the roots are
all real
if,
2
;
(ii)
least
;
(iii)
two roots are
x + x~
[If z
l t
then
z*
-f-
and two imaginary
real
pz + q 2 = 0.
Let
z l9 z 2
or
i(z 2
&=i(i>/z7^4)
if (fg-f
I)*
2 .
be the values of z, then N/z a 2
-4).
any value of a: is real, one and therefore both values of z must be real, and so p*>4(q-2). The conditions in question can now be obtained by considering the equation whose roots are z^ - 4, z a 2 - 4.] If
18.
// in
the last
all the roots
example
values of z are real, prove that
\
[For then 4(q 2)
-
l)*=a(x* -
1)
of the given equation are imaginary and the
p <4 and q<6.
\
and
etc.]
x^ 1, prove that '
x~
2(1 -a)'
Also if 0
but
20.
Show
that, if r
is
any
integer, the roots of
(4r+l)7r
are the values of tan
2fTT
21. If
yr = cos-~ show that o
,
and deduce the construction indicated figure
a
for
inscribing
a
regular
in the
pentagon in
circle.
In the figure
OH=\QA, and
22. Prove that the values of sin
and
cos-^ are respectively
-^^ and
A
(I
-v'17-f N /(34-2 N /17)}--i N /{174-3 N /17
[Referring to Art. 5, Ex.
1,
+ N/(34-2 x/17) + 2
sin=V(i-~iy2) and
{!
+A-
CHAPTER
XII
CUBIC AND BIQUADRATIC EQUATIONS
TheCubic Equation.
1.
The standard form of the cubic equation
is
w = ra + 3ta + 3cz + rf-0 ............................ (A) 3
If
x
y
bja this equation
where
= ay = ax + 6,
becomes
= ac-6 2
//
If 2
If a,
j8,
y are the roots
2.
3
is
+ 3//z +
6-'
=
................................... (C)
a -ft/a,
of (A), those of (B) are
fl
+ b/a, y + ft/a,
a/3 + 6, ay + b.
+ 6,
Equation whose Roots are the Squares of the Differ-
ences of the Roots. equations (A) and
(B).
The differences of the roots are the same Hence the equation whose roots are
(-y) 2 is
G=^a*d
,
the equation z
those of (C) are aa
2
obtained by putting
The
(y-) 2
>
= 3H{a 2 q
resulting equation
,
,
r = G/a*
(<*-)
for
2 ,
in Ch. VI, 17.
is
aV + 18a4^
2
-h81a 2flr2 2;-f27(G?2 -h4Jy 3 )==0,
...............
(D)
whence the important equality a6 (]3-y) 2 (y-a) 2 (a-j8) 2 = - 27(
2
4-4# 3 ) ................ (E)
It is easily verified that 2
.................................. (F)
A = a?d2 - 6a6crf + 4ac 3 + 463rf - 36 2c 2
where
4
2
2
,
.................... (0)
2
a (^-~y) (y~a) (a-j3) =->27J ...................... (H)
so that
The function A is
+ 4#3 = a 2J,
is
called the discriminant of the cubic u,
and
the necessary and sufficient condition that the equation
have two equal
roots.
its t/
vanishing should
=
CARDAN'S SOLUTION
180
Excluding the case in which two
Character of the Roots.
3.
roots are equal, possibilities are (i)
and remembering that imaginary roots occur
All the roots
may
One root may
and then by
be real,
and two imaginary
(H),
in pairs, the
J<0.
and, denoting the roots by a, A IIJL, we find that the product of the squares of the differences of the roots - A) 2 + /it 2 } 2 so that A >0. is equal to 4p,*{ (a (ii)
be real
;
,
Hence if J<0, all the roots are real and unequal ; ifA>0, one root is real and two are imaginary. Also it follows from equation (D) that J7 = 0, 6r=0 are the necessary and sufficient conditions for three equal roots.
Cardan's Solution.
4.
z=az + 6,
If
the equation
az + 3&z -h3cz-hd = 3
2
becomes
z z = w*~
Let
Comparing
(B)
3
+ 3#z + #=0 .............................. (B) z3
+ w*; then
and
(C),
.............................. (A)
- 3rwn*z - (m + n) =
m*n* = -H, m + n=
-G.
m and n are the roots of 2 + <#-#3 =0, and we may take m = J - G + >/(r2 + 4/73). Therefore
t
................ (C)
........................
(D)
(
If
Q
denotes any one of the three values of
the three values of
m*
are Q, ajQ,
<*)
2
Q, where
CD is
an imaginary cube root
of unity.
Also, because
m
3
w* = -/?, the corresponding values of
-H/Q, Hence the values
of
z,
that
is
ri*
are
-<*HIQ, of
ax 4- 6, are
Q-H/Q, O>Q+4H 8 <0, all the roots
NOTE. If C? 2 of the cubic are real, but Cardan's solution them in an imaginary form, which is very unsuitable for practical purposes.* In this case a solution can be easily obtained as follows. gives
5.
Trigonometrical Solution when
G 2 +4H 3 <0.
Taking the
equation z3
when
6r
2
3
+ 4Jff <0, a
+ 3Hz 4-0=0,
solution can be obtained
This
is
sometimes called the
................................. (A)
by using the equation
irreducible cote.
FUNCTIONS OF THE ROOTS If cos
30
is
known, cos
determined by
is
0-i cos 30 =
cos 3 0-f cos z = q cos 0,
Let
181
(B)
so that
cos 3
Equations (B) and
3ff
G
q
q*
+ ^-cos0+ 2
=
-^
(C) ' v
(C) are identical if
$
4
ff
= 2V-# and cos30-~-~= 3
2>/-/P
2
Now
G?2
equation.
ifif
a
is
3 ,
and so a
(D)
,___..
can be found to satisfy the last
real value of
any such value, the roots of (A) are ^27T
jcosl ^x.
1.
reduced
+ 3&# 2 + 3cx + d =0
3 // all the roots of ax
are reai, a/iow
tJiat
the equation
can be
to
3-j-f^=0,
2
where
27/x
<4,
xp
+ qt, where p and q are by a substitution of the form If z = ax + 6, the equation becomes z 3 - 7b + Q = 0,
A= -3#=3(6 2 -oc).
where
-aj
real.
2
Since aU the roots are real,
+4# 3 <0,
/.
#<0
and
h>0. Writing
the equation becomes
z=*Jh.t,
substitution
6
x~ NOTE. reduction.
The
XXIX,
a
t
a
M =a-f
2
co y,
an imaginary cube root
a> is
~- .t. + >Jh
3
1
=1,
of unity,
+ CO + C0 = 0, 2
TAe interchange of any two of the
M
CO
3
(0
2
a> /J-f coy,
and observe that
-^=1^3.
letters a,
into the other, functions L , Thus the transposition (a, j8) changes
/J,
y transforms
either of the
3
+ coa +
a>
2
y)
3 ==
(cua
+
i3 into
3 co ^8
+
2
co y)
3
= coW 3 = M3
.
It is the existence of functions possessing this property which the solution of a cubic equation depend on that of a quadratic. (2)
the
3.)
.L=a-hco/J-f
CU
(1)
= GI*Jh\ and
Important Functions of the Roots..
Let
where
jj.
roots of the cubic can be expressed as convergent series by using this
(See Ch.
Two
6.
where
t*~t+p.~
is
L and
M are functions of
the differences
1-f co-f
they are unaltered by writing a + A,
j8
a>
2
o/a>
j8,
y.
For since
=0,
+ h, y + h
for
a,
j3,
y.
makes
SUM OF TWO CUBES
182 (3)
We
have the following equalities
wM = 2j8 ~ y - a,
orL + a>
+
cu
2
M = 2y - a - 0,
by Ex. XII,
Also,
a>
2
:
L - wM =
a>L -
'
cu
W
-
(
(
a>
-
co
-
a>
oj
2 )
2
(y
- a)
- j8) ) (a
,
.
9,
Z 3 + M 3 = -27/a 3
Therefore
,
..............................
(A)
i3 -M 3 =-3(
iJf = 2k 2 - 2?j8y = -9#/a2 ......................... (C)
Moreover, (4)
From
equations (A) and (C)
follows that (-|a) 3 , (-JaM) 3 are
it
Ae
roofs of
which 7.
is
the auxiliary quadratic in Cardan's solution.
The Cubic
Sum
as the
of
Two
3 2 // u == ax -h 36x -f 3cx + rf, ^Aew constants
A
Cubes. ,
B,
A, p,
u^A(x-\Y + B(x-tf\ provided that u has no square factor
H
s=
(ac
Also
Eliminating A,
B
from the
equations,
coefficients it will be seen that the identity
\A + ^.B^-
first
-
2
a( A
Assume that
are the roots of
)
A+B = a
(A)holdsif
A, /x
........................... (A)
- 6 2 x 2 + (acZ - be) x + (M - c2 ) = 0.
By expanding and equating
of
.
can be found such that
2
A/u
jit
)
three
and from the
last three of these
+ 6( A2 -p?) + c(A -/x) -0, - A 2 ) = 0.
and that A^/x, and divide the these equations by A-/x and the second by Aju,(A~/t), then = 0, and aA/Lt + o(A+jLt)4-c
Eliminating
neither A nor
/u,
is
zero
first
/^,
Therefore A and, by symmetry,
^ are the roots of - (bx + c) 2 = 0, (ax + b) (ex + d)
which
is
the same as
H(ac~62 )x Also A,
B
are given
by
+ (od~6c)x + (6d!-c2 )==0 ................... (C) A + B =a, A-4 +fJ3 = - 6. 2
THE HESSIAN
183
// A = 0, then by equations (B), TJ 2 =A &i-c
so that also in this case A
// A
jit,
two equal
the reasoning roots.
i and
c o
and p are the roots
Ex.
w=
if
and only
)
A
has two equal roots, so also has
H=
UG the preceding
1.
if,
Let
2x*
to solve the
)
2x*
equation
A+B = 2,
and
A+2B=;-l,
2 -(#-7) =0,
so that
vice versa.
giving
A
= l,/i=2;
.4=5, 7?^ -3.
.
,
and the roots
.
_ 5"^ -2A;3^ _ (5Jfc3*)
,
has
+ 3#2 - 2lx + 19 =0.
Hence the given equation may be written 5(x- 1) 3 3(#-2) 3 2 by -^/5 (x - I ) = k /3 (a: - 2), where k 1, co, co
are given
=1
w=0
+ 3x*-2lx + l$ = A(x-X)* + B(x-iJL)* ........................ (A)
Then A, //, are the roots of (2* + 1)( -lx + 19) and, equating the coefficients of x 3 and # 2 in (A),
Since ^3
if,
For using equation (G) of Art. 2, we find that 2 - 6 2 (bd - c2 = 4 (ad 6c) (ao
Hence,
H = 0.
of
This case will arise
fails.
,
(
Hence the roots are |(
1
-
/75 -4/45),
The Hessian.
8. is
-
called the Hessian,
The roots
of
H=
will
hQ =
where
(1)
and
is
of great
H = ac- 6
2 ,
A,
ft,
and we
h^ad- be,
shall write
h2 = bd-
c2 ,
A 1 2 ~4A A 2 = A .................................. (A)
Using the identities of Exercise XII,
~18H = a (3)
importance in the theory of the cubic.
be denoted by
so that
(2)
H which occurs in the last article
The function
We
2
2
2
2
{(o;--a) ()8~y) 4-(a:--j3)
can find the values of
A, p, in
4, 5, 6,
we
find that 2
2
(y-a) -f(x~ y ) (a^j9)
terms of a,
/J,
y as
2
} .......
follows.
(B)
Let
M =a + eo j8 + coy, y, 2 = Z/ j8y + coya 4- a> aj8, M' = /Jy -f co 2 ya 4- a>aj8. L =
Then,
and
if
2
2
P = (z-a)(j8~y), Q = (a?-j8)(y-a), R = (x~y)(a-j8), P +
we have
AUXILIARY QUADRATIC
184
from (B) that one root A
It follows
that
2 A(> -
is,
(4) It is
by
+ (co 2 - w)L' =0,
^ = -M'/Jtf ...................... (C)
A = -L'/l and similarly
so that
easy to verify that the substitution
reduces the equation h^x?
which
<*>)L
H=0 is given
of
+ hfl + h2 =
to
the auxiliary quadratic in Cardan's solution. Thus the roots of the auxiliary quadratic are is
EXERCISE
XX
THE CUBIC (i)
Find to five places of decimals the real root of z s + 293-97=0, (ii) z 8 + 6a; 2 H-27a;-26^0, (iii)
2.
Find to
1.
five places of (i)
3.
decimals the real roots of
x* -3a;-Hl=0,
(ii)
- 3qx +r =0 and k = y are the roots of a: (0-; 3
If a, 0, (i)
>/3(4g
3
-r 2 ), show
tl
t
08 -;
(ii)
The equation whose
(iii)
roots are
j9
- y, y - a, a - j3
is
+ y + y a=|(r + Jfc), a y-f ^ a + y ^ = f (r-t). a 8 8 = 3 a y-hj5 3 a + y 3 )8= -9g 2 a (v) j8 + ^ y-f-y a If the are all roots real and a> j8> y, the difference between any two (vi) of the roots cannot exceed 2 N /(3g), and the difference between the greatest and least must exceed aa
(iv)
2
a
2
2
2
.
4. If a, 0, (i)
y are the roots the equation whose roots are
j8
y,
y
- a, a - jS )
(ii)
=
is
;
the equation whose roots are a 2 j8 + ]8 2 y -4r y 2 a, a 2 y + 2 a -f y 2 )9 a4* 2 - 3a 2 (acZ - She) x -f 9 (a 2d 2 - fabcd + 3ac 3 + 36 8d ) = 0.
is
In &e following examples ike letters L, M, L', M' 9 h , h l9 h 2 have the meanings assigned in Art. 8, and unless otherwise stated a, j9, y are the roots of
.5.
Show
that
L t + "Jf=3.M', and M* + a
a
L=3L'.
TYPES OF SUBSTITUTIONS If
6.
-
A (x - A)
u
A)
3
-f
B(x -
l jB
A/B=
k=
s/(
;[
y are the roots of u =0, assume that
^=
(1, 1, 1), (1, o>, o>
2
),
(1, o>
2
- JUT '/ Jtf
w),
,
(A)
.
add and use Ex.
5.]
+ i )>
^(*i
and
a 2 y + j3 2 a + y 2
+
3^
=
~
(/H-fc)^
Find the values of I, m, n such that
2^0^-^ + ^,
proving that
That
is
to say,
'
homographic ^ ^
which
jS,
A = - L'/L,
- L 9 /M\
y + J/? + ray -fr&=0,
9.
a,
- JJ), prove that
+ /8V4V + 8 =
8.
and
2
Hence prove that
If
/x)
185
+*(a - M )=a>4*(|3 - A) + J3*(j8 - ,x)=a) ^*(y - A) -f JS*(y - ft)=0.
[Multiply equations (A) by 7.
3
'
yoc
mh
2h
any iwo roots
+ Jy + ma-fn-0,
(a,
l
-k,
h
a/?
n=h
+ Za + w/? + tt=0 where &
29
o/ i^e cubic equation
j8)
u =0
;
iV( -JJ).
are connected by the
relation
Except when J =0, there are two substitutions of the form will transform the cubic equation
u=Q into
itself,
and these
are
(
where 3& 2 =4A ^ 2 ~^i 2 and ^ ^i> Explain why this fails if A ~0. >
[This
is
merely another
way
^2 are the coefficients of the Hessian.
of stating the last part of Ex.
m
8.]
f
10. Find substitutions of the form x~(ly + m}l(l'y + transform ) which will the equation x 3 - 3# 2 + 3=0 into itself, showing that these are
* = (2y-3)/(y-l) 11.
and
* = (y-
Find substitutions of the form x~(ly + m)l(l'y + m') which will transform 2 60; -f 1 ^0 into itself, showing that these are 4- 3#
the equation z 3
12. Show that any pair of roots a, j3 of the equation x 3 -3ic 2 4- 3s- 2 = are connected by a relation of the form pa. -\-qfi + r=- where p q, r are the same numbers whatever pair is chosen, proving that the relation is 9
13. If a,
j3,
y are the roots of x* -30#-f r=0, find Za
proving that,
where
2
+ ma 4- ft
if a,
j8,
y are
=
I,
m, n such that
n = y, Z /j9 + mj5 + real and a> > y, then 2
/?,
jfc
Hence show that with these values of I, m, n the substitution transforms the equation x 3 - 3qx
+ r=Q
hi to
itself.
1S6
BIQUADRATIC EQUATION
14. If
either
,
2 y are the roots of a?-21a; + 35=0, show that a + 2a- 14
jS,
15. If a,
y=x* + 2x~
(In other words, the substitution
or y. equation into ft
is equal to 14 transforms the
itself.)
w=0 and = = k, <**+p
y are the roots of the cubic
/?,
2
^
* prove that
where A
p=
36
a
A=
if
Further,
^=
tt=0
Prove that the equation y x* + px + q,
16.
6
3i=Jf(A-,*) =
-Jf'/Jf, then
2 8 (1, 1, 1), (1,
by
[Multiply
-L'/A
,
g= a hA-a
H =0.
a root of the Hessian
is
2c
,
-fA,
27/2
and add.] transformed into
is
36
.-,
p= a
if
where A
is
H=0
a root of
2c
N
6
N g= a + A-, a
+A,
and k
determined as in Ex.
is
Find substitutions of the form y ~x 2 + px -f q which
17.
by the
y*=k?
substitution *
15.
will
transform
=0 into the form y 3 = ^ 3 , giving the value of k in each case. 3 - 3 =0 and ( L) 3 , [A, p, are the roote of x* + 2x (JJif ) are the roots of
i.e.
of
t*
=4 3
If
.
we take A = 1,
fi
=
- 3, we must take
found that each of the substitutions, reduce tfc=0 to 18.
In Ex.
Explain
9.
J^^S
2
why it is not to
x = y - 6/a,
If 2= ay = ax + 6,
have
6/a,
j5,
3
-f
i/
= 8,
giving
ic
a
-f
6H 4G .OA +_ _^ !f+?
* This is
is
(A)
m
iC
2
G =a2d~3a6o + 26s
=a3e-4a26d + 6a62c-364
,
,
(B)
f
.
the equation becomes
+ &Hz* + Gz + K=Q
(C)
y, S are the roots of (A), those of (B) are
and those
4x + 3=0.
6c#2 + 4dz-f e=0
s
,
z*
>
It
becomes
this
iTac~62
where
If
will
The Standard Form of the Biquadratic Equation
-
-f
thus
6.
be expected that both roots of this equation satisfy u =0.
t ay +
8
is
y=# 2 -f-7,
.]
u=*ax* + 46x If
M=
and
3
y-x + 4x + 11, we may
17, if
L = - 6,
y~x + 4a;-f 11 2
of (C) are ooc *
+ 6,
a/J 4- 6,
ay
-f 6,
a + 6/a,
/?
+ &/#, y + 6/a,
aS + 6.
a case of Techirnhausen't Transformation.'
(See Art. 17.)
REDUCING CUBIC
Some Important Functions
10.
A = /Jy-fa8,
Any rearrangement
(1)
functions
A,
/i,
of the Roots.
= ya-f-/JS,
/A
of the letters a,
v into itself or into
187
j8,
Let
i>=a/J-fyS. y, 8 transforms
any one
of the
one of the other two.
found that the existence of functions having this property makes the solution of (A) depend on that of a cubic equation. It will be
(2)
We
The
cyclic substitution (a/Jy) changes
have
A to
/it,
/z
to
v,
v to A.
(j8-y)(a-S)=i>-/z
(y-a)(j3-SHA-v Hence if two
of a,
/J,
three of a,
j8,
y, 8 are equal,
then two of
........................... (D)
A,
//,,
v are equal,
and vice
versa.
Also (3)
if
The functions
A,
then A=/i = *>, and vice versa.
y, 8 are equal,
/x,
v are Ae roote
o/*
a*y*-&a cf + 4a(bd-ae)y-8(2ad* + 2eb*-3ace) = Q ......... (E) 2
For
2;A = 2:aj8 = 6c/a, 7a
.
ZajSy
+ aj3y 827a
- 4aj8y8 = 4 (4W - ae)/a2
,
2
~ (4d2 - See) + ^3 (4fc2 - Sac) '
a2
The second term
a
of this equation
can be removed by the substitution
and the equation becomes where
4^3_/^ + j = o, = / ae 4bd + Sc2
(G)
(H)
,
Expressed as a determinant a
&
c
c
d
e
bed
It will be is
shown that the
solution of (A) depends
on equation
(G),
which
called the reducing cubic.
The functions 7 and J are
of great importance in the theory of the
biquadratic.
K
B.C.A.
THE DISCRIMINANT
188
Boots of the Reducing Cubic.
(4)
(Q) corresponding to A,
Let t^t 2
^a = A-^-A-i(A+ a
/
t
s
be the roots of equation
Then by
v respectively.
/x,
,
(F)
+ v)=i{A- /t -(v-A)};
i
whence, by (A), t^^a{(y - a )(j8-8)-(a-jB)(y-8)}, with similar values for 2 2 tz If real,
y, 8 are all real, or if they are all imaginary, then
/?,
and
of the three
or
if
A,
a,
Also
t
/z,
jS,
t% is real,
a,
y, 8 are real
and
v
t%
v,
and therefore
y
y, 8 are of the
,
if
j8
are real
and two are imaginary
t
v
Z
3,
2,
lim,
forms
and y = / -f ^m,
are
V 8
all real if a,
im'
The Functions /?,
t
2,
t
3
are all
I,
Since
J.
,
then only one
t
l9
t%,
3
jS,
y, 8 are all real,
.
m,
== /
imaginary. statements are true. (5)
v
conversely.
also
Moreover, one of these cases must
of a,
t
conversely.
If two of the roots a,
For
.................... (L)
.
,
a,
............. (K)
is real and A, /x are and so the converse
then v arise,
are functions of the differences
y, 8, so also are /, J.
/ and
It follows that
J
therefore in equations (H)
same
are the
and
(I)
6 = 0, c-H/a,
a I= z
giving
K + 3H*
whence the important
for equations (A)
and
(B),
and
we may put d=
2
/a
e
,
= #/a 3
,
a*J^HK-G*-H*>
and
identities
= a 2/-3ff2 Another important equality
is
8-y)2(a-8)2 + (y-a)
()
G* + H* = a*(HI-aJ) .................. (M)
,
2 8-8) + (a-j3)2(y-3)2 = 24//a2 ....... (N)
2 (j
For
and
(6)
The Discriminant.
By
equations (D),
a(j8- y )(y -)( -/3)(-8)(
- <3 ) 2 (<3 -
= 256J by where
^ft -
< ) 2
2
by equation (K)
Art. 2,
............................................ (0)
J =/
-27,72 .................................... (P)
3
REDUCING CUBIC The function A ing
is
called the discriminant of the quartic w,
and
the necessary and sufficient condition that the equation
is
have two equal 11.
its
vanish-
w=
may
roots.
Character of the Roots.
J = 0.
this
If two roots are equal, then and remembering that if there are any
case
Excluding imaginary roots, then these occur in (1)
189
All the roots
may
be real,
pairs, the possible cases are
:
and then J>0.
Two roots may be real and two imaginary. Denoting the roots by a, /?, im, it is easily shown that the product of the squares of the differences of the roots is -4w 2 (a-0) 2 {(a - 1)* + !*}*{ (0 - 1) + 8 }*, and so J<0. (2)
I
w
(3)
All the roots
may
Denoting them by
be imaginary.
lim, /'iw',
we
find that the product of the squares of the differences of the roots
is
16w2m' 2 {(Z Hence
(4)
12.
- m') 2
+ (w + m')*}*{(l + Z') 2 +
2
}
,
and so
A>0.
// J<0, two roots are real and two imaginary.
(5) If
For the
2 Z')
J>0,
the roots are all real or all imaginary.
criterion distinguishing the
Solution
Ferrari's
two cases
of the
of (5), see Art. 15.
Biquadratic.
Writing
the
equation
u=ax* + we assume
bx*
+ &cx2 + 4dx + e^0,
(A)
that
au = (ax* + 2bx + s)*~(2mx + n)* Expanding and equating
coefficients,
2w = as + 262 ~3ac, 2
(B)
we have
wn = &s-ad, n2 = s 2 -ae
(C)
Eliminating m, n, 2
(s
- ae) (as + 2V* -Sac) = 2 (bs - ad) 2
,
which reduces to 53
-3c5 2 + (46d-ae)5 + (3ace-2ad2 -2662 ) =
(D)
The second term can be removed by the substitution
= 2t + c
(E)
&3 ~/* + J = 0,
(F)
s
and equation (D) becomes which
is
the
c
reducing cubic.'
i
(See Art. 10.)
= 2bt + bc-ad
Equations
I
(C)
become
(G)
FERRARI'S SOLUTION
190 Thus, if fj
a root of
is
m the equation
its
n t = (26^ + 6c - ad)/m v
b 2 - ac), l =\/ (at v +
u^=0 can
be put into the form
(ax
and
and
(F)
2
+ 26z + c + 2^) 2 - (2m 1 a? + n x ) 2 - 0,
roots are the roots of the quadratics
ax 2 + 2bx + c + 2^ =
(2m1x + 14).
It should be noticed tha^ the three roots of (F) correspond to the three of expressing
ways
u as the product
of
two quadratic
=x* + 3x 3 + x* - 2 = 0. u = (x 2 +px + s) 2 - (mx + n) 2 and equating coefficients, wo have Expanding
Ex.
Solve n
1.
Let
The
may
last
equation
is
a
and
u
[x*
-
if s -~
satisfied
take
and then
J-,
+ (p
-f-
77^) a;
?yr
-
-1+4, inn-
4.
Thus we
+ 5 -f n} {a; 2 + (p - ?/i) a; + s - n}
^(x*+2x-2)(x*+x + l). -1 >/3, w, a> 2 \vhero w ,
is
an imaginary cube root of
1.
Deductions from Ferrari's Solution.
13.
then
.
=-|,
=i,
=--j,
Therefore the roots are
Let
factors.
]8,
y be the
ax 2 + 2bx + c + 2^ + (2m 1 x + 7i 1 )=0,
roots of
2
ax + 26x + c +
a, S are the roots of
Further,
let (y, a), (a,
/?)
(H) by changing the suffix
1
2/ x
-
(2^0; + n x = )
...........
(H)
............. (I)
be the roots of the equations obtained from into 2 and 3 respectively.
= c + 2t l + n v a(3y
Now
a (/?y 4- aS)
therefore
which agrees with equation
(F) of Art. 10.
a(j8y-a8) = 2w
Also
= 2c + 4^,
1
,
a(]8-f
y-a-8)== -4m 1}
with similar equations given by the cyclic substitutions
................ (J)
(a,
j8,
y)
and
(1, 2, 3) of the suffixes.
Hence
(i)
the values of
m are
In connection with these functions
Ex.
3, (iii), p.
it
should be noted that by Ch. VI, 16,
96, ............ (K)
DESCARTES' SOLUTION by
Also,
This
(F)
and
The values of n are
(ii)
The values
(iii)
whose roots are
(G), the equation
an important equation, obtained
is
of
J^ZY^
___
and from
y + oc ~ p
o
Of.
- yS).
r(
ya-)8S 3+y
la(oc^
are
~
,
2m 92
in Art. 15.
way
- j8S),
|a(ya
-~- 2 >
2ml
another
in
\a(fiy -aS),
-~-
191
S
ot
+p
o
y
that these functions are the values of z found by
(F), (G) it follows
eliminating tfrom Z
+ bc~ad
2bt
1 __ _ ~
,, (M)
and
~2' 7
Biquadratic as the Product of Quadratic Factors.
14.
9
Let u
Descartes method.
ax*
4-
4foe 3
u = a (x + 2Zx -f 2
Expanding and equating l
+l =2
--
a
,
?/i
2 )
(x
coefficients,
m+m =6
a
~ 4a
-f
6cx 2
-!-
4c?o; -f e,
and assume that
+
we have k
d
:
,
a
,
,
-m?w
=-e
.
a
Now 1
1
I
l
1
m
1
l
m'
m'
I
Substituting the above values for
row by
If
we
a/2,
-f
Z
m 1
m + m'
2
|-
f
f
V
211'
m + m' lm +l'm m + m', etc., and + 1',
-0.
+ I'm 2mm'
lm'
multiplying each
we have a
b
3c-2all
b
all'
d
3c-2all'
d
e
f
=0.
\
write
the equation becomes
c
which when expanded
I,
+ 2t
a
b
b
c-t
d
d
e
+ 2t
c
=
0,
is
Corresponding to any root ^ of this equation, we can m, V, m' which will satisfy the conditions.
now find values
of
POUR REAL ROOTS
192
2 = given in the form u^-ofl + px + qx + r Q, we proceed as in Art. 15, using I instead of 2Z, to obtain the cubic in the form 2 2= 2 irl 2 0, which reduces to an equation lacking the second P(l +p) y
Again,
if
the equation
term on substituting
z
is
- 2p/3
for
I
2 .
In numerical work, unless this cubic has rational roots, the solution becomes by the methods described in this chapter the student can convince
NOTE.
very laborious himself of this
;
by verifying the steps of Cardan's method for solving the equation, 2 8 - 1 \z - 15 =0, which is obtained as above for the biquadratic, x* - 3# 2 - x -f 2 = 0.* Later, by one of the methods described in Ch. XXVII, it will be found that one root can be found approximately with very little trouble. This is all that it is only necessary to have one way of breaking up u into quadratic
15.
Four Real Roots.
In Art.
11, it
is
required, since
factors.
has been shown that
necessary condition that the four roots should be real rule (Ch. VI, 11, Ex. 1), if #>0, at least two of the roots
Hence, J>0, 77 <0 are necessary conditions that real but these two alone are not sufficient.
;
J>0 is a
also, by De Gua's must be imaginary.
the roots should be
all
;
The complete
may be found by using Sturm's theorem, the usual course This, given adopted in text-books, involves rather tedious reckoning the conditions can be found much more easily by the set of conditions
later.
;
use of either Ferrari's or Descartes' solution of the biquadratic. /
The necessary and sufficient conditions u == ax* -f 46x3 -f 6cx2 + kdx -f e =
Theorem.
aw
all real i*
(i)
or their equivalents
Let
z = ax
+b
;
v = z*
(ii)
J>0, J>(),
then by Art.
that the roots of
//<0,
and
12# 2 >
H<(),
and
2Hl>3uJ.
9, (C),
and Art.
+ 6Hz* + 4:Gz + K = O
t
2
/,
10, (M), z is given
K = a I-3H 2
where
by
2 .
v= has the same number of real roots as w = 0. For v = 0, using the method of Descartes, and supposing that - 2lz + v 55 2 + 2lz + m) m')
Now,
(z*
(z
we
get,
by equating
Eliminating
Hence, since
m
9
coefficients,
and m' from these equations,
3
(2Z -f
3HI)
2
- G2 = 1 2K, or
W + l2Hl* + (9H*-K)l*-G*=Q.
K=a*I -3H2
,
the values of
I
2
are the roots of 2
which, with the substitution, y +
H = at,
=0;
reduces to 4J
3
..................... (A)
-
*Thus, tt'+v^lS, ut?ll/3, t*-t?= 5-27397, ti=-2-16423, t>=l-0943, which ttS(*H2-42042s+l-63580)(s-2-42042a:+l-22265)=0.
-i+t>- 3-85845, from
FOUR IMAGINARY ROOTS Now
the three values of
I
2
correspond to the three ways of expressing
v as the product of two quadratic factors.
and
the roots of v =
If
(i)
i.e.
positive,
are
all real,
Hence,
the three values of
the three roots of (A) are
the roots of v =
If
(ii)
193
I
2
must be
all real
all positive.
imaginary, then, since their sum is zero, 2 Ait/i, -A-ti/i/, and thus the values of I are 2 -i(/A-/i') i.e. the roots of (A) are all real, but only one are
all
we may denote them by A2 is
2
-4(/4+/z')
,
positive If
(iii)
>
,
and conversely.
;
two
=
of the roots of v
are real
and two imaginary, we may denote
and then the values of Z2 A, JJL, A'i/*', where A-Fju-h2A' = 2 2 i.e. the roots of of which one is real and the other (A), are A' J(A -f X' t///) two imaginary and conversely.
them by
,
;
,
,
;
It follows that the roots of v if,
and only
if,
Now, the is
the case
the roots of (A) are
roots of (A) are real if,
and only
if,
Descartes' rule (Ch. VI,
negative and theorem.
if
12H 2 -a 2I
is
I2H
2
J=/ -27/
2
2
is
2
First observe that,
p + pq + q
2
but /
is
hence,
2
;
so that
,
J>0,
/>0
hence, since
;
//<0,
if
p and
q are
:
q according as
|
2HI\>\ 3aJ\;
any
2HI<3aJ.
and real
numbers, positive or negative,
hence, since
;
i.e.
,
2HI<3aJ.
2 3 3 2 p -q = (p- q) (p + pq + q ),
it
p g q*
we have 8/P/ 3 <27a3 J 3
2HI<3aJ, and J>0; and therefore
G 2 + I2H*
addition,
l2H 3 <.a 2HI, and
;
;
dividing by H, which Thus, the two sets of conditions are equivalent.
therefore
suppose
i2H 2 >a 2I.
4# 2/ 2 >9a 2
ff<0,
3
positive,
by
;
we have
>0,
= (p + %q) 2 + f 2 >0
follows that p* Then, since
this
also, the roots of (A) are, by and positive, if, and only if, H is which proves the first part of the
and
negative, therefore
Next, suppose that 2
and
;
HI<0
P>a P>27a J
HI
and, since
are real
of the second set of conditions,
//<0> 3
Then, since
all real,
all real
positive
that
Also,
and positive. those of 4 3 - It + J =
u = 0, are
9, 10), all real
follows that
it
therefore those of
A = / 3 -27e/ 2 >0
To prove the equivalence first
= 0, and
is
negative,
12H 2 >a 2 I.
~l * In 1 general, if n is odd, =*(p-~q)(pn- +pn-*q+ ... +gn ), where the last factor is the qn according product of pairs of complex conjugate factors, and is therefore positive hence p n
pnqn
^
;
&spr^q. But,
if
n
is
even,
and not as p :g q.
pnqn^ty
q)(p+q) (a positive factor), hence
pn^^n
as
|p|<|ff|,
EECIPEOCAL TRANSFORMATION
194
Alternatively, using Ferrari's method, the equation
whose roots are
s
which, when expanded, is equation (A) on p. 192. There are three possibilities. (i)
AH
the roots of u =
all three real (ii)
and
may be imaginary = /? A-
(iii)
Two
take a,
j8
;
I/JL',
of (A) are
-iaV-fO
three real, but owe owfa/
all
in that case the roots of (A) are
;
u=
a = A + i/z,
are
be real
positive.
All the roots of
and the roots
may
of the roots of
,
8 = A'
-iW)
I/A'
2
2
1
>
we may take ;
(A-
2
A')
,
which
is positive.
w=
to be the real roots
2
in that case
may
be real arid two imaginary 8 = A - t/i and then ;
and y = A + t/i,
we may
;
are the roots of (A), of which one only is real. The proof proceeds as before.
Another method Exercise
XXI,
of solving the biquadratic,
due to Euler,
is
given in
17.
Transformation into the Reciprocal Form. The substitution x = py + q transforms the equation w = into apY + *BrPy* + Wp*y* + 4:Dpy + E = Q, .................. (A) B = ag + b, C = aq 2 + 2bq + c, where D = (a, 6, c, <%, 1)3, E-(a, 6, c, d, efa I) 4 16.
.
= if ap* E, Bp*==Dp, 2 = 2^, p* = D/B .......................... (B) by aZ>
This will be a reciprocal equation that
is, if q,
p
are given
Suppose that the conditions (B) are satisfied, and that the roots of (A) where J/22/3 == J/iJ/4 :== l* Let a, j8, y, S be the corresponding !/i> 2/2> J/3> 2/4
are
roots of
w = 0, so that
It follows that
? H~?)(y-?) = (a~?)(S-?), 2
and
...................... (C)
therefore
Hence by
Art. 13,
(iii),
the values of q are the roots of the equation in z found
-
by eliminating tfrom
S' 2
i
LO
an d
TSCHIRNHAUSEN'S TRANSFORMATION
qp
The points corresponding to q and respectively the centre C and the foci NOTE.
F
F' of the involution determined by the
y
"p/
9
as given
COtfl
points a, j8, y, 8, in which (j8, y) and (a, 8) are corresponding pairs of points. That is
Find a
1.
substitution of the
form
x*
Use
into the reciprocal form.
o = l,
Here
4* 3
The equation One
and by
solution
is
(C), are
Y
fi"
F
C j8.Cy = Ca.C8=CJ2 ==JF C a /
f
xpy+q which will change the
+ x*-2x-l=Q
.
equation
....................................... (A)
this to solve the equation.
6
= J,
-/* + Jr=0
$=-J,
by
FIG 29
to say, the line segments are such that Ex.
195
and by
is
c
= 0,
4e 3
d=
c=-l;
-^,
-f^-^=0,
that
8
is
(4*)
+2 (40 -3=0.
(E) of the text, the corresponding value of q is
(B) of the text, the corresponding value of
p
2
is
-i + |~|_
*
Taking p = 1, one substitution of the kind required is * =y 1. 2 4 ~ becomes the + in 3t/ -f 1 =0. %!? 3y y equation (A), Substituting 1 2 Dividing by y and putting z=y-f-y"" we find that ,
2 1
Also
y -#s /.
17.
+ l=0,
a
- 3* + 1 = 0,
giving
z
= | (3
/5).
2/=i(2V2 -4), and aj=J(z~2db^ a ~4) a
.*.
*=i(- 1+^5^-2 + 6^5),
or
i( -1
;
-^^^2 -6^6).
Tschirnhausen's Transformation.
If
we
eliminate
x
between ~ + a^-1 + a2x n 2 + + a n = r "" 2 1 y=xr + p^" + ^2^ + + Pr
/(x)
and
we
shall obtain
=a
o;
n
an equation
. . .
of the
form
Thus for a single value of y corresponds to each of the n values of x. can be chosen so that r of the ... p r theoretically, in general p v p 2 , ^n are z ^ro.
coefficients bl9 62 >
Transformation
:
it
This process
is
called Tschirrihausen's
has been applied to the cubic in Exercise
XX,
16,
TSCHIRNHAUSEN'S TRANSFORMATION
196
Let
Case of the biquadratic.
a,
y, 8
j8,
be the roots of
u = ax 4 -f 46x3 -f 6cz2 + 4dz + e = 0.
We shall
prove that in general three if x is eliminated between
such that
u=
sets
of values of
p and
q can be
found
2 y = x + pz + q,
and
the resulting equation is of the form 4
Further, the values of p
and
is
+ =
.................................. (A)
<7
q are given by
ap=-Ab + where z
2
+/
2/
aq = 3c + 2bz,
2az,
......................... (B)
one of the three
aff-yS a + jS-y-S
ya-/3S
j8y~
'
'
y+a-jS^
J3Ty-a-S
*
the resultof y corresponds to each of the four values of x Let p, q be chosen so ing equation in y is therefore of the fourth degree. 3 and let are and of that the coefficients zero, y 2 be the values y l9 y y
One value
of y.
;
Then we may take 4-j>y
+ ?=
-~2/2
whence by addition and subtraction,
y-a-8) = 0, ................... (D) s 2 +^5 1 + 4y = 0, ................................. (E)
and where
sl
= 2oL,
sz
= Z
2
Now therefore
(j3H-y)
-(
where
-p^s.-Zz, *
2
we
a/
j8y-a8
=
p=
Hence Substituting in (E),
have, from (D),
46
+ 2z ..................................... (F)
find that
a
a
\
3c
a 2
giving
(
Moreover, is
of the
if
form
p, q (A).
have these values,
it is
}
obvious that the equation in y
SUBSTITUTIONS As
197
be seen in the next example, each value of y gives one and only
will
one value of x which* satisfies the equation w = 0. Ex.
1
.
Find a
4
to the
form y Here a = 1, we have
2
+/?/ b
substitution of the
+ g ~0. Find
=0,
c
=0, d = 3,
e
of
t
is
1,
-
5, so
that
where
4* 3
z~f and
giving
y=xz +px + q
the values off, g,
=
z^ One value
form
which
and use
will reduce the equation
the result to solve the equation.
7=-6, J=-9; and using Art.
13,
(iii),
+5*-9=0.
then
One such
p=3, q=0.
substitution
is
therefore
y
To
eliminate x from (A)
and
(B),
x2
O
flH\
\D j
H~ oiC.
we have
*
- 3x) =a;4 - 9x 2 - - 9x2 - I2x -H 5,
y(x*
xz + 43#
But
y
frcttn
(A)
;
;
which reduces to
- 200-0.
The
values of y are
Hence we
10*,
\/2,
and the corresponding values of x are given by
find that the roots of (A) are
-lN/2,
EXERCISE XXI
THE BIQUADRATIC Unless otherwise stated
1.
If
2.
If
+ y = a + 8, ( j8
-H
If
j8,
y, 8 are the roots of
y)8 = (-*- 8) j8y, show
[Deduce from Ex. 3.
a,
u^ax* + 4bx* + 6co; 2 4- 4d(a: + e =0. show that a zd + 26 3 - 3ofo = 0. that e 2 6
-f
2d 3 - 3edc = 0.
1.]
2 2 0y=a8, show that od =6 e.
L
4. If
show that the interchange of any two of the four 3 and into L 3
M
a,
j8,
* In the case of the general equation, tt=0, and the substitution, obtaining the second quadratic (G) is
y [ao? +(46 ~ap)x] * (
and
thus,
y, 8
changes
L3
into J/ 8
,
y**&+px+q,
the
method
for
FERRARI'S METHOD
198 5.
Show
that the equation whose roots are j8y
- a8
ya
/94-y-a-o'
-
-
a/?
/?S
yS
-/
is
where
B-az + b, D-(a,
[Dimmish the roots of 6. If a,
ft,
3
b, c,
u~Q by
y, 3 are the roots
I) #-(a, 6, and use Ex. 3.]
d$z, 2
,
of w==o; 4 -fg# 2
-f
d, e$z, I)
c,
ra-f s ~0,
4 .
show that the equation
whose roots are z3 -
is
and that
if 2 X is
2
- 4sz -f 4g# - r 2 = 0,
a root of this equation, then
where A ls A 2 are the roots of
A^Az^-s^O. 7,
Show
that the equation whose roots are 3
s
[Use equations 8.
Solve x*
23
(D), (G) of Art. 10,
- 2x* -\-a(2x -
1)
^0 by
and Exercise XX, putting
it
3, (Hi).]
form
in the
and choosing
s so that the right-hand side is a perfect square. the equation has two and only two real roots unless a = 1 or 0.
Hence show that
.
Solve by Ferrari's method 9.
:
* 4 + 12:r-5=0.
11. x*
- 3z 2 - 4x ~ 3 ~0.
13.
* 4 -f 12x3 4-54^ 2 -f 96x4- 40 = 0.
14.
4 4x 3 + lx z Express #
6a: -h
x*~2x- 1 ^0.
10.
a4
12.
x 4 - 4* 3
4
4- 5.u
4 2
= 0.
3 as the product of quadratic factors in three
different ways. 15. If a,
jS,
y, S are
the roots of x*
-\-
3x 3 4-# 2
-20,
prove that the equation
whose roots are
(j34-y-a-S) z
is
[Use equation (L) of Art. 16.
Show
and two
3
2 ,
(y 4-a
ft
- S) 2
,
(a
+ j8-y-S) 2
- 19z 2 -f 243z - 225 = 0.
13.]
that the equation whose roots are
similar expressions
is
3 2 4(*4-L) -e /(z4-L)4-eV=0
where L=ce-d*. [This follows from equation reciprocals.]
(L) of Art. 13
by changing
a, 0, y, 8 into their
EULER'S SOLUTION If
17. Euler's Solution.
and
= *Jl + \/m -f \/n,
z
(i)
(ii)
then
Z,
which (iii)
z
- 223
4
.
z2
Jl,
*Jm, *Jn denote either of the square roots of
is
.z
+
the same as
The complete
4(s-f//)
and(K)
6//z 2
4-
4s
3
-f
+ 4Gz + A" = 0,
127/s 2
~-K)s- G
2
4-
(9//
is
given by
2
- 0,
-a 2/(s + //)+a3
3
solution of the biquadratic u
aa -f b = -
*Jl + sjm + >Jn,
ay + 6 =
^Z + Jin -
Observe that the values of
18.
z4
identical with
a8
v/Z
.
-h
6
s/w
m, n are i\a
I,
=
af3 + b~
*Jn,
the square roots being chosen so that
(L)
m, n
l~- 22!tom =0.
m, n are the values of a given by is
/,
prove that
- &*Jl*Jm*/n
If this equation
199
.
=
*Jl- *Jm
-
Jl-~
*Jm -
(j3
>/n,
same
s/n has the
2
+ -s/n,
+ y-a-8)
2 ,
sign as G.
etc., see
equations
of Art. 13.
Find a substitution of the form
x=py + q
which
transform the
will
equation into the reciprocal form. to solve the equation. 19, In
Ex.
1
Show
of Art. 17,
if
that one such substitution
each value of y which y
is
substituted for y in y
*
is
satisfies
x = 2y -
1.
Use
this
the equation
+ 9% - 200^0 2
x + 3x, we get eight values of 2
x,
four of which satisfy
Without solving any equations, show that the other four values of x are the roots of x* f 12.r 3
[The left-hand side
+ 54x 2 + 96x
f
40 = 0.
(Cf.
Ex.
13.)
is
5H(z 4 + 12* - 5).] Find a substitution of the form y=x 2 +px + q which 3 2x - 1 to the form y4 -f fy 2 + g = 0. equation x* + x 20.
will
reduce the
that one such substitution is y =x 2 - x - 1, this reducing the equation 2 Use this to solve the equation. (Cf. Ex. 10. ) (2y) + 26 (2y) -11^0.
Show to
4
21.
Find a substitution of the form
y=x 2 +px + q
which
equation to the form
2 y*+fy + g=Q, showing that one such substitution
this reducing the equation to
4(1
is
will reduce
the
CHAPTER
XIII
Lower
class
A
Upper
O
class
A
P FIG. 30.
The system
of rationals is thus divided into
A
called the lower class
the rationals to the
two
parts,
which
will
be
upper dass A'. The class A contains all and the class A' contains all those to the of P,
and
left
the
A
is less tlutn any number in A'. Two right of P. Thus, any number in or does distinct cases arise, according as the point not, represent a does,
P
rational number.
P
to represent some rational, 3 for example. The Suppose contains every rational less than 3, and the class A' contains every rational greater than 3. The number 3 may be assigned to either dass. If First case.
class
A
3 belongs to A,
number For
it is
the greatest
number
of this class,
and there
is
no
least
in A'.
if a' is
A
supposed to be the least number in yationals exist which and greater than 3, and which therefore belong to the ',
are less than a' class
A
.
Similarly if 3 is assigned to -4', it has no greatest number. the class
is
the least
number
of this class,
and
A
Thus corresponding the
dass
A
to
has a greatest
any rational, the classification is such that number or the dass A has a least number. 1
either
DEDEKIND'S DEFINITION ticcond caw.
number.
201
P does not represent a rational be equal to the side of a square whose area
Suppose that the point
For example,
let
OP
square units. Classify the rationale according to the following rule The lower class A is to contain all the negative numbers, zero and every 7
is
:
number whose square
positive
The upper greater than
The
A'
is
is less
than
7.
to contain every positive
number whose square
is
7.
such that
classification is
Every
(i)
class
rational is included in one or other of two classes
For
A, A'.
exists whose square is equal to 7. number in A is less than any number in A. For if a is any (ii) Any number A and a' is any number in A', we have a 2 <7
no rational
,
a
so
The
(iii)
number.
and we
class
A
has no greatest number and the class
have to find b (greater than case
A
has no least
2 For, suppose that a is the greatest number of A, then a <7, 2 can find a rational b such that 6>a and 6 <7. To do this, we
that & 2
a), so
if
7
-a 2 <7
- <-s -a 2
7
6-flK-r
a2
.
This will be the
-a 2 .
2a
b -fa
Thus, 6 belongs to the class A, and a is not the greatest number in A. Similarly it can be shown that A' has no least number. This classification
by is
-s/7,
is
and we assign
said to define the irrational number which we denote a place on the scale with rationals as follows ,J1
it
to follow or to precede
:
any
positive rational a, according as a
iJJKDedekind's Definition.
2
<7
or
Suppose that a certain rule enables
us to divide the whole system of rationals into two classes, a lower class A and an upper class A', so that any number in is less than any num-
A
ber in A'
.
Two
cases arise
:
A
has a greatest number or if A has a least number, the tion defines a rational number, namely the greatest number in (i)
least (ii)
If
'
number
A
A
or the
.
no greatest number in A and no least number in A', the defines an irrational number which is to follow all the numbers
If there is
classification
in
in
classifica-
A and to
precede
all
those in
A.
number
is called a real number^ and the and is irrationals known as the system of real numbers. aggregate of rationals
Any
rational or irrational
EQUALITY AND INEQUALITY
202
The
be called the classification or the section (A, A'), and the real number defined by it may be known as the classification just described will
number (A, A'). The real number
zero is defined
A
by the section (A, A), where
A
contains
the negative rationals and contains all the positive rationals. A contains some positive rationals, the section (A, A') defines a
all
If
positive real number.
A' contains some negative rationals, (A, A') defines a negative
If
real
number.
We we
'
'
are not justified in regarding a real number as a have given fresh definitions of equality, inequality
'
'
number until and the fundadefinitions must be in
mental operations of arithmetic. Moreover, these agreement with those already given for the system of rationals.
equal If
Equality and Inequality. Two real numbers are said when they are defined by the same classification of rationals.
a and
precede
/J,
// a and
/J
are real numbers,
we say that a j8
is less
and some of the than j3 and that j8
are unequal real numbers, infinitely
to be
rationals
which follow a
is greater
than a.
many
rationals lie between
them.
For (B,
B
a
let
f
)
rational r lies between r belongs to
a and
jS.
each of the classes
Then r follows a and precedes jS, therefore A' and B. If r is the only rational between A'
ex
B Fio. 31.
a and
,
then r
Art. 2,
is
the least
by number r, Thus at least two, and andjS.
(ii)
in
A and the greatest in B.
therefore infinitely
(J5,
Hence,
B') defines the
many,
Each
Let
A and A'
be two classes of rationals such that
class contains at least one
Any number in
same
impossible, for a
is
(See Ch. II, 10.)
Theorem. (i)
number
(i), each of the classifications (A, A), and we should have
number.
A is less than any number in A
1 .
ENDLESS DECIMALS
203 .
(iii)
Two numbers a and
can be chosen from
a'
*
**
A
and A'
respectively, so
that a'
-a<,
is any positive number we may choose, however Then there is one and only one real number a such any number in A and a' is any number in A'.
where
is
a
that
where a
Divide the system of rationals into two classes according to the The lower class is to contain every rational less than any A The upper class is to contain every rational greater than
Proof.
following rule number a' in
any number a It will be tion.
small.
For
:
f
.
A.
in
proved that not more than one rational can escape classificatwo rationals Z and V (l
if
This contradicts the hypothesis, for a'
so that either a or a' lies between
we can choose a and
a! so
that
-a
and
I'.
one rational which escapes classification, then it is to be the upper or the lower class. to either Thus, by Dedekind's assigned a number defines real the a which satisfies the classification definition, If there is
Conditions stated above.
This theorem
may
also be stated as follows
:
If (a n ) and (a n ') are sequences of rationals such that (i)
and
(ii) it
a 1
is possible to
<
3 ...
e is
number 5.
...
find n such that <*n
where
...
-<**<>
any positive number we may choose, however small) then one and only one, exists such that # n ^a
real
'
a,
Endless Decimals.
Using the ordinary notation, a -a 1o 2 ...a n ...
let
be an endless decimal, where the successive figures are formed according to
some Let
definite rule.
dn = a
a^
The sequences (d n ) and theorem, and therefore a
We o
. . .
aw
and
dn
'
= d n + 1/10W
.
f
obviously satisfy the conditions of the last 8 is defined by real (d n
)
number
say that the decimal represents or
is
equal to this real
number
S.
B.O.A.
FUNDAMENTAL OPERATIONS
204
Conversely, any real number 8 can be represented by a decimal. For, using the same notation as in the preceding, if 8 is known, then for any
we can find d n so that d n <8
suffix n,
to any
That
'.
Of course, a terminating decimal
to say,
is
represents
we can
calculate
S.
be regarded as an endless decimal
may
in which, after a certain stage, all the figures are zeros.
Since
any
by a decimal which terminates or that a non-recurring endless decimal represents an irrational
rational can be represented
recurs, it follows
number, and conversely.
The fundamental Operations of Arithmetic. follows, a Greek letter denotes a real number, a small
In what
Roman
letter
Real numbers will be defined by using the theorem
represents a rational. of Art. 4, and when
we say that a
number a
real
is
defined
by a^a^a',
presumed that a, a' are any numbers in classes A, A! of rationals which satisfy the conditions of the theorem.
it is
(1)
Addition.
then a -f j8
is
If a,
j3
numbers defined by a
by a -f 6
defined
This classification defines a real number, for (i)
(ii) (iii)
There
Any
We
a
is
at least one a
-f
b
< any a' +
can choose
e is
and one
a'
+ b'.
b'.
that
a, a', 6, b' so
a<^e
a'
where
6
-f
and
b'
6
,
any positive number, however small, and then - (a 6) = (a - a) + (V - b)<. (a -f b') f
f
-f-
We
define
a -f /? + y as meaning
(a
-f jS)
+ y, and
it is
easy to show that
the commutative and associative laws hold good. .
It
is
Show
1.
obvious that a
and
j3
= a. b< 0<
-Ha are defined
defined
by
positive rational. Let a be defined
by
is
a< a<
&',
a',
since every 6
where
-f
6<
classification of rationals.
is
is
any negative rational and
defined
+ 0<
a'
b' is
any
by
-f 6',
X
is
negative and every 6 positive,
a + 6<
Thus a +
6
then a + a
and
by the same
Prove that a +
x. 2.
Zero
-f ft
a< a' -f 6'.
and a are defined by the same
classification of rationals
and are equal.
MULTIPLICATION OF IRRATIONALS If
.
3.
a
is
Prove that defined
by
205
a + ( - a) =0.
a< a< a',
then a +
(
a - a'< a
- a) 4- (
defined
is
by
- a)< a' - a.
every a< any a', therefore a a'< 0< a' a. Hence a + ( -a) and are defined by the same classification, and
Now
We
(2) Subtraction.
If a,
(3) Multiplication.
/?
are positive real
and
a^oc^a' aj3 is
Whence
define
follows that
it
for
(a-j8)+j8=a,
then
are equal.
defined
numbers defined by
6
These conditions define a real number,
by afe
for (i)
(ii) (i\\)
There
is
at least one ab
Every a6
a'b'
We can choose a,
6, a', &'
For
positive number.
and one
a'b'
.
.
a'b'
so that a'b'
-ab<,
-ab = a'(b' -b) + b(a'
a
a
6'
>
/}
where
is
any assigned
-a).
a'
'//
k
FIG. 32.
Choose rationals
a'6'
We
can
now
choose
For
a'6'
zero
- ab
a, 6, a', 6', so
6'
and then
-
^/7
Show
4.
that a'
-a
= = 0. a,
a(-j8H-aj8 = (~a) and (-)( -J8)=aj8. It is easy now to show is taken to mean (
that
defined by
that the
and associative laws hold good.
^1.^1 = 7.
a< N/7<
a',
where
rt,
a2
Hence *y7 ^7 is defined by a2 < 7< .
a'
negative factors, the definitions are
commutative, distributive
is
if
b)
and
b
a.
^ Ex.
and &>/J, then + k(a' - a).
- ab<.
and
Further, a/Sy
A>a
that
h, k, so
a' 2 ,
a' are
any
< 7< a'
and this
positive rationals such that
2 .
classification also defines the
number
7,
POWERS AND ROOTS
206 (4) Division.
If
a
is
a positive real number defined by
a^a^a', then
is defined by l/a'
I/a
There
(i)
We
(iii)
and one
at least one I/a'
is
can choose
Every l/a'< any I/a. so that l/a-l/a'<, where e is any assigned
a, a'
I/a.
(ii)
-
positive number.
To prove
this,
we have
show that
to
"i
a, a'
7*
can be chosen so that
a
;
^
p IG f
-a
choose a, a
First choose h so thaia'
"~~l
'
so that a>/i
and
a
h
NOTE.
Tliis
Consequently
JSfo.
it
5.
meaning
is
follows that
// a,
and
a'-a
reasoning depends on the existence of positive rationals
rio
j9
l/(
(a/j3)
.
less
than
a.
to 1/0.
assigned
Further definitions are
whence
h 2
we have
= - I/a and a/j8=a l/j8 for (a I//?) jB =a(l/jB /? =a -a)
.
.
;
are positive real numbers defined by
.
;
.
jB)
a< a< a' arw2 b< j8< &',
=a.
a/<
a' ft. defined by a/6'< This follows from the definition of
7.
Powers and Roots.
and multiplication.
l/j3
a positive integer and a a real numdefined by a n ==a a a to n factors. n m n m + follows that a .a =a and (aw ) w =amn
n
If
n ber, the nth power of a (written a )
is
is
.
.
. . .
m, n are positive integers, it That these formulae may hold for zero and negative values we must have a = 1 and a~ n = l/an
If
.
of
m and n,
.
Theorem
1
.
(i)
// a
x, x' positive rationals
is
a positive real number, n a positive integer and
such that
n
,
then positive rationals
xl9 #/
exist such that
For
if
We
can choose ^ 1 >x, so that
x
then
x x w - xn
The existence
of a
number x
~#<(a -xn )fnx' n
xx ,
so that
such that
~l :
and then
x^
a
can be proved in a
similar way. (ii)
//
is
any
positive number, however small, then
For as in the preceding, x/ n
~x1 w <(x1
/
-ajj)
.
n n x/ ~x 1 <, provided
nx^- 1
.
PRINCIPAL ROOTS Theorem number
,
207
a positive real number, there and one only, whose n-th power is equal to a. 2.
//a
If a = z n where #
is
is
a positive rational, there
is
is
one positive real
only one such number
x,
for the nth If
powers of unequal positive rationals are unequal. no such x exists, divide the system of rationals into a lower
class
A
and an upper class A' according to the following rule The class A is to contain every negative number, zero, and every positive :
',
number x such that x n
that x n =a. (ii)
Every #
(iii)
There
is
x',
for
sc
n
no greatest x and no
n .
least
x
f .
This follows from the last
theorem. Therefore the classification defines a real number Again, for every
x and every
.
x'',
xn
xn
and
.
Also every rational is an x or an x', therefore these may be chosen so x is as small as we like. Hence, as in Theorem 1, they may be where e is any positive number, however small. chosen so that x' n -xn
that x'
<,
n are defined
Thus both a and
by the same
classification of rationals,
a classification which satisfies the conditions of the theorem of Art. 4 and n
therefore
=a.
a positive real number, the principal n-th root of a defined as the positive real number whose nth power is equal to a. This Definitions.
is is
If
written J^a, and
a
is
is
generally called the n-th root of n
Thus
(;/a)
a..
=a.
It follows that
ya.y]8-y(aj8) If
n
is
odd,
we have (-
When n
is
WH^a^ y(.
and
an odd
simply the n-th
n==
(-
1 )n
)
w== -a.
we therefore define the -a as - ^a thus
integer,
root, of
(^a
principal n-th root, or
;
even, the 7^th power of every real number is positive, and therefore no real number exists which is the nth root of a negative number. If
n
is
SURDS
208 Indices
which are Rational Fractions.
If p, q are positive integers,
?
aq
is
p defined as *]a* or (!ja) This definition assigns no meaning to a x is
.
considered in Art.
Surds.
8.
If
a
when x
is
This case
irrational.
9.
is
not a perfect nth power, ^Ja
is
called a surd of the
nth order.
Two rational
surds of the same order are like surds
when
their quotient is
otherwise they are unlike surds.
:
The following theorems and examples depend on the number cannot be equal to an irrational.
Theorem
1.
If
x+Jy^a + jb
fact that a rational
x, y, a, b are rationals, then
where
x=a
and y = b,or else
y and b are the squares ofrationals. For suppose that x-^a and let x = a 4- z,. then z + Jy
Jb,
and by squaring,
Therefore ,Jy is rational, and from the given equation the other hand, if x = a, then y 6.
is
Jb
rational.
On
Ex.
1.
// a
+ b^/p + c*jq
where
a, 6, c are rationals
and Jp
t
^/q are unlike surds,
thena = Q, 6=0, c=0. By transposing and squaring we can show that 2 c*q ~a b~p. 2ab^/p
If
ab^Q, the left-hand side would be irrational and the right-hand side would
be rational ; which
is
impossible.
Therefore a6=0, and consequently a
or
6~0.
a=0, then b*Jp + CsJq~Q; and, if 6^0, then Vp/\A?~ -c/6; and which is not the case. */p *Jq would be like surds Therefore 6=0 and c = 0. //
so
that
;
7/6 = Ex.
2.
then a // a
-{-
~ c^q ~ 0, and therefore a
+ 6^/p + c^ 2 =0
wAere
a, 6, c,
and
#
c
= 0.
are rationals and
p
is
not
a perfect cube,
then a, 6, c are all zero.
Multiplying the given equation by %Jp
cp +a?Jp
and eliminating the terms containing Since ^/p
is irrational, it
therefore
c^Q we should have p 2 =(
fy
= ac and c 2p 3 2 2 2 c*p = a 6 = a c.
-
) \ c/
Hence c=0, and
+ 6Ap 2 =0,
follows that b2
If
we have
therefore also
,
so that ^/p 2
a=0
ab,
would be
and 6=0.
rational,
which
is
not the case.
IRRATIONAL INDICES Theorem
2.
Suppose
square, then if
p + *Jq
is
For we have
Since
Jq
is
Putting
a polynomial with
2 &ndf(x) = Q{(x-p) -q}
9.
Irrational Indices.
9
if
f(x)
is
identity of the form
x=p + Jq, we
irrational it follows that
=
a perfect
a root off(x)
we obtain an
are rationals.
is
0//(#)=0 where f(x)
is also
divided by (x-p)*-q,
S
root
that q is not
= 0. p-*Jq 2 = + {x-(p Jq)}{x-(p-Jq)} (x-p) -q, and
rational coefficients,
where R,
and
that p, q are rationals
a
209
have
Rp -f S = and R = 0.
p-Jq
therefore
The theorems
Consequently
a root of/(z)
is
of Art. 7 hold
= 0.
when a
is
a
number, the indices being rationals, and lead to the following Definition. Let a be a real number greater than unity, and let be an
positive real
:
irrational defined
by x
f
x belong to classes of rationals
x,
Then
satisfying the conditions stated in Art. 4.
a*
is
defined
by
These conditions define a real number, for (i) (iii)
there
at least one a x
is
'
and one a x>
we dan choose x and
x every a*
(ii)
,
' x' so that a x
-
ax
<
where
is
any assigned
number, however small.
positive
To prove
this,
ax Next, by Ch.
'
then a x
&>,
choose a number
-a x = a x (a xl -x - l)
II, 18, (7),
find a positive integer
n such that
l
an
NOTE.
The
'
-a x
At present
7/0
a*
is
defined
show that ax .
hold for
all real
it
way, is called its from other values which will be found
of will stand for its principal value.
by the
defined as (I/b)*, where 6 = I/a. able to assign a meaning to a* It is easy to
;
real positive value of a*, defined in this
principal value, to distinguish later.
.
x<.l/n, and then
Finally, choose x, x' so that x'
ax
-l
if
classification
We
ax '
define 1* as
when a
is
1.
negative.
a>0, the index laws
ay
= a*+* and
values of x and y.
(a
x y )
= axy
At
*
Or
present,
it
we
may
be
are not
OPEN AND CLOSED INTERVALS
210 10.
and
a^l,
a single real number
there exists
N
If a and
Theorem.
Logarithms.
are positive real numbers
such that a*
= N.
N
x and first let Suppose that no rational x exists such that a = x x '. n>l. Then, by Art. 7, rationals x, x' exist such that a Divide the system of rationals into a lower class A and an upper class A'
Proof.
by the following rule The class A is to contain every rational x such that ax
;
;
There
(iii)
is
no greatest x
the last article,
a*
we can
Similarly
;
for
find x2
if
>x
x
ly
supposed to be the greatest, as in so that a^ aPK^N i.e. so that is
-a^
can be shown that there
it
classification therefore defines a real
Again, for every
x and every
is
no
least x'.
number
.
#',
a x
and
a*
;
and since every rational is an x or an x', these may be chosen so that x -x is as small as we like. Hence, as in Art. 9, they may be chosen so that a x -a x <, where e is any positive number, however small. Thus both N and a* are defined by the same classification of rationals, f
'
a classification which If
a
satisfies
the conditions of Art.
find f so that (l/a)*
= l/N, and
4,
and so
a^
= 2V.
then a* = N.
any positive real number a is chosen as base and N is any positive real number, the number given by the relation a*~N is called the logarithm of N to the base a, and we write f = log a N. Definition.
Definitions. We say that between a and 6 forms the (open)
1 1
lie
.
(ii)
The
real values of
or the range (iii)
If
closed at '
a^x^b form
(x)
which
a
the closed interval
(a, 6)
the real numbers x form an interval
(a, 6)
open
at
a and
or the range (a <#<&). '
'
generally means open interval/ number x in the interval (a, 6) is often called a point in the interval.
Interval
Any
x such that
the set of real numbers
(i)
interval (a, 6) or the range
a
a
'
'
If
x l9 x2 are any numbers in the interval (a, 6) and/(x) is a function x such that f(xl )^f(x 2 ) when x 1
of
case/(x)
,
is
constant in the interval.
DEDEKIND'S THEOREM )
when x^x^
then
211
said to increase steadily in the
/(a?) is
stricter sense.
f(x l )^f(x 2 ) when #T <#2 f(x) decreases steadily; and
If
if
,
when x l <.x2 ,f(x)
f(x 1 )>f(x^)
decreases steadily in the stricter sense. 9
'
is the set of a real number, the neighbourhood of a real numbers in the interval (a-e, a + e), where e is as small as we like.
a
If
(v)
is
12. Sections of the
System of Real Numbers.
Dedekind's Theorem.
vx(i)
// the system of real numbers
is
divided into
A
and A' such that (i) each class contains at least one number, number belongs to one or other of the classes, (iii) any number (ii) every real in A is less than any number in A' then there is a single real number a, such that all numbers less than a belong to' A and all numbers greater than a belong The number a may be regarded as belonging to either of the classes. to A' two classes
9
.
Consider the rationals in
Proof.
and AI, which form a section
number
a
If
(1)
Two
a. is
A
of the
and A.
system
These form two classes
and define a
of rationals
A
l
real
cases arise.
rational,
it is
the greatest
number
in
Al
the greatest number in A v it is also the greatest suppose that there is a number ft in A greater than a.
or the least in A^.
number
in A. For Between a and /? there are rationals greater than a which therefore belong to A and also to A'. Hence j8 must belong to A', which is not the case. Similarly if a is the least number in A^, it is also the least in A. If
a
is
If
(ii)
a
is irrational, it is
greater than
all
the numbers in
Al
and
less
Also, a must belong to A or to A, and just as in the show we can that it is the greatest number in A or the least in A' preceding,
than
all
those in A^.
The theorem
just proved
section of the system of real
is
of great importance.
numbers
defines
It
shows that any
a real number.
Thus the con-
sideration of sections of this kind leads to no further generalisation of our idea of number. All this is sometimes expressed by saying that the system
of real numbers is closed, or the aggregate of real numbers is perfect. On the other hand, the rational numbers do not form a closed system, or in other words, the aggregate of rational numbers is not perfect, for a section of the system of rationals does riot always define a rational. (2) // the numbers in the interval (a, b) are divided into two A' as in (1), the section determines a real number a.
classes
A
and
9
For
if
we take with
A 1
all
all
the real numbers x such that
the real numbers x such that x'^-b,
of real
numbers.
we obtain a
x^a
and with A'
section of the system
ARITHMETIC CONTINUUM
212
1 that if we take any we in a straight line as origin and any length may choose as unit point of length, then to any point P in the line there corresponds a real number p.
The Continuum.
13.
from Art.
It follows
choosing suitable axioms regarding the characteristics of a straight line we can ensure the truth of the converse of this statement, namely, that to every real number p there corresponds a point P in the line. Thus
By
the correspondence between the points of a straight line and the
numbers If
that
p p
is
complete. the real number corresponding to the point P of the line, we say Hence the system of real is the measure of the length of OP.
is
numbers This
real
is
adequate for the measurement of the length of a straight
an instance of what
is
The aggregate
of real
is
known
numbers
as a continuous magnitude. called the arithmetic continuum,
is
of the points of a straight line is called the linear
the aggregate
line.
and
continuum.
RATIO AND PROPORTION NOTE. small
In
Arts. 14-16 capital letters will denote concrete magnitudes
and not numbers
;
denote positive integers.
letters will
Equality and
Inequality of Concrete Magnitudes. In how a magnitude or a quantity of any kind is to be measured, considering the at outset we must define what is meant by saying that 'A is equal to 14.
A
and B are quantities of the greater than or less than J5/ where of these statements depends on the particular kind of quantity we are considering. For instance, let A and B denote segments
J5,'
'A
is
The exact meaning
kind.
if A can be made to coincide with J5, we say A = B can be made to coincide with a part of B, we say and B>A. The student should reflect on the meaning of equal angles, equal
of straight lines if
;
;
A
A
velocities,
equal forces, equal quantities of heat, etc.
We
15. Ratio.
number
said to contain
This
is
B
able,
C
exactly
Definition.
(that
B
can be divided into any
Denoting any one of these parts by C, B is n times, and C is called the nth part of B. (exactly)
expressed shortly by writing
nth part of
a and
assume that a magnitude
(n) of equal parts.
we denote
C = ^B.
this
If.
by writing
A A
contains the
B.
A and B are magnitudes of the same kind, and if integers such that A contains the fcth part of B exactly a times
A~j BJ,
and the
times,
or
If
b exist is, if
m
B^nC
ratio of
the magnitudes
A
to
B
(written
A
A
and :
B)
B is
are said to be commensur-
defined as the
number
a/b.
RATIO AND PROPORTION If
no integers a and
6 exist such that
213
A = ^B, the magnitudes A
and
B
are said to be incommensurable.
Axiom
of Archimedes.
If
A
no matter how small
A may
always be found such that
mA>B
kind,
16.
Two
Ratio of
B
are magnitudes of the
same
be compared with 5, an integer
m can
and
or
^B
Incommensurables.
A
If
and
B
are incom-
mensurable magnitudes, the ratio of A to B (written A B) is defined as the irrational number determined by the following classification of the :
entire
the the
system of rationals
Lower
Class Class
Upper
:
is
to contain every rational a/6 for which
is
to contain every rational a'/b' for
This classification defines an irrational (1)
every rational
hypothesis, there
(2)
(3)
every a/6 there
is
hypothesis
is
falls
than any
no greatest a/6
bl
A>a
l
;
because
into one or other of the
no rational a/b
is less
number
;
for
a'/6'
;
which bA for
bA>aB,
which b'A
two
= aB
classes;
for,
by
:
^B
:
for suppose ajb^ to be the greatest, then
by
B.
Now by
the axiom of Archimedes, a multiple of a magnitude (however small the magnitude may be) can be found which exceeds any magnitude of the same kind, however Therefore an integer r can be found such great.
^-a
JS)>a 1 J5, and therefore rb l A>(r + 1)a 1 B. Let r61==62 and (r-fl)a 1 -a2 then b^A>o 2 B and a^-a^r-f l)lrb{> ajbl9 so that be shown that there is no ajbi is not the greatest a/6. Similarly it can number. irrational an defines Hence the classification least that
r(6 1
1
,
a'/6'
Theorem.
If
A, B, C,
D
are four magnitudes of the
same kind, the
ratio of A to B is equal to that of C to D if mA> =oi
mC^nD and A/B = C/D = n/m. If no such rationals m, A, B and also C, D are incommensurables, and the ratios
n
exist,
then
A/B, CjD are therefore are and irrationals defined by the same classification equal. = C:D, A:B and kind same the of If A, B, C, D are magnitudes then A, B, C,
D are said to be in proportion. * This
is
Euclid's definition of the equality of
two
ratios,
SQUARE AND OTHER ROOTS
214
EXERCISE XXII
IRRATIONALS is
1. Show that no rational exists whose nth power is equal to ajb, where a/6 a positive fraction in its lowest terms unless a and 6 are perfect nth powers. /x \** a - where x/y is a positive fraction in its lowest terms. [Suppose that ( j =
Then bxn =ayn where k
is
Therefore
Since y
prime to a whole number. Then .
is
a~xn and byn Jx + *Jy = */a + >Jb ,
that
is
y
x,
to say,
then either 2. If 9 Ja, */b are all rational or all like surds.
Hence a and b are .
xa, y=b
and *Jxy = */ab, unless
[Squaring, x + y=a + b, 3.
n must be a divisor of
akxn
^fa
where
a, b are
in terms of
b
= kyn
x
for
b,
Va6 are
y=a,
,
6.
or *Jx, *Jy,
rational.]
not a perfect square.
Also find x and y
b.
[Proceed as in Ex. 4. If
is
Let
exist such that
x, y may + ^/b = *Jx + *Jy,
given rationals and 6
a and
6.
a is prime to perfect nth powers.] Jk
or
*fxy,
Find the condition that rationals
= l,
2,
and show that x
a \(a dbVa -6).]
a + b*Jp + c*Jq + d*JpqQ where *Jp
9
*Jq are unlike surds,
then
a, 6, c,
d
are all zero.
[We have Hence by
Art. 8, Ex.
1,
ac=bdp,
and consequently,
if c
^0 and a ^0,
bc=ad, c*=dp, then p=c*.]
5. If x + $y=a + $/b where x, y, a, b are rational cubes, then x=a and y 6.
[Put
z= a; -a.
rational.
Show
Now use
that 2 8 H-y~6
+ 32\/6y =
a$lp* + b$]pq + c$q*Q that a, 6, c are all zero.
and neither of
7.
cients,
not zero, since pjq
2
2
j? ^, j?^
_+ _
3 z 2 [Show that C*q*-a*p -b pq = 3db(ap\/pq
is
.*.
;
y, b are
not perfect
$y=mg/b 2 where
m
is
Art. 8, Ex. 2.]
6. If
in brackets
and
is
bq\/p
2
q),
is
a perfect cube, show
and that the expression
not a cube.]
Show
that *J2 and v/3 are cubic functions of J2 -f *J3 with rational and that ^2-^/6 + 3 is the ratio of two linear functions of *J2 -f \/3
coeffi-
=
8 [For the first part, let y s/2 + \^3 ; find t/ , and eliminate \^3 and V2 in turn : for the second part, find the value of the product ( \/2 - V6 + 3) (V2 -f N/3 + a:), and show that this is equal to y + 5 when x 1.]
=
8.
root
Find the equation of lowest degree with rational
is (ii)
[(i) (ii) (iii)
If * =
&2 + 3^/4;
(iii)
2 N^ + %/SL a: -l=2\a;, etc. 8 if y=#2 + 3#4, i/ = 110 + 18(4/2 + 8 =5 + 34/6.2, and (28 -5) 8 = if z=4/2 + 4/3, ;
coefficients of
4/2 + 4/3.
which one
RATIONAL COEFFICIENTS 9.
If all the solutions of
as 2 + 2hxy + 6t/ 2 = 1,
a'x 2 + Zh'xy + b'y* = 1
then (h -A') 2 - (a -a') (6 -&') and - a'6) 2 - 4 (aA' - a'A) (W - h'b) are the squares of rationals.
and
are rational (oft'
215
a, A, b, a', h' 9 b' are rational,
[Let (a lf j8i), {a 2 , are the roots of
j8 2 ), (
-
i,
-
ft) (
-
-
2>
^2)
be the solutions. Prove that
~
,
,. /
y
show that a^, a a j3 8 are the roots of - 4 (a*') A6')} + 2z { (a - a') ( A6') - (6 - 6') (aA')} + (a - a') (6 - 6') = 0, - a r&. Hence show that =ab' where (ab') Also,
z 2 { (oft') f
(
aiaa
/ .
whence the second 10. If a, 6, x,
prove that
i )
/
-4(aA
f
/
)(A6
)
result follows.]
y are rationals such that
eiJAer
a;
a,
yb,
[Put*-a = -S:,y-&==r; If
(o6
or I .'.
-ab and
1
-cry are the squares of rationals.
(aY -bX)* + 4XY=Q.
X X and 7 are not zero, solve for ~
Next observe that the given equation y and 6.]
is
,
and show that
1
-ab
is
a perfect square.
unaltered by the interchange of x and a,
CHAPTEK XIV INEQUALITIES IN
this chapter, various
of the results are of
Many
of dealing with inequalities are explained.
fundamental importance.
Weierstrass' Inequalities.
1. less
methods
than
1
whose sum
Ij a v a 2 denoted by s n then
is
,
a n are positive numbers
...
,
+sn ), -s n ), where, in the last inequality,
For
(1
and continuing
-a
x ) (1
thus,
it
supposed that s n
is
- a2 ) = 1 -
(a l
+ a 2 + a x a 2 > \-(a l + a 2 ), )
we can prove that
(l-o 1 )(l-o 8 )...(l In the same
way
Again, 1 -a 1
and
-a,)(l -o.)
if s n
(1
2.
-a^,
...
(1
-a n )
...
(1
H-o n )
) ...
Many
inequalities
depend on
-00(1 -o,)
...
Ex.
2.
For
Using
+a n )
7/a>0 and 6>0,
// a,
6, c
(a
then
are positive
and
(a
(1
-)
the fact that the square of a real
positive. 1.
0
+ 0l )(l +az)
Ex.
(1
for
+ b)>*Jab.
not all equal, then
(a + b + c) (be + ca + ab) >9abc. + 6 + c) (be + ca 4- ab) - 9a6c
number
is
TYPICAL METHODS 3.
A
Ex.
1.
We
have
217
special arrangement of terms or factors is sometimes useful.
n
Show
that
[n>w
(|^)
and r(n~r + l)>n
8==1
if
r2
2 .
n 2(n-l) 3(n-2)
...
.
.
-r(n + l) + n<0
that
;
r(n-r + l)
is if
...
n
.
1,
~n)<0
(r-l)(r
or
if
l
Therefore
j&e. 2.
are positive numbers
6, c
// a,
any two of which are
together greater than the
third, then
Now,
since
a(c
+ a-b)(a + b -c)>0, we have
a(a + 6 -c-f c-f a-6)>2(c~f-a ~6)(a
if
which
the case.
is
greater than
4. If a,
Similarly, l/(a
2/6, 2/c respectively
x are
6,
For since
6 (6
according as ab
+ x)/(b + x)^a/b,
that
+ c - a) + l/(c -f a - 6),
l/(6
result is obtained
(a -f x)/( b
positive,
+ bx^ab + ax,
o 2 >a 2 -(6-c) 2 ;
or
is,
by
are
addition.
according as
a$b.
+ x)^ a/b
according as
ax^bx
or a $6.
Prove that
r. 1.
^
positive, then (a
-f x) is
+ 6-c)
+ a-6) + l/(a + 6-c)>2/a,
+ 6 - c) + 1/(6 -f c - a), and the
;
l/(c
8
*
1
..
(2*-D
,
A then
'
/o
AI Also
(2*
4 6 >- -.
+ ix !)^-.-...
^
Therefore
-
*^
and
'
5. If both sides of an inequality are symmetric functions of a y no loss of generality in assuming that
6, c, ... A, k,
there is
Ex.
"
1.
6, c
// a,
are all positive
and
n^O or n^
-
1,
then
a n (a-b)
an
let
w>0.
n
n
^b ^c
therefore
whence the //
On
account of symmetry, we may assume that a^6>c. and a n (a-6)(a-c)>6n (a-6)(6 -c), also c n (c-a)(ca n (a-6)(a-c)+c n (c-a)(c-6)^6 n (a~6)(6-c),
Hence,
result in question. 2;a n (a-6)(a-c)=Z(a-6)(a-c)=27a2
n=0,
27a 2
and -l,
let
easily seen that
n= -w-1,
-
-.
2:6c
<
-276c
so that
Ean (a - 6) (a - c) = -r-
m^O. 2?a
Let a = l/a, 6 = l/jS,
m (a -
j8)
(a
-
y)
^
c
= l/y,
then
by the preceding.
it is
INEQUALITIES INVOLVING POWERS
218 6.
//!, a2 ((!&!
-f
,
a n and bl9 62
...
let
a 2 b2 +
A=
bn are two sets of real numbers, then
...
2
2
2
-f a n ) (6 X #tA) 2:^( a i + a 2 + = = occurring only when a^Jb^ a^jb^ -f
. . .
the sign of equality
For
,
. . .
B = 2arbr C = 2b rz then 2 Z(ar + A6 r - A + 2 A
a r2 ,
,
,
have a r -h \br =
for every
is
the
7. If a 19 a 2
. . .
= a n/b n
. . .
-f
6n2 ),
.
+ A2 C.
the sign of equality occurs,
we must
so
2
of
,
also follows
sum
from the identity
\n(n -
of the
1)
squares of the form (a^g
a n are n positive numbers, not
...
all
equal
according as x and y have the same or opposite signs. First suppose that x and y have the same sign. Let
the numbers
-f
-f
The theorem
where S
and
62 2
A + 2XB + A C = are equal and B 2 = A 2A$ + A2 C>0 for all values of A, and therefore B2
Hence the roots Otherwise
r,
If
-f
for all values of A,
)
Therefore ^4-2AJ5+ A 2 C^O.
2
1, 2,
negative or zero.
...
n,
then a rx -a sx and arv
Hence a
x
(a r
x+v
to
- a 2 ^i) 2
-
one another, then
r, s
be any two of
are both positive, both
-a/ -a8x )(ar y -a8 v )^0, and
x y v v -h a*+ ^*a r*a 8 + a s a r
There are 2^(^-1) relations of this sort, not occurs in n - 1 of them and a r xa s v in only one.
consequently
.
all equalities.
Also a rx + v
Hence by addition
(n- \)Za r Therefore
nEa
x +v
x +v
>Z:a rxasy where s^r.
> Sa x +v + Safa* = Za x r
If x and y have opposite signs, then (a r be substituted for in the preceding.
x
.
2a rv
.
- a,*) (a r y - a/)<0, and
< must
>
8. The inequality in Art. 7 can be written
n Hence to
it
follows that if
one another, and x,y,z,
a^ a 2
...
,
...
a n are n
positive numbers, not all equal
are all positive or all negative, then
27a r +y++...
n
n
n
Ea ^
n
Zgv Zaf ^
n
n
FUNDAMENTAL INEQUALITIES
**-
9. If a is any positive number except then First suppose that
we have (a
,
_
l)/p
_
(aP
_
= (a - l){pa~ l - (a p
p, q are positive integers.
_
V)/(p
~
If
1}
4-a*~ 2
l
y are positive rationals,*
x,
x>y.
if
x~p, y = q where _i
and
1,
219
+
... 4-
l)}/p(p
- 1).
Now, the expression in the large brackets can be written 1 a*~ 2 ) + (a*- 1 - a*- 3 ) + + (a~ l - 1 )} (a*. . .
{
and each term
of this is positive or negative according as a.^\. in either case, (a p -l)/p- (a** 1 - l)/(p - 1 )>0,
Hence, that
p,
(a
is,
p
l)/p decreases as
if
y are fractions, we
may
one or both of
q,
d are positive integers and p>q----
TT~ d
P/
p>q.
(a-l)/p>(a<*-l)/q If
cc,
and therefore
decreases,
j>
> --TT~~ d
x,
take x=p/d, y = q/d where have to show that
We A
>
.
that, is
9
P
?/
-
,
9
i
where b = a d NOTE.
6>0 and
This follows from the preceding, for
.
It has
been shown that
a>0
and ^1,
passes through positive rational x values, (a ~l)/x increases with x. Putting I/a for a, it follows that (a- ~l)/x increases. x and as decreases Further, putting increases, consequently (l-a-*)/* if
as
a;
x
i
i
we conclude that x(a x - 1)
Ifx for x,
decreases
~ Ex.
Ifa>l and n^l.then
I.
i
i
n(a
n
and x(l ~ a x ) 1
-l)>l
.
i
__ji
If
w>l, n(an -
If
n = \ the inequality becomes a -1>1 --, or a (a-
1)
=an .n(l-a n)> a^(l
- a~ 1 )>l - a- 1
9
1)
.
> a-1,
10. If a and b are positive and unequal and x except
unless
1,
then
Proof.
6)>a _
^-i (a ,
*
in which case
xa
x ~l
6x
>x6
x-i
(a-b)
P
x
(a
is
any
i.e.
2 (a-l) >0.
rational
number
_ 6)f
- bx
.
First observe that, in either of the cases, the truth of one of
the inequalities involves that of the other. * It
increases as x increases.
can be shown that
this is true
when
x,
v are positive
Thus, real
if
for all positive
numbers.
(See Ex.
XXIII,
and
37-39.)
B.CJU
FUNDAMENTAL INEQUALITIES
220
x ~l x x we may interchange (a-~b)>a -b unequal values of a and 6, xa l x - a x and therefore a x -b x a and 6, so that xb*~ (b a)>b >xb x ~ l (a-b). Thus it is only necessary to prove one inequality in each case. ,
',
Let a/b = k, then
x
(i) if
according as
x
(ii) if
is
- bx ^xbx ^ l (a - b) according as kx - I $x(k - 1), or x (k l)/xg(A 1)/1, that is according as xgl (See Art. 9.)
x positive a
is
;
negative and
equal to
-
y then xa
x ~l
y
- b) ^a x - b x according (a
-l
(k-l)>k-v-I, that is according as -y(k~l)^k-k y + v +l - 1 5 k (y + 1 ) - y - 1 or according as according as k (*+'-l)/(y + 1)2 (*-!)/!; -yk~ v
as
l ,
or
,
that
is
Now y>0, and
y + l^l, or t/^0.
(by Art. 9) according as
xa*- l (a-b)>a x
so
-bf*.
This completes the proof of the theorem. 1 1
.
When
6
= 1,
the second inequalities in Art. 10 become
ax
-I>x(a~I) -l
ax
Writing n for x and
and
1,
1 -f
x> -1 and 7^0,
x for
,
0
n
>l+nx
if
n<0
(l+x)
if
0
-x)
(l-x) Ex.
or
n
>l -nx
if
n<0
n
if
0
// m, n are positive and
1.
>1,
any
rational except
that
m
>1,
x
again, changing the sign of x, if (1
it
or
is possible to
>1,
find positive values of x such
(l+x)< this inequality holds
Now (1
if
or
follows that, if n is
it
n
For
x<0
JAen
(l+x)
Or
if
(1
+ x)- m >l -mx, or
+7ix)(l
~mx)>l,
Ex.
// sn = a + a 2 +
2.
1
Denote the left-hand
if
if
x>0
and
hence the inequality holds
x{(n-m) -mnx}>0, . . .
side
+ an
where a v a 2
by un then ,
r=n4- 5
HH-Un-^^-j*
if
,
or ...
if
if
1
-mx>l/(l +nx),
0
an are
all positive t
n>l,
*_
^(l+aja+o.)...
then
or
if
ARITHMETIC AND GEOMETRIC MEANS Now
s r
n
-^ n Li> ra n ^n r^Tl
1 >
when r>l
and when
(Art. 10),
r
221
= l,
this
becomes
an equality, therefore
un - u n _ l >an u n ^
therefore
Now
M!
>0
and
n=2,
for
n >(l
+an )u n _ >u n , 1
l
.
3, 4, etc.
Arithmetic and Geometric Means.
12. are
0, therefore w n
If
(1)
a,
k
c, ...
6,
n numbers and
A^ then
^4
and
+ c+...-h)
arithmetic
n G= J (abc
...
&),
mean and G the geometric mean
of
&.
Theorem.
13.
not all equal
For
the
called
is
...
a, 6, c,
-(a + 6
n
A
let
T/?e arithmetic mean of n positive numbers, which are one another, is greater than their geometric mean. he the arithmetic mean and G the geometric mean of the n
to
numbers
positive
a, 6, c, d,
Suppose that
a^any
other
number
...
A* ..................................
(A)
and that fc^any other
of the set
number, then Let
&'
na>a + b + c + ... +k>nb and a>A>b. =a + 6 - -4 so that ?/">0, and consider the set A,b',c,d,...k,
obtained by substituting A,
Thus
A + b' =a
-f
b
the same arithmetic
b'
for a, 6 in (A).
and Ab'>ab.
mean
.................................
We
(B)
have
Hence the numbers
of the set (B)
as those of (A), but a greater geometric
have
mean.
the numbers in (B) are not all equal, then A is neither the the least of the set. By repeating the process we can therenor greatest fore obtain a set of n positive numbers containing two A's, with the same
Again,
if
mean
as before, but a greater geometric mean. we shall finally arrive at a set of n numbers each equal of which the geometric mean is greater than (?, and since is the
arithmetic
Continuing thus, to
A
A
geometric
Thus
mean
for the
of these equal
numbers
10, 1, 2,
7;
The arithmetic mean
numbers,
it
follows that
A>G.
10, 1, 2, 7, the successive sets are 5,
of the
increasing from left to right.
6,
2,
7;
numbers
5,
6, 4,
5
;
5,
5,
5,
5.
in each set is 5, the geometric
mean
FUNDAMENTAL THEOREM
222
Taking the
Alternative proof.
set (A),
we have
Proceeding in this way, we can show that abc
k^
...
(1
U
{
(a
-f
V
6+
n
is
k)
}* >
if
...
a power of
2,
then
,
'/
so that
n
is
where
A
If
not a power of
2,
occurs r times
and n-f r
a b
consider the set is
a power of
c9
9
9
k
...
By
2.
A A A 9
9
y
9
...
9
the preceding,
i
abc
The right-hand
G^A,
...
n +r
A n+r
side is equal to
Gn A r ^A n + r
and therefore
,
when
the sign of equality only occurring
the numbers
all
,
so that
a, 6,
...
k
are equal. Shoiv that n n >l
J&r. 1.
We have
(1
Now
AV
>\
/
1.3.5... ...
For
...
(2n
-
1).
(2w !)}/> +2n - 1 =
&{!
(2-l)
and
.3.5
l
+
a^ (n
-
There are n inequalities
- a 2 ) + (0 - a 8 ) + - a 2 ) (5 1) ^ {(5
. . .
...
-
(2
(2*-l).
2 1)} -r/
n n >l .3 5 .
a n are positive and (n - 1)* =
(*
therefore
.
+3 +5 -f ...
// a x , a 2 , a 3
2.
3 5
+ 3 -f 5 + ... + 1
.'.
.
...
^+a
2
;
(2- 1).
+ ...
-f
an
,
+ (* - a w = a lf )
8 ) ...
(a
-a
n )}
(or equalities) of this kind,
l /(
n~ l \
and the
result follo^a
by
multiplication.
14.
Theorem,
//a,
6, ...
k and
x, y,
...
w
are two sets of
n
positive
numbers, those in the second set being rational, then
+ ku
For in Art. 13 we may take x^x'lg, y = y'lg, ... w = w'/g, where x' y ... w' and g are positive integers, and then it is sufficient to prove that 9
+ ''
ax'
Now is
fy'
+ ...*M/ '
the left-hand side
the geometric
mean
b occurs y' times, the last theorem.
...
is
...
,
'" k
the arithmetic
of a, a,
r
6, 6, ... k,
mean and k
...
,
the right-hand side where a occurs x' times,
and k occurs w' times, so that the
result follows
by
MAXIMA AND MINIMA .15. all
Theorem. to
equal
// a, 6, one another, and
k are n positive numbers which are not
c, ...
m is any rational except
n
.,
a
b
c
k
x
y
z
w
let
or 1, then
n
V
m does not or does Iw between
according as
For
223
/
and
1.
a + 6-f c-f
...
+&
n
then 1
fl m ^ l-a m m m -a -(a + & + c + ...+ k )^ l-(a + + c + .:.+
and
xm + ym -f zm +
according as If
m does not lie
xm + y m +
Hence
...
and therefore If
0
and
between
-f
w-n) = Q,
. . .
Art. 11, the sign
>
must be replaced by +wn
xm + i/m + ...
inequalities, so that
Art. 11,
+ wm -n>m(x4-y-h ... rw xm -f m -f -f t^ > n. y
by
+ wm $ n.
. . .
by
1,
,
<
in all these
This completes the proof of the theorem. 1
6.
Theorem
.
Ifa,b,...k and x,y, ...w are two
sets
...
+wk m
ofn
rationals
and
m being any rational,
those of the first set are not all equal to one another, then, .
J
m
does not or does lie between and 1. according as * This depends on Art. 15 in the same way that Art. 14 depends on Art. 13.
17. Application to Questions of Maxima and Minima Val ues. In the theorems of Arts. 13-15 the inequalities become equalities when all the num'bers a, 6, c, ... k are equal. Hence we draw the following ;
conclusions
:
Suppose that then: (1) If
x, y, z,
x+y+z+
...
. . .
w are n positive variables and that c is a constant,
+w = c,
the value of xyz
so that the greatest value of xyz (2) If
...
w
is (c/n)
xyz...w = c, the value of x + y +
...
...
+ y+
... -f
w
is
is
greatest
when
n.
w i
so that the least value of x
w
ncn
.
is
least
when x = y=...
=w?,
APPLICATIONS
224 if
Again,
ra is
rational except
any
or
from Aft. 3 we conclude that
1,
:
between
x + y + z + ...+w = c then, according as m does not or does lie and 1, the least or the greatest value of xm +ym + ... +w m occurs
when x
y...w,
(3) If
the value in question being n l
~m
cm
.
.
m m m = c, then -f w (4) If x + y + according as m does not or does lie between and 1, the greatest or the least value of x + y + ...+w occurs when x = y=...=w, the value in question being n 1 " 1 /. c1 /. . . .
Ex.
1.
Find
the greatest value of
xyz for positive valuer of
x, y, z, subject to the con-
dition
Since yz + zx + xy~l2, the value of (yz)(zx)(xy) is greatest when yzzx^xy, that when x~ y~z-2. Hence the greatest value of (yz) (zx) (xy) is 2 8 and the greatest value of xyz is 8.
is
,
Ex.
2.
// the
sum of the
sides of
w given, prove that
a triangle
when
the area is greatest
the triangle is equilateral.
Let If
a, b, c
A
is
be the sides of the triangle, and let 2 - b) zJ s(s - a)
the area, then
(s
-
Now
(.?
Hence the value
of (s
Ex.
3.
Since
Find the x 2 y3 6.
o%3 = 6,
Therefore A#
if
-f (s
- a) (s - b)
Therefore the value of
condition
a)
A
is
/x
are
6)
(s
-I-
- c)
(s is
-
(s
-f
+ c ~2s.
a constant,
greatest b
6
- c).
s
c)
when a
greatest
when
s
-a~s -b~s -c.
= c.
3x + 4y for positive values of x and
least value of
A,
-
a
any constants, we have
+ Xx+^y -f/xy+^y
is
least
when 2
(
\x) (Xx) (^y) (py) (/z//)
A#=/u,?/
= (6A 2/z 3
)
+ 4t/
of 3z
Find the least value of x~* + y~ l + z~ l for positive values of x, the condition x + y + z = c. In Art. 17, (3), putting w= - 1, n~3, since m does not lie between
+ y~ l + z' 1
is
SV" 1
.
.
4.
value of x~ l
= 6A 2/x 3
y
3
Hence the least value of 2Az -f 3pty is 5(6A /i )^. Putting 2A = 3 and 3/x 4, it follows that the least value
Ex.
y, subject to the
is
y, z
which satisfy
and
1,
the least
or 9/c.
EXERCISE XXIII In
marked with an
the inequalities
*, all the
numbers concerned are supposed
to
be
positive. ... a 1. If ,, cr 2 n and b 19 b 2 ... b n are two sets of numbers and the second set are positive, then (a l -f a a + -f a n )/(6 1 -f 6 2 + + bn) lies between the greatest and least of a^/b^ an lb n ..., 2 /6 2 ,
,
. . .
.
.
,
2.
If a,
6, c
are (b
any
4-
c
-
real
a)
2
numbers, show that a - 6) 2 -f (a + 6 - c) 2 ^ 6c
-f (c -f
. .
-f
ca
-f
ab.
all
those in
TYPICAL EXAMPLES
225
3. If J + w -fn = l and J' + m' -f w' ^l, then 2 2 [E(mri - m'ny + (W + mm' -hnn') = El* Z7' .] 2
2
2
2
2
2
.
4. Show that a(x + y)* the quantities involved. 5. If any two of showthat
a>0
If
6.
+ bz + cz(x ~-y) cannot be 2
positive for all real values of
together greater than the third and x
a, 6, c are
and x>y, show that ax + a~ x >a y + a~ y
+ y + 2=0,
.
c, d are in harmonical progression, then a + d>b + c. [Let p 3g, jp f, 2>-l-#> ^>4-3^ be the reciprocals of a, b, c, c?.] ~ ~ 8.* If x^ty, then (axn + byn )l(axn l + byn l ) increases as n increases through
7.* If a, 6,
integral values. 9. If a, 6, c,
[2(ab
m* 10.*
-4-
dareall >1,
1)> (a +
+ 1),
!)(&
then
etc.]
--- + ^ a + b + c. c+a 2 2 [For, (& + c )/(& + c)^(& + c), etc.] 11. If x>0 or <-l, then r zn ' r ^ I +x [Show that x + + x -r &4-c
,
+
-
-
,
l
12.*
(i) (ii)
a(a-b)(a-c)+b(ba 3 -f 6 3 + c 3 + 3a5c ^ a 2 (6 + c)
[The second inequality 13. If a, 6, c
> 27a
15. If
[If
.
a>6, then
16. If
n>0
Za Ella^n* .
and not
p>l
Za 3
(ii)
;
all
.
El'.a'&nZa*.
equal, then
y are any two positive numbers such that
#,
\*
(^
-l
a?
+ y = l,
then
.]
anda?>l, then
> ~nx
x n ~x~- n 17. If
first.]
and Za(6-c) 2 >0.]
JEb"
a 9^6 and
+ a) -f c 2 (a 4- b).
merely another form of the
is
14. If a, b 9 c are positive
8
6 2 (c
are n positive numbers, then
... fc
(i)
[327a
-I-
and
r is
(p
any
4.
A
.
A
[Use Art.
ni 9.]
rational except zero, then
l)f+i
- r+i^( r + l)pr >-pr + l p
where the upper or lower signs are to be taken according as lie between and - 1.
Hence show that
r
(l
+ 2 a + 3 r 4-
...
+
r
)/w
r4
'
1
lies
r
does not, or does,
between l/(r-fl) and
[In Art. 10, let a;=r4-l. First put 0=73 + !, &=/>, Finally let p = l, 2, 3, ... n in succession, and add.]
and next a=|9, 6=p-l.
HARMONIC SERIES
226
19.* a*b
+ b*c + c*a^3abc.
21. If
x> I
22. If
any two of a,
and n
is
20.
a positive integer,
(#
n
-
1)1 (x
-
are together greater than the third, then
6, c
23. If all the factors are positive, then
abc^TI(b+c-a)
(i)
24.
The
and
25. If
x+y+z
of,
is
a, 6, c,
abcd^ TI(b + c+d -2a).
*
-ax -by -cz)
greatest value of xyz(d
factors are positive,
(ii)
;
4
4
provided that d are given positive numbers. is
/4 o&c,
all
the
6, c are positive rationals and x, y, z are positive variables such that a b constant, show that x y sf has its greatest value when x/a y/b zlc.
26. If x, y, 2 are positive
and x + y-f- z
then
1,
8xt/z
<( 1 -z)(l
-i/)(l
V
-z)<-2
[Observe that 27(1 -a:) = 2.] 27. If
*=!+-A + -+. ..+_ n o
and n>2, show that
[Consider the sequences \, \, ?,
^04:
...
!L^
and ?, ?,
*
1ZO
71
...
!tl
.]
71
H
28. Let a l9 a 2 , ... an be a sequence of positive numbers and let A,Q, respectively be the arithmetic, geometric and harmonic means of a^ and an , then n
29. (i) (ii)
,
an
a harmonical progression, nH
G
,
If a x a 2
...
.
is
n IfzjX&t ...xn =a where a
is
...
an <(? n
.
a constant, then
r The sum of the products of every r of the x's > CJ?a The least value of (x l -f (a?a + k) ...(xn + k) is (a 4- k) n .
ifc)
30. If a is
.
any positive number except 1, and x, y, z are rationals no two of which and which may be positive or negative, then a x (y - z) +a*(z - x) +a*(x - y)>0. let z be x, [First y, integers and let x> y> z (Art. 5). Consider the A.M. and the O.M. of the set of numbers a x a x a x ... a 2 a 2 a 2 ... where a x occurs (y -z) times and a? occurs (x y) times. If x, y, z are fractions, denote them by p/d, q/d, r/d where p, q, r, d are integers and d is positive. This case can then be at once deduced from the preceding.]
are equal
,
31.
,
,
,
Deduce the theorem of Art. 9 from Ex.
32. If
a l9 a 2
,
...
,
(a 1
[In Art. 15 put
a n are positive and + aJ + ...+an*)<
*
w = ?, a-a^, etc.]
p>q
,
,
30.
then
,
333
IMPORTANT GENERALISATIONS
-f-
16
a-f b
w= -1, n4, $(b+c+d)
[In Art. 15, put
227
+ c + d'
fora, etc.
Use Art. 16 to show that
34.*
a" 1 + b~ lc + cr l a
a+b+c
*)
35.*
[Use Ex. 34.] 36. If
a2
!,
...
,
OK and b l9 b 2 , then
...
two
bn are
sets of positive
numbers arranged
in descending order, (a l
+ a 2 + ...+an )(b + b 2 + ...b n ^n(a b +a 2 b 2 + ... +a n b n - 2a br Use the relation (a r ~a n )(b r ~bn )^0 r r r l
~na b
[Let u n that w n >w-n _!.] 37.
If a>
1
and p, q are a*
1
).
1
.
positive integers,
aa
-I
p [Let
)
.
~l
p-q, a 2
q
u^(a? - l)lp -(a*- - 1)1 (p ~ 1), 1
show
tfiat
.
,
to
2 -I) when p>q.
then by Art.
9,
... -f
-
p
show
1)};
~.
p(p-L) a p -l aq ~l
and
p 38. If
a>l and
-
q x,
y are
"
positive rationale, then
x
2
y
[Put x pfd, yqjd where p, Ex. 37, we find that
q,
d are positive
Let a lld
integers.
b
;
then, using
i
and by
Art. 9, Ex.
39. //
a> 1
ami
1,
x,
d(a
y are
d
- 1)> 1 -a- 1 .]
positive real numbers, then
(a*-l)lx>(av-l)ly
if
x>y.
[Let x' y' be rational approximations to x, y such that x'>x, y'
then
We
/(*)
can choose x', / so that when *>!/. See Ch. XIII, 7.]
|
-c'
-a" 1
2 )
,
and so
/(a?)
-f(y)>0,
CHAPTER XV SEQUENCES AND LIMITS (Continued from Art. 11 on p. 15)
Limit of a Function of the Positive Integral Variable
1
When
n
is
large,
1 4-
-
enough, we can make This as
n tends to
infinity.
I
further,
;
by making n large
l
as nearly equal to
1 -f
roughly what
is
nearly equal to
is
n
n.
is
as
1
we
meant when we say that
like.
1 is
the limit of
In this explanation, the meaning of
1
+-
'
large
enough/
'
as nearly equal to 1 as we like,' is by no means clear. If u n is a function of the positive integral variable n> the n tends to infinity is usually defined as follows
i
limit of
u n as
'
:
that we (1) If corresponding to any positive number no matter how a there is m such that small, positive integer may choose, for every integer n greater than or equal to m Definitions.
where
is
a fixed number, then
is
expressed by writing
I
This
lim u n
=
I
called the limit
is
tends to infinity.
lim u n = L
or simply
/,
ofu n as n
n->oc
We also say that
un
tends to
I
as n tends
u n -~>l
as
to infinity
n-+vo
M
corresponding to any positive number how great, there is a positive integer matter no
then u n
this
by
that
m
y
we may
choose,
such that for every
n greater than or equal to m, u n >M, is
said to tend
to infinity
wn
>
as n tends or
oo
,
,
to
lim u n
w n ->oo and v n = - u n we say that If u n -->l we say that the sequence (u n ) If w n ->oo or - oo to the limit L the If
and express
.
(2) If,
integer
;
infinity)
= oo
.
v n tends to is
and we write
minus infinity (-<#). and that it converges
convergent,
sequence (un )
is
said to bo divergent*
THEOREMS ON LIMITS u n does not tend to a
If
oscillate
un
If
limit,
It
if|a|
a n ->
is
u 2 ),
-
or
oo
a=-l,
if
according as n
oo
an
is
This
u3 ),
(3,
...
=r
1
even or odd,
often useful to represent the
graphically. (2,
it is
that, for
first
a n ~> oo if a - 1, a n = 1 for and a n oscillates finitely; if and so a n oscillates 1,
;
infinitely.
few terms
of a
sequence (u n )
done by plotting points whose coordinates are
is
It
.
said to
\
does the function a n behave as n-> oo ? II, 18, (5), it will be seen that if a>
How
1.
1,
,
Otherwise, u n oscillates infinitely.
said to oscillate finitely.
is
every n; -
oo
n,
Referring to Ch.
a<
nor to -
M
\u n
Ex.
,
and the sequence (u n ) is called oscillatory. can be found such oscillates and a positive number
every positive integer
then u n
+ oo
nor to
229
(1,
u
}
),
usual to join points representing consecutive
is
terms. Ex.
2.
un = ( -
//
n l)
I
1
+-
represent
,
j
Jew terms of the sequence and examine the diameter of the sequence
graphically the first
(un ), as
n->
oo
.
As ?i~> oo u zn -> 1 and u 2n ,!-> Thus un oscillates finitely. ,
2.
1.
FIG. 34.
Fundamental Theorems on Limits.
(1)
Ifu n -*l and
v n -*l\
then
u n + v n ->l + l',
(i)
l/u n ->l/l, unless
(iv)
Proof. (i)
and
un
(ii)
-v n ->l-l',
= 0,
1
(v)
Choose a positive number (ii).
By
w n ^ n ->Zr,
(iii)
u n \v n ->l\l' unless ,
I'
= 0.
no matter how small.
,
the definition of a limit
we can choose MJ and
w
2
so
that
\u n -l\<.l
and If
\v n
we denote
will hold for
-l'\<\t
the greater of
n^m, and
and |
Wn
mv m2
for
n^m
for
n^m%.
by m, then each n we have
for such values of
_ Vn _
(J
_ J')
Hence, by the definition of a
|
sg |
Mn
limit,
u + v->l + l'
l
and
of these inequalities
MONOTONE SEQUENCES u n = I + a and vn = T + Then
230 Let
(iii)
j9.
Choose any positive number
L
greater than
V + j8
|
then
1,
\u nv n -ll'\<\fll\+\\.L. the definition of a limit
By
provided only that
n^m. Hence
Let
tt n
= Z+a,
Z
is ftoJ zero
1
we can choose a
n
Now by
IV \
<.
This follows from
(iii)
number L
positive
and
than
1
+a
|
|,
and
|ifX'
we can choose
m so
that
also
* and therefore
11
i_
c,
less
<
-|
n>m. Hence
<
a
-
the definition of a limit
provided only that
(v)
|
then
A // then
u nvn -
also
u nvn -+ll'.
Therefore (iv)
m so that
we can choose
w
>T
.
Z
for
(iv),
when r Ifu n ->l and v n ~w n ->0,
(2)
(
Now we
can choose
m
|w n
Hence
-J|<|
vn
-un
\
+\ u n -l
\.
and
|^ n
-w n |
for such values of n, |
3.
w
n^m both
so that for
-Z|
f> |
vn
-1 1
<
c,
Monotone Sequences.
said to increase steadily, and (u n ) If w n +i
and therefore w n +i^ M
If is
un
v n -> Z. for a11 values of n,
un
is
called a monotone increasing sequence. is
called a monotone decreasing sequence.
said to decrease steadily,
and
(w n )
is
EXISTENCE OF A LIMIT Theorem less
231
// every term of a monotone increasing sequence (u n ) is than a fixed number k, then the sequence converges to a limit equal to, or 1.
less than, k.
Divide the system of rational numbers into two classes as
Proof. follows :
of
The lower class is to contain every rational a such that, for some value n (and therefore for all greater values), w n >a. The upper class is to contain every rational a! such that u n
values of n.
No rational escapes classification, and every a the classification defines a real number A.
We
shall
is less
prove that the sequence converges to
First observe that
every rational
which
A
than any
a'.
Hence
as a limit.
u n cannot exceed A. For if u n >A, u n would exceed This is impossible, for such lies between u n and A.
a rational would belong to the upper class. Next, choose a positive number e, no matter
how
small,
and
let
a be a
rational such that
A-
0
therefore
is
exceeded by some term um of the
;
if
limu = A.
Hence
k^A,
Again
for otherwise k
would be exceeded by some term
of the
sequence, thus
It follows that
or to
For
u n
-f
oo
if
if
un
increases steadily as
n->oo
,
then u n tends to a limit
.
u n does not tend to +
oo
,
number k
a positive
exists such that
for all values of n.
Theorem 2. If every term of a monotone decreasing sequence (u n ) is greater than a fixed number k, then the sequence converges to a limit equal to or greater than k. This can be proved by an argument similar to the preceding, or thus Let v n = -u n then (vn ) is an increasing sequence of which every term is Therefore (v n ) converges to a limit less than - k. Hence less than - k. :
,
(un ) converges to a limit greater than k. Hence also if u n decreases steadily as n->oo
to -oo
.
,
then u n tends to a limit or
EXPONENTIAL INEQUALITIES
232 i
Ex. If
1.
x
un
//
~0
un
1,
--ii(x
n -
1)
where
for every n,
#>0, show
that
un tends
to
a limit as n-^&
.
and so lim u n ~0.
x>I, u n >0, and by XIV, 9, un decreases limit, which we shall denote by /(x), as ?i~> x If
Hence un tends
as n increases.
to a
.
If
0<#<1,
(1) ///or all values of n* u n is positive and u n a constant greater than unity, then ?/ n ->oo
Theorems.
4.
where k
is
.
For u n >ku n _{>k*u n _2 (2)
y>\ and
x~l/y, then
let
all
Iffor
>k
...
values of n,
un
n ~l
u1
Now &>1,
.
therefore u n -+co
and u n i
is positive
is
.
a positive
constant less than unity, then u n -0.
The proof (3)
// u n
For
if
is
similar to that of
is positive
Theorem
(1).
and lim u n n /u n = l then w n ->o> ?//>!, and u n -+Q .
t
/>! we can choose k so that >&>!, and then hy the definition we can find m such that u n+1 /u n ^>k for n^-m. Hence by
of a limit
Theorem
(1), w n ->oo l<] we can choose k so that l
If
Ex.
.
9
1.
If p
is
a given
integer, positive or negative,
* (i)
un =
If
and the 5.
(1)
results follow
p
xn -^
->oo
if
x>l u
then
n
/
=(
-
~->0
(ii)
;
sJww
\v -_
if
find
m so that
that
|*|<1.
'X>x
as
rt-> oo
,
)
from Theorem
(3).
Exponential Inequalities and Limits. If n
For,
is
a positive
integer,
f
1
+-
)
<3.
by the Binomial theorem, ...
to
n -f
12 '(3
*
Or,
it'
n
^ m where M
is
a fixed number,
i.e.
after a certain stape.
1
terms
EXPONENTIAL LIMITS (2)
m and n are positive integers
//
1+
1
m
\
>
)
w/
/
n
1
+
1
\
I
1
\n
+m/
)
>
1
and
n/
ml
+
m
n
m>n, /
\
-
m.
11,
f
and therefore
,
then
1-
\
For since m/n>\, by XIV,
m
and
233
+
1
(
\
m
1 \ 1
m/
l\ n
/
>l-f-j. \ n/ n
therefore
Moreover,
0
Hence
l-
'
'
7
by J XIV,
1\*
f 1
11,
--
<1
)
n/
\
n
1
m
n
.
and
n/
(3)
The number
andu n = (l--) \ n
Ifu n = (l+-) n/
e.
'
\
u
through positive integral values,
and the
same for
the limit is the
,thenasn-+oo
'
tends to a finite limit, so also does
71
This limit
both.
is
denoted by
and
e,
is
un
'
;
one of
most important numbers in mathematics.
For
(i)
every n.
has been shown that u n increases with n and that u n <3 for Therefore u n as n->co where e is a fixed number less than 3.
it
-e
,
\~ n (f\ -~
' '
(ii)
Again,
Wfl
-*- - u = u fu n ( n \u n
\ 1
)
/
= un
f
0
therefore
'
- un <
n
u n '-u n -+Q
therefore
(by
2
V
n 2 /)
{\
l
Nowifn>l,
1
\
<-. Now
-
->0,
n
n
and
u n '-+e.
It should be noticed that
u n
can be shown that 6-2-7182818284
f .
....
EXERCISE XXIV 1.
State
how
the following functions behave as n->
(ii)
n + (-l) n
n
(v)
n 2 -f(
->
as n->
(i)
2
+ (-l) n
(iv)
2
+ (~l) n -.
.
xn 2.
Prove that
n
QO
.
.
(iii)
XIV,
-
n
1
1
.
\ j
11),
EXAMPLES ON LIMITS
234
Represent graphically the
3.
say
how each behaves
||
n->oo
%n
un ~
If
6.
-
A*
I
[The limit
.
W
~~
Show
x
is 1, if
\
|
and -
> 1,
1
otherwise.]
-H---- _H-------o+...-fr-, prove that f
tends to a limit as w-*oc 7.
*
n
-
then
^.fl
Find lim
5.
few terms of the following sequences, and
first
.
--*
(Os.i. 2 o 4. If
as tt~>oo
that
.
-n< -^ + o- + 4- -f
vr
... -f
,4
Hence show that
and that un
1 H-
J
+
4-
. . .
H
J
--->< n
/I
1
[Group the terms thus: -+
1\
^+-
)
+
as ?i->oo.
/I
1\
1
1
-4- -4-
)
+
... .]
^-4-
1
8.
If lim
71(0;^
-
prove that
1) =/(a;),
z>0 and
if
1
Ex.
llli [See Art.
xn y n -
3,
=yn (xn -
1
For the
1.
n
-
1)
lim yn ^
and
,
1
Q 9
T If
tt ^v
[If
/>0, then -
5 2
I
n
vn ->oo
n ->0,
/v
2
10. If
4
wn -
1
;
_^_^ and
i<
w n i' n
<
1
-
according as x
,
^
2n
"
^^'s-srn:' 4
Now
l
_____
_
.
1).
1
1
1) -f (y
1
we have f(xy)-\im n(xn yn -
last part
5 y, 1
hence
v< 2-^n
and
6
2w
46
,
-
''*
-
2n + 2
N/2/Hrl, prove that
between l/\/2 and 1 as n->oo [Show that w n ^
lies
wn
1
tends to a limit which
.
,
"
/2" has shown that this limit = A/ - .] ' 7T
9,
wn *=un vn and
^
Wallis
GENERAL PRINCIPLE n ^ 3,
11. If
n+ prove that \/(n +
1)
/
[For the
first
also
steadily;
part use the inequality
w>4
(
/
(r + 1)^
(i)
between
lies
7J 71
and
1
(
1 -f
+ l>0,
lim u n ^\l(r f
l
r
-f
2r
-I-
r-f-i
1
-
)
nj
\ (ii)
that
lim {(a n-
-
3 r -f
r-,--l
Corresponding small,
.
decreases
. .
-f
.
+
r
1
)
4-
l ,
n +
(a
-f
r
2)
-f ... 4-
(a
The
and add.
+ tt) r j/tt f + 1 = l/(r + l).
^-00
to
any
positive
must be possible
it
n r )/ti r
.
The necessary and
cient condition for the convergence of the sequence (w w )
how
oo
1).
General Principle of Convergence.
6.
n ->
|
,
Show
as
s'w
[In Exercise XXIII, J7, put 1, 2, 3, ... n in succession for ?;, necessity for the condition r~hl>0 in (ii) should be explained.] 13.
1
If Z>1, for ?*>3 we C'^>1, therefore "?i->Z where Z^l. n But J n /n->oo when J>1 (XV, 4, Ex. 1).] l /n
any rational except zero and
12. If r is
If r
Thus
<3.
4 ---I
1
(
that */n ->
n
]
^n>L and
should have
show that
235
to
number
m
find
we may
choose, no matter
so that, for every positive integer p,
\u m + ,-u m 1(
that
suffi-
as follows.
is
\<
(A)
say, the difference between u m and any subsequent term must be numerically less than e, so that if the terms of the sequence are repre-
That
is to
sented graphically, all the terms beginning with u m lie in a strip of width 2e. This statement, called the general principle of convergence, is of the greatest importance, (i)
The condition
and may be proved as For
is necessary.
|
um + p - u m
< \
The condition is points Wj, u 2 ... on the (ii)
,
so that
all
|
um + P -
I \
un
we can
>l,
find
u m+p - u m - (u m
Now
for every positive integer p.
Hence
if
follows.
+ u m -l\
,
]1t
and |
\
m
-l)
um
+ (l- u m ).
^ -um \<
e.
Represent the terms u v w 2 ... by the The condition asserts that m can be found
sufficient.
x-axis.
so that
,
the points,
on a segment of length with its middle point at um
lie
2e
FlG
35
.
By like.
choosing e small enough, we can make this segment as short as Hence it seems reasonable to conclude that as n->oo the point ,
tends to a limiting position somewhere near to um If the student satisfied with this reasoning, he may omit the proof on the next page. .
Q
B.C. A.
we un is
NECESSARY AND SUFFICIENT CONDITION
23G Given
for every positive such that
m
integer
e,
that,
no matter how small,
un - um for every integer n greater than m,
a positive
required to prove that the sequence (u n )
is
it
|
there exists
is convergent.
Using the given condition, we can determine a positive integer w 2 >Wj, u mi> is within the interval
Proof.
m
so that, for every integer
l
('U mi
Denote
this interval
by
-,
U mv
+).
(a v b^.
w >w2
Again, we can determine m 2 so that, for every integer 3 within the interval (u m
W 3 >m
2
Thus the whole or part
.
Denote the common part This //
^Hi.2
is
of length
The
.
(w ?Wo
-
Je,
of these intervals
lies
-e,
of
entirely within
figure illustrates the case in
um^ +
,
um ^
is
for (c^, 6 T ),
within (a v
b^).
and contains u n
tor
\t)
is
(a 2 6 2 ).
by
,
/^),
(c/ 1?
which the intervals overlap.
FIG. 36.
repeating the process, we can find a succession of intervals (a^ such that (i) each contains all that follow ... (a,., b r ) ...,
By (a 2
,
6 2 ),
(ii)
(iii)
;
for sufficiently large values of n, b n -ff
a n
^
*
.>, t -->
tl
Hence by Ch. XIII, that
b^),
1,
there
is
and
for every n,
ns
uu H
is
~~*
within each of the intervals
^
;
'
one real number
I
and one
only, such
this is the limit of the sequence (u n ).
NOTE. The necessary and sufficient condition for the convergence of the sequence (u n ) is usually stated (less simply) as follows. Corresponding to any positive number e that we may choose, no matter how small,
it
must be possible
to
find \
if
>w,/w
m
so that
",
-
"
.................................
(B)
every positive integer p.
The condition (B), though apparently more general than (A), is equiTo prove this, we have only to show that if (A) is possible, valent to it. then (B) is possible when n]>m.
BOUNDS OF A SEQUENCE we can
the condition (A),
By
all
the terms w wfl
m so that
U^^-Un
I
and then
find
,
I
um+z> m+z
wf 2 .
237
He in the interval
Therefore the difference between any two terms which follow u m so that Hence we can find
than
is less
m
.
|w n + p -tt n |<
H>w.
if
Therefore the conditions (A) and (B) are one and the same.
Bounds
7.
of a Sequence.
If
(1)
there exists a
number
M
such
that, for all values of n,
the sequence (u n ), and also the function u w are said to be bounded above (or
on
the right). If
number
there exists a
JV
such that, for
all
values of n,
then (w n ), and also w n are said to be bounded below (or on the left). A sequence (u n ), or a function w n which is bounded above and below, ,
,
is
said to be bounded.
Examples.
Here
(un ) has a greatest term,
namely w 2 ="2
:
^ has
also a least term,
These are called Ae upper and lower bounds of (MW ), or of u n (FiG.
namely
%
- 2.
37).
-1
FIG. 38.
FIG. 37.
1
2.
Although there
wn
,
is
no greatest term, we say that
for the following reasons (i)
(ii)
No term
of (u n ) exceeds
Infinitely
many terms
1
is
the upper bound of (un ), or of
:
1.
exceed any number
For similar reasons, we say that -
1 is
less
than
1,
for
u.in
-+
1.
the lower bound of (u n ) of u n (FiG. 38).
UPPER AND LOWER BOUNDS
238 // u n
Theorem.
no term of (u n ) exceeds h
(i)
at least
(ii)
This
bounded above
is
number
function u n
a number h such
there exists
;
one term of (u n ) exceeds any number h
is
that
less
than
h.
called the upper bound of the sequence (u n ) or of the
.
Divide the system of rationals into two classes. The lower class is to contain every rational a such that for some value
Proof.
of
n u n ^a. The upper
of
class is to contain
every rational
a'
such that for
values
all
n
u n
is
classification,
defines a real
bounded above, both classes and every a is less than any
number
exist. a'.
No
rational escapes
Hence the
classification
h.
Moreover, no term of (u n ) exceeds h, for in that case it would exceed a rational a' greater than h. This is impossible, for a' belongs to the upper class.
Also
if e
is
any
positive
number, no matter how small, we can find a
rational a such that
h-
:
(i)
(ii)
no term of (u n ) at least
The number function u n NOTE. h and
is less
than
one term of (u n ) I
is
I
;
any number
is less tJian
called the lower
bound
of the
greater than
I.
sequence (u n ) or of the
.
// (u n ) JLOS no greatest term, infinitely
many
terms of the sequence
lie
between
h-.
For every term is less than h, and at least one term u m is greater than h - e. Since ia no greatest term, infinitely many terms are greater than u m and therefore also greater than h e. there
Similarly, if (un ) has no least term, infinitely
and
,
many
terms of the sequence
lie beticc.cn
I
LIMITS OF INDETERMINATION Terms selected from u l9 u 2 said to form a subsequence. Ex.
Explain how
1.
to select
t/
,
3,
...
239
according to some definite rule are
a subsequence which will converge
to the
upper bound h
If a 2 ^h, let a 3 be the lirst If Ui^h, let a, 2 be the first term of (un ) greater than v term of (un greater than a.2 Continuing in this way we form the required subsequence The formal proof is loft to the student. ^i a2
.
,
,
Lower Limits
Upper and
8.
.
.
)
(of
Indetermination) of a
Sequence. Let (u n ) be a hounded sequence and let h l9 h 29 A 3 ... and l v J2 ^> the upper and lower bounds of the sequences u 19 u 2 u% ..., u 2 u 3 w 4 (1)
,
,
,
,
u39 w4 u 5 ,
If
fi
...
,
...
,
etc., respectively.
then u v
l=sUl9
,
* >e
h^u
h^h^.
If
hl = h 2
Thus, in
not
is
ly
than any term of
less
then Aj
is
t/
2,
^h
%, ^4
,
...
u 2 u3 w4
the upper bound of
,
,
>A
,
;
therefore
...,
and so
and so on. 4 Similarly A 2 ^/fc 3 A 3 any case, h l 2 and Hence h l9 h 2 A3 ... is a steadily decreasing sequence every term is greater than or greater than l v Therefore this sequence tends to a limit .
.
,
,
,
,
H
equal to l v Similarly
than h v
I
l9
1
2,
Z
3,
Therefore this
The numbers
H and
an increasing sequence and every term is less sequence tends to a limit L less than or equal to h r is
...
L
are called the upper
and lower
limits (of indeter-
mination) of the sequence (u n ).
lim u n = //
It is usual to write
(2)
and
lim u n ~L.
Examples. Here h 1
= h 2 =h3 = ...=H.
(Fig. 39.)
FIG. 40.
FIG. 39.
1\
Here
^ = ^3-
;
h^h^u^;
=l z ~Uq-,
etc.
etc. .',
/.
//=lim u2n ^l.
L~tim w 8n+1 =
-1.
(Fig. 40.)
GENERAL PRINCIPLE OF CONVERGENCE
240 (3) It is all
easy to see that,
terms of u lt w 2
...,
,
wn _1
After a certain stage,
(i)
that
if
A n >Aw+1 then h n = u n and h v A 2 ,
As w->oo two cases
.
arise
the A's are equal, that
all
7> /> /> ~ "'m+2 ~ n m ~ Mrn+l
:
m exists
to say,
is
h n -\ are
...
,
so
H "
-
In this case, (u n ) has no greatest term. None of the terms from u m onwards - e. can exceed H, but infinitely many terms exceed
H
There
(ii)
no such stage as that described in (i). That is, no matter h n >h n+l be, we can find w, greater than m, so that
is
m may
how
great Therefore every h
.
is
a term of (u n ), and in
cases //
all
H
such that, after a
is
certain stage, every term of (u n ) is less than + e ; Hence in all cases, infinitely many terms of (u n ) are greater than a in other than are certain all the terms less + also, after stage of (u n ) .
H
H
words, only a finite number of terms exceed II
In a similar
way we can show
}
+ e.
that infinitely
many
terms of
un
are less
L+e
than
and, after a certain stage, all the terms are greater than L~-e; in other words, only a finite number of terms are less than L-c. It follows that, if
H=L=
then the sequence converges
l,
For, after a certain stage, all the terms of (u n ) //
-f
that
,
is
between I -
and
l
lie
to
I
as a limit.
between
L-e
and
+ e.
We can at once deduce a proof (4) General Principle of Convergence. that the condition in Art, 6 is sufficient for convergence. Given that
m
can be found so
that
\u n
-u m \
nl>m,
for
it
is
required to prove that (u n ) is convergent.
u n cannot tend to
First observe that
-f
co
or to
oo
.
Therefore
We can therefore choose m (u n ) has upper and lower limits // and L. u n - um and n (n> m) so that un 3-6 Je, Je, \u m L |
and since
H
H
-L
H
- u n + um
\
<
-L + u n - um
\
<
\
|
<
;
,
\H-L\^\H-u n \+\u n -L\+\u n - u m
|
H and L is H = L and
to say, the difference between the fixed numbers Therefore less than any positive number c, however small.
That is
(u n )
9.
is
convergent.
Theorems.
(1)
certain stage,
where k
is
But
un
if
In
// (u n )
u
a positive constant -u n ^ ;>w n - u n }
the first case,
,
is
an increasing sequence and
if,
after
a
u less ^.
2
,
than unity, then the sequence the sequence is divergent.
suppose that the conditions hold for
is convergent.
Then
if
therefore
IMPORTANT THEOREMS w>s, u m+l -um
241
M
.
Also
Hence by addition
Choose any positive number can find
,
no matter how small then, since &<1, we and since 'Umirn >um it follows that ;
m so that &m
tWn
,
- um
|
for every n.
Therefore the sequence (u n ) converges. In the second case, if n>s, u n+l - w n >w s
Now (2)
us
w s+1
is
^
t
~ w 6-
a fixed positive quantity, therefore obviously
// M M >0 /or a^
m^.9
o/ n awrf w n+1 /?/, n ~>/>0 where
I
w n ->oo
is
.
a fixed
i
n number, then u n ~+L Because Z>0, we can choose a positive number like. Then we can find m so that
e< and
as small as
we
and by multiplication,
Since
Z-e>0,
it
follows that
^u n
^u n therefore
,
(Z-)
n
<
J m <(Z + e) n
i
/
\
n
1
'
and
i (t
m
')
.
(I
/
-)
j V
m 1
Now
u m fl m
is
a fixed number, therefore (u ni /l m ) n
1
>]
and
?/
n n
>i!.
i
The converse of this theorem is not true. That is to say, if ^n n -> follow that w n4l /w n ->/. For consider as a special case the sequence 1,
NOTE.
/,
it
does not
2, 1, 2,
...
in
i
which the terms are alternately a limit.
1
and
2.
Here w n n ->
1,
but n n
fl /^ n
does not tend to
IMPORTANT THEOREMS
242 (3)
We
//
,
= u + uz +
w
.
.
.
l
can find
w
-f
u n and Urn u n = I,
so that for
any
Urn
then
n-*x
n
positive
>
-
oc
^
sn
=
/.
c,
Hence by addition, - m)(l - e)
for
and
Now let m and n lend to oo in such a way that w/m->oo we may suppose that m~>cc and n>m 2
(For example,
.
.)
We
show that under these circumstances s m /n~>0. Since u n -*l we can find a fixed number k such that u r and then m < WA', 8m < % + h/ 2 4- ... 4shall
|
r,
I
I
I
1
I
so that
m
<
I
.v
Thus
(4)
//
Let
like,
and
/
s
n
P n = a n & +a n _ 1 6 2 + a n _ 2 1
and
bn
';
3
+
= + f3 ni l>
---
+a
(3),
We
shall
+a 2 + ...+a n )->0 and
l(a 1
e
can betaken
j8 n
->0, then
4-a^). i(j8 1+
j8 a
+
...
+j8 n )-^0,
Te-
prove that $,,->0.
have |
and since a n->0 and
jS n
can also find |
m
as
|
A-m
Q n |< ~n T| r =i
->0 we can find a |a f
We
and
to zero
so that a n -~>0,
?Z
and we
/
n->(>.
A
Q = ~(^i+n-ift +
where
by
for every
therefore s n jn->l.
nfy^^a-fay,
Now
I
m/n and 5 m /?i tend
in the inequality (A),
we
1
7
A:
n
7i
as small as
I
|
|
1
\
and
|<*
fixed
I
,
number k such that for every s
|ft|
so that
<
/k
and |
f38
\
< c/A
for
Hence
|a r ]9 n _ r+]
and
COMPLEX SEQUENCES for n-r + l>m, |<*| j8 n-r+1 |< * r
n>2w,
10.
zn
If
(1)
i.e.
= xn +
corresponding to any positive such that integer
m
-z\
\z n
where
we
a fixed number, then z
z is
zn = z
lim
write
said
(z n ) is
is
a positive
n^m,
for
called the limit
is
limz n = z
or
however small, there
e,
r^m,
where x n and y n
iy n
are functions of a positive integral variable n, the sequence to be complex. If
for
B)
P n-~>ab.
|Q n |< e: consequently Q n->0 and IV
Complex Sequences.
(
l^r^n.
|
Therefore
(A)
r>m f^>r r
|r0n-r+l|<*| a r|<
the inequality (A) will hold for so that, with this^proviso, oc r n _ f3 r+l \
243
or
ofz n as n->oo
,
and
z n ->z.
n->oo
Under these circumstances the sequence limit
(z n ) is
said to converge to the
z.
In order
lim z n
that
z
=x+
it is
ty
limx n = x
and
necessary and sufficient that
limy n ~y. 2
K-H=-y{(z-z) + (2/--?/) 2
For Therefore |
zn
z \
^
|
First suppose that z n -*z.
and
x
xn
|
\
By
l^n" !"^^?
z \
|# n -#|
therefore
~z\-> 0,
All this is evident geometrically, for \z n
points z n
and
yn
|
and
tend to zero, so that xn +x and //>?/. Again, if x n ->x and y n ->y, from equation (A) \z n
(A)
}
^
~
y
. \
definition,
therefore
2
zn
-z\
\y n ~y\
it
follows that
zn
-> z.
is
the distance between the
z.
The theorems of Ch. XV, 2, regarding the limits of the sum, difference, product and quotient of two functions of n, continue to hold when the (2)
functions are complex. (3)
Limit of
zn
as n->oo
Otherwise z n does not tend to
For
let z
(i) if
(ii) if
= r(cos 6+ i
=l
to a limit.
then z n
z is
complex.
a limit except then z n = r ri
sin 0),
|z|=r
when
r n ->0
= l, and
and z
if |
zn
->0;
1
then
|>
7/|z[
then z n ->0.
when z = l, and then limzn = l. (cos n0+ sin nd), and i
r n ->oo
and
zn
does not tend
SUBSEQUENCES OF PROPER FRACTIONS
244
11. General
condition for the convergence of a complex sequence
sufficient
as follows
so that \
to
(z n ) is
Art. 6,
any
e,
positive
-z m \
zm + p
If
Proof.
and by
(z n )
is
:
Corresponding
m
The necessary and
Principle of Convergence.
however small,
convergent, then (x n )
we can therefore choose m \
\
N W
X ^m+y
__
Jr'm
it
must be possible
to
find
every positive integer p.
|<-1<: ^ 2 fc
and
(y n ) are
convergent sequences
;
so that for every positive integer p,
MTU] *iu-i
< 2f e Mm \^l
lay |
\
__,,/
.7^40
->.
-
\
htn + p-^ml
therefore
Hence the condition Next suppose
that \
Therefore (x m ) (y n )
is
As
any
,
^ - z m \
then |
convergent (Art. (z n )
is
6),
however small,
e,
and
x w +p - x m
|
it
similarly
<
z |
m
-
f
it
may
be stated thus.
must be possible
to
find
p
can be shown that
convergent, and the condition
in Art. 6, the condition
positive
if
necessary.
Hence
convergent.
NtfTE. to
is
is
zm
is
sufficient.
Corresponding
m
so that
for every positive integer p.
EXERCISE XXV 1. If u n is a decreasing function of n, then, after a certain stage, all the terms of u n have the same sign. - QO First let u n -> I. If / ^ 0, u n is positive f Kithor u n tends to a limit or to for all values of T?, for un tends to / from above. If I <0, we can choose so that .
m
l
n^m.
for
Therefore the terms of (u n ) beginning with u m are negative. If u n ->--<*), we can find so that u < - k for
m
.
2.
If u n
is
n^m.]
n
an increasing function, the statement
in
Ex.
3. Show that the positive proper fractions can be arranged in a definite order, so as to form a sequence. Represent the first few terms graphically indicate terms which form subsequences converging to the upper and lower bounds as a limit. [The sequence, omitting such fractions as |, is
1 is
true. ;
'
|
I
!
;
1
1
1
2'
3'
I
9
3
1
2
4'
5'
5' 5'
The required subsequences dotted
lines.]
3
5'
6'
6
arc indicated
by the
-Q
FIQ. 41.
GEOMETRICAL CONVERGENCE 4.
un >0
If
[See Ex. 5.
6.
n and un+l /un ->l then (nun
for all values of
XXIV,
and Art.
11,
}
\+1+ 2 n (\+ V 3
un ^
,
)
n -+l.
9, (2).]
Prove that lim -
If
245
. . .
-
+
)
-0.
[Art. 9, (3).]
n/
show that
->
and consequently
e,
(n
e.
4-
"n-l
7.
n
- .^
z~
,
"~l.n
5-
+
71
1
~w
-7i
prove that
1
1
oo
[Show that
r(7i
k>0
prove that
(ii) if
>
where
.
for
and
(j) if
.
+ -^T;,
-zr*+
prove that un -*
If
1
1
If
9.
,
i .
+l
1
8.
+
...
3(71-2)
a,
r
= l,
2, 3,
...
n.
Hence show that
and negative roots of x 2 and %>0, then ^ ->a; and v 1
-/? are the positive
= */ (k + u^) vn = */ vn
un
__ l )
(A:
-x-k
rt
Illustrate geometrically. [(i)
Here
t*
n
2
= ^ + w n -i ^n
-
1
~&+w
'
f2-2
wn 2 ~ wn - 1
~ u n~i -Un-zi
hence u n '^u n _ l9 according as u n _^'u n _ z therefore (u n ) is monotone. 2 2 u t ~ a) (u l + /?), J^ir5< suppose that u t > a, then u l >u 1 +.k for 'i^! - u t - k ( 2 = 2 a u and is /. /. u 2 2
i/,
,
)
.
f
^n -> Z ^ a.
The
Thus wn
case in which
i^ n_!
and
->
follows that
u l
of y x and (ii) Draw graphs meeting at A. In Fig. 42, we have
in the
same way.
y~
PrPr ~ u r ~ u r+\*
and thus
The
it
figure illustrates the case in
which
Ui = ON = N p > a, l
1
l
and shows that (un ) is a decreasing sequence. Thus J\^rPr > Nr pr for every r, and so Nr is always to the right of N. Thus un -> I ^ a and
Hence
N
r
tends to
N
as a limiting position
^
and un -> a.]
NN r N 2 FIG. 42.
N,
CONVERGENCE TO ROOTS OF EQUATIONS
246
Wn=/( wn-i) where f(x) is such that (i) f(x)>0 and (iii) the equation x=f(x) has a single w that ->a. prove n [It is advisable to draw a figure.] 10. If
increases with x,
11.
Show
number
that as the
tend to the roots of x 2 - x - 7
=0
if
*>0,
positive
f(x) root a,
(ii)
of roots tends to infinity, the expressions
as limits.
3 = v7-v7+y, prove that 2/ ^( a; 2_7)2_7 last the Draw a graph of equation, and prove that 12. If
%
un ^
\7-V7-f \/7-N/7T...
%, w 3 u5 u Z9 u^ u$
then
,
to
if
2n
roots,
a decreasing sequence,
... is
... is an increasing sequence, and that both sequences converge to 2 as a limit.
Prove also that
\7-fv7- V 7 WT"^"... 13. If
to oo ^3.
u l9 v1 are given unequal numbers and
n^^
for
u n decreases, and vn increases as n increases, (ii) Each of the functions u n and vn tends to a limit, which is the same for both. This limit is called the arithmetico- geometric mean of u and i\ (Gauss). ~ v = i ( \^w_i - N/v -i ) 2 > u > v for ever ^ prove that
K
(i)
*
n
>
n
Un - I u n~i + vn-i) < Wn-i ...
.'
Thus v^^ Hence u.n tends
l
t;
Vn
.
t*
i
= s/(^n -i^n-i)> ^n-i-
.
and so does vn Also lim vn = lim n_ 1 = lim (2un - Mn^) = lim to a limit,
y
n
and
(
n .j
CHAPTEK XVI CONVERGENCE OF SERIES 1
.
Let u n be a function of n which has a definite value
Definitions.
An
for all positive integral values of n.
which every term
in
is
R Ui v
by
,
sn
expression of the form
followed by another term,
This series will be denoted by En^\u n
n terms by The sum
(1)
,
or
by
is
27w n
,
called
an
infinite series.
and the sum
of its first
.
of the
p terms immediately
following the nth term
is
denoted
so that **n, v
As n tends
~ w n-f 1 + u n+2
~^~
~^~
u n+j> ~ s n+v~~ s n-
to infinity, there are four distinct possibilities
to a finite limit, to infinity or to
minus
infinity, or it
:
sn
may
may do none
tend
of these
things. If s n
tends to a
called its
sum
finite limit s,
to infinity.
Thus
lim
s n = s,
the series
is
said to be convergent and s
is
defined by
s is
or briefly,
lim s n
= s.
n->oo
This
is
also expressed
w t + w 2 + w3 + If s n
tends to
oo
by writing
=s
to oo
...
or to -
oo
,
or
the series
27fw n = s, is
or
2u n = s.
said to be divergent.
tends to no limit, whether finite or infinite, the series is said to oscillate. We say that the series oscillates finitely or infinitely, according as s n oscillates between finite limits, or between + oo and - oo If s n
.
Series
which diverge or
The sum not a
sum
be non-convergent. essentially a limit, and
oscillate are often said to
(to infinity) of a
convergent series
is
in the sense described in the definition of addition.
We
is
are
therefore not justified in assuming that the sum of an infinite series is unaltered by changing the order of the terms or by the introduction or
removal of brackets. In fact, changes of this kind
may
alter the
sum, or they
may
a convergent series into one which diverges or oscillates. Thus the series (1 - 1) + (1 - 1) + (1 - 1) + ... is convergent and zero,
but the
series 1
1
+ 1-1 + ..
.
oscillates.
transform
its
sum
is
ADDITION OF SERIES
248
The Geometric
2.
converges
l^l^l and
if
oscillates finitely if x== - 1, For sn (l-z n )/(l-z) if
=
5 n~>oo
If
x^l,
If
x< - 1,
.
s n->oo
If
x=
or
-
Theorems.
3.
Uj
The
Series.
series
+x + x* + ... +xn - 1 -t-...
1
its
sum
and
oscillates infinitely if If |z|
is
It
l/(l-x).
diverges if
x< - 1.
x^l. -1, sn = l
and
or 0, according as n oo according as n is odd or even.
(1)
m
If
+ Wg + tia-f
is
...
a given number, the
is
* n->l/(l
-x).
odd or even.
series
um+l
and
converge, both diverge or both oscillate.
6
For
let
Then
5n
n = sm 4. n -Sfn
(i) if
(ii) if
(2)
= w 1 + w 2 -f...+w n
sm+n
an d
tends to a
5
m
is
,
*n
a fixed number, hence
finite limit or to
oo
so also will t n
,
;
sm + n tends to no limit, whether finite- or infinite, neither will t n
If k
.
a fixed number and Eu n converges to a sum s, then Eku n is its sum is ks. Also if the first series diverges or oscillates, so
is
convergent and
does the second, unless & = 0.
-oo
if s n ->s, then ks n ~+ks. If k 9^0 and s n so. does ks n tends to no fixed limit, finite or infinite, neither does ks n
For $n
,
.
If
.
4.
Addition and Subtraction of Series. wx
converge and (i) (ii)
(Wj
4-
(MJ
their
Vj)
Vj)
+ (u 2 -f v 2 ) + -f-
(ii)
v2 )
(u 2
-f
NOTE.
-f
sums are
w2 -f
In the proof,
The proof
is
-f
. . .
all
w2 + s
. . .
. . .
and
- . .
and
t
vl
+ v2 +
If the
series
. . .
respectively, then
converges,
and
its
sum
is s
converges,
and
its
sum
is s
+ w n ) + Iim(v 1 -f v 2 4-
the series in brackets are
. . .
+1
;
t.
+ v n ) = s 4- 1.
finite.
similar to the preceding.
SERIES OF POSITIVE TERMS
We
first
heading we
consider series in which all the terms are positive. Under this include series in which all the terms are positive beginning with
some particular term. 5.
Introduction and Removal of Brackets. Let Zun be a series and let the terms be arranged in groups without altering
of positive terms,
SERIES OF POSITIVE TERMS Denote the sum
their prder. (i)
// Su n
For
converges
to
a sum
let
Then
m tends
where
Now
lim
s
n we can
terms in the nth group by vn then ,
so does
s,
2v n
.
=
s
for every
of the
find
m and p,
to infinity with n. therefore
m = lim sm + 9 = s,
so that
lim t n = s.
If Zv n converges to a sum s, so does Zu n For corresponding to every n we can find (ii)
.
m
But lim tm = lim tm+l = s,
y
so that
lim s n = s.
therefore
Changing the Order of Terms.
6.
249
// the terms of a convergent remains convergent and its
series of positive terms are rearranged, the series
sum
is unaltered.
Proof.
manner. is
a
Let
Eu n -be
Denote the
.
first
n terms
Further, suppose that the
(n +p) terms of
Zu n
of
first
Zu n
m
are,
among
the
first
n^m^n+p,
m tends to infinity with
Also
n. sn
< >m
.
lim 5 n ==lim 5^ +J> ==5, therefore lim m = s. Hence Zvn is convergent and has the same sum as
Now
NOTE.
The argument
fails for
MI
Zu n
.
such a derangement as
+ wa 4- u6 -f
.
..
+ uz + w4 + w e +
. . .
,
where Zun is broken up into two (or any finite number of) infix For here we cannot find m so that the first n terms of Zun m terms of Zvn Thus u2 does not occur till infinitely many of ti have been placed. The rule for the addition of aeries applies in cases of this kind, but Sun is broken up into infinitely many infinite series. Derangements of the last sort are considered in another volume. .
m
tsrms of
terms of 2v n are among the
.
Thus
and
,
Let
u.
Suppose that the
Zv n
convergent, and let the terms be rearranged in any v n so that every u is a v and every v series by
new
first
COMPARISON TESTS
250
Theorem.
7.
of positive terms cannot
series
and
oscillate,
than some fixed number k, the series converges and
less
always
A
if s n is
sum
its
is less
than k.
For
XV,
(Ch.
and therefore tends to a
increases with n,
sn
Hence the
3).
every n, then
s n ~>s,
s is
Comparison Tests. Ua n is known to be
8.
terms and
* (i)
(ii)
u n ^a n for
* or if or if
(iii)
Proof,
u n/a n
to
+
...
a + a2
If
(i)
u l + uz
and since For
(ii)
is
n
less
to oo
-r ...
be
Eu n and Za n are divergent, then Uu n will (ii) (iii)
3).
=
we have
,
...
a n
un
follows that
it
;
is
;
convergent.
Now
.
ka n
is
convergent,
convergent.
//
(i)
for
limit.
+ u n ^.a 1 +a 2 -f
This follows at once from
*
XV,
than some fixed positive number k
for
(ii),
if
lira
a positive constant k, such that u n /a n
;
we have u n
values of n,
all
than k (Ch.
//
a finite
a fixed number,
is
un
therefore (iii)
t
sn
if
Further,
less
2u n and Sa n are two series of positive convergent, then Zu n will be convergent if (1)
always
u n/a n tends
oscillate.
a fixed number
values of
all is
cannot
series
where
finite limit or to
u n /a n
all
we can
is finite,
find
values of n.
two series of positive terms and
Za n
is
known
to
be divergent if
^a
un n for all values of n ; * or if u n /a n is always greater than some fixed positive or if
u n/a n tends
to
a limit greater than
/. :
s
;
zero.
TV be a positive number, no matter (i) Let Proof, Ea n is divergent, m can be found such that
how
great.
a l + a2 +
. . .
-f
>N
provided that
n> m
t^-f w 2 +
...
+w n >JV
provided that
n>m,
an
number k
Since
;
divergent. values of n,
u n >ka n
% therefore Eun
is
.
Now Ea n
is
divergent
;
therefore
divergent.
at once from (ii), for if lim u n /a n is finite, a positive ound such that u n/a n >k for all values of n. .y these tests, we require certain series which are known or divergent. The geometric series and the series of the .hose which are most generally useful as test series. flcient
that these conditions should hold for
all
values of
n
greater than
some
METHOD OF COMPARISON Theorem.
9.
The
series
p>l and divergent if Denote the given series by convergence or divergence
any way we
in
First
let
is
+ H^ + O +
TJ
S
Group the terms
is
Now
the
(B)
unity, for
Next, if
thus:
as follows
:
1
1
,
2, 4, 8,
and the
terms, respectively. than the corresponding term in
is
/ 1
1 \
!+_+_ +
same as
1
1
1
!+_+
or
...,
a geometric series whose
common
+ ....... (B)
ratio 1/2 P-1
is less
than
.
last
Each term
,
terms in the brackets are 1/2 2
brackets contain
Now
etc.,
,
in (A) after the first is less
(
is
convergent if
p>l. Hence (B) is convergent, and therefore S is convergent. p = 1, the series S becomes 1+J-f^ + J-H... Group the terms i i+ i + i +i + ... .................. (C) i + + + i) +( i)
where the
which
^s
is
,
...
I
which
.
terms in the brackets are l/2 p l/2 2p l/2 3p
brackets contain
Each term
+
a series of positive terms, its not affected by grouping the terms in brackets Since
S.
1
first
.+-
please.
p>l.
where the
.
251
2, 4, 8,
of (C)
...
4
3 ,
1/2
,
1/2
,
and the
etc.,
terms, respectively.
greater than the corresponding term of
is
the same as
+i + f + f + ...,
1
l+ + i + J + .......... (D)
or
and therefore S is divergent. Lastly, ifp
is
divergent,
;
fore
S
is
10.
divergent.
Examples on the Comparison Tests.
1.
Is the series
i
- -f
~
= .
K
Q
1
Ex.
Z
.
3
j.*>.4: let
+ o~~l
an
r o 4 o .
=
+ ~*
converge^ or divergent
*
Then we have
=
Now Zan
is
convergent
;
un
therefore
*
is
?
.
2
.
convergent.
The nth tenn a w of a suitable test series is found thus Keeping only the highest powers of n in the numerator and denominator of u n we obtain 2n/n8 or 2/n*. Disregarding the numerical factor 2, we take !/ for e* n :
,
.
B
B.C.A.
D'ALEMBERT'S TEST
252 Ex.
Test for ro nvcrgence or divergence the series x xn ~ l 1 x2
2.
<1,
If
a;
n
l+#2
,
l+o: 3
->0 and vn lxn ~ L
->
a:~
n ->
and
w^ 71
#n
=1 1
a:
= 1, u n = ^ and
If
}
the series
is
+-
...
l+zn
+
Now Zxn ~
1.
convergent. If a;>l,
+
_
4.
^
1+z
+ xn
=1
l
...
where x>0.
,
convergent, therefore
is
1
1 --
x x~ n +
->
Hence
.
x
1
ZV-, n
is
divergent.
divergent.
TESTS DERIVED FROM THE GEOMETRIC SERIES
A
D'Alembert's Test.
11.
u n+l ju n
vergent if
9
where k
Zu n
series
is
of positive terms is eon may have an\
a fixed number and n
value.
The
series is divergent if
For
if
u n+} fu n ^l for
u n+l /u n
Thus w w
71
.
2^7/j
.
all
values of n.
n, then
"1
A-
convergent, therefore
is
27w n
ii
convergent.
Again,
if
u n+l/'u n ^l for
+u n ^nu
w t -f w 2 +
therefore
values of n, then
all
1
-*cc
In particular, if u n ^/u n tends
l<\ and
to
.
Hence Su n
a limit
is
divergent.
then 2!u n is convergent whei
Z,
when />!.
divergent
For if Z<1, choose k so that l
of a limit
?
Wi + is
convergent, and therefore
Again,
if
/>!, then
um+2 + NOTE,
(i)
is
m
Zu n
tt
is
m+2 +... convergent.
can be found so that u n+ ^>u n for
divergent, and therefore
It is sufficient that the conditions
un
is
n>m.
Henc<
divergent.
should hold for
all
values of
n
greate
than some fixed value.
Nothing is said as to the ease in which Urn un+l lun ~l. In this case the be convergent or it may be divergent, and we say that the test fails. may For example, consider the series (ii)
1111 + + + +
i
The If,
first is
2
3
divergent, the second
however, u n ^^u n tends to
the series
is
divergent.
The
_
"'
4
series
and
1
r
+
1
+
2i
1 i
+
1
4*
+
is
convergent, and in both
1
as a limit
from
1+2+3 + 4 +
...
seriei
--
we have lim un +Ju n = 1 u n+l >u n
above, so that always is an instance of this .
CAUCHY'S TEST 12.
A
Cauchy's Test.
Eu n
series
253
of positive terms
is
convergent if
i
w n n <&
is
a fixed number and n can have any value. i
The
series is divergent if
u n n ^l for
many
infinitely
values of n.
i
u n n
if
series
Zk n
many
values of n,
,
con-
is
i
w n w >l,
If
Zu n
then
is
and consequently w n >l,
for infinitely
obviously divergent, i
In particular, if u n n tends and divergent when l>l.
The proof
Zu n
a limit L then
to
is
convergent when l
similar to that in the last article.
is
It is sufficient that the conditions should hold for
n>m,
where
m
is
a
i
number.
fixed
but
Also nothing
the limit
if
is
Zu n
approached from above,
Comparing these
13.
said as to the case in which lim u n n
is
useful than Cauchy's.
concerned, u n+l ju n
is
is
divergent.
more generally with which we are
tests, that of D'Alembert
For
most
in
the series
of
\,
a simpler function than u n
is
.
On
the other hand, Cauchy's test is more general than D'Alembert's. That this is the case will be seen by noting that
Cauchy's condition for divergence
(i)
is
greater than a fixed value,
and
this
is
For
wider than D'Alembert's.
in D'Alembert's test the condition has to be satisfied
by
all
values of
n
not the case in Cauchy's. i
(ii)
It
has been shown in Ch.
XV,
9, (2),
that
if
u n+1 /u n->l then u nn ->l.
i
But
if
u nn ->l,
So we
does not follow that u n+l /u n tends to any limit. expect that Cauchy's test will sometimes succeed when it
may
D'Alembert's
fails,
as in Ex. 3 below. 2
Ex.
Show
1.
that the series
ic
+
+
#3
+
yA
+ ...
is
convergent if
0<#<1
gent if
un =
Here Hence, If
and
x
is
if
1,
xn -
n
,
un n+1 =
0<:r
~;
n+1
.*.
lim
n w n41 =lim T un n-fl
series is convergent,
D'Alembert's test
divergent.
xn+l
fails,
and
if
but in this case the
x>l, series
xx. it is
divergent.
becomes
and
diver-
EXAMPLES OF SLOW CONVERGENCE
254 Ex.
Show
2.
that
series
the.
+x + x2 /
1
+ x* f
12
+ ...
convergent for all positive
is
(3 "~~
values of x. />.n-i
TT
~n
7?
lim
II
-= 1' lim
iv
=i
-
series is convergent.
+ 6 + a 2 4- 6 2 + a 3 H- 63 -f
a
Test for convergence the series
3.
*
,,,
!
'
.
Hence the Ex.
=
=
Horo
. . .
.
-^
Here
according as n divergent
is
a^l,
if
test
is
odd or even, the or if b^\.
series is
0
if
convergent
w n n ->a a or 6 2 and 0<6<1, and
test, since
Using Cauchy's
inapplicable.
i
i
~
and D'Alerabert's
EXERCISE XXVI 1.
State the character of the series
showing that
for the
second series lim$o M ~-l and lim
2.
Show
that the scries x
3.
Show
that,
if
ci
l -\-
-?>
+ x ^ + x2
+ 3 + a t ) 4-
a2
(a,
ft
...
-\-
is
5 2n _
3
x~ 2 + x + x~* +
-}
)
I
3
(
4-^4)
Criticise the following 2(-J
+
i-
+
i
+
J-f-.-)
which 5.
(iii)
is
absurd, for
If
n
is
11
J
<1,
^-i-*-l-i
+ l-+-J +
J
1
<,
v
'
(ii); v
>
some
fixed
number m.
=(i + 4 +4 +
...)
+ (* + ! + i +
)
= n-i+4 + ...,
,..11111
1
first
1
million terms of the
. .
that
(m +
:
The sum lies
(iii)
is
....
4-
Given that 2 20 > 10 6 4- 1, show that the sum of the series I -f i f 1 4- j -f is less than 20.
Show
never convergent.
and so on.
1
.
6.
-..
...
a positive integer, show that 1
is
...
:
=i+ .-.
2.
1
convergent, so also
J (a 2 + a 3 +
Prove also that the converse is true if a n >0, for n [Consider the sum to n terms of the second series.] 4.
. |
between
Hence show 4-
m-fl
and
m+1 ,
. .
.,
must be taken.
with an error
-~
(m-fl) -f(ra
that, in order to find the
^ + 4~r +
o
-
to infinity J of the series
less
2
+
/rr2 + : ~^r2 + (w-f2) (m-l-3)
rri.
+ 1) 2 sum to
than
infinity of the series
001, at least 1000 terms
CONVERGENCE OF SERIES Determine whether the n
1
,
9.
-
a+6
1
+
+ 26
+
a + 36
3.4
.2
1
Exx. 7-14 are convergent or divergent.
series in
o.lll
1
a
+
.
10
.
. .
255
5.6
3
1
7
_5_
~ 13.
7
J_ + __ + _?_ + ....
-A-
H.
Test for convergence or divergence the series whose nth terms are 15.
19.
-^_.
is.
-JL_.
17.
/t
-. +a
20.
n-
21.
N/rc
2n
.
/_4.
is.
_______ 2 + l-n.
22.
_
V^-fl-s/n 8
1
.
|
For what positive values of x are the what values are they divergent ? 23.
25
'
l+2x + 3x* + 4x* + ... .
Z
1
29.
32.
terms
--
x
2
30.
.
-
-
31.
.
:
refer to series of positive terms.
Zun
is
34. If
Hun
is
35. If
Eun
is
-2! ^ -^t-
1
convergent and
convergent, so
un
vn
^t-1 ^ ^=^ vn un
divergent and
[After a certain stage
un z
is
un
for
,
37. If
2un
z
is
convergent, so
Find the sum to n ^ 1.3.5 ,
+
ww ->0
2
.
2.5.6
then Zt'
is
is
convergent.
divergent.
un 2
/.
;
).
+ ~~^-~ -f ... 5.7.9 4
3 i
1.4.5
-
n ^ m, then Zvn
Zun f(\ +un is Sun ln. [For ww /
infinity of the series in
3.5.7
for
.
36. If 27wn is convergent, so is
A 41)
2
Explain why D' Alembert's test cannot be stated thus A series of positive convergent if the ratio of each term to the preceding is less than unity.
33. If
-
a;
is
Exx. 33-37
38.
-
..... 26
i
3a
2a
1
7.8.9
Exx. 23-31 convergent, and for
+ */2 2 4- x 2 /
24. 1
.
,
4.5.6
1.2.3
series in
_i
i
3.6.7
.
Exx. 38-41. 39. -
-
1.4
-f :r~
2.5
3.6""
.,12
41
3
-i
1
.3
.
7
3 .6
9
u
--
i
.
5
.
7
.
11
ABSOLUTELY CONVERGENT SERIES
256
TERMS ALTERNATELY POSITIVE AND NEGATIVE
Theorem.
14.
alternately positive
The
and
u^-u^ + u^-
series
negative, is
...
,
in which the terms are
convergent if each term
is
numerically
less
than the preceding and lim u n =^0.
We
Proof.
and
have
can be written in the form
this
= w l-( w 2- w 3)-( w 4- w 5)----- w 2n
*2n
Now
Wj-i/a, w 2 -i/ 3?
From
(A)
it
3
-w 4
,
...
are
all
therefore follows that s 2n
positive, for
is
and increases with
positive
n,
(B) always less than the fixed number u v Hence, s.2n tends to a limit which is less than MJ (XV, 3).
and from
it is
seen that
2n is
$.
Again, S 2ni l ~s 2 n + u 2n+i an( l l* m U 2n+i same limit as s 2n The series is therefore convergent, its sum .
therefore
\
tends to the
s 2n+l
.
Ex.
Show
1.
that the series
1
-
^ 4- -3"--
4
+
...
is
positive
and
less
than u v
is convergent.
Here the terms arc alternately positive and negative, each term is numerically than the preceding and lim l/--0 hence the series is convergent.
less
;
SERIES WITH TERMS EITHER POSITIVE OR NEGATIVE
Absolutely Convergent Series. The series 27w n containing positive and negative terms, is said to be absolutely convergent if the series 15.
2
1
un
,
is
convergent.
|
Hu n
convergent and
E
u n divergent, then or conditionally convergent semi-convergent. If
is
Thus the series 1
-f
series
f+
\
;
+
\
l-| + i-i + --+
ally convergent, for
An
Theorem.
Zu n
Let
|
.
.
.
it
is is
|
Su n
said to be
is
absolutely convergent, because the
is
But, 1-^4-
convergent.
convergent and
1 4-
+
1
3 4-
3-
1
-$+... 4-
. . .
is
is
condition-
divergent.
absolutely convergent series is convergent.
be absolutely convergent
;
then by definition
Z
\
un
is
con-
|
vergent.
Now
i/
n
4j
un
=2u n
\
or 0, according as u n
positive or negative.
is
Therefore every term of the series (u n + u n \) is term of the series 272 u n |. corresponding convergent \
>0
and
is
< the
1
Hence Z(u n + u n \
convergent.
\)
is
convergent and consequently, by Art.
4,
Zu n
is
PRINGSHEIM'S THEOREM NOTE.
if
we say
is
absolutely convergent, and not that of 2u n itself.
,
\
Zu n
that
another series, un From Arts. 11 and 12, \
it
a fixed positive number
is
absolutely convergent
less
than unity.
Since s n
of the positive integral variable n,
Zu n
sufficient condition for the convergence of
Corresponding it
to
must be possible
//j/7/ HUM
any
to
-jo t-o,
j
Thus
Zu n
if
is
positive
m
find
7? it n>
Ex.
is
p 1.
to
tend
if
y
-4^
convergent and
allowed
Apply
Mft-fi
I
I
and in particular u n ^0. But the condition (A) is unless
so that
-4-
'*n~f 2
is
'
*
in
to oo
-U II
* *
'
<^~ ~-
f
ensure the convergence of
Hu n
any way whatever.
11 4 + '"~
2 "3
7f w
'
p ~
n+
1
f "'
h
4-
1
n+2
+...+
n+p
For any fixed value of p, p/(n H />) -> as ?i-> oo hence, Rn p cannot tend to zero, and so the series
>
n~\-p
But
.
is
t
Pringsheim's Theorem.
'
lp
/i'
1
1
Here
tive
I
n-f j?|
the general condition to the series
+
17.
and
:
any integer,
sufficient to
'not
a function
e that
?/
p
as follows
is
is
(B), the necessary
we may choose, however small, u^in^for every positive integer p,
number
'}/
j,
a
after
|uj<*,
or if
General Condition for Convergence. by Oh. XV, 6,
16.
if,
i_
,
?p is
Zu n
follows that the series
certain stage,
where k
statement of this theorem, he should we assert the convergence of
If the reader fails to see the point in the
note that,
257
// Su n
if
2
ri
p
,
then p/(n
-f
p) ~>
1
;
not convergent.
is
a convergent series of posi-
decreasing terms, then nu r ~>0.
For
if
p
is
any
positive integer
as W->OQ P u m+v< u m+l+ u m+2+ + u m+jT> = Let p m, therefore mu 2m >0 and 2mu 2m->0 i.e. nu n -*Q, when n = 2m. and Let p = m 4- 1 therefore m + 1 u 2m +l~-> ;
,
(2m +
1)
u 2m+l
(
)
= 2(m + 1) u 2m+l - u2m+l ~~>0
This proves the theorem, whether n Ex.
1.
Use
Hero nun and, since
this
n-
it is
;
-
n
theorem
to
show
=1, and does not tend to it
nu n
when n = 2m -f
1
.
Hence E- cannot be convergent n
;
i.e.
>Q,
odd or even.
is
that the series
a series of positive terms,
;
27
zero.
-
n
is divergent.
must be divergent.
REARRANGEMENT OF TERMS
258
of
Zu n
sum
Removal of Brackets.
Introduction and
18.
Denote the
be arranged in groups without altering their order. terms of the nth group by v n then
of the
,
2u n
If
(1)
Let the terms
is convergent, so also is
Sv n and
these series have the
,
same
sum.
For
Then
= !*! +
$
let
for
any value sn
of
= m
+ ...+M n
2
n we can find
or
t
,
sn
t
= v l +v% + ...+vn
n
.
m so that either
= tm + (ww4l + w m+2 +
.
..u m + 9 ),
where the terms in the bracket are included in the group vm+1 If
Zu n
is
convergent, fi
n,p
w n+1 + t w+2 +...ttw4 ->0 as n~oo, therefore s n - Z m-> 0. as n ~> oo
:=s
.
m~> QO
also
J>
,
Hence Svn converges to the same sum as Su n If
(2)
Sv n
is convergent,
Zv n
For, given that for all values of p.
Denote
(3)
as in
Ev^
.
|
un
is
Eu n
is
.
not necessarily so.
convergent,
we cannot conclude that
by u n \ and suppose that the terms of Zu n
\
f
Rnjtf->Q
are grouped
'
Let v n be the sum of the terms of the nth group. (1). is convergent, so also is u n (Art. 5) and, as n -> oo
Then, if
'
,
;
Therefore, as in
(1),
Zv n
Rearrangement
19.
converges to the
of Terms.
same sum as Zu n
.
The following example shows
rearrangement of the terms of a semi-convergent series may alter its sum. and a rearrangement of the signs may change the series into one that
it
which Ex.
is
1.
divergent.
With regard t 1
I
sum
Lft's be Ihe
to the, series I
I
oj the tern i -convergent series
-5-4*3-5
I
I
1 + 5-
/ri
i
1
1
1
i
1
^r^rr^r-
,
,
........... (B)
......... (C)
obtained by rearranging the terms and the tigns of (A) respectively, prove thai (B) is convergent with a
(i)Let
then
w
sum
- 1 -5
{s
and
(C)
w
divergent.
-r- ---,-... to M terms.
^
or
w
= l -~ --H-- -- -~ + ...
to
n terms
:
SUM ALTERED BY *. 1
1
1
--lim9 2n =-*. Hence, the second (ii)^t
n
*>(
then
converges and
series
<
Also
= i + l_l + l + l_l
-+-
-
}+( -+- -O
D
I/,
Now
the series
+- +-
1
t
sn +%
&U tend to
oo
-f
is
. . .
3
2i
Hence the
.
1
divergent
;
series (C) is
Theorem.
// the terms of an absolu remains convergent and its su?
the series
Denote the new
We And (u n
have un since E\
un
-f |
Let
E\u n
\
|),
-f
\
|
un
v n) so tha series by u n = 2u n or 0, accor is
\
a convergent se
for its terms are
<
the c
Z\u n \=s and Z(u n + \u n \)^ and (u n + \u n \) -are series o
unaltered by any rearrangement of tern
27(^ + 1^1) =
2\v n \=s> Ex.
Find a range
1.
^
of values of x far whicJ 2 z l+(x-* + (x-x*)
-
)
can be arranged as a convergent
series of ascendi
Expanding the terms of (A) and removing h
Now,
|a?-a^|
if
justify either the
However, by Arts. convergent
where
/
a?
;
16, 18 (1), 19,
and, by Art.
=| x
|,
is
convergent
is
NOTE.
be the case
that
;
x'<\(*J$-l) which
each of these
5, this will
t
or
is,
if x'
+
-tON/5-
the required range.
What
has been said does not preclude
can be made over a wider range.
ERROR
7
^arching for the sum of an infinite content with an approximate value.
and >,ries
approximate value of
i
Rn
sum, denoted by Thus we have (A). its
,
called
is
It should be
5.
s.
)er
limit to the numerical value of the
number which
o find a
|
Rn
cannot \
t series
w 4-...+(-l) W w +...
,
Here
ive terms.
n+ 3 Brackets
Hence lij less
is
positive
the
than w n
less
than u n ^
error in taking s n as _ _ i
ror in taking x
1
an
.
- x2
/2 as the
sum
of the
than x 3/3.
3ss
.n
and
which
erms are
all
the terms are positive or in the same sign. This is
all of
n often proceed as 3
--K"H 3
xn
n
+... where
follows.
0<^<1.
The
series is
(
n +1 numerically less than x / (n
le of s is
;ence or Divergence, lue of
Rn
+1
) (
1
- x).
For a con-
for a given v&lue of n, the
more
ge.
ilue of r y the
2
+
more rapid is the convergence
convergent, but in order to find mals, at least 1000 terms must be taken /4
. . .
is
fore say that the series converges slowly.
SERIES OF
-v"
Next, consider the of a million terms
series
is less
1 -f
|-
f
_
than 20
(st
therefore said to diverge slowly. In prat, the reckoning so that we have to deal
w
For instance, the series, 5 = 1 - J J By grouping the terms, we find '
1
This series converges more rapidly 8 5 *-2[t + i(i) + i(i) +.
find that
rapidly than a geometrical series '
22. Series of
Complex Zx n and Sy n c
that the series sn ,
r n be the sums to n
cr n ,
Ch.XV,10,
tei.
lim Sn
= lu
and we say that z n converges ti Thus Ez n is convergent if Zx n a
(2)ByCh. XV, U^t of Zz n it
is
as follows
must be possible \s n+v -s n (3)
to
Definition of
As
zn
find
tha
\<,
terms and 2\
Corrcs
:
P
is \
for real sei
convergent.
then 2\
zn
is c \
Consequently (4) It is
impo.
2x n and Zy n9 an The
first
Hence
if
part
Z \
c
xn
\
The theorems c brackets and the c
(5)
of
complex
series.
This
1
T
;
'
,
FOR SERIES
is
..+zn
complex, the series
~l
and when
1
convergent,
z
if |
+ ...
|>
0, (3), 3
sum
is
1, z
its
n does
sum
is 1/(1 -z).
not converge to
*->0 and
s n -*l/(l-z),
1/(1 -0).
0
re
- r cos
+
ir sin
1
- r cos 6
that these results are also true
1
1-5
FIG. 43.
then
w n->0.
then we can find
ULTIMATE SIGNS OF TERMS
Now
m m+I m+2 + +
the series
-
H
. . .
The Binomial -
1
is
~ 7 -~ + n(n-l)(n-2) When n
called the Binomial Series. its
sum
is (1
which we denote by
+ x)
n
In
.
+ ti 1 +
tt
2
+
all ...
stage, the terms are alternately positive
'
if
+ t< r +
...
.
1).
which
1, 2, ...
,
,
Convergence, etc. We shall prove that the series 1 ; and when x = l, the series converges if vergent if x (2)
\
\
Denote the
Proof.
n-r r+1
ur
and the
<
n = -1 and
finitely if
oscillates infinitely if
is
-.x,
r->oo
as
series is absolutely
;
convergent
x = l.
. . .
is absolutely
n> - 1,
con-
oscillates
then
,
therefore lim
x
if
greater than n, negative, or all
w< - 1.
by UQ -h u + u2 4-
series
a certain
negative, or they all have the
um+2 ... are alternately positive and have the same sign according as x^Q. ,
+
the beginning, or after
and
m is the least of the numbers 0,
the terms u m um+l
x
a positive integer, the series terother cases, it is an infinite series,
same sign according as x^O. For w r +i/ M r = (n-r) xj(r +
Hence
of n, the series
+ 1) (n-r *
...
is
From
Ultimate Signs of the Terms.
(1)
w n->0.
For any given value
Series.
9 + nx + n(n-l) -Vs ' ^ +
minates and
divergent, therefore
and
uju n->
is
263
|
\
<
*r+l
1.
^
and u r have opposite signs, so that, after a If r>n, ur Next let certain stage, the terms are alternately positive and negative. Also
r-n (i)
Also |
n>-l,
If
u r+ i
|
<
|
ur
(ii)
//
n= - 1
(iii)
//
n<-l,
as |.
-si
\
r
r~>oo,
a,,.->n
Hence, by
the series let
is
n + l=
Hence u r -> oo and the The case in which x~ - 1 |
+-T where a r = Hence, by Art. 24, w r-0.
+ l>0.
Art. 14, the series
1-1 + 1-1 +
-m
so that
..
.
is
and
w>0,
convergent. oscillates finitely,
then
series oscillates infinitely. is
considered in Ch.
XX,
6.
CONVERGENCE OF SERIES
264
EXERCISE XXVII
RANGE OF CONVERGENCE Show
3
. K
that the series in Exx. 1-5 are convergent.
L
_J___ a+ , |
1
a+2 a+3
1
I
L__JL
,
1
1
1 1 L _____
__ ____ __ 22 32 42 52 I
I
.
6.
Prove that the
7.
Show
V'2.4 3
"
,
+4
a
4
J
2
1
-
3
'
5
2.4.6
,
'
62
series
that the series
X -l+x
#2
-
1 4-
,
*
x + x2
j
2
+4-
X^
-
-
...
*
^3/ 3
4-
. . .
is
1
|
convergent 6
is
if
values
a divergent
series.
convergent for
of x t real or complex.
8.
9.
Show Show
that - +
a
-
-a+2
-
- -f
~
a-f-1
a
+3
+
-+ a-f5 -
a+4 ;
...
is
that the series
3
is
2
convergent if /n <4/27. [Consider separately the 10.
is
(i)
sum
odd and the sum of the even terms.]
of the
Find a range of values of x
for
which the
series
convergent.
(ii) Find a range for which the series can be arranged as a convergent series of ascending powers of x.
11. If
0
0-4,
show that the series 2 1 4- (2x cos 6 - x + (2x )
cos 6
- a: 2 ) 2 -f
. . .
can be arranged as a convergent series of ascending powers of x. 2x cos [It is enough to show that with the given condition
\
|
4-
x 2 < L]
12. Let the terms of Su n be arranged in groups, without altering their order, and denote the sum of the terms of the nth group by v n Given that 2vn is convergent, prove that Eun converges to the same sum as .
Evn
in the following cases
(i) (ii)
If the
:
number of terms
in every bracket
If all the terms in each bracket have the
[See Art. 18. In case (i), J?n , j,->0. v n tends to zero, and case (i) applies.]
In case
is finite
same (ii),
and w n->0.
sign.
v n -+Q
;
hence, every term in
SUMMATION OF SERIES Show that the series l/(l+i), and find its sum.
l+z + z~ + ...
13. 2
14. If z n
=
+-~i n
n
show that Ezn
L>
is
is
265 convergent
absolutely
when
convergent, but not absolutely con-
vergent.
Rn
Prove the results in Exx. 15-17, where the given series.
the remainder after n terms in
is
^
15.
For the
series 1
16.
For the
series
+ry-
1
+ 7-5 - - +
r
1
1
. . .
\R n \
,
o
I
that
I
X 17.
For the
series
1
Q
18
AnL--
27
+-J +
-.
20. If
2L
l_
1^:
Exx. 18-23
in
X^ i
2
that ~ x provided
--
tOQ0
^ ]
or
!^
.
as
TI~> ^cording
|
x \$l.
then
a(a
6-2>a>0,
Show
l
|n
+ l)
+
a(a-f l)(a
+ 2)
6-1 ^ ^^
f "
"
'
then
___ __
o
24.
r-
""' i
6-l>a>0,
6
+1
19 ii/
/-
+
tl/
n+
:
L -i- + -l- + _i- + ...to*=- T ? I-x* x-l
|*|>1,
a
23. If
Rn <
1
li + l-i4 + ri~;8 +
22. If
X^
!_
IJ
Prove the results stated JLU<
X^
+ r^ + |- -+...,
+r
6(6
+
that the
+
6(6
1)
sum
l)(6
cos I
^(6"-a-l)(6-a-2)'
n terms of the
to
is
+ 2)
{sin
series
+ cos 2^ + cos 30 + ... - 1}. (n + 0) cosec \Q
Hence show that the series oscillates of TT, in which case the series diverges.
finitely
except
when
6
is
an even multiple
n
[2 cos r6 sin
25.
of
TT,
26.
Show in
(r
+ 16) - sin (r - 10).]
that the series Z'sin nO oscillates finitely except it converges to zero.
when
6
is
a multiple
which case
Show
convergent, onvergent,
by
-sin
that the series z
when
z
= cos + -
1
+ |2 2 +
sin -
4
plotting the points 8 19 s 29
4 ...
$7
.
z
3
+ ...
is
convergent, but not absolutely
Illustrate the slowness slownc of the convergence
on an Argand diagram.
CHAPTER XVII CONTINUOUS VARIABLE 1
it
We
Definitions.
.
passes once through
cumstances x
is
say that x varies continuously from a to b when real values between a and b. Under these cir-
all
called a continuous real variable.
Limit of f(x).
Consider the function
When x = Q,f(x)
f(x)
= {(1
has no definite value, for
-
3 -f-
x)
l}/x.
then takes the form 0/0. In other words the function f(x) is undefined for the value zero of x. For 3 2 2 every value of x except zero, we have f(x) = (3x + 3x + x )/^ = 3 4- 3x -f x it
.
Thus, if x
is
nearly equal
to zero,
f(x) is nearly equal to 3 : further, by the difference between f(x) and 3 as
making x small enough, we can make small as we please. This is roughly limit off(x) as x tends to zero.
what
The matter may be put precisely number e, no matter how small. If x
|
f(x)
Thus, as x tends positive
number
-3 1
meant by saying that 3
as follows
|
therefore
is
\
< 1,
is the
Choose any positive
:
then
x
provided that
|
\
<
\
|/(z)-3| becomes and remains less than any we may choose, no matter how small. This is precisely
to zero,
that
meant by saying that 3
the limit off(x) as x tends to zero. The usual It should be observed that f(x) never attains the value 3.
what
is
definitions are as follows (1) //,
I
any
small, there exists I
where
:
to
corresponding
how
matter
~ f( x ) 1
1
is
<
positive
a number
number
rj
provided only that
and a are fixed numbers, then
I
c
which we
may
choose,
no
such that
x-a |
\
<
is called the limit
TJ,
off(x) as x tends
to a.
This
is
expressed by writing lim f(x)
=
I.
z-+a
Or we may say that /(#)-> I as x-^a, where it is understood that x may approach a from either side, i.e. x may tend toa+Oora-0. Observe that nothing is said as to the value of f(x) when # = a. Such a value
may or may not exist.
In any case, we are not concerned with
it.
THEOREMS ON POLYNOMIALS corresponding
(2) If,
exists
to
any
number
no matter how small,
c,
there
N such that - < provided only that x> N, f(x)
a positive number
1
|
where
positive
267
I
This
c,
1
off(x) as x tends
is
a fixed number, then
is
expressed by writing lim /(#)== Z or f(x)->l as x->oo
is called the limit
I
to infinity. .
'x-+
,(3)
If
any positive number number N such that
a positive
or
no matter how
and we write /(#)-> oo
that f(x) tends to infinity with x,
correctly) lim f(x)
(less
oo
great, there corresponds
x>N,
provided only that
f(x)>M, we say
M,
to
as
x->oo
.
3->oo
The reader should be able to frame for himself corresponding for the meaning of f(x) tends to minus infinity as X-+CG
definitions
'
.'
Fundamental Theorems.
2.
tend to the limits
lim (u v) = I + V, liml/w = l/Z unless
(iv)
infinity,
lim (u - v) = I - 1',
(ii)
(v)
Z--=0,
(iii)
Two Theorems
(1)
= aQ + ax -f a 2 x 2 4-
is
If
number
17
any
. . .
f(x)
a 1
and the
how
small,
provided only that |
Z'
make
for
said in this article coefficients are real
we can find a
positive
|
'
f(x) a Q
then
kx'f(I
\
<
al
|
|
/(k+*)
is
f(x)
\
|,
-f |
,
\a n
-a 1
f
),
<
that c,
<
77.
;
1
that
is if
x
if
a suitable value of
\
\
a 2 x' 2
-x')<, provided
kx'<(l-x Therefore
|a 2 x'
...
x \
k be the greatest of the
let
\
(a^,
then
= x' and
x
let
numbers
so
is
easily
= IV, = 0.
a n xn , then
positive number, no matter
For shortness,
Proof.
and
-f
lim uv
so that
|
If
What
on Polynomials.
holds good, no matter whether the variable x or complex.
Let f(x)
then
limufv = l/l' unless
Apart from some verbal changes which the reader can himself, the proofs are the same as those in Ch. XV, 2. 3.
x which
u, v be functions of
V as x tends to a or as x tends to
I,
-f-
(i)
Let
|
\
<
/(k
+ e)
;
77.
B.C. A.
QUOTIENT OF POLYNOMIALS
268 (2)
//
number
M
is
a positive number, no matter how
m so that
\f( x )
The
letters
\>M,
provided only that
having the same meaning as in
|/(*) !=*">.
great,
-
n -l
+n-2
we can find a
positive
\x\
we have
(1),
^2 1
i
1
X
Choose any number
e
and a n
between
|
1
-
X
For such values
of
x we have
= i|a w
if
theorem
#'>
is if
f(x)
\
> x' n
{|e&
.
n
\
- e}.
then
|,
K
|/(x)|>t|a B
Hence
last
<
that
-
|
In particular, put
Then by the
1
< K+
if
. \
if
x':
m is the greater of the two numbers and
then
f(x)>M
From Theorem
(1) it follows that
if
\x\>m.
lim f(x)
= aQ *
a;->0
= Again, if x y -fa, then y->0 as x-^a and /(a) Hence of y in the polynomial f(y + a). -
lim /(x)
.
1.
// f(x)
_3
=
the term independent
lim f(y -f a) =f(a).
Km /(a;) -3
is
;
(i)
0,
(ii)
,
(iii)
cw
a?-
3
^5' (ii)
Let
a;
= 1/y,
then y ->
*-~
lim
->
as
lim /(X) (iii)
*
Let x = 1
-f y,
then
as x -+
t/
The reader must understand that
a?
oo
->
1
and
and
this is not
a roundabout way of saying that f(x}^a Q when
CONTINUOUS FUNCTIONS 4.
269
Continuous and Discontinuous Functions.
We
confine
ourselves to one- valued functions of the real variable x.
We may say that/(x) if
is
a continuous function of
the curve whose equation
where
xa and x =
yf(x)
is
is
or
in the interval (a, 6)
continuous between the points
6.
This supposes that we know what is meant by a continuous curve/ representation of the curve in Fig. 44 can be drawn without allowing '
A
the pencil to leave the paper, and we say that it is continuous. In Fig. 45, the curve cannot be so drawn near the points where x = x and x~x 2> and
we say that
the curve
is
discontinuous at these points.
FIG. 44.
We curve
:
FIG. 45.
choose the following as the distinctive feature of a continuous As the point P (x, y) passes along the curve, any small change in the
value of x is accompanied by a small change (or by no change at value of y. is true everywhere in Fig. 44, but xl and x=*x% in Fig. 45.)
(This
x
More
precisely thus
P(x, y) any other point on c, no matter how small. values of |
Further,
X~XQ
PQ (xQ
let
:
on
it
|,
is
point on it. Thus the curve
AB is
(1)
(or at the point
XQ
,
for all sufficiently small
continuous
is
if it is
P
.
continuous at every
x=xt
and
x=x2
it is
discon-
.
single-valued function f(x) is continuous at x = x or for # ), if for any positive number e that we may
The
however small, a number provided that x XQ
\
.
\y-yQ \<
if
continuous in Fig. 44, but in Fig. 45
tinuous at the points where Definitions.
yQ ) be any point on the curve and Choose a positive number ofP
said to be continuous at
we say that a curve
in the
not true near the points where
is
either side
Then
the curve
,
all)
77
exists
such
that \
f(x) -f(xQ ) <<=, \
\
This is equivalent to the following f(x) is continuous at x iff(x)~*f(xQ ) as X-+XQ, where x may tend to XQ either from the left or from the right. :
TYPES OF DISCONTINUITY
270 It
important that the student should fully realise all that If /(#) is continuous at z then
is
the definition just given. (i)f(x) has a single
is
implied in
,
definite value
when
x~x
;
h tends to zero through positive values, then /(x + A)-> a number L 19 and/(x - A), a finite number L 2 (ii) if
finite
;
i x =/(# =
and we must have
(iii)
)
."#. 1. A simple case, in which f(xQ ), given by Goursat.
If f(x)
zero,
= x2 +
x2 ...
^-f 1 ~}-X"
and so
f(x)0
x2
+7
:
4-
(1
,^i T* ar)
+
...
2
-
L 19
to oo
,
L%
but are not
all
equal,
is
when #=0, every term
of f(x)
is
all exist,
then,
but, for all values of x except zero, f(x) = 1 -f x lira /(*) = 1, so that f(x) is discontinuous at
2.
;
Thus, /(O) =0, and
# = 0.
*->0
TAe function f(x)
(2)
is
continuous ih any interval if
every point of the interval. If the interval (a, b) is closed (a
f(x) as x-+b-Q.
continuous at
end points are included,
x->a + Q and
continuous at a if f(x)-~>f(a) as
is
it is
at b if f(x)->f(b)
b X
FIG. 45.
Types of Discontinuities, 7
C,
C
as
x->x l from the
,
then y-^y^ as x-^x l i.e.
right,
lim JC
(ii)
In Fig. 45 let y v y/ be the ordinates of from the left, i.e. as x-^xl -0 and y->y%
(i)
^ 7
'^^ 7 JL
~
^ fl/j
and
lim X
At the point where x = x2
Thus y
This
as x-^a^-hO.
f
2 example, 1/x and 1/x
expressed by writing
/
^ 3/1 ~H
supposed that y is infinite. For x = x1 and at # = #2 are discontinuous at x = 0. in Fig. 45, it
is
a discontinuous function of x at
is
is
v = Vo. 7a
,"
.
that f(x) has real values only on one side of the In such a case the curve y=f(x) stops abruptly and/(x) is discontinuous at # (iii)
It
may happen
= point x x
.
.
Taking x,
we
see
(as usual)
Jx
that v x - x
to denote the positive square root of the is
discontinuous at x
.
number
FUNCTION OF A FUNCTION
Theorems on Continuous Functions.
5.
(1) It follows
x=x
from the
called
a,
and product of these functions. = 0. except when
// f(x) tends lim tinuous at x^lythen
y=
If
function of a function of
Theorem.
and
x,
is
is
{f(x)}.
and
as x->a,
I
then y
w=/(x),
denoted by
= cf>{lim
if
where
(f>(u)
a finite limit
to
= lim/(x) = Z, then /(z) Z = continuous at x I, therefore
For
are continuous at
is true for the quotient f(x)/
Function of a Function.
(2)
and
definition that iff(x)
so also are the sum, difference
,
The same
is
271
con-
is
f(x)}.
+ 7?, where
77
-->0 as
x->a;
also
lim
Continuity of Rational Functions. If/(x) is a polynomial, by Art. 3, lim/(x) =/(&), where a is any real number.
6. then,
x >a
Therefore the polynomial
continuous for
is
values of x. a polynomial or the quotient of
all
Hence also any rational function of x (i.e. two polynomials) is continuous for all values of x except for such as make the denominator vanish.
The Function x n
7.
// n
.
is
rational,
xn
is
continuous for
all
values of x for which x n has definite real values.
Suppose
(i)
that
x>0.
x
If
is
any
positive
number, we can choose
a, b
0
so that
;
Now
~ both x"- 1 and x n 1
greater of the last If
then
c is
(ii)
|
7/x<0,
At
<,
|
and
bn
let
~l ,
hence
|<&.|w|.|x~rX
provided that
continuous at x = x
x-x |
is
the
|.
small,
=-.
1
k
if
r
.
n I
I
.
x= -y, so that xn = (-l) nyn where it is supposed that Now yn is continuous for positive values of y> therefore
xn has a real value. xw is continuous for (iii)
x
w
~1
any assigned positive number, however xn - x n
is
xn
two numbers,
|
Hence xn
between an
lie
the point
negative values of
x = 0.
If
n
is
x.
positive,
limx n = 0.
If
n
is
negative,
s~>0
xw ->-oo or to tinuous at
x=
-oo, according as x->4-0 or to -0.
Hence xw
for positive, but not for negative values of n.
is
con-
CONTINUITY OF A FUNCTION
272 8.
Fundamental Theorems.
and f(a)^Q, then f(x) has the neighbourhood of
the
(1)
is
continuous at x=*a
same sign as f(a) for
that is to say,
a;
If f(x)
if
values of
all
a-r]
r]
where
x in 77
is
arbitrarily small.
For since /(a)^0, we can choose e so that f(a)-e and f(a)+e have and since f(x) is continuous at x a, we can find
the same sign as /(a) so that 77 f(a)
-
:
For such values
provided that
-f e,
of x, f(x) has the
\
<
77.
same sign as f(a).
// f(x) is continuous for the range the value a to the value 6, f(x) assumes at (2)
and
x-a |
then as x passes from least once every value between f(a)
a^x^b,
f(b).
From
the graphical point of view the truth of this statement
is
obvious.
Let A, 6,
and
We
let
B
be the points on the curve y=f(x) whose abscissae are a and k be any number between /(a) and f(b).
assume
that
any
straight line
which passes between the points A,
B
cuts the curve at least once. It follows that the line y
varies
=k
cuts the curve at least once.
from a to b,f(x) takes the value k at
Hence as x
least once.
y-k
x
a XQ
l
x
f
FIG. 46.
Proof.
Suppose that
tween /(a) and
f(b).
f(a)
of x in the interval (a, 6) for which f(x)
there are values
because f(x) (a, 6) for
is
a
which f(x)>k.
DERIVATIVES Divide the real numbers in the interval
The lower class
A
is
to contain every
273
(a, b)
into
two
classes as follows
:
number x such that f()
any number
in the range a
A
(#!<#' <#2
some
) belongs to class A', for although f(x')
It has been
shown that both
(a, b) is
included,
Therefore the classification defines a real number. x
A
in
for
classes exist.
Also every number in the interval than any x'.
number
f()>k
or the least
number
A
in
,
and every x which
is
is less
either the
greatest It remains to prove that /(# ) = k. we can find If f(x )-k<0, by reason of the continuity of f(x) at x so that f(x)-k<.0 if # Thus f()
,
-<<#
<
Hence x +
belongs to class A, which is impossible, as it is greater than x If f(x )~k>0, we can find e so that/(x)-fc>0 if XQ Q + c. Thus XQ e belongs to class A', which is impossible, for it is less than x .
-^X^X
.
Hence it follows that/(# ) = case when /(#)
It
.
is
NOTE. If f(x) =k for more than one valuo mined as above, is the least of these values. 9. Derivatives.
If f(x) is
easy to modify the proof to suit the
of
x in tho interval
(a, b),
is
and
by the equation
denoted byf'(x).
o:
,
deter-
a function of x such that
tends to a limit as h tends to zero, this limit is
then
Thus/'(x)
is
defined
called the derivative off(x),
The function f'(x) is also called the differential coefficient of f(x). It is possible that for some values of x, this limit does not exist. If x is such a value, f(x) has no derivative at X = XQ This is certainly the case when .
discontinuous at X = XQ
For in order that {f(xQ + h)-f(xQ )}/h f(x) tend to a limit as h-+Q,f(x + h) must tend to/(x ). is
Ex.
1.
Wo
have
// f(x)
=xn
,
where n
f'(x)= lim
is
.
a positive
integer,
prove that f'(x)
may
GRADIENT OF TANGENT
274 10.
to a Curve.
Tangent
Q
be a point which
P
from either
Let
P
supposed to move
is
be any point on a curve and let along the curve, so as to approach
It is supposed that the curve continuous near P, so that Q may be as near
is
to
P
as
we
side.
like.
T'PT
If a straight line
approaches
PQ
P
exists such that, as
from either
makes with T'PT
tinues to approach P) remains less than
we may
Q
the angle which becomes and (as Q conside,
choose, however small, then
any angle
T'PT is
called
47.
the tangent to the curve at P.
This at
P
is
sometimes expressed by saying that
is the limiting
position of the chord
PQ
as
tangent to a
the
Q
curve
tends to coincidence
with P.
Notice that
P,
PQ
The curve
(ii)
is
(i)
Q
would cease
direction
at B,
supposed a chord,
in Fig. 48
thrown from A,
ball hits the
is not
to be
is
to
reach P, for if
Q
were
to
coincide with
supposed to represent the path of a ball which ground at B and proceeds to C. Just as the
strikes the
it is moving in the direction TB and, just after, in the We may, if we choose, say that the curve has two tangents BT and BT Strictly speaking, according to the definition,
ground,
BT'
namely
.
'.
the curve has no tangent at B.
This at
is
an instance
of a continuous curve which has
no
definite tangent
one point.
B FIG. 48.
Gradient of Tangent. on the curve whose equation
1 1
P
.
curve has a definite tangent
PT
Let
be the coordinates of a point is y=f(x). We shall suppose that the at P, and that this tangent is not parallel (x, y)
to the y-axis, as in Fig. 49. Let Q be the point on the curve whose coordinates are (x
+ A,
y-f k).
DIFFERENTIAL COEFFICIENT Draw
QM, and draw PR
the ordinates PiV,
R.
Then
and
since
P and
($ are
Now,
as
PT and h
y + k =/(x 4- h)
gradient of chord
.'.
Q tends
PQ = n\
J^^M
.
h
;
gradient of
is the
;
PQ tends to the limiting position
Pr =
lim^4^W'(*)^-~>;o
Hence /' (x)
OX to meet QM in
k =/(* + h) -f(x)
/.
;
to coincidence with P,
tends to zero /.
parallel to
on the curve, we have
and
y =f(x)
275
fl
gradient of the tangent to the curve y
f(x) at the point
fey).
P
PT
tends to is nearly parallel to OY, f'(x) will be large, and if If coincide with a point where the tangent is parallel to the y-axis, we may expect that/' (x) will tend to infinity. Ex.
1.
Find
the equation of the tangent to the curve,
point whose abscissa
whose equation
y = x?,
at the
is a.
If f(jc)~x^, we have /'(#) the gradient of the tangent at the point 3# 2 .*. 3a 2 , and the equation to the tangent is 3 2 -a) or y = 3a z 2a ;
is
is
(a,
a3 )
.
12.
Notation of the Differential Calculus. In the differential is given a different name and a different
calculus the derivative of f(x) notation is used.
Any number by
z-f 8x; and,
nearly equal to x is denoted if
responding to x thus we have
+ Sx
is
denoted by y + 8y
2/
This limit
is
denoted by
(with regard to x). to x,
In finding
and the process
We
often write
-=^.
,
is
of differentiating f(x)
:
is
-=-
we
called 2Ae differential coefficient of y
are said to differentiate y with regard
called differentiation.
-~ in
dx
and
ctx j
;
where ^~ denotes the operation the form ~7 -/*(o:), v ' *
it is
dx
also written shortly as -~-
ax
.
DIFFERENTIATION
276
Rules of Differentiation.
13.
(1) If u,
v are functions of
d
,..
(l)
.....
(Ill)
dx
,
v
v
'
(.
.
and
in (v)
Proof.
dv
iP\
dx
dx
dx
and
-=-
ax
have
(i) (iii)
and
v
du dv T"""T"> dx dx
,
'
I du 5 -Tu ax
d /1\
.
-
i
i
>
definite values, also that in (iv)
Let
-5-
ax
(ii).
8 (uv)
Sv> 8(uv), etc.,
Su.,
,
^~ are not ax These are
= (u + Sw)
be the increments in w,
x.
infinite,
left to
v, wv, etc.,
Sw and Sv tend to zero as 8x->0.
the student,
( v + Sv)
d
^+
lim(fF
dw dx
dv
dx
+ fc)lim
s
'
l^___Sw_^ du dx
A^V ^ v ^ dx\v/ dx dx 1
N
v
/
^
1
1N .
dx v/
du
v dx
u dv dv_l1 / du 2 v dx~^v*\ dx
fo J3 + l)2-(2-l)8_
'da:
.
w
N
v
d /^N /u\ dx\v/
(i) (
rf
(Sar+l)"
-l
dvN dv\
dx/
5
~(3a: + l)*'
.
du
1
wlim(u-fSw)
u(w-fSu)
We have
u
v^Q.
responding to an increment 8x in
Because
.
dx
ax-u/
du
-=-
~
.
^
..
d
T-""
'
(iV) -y-
,
/
1
assumed that .
,.. % ^
dx
.
.
dx\v/
j
dv -J-
d dv du T- (UV) = U -7- + V -7dx dx dx
d /wN
It is
du Tdx
then
x,
-_*._
2
cor-
POWER OF A FUNCTION Ex.
Slww
2.
277
a product and a quotient can
that the rules for the differentiation of
be written in the forms
du dv dp /!)___+_ p dx u dx v ax ...
where (iii)
1
1
1 dq = 1 du---1 dv u \_^ qdx u ax v ax
.... /
,
an
q = ujv.
p = uv and Deduce
1
that
2
if
= (u1 w2
un )l(vt v2
...
u
wtare
v m ),
...
u 29
t> , 2
..., t^,
...
are functions of x; then
\- Z zdx (i) ,
If
p
uv,
then
,
-V-T- + u-=-
~f-
ax
dx
and,
;
dx
each side
if
divided by
is
p(=uv) the t
result follows. (ii) (iii)
This follows in a similar manner.
Let
7
UjU2
...
^n V
v^
,
...
__...
/ Z_ \J
and by repeated application
I
'
= !^+! du*
u 3 u^
= Z ~ m(-~}; ^^ V dx ^\v dx 8
i ssc
Prove
that,
dx
I
rr U
7
,
*j^
j:
V dx
dx
.
2
un)
w3 55"
dx
... i*
n
and the
a function of
is
rs=1
dx
\ur
result follows.
J
s
3.
I dV dz _ _ dU ~_ _ _
z
'
!
dx
Ex.
1
of the rule for a product,
dU _ 1 d% 1 rf(w U dx ~u dx u2 uB ... un 1
SimUarly J
then
;
onH cHlU.
[/ r
/ /
vm
and -~
if
y
let
n be a positive integer
x,
then for
exists,
all
rational
values of n,
Let
w
y
n \
and . f t
(ii)
Let n
(i)
Idw == 1 dy 1 dy _. _^__^. w dx y dx y dx
-m, where
m
is
<
Hence, by^
wy = 1
.
and
m dy 1 ^w ~ y- +- dx = w dx y
,
,-r
(i),
dy __j____ y dx
a positive integer _
/.
,.
then
;
1
/
w~y
y
.
y^
.
n terms) = ndy T -
...
y (n factors)
;
.
y dx
;
then w=y~~ m ,
m Idw -~ 1 dy + m -|~ =0. w dx y dx - -=-
~1 rfw -r- =-
or
.
w dx
y dx
then
wy*
P (iii)
Let n=p/q where p and q are integers 'u
cr
Hence, by J
That
is,
we have
/-\ (i)
w or
/--v (11), ' v
in all cases
d2
^ c?y i-^^-fl-^ wdx zdx ydx
g
-
1
rft^
-=-
=-
*f oo?
NOTB.
;
~
;
* (
oo;
and
and
t)
9
say.
- dw ndy --. 1
w
dx
y dx
and multiplying each
^ )=ny
wQ =y =z
side
by w(=y
n ),
-i.*
It should be carefully observed that there
00; is
always a factor
y-,
unless
y=s.
FUNCTION OF A FUNCTION
278
Function of a Function.
14.
is
If w=/(x)
(1)
and y~
a single-valued function of
x, continuous for
a single-valued function of u, continuous for the correvalues sponding range of of w, then yisa continuous function of xfor a^x^b. Let x be any value of x in the given range, and w , yQ the corresponding is
u and
values of
y.
Because y =
for
,
any assigned positive number
e,
provided that \u-uQ \
and because /(x)
,
and consequently
\u~UQ\
\y-yo\
if |
x-x
]
This proves the theorem.
If the derivatives
(2)
values of u,
and
if
f (x) and
a
(u) exist for
F(x) =
fW-f <) ./<>,
then
ttati,,
|=|.g.
Let 8w, Sy be the increments u and y corresponding to an increment 8x in x, then
Sw^0,
if
8y^8y Su Sx
du ax
8u
.
--->- and
As o#->0, fi
oo;
~8u'
du -r~ dx
since
and therefore
ou
w
Sx
.
is
.
a finite,
Sw->0
-+--. du
Consequently
dy = .. 8y _. 8w dv dw r ax lim/-lira^--^-ow ox du dx
,^ v (B)
/
It
may happen
that Su =
for
some value
_
In this case we must have 8y = 0, for otherwise 7
7
= 0.
hence
and then equation (A)
of Sx,
does not hold.
Also
-^-
= 0,
8w = 0.
for
~ would not be
finite
U/U
Thus the equation
dy ^dy du dx du dx is
true also in this case. Ex. (i)
/Y (11)
1
Find
.
Let
,
u=ax + b,
*
Let
^
ti
a
(i)
then
70.1
-f-
when y =
(ax
y=u n
,
1
ox% then y =u
+
and i
,
n fc)
and
,
dx <^V -
dx
(ii)
=-~ du /
=( \
when y = .
dx 1
=nu n~ l .a=na(ax+b} n~ l
\ )
u*J
'
v
~,
.
26x =
26a;
(a+bx*)
2
.
.
:
SIGN OF FIRST DERIVATIVE
//+,!, a 6
&.*.
We
279
'
2
2
regard y as a function of x, and differentiate both sides of the given equation
with regard to x thus 9
2k 2y dy ~ + 7? -5? =0. a j cr eta
15. Derivative of
xn
// y = x
.
n wfore
w
is
any
rational, then
dx x
If
Proof. of h,
is
positive, x
and by Ch. XIV,
Also
a;
71
"1
is
x
is
positive for all sufficiently small values ~ 1 -1 'and nhx n l . lies between
is
nh(x + A)*
a continuous function of
dy -ax If
+h
+ h) n - xn
10, (x
Therefore
x.
--
..
(x
hm
v
+ h) n -xn '
h
h-+Q
- u, then = ( y
= negative, let x
l)
n wn
and
r~
n n l when n is a Or,* having proved that the derivative of x is nx positive integer, we can prove that this is also true when n=p/q as follows.
p
j
Let y = a^, therefore y* = x p ,,
f
therefore
16.
qtf^
.dy 1 1 /- = px*>dx ^
.
Now _
.
and
Meaning of the Sign of
7
q
~T-y
= qy9 " 1 ~/-
-
~ dy = px p l = pdx qy- 1 q -/
.
,
xp ~l
Z^v
xq
Iff(x )>0
f'(x).
9
--*
= -p x ^ q
then for all values
^
x of x in the neighbourhood of x /(x) 0. ,
sufficiently small values of /
Similarly
t//
(x
A,
Hence
/(x + A) ~/(x
)<0, /or every x in z
.
S
JAe
a5
has the same sign as ) neighbourhood of x
for
A.
,
5x
.
NOTE. If /'(a;)>0 only for the single value # of #, it does not follow that f(x) increases steadily as x increases through XQ For if x t
that f(Xi)
17.
and f(x2 )
)t
and we cannot conclude that f(xi)
Complex Functions of a Real Variable is
See also Art. 13, Ex.
3.
+ i<(x), + i
x. If y=/(x)
called a complex function of x and its derivative rules of Art. 13 obviously hold for such functions.
then y
The
)
is
MAXIMA AND MINIMA
280
Higher Derivatives.
18.
the derivative oif'(x) exists, it is called the second derivative or the second differential coefficient of f(x). If
this is written in
y =/(#)>
If
of the forms
any
So by n differentiations (when possible) we obtain the nth derivative or the nth differential coefficient of f(x) which is written in any of the 9
forms
The, following are important instances. (i)
If
(ii)
If
t/
= zn
-t = nxn - 1
,
,
= a^xn + na^*- 1 + - ~^~1
a 2xn ~* +
(
f(x)
4-
. . .
a n)
L?
then /' (a?)
=n
a^f
1
+ (n - l)a^- 2 + (n
Or, using the notation of Ch. Ill,
= n(a f'(x)
then
and by successive
,
a l9 a 2
,
...
2)
1)
^" f(x)
a 2^~ 3 +
= (a Qy a v
a n _Jx, I)*-
a2
,
...
. . .
+ an
a n $x,
w l)
,
...3.2. (a^-f a x ),
Maxima and Minima. We say
that/(z) has a
maximum
XQ when f(xQ )>f(x) for all values of x in the neighbourhood In other words, if a number exists such that f(xQ )>f(x) when
at X
then /(x replaced
)
is
by
a
<
maximum ,
,
1
differentiation
Hnfn- 1) 19.
1, if
"
value of /(x). If in the preceding, the sign then f(xQ ) is a minimum value of (fx).
Maxima and minima
value of
>
z
.
is
values of a function are sometimes called stationary
or turning values.
A
necessary condition that f(xQ )
may
be a
maximum
or a
value of f(x)
is that f'(xQ ) =0. This follows from Art. 16 but the condition ;
is
not
sufficient.
minimum
TEST FOR MAXIMUM OR MINIMUM If f(x
)
a maximum
is
281
then for sufficiently small values of
value,
A,
Hence by Art. 16, as a increases through x the sign of ). from -f to -. f'(x) changes Suppose now that /"(# ) exists and is. not zero. Then since f'(x) decreases as x increases through rc f"(xQ ) must be negative. /(z + A)
,
,
Similarly sign of f'(x)
be
minimum value, as x increases through z the f(xQ ) must change from - to + and if f"(xQ ) is not zero, it must a
is
if
,
,
positive. If the sign of
/
-
(x)
does not change as x passes through x
is
f(x)
,
neither a
maximum nor a minimum. The /" (x = will be considered ma 7 be a maximum, a minimum or neither of these. later. In this case /(# To illustrate this geometrically, in Fig 51, the tangents at A, B C to case in which
)
)
9
the curve y ~f(x) are parallel to
OX.
FIG. 51.
The
sign of f'(x) for different parts of the curve
has
f(x)
maximum values
Fig. 52 represents the
C and
graph of
t/
a
minimum
= x3 -f 1. Here
as indicated,
Thus y 1.
Here
Thus
-=^
= 3z2
dx
is
neither, a
Search for
maximum
nor a
,
so that
-~=0
ax
maxima and minima
/'(*)=6*
a
minimum when x=0. values of
-6*-12=6
/'()=0
if
*=-l
or
2.
As x increases through -1, the sign of f'(x) changes from increases through 2, the sign of /'(#) changes from - to 4Thus /( - 1) is a maximum and /(2) a minimum value.
+
to
-
.
Or
and
value at B.
dy =0, but as x passes through 0, the sign of -~ does not change. ax
when
Ex.
at A,
is
thus,
f"(x) = 12s
- 6 =6(2* -
so that f"( - 1)<0 and /"(2)>0, leading
to the
1),
same
results as before.
;
and as x
CALCULATION OF LIMITS
282
20. Points of Inflexion.
APE
In Fig. 53
is
supposed to be part
of a curve represented
by y =/(#). moves along the curve from A to B, the gradient increases along the arc AP and then decreases. If a point
Thus
~
maximum
has a
ax
of the tangent
value at P,
,
2
and therefore
-~=
at P.
dx* If
Q
is
the secant
a point on the curve near P, QP cuts the curve at another
point Q' near P, and we may regard the tangent TPT' as the limiting position of the line QPQ\ when the three points Q, P,
and Q' are very Definition.
minimum
is
At such a
Q FIG. 53.
close together.
A
point
(x, y)
on a curve at which j-
is
a
maximum or
a
called a point of inflexion.
dty -~
point,
dx*
and the curve
0,
crosses the tangent.
EXERCISE XXVIII Find the limits of the functions in Exx. 1-3 tends to oo . 1.
(i)
as x tends to zero
;
3.
'
(1
+*)- (l-
Find the values of 4. lir
Prove that
(i)
10.
Prove that
sin
[sin (x -4- h) -sin
a;
8.
Iim
1;
a;
hm ,.
6.
(ii)
Iim -
a:
= 1.
x and cos x are continuous functions of
x.
=2 sin^A cos
For what values of x
12.
Prove that
r
Iim
Iim cos
11.
L
5.
sin
9.
-7-
sin
is
tan x discontinuous
x = cos x 9
Bm(x + h) -sin a; = 2 sin _ %h
(ii)
-=-
,
.
cos x ,,
v
?
- sin x
y
CO8 (x -f \h) ~> cos x as
-=-
tan x = sec 2
x.
as x
DERIVATIVES AND TANGENTS 13. If
f(x)=z(x-a)
Hence, 14.
m (x-b) n
if
m, n are
prove that
,
positive, f'(x) vanishes for a value of
Prove the following d \r
283
x between a and
b.
:
(i)
dxj
dx (m) 15.
\
(d
n
_
1
_1 / jl l-x*-2\"((l-
(dx)
1
+ ~
}
'
l)
Find the derivatives of (iii)
n (iv) sin z. 16.
Find -%
n (v) cos #.
if (i)
(vi)
ax 2 + 2hxy + by* + 2gx + 2fy + c=Q,
tan n tf.
(ii)
if
x 3 + y* =
(Z-iC
,-
17.
_
2
cZ o;
.
12 =
Prove that
rfy
- d*y //dy\*
-^ / aa: / \^/ 2
.
(
/
^
2
18. If
y=A
19. If
y = cos x -f
20.
cos
Prove that
if
|Lia;
+ J5sin
sin
21. is
r7 y then T-O
"
M
2
2/-
then -~ = iv.
or,
dx
= tan-
[Consider the values of
IJLX,
1
l
/
a/a:,
(x
-
then
I /
10)
(x
+ ia).]
Find the equation to the tangent at the origin to the curve whose equation
y
on the parabola y ~px 2 + qx + r whose abscissae are a - h, a + k respectively, prove that the tangent at P is parallel to the
22. If P, Q, Q' are points a,
chord QQ'. 23. 24.
x(x25.
Prove that the function 2x 3 - 3x 2 + 6x - 5 always increases with Sketch roughly the graph of y = x(x~ 2 for values of a; between and 1.
2
I)
,
and
x.
find the greatest value of
I)
(i)
If f(x)
=(x-l)*(x- 2),
for
what values of x does f'(x) vanish
?
= to x = 2, and show that one (ii) Sketch roughly the graph of y=f(x) from # of the values of x found in (i) gives a minimum value of f(x) and that the other gives neither a maximum nor a minimum. 26. Prove that aa; 2 -f
T
x~
-b/2a
bx -f c according as a
^ 0.
gives
a
maximum
or
a
minimum
value B.C. A.
of
INVERSE FUNCTIONS
284 27.
When
(i)
does the function 3x* - 4# 3 -
36a; 2
-
1
increase with
x
?
Find the turning values of the function, and state the character of each.
(ii)
Show
28.
that the turning values of (x-l)(x -2) (2 -3) are approximately
0-38.
Four equal squares are cut off from the corners of a rectangular sheet of 8 inches long and 5 inches wide. The rest of the sheet is bent so as to form an open box on a rectangular base. Find the volume of the box of greatest capacity which can be formed in this way. 29.
tin,
30. The regulations for Parcel Post require that the sum of the length and girth of a parcel must not exceed 6 feet. Prove that the right circular cylinder of greatest volume which can be sent is 2 feet long and 4 feet in girth.
Show
31.
that the height of the right circular cylinder of greatest volume ball of radius r is 2r/\/3.
which can be cut out from a spherical
21.
Inverse Functions.
Let f(x) be a single-valued function of
x, continuous for the range a
shall
prove that
The equation y=f(x) determines x as a
(1)
function of y. This function
x=/~
1
(y)
denoted
is
mean the same
by/~
thing.
1
(j/),
single-valued
so that the statements
Also the function/"
1
is
continuous
y=f(x) and
called the inverse
of/.
Since f(x) is continuous, as x varies continuously from a to 6, Also f(x) takes f(x) takes every value from /(a) to f(b) at least once. Thus, correevery value once only, for f(x1 ) =f(x2 ) only when x 1 = x2 Proof.
.
sponding to any value yQ of y between /(a) and/(6), there XQ of x such that yg=f(xQ ). l For, taking Again, f~ (y) is continuous. the case in which y increases with x, let yQ be any value of y between /(a) and /(&) and
is
a single value
I
XQ the corresponding value of #, so that
For
+
sufficiently small values of c,
x -
and
Let yQ -k
X Q~6 X Q X^ X and y + k be the corresponding values of y, FIG. 54. and let 77 be the smaller of the two k, k'. As y varies from y -7? to y + 7?> x ^ es between # -e and XQ + C, for a; are within the interval
(a, 6).
f
increases with y.
x~x
Hence |
\
-1
and therefore x =/
<
(y) is
,
provided that
continuous at
j/
|
.
y
PRINCIPAL VALUES
285
- k' as the values of y y decreases as x increases, we take yQ + k and y ~ To show that/~ 1 (y) is continuous at and XQ + corresponding to x If
.
y=zf(a) and y =/(&), we consider the variation
(2)' ^
// J
For ~ax
if
-~- exists
dx
and
is
then ~-
a
dy
Sx and 8y are corresponding increments of x and
is finite,
as Sz->0,
8y
dx T dy
ox
,.
-
22.
never zero Jfor the range y
of y for the ranges
and
.
by
Sv
/,.
1
Si"
I
I
~-
.
dx
then, since
Therefore
vice versa.
Joi,
,.
^-
y,
=1
,
lim^ = l
Idy --
.
dx
The
value of
Inverse Circular Functions. The numerically smallest = y which has the same sign as x and is such that sin y x is called
the principal value of sin~ l x. A similar definition applies to tan^ 1 x. The smallest positive value of y such that cos y = xi$ called the principal value of cos~ l x.
Y
Y
Y
l
jy=s//i"
jr
FIG.
Thus the principal values ,
and that
of cos" 1
x
lies
of sin" 1
between
5^>.
x and tan" 1 x
and
lie
between - - and
n.
Unless otherwise stated, sin" 1 ^, cos" 1 ^ and tan"" 1 ^ will be used to represent the principal values of the functions for these, the derivatives of ;
sin*"
1
a;
and cos" 1 ^ are
positive
and negative
respectively, as can be seen
from Fig. 55. Ex.
I.
Show
Let y = tan~ 1
that
-=-
dx a;,
then
tan~ l x =
l+x*,
x=t*ny and
.
dx -=-=8
dy
'
dy te
=
1
r+?'
BOUNDED FUNCTIONS
286
Bounds of a Function.
23.
Let f(x) be a function which has
a definite value for every value of x in the interval (a, b). If a number exists such that/(z)
M
is
(a, 6),
then/(x)
said to be bounded above. If
a number
N exists such that/(x)>2V for every
a;
in (a, b),f(x) is said
to be bounded below.
A function
which
is
bounded above and below
said to be bounded.
is
Suppose that f(x) is bounded above in (a, 6). Let S denote the aggregate of the values of /(x) as x varies continuously from a to 6. applying Dedekind's theorem (Ch. XIII, 12) to the set S, exists such that
By
we can
show that a number h
no value of f(x) exceeds h
(i)
at least
(ii)
one value of f(x) exceeds any number
number h
This
;
is
than
h.
called the upper bound of f(x).
Similarly, if f(x) is bounded below, a (i)
less
no value of f(x)
is less
than
I
number
I
exists
such that
;
one value of f(x) is less than any number greater than (ii) This number I is called the lower bound of f(x). at least
Ex.
Consider the function f(x)
I.
=
lim
.
n-^-oo 1
If
+nx 2
3=0, f(x)=0.
If
I.
x^O.
Thus f(x) has a definite value for every value of x. But f(x) is not any interval including zero, for by making x small enough, we can make exceed any number we may choose.
f(x)~ljx.
bounded I/a;
in
Theorem bounded in
1.
If f(x)
is
continuous in the closed interval
(a, 6),
it
is
(a, b).
Suppose that f(x) is not bounded in (a, b). Because f(x) is continuous at x = a, for sufficiently small values of x - a, f(x) lies between f(a) - c and /(a)
x in
+ e, where
is
any small
such that f(x)
(a, 6)
is
positive
bounded
possible) not bounded in (x, b). Divide the real numbers in (a,
A' as follows f(x)
is
is
or
is
:
The number x
not bounded in
b)
is
number. in (a, x)
Hence there are values and (if our supposition
into a lower class
to be placed in
A
A
and an upper
of is
class
or in A' according as
(a, x).
According to the supposition, both classes exist and every number in A less than every number in A'. The classification therefore defines a real
number
a,
which
Now f(x)
is
the greatest
number
in
A
or the least in A'.
continuous at x=a, therefore for sufficiently small values of x-a, f(x) lies between /(a-e) and /(a-f e), where is any small is
ROLLE'S positive
belongs to A, which
Theorem
2.*
is
287
bounded in (a, a-fe), and so impossible, for a + >a. This proves the theorem.
Hence f(x)
number.
THEOREM
// f(x)
is
is
continuous in
lower bounds f(x) takes the values h 9
and
I
(a, 6)
and
h,
at least once as
I
are
its
x wries continuously
from atob. For if c is an assigned positive number, however small, there one value of x in (a, b) for which f(x)>h -, so that
h-f(x)<
and
1/{A
upper and
is
at least
-/(*)}> 1/c.
Hence l/{h-f(x)} is not bounded. But if h-f(x) does not vanish for some value of x in the interval, l/{h-f(x)} is continuous in (a, 6). Therefore h=f(x) for some value of x in (a, 6). = Similarly there is a value of x for which f(x) I.
Theorem. Suppose that f(x) is continuous in the closed and has a derivative f (x) for every x such that a
24. Rolle's interval (a, b)
If /(a)
=
a and 6.** This theorem is of the greatest importance. A strict proof (as given on the next page) depends on the rather difficult considerations of the last article, but from the graphical point of view its truth is obvious.
Let the curve y=f(x) cut the x-axis at A, B. Suppose that it is conA to B, and that at every point it has a definite and finite gradient. The theorem asserts that there is at least one point on the curve
tinuous from
between
A
and
B where the tangent is parallel to the x-axis.
FIG. 56.
FIG. 57.
= l -x$. Here f'(x)= -GO at j/ not zero for any value of x between - 1 and + 1.
Fig. 57 represents the curve
f'(x)
is
* This proof
is
and
taken from
** If f'(x) exists for a
must be continuous
in the
open interval
(a, &),
but not neces-
MEAN-VALUE THEOREM
288
Because f(x) is continuous in (a, b), it is bounded in (a, b) and attains at least once its upper bound h and its lower bound /. Either (i)/(z)=0 for all values of x in (a, 6), in which case /'(z) = Proof.
x
for every
;
or
some values
(ii)
of f(x) are positive or negative.
positive and unequal to /(a) or /(&). Moreover, f'(h)=*Q, for otherwise there would be values of x in the neighbourhood of h for which f(x)>h, which is impossible. If positive values of f(x) exist,
h
is
there are negative values of f(x), then I is negative, unequal to /(a) or /(6) and /'(Z)=0. For if f'(l)^Q, there would be values of x near I If
impossible. This proves the theorem. simpler proof, for the case in which f(x) is a polynomial, is given in
which
for which f(x)
A
is
the next chapter.
25.
The Mean-Value Theorem.
and f'(x) where
exists for
is
a<#<6,
m
-f(a)
= (b-a)f (),
some number between a and Let
Proof.
then
(x)
=/(&) -f(x)
is
Iff(x)
continuous for
a
then ........................... (A)
b.
-
--
= -f(x)
-/(a)},
{/(&) -/(a)}.
Also ^(a) = and
Let A, B be the points on Geometrically. the curve y~f(x) where x = a,b respectively. The theorem asserts that there is a point C on the curve between
A
and
B where the
If P(x, y) parallel to AB. the curve, and the ordinate is
NP
duced) meets
AB in
Q,
it is
tangent
any point on
is
easily
(or
NP
pro-
shown that Fia. 58.
Corollary. (i) (ii)
//,
throughout any given interval,
f'(x) is always positive, then f(x) increases with if f'(x)
is
(ill)
if
f
(x)
All this follows
two values NOTE.
of
;
always negative, f(x) decreases as x increases
f
is
x
always
zero, then f(x)
from equation
x in the
(A),
is
where
it is
supposed that
interval.
It has not been
assumed that /' (x)
is
;
constant throughout the interval.
continuous.
a, b are
any
AREA UNDER A GRAPH 26.
a given function of x, and that Such a function is '(x)~f(x).
Suppose that f(x) we try to find a function ^>(x) such that called
If
and
I
ntegration
an
and
t/j(x)
differ
Hence
if
(x) are
denoted by \f(x) dx.
is
that
any value
of
(iii),
<}>(x)
is
(f>(x)
-i/f(z)
=a
+ C.
Thus
equations which Ex.
1
.
NOTE.
mean
^ dx
Since
xn
denoted by
j
Integration
Ex.
2.
//
n>0,
+ 1)
j-(x) =/(#),
and
(x)
=
r
f(x) dx
+ C,
are
two
the same thing.
^^
( v
n+
1
) '
x n , therefore (x n dx -
...
dx
is
*-
n+1
+ C.
to bo regarded as denoting the operation of integra-
These are inverse operations.
denotes that of differentiation. '
where
J
-j-
4
then
/(x)rfx, the general value
J
The symbol
tion, just as
= $'(x),
constant C.
/i
f(x) dx
\l(n
and
by a constant.
r is
or),
any two functions such
Art. 25, Cor.
by
therefore,
iff
is
.
integral of f(x) (with regard to
289
PN
can be applied to find the area of a curve as in the next example. ordinate of
is the
prove that the area
any point P(x
ONP
of the rectangle contained by
9
y)
on
bounded by ON,
the curve represented
NP
and
the arc
OP
by
y~kxn
is
equal to
ON, NP.
FIG. 59.
Denote the area by S and let #, y be the coordinates of P. Let Sy, 8/9 be the increments in y, S corresponding to the increment ox in x. If Q is the point (x -f oxt y + Sy) and QM its ordinate, then rect. PM
therefore
y<
JO
Of
-
< y + Sy
!f y rfa: =
Now S=Q when
a;
=
I
;
and
since Sy -> 0,
.*.
y=
-
.
11 "1" 1
A:a;
n
/b: = --
-j-
(7,
^0 and 5=
where
C
is
a constant.
TAYLOR'S SERIES
290 27. Taylor's Theorem. is
If f(x)
tives f'(x),
Theorem,
known
also
as the General
the first
n
values of x between a and
b,
continuous in the closed interval
f"(x) .../
(n)
(x) exist
for
all
=/() + (6 - a)f () + -(&-
f(b)
/" (a) +
.
.
.
1
i^
some number between a and is
then
"
|>*-1
The proof
deriva-
2 )
f/
is
and
(a, b)
1
where
Mean Value
b.
similar to that of Art. 25.
Let
and then
F' (x)
and
f (x)=
Now
= -
1
p
(b
- x)"-
-
= Q and
^(6)^=0, therefore by Rolle's theorem for some value ^ of x between a and 6, hence
which
is
Let b
'(#)
=
equivalent to equation (A).
= a + h,
then equation (A)
may
be written
..+
...(B) /^
where
J2 B
=
n
pA /<
w >(a
+ 0A)
and
i.
0<0<1.
For any number between a and a-h/* may be written in the form a + Oh where 0<0<1. In general, the value of depends on n. Taylor's Series.
(a-c, a + c),
and
Suppose that f(x) is continuous in the closed n derivatives f'(x), /"(#), .../(n) (^)
the
a/wi /Aa< |
h |
Then equation
(B) holds.
interval
FUNCTIONS OF A COMPLEX VARIABLE
Rn =
Further suppose that
-.-
h"f
w (a
-f
n-> oo
as
9h)->Q
291 then
,
vL "
For
sn
if
is
the
sum
/(a + h)
The :he
to
n terms
...to*>.
of the series,
s n = 7? n ->
~
so that
When
V(0) +
a
Rn
A) .
-4-
is
Lagrange's form of
equation (A) becomes
0,
AT
1
s n ->f(a
and
series in (C) is called Taylor's Series
remainder after n terms.
...(C)
/
(0)
+
...to
................
(D)
11 This expansion
is
known
as Maclaurin's Series.
THE COMPLEX VARIABLE
The Quantity
28.
x-f ty, which we denote by
z, is
called the
com-
Here x and y are real variables, independent of one variable. oo to +00. another and each capable of assuming any values from As z passes from a value z to a value z v we suppose the variation to be plex
By this we mean that the point z is to move along a continuous curve from the point z to the point z v Thus the variation of z is continuous if the variations of x and y are continuous. continuous.
29. Function of a
X
Y
and
Complex Variable
are functions of x
and
y,
z.
If
we say that Z
is
Z=X+
iY, where a complex function
of the real variables x and y. If z is known, so also are x and y. Consequently X, Y and Z are known. Hence it would be perfectly reasonable to call Z a function of z.
But
it is
usual to restrict the meaning of the term
'
function of z
'
as
follows.
We
use the equation Z=f(z)
to indicate that
Z
is the result
operations performed on z, the variables x and y not occurring we say that Z is a function of z.
Thus 2x -f 3iy
is
a function of z in the
first sense,
of definite
explicitly,
but not in the
and
restric-
ted sense.
For
if.
terms of
z is
known, so
z alone, i.e.
It follows that
is
2x + 3iy
;
but 2x
-t-
3iy cannot be expressed in
without the explicit use of x or
y.
if
one series of operations determines
X and Y in terms of x and y.
CONTINUITY OF RATIONAL FUNCTIONS
292
30. Definitions relating to Limits (1)
way
and Continuity.
To say that z tends to zero (z->0) is to say that z varies in such a that z becomes and remains less than any positive c we may |
\
choose, however small.
a
Thus, after a certain stage, the point z (on an Argand diagram) is within which may be as small as we like. with centre and radius
circle
,
To say that z tends to a (z->a), where a is a fixed number, is to say that 2-a->0. After a certain stage, the point z is within a circle with centre a and radius c, which may be as small as we like. (2)
We
say that /(z) tends to a limit I as z tends to a if, for any posithat we may choose, however small, 77 can be found so that
tive
|
/(z)
-1 1
<
provided only that
e,
z
-a
|
\
<
rj.
In other words, as z approaches a by any path whatever the distance from the fixed point I becomes and remains less than
of the point f(z)
any length e that we may choose, however small. = z or at (the point) (3) The function /(z) is continuous for z e that we however choose, small, we can find 77 may any positive z -z provided only that f( z ) ~f( z o) < <^. I
>
I
z
if
for
so that
1
|
the same as saying that as z tends to z along any path whatever, /(z) tends to /(z ) as a limit. It is assumed that /(z) has a definite value at every point in the
This
is
neighbourhood of the point
z
.
A
The function
if it is continuous /(z) is continuous in a region of the region. at every point Hence if z describes a continuous curve in the region of continuity of the function /(z), the point /(z) will also describe a continuous curve.
(4)
31. Continuity of Rational Functions of z. All that has been and 3, with regard to the real variable, holds for the complex
said in Arts. 2
In particular
variable. (i)
(ii)
(iii)
lim (o
z-o
:
+ az + a^
2
-f
. . .
+ a wzn ) = a
.
A polynomial in z is continuous for all values
of
A
values of z except for
rational function of z
those which
make
is
continuous for
the denominator zero.
all
z.
EXPANSIONS IN SERIES
293
EXERCISE X3TTX 1.
Prove that
(i) (ii)
(iii)
x-x always decreases with x. tan x-x increases with x in the intervals
sin
(-lU,
-
ITT),....
(71-,
|TT),
decreases as x increases from
x
to
TT.
2 3 2 1 2. If a, ax + b, ax + 2bx + c, ax + 3bx + 3cx + d and ax + 4bx* + bcx* + 4
F=^o w 4~4ii 1 w 3 + 3% 2 then
=0
-7-
,
ax
and
ax
=0, so that
U
and F are indepen-
dent of a;. 3.
/ -l-
cos,
If
re 2
and dun
dvn
^=- -
xi_ x prove that
^=
v
Hence show that un and vn are
M
'
positive or negative according as
n
is
odd or
even. [If
#=0, then wn =0 and dui -7
Now
= ~v >0,
.*.
= 0;
vn
%
is
^
also
= cos x- 1<0,
an increasing function,
v
= sin x - x < 0.
%>0;
.*.
rfv
and
^-=w
and so 4.
1
>0,
/.
Prove that for (l)
(ii)
[(i)
Let
t
3, for
all
r
an increasing function,
t;
2 ),
(w a v3 ), ,
sina;
/.x
... .]
y*
3 X* = x- X + -
QO;
to oo
.
r^
be the rth term of the series arid sr the x, s zn
x as
< cos x < s 2n+l and s
r -> oo
2
+i
- s zn
sum
tzn+i
to r terms. ~> as ?i ->- oo
Then by .
There-
.]
-
/
sin (x
rv =sm x + hT + h) -
cos
x
W- sin x - & cos x r^ o
2i
(ii)
= cos - A sin a: - h
cos (# -f- h)
a;
2 -
cos x
+
2i
6.
^!>0;
Use Taylor's theorem to prove that (i)
[(i)
/.
values of x
COSX = l-+r-... tO
every
fore sr -* cos 5.
is
on, in succession for (u 2 ,
X2
Ex.
vx
If f(x) = sin x,
|^n
|<-|^n |->0as
+ ...
;
a; -f- ...
.
h3 sin
3
n-*oo.]
Deduce the expansions of sin 2 and cos x
in
Ex. 4 from Taylor's theorem.
CASES OF BINOMIAL
294 7.
Prove, by means of Art. 22, Ex.
tan" 1
;*;
=# -
o
x--5 -f
x'
5
7
Sr =x-~ + ~-...+(-l) r ~i
{Let
o
o
/(0)=0, and/
/
(x)
0
that, if
1,
x*
THEOREM
;
tooo.
4-
^,
and
/(s^tan- 1 *-^;
1
JLf
= l/(l-fa: 2 )-(l-x 2 4-a:*~...4-(-l) r-1 x 2r ~ 2
so that
,
which is positive or negative according as r is even or odd. Hence, tan~*aj lies between Sr and Sr+1 and 8r - Sr+l so long as x is positive and not greater than unity.] ,
8. If f(x) is
a function whose
first
= [Proceed as in Ex. 9. If
1
|
derivative
show
is I/a;,
then
as r ->
-> 0,
that, if
oo
,
0
x3
x2
x-j + ---...
to QO.
7.]
0<0<7r/6, prove that tan
+ i0 3 and
between
6 lies
+ f0 8
= + a03 -tan
.
show that /(O), /'(O), ; [For the second /"(O), are all zero, and that /'"(0)^0 throughout the interval 0<0^7r/6 if 9a>8 hence, in succession, / // (0),/ (0),/(0), are all positive.] inequality, let /(0) /
;
The Binomial Theorem a positive integer, show that the expansion of (a+h) n by the Binomial theorem is given by Taylor's series. 10. If
11. If
n
n
is
is
any
rational
number and -
n(n-l)^ z 2 + ^
1
.
-
...+
-- -
(i)
show that
n(n-l) ...(n-r + 2) x^T-* '
,
-.
-,
IT
1
-
where (ii)
If
0
that
J r
->0 as
r -> QO
,
so that in this case ...
[For r>n, (l+0x) (iii)
-
If
n - r
and
"'
to oo ...................... (A)
xr ->0 as r~>oo.]
<#<(), explain why we are unable to say that + Ox) n ~r > 1 for r>n, and so far as we can tell, 1
ftr
-+Q.
Rr may not tend to the expansion (A) in this case, another form (named after Cauchy) of the remainder after n terms of Taylor's series must be used. See any treatise on the Calculus.'] [Here
zero.
(
1
To deduce 4
CHAPTER XVIII THEORY OF EQUATIONS
(2), POLYNOMIALS RATIONAL FRACTIONS (1)
Multiple Roots.
1. is
a polynomial, then
(x
For /(#) = (#- a) r
*
one real root of
Let
Proof.
If
Theorem
2. Rolle's at least
.
a,
j8
where
is
a polynomial and
for Polynomials.
f (x) =
lies
// f(x) is a polynomial, between any two real roots of f(x) = 0.
be consecutive real roots of f(x) = 0, so that
f(x)=(x-*r(x-p).(x) integers, and
where m, n are positive
Then
#-/?.
(2),
is
divisible
of invariable sign for oc^a^jS
by x-at or
and
/'() = (x a) where
Now
0(a)
= m(a-j8)<(a) and
Therefore 0(a) and 0(jS) have unlike signs where continuous.
Hence
and consequently /' (x)
\fj(x)
between a and
3. (
1)
is
;
moreover,
ift(x)
is
every-
equal to zero for some value of x
/?.
Deductions from Rolle's Theorem.
= // all the roots of f(x)
are real, so also are those of
f (x) = 0,
and
the
roots of the latter equation separate those of the former.
For if f(x)
is
of degree n, /' (x) is of degree n-1, and a root of /'(x) n - 1 intervals between the n roots of f(x) = 0.
=
exists in each of the (2)
//
all the roots
f" (x) = 0,
= f'" (x) 0,
of f(x) . . .
,
and
those of the preceding equation.
This follows from
(1).
=
are real, so also are those of /'(x)=0,
the roots of
any one of these equations separate
DEDUCTIONS FROM ROLLE'S THEOREM
296
Not more than one
(3)
(ii)
be the real roots of
j8x , /J 2 , ... ]8 r
a x =a 2 then a x ,
If 0^:56 a 2 ,
Hence
by if
(i)
if (ii)
;
(iii) if
or
jS1
= 0, any
/'(o;)
Let
.
of which o^,
may
a2 be
real
= f(x) 0.
roots of
roots
off (x) = 0,
be greater than the greatest of them.
be multiple roots, and suppose that If
can
be less than the least of these, or
(iii)
let
=
between two consecutive roots
(i) lie
For
root of f(x)
j8 r
is
one of the set
=
j8 1? j3 2 ,
(Art. 1.)
j8 r .
between o^ and a2 cannot be consecutive
Rollers theorem, a root of f'(x) lies
j31
a 1
then
,
then
,
/J 2
^
cannot be the least root of f'(x)=Q cannot be the greatest.
then
,
and
j8 x
.
j8 r
Thus, not more than one root of f(x)
can
;
in any one of the open
lie
intervals
(4)
// /'(x)
=
Aas r razZ roote,
f(x)=0 cannot have more than
r
+1
has no multiple roots, none of the roots of f'(x) = is a root of /(x)=0, and the theorem follows from (3). If f(x) = has an in-multiple root, we regard this as the limiting case in
If /(#)
=0
which
m
(5)
If
f
(r
is
\x)
It follows
from
(4)
= f(x) that
true in
is
and
all cases.
(r)
the equation / (#)=0 has as many. has at least as many imaginary roots
derivative of f(x)
any
roots, then
imaginary
Thus the theorem
roots tend to equality.
has at
/(#)=0
least
as/'(z)=0. (6) If all
number of
A
the real roots real roots of
...
j8j, j8 2 ,
f(x)=Q by
Ex.
and 1.
have opposite
/(/J 2 )
Find
the character of the roots'of
* /(*)
and /(#)=0 has a
real root
f(x)
same
-1
+
-
1,
j3 x
and
/? 2 ,
faw a4
2.
a // 6
tecw*
oc<
0,
2
+
-
are
1
one between
(uw imaginary
For
and the equation
then the equation roots.
1, 1, 2.
00
+ + -1 and
oa; -f-26a;
= (a,
6, c,
See Ch. VI, 11, Ex.
/(->(*) = ?i(tt-l) 2
f(x)
...3.(a
according as
sign.
and two imaginary
1,
roots.
Ex.
....
= 3a;4 - &E3 - 6* 2 + 24x + 7 = 0.
roots of /'(*)-=
-00
<-
between
lies
signs, or the
Here f'(x) = 12(x*-l)(x-2). The
find the
considering the signs of /(&), /(/J2 ),
single root of /(x)=0, or no root,
f(Pi)
f'(x)=Q are known, we can
of
3i
-f26
+ c =0 has two imaginary
. . .
1.
+ c) roots.
k $ #,
1
w )
~
NEWTON'S THEOREM
Sums
Powers Newton's Theorem. 4.
Let a,
j3,
of
n
let s r ==a r
+ j8r -h... + K T
n ~l
(r
+pzxn
= 0,
-2
+
...
1, 2, ...),
+p
(fi)
*r
+Pi*r-i+Px*r-*+- +Pn*r-n=<>
(i)
sr _ 1
We
..................... (A)
= n,
have identically f(x)
- (a? - a) (x - j8) ... (x - /c),
and, assuming the rule for differentiating a product (Ex.
(B)
2, p. 277),
synthetic division, 2 - a) = xn ~ l + Atf"- -f
f(x) / (x
where the
A
l
coefficients are given
by
A2=a?+p
=
in succession
,
l
;
,
other quotients in (C) are expressed in a similar way,
If the
Q l - s l + np v
and so on
;
and
Q
- ^ 2 -f p^ L -f np2
Q n ^ - * n-1 -f jvn _ 2 -f
Now f'(x)nxn ~ + (w-l)p l
comparing
this
with (D),
-*
"-*
-*
'
1
xn
*~ 2
^ - (n -
on,
it will
be
. . .
+ ...+Q n
................. (D)
,
Q3 = s3 + p^ 2 + p^ -f np3
,
,
+ ;v-2*i + w ^n- 1-
-h(n-2)^2x 1
and so
~"
~*
seen that
where
.........
x~
x-oc
By
then
+p2 s r _ 2 + ...+slpr _ l +rp
sr
1
+;p n =0,
so that $
(i)
Proof,
Equation
be the roots of
... AC
f(x)^x +p lx and
Roots of an
the
of
297
n -3
4-...
+|> w _!
^^ Q 2 - (w - 2)^2
,
. . .
,
;
and
Q n =p n - V
Therefore
2
(ii)
+ --+^--
1
)Pn-l
=
Let r^zn, then multiplying each side of /(a) "- 1
a r -h fta 7
+ p 2 a *- 2 -f 7
. . .
-f-
=
r-w = 0, ^ a rt
and similar equations hold for each root. Hence the second result follows by addition.
(E) r by a
~n
we have
ORDER AND WEIGHT
298 Ex.
an
For equation (A) of
1.
the last article, find the value of
even number, in terms of S Q s l9 ,
If q, c 2 ,
(a
-
r x)
In this equation, substitute
= a - qxa r
a,
/?,
...
y,
in the expansion of (1 4-x) r ,
...
7"' 1
+ C 2z V~ ~...+x 2
is
then
r .
K in succession for x and add, therefore,
Similar equations can be deduced from the expansions of (/3-#) r ,
(y~ z)
r >
etc.
we have
hence, by addition,
5.
where r
,
....
s.2
are the coefficients of x, x 2 ,
...
r
27(a-/?)
Order and Weight of Symmetric Functions.
Leta,
/J,
y
...
be the roots of xln
+p
1
xn
-l
+p&n -*+...+p n = Q ...................... (A)
Consider any function of the coefficients
;
for example, let
Let ^ be expressed in terms of the roots by means of the equalities
Consider Jfo highest power of any root a which occurs in u, and denote it by a*.
Each of p 19 p 2 ... involves a to the first degree from the highest term Pi 2p% and /row. /i/s alowc ,
:
degree of this term, namely
The
order of a
this expression for
5 only, therefore a arises also the index s is the
3.
symmetric function of the roots of an equation
is
defined
as the index of the highest power of any root which occurs in the function. Thus the order of 27a 3/8 is 3, and that of Za/3y is 1.
From what function of
the coefficients
If u
is
has been said
it
will
be clear that
the roots is the degree of the
p l9 p^
...
pn
a symmetric function of order atfv
+ al
xn l
s
of the roots of
+ ...+a n = Q,
s
then a Q u is a homogeneous function of # a l9 a 2 For if the equation is written in the form (A), u ,
which the highest term
The in a,
is
of degree s
,
and Pi =
........................... (B) ...
is
of degree s. a function ofp l9
i/a
>
weight of a symmetric function (u) of the roots a, |8,
...
.
Thus the weight of Za3/?
a symmetric
.
n
in
the order of
function ivhen expressed in terms of
is 4,
and that of ZajSy
is 3.
?2 ==a 2/ a o j8,
...
is its
j> 2 ,
e^ c
...
-
degree
PARTIAL DIFFERENTIATION Theorem.
// a symmetric function
expressed in terms of the coefficients term,
when
299
(u) of the roots of equation (A) is
pv pz
,
sum
the
...
of the suffixes of every
written at full length,* is equal to the weight (w) of u.
Denote any term in u by Pi hp 2 kp^ and in equation (A) write n 1 n The equation becomes y -f A^y"" + A2j92 y n ~ 2 44- A p n ... 0, 1
Proof.
-
x = yl A.
. . .
and the roots of If then (i) (ii) (iii)
;
this are
Aa,
we multiply every
to multiply
A/3, Ay, ....
root by
\w
u by
the effect of this
is
;
to multiply piSpz, p^
to multiply
A,
pfp^pz
1
by ...
A2 A 3
A,
,
,
...
respectively;
by
/
which proves the theorem. down the literal part of Za 2/? 2 expressed as a function of the coefficients. of 2 and weight 4. The only terms of weight 4 are such as contain order p 2 z The first two of these are of too high an order. Pi*> Pi Pz> PiPv Pz Pv Ex.
2
oc
1.
2
Write
,
is
>
2 2 2 Thus we have 27a j8 = ap\p$ 4- bp 2 -f cp^, where a, 6, c are independent of p lt p 2 p 3 .
,
6. Partial
Derivatives.
Suppose that u=f(x,
y, z, ...)
is
a func-
tion of the independent variables x,y,z, ... The operation of differentiatu with on to the x alone varies, is denoted that x, ing respect supposition .
by the symbol OX ^~ respect to x.
,
and ^~ CfX
called the first partial derivative of
is
u with
-.
Sometimes the ordinary symbol -p
the context implying that the differentiation
_ 2
d u ___
,
used in the same sense,
is partial.
_3 fdu\ _ .
.
dxdy
is
I
1
-
We
also write
d*u
dy\dx/
with similar notation for higher derivatives.
For example,
if
u = xm y n
j.nereiore
,
(xx)
^ ^ ox oy
oy ox
It follows that equation (A) holds
when u
*
U
Thus the term
2
2>i j> 2
3
is
is
a polynomial in
x, y.
written PiPiPzPzPzB.C.A.
TAYLOR'S THEOREM
300 It is
proved in books on the
represented by (A) holds
'
when
'
that the commutative property the functions in question are continuous. Calculus
A strict proof is difficult, and here we are only concerned with polynomials. Again,
we
often regard
h^ + k-^-
as
an operator, and we write
du da ( d 9 \ 7 7 ^ h-~-+k-=- = (h-r-+k-^-}u = Du, ox oyj oy \ ox
^^=^^^--5-^
dx
if
h k are constants y
a result which
;
dx 2
\
dyJ
\
dxdyj
we may
dxdy
express as follows
:
u.
dxdy The value of
D n u can be written down in the same way
by the Binomial
theorem.
7.
Taylor's
Theorem
nomial of degree n in
for Polynomials.
//IN
2
7
-r^r
+
:
av
...
a n are independent of
A,
+ TT. a n
'
\^
and
if
X=x+h
Hence, by successive differentiation with regard to
f (x + h) = aj + ha
The
result follows
h
2
by putting h =
2
+ :-- a3 +
(A)
we may assume that
[L
,
a poly
...+~fM(x) n
This follows from Ch. XVII, or thus
f(x + h) a + hdi + where a
is
x, then
~ f(x + h)=f(x) + hf'(x) + f"(x) + & Proof.
// f(x)
(1)
. . .
+
h"-
h,
1 -
-^
we h&ve
an
in these equations
;
,
for
we
find that
EULER'S THEOREM If u z=f(x, y)
(2)
is
a polynomial of degree n in 1
f(x -f A, y + k)
For we
where a
may assume
301 x, y, then
_+*-Dn u
u + Di
(
that
... are independent of A, k. = k A we have a = u', and differentiating 0, Putting with regard to h and r times with regard to k, we find that ,
6
,
^T) /(# + A, y +
Then,
if
d
f
\
X = x + A, Y = y m -r
( d
-f
yfc,
k)=pr + terms
m-r
times
k
A,
(D)
equal to
is
\r
and putting A = 0, A; = 0, we have which proves the theorem. // u?Ef(x,
involving
the left-hand side of (D)
= \SV/ /(^ ^) /tn-r,r(^
^3T/
(3)
y, 2, ...) is
^)> sa 7 ^ e fm~r -
^ r =/m _ r
,
r
(x, y)
=^
a polynomial of degree n in
^
v^
x, y, z,
ZAen
...
1
1
(E)
n where
D denotes the operator
The proof
is
8. Euler's x,
B )*
dx and, in general,
+ l~- +
...
.
oz
similar to the above, using the Multinomial theorem.
Theorem.
y and of degree -~-
h -- + k^ox oy
Let
(1)
u
be a polynomial
homogeneous in
n, then
J ^~
dy
=,
^9 dx 2
y
^^-
dxdy
+ y-^-) w = n(n-l)(n~2),..(n-r=f (x^ \ ox oy/
Proof.
Let u=f(x,
y),
then by Taylor's theorem,
f(x + \K,y + Xy)
du
+
---
>
and
since
u
is
homogeneous and
of degree
n
6.
}
........................... (B)
The coefficients of the powers of A in the expressions on the (A) and (B) must be equal, whence the results follow, end of Art.
r
in x, y,
/(x + As,y+Ay) = (l + A)t
* See
/AX (
right of
PARTIAL FRACTIONS
302 // u=f(x,
(2)
a polynomial homogeneous in
is
y, z, ...)
x, y, z,
...
and
of degree n, then
du du dn z *x a" +y~*-+ y oz ox dy
D u = n(n-l)(n-2) r
wAere
D r tfawdx for J
(x \
^dx
4-
y J
^~ dy
9. Partial Fractions.
.
. .
...
Let
in
)
form. J expanded r
its
/
in equation (E) of Art. 7, the proof
is
be a rational proper fraction // x a is a non-repeated
f(x)l(f>(x)
let
x-a
the fraction corresponding to
(f)'(a)
= (x-a)^(x) /r
'x-a
^
and
is
.
=
x-a
^>(x)
From
l)w,
to be expressed in partial fractions.
is
factor of
For
4-
(n-r +
...
dz
= Putting h = Xx, k Xy, /-Az, similar to the preceding.
which
2 -5-
4-
+-^nu>
4-
the second equation f(x)
and (rom* the
first,
by
= At(x) + (x
differentiation,
Putting x = a and observing that 0(a)^0, we have
and
f(a)=Aif>(a) Ex.
and If
1.
^(a)=0(a),
therefore
A=f(a)/
m - 1 /(x n ~l) in partial fractions, where m, n are positive integers Express x
m^n. x-a is any
factor of x n
xn ~
The imaginary roots of xn cos
according as
n
Let
-
1,
then a n
na n ~ l
I
=
1
= 1 and
x-a
x-a'
are
sin^-^ where r = l,
i
n
n
n
2,
...
i(w-l) *
or ^(n-2),
odd or even.
is
a = cos
n
and denote the sum of the
+
*
sin
u
,
then a" 1
= cos
i
n
fractions corresponding to x - a
sin
and x - a"
the preceding,
wr = -
-
# cos rmoL
2
/.
.
a;
r 2
-
tr - 2a:
cos
r(m - l)a where a = +1 ,
-
cos ra
;
n
2?r *
1
by u r then by ,
SYMMETRIC FUNCTIONS OF ROOTS Hence, according as n 1
1
n
x-l
r-*i
Theorem.
10. 7/a,
...
j8,
are
x
is odd, or even,
+r =J(n-
m -l
l(x
n-
equal to
is
1)
303
1
1..J .(zl) n x-l
'
r'
n
+
!. a;
4-1
1--K-V T
r--i
The following theorem is often useful. n w==(a a p ... a n $#, l) =0 awd
roote of
f/te
,
t;s^(a ,a 1 ,...a n ) is
a symmetric function 0/a,
...
j3,
involving only differences of the roots,
,
then v satisfies the equation '
dv
dv
dv
For the substitution x = X + h transforms w =
A l = a l + a^i J 2 = a 2 + 2a
where are
9
a -A, ]8-
A,
...
,
1
A-f a A 2
etc.,
,
into
and
since the roots of [7
=
we have
v = (f)(a^
a l9
= (f>(a^ A^
an )
...
... ^[
n ) ................... (B)
If the right-hand side of (B) is expanded in powers of A, every coefficient must vanish, and by the extension of Taylor's theorem the coefficient of h is the same as that in
dv
dv
dv
a Qh
2 2 ^- + (2a x A -f a Qh ^~ + (3aji + 3a x A + ajh?) ^ )
dv
.'.
^or
J&x. 1.
^
dv
+
. . .
;
dv
a,Q^~ + 2a l ^~ + 3a 2 ^ +...=0. ^a x ca 3 aa2
biquadratic (a
,
a lf a 2 a 3 a$x, ,
,
4
1)
~0, find in terms of
the coefficients
the value of
This function
where p of
a,
j8,
9
q, r
. . .
,
is
of order 3 and weight
are independent of a
,
a lt
...
3.
.
We may And
therefore
since v
is
assume that
a function of the differences
we have l l5
+ 3a 2^
+ 4a 83 ^
=
;
=
;
4 2
Q
This
is
true for
all
values of a
,
a 1$ a 2 therefore ,
= 0,
giving
q-~3p,
r-2p;
The value of 3? may be found by giving special values to a, y, 8. Thus by taking = 1, y=-l, 8=4, we find that ^=32, which agrees with Ex. 3 of VI, 16. ,
APPLICATIONS OF ROLLE'S THEOREM
304
EXERCISE XXX 1.
Prove ating
The equation f(x)==x* + 3qx + r=:0 has two equal this (i) by applying the II.C.F. process to f(x) and x between /(#) and f'(x)=0.
The condition that the equation
2.
equal roots
Show
that this
may
Find the values of a and solve the equation in one
bx 2
and
by
elimin-
may have two
+ 2cx + d~Q.
which ax3 -9x 2 -f I2x -5 =
for
the
5. All are real. .
roots
Ii
!_?
[Use Ex. VII,
Show
if
-
of
roots
has equal roots,
case.
~ The equation xn -qxn m + r=Q has two equal
14.]
that the equation x 3 - Ix + 7
and one between - 3 7. If
(ii)
be obtained by eliminating x from
3.
6.
+ 3bx* + 3cx + d =
if ;
is
ax 2 + 2bx + c=Q
4.
ax*
roots f'(x)
=
has two roots between
1
and 2
- 4.
arid
~ ~ the equation x n +p 1 x n l +p 2 x n 2 show that a is a root of n~2 + n _ n txn-i + n _
+ ... +p n ~0 has
each
three roots,
equal to a,
n ~*
-f ... +Pn -i =0. p 2X 2 equation x -4a^-hlOx + 7a;-5-0 has one (
l)*p lX
2)
(
2
8. Show that the negative and two imaginary roots. 1
9.
The equation 3x 4 + Sx3 - 6# 2 - 24x 4- r = 7 13
has four real roots
positive,
one
if
-8,
real roots if - 8
two
if
r> 19.
The equation f(x)=(x - a) 3 + (x - 6) 3 + (x - c) 3 -0 has one real and two imaginary roots. 10.
11.
The equation x 6 -5a# + 4&=:0 has
three real roots or only one, according
as 12. The equation x 5 + 5o (i) if a>0 ; (ii) ing cases :
3
+b
if
a<0
has one and only one real root in the followand 6 2 -f 108a 5 >0.
13. The equation x 6 + 5ax 2 + b has one and only one real root in the following cases (i) if a, b have the same sign (ii) if a, b have opposite 5 signs and b(b* + 108a )>0. [Deduce from Ex. 12 by putting I/a? for a?.] :
14.
Show
and that
that
;
if
36 2
a>0
these values are all positive if and a <46 . 2 [x 9 y, z are the roots of an equation of the form 0*-a8* + b + 2 2 3 this reduces to one of the form +<(6 Ja ) + fc'=0.] B=
<
2
#=0
and
if
FUNCTIONS OF ROOTS OF EQUATIONS 15. If
A2
Aj,,
A 3 are the values of A given by the equation
,
jc*^ +
^A prove that these are
all real
order, then
;
y &2
2
z2
^+
^A ^-A~
also, if
6,
,
'
and A lf A 2 A 3 are
c
,
...
+
a-n-i
+
x n- 2
!
ZV^O
show that
+ "- +_____
I
for
U
_i
'
h*~~
|2
|j.
r
= 2,
3, 4,
...
n.
17. If f(x)~0 is a cubic equation whose roots are harmonic mean of the roots of /'(a?) ~0, prove that a 2
The sum
19. If s n
a,
the
is
sum
and a
y,
/?,
is
the
/?y.
of the ninth powers of the roots of # 3 -|~3#4 9
of the nth powers of the roots of
that
x-
is 4
-x
2
4
zero. 1
-~0,
show
H7r s 2 tt-i^0,
a,
/?,
...
equation whose
m and n
according as n
is
* 2n -=-4cos
-. 3
are the roots of (a a l9 a 29 ... n ^/, l) w ---0, are ay-x, py-~x etc., is (a u l9 u 2 ,
roots
where ^^a^x-^a^, 21. If
descending
are the roots of the equation 'vn j______ _________i_
20. If
in
a 2 > A x > 6 2 > A2 > c2 > A3
16. If a, 0, y,
18.
305
U.2 ~~a
x 2 + 2a 1 xy + a 2 y*,
are positive integers
odd
or even,
and
,
show that the un $t, l) n ^0,
...
etc.
m^n,
prove that
where
__-to
#cos(2r-f l)wa-cos _ __ ___ // ~ _ ___ " r j
rc*
22. Let k be given
,
y
^^
__
cos (2r
(2 \_
+
by the equation A J'^ a-fc ^ 6 -
/
f7
If this equation has a rejx?ated root
^, prove that
k^a-gh/f^b- hf/g - c -fg/h, provided that none of another (h), and then
/,
&!
g,
h
is
zero.
= a and
(a
If one of these (g)
-b)(a- c)
/
is
zero, so also is
2 .
Under these circumstances, show that ax 2 + by 2 + C2 2 + 2/yz + 2gzx + 2hxy - k t (x 2 + y* + z 2 ) a perfect square. [Denoting the cofactors of the elements of J' by A' 9 F', etc., the equations are qiiadratics in ^, and the roots of any one of them A' =0, B' 0, C" = Hence if k~kn then A\ B', C' and consequently separate those of J' 0. See Ch. IX, 21, (4).] F', G', H' are all zero. is
CHAPTER XIX EXPONENTIAL AND LOGARITHMIC FUNCTIONS AND SERIFS Continuity of a* and log a
1.
tinuous for
all real
XVII,
(Ch.
2. Ch.
21.)
Exponential
XV,
(1)
The reader
log a x
and Limits.
where
and
(l--}~*-*e. xj
\
m is
a positive integer, then
11 + ~
1,
1
x
l
therefore
If
z->oo
(l
,
~
+
<(
}
1 -f
-
1+
\
7
1
/ /
t
\
XV,
m
5, (3),
in -f-1
"
*
=fl+
<(l +
}
xj
then m->oo and by Ch.
1\ m
/
\
/
1 A
i
-7-I1+-
7
,
m + lj
\
'
v
_
m+l
.
1+1) (\m m/ .(l+j^
e;
therefore
Next
let
x=y +
l,
then as
x->x
,
?/->co
and v
=(i+ii .fi+h V \ y/ yj It follows that if
x->x
or
-
x
,
= lim(l+^)*
,
i
and consequently
9.
(Continued
through real values,
m
lim
con-
continuous for positive values of
is
Inequalities
(l+-)%c \ xJ let
is
reminded that log x stands for log e
is
5.*
As x->oc
For
a>0, the function a x
If
This follows from Ch. XIII,
values of x.
Hence the inverse function
x.
1 -f x) (
x
6.
x.
x.
from
EXPONENTIAL THEOREM If z
(2)
is
any
real
number, z \x
/
= ez
lim (14--) X/ a^oo \
This
is
obviously true
or -
oo
z
if
= 0.
*Aew
\
a*_i
.
hm
is
= obviously true jf a
Now by
Ch. XVII,
=
1/2;,
then as
x
L
y
a^l,
If
ax = l+y, then as x->0,
let
5, (2), 1
lim log a
a;
= loga.
------
x-*Q
This
let
In either case, by the preceding,
y/
(3)// a>0,
.
z^O,
If
according as z.50.
,
307
(1
1
+ y)* - log a {lim
-
y~>o
(1
+ y) y } - loga e = r
a*
lim
1
X
ic->0
;
Aoga
i/->o
therefore
1
= log a.
Here x may tend to zero through positive or negative values, so that x 1 - a~ ax 1 and x x
-
--
each of the functions
tends to log a as a limit as #->0.
been shown in Ch. XIV, 9, that the first of these functions increases and the second decreases as x increases from zero. Hence It has
.,
if
A
x>0,
.
then
l-o-* ---
The Exponential Theorem.
3.
x
2
x
3
.
<
a*-l
all real
-
values of x,
xn
'
I
J
Proof.
The
series (A) is
function of x which
x, n
we
Convergent for all values of x,
shall denote 2
x
by E(x).
If
- --
n
is
where
sum
is
a
8
y
I
its
a positive integer,
, n(n~l)/x\ w(n-l)(n-2)/a\ ~ =l+n.~ + + (1+-) f-) \ 3 n --Vo2 ^(-) \n/ \n/ n/
/-
and
+...
ton + 1 terms
IRRATIONALITY OF
308
Now
jo r
since
E(x) are positive
1/n, 2/n, ...(n-l)/n
than unity, for the values
- 1)
2, 3, ..., (n
of r
we
numbers
have, by Ch.
less
XIV,
1,
n .;
Let
be the
sn
sum
0
n terms
to
I
is finite
let |
x =^x 1 \
I
2.3
Now
and
of (A),
3 -4
>-D
,.
for all values of
then
i.
TT a!l+ - +
+ ,
Xl
;
g Xl
n
^.\ j
x ly therefore
/
x\
Iim^ n4 1 = lim( 1-f
n
.
n '00^
>QO
71
but W-^OO
This result
is
called the Exponential Theorem.
In particular, when x = 6
1
we have
11 + +-+
- ii 1+1+
[2
from which Ex.
it
can be shown that
|3 e
1
5
+
= 2-7182818284
...
$A#tt> JAaJ e i$ irrational.
1.
Suppose that e=pjq
where p and q are integers
t
then
;
where ?
If the equality (A) holds,
account of
4. //
(B).
a>0
Hence
e
and x
we should have
|
is
integer,
which
is
impossible on
cannot be a rational number.
is
any
real
a* = e*
that
R y=an
+2
number. lo 8
a
.
= jE(zloga),
........................... (A)
to say 2
1
3 )
+
.................. (B)
GRAPH OF EXPONENTIAL FUNCTION
An
6.
When a>0 and x
Irrational Index.
309
is irrational,
we take
ax
~E(x log a) as defining the meaning of a*. This agrees with what has been said in Ch.SXIII, 9, for it will be shown later that the sum of the series denoted by E(x) is a continuous function. the equation
6. (1)
Derivatives of a, logx and xf. Suppose that a>0, then by Art. 2, (3), x
-j-a
r lim
=
ah -l
lim a x
r
ax
A
In particular,
NOTE.
e
= a x log a.
x .
Tx be observed that,
It should
a<0, ax has not been
if
defined
for all real values of x.
y = logx, then
(2) If
^ therefore
and
dx -j~
= ev = x
d
-,-
-=-
is,
dx
!
=log G x
i
.
x
has been shown that when n
(3) It
prove that this
is
true for
_d NOTE.
Since
is
y log a,
all real
d y,n
dx it
en log a;
_n __
.
if
rational,
en log a?
instance,
the curve
(1) that, for
TN
y
,
TN
P
:
i.e.
y=ax
until it
at B.
9
the gradient
OX
the subtangent is constant.
l,
A,/'
mark one of the rulings as the axis of y. On this take OA, equal to, say, 4 intervals, as the unit; and on OX, to the left of O, lake Oa OA. Draw
OY
To
:
on the graph meets
sheet of ruled exercise paper and draw a line perpendicular to the ruling as the axis of x ; and
Aa
.
--nx n
the tangent at a point
PN is the
xn = nxn ~' 1
x
= l/log a ordinate of P, then This fact gives a rapid construction for the graph of y=ax , with considerable accuracy. = ex where For for take a and
in T,
ax
we proceed thus
_n _ xn
x
dx
follows that,
is
values of n,
has -been shown in
it
9
ay
fy = 1 xi. * -; that dx x
x
dyjdx
x = ev
meets the first ruling to the left of Without' removing the pencil-point
da
O FIG. 60.
B9 rotate the ruler until it passes through 6, the foot of the ruling next to the left of or, and draw bB until it meets the ruling next to the left of A ; and so on. Find .points on the curve to the right of Y in
fr6m
the same way, and replace the broken line by a If,
in Fig. 60,
Also,
if
'
fair curve.'
OX and OY are interchanged,
the paper
is
turned through 90 in
its
the curve
is
then the graph
own plane, and then turned
over, the curve as seen through the back of the paper y = log x, with the axes in the usual position.
is
the graph of
EXPONENTIAL AND LOGARITHMIC LIMITS
310 7. (1)
For
For
all real
let
f(x)
so that /' (x)
from (2)
values of x other than zero
= ^-1 -x,
(3)
then
according as x 5:
:
//
- 1 and ^0,
x>
//
t/
For
x>0,
then
Hence, if x increases or decreases from zero, and is therefore positive.
0.
x>log
ix2/(l + x)
-l
-|x (1
2
i -*
^
if
Hence
0(#)
x
+x)
/(l
+ &).
T2
_
increases
$
according as x
= a;~log(l + x)-%-
from -1, and since /(0) = 0,
it
0.
x2 X
1 ~r
then
,
and
increases as x increases from -1,
,
2
I+x
Hence f(x) decreases as x
Again,
(1
+ x)<-|x2
,
1
$
(1
- |-x 2 then
+ x)
f'(r\ Ji vv**'/'
follows that f(x)
+ x).
(1
by taking logarithms.
(1)
then
e*>l +x.
and /'(z) = e*-l,
=
/(0)
zero, in either case f(x) increases
This follows from
and
and Limits (continued).
Inequalities
since <(0)
= 0,
it
^
0. according as x
follows that
(4)
//
x>
greater of the
"2
-'1
^0
and
numbers
and
and
1
I,
1 -f x,
h are respectively the smaller and the then
x2
This
(5)
is
merely another
For any
x2
way
of stating the inequalities in (3).
positive value of n*
(i)
lim a;->oo
(i)
For
all
positive values of
logx _2__ xn Choose
m so
that
0
(ii)
<-
n - w ->ao
m
we have by Art. 1 _1n x__m -l__ ^ ______ x mxn ~ m m
_
and
(log x)/x
2, (3),
,
~>0.
and
= ~ (log y)/yn ->0. log x
lim (x n logx) x >o
*
n
J/->QO
(ii)
.
then a^ x~>cx)
Put x = l/y, then as x~>0, xw
-~- =0, X
= 0.
EULER'S CONSTANT (6)
For
This
is
all values
of
lim x
r,
obvious when r<0.
x
r
/e
= 0.
r>0,
x let e
y
\
If
311
= y,
then, putting -
=n,
'
e*
Now lence 8. >e
as it
x-^oo, y->oo;
and, by follows that x r /e x -+0.
The Manner
which e and log x tend
any positive number. First suppose that #-^oo
Now, by n ;
in
Art. 7,
however great n
,
xn
(6),
w = ->0, (logy)/yw ->0.
since
(5),
may
,
then
e x ->oo
Let n
to Infinity.
and logx->oo.
x
/e -+Q, therefore e* tends
to infinity
faster than
be.
n Again, by Art. 7, (5), (log x)/x ~>0, therefore Zo<7# tends n nore slowly than x no matter how small n may be.
to infinity
,
n If #->0, then -logx->ao., and by Art. 7, (5), (log x)/x~ ->0; hence n log x tends to oo more slowly than l/x , no matter how small n may be.
9.
// M n = l+-g- + J
Theorem.
...H----
+
logn, then as n->oo
,
w n->y
Tt'
y
a fixed number lying between
is
Art. 7, (2),
By
Proof.
n-l
1\
/i
i
log
1+-
)
w/
V
log f
log(n-f
, ,
that
n-
<1
1,
n - 2,
...
2, in
7, (2),
n+1
is,
log
n
1
<-w
n
,
succession for n, we. have
and by addition,
;
n
Art.
.
I
(
& \
.
\
Substituting
Also
,
I
n _l
Again, by
1\
/, < n <-log& (1-n/
6
n
that w n decreases as n increases.
1.
n
1
n
1
and
it
l)
follows that ;
i.e.,
log (w
But log(w + l)-logw>0, hence 0
V>y NOTE. /
where y
is
a fixed
The number y
0-577215
....
It
is
is
number >0; and, known
as
since
Euler's Constant.
w w
can be shown tnat
often convenient to state the theorem as follows ,
where
->0 as n-^oo
:
,
EXPONENTIAL FUNCTION
312
10. Series for log 2.
Let
i_
l_-i
(-n-i!
,,11
.,
then
1 N
= (y + log 2n + Therefore
2B )
-
(y + log
the
z2
)
- log 2 + e2n - en
~s2n --0, hence
.
s 2n+1 ->log 2.
l+z +
series
r-
+-- +
r
jE
Proof.
It has
fv hen T
...
is
is
sufficiently
is
z.
real,
first of
these equalities
and exactly the same proof holds when
z is
complex.
II
+- = 1 +- -M - =p(cos n n n
For
sum
z n
(
3T
1
real
z,
Its
convergent.
been shown in Art. 3 that the
Let
values of
x (z)=lim (1+-) =e (cosy + 1 siny).
z is real, Z
all
denoted by E(z), or by exp
// z = x + ly where x and y are
Theorem.
For
!
and
called the Exponential Function,
(1)
z3
I
holds
1
w+
The Exponential Function.
or complex,
(2)
1
1 -+... to oo. log2 = l-| + |-... + (-l)"- 71
Thus,
11.
to r terms of the series, to oo
/I
Also s 2n+1
2n --log2.
sum
be the
sr
+
i
sin
where
p>0 and ~7r<^<7r.
cr
values
large
of
n,
1+->0; n
P>0,
hence, since
and />cos< = l+-, we have cos ^>0, and -\-n n
^ = y/(n +
Also tan
x \2
<- 0, and ^/tan ^ -> 1 = lim {ny/(n + x)} == y. <^)
)-> 0, therefore
lim 7^^ = lim y2
(n tan
x \2
/
n/ +^i=(l+-) nz \ ( 1+-) n/
w therefore p r /
and,
lim see"
since
we have
lim p n = e*
and
;
- lim
1
sin
(
1
+ tan 2
hence, JE(z) jE (z)
(3) It follows
nor cos y +
\n
/
sec 2 <; T
n = lim
f
x = (l-f-j
+^
)
=
sec"^;
n/
\
w2\in
1
hence
;
JL 2n
2
1
(^
)
=1
>
= lim p n (cos n
= e?(cos y -ft sin y).
that E(z) cannot vanish for any value of y can vanish.
z,
for neither e*
AND COSINE SERIES
SINE
313
The function E(z) is periodic, and its period is 2i7r. For if k is any integer, positive or negative, the values (4)
siny are unaltered by adding 2&7r to y. Therefore E(z) < adding 2ik7r to z, thus E(z + 2 ikn)^E (z).
is
and
of cos y
unaltered by
t
A Complex
Index. When z is real, e z is a multiple-valued The real positive value of e z is function, except when z is an integer. called its principal value, and it has been shown that 12.
-
E(z)= the
z principal value of e
.
We take this equation as defining the principal value ofe when z is complex. z
This implies that for imaginary (as for real) values of garded as a multiple-valued function.
For
the present,
Thus e
if z
z=x 4
we
and %' = x' 4-
iy,
= E(z) = ex e
thus,
z .
e
take e z as denoting
(cosy-f
i
f
iy
,
e z is to
be
re-
principal value defined by E(z). we write
its
and
sin y),
z,
ez
'
= ex
'
(cos y'
+
tsint/')
;
= ex e x (cos y + i sin y) (cos y' + i sin y) = ex + x> {cos (y + y') + i sin (y 4- y')} = e* +2
z'
'
.
'.
Hence
the
Again,
if
same index laws hold for imaginary as for real indices. a is real and positive and z is real, a z = ezlo ^ a = E(z log
In agreement with
E(z log
this, if
a) and, at present,
we
a>0, we
take a
z
a).
a z as
define the principal value of
as representing this value.
and Cosine. For all real values of x and y, it x E(x + ty) = e (cos y + i sin y). When x = 0, this gives
13. Series for Sine has been shown that cos y
Equating
real
+i
sin
and imaginary
-
14.
mean
yE(iy)
v
2
parts,
we have,
4
t/
-
for all real values of
-*y
.
5
3 ?/
-
Exponential Values of Sine and Cosine.
Using
exactly the same as E(z)> by the preceding, cos y
therefore
+i
sin
cos
y = E(iy) j/
=
(c
= &u
9
ty 4- e"~ l *0,
cos ?/-
t
sin ?/=
sin
y~E( - iy)~e~~ iy
(^ rJL~^li ""*"JWK*"
;
e*
to
SUMMATION BY EXPONENTIAL THEOREM
314
which can be summed
15. Series
by the
Exponential
Theorem. (1)
The
Zu n
series
xn
where u n
,
is
a polynomial in
n.
'-
We
can find a
un = a
av
,
-f
a 1n
...
ar
independent of n, so that
,
+ a 2n(n -l)-f
...
+a r n(n~ 1)
...
(n-r-f
1),
and then s^n
/v>tt
1
spf*
n=rr
= (a + a^x + a2x 2 4Ex.
1.
w
Here 2
in
s
-f
the value of
(n*
n-r
a rx r ) e?.
5)/ n. i
6=n(n-l)(n-2)-f 3n(n- l)+n-f5;
TAe
n 5en*65 2/M n x /(w
n and
a, 6,...
+ a)(nH-6)...(rn-A;)
A are unequal positive
Series of this kind can be Ex.
-f
-f
2
hence,
putting
a
5,
a8 = l, and * = 1 in the result above, we have
= 3,
(2)
Find
. . .
T**
2..
Find
summed
|n w;Aere
t/
n is
a polynomial
integers.
as in the following example.
the value of
Denote the sum by
$, then
Now where
a, 6, c,
d=l.
Thus
rf
can be found by division, their values being
XT Now
=->-'-*)> _ -"' 6
a=
-5, 6 = 5,
c=
-2,
LOGARITHMIC THEOREM
315
EXERCISE XXXI X
Show
1.
lim fl -
that
Find the sum to
}
= 1. Exx. 2-9
infinity of the series in
3579
0,234
3579 +
n
f 4 f ji" [6 "L?"
1*1 23 1
+
Prove that
33
1+2
"" + '-
7
'
s " 4 + ""
^
A
*
A
23
+2+3
ni~
38 4"
a;
,
1
48
4""
" "
1
""^" 2**
3a:
8
:
-=t(4S-16e).
11.
+
~ii"
1
1
:
12.
(
cos 6 ox
13. If
= tan" 1 -
where
16.
-f
^^ + ...4- wa:n + ...
//
Suppose then
(i)
yn = log
rz^
n
(
1
- !<#
+ x) - s n
^
^ 0,
0
that
dx
n terms of
and
1-hx*
" " l
s n~>log (1
is
even or odd.
and s^^ 1 /( w f i)"^ -
+x)
this case of the theorem.
in
to
9
n
The case
sum
x = 0,
when
1+a;
a;
sn
which proves
be the
sn
__ 1 -f
according as
therefore
Let
yn =
then
;
Hence log(l-fx) lies between n n But n+l-n = (~l) ^
Therefore y n
X
i
.
Theorem.
series,
(ii)
then
,
a
Proo/.
the
cosfea;=w
as (Of.
for every n.
as W-^QO,
n->oo Ex.
,
XXIX,
which x = 1 has been dealt with in Art.
8.)
10. B.C.A.
LOGARITHMIC SERIES
316
When x<0, changing
(iii)
the sign of
has to be shown that
x, it
3 Xn X ~ X + -x) = x + + -+...+ 2
0
if
Let y n = -log
therefore y n
>0
-log
(1
-x)-s n9 where
(1
yn =
then
series;
then
sn
x = 0,
when
n
and
~r
n _ x ~ =
(i
then
'
~5
i
xn
|
i-x
~dx
-x
)_ ~
'
(T7*)
r^Ti
2
~
/
0
Logarithmic Series.
^i
Changing the sign
X
and, since
0
s n~>-log (1
-
+# =
x
x
log (1
+ x) - log
/
- x),
(1
x3
./
1+x = n-f --1 1-x
-
,,
.
so that x
-
+
1
=^
4-
(A)
follows that
it
x5 5
\
+
-;
/nv
......................... (C)
,
7,
we nave
2n-fl
7^
.....
3
L
//
0
o: and sn ~x+ o
a;
4-
If
the,
Rn
value of log
"
is
is less 1
~~
x
than
...(D)
5
- +... to
1-fa;
2s n as
-x),
have
/-tv~,i** 71 - 1
3 .
;
of x,
1+x
Writing &
x
;
-...+ (-I)
1
1
("r-lo^
;
x
1
-l
If
X
log
--
'
thus y n->0 and
;
since
__ 1x =
x n+1
1
-
-------
the last expression ->0 as n->oc which proves the theorem in this case.
and
x
this
x" +1
1
17.
n terms of
when x = 0, and
=
2n
x n+1
1
*
Hence
z n <0.
therefore
to
for every n.
n+i x n+
ofe n
=^
1
cfx
Zn^yn-"*
Let
sum
the
is
....
-
2ii\>
n
terms,
---
-
-
+1
1
x
the remainder after n terms of the series,
.
show
that the err.or in taking
CALCULATION OF LOGARITHMS 18. Series
317
which can be summed by the Logarithmic
Series. Ex.
1.
Ex.2.
The
If k
If
is
a positive integer and x < 1, then n xk f x2 1 ^n=oo' x \
\x\
series is
\
find the value of
convergent when \x\
< 1.
*-*
~~~
Let
8
be the sum, then since 3
1
1
*"
1
wehave
TAe method of Ex. 2 can term
sum any
6e applied to
series of
which the general
is
where
P
is
a polynomial in n and
a, 6,
19. Calculation of Napierian
k are unequal positive integers.
...
Logarithms.
The number
e is
chosen as the base of the system of logarithms used in theoretical work. Such logarithms are called Napierian after Napier, the inventor of loga}
rithms.
In theoretical work, log^V means log e
reckoning, log^ means log ]0 jV. The method of applying the equations
Napierian logarithms Ex. T
x
Let
1.
N;
just as, in practical
of Art. 17 to the calculation of
exhibited in the following example.
is
Calculate Iog e 2 to seven places of decimals.
i+*
i
,
.-.=-;
=2;
rt
.-.log, 2
-2
Carrying the reckoning to nine places, we have 333 333 1/3 =0-333 1/3 1/3
8 5
1/3=0-333 3 8 ) =0-012
333
345
679
3 5 ) =0-000
823
045
=0-000
065
321
=0-000 3 11 ) =0-000
005
645
000
513
=0-037
037
037
l/(3
.
=0-004
115
226
l/(5
.
7
=0-000
457
247
I/ (7
.
9
=0-000
050
805
I/ (9
.
=0-000 18 1/3 =0-000
005
645
1/(11
.
000
627
I/ (13
.
000
070
1/3
1/3 1/3
11
15
1/3
=0-000
/.
log e 2
3') 3')
3 13 ) =0-000
1/(15.3
15 )
333
000
048
=0-000
OOP
005
=0-346
573
589
0-693
147
178
=0-693147178 nearly.
BORDA'S METHOD
318
The (i)
this
error in the value obtained for log e 2
We
have taken 2S 8
account
is less
for log e 2.
JL
.
The
Ex.
:
the error on
1 of Art. 17,
than
17 (ii)
By
due to two causes
is
1
>000 000 002.
9
2SQ
error in the calculated value of
is
numerically
2x8x0-000 000 000 5 = 0-000 000
than
less
008.
Adding the results of (i) and (ii), we see that 0-693 147 178 is an approximate value of log,, 2 with an error numerically less than 0-000 000 01. V. 0-693 147 168
can be found in succession by using formula of (D) Art. 17. The logarithms of successive prime numbers may be calculated by the following method, due to Borda, in which the series employed converge very rapidly.
Using the identities
x3 -3x + 2 = (x-l) 2 (x-f2) it will
and
z3
-3z-2
be seen that
(x
+ l) 2 (z - 2)
"
x3 - 3x - 2
~ 1
- 2/(z3 - 3z)
-f ...
Putting z = 5,
6, 7, 8,
log -|
it will
+ -
+ log 5
log 7
Hence
it
log 3
+i
log 5
+
=0-0181838220
...
,
=0-0062112599
...
,
= 0-0040983836
....
= 0-0101013536 log 7
-2 log 2 + 2 log 3 -i log 5 ~I
.
be found that
2-f log 3
log 2
'
log 7
can be shown that
= 0-693147 180 = 1-609437912 log 5 log 2
The logarithms
= 1-098612288 7 = 1-945910149
...
log 3
...
log
of successive primes 11, 13, etc., can etc., in the formula.
....
now be obtained by
putting z + 2 = ll, 13,
The method
of the
example which
follows, in
which logarithms to a base
10 are calculated without the use of the logarithmic interest, being one of the methods given by Briggs.
series, is of historical
HYPERBOLIC FUNCTIONS Ex. 2. Obtain 5 decimal places.
319
by the use of square-roots only,
the logarithm of 7 to the base 10,
to
In the reckoning below, when a square-root has been obtained which is greater than the square-root of the product of it and the last square-root obtained that was less than 7 is found next and vice versa. Thus we have 7,
;
= log ,4 =0-5, log */T6I=log 5-623413 ... = log J5=:0-75, =0-875, Jog ^105 = log 7498942 ... = log log >/]&;= log 6-493816 ... -log D = 0-8 125, ^0-84375, log s/'OD^log 6-978305 ... = log = 7-233941 ... ^ 0-859375, log */(7JE=log =log log
and so on
;
until
we
VT6-log 3-162277
arrive at
= 0-845100 = 0-845096
log 7-000032 log 6-399968 thus, Iog 10
...
7=0-84510 to
five places,
and probably
Common
20. Calculation of
...
is
j
equal to 0-845098 to six places.
We
Logarithms.
have
logjo.V-log.tf/log.10.
The expression of logarithms and
is
l/(log e 10) is
modulus of the
called the
denoted by /*.
It
common system
can be shown that, to
fifteen places
of decimals, /*
Thus Iog 10 tf =/*log
21
.
tf
0434294481903251. where
tf,
has the above value.
ju,
The Hyperbolic Functions.
hyperbolic cosine, sine and tangent and tanh x, are defined by
The functions known as the and denoted by coshx, sinhx
(1)
of x,
--* ,
The
reciprocals
secant, cosecant
(2)
x .
= .
cosh x
.
,
cosech WQViVAU.
x = -r-i */
-7-
ax
.
i
sinh x
,
T- cosh x = sinh x,
and we write
of x,
Hence the following equations cosh 2 x - sinh 2 x == 1 ax
,
cosh x, sinh x, tanh x are called the hyperbolic
of
and cotangent
sech x =-
*~
A tannx
,
sinh x
,
coth VVVJLX
x rf/
--
1
i
.
tanh x
:
1
- tanh 2 x = sech 2 x,
cosh
x,
-7-
ax
tanh x = sech 2 x.
ADDITION FORMULAE
320
cosh ( - x) = cosh x, sinh ( - x) = - sinh x, tanh ( - x) = - tanh x, so that cosh x is an even function, and sinh x, tanh x are odd functions of x. Graphs of these functions are shown below rough sketches of the Also,
:
two can be constructed quickly from the curves y = ex y^e~ x (see p. 309), by observing that, in Fig. 61, C is the middle point of AB, and ED = BC. The ordinate of y = tanh x is equal to the ratio ED/EC
first
,
;
this
can be found numerically or by a geometrical construction.
61.
(3)
We
Addition Formulae.
2 cosh x cosh y
have
= \(ex + e~ x -f-
)
(e
y
+ e~ y
)
e
= cosh (x -h y) -f cosh (x-y). Similarly,
= cosh (x + y) - cosh (x-y), 2 sinh x cosh y = sinh (x + y) + sinh (x - ^ 2 cosh x sinh y = sinh (x + y) - sinh (x
2 sinh x sinh y
-
Therefore
(xy) = cosh x cosh y sinh (x y) = sinh x cosh y
cosh
sinh x sinh y,
cosh x sinh
y,
and consequently A
,
tanh
,
.
(x v dby) y/
=^
tanhxtanht/ T
r
T
r^-
ltanhxtanhy
where in each equation the upper or lower sign
is
,
to be taken throughout.
INVERSE HYPERBOLIC FUNCTIONS (4)
321
Using the exponential values of sin x and cos x (Art.
14),
we have
cos x =
2
x=
sin
= ~ sinh
tan x=
tanh
ix,
ix,
i
cosh x = cos
and
tx,
sinh x = i sin *x, tanh x
Inverse Hyperbolic Functions. Let x e -x _ and e2x - 2e*z/ -f 1 e
22.
^ = ?/v/(2/
therefore
Thus
if
2
there are
#>1,
i/ ==--
2^
_j_
and
-l)
two
tan
i
= coshz,
positive
x~log{y,J(y
2
-l)}.
real values of x, equal in
which correspond to any given value value is denoted by cosh" 1 y, so that cosh" 1 y = log {y + J (y2 ~
Next,
let
?/
values of #,
= sinhx,
1)}
6
magnitude and
of y.
.
-2e /-l=0, and since we have eF~y + ,J(y 2 + l), so that # = log then
2a;
then
0,
of opposite signs,
The
ix.
a?
ex
{y
>0
for real
+ J(y2 + 1)},
and we write sinh- 1 y = log {y *
A
Again,
.
it
/
e2
then
*
=-
= tanhx ~
+ Jy2 + 1}.
e^-e-* =
and
Hence we write ,,
Thus cosh" 1
/
is
1
a single- valued and continuous function of y (see
sinh"" 1 y is a single-valued and conCh. XVII, 21) for all values of j/>l tanh"" 1 y is a single-valued tinuous function of y for all values of y and continuous function of y over the range -l
;
EXERCISE XXXII The notation of Art. 9 1.
Show
2.
Prove
that
that
is
used in Exx. 2 and
-~H--- -+
1 -f
t+
---
+ ...
-J-^71
-f
. . .
3.
-f ur-
2n
~"=log 1
-> log 2 as
n->oo.
+ Jy + J
n+
2
log
2n
~
n,
so
that
LOGARITHMIC SERIES
322
1-J-fJ-J-f... be deranged as follows. Write down the by the first q negative terms, then the next p positive terms followed by the next q negative terms, and so on. Prove that the series so deranged is convergent, and that its sum is 3.
first
Let the
series
positive terms followed
p
log 2
[Let ar be the sum of the
first r
+ | log plq.
groups of
p
positive followed
by q negative
terms, then 1
1
\
I/
1
1
1
= (log 2 + Jy + \
- i + log qr + c ) -+ log 2 + \ log plq. log pr + 2pr (y ar pr ) Also, if sn is the sum to n terms of the deranged series, r can be found so that Thus ar -sn is the sum of a all these terms are contained in oy, but not in a r i. Also r->-oo as w->oo, and finite number of terms each of which ->0. consequently sn -> log 2 4- i log p/q.] 4.
Use the (i) if
if
(ii)
[For
(ii)
series for log
(
1 -f x)
to
show that
:
0<#-log(l + x)<\x* 0<#-log(l +x)<\x*l(l + x). x= -y, then 0
0
then
let
;
then
...),
5.
etc.]
(i)Showthat (ii)
Hence
calculate Iog 10 7 to five places of decimals, given that
Iog 10 2 =0-301030, 6.
(i)
(ii)
7.
(i)
(ii)
Showthat
lo glo
1-1=2,1
|l
+i
g=2 M {^ +
that log,
-
=0-434294.
i+i
-
-
21-.+ ...} 11 to five places of decimals.
Given ^=0-434294, find Iog 10
Show
p,
1 .
^.
+
L
^+
.
...}
Given that Iog 10 2 =0-301030, Iog 10 3=0-477121,
^=0-434294, find
log lo 13 to five places of decimals. 8.
(i)
(ii)
x + z 2 ) = log e ( 1 - x9 ) - log e ( 1 - x). x x2 ) to six terms, and show that the coefficient of + Expand log c (l-f xn in the expansion is -2/n or Ijn according as n is or is not a
Show
that log e
(
1 -f
multiple of 3. 9.
Find the
Sum
coefficients of x*
m and
x* m+ l in the expansion of
to infinity the series in Exx. 10-14. 1
1
1
'
IT2
+
37*
+
6~6
APPROXIMATE FORMULAE 15.
323
In the expansion of (I +bx + cx*) log e (1 -fa;), (i) Find the coefficients of x* and a*, and determine the values of b and c so that each of these coefficients may vanish, Giving to 6 and c the values so determined, expand the given expression as far as the term containing x*.
(ii)
-
(iii)
Hence
an expression of the form
find
-
r-
the best possible approximate values of log e 16.
(6
( 6 J .
---
Denoting the
(ii)
--
(
by u v
series in brackets
un
(in- 1)
=
will give
+ x) when x is small.
(1
- (6x + 3x*)
+ 6x + x*) log e 1 -f x) 1.2 2.3 3.4 t2 x xH x 5.6. 7 [3.4.5 4.5.6
Prove that
(i)
which
-
}
... V
w a -fw 3 -
-
,
/' ...
prove that
,
2
*n-i (n-l)( + 4) If 0
(iii)
-i
(iv)
17.
Use Ex. 16
759
(iv) to
with an error in defect
show that Q0^AO
less
than 5/10
11
is
an approximate value of log e
Hence
.
find Iog 10 1-024
and
to ten places of decimals. 18. If
a-b=h, where h is a-6/1 l\h a-b\
small,
h2
show that h
2/a-6\ 3 _^
h*
" +
+
+
A8
i
(iii)*
Hence show
that, if
a
,
a
*
+ "'"
is^nearly equal to 6,
a-bfl
l
the error being approximately equal to (a -6) 3 /6a 3
Sum in
the series in Exx. 19, v"*
83*
00
n
~*~*
21. If 8n is the
on
~n
sum
to
n terms of the 1
1
1. 2.
20':
3*5.
6. 7
+
v WamQO
(
n + *-J
xn
series
1
1
9. 10. 11
13.14.15
'
prove that a
a
8^= Hence show that the sum to
1
1 i
,
..,
infinity of the series is
* This is Napier's formula.
,
|
J log
2.
.
1-024,
Iog 10 2, each
LOGARITHMIC SERIES
324 22. If sn
is
the
sum
to
n terms of the
series
1
2. 3.
1
45. 6. 78. 9. 10
(3w l)3n(3n +
1)
111^3+".+3^T- L 2^^+^
prove that
1
+
Hence show that the sum to
infinity of the series
is
(log
3-1).
23. Prove that
+++ 24.
2
By expanding (i ,
log(l-f-o;-fa;
n
as ,..,
,
is
of the form 3r or
n a (?i 2 -l 2 ;)
w2
1-^+-^ 1^ Li n(n +
,
g)
[3
cording as n ,...,
in various
i- + ^Ll)-(-f)(|J8
(u)
)
-
-
ways show that
+ ... =( -i r
-
.-ii
is
.
+
n
is
+1
1XM
=(-!)"
.
or
of the form 3r or
the form
6r,
6rl, 6r2,
[These are obtained from the identities
-, -
or
71
-
according as n
71
6r~3. :
2 -^)-21og(l -x ) + log(l -s)=log
identities
^-^ to show that
/
This formula
is
.
ac-
3rl.
(ffi
Use the
lmn - 1
(-|)
or
7Z-
25.
...
3rl.
of the form 3r or
-=0,
log(l
according
(n-
l)
according as
(iv)
(-ir-
l_z
2n
(ii)and
2 or
.
3rL
n 2 (n 2 -l 2 )(n 2 -2 2 ); i
-=
"- to
72
1
/
3
U
72 4
-
2/
+ ""
"
J
useful in finding the logarithms of large primes.
is
of
CHAPTER XX CONVERGENCE
(2)
SERIES OF POSITIVE TERMS (Continued)
Cauchy's Condensation Test.
1.
and a
creasing function of n, n and 2a n
f(a
Group the terms
is
any
// f(n)
positive integer
is
>1,
a (positive*) dethen the two series,
are both convergent or both divergent.
),
of Zf(ri) as follows
:
{/(I)
sum
v n denotes the
if
where,
v n =f(<*
The number
n -1
of the
+ 1) +f(a n ~ l + 2) + ... +/(a).
of these terms
function, n (a /.
- aw ~ 1
terms of the nth group,
is
n
)/(a
~ a n - a n l and since f(n) ,
< vn < (an - an ~ )f(an l
)
1
)
-(a-lJaYKXvn^^- 1 )^" /^'
a nf(a n )
verges (Ch.
XVI,
;
1 )-
therefore
2.
5).
vn and therefore Sf(n) conconvergent, so also is a nf(a n ) is divergent, so also is vn and therefore i
If
But a
An important p>l
gent if
a
;
terms must converge or diverge, n n are both and a f(a ) convergent or both divergent. f(n)
2f(n) diverges. .
is
a decreasing
1
a
Now, if
is
and
series of positive
Test Series.
The
-
2r
series
}
^
is conver-
divergent if
positive integer, this series is convergent or divergent according as the series Zv n is convergent or divergent, where If
is
any
n
~
a n (log an )*
~ (log
ay n p
"
Since (loga) p is constant, the given series is convergent or divergent according as 21/n p is convergent or divergent, that is according as p>l or *
In any case, the terms of the decreasing function /(n) are ultimately of the same sign (Exercise
A-A.V,
1}.
RAABE'S TEST
326
Kummer's
3.
and
terms
Test. that
suppose
Vn = dnUn/Un+i-dn+l ;
for
vn >k>0,
n^m, if v n
(ii)
Proof,
condition,
...
that
so that after
a certain
stage,
say
is convergent ;
n-1
,
the
s r is
if
sum
. . .
in succession for
n
+ u n < 7 (dm um - dn u n ) < 7 dm um
in the given
Su n
is
,
a fixed
convergent (Ch. XVI, -
dn
Since
.
to r terms of
m + dm u m
(ii)
suppose
is divergent.
m + 1,
Putting w,
found
Zu n
the series
um+l + ww + 2 +
therefore
also
divergent;
we have
whence
Thus
be
<0 for n^m, Zu n
(i)
is
2l/dn
be two series of positive
then
number k can
if a fixed
(i)
Su n and 21/dn
Let
dn4 1 <0 .
number
7).
n^m, we
for
;
have
"n+i d therefore
u n >dmum
divergent, therefore
Now
-5-.
an
2u n
is
d mw m
is
a fixed number and
l/dn
is
divergent.
NOTE. If t>n ->-0 we cannot find k so that t;n >&>0 for n^m, and the test fails. In part (i) of the test, S\jdn need not diverge in fact, dn may be any positive function of n. ;
Raabe's Test.
4. (i)
Let
Zun
be a series of positive terms, then
if after a certain stage, say for
where (ii)
"k
If
is
a fixed number, then
n(
~
Xu n+l
1)<1, '
then
Zu n
Zu n
is convergent.
is divergent.
This iollows at once from Kummer's test by taking d n = n, giving
GAUSS'S TEST In
This test
^u
1
n+l
may
TT
*,
un+l -~-^ (2-l)
D'Alembert's fails
test, 27wn
with Raabe's
;
-'
.
.-.
; '
2un
NOTE.
From 5. that
is
* -f
4 o
hm
n 2wn+1 x92
<
l
.3 x 5
5 ^2
.
\
=-
+
.
1 and diverges converges if x we have however, |
-r
4 o
if |
x >1.
x
If |
\
\
= 1,
test,
-n
3
.
convergent.
This series
is
the expansion of
he deduced the series for sin
it,
fails.
--
x+-
..
-~
*a
6n 2 therefore
when l> 1 and
converges
l
.
,.
-T5L
Here
By
Zu n
then
Z,
the convergence or divergence of
1.
the test
=
:
.,,-,.., Consider
Ex.
'J
but if Z = l, the test fails. be tried when D'Alembert's test
when l
diverges
-
n(
particular, if Urn
327
sin" 1 x,
and was discovered by Newton.
x.
Gauss's Test. Let Zu n be a series of positive terms and suppose u n/u n+l can be expressed in the form ^n-fi
where
and
p>l
\
limit as n->
\< a fixed number k or
bn
then
,
Zu n
First suppose that
Proof.
U n( n \u n+l n ,, |
bn
|
and diverges If
a=
in Art. 3
1,
bn
j=a+ n p^ --*r ,
a>l
and
diverges if
><
as
/
and p>l. Therefore, by Raabe's if
a finite
then
a^l,
l\ i
(
for
converges if
(in particular) b n tends to
test,
Su n
converges
if
a>l
a
Raabe's test
fails
Taking dn = n log n
and we apply Kummer's.
we have -
(n
+ l)
log (n
log
n-
(n
+ 1)
^n+l
=n( 1
Now and
(n-f-l)log
i^.6n->0,
Therefore v n-> divergent.
l
4-
~
for
-
+ -j J
= (w-f l)logf 1 jp>l (Ch. XIX,
+ 1)
log (n
+1
r)->-l
and, after a certain stage,
as
and
7, (5)),
t>
)
n <0.
|
Hence
,^
27w n is
BINOMIAL SERIES
328 In
cases
many
u n/u n+l can be expanded
and we have the following
rule
in a series of
:
b u n _ ^I j__a j___ i -r -r o ~r n n2 U
r* Tf
.,_
i
M
// J
,
valws of
the series being convergent for sufficiently large
convergent if
For
a>l and
sum
Zu n
n, then
is
a^l.
divergent if
in this case b n is the
powers of 1/n,
of a convergent series.
the most generally useful test for series of positive terms, and in cases where it applies, it is better to use it at once instead of beginning
This
is
with D'Alembert's and Raabe's Ex.
_.
By
P -
Discuss the convergence or divergence of
1.
_
tests.
.
.
.
division
n
we can show Athatx
where
bn
,
=^1
^
-r
1
\2
32.52 42 6 2
+
'
"
'
n n->oo.
aa
j
)/(l 4-^) is
+n
v- - i
i
32
l%
+ W~T* * 02
-
I/, + -i\ //,
-
Therefore by Gauss's test the series
Raabe both
-
=[
2
Here the
divergent.
tests of
D'Alembert and
fail.
CONTINUED FROM CH. XVI, 25 6.
Binomial Series.
When x=-l,
(n-l) v
2
the series
n(n-l)(n-2)
-
/
L
L converges if n>0 diverges if n<0. When x== -1, after a certain stage
and
+ Wj -f w 2 +
Denoting the series by u
M-
r
*
n-r
w r +i & r = n(n
where
Hence, by Gauss's
5 1,
+ 1)
test,
+1
_
l( I
the terms have the same sign. division that
all
. , .
we can show by
,
n-fl
.,
i
b~ *
i
t
~j
)
r
--)->n(n + l)
-
rz
r~>oo.
as
the series converges or diverges according as
^
0. according as n the that the test says series diverges when n Apparently the reasoning fails because u r/ ur+i * s f the form 0/0.
n+1
7.
that
is
The Hyper-geometric X+a
1+ rT^ is
known
Series.
The
0,
series
xi
+
"-
r.2.y(y+V
as the hyper-geometric series,
and
is
of great importance.
but here
HYPER-GEOMETRIC SERIES If none of the constants a, and we shall prove that (i)
j3,
is
y
329
a negative integer, the series
the series is absolutely convergent
|x|
if
is
is
endless,
divergent if
\x\>\; (ii)
when $
1,
(\\\)
when x
-1,
the series converges if y-f
Denoting the
Proof.
,
and not otherwise;
the series converges if y>a-f-/8,
by UQ + ^ +
series
. . .
-
=
l>a + /3,
-f
un +
. . .
and not
otherwise.
where
...
1.2...n.y(y + l)...(y + n-l) .
we have
Hence the
(i)
series
converges absolutely
if
|x|
if
|x|>l. (ii)
When x = l, w n /w n+1 ~>l.
have the same
Hence, after a certain stage, the terms
sign.
Applying Gauss's
we can show by
test,
division that
n
n+i
where n-
\
and 4,
5
are fixed numbers.
verges according as y-f
1
Thus
-a->l
y>a-f/J (iii)
If Aen
x= - 1,
or or
w
w/
\
b n -+A
and Z"w n converges or
<1, that
is
di-
according as
y^a-f/?.
after a certain stage, the terms are alternately posi-
and negative. Also |w n +il
tive
(a
+ n)(]8 + n)<(l+n)(y + n)
;
that
is,
if
(a + j8)
or
l+y~a-j8>(ajS~y)~.
if
This
is
true for sufficiently large values of
n
if
--!*_-.! +-5
y 4- l>a -f /?.
where a n->y-f 1 ~a~)3 as w n+1 n n->oo. Hence by Ch. XVI, 24, u n->0 if y-f I>a4-j8, but not otherwise. Therefore by Ch. XVI, 14, the series converges if y-f l>a+/}, but Again, as in the preceding,
not otherwise.
ABEL'S INEQUALITY
330
De Morgan's and Bertrand's
8.
positive terms,
and suppose
Test.
u n/u n+l can
that
Zun
Let
be
a
series of
be expressed in the form
~~
n
u n+l
a certain stage an >l + k where k
then
(i)
Zun
is convergent.
if after
// a n
(ii)
Zu n
Using Rummer's
Therefore v n ->a n diverges If
is
a fixed positive number,
is divergent.
test,
we take dn = n log w. Then,
v n ~n log
if
nlogn'
-l
n
u u n+l
.
as n->oo
(n
as in Art. 5,
+ 1 ) log (n + 1
)
Hence Zu n converges
.
if
an >l
+ k and
a n
a n = 1, vn <0, for log ( 1 \
-n
r
-f-
-n+
)<
=
1/
,
Zun
therefore
1
is
divergent.
SERIES WITH TERMS POSITIVE OR NEGATIVE 9.
// Eu n is a convergent series of positive terms and a sequence of real numbers, positive or negative, and all than some fixed number k, then 2a n u n is absolutely con-
Theorem.
av a2>
t5
39
numerically
less
vergent.
For |
Since 27wn
where
is
is
^
\
+ a m+2w 1
convergent,
we can
find
m
such that for
any positive number, as small as we choose. |
Hence
am ^um
am+lum+l
-f |
|
am+2um+2
-f ... -f 1
|
values of p,
Therefore
am+ ,um+ ,
27a n w n is absolutely convergent.
^a:. 1. // Eun is a convergent series of positive terms, are absolutely convergent for all values of 0.
10. Abel's
numbers
all
swh
for the values
Inequality.
// uv
ti
t^,
2un cosnO and Sun ainnB
8 , ... t*
a sequence of
real
that
nofr, and if al> aa a3 ... is a a j < Q,^ + a + + aj^ n < a
1, 2, 3, ...
of positive terms, then
,
^
m . .
,
decreasing sequence
^
DIRICHLETS TEST For
let
tf
n
~ u l + u^ 4-
-f
. . .
un
331
,
- a Wj 4- a 2 n 2 + + a n w w n = Then u l s v u^s z -s v W 3 = s 3 -.s 2 etc., therefore = 1 5 1 + 2 5 2 ~ S l) + + a n (*n ~ S n~l) * = (a l - a 2 }s 1 + (a 2 - a 3 )* 2 >+ (a n ^ l -o n n _ 1 + a n * n
and
t
.
.
.
.
3
,
' ' '
(
-t-
Now and
the factors a l
sum
their
Also
52 ,
sv
-a 2>
2
~a
.
av
...
5 n are all greater
.
).s
.
a n-i~~ a ni a n are positive or zero,
3>
is
.
than
and
I
than
less
h.
Therefore
Dirichlet's Test. // Zu n converges or oscillates between finite and a l9 a2 a 3 ... is a decreasing sequence- of positive terms which tends to zero as a limit, then Ea n u n is convergent. 11.
limits
,
,
Let
sn
= u l 4- w 2 +
+ u n-
we can
Since 27w n converges or oscillates finitely, and p, that, for all values of
find
numbers
A,
I
so
m
l that
is
to say,
l< um+I + w m+ 2 + by Abel's inequality
Therefore,
Now
m+1 ->0 as w->oo
w
// (an )
limit, both the series
which case
For
therefore for
,
^a nu n
hence j&ar. 1.
if
Ean is
+
cos
n6
*s
any
e
we can
find
m
so that
convergent.
a decreasing sequence of positive terms which tends to zero as a an sinn& are convergent unless 0=0 or 2for, in an cos nQ and diverges.
not a multiple of
2?r,
ScoBnO and
tho series
27 sin
n^
oscillate
between
(Exercise XXVII, 24, 25). the result follows by Dirichlet's test.
finite limits
Hence
12. Abel's Test.
n n converges and a 1? a 2 then Za nu n is convergent. sequence of positive terms, For as n tends to infinity, a n tends to some limit
Hence by
Dirichlet's test,
a n u n - lLu n
But Zu n y
If
Z(a n -l)u n
is
,
a3
I,
,
...
is
a decreasing
therefore
convergent; that
is
to aay,
is
is
convergent. convergent, therefore
a nu n
is
convergent. B.C.A.
CONVERGENCE OF POWER SERIES
832
POWER
A
series of the
x may be
real or
type
SERIES
Za n xn is called
a Power Series.
In Arts. 13-15,
complex.
Za nx n
converges (absolutely or conditionally) for x x x1 verges absolutely for all values of x such that
13. //
\
For
a n x^
since
convergent, a n a: 1
is
k exists such that |
a n xf
\
n ->0.
\
<
= x ly
con-
.
\
|
Therefore a positive
Hence
for every n.
it
x
if |
r
\
<
|
xl
number |,
In
4 Therefore every term of 27 a n
xn
is
less
\
1
than the corresponding term
of the convergent series 7
,
k+k
Z
Therefore
\
Za n x n
14. //
x such For
that
is
is \
7
i
H-An I
X T ^l
convergent and
non-convergent for
Ea nx n
x^x v
is
it is
absolutely convergent.
non-convergent for every
\x\>\x\. x 2 where I^l-H^il* by the last which is contrary to the hypothesis.
the series converge for x
if
theorem
an
xn
X Tl X
converges for
it
x^x^
and 15. With regard to the series Za n xn either (i) it converges for x = it all other or values no value x, (ii) converges absolutely for of x, for of a n x n converges absolutely or (iii) a positive number R exists such that ,
when
x |
\
and
is
non-convergent when
\x\>R.
Suppose that there is a value of x other than zero for which converges. Also, suppose that there is a value x' of x for which x' non-convergent. Choose a positive number a greater than
Proof.
2a nx n it
is
,
|
then,
by Art.
14, the series is
Divide the real numbers in the interval
The lower
class is to contain
(0, a)
every real
into
number
two
classes.
such that
r
Sa n x n
The upper class is to contain every real number r', does not converge if |x|=r'. Both classes exist, and follows from Arts. 13 and 14 that every r is less than any r'. Therefore the classification defines a real positive number R which
converges such that it
\
non-convergent for x = a.
if
|
x\ =r.
Za nxn
R
This number separates the classes and may be assigned to either class. such that a n xn converges absolutely if x
is
|
if |
x \>R.
Nothing
is
said as to
\
what happens when x = R. |
\
RADIUS OF CONVERGENCE
333
= = writing R Q, R
By
All that has
been said applies to any power
16. For a real series of convergence. according as x
Za nxn may
The is
Za nx n
,
the interval
series, real
(~R,R)
and
(i)
is
(ii)
respec-
or complex.
called the interval
converges absolutely or is non-convergent within or outside the interval. At either end point series
converge, diverge or oscillate.
Fdr a complex series Za n zn R is called the radius of convergence. The circle (0, R) with centre at the origin and radius R is called the circle of convergence. The series converges absolutely or is non-convergent according as the point z is inside or outside the circle. If z is on the circum,
ference, the character of the series
Sa n z n
17. Let
be any
not determined.
is
series, real or
complex, and suppose that
lim \a n+l /a n
Then and
it
\=l
= l\z\,
lim
R = l/l.
follows from D'Alembert's test that i
i
Again,
if
lim |
a n n = l, then lim \
|
anz n
n \
^l
z \
|,
and by Cauchy's
test
18. In considering the behaviour of a complex series at points on of convergence, the following theorem is often useful.
its
circle
// (a n ) is a decreasing sequence of positive numbers such that a n+l/a n->l and a n ->0, then, (i) if Za n converges, 2a n z n converges absolutely at every n point on its circle of convergence, (ii) if Za n diverges, Za n z converges = 1, (though not absolutely) at every point on the circle, except at the point 2 n n where it diverges, (iii) the same statements hold for the series (-I) a n z ,
except that
For
it
diverges at the point z
= lim 1/72 an
Moreover,
(i)
a n+l /a n = zn if
l,
= - 1, when Z( -
and at a point on the
l)
na
n is divergent.
circle of
convergence
= a n (cos + i sin d) n = a n (cos nO + 1 sin nO). Za n
converges, both
absolutely convergent (Art. 9,
Ex. 1);
Za n cos nO and Za n sin nO are therefore Za n z n is absolutely
a n cosnO and convergent when |z|=l. (ii) If Za n diverges, both = or 2&7r; in which case 2a n sinnd converge unless a n cosn0 Ex. the Hence statement follows. diverges (Art. 11, 1). (ii) (iii)
This follows from
(i)
and
(ii)
on substituting -z
for
z.
COMPLEX VARIABLE
334 19. If
Za n
n z
vergent when
when
converges
z |
|
<|
zx
z=*z lt
Denoting
|.
Art. 13,
by
sum by
its
/(z),
= a + a 1z + a 222 + --- + (a n + n)z n where /(z) z n = a n+l z n + l + a n +2* n+2 + We have >
|
As
|<| a n+l z+
T?Z
we can
in Art 13,
\
+ a n+2z+
2
find a fixed
number
+
1
1
as
|z|->0.
>
T?
and
absolutely conprove that
shall
|^| >0
f
l
it is
we
...
.
such that for every
k,
r,
20. Criterion for the Identity of Power Series. // Za nzn =Zb n z n = for every value of z whose modulus is less than some number /z, then a n b n for every value of n.
For by Art.
where
77
and
19, for
every
n,
tend to zero as
17'
z |-->0.
|
- 6
be seen that
z |-^0, it will
1
|
.
,
can divide by
z,
therefore
+ a n + 'V)^n ~ 1 = ^i +
a l + a2? +
Making z->0 as a n = 6n
21
.
(
before,
^+
'
+ (^n + ^)2;n "" 1
*
we have a l = bv and by continuing the
process,
for every n.
We
Binomial Series. (n-l)
consider the convergence of the series
2
n(n
...
1)
(n
r
+ 1)
i
where w
is real
Denoting the
and
z
series
complex.
by
...
n Therefore, (i)
it
,
we have
=1 r
follows from Art. 17 that
The series converges absolutely
if
z |
|
<1, and
is
non-convergent if
z>l.
MULTIPLICATION OF SERIES
When shall
|=1, the point
2; 1
z
on the
is
circle of
335
convergence, and
we
prove that
(ii)
// n>0, the
series converges absolutely at every point
on
its
circle
of convergence.
l
'
|
/
ur
n
r
4. i
"
r
u/ Therefore
If n
^~
u'r+l
1?
ur
=u
r ',
+
1
according as r
increases with
r,
or
is
n^r + l,
that
is
according as
constant and the series cannot be
convergent.
- 1, we If ?i> apply the theorem of Art. 18. After a certain stage, u r a decreasing sequence, u'r /u r '- >l and, as shown in Ch. XVI, 25, '
is
<->0. These are the conditions required in Art. 18, to apply the theorem of ur and we have to consider the convergence of '
this article,
'.
After a certain stage, the terms of Zu r are alternately positive and r - l) ru r negative, so that Zu r is convergent or divergent, according as Z( is convergent or divergent, that is according as n^O. (See Art. 6.)
Hence the statements
(ii)
and
(iii)
follow immediately from Art. 18.
MULTIPLICATION OF SERIES 22. //
t/j
sums being
... and V ^v 2 + v + ... are 1 3 convergent series of posiare absolutely convergent real or complex series, their respectively, then the series
+ Wg-f w3 +
tive terms, or if they
s
and
t
U l v l + ( U IV2 + U 2V l) + ( U 1 V3 + U 2V 2 + U 3 V l) + is absolutely
convergent
and
Multiply the terms of products as below.
its
sum
is st.
%-f w 2 -fw3 +
...
by vl9 v2
,
v3
,
...,
and arrange the
This array extends to infinity on the right and below, and we shall consider two ways of arranging the terms so as to form simply infinite series.
PRODUCTS OF SERIES
336 Let
an
is
sn ,
the
t
sums
n be the
sum
to
n terms
n terms
of the first
of 27w n
Evn
,
of the first
FIG. 62.
Draw
respectively
n rows
then
;
if
we have
of (A),
Fia. 63.
a set of squares, as in Fig. 62, where the terms are represented by cr Then the terms n
dots and the nth square just contains the terms in in (A) can be arranged to form the infinite series
u i vi + ( u i
2
.
+ W 2v 2 + uflj + (ttjVa + U 2v3 + u3Vz + U 3 v2 + u^) +
where the nth term
is
the
sum
of the terms of (A)
. . .
,
......
(B)
between the nth and
the (n - l)th squares.
The sum
to n terms of this series
and
is cr n ,
its
sum
to infinity
is
st.
Again, the terms of (A) can be arranged to form the infinite series
U V l + (M l Vt + U 2V 1
where the nth term
the
is
1 )
+ (tijVs + Wg^ + tVi)*...
sum
of the terms
(n-l)th diagonals, drawn as in Fig. 63. If the brackets are removed in (B) and differ
only in the
convergent and series
Ex.
The
(Chap. 1.
sum
and
= l +S +
+
5,
series equivalent to
~n
where
is st
6
XVI,
// E(x)
arrangement
its
= r- + \!L
it
we have two
The
series
first series is
which
absolutely
this is therefore also true for the second
;
19).
-
+ ...
,
prove
that,
for all value* of x
and
y,
E(x) and E(y) are absolutely convergent, therefore ~n~i
l^Lil II
In Arte. 23 and 24, the series dn
............... (C)
between the nth and the
(C),
of terms.
,
= uvn -f
i/
!
n ii2
(C) will
l?_
!
be denoted by
w s l -2 + 2 tv_i -^
should be noted that, in every term of d n the ,
Zdn where
+ w w Vi
,
;
sum of the suffixes is n + 1.
MERTENS' THEOREM
337
23. // Su n and Ev n are convergent with sums s and t respectively and one of these series, say Zu n is absolutely convergent, then Zd n is convergent ,
and
sum
its
Let
sn ,
is
tn ,
.st.
(Mertens.)
Dn
Considering the array (A) in Art. 22, 2
wl = v n
w2 -
,
3
l
t;
2
+ vn
w-1
4
wn -i = ^2 +
lastly
?;
+ ...+u n w n _ l
............... (B)
,
w3 = vn _ 2 + t;^ + vn
,
wm - vn _ m+1 + v n _m+2 +
and generally
3
,
,
be seen that
it will
-Dn = u w +u w +u w
sn tn
where
u n 2vn Edn respectively.
be the sums to n terms of
.
+ vn
;
+ vn-
+
s
. .
,
The series Zvn is convergent; and hence (i) we can v r +v r+l +v r+2 + ... number p such that v n |< for is an arbitrarily small number therefore and positive -f
|
find a positive
r>/x,
where
c
;
l^l'
l^il*
l
w'm|j
the sequence number k such that
is
(t n )
(ii)
t |
n
are
\
all
bounded
hence we can find a positive
;
for every
/?,
and consequently
We
therefore consider separately the sum of the first series in (B), and the sum of the remaining terms, where m
p=u
Let
chosen.*
^ + u#o + 2
+ u m+l w m
. . .
...(C)
m is
terms of the
to be properly
,
Q^W then
with the condition 2?|t/ n
|,
then
if $'
is
the
sum
of the convergent series
\P\
Q
Also
and
Hence,
(C).
since
27 1
that |
vm f 2
un H 1
is \
1
convergent,
u m +% +
. . .
|
-f |
nn
and then |
Q
we can
<,
|
find a positive
provided that
number
m
4-
pf such
2>//,
. .
.(D)
The conditions (C) and (D) are satisfied if n>2m and m> either of the - 2. Hence, as ?/?, and p 1 and consequently w, tend to infinity,
two,
/u,'
\Q\->0
|^|->0,
and
sn tn
-Dn ->0,
which proves the theorem. *
This artifice
is
often useful.
THEOREM
ABKIAS
338
24. // Eu n and Zv n are convergent with sums if
Sd n Let
it
sn,
convergent, then t
Dn
n,
sum
its
is st.
and
s
t
respectively,
and
(Abel.)
he the sums to n terms of
Zu ny ZV n Ed n ,
respectively,
then
+ D 2 + D3 +
Hence
D
Now
and /->/,
s n ~->s
i
/r Also,
if
D
the
is
sum
+ D n - ,Vi + n-Va ^ *n therefore by Ch. XV, 9, (4), . . .
s n t,1
+$ /<^1.L2 -f
of
ZWW which
-
...
is
,
convergent,
by Ch. XV,
9, (3),
1
D=
Therefore l/^. 1.
//
.s/.
-l<.r^l,
-prove,
(log
(
dn
that 1
+ x) 2 ^ rfrr2 - ^.r3 -f c?3 r4 -
=
"-, '/*
Under
4-
1
(
is
. .
+
:
-f
-
.,
^
+
...
4--
,
).
W- /
suitable conditions, the rule for the multiplication of series gives
easily
applies,
1
V
// -1<#<1, the series for log ( 1 + question holds by Art. 22. - + ~ + ~ 4 2 "a // x 1, the series 1 it
.
J
shown that dn
and the statement
is
a;)
is
absolutely convergent and the result in
and so is d l - d% + d3 is convergent for , and dn ->0 by Ch. XV, 9, (3J. Hence Art. 24
^
. . .
;
true in this case also.
EXERCISE XXXIII Prove that the .
t
!
1
+
1 H-
1.3 --
series
+
:
1.3.5 -
.
r T-^-f...
^
is
,.
divergent.
-...
la + 1.3
2
6
2T4
a(o + '
l)
is
convergent
1.3.5 a(oH-l)(o+2)
if
6-l>a>0,
its
sum
-
MULTIPLICATION OF SERIES un
rr If
A 4.
=
-
nP +
a - a' > 1 and divergent
series 27#n cos
The
5.
.
a - a' <
if
+
+k
. .
M '-*
,,
,
,
v~ z
.-
2/w n
is
if
convergent &
1.
Exn sin n0
n# and
show that
=-> ' >
339
are absolutely convergent
x
if |
\
< 1.
4
2n
6.
The
series
7.
The
series 27 - cos
a 2
and
cos nO
diverges
The
8.
sin
n6 are convergent excepting that the
71
when
an even multiple of
6 is
27 - cos nS
series
are absolutely convergent.
n2
nO and 27 -
71
first series
~ sin nO
27
and 27
-
IT.
n6 are not absolutely convergent
sin
71
71
excepting the last series, [Use the following
when
a multiple of TT.
is
:
|
cos
n6
|
>
cos 2 nB&%(l+ cos 2n0)
9. If u l9 u, u 3 , ... as a limit, the series
is
is
and
|
sin
nB
\
^
sin 2
^
nO
r
-J
(
1
- cos
2n0).]
a decreasing sequence of positive terms tending to zero
convergent.
[Deaote the *>i9
v z> ^3 ,
...
is
series by vl - v% -f v3 Using Ch. XV, 9, a decreasing sequence tending to zero as a limit.]
convergence of
10. Discuss the
If.
Show
11.
that, if
Q
Zun
where
1
1
show that
(3),
1
the rule for multiplication
fails
to
obtain a
convergent series for
/i__L_j__ [Show that the
series is 1
- d^
-f
d2 -
. . .
L_j_
where 1
n
12. If
/ an = ( 1
-f
^\ -
-l=-2
is
13. If
that
i)v
-~ e
x
J
absolutely convergent for
27l/n
_ r+
r p( n
r(n-r+l)^%(n + l ) 2 .]
and
[an
r=I
show that the
,
all
series
whose nth term
is
an -
1
is
values of x.
2 2 -t-r~-...; therefore n .|an -l|-*-|2;| as n->oo.
Also
convergent.]
+ FfrlKX^l+x *Y L
yJP(a,j8 > y,*)-(y-a)JP(,j8-f
Under what circumstances
is
1
1,
-^-y(V + y+
this true
1,
**^ 1
and \x\
)
*)=a(l -*)F(a-f
(i)
when z = l, and
1
9
p + l,y +
(ii)
l 9 x).
when a= -
1 ?
CHAPTER XXI BINOMIAL AND MULTINOMIAL THEOREMS
A
1.
the
sum
General Statement.
The Binomial theorem
asserts that
of the series
n(n-l) _J_^_J x 2+_
when convergent,
is
L\(w-r - +Jx
n(n-l)
+__V----
one of the values of
...
(1
-f
1)
n x)
.
^ r
+
mtm
(Of course, in a com-
must be specified.) and only if, n is a positive integer
plete statement, this value
The the
series
sum
terminates
is (1
+x)
if,
;
in which case
n .
This leads to the following theorem, which enables us to deal with the case in which
is
rational number.
any
Vandermonde's Theorem.
2.
used
n
;
we
The following notation
x r = x(x-l)(x-2) where x
is
often
is
write
any number whatever.
...
(x-r +
1),
Vapdermonde's theorem
asserts that
if m and n are any numbers whatever, then
Proof.
First suppose that
m and
n are
positive integers
1
;
then
+r n *+ro*+r Z
:
1
x r in the product of these series the coefficient of x r in (1 -f x) m + n that is to say,
The
coefficient of
is
therefore equal to
,
r
m
r ._
^r-1^1
^r-2^2
.^r
Multiplying by r, we obtain equation (A). This equation holds for all positive integral values of since each side
is
a polynomial in m, n
it is
true for
all
'
m
and n
;
values of m, n.
and
BINOMIAL THEOREM Case of a Rational Index. !, the sum of the series
3.
n(n-l) is the real positive
We
value of
// n
nln-l)
2
(1 -f x)
is
...
341 rational
any
(n-r + 1) '
number and
r
n .
give what is essentially Euler's proof, although in the we have to assume a property of infinite series which is not proved
The series and let
Eiderjs proof.
sum by
its
last step
first
f(n),
is
absolutely convergent
M fM n r ~n(n
1 \
Art
O\ &)
///.
i)\n
...
f
M
A*
r +
\n
I
x
if |
\
<1
till
;
later.
demote
1 \ i j,
so that
n2 For
values of
all
m
and
wr
2
r
n, the series
absolutely convergent, therefore
corresponding to f(m), f(n) are the rule for the multiplication of by
series,
f(m) f(n) = .
dr =
where
d vx
1-1-
-I-
dzx2 -f
. . .
-f-
A^x
r
-f
. . .
,
^+ -^l + ,-^A + - +^-
Hence, by Vandermonde's theorem,
and
/(/).
/"(/A)
/(m)./(n)=/(m + n) .............................. (A) ./(p) =/(m 4- n) .f(p) =/(m + n +p),
so that
Hence
=
/(m) ./(n)
a similar result holding for any number of factors. Positive index.
If
n=p/q where
jo,
are positive integers,
by the
preceding,
Hence
/(~j
by
(1 -fx)
,
we
With regard
is
a
q-th root of (I
+ x)
......................... (B)
the real positive gth root of (l+x) p ? " /p\ = (1 +x)* that is f(n) conclude that
Thus, x being ?
*
real, if
/(
to sign,
it will
a continuous function of x in
)
,
is
denoted (1 -f x)
be shown in another volume that f(n) the interval
-
n. is
BINOMIAL THEOREM
342
Now
f(n) does not vanish for any value of x in the interval, and consequently its sign is the same throughout the interval.
But when
cc
= 0,/(n) = l,
therefore the sign
to the real positive value of (1
-f
x)
is
positive
is
equal
.
Let n be a positive rational.
Negative index.
and/(n)
n
Putting
m~ -n
in
equation (B), ............................. (0)
l; therefore
/(
- n) =
,
= (1 +
NOTE, (i) Equations (A)-(C) hold when x is complex, provided that x |< 1. Hence x is complex, \'x\
if
.
a positive integer, /( -n)~ (1 +x)~ n x |< 1, it can be shown that f(n) is continuous for all n x are real and and (ii) // values of n. Hence if (l+x) n is defined as in Ch. XIII, 7, when n is irrational, we have (1 -Hx)n f(n). is
if n.
particular,
.
\
Second Proof for the Case of a Real Index.
4.
The following
proof depends on very elementary considerations and, although it is unsuitable for reproduction in toto, the reader will find it a very useful exercise to work through the details. We shall prove that if n and x are real
and \
x |<1, then
n(n-l)' v
-
n(n-l)
'
2
...
lo
T*
(1
+ x) n
Proof.
Let
where
Also
series.
has t
its real
...
,
positive value.
be the rth term and
nt r
-
(w-r + 1)
...
s nt r
sum
the
to r terms of the
let
Q
First suppose that 1/n,
Also
~j^
If for
= ny n ^
any assigned values
of invariable sign for sign as ny n _ lt r-1
r
according as
^
and y n
of
n and
0
,
r
n^O ................ (A)
when x =
=
................ (B)
can be shown that yn -\, r-i i Q (B), y nt r must have the same
r it
then by
.
Repeating this reasoning when n-\, n-2,
and observing
that,
by
(A),
y n - r +i
t
i
^e
^ as
...
are substituted for
same sign as n-r-f 1,
n it
follows that the three expressions
yn
,
r
,
n(n-l)...(n-r-f-2)2/n ^ r4 1>1
have the same
.
sign.
Therefore y n
,
After a certain stage, say for r
r
,
and
n(w-l)
has the same sign as
> some
fixed
(n-r + 1)
...
number
t
nt
k,
r+v the signs of
ALTERNATIVE PROOF ^n,
an(i consequently those of y nt
r+i>
Therefore (l+x) n lies
increases.
But
are alternately
r
between
since the series is convergent, s n> s W| r-> ( 1
Therefore
which proves Next
let
0
/i
(l~x)
Here so that
+ x)
n
343
and
s Wjt
.
r -> oo
as
* Wff+1
s n> r~^0 as
y^
4-
and - as for r>&. ,
r->oo
r
.
,
this case of the theorem.
x<0.
Changing the sign of
x, it
-
then
-
has to be shown that if
(n-r-f 1) r -^ = il~nx + n(n-l) x 29 ---+ (- 1 _ n(n-l) -x + ... ^T-* -~2~~ ^^ = (1 ~^) n ~n,n n y njl = (l -cc) -1^0 -according as wgO; ............... (C) x
)
r
also
-
,
-^-
= -^-i,*.-i and
y w ,r =
...
^=
when
.
............. (D)
Reasoning as in the previous case and observing that by (C), yn _r+l> and n -r ~h 1 have opposite signs, it follows that the three expressions
r
and
(~l) n(w-l)
have the same
...
(n~r-f
1)
Hence, as before,
sign.
yn
has the same sign as
r
^
t
r,
t
n>
r+l is invariable,
and
:
n
For every
r+1 ..................... (E)
nt
After a certain stage, say for r>k, the sign of
we proceed thus
t
n
r
,
,
r
-
-
n r+1
.
,
we have
+^- r Therefore
(l-x)* n>r = * n+1 '
Hence
(1
Hence by
(E), 2 Wt
r
,
r
-x
- x) z w
by
,
r
and
(1
-x)y n
,
r
= y n + lff + 2* n>r ....... (P)
= (1 - a?)y n r - tn> r+1 ^ 2/n+l, r + x^n, r ~ ^n, r-fl ,
has the same sign as
,
so that
ntr
t
n+lj r+2 .
Now
r+2
(E) z n>T
has the same sign as
(w-
+ l)y nif ................ (H)
PARTICULAR INSTANCES
344
n>
If
(i)
fore
lies
yn>r
follows that y nt
SL8
cession
n-h it
Hence
r-^
since
m> -1,
and
,
-x)
z nt
t
-x)
n ,
r
.(1 -x)""
r +i
have opposite 1
But
.
We shall prove that
if
n
where
by
case
from
(F) that
yn + m
(i),
,
a negative
,
21)
and
(compare equation
J
n>r ->0
as
r->o>
.
yn> r->0.
r->0;
Hence
^nd
to zero.
complex and
in suc-
z \
\
<1,
2
Using the same notation as before, we see that the
XX,
,
in all cases.
integer, z
n(n-l)
(Chap.
r+1 ->0 as r->oo
u positive integer. If
is
//*
a ll
is
,
.
r yn +m-2,r> 2/n,r and the theorem is proved ,
^n
There-
signs.
n
ntr->Q it follows
therefore,
follows that y n + m _ 1 s nt r->(l
*w
s Wjr ->(l
and
,.
-M-l
//)/<-!, then
Now
and
between
,
yn+i,r~*Q
5.
1 it
yn r->0 and
therefore (ii)
-
Also
(1
series is
+z)ynj r =
convergent
\
(F)).
Hence as before, if yn+l} r->0, then y n> r~>0. Now when n= -1,
Hence
in succession y n
,
r
~>0
for n==
-2, -3, -4,
...
.
So that
nir ->(l+z)n
when w is a negative integer. The theorem is obviously true when n is which n is a fraction is considered later.
a positive integer.
The case
in
6. The reader should be familiar with the following particular instances is assumed that x 1
in each case it
|
and
\
<
:
in a similar
way 2 (l-z)- = l+2z 3.4.5
3 4 -
n(n-f-l)
1.3
.
1.3.5
3.4.5.6
;
NUMERICALLY GREATEST TERM
345
2
Ex. 1. Expand l/(3 + 2a:) to three terms (i) in ascending powers of x; scending powers of x ; (iii) say for what values of x each expansion is valid.
(ii)
in de-
27
The
(iii)
first
The second
Ex.
To
see
if
valid
if |
-|ar
|
<1, that
is if |
x |
valid
if
this
that
1,
1
+ +
is if
~
can be done by the Binomial theorem, take some particular term, say
the fourth, and proceed thus
5.8.11 ._..
:
o f.f.^s o 3
/5.\3
__.
J !-! J
1 ^
O
.
_______
* \4 5.
".
'
sum
is
equal to 2n(n
of the series
;
+ l)(n + 2)(n + 3)x*/\
4,
where
treating every term as above,
Numerically Greatest + x) n where |x|<1.
7. (1
1.2.3.4
2.3.4
8.12.16 which
.
3 is
Sumtheterie*
2.
is
expansion
w~|
and
a;
= -|.
Let
5
be the
we have
Term
in
the
Expansion
of
Let u r be the rth term, and let accents indicate numerical values so that x' = x and u r = t/ r t/ n is positive, f
|
|
|
|
;
wr Let
ifc
be the positive integer such that
therefore ^4-1, greatest term If r^.k, .,
,
that
.
is
r
is
% +2j among
-
is
the
a decreasing sequence, k 4- 1 terms.
and the numerically
first
then wr _}_iw/
,. ^ -according & as r dr. -* 1
i
according as
(n-f 1
r)x' .-
r,
NUMERICALLY GREATEST TERM
346
-
~
If
(i)
r~ ~i+f, where
~\~
proper fraction, the
I
tth term
Next,
is let
~r~
(i
+ l)th term
= i, where
-
//
(ii)
X
/
is
a positive
numerically the greatest term.
is
a positive integer, then, numerically, the
t is
+ l)th, and each is greater than any other term. and equal to - m, then
equal to the
n
a positive integer and
i is
x
j
1
(i
be negative
l '
wr+1 ~w/ according
therefore
,,,.,. according 8 as
that
r
is
It follows that
(ii)
If
= i +/, where
i is
a positive integer or zero and
(m-l)x' = "*~~
X
to the (i-f l)th,
We
.
first
^
JL
7 '
r,
the
(i)
//
positive proper fraction, the (iii)
<- (m-l)x' ~ -:--
=
m^l,
:
If
(m + r-l)x'
as
f
(i
where
i,
and each
term
+ l)ih term
i is
is
is
the greatest,
is
greater than
any
other.
have thus proved the following important property n // |x|
series:
is
a
the greatest.
a positive integer, the iih term
of these
/
is
equal
.
of a binomial either decrease
numerically from the beginning of the series or else they increase numerically up to a greatest term (or to two equal terms) and then decrease numerically. ""
Ex,
when
Which
1.
numerically greatest term (or terms) in the expansion of (I
is the
+ a;)"2
x%?
Let ur denote the rth term
;
ur or
if
r
^
14.
5
r
First consider the terms for which
4r>54
54 - 4r
%^ -r + l 4
u r+i
uril jur
For such values of M r+i
(A)
5r is
negative.
This will be the case
if
r,
ur
Hence the
greatest term or terms must be included among the
first
1
4 terms of the
expansion. If r
^
13,
u r+l /u r
is
|
positive,
u r+1
rr 1
and from (A) \u r
that
Hence, the sixth term other term.
is
|
is
it
follows that
according as 54 4r according as r
~
^
5r,
6.
equal to the seventh, and each of these
is
greater than
any
CALCULATION OF ROOTS The Binomial
Approximate Values.
8.
tain approximate values, as follows Ex.
347
series
can be used to ob-
:
Calculate */2 to six places of decimals. perfect cubes are 1, 8, 27, 64, 125, .... Search for have roughly twice the other. 1.
The
two of these such that one
We
is
.
*
3* 5 s 1
L
1.2.3
1.2
59
+ ""/
L
(
Denoting
this series
by u t + u 2 u 3 + u t - u 5 +
. . .
'
we have
,
=1*25,
^ -
*.=
-0-01,
--
= 0-000
07
001
8
=
08
=,0-000 Jj
~= 4
0-000
OOP
02
0-000
080
02*
'
'2
Therefore,
07
001
1-260
= 1-260 = 1-259
001
07-0-000
921
05 nearly ......................................... (B)
02 nearly
080
possible error in this result, let s n be the sum to n terms of the series R and after n terms. The error arises from two causes the remainder (A) n The error caused (i) by taking s 5 as the value of '^2 is R 5 Now, in the series (A), the terms beginning with the second are alternately positive and negative, also each is therefore J? 5 is positive and less than w e numerically less than the preceding
To estimate the
:
.
.
;
u= -^5
Now
.'.
The
.~-u 5 = ~~u 5 < 0-000 000 000 4; 3
the error
5
5*
is less
than 0-000 000 000
4.
from the errors in the calculated (ii) values of w 4 and u 6 These cannot exceed 0-000 000 005, hence the error in the calculated value of s 5 cannot exceed 0-000 000 01. Adding the results of (i) and (ii), we see that the error in the result (B) is numerically less than 0-000 000 010 4. Thus, to six places of decimals, error in the calculated value of S B arises .
/2
= 1-259921. i
Ex.
2.
p + qx -,
,
We
and
If x
where p,
is
small and
q,
p' 9
q'
n>l,
obtain
an approximation
to
(l+x)
n in the form
are independent of x.
have
(B) B.C.A.
NEWTON'S APPROXIMATION
348 Choose a and
b so that 2
and
a-b-l/n
-&)&-=(&- l)/2w and a = (n + l)/2/i.
(a
b-^(n-l)l2n Substituting these values for a and 6 /.
n
-l)x
we have
in (B),
2
;
2
\n J both convergent, and since they
\nj
\
(C)' v
J
If x is small, the series (A) and (C) are as far as the terms involving x 2 it follows that
are identical
,
.
+x
(1
9.
Theorem.
n
)
^
_
----
-
,
- rf- nearly.
}
n>l and 0
// ^
~ n2 -
1
proximate value of
By it
(1 -f
n
x)
w/A an
,
1
error in defect less than -ys~3
^3
-
multiplying each side of equation (A) in the last example by
2n + (n- l)x,
_
can be shown that
__
i
-x-x .x)
-
X
2
on"5
Denoting the
< 1
--
-
series in brackets
MI
-h
u r-
we have
w r _!
=7
OX
_
.
4w .on r
+ (-
rn
l) '^r
r
1
-
--,
-
(r-f-2)n
-
-
r-1
+
. . .
i-
.
J
,
x.
(rn-l)r<(r-f2)(r-l)?t provided that r>2, for _ 1 r _ ( r + 2) (r - 1) n - - r ~ (r - 2) n<0 (
m
...
by
-^ + 2 ^3
,
Now
LX
4n
^
)
and since 0
;
.
and, since
n> 1
and x
is
Ex.
2
-l)rryi2n
v2
than
3 ,
that 63.5/504 is
an approximation
\2
to
with an error in defect less than
0-0000005.
Here
is less
;
result follows.
Show
1.
than unity
positive, the last expression
(n
whence the
less
*
1
(
+
J
j
henco, an approximation
5750-4-12 4
750
i
6
635 r
'
504'
--
... r * an error in defect with i
4
12.27
-
liM 3
10 8
10 7
is
KXAMPLES OF APPROXIMATION
349
EXEECISE XXXIV If a:
1.
is
positive,
which
is
the
first
negative term ih the expansion of I
x
2. If
is
(
1 -f
x)
*~ ?
9
terms in the expansion of (1 -x)~* beginning with a Prove this, and find which is the first of this series
positive, the
,
certain term, are all positive, of positive terms.
Which 3.
(1+x)*' when ar^J
5.
(l~x)~
7.
Between what numbers must x _
(I
the numerically greatest term or terms in the expansions of 35 _3
is
-x)
(l-x)
when x = ]l
6.
(l+x)~* when *---{??
2,
the third term
x> 0,
9.
If
0<#<1, show
(ii)
lie
order that, in the expansion of
in
3
If
than
when *-:&?
4.
8.
and
defect
2
?
less
which
than
~x~ \ ~\x
\/l
may the
is
be the greatest
first
that
(i)
?
negative term in the expansion of 1
Vl 4-x
1
+lx-{x-
nearly, the error being in
^x'^. -
.r-
nearly
*
the error being in excess and numerically less
-
x
10. If
is
large
and
positive, s/a;
3 -t
show that ~
\~x-\3
^2
the error being in defect arid less than 5/81 a11.
IfO<
1/(1 -ir)
2 ,
and show that
/? n
is
x5
9
nearly,
8 .
the remainder after n terms in the expansion of
0
0<#<1, and Rn
13. If
1/(1
-x)
3 ,
14. If
1/(1
-x)
(ii) 1
less
3
is
the remainder after n terms in the expansion of
show that
0<#<0-01,
show that
with an error
in defect less
-3#
f fix* is
(i)
l+3o; + 6x 2 is an approximate value of than 0-000011.
an approximate value of
1/(1 +x)*
with an error in excess
than 000001.
15. If
is
small,
show
that, neglecting cubes
and higher powers of e,
APPROXIMATION TO ROOTS
350
16. // we have tables giving the square roots and cube roots of numbers to f signicant figures, show that the roots can be found by division only, to 2f significant figures as follows.
If *Jx~a
to
f significant figures,
then *Jx
f significant figures,
then
-
2 (\a 1
If
$x-a
Show also
Given
Given 4/10
19.
Show
x
is
20. If
(i)
(iii)
1
\ to )
2f significant figures.
/
that
x
is
by
= 141 421356237
division, that
(approx.).
2-1544 (approx.), show, by division, that $ 10 = 2- 154434690 (approx.). if
(i)
x
is
small,
v/(z
large,
2
*J(x*
+ 4) - *J(x* + 1) ~ 1 - i* 2 + 3 /
s ~ +4) -V(x + 1)= 2
1
~
(
5
7 4 6 4~z
~
nearly
;
21 \
^ + ^)
nearly-
show that approximately
small,
V(4 - 3*)
(
/x
$x = o-(\a~ + 2a
1-41421 (approx.), show,
v/2
18.
if
2f significant figures.
use Ex. 15.]
>/2
(ii)
to
/
that in each case the error is in excess.
*Jx-, and
[Put a 17.
to
/x - + a\ )
=1
.
4-
# (8 + 3*) -r (9 - |z)
+ 4a; -f 10x 2
* )
= 1 -f 2x + 3z
--,
7
(1
-
-iVr)
;
2
(iv)
(v)
(
I -fa:
+ * + *)* = l+ix-h A* 8 + ^5**.
In each of the examples 21-26, (i) prove the given equality ; (ii) by expanding the right-hand side to four terms, deduce an approximate value of the given surd ; (iii) find an upper limit to the numerical value of the error. 21.
^1001 = 10(1+0-001)3.
22.
V2 =-J(l -0-02) "i
-f(l+ 0-024) "i
24.
#5 =(1+0-08)*.
26.
#4=f(l
23. 4/4 25.
*/3=i(l-0-01696)*.
27. If
a
is
nearly equal to
b,
show that ~
^ and that
if
a>6,
the error
is
in defect
and
is less
3 tt -! /a-6\ "12^ V~T/
than
2
[Assume the result of Art. 28. 10*. 3/10*.
Show
that
/543
9,
2709
^^ = 2706 ^TTO ^ 54 5
J
'
C
'
and put x = (a - b)/b.] '
nearly, the error being in defect
and
less
than
SUMMATION OF SERIES 29. If a -b=^h,
whore h
a
I
+
ft" 1
2 (iii)
h
6~a
2 ft~8 1
A2
4 ft" 8
8
1
+
small,
is
I h'1
7i
+ "
Hence show that
4
a
show that 5
I 3
T6
ft~
1
&3
16
ft"
h*
~T28 b** 1
&*
"
ft"
351
3
+
'*
""
"
""32 6*
nearly equal to 6, then I a +b = + nearly b a+b 4~'b'-' if
is
a
Va
'
the error being approximately equal to (a-ft) /128ft 4
Application of the Binomial
10.
4 .
Theorem
Sum-
to the
mation of Series. (1) Many series may
be summed by expanding some function of x as a convergent power series in two different ways and equating coefficients. Ex.
1.*
If
prove that
It will be easily seen that
w2 -|?.(g + l),
wl
therefore
and
+i/ 2 o;
+ w3a;2 -f-... /
-f
wn _!a;n<" 2 + ... =
-
%--f v
Hence
if
8
is
the
sum
of the series in question, of n 2 in (1
[g=coefft.
=coefft. of
(2)
|g (1
Theorem.
where k
// f(x)
-a
-f
xn ~* in
-*)-(-).
(1
(1 -a:)-<
a^ + a2x2 -f a3 x2 +
. . .
+ a nxn +
positive number, then a + aj-f a 2 the coefficient of x n in the expansion of f(x)/(l -x).
For
is
any
~
-f ...
. . .
+a n
for is
\
x
equal
to
convergent for sufficiently small values of
x.
fix)
z=
x
and the
-
x
series are absolutely * This equality
is
required in the Theory of Probabilities.
SUMMATION OF SERIES
352
In what follows, we suppose that x <1 except when n tive integer, so that all the series which occur are convergent. (3)
|
We
shall write
,,
w
then,
(1
+ x) n = c
/i ,n- * , 1 H- ;r) (
4-
4-
c^x
c 2x 2
a posi-
is
\
c3x
4-
3
-h
-f
. . .
c rx r
-f
. . .
n(n-l)^ x ~n(n-l)(w~2) --~~----- x2a + = w -f -^r 4-
l
;
. . .
r
!
= c x 4- 2c 2x
-f
3c 3 ,r 2
-f
+ rc^ r
. . .
'1
-f
. . .
,
and, by a succession of similar steps,*
=l
2
and so
.
+2
2c 2
3r3 :r
.
+
... -f
r(r
where u r
is
-
on, as far as required.
Hence we can sum
*_ Q u r c r x
the series
r
a polynomial in
For, dividing u r and successive quotients by r, r-1, we can find constants A Q A L A 2 A 3 ... such that
order
,
Z^^u
then,
Ex.
c
r r
Find
2.
x r ^At(l+jc) n +
Z*^
(r
Denoting the sum by
+1 *V
3 )
c
r
xr
,
A
v
S=-(l+x)
= (1 + (4)
n
-f
n -8 a;)
nx(l +x)
n ~l
+ A 2 n(n- })x 2 (l +x) n
'2
.
and proceeding as above, we find that +7/' + 6r(r - 1) +r(r - l)(r -2)
1nx(l +ar) }l -f (7
in this
...
,
,
3 (r + l) -a .'.
r-2,
r.
n -1
;
+ 6n(w-
l)x~(l +x)
n
~2
+ n(n-l)(n
4 3) a: -f (6/& 2 + 8w + 3) x 2 4 (w + 1)V}.
Again, by expanding
+x)
(1
1 (Ijfx^^^ " " "V" n+1 ?i + l ^
i
w+1
^n^
we can show
,
GI '
*^
r
2
2
that
i. ..i" _c, -i~X
l "r~
r-fl
and, by similar steps, "
(n
+
l)(n
+ 2)
(M
+
l)(
+ 2)'n +
n 3 (l+x) +
"
^,2
.
-,
Jv
~~~^,
i
(r 1
^.3
__
j
1.2.3 on.
i
\
1
.
and so
i
1.2
l
+
l)(r
+ 2)'
_
_ X+
?r
'
2(n + ,v.r+3
l) ,
"
These results can also be obtained by integration,
* Tlirse results
can also bo obtained by differentiating with regard to
process has not been justified in ease of an infinite series.
x,
although so far the
SPECIAL TYPES OF SERIES Hence, if u r
is
a polynomial in
we can sum
r,
353
the series
r r
and
For, dividing u r
we
in this order,
successive quotients by r -f k, r can find an equation of the form
A
ur r
where vr
is
^4^, -4fc_!,
...
Al
l
~~~
"*.
":;
i
r
r
+1
A% \ ^rr (r + l)(r + 2) '
..
/
The value
z r~~
'"
1,
. . .
r
+ 2,
sum
-f
1
Ak
.
(r +
l)(r + 2)
...
and
of the series is
-
r+1
r
r+2
+
x
Zv r c r x r
of
i~
+k-
are the remainders in the successive divisions
the final quotient, and then the
zvx rr +
/
'
is
found by
(3),
and the other sums are given by
the above equations.
Further, we can
sum
the series
k are unequal positive integers of which k is the greatest. can be reduced to the preceding by multiplying numerator and denominator of the fraction by such factors as will convert the where
a, b,
For
...
this
denominator into
Find
Ex.
2.
We
have
hence,
S = 2^1 -
r
(r -f 1 ) (r
-
+ 2) (r + 3)
. . .
(r 4- k) .
f -s
-
+3
*"-**
(MMT,
(7TIj(iFT2MF?S,
z2
n -Hi
n (i+x) +*-i
v.
d+*) n+l
i
i
a;
n-fl
2
r
i
*
x
MULTINOMIAL THEOREM
354 yiOO
^r^O 7+3
Wo
tr
""
'
~ =r -
have r
and
4-
3
H
3
r
+3
27(r
where
f,
If 6, c,
...
i are any
m
f(x) (i)
2.
numbers, n
= (1
&#
4-
and
is real,
4 ex 2 4-
fcc
... 4-
m
n )
,
/(x) can be expanded in the form 1
4 ux + u 2x* 4-
where the series terminates if n
is
urx r 4-
... 4-
a positive
. . .
The
u r of x r
coefficient
n(n -
~
1)
(
is the
sum
and in
other cases is
x.
of all terms of the form
+ 1)
H--2) .^(a
,
integer,
absolutely convergent for sufficiently small values of (ii)
the same as in Ex.
Multinomial Theorem for any Real Index.
11.
then
is
, /B
y
c
""
^
'
B'Ty/'-E where
/?,
y
,
. . .
K:
Aave any positive integral or zero values such that
and the corresponding values of a are given by a-f/3-fy-f-...-f/c = w then For let y = 6x + ex 2 -f 4- fcr by the Binomial theorem . . .
,
* provided that |y|
By
substituting bx
+ cx 2 +
...
for
(1),
we can
find a positive
y in (A) and expanding the terms, we
obtain an expansion of /(x) in powers of x which is convergent if x\
are
made
where
This will be the case
positive.
x', b', c', ...
stand for \x\, |fej, this is the case if
if
c |
number a such that
|
|,
+ k'x' m
b'x' ...
;
. . .
x
This concludes the proof of the statement (i). r Again, s being any positive integer, by Ch. Ill, 16, the coefficient of x 2 s m in is the sum of all terms of the form (bx + ex + ... 4 kx )
EXPANSIONS where
. . .
y,
j8,
ic
have any positive integral or zero values such that jS
and
+ 2y + ...+w/c = r, K
y-K..+
/?+
from (A) that the coefficient the sum of all terms of the form It follows
is
n(n-l)...(n-* + T-7f
where the values of of s
by
Ex.
1.
Here
j8,
y,
JPt'nrf f/ie coefficient
n~
--J,
j8,
y, 8
x r in the expansion of f(x)
of
l)
fl
c/
A;
. . .
,
j
of X* in the expansion of (\
r
^4, and the
( v
~i)( -i--l)( v -i ~2) I _JLI *_ -_ _z
coefficient is the ...
(a
sum 1
.
2#(
3
8
terms of the form
- 4) v ( - 2) 6 ,
have positive integral or zero values such that
and the corresponding values of a are given by
~~
+ 2x - 4# 2 -
of all
+ 1)
i
-
where
(B)
S ............................... (C)
c; i
..............................
are given by (B) and the corresponding values = n-s. (ii) follows on writing a
...
The statement
(C).
355
/J
+ 2y + 33 -4,
a+/?+y + 8=~-|-.
The possible values of a, /S, y, 8 are shown in the margin. The term corresponding to the first set of values is
a
y
ft
2>
>
-3i,
2,
1,
-4,
4,
0,
-2j,
0,
2,
S
'
j
Similarly it will be found that the other terms are ^, 6, so that the required coefficient is
15,
-3 + 15+4f + 6=22-|.
EXERCISE XXXV 1.
By
n considering the expansion of (1 -a:)~
A ~r H H
n(n +
n(n +
1)
,
show that
l
r
r^,:
2
L 2.
Show
that, if
z
= ~^, and x+y
2 |
|
<1, then n-f2);
3. If n, r are positive integers (including zero), show that n f| - 1 2 . t he expansion of (1 2r)2 x) /(l -a;) is (n
n+r-i
^
+
+
n - 2 [Expand {2 - ( 1 x)} l ( 1 a;) .] 4. If w is a positive integer and (1
show that
+ c*i + a, +
+x) n .
+ an_! = $n (n 4- 2) (n + 7) 2n
-4 .
the coefficient of
COMPARISON OF COEFFICIENTS
356 5. If
n
a positive integer, show that
is
4
3.4n(n-l)
= coefficient of z n 6.
If n
Prove that
Show
n
if
!?L 8.
{1
p
in
{2
-
-
(1
n x)} l(l
(n
+ l)(n + 2)
- x)*=(n z + In 4- 8)2W ~ 3
"~~ ""
1)
(
"
f
-# (2 -x)}~ 1 = (I -x)~ 2 .]
'
JL??_
that, for all values of
p 4-^ = 1 and
n(tt-l)(n~2)
+" ~
'
*
m and n
9
coefficient of
x r in the expansion of
n, r are positive integers (n>r),
put 1 q for p, and show that the hand side when expanded is
(1
+x)
n~m
.
show that
p
r
,
coefficient of q k
^coefficient of z fc in (1 +3;) n x (1 +a?)-
+0^+
value, find the
sum
2 r==Q mc x
of the series,
r
IL^J; 14. If
'
n
[Divide by
any
.
a positive integer,
is
n n
= the 9. If
5 n(n - l)(n -2)
a positive integer, show that
is
[From the expansion of 7.
.
12.
r 9
when
on the
left-
- r - fc+1 >.]
n may have
m is
,-,!-=,;
13.
ur =Ao + A lr + A s r(r-I) + ...+A h r(r-l)...(r-h + l) and ~ h~1 h 4- A l nx + A 2n(n- l)xh 2 -f + A h n(n ~I) ...(n-h + l),
(x)-A Qx
. . .
prove that
15.
Prove that
CfCfJlJI
[Consider the product 16.
Show
(1
+ Cf^Cfilf H-(7J +1 C;Si; + ... T +^ x (1 - X )~^- T \] -x)~(
that
3 2n + a
1
to 40 terms -6' JJ.
-3
2714-3*
the summation being taken for n. 9 r such that p + q + r [Consider the expansion of
all
positive integral values, including zero, of
p q y
(a 4-6
+c)
2n+3
- (6 *
-f
c
-a)
2n + 3
- (c 4-a - ^>) 2n+3 - (a 4-6 -c) 2n + 3 .]
Check the answer by putting
nl.
CHAPTER XXII RATIONAL FRACTIONS
RECURRING SERIES AND DIFFERENCE
(2),
EQUATIONS 1.
Expansion of a Rational Fraction.
fraction in x,
Any
rational proper
whose denominator does not contain # as a
factor,
may
be
written in the form
P
+a _ - - -+p
+ a 1 a; + a 2z2 + ... 2 l+piX+p^x + ...
a _
Q
mx rx
r
Various ways of expanding such a fraction in a series of the form
Zu n x n
are given below.
can be shown that the
obtained by any process is convergent for all values of x in any interval including zero, then by Ch. XX, 20, any other method of expansion yields identically the same series. If it
2.
Theorem
.
The
series
rational fraction
P/Q can
be
expanded in a convergent
series of the form
U + u l x + u 2x2 + if
and only
if
For, by the
x \
\
<
A |
method
of several terms of the
|,
where A
+u n xn + ...
...
is the root
of partial fractions,
of
,
$=
with the least modulus.
expressed as the sum a constant, r a positive
P/Q can be
form A(x-a,)~- r where ,
A
is
integer and a is any root, real or imaginary, of the equation Q Q. All of these terms can be expanded in a convergent series of ascending
powers of x if, and only the least modulus.
When
this
the various Ex.
1.
is
if,
x |
\
<
A |
|,
where A
the case, the expansion of
is
the root of
the fraction
be expanded in a series of ascending powers of x? Equating the denominator to zero, the roots are
-iV2
and t(i
and their moduli are -s/2il, >/5, ^5. Hence the expansion is possible if and only
with
P/Q can be obtained by adding
series.
For what values of x can
$=
if |
x \<*J~ -
1.
RECURRING SERIES
358
Methods of Expansion.
3.
where the Art.
1.
(1)
Suppose that
assumed to be convergent and P,
series is
are as described in
Q
Multiplying by Q,
Expanding the right-hand side and equating uQi u v u2 ... are determined by the equations
coefficients, tjie values of
,
Hence, after a certain stage, any r successive coefficients are connected by a linear relation. A series having this property is called a Recurring Series. (2)
The process
equivalent to synthetic division carried Ch. Ill, 3.) (See
just described
out as in the next example. Ex.
Show
1.
that,
The reckoning
is
for sufficiently small values of x 9
1
+ X + X 2 + X3
is
as follows.
1
-
-3
(2
4-
4)
1+1+1+1 + 4 + 12 + 16 + 16 + 32
-3-9-12-12-24 + 2+6 + 8+8 + 16 1+3 + 4 + 4+ 8 + 20 The
first
term
diagonally in
(c),
in the quotient (d)
2
(a);
(6),
+ 1-3,
is
1
;
(a) (b) (c)
(d)
2, -3, +4 3(2-3+4)^6-9 + 12;
1(2-3 + 4)^2-3+4; put
the next term in
(d)
;
- 9, + 12 6 - 3 + 1 =4, the next term in (d). diagonally in (c), (6), (a) do this example by the method in Art. 3, (1), he will see that the reckoning is essentially the same as that just described.
put
6,
;
If the reader will
(3)
The method of Partial Fractions.
Ex.2. If We have
where
Hence
Now hence
a,
j8
1/(1
-2ar+5x 2 )=w
a;
1
1
are the roots of i
+w1 + ...+Mwarn + ...
x*
find the value of un
_
= _1 /__?_ _ _J )~a-8 \1 -a* 1 -j -j8*)~a-j8
- 2x + 5 =0, so that we
may
take a = 1 + 2*,
i
l a=v/5(cos0 + isin0), 0=V5(cos0 -tsin0), where 0=tann +1 n ^ 1 =5*(n + 1 /. a ).2tsin(n + l)0; -/3
w n =i.5*(n fl )sin
.
-
(n
+ l)0, where 0=tan"-
!
2.
2,
ft
= 1 - 2i.
HOMOGENEOUS PRODUCTS
A
(4)
fraction can also
expanded by the Multinomial method and that -of Partial Fractions and equating
rational
Theorem.
359
Using this we can obtain
be
many important
coefficients,
results, as in
the following
exercise.
EXERCISE XXXVI Use synthetic division to expand
1.
w (l-a;)
v
2
'
1 (l-2*)' as far as the terms containing a; 6 Find also the .
Show
2.
that, if
p and
coefficient of xn in
each expansion.
2 q are real numbers, the condition that 1/(1 4- px + qx ) series of ascending powers of a; is as follows
can be expanded in a convergent 2
p >4#, then
(i) if
<
x
|
\
:
- Vp 2
\
(p'
*q)/2q
where p'-\p\\
|,
(ii)ifp*<4q, then \x\
x lies between a and show that the fraction l/(x -
|
|
|
|
. . .
[Let If
The
|a|
,
= (a- ]8){l/(a;- a) -!/(*- J8)}.
fraction
|a|<||<|j8|, prove that
and
H
[Hn i.e.
is
sum
of the homogeneous products of n dimensions which can show that dn +*(b-c) + b n +*(c-a)+cn +*(a-b)=-Hn (b--c)(c-a)(a--b). the coefficient of x n in the product
4. If n is the be formed with a,
6, c,
The
in the expansion of 1/(1 -ax)(l -bx)(l -ex). fractions.]
result
obtained by the
is
method of partial 5.
Prove that, for
sufficiently small values of x, 1
1
a n =p n
where
n
-(n-
""
2
l
-px + qx*~~
~
+
l)p ~*q
n
-p
n
~*q*
-
. . .
,
I
the (r+ l)th term being 6.
(
T
l)
r
C^- p
n - zr r q
.
use the identity
If a-f/?=jp, aj9=5 , f
/
to prove that
an + p" =pn - pr
n~2 j3
Li the (r + l)th term being
[Expand both
sides,
(
I)f
g
+
1
-
__ *>\ ,
p
f-
w~ 4
g
1
-
2
-
1
fix
. . .
-px + qx*
,
!j_
^~>-l)^-^-2) - (n-2r + l) yn _ ifgr
equate coefficients and use Ex.
5.]
SCALES OF RELATION
360 7.
+ p=p, 4=9,
If a
to prove that
-
a
p
the (r + l)th term being
[Expand both 8. If
%
is
sides
(
(-1?
^n
*~ r "
and use Ex.
~^
(r
(ii)
+ 1) th term
552* sin
-*V-- y
.
5.]
n 0)
-
~ (2 cos B) n ~
2
+
r
Li
the
"' (n
a positive integer, prove that
2 cos nB = (2 cos
(i)
-
use the identity
being
(2 cos
-
r
0r- -
-
the (r+ l)th term being
(
n n~r~ ^
^
2
I
n (2 cos 0)
~4 -
...
,
_
l)
(
^r~
(2 cos
1
(n
*) "' (*>-
^
~ T ~ l)(n
(
(2 CO8
"- 3
4)
y-
"'
(
-
(2 cos *)
12
~*~ 2)
)n -sr.
"~ 2r)
1)'
(2 cos
...
,
0)-->.
7"
[In Exx. 6, 7 put a 9.
(ii)
Show The
that
(i)
= 2, p = z~
l
where
z~ cos
0-fi sin
0.]
the coefficient of x m in the expansion of -~
coefficient of x* n in the
expansion of
-- -
--
-
(1
-x)(\
-x
;
2
)(l
- x4 )
1
+#
is
(n f
2 Recurring Series. (1) Let u Q + u 1 x + u 2x + ... be a series which any r + 1 successive coefficients are connected by the equation
4.
u n +PiU n -i+P2U n _ 2 +...+p r u n _ r = where r
is
a fixed number and
p l9 p 2
,
...
pr
)
is
-, ''
1 )
2 .
in
.................... (A)
are constants.
Such a
series is
called a Recurring Series of the r-ih order.
Some authors
equation (A) the Scale of Relation of the series 1 +p 1 x+p 2 x 2 + ... + p r x r others use this term to denote the polynomial We shall take it to mean either of these things. call
u l9
;
.
u r _^ are known, the subsequent coefficients can be found in succession by equation (A). Thus a recurring series of the rth order depends on 2r constants. Hence if the first 2r coefficients are given, in general the series can be continued as a recurring series of the rth order, and in one way only. Also If the coefficients
it
u
,
...
can be continued as a recurring series of the (r + l)th order in a doubly number of ways. For to do this we can give ur and ur+l any values.
infinite
2r terms belong to a recurring series of necessary that this may be so, and then the series can be continued as a recurring series of the (r - l)th order.
may happen that the first order r-1. Two conditions are It
CONVERGENCE OF RECURRING SERIES Ex.
361
Discuss the question of continuing 2un x n as a recurring series when the
1.
first
six terms are known.
un +pu n _ l + qu n _ 2 + run _ 3
Let the scale of relation be
Then
are given
q, r
p,
u3 4- pu2
4-
qu
0.
by
4- r u
w4 4- pu9 4- #1*2 4- r^j
0,
Now Sunx n
can be continued as a recurring definite values and rr^O. This will be the case
w e 4- pw 4 4- g?/ 3
0,
4-
ru 2
order
series of the third
if
0.
have
p, q, r
if
and
(A)
"3
In case either determinant then
is
zero,
suppose that the scale
%
u & +pu^ 4- gw 3 =0.
That these equations may be
one of the
u n ^-pu n _ l
W 4 4-^ 3 4-
first
+ p W2
4-
^ Wj = 0,
consistent, each of the determinants in (A)
vanish, so that two conditions are necessary that the second order. If it is
is
u2 4- p^ + qu = 0,
order, the scale being
Eun xn may
must
be a recurring series of
un +pu n ^ 1 =0, then
Ut+pUQ Q, +pu 1 = 0, ... uB +pu4 0, 2 conditions u l -u^u 2 = Q u 2 2 -u u 3 =0 w 3 2 -w 2 w 4 u2
so that the four
must be
(2)
9
9
1
u4 z
0,
-u 3 u5 =Q
satisfied.
Any
u n xn
recurring series
is
convergent for sufficiently small values
of x.
For
let
the scale be
u n +p l u n _ l +p 2 u n __ 2 +...+p r u n _ r = 0,
and let U Q Then by (A),
|,
|
|,
... |
pl
|,
p2
|
|,
be denoted by UQ ',
...
u,
p, p%',
...
(A) ....
'
the greater of the numbers Pi +p% + +p r an d 1, and un ^ the same way the greatest of the set u' ... In __ n 2 l9 UQ.
where # is
u |
..................
is
u^
where
w^
is
,
the greatest of u^_ s _ l) u'n _ 8 _ 2
Continuing thus, by multiplication
where
A
is
the greatest of Wn-i> Wn-2>
Thus |
,
...
u$.
we can show that |
...
2u n x n
\
u \
u nx n \^\gx\ n .A,
n convergent if x |
and
un
is less
is
|
|
|
than the
GENERATING FUNCTION
362
n
when convergent, of any recurring series Zu n x a rational proper fraction, which is called the generating function of The sum
(3) is
to infinity
,
the series.
Let the scale be
U n +PlU>n-l+P2U n _ 2 +...+pr Un __ r = Q, and
..................
(A)
let S
= w + w 1 x + w 2 x2 -f ...
to QO ......................... (B)
p^, p2^> p rxr and adding, we find that + p& 4- p 2x2 4 4 p rx?) = a 4- djX 4 a 2x2 4 + a r _ 1 xr - 1
Multiplying (B) by s(1
where
. . .
a
=w
al
,
=w
. . .
4r
,
......
(C)
a 2 = t/ 2 4 jo^
^WG,
the coefficients of powers of x higher than x r
~l
vanishing on account
of equation (A).
Thus
s is
It follows
A
ti;Aer6
i
equal to a rational fraction, of which the denominator
from Art. 2 that
// w n
(4)
i
a recurring
is
Q=
iA6 root of
To prove
a
of which the scale
is
(1
-x)
n of r+l
is
r,
then
Zun xn
. . .
Let ,
1
so that v n is a polynomial in
,
r steps of this kind,
where &
degree
n 2 - x) = w 4 v^ 4 1 2^ 4 ... 4 v nx 4
s (1
where vn = u n -u n _ l
|,
.
we can proceed as in Ch. VIII, 7. s = u 4 ux 4 u%x 2 -f 4 u nx n . . .
then
if\x\<\ A
wjAicA Aas $Ae Zeas^ modulus.
positive integral function of
series
this,
2Ae recurring series is convergent
is
we
. . .
,
n of degree r-1.
By
find that
independent of
x,
and therefore
s(I-x)
r+ l
~u +(k-u Q )x,
which proves the theorem, Ex.
2.
Find
the
sum l
This
by
is
to infinity 2
of
42 2*43 2a: 2 44V4..., when |*|<1.
a convergent recurring
series,
and the
scale
is
(1 -a;)
8,
= =
-3
.
l*x
-3
.
2 2 *2 - 3
.
3.
and
S--
8 .
Denoting the
sum
FINITE DIFFERENCE EQUATIONS The sum
(5)
n terms of a recurring
to
similar to that in Ex.
Find
3.
the
to
can be found by a process
or in the last example.
(3),
sum
series
363
n terms
+ u^x 4- u 2x2
of w
+
. . .
,
where u n
+pu n _ t + qu n , z =
n>l.
for
=u + u lx+ u2 x 2 + ... + u n _ xn ~ 2 + pu n _ 2 xn - + pu n _^xn pas n = 2?tt # -f >Wi# +
Let
l
6n
l
1
then
. . .
Since w r +7>w r _ 1
+ ^w r _ 2 =0
for
::
*n (1
+ JP^ +
^
2 )
t
r>l, by addition,
~ uo + w i +P ?/o) x ~ u n xn (
which determines the value of sn unless a; is a root a of the equation 1 In this case the sum may be deduced from the general result by supposing that #,
Given the
(6)
uv
w n -i-^ 1 w n _ 1 +
scale
ur _ v we can
...
find the value of
4-j? r w n _ r
= 0, and
the values of w
,
u n by
(i) finding the generating function (ii) expressing this as the sum of partial fractions of the form ~ ~ m where a is a root of xr A(l-ocx)" +p 1 x r l + p 2xr 2 + ... -f 7? r = 0; and n (iii) expanding the partial fractions and collecting the coefficients of x ...
;
}
.
Examples
Examples of Finite Difference Equations.
5.
any
are given in the next article.
r
+1
consecutive terms of the sequence (u n )
u n +PiU n ~i+P2U n _ 2 +
equation
where p l9 p2 ...pr are constants. ,
is
From
...
Suppose that are connected by the
+;p r w n _ r = 0,
................. (A)
this point of view, equation (A)
called a Linear Finite Difference Equation with constant coefficients. So far as equation (A) is concerned, the first r terms of the sequence
may have any
values whatever, that
is
to say they are arbitrary constants.
Thus the process described in the last section enables us to find the general value of u n which satisfies equation (A), and shows that this value involves r arbitrary constants. The general value of u n is called the general solution, and the process of finding it is called solving the equation. The process leads to the following results. The general
(i)
(A
-f
wJ5)a
n
where a,
,
u n + pu n _ l + qu n _ 2 ^=Q is wn = ^4a n -f J3/?" or are the roots of x 2 +px + q = Q and A, B are arbitrary
solution of j8
constants, the first or second form being taken according as (ii)
The general
where a,
j3,
y
solution of
//a^^y,
then
2 +px + qx + r = Q and A,
no two of
the roots are equal. n a n -f
u n = (A+ nB) u n = (A+nB + n*C)oL n
If a = P 7^ y, the solution 2A
u n +pu n ^ l + qu n _ 2 + ru n _3 = Q
are the roots of x 3
constants, provided that
a^/?
is
Cy
B,
or
a = j8.
is
C
are arbitrary
.
.
B C A -
-
-
GENERAL SOLUTIONS
364 For any
(iii)
other value of r the general solution of equation (A) has
similar form.
We If
give the proof of
w
,
(ii).
%, u 2 have any values, then
+ u 2*x2* + + u^x l
UQ
where
a, 6, c
where
a,
j3,
. . .
+ u n xn +
are functions of ^
y are the roots
. . .
=
-a + bx + cx2
_
u v w2
,
~ 2
==
3
P
a
.
J 7: (say),
Q
Also
.
of
# 3 + px* 4- qx 4- r = 0. If
(i)
no two of
where A, B,
a,
j8,
are equal,
y
C
are arbitrary constants. the fractions on the right and equating coefficients, Expanding
(ii)
If
a = /? 9^ y, then
P/4 (iii)
If
w
JJ
u n = (4
and
f^
JD
^L
-f
nfi)a
n
+ Cy w
.
a = j3 = y, then
B
P__A__ ~ + T - ocx
Q
(^
+
(l^axp
(1-a;
u n = ^4o
and
where
JB',
C' are arbitrary constants.
The first term of a sequence is 1, the second is 2, of the two preceding tertrw. Find the n-th term.
j&r. 1.
sum
Denoting the sequence by u l9 w 2 , ... we have u -u and n -2^Q
Hence
Eliminating A,
,
B from these (a
-
equations,
a
-a
we
0K = (2 - )3)a" -'
Now a+0 = l, a/3=-l and 2
1,
other term is the
U 2 ^2.
2-Ao^ + B^, l=.4a + ^, where
1
Un^AoP+Bp
^=
n~i-u
n
and every
a,
j8
are the roots of
find that
-
(2
- )"-'.
a>/8, a- 8=v'5, Therefore -1=0), and similarly 2-a=/?
hence 2-y8
if
J
2
.
I
-d
-
= l+a=a 2
(for
GENERAL TERMS Ex.
Find
2.
a recurring
the n-th term of
which
series of
Find also the sum of the first n terms when x ~ 1 and Denote the series by w -f u^x + u 2 x 2 + . . .
,
365 the first
.
let
the scale be
7+2p+# = and 20 + 7^4-2^=0, giving p= -2, un -2u n ^ l -3u n _ z -Q. The roots of x*-2x-3^Q are 3, - 1, and therefore Then
is
un =A.3n + B(-l) n
A+B^\,
where
Thus the nth
-^2.
3.4
n
term-w^a;"- -i-[3 + ( 3
.*.
sum to n terms = -
1
l-3#
4
---h
--
1
Now the
sum
to
Ex.
3.
The
solution
n terms
are arbitrary constants.
The values
of a,
ft
Hence the values
~iin
a,
==
;
...
n-1 + u n _ 2 =0 ^w a rea/ /orm. are the roots of x*-x + \
and
j8
Changing the constants,
this
^3
.77
may
(7,
be written
are
an
of
,,!.. solution
;
.
1
-w
scale
n ~l
-2 + 7- 20 +
1
of un + />^3n where
1
and the
v\n
of the series
Can
un
]x
1+s
J'iwrf the general solution is
l)
!-(-*)" _
/
and the
-3,
,
iZL-^^ limLJL
lim
q=
B=^\. n ~l
--
l-3w z n
4
/.
A-%
Hence 1
four terms are
,
j3
n are cos
t
sin
-^~, o
o
AA
un
is
7T
r + B am -
cos u
and so
~u
EXERCISE XXXVII 1.
Find the wth term and the sum to n terms of the recurring (i) 1
2.
is
Show
+2+5+
Show
vergent
.
;
1+2 + 5 + 12 +
(ii)
x |
|
-
<
sum
to infinity
..
.
series
:
.
is 7J-.
un -2un _ 1 + 4un _ 2 -3un _ 3 = Q
that, if
if
infinity is
+ ..
that the recurring series
convergent, and that the 3.
14
;
and
if
the
first
the series
9
three terras are
1
2unxn
+ 2x + 3#
2 ,
** ( i
^ ~\
9
_1_ 1*2^i ~^~ fjjL,
//Ix
; i
_
f?l* 4Utfe
I
n^
4.'/*^ xC/ <
-
^f*^ O*<
!
3 2 [The moduli of the roots of x -2# + 4x-3=0 are
1, 1/^/3, l/>/3.]
the
is
con-
sum
to
EXAMPLES OF RECURRING SERIES
366
3un -lun_ l -{-5un _ 2 -~u n _ 3 =0 and
4. If
%=
u 3 = 17,
u 2 = 8,
!,
the
find
un and u l
values of 5.
Find the nth term of the recurring
6.
If u n
series
l-f2-f34-54-74-94-...
.
the nth term of the recurring series 14-24-34-84-134- 30 4-... , find the sum to infinity of the
is
show that un ={(3n-4)( -l) n -f 2n + 2 }, and 8eries u If w n
7.
-^~w _ n
un ain6~u 2 q
+ gw n _ 2 =:0, where ^ 2 <4g, and -^7r<6<^rr show
1
9
2
2
?L:J
sin (n
- 1)0
2
-u^
sin
tan#=
where
(n- 2)0,
that
\/(4
2 ) .
J
Zu n x n Er n xn
If
8.
and
1
,
+p'x + q'x
2
then
,
are recurring series of which the scales are 1+px + qx 2 2(un + vn )xn is a recurring series whose scale is (
ZW n ZV n
9. If
1 -\-px 4-
2 flo:
) (
1 4- 2>'z 4- q'x*)-
are recurring series of which the scales are and vn + p'u n 4- q'u n__ 2 = 0,
,
un + pu n _^ 4- qu n _i ^
^
7^ 4^', then Sun vn is a recurring series whose ^ 4<^ and un ~PP' un-i + (PV +P' 2<1 - 2^0^-2 -pp'OU'lln-z +
where
2
>'
^>
2
7
r/
scale
is
respec-
'
,
(
10. If 1
2un
- 2kx 4- A: 2x 2
,
xn
Evn
are recurring series whose scales are then Eun r n xn is a recurring series whose scale ,
Hence show that Ununxn
Eun
(
xn
xn
is
a recurring
series
whose
scale
l+px + qx 2 and is
is
(1
a recurring series whose scale is 1+px + qx 2 whose scale is [1 - (p 2 ~ 2q)x 4- g 2 # 2 ] ( 1 - qx). are the roots of z 2 4-pz4-?=-0, then u n * = A(ot*) n [If a, fi where A, B, C are constants. Hence the scale of Eu n 2 x n is 11. If
is
a recurring
12. If
-
is
-
l+px + qx
13.
,
then
Uun
2
xn
series
+ B(fi*) n + Cq n
,
(l-a 2^)(l-^)(l-^).] , 2
= ^r u xn n==0 n
Prove that each of the COS a sin
use Ex. 11 to show that
series
4- X COS
oc + x
,
sin
(0
4-
a)
4-
X 2 COS (20 4- a)
4- ...
,
2
(04-a)+# sin (204-a)4-... 2 being 1 2x cos 4- x in each ,
Show
also
and that the sum to n terms of the second series may be obtained from expression by changing cos to sin in the numerator.
this
a recurring series, the scale that the sum to n terms of the is
case.
first series is
cos a - x cos (a - 0) - x n cos (a -f nO) + xn + l cos (a _____._____
+n -
Id) 9
FORMATION OF DIFFERENCE EQUATIONS Finite Difference
6.
367
Let (u n )
be a sequence in u n ^ r are connected by
Equations.
all of the terms u n u n __ v u n _%, ... an equation which holds for all values of n^r, where r is a fixed number. Such an equation is called a Finite Difference Equation.
which some or
An
,
important class of these equations has been considered in Art.
5.
Here we give methods which apply to various cases which occur in ordinary algebra. The general theory belongs to the Calculus of Finite Differences. To solve a difference equation is to express u n as a function of n in the most general form. The result is called the general in Art. 5
it is
clear that this
and from what has been said solution must involve one or more arbitrary solution,
constants.
A
particular solution
obtained by giving these constants special
is
values.
un
a given function of n containing one or more arbitrary constants, a difference equation may be obtained as in the following examples. If
Ex.
is
// u n
1.
~-Gn-2
where
C
an arbitrary
is
constant, obtain the corresponding
difference equation.
The required equation
is
found by eliminating
un ~Cn-"l
- l)w (n n
giving
C'
from
^n-\~ V(n -
and
1)
-2,
-nw n- 1 = 2. .
n n where A, B are arbitrary constants and // un ~AoL +Bf$ numbers neither which is zero, prove that unequal of u - (* + $)u -=0. fl _
Ex.
2.
The
result is
n _i +aj8t*
n
found by eliminating A
Solution of
7.
,
a,
ft
are given
8
B from
some Elementary Types.
n // u n = au n _ l9 then u n = Ca where C is an arbitrary constant. For w n /w n _ 1 ==M n _. 1 /w w ._ 2 ==... =w 2 /w 1 = a, and u v may have any value. (1)
// u n = a n u n-i where a n
(2)
where
C
is
For *or
an
Wi/a i
n> then
arbitrary constant.
Un ~ l ~Un un-\ u n-Z *
un =
therefore
and
a given function of
is
mav have any
value
...
al
we
^-aa --a na n -i ul
a^
. . .
choose.
a ny
a
<*2i
w n = Ca xa2
...
an
SOLUTION OF ELEMENTARY TYPES
368 (3)
= bn
// u n a n u n _i
where a n and b n are given functions of n, the
general solution is
where
C
is
an
arbitrary constant.
Let vn = a 1 a 2 ...a n
The
result
,
and divide each
side of the given equation
by v n
.
is
Un/Vn~Un-lK-I = b n/ v n' In
this,
writing n
Whence by
n~
1,
2 in succession for w,
...
we have
addition,
wK This
2,
the
gives
^1/^1
= &iM + 6 2 /v 2 +
fe
the value of
question, for
in
revsult
+ & nK - iKu l lv l -bl jv l
is
arbitrary.
In examples of this type, it is better to apply the rather than to use the actual result. -l)u n - nu n _ 1 ~2. we have Dividing by n(n-l), u n \n - u n ^l(n - 1) =2/ Ex.
1.
1,
nn
-2,
...
2 in succession for
U.2
I
Hence un ~ Cn-2 where C. (See Art. 6, Ex. 1.) Ex.
2.
Find
(4)
we
2.3
C~ 2 + u v
the general solution of
(n
-
1).
n and adding,
(n-l)n) and
since u t
u n - nu n _ { =
I
is
an arbitrary constant, so
also
M.
and proceeding the equation becomes un j\n--u n _ l l\n-\ l, find that wn + (7) where C is an arbitrary constant.
Dividing by as before,
just described
Solve the equation (n
Writing n
is
method
n,
|
The general
~[n(n
solution of
^a n + c according as a and
?/
n
n
~aw n _ 1 = cjSn
+Y03--a)
is
or
are unequal or equal.
j3
For dividing by an we have ,
Writing n-1,
n-2,
...
2 in succession for n and adding,
a
LINEAR DIFFERENCE EQUATIONS u n = coc n
Therefore
=u
where B( Hence,
l
#
-
gn r
a
if
+ -^2 + [a a -
{
.
.
.
n
369
l
+ r-n \ a
j
an arbitrary constant.
is
/ot-cft/oi)
B
(B
2
_ an
-
n
n
,
/3-a which-
may
be written
where A(^B-cft/(ft~oc)) If a = ]8, then i/ n = cna n
The general
(5)
is
an arbitrary constant.
-
w n ~(a-f j3)w n _ 1
solution of n
u n ^AoL + Bft according as a and
/?
n
-faj3w n _ 2
=
u n ^(A + nB)oc.
or
are unequal or equal, where A,
i
n ,
B are arbitrary constants.
For the equation may be written
^-atV-i^K^-a^o). Hence by where C
is
u n ~ocu n
(1),
^^C^
t
an arbitrary constant, and therefore, by (4), w n ==^a w + Cj8 n+1 /()8-a) or (4+Cw)a n ,
according as a 7^)8 or
Cft/ (ft -of)
is
a = /J.
This establishes the result in question, for an arbitrary constant if a^/3. (See also Art. 5.)
Linear Difference Equations.
8.
au n + bu n _ l + cu n __<>+ is
With regard (i)
it
An
equation of the form
ku n , r = l
r is a fixed
n or constants.
to equation (A),
The general
... 4-
Here
called a linear difference equation.
k, I are given functions of are of this type.
(1)
.....................
number, and
(A)
a, 6,
...
All the equations in Art. 7
should be noticed that
u r+l9 wr4 2 >---
solution involves r arbitrary constants, for
can be found in succession in terms of u l9 u^
...
u r) which
,
may
have any
values. (ii)
// v n
is (he
general solution of
au n -f 6w n _ x -f cw n _ 2 +
and
wn
is
solution
any particular For it of (A).
. .
.
+ ku n , r = 0,
solution of equation (A), then v n is obviously a solution, also
+ wn it
is
is the
general
the general
solution, because v n involves r arbitrary constants. (iii)
The method
of dealing with such equations as
au n -f bu n ^ where
a,
fc,
...
k, r
-f
. . .
are constants and
trated in the next example.
-f
ln
ku n _ r = /, is
a given function of n,
is illus-
GENERAL TERM OF A SEQUENCE
370 Ex.
-Su^j-f 6w n _ 2 =n
Solve the equation u n
1.
The general
-5%^-f 6w n _ 2 =0
un
solution of
A
*^tl
Next search
Assume
w n =a + fru + en 2
a + bn+cn -5{a + b(n-l)+c(n-l)
powers of
coefficients of
6
-J-,
+ 2n
.
is
^
a
J,
-^
n,
we
then for
,
2
}
A
/
all
values of n,
+ 6{a
find that
2a
26-14c=0,
2c^l, c
5n
+/J.O
z
giving
-f
for a particular solution of
as a trial solution
Equating
2
~ n d so a solution
is
15) ...................................... (B)
Again, take the equation
and assume as a
a(5
Thus a solution
n
a 5n
un
trial solution
-5.5n " 1
.
,
.
then
+6.5n - a
)=5
n ,
? giving a = 6^.
is tS.
Finally, take the equation
Since 2
un
is
a root of x 2
an. 2 n ,
-5x + 6=0,
it is clear
we have a{n
- 2. giving a =
.
2n
-5(w -
Thus a solution
l)2
n -x
that a
.
2W
is
not a solution.
+ 6(n -2)2n ~ 2 } =2n
Trying
,
is
u n = -n.2n + l .......................................... (D)
The general
solution of the given equation
sides of equations (A)-(D),
and
is
obtained by adding the right-hand
is
EXERCISE XXXVIII The first term of a sequence is 1, and every other term the preceding terms. Find the nth term. 1.
2.
If
3.
The
2ttn
-tt n _ 1
first
the arithmetic
is
the
sum
of
all
=na, show that
is a, the second is 6, and every other term of the two preceding terms. Show that the nth terra is
term of a sequence
mean
thus proving that the nth term tends to J (a
+ 26)
as a limit as
n ->
oo
is
.
4. The first term of a sequence is 1, the second is a, and every other term the geometric mean of the two preceding terms. ~ Show that the nth term is ax where x ^i:{2 + ( - i) n 2 }.
is
DIFFERENCE EQUATIONS
371
5. If ww = />_! gw n _ a and un tends to a limit other than zero as w-> oo are 1 and -~q; also show show that |g|
^i
,
x
6.
i-
Show
+ qui
Ui
hm unn
that
.
,
1+q
that the equation
un - (a 4 ft + y) U n _i 4 (fty
4-
ya 4 aj8)
tt n _2
~
can be written in the form
Hence prove that
Un-oMn-i^Bp + Cy" fi^y or j9 y, where J5, C
OT
are arbitrary constants. according as as of the in Art. 5, (ii). solution equation given general 7.
If
un
nu n _ l + (-l) n and w ~l, show that un j\n-*e~* as ^->oo.
K/|n_= Wn_ /ln_-J-K-D 1
8.
If un = ?i(u n _ l
+ un _ 2
[Write the equation 9.
If
wn
(^
t*
n
)y
n
/[n..]
un
find
- (w 4-
in terms of
w and ^, and show that
u n _ l ~ - (^n-i ~ nu n-2)-]
1)
- l)(^ n ~i + ^n-2) an(i
V 3
j
l
M2 = l
= 0,
"
1
"
(
prove that
1)n 1
,
li
/'
L?
[Write the equation in the form
un - nun ^ = - [u^.! - (n 10.
Obtain a particular solution of
by assuming un =an + b. Hence find the general solution of the equation. Find the general solution of un -5u n _ l -i-Qu n _ 2 n [Assume un ~a .l as a trial solution.] 11.
12.
Find the general solution of
[Take as a trial solution 13.
wn -wW -i-Wn_2=ft 2 un ~ an 2 + bn + c.]
.
Find a particular solution of
un + u n-i sinna by assuming un a cos no, -f b sin noc. Hence show that the general solution is
~~ 2(1 14. If
Deduce the
u n un _i
=-
[nu n = (n l)?/n_ 2
un
in terms of
-f
cos
)
,
find
;
no^e that there are two forms.}
u^
ln
.
GENERAL SOLUTIONS
372 15.
Show
that the general solution of the equation
Vn-l + aun + bun-l + C = can be written in the form
where
A
an arbitrary constant and a, /3 are the roots of 2 - ab =0 a: - (a - b)x 4- c
is
;
unless
a.
=
fi,
in
which case
[Write the equation as ^(^ n _i4-a)4-&w n _ 1 4-c^0. v The equation then reduces to so that u n 4 a /
.
v n-l
vn
unless vn _j 16. If
4-
(6
- a)v n _ 1
4- (c
- ab)v n _ 2
= 0.]
Vn-i + 3ww - 4
w_ 1
-2=0, prove
that
>l.2 n 4-5 n ^
/I
17. If
<
wn w w _ 1
4- 5itw 4-
WTJ_I 4-
9
0,
2w
_,, ""
1
4-5 n
~1
<
then 2
,
18. If
tt
n+1
=-~
-'
2n-wn
and w l = a, prove that
a4-w(l-a)
and [Put
wn ^nvw .]
0,
Put
CHAPTER XXIII THE OPERATORS 1.
The Operator
variable n, (i)
J, E, D.
INTERPOLATION
Let u n be a function of the positive integral
A.
and consider the sequence u v u 2 u3 ,
The meaning
symbol A
of the
is
,
...,
un
...
,
.
defined by
^n^tVrt-f^, and A
regarded as denoting the operation which when applied to u n the Here n may have any value, so that difference u n+l - u n produces is
.
Jw 1 = w2 -^^ 1
Au 2 = u3 -u 2
,
etc.
Au n = v n *Aen Vj + Va + Va + .-.+Vn^Wn^-ti!. = w a - tij, = v n = w w+1 - ti w Vj *?_! w n !*__!, ...
(ii)//
,
For
,
whence the (iii)
,
result follows
by
addition.
d(u n + v n )=Au n + dv n A (u n + v n ) = (w w+1 + v w+1 ) - (u n + w n ) ~ *n) + (V i - V n =A
// u w v n are functions of n, then
For
(
In this connection
then As n
is
it
n-fl
w+
not equal to
.
)
should be noted that
.
if
Au v +Au% + ...+Au n
unless
w 1 = 0, for
and
Following the index notation of Algebra, the symbol that the operation A is to be repeated, and we write (iv)
,
and so on
;
thus
J2
denotes
TABLE OF DIFFERENCES
374
This process may be continued to any extent and, from the way in which the successive differences are formed, it is obvious that the numerical 3 coefficients in the expressions for J 2w n , J w n ... are the same as those ,
in the expansions of (1
-x)
2 ,
(1
-x)
3 ,
Hence we conclude that
....
4 ru n = u n+r -Clun+r _ 1 + C'2 un+r
(A) 2 -...+(-I)nu n in Art. 4. more concisely proved For the sequence uQ u l9 w2 ... a table of differences is conveniently
a result which
,
is
,
written thus:
,
UQ
u2
Ul
2 Thus, for the sequence O ,
u
I2,
2 2 , 32 ,
Au
...
W3
,
4
1
1
A*u A*u
(v) If 'u x is
_.
where u n 9
222
a function of any variable
n2 we write ,
16...
5
3
...
7
x, the
meaning
A applied to u x is defined by Aux ^u x ^-u x All that has been said about the symbol A still
of the'operation
.
successive differences of II u x>
I/
holds, for in finding the
u x we are merely concerned with the sequence
07 x+V ux+2>
Illustrations,
(i)
u n = n(n~ l)(n-2), then
If
J rw n -0
and (ii)
If
w n = n3
2. Special
,
since
for
r>3.
w 3 = n(n~l)(w-2)-f 3w(w-l)-hw, by the preceding,
Cases,
(i)
The following notation
xr =x(x-l)(x-2)...(a;-r-f 1) and 3_ r = If
x
that
is
is
variable
and J
is
-
often used
-^-^-
x(x + l)(x-f 2)
...
;
we
write
-
applies to x, then
Jx r = rx r _
1
........................................
(A)
THE OPERATOR & Similarly
we can show that Ax^ r
-
=
ra( f +D,
375 ...........................
(B)
being assumed that no denominator vanisJies. These formulae may be compared with
it
x r = rx r
-r--
ax Ex.
Shaw
1.
~l
and
-_-
ax
x~ r = - rx~ (r + 1 \
that
n(-l) + (n-l)(n-2) + ...+2.1 w(7i-l)(/i-2) + (n-l)(n-2)(n-3) + ... +3
.
2
.
=(/t + l)?i(n l=(n + l)n(w
-
1),
and so on.
8
If
and
is
sum
the
of the
Jn3 = 3wa by
since
>
first series,
Art.
1, (ii),
S=^ (n + 1) 3 +
vhere
C
is
a constant.
Putting n =2, we find that (7 = 0. Hence the result. Similarly for the second Of course, the rule in Ch. VIII, 3, might have been applied.
A
(ii)
by
Polynomial.
a?
x,
dent of
1,
#,
a? -2, ... such that
a polynomial in x of degree r, dividing in succession, we can find a a^ ... ar indepenIf
ux
is
,
ux = ao + a i^i + a 2 x2 +
J w^ = a l + 2a2 x x + 3a 3#2 -f
and then
and so
on.
series.
Finally
A ru x = a r
+ a rx r
,
. . .
J r ^wx = 0.
and
r
.
. . .
,
|
2.
?x.
For
//
A mu
x
w
as a factor.
\
The Operator
3.
.
when applied
to w n ,
is
E.
The
the operation denoted by E by unity. Thus for all values of n,
effect of
to increase
n
Eu n ^u n+l a+
.
a constant, the following equations define the meaning of E, Ea and a$, regarded as operators.
If
a
is
(E + a) u n = w n ^! + a?/ n = (a + J0) w n Further,
and so on
;
we
write
thus,
if
r is
y
a positive integer,
,
E + a,
LAWS OF OPERATIONS
376 Hence,
p, q are positive integers,
if
E*E*u n = E* (Eu n ) = Eu n ^ =
E*E*u n = E*>+
so that
Again,
.
a, 6 are constants,
if
(E + a)(E + b)u n = (E + a) (u n+l + bu n )
+ abu n Thus, so far as addition (subtraction) and multiplication are concerned, the operator E combines with itself and with constants according to the laws of Algebra.
The same
for A, for
is true
4* n = ti w+1 -w n = (^-l)ti n More defined
4.
x
9
generally,
if
ux
is
a function of any variable
x,
the operation
E
is
by
Fundamental Theorems.
and
.
r is
a positive
// u x
is
a function of the variable
integer, then
(i)^ux ^ux+r -C[ux+r ^^C 2 ux+r ^-...+(-lYu X) ......... (A) = ux + C[Aux + ClA 2 ux +...+A r u x ....................... (B) (ii) ux+r r
The first theorem Proofs. on writing x for n. More easily thus
follows from the reasoning in Art.
1, (iv),
:
(i)
J'.-(-l)X = {Er - CJ^- 1 + Cj'- -...+(- l) r}ux
Again,
(iii)
(ii)
If u x
u x+r =Eru x = (l +A) ru x and therefore ,
is
a polynomial in x of degree
For the left-hand (iv)
//
s n is the
side
sum
=4^ =
to
n(n-l)
l,
then
(Art. 2,(ii)).
n terms of the A
r
series
n(n-l)(n-2) '
u
-f
u 2 -f u3 -h
. . .
,
then
METHOD OF DIFFERENCES For w
=
n
l+^) ~X>
377
therefore
Similarly
Adding and using the Ex.
4
3,
Find a
1.
results of Art. 2,
n which has
cubic function of
Ex.
the result follows at once.
1,
the values
-3, -
1,
1,
13 when n
= \,
2,
respectively.
Denoting the function by u n
and A r un =
,
we have
for.r>3, since u n
Also
tt
is
a cubic function.
= ^ w -X =
-
_
+6
j&a;.
Sum
2.
the series
2.3+3.6 + 4. Here un ~ (n + l)(^ 2 +2). Writing down the as
on the
first
right,
This
four terms,
that
is
ll
+ ...+(
a cubic function of
we
un
find,
^=6, 4^ = 1
?i,
A r un
for
44
18
Aun A 2u
A^u^U, J 3 u 1 = 6.
so that
6
26
12
n>3. 90
46 20
14 6
Hence by
(iv),
i9 12
.
Ex.
3.
//
positive integers,
For by Art.
if
2,
A
(Cf.
Ch. VIII,
5 = *n -C7J(a: + l = (-l) n \n then 8 if r>n and S
applies
(ii).
,n(n-l)(n-2)(n-3)
1A 14
-
to
x,
S=^(l
4,
Ex.
1.)*
r,
if r
- E) rxn ^(-l) r A rx n f
are
= n. and
the
result
follows
BERNOULLI'S NUMBERS
378 Ex. 4.*
(Bernoulli's Numbers.)
For nr =tf n O r =(l+J)n O r and
If
Br
is
defined as in Ch. VIII, 8, (5), then
-l+$-...
+ (-ir
m>r;
if
i
}r
1)
...
......................... (A)
...
Li
II.
#r = l f + 2r + ... +nr we have f(n + l)n (n + l)n(*-l)
Hence
if
,
l)n(n-
(n-f
(*~r +
l
'
*'~\
LT+JL
[3
[2_
Now, by
definition,
Br
coefficient of
n
in
Sr
hence
,
Again, equating the coefficients of w in (B), A3
J "T + {AZ Hence
~- +( ~ 1)f "
Ar^
^ =i
and (-l) rJ5r =J5r Hence equation (A) holds for r>0.
5.
5.
J0r =l,
Apply equation
in
Operators '
'
following
"
r>l,
if
Moreover,
Ex.
r=
r}
proof
(A) of
for
r>l
Ex. 4 to show that B^
a Fractional Form.
of the
theorem in Art.
4, (iv)
(Ch. VIII, 8,
(5)).
^.
Many
writers give the
:
En -l n(n-l)
.
n(n-l)(n-2)
i
Here the operator is written in a fractional form in order to transform a polynomial in E into a polynomial in J. This result was obtained his collected works).
by Cayley by a much more
difficult process (Vol.
IX, pp. 259-262 of
LAGRANGE'S INTERPOLATION FORMULA The
when
result,
obtained, can be verified by multiplication
379 and addition ;
this fact justifies the process.
But
Thus the
at every stage the operator must be regarded as a whole.
E -l n
equation Sn
(E -
does not imply that
In fact the last statement
is
~~E^T Ul 1) s n
= (En - 1) %.
not true, for
(E-l)s n = s n+1 -s n = u n+l
(E -l)u l ^un ^ l -u l n
and
.
We
conclude that, in such transformations, the operator can be put in the form /(J)/^(J), where f (A) and
f(A)
is divisible
6.
by <(J), andf(A),
are not regardedas separate operators.
Suppose that y
Interpolation.
a function of x whose value
is
has been determined by experiment, or otherwise, for x = a,
The problem
of interpolation intermediate values of x.
The graphical method is to plot the points draw a smooth curve through the points.
(x, y) for
We
'
*
6, c, ...
.
to find approximate values of y for
is
x = a,
6, c, ...
assume that
,
and
this curve
represents as nearly as possible the graph of the function y. Let us suppose that the equation to the curve so drawn is y=f(x). It is natural to take for f(x) the simplest function of x which satisfies the conditions.
If
y
is
known
for
n values
we assume that
of x,
where the n constants p p^ ... are to be determined by making the curve go through the n points. If only two values of y are known, the curve is the straight line joining ,
the corresponding points, and we have the rule of proportional parts as used in working with tables of logarithms and trigonometric functions. 7. ively,
Lagrange's Formula. we assume that b)(x
(x
y=syi
For y
is
when x = a, Similarly
c)
(a-b)(a-c)
If
+y*
(x
y=y
y2 y$ for x=a,
-a)(x
6, c
,
l9
(x
c)
(b-a)(b-c)*
y3
a)(x
respect-
b) *
(c-a)(c-6)
a quadratic function of x of which the values are yl9 y& y3 6, c. if
y = j/1
y2
,
y s- y
>
-
^3
when x = a,
-^4
(x-b)(x-c)(x-d) \
'
_L^
'
b,
.
+ three
c, .
d we take ..
similar terms
;
*(a-b)(a-c)(a-~d)
with similar formulae when more than four values of y are given. B.O.A. 2B
INTERPOLATION
380
8
.
If
the values of y are
known
of applying Lagrange's formulae,
we
For example, suppose that y
Assume that y is
a positive integer,
at equal intervals, instead
use the operators
y&
y\,
a cubic function of x
is
x
for values of
;
y%,
then
E
for
j/3
J ry =
and A. x v x2
X
XQ
for
r^4, and
,
x$.
,
if
n
= (l+A) n y0)
n
y n ^f! y therefore
Hence the equation
represents a cubic curve which passes through the four points (# ?/ ) (x l9 T/J), ... and gives an approximate value of y corresponding to any value ,
of x within the specified range. Ex.
1.
Given
tJiat
sin 45 =0-7071,
sin 50 =0-7660,
in 55 =0-8192,
A-rn
60 =0-8660,
n62. Let
1*3
= 10 4 sin (45+5x).
Construct a table of differences as below.
Au
u
A 2u
A*u
589
-57
Mj-7660 532 "
7
-64 468
J^ = 589,
/.
Assuming that
r
J w
J 2 w = - 57,
-=
A92
7.
-
M x(x-l)(x-2) _j
x = 1-4 and
so that
u
=7071 D
= 1-4x589 MA
=
2
=
824-6
=
- 0-7 x 0-4 x 57
|
ILl-JpZJ
A 3 u Q ^ 1 -4
x 0-4 x 0-7
=
0-392
7895-992
I
15-96
u x = 7880-032 .'.
The
-
r>3, approximate values of w^are given by
for
x(x-l) -L_J
52=45 + 5z,
J 3w
sin 52 =: 0-7880 to four places.
correct seven-place value
is
0-7880108.
- 15-96
BESSEL'S INTERPOLATION If
wo only
FORMULA
381
use the values of sin 50 and sin 55, the rule of proportional parts would
give sin 52
=0-7660+f
x 0-0532 =0-7873.
If UQ and several other terms of the sequence U Q u v u%, ... are known and approximate values of the missing terms are required, we may proceed ,
as follows.
= 4-7046, w 3 = 5-6713, w 5 = 7-1154, find approximate w = 4-3315, and w 4 Four points on the graph of u x are known. We therefore assume that it can be 4 represented by a cubic function of x. This is the same as assuming that J wa = 0. Ex.
2.
^
//
values of u
.
.
Hence we have approximately 0,
/.
+ ^i) ^11-4619. = w 4 6-3199 (approx.).
w 2 ^5-1420,
NOTE. In this example w ~tan77, w 1 -tan78, w 3 tan 80, W 5 ~tan82, we have found that tan 79 ^5-1420, tan 81 ^6-3199 (approx.). The correct values are 5-1446 and 6-3138. It should be noticed that the
approximations just found are
given by )
9.
=5-1879,
much
closer
so
than those
w 4 = \ (u 3 + u 5 )= 6-3933.
Bessel's Formula.
second differences
In using Mathematical Tables, the effect of be easily and accurately allowed for by using a
may
formula due to Bessel.
Let u_ l9 u
,
u lt u
be four consecutive values of a function u XJ as
and suppose that values and 1.
given in the tables, of
x between
of
u x are required for values u -i
Denoting ences by a
9
on the
first 6, c
a
and second differand d e, as shown
,
b
9
right, Bessel's
formula
u*
is
f e
c
\
x(x-l) d + e which
is the
same as a) ............................ (B)
The values little
labour
is
of
uQ9
a, 6, c
can be taken from the tables, so that very
required in using the formula.
THE DIFFERENTIAL OPERATOR
382
if
To show that this gives a good approximation, and to estimate the error, we neglect differences of the fourth and higher orders, we have
Now
e-d^f; and
w -w_;i=a, b-a=^d,
x(x-l).
+ l)x(x-l),
(x
'
V
e-d\
x(x~l) fd + e so that
if
?'
x is
Hence the
(x
+ l)x(x-
1)
the approximation given by (A),
error
C/5^ to6/65
Ifo. 1.
therefore
is
approximately izx(x- l)(2x~
of square roots
to
I)/.
find
Using Barlow's tables, we take
045020= a -:
11 1-085553
045001=6 M1
= N /12350 = ............... 044983 -c.
^2-^/12360 = ................ (The differences
a, 6, c are
z^-4,
given in the tables,)
For >/12344, we have
c-a-^ -
-000037. -J*(3?-l)=--06, w = 111-085553
Ja?(a?
-
1
)
(c
-
6= a) -
-0180004
-00000224
~
N /12344=!ll-1035o7>
The
result is correct to the last figure.
10.
The Operator D. du
the symbol
-r-
is
When we d
dn
write
u/ d
\
n
regarded as an operator, which
is
sometimes denoted by
118
If
a+
a
is
a constant, the following equations define the meaning of
A Da and aD, regarded as operators.
D + a,
THEOREM
LEIBNIZ' Hence
(D + a) (D + 6) u =
a, b are constants,
if
383 -I-
(Z)
a) (Dw
-f
6w),
= Z)2 w -f 6Z)w -f aDw -f afrw = {D2 + (a + b)D + ab}u. (/) 4- a) (D -f 6) u
and
Thus, so far as addition, subtraction and multiplication are concerned, the operator D combines with itself and with constants according to the laws of Algebra.
The n-th Derivative of a Product.
11.
d
du
dv
dx
dx
dx'
x
of
D D
Let the symbols
l9
If w, v are functions
denote differentiation with regard to
2
x,
D
suppose that only operates on u and its differential coefficients, that 2 operates only on v and its differential coefficients, so that
and and
D
Dn
D
As
/
l
\
(uv)
=
ax
and addition and / ^
&u
t
D
dv
D 2 (uv) = u jr\
v,
i
\
ax
\dz/^ that
WV '"~'
is,
-j
1+
*'
(uv)
=^
This result
Ex.
is
(
1+11 -
uv >~\
known
If ij~u
+v
and
-
(uv)
n -f D n = (D l 2
.
)
uv.
where u m
=x + (xz - ^)
vm
,
=x -
-f
uv n
,
x.
- k) ",
/^
W = 0,
1, 2, ...
.
we have
Differentiating with regard to x,
_ i +x Wi
'
2
)>
often written in the
is
(
2)( u
2
with regard to
=
tn-i
-
. . .
suffixes indicate differentiation
ww
ax
21
2
as Leibniz' Theorem,
=
1.
~=
are independent of each other, they obey the laws of multiplication, therefore
form where the
d
,
2
n \
,
and hence
;
_ jt)"^ = tt w (a; 2 - Jt)"" (x z
;
hence (# a - k)^u l =u/m.
Differentiating again,
.'.
Similarly
~*
*-
*
(x
-
2
(#
2
k)
-
wa
!
-f-
arM!
k) v^-^-xvl
= (x* - kf*
- w/m 2
0.
- v/m?
0,
.
ujm*
;
- y/m 2 =0. whence by addition, k) y 2 -f ar^ (x Differentiating w times by Leibniz' theorem, 7 2 ^-i (* k) yn +* + n.2x. t/w+1 + 2
-
-
This example leads to a remarkable theorem on cubic equations, given
in
Ch.
XXIX.
USK OF OPERATOR A
384
EXERCISE XXXIX 1.
Find the polynomial of the third degree in n which has the values when w = l, 2, 3, 4 respectively. (Use Art. 4.)
2, 11,
32, 71, 2.
Use the method of Art.
P+3 +5 + 3
(i)
3
u n = xn and A
3.
If
4.
Show
(ii) I
;
apph'es to
/-l, 2 or
S be = u r (* n/) r
the
(ii)Lct
.
show that
0,
according as n
of the series .
(
r
~y)
n
n~2
or ?z->2.
n(n +
l),
I,
2
sum
n~ 3
if
n>2
of the series.
S=
then
',
?/
.
;t,
2
[In each case let
Also J
+ 22
(w-l) C?4-(n-3) r^ + (w-5) Ct4-...^2
(iii)
r?
.
xn -C'i(x-y) n
2
Let
22
2
that
(i)
(i)
sum to n terms 3 2 + 3 a 4 2 + ...
Ex. 2 to find the
4, ...
(
1
=
- E) n ii
-
1 )
(
nJ w
.
etc.
n, |
w r = (a:-r)(y-r), then S = ( -
l)
nA n
Also
u^.
w if n>2 Aut = (x-l)(y-\)-xy=-x-y+\ J M = 2, J w r = n w = r then S = i-{(^+ l) ~(E I) )w (iii) Let w r n n Also w = 0, Jw = l, ^ w = 2, and JX = 0if /. S-=l-{(2 + A) -A }u 9 8
9
2
,
;
2
.
5.
^x(x-l)(x~2)...(xr + l), show that - 1 )a:r_ 2 -h (x -h n) r = a- r + ( ?rar r _ l + CJr (r
If x r
the last term being r(r-
Verify that,
6.
If
if
n
1)
...
(r
3 and r
xr = x r -f o^^!
-}-
~n-\- l)x r _ n if
. . .
a^l-r(r-l),
[Equate to 7.
Show
,
,
if
prove that
6=A-(r-l)(r-2)(3r-5).
the coefficients of x r
JV+^fc+JrJ'rj-JL
(ii)
8.
.
~1
" and x r 2 on the
right.]
that
(i)
[Use Ex.
.
the identity becomes
2,
6a:r _2 -f
.
r>n and C^jr
J rx r + 2 =
J
[x
2
-f 7-a:
4-
;
^r (3r -f 1)]
. |
rjh
2.
6.]
Prove that xn -Ci(x+l) n + Cl(x + 2) n -...+(-\Y(x\-r) n
+\
= - Yl {& + r-r + iV(3r -f 1)}
and [Use Ex.
(
7.1
1
|
when
n~r-\
when n = r 4- 2. r+_2
1,
EXAMPLES OF IDENTITIES 9. If
385
none of the denominators vanish, prove that 1
x
(-ir\n
= __
Deduce that
4- C? ^-fC? ^-"'+ 1
+
(
1)n
1
1 -
n(n-l)
1
n
-
-
(x+l)(x + 2)
2
n(n-l)(n-2)
I
3 10.
Prove that 2
1 *
~r 71
r-
22
1.2
m+2
m-f-1
n.2.3 n ~ l
n
^3
m(w-f-l) 11. If u n
2W
yt(n-l) T
nt/^x/.^-.um(m + l)(m + 2)
,
'
/
i\n.
m(m-f 1)
...
(m-fn)
~\l(an + b), show that
(i)
A ru ~( -
r
l)
r
111
(an
+ b)(a .n 1
n ' n(n-l)
a '
n(n-l)(n~2) "
a2 '
"3
= l/a;(a: + l)(a; + 2)...(a: + r-l), show that + n)_ T = a?_ r -6 ?ra- r+1) +(7?r(r+l)a;_(r +8)-...
12. If a:_ r
I
(
n+ 1
to
(
13. If n is a positive integer vanishes, prove that l
+
n
n n ~l) (
and y
is
any number such that no denominator
n(n-l)(n-2)
n. ^n(n-l)
terms.
_
y+
l
n(n-l)(n-2)
/.x (in)
each
series being
continued to n 4-
[These results follow
1
terms.
from Ex. 12 by putting successively
r
= l, x=
-(t/4
1)
;
USE OF THE OPERATOR J
386 14.
Show
tfcat
ln + :
1
+ 2)... 1
V
/
,*//*.
i
I\/,M
i
* //
o\
IW/-M
i
-...to n +
-.
_
15. If
>
...
'
16. If
(a-n-f
a(a-
a-b a~6-fl
a-6-fr-l
tti=7
^2=7-77 6(6
b
_
^ rWn=
.
1)
...
(a-n-f 1)
a(a-
...
1)
(a-n-f
1)
e tc., as in Ex. 15, prove that
1)
a-6 a-6-fl a-6-f2 ^ T 6^1 6^2~
[Putting n = l in Ex. 15, find
17.
T;>
+2
prove that
66-16
1
ln '
1)
a-6 a-6 + 1
w
'
o\
..
___
__a~b ~~ a(a- 1) U==
i
terms=
1
--...-
^^
~w
o\/ M
i
Au 0t A*u
a-6-fr-l *
6-r-fl
etc., in
,
succession from the equations
Prove that fI
_? )
__
"*
a
M n
a 6
+
a
)(1
6
6/\
\
j
...
?L 6
Ti-
n(n - l)(n-2) a(a- l)(a-2)
n(n-l)a(a-l)
JT~
(1
6-2/"\
1/\
6(6-1)
6(6-l)(6-2)"
[3
to n-fl terms.
[Using Ex. 16, the left-hand side = ( 18.
Show
that the
sum
n J nw
n =(l -E) u
.]
of the products two together of
1
x is
l)
9
JL JL x + l'
i
x+2
equal to
n(n-l)
(
n-2
"--
I
~+ (n-2)(n-3)
1
"-
[2
[Equate the
coefficients of a* in
Ex. 17 and put
#=
-6.]
to
n ~ 1 term8
1
r
INTERPOLATIONS 19.
In Art.
4,
Ex.
387
has been shown that
4, it
Prove also that
and use each of these formulae to show that B^ = [Equation (D) of Ex. 4 can be written
and
(
^12500 = 1 1 1-803399, ^12510 = 111-848111, ^12520 = 111-892806,
20. Given that
^12530= 111-937483, ^12516 = 111-874929.
show that 21. Given that
show that
sec 89
4'
sec 89
5'
= 61-3911, = 62-5072,
sec 89
6'
=63-6646,
sec 89
7'
= 64-8657,
Bessel's formula gives
sec 89
22. If
Sn = l n +
on
on *
/
L
w
+rz+.-+ ^ 1 i
2 fl .l f
3n
40" = 63-2741.
5' i
i
\n
+
to oo
971
n
r
r
.2 4 .3 + WJ^^ +--+_ ,.,
A
IJL
J"5. = F-i +
or-i
+ LL
(iu) (iv)
(v)
and
. . .
= 5, _,.
J
Use
(iii)
(iv),
S19 SZ9
or (iv) to find 2,
4,
3,
)
S19
showing that for
6,
-T
B
+-
i
L_
(iii)
5,
Qn
rn = l n -"r? + r5-- + ('" I m
(u)
...
7,
^n /e=2, 5, 15, 52, 203, 877, 4140. we have SW =(J + l) n and Sn ^ = (E - l)n S on
23. If
+
L_
Sn = -Sn_ 1 SB = 5 ^ 1 n = l,
[(iii)
....
d
I
or~i (ii)
prove that
,
I
i
= Tn.!
Using
(ii),
show that
for
n=l,
rn .e=0,
2,
3, 4, 5,
-1, -1,
2, 9.
to
.]
<, prove that
CHAPTER XXIV CONTINUED FRACTIONS 1
Definitions.
.
Any
(1)
expression of the form .................................
i-: called a continued fraction,
is
The
and
-
Oj
may be supposed to be attached fraction may be expressed in the form
- Ui ~ F-a + JU
where the
and
a's
ai
^2/a 2
by
stopping at ,
b n /a n
.,
.
6's ...
&3
&2
~
"
is
,
convergents are
av +
,
1
is
.............. (B)
and so any
6's,
........... .......
called a convergent.
flu
. . .
~
..
........
(C) VI \
be positive or negative numbers. The quantities are called the elements of jF, and the fraction obtained
,
second, third,
3
may
any particular stage
,
bn
"...
1
--
to the
signs
continued
It
a
written
is
obvious that the fraction
is
6<> "
a:
,
H
6<>6q----
. . .
,
#2
unaltered
Thus the
first,
.
2^~ ^3
if
6n
,
an
,
b n+l
(u
= 2,
3, ...)
by any number k or if the signs of 6 n a n and 6n fl are Thus any continued fraction may be written in the form
are all multiplied all
(C)
changed. where a 2 a 3 ,
Ex.
1.
Prove
,
are
...
,
that, if
all positive.
JLJLA
2.
fraction
on the
left is
*
elements,
*
l
1 .
4+ 4+
4-f
The
n
each fraction contains
2+ 4+
to s equal ^
4+
2-h
1122 ---- -
-
2+4+4+44-
Formation of Convergents.
(1)
...
""
=x
1111 -
~
x ^ 2+4+2+4 +
Let u n denote the nth con-
vergent of the continued fraction, _.
62
60
a 2 + a3
We
shall write
+
bn
.
w=^) 1 /
a,
J
an + 1,
,
so that
w=^22
fi,=r 1 1
where
J a2
a2
THE RECURRENCE FORMULAE Observing that u3
389
be obtained from w 2 by changing a 2 into
may
a2 4- 63/03, we have u3 =
(
and so u$ = psjq& where y3 - a3j> 2 4- 63^ and Proceeding thus we can show that, if p n and values of n by the equations
? =
?-! + &nPn-*
Vn
?3
==
q n are defined for successive
= anqn -i + b n qn _ 2t
.................
(A)
the nth convergent of jP. p n/q n Equations (A) are called the recurrence formulae. It will be found convenient to write = 0, ............................. i and
then
is
yo==
and
it
will
Pz^^Pi^^Po an d
be seen that
important to state
For
Definition.
#Ae quantities
?2 ==a 29 i r
rather more precisely
all this
with the initial values
If F. convergent of
pQ = l 9
p n and
qo
=
Pi
9
= ^i>
q\
:
q n are defined in this way, then
,. .
.1
II
'
I
J.
I JL
I
............... (A)
,
=1
Assume that this holds for the values 2, 3, Proof. the nth convergent by u n , then
Now p
so
the continued fraction
p n and q n are defined by the equations = a n?n-l + 6wyn _2 Pn = ttfiPn-l + ^n-2 ?n
Theorem.
+ ^25 o> r
n^2.
equations (A) hold for (2) It is
(B)
?0
...
p n /qn r of n,
is the n-th
and denote
f,
also u r may q r _ z are independent of a r and b r be transformed into u r+1 by writing ar + br+1 / a r+i for ar therefore r
_v
gv_
j,
pr _z
,
;
,
(
. Ur+1
r
+
Hence
br+l
^ \
w^
,
,
Thus the theorem holds therefore for
w = 3,
4, 5,
TPr ~ Z
P '~ l+
...
=
-
_ 5^i (r^i
-
r +-'
for
n = r-fl;
in succession.
but
it
holds for
n = 2,
and
INFINITE CONTINUED FRACTIONS
390 (3)
For the fraction
p n and
q n arc defined
b
by equations
Again, the fraction
b2
ft
is
bn
b
(A) with the initial values
b3
obtained from the fraction in
by changing the
(2)
signs of the 6's, so
that the recurrence formulae are
Pn = fln^n-l
~ &nPn-2>
?n
= a n?n-l - 6 n?n-2'
In the following articles we shall call the reader's attention to certain types of continued fractions which will be considered in detail later. Ex.1.
Prove that (I-
x)-^l+~ |j 3^ y
Calculating the convergents of the continued fraction,
we have
2.1-3&.1 6-8o;~
which provea the statement in question.
Notice that no fractions are reduced
to
lower
terms, until the final stage.
3.
Infinite
Continued Fractions. ,_ If
JL'
/I t*i
I
T"
694
"
With regard
__6q?_
to the fraction
6n )
-
if
the
number
of elements is finite,
F
is
called a terminating continued
fraction.
We
may, however, suppose the b's and a's to be determined by a rule In this case F stands for an is no last element.
of such a kind that there infinite
continued fraction, which
regarded as defining the sequence of
is to be
convergents Pl/
?2/?2>
>
Pn/
>
and we say that
the continued fraction is convergent, divergent or oscillatory this as according sequence converges, diverges or oscillates. is defined as If the sequence converges, the value of the fraction
F
lim Pn/q n
,
and we write jF
= a 1! +
-
--
a2 +a3 +
...
SIMPLE CONTINUED FRACTIONS Let x stand for ^ 2
Illustrations.
after the first is
Again,
meaningless.
,
2
may
x=lJ2i,
leading to
where every element
be treated as a number,
a result which
is
clearly
x stands for
if
333
t000
2T2T2T"* and we assume that x
to oo
...
-=
^ 2
// we assume that x
-3/2.
we have # = 3/(2-x)
333
391
is
we have
a definite number,
x = 3/(2+se);
z2 + 2z-3 = 0;
/.
We
have thus proved that unity, but we have not shown
.'.
that
a value
z=*l
or
has a value,
the fraction
if
'
its
-3. value must be
exists.
In neither of these examples can the first step be justified without knowing that the continued fraction is convergent. 4.
An
Simple Continued Fractions.
_
1
a i T^___ i
_ _ 1
1
a 2 + a3
expression of the form
..."
+
>
an +
where every a is a positive integer, except that a l a simple continued fraction.
may
be zero,
is
called
Theorem. Any rational number can be expressed as a simple terminating continued fraction which can be arranged so as to have either an odd or an even number of quotients. Let
A/B
be the number,
of finding the G.C.M. of r2 ,
...
A
and B being positive integers. In the process and 5, let a v a2t ... be the quotients and rl9
A
the remainders, so that
A^a^B + r^ and therefore
IB
A= #1 + B l ^prrBr
-^
JS==r 1a 2
,
r
Again,
if
>!, we can
r1
,
1
=flo"f -
This process terminates, and
A B
+ r2
g,^
if
an
r,
7-
rr is
,
r
111
-^
write 1
=O+ r 3
etc.,
,
1
7-, etC.
the last quotient,
1
a n ~an -l+
andifa n
= r2a3 + r3
1
RECURRING CONTINUED FRACTIONS
392
Thus matters can always be arranged so that the continued either an odd, or an even, number of quotients, as required. Ex. 1. Express ^g$- as a simple continued fraction with add number of quotients. al
157
.
i.e.
(i)
fraction has
an even and
an
I
JL
=2 +
(ii)
-gg-
a4 =5
i
i
i
number of quotients can be made odd or even as required For instance, the penultimate convergents to 157/68 in the two forms above are 30/13 and 127/55; and it will be found that (127, 55) and (30, 13) are The
NOTE.
is
fact that the
of importance.
the least solutions in positive integers of
68* - 157y = + (See Art. 9
respectively.
Ex.
Use
2.
The
z -x)~ as a continued fraction.
is
giving
*
- x) 2
(1
^ l/2a?
-
1+
"^4/3+"
1/3
-
L
^9/2*+
1/3
2x
2s 3*
+
1
-3/2
1- 2+ -9/2* + l/3 x 2x
2x 3x
~ 1+ last three steps, the
+ - 4/3 + -9/2* +
??.
""
In the
1
3.)
the H.C.F. process to express (1
H.C.F. process
68* - 157y = -
and
1
and Ch. XXV,
ri2 + 3^T*
statement at the end of Art.
1
has been used.
Recurring Continued Fractions. If after a certain stage, the elements recur in the same order, we have a recurring continued fraction. 5.
,
'
The recurring elements form the the non-recurring elements,
'
'
*
recurring period
such exist, form the
if
or the
*
9
cycle,
*
'
acyclic part
and
of the
fraction.
The
cycle is usually denoted
last of the recurring elements, _____ _____ _ _ ~~~ _
.-
-
-
_____ _
-.
~.
1+2+3+4+3+4+ A
by putting
asterisks
under the
* . .
1Q AD
/l ATir'I'f'.Afl VA.\3JLlv Uv7vA
MY J
r4"\7
-
._,....
_.
.
_
1+2+3+4 *
Here 3-f
-
4+
is
first
thus
and the cycle J
-
1+2s
^e acyclic part. J *
*
.
and
SPECIAL TYPES Ex.
is
1.
equal
to
(a
n+1
are the roots of x* -00;
ft
-^n+l )l(
1
Hence show
(ii)
is the
// a,
(i)
(l
+ l/n)a
- ax + 6 of x 2
according as ot^fi or a=j5.
a 2 ^ 46, and that
value
Here pn and qn are given by
together with
p = l, p^^a, qQ
where A, B, A',
E
f
= aq
4n
^ = 1.
0,
a^jS, then by Ch. XXII,
First suppose that
p
(i),
B^a=a+^
=A +B
gvng
5,
Therefore
are constants.
p 1 =Aoc +
=1,
^l/= yw =
thus leading to the
Next
let
n+1
-]8
(
w+1
and
)/(-]8)
fc
n (
n -)8
)/(-^
= a=)S, so that a 2oc and 6=^2, then by Ch. XXII,
A~B~\
,
If
=
first result.
Proceeding as before, we find that n -1 n yn = na ;>n -(n + l)a a 2 >46, then
a,
j3
>
^n /gn = (l + l/n)a.
,
l-W
^
5, (i),
A' = Q, B' = l/a., hence
n
Let |a|>|/?|, then
are real.
qn
j8
/a
n ->0 as n->oo and
'
2
a 46, then a~/? and pn /qn (l+I/n)a-+oc. Hence, in both cases, the continued fraction is convergent and
If
Ex.
2.
//
pn lqn
is the
11 prove that
and
n
is
its
value
is
a.
n-th convergent of the recurring fraction 1
aT 6TaT
If
its
= 0.
Pn = <*
(ii)
prove that the n-th convergent of
that the continued fraction is convergent if
numerically greater root
(i)
or
)
+ 6=0,
393
even, the recurrence
1
'"
'
6-f
p n - (ab + 2)p n _ 2 qn -(ab + 2) qn formulae give
Pn
=t>Pn
Pn-2 = pn
whence eliminating p n -i and pn -3> we have n +2)pn -2
p
(ab
+Pn-t =0.
If n is odd, we have the same equations as before, except that a and b are interchanged, leading to the same result. And so for the g's.
BROUNCKER'S FRACTION
394 Ex.
3.
//
pn lqn
n4h
ta the
convergent of
3a
52
1+ 2+ 2+
2+
1
I
(2w-3)
1)
<
Hence show
this holds
and
that the fraction is convergent
If vn denotes either
if
n
Replacing the
we
n-
write
ft
1,
- 2,
-(2- lK_i=(- l) t;'s
by
p's
n ~1
'
2 ^Ti
that its value is w/4.
we have
or qn ,
pn
"
2+
Pn l3n = l ~ 3+5-- + ~
prove that
and
"
2 in succession for
...
n -1
(2-3)(2n-5)
and observing that
=
jt)
3
...
and
n, therefore
.
^=
!(,-,). 1, we have
^n-i
2n-l'
,
" where
^4 is
"" 4
independent of
Next, replacing the
v'a
n.
by
find that qn
Again, the series fraction converges
-
1
and
=1 1/3
its
n~ 1, we find
and observing that ?n
whence we
1
3
1
Putting g's
1/5
value
A =0.
that
an(^ 5 i
l
(fo
F
we
== l
-(2n- l)7n-i-0,
.3.5... (2n -f-
""'"" ,/. "*
.
,
1.3. ..(2-l)""
- ...
is
1),
leading to the
convergent and
its
first result.
sum
is Tr/4
;
therefore the
is Tr/4,
EXERCISE XL 1.
Find the
first
four convergent (unreduced) of
w JLJLJL* 2+3+4+5 r\
2.
W JLAJL! l-2-3-4' /-\
f
Without calculating the convergent^, prove that 2
3
2+ 3T
5_ 1 1 23 T 5^1+ 2T 3T 4
4
'
3. Express fff as a simple continued fraction having odd number of quotients.
4.
The reckoning on the 85
5.
3- 4-
52 57 5
an
85 104 19
I
this.
Express
(ii)
20
5*
^
in the
form a l
greater than unity. 6. Express (2a -a- l)/(2a*-3a) calculating the convergents. 2
by
an even and
right shows that
1_J_1
52""
Explain
(i)
a * ~ a* in the
...
,
where the
"""
form
-
1+^ /i + /a +
a's are integers
...
,
and verify
INFINITE CONTINUED FRACTIONS Show
7.
that the nth convergent of
JL_1__!
2-2-2- -
2 1 is
+ l)/n, and
(n 8.
Show
395
that the value of the infinite fraction
is
unity.
that the nth convergent of 1
1
1
1
rr^r^r^r is
n,
and that the
9.
infinite fraction is divergent.
There being (n +
1)
elements in
2~2- 210. If
show that
all,
1
'"
2^"
l_(n + l)a?-n nx-n + T
*
a?""
a z >46, the numerically smaller root of x 2 -ax + 6=0
A JL J A a
a
[This follows from Art. 5, Ex. 11.
For the fraction
b
is
equal to
a
#
1.]
b
b
b
a+ a+ a+
"
*
*
a
+
"
'
pn = - a -)/(- 0), gn -( n+1 - j3n+1 )/(a - ]3) where a, ^ are -a# -6-~-0. Hence show that if a, 6 are positive, the fraction is convergent and equal to the positive root of x~bj(a + x).
prove that
(
an
the roots of # 2
12. If
p n jqn
IL
the (unreduced) nth convergent of
and the number of terms tively,
then u n \vn
is
in
>
n,
qn
respectively are denoted
by un
,
v n respec-
the nth convergent of
Hence show that
13.
Prove that
,
x
l
fi\ v
14.
;
_z_ z = _ 1
(
_o/JLJL-i-
1^4^1-4^'
^
\2~2^2-"V"
Hence show the nth convergent of the
first
fraction reduces to 2n/(n
+ 1).
Prove that 1
showing that the 2o
9fy.
Prove that, each fraction containing n elements,
J_ J_ JL JL
15.
_
1- 1+ 1- 3+ 2-
I2
infinite
22
2
J3_
(^-1]
_1
continued fraction
is
11
1
divergent.
B.C.A.
SIMPLE CONTINUED FRACTIONS
396 Prove that
16.
~
' ' '
2a + 2n -I+5= k n a (a + 1) (a + 2) (a + n)
......
'
2a
if
showing incidentally that,
Use
show that the
this to
fraction in Ex. 15
Putting
a~ -m,
where
in Ex. 16,
m J-itJli
a
O
iftl
m is
~~ ju'tTl
~~
9
then
divergent. l)
a positive integer, show that 2
I
~~
x
(w--2)
jw-l) i.
is
!){?! -(a 4- w -
[Here gn -(a + H)g n _ 1 ^=(a + n17.
. .
.
2
Also prove this result by induction. 18.
For the fraction
-
+
1
Hence show that the n
...
prove that r
,
n+
^1_1_
fraction
12
is
13
~ convergent and equal to l/(e
1).
a positive integer, prove
is
n
by induction that 2 11 ~ n 1 "" ?T-~ "2+ w+~2 2"+ 1 + 2 n - -2
[
w"+ w - 1*+
-t-
'
For the fraction I
2
22
1_
r+ 7)
n
-
prove that
1+1+
1"*-
^
1
3a
11 + -
-
fraction
is
. . .
(
where the
,
...
l
_
,
...
3/3
>
.
2.
l *
*
'
~j
except that a x
1
1
then x2) ^3
"1 1 -
are called the (partial) quotients
X2 =
'
are of the form
+ ~~"~" r~T
a's are positive integers,
Here a l9 a 2
71
I)
""
convergent and equal to log
1
at
2
r+
+ -
Simple Continued Fractions xl
(?i-l)
"*
t>
Hence show that the 6.
...
\n-i I
n
20.
-
+
?n
^j
J>n_
in -hi
19. If
~^- _ 2+ 3
Gt
3
+
-
-
a4
be zero.
may and
if
11 ~ --
+ a5 +
are called the complete quotients.
Thus we have Xl
that
is
"~
JL ai f "
a2
_L_JL.
+ *"a n ^+ x n
to say, the continued fraction x l
is
l
obtained
its
from
by substituting the complete quotient x n for the quotient a n
.
n-th convergent
FUNDAMENTAL PROPERTIES
397
PROPERTIES OF THE CONVERGENTS
As
7.
with the
in Art. 2,
initial
p n and
values
mi
!,
)
U/
be seen that q n a2 ... an _ x
by the equations
p = a 1?
,
1
-
,
,
and so
the
*'or
the case of
an
From
9.
and
a n as p n ^
...
,
etc.,
is
.
,
We have p2 >p v and if n>3, p n ~ pn ^ = (a n -
8.
. ,
1
-4-
#o#3
same function of a 2 a3
is the
.
!
,
a2
1
=1
g^
a^a., ~*--3 ~f-o\ -ha, -
a^o-fl -2--
0i1 ,
= 0,
l
/1\ --
The convergents are
of
=1
jo
,
it will
q n are defined
since
y's,
thus
q^q^,
1
)p n
,
-f
pn - 2 and
j) w a/?d g n increase with n, .
infinite continued fraction, they tend
to infinity
in
with n.
we have
the recurrence formulae (A)
Pn ln-l -Pn-dn^fanPn-l +#n -2)^-1 ~ Pn-l (n?n-l +
= This holds
if
we
write n
therefore
Again,
(B)
-(Pn-i9'n-2-? n-27n-l)-
-
1,
n-
2,
.
Pnfr-1-.Pn
..2
in succession for n, also
-i7n
= (-1) w ............................ (B)
we have
P1n-z-Pn-&n =
therefore
From
and
(C)
^} n " ^n ........................ (0) l
(
we obtain the formulae
10. ^uer?/ convergent
pjq n
is
in
its
lowest terms, also
Pn-\ and q n to q n _ v For if the numbers belonging to any (q
7n-i)
have a
would divide 1 1
.
3)
>
From
common
pn ,
the recurrence formulae (A), -
n
prime
to
of the pairs (p n q n ), (p n p n ^) factor g, then on account of equation (B), g ,
1.
Pn-l
is
- --
Pn-l/Pn-2
we have
and
-
=
n
?n-l
The first of these holds if we write n - ] n ~ 2, The second of these holds if we write n - 1, n - 2, ,
---;
-
2 for n.
3 for
n.
.
9
ODD AND EVEN CONVERGENTS
398
1111
and
Also PI/PQ^^I
therefore
JaAh^^J
/n
-,
(E)
(F)
11,
It follows that if
pr/qr pr '/([r ,
a*2
are respectively the r-th convergent of
and
+
an
-r
an +
i*
11
'
a-, n-l +
a,l
where in the second the quotients occur in the reverse order, then
p n '/q n ~p n lpn -i and pn ^ 1/?n-1 = ? n/?n-i '
For by the preceding,
by
and
Art. 10 these fractions are in their lowest terms.
12. It has been shown that
if
p'q', p/q are consecutive convergents
of a simple continued fraction, then pq' ~-p'q =
1.
where q'^q, and suppose that p/q is Conversely, let as a fraction continued with an even or an odd number expressed simple
pq-p'q=l
of quotients according as the arbitrary sign
that p'jq'
For
is the
p" jq"
if
is
-f
or
-
.
We
can prove
convergent immediately preceding p/q.
is
the last convergent but one,
by hypothesis and equation
(B) of Art. 9, f
pq
-p q = pq" -p"q, and f
r
therefore
p(q
-q")q(p' -p")> .
Now p else
is
prime to
q,
for
pq -p'q=
1,
therefore q divides
q'
-q"
or
= q"q'
Also
5'<<7
sequently
13.
and q"^q, therefore q cannot divide q'-q", and con-
= q" and p'=p", which q'
proves the theorem.
Sequences of Odd and Even Convergents. Let w n
then, from equations (D)
of Art. 9, it follows that
Wn-w-i
and
u n -u n _ 2
Therefore u n lies between w n _rl and w n _2 between the two preceding convergents. say, every convergent
are of opposite sign.
>
that
is
to
lies
Now W!
consequently W1
on, therefore
Thus
the
odd convergents form an increasing sequence, the even convergents
form a decreasing sequence, and every odd convergent convergent.
is less
than any even
CLOSENESS OF APPROXIMATION
399
Consequently, as n tends to infinity, both w 2n _! and u2n tend to limits, and these limits are the same, for by equation (D) of Art. 9,
finite
W2 n 7 U*n-l == l/?2n?2n-l-> 0.
Hence un tends to a
limit x such that for every n,
In other words,
and
(ii) its
continued fraction is convergent, (i) every infinite simple value is greater than any odd convergent and (iii) less than any
even convergent.
4. Each convergent is a closer approximation fraction than the preceding. 1
For
let
so that
z'
is
the (n + l)th complete quotient, then
= a n+lPn + ^n-l)/( a n+l?n + ?n-i) z = (z'p n +p n -.i)/(z q n + ?n-])
Consequently
2;'
and
(zg^
and
>? n _i
- j> n ) = "
2
r
'
(
f
and therefore
gn
and 2'=a
...
Pn+l/
Now
value of the continued
1
1
1
= %-{-
z
to the
(2;<7
- y n _!)
n _1
=
2;'^!, therefore
z-that
is
to say,
is
nearer to
^an to
p n/
pn -\l
5. ulwy convergent p/q of a simple continued fraction is a closer approxito the value of the fraction than any rational fraction r/s with a denominator less than q. 1
mation
For
let
z
al
preceding p/j. to z than p'/q'. lies
-\
--
By
...
and
,
Art. 14,
if
let
Moreover, between y/g and p'jq', and
that
?
|
ry'
nearer to z than ^/j, it is also nearer between p/q and p'/gr' therefore r/s ;
1
P'i
r_ Therefore
be the convergent immediately
r/s is
z lies
\~8
p'jq'
i.e.
9
q'
- ^jp' <^ |
;
9'
and, since
ry'
- sp'
is
an integer>
it
follows
16. If
is the
numerical value of the error, in representing the value z
of the continued fraction, by
and within wider For
then
,
limits
u n =p n /q n
if
p n/q n
then by Art. 13,
,
w n> is
ERROR
LIMITS OF
400
U n+l9
^>
^n+2>
an increasing or a decreasing sequence, therefore ~ ~ t*n Un+2 <\ W n *
I
I
Also by equations (D), I
W n - W n +2
whence we have the
I
=nW?n7nf2 and
I
w n ~ w n-h
inequalities (H).
Again, g n + a = a n +29Wi
+ ?n>
therefore
^ ?n +2/a n+2 Also qn+l >q nJ and therefore l/9 n ?n Hence the inequalities (I) follow from (H).
17.
fraction,
Let then
and al
z
l/z 2
that in only
a3
is
Z2
,
one way.
+ -- where a l and Z2
where a 2
can be expressed as a simple continued
positive irrational z
Any
,
...
Z3
,
-a 3 -fl/Z4
z a >l,
it
Zn
...
,
,
be zero),
let
r,
...
,
11
1
may
l,
follows in succession that a 2 a 3
and we have
(which
= <*n+l/Zn+l>
are integers such that, for every
a r <2 r
z
Continuing the process,
z 2 '>l.
= a 2 4-l/03
the integral part of
are positive integers
The process can be continued indefinitely, and leads to the infinite continued fraction ] j =O I + --
F
S
0-2
Again, we have a 2 <2 2
,
# 3
,
-
an
etc.,
j and
al
+
a2
>z,
11
-f --------
a + a3
.
+
therefore 1
1 !
----- ...
...
+
a2
+-">2 2 a3
~ 1
a 2* +
-
a3
-
H-
,
etc.,
1
a4
<2o, -
etc,
EXPRESSION FOR A SURD Continuing thus, F, then
Now
it
can be shown that
ur
if
401 the rth convergent of
is
n tends to infinity, u 2n _ l and u 2n tend to the same limit, which must therefore be equal to z. That is to say, the fraction F converges to z as a limit, and we write as
JL '" JL
1
It remains to
show that
a 2 -h
this
+
an
mode
of expression is unique.
Suppose
that
_JLJ_ The
1
1
_A
1,1
integral parts of these continued fractions are equal, therefore
j and
= 11
a,
i
6,
------
a2
H
a3
+
.
.
= 62 + T
.
Hence, by similar reasoning, 1
a 2 -62
and
... ,
and so on
When
Thus the continued
indefinitely.
z is
=6t 3 + r
1 >
v:
C/4 -r
C&4 -r
fractions are identical.
expressed in this way, any convergent of the continued
fraction will be called a convergent to
18. The value of any number.
infinite
z.
simple continued fraction
For any rational number can be expressed as a
finite
is
an
irrational
simple continued
fraction. Ex.'l.
The
Express (/v/37+8)/9 a* a simple continued fraction.
integral part of this surd
N/37+8 9
"~ -- ~ 1 1 -f ^37-l_, iT "9"
and
is 1,
4
7m
~\
s/37
-
s/37-4
3
4
.
^2+
\/37~3 ~'
:
7 1
4
-
~ "f*
V37+3'
4 N/37~5_ o -r -- + 4 ___ud H--H -7=^ 3 3 V37+5 -i
7
,
"
1 H A
:
4 ------
>
^37-3 =2+-, 4
--
\/37
-7=
V37+4 V37+5
+1
N/37 ,
+r
_
tj
.
=l+eto.
-
7
v/37+3
The complete quotient (s/37+3)/7 has occurred
before,
and the subsequent work
consists of repetitions of the last three steps, thus, with the notation, of Art. 5
is/37+8 9
1
IH-
>
J_ JL_ 1+ 3+ x-
the continued fraction being a recurring fraction.
1
2' -ti-
:
INTERMEDIATE CONVERGENTS
402
a given number x is expressed as a which pr/qr is the rth convergent, then pn/q n is an approximate value of x in defect or excess, according as n is odd or even, and the error is numerically less than I/qn q n +v an(i a fortiori
19.
Approximations.
If
(1)
simple continued fraction of
less is
than l/q n 2
Now
.
2n+i
= 0n+i7n + ?n-i>
hence if a n+1
is large,
p n/qn
a specially good approximation.
Or again, if we require an approximation with an error numerically less than a given fraction I/a, we calculate the convergents until we find one p n/q n such that q n q n+ i ^a. This is the required approximation. Of course, the conditions are satisfied
if
q n ^*Ja.
Ex. 1. It is required to find good approximations to the value of vulgar fractions, given that 77
The convergents
_q
,
TF
in the form of
JL _JL _L _JL. L JL 7+ 15+ 1+ 292+ 1+ 1+""
are
3
22
333
356
103993
I*
T*
Toe*
m'
33ioT'""
The denominators 106, 33102 are large compared with the preceding ones, therefore 22/7 and 355/113 are specially good approximations, and they are both in excess. For 355/113, the error is less than 113x33102
3740000
20. The problem may be of such a kind that no suitable approximation can be found among the convergents (as in Art 22, Ex. 2). We may then proceed as follows Choose two convergents p/q, p'/q, :
one an odd and the other an even convergent. integers, the fraction
;
mq + m'q'
and p'/q'. Also x lies between these convergents, a closer approximation than p/q, and perhaps also than P/Q and it may be possible to choose m, m' to suit the conditions of
between
therefore p'/q'
m, m( are positive
p mp + m'p' Q
lies
If
p/q is
the problem in question.
21. The most important fractions of this kind are the so-called Let p n/q n be the nth convergent of
mediate convergents.
1
Consider the sequence
Pn
Pn+Pn+l
the (r-hl)th term being
PrjQ r
where
Pf
inter-
APPROXIMATIONS The terms
of this sequence (excluding the first
p n /q n and
intermediate convergents between
possesses the following properties (i)
It
or even, (ii)
(iii)
and
last) are called
Every
Any
PT jQr
fraction
fraction
which
is
the
The sequence
;p n +a/?n+a-
:
is an increasing or a decreasing sequence according as n and the denominators form an increasing sequence.
is-
odd
in its lowest terms.
lies
denominator greater than either (iv)
403
Pr/Qr
between
Qr
and
Pr+1 /$r+1
has a
Pr/Q
as the
or Qr +v
Limits to the numerical value of the error in taking
r
value of x are given by
For we have
P r+l Q
r
whence the statements
- PrQ
(i)-(iii)
PrIQr> is
either
r+l
=pn+l q n -p n q n+ - ~ l) w+1 i
follow
:
Pn+2/q n +2>
Now
x
2W]/?n+1>
>
an increasing or a decreasing sequence, therefore
r
Qr
(iii)
all
by writing down the odd cqnvergents
the intermediate convergents, then
an increasing sequence of fractions in their lowest terms, and a closer approximation in defect to x than the preceding. The denominators form an increasing sequence. This
is
each fraction (ii)
this proves the statement.
,
consider the sequence formed
and inserting (i)
,
and because
Qr Since
(
If
is
P/Q, P'/Q' are successive terms, any fraction which greater than Q or Q
lies
between
f
them has a denominator
.
Again, the sequence
formed by the even convergents and their intermediates, has the same properties as the preceding, except that it is a decreasing sequence in which each term is a closer approximation in excess to x than the preceding. 22.
Problem
denominators
less
It is required to find the fraction which, of all those with than a given number a, is the closest approximation (in .
defect or in excess) to
a given number
x.
APPLICATION TO CALENDAR
404
To do
we
express x as a simple continued fraction, and calculate the convergents p^q^, pdq& until we find p n/qn such that this
q n ~a, then
If If
we
q n <.a,
is the required fraction. consider approximations in defect.
p n/qn first
Taking the
sequence (A) formed by the odd convergents and their intermediates, then P/Q P/Q, P'jQ' such that Q^a
find successive terms
;
the closest approximation in defect of less
than
all
we is
fractions with denominators
a.
For if h/k is nearer to x than P/Q, then h/k must coincide with a term to the right of P/Q in the sequence, or else it must lie between two
and
suc.h terms,
in either case
k^Q'>a.
In the same way, by considering the sequence (B) formed by the even convergents and their intermediates, we can find the fraction with denominator
two
Ex.
As
less
than a which
fractions, the
Find
1.
Art
in
18,
the,
is
the closest approximation in excess. Of these is nearer to x is the number required.
one which
fraction with denominator less than 500 which is nearest to s/14.
Ex.
1,
we can show that
N /14
=3 +
--
-
-
- -
1+24-1+6
,
and we
<,
.
,
t
P!
,
find that
~ 333
q7
89
p B - 449 , '
q8
120
p-9 = 3027 -
,
qg
,
*
* .
809
For the approximation in defect, we insert intermediate convergents between p 1 lq1 and p 9 /q9 These are of the form Pr /Qr -= (333 + 449r)/ (89 + ] 20r). .
The conditions $ r ^500<(> rfl give
r---3, and 2\IQ 9 -- 1680/449. For the approximation in excess, we consider the convergents 8 /io/
,
?/& =449/120. -
2
Finally, we can show that 14 (1680/449) <(449/120) so that the required approximation is 1680/449.
Ex.
2.
The mean tropical year
convenient method of correcting
We have
is the,
-
14,
It is required to find a practically calendar, the year being taken as 365 days.
365-2422642 days.
*
<
2
1
'
'
and the convergents are 1
1*
4'
7
29*
j*
33'
39
j47 f
161"
194'
321
1325""
Take 1/4 as a first approximation. This, amounts to adding 1 day every 4 years, and is an over -correction. The fraction 7/29 is inconvenient but 8/33 suggests an addition of 24 days in every 99 years, i.e. make every fourth year a leap-year, except at the end of a century. This is also an over-correction. ;
CHARACTER OF APPROXIMATION
'
405
None of the other convergents give convenient approximations ; and the same is therefore proceed true for the intermediate convergents in the first few intervals. choose the convergents 1/4 and 39/161, the first an oven and the as in Art. 20.
We
We
second an odd convergent
if
then,
;
m,
m
f
are positive integers,
m. l4-m
1
x
.39
39
ra.4 + m'. 161
161
'
4
The values
m'^2
??i~19,
give the approximation
19.1+2.39
97 *
+ 2. Tel "400
19. 4
This gives one more day to be added in 400 years than the fraction 8/33 i.e. it makes every fourth year a leap-year, except those at the end of a century, with the further ;
correction that every fourth century is to be counted as a leap-year. The fact that this approximation, 97/400, can be obtained from the convergents 1/4 and 47/194, which are both in excess, shows that this also is an over-correction.
The
error
is less
than a day in 4000 years.
MISCELLANEOUS THEOREMS
23 Let p/q be a fraction in its lowest terms, which is an approximate value some quantity z. It may be important to know under what circumstances .
of
p/q
is
a convergent
The answer to
to z.
this is contained in the following
Let p/q be expressed as a simple continued fraction with
number of quotients according as p/q
~g
and
z,
an even
or
:
an odd
f
let
p
/q' be the last convergent
but one. !
z
If J
~P
I
\
<
q\ then p/q is
------
^
q(q + q)
a convergent
For suppose that
1
V
q
\
to z.
11
p= a l + ---
...
an
#2+
even or odd according as p/q
is
i
or a fortiori if J
9
q
where n
P -
1
"7
g
1
^a-i
,
z,
1
and
let z'
be determined by
1
a
prove that z'>l, for then z' can be expressed as a simple continued fraction in which the first quotient is not zero. We have It is sufficient .to
*=& +q
andthcrefore
Now
f
pq -p'q^I
f
q(qz
4- q')
is
positive
or
1,
according as p/q
qz'
q(qz L
z
;
+q)
therefore in both cases
and
____ q(qz'+q')
Hence
'-*-&-:** q
qz
\q
+ q'>q + q' and z'>l, which proves the theorem.
SYMMETRIC FRACTIONS
406 24. If
pr/
which the value
For
if
k
is
Therefore
is th6
r~th convergent of the simple continued fraction of
then z 2
is z,
+ Pn-l? ~PnPn-l (*? + ?-l) 2 ^ left is
25. // p/q is an even convergent, and continued fraction whose value is z, then according as p/q precedes or follows
X/Y.
ately precedes
than p/q or
else
X/Y
is
even or odd, for
let
X'
-
p~
p'/q' precedes p/q. with or coincides else X/Y
is it,
pX
9
be the convergent which immediately it occurs later than
let p'/q'
// p/q follows X/Y,
an odd convergent, and therefore
x _/
7 since p/q is
in the sequence of convergents.
an odd convergent immediately preceded by X'/Y',
Then
p
|<*
an even convergent immediately preceded by
p'/q',
and
z*>
qY
qq
26.
is
Then X'/Y'
XX'
and
n
X/Y an odd convergent, of a simple
X/Y
X
(ii)
0.
X'/Y' be the convergent which immediis an even convergent, and it occurs coincides with it, therefore
// p/q precedes X/Y,
since
even or odd.
and qn >qn -i, which proves the theorem.
Pn>p n -i,
and
is
equal to
this is positive or negative according as
later
n
according as
z^^n-i/g^n-i*
The expression on the
(i)
according as
the complete (n-f l)th quotient,
Jrfn-l ( kPn
and
5 PnPn-il^n-v
Symmetric Continued Fractions.
A
simple finite continued
which the quotients equidistant from the beginning and end are equal, is said to be symmetric. For instance, the fractions, fraction, in
77
^
.^:
4+l-h4+3'
132
are symmetric, the former having an odd latter
an even number.
_L__
4+1+IT4T3' number
of quotients,
and the
SUM OF TWO SQUARES The
is
following
407
an important property of a fraction of
with
this kind
an even number of quotients. /n Let r , P - = #!
H
(1)
j and
...
1
^=
ar + a r + a r _ +
q the fractions
P/Q
p/q,
being in their lowest terms,
the convergents immediately preceding p/q
P=p*+p'2, For by Art.
11,
therefore
P/P'=P/Q,
f
let
p'/q
,
P'lQ' be
and P/Q, then
= q* + q'* and
Q'
and
ax
Q* +
P ~Q,
l=
also
therefore
p
pq+p'z" /
ar + q
X=p* + p' 2
If
therefore
and
Y^pq+p'q',
any common factor
to one another, therefore
therefore
##.
1.
X/Y ,
PQ -P Q = (~l) 2r /
//
pr /qr
/
is the r-th
r
+ q'
we have
X and Y divides p and p', which are prime
of
is
P = X and Q = Y and
Moreover,
= pq+pq2
in its lowest terms.
ff
Q~
consequently
and
P/==Q,
therefore
1111 _____
i
"
._
at+'" ar + ar +'" a^
prove that the remainders in the process of finding the G.C.M. of Q,
P=p 2r
is
convergent of the continued fraction
P
For
So also
P are
Q=p* r -i
,
with sinaikr equations, the last being p 2 =a jp l + l. Hence the G.C.M. process is that shown on the right.
&r.
2.
Given that 218 2 -f
1
=25
.
1901, express 1901 ow fe
The reckoning shows that 1901
+ _LJL_L J_ JL J_I 1+ 2+ 1+ 1+2+ 1+ 8*
218 Hence,
if
1901
p r /qr
=^a
2
is
the rth convergent,
+z> 4
2
= 26 a + 35 2
.
(See Ex. 1
and Art.
26.)
*ww
of two squares.
P/Q,
SUM OF TWO SQUARES
408
27. Application to the
any factor of
positive integer,
Theory of Numbers. // Q Q 2 + 1 can be expressed as the sum
is
any
of two
squares.
For
let
Q 2 + l=PR
P>R,
where
p -=
l
Therefore
P>Q
is
P'fQ'
divisor of
p P= a 2r 4
Consequently ^ J 7;= P
Q
a 2r _ 1 = a 2
;
/A* (A)
the convergent immediately preceding P/Q,
P(R-Q') = Q(Q-P') Q-P' or else Q = P'.
.1
...
l
and P>P', therefore
n
..............................
and
l
0,+ ----
Q if
is
P>Q
Let
prime to Q.
then
P
then obviously
P
and, since
The
-1
a 2r _ t
...
11 --a 2 4- a l
4-
and the continued fraction
is
is
it
a
impossible, for
R = Q'.
and
Q==P'
prime to Q,
is
alternative
first
therefore and1*1.*
,
a 2r = a l9
is
symmetric. Denoting the mth convergent of (A) by p m /qm we have, by Art. 26, ,
etc.,
,
which proves the theorem.
28. Conversely integer
Q
--
if
x2 + y~ where x
can be found so that Q*
z 1 T Let - = a r 4 a y r-1 then,
P
if
9
pml^m
is
1 .
.
the
.
.
ax
-h
,
and
+1
X = a, --
4-
.
a r-1
.
.
ax
4-
=p
Using Art. 28, we have
2
?
= 46 2 4- 39 2
o9
Q
therefore
,
2
an(j
4-
i ^ ^ 3 54-14-14-
,
4-"
1027 -
is
LA
l
1 -f 1 4-
3637 -
f
3
we have 3637
,
x 200 - 1027 2 =
1,
10/27
(1027, 290).
the last two convergents of 3
290
and writing down the continued
,
<
and the required solution therefore
^=^2^.
find a solution in positive integers of the equation
= l4-r
of which the last two convergents are
;
= Pq^r-v
1
JL _ X -1 JL _L 1 1 4- 1 5+ 1 4- 1 + 5T
:
ar
2
fraction
Or thus
4-
to the last continued fraction,
Given ^Aa^ 3637
.
1
----
ar
mth convergent
Also ^L^r-i ~ Qpzr~\ == 1
,
111
. . .
a2 +
Q
r
1.
by P.
1
4------
y and x>y, then an
to
prime
is divisible
Z=y 4-^ i==a;2_h2/ Ex.
is
+ 1-
F
-
--
are TT ancl
and x = >/(3637 290- 1)=: .
i
o
Therefore
RECURRING CONTINUED FRACTIONS
409
29. If x is prime to y, any factor of x* + y* is the sum of two squares. For we can find an integer Q such that x2 -f y 2 is a factor of Q2 + 1 and every factor of
Q2
-f-
1
is
the
sum
two squares.
of
30. Simple Recurring Continued Fractions. grouped as follows (i) (ii) (iii)
These
may
be
:
Fractions which have no acyclic part.
Those with an acyclic part consisting of a single quotient. Those with an acyclic part consisting of more than one quotient.
The reasons
for this classification will
and we proceed
of fundamental importance,
later.
appear
The conclusions are
to consider these types in
order.
31.
No Acyclic
iT
Part. U-t
(X>
T~ 7
1111 .
7~~~
&2+ b3+
................
7
7
b n+
&1+
&2
the fraction having no acyclic part, then a with integral coefficients.
1
7
Moreover, if
j8
+
(**)
a root of a quadratic equation
is
second root of the equation ,
is the
then
where the order of the quotients
is reversed,
whence
it
follows that
-1
l/i
T" 7
'"
62
therefore
and
a
in the
is
~"
-fractions (A), (B)
by p r/q r and
'
.
+
T
b n +oc
the positive root of
same way the
fraction (B)
is
the positive root of
Jn'zMK-CJ*-^--!^ which by Art. 11
is
'
the same equation as
Now
equation (D) can be transformed into equation (C) by writing Therefore - l/j3 is the positive root of (D), and is conIfx for x. sequently equal to the fraction (B).
-
'Also, from (B),
/?
is
negative,
and
-^>1
;
from which
it
follows that
STANDARD TYPES
410
A
32.
single Acyclic Quotient. l
siw/Ze quotient
l
]
a l does not recur, then a
equation whose second root cannot
and -
between
1,
11
11
then x-a^^I/y, and
lie
a root of a quadratic
is
if
pr/gv
is
1
1
the rth convergent of the second fraction,
as in Art. 31,
^-(Pn-ffn-Jy-Pn-l-O,
(
Denoting the
a'^ + l/a, The root
a'
(B)
pn -i(x-<*i) + (Pn-q n -i)(x-<*i) -?n = ................. c ) roots of (B) by a, j3 where a is positive, and the correspondby a', /?', we have 2
and therefore
ing roots of (C)
........................
is
'
= 04 + 1/0.
the value of the fraction (A), and by Art. 31
where /is a positive proper fraction. Now a l and b n are unequal, for
if
we have
a l = 6 n then a a would belong to
the cycle, whiqh is contrary to the hypothesis. Consequently /?' = /--/ or -I -f where 7 is a positive integer, and therefore /J' cannot lie between
Oand -1.
1111111
More than one Quotient
33. // XI
OC
_ -n
l
Uf-l
"1
Acyclic.
in which at least two quotients do not recur > then a is a root of a quadratic equation with rational coefficients whose second root is positive.
- -111 +6 +6
,,11-
1
Let x = a t +a2 +
and
Pr/Q r
let
...
=
am
>
2
1
+
...
and
t/ y
= 0i +-
7
62
and p r/q r be the rth convergents
+
...
=
bn
of the first
+
...
,
and second
fractions respectively, then
2
and
?n!/
If a,
/J
are the roots of !
A ___ _ t'lii /.
i
1
__
11
_____ i
-(F-?-i)!/-yn-i =
......................... (C)
_
a being positive, then by Art. 31,
(C), O
j
Tlfi
<*l-l.vl
1 -.
^
i
A) t/
n "
~ i
1
_
7
.
. .
_11_ ,
r
...
.
I II l ...IJL/I v '
SUMMARY OF RESULTS
411
eliminate y from equations (B), (C) we obtain a quadratic in x with rational coefficients, whose roots a', /?' are found by substituting Thus a' is equal to the fraction (A) and a, ft respectively for y in (B).
Also
if
From
we
equation (D)
it
+ l/p=a m -bn -f, where / is a not equal to b n for if a m = bn
follows that a m
Now a m is positive proper fraction. then aw would belong to the cycle, which It follows that am + l/j8-/-/
where I
is
_ 1
1
.
and
since a /m __ l
fore also
/?',
is
is
or
contrary to the hypothesis.
-/-/,
Thus
a positive integer.
Um - i+
,
is
1
__
ft
m - i+ I
1 fYf or
________
ft
-1
a positive integer, the expression on the
NOTE. The argument depends on the existence of a m _ l9 and than one quotient in the acyclic part.
34.
left,
and
there-
positive in all cases.
Summary.
It has
fails if
there
is
not more
been proved that any simple recurring con-
tinued fraction is a root of a quadratic equation with rational coefficients. Also the second root (3 of the equation is restricted as follows : (i)
If there
is
no acyclic part, then
1
(ii)
If the acyclic part consists of
a single
(iii)
If the acyclic part contains at
least
Ex.
1.
Find
the value of 1
JL
~ o
-
+~f~
JL
~f"
quotient, then
/?<
-
1
or j8>0.
two quotients, then j3>0.
-
.
~}~
*
*
Denoting the value of the fraction by xy we have
^l + _. i
1
1
where
^ = 1+ ^
and y = (s/37
giving
+ 3)/7
and
__.. 1
1
1
;
y=
*=
EXERCISE XLI 1.
provided that 2
D
---
For the fraction
|
x
\
<|
(
...
- a +
,
prove that
(See Ch.
pn
XXII,
q n -i and
2.)
B.C.A.
COMPARISON OF PHYSICAL UNITS
412
----
For the fraction
2.
(ii)
3.
l>n?n-4
-S /iJPn-4 =
-
f
w l)
(
...
-JT4-1)^ CL '
1
prove that
~Mnn-ian.-2
4
2+
:
1
1
2a
+ 2+
1
-
...
~
.111
/-T\<2(
.
'
1
1
1
1
Prove that /
n
1
A
,~
2a -f 2a
va 2 ~-2 5.
,
Prove that
Deduce the following
4.
... ------
2a -f
-f
1
a-
1
1
Prove that (i)
(n)
6.
Ex.
Show
a -f
f
1111 r~ *T
&4-c-f6+2a
that 449/120 differs from
//
A/
V\
a
1\/
+7
(
c
a
6/\
^14 by
less
+ r~"7i
\
6c-{-2/
than 1/90,000.
(See Art. 22,
1.)
A
7. straight ruler is graduated in inches and centimetres, the ratio of an inch to a centimetre being N. If is expressed as a simple continued fraction and p/q is any convergent, show that the distance between the graduations p centimetres and q inches is less than the distance between any two preceding graduations corresponding to a convergent of the fraction.
N
4
'
'
'
8. Find the fractions which, of all those with denominators less than 500, are the closest approximations (i) in defect and (ii) in excess to 0-2422642. (See
Art. 22, Ex. 2.) 9. Given that 1 metre 3 -2809 ft., obtain the two approximations, 8 kilometres 5 miles, 103 kilometres 64 miles and find an upper limit of error in each case. ;
Given that 1 kilogramme = 2 '2046 pounds, show that 44 kilogrammes slightly greater than 97 pounds, the error being less than half a grain. Obtain also the usual approximation, 7 grammes 108 grains. 10,
11. If p/q,
p'lq',
is
p"lq" are consecutive convergent^ to a simple continued F lies between *Jpp'/qq' and *Jp'p"Iq'q"-
fraction F, prove that
SPECIAL TYPES
F=
For the fraction
12.
Also
-.
------
...
-i
a+ b+ a+
b
2 p are the roots of x - (ab 4- 2)x -f
if a,
ap 2n = bq 2n __ l ~abd n /d l
(iii)
e^a& + 2, and
(v) if
cc
2
1
(iv)
;
< |(
that prove r
,
+
413
~
and dn = a n - /_*", then
^ 2 n4i = _/2n
:r::
(^n-|.i
-dn )/d
1
;
-c-f \ 7c 2 h4), then
2
l+frr-o: _ - CX 2 -f iC 4
1
1 -f fix
[If w n
stands for
un =-p 2n
If
or for
y; 2n
we have
,
- ^2
.4 n-
>
2 n-i>
then, by Art. 5, Ex. 2,
B~ p ~ 0,
Ay.\Bfi~p 2 ~b,
Q
-4~ -B=6/c/ lf
giving
w~^? 2 n-i> since c^a-f-^, then Ja
If
^=(l--i)/^,
giving
Finally,
it is
and then the 13. If
p^n^d^d^
/.
B~-(l-fi-*)ld l9
results in (v)
# n /gn
is
can be obtained by equating
the ?ith convergent to ------
+
2
the numerator of the (n~5)th convergent to
rt
that
the
fraction
a
----a- a-
then
...
n
n
Prove
coefficients.]
--->
-f
Pw-i^s-i-gn-iPs-i
14.
1,
d fi-i)/ d ij) 2w -i-=(^n-
:.
easy to show that
i
is
Aa. ZjrBfP~p z -=c-
1,
\~lJf3=pi
-
+ a n-l +
----
a- x
,
n_2
+
....
which a
in
is
equal
- 1
and is repeated any number of times, must have one of three values, to and that if x satisfies the equation 2# 3 -f 3# 2 - 3# - 2 ~ 0, the fraction satisfies this equation. 15. If n L
H
--a2
. . .
-f
terms, prove that
---#r-i
$2 -
1
.
. .
+
a r + a r-i
+
is
divisible
by P.
2
+ a
i
P, Q, R are positive integers such that can be expressed in the form
16. If
P __ _ Q and
if
Verify
pm jq m
is
1 i
^,
x
a,+
'*'
111 _
__________
a r_t
=~ V
and P/Q
Q~-l=FR
...
+ a r + a r _ t -f
'"
11_ _.'
the rath convergent, then
^ =Pr (Pr +Pr-i)> G =.Prfr-i +Pr-i
R = V2r = 3r-JL
in its lowest
and JP>JR, then
_______
cu-f
is
(
CHAPTER XXV INDETERMINATE EQUATIONS OF THE FIRST DEGREE In
this chapter a,
6,
stand for positive integers.
c, ...
Underthis heading we
consider integral solutions, and more particularly positive integral solutions, of a single linear equation in two or more variables or of a system of linear equations in n variables (m
m
2.
that
A Single Equation of the Form we need only consider the forms ax
If either of
any common
by
by
have a
b
a,
division.
prime
to
common
Hence and we
b,
(i)
To find a
by c.
integral values of #,
Hence
if a, b have
t/,
then
a common
no solution in integers exists. divisor which divides c, this can be removed no
assume
a
of generality in supposing that
loss
is
this to be the case.
The Equation ax~by = 1
3.
obvious
c,
there is
shall
It is
ax -f by = c.
and
these equations is satisfied divisor of a, b must divide
divisor which does not divide If
=c
axby=c.
(a prime to b).
solution in positive integers, express
as a simple
a/6
continued fraction with an even number of quotients (Ch. XXIV, 4). The last convergent is a/6. Let p/q be the convergent immediately preceding (Ch.
a/6.
XXIV,
9),
Then since and therefore
is
a/6 (q,
p)
is
an even convergent,
(ii) To find the general solution in integers, suppose that solution in integers, then
ax - by = Therefore since a
where
prime to
is
t
a(x~~q)
is
This
is
6,
x-
q
Hence a(x-q) must be divisible by
zero.
y~p + at
is
(x, y)
any
=aq - bp.
= b(y-p).
an integer or
x = q + bt,
1
aq-bp~l
a solution.
is 6.
divisible
by
Consequently
Hence where
= 0,
called the general solution in integers.
1,
6,
2,
etc.
and
GENERAL SOLUTIONS Since
(iii)
that (q,p)
We
a convergent which precedes a/6, it follows that and from the form of the general solution we conclude is
p/q
and q
p
only solution in positive integers such that least solution in positive integers.
is the
shall call this the
Since
(iv)
integers of
'
p
and q
'
a(b-q)-b(a~-p)= 1, the ax -by = -1 is (b-q,a-p).
The Equation ax-by = c
4.
415
'
'
solution in positive
least
(a prime to b).
(i) To find a solution in positive integers, express a/b as a simple continued fraction with an even number of quotients, and let p/q be the
convergent immediately preceding
aq-bp = l Hence (ii)
(qc,
pc)
To find
is
Then
a/b.
aqc~bpc = c.
and
a solution.
The equation may be written prime to 6, x-qc must be divisible
the general solution in integers.
a(x-qc) = b(y-pc) and
since
9
a
is
Therefore (x qc)/b = (y-pc)/a by Hence the general solution is b.
x = qc + bt, Let
where
t,
y = pc + at
= 0,
(
is
t
an integer or
zero.
2, ...).
1,
m
be the integral part of the smaller of pc/a, qc/b 9 then the in positive integers is (a, /8) where a qc- 6m, ft =pc - am, the general solution in positive integers is
(iii)
least solution
and
x=oc + bt,
Thus Ex.
1.
the equation has infinitely
Find (i)
We
y~p + at many
(J
= 0,
1, 2, ...)
positive integral solutions.
the general solution in positive integers of
13.r~l7y = 5:
(ii)
13z-17y= -5
(iii)
;
13*-172/=996.
find that
l^o+J--^^*-!---^-!-! 17 1+3+4 1+3+3+1* The convergents
of the continued fractions are
f
1
3
13
i'
I'
n
13.4-17.3 = 1,
and
Hence for
the first equation
* = 20 +
The
1
P P
;
a solution 17*,
3
10
13
4'
Is'
IT
13.13-17.10=-!. is
The. general solution in integers is
(20, 15).
y = 15 + 13*
(*=0,
1,
2, ...).
-
solution in positive integers is x = 3 -f 17$, Similarly for the second equation a solution
1) is (3, 2), and the general by y = 2 + 13* (t 0, 1, 2, ). is given by # = 13. 5 = 65, y = 10.5=50,
and the general
is
least solution in positive integers (given
solution in positive integers
t
. . .
NUMBER OF SOLUTIONS
416 In case of
the third equation, the
be shortened thus
may
reckoning
:
Dividing 996
by 13 (the smaller of the coefficients), we find that 996 = 13 76 + 8 = 13 77 5. Hence the equation may be written 13 (# 77) ~\ly -5, and by the preceding the least positive integral solution is given by x - 77 = 14, y 11. .
Therefore the general solution in positive integers
s = 91+17*
is
y = ll + 13* (=0,
f
.
1,2,
...).
c can be made to depend on that NOTE. In this way the solution of ax - by of an equation of the same form in which c is less than or equal to the smaller of $a, \ b.
The Equation ax + by = c
5.
Let
General solution in integers.
(i)
prime to
(a
continued fraction with an even number last
Then aq-bp =
convergent but one.
b).
be expressed as a simple of quotients, and let pfq be the a/b
\
and the equation may be
9
written
ax + by = c(aq -bp) a
Since
prime to
is
a(cq-x) = b(y + cp).
or
cq-x must be
6,
t is an integer or zero. Hence the general solution in
divisible
by
6,
and therefore
where
x = qc-bt (ii)
integers
y^at-pc
9
(2
is
= 0,
The values
Positive integral solutions.
2, etc.).
1,
of
t
(if
such exist) for which
x and y are positive are given by the conditions, pc/a
then
y
x,
Hence
m+f,
qc/b~n+g,
be positive integers if ... n. is the number of solutions in positive integers,
will
if
N
b
Now
0
where
-1
,
a
ab
and therefore
ao
-..
ab
not divisible by ab, the number (N) of solutions in one of the integers next to c/ab 9 and is greater than or positive integers than less c/ab according as f^,g.
Hence, if
c
is
is
If
c
is divisible
by ab
9
/=0,
then pc/a and qc/b are integers
= 1 and N--.-L ao
;
therefore
THREE UNKNOWNS Ex.
Find
1.
13*
and the
+ 17^3000(13.
4
4 - 17
.
-17.
_
Moreover,
3
.
= 1, and
may be written = 13(12000 -*) 17 (y + 9000)
or
3)
the equation
:
x~ 12000 - lit, y~l3t- 9000.
by
_
9000
15
_. c =706
^,
3000
13
any
also
4
15
/=Yo LO
^"i^ Li
-693 +
1
13,
^ nat
so
= 093,
by
694,
...
These
705.
hence the number of
/<#
this result agrees with the preceding.
Equations with Three Unknowns.
may
exist,
1.
<14;
Since 705
13.
The general
(1)
Ex.
is
Two
6.
-
XO.JLV
solutions
if
<~'--
4
^692 ^
therefore the positive integral solutions are given solutions are (15, 165), (32, 152), ... (219, 9).
Now,
3000.
that there are 13 such solutions.
integral solutions are given
HT
+ 17/
13#
the positive integral solutions of
Use the rule just given, to show As in Art. 4, Ex. 1, we have 13
417
Find
solution in integers, and all the positive integral solutions, be found as in the next example.
all the positive integral solutions of the
2x -
= 7, Zij + 3z
pair of equations,
- 4# + ly + 3z = 1 9.
4 ; and tho general Eliminating one of the unknowns (z), we have 2x 3y = solution in integers of this equation is #~3w-2, 1, 2...). y--2m, (in~0, Substituting these values of x, y in one of the given equations, say in the first, we
-
find that
2m + 35 = 11 and the general solution in integers of this equation is z = l+2, and therefore a;^=3/-2 = 10-9, y~2w = 8-6*. ;
M = 4-3t,
Hence the general solution in integers is x 10 - 9J, y = 8 - 6^, 2 1+2^. and 1, therefore the only The only values of t for which x, y, z are positive are positive integral solutions are (10, 8, 1) and (1, 2, 3). (2)
//
(a,
j3,
y)
is
a solution in integers of the two equation
ax -f by + cz = d, it is
required
(i)
to
a'x
4-
b'y
f
the three
(be'),
the equations are equivalent to
(ab
(ca') y
/
is
a
common
be integers for
factor of
(ca') 9
t
all integral
(ab
by Oh. X,
6,
...............................
r
(bc') t
zero,
(4),
v
;
(aV)
Denote each member of (A) by t/f where and / is an integer to be chosen later, then
will
is
)
-=
(te'P(ca')
x, y, z
',
find the general solution in integers.
// none of
Thus
+ c'z = d'
).
may
be any integer or zero
values of
t
if
and only
if
SINGLE EQUATIONS
418
Now hence
common
every
and
is the
where g
any
The only
G.C.M. of
is the
l
A
Ex.
(ab
(ca'),
)
and
(
= 0,
easily seen that the general solution in integers
(ca'),
2,...
1,
.
exist)
Find
1.
are
...
a, 6, c,
may
all
is
(ab').
Form ax + b
Single Equation of the
(1) If (if
(be'),
If (6c')=0, the equations are equivalent to
it is
7.
G.C.M. of
a factor of their G.C.M.,
(ca') 1
where g
is
are given by
all integral solutions
(be')
(ii)
numbers
factor of these
then
positive,
be found as follows
the positive integral solutions
all
:
lx + 1 1?/ 4- 26z = 123. + 11 +2(>2< 123, hence z<4.
the positive integral solutions of
greatest coefficient is
2(>,
and
7
Therefore the
possibilities are 2
+ \\y~ 97, + llr/=:71, 7z + ll?/ = 45, 7# + !!?/:= 19, lx
1,
2-2,
= 3, = 2 4, 2
Thus there are two solutions Ex.
Find
2.
Here
Il:r4
~(3-f 4
+ 7),
possibilities,
are 28 solutions in
(2) If
(a,
namely
(0, 5),
(7,2),
no
therefore
x3 = 4,
one solution
one
no
all the positive integral solutions of
^67
is
in positive integers,
~ 2,
The other
of which there
7.r
3arj,
a:
4
x^
2
t
1,
(6, 5, 1)
and
3^ + 4x 2 -f 7# 3 + 1 1.r4
2
= 6, may
no solution. be treated in the same way.
^ y)
w
There
all.
is
an integral solution of ax + by + cz = d (where
xa + bw-cv,v,
07.
no solution,
are positive or negative integers), other solutions are given by
where u,
(7, 2, 2).
<5, and we may have
+ 4#2 = 9,
^xl + 4
namely
y = /3 + cu-aw
9
have any integral or zero values.
z
a, 6, c
GENERAL SOLUTIONS This
obviously true.
is
a, 6, c are
For
prime rr-a,
if
we can show that
Further,
each other, then every integral solution can be so expressed. y~fi, z-y are denoted by X, Y, Z respectively, the
a-Y
Let a be prime to
6,
Z may
+ ftY + cZ-O ................................. (A)
then integers
u, v
and
X, 7,
then,
since a
w
where
is
is
Z
can be found so that, whatever
have,
av-bu If,
if any two of the three
to
equation becomes
integral value
419
Z.
are integers which satisfy (A),
we must have X + cv == biv, Y - cu = - aw,
prime to
6,
some integer or
This proves the statement in question.
zero.
EXERCISE XLII Find the least solution
1.
in positive integers of
=l
(ii)
;
68s
In Exx. 2-7, find the general solution in integers and the least positive integral values of 2. 5.
3.
27a:-8y = 125.
4.
17a;-41y^l.
6.
12a?-7y = 211.
7.
Find 8.
11.
x, y.
8z--27?/:--125.
all
7* +
1.
the positive integral solutions of equations 8-10. 8//
= 50.
9.
lLr + 7y = 151.
10.
Express 68/77 as the sum of two proper fractions whose denominators
and
are 7
29* - 130 =
11.
12. Show that I/ (13x17) can be expressed in two ways as the difference between two proper fractions whose denominators are 13 and 17 ; and express
the given fraction in these ways.
AB
AB
CD
are two rods, each one foot in length. is scaled off into and 13. 13 equal parts, and CD into 17 equal parts. If the two rods are placed side by side, with the scales in contact, and the ends of the rods in alignment, show that and one of CD can never be less the distance between one scale division of than I/ (13 x 17) of a foot, and that there are two cases in which the distance has this value. Which are the pairs of divisions in these cases ?
AB
The sum of two positive integers is 100. If one is divided by 7 the remainder and if the other is divided by 9 the remainder is 7. Find the numbers.
14. is 1,
15. If c is increased
ax + by=c
is
by
kab, then the
increased by k.
[For c/ab
number
is
of positive integral solutions of increased by k and/, g are unaltered.]
SOLUTIONS IN POSITIVE INTEGERS
420 16.
The
least value of c for
which the equation ax + by
[This follows from Ex. 15, coupled with the fact that a of c for which the equation has one solution.] 17.
c
has
+b
is
N solutions
is
c^a + b + (N -l)ab.
given by
Find the
least positive integer, a, for
the least value
which the equation 3x + y
a has
six positive integral solutions. 18. Find the least value of c in order that the equation l%x + Yty have 13 solutions in positive integers, showing that the required value Verify by finding the number of solutions for c-2682, 2681. 19.
The number of
the integral part of
Find
all
20.
x + 3*/-4z-
solutions in positive integers of the equation
may 2682.
x + by~c
is
c/b.
the positive integral solutions of equations 20-23.
Ix + lly + 133 = 201,
21.
S,\
7z 22.
9x+I6y + 25z = IQ9.
24.
Find the
remainders
c is
23.
number which when divided by
least
1, 5, 9,
= 209.
respectively.
7,
11,
Find also the general form of
all
20,
leaves the
such numbers.
a collector has 25. A certain set of stamps is made up of four different sorts one of each sort and some duplicates. The values of the stamps are 2d., 3d., and the lot are worth 3s. 9d. Find all the different ways 9d., Is., respectively in which the batch could have been made up. :
;
A
man has in his pocket three half-crowns, seven other silver coins which 26. are either shillings or sixpences, and eightpence in coppers he paid a bill of 9s. 7d. Find the possible number of different ways of doing so. ;
27. Find the sum of money which is expressed in farthings by the same digits written in the same order as those which are required to express it in pounds, shillings
and pence.
How many
solutions are there to this problem
?
28. If (p l9 2h> P& P*) i g an integral solution of ax l that other solutions are given by
+ bw -cv du', x 2 =p 2 + cu - dv' - aw, ~ #3 ~p3 + dw' + av bu, XIPI + an' + bv' - cw'
+ bx z ^-cx^ + dx^
e9
show
x i ~P\
where u
9
29. If
v,
w, u' 9 v' 9 w'
ax + by + c~Q, a'x + b'y + c'
integer, unless both b
[For a(bc')
9
have any integral or zero values.
and
b'
Q and x
are divisible
+ b(ca') + c(ab') = Q
and
by a'(
an integer, prove that y - a'b.
is
ab'
is
an
CHAPTER XXVI THEORY OF NUMBERS 1
Congruence.
.
(1)
Let n be any positive whole number, which we any numbers, positive or negative, such
shall call the modulus.
If a, b are
that a
by
~~
b
is
divisible
respect to the modulus n,
This
is
expressed
(2)
then a and b are said to be congruent with and each is said to be a residue of the other. n,
= by writing a ^b (mod n) or a 6 an abbreviation for is congruent with.''
(mod
n),
When no where the symbol doubt exists as to the modulus in question, we simply write a E=&. Any statement of this kind is called a congruence. '
==
Every residue
is
a to the modulus n
of
is
form a -f qn, where q
of the
may be positive or negative. If r is the remainder when a a = r (mod n) and 0^r
is
divided by n,
;
thus
r is the least positive residue of a.
Again, a
= - (n - r) (mod n), and either r or n - r^.^n. Hence we can and |/|<^n, where r' may be r so that a = r'(modw)
== r
always find
This
positive or negative.
For example,
if
5
is
number
the modulus, 34
least residues are (2) (i)
1
and
^6 + 6'.
(ii)
4
s - 1 and - 34 = -4 = 1. Thus are 4
and
and the absolute
1,
1 respectively.
Fundamental Theorems.
a -fa'
==
-34
the least positive residues of 34 and
-
called the absolute least residue of a.
r' is
f
a^b(modn) and a ^b'(modn),
//
a -&==#' -6'.
(iii)
then
aa'^bb', and in particular uc^bc.
d, a number prime to n, then a/d = b/d. For by hypothesis a b and a' - b' are divisible by n, hence - (b + b') and (a - a') -(b-b ) are divisible (a + a) by n. = (a - 6)a' 4- (a' - 6') 6, aa' (iv)
// a and b are divisible by
f
W
Again,
therefore aa' -66'
divisible
by
- = Lastly, a 6 qn, where q
is
is
by d, so also is qn. Now number, and since a/d-b/d = q/d
divisible
n. This proves the theorems (i)-(iii). a whole number, then since a and 6 are d is prime to n, therefore q/d is a whole .
n>
it
follows that a/d = b/d (mod
ri).
In particular, if a == 6 (mod p) wAere p is a prime and if d is any common divisor of a and b, then afdz=b/d(modp), except when d^Q(modp).
Thus a close analogy and equalities/ *
exists
between
'
congruences to a prime modulus
'
TEST OF DIVISIBILITY
422
The following
(3) (i)
ab
...
f
then
,
(2)
:
same modulus
to the
.
(ii)
and
are immediate consequences of the theorems in
If any modulus as=a', 6 = 6',... k~k &==a'6' ... &', and in particular a m z=a' m to
// f(x\ y, 2, ...) is a polynomial in x, y, z, ... with integral coefficients x = x', y^-y', z^z, ... to any modulus, then to the same modulus
f(x,y,z, ...)=f(x,y',z, Ex.
1.
Give a
test
as
...).
to the divisibility
number by
of a
N may be written in the form N=a +a + a. + +an n where
11 or 13.
7,
Any number
l
t
i
t
2
t
...
f^lOOO
0^a r
and
s-a -a 1 + 2 -... + (-l) n an then we have + l =1001 =7 t~ -1 and r =(-l) r with regard to each of the moduli Let
rt
,
t
,
N~s
Consequently
J
(mod
7,
.
11
.
13;
11
7,
hence
and
13.
11 or 13).
Thus if 2V = 12345671, then 5 = 671-345 + 12 = 338. Now 338 is divisible by 13 but not by 7 or 11 therefore 12345671 is divisible by 13 but not by 7 or 11. ;
The Numbers
2.
less
than a Given Number and prime to
n is any number greater than 1, the number of positive integers than n and prime to it is denoted by
it.
less
:
value of <>(1) as 1. Here, it should be noted that 1 is to be regarded as prime to every other number. Thus the numbers less than 12 and prime to
it
(1)
are
and
1, 5, 7, 11,
If a
is
prime
For
if
to
n, so also
n
is
by
in a certain order (Ch. its
is
remainder.
gression are prime to
If
number of terms of the arithmetical progression x + 2a, ... x + (n-l)a x-i-a,
these numbers are divided
w-1, taken
(2)
= 4.
to n, the
x,
which are prime
gp(12)
I,
n, the remainders are 0, 1, 2, 10),
Consequently, as
n as there are numbers
m is prime to n,
and
then cp(mri)
(p(m)
.
less
if
a
number
many
...
prime to terms in the prois
than n and prime to
it.
(f>(n).
To enumerate the numbers less than mn and prime to it, we arrange the numbers 1, 2, 3, ... mn in n rows and m columns as below :
1
...
w-hl 2m-fl
...
...
(n-l)w-t-l
m
2
m-f2 2m + 2
...
k
...
...
m+k
...
...
2m + k
(n-l)w-f2...
....
(n-l)m+k...
m 2m 3m nm.
is prime to n, the numbers in question are those which are to both and n. Now member of the &th column is or is not prime every prime to m, according as k is or is not prime to m.
Since
m
VALUES OF
423
p(n)
Thus the numbers in question, being prime to w, are members columns headed by numbers prime to m. 99 (w) ft,
of the
Moreover, every column contains cp(n) numbers which are also prime to for the numbers in each column form an arithmetical progression of
n terms with a common
It follows that there are
m and n,
to both
(3)
//
that
m v m w3
cp
(m)
m
.
cp(ri)
are prime
r
prime to n. numbers less than is
and so cp(mn) =
to mn,
is
...
2,
m which
difference
.
mn and
one another, then
to
cp^m^m^ ... w r = 9?(%)
)
w2
,
therefore
and so on, (4)
If p
for is
it is
m m2
prime to
any number
)
.
)
.
)
)
and
1
.
prime
both m^
Hence
of steps.
a prime, then q>(pr )
p
r
(l-l/p).
when r = l,
The proposition is true If r>l, since p is a prime,
for cp(p)=p-I. numbers 1, 2, 3, ... p r those which ~ are not prime to p r are p, 2p, 3p, ... p r and their number is p r l ~ r l which All the rest are prime to p r and their number is p r -p of the
,
.
,
y
,
proves the theorem. (5)
If
n=pa
.
r
.
...
where p,
p/\ For pa (f, (3) and (4). ,
Find
Ex.
1.
We
have
r
q/
are prime to one another,
...
the
are the prime factors of n, then
... q, r,
number of
integers less than
and the
n and prime
to
result follows
it,
from
when
w = 1024,1025, 1026.
1024=2 10
,
1025 = 5*. 41 1026
= 2.
= 1024 (1 -|) =512. and 9? (1025) = 1025(1 -J)(l -^-)=800. 19 and
hence
33
.
(p (1024)
This illustrates the irregularity in the variation of the function
Ex.
2.
This
is
prime to
//
n^2,
the
sum of the when n
obviously true it,
so also
is
Hence the numbers each pair has a
sum
integers less than 2.
If
n>2
q>(ri).
n and prime to it is \nq>(n). is any number less than n and
and x
n - x. less
than n and prime to
it
can be arranged in pairs such that
n.
Thus
pairs is
fyp(ri)
and the sum of the
FERMATS THEOREM
424 (6)
Theorem.
number
// ^(
= 1),
d3
2,
dr (=-n)
...
,
n^p^r
+ y(d2 + 93(^3) + )
where p,
...
q
l)
then every divisor of n
0
...
)
form p x qVr z
of the
is
and
let
where
...
Q^x^a, Q^y^b,
so that
etc.,
S = Zcp(p*qyr*
S
have
...
x, y, z,
= 27{y(p*)
...)
.
99(7")
.
Hence
values subject to the above conditions.
all
equal to the product
is
Now
1
+ (jp(p) +
2
cp(p
-f
)
.
.
.
4-
a (jp(p
)
= l+(p-l) + (p a -ji) + ...+(^and
any
= n.
(p(d r )
are primes
r, ...
y
+
...
S =
where
are the divisors of
n, then
Let
rf
similarly for the other series in the brackets.
S-=p For instance, the divisors
~l to p, then a p
For
a
1
is divisible
p
)=
Therefore
.r ...=n.
.q*>
//
-1
p
of 24 are 1, 2, 3, 4, 6, 8, 12, 24
Fermat's Theorem.
3.
a
fl -
is
a prime and a
is
;
and
any number prime
by p.
-
l)a are divided by p, the a 1 in taken certain order. p Also the product of the numbers a, 2a, 3a, ... is congruent (mod p) ~ with the product of the remainders, therefore a p l \p-I = \p-I; and since
is
prime to p,
if
a, 2a, 3a,
remainders are
1, 2, 3, ...
since
prime to p, therefore
1
\p
is
...
(p
a^^
^mod^), which
proves the
theorem. Corollary 1. is divisible
If p
is
a prime and a
is
p any number whatever, then a -a
by p.
"" ~~ For a p -a = a(a p 1 -l), and if a is prime to p, a p l - 1 is divisible p - a is divisible by p which is a prime number. Hence in all cases a by p. y
Corollary 2.
If p
an odd prime and a
is
a i (11-1)52
For a therefore
number.
p -1
p
is
prime
is
a divisor of
I>
a^^~ 1^
p, then
l(modp).
-l = (a*
to
1
or of
a"- 1 -!
a^/>-i)
+1
?
is
divisible
since
p
is
by
p,
a prime
THEOREM
WILSON'S Ex.
The fifth power of any number N has is divisible by 5 and by 2.
1.
For A75
-N
Ex.
The ninth power of any number
If
2.
N
prime to 19, since
is
\r(19
425
same right-hand
the
N is of one of the forms Cor. 2
-1)=9, by
19w, 19m
Euler's Extension of Fermat's Theorem. and a is prime to n, then afW ^ 1 (morf n).
^( =
a2
1),
,
a3
than n and prime to
...
,
9
19).
If n
a
divided
is
any num-
be the numbers, in ascending order, less
and consider the products
it,
aa l9 If these are
1.
N ^l(mod
we have
4.
Let
N.
N = 19m.
Otherwise we must have
ber
digit as
by
aa 2
aa^
,
n, the
...
aa9(n}
remainders are
(A) all different.
For
if
we
suppose that two products as aa r aas (r>s) give the same remainder, then a(a r -as ) would be divisible by n. This is impossible, for a is prime to n and a r a s
Also the remainders are
all
prime to
n, for the factors of
any product
are both prime to n.
Hence the remainders are the numbers a v a 2 ... a 9 n taken in some Now the product of the numbers in the set (A) is con,
(
)
order or other.
gruent (mod n) with the product of the remainders, therefore a*( n >
.
and dividing by a x a 2
a xa 2 ...
...
a
a
^a^
which
is
...
a9(n) (mod
prime to n,
n),
we have the
result in
question.
Wilson's Theorem.
5.
divisible
1,
is
.
is
by p. a is any one of the numbers
a prime number, then
\p-l +1
is
p 1, and if the products the remainders are the numbers a, 2a, 3a, (p-l)a by p, >~1, taken in some order or other. Hence, for every a, there 2, 3, ... one number a' and one only, such that aa' =zl(modp).
For 1
If p
if
If
1, 2, 3, ...
are divided
...
2 a'=a, then a -!
must be
divisible
by
p,
and
since
p
is
a prime
\p
or a -1=0. and a
such that the product of each pair is congruent with 1. 2 3 ... (p~2)~l (mod p) and \p- 1 ==j?- 1 (mod^).
Therefore
Hence NOTE. the other.
\p
.
-
1
+ 1 ^0 (mod p), which
The numbers (Euler.)
a, a'
is
the result in question.
are called associated residues, each being the associate of
LAGRANGE'S THEOREM
426 Conversely, if
p-1
\
is divisible
4- 1
p would have
For otherwise
by
p
then
9
p
is t3
prime number.
a factor q which would divide
and
[p-1
[p-1 + 1. 28 4-233
Ex.
1.
We
have 899 = 29.31,
Prove that
by 899.
is divisible
|
233-1 (mod 29) and 233= 10 (mod 31). By Wilson's theorem, 28 + 1-0 (mod 29), therefore |j|8 + 233=0 (mod 29). 30 + 1 = 0, therefore 30.29. 28 4- 1 = 0, Again, to the modulus 31, we have also
|
1
|
and
so
(
-
1)(
28 4-32=0.
2)
Since 2
prime to
is
31, it follows that
[28
|
and therefore
[28
Hence 128 4-233
4-233=0.
is
divisible
by 29 and
31,
4-
16=0,
and therefore
by 899. 6.
than
Lagrange's Theorem. If p p-I, the sum of the products of
a prime and r is any number less numbers 1,2, 3, ..., p-1 taken
is tJie
r together is divisible by p.
Let f(x)
= (x 4~ 1
products of
(x
)
4-
...
1, 2, 3,
,
2)
p
.
.
.
-
p - 1)
then
4-
1
taken r together,
;
a r denotes the sum of the
if
(x
= tf>- +a l x*- 2 + a&*-* + ...+a P _ l ................... (A) = (x + I)f(x + l), that is, identity (x+p)f(x) l
f(x)
Also
we have
the
(B)
Equating
~ the coefficients of x p 2 ,
etc., are all all divisible
,
...
x,
we
find that
~2
1
a prime, Cf is divisible by p, for r
since
Now,
~ zp 3
p
is
>
;
,
by
which
p,
is
the theorem in question.
For another proof,
see Art. 11, (6).
We
NOTE. (i)
and is
can immediately deduce the theorems of Wilson and Fermat, thus
Equating the terms independent of
since
divisible
av a2
,
...
,
by p which t
re
a p _ 2 are divisible by is Wilson's theorem.
p
9
so also
is
a^^ + 1,
that
is
any number prime to p one of the numbers x 4-1, x 4- 2, ... by p, hence f(x)=0 (mod p). ~ ~ Also /(#)= x p l 4-ap^ x p l - 1 (mod p) therefore x3*- 1 -1=0 (mod p) (ii)
If
must be
x
is
:
in the identity (B),
9
\p
,
-1
4-1
x+p - 1
divisible
9
Format's theorem.
9
which
is
EXAMPLES OF DIVISIBILITY Ex.
If p
1.
is
greater than 3, prove that
awrime, '1
1
1
+ '"
^2 in the identity (A),
If,
(p
and
N6w,
p
since
divisible
Hence,
is
-
(2p ...
1)
a prime greater than
a 3) _ 2
.
divisible
is
If p
divisible
is
a prime, greater than
by p
m
~\
3,
mp
r
~l,
2, 3,
since 3
divisible
*s
prove
-
m
I
.
we have
-f ...
7>
;
and, by Lagrange's theorem,
prime to p,
is
by p
that, I
(
3
it
follows that
2 .
for all integral values of m,
p\m ~
.
Substituting rp for x in the identity (A),
where
2 .
subtraction,
a/p-s exists
3,
by p*\ and,
1
is
P+^-s
ap^ 2
j)~i-f
in succession,
by p.
3_p
2.
p and - 2p
+ l)~2>_i ~%_2 2; are identical, we have by
a^_ 2 Ex.
to
(p
since the left-hand sides
it is
p~-}
we put x equal
+ l)(p + 2)
(2p -l)(2p -2)
427
...
m - 1,
,
Hence,
|
(r
and, by giving r the values
and p -f-
1)
p
I
I
(
1, 2, ...
hence,
is
I
mp -
j
as in Ex.
prime and greater than
rp
.
I
/;)
m - 1,
,
we have,
ra(|
~r+1 it
(mod
1,
3.
3 )
jt?
;
follows that
p\ m =0 (mod
p
m +s ).
EXERCISE XLIII 1.
N
If
ao
+ a t + a 2 t* + l
...+a n t n
,
... -f (
-
I)
n tn
9
T
Q^a r
where Z~ 10000 and
S^aQ-^^a^ -
and
prove that J\ ~/S(mod 73 or 137). Hence state a rule for the divisibility of a number by 73 or 137. Apply it to 30414 and 81103. 2. If
and
p
is
a prime and x l9 x 2 , X Q
in particular (ax) p ~ax p
Prove that,
4.
Prove that,
if if
n
n
is
...
xa are any numbers, show that
(mod p), where x is any number. and if a is not divisible by p, a p~~ 1 == I (mod p)
p Consequently, a ^= a (mod p) which is Fermat's theorem. 3.
,
9
a prime greater than
is
then n 6 -
7,
1
is
a prime greater than 13, then n 12 -
divisible 1
is
by
9
504.
divisible
by
65520. 5.
If
p and
6.
If
p
is
q are different primes, then
a prime of the form
pq
~l
4m + 1, show
-f
q
p~ l
that
I
1
is
%(p -
+ 1 = (mod p). 2ms (4m) (4m -1) (4m -2) 1.2.3
divisible 1)
is
by pq.
a solution of
the congruence, x 2
[Show.that 2 E
.
.
...
(2m -f l)(mod^).j B.C.A,
ROOTS OF A CONGRUENCE
428
4m -
7. If p is a prime of the form the congruence, x 2 - 1 = (mod p).
the congruence,
8. If
-
(i)
9.
\p
Show
10. If
11. If
1 =
that
-
2
^
l)fc
(
118
a'
+1
=s
-
p ~^
is
1
divisible
-3 + 1
|
(modp), has no
and
;
show that
1,
(ii)
by
p
a prime, 2 \p
n
is
any odd number, show that
divisible
\n-\\ I+S + QT--- + 11 I
12.
Ad
I
Arrange the numbers aa ~\ (mod 17).
...
2, 3, 4,
,
p
-
1)
is
a solution of
show that of the form 4m -
solution, is
1.
437.
is
is
that
\(p
by
p,
T h=0(inod7i). l_j
15 in pairs, a and
a',
such that for each
f
pair,
13.
Let
n~a pb Qc r
...
,
where
a, 6, c,
...
and consider the
are different primes,
groups a, 2a, 3a,
...
7i/a
.
a
6, 26, 36,
...
n/6
.
b
1
a&, 2a6, 3a6, .
(
f
A
)
6c, 26c, 36c,
...
n/ab
...
n/bc
.
.
a6 be
1 f
(
B
)
together with groups (C), (D), etc., formed in a similar way from the combinations of a, 6, c, ... taken 3, 4, ... together. If s l9 s 2 5 3> denote the sums of the rth powers of the numbers in the groups (A), (B), (C), ... respectively, show that the sum of the rth powers of the numbers not greater than n and not prime to it is ,
14. If less
# 2 is
the
sum
of the squares, and it, show that
$ 3 the sum
of the cubes, of the numbers
than n and prime to
where
a, 6, c, ... are the different prime factors of n. [Use the last example to show that the sum of the squares of the numbers not greater than n and not prime to it is
U 3
7.
\
a~
ab
+ n +
+ '")
Roots of a Congruence.
degree in x, condition
with integral
f(x)=Q(modn)
2
6^
a~
a+ ~->-l
is a polynomial of the rth value of x which satisfies the any called a solution or root of the congruence
If f(x)
coefficients, is
f(x) =0, which is said to be of the rth degree. If xQ is a root of the congruence f(x) ^0(mod n), so also congruent with x to the modulus n.
is
any number
Two roots will be regarded as distinct only when they are incongruent with respect to the modulus, and when we say that a congruence has r roots we mean that it has r distinct roots.
THE LINEAR CONGRUENCE Ex.
from Format's theorem that, (modp) are 1, 2, 3, ... p 1.
It follows
1.
1 congruence x^"
1
if
is
p
429 a prime,
the roots of the
E\rery square number is of the form 5n. or 5wl> thus no square can be z congruent with 2 to the modulus 5. In other words, the congruence x ==2(mod5) Ex.
2-.
has no solution.
NOTE.
It
13x=
congruence
-
z=
1
obvious that, in dealing with the congruence /(#)=0 (mod n), we may by any number congruent with it to the modulus n. Thus the
is
coefficient
any
replace
(mod
17
(mod
5)
is
3#=2
identical with
or with
3a;=
-3
or
with
5).
8. The Linear Congruence. (1) If a is prime to n, the congruence ax H= b(mod n) has just one distinct root. For if the terms of the arithmetical progression 0, a, 2a, ... (n - 1) a are is prime to n, the remainders are the numbers 0, 1, some or other. Hence there is just one value of x order 2, n-1, such that ax^b(modn) and 0^x
divided by n, since a in
...
be the last convergent but one.
Then ax - nyQ = 1 and since a is prime to ,
a
If
so that n,
x^=bxQ
axQ == ,
1
(mod
which
is
n).
,
the required solution.
small, or the product of small primes,
is
ax E= 6 = abxQ
Therefore
it is
generally easier to
proceed as in Exx. 1-3 below. (2)
// a
is
not prime to n,
and g
is the greatest
numbers, there are g incongruent solutions of is divisible
by
common
ax^b(mod
n),
divisor of these
provided that b
If b is not divisible by g, there is no solution.
g.
a~ga' n=gri, so that a' is prime to ri. We require values of x such that ga'x - b is divisible by gn'. No such value exists unless 6 is divisible by g. If this condition is satisfied and b=gb', then a'x-b' is divisible by n' and the given congruence is equivalent to a'xs-fe'fmod n'). Since a' is prime to ri, the last congruence has one solution a
(3)
let
,
The expression 6/a(mod
n)
is
used to denote any solution of
ax = 6 (mod If
n).
common divisor which does not divide 6, then 6/a(mod n) no meaning in every other case it has infinitely many values. If a
a and n have a
has
:
if a is not prime to congruent (mod n) is the where congruent (mod n/g) g greatest common divisor
is
prime to
n, these values are all
n,
they are
all
of a
and
n.
:
>
SIMULTANEOUS CONGRUENCES
430
Such expressions possess many of the properties of ordinary and it is easy to prove the following
fractions,
:
a = #'
If
(i)
and
a7&'(modn) ak/bk(mod
(iii)
Ex. (i)
(ii)
equivalent to a/6 (mod n). equivalent to a/b (mod n), if k
am/bm(mod mn)
(ii)
and
a/6 (mod n)
is
is
n)
the expressions
bz=b'(modn), are equivalent.
is
1. Solve (i) 5ar=2 (wod 7) (ii) ISzssG (mod 21). Hero 6#==2-7== -5 (mod 7), therefore #~ - l==6(mod
prime to
n.
;
Since 3
is
a divisor of
so that
z==6(mod
Ex.
2.
Solve
We
have
and x~G,
7)
3.
We
have
?=
12*==
Sotoe -)-
1
20 (mod
13,
7z=
1
~ 2 4-
=43
.
- 50,
^|?== o 11
-
(mod -
-
2 - 17
.
5~
7).
equivalent to
as -Hf?=
therefore
5#=2(mod
7),
IA
^s ~ O
==
O
22 (mod 157).
43).
and the third convergent - 17
.
5(mod
Simultaneous Congruences,
have x = a +ocy where y
is
a + ocy = 6 (mod
and
43)
(i)
is
5/2, thus
&(mod 43 )~
~5
'
55= 31 (mod 43). find x so that
It is required to
and
x-a(moda)
We
is
21).
x= -\^- (mod 43)~ -
Therefore (4)
the congruence
36* = 7 (tnod 157). O
Ex.
15, 6, 21,
x==b(mod
/?).
given by
j8)
or
a2/
= fe~a(mod j8).
Let ^ be the greatest common divisor of a and j8. If 6 - a is not divisible by g there is no solution. Otherwise, there is just one value y l of y less than /?/
+ f}/g
.
t,
so that the general solution is
x = x l + afilg
.
t
x t = a + at/j.
where
and only if, 6 - a is divisible by g, the greatest common divisor of a, j8, and then the conare gruences equivalent to the single congruence x = x (mod I), where I is the L.C.M. of a and j8. Thus a solution
^
of the given congruences exists
It should be observed that solutions (ii)
To find x
so as to satisfy
xz=a(mod
always exist when a
a number of
x = b(mod
if,
j8),
relations of
tJie
x^c(mody),
is
prime to
/?.
form
...
replace the first two by x = x 1 (modi!) as in the last section. this with the third, the first three congruences are equivalent to
we
Taking
x = x2 (mod m),
where x2
is
any
solution (found as before)
and
m
is
the L.C.M. of
I
and
y,
THEOREM ON FRACTIONS and
therefore of a,
congruences are together equivalent
and L if
is the L.C.M.
A
<*>
Ex.
We
35x^4 (mod
Driven that
1.
x=~
have
oo
Hence
.
Thus y = l+4t and
a;
and
9),
~ (mod 12) - 1-- 2 (mod 12).
x=-
&o
3
= 5 4- 9 (1 + 4$)
+ 362, which
14
solution
any
12), find the general value of x.
1
12) =s--
y^-Wmod y
giving
12),
is
the given
Moreover, a solution always exists
and 55x=2(mod
5 (mod (mod 9)==-^r== 1
= 5 + 9ys 2 (mod
a;
9)
where
xz==g(mod L)
of a, /?, y, are prime to one another.
...
y,
to ...
be seen that
will
it
Continuing thus,
y.
j8,
431
o
is
(mod
4)
si (mod 4).
the general solution.
We
can apply these methods to solve a linear congruence when the modulus is a composite number, as in the next example. (5)
Ex.
2. Solve 19*== 1 (mod 140). and have 140 4 .5.7, and therefore 19#==1 for the moduli 4, 5 and 7 since these numbers are prime to one another, any value of x which satisfies these conditions is a solution. Thus we have
We
:
#==
17
=
-
(mod 4)==
1
- l(mod
x=
-
1
5),
7)
= -=-4 (mod 7).
- l(mod 5), so that y = 5z. - 4 (mod 7) z= and -f 202=
+4t/==
x= -
Therefore
(mod
1
8
z==y-(mod
Hence
:==-- (mod 5)= 1 t/
1
and
4),
1
7)= 3 (mod 20 (mod
-^r-
Thus
7).
and x = 59 In
this case the solution
_. = 7 + 2
1<7
Therefore
-
-
+
1
2*
~\r
4"
2
-f 140J, which is the general solution. can be found more easily by the method of
The fourth convergent
.
is
~ o
,
and
59
.
(3),
Ex.
19 - 140
3,
.
thus 8
= 1.
x~ ~ (mod 140) == 59 (mod 140). 19
A Theorem
9.
on Fractions.
If n
is the
one another, and Z, m/n can be expressed uniquely in the form
c 9 d,
which are prime
...
to
m = a- B y A + y + + ...+ 7 n
where a,
Let
a'
)8,
...
A,
b
c
is
product of factors a,
prime
6,
to n, the fraction
...
7
..............................
fc,
v
A)'
I
k are positive integers and a
etc.
and therefore integers x, x' can prime to a', for any common these two would divide m, which is prime to a'. Thus we have bed
then a
... I,
be found so that a'x factor of
a
m
is
prime to
a',
+ ax' = m. Moreover,
m _x
x
aa
a
f
a
'
. -I
1
x' is
OUJ.VA
w_x n
a
T
7
:i
7
bed ...I
THE GENERAL CONGRUENCE
432
Proceeding in this way, we can obtain a relation of the form
m-x
._
n where
u I
_. . .
I
j
a
w
z L ~___ J
__ _i-
b
c
w are integers, and then mfn x^a + y^/3 + kJ), etc.,
x, y, z,
by writing
j
,
I
frjcr,
can be put in the required form where a, 'j8, ... are the least
with regard to the moduli a, 6, ... respectively. we multiply each side of the equality (A) by n, it will be seen
positive residues of x,
Again, if that a he
...
.
?/,
...
Z^m(moda),
ac
.
j8
= w(mod
...
6),
etc.
Since
6c...
i!
is
a has just one positive value less than a. Similarly j8 has just one value less than 6, and so on. Hence the expression found for m/n is unique. prime to a,
Ex.
}
Express
.
^ e form
Yr~~7\
- + -+- + - + A% 4
O
/
J.
the fraction* being positive
1
proper
fractions and k an integer. Here x, y, z, u are determined by
5.7. liars
4.7.
1
101
(mod
= 101 (mod ly
1
4), 5),
(
4.5 Ilz=sl01(mod7), With
a?
.
5
.
= l,
lu
~ 101 (mod
y = 2, = 1,
it
1)
(
2
.
1
.
.
= 1,
== 1
= 1 (mod
a?
-4,
4);
y^2(mod5); ZEE! (mod 7);
s 1= 12,
w
mod
== 3(
i + f +y+^r = l^^;
find that
The General Congruence.
10.
y
4?t
11),
= 3, we
- l)x
1)(
(-3)(-2) .4^-3,
.
4
.
In this article
11).
so that jfe=-l.
/(re), 97(3),
...
will
denote polynomials with integral coefficients. (1) If
every coefficient of a polynomial /(x)
is divisible
by n
for all values of x,
gruent with zero to the modulus n.
/(x)
r=
divisible
by
n,
then /(x)
and we say that f(x) is identically This is expressed by writing
con-
said to be identically congruent with (p(x)(modri) 0(mod n), which is also written as /(x) H^ 99 (x) (mod n).
Again, /(x)
-
is
is
if
If /(x) is divided by
whore the remainder R'(x)
is of lower degree than
is called division
(mod
n),
f(x)
where R(x) If
72 (x)
is
and leads
= Q(x)
of lower degree
^0 (mod w),
.
to
an
identical congruence of the
form
(p(x)+R(x)(modn),
than
then f(x)
is
said to be divisible (mod n)
by
(p(x).
CONGRUENCES said
is
Ex.
divisible
is
// f(x)
have r roots equal
to
(mod
shown on the
,
the congruence
7)
right.
f(x)==Q(modn)
to a. 2
2, 4,
-6 by + 1. In
and
1
r
G are roots of 5x* + 3# - \x - 2===0(mod - 2 and x 4 in succession is x by In the third line 13 is replaced by
Prove that
1.
Division
-
(mod n) by (#-a)
433
the last line 19
is
replaced
0. The reckoning shows that + 3z2 - 4x - 2~ 5(x -2)(x -4) (x + l)(mod
by 5
7).
12
5+342
\ __4
5-1+1 + Q
10-2+2
and 21 by or*
when x ~2, 4
so that the congruence holds
or
20 + 9 7),
6.
Congruences to a Prime Modulus.
The analogy between a step further by the following
11.
congruences and equalities is carried theorems, in which the modulus p is a prime.
or is
N ^Q(mod p).
For since
MN
is
M
= Q(modp), a prime, then either or N divisible by the prime p, either is
M
so divisible.
If a
a root of f(x)z=Q(modp), then f(x) is divisible (mod p) by For by di vision (mod p) we have f(x) ^Q(x) (x a) -f R, where
(2)
x-a.
R
where p
MN-O(modp)
//
(1)
is
.
is
of
independent
therefore /(.r)
then
it
of x
- a and x
(mod
p)
divisible (mod p)
divisible (mod p)
is
/?
is
x=oc
Putting
divisible
is
is
Iff(x)
(3)
x.
x-oc.
by
r
and by (x-fly where a For any common divisor which is independent of x.
by
(x~a)
-
r
by (x
we have 72=
- ft)* (x /?
.
(4) If f(r) is a polynomial of degree n, the congruence f(x)^Q(rnod p) cannot have more than n roots, any root which occurs r times being counted
as r distinct
This follows at once from the preceding theorems.
roots.
// f(x)'^Q(modp) is a congruence of the n-th degree with n roots and a factor of'f(x) of degree r, then the congruence (p(x)^0(mod p) has
is
(6)
p
//
is
a prime,
p congruence x
Hence it
if
~l ~
(Art. 6)
Ex.
1.
;
it
follows
ar
a^"-
^0(mody)
also
I
p
-
1
+
sum
-a
I
2
from Fermafs theorem that the roots of 1, 2, 3, ... p - 1, and therefore
are
^0(mod p)
a r denotes the
follows that
Hence
1
of the products r together of
p -3 ar
for
4- ... 4-
a p _,
r
=0(mod
p),
the
,
9
x2
{
I
p
-
which
which
is
1 -f 1}
1
,
2, 3,
...
p-
1,
^ 0(mod p).
Lagrange's theorem Wilson's theorem. is
Solve
Here 4z 2 4-43;-f4==0, or (2^-fl) 2 ^
-3s 4;
and2.r-f IE=
2.
Hence x~ 2 or 4 (mod
7).
ROOTS OF CONGRUENCES
434
In treatises on the Theory of Numbers the notion of congruence is extended to complex numbers, thus completing the analogy referred to at the head of this
article.
Let /(a;)^0(mod p) be a connth degree to a prime modulus p. Let the roots of the congruence be oc v 2 ... a r where some of these may be multiple roots. By a modified H.C.F. process we can find a congruence R(x) ^0(mod p) of 12.
Reduction of a Congruence.
gruence of the
,
,
degree r of which the roots are ot v
,
1, 2, 3, ...
~ xp l
-I ==0(modjp). /(#)=EO(mod^) is a solution of ~ Let the H.C.F. process be carried out for the functions f(x) and x p l - 1 with the modification that any coefficient may be replaced by any number congruent with it (mod p) then the last divisor R(x) is the highest common solution of
.
p - 1, hence every
;
di visor ( mod p)
and x p ~ l -l.
of f(x)
The truth of this statement depends on the following. Let u, v be any two consecutive remainders in the process. If w is the next remainder, w au + bv where neither a nor 6 is divisible by p. Hence the common divisors (mod p) of u and v are the same as those of v and w. It follows that
Ex.
1.
SJunv that the congruence
distinct root
is
R(x)^0(modp)
and
solve
is
the congruence referred to above.
3# 4 -2x* ~4x 2 -x-l==Q(inod
7)
has only one
it.
6 Carrying out the modified H.C.F. process for the left-hand side and re -!, it the x 2 is the distinct found that last divisor is root. + 2, showing that only we find that 3z4 - 2#3 - 4z2 - x - 1 == 3 + 2 x~ division
(mod 7) By Now afe3{mod7)
(x
-2
has no solution, hence
is
2)
(
3) (mod 7).
the only distinct root.
EXERCISE XLIV 1.
Find the form of x, (i)
29x=
1
if
(mod
13)
;
(ii)
2:c==3
(mod
7)
and 3*^5 (mod
2.
(iii) 8z= 1 (mod 7), 3z=4 (mod 11), and 7z~3 (mod Solve (i) 78z= 1 (mod 179) (ii) 78a;= 13 (mod 179).
3.
Solve
16 ^^ 31
.
(mod 1217) (iii) 15*= 28 (mod 1009) #= a (mod 22) and x =: b (mod
(H)
;
4. If
20).
If
x~a (mod
b
16)
40),
(mod 5)==c (mod
30#=~31 (mod 1861)
If x==a
(mod
7)== 6
z==
;
(iv) 26#=35 (mod 1901). show that a -b is even and
11),
prove that
x- 385a + 1766 - 560c (mod 880). 6.
;
;
(i)
5.
11)
(mod 11) c (mod 13), prove that - 286a + 3646 - 77c (mod 1001).
BICYCLE-GEAR PROBLEM 7.
If x = a
8.
Express
integers 9.
and
Express
integers
435
(mod 3)^6 (mod 5)=== c (mod 7), prove that x== - 35 4- 216 4- 15c (mod 105). --
1
/
form
in the
-
OU
< 32,
?/
-
< 5,
2
< 11
k,
where
x, y, z,
k are positive
k,
where
x, y, z,
k are positive
11
-
form
in the
~
.
1001
and #<7,
4-
O
11
7
2<13.
2/
10. Find tfie form of numbers, of which the first, second and third powers are of the forms 3^4-1, 4^4-1 and 5>i4-l respectively. 2
-2
divisible
11. If
?i
12. If
# 2 4-4oH-2
is
13. If
n 2 and
+
is
(n
by
divisible 2
l)
and n 2 ~ 3
7,
is
divisible
23, find the
by
are of the forms
11, find the
by
form of
form of
n.
x.
llm 4- 4 and 12m, 4- 4,
find the
form
of n. 14.
Find four consecutive numbers
divisible
15. 'Prove that the congruence 3# - 4s - a; incongruent solutions, and find them. 3
4
by
2
4-
5, 7, 9, 11 respectively.
3# - 4^
(mod
16. Prove that the congruence incongruent solutions, and find them. 17.
7)
(mod
~0 (mod 29)
Prove that the congruence and find it.
has only two
11)
has
four
has only one distinct
solution,
18. Bicycle gear as
A
a revolution
counter.
bicycle with no free wheel has
,,-
19 teeth in the back chain- wheel, 62 teeth in the front chain- wheel, 121 links in the chain,
^
c
and the circumference of the back wheel of the bicycle is known to be almost exactly 7 feet. Prove the following rule to find
p IG>
^
the number of revolutions (R) of the back wheel (and so the distance travelled) in any journey of not more than 10 miles. Mark a tooth .4 in the back chain- wheel, a tooth in the front chain- wheel, and a link C in the chain. At the start count the teeth or links from A to B and from A to C in the direction of turning. At the finish count these again. Take the increase from A to B or 62 minus the decrease, and denote it by a. Take the increase from A to (7 or 121 minus the decrease, and denote it by 6. Then the number E is the remainder in the division (3025a 4- 21086) 7502. [If x is the number of revolutions of the back wheel,
B
19z=a(mod Hence show that and therefore
62)
#==49a(mod
and 62)
u;= 3025a
-f
19#==&(mod 121). and #=51&(mod 121),
21086(mod
7502).J
CHAPTER XXVII RESIDUES OF POWERS OF A NUMBER, RECURRING DECIMALS Residues of Powers of a Number. 1
// a and g are prime
.
to
any modulus
n, the least positive residues
successive terms of the geometrical progression
number of terms in
is the
t^cp(n) and
the recurring period,
and
recur ,
...
a, ag, ag~,
is
of
if
t
independent
of a.
Denote the residues by r r l9 r2 etc. Each of these is one of the
,
,
,
;
,
prime to n, hence ag
i+2
~ag
2 ,
etc.,
Moreover, e
<7
~l(mod
the
= l(modn).
^*
and so r rQ r t+l r^ f^^^* number of terms in the period is
rM
If
/A==v(mod
t
~a
t+l
^ag, where t^
ag
9
e ^ c ->
,
t
the
n).
in (i) No two residues = and 0(modJ), conversely, /z
It should be noticed that (ii)
that
It follows
= l, J),
then
:
and conversely,
(iv)
Thus for
the
(modn) with
modulus
~gr,
2
-g
,
,
0,1,2,3,
Residue.
1,2,4,8,16,13,7,14,9,18,17,15,11,
7,8,
= /v,
rM
then
n-l occurs w-r3 etc.; for
,
,
etc.
,
1 2 2, 2 2
Index.
5,6,
3
If
(v) If
n-rv n-r2
~g
19, the residues of
4,
(iii)
^^^^(mod n).
as a residue, the subsequent residues are
they are congruent
the period are equal,
,
...
are
9,10,11,12,13,14,15,16,17,18.... 3,
6,12,
5,10,
1
...
.
Having found r4 = 16, we have r 5 = 16 x 2 = 13 (mod 19). Again, r = 18 = 19 - 1, and the sul>sequent residues are 19-2, 19-4, etc. Here J = 18 =
For
the
modulus
13, the residues of
1 2 5, 5 5
Index.
0,1,
Residue.
1,5,12,8,1,5,12,8,1,5,12,
Here
2,3,4,5,
= 4, which
is
,
,
6,7,8,9,10,11,12. 8,
a divisor of 99(13)
1.
~
12.
...
are
RESIDUES OF POWERS 2. // g
is
g ^\(modn}
to
prime
i
then
9
For #o (n = l(mod )
If
ceding t^(p(n).
n and
t = (p(ri)
t
r
for
;
for
b
Hence,
=g mod
is
not the case, for r-s
S
denote the set of numbers
S
the numbers in
b of the set
the residues of bg, bg l bg 2 ,
,
6
if
included in the
"l 9
&A
g
,
than n and prime to where s
r~5
=1
,
than n and prime to t
it.
it.
If the set
q>(n).
does not occur in the set (A), consider
bg ,
less
then
9
S t
...
and
n)
(
9
less
a r ~as
if
s
Let
some number
numbers
q>(ri)
which
all
= 0,
9
g
These are
6
q>(n).
of these is one of the
all
+ &, where 0<6
and consequently
9
Also no two of them are equal
If
be equal to q>(n), and by the pre-
may
t
let
Let the least positive residues of g^ g l g 2 ... g*~ l be a t-i ............................... ( A ) a i> a 2> o>
Second proof.
(A) includes
index other than zero for which
or is a divisor of
n), so that
Therefore g b ^l(modn) t
Each
is the least
t
437
,
denoting them by
62 ,
set S, for
b
bt
...
is
^ ............................... (B)
prime to
n.
Also they are different
from one another and from those of the set (A). For if br = bs where s
9
9
)
which
is
contrary to supposition. and (B) include all the numbers in
If the sets (A)
/S,
then
2t
t
=
>
Otherwise 2t<(p(n)> and then some number c of the set S does not occur We then consider the residues of cg cgl cg 2 ... cgr*-1 ,
in (A) or (B).
denoting them by
As
be'fore,
CQ>
we can show that
^
9
^^
...
these
numbers are
9
9
................................ (C) all
included in
S and 9
that they are different from one another and from those in the sets (A)
and (B). Thus 3t^
(p(n).
Incidentally, this proof establishes the truth of Euler's generalisation of
FermatVtheorem.
PRIMITIVE ROOTS
438
An Odd Prime Modulus:
3.
prime, g any
g
t
s 1 (mod p),
number not then g
is
Primitive Roots. Let p be an odd
divisible
by
p,
and
t
the least index for which
said to belong to the index
t.
Since cp(p) = p-l, t is equal to or is a divisor of p-1. If tp-l, so that the period of residues includes every number less than p, the period is
said to be complete
is
a divisor of
and g
is
called a primitive root of the
the period
p-1,
is
modulus p.
incomplete and g may be
It
t
called a sub-
ordinate root.
Hence if g is a primitive root of p, every number prime to with some power of g.
p is
congruent
proved that, for any odd prime modulus, primitive roots exist for the present we assume this to be the case. When one has been found, It can be
;
the others can be found as in the following illustration. Illustration.
In the margin, for the modulus 13 the least positive U are written down round ... 2
residues of 2, 2 l 2 2 ,
,
the circumference of a
Every number
circle.
than 13 occurs, and so 2
loss
is
a
primitive root of 13. Hence, without further reckoning, we can write down the residues (mod 13) of powers of any number. 5 6 2 s2 10 etc., powers of 6, since 6-2 1 2 the residues of 6, 6 6 ... are found by starting with 1 and taking eveiyffth number. The residues are
Thus
for
,
,
,
,
1,
6,
10, 8, 9, 2, 12, 7, 3, 5, 4,
11.
Every number less than 13 appears, the reason being that 5 is prime to 12. Hence 6 is a primitive root; and, in the same way, if is any number less than 12 and prime to it, 2 M is congruent (mod 13) with a primitive JJL
root.
Since
with
2
2
5 ,
= 4, 27
,
there are four primitive roots of 13; they are congruent and are equal to 2, 6, 11, 7, the first, fifth,
2 11
,
seventh and eleventh
from
numbers
in
order
round the
counting
circle,
1.
9 and the residues of 5, 5 1 5 2 ... are obtained Again, we have 5 = 2 by counting every ninth number, or every third number in the reverse ,
,
,
order.
Since 3
is
four counts
Thus 5
is
the greatest
common
divisor of 9
we reach the number, a subordinate root
is 1, 5, 12, 8.
;
5,
it
and 12 and 12 = 3
from which we
.
4,
after
started.
belongs to the index 4, and the period
RECURRING DECIMALS
439
an odd prime, g any number >1 and not divisible by p, t the r v r^ ... the least positive least index for which #' == 1 (mod y) and r residues of g, g l g 2 ... then 4. If
p
is
fl ,
,
that
,
sum
to say, the
is
Thus
,
= 13 and jo
if
of the residues which form the period is divisibk by p. = 5, then 2 = 4 and
Again, suppose that e
with any residue
rm
and
= ef. Starting (not t itself) and taking every eih residue, we have the sequence is
a factor of
^m>
^m+e*
t
^m+2e>
These residues recur, and they form a sub-period containing f residues for g x ^l{modp) if x = ef, and for no smaller value of x. Also
^ + W.+W2
+
-.-+W(/-i)=^
Hence the sum of the residues forming Thus if p = 19 and # = 2 (See Art. 1,
5. (1)
Recurring Decimals. If n
is t
period of
prime
;
to 10,
figures, where
m/n t
Let
is
is
equal
the sub-period is divisible by p. (v)),
m
the least
then
*
= 18 and
and prime to
n, .then
a pure recurring decimal with a index other than zero for which
10* =
1 (mod n) ; so that t=*(p(n) or is a divisor of cp(n). For the remainders rv r 2 3 ... in the process of expressing m/n as a 2 decimal, are the least positive residues (mod n) of m, 10m, 10 m, .... Hence, by Arts. 1 and 2, the remainders recur, with a period of t figures. ,
Also all
m/n = O-a^
if
...
/*
dt) then
the fractions m/n, rjn, r^n, (i)
If
t
If
groups of
...
r8 /n = 0'ds+l as+2
...
a ta l
have the same period.
...
as
,
so that
It follows that
= y(n) and m is any number
m/n have the same (ii)
,
,
(jp(n)=et t
period.
where e>l, the fractions m/n can be arranged in e such that the periods are the same for fraction in
fractions,
the same group and different for those in different groups. This follows from the reasoning in the last part of Art. 2.
DECIMAL PERIODS
440
np
If
p
any prime except 2 or
according as 10 is or
p-l, r ~l
np,
//
(iii)
r
then
,
The case
(p-l).
is
t=p-I,
or
not a primitive root of p.* 1) and t is equal to or (p
n=p
2
a divisor of
is
t
r ~l
which
in
5, then
is
a divisor of
deserves further consideration.
5, and l/p has a period of t figures, then If p t l/p has a period of or pt figures. For let T be the number of figures in the period of l/p 2 and let
a prime, not 2 or
is
(iv) 2
,
Then F(t)^0(mod p),
lO'sl(mod^), and therefore
for
Hence T is pt, or a divisor where r is t, or a divisor of t.
r<, we
If
1
for 10
T
+ 10 + 102T +
...
+ lO(*-V* = (lQ(p-Ur _
a prime,
T=t
or pr,
Consequently r
T
1)/(10
-
1)
E=0(mod
p),
Hence F(T) = 10<*-V = l(modp). 10 PT - 1 - (10 T - l)F(r),
and therefore 10* T -1
JJL,
is
T
isnot si.
is
is
not ^O(modj) 2 ).
T=
not
The only known
NOTE.
p
have
T
Now
p =3
of pt, and, since
cases in which
t
or pt. Ijp
2
has a period of
t
figures are
when
or 487.
(2)
//
v,
then
.F
iyy\
=
<1, where
^
F
is
For 10*F
is
m is prime to
10,
and k
is the greater
of the two
a terminating decimal with k figures. an integer with not more than k digits, the last of which
is
not zero. (3) is
//
m/
F=
<1, where m, n and 10
a mixed recurring decimal, with a period of t figures, where
of figures in the period of l/n. For let k be the greater of the
prime
n and
to
integer of not
period of (4)
//
t
than
more than k
jit,
v,
t
is the
then IOkF = m'/n, where m'
Therefore IQ kF = I+f, where /
10*.
digits
is
is
an
and /is a pure recurring decimal with a
m
where p,
q, r, ...
are different primes other than 2
prime to p, q, r, ... , then if t, t', t" ... are the a figures in the periods of the decimal equivalents of F, l/p , respectively, *
number
figures.
F = -~j~
and 5 and
less
two
F
are prime to one another, then
The primes
t
is
is the
less
L.C.M. oft', t",
than 100 of which 10
is
numbers of b
l/q
,
etc.,
t'", etc.
a primitive root are
7, 17, 19, 23, 29, 47, 59, 61, 97.
USEFUL THEOREM For
n = p a (f
let
and, since
t'
Therefore
;
t
similarly, is
t
then
,
s 1 (mod n),
10*
a multiple of t",
is
common multiple of any common multiple
a
t' ,
Again, if r is each of the moduli p*, f, It follows that
t
the
is
...
,
least
and
since
common
10 e s t
a
1
(mod p ) must be a
t'" 9 etc.
t", etc.
of f,
p
therefore
10*' = 1 (mod p a ),
the least index such that
is
$'
multiple of
. . .
441
a ,
(f,
...
we have 10 T ^1
...
",
for
are prime to one another,
multiple of
t"', etc.
t',
Given a table showing the periods of decimals equivalent to fractions 1/n, where n is a prime or a power of a prime, we can easily express any fraction as a decimal by using the theorem of Ch. XXVI, 9. 5. In practice, the following theorem is useful. Let m/n=0' a^a^ where n is prime to 10 and . . .
r steps of division,
9-a
decimal are
For
let
m,
9-a 2 9-a3
x,
,
m m2 ly
...
,
n-m
remainder
tfie
r
a r+l
is
is
,
-m
-m2
l9
,
etc.,
etc.
a r+l = 7(10-0^ +/)
/
...
.
are congruent with ,
the greatest integer in
hence
where
the period contains 2r figures.
m (mod n)
n- m =
Hence the subsequent residues and are equal to n m l9 n w2 Moreover,
and
,
//, after
be the least positive residues of m, 10m, 10 2 w,
m
then
...
,
m prime to n.
occurs, the subsequent figures of the
10(n-m)/n;
= 9 - %,
Similarly a r+2
a positive proper fraction.
= 9-a 2
,
etc.
Ex. 1. To express 1/73 as a decimal, the division need not be carried beyond the stage shown in the reckoning, for since the remainder g \ ^AQ / .9136 72(=73-l) occurs, the subsequent remainders are 73-10,
73-27, etc., the corresponding 9-0, 9-1, etc. Thus
figures in the decimal being
270 ~510
-^Q
1/73-0-01369863. Again, the number of figures in the period is 8 and (p (73) 72. Hence for values m less than 73, the fractions ??f/73 can be arranged in 72/8=9 groups such that the period is the same for those in the same group. By the preceding, the complete of
set of remainders for
1/73 1,
is
10,
27,
51,
72,
63,
46,
22, .......................... (A)
and one group is obtained by giving m these values. To get a second group, choose any number loss than 73 not included in the set (A), say 2. The remainders for 2/73 are congruent (mod 73) with 2x1, 2 x 10, 2 x 27, etc., and are therefore 2,
A
20,
54,
29,
second group is obtained by giving in the same way by choosing a number (B)
and
so on.
m
71,
53,
19,
less
44 .......................... (B)
A
third group can be found than 73 and not included in the sets (A),
these values.
RESIDUES
442
EXERCISE XLV The prime
NOTE.
the following table
/(4)-3 /(7)-3
2
11
.
2
/(8) = 3
4649,
.
= 10*-1
for
{
= 1,
2, 3,
9
...
are given in
3 /(6)-3 .7.
2 /(5)-3 .41 .271,
101,
.
239
.
factors of /(0
:
2 .
11
73
.
101
.
/(9)^3
137,
.
4 .
37
11 .
.
13.37,
333667.
1 2 etc. given the least positive residues of 2, 2 2 2 2 1 1 are etc. What 7 7 of and write down those 10 10 7, 10, , etc., (Art. 1), the primitive roots of 19?
1.
For the modulus
19,
,
,
,
2.
For the modulus
are the primitive roots of 17 3.
If
p
is
the least positive residues of 3, 3 1 3 2 , etc.
17, find
,
(ii)
p
an odd prime, g belongs to the index
pccurs as a residue
1
v^o
rt
...
_^
when #^13, g^2, (iii)
If g
-
and only
if,
t,
and
r l9 r 29 etc., are
,
the
even.
is
if, t
according as
t
even or odd.
is
Verify
3, 5.
a primitive root of p, show that theorem
is
r
prove that
+1 (modp)
or
1
What
?
least positive residues of
,
,
(ii)
becomes Wilson's
theorem. [(i) (ii)
If
is
t
even, g ...
/wr.,
rt_ l
f
!2
~
-
I
(mod p).
s
^g (modp)
where
9
= + I +2 + ... +(t -
8
1).]
RECURRING DECIMALS Here n
is
(modn
etc.
4.
O-a^
etc.,
1
...
...
1
at
aA
t
a **./ ...
l/n= O'(iia 2
,
= 0'a 1aotf 3
a2
at
...
and
in the division,
...
,
^/(lO*
7^-1
r is
then
r
t
+
-
where
1)
. . .
a 1 a 2 ...a t
is
the
.]
J
= 0-6i6
n
....
.
+ ---^ +
and
r ...
...
a^a^
8uchthat
j.
6 rf
2 ...
...
y,
prove that j.
the remainder immediately preceding the
and a t are the
least positive integers
x y 9
lte-ny = l
and where the
5, t is the number of figures in the are the least positive residues of 10, 10 1 10 2 ,
p
digits are a l9
= a& 2
---O-atGU
V) n
If
a prime not 2 or
n
Prove that
remainder
is
r 19 r 29
,
or p), and we write - or ~ *
[The decimal
6.
r
;
number of which the
5. If
p
to 10,
prime
period of l/n or l/p
a^z
...
a's are digits of the
the figures in the period
[l/ft^O-a^ to be carried
...
a^!
when a l
may
...
is
,
at x
r
a^a^
. . .
a^_ l9
numbers
indicated. be found as in Ex. 8.
rln
Q-a t a l
multiplied by
...
r.]
a^
...
;
Having found explain
r
why no
and a t9
all
figure has
RECURRING DECIMALS 7. If
p
is
a prime and 1/p
=
a^
. . .
and give an example to show that such number. 8. (i)
arar+l
. . .
relations
a 2r
443 ,
prove that
do not hold
if
p
is
a composite
= Express -^ as a decimal by using the equality 10 x 4 13 x 3 1. Put down 3 (the last figure of the period) and multiply by 4 as follows
:
076923 4
If the process
4x3 =12, 4x2 + 1= 9, 4x9 =36,
put 2 to the
left
put 9 to the
left
put 6 to the
left
4 x 6 + 3 =27, 4 x 7 + 2 = 30,
put 7 to the
left
is
put
of 3 and carry of 2.
1.
of 9 and carry of 6 and carry
3. 2.
to the left of 7.
continued, the figures recur and tV=0'076923.
in Ex. (ii) Or thus Hence the following
a ta^a z ... a^_i/r=a 1a 2 ... a t Put down 3 and divide by 4 as below
7, :
.
:
"076923.
which is the first figure of the period. and 2 over. Put 7 to the right of 0. 4 into 27 is 6 and 3 over. Put 6 to the right of 7. 4 into 36 is 9. Put 9 to the right of 6. 4 into 9 is 2 and 1 over. Put 2 to the right of 9. 4 into 12 is 3. Put 3 to the right of 2. 4 into 3 is 0, and the figures recur. 4 into
3
4 into 30
is 0,
is
7
9. Cases in which l/p can be quickly expressed as a decimal by the methods of Ex. 8 are p = 17, 19, 23, 29, 43, 59, 79, 89.
10. Justify the following
:
A
If
C=
~0
9408
_
then
tV = -0588235294
where jB=4J, (7=4J5,
etc.
. . .
[The division shows that 10* =588 x 17 + 4
11.
By
the
;
method of Ex. 10 show that j& =0-032258064516129.
[After 6 steps of division, the remainder 2 occurs.] 2F
B.C.A.
BINOMIAL CONGRUENCES
444
12. Show that for values of m<41 the fractions r/i/41 can be arranged in 8 groups such that all the fractions in any group have the same period. What for the group to which 2/41 belongs ? are the values of
m
13. Prove that gV^O-0 12345670, and show that the values of m/81 has this period are given by m 1 (mod 9). 14. If n is any number prime to 10, prove that a that every digit in the product nn' is 1. Find n' when n~ 41 and when n~123.
~ or ~- as a decimal, according as n
[Express
fit
is
m
for
which
f
number n can be found such
or
not prime to
is
3.]
\jTt
15. If n is a prime or a power of a prime, prove that the values of n for which the decimal equivalent of 1/n has a period of t figures are as follows :
1
3
4
5
3,9
27,37
101
41,271
6
9
|
n
|
7,
13
239, 4649
73, 137
333667
[The values of n for which 1/n has a period of t figures are included among the divisors of 10* - 1. See note at the head of this Exercise.]
Show that the only prime p for which the decimal equivalent of l/p has a period of 10 figures is 9091, and (ii) a period of 12 figures is 9901. Also express 1/9091 and 1/9901 as decimals, explaining why only 5 and 6 16.
(i)
steps of division respectively are necessary.
Primitive Roots (continued). Theprocessof finding a primitive one of trial. The orthodox methods of shortening the reckoning given in another volume but in the example below we give a method,
6. root are
is
;
founded on the assumption of Art.
3,
which
is
convenience in solving the congruences in Ex. least primitive root,
= 2, = 3, 0-5, Ex.
Find
1.
g,for any prime modulus, p,
j0-3,
5,
usually successful. For we here give the
XL VI,
less
than 100.
11, 13, 19, 29, 37, 53, 59, 61, 67,
p = 7, 17, 31, 43, 79, 89; = 6, y = 23, 47, 73, 79;
p=
I
;
= 7,
83;
p = ll.
the primitive roots of 67.
converted into a decimal fraction, the reckoning shows that 10 is a subordinate root of modulus 67, with a period of 33 figures. Art. 3 suggests that this If 1/67
is
period consists of every alternate figure of the period of some primitive root, g 2 10(mod 67) will lead to a primitive root. consequently, that g
;
and,
\Ve find that a value of g is il2 and, converting 12/67 into a decimal fraction, obtain, as remainders in the process, the residues of the odd powers of 12. When ;
we
these are interpolated between the remainders in the reckoning for 1/67, we find that the numbers less than 67 are included ; thus 12 is a primitive root, and its period is
all
1,
From
12, 10, 53, 33, 61, 62, 7, 17, 3, 36, 30, 25, 32, 49,
this, as in Art. 3, all
U,
21, etc.
the other primitive roots can be immediately written down, together with their periods thus we find that other primitive roots are 61, 7, 32, 51, 41, 13, 2, etc., these being the residues of 12 fc where k is prime to 66. ;
,
OF PRIMITIVE ROOTS
tFSE
The Congruence x n ^a(mod
7.
It is
p).
445 supposed that
p
an
is
odd prime.
Ifn
(1)
if
For,
is prime to p- I, there xn z=a, then x nA = aA
is
3, a single
value of
A has this value, xnX = x (x v which is the only solution. If
~l
k
.
Ex.
x* ~ 21* (mod 23) z:EE21
// n
(2)
)
A, less
prime to j>-l, there than p-l, such that
= x (mod p
)
is
and
;
is,
x s= ax (mod
thus,
p),
Solve the congruence, x 13 == 21 (mod 23).
1.
Here
and, since n
;
by Chap. XXV,
a single solution.
is
17
and 13
;
17 E=
.
17
~(~2) ~(-2)
not prime to
.
1
(256)
(mod
a
22)
= (-2)
;
hence we have 2
.
(3)
s5(mod
23).
there are several solutions, or there is
p-l,
no
solution.
common
Let d be the highest
n and p - 1
divisor of
and- let g be a
;
Then, since the period of g contains all the
primitive root of p.
numbers
r
than y, we can find r, such that g z=a(modp). s r x^g (mod p); then g ns = a^g (mod p), and therefore
less
Let
ns^r(mod p-l) ................................ (A) The congruence solutions, or
no
and therefore
(A),
also the congruence
solution, according as r
is,
or
is
a factor of
is
by d
not, divisible
n and p 1 (Chap. XXVI, 8). n # E=a(mod p) are given by x^gs (modp), where
d
xn ^a, has d ;
for
Also, the solutions of 5
has any value which
satisfies (A).
Ex.
A
Solve the congruence, # 15 =31(wod 37).
2.
primitive root of 37
2 15S ==2 9 ,
is
and 15$~9(mod
and we
2;
36)
;
z^2 3
2 15
,
,
2 27
9
2 =;31. Hence, 27 (mod 36), and
find that
hence 5=3, ,
15,
i.e.
8,
if
x=2 s
,
then
23, 6.
EXERCISE XLVI Solve the congruences in Exx. 1-6 1.
z 3 =l(mod 11
19).
= 5 (mod 31).
:
2.
x 5 =2(mo&
5.
^=31(mod41).
17).
3.
*7 ==8(mod
6.
a:
21
a;
7.
Find the general solution
8.
Find three solutions in integers of # 8 = 73t/4-3.
9.
Show, without calculation, that a primitive root of 237
in integers of
least of the congruences, <7== 10, g a
= 10,
4
13).
= 2 (mod 31).
4.
# 2 = 19z/-f-5.
s 10.
will satisfy
one at
CHAPTER XXVIII NUMERICAL SOLUTION OF EQUATIONS 1
The Problem
.
mate values of
under consideration
the real roots of
is to
witli
find the actual or approxinumerical coefficients.
any equation This question is quite distinct from that of finding an algebraical soluIn fact, no algebraical solution of the general equation of the fifth tion. or higher degree has been discovered. If any rational or multiple roots exist they can be found and removed from the equation (Ch. XVIII, 1, and VI, 14). It may be convenient not to
is
remove the multiple roots. Thus we are only concerned with as follows
We
(i)
irrational roots.
The usual procedure
:
find
an interval which contains
all
the roots.
Ways
of doing
have been given in Ch. VI, 12. A method due to Newton which involves more calculation, but which yields closer limits, is given in the this
next
article.
We
separate the roots, that is to say a contains single root or a multiple root. (ii)
we
Taking any interval which contains a
(iii)
find intervals each of
which
by a process of which contain the
single root,
approximation, we find smaller and smaller intervals root.
Newton's Method of finding an Upper Limit to the
2.
This depends on the following theorem a number such that f(x) and all its derivatives are positive when then h is an upper limit to the roots of f(x) =0.
Roots. // h
x = h,
For
:
is
if
Hence that
is
NOTE.
power
of
n
is
if
the degree of f(x), then
f(h), f'(h),
to say
/(x)>0
...
for
fM (h)
are all positive, f(x + A)>0, for which x^h, proves the theorem.
In applying this theorem, we suppose that the x inf(x) is positive, so that
coefficient of the highest
SEPARATION OF ROOTS
We then
^ such that
find the least integer
of x does not
make
Continuing thus,
n f^
~^(x)>0
we can
9
we
find
by
f^
n
xh^
"^(x)>0 when
trial
find the least integer
447 If this value
a greater integer h2 such that
h which makes f(x) and
all its
derivatives
positive.
It should be observed that if
are
all positive,
above theorem, Ex.
1.
Find
we have found a number h r such that This follows from the positive for x>h r n (r+l) are (z), .../ -^(), f^(x). / f^(x)
then these functions are for the derivatives of
all
.
the integral part of the greatest root of
f(x)=x*-10x*-Ilx -100 = 0. Here
The
2 f'(x) -3rr
-20*-
11,
least integral values of x for
(Cf.
Ex.
1, p.
91.)
/"(a) =2 (3s -10),
/'"(z)=6.
whieh
and
f"(x), /'(a*)
f(x)
are positive are
respectively 4, 8 and 12. Thus 12 is the least integer which exceeds the greatest root, the integral part of which is therefore 1 1 .
EXERCISE XLVII In Exx. 1-5 find an upper limit to the roots by Newton's method. 3 2 2. z 3 ~2z 2 -51z- 110=0. 1. z -20z - Six + 1609^0. 3.
* 4 -4x3 -3* + 23-0.
4.
5.
Separation of the Roots. Rules for this purpose were given and by Sturm. Fourier's rule, though often convenient, is Fourier by incomplete, while that of Sturm definitely separates the roots, but its application may be very laborious. We require the following theorems. 3.
Subsidiary Theorems.
(1) If
a
is
a root of /(x)=0, as the variable x
increases through the value a, the functions /(x) and/'(#) have opposite signs just before x=
For
since
/(a)=0, we have 7>2
/(a + A)-A/'(a) +
7,r
/"(a)
a)
+ ... +
/W(a) + ...,
+ ...+ 1
/0(a) +
........... (A)
................. (B)
Suppose that /^(a) is the first term of the sequence /'(a), /"(a), ... which is not zero. For sufficiently small values of h, the signs of /(a -h h) and /'(a + A) are respectively the signs of the first terms in (A) and (B) r r) they are therefore the same as the signs of A /* (a) and h f^((x). Hence /(a + h) and /'(a -f h) have the same or opposite 0, which proves the theorem. signs according as h
which do not vanish
;
r ^l
^
STURM'S FUNCTIONS
448 If
(2)
an
is
a.
r-multiple root of
then, as the variable
f(x)=Q,
x increases
through the value a, the signs of the functions /(*),
/"(*), .../
/'(*).
- 1)*,
(1
/'>(*)
(C)
x = a, and all
- and
the functions + and -, or -f, just before have the same sign just after X QL. For, when z = a, all the terms of the sequence (C) except the last vanish, and the result follows by applying theorem (1) to every two consecutive It follows that as x increases through the value a, r changes of terms.
are alternately
in the sequence (C). important to observe that since
sign are It is
lost
#=a,
at
is
continuous
just before and just after X
derivative.
Let f(x) be a polynomial and fi(x) its Let the operation of finding the H.C.F. of f(x) and fi(x)
be performed with this alteration
The sign of each remainder the sign of the last
remainder
:
is to be
changed before
is also to be
it is
used as a divisor
\
known
:
changed.
then Denote the modified remainders by /2 (z), /3 (#), ... fr ( x ) fi( x )> fz( x )> "-fr( x ) are called Sturm's functions and f^x), f2 (x), ... are
OL.
Sturm's Functions.
4. first
/^(aJ^O and f^(x)
same sign
this function has the
/(#),
fr (x)
as the auxiliary functions.
In the ordinary H.C.F. process we can multiply (or divide) any remainder by any constant. In the modified process it is essential that such multipliers should be positive. Sturm's functions are connected
by the equations
.(A)
where q v q2
are the quotients in the process just described, these quotients being functions of x. It is important to notice that the relation between f(x) and fi(x) is ,
...
essentially different
from that connecting f^x) and/2 (x)
;
fz(x),f3 (x)
;
etc.
5. Sturm's Theorem. Iff(x) is a polynomial and a, b are any real numbers (a<6), the number of distinct roots of f(x) = which lie between a and b (any multiple root which may exist being counted once only) is equal to the
excess of the
number of changes of sign in
the sequence of
Sturm's
functions /<*),
when
x=a
over the
A(*),
MX),...
fr (x),
number of changes of sign in
the sequence
(S)
when x~b.
STURM'S THEOREM
449
which f(x)^0 has no repeated Equations (A) lead to the following conclusions. the case in
Proof for
(i)
factor, (ii)
Since /(x)
=
has no repeated root,/(z)
(iii)
If
any term
and follow
it
if
this
happened,
Thus
if
have opposite
signs.
if
(x)
x
(S)
have no
common
can vanish for the
the subsequent terms including
is
remark applies
similar
all
of (S) except the first
/.-i (a)
A
and/
and consequently fr (x) is independent of x. No two consecutive terms of the sequence
same value of x, for, fr (x) would vanish.
root.
terms which precede /s (^)==0, then zero, the
=-/,+i (a).
one of the
y's is zero.
Thus
if
<7 S
= 0,
then
/s-i(*H -/mfc). Suppose now that x increases continuously from x a to x b. As x varies, no one of Sturm's functions can change its sign unless x passes through a value which makes that function vanish, for these functions are polynomials.
Suppose that x passes through a value a which makes just one
of Sturm's
functions vanish. (i)
If
a
a root of /(x)
is
= 0, one change
of sign is lost in the
sequence
For f(x) and f^x) have opposite signs just before x=a, and they have the same signs just after x=a.
(S).
(ii)
is
If
a
a root of fm (x) where
is
gained or
lost.
For
m = l,
/rn (a)=0, and
2,
...
or
therefore
r-1, no change
of sign
fm -i(oc)=-- -/w+iM-
= l, /
Moreover, fm ^(x) and fm+ i(x) are (a) stands for /(a). continuous at x=a, so that each of these is of invariable sign near #=a. If
ra
Hence, just before and also just after #=a, the signs of /m _i(^),/m (x), - or + or h-f or 1-, showing that fm+i( x ) are either -f -f
no change
of sign is lost or gained.
If x passes through a value a which makes more than one of Sturm's functions vanish, no two of them can be consecutive functions. If f(x) is
one of them, then by the preceding, one change of sign
is
lost
and/1 (a?), and no change is gained or lost in the sequence ... /1 (x),/2 (x), fr (x). Thus one change is lost in the sequence (S). If f(x) is not one of them, no change of sign is gained or lost. between f(x)
Thus, as ar increases, a change of sign in the sequence (S) is lost whenever x passes through a root of f(x)=0, and under no other circumstance is ;i change gained or lost. This proves the theorem.
ESSENTIAL PROPERTIES
450
Remarks.
In applying Sturm's theorem, labour
be saved by the
may
following considerations. If there is
(i)
of
x and
its
no repeated
sign only
is
root, the last function
the last three functions are
then
a'"
and from (ii) If,
any
stage,
fr -i(x) = a'x 2
9
then
+ b'x + c',
we
fr (x)=
a!'x
= - (a' -^ - V \
this the sign of a'"
at
is
independent
required.
Let x have a value such that if
fr (x)
+ 6",
Thus,
-fr-zfa)a'",
+
may often be found
without
much
calculation.
arrive at a function fs (x) such that all the roots
/s (.r)=0 are imaginary, the H.C.F. process need not be continued, and we can use /(x), /j(x), ... fs (x), instead of the complete set of functions. of
For the
essential property of the last function
of invariable sign for all real values of x,
then, in counting the
a\
number
that
and fs (x) has
Suppose that one of Sturm's functions as
(iii)
x
is
it
this property.
fm (x)
may regard the sign of/rn (a) as either + /TO (a) = 0, then/TO-1 (a) and /m+1 (a) have opposite signs.
Ex*
1.
Find
the
number and
when
vanishes
of changes of sign in the sequence
/(),/i (), .../r (^)> we if
should remain
position, relative to
numbers on
or
-
.
For
the scale of integers,
of the real roots of
Here fl (x)
4x* - 9# 2
-4z + 7, and
the modified H.C.F. process
is
as follows
:
43
-99 2237
/(*) =
Thus /2 (aO-43a;2 -72:r-69, fs (x) = S3x-191, /4 (*)= +48074. The sign off^x) is found more easily by Remark (i). If /3 (#) =0, then/4 (s) = -
When x=-^> /2 (x)
is
negative.
Thetefore /4 (x)
is
positive.
MODEL SOLUTIONS The number values of x are
of changes of sign in the sequence of Sturm's functions for various
shown below. x
-oo
-2
-1
-f
4-
2
/i(*)
Number ,
o*
,
.
.
} changes of signj
-
451
4
-
+
+
3
Hence the equation f(x)=0 has one root between 1 and and two between 2 and 3. Ex.
(i)
Find
2.
the
number and position of
Taking f(x) ~x* - x -
Also,
1,
For various values of
x,
-f
-f
-2 and
-1,
one between
the real roots of
we have f^x) 5**-l
when /2 ()=0, fs (x)^
oo
2200
.
4
-f
3
5x* -
and the modified H.C.F. process is
1,
-fi(x), therefore
the signs of Sturm's functions are
-oo
2
1
:
oo
/(*)
Number
/>(*) of
"
'
changes of sign/
showing that there
is
one real root between
1
and
and that the other roots are
2,
imaginary.
x.
The
roots of
/2 () =0
are imaginary, Jfor 15 2 <4
.
8
24,
.
and we need not proceed
further.
Thus there
is
one real root between
other roots being imaginary.
- 2 and -
1,
another between
1
and
2,
the
ALL ROOTS REAL
452
Sturm's Theorem (Multiple Roots).
6.
the case in
which the equation f(x) = /(*)
where p,
and
a,
are
...
y,
/?,
Suppose that
-
= (*-)"(* -W*-y) r
are positive integers
...
q, r,
It remains to consider
has repeated roots.
all different (real
or imaginary), then
u = (x-a)*- l (x-p)*- l (x-y) r - 1 ...
where
Hence u
is
the H.C.F. of f(x) and /^z), and /(*)>
are divisible
u.
by
all of
Sturm's functions,
.......................... (S)
/a(*),-/r(*),
fl(*)>
.
Denote the quotients by
so that
= (x-
...
,
Also for any value of x, the number of and r (x) changes of sign in the sequence (S) is the same as that in the sequence (S'). Dividing equations (A) of Art. 4 by u, we have the identities is
tfj
independent of x.
(x)
=q
l i/J l
(x)-ifj 2 (x),
tl*i(x)
= q.j/i2 (x)-fa(x),
etc.
no change of sign is gained or lost as x a value makes which one of the set ^(x), $%(x}, ...iff r (x) passes through
Whence, as
in Art. 5, it follows that
vanish.
^ (a) =p(oc - /3)(a - y)
Again,
...
-^(a),
have the same sign when x = oc, and since these functions are continuous, they have the same signs for values of x near therefore
x=oc.
^(x) and
L (jr)
Hence, by Art.
change of sign
is
5,
as x increases through
lost in the
It follows that the
sequence
number
any root a
of
a
(S').
of real roots of
(f>(x)^0
in the interval
number of changes of sign a
is
equal to the excess of the
Conditions for the Reality of all the Roots. Let/(x) be a polynomial n with its leading coefficient positive. In general, the number of Sturm's functions is n -f 1.
of degree
The necessary and sufficient conditions that the roots of f(x) = Q may be and different are : (i) the number of Sturm' s functions must be n -f 1.
all real
and
(ii)
the leading coefficients of all these functions
For, as x passes from - oo to sequence of Sturm's functions if
+ oo n changes ,
and only
if
must be
positive.
of sign will be lost in the
both of these conditions hold.
THE BIQUADRATIC 7.
where
Application of Sturm's Theorem.
K=a2 I-3H
2
(Ch. XII, 9; 10
(6)),
453 For the biquadratic,
Sturm's functions are /() and
2 fz (z)=-3Hz ~3Gz-K, f3 (z)=-(2HI-3aJ)z-GI,
In proving this
we
require the identity (Ch. XII, 10, (6)) 2
Sturm's process
+ 4#3 = a2 (#/-aJ) ............................ (A)
is
-Hz G
Again,
if
f3 (z)=-Lz-GI where L = 2HI-3aJ. /3 (z) = 0, then /4 (z)= -/2 (z), therefore = 3H(GI) 2 - 3G(GI)L + KL2 )
= (a2/ - 3H2 (3aJ - 2HI) 2 + 3G2I (3aJ - HI) = 9a4 /J2 - 12a 3HI*J + a2H2 (U 3 - 27J2 )
)
+ 3I(G2 + 4:H3 )(3aJ-HI)* = a2 H2 (4/ 3 - 27 J2 ) - 3a2#2/3 Disregarding square factors,
we may
therefore take
3 2 /4 (z)=/ -27/ Hence if A is positive and both H and 2HI - 3a J .
are real
If A
is positive
and
are negative, all the roots
at least one of the two,
H and
2HI - 3aJ,
is
are imaginary. - oo to -f oo , in the first case four changes of sign as x from For, passes are lost in the sequence of Sturm's functions, and in the second case no
positive, all the roots
change of sign If A
is lost.
two roots are real and two imaginary. All this has been proved more easily in Ch. XII. is negative,
*
The
identity (A)
is
used in these steps.
FOURIER'S THEOREM
454
Fourier's Theorem.
8.
Let f(x) be a polynomial of the n-th degree
and let fi(x), /2 (x), ... fn ( x ) be its successive derivatives. Let R be the number of real roots of f(x)~0 which lie between a and b, where a, b are any real numbers, of which a is less than 6, and an r-multiple root is counted r times. Let N, N respectively denote the number of changes of sign f
in the sequence
AW,
/(*),
when x
a
and when x = 6,
/.(*),../(*)>
........................
(?)
then
N^N
f
R^N-N'i
and
(N - N') - R is an even number or zero. In particular if N - N' = 1 there is just one = no real root lies between a and b. if N N'
also
real root in the interval,
,
In this connection /(x) and
its
and
derivatives are called Fourier's functions.
Let x increase continuously from x = a to x = b. As x varies, no change of sign can he gained or lost in the sequence (F) unless x passes through a value which makes one of its terms vanish. Proof.
(1) (i)
Suppose that x passes through a root a of /(x) = 0. If a is an unrepeated root, a change of sign is lost between /(x) and This has been proved in Art. 3.
fi(x). (ii)
If
a
an r-multiple
is
f(x),fi(*)..:fr(*)-
changes of sign are lost in the sequence
Suppose that x passes through a value
(2)
of the functions /j(x), (i)
root, r
(See Art. 3.)
If
/m (a') = 0,
/2 (x),
a'
which makes one or more
.../n (x) vanish.
but neither
fm -i(
nor
fm +i(a)
is
zero, there is
no
gain or loss of changes of sign when /m _i(a') and fm+ i(at') have opposite but if these have the same sign, two changes are lost. signs :
For fm (x) and fm+ i(x) have opposite signs just before x=a', and they have the same sign just after x=a'. (ii)
all
Suppose that when x=a' the
vanish, but neither
Then as is
if
fm -i(x)
/m + r (x)
is
zero,
r-
odd there is a loss of 1 or r + 1 changes of sign according and fm + r (&) have opposite signs or the same sign, while if r
r is
/m -i(a')
nor
r functions
even there
is
a loss of r changes of sign in either case.
Thus, under this heading, no change of sign are lost, their
number is
even.
is
ever gained, and
if
any
DISADVANTAGES Ex.
Apply Fourier's method
1.
to
455
separate the roots of
2 ~ f(x) =2^ + Ix* 4Qx* 23* + 38* 4 -0. 3 2 /x (x) = 2 (5^ -f 14* - 60* - 23z + 19), 2 120* +42* ft (x) =2(20** -23),
Here
/4 (a?) =8 (30* + 21),
/6 (*) = 240. By Newton's method we find that
all
the roots
lie
in the interval
(
7, 4).
For
various values of x the signs of Fourier's functions are as below.
These results show that there is one root in each of the intervals ( - 7, - 6), ( - 2, - 1 ), (3, 4), and either two roots or none at all in the interval (0, 1). Subdividing the last interval, we find that when x =0-5, f(x) is positive. Hence there
a root in each of the intervals
is
It should be noticed that
for it lias
(0, 0-5)
and
we need not find the
(0-5, 1).
signs
oifl (x), /2 (#), e te when re =0-5, and 1. all, lie between
been shown that either two roots, or none at
Remarks.
In applying Fourier's theorem, as stated above, the what sign is to be attached to one of Fourier's functions
(1)
question arises as to
which
may happen
number
roots
vanish when
x=a
or
b.
met by observing that, for sufficiently small values of A, between a and b (a
This difficulty
the
to
is
of roots
and b -h instead
of
where A->0.
6,
Disadvantage of Fourier's Method. Suppose that (as in the last example) all the roots of an equation are accounted for except two, which, (2)
if
real,
must
it
may
require a large
lie
in a certain interval (a,
number
this the possibilities are (i)
The
roots
separated by
it
may
further
6).
If these roots are nearly equal,
of trials to separate them.
we
may
to do
be very nearly equal, in which case they can be for equal roots
The roots may be imaginary, and no matter how many
made, we
fail
trials.
if we begin by testing (ii) The roots may be equal, and would be better to use Sturm's process. (iii)
If
:
not be able to discover whether this
is
or
is
trials
are
not the case.
NEWTON'S METHOD
456 (3) If
N
so that is the
the roots of f(x) are diminished
is the
by
the transformed equation
a,
number of changes of sign in
is
Similarly N'
this equation.
number of changes of sign in the equation obtained by diminishing the 7 f(x)=0 by b. If N and A arc defined in this way, we obtain '
roots of
Sudan's statement of Fourier's theorem.
from -
oo to 4- oo n changes of sign If it is known that a are lost in the sequence of Fourier's functions. = no of root certain interval contains f(x) 0, and that, as x increases through
Imaginary Roots.
the interval,
than n - s
s
As x
increases
changes are
real roots,
lost,
,
then the equation cannot have more
and must therefore have at
least s
imaginary roots.
EXERCISE XLVIII 1.
From
Fourier's theorem deduce (i) Descartes' rule of signs of finding an upper limit to the roots.
method
In Exx. 2-11, find intervals
(a, b)
where
a, b are
;
(ii)
Newton's
consecutive integers which
contain the real roots of the following equations.
In Exx. 2-4 use Fourier's method; and in Exx. 5-11 use Sturm's method. 2.
a^-3a; J -4a:-f 11=0.
3.
z3 -20z 2 - 3 Ix +1609 = 0.
4.
2z3 + llz 2 - 10z + l=0.
5.
s3 + 2z 2 -51z + 110=0.
6.
2z 3 -15z 2 -8a; + 166=0.
7. a;*
8.
3z 4 -10z 2 -6a;+16=0.
9.
10.
12.
z - 3z - 24z 5
4
3
+ 95s - 46* - 101 =0. 8
11.
~4* 3 - 3x-f-23=0.
x*~x*-4x* + 4z + l=0. x 5 + 3x4 + 2s3 - 3x a - 2x - 2 =0.
Use Sturm's theorem to show that the equation xz + 3Hx + G=Q has and distinct roots if and only if G* + 4# 8 <0.
three real 13. If
real
e<0-55,
4 8 2 prove that the equation # -f 4# -f 6x
and two imaginary
9.
roots.
[/ = e
-4x + e=0 has two
+ 7, J = - 4.]
Newton's Method of approximating to a Root.
Suppose
equation f(x)~Q has a single unrepeated root in a small interval (a, jS), so that f'(x)^Q within the interval. Let a 4- h be the that
the
by the second mean value theorem, /(a + h) =/(a) + hf (a) + &*/" (a + 0h) - 6,
actual value of the root, then
where
0<0<1.
Hence
FOURIER'S RULE It follows, that
if
k
the greatest value of f"(x) in the interval, and
is
=a -/()//'(),
,
the value of the root
is
457
ocj -f
e
where
Thus, if /? - a is a small number of the first order, the error in taking o^ as the value of the root is of the second order of smallness, unless k/f'(
approximation can be carried to any
By repeating the
process, the of accuracy. required degree
is
large.
NOTE.
In the preceding, f(x) may be any function of are continuous in the interval (a, jS). /"(a?) i
x such that
f(x), f'(x)
and
.
Find an approximate value of a, such that a A/a = 3. Put a = x*; then f(x)=x 2x -3=0; and f'(x)-x zx (2 + 2 log e x). Also /(ar)=-2 when z = l, and f(x)- +0-375 when 3 = 1-5. Starting with a - 1 5, the work proceeds as follows.
Ex.
I.
/(a) =0-375, /'(a) =9-487, /(a)//' (a) =0-04, /( ai ) =0-0193, /'( ai )
/(,)= 0-000255,
/'(a a )
hence, the value of a required
This method
is
often
is
and
ai
/
= l-46;
and
/(a 1 )// (a 1 )--0-0023,
= l-4577;
=8-262, f(*jlf '(**)= 0-000031, and a 3 = l-457669; (1-457669)*
employed
= 2-12480.
in cases
where the interval
very small, important to know under is a closer approximation to the root than a.
and then
11
= 8-324,
(a,
is
jS)
not
what circumstances a t
it is
We shall assume that neither f'(x) nor f"(x) is zero in value x the interval (a, /?), so that each of these functions reof any tains the same sign for all values of x with which we are concerned. Now Fourier's Rule.
for
where
0<0 <1, 1
so that
*=-/()//'( + Hence
Now
if
oq
t
is
W
-/(a)//'(a), it follows that t has the certainly a better approximation than
and a + A, and
since
This condition
a1
=a +
/,
this is the case
1 |
7 '
is
if
same sign as h. a if o^ lies between a
satisfied if /(a)
and/
(a)
1<| h
|,
that
have the same
increases with x or decreases as x increases according as
is if
sign.
.For
/"()0,
f(x) and h 2^0 according as /(a) and /'(a) have different signs or the same sign. Thus we have Fourier's rule, which is as follows // neither f (x) nor f"(x) is zero near the point a, and /(a) and f"(a) have the same siyn, then :
ax
is
a
better
approximation than
a.
VARIOUS CASES
458
The various cases which can arise are illustrated in Fig. 66, (i)-(iv). It is to be noted that the tangent to the graph of f(x) at the point where #=a cuts the x-axis at the point a x .
In the figures on the left, /(a) and /"(<*) have the same sign; in those on the right they have opposite signs and Fourier's rule does not apply.
+./(*)-
a
a a
a cn+h
FIQ. 66.
We
now show
shall
that a a
is
a better approximation than a, provided
that the interval is sufficiently small.
The true value
of the root
approximation than a JLA2
if
= a 4- h =! +
|J<|A|,
/f"(\
that
or
and therefore
,
/" (x) ^t
such that
1 1
If neither
is
a
in the interval (a,
\<.\f"(
f (x)
better
+ 0h)
\,
nor f"(x)
^3),
x
is
a better
is if
if |
If
<2
we can
find a positive
number
and we draw the following conclusion is zero
in the interval (a,
approximation than
a.
/J)
and
:
I
EVALUATION OF ERROR This
is
certainly the case
if .|
j3
-a |
<
'
^
459
'
,
.
Moreover, in cases where Fourier's rule does not apply,
at
a closer approximation than a
is
that
,
|
h \<\
t\,
and
if
is, if
2
Z|/(a)|<2{/'(a)}
............. (C)
In this case, by (A),
J5te.
2.
Let
Find
/(x)r,x -2a;-5,
Since /(2)<0
/"(2)
Xs = 2x + 5
the real root of
3
to
nine significant figures.
th-n
f'(x)=3x* -2, /"() -60:. / the root lies between 2 and 3. Also /(2)= -1, / (2)=:10,
= 12.
Taking a
2 and applying Newton's method,
ai
=2
-=^-=
2-1.
Here /(2) and /"(2) have opposite signs, and Fourier's rule does not apply. /i<0-l, ;and the least value of |/"(#)| in the interval (2,3) is 12. Hence
But
t
and
KI is a better approximation than aMoreover, /"(#) increases with a:, and therefore
/"(a
,.,. Next, taking a -2-1,*
,
.
we have
Hence
ai ---2-l
/(2-1) =0-061,
-Pl -^?
/'(2-1) -11-23,
//
/ (2-l)-12-6.
=2-094569
....
in the first
approximation, so that
J jL'2ij
Also the value of h
is
equal to that of h |
and consequently
|
|
<~
|
< 0-007,
(0-007
)2^|~|<
0-00003.
Moreover, e is negative, and so the root lies between 2-09457 and 2-09453. the error in taking 2-09455 as the root is numerically less than 0-00002. Again, taking a
find that
.
Also
|
c is negative,
*
2o
we
/(a)- -0-000016557... and /'(a) = 11-1 1607, ai =a -/(a)//' (a) =2-094551483 ...
giving
Moreover, 2-09455148,
2-09455,
Hence
\
2
n ^<10~ 11
and BO the value
.
24.
of the root to nine significant figures
This approximation might have been obtained graphically. B.C. A.
is
NEARLY EQUAL ROOTS
460
Ex. 3. A donkey is tethered to a point A in the circumference of a circular field of radius a by a rope of length L Find the value of I in order that he may be able to graze over exactly half of the field.
In Fig. 67
AB
Thus
2a'
is
LAOB=x, LOAB^x'
the stretched rope,
+ *= w
measured
in radians.
J
,
Area donkey can graze over segment A BB' + segment CBB'
= a 2 (x - \ sin 2x) + l*(x' - \ is
JaV,
to
will
Equating
this
given by
f(x)~x- tan x -f r.
Hence,
/' (x)
it
r
( I
be found that x sec
x-
n
f/
f (x)~
} ,
(
\ **
and
0.
1)
x - tan x
- sec
tan x
sin 2x').
sec x(\
j
+ 2 tan 2 x)-2 tan x sec
a:.
In a question of this kind it is important to start with a really good approximation, to be found graphically or otherwise. draw the figure so that the shaded areas are nearly equal, and then by measurement
We
Z_^O# = 71^l-239 Taking a 1-239, we find and the next approximation ^=
i
If
x increases from
positive.
radians (nearly).
that, approximately,
/(a) ^0-018,
^
7
-/(a)// (a)
= 1-239
--
0-003
-=
1-236.
therefore /"(a) is f"(x) increases from ?r/2 to oo have the same sign and, by Fourier's rule, a z is a closer
to ir/2,
Hence /(a) and /
"
(a.)
;
approximation than a. Further, we can show that /"(a)<34 if x
Thus the following
therefore / '(x)<34
a
radians
Hence
/
;
|
results are correct to the last figure
L AGE = 1-236
/'(a) ^-5-578,
is
- 70 50
X ,
and
/7a/6, very
Ifa ^2
:
sin 35
25'
= 1-16.
nearly.
10. Nearly equal Roots. Suppose that the equation /(z) = has two nearly equal roots within the small interval (a, j8), and let oc + h be the true value of one of them, then
where
a=/(a),
6=/'(a),
(?-/"(),
etc,
Neglecting squares of /?, we have a + 6A==0, giving Newton's rule. But the roots are separated by a root of /' (x) ~ 0, so that /' (x) is zero for some 2 Consequently /'(a) is small, and the term ch cannot be neglected in comparison with bh, so that Newton's rule does not 2 apply. Neglecting cubes of h, we have a + 6A + cA = 0.
value of x in the interval.
HORNER'S METHOD This equation has nearly equal roots, and
same as
(a
+ |6A)
2
= 0.
This result
is
therefore approximately the
is
Hence we have 2a
b
T = "2c
i
value of
461
.
f
"
fcPP 1
*-)'
used in Homer's method (Art. 11, Ex. 3) to suggest a
when
trial
2a/b and
b/2c are nearly equal. only applicable Jn the same way, if the equation has three nearly equal roots in the interval (a, j8), the value of one of them being a-fA, neglecting A 4 h.
It
is
,
we have a -f bh -f ch 2 + dh 3 Hence,
1 1
.
;
3a
7
*
which
is
b
nearly the same as c
--y---- -35
.
.
,
(nearly).
Homer's Method. The
Horner,
is
process described in this article, due to the most convenient way of finding approximate values of the
Any
rational roots
which
may
=
where f(x) is a polyexist can also be found in the same
irrational roots of equations of the type /(z)
nomial.
(a
+ |M) 3 = 0.
way. One great merit of the method is the very concise and orderly way in which the reckoning can be arranged. The root is evolved as a decimal, the figures of which are obtained in succession. In what follows, we are only concerned with positive roots. To find the negative roots of f(x) = 0, we find the positive roots of /( x) =0.
Consider the equation
and suppose that
f(x)
= a x n -f a^~
this has a single root
a
v
4-
.
.
.
a n ^x
-h
in the interval
-f
(p
an y
p
(A)
0,
f
1
),
where
a positive integer. (The case of two nearly equal roots will be conp sidered later.) Let (x^p-qrs ... where q, r, s ... are the figures in the decimal part of the root then p can be found as in Ch. XXVIII, 2, is
:
or preferably figure by figure as in Ex. 2 below. Decrease the roots of (A) by p, and let the transformed equation be n n -l + ...+b n _ l x + b n = a
(B)
This has a single root between and 1, namely 0-qrs .... In finding the value of q, we avoid the use of decimals by multiplying the roots of (B)
by
10, obtaining the
equation
= aQx n + lOV?
71
-1
* ...+I0 n - l b x + 1
W
n
bn
=Q
(C)
This has a single root between and 10, namely q-rs .... Moreover, is the number that such the roots of (C) are decreased by q the greatest q if sign of the last term of the transformed equation is the same as that of th>e last y
term of q and and
(C).
q+
This follows from the fact that
l, and there is no root between have the same signs.
<(#)
=
and
has a root between q,
so that
<(0),
HORNER'S METHOD FOR CUBE ROOT
462
In practice the value of q given by Newton's method. For, regarding 0-qrs
p ...
is suggested
and, after a certain stage,
as an approximate value of a,
-&/&_!
-f(p)!f'(p}
Having found the correct value of
is definitely
we have
(approximately).
we
decrease the roots of (0) by q, multiply the roots of the resulting equation by 10, so obtaining an equation of which r-s ... is a root. y,
Continuing as above, we can find as decimal representation of a.
When
a certain
number
many
we choose
figures as
have been found, about as
of figures
in the
many more
can be found by a contracted process explained in the next example. Ex.
I.
Find
to ten significant figures the real root
/(*)=
3
of
- 111=0
(A)
The work is arranged below in a form suitable for explanation. In practice be arranged as in Ex. 2. The explanation is given on the opposite page. 1
0+ 0-111(4 4
47.
48
it
would
(a)
+ 16- 47
8+48
_ uO .Qy
12 (6)
120+4800-47000(8 128 + 5824408 136 + 6912 144 1440
+691200
-
408000(0
14400 +69120000 -408000000(5 14405 +69192025 - 62039875
(c)
(d)
14410 +69264075 14415
11=0-8,
144,15
+ 6926407,5- 62039875(8 6927560
-
()
6619395
6928713 1,44
+ 692871,3 692884
6619395
9 (
(/)
383439
692897 69289,7 6928,9 692,8 69,2 6,9
383439 (5 36990 5
(j
(
2345 267
3 (
3 (
59
8 (
4
Hence the required root
is
4-8058955338, with possibly an error in the last figure.
EXPLANATION OF PROCESS NOTE.
Denoting the root by
Also that for the values f(x)
=
4-80
4-8
4-8,
4,
-47,
the successive steps show that
a,
4
463
-0-408,
4-8058955338 of
...
-0-062
...
...
,
etc, x,
-0-000000004
....
EXPLANATION OF THE SUCCESSIVE STEPS (a)
Decreasing the roots by
the resulting equation
4,
is
(B) If
(b)
by Newton's method
a4-fA,
Multiplying the roots of (B) by 10, the resulting equation (f)
^o;
(a?)
3
is
+ 120a; +4800* -47000-0 ............................ (C) 2
Taking 9 as a trial figure, we find that this is too large, for decreasing the roots of (C) this shows that
dgns,
and thus a root of
Trying
8,
we
(f>
(x) lies
find that this
resulting equation
is
and
between
9.
the correct figure, and decreasing the roots by
the
8,
is
s3
+ 144s 2 + 6912&- 408 .=
Denoting the root of (C) by 8
-f k,
................................. (D)
we have
Hence we may expect that the next two
and
figures of the root are
5,
and
this
proves to be correct. (c) Here the roots of (D) are multiplied by 10, and those of the resulting equation iecreasod by if they were decreased by 1, the sign of the last term would be changed, is the next figure of the root. 30 :
(d) is
The
roots of the equation obtained in
(c)
are multiplied by 10,
and we
find that 5
the next figure. (e)
The equation found ar
and the next
3
in (d)
is
+ 1441 5.T 2 + 69264075z - 62039875 = 0,
figure of the root is 8.
We
...
..................... (E)
have to multiply the roots
decrease those of the resulting equation by
of (E)
by 10 and
8.
In doing this the important figures will retain the same relative position if, instead of introducing O's, we write down the last term of (E) as it stands, cut off one figure from, the 3 2 coefficient of x, two figures from, tJiat of x , and neglect the coefficient of a: .
In
an equation of higher x3 and so on.
the case of
coefficient of
NOTE.
In
degree,
we should
cut off three figures
from
the
,
the multiplication allowance should be
(f)
This step
(g)
Only the
is
similar to
last
mod J 1
r the figures cut off.
(e).
two terms remain, and the process
is
simply contracted division.
TWO NEARLY EQUAL ROOTS
464 Ex.
Find
2.
to
nine significant figures the real root of
It is easily seen that the root lies
the roots by 10.
x*
- x* - x - 2000 =0.
between 10 and 20, and the
In practice, the work
is
arranged thus
first
step
is
to decrease
:
|f =0-7.
The required NOTE.
which
3.
lie
is
12-9686724, to nine significant figures.
It will be seen that
figure until
Ex.
root
we come
Find approximate
is
more than the correct
1
values of the two roots of
3s4
- 61X3 + 127z2 + 220*
as follows,
and the explanation
between 2 and
The reckoning
Newton's approximation gives
to the 8.
-520-0
3.
3-61 -55 -49 -43 -37
is
+127
+220-520 (2
+
+ 254+ 92
17
- 81 -167
3-370 -16700 -364 -17428 -358 -18144 -352 -18848 -346
on the opposite page. (a)
12
+92000 +57144 +20856
- 120000 - 5712
2 (
3-3460-1884800 + 20856000-57120000 (5
+ 1 1345875 -3430-1919175 + 1750000 - 3445 - 1902025
,.(6)
(c)
390625
-3415-1936250 -3400 -3,4
-19362,5 + 175000 -390625(4 -19376 + 97496 641
- 19390 -19404
+
19930
.(d)
SEPARATION OF ROOTS For the smaller root
For the greater root
:
-194,04 + 1993,6-641 (0
-641
199,3 193
- 62
18,7
- 62
+ + 187
465
(e)
(/)
(3
-149,9 +1582 (9 - 167 79 f
-1,9'
-
1,8
(3
(g)
6(3
...... (h)
:
-194,0 + 1993,6- 641(9 + 247 +1582 -1499
-18,4
79
-1,8
Thus the case
is
and 2-2549942..., where the
roots are 2-2540333...
(4
(')
....(/')
....(
5(2 last figure in
each
not to be relied upon.
EXPLANATION OF THE SUCCESSIVE STEPS (a)
Decreasing the roots by
2,
the resulting equation
3a^ - 37z - 167z 3
is
2
+ 92z -
By
the question, this has two roots between
(6)
Suspecting two nearly equal roots,
12 =0.
and
we apply
is
1.
the rule of Art. 10, and the figure 2
suggested by ...
Decreasing the roots by 2, the sign of the last term of the resulting equation is If the roots are decreased by 3, it will be found that again the sign of the last term .
-
So far, then, we have no assurance that 2 is the proper figure. But when the roots arc decreased by 2, the signs of the terms are + - - + so that two changes and when they are decreased by 3, the signs are of sign are lost. Hence, by Art. 8, the two roots lie between 2-2 and 2-3. is
.
,
+----,
(c)
and
(d).
Similar remarks apply to these steps.
~
2(-5712)
Thus the (e)
and
matter of
roots f
(e
).
trial.
of the terms are
lie
__
20856--'
figure 5 in
20856
.
and
The
(c) is
suggested by
nr5 -'
2rT8848T
between 2-254 and 2-255.
Here the
rule of Art. 10 fails,
If at (e)
+--
+
and the separation of the roots
is
a
we
decrease the roots by 1 (instead of by 0), the signs +, one change of sign being lost. The same is true if,
we
decrease the roots by 9. Thus one root lies between 2-2540 and 2-2541, (e'), and the other between 2-2549 and 2-255 ; and the roots are separated.
as at
Immediately after the roots have been separated, is suggested by Newton's rule. (/'), the proper figure
i.e.
beginning at the stages
(/),
APPROXIMATION TO ROOTS
466
EXERCISE XLIX Find by Newton's method to five significant figures 1.
The two
2.
The
positive roots of x*
3 positive root of 2x
:
- Ix -f 7 ^0.
- 3x - 6 =0.
Find the roots indicated below correct to seven places of decimals by Horner's method. 3
+ 29* - 97 = 0.
3.
The
4.
The cube root of 4129,
5.
The three
6.
The two roots of 2* + ll# - 10z-f
real root of
a;
real roots of x*
- 3z 4-
3
1 =- 0.
2
1=0
7.
The two roots of # -2(Xr - 3 1#+ 1609 =
8.
The two
9.
The
3
roots of 2x* -
greatest and
in the interval
2
Wx
z
(0,1).
in the interval
- Sx + 166 =0
in the interval
(13,15).
(5, 6).
least roots of
3s 4 - 6Lr 3 + 127* 2 -f 220* - 520 = 0.
The two
roots of 3z 4 - 10z 2 ~ 6x
+ 16 =
in the interval (1,2).
11.
The two
roots of 3x4 - 2x* - 34x + 40 =
in the interval (1,2).
12.
The two
roots of Ix* - llx 2 -32x-f -40 =
13.
The two roots of x* - 2x 3 - 13s 2
14.
The two
roots of
15.
The two
roots of 2z 4
16.
The two roots of
17.
The two
10.
3a;
x*
4
-4^ 3 ~
+ 4z
18.
The
positive root of 2z
19.
The
greatest root of the following (i) (ii)
(in)
-f 5a;
2 -f-
(4,5).
in the interval
f2, 3).
3x = 8002.
:
x + 4# - 2o;3 4- 10* 2 - 2x rr 962. x 5 + toe* - 10r3 - 1 12x 2 - 207* x5
in the interval
in the interval (2,3).
-27z 3 + 25z 2 -j-179:r-275---0 3
(3, 4).
12^ + 9^0 in the interval (1,2).
2
12o; -f
-16z 3 -17a a + 392- 782 =
4
(1,2).
llx + 133 --0 in the interval
+ 5$x*-388x + 511=0
roots of 2* 4
5
in the interval
4
4-
12z 4
4-
1 10,
59z 3 + 150* 2 -f 201* = 207.
x 20. If f(x) = e - 1 - 2#, show that the equation than other zero, the solution being a? =1-260 .... 1-260. Show also that f(x)<0 if
f(x)-Q has
just one solution
0
21. If
f(x)~e
x
-e~ A
positive solution given
-4x, show that the equation ,/(x)=0 has just one by x = 2- 177 ... and that/(z)<0 if 0
Find an approximate solution of (nearly), where the last figure is not to be 22.
of x, x*
a:*
=10,
relied on.
showing that
x -2-50616
Verify that for this value
9-9995 (approx.). l is given by f(x)~x~ ~log, ;r--0. By drawing the curves y~xr l and Five applit/=log 10#, it will be found that x 2-5 is a good approximation. cations of Newton's process give x 2-50616.1 -=
[Here x
CHAPTER XXIX IMPLICIT FUNCTIONS, CURVE TRACING 1.
Implicit Functions.
If
an equation represented by /(#,
y)
two variables x and y are connected by =0, we say that y is an implicit/unction
of x.
In particular, is
called
is
f(x, y}
an algebraic function
Taking
this gives
But
if
a polynomial in x, y
and
/(x,
y)=0, then y
of x.
as a simple case the equation
y=x
-
j(1i
t
a;
2 ),
and so defines
?/
as a two-valued function of x.
we
are unable to express y explicitly in if, terms of x, we are not justified in assuming without further enquiry that y is really a function of x in the sense hitherto understood. as generally happens,
The general question involves
difficult analysis,
and cannot be con-
Some exercise in tracing curves from their equations will, convince the student that, at any rate in a large number of however, does define y as a single- or multiple-valued cases, the equation f(x, y) =
sidered here.
function of
x.
The general form
of a curve represented by an algebraic equation can be found by pure algebra, and affords excellent practice in finding generally
approximate values. 2.
A
Rule for Approximations.
The following process
is
of
great practical use.
Suppose that
J/
= ^ + ^/(?/)
*~
x2 (y) +x?i/j(y)
+
...
,
small and/(y), <(?/), etc., do not involve approximate to the value of y, in ascending powers of x.
where x
Let
is
f(y + h)
=
f(y)
+
A/, (y)
.................... (A)
x.
It is required to
+
................................................. J
The first approximation
is
obtained by putting x=0, giving
y-
....................................... (0)
INVERSION OF SERIES
468
For the second approximation, y = a + xf(a+
2
y = a-fa where a
let
+ ...
<(a-fa)
= a-f x {/(a) -fa/! (a) +
+ x2 {(a) +
...}
0(x),* then
is
and neglecting small quantities of the second 2 etc., the second approximation is XOL, x
order,
+
...}
...
,
terms containing
i.e.
,
y = a + xf(a), so that
Next
oc
xf(a).
let
y = a-ha-fj8 where
j8
y = a + xf(a -fa -f j3) +
and neglecting small quantities
is 2
0(x
2 ),
...................................
then
<(a -ha + ]8)
Similarly,
we
if
y
is
0(x
3 ),
. . .
(a) ,
...........................
(E )
namely
2 jS-x {/(a) ./x (a) +
Thus
4-
of the third order,
2 y==a + xf(a + a) + x
which gives
(D)
^(a)}.
neglecting small quantities of the fourth order,
shall find that ,
which gives
the fourth
.................
(F)
approximation.
Hence the following Rule
Having found
:
r successive
approximations
to the value of y, to find the equation giving the next approximation, in equation (A) substitute the result of
the rth
approximation in xf(y),
the (r-l)th approximation in x 2
the
first
r approximation in the term containing x
and neglect terms containing powers Ex.
1.
Newton used the expansion of sin~ l
x=x-
sin
?/,
of
x higher than x r
2 (1 ~a; )~i
a
1
8
-
g
Putting x
y
1.3
-f
3
^
,
.
to prove that
z -
5
.
+
(A)
he deduced that
1^,U'* This
means
'
of the
same order
as
a;.'
(Ch. II, 7.)
ROOTS OF A CUBIC To do
469
write equation (A) in the form
this,
7
If
y
is
small, so
is
sin y,
and the
first
2/- ......................... (C)
is
approximation
Substituting y for sin y in the second term of (C) and neglecting higher terms, the second approximation is
y-%y* ........................................... (E) Following the rule, the third approximation
is
given by
siny^y-^^ and neglecting the powers of y higher than (F)
The fourth approximation
is
given by
Expanding and neglecting powers
than y7 ,
of y higher
this gives
and so on. Ex.
2.
If
is
fj,
If fj,
is
small, so
small, find approximate values of
is
y(y
- l)(y + 1), therefore y
y which
satisfy the equation
nearly equal to
is
or
1.
Write the
equation (A)
(i)
If y is small, the first approximation is 2/ = /x. For the second, substitute p? for y3 in (A), giving
For the
third,
y = /x -f (p, + /r ) 1
3 ,
^+
For the fourth, y =fi + (p. 4-
giving 6 3
3/z
) ,
y=
//,
4-
2/=/
p? + 3/A
y -p +
giving
^ + 3^
5
+ 12/z 7
.
Further approximations can be found in the same way.
(ii)
If
The
y-
1
first
is
approximation
The second For the
small, put
then equation (A)
is
Y = - ^/LI. (
F=
-J/i-|( -i/x-f/x 1
(iii)
is,
Similarly
a 2 )
that
if
y+
1
is
zero.
3 ,
.
small, then approximately
This result also follows from the preceding, for the is
.
-^
2/
we can show
be written
-i(-^)
y^-iM-lf* = 1 - i^ - f M 2 - ift8
giving
that
may
y = - J/i -f - Jft)= - J/t -f /x 2
is
third,
Y ~y - 1,
sum
of the roots of equation (A)
ROOTS OF A CUBIC
470
Cubic Equation with Three Real Roots. If all the ax -h 36x2 -f 3cx + d Q are real, the equation can be reduced to 3.
roots of
3
y -t/+/x = 3
by a transformation (Ch. XII, 5,
Ex.
1).
last
/x
x=p + qy,
form
of the
The
2
where
2
//
/^
+
'"
the roots of
<4/27,
....................... (A)
where p, q are
real
numbers
example suggests the possibility of expressing
the roots of equation (A) as power series in
Theorem.
<4/27
y
//,.
-y+^ =
3
are
where
+
V2
\n
y*-y = 2x,
Write the equation in the form
where
x=
-/z/2
..............................
of y can be expressed as power series in values these are given by theorem, Taylor's
Assuming that the values
/
and ^)o
(^i)o
x,
by
xn
x2
where y n = 'r: ax
(A)
and x2
are the values of y, y l9
(^2)0*
j/ 2 ,
...
n
when cc = 0. (Of course, x is to be made zero after differentiation.) The assumption will be justified if the series is convergent. Let y = u + v, then y 3 - 3uvy = u* + v 3 with (A) if w3 + v 3 = 2x and
w = x + (x2 - fc)i, 8
Thus we may take un =
Writing
dn u -=-^
vn
,
=
dnv -7-^
and
,
This equation will be identical
.
3uv v
3
1.
= - (a; 2 - &)*, where a?
& = 1/27.
differentiating with regard to x,
we have therefore
2
(x
= u/ 3
- fc)
Differentiating again, 2
giving
(x
;
jfc)
Since
y *
An
u + v,
- &)% = - v/3.
- ^)^w + 2 xu^x* k)~% = Wj/3, = w2 4- x^i = (x2 - &)i 2
(x
.
(x it
(x
similarly
we have
2
Similarly,
2
and
i) v 2
follows that
alternative proof of this result
2
(x is
^/S
w/9.
+ xt;j = v/9.
-k)y2 + xyl ~y/
.................... (C)*
indicated in Ex. 6 of the next Exercise.
EXPRESSION AS POWER SERIES Differentiating
n times by Leibniz' theorem (Oh. XXIII,
471 11),
(^-k)y n+2 + (2n + l)xy n ^ + (n^^)yn ^O ................. (D)
#=0
Putting
The
possible values of (y)
(i)J/ (yJo
-
885
Hence from
and
= 1/27, we have for n = 0, 1, 2, ... = 3(3w + l)(3M -l)(yn ) ........................ (E) (yn +2 )o
and
1*
are 0,
yi* 2
since
/^
where
3
x
4.6
a
(n
, 3
such values of x as make the
Denoting the series
Hence the convergent //
u 2n
series
%
series convergent.
u z + u3 +
=l, then
(y)o
Since the
O
is
Putting
A
we have
,
consequently also
un
(y 1 )
=l-
i
sum
Hence the numerical value of (y n Q is + or - according as n is odd or even. )
-F(-x).
of the roots of (A) is zero, the root corresponding to
y = F(-x)-F(x). -/x/2, we have the results stated above.
given by
x=
more general theorem, which includes
given in Ex. L,
this, is
7.
EXERCISE L Having given that x = tan x - J tan3 x + -| tan 6 a: - ^ tan 7 as + tan x - x + fx3 + ^-a; 5 4- ^^x 7 + prove that x + ayn where x, y are both small, prove that approximately 2. If y ~ y=x + ax n + na 2x m l + f n (3n - 1 ) asa: 3n 2 1.
. . .
. . .
,
.
.
3.
If
are
,
x
y
ss
. . .
Eu 2n+v and
,
-
-
+ l)(n + 3)...(3n-3) ^
2
if
(y)
by
4-
the same as before, the sign being Therefore =l (iii)
we have (y^o^l.
1)
(E) in succession,
y=-l+F(x) ,
(ii)
-
n
if
Thus
for
1. 2
are both small, then .approximately y=x + ay* + by* and = x + a* 2 + (6 + 2a )x3 + 5a (b + a 2 z4 a;, ?/
2
?/
)
.
APPROXIMATIONS AS POWER SERIES
472
y=l+xav
4. If
and x
small, then approximately
is
z 2 3 2 y = 1 -f xa -f x a log a + fa^a (log a)
5.
If
are positive integers (>s>
8, t
and A
t)
is
.
small, the equation
y*-y* + (-OA = has a root nearly equal to
l-X-(-l+8 + t)-(-l+2s + t)(-l+s + 2t). ii
y*-y=2x, show
If
6.
2
(ii)y,=
where y1 ~-~, 7.
yt =
y=u + v
If
li
that
-l) = (32/ ~l) 2
24*,
-
3
12 (t/-
Deduce that
~.
v m ~x-*/(x 2 -k),
um =x + ^(x 2 -k),
where
m
being
a
positive integer, then (i)
x and y are connected by the equation '
~m
~
'
2
li (ii)
vhere
If
a;
(t/)
is
sufficiently small,
is
a value of y obtained by putting dii d 2y
are the corresponding values of (iii)
Using Ch. XXIII,
11,
,
u>x
Ex.
1,
-^~ dx
m is odd,
one value of
. . .
in (A),
and (y^,
(t/ 2 )
,
...
.
show that
Hence prove that equation (B) holds (iv) If
,
#=0
if
(y) Q is 0,
x 2
and then
L* + ?3 VP Mn-^16'F "^^ L- 'H
K
|
I
(v) If
.
m=5
and
A;
6
=^, equation (A) becomes 6 y -
and the values of (y) Q are
Thus
all
I
0,
A/f g
the roots are real, and they can be expressed as power series in x
! &' (
vi) If
m=7
and
Ff = |,
the equation
is
y
1
- y 5 -f fi/ 3 - ^i/ = 2x.
if
DOUBLE POINTS Tangents and Asymptotes.
4.
Q
a point on
of the secant of
Q If
is
P
P
a point on the curve and defined as the limiting position is
near P, the tangent at P is PQ as Q moves up to P. Thus as it
.
from the tangent at
curve
If
P
473
Q moves to
P, the distance P the
becomes indefinitely small, so that near
indefinitely close to the tangent.
moves to an
the tangent at
infinite distance
along a branch of the curve, and
P tends to a limiting position,
this limiting position is called
an asymptote to the curve. It may happen that an infinite branch has no asymptote. We then find the simplest equation which nearly
and say that this is the equation to a represents the branch at oo curvilinear asymptote. In the examples which follow, tangents and asymptotes are found by a ,
process of approximation.
Intersections of a Straight Line and Curve. equation to a curve of the nth degree and y =/*# -f v that 5.
line,
-+
the abscissae of the points of intersection are given ...-fa n
= 0,
w=0
If
is
the
of
any straight by an equation of
.....................
(A)
obtained by substituting px + v for y in u = 0.
*
(i) If this equation has two equal roots, the line meets the curve in two coincident points and, except in special cases (see (iii)), touches the '
three roots are equal, the line meets the coincident points. Except in special cases (see (iii)), the
curve.
If
curve in three
point of contact is a point of inflexion (Ch. XVII, 20). At such a point, the curve crosses the tangent.
If
a
= 0,
a
If
(ii)
and the
line
=
one root of equation (A) meets the curve in one point at
and
^ = 0,
is infinite, '
F
infinity.
68
the line meets the curve in
two points at infinity and, except in special cases (see
where a curve crosses
(iv)), is
an asymptote,
a double point, or node. At such a point there are two tangents, each meeting the curve in three consecutive points. Any other straight line through (iii)
Any
point- (D)
D meets the curve in two consecutive points. If
(Fig. 69.)
the double point is at infinity, the
parallel straight lines.
two
itself is called
tangents are So in this case the curve has
parallel asymptotes, each
meeting the curve in
three points at infinity. Any straight line parallel to the* asymptotes meets the curve in two points at infinity.
FIG. 69.
CURVE TRACING
474
tangents at a double point coincide, the point There are two kinds of cusp, as shown in Fig. 70. If the
called $ cusp.
is
FIG. 71.
In tracing a curve, it is important to be able to account for all the n points in which any straight line meets the curve, remembering that imaginary points occur in pairs. (iv) Fig. 71, (i), is
a sketch of the curve whose equation
is
= (x-a)*(x-b). y* The
axis of x meets the curve in
A (a, 0)
and B(b,
2
(x
a)
The equation
0).
to
any
AP through A is y = m(x - a) and, on eliminating y we have ~m = hence, AP meets the curve in two coincident points (x -b
straight line
t
;
2
)
;
at a point P, given by x = 6 -f m 2 Therefore, A is the only point on the curve to the left of the line y = b and thus may be regarded as a double point- at which the tangents are imaginary.
at
A and
.
,
A
;
A point on a curve, in its
such as A,
is
immediate neighbourhood
is
The conjugate point at A, in Fig. form, when a a', of the oval in curve whose equation is
Curve Tracing.
6.
To
called a conjugate point, if
no other point
on the curve. 71,
(i),
may be considered as the limiting
Fig. 71,
(ii),
which
is
a sketch of the
find the general shape of a curve represented
by a given equation, proceed as follows, (i) Consider any symmetry which the curve may possess, e.g. symmetry with regard to either axis, or in (ii) Mark any points on the curve which can be opposite quadrants, found easily, e.g. the points (if any) where it cuts the axes, (iii) Find the approximate form of the curve near some of these points. Near (0, 0) (if on the curve) x, y are small, but not necessarily of the same order. To find the form near any other point, move the origin to that point. The quickest way of finding the tangent at any point is to find the value of T-
But, without further different!
of the tangent the curve
may
possess.
lies,
or discover
So, for a/etu points,
mation, as in Ex.
1.
ition,
it is
any
we cannot say on which peculiarities
side
which the point
better to use a process of approxi-
APPROXIMATIONS TO FORM (iv\
Find any regions
Find the form
(v)
which there
in
is
no part
of the curve near infinity,
475
of the curve. i.e.
(Exx.
1, 2.)
where one or both of
x and y are large, but not necessarily of the same order. In doing this, we find the (linear) asymptotes (if there are any), and the approximate form of any parabolic branches. '
'
The reader should be able to draw the 'parabolas' represented by m and n are positive integers. For example,
m n where y =x ,
y x
x
f
O
O
y-
x
y FIG. 72.
By
(vi)
are parallel to the axes.
(vii) It
or again,
We
~
considering the value of
may be
,
dx
find the tangents
(if
any
exist)
which
70 JS)
d y dx 2
Notice that, at a point of inflexion,
''
possible to find y explicitly in terms of x or x in terms of y able to express x and y as functions of some variable /. ;
we may be
can then find any number of points on the curve and the gradients at '
these points.*
Examples on Curve Tracing.
7. Ex.
1.
Trace the curve
The equation
is
(
The points
is
= (x2 - 4) x
~x,
~y
y
1),
(A)
Hence there
for x, y.
(2,
0),
(2,
is
are
symmetry
in
on the curve. (A)
y
y~4=x.
the equation to the tangent at
2nd approx.,
1)
- x3 - y + x
x and y are small and of the same order.
1st approx.,
This
1 )y
(0, 0), (0, 3
Write the equation (0, 0),
-
unaltered by writing
opposite quadrants.
Near
if
~ 3 y 4x + 63^
x
(0, 0).
,
obtained by putting (4#) 3 for y3 in (A), showing that the curve is above the tangent in the first quadrant and below it
in the third quadrant.
Thus
(0, 0} is
a point of inflexion.
FIG. 73.
* The examples given in Art. 7 and in Exercise LI have been chosen so that one of these methods can be used. Thus in each case the curve can be plotted to any degree of accuracy required so that the student can verify the conclusions arrived at by the methods just described, and gain confidence in using them. However, when only the characteristic form of the curve is required, the advantages of the above method over the method of plotting, even when the latter can be done, will be obvious. [See Ex. 22 of Exercise LI.] '
'
;
2
H
B.C. A.
APPROXIMATE FORM AT INFINITY
476 Near
(0, 1
2 J = - 4x
1st approx.,
which
is
Y = y - 1,
putting
),
y=I
or
27
equation (A) becomes
- 2x,
the equation to the tangent at
+ 3 Y 2 + 7 s = (x* - 4) x.
y
(0, 1).
2 7 3 - 2#) 2 - - 4z, 2nd approx., giving - 2# - 6# 2 which shows that the curve is below the tangent. -f-
y=1
(
,
\
Near (2,0), putting X=x-2 we find that = - 8z - Gz 2 (approx.), showing that the t/ -S(x -2) and that the curve equation to the tangent is y
FIG. 74.
9
Again,
-
dyjdx
(3x
2
gradient of the tangent
and the tangent
is
2 - 4)/ (3?/
1
parallel to Oy,
3 2/
From
is
it.
1 ), the Therefore, at each of the points ( 2, is parallel to Ox where x=2/\/3=l-16;
).
where
y~
large
-a^~0,
dil/s/3-
and
of the
giving
0-58.
same order
;
and approximately
y~-x~Q.
y - x=
(A),
This shows that, as :r~>
which
below
The tangent
is 4.
Near (00,00), x and y are both
2nd approx.
is
oo
,
the curve becomes indefinitely close to the line y - x
0,
therefore an asymptote.
Again, on the curve that the curve
is
Thus the curve
y~x--
(approx.),
and on the asymptote y = x; showing
below the asymptote on the right and above is
as given in
-Fig.
it
on the
left.
75.
y
Putting y ~tx in equation (A), it will bo seen that any number of points on the curve can be found by giving any values to t in the equations y = tx, x 2 = (4 - *)/(! - 13 ). If
l
in the space
the corresponding point is imaginary, thus no part of the curve lies from the position y = x to the swept out by a line turning round
position y = 4x.
ASYMPTOTES PARALLEL TO AXES Ex.
z 2 (x-l)(x~2)y = 2x
Trace the curve
2.
477
-y
(A)
henco the curve for y symmetrical The equation is unaltered by writing hence no part of the curve lies with regard to Ox. Also y is imaginary if 1
;
;
* 2 )=0
N ear
the curve
(0, 0),
by
y
z
-x 2 =
is
(A)
approximately given
y~x.
or
Taking y =x as a first approximation, from (A) 2 (y - x) = (3*2/ 2 - re j(y 4- x)
V
;
)
giving as a second approximation
showing that y -x Similarly y + x~Q
Near
(1, GO
),
is
x-
1
3z 3
3
2
2x
4
FIG. 76.
.
is a tangent at (0, 0) and that the curve a tangent at (0, 0) which is a double point.
2^2 1
;
-..
^ear
(2, oo
),
9^2 "
r-2{i
-
.-,
(#-")?/
1} y3
~r~
-2)#
(1
O
-
^
92
~-
showing that x 1 =0 and x 2 =~0 are asymptotes, the curve approaching the first from the left and the second from the right. x 2 (y z -
Again, writing equation (A) in the form
near
We
3
'"
(oo
>s/2),
y
above the tangent.
I2
2 "
~
is
and near
(oo,
2)
FIG. 77.
=
-
I
-^2),
y
- 2y 2
,
it
+ N/2 ^ -
will be seen
that
3
can now sketch the curve, as in Fig. 78.
parallel to Ox can be found by equating to zero the coefficient of the power of y in (A) similarly for those parallel to Oy. Also, for all values of k 1, the line y - k meets the curve in two finite points and in two points at oo except The same is true for the line a; /, where /2.
NOTE. The asymptotes
highest
:
.
FOLIUM OF DESCARTES
478
8. In the next example we consider the possibility of finding the approximate form of a curve near the origin, or near infinity, by retaining only certain terms in its equation.
Ex.1.
Trace the curve
+ y*-3axy-Q
x*
(a>0) .............................. (A)
3 term, we have y -3axy=0. the we the equation factor consider y, Disregarding
Omitting the
If
y
is
first
small, x
is
small and 0(y 2 )
so that r*
;
t/
- Sax =
),
and can
2 6
is
0(?/
................... (B)
therefric be neglected in comparison with the other terms in (A). Hence (B) approximately represents the curve or part of the
curve near
(0, 0).
by omitting the second term, we find that part x 2 - Say = 0. (0, 0) is represented by 3 3 0. Omitting the third term, we get a; -)-?/
Similarly,
of
the curve near
O
FIG. 79. This represents the curve near (oo oo ) ; for if y is O(x), then 2 Saxy is O(x ), and can be neglected in comparison with the other terms of (A). Thus near ( oo, oo), as a first approximation ?/-ha: = 0. ,
For a second approximation,
A
third approximation
is
-f
x
3a
(the last step being obtained
-a
a3 4-
approaches
;
it
,
-
z
-
-
-
=
- x - a for
y,
*
f
~x(x-\-a)
by long
division).
showing that y -f x = - a
is
/V.2
3o -^-^
;
obtained by substituting
xy
~ y+x
_
SAJ
y
Hence the
- a. thus "2
third approximation
is
an asymptote, and that the curve
from above at both ends.
y
o
-a \
\
\ \
-a
FIG. 80.
The curve
is symmetrical with regard to the line y x, for its equation is unaltered x and?/ are interchanged. Further, if y tx, then s = 3a//(l -f J 8 ), so that any number of points can be found by giving different values to t e.g. < = 1, = 3a/2.
if
;
xy
ORDER OF MAGNITUDE OF TERMS Newton's Parallelogram.
9. t/ =
A
where u
is
a polynomial in
479
Consider the curve represented by
x, y.
Let v be the expression obtained by retaining only certain terms of u. rule which goes by the name of Newton's Parallelogram enables us to *
'
choose these terms so that v =
may
near the origin or near infinity,
when such
represent the curve,
or,
part of
it,
parts exist.
axyn
Let
cx r y s
1
bx^tf
,
,
,
...
be any terms
of u. Represent these by points A, J5, C, ... whose coordinates referred to any axes are
(m, n), (p,
The
q),
(r, s), ...
formed
figure so
.
H is
m n Suppose that ax y
,
'Newton's Diagram.' bxVy* are small or
K Fio. 81.
~m 0(x p ), and therefore y is O^te-wO/fr-fl)). Join AB, cutting Ox in H, and let the angle xHA=0; ~ ie and axmy n is 0(xm ncoie ). MHO, w-ncot0 = 0#; then y is 0(x) therefore the terms represented by A, B are 0(x H ). Draw CK parallel to AB to meet Ox in K then, for similar reasons, Thus the term C is of an the term cx r tj s represented by C is 0(xos ). order lower or higher than the terms A, B according as C is on the same side of AB as 0, or on the opposite side. Hence the following rule. large
numbers
of the
same order
;
then yn
~~
(l
is
;
Rule. Mark on Newton's diagram points A, B, C, ... representing all the terms of the polynomial u. Choose any two of these, say A, B. Let v be the expression whose terms are those represented by A, B and
any points which may happen to then v =
lie
on the
line
AB.
AB
remote from the origin 0, = (or part of it) near approximately represents the curve u
If all the other points lie
on the side of
on the curve).
(if
v=
AB as 0, then
the other points lie on the same side of sents the curve u Q (or part of it) near infinity. If all
NOTE. If the straight line AB passes through 0, the rule terms cannot be of the same order. In the following examples the points representing the the equation are marked 1, 2, 3, ... in the diagram. Ex.
fails
1st,
;
repre-
for the corresponding
2nd, 3rd,
...
terms of
1.
The terms
1,
3 and
2,
3 give the form of the curve near
the origin.
The terms Art.
8.)
1,
2 give the form near infinity.
(See Ex.
1,
Fio. 82.
EXAMPLES OF CURVE TRACING
480 Ex.
2.
- by)
(ax
2
= ax 2
ij
-f
4.
y
Write the equation in the form, a 2 x 2 - 2abxy + b 2 y 2 - ax 2 y -y*^0.
The form near the
by terms
origin is given 1, 4 and 4, 5.
by the terms
infinity
Thus near (0, 0), (ax - 6//) 2 =0. Near infinity, a z x 2 ~ax z v-a=0 and
yQ
ax 2 y-l-y*
and
and near
1, 2, 3,
~^
giving
0,
EXERCISE LI Trace the curves represented by the following equations and work out the accompanying details. (For hints for plotting, see Ex. 22.) 1.
y*-y + x=Q.
Near
(GO
,
2 ?/
3.
a;
4.
a;
3
3
Near 5.
(i).]
3
Near
.
oo
),
The tangent 2.
[Fig. 84,
y ^x + x y~ -x* -
Ne,ar (0, 0),
[Fig. 84,
1).
+ 2/ 3 ^a 2 x.
+ 1/ 3 - 3ax 2 3
(0,0),
t/
[Fig. 84,
= 3aa; 2
(a, oo
a
passes through
Near
(ii)].
(0,0),
Near
(iii)].
a
(a;
no
=
),
8.
x(y-x)
Near
Near
(a, oo
a
2nd approx.,
2
O
.
2
near
) ;
z y*=a x;
a),
closer
Ox where
Near
oo
2 ,
?/
^a; 3 .
near (00,00),
y=x*Jax + a.
dy -~
on the curve, but the curve
0,
(0, 0),
a (
x
O
is
y
*
a
conjugate point.'
-,
2-a(>/5
x*^a*y.
approximations being
[Fig. 84, (viii)].
.
y=
consequently If
[Fig. 84, (vii)].
x = a-{
:
\
-1.
2
),
!#
.
(0,0),
is
[Fig. 84, (vi)].
y~xa,
=;aij
(a; -f
(0, 0),
near (oo,
;
+ a).
Tangents are parallel to 2
^
y=
/
real point near
2 2 x(y-x) =a y.
Near (00,00),
.
near (00,00),
;
Near (00,00), y 7.
- \x - |z 2
at the points
[Fig. 84, (v)].
,
y (# ~a)~tf
1
1
Ox
y 6.
y ==
),
[Fig. 84, (iv)].
.
(x-a)y*=a*x. near
1
3~.
perpendicular to
is
^a; 2 (2;-f
$jc~
(0,
+ 1)
Near
I'D approx.
(0, oo
),
x=?* y
y=
3x.
Near
Near (00,00),
(0, 0),
a3
-
1st approx., la*
(A).
3rd approx., y
x
V^T"
This shows that the curve becomes indefinitely close to that represented by (A), to the parabola (y-x-a) 2 ~ax. This is therefore a parabolic asymptote.
i.e.
EXAMPLES OF CURVE TRACING
482 x*
9.
Near
- 3*y 2 + 2y* = 0. a;
(0,0),
3
-3i/
Near (00,00),
2
[Fig. 85,
^0
and
(i).]
t/^fz-fz
1st approx.,
# 4 -f2i/ 3
2nd approx.,
y-^
2 .
i.e.
0,
r ..................... (A)
y
2^ y3
-
T + &- ..................................... ( B )
03
-----
in the diagram ; it gives the general Equation (A) is represented by trend of the curve, but (B) shows that it is not a parabolic asymptote. If y tx then x = t*(3-2t). Mark points on the curve corresponding to < = 1,2, -A -1. %
10.
2 ax(y-x) = y*.
[Fig. 85,
Near
(ii).]
and
y-x=Q
(0, 0),
ax^y*.
3'
With y = #
as 1st approx., the
2nd
x is
z
?/
z .
^
Near (00,00), ax*-y*
0.
a2 a2 11. x'A
x'-a'sy-iy^O.
-a zy=0.
powers of
x,
[Fig 85,
Near (00,00), by solving show that approximately
The equations y =
(0,0) (0,
2nd approx.,
y~^2
beuig 3x 2?/
0,
Near (oo,3),
2nd approx.,
?/
3
and
-
-^
in descending
a
by -------
in the diagram.
...................................... (A)
[Fig. 85, (iv).]
are on the curve.
Near
O -
3x~2y^0.
(0,0),
xJ, showing that the point
O
is
a cusp, the tangent
2*^/2
12 -
.
x
The asymptote
2/
viz.
(
- f, -
3 cuts the curve again at
3x 2 + y*~Q (represented by
-----
)
(
-J).
-f
,
3).
................... (B)
y~ -3 s X s -
3# 2 ~ I) (y If 3z 2 + ?/ 3 ^0, then - f - f ). the ( 3
-f
y and expanding
which meets the curve at one other point,
Near (00,00), that
2)
O
for
a*x
are represented
(3x~2y)*-3x*y-y*^0.
The points
x3
y=-r$--i D
(0,0),
tt
x2
12.
Near
(iii).]
-4
I, showing that the curve (A) is below (B) and a parabolic asymptote. 2 (3.r -2*/) -0, showing that the curves (A), (B) touch at
is
point solving for x, show that ,
By lines
y=
-
1
and y
Fig. 85, (v), (vi), 16.
y< - 1 or 4 touch the curve.
(vii), (viii),
else
0
Also show that the
are curves whose equations are given in Exx. 13-
These are of the form xy*-2ky~px* + qx* + rx + $* (See Exx. 2
- I2y = x- 15x 2 4- MX - 85.
14. xy*
15. zz/ a
3 2 -242/^:c -10:r ~5a:+150.
16.
13. xt/
17, 18.)
- IQy^x* - 14z a + 60x -
104.
a 8 8 a^ -8y = x -8a; -3aJ-8.
17. Show that every curve whose equation is xy z - 2ky = px* -f- qx 2 -f rx + s has three asymptotes, or only one, according as p is positive or negative. * Curves having an equation of this form belong to the first of four classes into which divided cubic curves (Enumeratio Linearum tertii ordinis, 1706).
Newton
/
FIG, 85*
(viii)
POINTS OF INFLEXION
484 18.
Show
(i)
that, if
of x other than zero, y
xy
2 2ky px* + qx + rx + s, where p^Q, then, for values given by an equation of the form
2
is
(xy
-
2
k)
=p (x -a)(x- b) (x -c)(x~d).
(ii) In Exx. 13-16 employ the methods of the previous examples to find first and second approximations to the curve at infinity and, by the help of the points, A, J5, C, Z>, which correspond to x=a, b, c, d, and of the points where the curves ;
cut the asymptotes, trace the curves.
NOTE. Diagrams (v) and (vi) show that a cusp ia of a higher order of singularity than a node ; while (vii) and (viii) show the change of partners that generally '
*
occurs when two of the points, A, B, C t D, approach one another, coincide, and then separate again, owing to a change in the coefficients, p, q r, s. t
Points of Inflexion 19.
Show
that the points of inflexion of the cubic curve z
y(a'x lie
on the straight
line f
%y(
[From
(i),
by
+ 2b'x + c') = ax* + 3bx z + 3cx + d
whose equation is - 6"2 - 2bb' + ca') x + (be' (ac' -:=
)
2cb'
....................... (A)
+ da').
differentiation,
(B)
Differentiating again,
and putting
d 2 )/
~0, we have
ct>x
(C)
The coordinates of a point of
inflexion satisfy equations (A), (B),
and
(C).
Multiplying equations (A), (B), (C) respectively by a',
the result
is
-2(a'x + b')
9
a'x 2
+ 2b'x + c',
obtained by addition.
20. Show that the curve (p. 475), whose equation is (y*-l)y~(x 2 -4:)x 9 has three points of inflexion which very nearly lie on the line x 64y. 2 = 6 4 6 [If p=dy/dx 9 show that, at a point of inflexion, # (4p -p )/(^ -l), and
y* = (4p*-l)/(p*-l); so that
21. Prove that the points of inflexion on the curve y*~x z (x 2 + 2ax + b) are determined by the equation 2x*-'+ Qax 2 + 3 (a 2 -f b)x + 2ab =0.
[At a point of inflexion
22. Plot
For For For For
+ 3ax 2 + bx, and y~-~2x* dx
some of the curves
in
Examples
-^
( ) \flte/
1-12, as follows
= 6z + 6o# -f 6.1 2
:
Ex. 1, plot x~ -t/ 3 , and # = ?/, and add corresponding abscissae. Exx. 2, 5, 6, 7, 8, 11, express y in terms of x. Exx. 3, 4, 9, 10, use y=te, or x=ty. Ex. 12, express x in terms of y.
CHAPTER XXX INFINITE PRODUCTS a v a2
1. If
product a x a2
an
...
,
...
an
is
...
will
a sequence of numbers, real or complex, the
be denoted by
Pn
or
by II^ia r .
Definition.
n-> oo a 1 a 2 a3
we
,
Pn
If
tends to a
except when a factor of
...
is
convergent and
Pn
Thus we write
.
limit P, different
finite is
an
.
from
zero, as
we say that the infinite product
zero,
that its value is P, or that it converges to
P, and
write
P=a Pn-> oo
If
infinite If
Pn
or
product
is
oscillates,
a 2 a3
1
-
oo
...
,
to oo
or
= /7"~5a n
Pn->0
if
or simply
,
when no
=
factor
Uan of P n
.
is
zero, the
said to diverge or to be divergent.
the infinite product
is
said to oscillate.
The reader may ask why we say that the product diverges (to zero) when Pw ->0. The reason is as follows. According to the definition, if a a2> ... are real and positive, the infinite product Ua n and the infinite >
series
2 log a n
both converge, both diverge or both
For logPn^logaj-f log a 2 +
...
-flog
oo
,
an
,
and
or oscillates according as log , to zero oo or oscillates. to
to oo
If
Pn->
a
finite limit,
then
Pn
..
1
Pn
Pn
oscillate.
tends to a
tends to a
~> the same
limit,
and
finite limit,
finite limit, to
this limit is
if
not zero a n = Pn/Pfl _ 1~>l. Thus if Ua n is convergent, then a n->l. It is generally convenient to write an infinite product in the form
A necessary (but not sufficient)
condition for convergence
that w n->0,
is
and so, after a certain stage, un |<1. The character of P, as regards convergence, will not be affected by removConseing any number of factors from the beginning of the product. quently there will be no loss of generality in supposing that u n |<1 for |
|
every
n.
The simplest and at the same time the most important case the us have the same sign.
is
when
all
CONVERGENCE
486 2.
Theorem.
// u l9 u2 w3 ,
and
are positive
...
,
than unity, each
less
of the infinite products
is
convergent or divergent according as the series
The
Proof.
series
Suppose that found so that
Zu n
2u n is
is
convergent or divergent.
convergent and that
um + u m+l + um+2 + Also the omission of the its
first
m-l
sn
quently
By
Ch.
is
XIV,
1,
no
;
then
m
can be
factors of a product does not affect
Also
Pn
increases
1
)(l-fw 2 )...(l+w w )<
-U,)(l
-,)
...
Q n >0.
3.
T1
-<,
1
-
(!-)>
By
Ch.
:
1
XIV,
-*.
u n ) diverges to
to
any
so that 77(1
+u n
)
zero.
is
convergence of the
Corresponding
1,
and Q n ->0,
complex, none of which
sufficient condition for the
as follows
>
,
Pn->oo
Therefore
and 77(1
real or
1~6
6n
General Condition for Convergence.
numbers,
conse-
and Q n decreases as n increases. Therefore P n and and both of the products 77(1 -f w n )
and 77(1 ~u n ) are convergent. Next suppose that Zu n is divergent.
diverges to oo
s
for every n.
tend to positive limits as n->oo
Moreover,
assuming that
we have
g n = (l
is
is s
<1.
to oo
...
loss of generality in
Pn = (l+u
of
sum
its
convergency.
Hence there
Qn
Eu n is convergent or divergent.
positive
Let
zero.
infinite
number
(a n )
be a sequence
The necessary and product
6,
however small, an
integer m must exist such that for all values of p
that is to say,
Proof.
which
is
First suppose that
< P
is
for
^ = 1,2,3,
convergent.
not zero, and consequently a positive
(A)
Then P n tends to a limit number k exists such that
for every n.
ABSOLUTE CONVERGENCE by Ch. XV,
Also,
6,
we can
find
m such that y = l,2,3,....
for _rn+p
Therefore
487
fa
~i
so that condition (A) holds.
condition (A) holds, then
if
Conversely,
l-
and Taking e
I
I
I
~~
I
I
\
|
XV,
so that
|,
|P n |>some
\
IP m\ *
fc
\
I
\
is positive and as small as we choose. is convergent. zero and the product
P
6.)
Absolute Convergence.
numbers whose absolute the infinite product
is
convergent. Hence the product is
u l9 u 2 w3 ,
...
,
are real or complex
by w/, u 2 ', u3 ',
...
,
P = (l +*!)(! +^(1 +113)...
said to be absolutely convergent
Zu n
If
values, or moduli, are denoted
is
the series
|
IP * f m \< ^ m+p -IP
Pm+p Pm \
(See Ch.
Pm
.
Consequently Pn cannot tend to zero. Pm+p and Pm have the same sign, and therefore
IP L m * m+p -P Hence Hence
-e)
................ (B)
number when n>m.
Also by (B),
4.
Pm+J) |>(1
|
77(1
+u n
)
when
is
the product
absolutely convergent
if,
and only
if,
absolutely convergent.
Theorem. An absolutely convergent infinite product (P) is convergent. Let P and P' be defined as above and suppose that P' is convergent. Then, by Art. 3, we can find m so that
|(l+t4+i)(l+t4+a)-..(l+t4+p)-l|<6
Now
for
^ = 1,2,3,....
(1
This follows on expanding each side and remembering that for any
numbers
a, 6, c,
...
Therefore
|(l+t/m+1 )(l+wm ^2 )...(l+wm+P )-l|<
Hence, by Art.
3,
P is convergent.
for
2>
= 1,2,3,....
EXPANSION OF PRODUCT
488
We
Derangement of Factors.
6.
shall
prove that if the factors
of an absolutely convergent infinite product (P) are rearranged in any way, remains absolutely convergent and its value is unaltered.
the product
Consider the infinite products
where
P is absolutely convergent and Q is formed by rearranging the factors
of P, so that every v
Since
is
a u and every u
a
is
v.
P is absolutely convergent, so is the series Su n when
absolutely convergent
This series remains
.
Zv n
Therefore
terms are rearranged.
its
absolutely convergent, and consequently the product
Q
is
is
absolutely
convergent.
For any
Pm =a and
is
a2
...
w,
am
it is
q,
...
|
Pm
are
s
a p aa
all
>w.
follows
it
. . .
as -
1
Also, as
n
= 1 + vr
br
,
.
so that all the factors of
Q n = b^b2
the factors of
...
bn
,
a*
-
m tends to
oo
so does n,
,
and
since
from Art. 3 that
and therefore
1->0,
and Q n tend to the same
Expansion of an
6.
among
QJPm = a vaQ
convergent,
Thus
possible to find
are included
therefore
where p,
P
1
suffix
Let a r = 1 + u r
P = Q.
shown that
It remains to be
Qn /Pm -* 1
.
P = Q.
limit, that is to say,
Product as a Series.
Infinite
Consider
the infinite product
where ul3 w 2 w3
... and x are any numbers, real or complex. Let p r be the sum of the products of u ly u 2 ... u n taken r together, so ,
,
,
that
Pn = (l +xu We
(ii)
)(l+xu%)
prove that if
shall
(i)
1
As n->oo For
,
...
(1 -f
limit
so that j>/
Each
is
|
,
the
lr
+px + p2x* +
...
+pn xn
is absolutely convergent,
= for r
l, 2,
...
.
then
n.
of x,
P = (l +zw 1 )(l -f xuz )(l +xiiz) Let w/ = u r x' = x and |
Eu n
the series
pf-> a finite
all values
xun )^l
...
to oo
= 1 + ^x +
Z
2
x2
-f ...
to oo
.
,
|
sum
|
w2 ', ... w n ', taken r together. r occurs in the expansion of
of the products of W/,
of these products multiplied
by
1
GENERAL COEFFICIENT '
therefore
pr
.
Eu n
Since
occurs in the expansion, and
r
I
is
erefore
Also
If
pr '
Zu^
absolutely convergent,
r->oo
then
,
s'
r
/
I
follows that
it
converges to a
sum
s',
and
r
j\r.
increases with n, therefore, as n->oo
pr
489
r->0, and consequently
,
#>/-> a limit
Z
r
'
l
'->0, so that
such that
r
f
lr
is finite,
however great * may be. Thus as n->oo p r becomes an infinite series, which is convergent and pr becomes an absolutely convergent series. Hence pr -> a finite limit l r) and we may write Z r = (l +e r )Pr where e r->0. f
;
,
Let
where
Qn = l +
m
I
l
x+
I
2
x2 +
...
the greatest of
is
Q n -Pn ->0,
Therefore
+l n xn
|
and
l |,
then
,
|
c2
I?
e^c
-
Pn->P,
since
...
JE'ar.
1.
^e
^t'nc?
coefficient
In this identity put zx for
therefore
(1
+zx*)(l +zx?) +zxn+l
2,
therefore
... (1
)(l -\-p l
+z*n + 1
z+p 2 z 2 + ...
-f-pn 2"
coefficients of z r ,
Equating the
1
-xn - r+1 ^
Putting
r
~-
1
,
r
-
2,
...
2,
1
in succession for
n ~\-x Whence by
multiplication
r
r,
follows that, for
to oo.
r of z in the expansion of
Assume that
(1
it
all
values
LOGARITHMIC METHOD
490 Ex.
2.
//
|
x <1, show
that for all values of z,
|
...
where
= l+f
00
to
I
x 4- x 2 -f jr3 and denoting convergent,
The
where
series
Z
its
pr and pr
lim n *>
r
is
. . .
tend to zero as
absolutely convergent, therefore the infinite product value by P, by the theorem just proved,
is
the same as in Ex.
therefore
?i->oo,
Another Method.
Z
r
Also
1.
xn
xn ~\ ...xn
,
~r +l
is
all
has the value stated above.
When
the factors are real and positive, the theory of infinite products can be made to depend on that of infinite series by the use of logarithms. Let P w -(l + wi)(l +^)...(1 +u n ), 7.
where u l9 Art.
1,
t/
2 , ...
~I
are real and
Hence the
Pn = log
(1
+1^) + log
P
infinite
if
I
is
z>-l
+w2 + ... +log )
1 ~
um+l
x~
y
Putting
(1
ti^+i,
,
+w m+1 + log )
w m 4.2>
...
ww
4.
(1
^w_,jn ^- r
L is l+u n
(i)
the least and
_L 77
+ti ro+2 )
+
for x in (A),
2 "if
where
numbers
of the
H the
m>n
2
both
)
1
and
.........
,
(A)
1 -f x.
,
~ sm^ <%
... -f
log (1 +ti w )
by addition we
_ 4 m n ^-^ ?i5
2~L
>
+u n
7, (4), 1
m+I^m+2^~'^ U n
log
XIX,
in
).
27 log (1
r-
a*
+uM+2 + ...+u n = rm n
4-7/2
7/2
Ch.
By
+w n
(1
2
and h the greater
Let
infinite series
oscillate.
and ^0, then
tV> smaller
(1
and the
product converge, both diverge or both
...,
As explained Then
r.
the last assumption involves no loss of generality. log
where
for every
r
greatest of the
= Jmfn
.
find that
'
numbers
1
,
1 4-
uly
1 -f
w2
,
.
Suppose
that
u n2
convergence we can find
is convergent,
m
so that for
then by the general principle of all
values of n,
5 m>n <2Lc,
and
therefore
c is positive and Hence by the general
where
as small as
please.
principle of convergence,
diverges to 0, or oscillates, as
or oscillates.
we
un
P converges, diverges to oo
converges, diverges to oo
,
diverges to
-
oo
,
,
EXAMPLES OF INPINITE PRODUCTS (ii)
it
Zu^
that
Suppose
then since
diverges,
491
r,n,n~^,n^**Vn
follows that, for sufficiently large values of n,
for
any
T
A
positive value of
however
,
(a)
If
u n converges or
oscillates
(b)
If
u n diverges
+
then
P may
Hence
between finite limits, then
or oscillates so that
o>
P diverges
upper limit
its
is
to 0.
-f-
GO
,
converge.
possible cases, and when the logarithm of a complex has been defined and its properties investigated, the same method
This includes
number
to
great.
all
can be applied when the factors of
P
are complex numbers.
EXERCISE
LII
Obtain the results in Exx. 1-5 by considering the value of 1.
d+t)(H
2.
(i-
4.
5.
(l+i)(l
6.
Show
\P*ni =
l
Pn
in
each case.
and
that .
A
Hence
x
if |
\
(l+x)(l+x z )(l+x*) 7.
Show
x = 2n
sin
---
Remembering that Jim 0->o
^
~ cos
1,
that
Show
^v. A
C
+ a:) (1 +-)( 1 V 2 /^
|
cos
^
...
cos
^
.
x
sin
a;
diverges to oo or to 4-^ 3 /)...
according as
that the infinite products
are convergent for ~
(1
^
O. 9.
to oo
deduce that
xx
Show
...
that sin
8.
X
< 1,
all
values of x. B.C.A.
EXPANSION OP PRODUCTS
492 10. If
|a;|
_(* a ? '~
where CTe
[Proceed as in Art.
-*")(*
1.]
11. If n, r are positive integers,
exactly divisible by
is
12. If
(
1
)
(!-*)(! -*)...(l
Ex.
6,
~* +i - d
show that
- x) ( 1 - x 2 )
.
.
.
(
\x\<\ and \zx\<\, show ...
where
*
[Having explained
r
why
^zr/(l
~ xr ).
1
at that
~ 1 + *iZ +
to oo
2 -x)(l -a: )
...
such an expansion
is
* 2Z
2
+
-
H-
tf ^
r
+
. . .
to 00
,
r (1 -a; ).
possible,
put zx for
z
and show
that
(l-zz)(l+tl z + t 2z* +
Hence by equating the
...)
coefficients of z
= ! + tjzx +
t 2z
z
xz
-f
. .
.
.
r ,
(l-X^t^X.t^.] 13. If
(l+zx)(l+zx*)...(l+zx*
x
14. If
|
|
< 1,
show that
15. If
a; |
|
<1
and
zx |
5
2
l
\
< 1,
,
z,
)... ==1
--____
^
/
)^l+p z+p z*+...+pnzn show
for all values of
(l+zx)(l+zx*)(l+zx where
n- l
.
x r\
show that
,
where
(ii)
where
- - --
3 (1 -sa;)(l -2o; )(l
^
r
-zx 5 )
...
to oo
=l+^2 +
=a; r /(l -a: 2 )(l -a; 4 )
J
^2
2 2 -}-...tO 00 .
...
(1
-x* r ).
,
that
CHAPTER XXXI PERMUTATIONS, COMBINATIONS AND DISTRIBUTIONS 1.
Combinations with Repetitions. The number n things r together, when each may be taken as
tions of
is
please,
denoted by
H,
and
number
the same as the
is
of
of combina-
often as
we
homogeneous
products of degree r which can be formed with n letters a, 6, c, ... k. The sum of these products is the coefficient of x r in the expansion of i
When
a, 6,
...
k are
all
equal to unity,
we obtain
the number of the
Thus
products.
therefore
#? = coefft.
of
x r in
(1
+x+
= coefft.
of
x r in
(1
-x)~
//
-
2
-f ...)
n
n ;
"^
Combinations and Permutations of n Things, not all different. LetC be the number of combinations, and P the number of 2.
permutations, of n letters aa (1)
C
is
equal
...
to the coefficient
bb
...
cc
is the
ofVs, r
the
taken k together, then
k of x in the product
number of different letters, p number ofc's, and so on.
where f
...
is the
number of cCs,
For the sum of the combinations in question
is
q the
number
the coefficient of x k in
the product
...
and
their
number
is
(1
+cx + c 2 z2 -h ... + c rx r )
unity.
-h#
+
2
+...
2 +*)(! +Z-fZ -f
obtained from the preceding by putting Whence the result follows.
is
to/factors
;
the coefficient of x k in the expression
(l+X+X 2 + ... +X p )(l which
...
... -f
a, b, c,
Xr) ...
equal to
DISTRIBUTIONS
494 (2)
P
k \k times the coefficient of x in the product
is
x\ L
x2
x
/
For the number
x
x2
tf\
of permutations involving Aa's,
re's,
//,6's,
...
is
Hence where the summation
is
to include every set of values of A,
(with repetitions) from 0,
1, 2, 3,
A Therefore Ex.
1.
We
+v+
. . .
= k.
xk in the above product.
number of combinations (C) and
the
the letters of the ivord parallel taken
four
chosen
such that
\k is the coefficient of
P/
Find
-f/z
...
... /x, v,
number of permutations (P) of
the
together.
have by the preceding,
PI whence we
3. result (i) (ii)
1
C
coefficient of x* in (1
4
coefficient of x*
find that
C22, P
Distributions, is
Some
286.
If
of the things
5
i
n things are divided into
called a distribution.
The
~x)~
may
A number be
In the
first
of
classes, the
of different cases arise,
alike, or
they
order of the things in each class
considered.
r lots
may
may,
or
be
thus
:
all different,
may
case the classes are called groups
not, have to be ;
in the second,
they are called parcels. (iii)
When a
distribution has been
made,
it
may, or may not, be necessary
to consider the order in which the classes stand.
In the
first
case the classes are said to be different (meaning that they in the second case they are
have to be distinguished from one another)
;
said to be indifferent. (iv)
Blank
may, or may not, be admissible, a blank
which contains none
Ex.
1.
How many
lot
'
being a
lot
of the things.
For example, consider what
three masts ?
'
'
lots
is
different signals
implied in the following questions. can be made by flying
five flags a, 6, c,
d
y
e
on
ARRANGEMENT Denoting the masts by
(A)
495
the following arrangements give different signals.
1, 2, 3,
123
123
123 d
IN GROUPS
*
e
<
< {C ,
:
c
c
; c
Because such arrangements as (A) and (B) are
different, it is a case of distribution
in groups. (A), (C) are different, the groups are said to be different.
Because
Further blank
two masts.
lots
are admissible, for all the flags
(See Art. 4, Ex.
may
be flown on only one or only
1.)
Ex. 2. In how many ways can five books a, 6, <% rf, e be divided among three people denoted by 1, 2, 3 ? Such distributions as (A), (B) in Ex. 1 are identical, so it is a case of distribution in parcels. Also the distributions (A), (C) are different, and so the parcels are, said to be different.
Further,
it is
Ex.
implied that each person gets at least one book, and so blank lots are
(See Art. 5, Ex.
not allowed.
1.)
In how many ways can jive books
be tied up in three bundles ? Here the order of the books in a bundle does not matter, so the distribution is in Also the bundles are not to bo distinguished from one another, and so the parcels. 3.
parcels are indifferent.
4.
(See Art.
Arrangement
different things
in
5,
Ex.
2.)
Groups.
The number of ways in which n
can be arranged in r different groups r(r
+ l)(r + 2)...(r + n-l)
or
is
|wC*l},
according as blank groups are or are not admissible. Proof,
(i)
Let n
letters
k be written in
...
a, 6,
a,
row
in
any
All
order.
the arrangements of the letters in r groups, blank groups being admissible, - 1 marks of can be obtained thus partition, place among the letters r :
and arrange the n + r-l things possible orders.* Since ferent arrangements is
r
-
1
\n-\- r
(consisting of letters
and marks)
of the things are alike, the
-
I
/
\r -1 =r(r + l)(r + 2)
...
in all
number of (r + n- 1).
dif-
arrangements of the letters in r groups, none of the groups can be obtained as follows (a) Arrange the letters in all being blank, possible orders. This can be done in In ways, (b) In every such arrange(ii)
All the
:
ment, place (r-1) marks of partition in (r-l) out of the (n-1) spaces between the letters. This can be done in C"~\ ways.
Hence the required number Ex.
1.
How many
n C^l*.
is
different signals
|
can be made by flying
five different flags
on
three masts ?
The required number = 3.4.5.6.7= 2520. *
Thus one arrangement of
7 letters in 5 groups, the
second blank,
is
indicated by a
|j
bed \ef\g.
PARCELS OF UNLIKE THINGS
496
Distribution in Parcels.
5.
(1) The number of ways in which n different things can be distributed into r different parcels, blank lots being admissible, is r n For each of the n things can be assigned to any one of the r parcels. .
The number of ways in which n
(2)
r different parcels, there being
which
different things
no blank
can be distributed into
lots, is
In times the coefficient of xn in the expansion of (ex
is
-
In any distribution, denote the parcels by a r a 2 Proof. consider the distributions in which blanks are allowed. ,
The total number of these is The number in which a x is blank
number
Therefore the
not blank
Of these
in
which
rn is
(r
the
last,
number
in
,
1
,
blank
Therefore 2>
i>
which
in
-
1 )
(r-1
a.*,}
n > r
1
the
last,
,,
.
is
n .
n
-(r-2".
is
1W| -2ft/ r-l) + r-2 tt
blank
is
number
in
which]>
.
n .
(T
Z(r
1)
4-
A)
o)
(r
.
J
the
number k
blank
in
which^
is
J it is
obvious that
number of distributions in which no one of x assigned parcels r n _ (j*(r - l) n + Of (r - 2) n -... + (- l) x (r - x) n
is
.
When x = r,
A
.
J
is
a a are n
the
ft
,
This process can be continued as far as we like, and the coefficients are formed as in a binomial expansion.
Hence
and
J
! a 2 are not blank
a3
,
.
,
>
number
Therefore the
Of these
ar
...
o^ is]
which]
U11 blank is
is
,
.
is
.
a2
r
l)
this gives the result in question.
complete proof
is
as follows
:
Let ux denote the number of distributions in which no one of the parcels a t a2 ... a x is blank. Of these, the number in which ax+1 is blank is ,
,
obtained by changing r into r -
number
is
changing
represented by
r into
in
1
ux (which
E~ ux where E~ l
,
l
a function of
is
blank
is
number of distributions in which no one of
Also we have
This
denotes the operation of
r-1.
Therefore the is
r).
.
,
u { - r n - (r - l) n = (1 - E~ l )r n
.
PARCELS OF LIKE THINGS w x = (l
Hence
-E~
l )
xr n
~2 -
= (1 - Cf E- 1 + Cf and when x = r, Ex.
. . .
+ ( - 1 )*E- X
)
rn
this gives the result in question.
In how many ways can
1.
497
five different books be distributed
among
three persons,
if each person is to have at least one book ?
The required number ^3 5 - 3 Ex.
25
+3
In how many ways can five
2.
The required number ~ (3)
.
I
(3
= 150.
5
different books
- 3 2s + 3
5
O
.
.
.
1
-
--
5 )
be,
bundle
tied HJ> in three
?
= 2,~>.
v)
The number of ways in which n
1
tributed into r different parcels is
same
things of the
can be dis-
sort
or C^Ii, according as blank lots
C>i{~
are or are not admissible.
If there are no blank lots, any such distribution can be effected as follows place the n things in a row and put marks of partition in a 1 out of the n l This can be selection of r spaces between them. which is therefore the number of such distributions. done in C n ways, Proof.
:
~~^
// blank of
lots
are allowed, the
n + r things
of the
same
number
of distributions
sort into
r
parcels with
such a distribution can be effected thus
and
into each of the r parcels, parcels, blank /Tn-fr O r _i
lots
:
is
no blank
put one of the
Hence the number
lots.
n
h r
For
things
in
question
r is
i .
These theorems can also be proved as in Ex. 2 of the next (4)
as that
n things into
distribute the remaining
being allowed.
same
the
The number of ways in which n things of the same
into r different parcels, where no parcel is to contain there are
no blank (x
lots, is
the coefficient of
+ x 2 + x* +
...x k ) r
or of
sort
section.
can be distributed
more than k things and
x n in the expansion of
(l-x
k+l
r )
(l-x)-
r .
r For the product of r factors, each of which is x x 2 -f x 3 -f 4- x may x v A v x^x where X x^, x ... are any terms selected be denoted by 2x from x, x2 ... x k the summation to include every such selection. The coefficient of x n in the product is therefore the number of ways in -f
. .
,
which
r
numbers
.
,
,
.
.
.
,
,
,
A,
/x, ...
can be chosen from
being allowed, such that their
sum may be
n,
1,
2, 3,
and
...
this
repetitions the required
A:,
is
number. Distributions of things of the sidered in Art. 8.
same
sort into indifferent parcels are con-
DERANGEMENTS
498 Ex.
In hoiv many icays can
1.
total score
three persons, each throwing a single die once,
make a
1 1 ?
of
The required number
is
>3 2 11 equal to tho co. of x in (x + x + x f
^co. of x
11
3
-^)
3
in
u;
of x 8
in
(l-o:)- (l
^co. of* 8
in
(^*
^c',o.
(l
-a;)-
(1
.
..
+ a; 6
3 )
3
3
2
~*
+
8
6
)(l"3z
9
)
= 45 -18 -=27. Ex.
2.
Apply
this
method
to
The number of distributions r or in in (1 +x + x z -f ... to GO
prove the theorems of the la^t section. of n like things into r parcels is the coefficient of x n (x
}
-f
x2
-f a;
3 -|-
.
,
.
to oo
r )
according as blank lots are or
,
are not allowed.
These expressions are respectively equal to
Derangements.
6.
-
(1
in
Any change
x)"
r
and x r (\ -x)~ r
,
etc.
the order of the things in a
group is called a derangement. If n things are arranged in a row, the number of ways in which they can be deranged so that, no one of them occupies its original place is
Denote the group of things by a j? a 2 arrangements is \n\ and of these
Proof. of
The number
in
which
stands
a,l
first,] I
in situ,
i.e. is
not in situ
Of these .
.
is
last,
in
which a
l
number
1M-
is
n
In
1.
is
the
number
in
which a 2 l \
.
.
in situ
total
is
number
Therefore the
The
a n-
>
is
i
\n~\-
i
!tt-2.
I
J
number
the
Therefore
.
neither a l nor a 2
Of these is
last,
in situ
the
is
which] I
in situ
,
,
|w-2 74-1 +|tt-2.
is J
number in which
a3
*\
is
Therefore the
one
in
.
number
....which
,
or aj, a 2 , a 3 is
?,?i
in
no] y
,
\n--3
^ w-1+3 ln-2,
,
in-
,si^/ is
This process can be continued as far as we like, and formed as in a binomial expansion.
it is
obvious that the
coefficients are
Hence inaitu If
the
number of arrangements in which no one of x assigned
is
x = n, the
n-Cf JB-l+Cfjn -2 -...+(-!)* last
term
is
|n
letters is
-a.
(-1)", and we obtain the result in question.
THEOREM OF WHITWORTH A
499
complete proof is as follows Let Vy denote the number of arrangements in which no one of a ...
:
ax
Of these, the number into n - 1 in uj. (which This of
number
is
changing n into n in situ
vx
is
which a x ^ l
in
,
a%,
obtained by changing n
is in situ is
a function of n).
is
l represented by E~~ v x
where E~~ l denotes the operation
,
1.
number -E~*v x
Therefore *he is
}
in situ.
is
,
of arrangements in which
that
is
no one of a {9
fl
2,
...
ax
to say,
Also, as in the preceding,
Hence
1 ,
it
follows that vx
=-
(\-E~iy\n = (i c 'f A'- + <7/?- 2 -... + (- 1 yE-*) n - j>i-Cf l'/&-l+0 |n--3-...+([)*~[n-.r. 1
-
|
=
If
NOTE,
by
n.
1
1
(ii)
??.,
(i)
It
is
the last term
is
(
Tho expression (A) easily shown that
Tho values
n
of
n
for
||
-I) is
1
77 ,
and we obtain the
sometimes called
n
is
1,
2, 3,
^n(>factorial n,
the integer nearest to \n
1
result in question.
can be calculated thus
...
and
,
I
a.
I
-1^0,
|
denoted
Begin with
:
and multiply successively by 1, 2, 3, ... increasing any even product by ing every odd product by 1 before the next multiplication. Thus ||
is
Ic.
1
1
and decreas-
2=0.2 + 1^-1,
j
3=3. |
I
-1^2,
1
[[4^4.24
-9.
A
7. General Theorem. In Arts. 5 and 6 we have instances of a * as follows theorem, stated by Whitworth :
If there are letter
N sets of letters and, if out of r assigned letters a, N sets, each combination of two letters occurs in N
occurs in
,
each
sets,
each
6, c, ...
t
combination of three
letters
occurs in
N%
and
sets,
so
on; and finally
all the
N sets, then the number of sets free from all the letters is N ~ C[N + CIN - GIN., + + - YNT Or more generally * If there are N events and r possible conditions such that every single condition is satisfied in N of the events, every two of the conditions are simultaneously satisfied in N of the events, and finally all r letters occur in
r
t
.
Z
.
.
]
(
.
:
l
...
2
the r conditions are simultaneously satisfied in
N
r
of the events, then the
number of events free from all the conditions is as stated above. The proof is exactly similar to that in the last article. *
Whitworth, Choice and Chance,
p. 73.
CONSTRUCTION OF PARTITION-TABLE
500
Numbers.
Partition of
8.
In this section we consider a num-
(1)
ber as formed by the addition of other numbers. Any selection of the numbers 1, 2, 3, 4, ... (with repetitions), of which if the selection contains p numbers, the sura is n, is called a partition of n ;
it is
called a partition of
n
into
p parts
or, shortly,
Thus 7=34-2 + 1+1, and we say that 3211 Ex.
Siij}posing that all the partitions of
1.
explain how to find all the partitions of Take first 6 then 5 followed by ;
a p-partition of n. a 4-partition of 7.
2, 3, 4,
5 have been written down,
6. 1
;
then
4,
followed by
all
the partitions of 2
then 2, followed by all the partitions of 3 (that is, 3, 21, 111) followed by the partitions of 4 which contain no part greater than 2 (that is, 22,211, 1111); lastly, 111111. The complete set is (that
11)
is, 2,
;
6,
(2)
//
Up
then
1,
is
3,
;
51, 42, 411, 33, 321, 3111, 222, 2211, 21111, is
the
111111.
number of p-partitions of n* then
n$=n$-i + n$To make the p -partitions
of
n
is
v
..............................
to distribute n elements into
p
(A)
indifferent
parcels, blanks being inadmissible. This can be done by placing one of the elements in each of the
and then distributing the remaining n
p
1
p parcels or 2 or 3 ... or
Therefore
parcels.
Changing n into n
and p
1
n
TTH-I
n -p
llp^l^lll whence the
p-1, we have
into
rjn-p -Til*
-r
rjn-v +
11$
-
rjn-pij + 11 p
-
result (A) follows.
If P>ln> then
II%= n"1\.
For
p^np,
(3)
~p
elements into
in this case
and therefore 77^^
Construction of a table of values of 77^. 77? =
1,
77S
= i(w-l)
or
0.
We
have
\n,
according as n is odd or even. Also 77^ = 1, 77^-1 = 1. In the table on page 501, denote the nth space of the mth row by (n, m) call the set of spaces (n, 1), (n + 1, 2), (n + 2, 3), ... the n-th diagonal.
We
can
mid using
up the first and second rows and the (2), we have
fill
n
i 77r -77 1
f2 ,
77?
Hence the numbers three,
...
^
l
77? +77'2
in the (n
,
first
and second diagonals,
77^= 77? + 77^+ 77*,
+ l)th diagonal are the sums
terms of the nth column, starting from the top. *
Hp
is
subsequently denoted by P(n, p,
:
*).
etc.
of one, two,
CLASSES OF PARTITIONS
1, 1,
The second column is 1, 1, 0, 2, 2, 2, .... The third column is and so on to any 2, 3, 3, 3, ... ;
0,
...
1, 1, 1,
therefore the third diagonal is and the fourth diagonal 0, 0, . . .
,
extent.
TABLE OF ^-PARTITIONS OF n
Various classes of partitions of
(4)
,
501
(i.e.
VALUES OF P(n
n may
y
p, *))
be considered
:
the
num-
ber of those in any particular class is denoted by a symbol of the form P(n, , ) where the number and the nature of the parts are respectively indicated in the first and second spaces following n.
In the
the first space,
that there are
p
parts
;
and
means that
number
In
and
p means
the <;<7
An
of parts does not exceed p. second space, q means that the greatest of the parts means that no part exceeds q.
asterisk
means that no
restriction
nature of the parts, as the case If the parts are
P(n, p,
q) is
to
may
be unequal, then
the
number
of which
is
is
q
;
placed on the number or the
be.
Q
is
used instead of P.
of partitions of
n into p
Thus
:
parts, the greatest
is q.
number of jo-partitions of n. P(n, ^p, *) is the number of partitions of n into p or any smaller number of parts. number of partitions of n into unequal parts, none the is Q(n, *, ^q)
P(n, p,
*) is the
of
which exceeds
q.
CONJUGATE PARTITIONS
502
The values
(5)
of
P(n, ^p,
For instance, the number is
sum
the
the 4th
That
can be found from the table on
of partitions of 10 into
of the first four
number
*)
numbers
and
therefore
is
23.
<4,*HP(14,4,*),
we have
and, in general,
P(n,
We
is
to say that
is
P(10,
(6)
not more than 4 parts
in the 10th column,
in the llth diagonal, that
p. 501.
= P(n+p,p,*) ......................... (C)
have the following relations
:
Q(n,p,q) = Q(n-q,p-l,
and
For consider any partition of n into p parts, of which the greatest is q. l parts, If we remove the part q, we have a partition of n q into p none of which exceeds q. Also from every partition of the latter kind we can derive one of the former. Consider a partition (7) Conjugate Partitions. which the greatest is 4, say 431. This
may
be written as in the margin, and
columns instead of the rows, we have 3 2 2 into 4 parts of which 3 is the greatest. Clearly
431
can be derived from 3 2 2
1
1,
if
of 8 into 3 parts of
we sum the
i
]
1
1
111
a partition of 8
1
by the same
process,
and these
are caljed conjugate partitions.
Thus
to
every partition of
n
into
p
parts of which the greatest
corresponds a partition of n into q parts of which the greatest
Consequently,
is q, there
is p.
we have P(n,p,q)^P(n,q,p),
............................. (E)
P(n,p,*) = P(n,*,p), P(n,
g)
= P(n,
q,
.......................... (F)
P(n,
5,
,*) ........................ (H) (8)
Euler's use of Series in the Enumeration of Partitions.
(i)
For
Q(n,
*,
is the coefficient
this coefficient
adding a selection of
n of x in the expansion of
number of ways in which n can be obtained by the numbers 1, 2, 3, ... q. is
the
EULER'S USE OF SERIES P(n,
(ii)
*,
n of x in the expansion of
is the coefficient
^q)
(1
503
- X )~ l (l -x*)- l (l -x 3 )- 1
...
(1
-a*)-
1 .
For consider the product
terms out of the
Any by
ce,
x2/3 x3y
...
,
a,
j8,
a+
y
2/3
...
factors
are
any we have
...
n is x ,
the product of these terms
If
second,
first,
x q& where
+ 3y 4-
. . .
-f
may
be represented
numbers
of the
0, 1, 2,
...
= tt.
qO
Hence the coefficient of x n in the product is the number of ways which n can be made up by adding any selection of the numbers 1, 2, ... Hence the result follows. q, repetitions being allowed.
Q(n
(iii)
p,
9
is the coefficient
^q)
(1
and
For the in
zpx
coefficient of
n
in 3,
p n in the expansion of of z x
3 +zx)(l +zx*)(l +Z3 )
...
(1
+z&),
n -iP(p-fi) in the
therefore the coefficient of
is
y.
in the first
expansion of
product
number
the
is
which n can be obtained by adding a selection of
of the
p
of ways numbers
1, 2, 3, ...q.
The
rest follows
follows that
(iv) It
of
expansion
Hence by
(i
L
,
9
3,
...
,
P(n, p,
0(17, total
4,
number
-x
3
1
)-
...
into
(1
i), *,
1),
p unequal
~x
p
~W+
l
>
in the
l
)~
.
p, *)
;
parts can be found from the
*) in (3). be
made up by adding four
of the
*)=P(17 -^4.3, 4,*)=P(11, of
of forming 17
numbers
ways
4, *)--=ll.
by adding four of the numbers
this
number we must subtract the number
1,
2,
of partitions with a part greater
9.
These partitions are
Or
the coefficient of x n
is 11.
From than
1.
?
havo
Thus the
is
Ex.
*)=P(n~^p(p-
In how many ways can 11
...
6,
- x )- l (l -x2 )~ l (l
Q(n,p,
(5),
table of values of
Wo
Q(n,p,*)
number of partitions of n
/Au^ ^Ae
Ex.
XXX,
Q(n,p, *)=P(n~l-p(p +
(ii),
whence by
1,2,3,
from Ch.
thus,
and similarly
10, 4, 2, 1
G(17,4,
;
11, 3, 2, 1
;
and
10) = C(7,3, <10)=0(7,
<2(17, 4, 11)
1.
their
3,
number
= P{4, *)
Hence the required number
is 2.
3, *)
is
= !,
11-2 = 9.
EXAMPLES OF SELECTIONS
504
The reader should have no
(v)
-zx)-
(1
l
(I
-zx2 )~l
...
difficulty in verifying the following (1
1 -2^)- = l + EP(n,
l
-Z)-
1
1~ZX~ \-ZX 2 )- 1 1
to oo
...
)~
...
tO 00
EXERCISE
=l
=
LIII
Find the number of ways of displaying 5 flags on 4 masts if more of the masts need not be used
1.
are to be used, and (i) one or masts are to be used.
Find the number of ways
2.
in
which n things, of which
all ;
the flags the
(ii) all
r are alike,
can be
arranged in circular order. 3. Show that a selection of 10 balls can be made from an unlimited number of red, white, blue and green balls in 286 different ways, and that 84 of these contain balls of all four colours.
4. Show that the number of made from five a's, four 6's, three
Show
5.
which n are sort, in
n
.
which can be
different selections of 5 letters c's,
two
and one
cf s
e is 71.
that a selection of (n - 1) things can be made from 3n things, of n are alike of another sort and n are alike of a third
alike of one sort, ~ 2 n 1 different
ways.
O
w where n is In the expansion of (a x +a 2 + + any positive integer not the that coefficient than ofany term in which none of the numbers greater p prove a lf a 2 ... a v occurs more than once is In. 6.
9
,
1.
in
In how
m different 8.
many ways can a bags, if one or
white
balls, 6
black balk and c red balls be put
more of the bags may remain empty
?
In an examination, the maximum mark for each of three papers is n Prove that the number of ways in which a candidate it is 2n.
;
for the fourth paper
can get 3n marks
is
i(w +
[Number = coefficient 9.
10.
of
z 3n in
8 + 10n + 6). - xn+l ) 3 1 - x 2n+1 )
l)(57i (
1
(
(
1
- x)~ 4 .]
In how
many ways can 5 rings be worn on the 4 fingers of one hand ? In how many ways can an examiner assign 30 marks to 8 questions,
giving not less than 2 marks to
any question ? 11. The number of ways hi which 2n things of one sort, 2n of another sort and 2n of a third sort can be divided between two persons so that each may have 3n things is 3n 2 + 3n+ 1.
[Number of selections of 3ft things = coefficient of x? in (1 -a;2n+1 (l -#)~ .] 12. The total number of permutations of n things taken 1, 2, 3, ... or n together n
3
)
is
the integer nearest to e in [Total L
number=e n I
L_
I
-
1.
/I [
\n +
x
1
+ l
n + l.n + 2
-f ...
)
/*
1 J
3
EXAMPLES OF ARRANGEMENTS
505
13. There are 5 letters and 5 directed envelopes, (i) In how many ways can all the letters be put into th wrong envelopes ? (ii) In how many ways can 2 letters be rightly placed and 3 letters wrongly placed ?
The number of arrangements of the n terms a l9
14.
of r sequences
a^,
2
a ija r+i occurs
a 3>
...
an
in
which no one
is
r r C [n-CT [w-l + r:; \n-2-. + (-l) r \n-r. number of arrangements is \n. Any one of ..
[For the total occurs in \n-l of these
Show
15.
that the
any two
;
number
none of the sequences
in
I
n-2
and so
;
the sequences
on.]
of arrangements of the letters abed which involve Verify by writing down the arrange-
ab, be, cd is 11.
ments.
The number
16. c,
d come together
Explain
this,
of arrangements of the letters abed in which neither a, b nor is
= 8. 3 + Cl .2* [4-CJ.2 [2 down the arrangements in question.
and write
A
party of 10 consists of 2 Englishmen, 2 Scotsmen, 2 Welshmen and of other nationalities (all different). In how many ways can they stand in a row so that no two men of the same nationality are next one another ? In how many ways can they sit at a round table ? 17.
4
men
18.
n are
The number of combinations n together ~~ and the rest unlike is (n 4- 2) 2 n 1
b
of 3n letters of which n are a
[Number == coefficient of x
n in
(1
~x n ^)
= coefficient of xn in 2 n (l
2
(l
+ x) n (l ~x)~\
-x)~ -n z
.
~ 2n 1
(l
or in (2 - F"-a-) w (l
~x)~\
nm
.
,
Show
also that the
number
-x)~
z
etc.]
Show that if a l9 a 2 ... a m arc distinct prime numbers other than number of solutions in integers (excluding unity) of the equation 19.
is
and
.
of solutions in which at least one x
unity, the
is
unity
is
[Solution follows from the general theorem of Art. 7.] 20. Given n pairs of gloves, in how many ways can each of n persons take a right-handed and a left-handed glove without taking a pair ? 21. (i) If n things are arranged in circular order, the number of ing three of the things no two of which are next each other is
ways of
select-
ira(n-4)(n-5). (ii)
22.
n things are arranged in a row, the number of such sets of three is -J(n-2)(n-3)(n-4). Find the number of positive integral solutions of x + y -f-z-h w>~20 under
If the
the following conditions (i) (ii)
:
Zero values of
x, y, z,
w
are included.
Zero values are excluded.
(iii)
No
(iv)
Each
(v) x,
y
variable
y
may
variable
z,
w
is
exceed 10
;
zero values excluded.
an odd number.
have different values (zero excluded).
PARTITIONS
506 23. Prove that the of the equation
number
of positive integral solutions (zero values excluded)
x 4- 2# + 3z~n i(*--l)("
is
where c~--H
or
tt-9, 10. [The required
24.
according as n
J
number
is co.
or
is
not a multiple of
is
3.
when
Verify
of xn in
The number of independent
solutions in unequal positive integers, zero
included, of
x is
4-7/4-z \-u-
the integer nearest to -&$p*(Zp ~ 25.
Show
that
TIlj'^S/*
2p
3).
2 .
26. A necklace is made up of 3 beads of one sort and tin of another, those of each sort bein# similar. Show that the total number of possible arrangements of the beads is 3n 2 -I 3/i -f 1. [Having put on the three beads, the number of arrangements of the CM, beads
between the three (which are
in the spaces
27.
Find the values of
28.
In how
integers 29. total
30.
/>(18, 5 *),
(ii)
P(12, <5,
ways can 20 be expressed as the
many
is
*),
sum
P(12,
(iii)
*, 4).
of four unequal positive
?
Show that the total number of number of partitions of 10. Show
Art. 8, 31.
(i)
indifferent)
3
that
7 (2ri-fr,
(3), to tind
P(25,
n
partitions of
+ r, *)=P(2n*
r?,
*).
n
is
Find the
P(2/i, n, *).
Hence
the table of
use
18, *).
Show that P(n, p, ^q)
is
cqiioi to the coefficient of
~ xn p in the expansion
of
Hence show that Q(*,P, 32.
Show
that
EXERCISE LIV * 1. There arc n points in a plane, no three being collinear except p of them, which are collinear. How man}7 triangles can be drawn with their vertices at
three of the points
?
There are n points in a plane, no three being collinear. number of n-sided polygons with their vertices at these points is
Show
2.
n-
-J j
if
n=4. *
The
results hi
Exx.
i>-12
were given by Dudeuey
~~
that the
1.
Verily
GEOMETRICAL PROBLEMS
507
m
points in a straight line are joined in all possible ways to n points in line, then, excluding the (ra-f n) points, the number of points of intersection is If
3.
another straight
^mn(m- l)(n- 1). n straight lines of indefinite length are drawn being parallel and no three meeting in a point, then 4. If
(i)
the
number of points
(n) the plane [(ii)
If
un
Show
5.
is
the
is
of intersection
+ n-f
divided into \ (/t 2
number
of parts,
that, in general,
show that
is
2)
in a plane,
^n(n
1)
no two of them
;
parts.
u n ~-u, _ l
<
l
n planes divide space into
4-
n.] 3
J-(?i
+ 5?j.
\-
0)
rerions.
What
are the exceptional cases ? [If no three planes have a common line of intersection and no four meet in a v n ~\ + u n-i where v n is the required number and u n is the same point, then v n as in Ex. 4.] 6.
lines
The number of squares formed on a piece of squared paper by and n vertical lines (m
%m(m -
m
l)(3>i
-i(a
according as a (ii)
is
+ l)(a-l)(a-3)
or
horizontal
l).
n are positive integers such that m^n and m + n (i) constant, prove that the greatest value of m(m ~ l)(3w- -m - 1) is If M,
7.
m
where a
a,
is
ia(a-l)(a-2),
odd or even.
If a straight rods of indefinite length are placed so as to number of squares, prove that this number is
form the greatest
possible
Ar(a according as a
is
+
l)(a
-])(- 3)
or
Aa(a-l)(a-2),
odd or even.
8. Show that the number of ways in which three numbers in arithmetical - I) 2 or \n(n - 2), according progression can be selected from 1, 2, 3, ... n is i(n as r? is odd or even.
The
9.
If c
sides of a triangle are a, b, c inches
is
given, according as c
where
a, b, c
are integers and
show that the number of different triangles is J (c + is odd or even.
1 )2
or
-Jc (c
+ 2),
10. Of the triangles in the last question, the number of those which are isosceles or equilateral is i(3c - 1) if c is odd and -J-(3c - 2) if c is even.
Each
side of a triangle is an integral number of inches, no side exceeding Prove that the number of different (i.e. non- congruent) triangles winch can be so formed is 11.
c inches.
or
according as
c is
odd or even.
12. Of the triangles in the last question, the number of those which are isosceles or equilateral is J(3c 2 -fl) or Jc 2 , according as c is odd or even.
2K
B.C.A.
CHAPTER XXXII PROBABILITY First Principles. In speaking of the probability of an event, doubt is implied as to whether it will or will not happen. The degree of 1.
doubt depends on our knowledge of the controlling conditions. knowledge
of these enables us to say that the event
or certain to
is
A complete
certain to
happen
fail.
Measurement of Probabilities. We are constantly forming rough estimates of probabilities, saying that some event is unlikely, likely, very This implies that (at any rate in certain likely or certain to happen. can be measured and can be compared with certainty, cases) probability which must be regarded as a degree of probability. Thus there is nothing to prevent us from choosing certainty as the unit of probability, and this is always done. In what follows the words probability and chance are synonymous, both of them being used to denote measure of prol)ability.
Suppose that n balls A, B, ... K> all of the same sort, are placed in a bag, and that one of them is drawn. There is nothing to favour the drawing
more than another,
of one ball
A, B,
...
Thus say 5,
or
if
is
x
K will is
the chance that
drawn
is
the chance that either
n
words
it
equally likely that
is
A
is
drawn, the chance that any other
ball,
also x.
Our notions with regard
so
in other
be drawn.
A
or
to probabilities therefore lead us to assert that
B is drawn is
2x,
and the chance that one
of the
drawn is nx. But it is certain that one ball will be drawn, and nx = 1 and x = l/n. Thus the probability of drawing A is 1/n. The following definition is a statement of Laplace's First Principle.
balls is
Definition,
If
an event can happen in a ways and
fail in b
ways, and
nothing to lead us to believe that one of these ways should occur rather than another, that is to say, all the ways are equally likely, then
there
is
the chance that the event fails is
+ 6), + 6/(a 6),
the odds in
a
the chance that the event
happens
is
favour of the event are as
the odds against the event are as
a/ (a
b
:
:
b,
a.
EXCLUSIVE EVENTS If b = 0, the event
a = 0,
certain to
it is
is
fail,
If a = 6, the event is there is an even chance of
509
certain to happen,
and
and the probability as likely to happen
of its
its
as to
probability
happening
1
is
;
if
is 0.
and we say that
fail,
its
happening. the respective chances of the happening and failing of an
If p, q are
event, then
Events
of a type to
which
this definition does not directly apply will be
considered later. Examples, (ii)
(iii)
(i)
// a coin
// a six-faced die // a ball
is
is tossed,
the chance of
thrown, the chance that
drawn from a bag containing 3
is
'
heads '
ace
;
It is easy to
Ex. toss
1.
A man
i.e. all
heads or
is
J-.
turns up
is
ace of spades
is
the chance that the card
is
a spade
is i
the chance that the card
is
an ace
a mistake as to the meaning of
much
of
-^>-
|
simultaneously,
is
an even
-J-.
the chance of
;
;
'
equally likely.'
practical experience in the laying of it
is
balls,
2.
the chance that the card
make
three pennies
:
'
and 2 black
white,
drawing white is J the odds against black are as 3 (iv) // a cord is drawn from a well-shuffled pack,
'
odds said
cha?ice that they
'
// you
:
will fall all alike,
all tails.
1
For they must fall all heads, or all tails, or 2 heads and a tail, or 2 tails and a head. So they fall all alike in 2 out of 4 possible cases.' As a matter of fact, the odds are 3 to 1 against their falling alike. For, if the pennies are denoted by a, 6, c, then a can fall head or tail, and so for b and c. Hence there are 8 equally likely cases of which 2 are favourable.
Events are said to be mutually exclusive 2. Exclusive Events. when the happening of any one of them precludes the happening of any other.
In the case of mutually exclusive events,
the ctiance that one
sum of the chances of the separate events. known as Laplace's second principle. It can
or
other of them occurs is the
This statement
is
also be
If an event can happen in several different ways, one only of which can occur on the same occasion, the chance of its happening is the stated as follows
:
sum
of the chances of its happening in the several different ways. For consider n events of which the chances are ajg, a,2/<7, . . .
,
a n /g respec-
denote integers. Out of g equally likely ways, tively, the events can happen in a ly a 2 ... a n ways respectively, and since the events cannot concur, one or other of them can happen in a l + a 2 -f -f a n
where the
letters
,
.
ways.
sum
The chance
of this
is
therefore
of the chances of the events.
(a4
-f
a2
-f
. . .
-f
a n )/g,
which
.
.
is
the
CALCULATION OF PROBABILITY
510 Ex. (ii)
1.
In a
Any
face of either die
may
These are made up of the thirty throws (1, 2), (2, 1), ...
We
count
what
single cast with two dice,
doublets, (ui) five-six (i.e. one die turns
(i)
two aces,
the other six) ?
turn up, so theie are 6x6 equally likely cases. six throws (1, 1), (2, 2), ... (6, 6), together with the (5, 6), (6, 5).
as the
(5, 6), (6, 5)
chance of throwing
the
is
up five and
same throw, and
aces^g
the chance of throwing doublets the chance of throwing
;
3%
five-six:
In a single cast with two dice,, what are two numbers of which the sum is 7 ?
Thus
call it five-six.
the chance of throwing two
;
-3%.
Ex.
2.
Out
of 36 possible cases, those in favour of the event are and the number of these is 6, therefore
odds against throwing
the
7, i.e.
against
(6, 1), (1, 6), (5, 2), (2, 5),
(4, 3), (3, 4),
the chance of throwing 7
and the odds against throwing Ex. (ii)
3.
eleven,
-/
J~,
1.
:
In a single cast with three dice, what is the chance of throwing (iii) less than eleven, (iv) more, than ten ?
The number (i)
7 are as 5
-
is
Among
of possible cases
is
63
1
3 times
;
3
(ii)
Let
s
of cases in which 11
is
-^ o
~Q^ t5u
thrown, then
s^coeflt. of x 11 in (x f x'2 +x*
+
+ :r 6 3 ^27
...
)
27 .*.
1
i
chance of throwing four-five -six
number
be the
four -five-six,
.
those four -jive-six occurs .'.
(i)
chance of throwing 11
-.
-
is
;
1
These cases are (iii) Let / be tho number of cases in which less than 11 is thrown. those in which exactly 3, 4, 5, ... 10 are thrown. Therefore t is the sum of the coeffi4 10 cients of 3 x~ a; in the expansion of }
,
,
.
,
.
.
(x
+ x- + x*
t- coefficient
hence,
= coefficient
of x of
+ #6
...
3 ;
)
x 3 (l -o; 6 ) 3 (l -x)~*
in
x 1 in
(1
-z)~
^coefficient of x 7 in ( 4x
4
(l -3rc
+ -, I
\
.
8 )
6V6
2
.
= 120-12^108; chance of throwing
.'.
(iv)
are
The
total
number
thrown and those .'.
(In the old
Ex.
4.
game
^4 parti/
in
of possible cases
which
less
less
is
made up
chance of throwing more than 10
of ten take, their seats at
B)
=-
is
^-
.
s>
of those in
than 11 are thrown
of passe-dix, a player stakes
specified ptrsons (A,
than 11
which more than 10
;
1
-J ==i.
on throwing more than
a round
table.
What
10.)
arc the odds against two
sitting together ?
A having taken his place, B has a choice of 9 places, 2 of which are next to A, hence the odds against B sitting next to .4 are as 7 2. :
INDEPENDENT EVENTS
511
Four cards are drawn from a pack of 52 cards. What is the chance that one of drawn ? Four cards can be selected from 52 in C% z ways and this is the number of possible One of each suit can be selected in 1.3* ways and this is the number of cases. Ex.
5.
each suit
is
;
;
favourable eases
;
28561
1.3*
Ex. is the
Five, balls are
6.
chance that 3
white.
drawn from a bag containing G white and 4 black and 2 black balls are drawn ?
What
balls.
Five balls can be selected out of ten in CJ ways. This is the number of possible Also three white balls and two black balls can bo selected in C\ C% ways,
cases.
which
.
the
is
number
of favourable cases, therefore
-iv
,i
the required chance
3.
C
T
^L
5
Events are said to be independent when
Independent Events.
the happening of any one of the others.
- 10 = C ^*'t ^p ">T-
them does not
affect the
happening of any of
The probability of the concurrence of several Laplace's third principle. independent events is the product of their separate probabilities. It is sufficient to consider
two independent events
A
A2
and
t
,
of
which
the probabilities are p l and p 2 respectively. Suppose that A L can happen in a L ways and fail in 6 t ways, all of which are equally likely. Also suppose that A 2 can happen in a 2 ways and fail in 6 2 ways, all equally likely.
Then
Pi
= ai/(<*>i + b
^ 2 -=a 2/(a 2
1 ),
f 6 2 ),
and there are the following
:
possibilities
A and A 2 may both happen, and this may occur in a^a* ways. A may happen and A 2 fail, and this may occur in a 6 2 ways. A may happen and A fail, and this may occur in aj> ways. A and A may both fail, and this may occur in 6 6 2 ways. The total number of ways in which both A and A 2 are concerned }
1
l
l
}
2
l
t
l
(a l
+b 1 )(a 2 -\-b.2), and
both
A
1
and
A2
occur
these are is
all
equally likely.
is
Hence the chance that
# 1 a 2 /(a 1 + 6 1 )(a 2 4-6 2 )j which
is
equal to Pip%.
It should also be noticed that
A that A 2
the chance that
the chance
l
happens and happens and
the chance that both
Hence
and
also the chance that one
PiO -ft) Also the chance that at fail) is
Al
1
-(1
A2
A 2 fails is p (l -p 2 A fails is p 2 (l -pi) t
)
1
fail is
(1
and only one
-
Pi)(l
-
p
;
;
2 ).
of the events
happens
is
+^ 2 (1 -PiH Pi +Pz- ZpiP*
least
one of the events happens
-^(l -pzl^
(i.e.
both do not
INTERDEPENDENT EVENTS
512 Ex.
What
I.
is the,
chance of throwing ace with a single die in two
trials ?
The chance
of not throwing ace in one trial is -|, the chance of not throwing ace in two trials is () a
the chance of throwing at least one ace
.'.
Or
1
-
;
2
(^ )
= ^Q.
thus,
Chance Chance Chance Chance Ex.
of success in both trials
3^. of success in the first trial and failure in the second
and success in the second one success yg + 2 ^ f J^~.
of failure in the of at least
2.
Referring
f
first
to the last
example, explain
The chance of throwing ace in the first is J, therefore the chance of ace in two
why
\
.
-|.
the following reasoning is incorrect
:
^ and the chance of ace in the second trial
trial is
trials is
-|.
The throwing
of ace in the first trial does not exclude the possibility of throwing ace in the second trial. Thus the events are not mutually exclusive and Art. 2 does
not apply. It should be noticed that, if the reasoning were correct, it would follow that the chance of throwing ace in 6 trials would be -f i.e. that ace would certainly occur in 6 trials which is obviously false. ,
;
Ex.
In how many throws with a
3.
single die will
it
be
an even chance
that ace turns
at least once ?
up The chance against ace turning up in n throws is (|) n and the chance that n If then n is the required number, we have up at least once is 1 - (|) ,
it
turns
.
l-() n =
/.
;
w (f)
=4;
/.
n = 3-8
nearly.
Hence it is less than an even chance that aco turns up once than an even chance that it turns up in 4 trials.
in 3 trials,
and more
Interdependent Events. If two events A v A 2 are such that p the probability of A ly and p 2 is the probability of A 2 on the supposition
4. is
A
that is
p p2
has happened, then the probability that both
l
A
and
A2
happen
.
}
Also
p3
if
is
the probability of a third event
happened, then the probability that A v A 2 and and so on for any number of dependent events.
For the reasoning Ex.
One bag contains
1.
.and 3 black
chance that
If a bag white ?
balls. it
is
The chance that the from
of the last article holds
it is .".
first
3 white balls is
A 3 after A l and A% have A 3 will happen is Pip^p^
good under these conditions.
and 2 black
balls
chosen at random and a ball
bag
is
chosen
is
i,
;
is
another contains 5 white
drawn from
and the chance
of
it,
ivhat is the
drawing a white
;
the chance of choosing the
first
bag and drawing a white
Similarly the chance of choosing the second bag .'.
is
\ \
and drawing a white is \ random is
the chance of drawing a white from a bag chosen at
.
f-
;
ball
FREQUENCY NOTE. its
Ex.
were put into one bag and a is not the same as before.
If all the balls
being white
A
2,
person draws a card from a pack, replaces
at least three trials,
The chance of
(ii)
success at
any particular
second caw, he must
What
balls.
is the
First method.
chance
It
In
trial is \.
the first case,
he will have
;
2
at the
first
(J)
~-fij.
and second
2 required chance = (2)
trials
and succeed at the third
;
~~Q.
are drawn, one by one, from a bag containing 6 white and 4 black
Five balls
3.
the pack. He conhe will have to make
and shuffle
it,
is the chance, that
required chance
fail
.'.
Ex.
were drawn, the chance of
exactly three trials ?
.'.
the
What
a spade.
to fail at the first anct second trials
In
ball
-^, which
is
tinues doing this until he draivs (i)
513
3 white
tliat
makes no
and 2 black
balls are
difference to the result
drawn
?
we suppose the
if
5 balls
drawn
Hence, as in Art. 2,
simultaneously.
the required chance -=C'*
Second method. Suppose the balls drawn chance that this occurs is easily seen to be -
_6_.
10
.
9
4 8
.
C*/C\ ~iy.
.
in the order indicated
4
.
7
by wwwbb.
The
3.
6-
Also the chance that 3 white and 2 black balls are
drawn
in
any other particular
obtained from the preceding by merely altering the order of the numerators. The number of orders in which this can occur is the number of permutations of the 10 5 / 3 2 letters wwwbb, and is therefore order
is
1
;
j
.*.
which
5.
is
the required chance
10
-^ f
-f
-f
'5,
equal to the result previously obtained.
Another
Way
of estimating Probabilities.
It
is
obvious
that the definition in Art. 1 only applies to events of a restricted type in particular, to games of chance. There are other cases in which probability ;
is
estimated by considering the frequency with which the event occurs, or believed to occur, in the long run/ For example '
is
:
(i)
It
is
an observed fact that about 51 per cent, of the children born in Hence we say that the probability that a child, about to
Europe are boys. be born, will be a
boy
is
about 0-51.
'
According to tables of mortality,' such as are used by Insurance Companies, out of 81,188 men 50 years old, 70,552 live to the age of 60. We therefore say that the chance that a man of age 50 will live another (ii)
10 years
is
70552/81188 = 0-869.
In general, trials, where
an event has happened in pN trials out of a total of N a large number, we say that p is the probability that happen on a fresh trial.
if
N
the event will
is
EXPECTATION
514
This notion of probability
is
not at variance with that contained in
Laplace's first principle (Art. 1). If a six-faced die is thrown a large
number
we expect one
of times,
face
up about as often as another, and this has been verified experimentally. Thus in a large number of throws we may expect ace to turn up in about J of the total number of trials, and from this point of view we
to turn
say that the chance of throwing ace
is |.
Expectation. Suppose that a person (A) has a a. lottery which gives him a chance p of a prize of 6.
ticket in
a
N
N
is a large number, we may the lottery were held times, where him to the about expect get pN times, receiving pNa. Thus we prize may say that on an average he receives pa for a single lottery, and this
If
is
called his expectation.
Next suppose that
him a chance p t
ticket gives
/1's
chance p 2 of receiving a 2 and so on. If the lottery were held times, where
of receiving
ap a
,
N
N
is
we may expect him
large,
pfl occasions, 2 on about p 2 N occasions, and so on. Altogether he may be expected to get about N(p a l -f p 2& 2 + ) f r a Thus we may say that on an average he gets ) (p l a l 4-jt) 2 tf 2 -f his from sum is sum of This the the single lottery. expectations arising a l on about
to receive
<2
l
chances of securing the separate sums, and
is
called his expectation.
The average value of a quantity P, subject to risk, is the value P assumes in the long run. This value is also called which average the expected value of P or the expectation with regard to P. Definitions.
If
that
P
a quantity can assume the values P l9 P2 P3 it has these values are respectively p v p 2 jo 3 ... ,
,
,
expected value of
This
is
P
merely a
,
^1^1+ j>2 P 2 + p 3 P3 ... generalisation of what has been is
...
and the chances the average or the
,
.
-f-
said in the case of a
lottery.
It should be observed that every p denotes the chance that corresponding value, so that events need not be independent.
P
has the
1. A man's expectation of life, is usually taken to mean tho average number of which men of his age survive. years If p r is the ehanre that he will survive r years, dying before the end of the next year,
Ex.
his expectation of life
is roughly given in another volume.
Ex.
2.
A
pl
.
1
+ p2
.
2
+p 3
.
3
4-...
.
A
closer estimate will be
person draws 2 balls from a bag containing 3 white and 5 black balls. If he for every white ball which he draws and Is. for every black ball, what is his
is to receive 10s.
expectation
?
The number
of
ways
in
which 2
balls
can be drawn
is
C\
28.
SUCCESSIVE EVENTS Of these the number of ways in which 2 white balls can be drawn 1
white and
2 black balls can be
Hence
expectation
= (/g
Successive Events.
7.
drawn
his chances of receiving 20s., lls. .*.
happening and of the
For the chance
Let
failing of
specified order is
of its r
drawn
2s.
3,
= 15, Cf = 10.
3
is
5
.
is
and
20 + |f
-
are respectively 3/28, 15/28, 10/28;
11
+ Jf
=88. 9d.
2)
chauces of the
p, q be the respective
an event at a single
The chance of its happening
(1)
C\
is
black ball can bo
1
515
exactly r times in
n
then
trials ix
times and failing
r
happening
trial,
n-
:
C"p r
n ~r
r
q
.
times in a
n ~r
p q Now the number of different orders in which these things can occur is the number of different arrangements of r things of one sort and n - r n - r, which is equal to C r things of another sort. This number is n/ \r \n .
:.
I
Thus the event in question can happen in C ways, which are mutually r n ~r exclusive, and the chance that it happens in any one of them is p q ;
therefore the required chance
We
(2)
n
exactly
have times,
C^ = C^ r
Cp q r
is
(n~l)
The greatest term
(3)
part of
-
that of
n-r
If
+ q) n =p n +
,
nq
is
exactly
is
and
of
C> ~V
(n
+
r failures
integer, then
r
])q.
Hence
where
r is
= nq and n
p
:
times,...
2)
are
.
.
.
.
which
the
r is
the integral
most probable case
is
the integral part of (n + 1) q. = np, so that in the most
r
probable case the ratio of the number of successes is
(n
...
in the expansion is that in
that
successes
an
times,
terms in the expansion n~l n + C^p q 4-
respectively the first, second, third,
(p
.
therefore the chances that the event happens
,
exactly
n ~r
to the
number of failures
q.
may be useful to state an important theorem due to Bernoulli chance of an event happening on a single trial is p and if a number of independent trials are made, the probability that the ratio of the number of successes to the number of trials differs from p by less than (4) It
:
If the
any assigned number, however small, can be made as near to certainty as we choose, by making the number of trials sufficiently great. The proof of this is beyond our present scope, and requires a use of approximate values of large factorials. For instance, we may use Stirling's theorem, which states that
PROBLEM OF POINTS
516 If unfT
(5)
we
is the
chance that the event happens at least r times in n
trials,
shall prove that (i)
u nt r = p n + C% p n ~ l
For if the event happens at least r times, then (i) it must happen exactly n times, or exactly (n-1) times, or exactly (n - 2) times, ... or exactly r times, and the chances that these things occur are the successive terms of the first expression. (ii) it
Or,
must happen exactly (r + 2) trials, ...
or in just
trials,
r
times in just
or in just
n
r trials, or in just (r
Now
trials.
happen exactly r times in just (r + s) trials, it must happen these and also in (r - 1) out of the preceding (r + s -I) trials. r 1 1 of this is p C and therefore j;+*- ^r-igS^Cy+*~ ^ r/
that
it
+ 1)
may
in the last of
The chance
f
;
Un,r
-f + Vr-lP'q + ^iP^ +
-
+ Cr~l
P^f^
the second of the expressions in question. That these two expressions are equal may be proved as in Exercise XXXV, Ex. 9.
which
Ex.
is
// a die
1.
3 timej,
at least
(ii)
is
thrown 5 times, what
3 times
is the,
chance that ace turns
up
(i)
exactly
?
Let x and y be the required chances, then
and or by the second formula,
46
fe. 2. A and B throw alternately with a single die, A having the first throw. The person who first throws ace. is to receive 1. What are their expectations ? The chances that the stake is won at the first, second, third, ... throws are i
5
6'
6*6'
i
5 \4
Hence, the expectations of
i
/sy !
*6'
+
-
..
A and 5
i
/r>y
\Qj
;
\Qj '6""
!
!
=
'
are
-ff
and
G
^B r,,
;
^-
,65 =
chance==1 i
-
respectively.
The Problem of Points.' A and B play a scries of game* which cannot be J^a;. 3. drawn, and p, q are their respective chances of winning a single game. What is the chance that A wins m games before B wins n games ? '
Here
A
must win at
least
m
games out
obtained by substituting these numbers for
of r
m + w-1, and n
and the required chance
in either of the formulae in (5).
is
CHANCE UNDER DIFFERENT CONDITIONS (6)
So
Several Alternatives. the event
alternatives
at each trial there have been
far,
succeed or
may
517
may
it
two
fail.
Consider the case of an experiment which at any trial must produce one of three results denoted by A, B, C. Let the chances that A, B, C occur at
B
be p,
trial
any
therefore
occurs
q,
P
One
r respectively.
p + q + r = l.
C
times and
of the three results has to occur,
Also the chance that in n trials occurs y times (where a
A
occurs a times,
+ p -f y = n)
is
Jl [*[(/ that
is
to say, the chance
For the chance that in a specified order
p
is
the term containing
is
A may a
qPr
Y
happen a times,
p
a
q^r
B
y in
the expansion of
times and
j8
C y
times
.
Now the number of orders in which these things can
occur
is
the
number
arrangements of a things of one sort (A), j8 things of another sort and y things of a third sort (C), placed in a row. of
This Ex. this is
number
4.
A
card
In/ la
is
I
Hence the
\y.
j8
(J5),
result follows.
the card is replaced and the pack shuffled. If chance that the cards drawn are 2 hearts, 2 diamonds and
drawn from a pack,
is
done six times, what
is the
2 black cards ?
At any
trial,
the chance that a heart or a diamond
chance that a black card
is
drawn
is
is drawn Hence by the preceding,
^.
is
^=4
;
also the
=
512* Ex.
5,
chance that
A it
'
hand
'
of six cards
from a pack in the ordinary way. diamonds and 2 black cards.
is dealt
consists of 2 hearts, 2
Find
the
The chance that the card dealt
is
a heart
is
second card dealt
is
a heart
is -^-f ,
card dealt
is
a diamond
is
fourth card dealt
is
a diamond
is
^f
fifth
card dealt
is
a black
is
ff,
sixth
card dealt
is
a black
is
f-f .
first
third
Hence the chance that 2 in any specified order
hearts, 2
,
,
,
diamonds and 2 black cards are dealt in
this or
is
j^3 5~2~
is
^f
The required chance therefore equal to
is
. *
1,
2
5 1"
*
3,
3
50
. *
12 40
*
26 48
, *
25 4 7"
obtained by multiplying this number by [6/[2 [2
1_2
;
and
INVERSE PROBABILITY
518
Ex. 6. Find the. chance that in a game of whist the dealer (A) has exactly two honours. The card turned up is, or is not, an honour and the chances of these possibilities are ;
respectively ^3
A
an d Y^.
In the first case A certainly has one honour. Of the remaining 51 cards, 12 belong to and 39 to the other players. Let p be the chance that, of the 3 remaining honours, 1 is among the 12 cards and 2
are
the 39, then
among
^_/T3 P ^I In
the second case,
2 honours belong to
.
12.
51
39.
.
where the card turned up
A and
49-
not an honour, we have
if
p'
is
the chance that
2 to the other players, '
P
__
p4 U
12.
.
.
51
2
1 1
3j9
.
50
the required chance^
/.
is
3J.
.
50
49
.
38
.
48
>
-$p +-sp = f
8. Probability of Causes. (1) Suppose that an event has happened which must have arisen from one of a certain number of causes C l9 C2 ... ,
What
This question Ex.
bay
is
Each of three bags A, B,
1.
are as follows
A
a specified cause, C l9 actually led said to be one of inverse probability.
is the probability that
is
C contains
v)hite
and black
.
to the event ?
balls, the
numbers of which
:
A
B
C
white
-
-
al
a2
as
black
-
-
bl
62
63
chosen at random, a ball
is
drawn from
It is required to find the probabilities
Q l9
it
and
is
found to be white. came from A, B,
Q%, Q% that the ball
C
respec-
tively.
If
the numbers of the balls are altered as below, the probabilities in question remain
ABC
unchanged:
when
#,
?/,
z
are
white
-
-
atx
a^y
a3 z
black
-
-
bvx
b^y
b^z
any numbers whatever. x(a l
Choose these so that
+ 6j) ^y(a 2
-t-
62 )
^ z(a 3 + 6 3
).
The three bags now contain the same number of balls, therefore any one of the a^y + a 3 z) white balls is as likely to be drawn as another. If the ball which was drawn came from A, then it belonged to the group a x x of
(a^
-|-
white
balls,
and so
for the other possibilities.
Therefore
Pi^ 1 /(a 1 + 6 1 ),
where
Now
Qi'.Q 2 :Q3
a white ball
is
It is to
:
a 3 z ^p l
'
'.p z
Ps>
P2 =
+ e 2 +
#2 Qv p is
be noticed that
thai the ball
az y
:
drawn, therefore
Ci with similar values for
= <*>&
l
and
the probability that, the event will occur
comes from A, and so for p^
pz
.
on
the supposition
PROBABILITY OF FUTURE EVENTS Ex.
The same as
2.
a x white and
m
z
such as B, and
from an
A
y
m3
such as C, as in Ex.
C
B,
and as
is
m
l
1.
containing
Q lt Q2t Q3
bag respectively.
Then any one
Alter the numbers of the balls as in the preceding.
white balls
A
Also
the preceding, except that there are ra x bags such as
b t black balls,
are the chances that the ball came
one of the
519
of the
as likely to be drawn as another, and Q l is the chance that balls, so also for (? 2 Q 3 , therefore groups of
ax
Ql + Qz -f
before,
therefore
with similar values for
it
comes from
,
Q Qs 2,
3#j)
.
A
NOTE. If PI is the chance estimated before the event that an and P2 P3 have similar meanings, we have
bag
will
be chosen
,
P2 PS^W! W 2 m3 l Q x Q 2 Q3 =Pl p l P2P 2 PsPa Qi^PiPiKP&i + PsP* + ?&*), etc. P
and Observe that (2)
A
:
:
:
:
:
:
;
*
therefore
Pp J
l
is
:
the chance that a white ball
drawn, and that from an
.4
bag.
Suppose that an event has occurred which
General Statement.
must have been due to one
is
of the causes,
Cl9 C 2
,
...
Cn
.
Pr
be the probability of the existence of the cause Cr estimated the event took place. before Let p r be the probability of the event on the assumption that the cause
Let
Cr
,
exists.
Then
the probability
Qr
of the existence of the cause
Cr
,
estimated after
the event has occurred, is given by
For an argument similar to that in the Qi
and
:
%
:
3
:
last
two examples shows that
=
since the event has happened,
Gl + 02
whence the NOTE.
result follows.
It is usual to call
the causes and
P P 19
2 , ...
Pn the
a priori probabilities of the existence of
Q l9 Q 2 ... Q n the a posteriori probabilities. The product Prp r is the antecedent probability that the ,
event
will occur,
and that
from the rth cause.
The argument depends on the assumption
P2Pz>
(1)
that
Qlf $2 >"'
ar ^ proportional to
which is justified in (1) PiPit In particular, if an event is due to one of two causes, the odds in favour of its having occurred from the first cause are as P1 p l P^Pf The way in which these principles are applied to determine the Probability of Future Events is illustrated in Ex. 4. :
A POSTERIORI PROBABILITY
520 Ex.
A
1.
A
white.
white ball
bag contains 5
ball is
drawn and
and of
balls, is
these
What
is the
only
-
5
i
.
1+2 + 3 + 4 + 5 = 1 ~5~~~" 2
1
g /.
the required chance
2 5,
p2
,
,
73
may
= Ql
P
l
pl
---
// ut
be
0, 1, 2, 3, 4, 5.
a notation similar to the
,
,
2.
5 are
1, 2, 3, 4,
chance that this
of white balls
C lt ... C5 and using - u, p l _,i5 nm p Q ^n ana
Y
1
number
the
;
Denoting these possibilities by 6 _i above, we have 2^ p ~~JJ _ _p ^
$#.
is the
?
There are 6 possible hypotheses
that
equally likdy that 0,
it is
to be white.
found
^
\
.
.
,
^5
_5.. 5
>
;
.
is replaced,
what
is the
chance
a second drawing
~6.5 2X Hence the required chance Ex.
3.
the
A
pack of cards
Two
missing.
cards are
is
The
event
'
is
counted, face downwards, and to be spades. ?
that two spades are drawn. (Cj)
(C 2 )
is found that one card is What are the odds against
it
drawn and are found
missing card being a spade 4
li-
The missing card The missing card
There are two possible hypotheses is is
:
a spade. not a spade.
The a priori probabilities P19 1\ of C19 C 2 are P l J, P z = %. The chances p l9 p 2 of the event under the hypotheses C\, C 2 are
Mil *
5TO>
1312 = 5~l 5~0 '
P'2.
The odds against the missing card being a spade
are as
m
balls which are either white or black, all possible numbers Ex. 4. A bag contains being equally likely. If p white and q black balls have been drawn in p + q successive trials without replacement, the chance that another drawing will give a white ball is
The
possible hypotheses, all equally likely, are that the
m Denote these by If
pL p2 ,
,
we have
.
. .
m
q,
C lf C 2
,
...
,
lf
...
Cm_q
m-g-rH-1,
If
A
is
r-r+l
p.
r(r
nq \-r-\\rm _/j ^ 7/ur,.vr> '^q l^p^q
+ l)(r + 2)
...
(r
independent of r. ... are the a posteriori probabilities of
Q19 Q&
where
...
of white balls are
_^ r
+ q- 1),
...
and
number
are the antecedent probabilities of the event under these hypotheses,
^ /?m Pr'-^p ur
where
q
Cr
...
a^irVr-i
\p \q
(n-r+p-l),
C C2 lt
,
...
,
then
Qr ur vT jS
\m-\-\
^
(by Ch.
XXI,
10.
Ex.
1.)
t
VALUE OF TESTIMONY If
pr
'
is
the chance that, with hypothesis , f
n^
thus
pn
l
r
-.
<7
521
_
another drawing gives a white
r,
m-q-p-r+l _ __ _ _. n-l -r m-q-p* m-q-p *
*
__
.
=0, and the chance that another drawing gives a white
r~r*QrPr' = S'l(m-q-p)S, where S' = 2 /S" can be obtained from S by writing n r
r r
Thus
T"T
ball,
^\(n-l ~r)(n-r)
-_
j
and
ball is
;
n and p
for
1
the chance
^-
for
...
1
;
(n-r +p). and hence
=
9. Value of Testimony. The theory of probability has been used to estimate the value of the testimony of witnesses. Such an application is open to adverse criticism. It rests on two assumptions which can hardly
be
namely (i) that to each witness there pertains a constant p credibility], which measures the average frequency with which he speaks
justified,
(his
:
the truth; (ii) that the statements of witnesses are independent of one another in the sense required in the theory of probability. If
we are prepared to make these assumptions,
the procedure
as follows.
is
probability that a statement made by A is true and p has a similar meaning for B, what are the odds in favour of the truth of a statement which A and concur in making ?
Ex.
1.
If
p
/
is the
B
4
The
the agreement of A and the statement is true
'
event
is
possibilities are
:
(i)
;
B
in
The
making a certain statement. false.
(ii) it is
chance that they both say it is true is pp'. the chance that they both say it is true is (1 -p)(l -p')> Thus the antecedent probabilities of the event on the hypotheses (i), (ii) are
If it is true, the If it is false,
pp' and (\-p)(\-p') and the odds in favour of the truth of the statement are t
as pp'
:
(1
-p)(l -p')
2. A bag contains n balls, one of which is ivhite. The probabilities that f speak the truth are p, p respectively. A ball is drawn from the bag, and A and assert tJiat it is white. What are the odds in favour of its being white ?
Ex.
B
The are
'
(i) it is
The a If
'
event
is
true,
the agreement of
(ii) it is
A
and
P P
priori probabilities 2 , of (i), (ii) are 19 are the chances of the event under (i)
In the case of
(ii),
(n
The
and both
possibilities
false.
pv p2
-
1)
balls
remain
P
1/n,
t
and
(ii),
in the bag,
(n-
P^
l)/n.
we have p ~pp'.
and one
of these
A
is
white.
The
should choose this ball and wrongly assert that it was drawn from the (lp)(n-l). The chance that B should do the same thing is (1 p') (n- 1) ;
chance that
bag is hence
B in making a statement.
A B
and the odds
Pt =(l-p)(l-p')(n-l), in favour of
a white
ball
having been drawn are as
RANDOM
POINTS TAKEN AT
522
axiomatic
that
:
If a
(i)
The following statements are
Geometrical Applications.
10.
taken at random on a given straight line AB, the chance on a particular segment PQ of the line is PQ/AB.
point
it falls
is
Or we may say that the total number of cases and the number of favourable cases by PQ. (ii)
a point
If
the chance that the point
L
Ex.
Given that x +
and 2a are equally
AB
Let
likely,
on a
falls
y-2a show
where a
that
a/S.
is
constant
and
an even chance
is
it
is
P
at
random
in
AB,
includes an area
that all values of
o-,
x between
that
and
be a diameter of a circle with centre
Take a point
radius a.
represented by
S which
taken at random on an area
is
is
AB.
AP = x, PB~y,
then x + y~2a, and all values and 2a are equally likely. Draw the ordinate PQ, then PQ 2 = AP PB^xy. Let
of x
between
.
If A',
B' are the mid-points of OA, OB, the ordinates
at these points are equal to a
.
^f
O P
.
Hence PQ>cuJ% if, and only if, P lies in A'B'. Hence the chance that z//>f a 2 is A'tt'/AB, that Ex.
Two
2.
The points
We may
Q PQ>b,
points P,
that the chance that
are taken at
random on a
where b
are as likely to
is
(a-b)
is
therefore suppose that
Q
is
B
\.
straight line
2
OA
of length a.
Show
z .
ja
the order 0, P, Q,
fall in
B'
FIG. 86.
A
as in the order 0, Q, P, A.
to the right
of P.
Draw OA
'
at right angles to
OA and
equal to
it.
Complete the figure as in the diagram, where
OL-^PR^b. If
8x
is
distance of is in
PA,
is
small, the
P from
number
lies
of cases in
which the
between x and x + $x and
Q
by the area
of
represented by
Sx.PA,
i.e.
the shaded rectangle.
Of these, the favourable cases are those in which lies in EA, and their number is represented by the upper part of the shaded rectangle cut off by LM.
Q
Hence the the total
total
number
number
of cases
.'.
represented by area of the triangle OAA', and by the area of the trianglq LMA' ;
is
of favourable cases
FIG. 87.
the required chance
=
A OAA'
CALCULATION OF CHANCE
523
EXERCISE LV 1.
balls,
two balls are drawn from a bag containing 2 white, 4 red and 5 black what is the chance that If
they are both red ? one is red and the other black
(i) (ii)
A
?
white and n black balls. If p+q balls are drawn bag without replacement, the chance that these consist of p white and q black balls a rim finisim+n is C p .^/C p+g 3. If four balls are drawn from a bag containing 3 white, 4 red and 5 black balls, what is the chance that exactly two of them are black ? (i.e. two are black and two are white or red.) 2.
contains ra
;
.
4. In the last question what drawn from the bag are black ?
is
the chance that at least three of the four balls
5. Out of 20 consecutive numbers, two are chosen at random chance that their sum is odd is yf 6.
alike,
In a throw with three dice what
two
(ii)
Explain
alike
the
why
and the third
sum
is
different,
;
prove that the
the chance that the dice (iii) all
of these probabilities
is
different
unity,
fall
(i)
all
?
and
verify the truth of
this statement. [(ii)
Any
there are 6
particular throw of this sort, as five-five-six, occurs in 3 cases, 5 throws of this kind.]
and
.
7.
What
8.
In a single cast with four dice what (ii) exactly 12, (iii) less than ^2 ?
is
the chance of throwing 10 in a single cast with three dice is
?
the chance of throwing
(i)
two
doublets, 9.
How many
times must a pair of dice be thrown that may occur at least once ?
it
may
be at least an
even chance that double six 10. If four cards are
king, queen,
knave
A letter is
11.
out of
i
are ace,
'
'
chosen at random out of assinine and one is chosen at random Show that the chance that the same letter is chosen on both
assassin.'
occasions
is |.
[The number
of possible cases
12. If the letters of
that
drawn from a pack, what is the chance that they same or of different suits) ?
(of the
(i) all
the
'
'
attempt
are together,
/'s
is
(ii)
7
.
Of these 14
8.
are written
no two
's
are favourable.]
down
at random, find the chance are together.
13. A party of n men of whom A, B are two, form single rank. What is the chance that (i) A B are next one another, (ii) that exactly m men are between them, (iii) that not more than m men are between them ? 9
m
balls arc drawn simultaneously and 14. From a bag containing a balls, then n balls are drawn. Show that the chance that exactly r balls replaced are common to the two drawings is 'C'^~/( '^. Prove also that this ex;
m
and n. by interchanging a the second There are C n possible cases. The favourable [Consider drawing. cases consist of selections of r balls out of the already drawn, with selections n-r out of not already drawn.] pression
is
inialtered
m
a-m
B.C. A.
INDEPENDENT EVENTS
524
15. A squad of 4w men form fours. are next one another in the same four. (16.}
two squares are chosen at random on a chess board, the chance that they have a side common is y\-
If
(i)
show that
;
the chance that they have contact at a corner
(ii)
[(i)
Find the chance that two specified men
Out of 64 x 63
possible cases, the
number
is
yj^.
of favourable cases
is
4.2 + 24.3 + 36.4.] 17. One bag contains a white balls and 6 black balls another contains a' white and b' black balls. Let p be the chance of drawing a white ball from a bag chosen at random, and let^/ be the chance of drawing a white ball from a bag containing (a + a') white f and (b + b ) black balls. r Find p, p', and show that p ~", p according as a-\-b ~a' ~b' and ab' a'b ;
have the same or opposite
signs.
18. If v4's chance of winning a single game against winning (i) two games (at least) out of the first three out of the first six.
B
f find /Ts chance of four games (at least)
is
,
(ii)
;
the card is 19. A draws a card from a pack of n cards marked 1, 2, 3, ... n replaced in the pack and then B draws a card. Find the chance that A draws (i) the same card as B (ii) a higher card than B (iii) a lower card than B. Verify that the sum of these chances is unity. ;
;
[For
n 112 + - + ...+
1
-
chance
(ii),
;
{--
n ^n
n
1.1 } J
n
bag contains n cards marked 1, 2, 3, ... n. If A is to draw all the cards and is to receive 1 shilling for every card which comes out in its proper order, prove that his expectation is 1 shilling. 20.
A
in succession
[The expectation arising from each ball
~ of a shilling, etc.
is
The
fact that
Ti
the probabilities, on which the several expectations depend, are not independent does not affect the argument.] 21.
A bag contains n tickets numbered
at once, and
drawn. 22. (ii)
to receive a
What
What
at least
[For
is
(ii),
is
is
his expectation
1, 2, 3, ... n.
A person draws two tickets
of shillings equal to the product of the
number
p3
pz< (i)
,
numbers
?
the chance that a hand of five cards contains
two aces
23. If p l9 chance that
number
(i)
exactly two aces
;
?
of favourable cases
8
-C\ C
-\-C\
.
.
6*2
+48.]
pi are the probabilities of four independent events, find the (ii) at least
two and not more than two of the events happen
;
two of the events happen. 24. Two persons each make a single throw with a pair of chance that the throws are equal ? [If
and
(x
+ x2 +
this is
. . .
+ x6 2 - a 2x 2 + ...+ a 12x 12 )
dice.
then the chance = Q A~2 ( a * 2 +
,
t>D
equal to
{coefficient of
a;
10
in
What
f\-
(
\
JL
X*
4 \
2
-
x
j
/
}/36
.]
is
the
+ an2
)
:
GAMES OF SKILL
525
25. The decimal part of the logarithms of two numbers taken at random are found to seven places. What is the chance that the second can be subtracted from the first without borrowing ? '
'
26. In a game of whist, find the chance that each of the four players should have an honour :
(i) if
the card turned
(ii) if this
is
up
is
an honour
not an honour
before the last card
(iii)
27.
card
;
;
turned up.
is
In a game of whist, find the chance that each party has two honours the card turned up is an honour (ii) if this card is not an honour (i) if
:
;
;
before the last card dealt
(iii)
g.
is
turned up.
Out of n persons
sitting at a round table, three, A, B, C, are chosen at Prove that the chance that no two of these three are sitting next one
random. another
(n-)(n ~
is
5)/(n ~l)(n-2). For 2 of t^em, C can [Relative to A, 13 can sit in ri-~3 places. for n - 5 of them, C can sit in n - 6 places, Hence, Ibhe places favourable cases = 2(n - 5} + (n - 5) (n - 6).]
sit in
n-5
number
;
of
29. Three cards are drawn from a bag containing n cards marked 1, 2, 3, ... n. Find the chance that (i) the three cards form a sequence (ii) they contain a (iii) they involve no sequence. sequence of two ;
;
between A and 7J, the chance that the server wins a game always p, the chance that he loses is q. The score is five all, they proceed to play deuce and vantage games and A has the service. Denoting /Ts chance of 30. In a set of tennis
is
2 winning the set by x, show that x=pq + (p + q*)x, whence and B have equal chances of winning the set.
31.
it
follows that
A
A and B
before
B
chance
is
play with two dice on the condition that A wins if he throws 6 throws 7 and B wins if he throws 7 before A throws 6. Show that A's to J5's chance as 30 31. (Huyghens.) :
Three players A B, C of equal skill engage in a match consisting of a games in which .-1 plays 13, 13 plays O, C plays A A plays B, and so on, in rotation. The player who first wins two consecutive games wins the match. Prove that the chances which A B, C respectively have of winning the match are as 12 20 17. [First show that if ^4 loses the first game, his chance of winning the match is ^ if he wins the first game, his chance is JJ.] ^2
,
f
series of
,
,
:
:
:
A
33.
white. (i) (ii)
34.
bag contains 3
A
ball is
What What
is
is
balls,
drawn and
and
is
it is
equally likely that
1,
2 or
all
of
them are
found to be white.
the chance that this
is the only white ball ? the chance that another drawing will give a white ball
The same question
as the last, except that each ball
is as.
?
likely as
not to
be white.
[Proceeding as in Ex. 3 expansion of (i + i)
.
1,
p. 520,
The
we
see that
rest proceeds as in
P P P ,
Ex.
l9
2
^3 are ^ ne terms in the
2, p. 520.]
GEOMETRICAL PROBLEMS
526
A
bag contains 5 balls, and it is not known how many of these are white. balls are drawn, and these are white. What is the chance that all are white ?
35.
Two
36. The chances that A , B, C speak the truth are respectively p, p', p". are the odds in favour of an event actually having happened which (i) all (ii)
A,
three assert to have happened ? assert to have happened and C denies
B
What
?
Geometrical
The sides of a rectangle are chosen at random, each than a given length a, all such lengths being equally Show that the chance that the diagonal is less than likely. a is 7T/4. [Draw a square, OACB, whose side is a and the arc AB. If is one of the rectangles, the favourable cases are 37.
less
;
ONPM
when 38.
P lies in the quadrant GAB.] A floor is paved with rectangular
a and breadth the
A
b.
Show
floor.
bricks, each of length circular disc of diameter c is thrown on
that
the
chance
that
it
falls
entirely
on
one
brick
is
(a-c)(b-c)lab. 39. A point P is taken at random in a straight line AB. Show that the chance that the greater of the parts AP, PB is at least k times the smaller is 2/(k+l).
40. If two points are taken at random on the circumference of a circle, the chance that their distance apart is greater than the radius of the circle is -|. [The unfavourable cases are represented by the thick arc in Fig. 89.]
FIG. 89.
random on the circumference of a the chance that they do not lie on the same semicircle is J. [Choose P. Choose Q on one semicircle on PP' E must lie on P'Q'. (Fig. 90.)
41. If three points, P, Q, R, are taken at circle,
;
If
PQs,
number
of favourable cases-
total
number
ic
.
c,
etc.]
Jo
42.
On
AB of length a + b, two segments PQ P'Q', of lengths measured at random, (i) If a>6, the chance that P'Q'
a straight line
a, 6 respectively, are
9
PQ is (a-b)la. (ii) If c is less than. a or of PQ, P'Q' is less than c is c*/a&. part Let AP==x. If [(ii) <2=c, ^g=rz, and P'Q' overlaps PQ towards B, P' must lie in LM. Hence,
lies
entirely within
the
common
the
number of favourable
cases
=
I
(c
- x)dx =-|c 2
b,
the chance that
kP
L
c
M
Q
FIG. 91.
;
Jo total
number
of favourable cases
c
2 .
Total
number
of cases
ab,
etc.]
CHAPTER XXXIII CONTINUED FRACTIONS
(2)
EXPRESSION OP A QUADRATIC SURD AS A SIMPLE CONTINUED FRACTION
Surds of the Form
1.
(JNb
l
)/r v
follows that
From
any simple recurring continued fraction surd. In other words, its value is of the form (JN are positive integers, except that 6 t
may
is
Oh.
XXIV,
33, it
a quadratic equal b^)/r v where N, b l9 r } to
N
be zero, and
is
not a perfect
square.
We
shall
prove that conversely any positive number of
UNb
1
)/r 1
can be expressed as a simple recurring continued fraction. In considering this theorem, it is to be observed that There
(i)
is
no
loss
the form.
of generality in assuming that
:
N - b^
is divisible
by r v
For
(JNb
1 )/r l
= (jNr
1
*b r
1 l )/r 1
*
and Nrf-farJ*
is
divisible
by
2
hence (^/N b 1 )/r 1 can always be replaced by an expression of the r^ same form for which the above condition holds. ,
We
(ii)
For
if
need only consider surds which are greater than unity. 2 where r2 is a positive integer, then ^V-^i 2
=^
It follows that every positive quadratic surd or its reciprocal is of ,four types considered in the next article.
2. (i) (ii)
(A)
Types of Quadratic Surds. The surd in question
N ~b^
is divisible
The type
it is
assumed that
greater than unity.
is
by
In every case
one of
rv
(JN + bJ/Tt
where
b l *
This
will
normal type, and is of special importance in the theory forms \/AjB and JN.
:
be it
called
the
includes the
SURDS OF NORMAL TYPE
528
To express
(^N + b^)/^
as a simple continued fraction,
we form
the
equations -
where a a
is
-a
_n f
,
1
U-1
W"i
the integral part of (^/N -i-b l )/r
shall first
Since
^
is
show that
62
are positive integers. + b l )/r l
r2
,
and
li
2 r I r 2 = N-b^ ......................... (B)
kj^'Vi-fri*
We
,
the integral part of (*JN
,
&!?!<*/ N + b^af.L +
and therefore
b2
we suppose that
If
62
<0,
it
2 4-
........................... (C)
r!,
r x .................................. (D)
follows that
N/A
r
1
,
but
61
< N/^
r >
and consequently 6 1
therefore 6 1
Again,
62
is
.
a positive
and consequently
r 2 is positive.
N-b2 and
since
N -b^
positive integer.
for
n = 2,
3,...
2
number
*
is
= N- (a
x
fj
divisible
less
than
Further,
is
N
62 2 >0,
we have
- ^) 2 = N-b*-r l (afa - 2a 1 61 ), by
rp
N -b
2
Hence we form the equations
so also
Continuing the process,
where a n
therefore
^/-ZV,
is
2
.
r2
is
a
an integer such that (F) .................... (G)
leading to
steps similar to the above, we can show that every a, 6, r is a positive For taking n = 2, by the preceding (*JN + b 2 )/r 2 >l, therefore a 2 integer.
By
is
it
N
- b^ is divisible a positive integer. Also by r 2 Reasoning as before, will be seen that 6 3 andr3 are positive integers, and so for n = 3, 4, 5, ... .
.
Finally, the fraction is periodic, for the
and
for every n, b n
y
therefore
rn
nth complete quotient
is
= (b n + b n+l )/a n <2JN.
Hence the fraction (JN + b n )/r n cannot have more than 22V distinct values, and one of the complete quotients must occur again. From this the stage, all the succeeding a's recur and in the same order for the fe's and r's. Thus the continued fraction is periodic. :
same
is
true
REDUCTION TO NORMAL TYPE
529
It will now be shown that in the process of expressing a positive quadratic surd of any other type as a simple continued fraction, a stage must occur at which the complete quotient is a surd of normal type. Whence it ivill follow that in every case the continued fraction is periodic. 2
The lype (*JN -ftj)'?*! where quotient is a surd of normal type.
/>
(B)
-
N/57-7
T.
X or instance,
/
r
~~p
The form
_
/>/57 + 7
,
_
__
b
i-t
.
where a x
a2
,
1
1
1
%
-1
form the equations
rz
__.,
are the integral parts of the fractions on the left and,
...
,
.1
--i
>N. We 0l+
1
3+l-t-l+l+l4-7
4
/
2
1
1
2
+ JN)/ri where
(bi
Here the second complete
.
/
s/57-7
2
(C)
T
=2- --- =
/
= -/ 1
dim and
1
for every n,
"
"w-f-l
As
in (A), every r
hypothesis
}
~~
n
a n rn>
r n r n+ 1
~ ^n 4-1 ~
an integer and every b
is
an integer or
is
zero,
and by
> S/JV.
b^
Suppose that are
*
6 15 6 2
,
6r
...
b l9
all positive, arid
b2
,
are
,
...
all
6 n41
^N
then r ly r 2 ... r n greater than is a decreasing sequence of integers. ,
;
Hence a stage must occur at which This being
and therefore Hence n is since also is
so,
n cannot
b n+l
odd,
JN
-
be even
;
for in that case
we should have
>JN. and
bn f t
+,JN>a n rn
bn
> 0,
it
x /-ZV
+ 6 n+l >0; and
Ar ~^ 4 >0, and
therefore r n+1
Consequently
.
follows that
.
1
negative.
Now
(n
+
being even, the
1)
(n
+ l)th complete
quotient
may
be
written
and
since
7
^w+i^^
according as 6 W a. 1 (D) The type
quotient
is
(b l
?
^i s
or
^^ ie
next quotient
is
a surd of normal type
0.
~JN)/r 1 where
a surd of type
(C).
6 1 2 >^V.
Here the second complete
MENTAL PROCESS OF RECKONING
530 3.
Theorem.
Let the surd
ing fraction, then
if
(i)
(JN + b)/r
b<>JN
be expressed as a simple recurr-
has no acyclic part;
the fraction
if ,JN>b + r, it has an acyclic part consisting of a single quotient; b > JN, it has an acyclic part of one or more quotients. (iii) if For if x = (JN + b)/r then (rx-b) 2 = N and the second root of this - /iV + hence, from Chap. XXIV, 33, it follows that quadratic is &)/r + (i) t/ -1<( N/2V 6)/r<0, the fraction has no acyclic part and these (ii)
,
(
;
conditions are equivalent to 6< N/2V<6-hr; (ii) i/" ( - N/A + 6)/r< - 1, that is, if and 6 -f r, the acyclic part consists of a single quotient that if contains one + the is, (iii) if b)fr>0, acyclic part b>JN, r
JN>
;
(~JN
or more quotients.
In particular, if
v A/B has a
A>B,
simple continued fraction equivalent
the
to
single non-recurring quotient.
4. Method of Reckoning. In practice we replace the written work involving surds by an easy mental process, as in the next example. Ex.
I.
Express (\/37
-f
as a simple continued fraction.
8)/9
Here a i integral part of the surd = l, also are now found in succession from the equations fe
n
=
fl
-&-n
n.-i r n-i
giving the table
rn
= N~ (
W
r n-i>
^ n
if
continued,
part of
a repetition of the part between the dotted
is
s/37+8 " 9
+
1111
lines,
and
i+I+3V2+"" Jf
The
= integral
various quantities
:
The reckoning,
NOTE.
^=9. The
8,
-X-
third complete quotient, (\/37+3)/7, marks the beginning of the 3<
recurring period, for
chapter that, from
5.
this stage, the above reckoning
can be replaced by a G.C.M. process.
>/A/B as a Continued Fraction.
(1) It is
supposed that A,
have A/ -= = \ />
-5 jD
=
l ,
B
are positive integers
where
N = AB,
6a
and that
= 0,
rl
= B.
A>B. We The surd
is
7*j
therefore of the type discussed in Art. 3, and it is expressed as a simple continued fraction by constructing equations of the type
,
for values of
n = l,
2, 3,
...
;
CYCLE OF QUOTIENTS where a n
is
an integer such that a n
an(i
b n+l
leading to
As
< (JN + bn )frn < an +
1,
= =
A/"S r is a positive integer and the continued Also, it has been shown in Art. 3 that a 1 is the only
in Art. 2, every a, 6
fraction
631
is
periodic.
and
non-recurring quotient. (2) is
The following
greater than
inequalities are required.
JN + bn >rn
therefore
1,
Every complete quotient (A)
.
= (# - bn*)/rn = (JN - b n (JN -f b n )/r n >JN- b w therefore ,/#<& + rnHl Again, if n>l, the continued fraction equivalent to (*JN + b n )/rn no acyclic part, therefore, as in Art. 3, 6n <\/^ <^n + rn n ^2) Also
r n _!
)
(B)
has
r
(
For any
suffix
w, 6m
(3)
hence from (B) and
bm <*JN,
-6 n
and
111
/-^
*
Now JNfr^-a^ is a root equation being V#Ai ~
Therefore by Ch.
Hence
ac
,
K -f
-
a 2"^" a c-fl
and therefore It has
recurring fraction in
+
from
2a l
;
(iii)
the beginning
etc.
for the rest of the
and end are
B
rl
l
_L_JL a 2 -i- a 3 -I-
"
*
_L
_J
a3
a2 4- 2a x
-f-
*
and we
,
cycle, the
equal.
be expressed by writing
M_N/A[ \
a3
<._!==
been shown that \I^AjB can be expressed as a simple (i) there is a single non-recurring element, a l ;
partial quotients equidistant
may
31,
which
the last partial quotient of the cycle is
All this
XXIV,
J
= o2a t -f
. . .
2a l9 a c = a%,
a c+l
Summary. (ii)
1
...
^c+
the other root of this
111
l
I
TT
*
r^ (x + a^^N,
of
i-
(D)
Let c be the number of elements in the
The Cycle of Quotients.
cycle, then
(C),
-6n <^n (n^2)
6m
(C)
shall call the sequence, a 2
cycle of quotients.
!_ +
*
"
'
* ,
a3
,
...
a 3 a 2 the reciprocal part of the ,
,
CALCULATION OF QUOTIENTS
532 (4)
The b and
Let
r Cycles.
cycle, so that
be the number of elements in the
c
jjf
\
j
*
*
The recurrence
due to the fact that the second complete quotient
is
From
appears again as the (c + 2) th. recur in the same order hence,
and therefore
bc + m
Hence, bc
63
,
r c +m
= 2a
complete quotients
ci
1,
rm
,
^^2.
f r
m = 0,
c _ m ^a 2+m for
1, 2, ...
,
we have
.
(i)
the cycle of
etc. ;
?'
=
c
The character is
Vs
6 2 , 63
is
,..
,
the cycle of r's is r v
(ii)
first term, i.e.
sequences
and
that a c+1
Kemembering
and so on
bm
this stage all the
(JN + b c+m )/r e+m = (JN 4- b m )/rm
;
2,
f c -i
of
the
/'
= r^
bc
r2
^ and is
,
i.e.
reciprocal,
and
rc ,
...
b c+l
= b2
,
is reciprocal after the
etc.
and the reciprocity of the three
recurrence
exhibited below, where the recurring periods are enclosed in
brackets.
Q
T
al
a2
&3
It should be noticed that the a
parts of the a
and
i
,
...
and
tt
,
a2
a
2
o
.
6 cycles correspond,
and the
reciprocal
r cycles correspond.
Calculation of the Quotients. In finding the values of 6 n r w a n teH w/ien middle the the b z/ of cycle has been reached, no further calculation is necessary. This matter is settled by the following theorem. (5)
,
,
,
we can
If b m r rn a m are respectively equal to are equidistant from the ends of the b cycle. ,
,
For
rn+i
and
:=
(6 n42
- bm _
the
l )/r
m
inequalities
__ l
less
rn ,
n+1 r n+1
Subtracting and using the hypothesis,
Now by
,
an
,
and
b n+l
+ 2 )/r n+l
and
then b m
= (N-b$ + i)lr n = (N-b m *)!rm = rm _ r
^m-i rm-i =b m _, l + bm
Also
b n+}
in
.
.
1
find that
(D),
(2),
are less than unity.
we
= 6 w+2 + 6 n4
both
(6 rn _ 1
- bn
Therefore the right-hand side of (A)
is
than unity, and consequently a w _i = a n+ i and bm _ l = b n+2
numerically We can show by a similar argument that 6 TO _ 2 tively equal to 6 n+3
,
r n4 2 , .
n+2
,
and
so on.
,
^m _ 2
-
a m-2 are respec-
CONVERGENTS
m=n
Putting
m=n+ 1
and
533
we have
in succession,
the following
:
If two consecutive b's, namely bm and bm + v are equal, then rm and a m are the mid-terms of the reciprocal parts of the r and a cycles. In this case (i)
c is even.
= // r n rn+l and a n =^a n+l
(ii)
two consecutive
(so that
likewise the corresponding a's), then
rn ,
and a n
rn+l
r's are
equal and
an+1 are
,
the
mid-
terms of the reciprocal parts of the r and a cycles, and bn+l is the mid-term of
In
the b cycle.
Ex.
Express ^/^i ana V^l as simple continued fractions.
1.
= V.
(i)
this case c is odd.
......
we construct the
N
Henco
.
A/T^J-
following table
^ = 11,
187,
al
by a mental process
[11
13
r
[11
6
3
22
a
1
[4
8
1
6
~l
11
and proceeding
t
as in Art. 4,
:
11
4
8
2].
Because 6 4 = 6 5 11, therefore a 4 is the inid-term of the reciprocal part of the a cycle. This cycle can therefore be completed, for its last term is 2a x = 2, thus 17
V
111111
+
II"
4~+
8+ r+
8"+
IT 2* *
x-
(ii)
For
/s/61
r
[I
60 a7
we have
= l,
=7 and
the
6, r, a,
table
is
[1431221341
12
4
3
a 7 and r 6
Seeing that a e
r7 ,
reciprocal part of the a cycle.
Therefore
ax
[757546 rj_
v
5
9
5
we conclude
Also the last quotient of the cycle
-
---
-
J+
+
j-
2+ 1+
where
*
Relations connecting the p's and q's.
N = AB, VIN -
rl
=
^B- and J
let
.-
-}-
^ B t
o-f
^i
p r /qr be
11~
1 05,-
an +
?i
zq n
Suppose that
the rth convergent, then
t.
wHere
z
= *PnPn=i
therefore
.
J+
* (6)
is 14.
11111111111 _ 2+ 4+
,
14].
that a 6 , a 7 are the mid-terms of the
+ ? w -i
and Equating rational and irrational parts,
21
==
7 V/A + h W-Hl ~~~
r^,.
i
CONTINUED FRACTION FOR
534
Solving for b n+1 and rn+1 and putting that
N = AB
and
r^B, we
find
=(l)rw+1 ........................ (B)
Bp n *-Aq*
2 2 Integral Solutions of Bx -Ay = M. If for some value of n it n happens that (- l) r n+l = M, then, by equation (B), (j? n q n ) is a solution
(7)
,
of
Ex.
1.
Find two
Expressing ^/^f- as a continued fraction
= 5/4, lqi 6.
^N
and therefore
1^4/^4=46/37,
In
(
(5),
(5,4),
Ex.
the cycle
2
3.
17?/
1),
we
find that r3
r6
= 3.
Also
(46,37) are solutions.
as a Continued Fraction.
Here
(1)
rx
= l and
the
all
Additional theorems are the following.
conclusions of the last article hold. (i)
-
z positive integral solutions of 11 x
o/Vs, one member and only one is equal to 1, namely r, r m = l, then, remembering that 6 2 = a 1 and using the
For suppose that
_r
ar
equations
,
+y
we have a m = integral part Also o
= &H-&,
Therefore
beginning of a (ii)
//
r n >2,
cycle of
^n
- 1V /V ~
(JN + 6J/r TO - % + &m = 6 2 + b m
of
/^H-6 m+ i)/ m ^i =
new
mrw+l
.
therefore
i
(N
f
--J
(
and
N /^-r62)/r 2>
marks
^tn^l
the
r's.
a n ^b n
.
For
in this case
n>l
and
so that
=
// r n 2, JAe number (c) of elements in the cycle belonging even. Also a n is the mid-term of the reciprocal part of the a a n = a x or a l - 1. (iii)
'
For a x Also
an
is is
cycle,
the integral part of
.
,
therefore 6 n = 6 w+1
is
and
a^bn
,
,
a w >(a 1 H-6 )/2-l>6 n -l>6n But by the preceding, a n <6 n therefore a n = bn
Hence by
JN
to
3
.
we have
.
Also
2a n = 6n
+ 6n+1
,
.
even and a n is the mid-term of the reciprocal Finally a n is the integral part of (^i + n )/2, so that
(5), (i), c is
part of the a cycle. a n = (a 1 + a n )/2 or (a l ^ a n - 1)/2, and therefore a n ^a l or a^-l. (2)
Equations (A) and (B) at the top of the page become
^?n?n-l-^n^n-l = (~l)
p n *-Nq n
*
W
&4-l
....................... (A)
=(-l)rn+1 ......................... (B)
CALCULATION OF CONVERGENTS (3)
x2 -Ny 2 = M.
Integral Solutions of
(-l) n+1 = M, nr
found that
In particular,
if
c is
If,
for
535
some value
of w, it is
then (p n q n ) is a solution of x -Ny* = M. the number of elements in the cycle belonging to 2
,
the necessary and sufficient condition that r n+l may be equal to 1 that n = tc where = 1, 2, 3, ... hence, it follows that
JN, is
;
x -Ny = l,
(i) for the equation and- (PM C ?2fc) when c >
2
is
(ii)/or the equation x* c is
odd
2
odd
Find a
Ex.
1.
We
find that
Ny =
c
= 5.
>
18,
6
x2 -
2
13^/
-
1.
* q5
~ 5 and (18,5)
is
a solution.
The following formulae are
Calculation of Convergents.
(4)
we multiply equations (A), (B) firstly by and secondly by p n and p n _i> we obtain by addition
this connection.
q n -i>
even
when
solutions are (p(2*-i)c> ?(2*-i)c) solution when c is even.
positive integral solution of
Also
c is
I,
* so that
when
;
2
the method giving no
;
solutions are (p tc yjtc)
If
useful in
and
qn
Again, changing n into n 1 in equation (B) and using equation (A), find that p n ~i = r n q n ~b n+l q n and Nq n _ l ^rn p n ~-b n ^p n _ l ..... (D)
we
^
In general, the p's are considerably larger than the
The Convergent p c /q c
(5)
elements in the cycle belonging
.
(i)
11~ ?c
f
i
where
~
m + n = c,
where c
is the
number of
to
.
For
//
_..._ a
2+
111
m+ f
-- "
~
q n+l
therefore If
^
is
the numerator and
Y
the denominator of the last fraction,
m Xq m -Ypm = (-I)-iqn and Xft^- Yp m .^(- l) qn+l Hence any common factor of X and Y is a factor of q n and of
*
q n+l
,
and since these numbers are prime to one another, so also are X and Y. Thus the fractions pjqc and X/Y are in their lowest terms, and therefore pc = X and qc = Y, which are the results in question.
536
HALF-CYCLES In particular, if c
(ii)
odd and n = |(c -
is
l>c=P*!7n+;W5Wi // c
is
(iii)
//
even and n
m
w
equivalent
to the
m
-qm p c -(p m qm ^-q m p m ^)q n -=(-l)q , r
^
p m -Nq m ^(-lY"rm V-AV-f-lJ'VHi; mtn = c, we have r^^^r^^; and therefore ~ N
-Pn
remains to consider the sign of p mp c -
The convergent
m
.............. (H)
)
(PmPc It
),
2 2 %n?c) - # (PmVc ~ VmPc? = (?> w ~ ^?m2 (^ 2 ~ W)2
since
and,
~
(P*-JNqm ){p e + JNq e
two equations
For using equations (E), p m q c Also we have the identity
Now
= ? n 2 + ?JM-i ................. (F)
n = c,
-f
(PmPc
?c
\c,
Pn + JNqn = (-I) icA
and
1),
pjq^
precedes
pjq c
.
2 -
Nqm q c
.
First suppose that c
is
even.
and p m p c >Nq n q e If m is odd, it follows from Ch. XXIV, 25, that p m P c <^r Tlie samc results follow q mq C w in a similar way when c is odd, and thus pm p c = I) 7> n Nq m q c If
is
even,
JN
.
c
'
.
(
It should be noticed that these results hold if c is replaced by (iv)
N is a prime and c is even and n = ^c
//
>
tc.
then
Pc + JNq = l(p n +JNq n )*, ........................... (J) 2 2 and q c - p n q n p c = (p n + Nq n e
which
is
Also
a cycle
equivalent 2
jt;
n
-
is a^
to
-J-
= 2Vyn or a, - 1. (
For by equations
(I),
therefore /T) 2
and
M
2
_ JV0
2
2
1)
.
2,
.
)
and
the
mid-term of the reciprocal part of the
p np c - Nq nq e = - l)p n and p n q c ~ q n p c = - l) n qn = (Pn* + ......................... (K) Pc/Vc n
/-n
V
21 NqM>
(2s_^L) ^(A) -A^-,; V 2p w 7 n / ?c 2
,
(
hence
^2 * ~ Nq " \7
(
pw ?n
L
9
2\
(
-1)* ~, ?c
the sign being determined by the fact that p n <>JNq n according as n is even or odd. Since is a and is to prime pn prime q n the last fraction is in its lowest terms. Hence p n 2 - Nq n 2 = ( - l) n . 2 or ( - l) n and the >
N
,
,
latter alternative
is
impossible.
MULTIPLE-CYCLES * ^ p n - Nq n = -
Therefore
Vfi^
Hence, NOTE. and noting
(
When N that p c /q c
Thus we
and the
;
in
is
find that
its
>/21 ^
and
2
.
rest follows
p c and
not a prime,
is
n l)
537
q c =p n qn
by
(1), (iii), p.
may
be
qc
c
= 6,
= 4 + -i- -i-
L
*
_L I
1+1+2+1+1+8
9,
713
o
C^Q
(6)
and
.
n o y ^
The Convergent
c is the
number of
quotient
is
j9 6
_r t>5,
g6
= 12.
V Pn/q n the n-th convergent of
is
(i)
quotients in the cycle, then
Pn+tc =PnPtc
For the
-
anc* therefore
fo iw
.
and
2,
#3
,
*
p ~ 81+21.4 = 55 2/
found by using equation (K)
lowest terms.
# so that for \/21,
534.
+ N(lnqtc
(c + l)th quotient a x + N/A therefore
is
and 2a x
q n + tc
=p n q + q np tc
(L)
tc
and the corresponding complete
r
,
and equating rational and
fractions
Removing Again,
11
1_
irrational parts,
^_ %
the last quotient being the
(te
we have
From
+ l)th.
this,
H- _p n
by means
?n
of equations
(I),
Pn+tc
If JC is
(
a l + Pn/
the numerator and
Y
X-N Y= Pn Ptl?-Nq
PnPtc
+ Afyngtc
the denominator of the last fraction,
and
PtcY -q teX = q n ( Pt
-Nq *). Now Ptc 2 -^9tc 2=: il ^7 (5)> (iii), therefore any common factor of -X and y divides p n and y n which are prime to one another. Hence Xj Y is in its lowest terms so also is p n+t c/
qtc
(
t
*)
t
,
:
tc
which are the equations in question.
As a
special case
we have
P(s+t)c=PscPtc
in particular,
+ N qsc q tc 2
P 2 tc-Ptc + Nq tc
and 2
and
q( S + t )c=Psc
q^ ic
+ y8cPtc'>
......... (
N
)
^^p tc q tc ..................... (0)
ABBREVIATED CALCULATION
538 (ii)
JN
Since
is
irrational, equations (L) are equivalent to the single
relation
Putting
t
= 1 and n = c,
In Ex.
Find a 1, p.
t,
)
533,
solution in positive integers of x*
^61
- Sly 2 = 1.
Hence the smallest solution
The mid-elements of the cycle are the 5th and 58. Hence by equations (F),
(p 22 , q 22 ).
;
) (p c + JNq c y ............. (Q)
expressed as a continued fraction, the number of quotients of the equatiort in question is
is
in the cycle being 11.
2? 6
+ V-%c = (Pc + J^Vc) 3 -, .......... (P) p tc + JNq tc (p c + JNqc P3C
p n+tc + JNqn+tc = (p n + JNqn
and therefore
find that
(
generally, for every positive
Ex. L*
,
we
in succession,
...
+ >/#fee = (Pc + V^c) 2
Pte
and
2c, 3c,
6th,
and p 6 ~164, q 6 ~2l,
^453, # e
Pu -Ms +Pe?6 = 164
.
+ 453 58 - 29718,
21
qn
.
Using equations (0) and observing that p n
*
= - 1, q 22 = 2p n qu = 226153980.
Pn ^Pn + 61 ^u = 2^ii + 1 = 1766319049, 2
2
Thus the 7.
where
2
least solution is
(1766319049, 226153980).
The Cycle belonging is
r'
= q* + g c2 - 21 2 + 58 2 = 3805.
- 61g n a
to z
=
(
N/N
+
Suppose that
b)/r.
a positive integer, then the simple continued fraction correno acyclic part (Art. 3), and the cycle can be calculated
sponding to z has as follows.
Rule.
Let
(x> y)
be a solution in positive integers of
x
equations
2
-Ny*=
the latter pair being used only
when
or
1
r,
r'
4,
any one of
the
...........................
(A)
are both even and N,
are odd.
b
x/y for .JN in the given surd. Express the result as a simple continued fraction with an even or an odd number of quotients according as Substitute
on the right of the chosen equation is + or or more of the cycles belonging to (JN + b)/r. one of the sign
Since
Proof.
N =6
2
-f-rr',
the surd rs 2
(1)
We
shall
such that
where y *
may
the positive root of
......................... ....... (B) ,
f
y
pq'-p'q=l
For equation if
is
This fraction consists
prove that this equation can be written in the form
z~(pz f p')/(qz + q'), where p
with (B)
(JN + b)/r
-2fe-r' =
.
(C) is
the same as qz 2
= q ry,
q p', q are positive integers y
and (p
p-q' = 2by,
......
(C)
q'^q .......................... (D) -
q')z
p'
= r'y, p'
= 0, which
is
identical
........................ (E)
be any rational.
Thi8 problem was set by Fermat as a challenge to the English mathematicians of his time.
GENERAL RULE From
we have -Kp + q')
(D), (E)
where x
^Ny
2
I
=,r
2 (A) x, y are positive integers such that x
(B) x
/2,
539 2
........................... (F)
,
All possible cases are included in the following
rational.
is
2
y = 77/2 where
,
77
are
2 ~-Ny =
1.
odd integers such that
2
- A7 2 -= ??
N
'
In case (B), divisible
by
4,
r
r,
and
f
are both even, for T
if
A
is
odd
so also
In both cases, from (E) and
(F),
we
p = x + by, and
p/q
It remains to be
(i)
2 // x
(ii)
Ny
2
iind that
=x-b\j .............................. (G)
q'
This will be so
if
x/y^.b +
r.
:
the least value of y
,
therefore
N**:(b
x/y^\/N -i-l^b + r where
~/2, y^rj/2
//
is
- Afy 2 = - 4, then or xly<^JN
Now N<(b + r) 2
(iii)
b2
*
1
^=I,
Hence
r'y.
= (x/y -f b)/r ................................. (H)
Three cases must be considered
- Ny 2 = ~ // x
p'
ry,
J-/1.
is 6,
shown that q'^q.
2
g=
:
z
-Nr)
and
is 1,
+ r) 2 and
2
--^i,
1.
Hence
(f^q* the least value of
77
is
land
xA/
Hence (2) It
r
and
x/7/
follows that
Up
q'^q.
11
-=a x + -----
..-
an
2 -f
(7
............................... ( J )
,
even or odd according as pq -p'q = the convergent immediately preceding p/q. (Ch. be determined by z = al + ... -7 a n -f z 2 +
where n
f
is
-111
l
-1, then
or
XXIV,
12.)
is
p'/q'
Now
f
let
z
,
'
then
z
= pz + P' ? 2/
Hence belonging
2;'=^
and
+ 2'
and
^jtP'^^tP', '
J2
'
+ ?'
2 21
the fraction (I) consists of
+ ?'
one or more of the cycles
to z.
The rule can be used (as in Ex. 2) to express any quadratic surd as a continued fraction, beginning at the stage where recurrence commences.
2M
B- C
-
A
EXAMPLES OF USE OF RULE
540 Ex.
Express (*JQ2l +21)/18 as a simple continued fraction.
1.
21< s/621<21 +
Since
there
18,
no acyclic part.
is
Also
18r'
= 621 -21 2 = 180,
r' =. 10.
giving
Thus
r,
r'
are even
and
6 is odd, also
25 2 -621.
I
2
^4.
Using the rule, we substitute 25/1 for ^/621 in the given surd, and express the result as a simple continued fraction with an even number of quotients, thus
+ 21
25
23
1
_9 +
1
1
~~T8~~""" r+ r+ s/621 ^ -
and
+21 -
18
2.
1 -
-
v 1+1+4
* Ex.
1
1
= 2n +
4' .
#
Express (^13-+ 7)/4 as a simple continued fraction.
Since 7> N /13, there is an acyclic part. The reckoning on the right shows that the second complete quotient is (/s/13 + l)/3, and since l<*yi3
3
Since
r --3,
-13.1
2
^-4
2
first
+
l)/3,
jy/13
fraction with an odd
number
in (;s/13
18
\1._
5/3
is
I
]
23 j^
_
5
.
and express the
~
, 1
15
-
2
-
7,
1
4,
3
a
2,
1
1.
Taking the second, we
useless.
is
equation
of quotients, thus
___L This
and
an odd number, the
substitute 18/5 for
18 - 13 2
b r
result as a simple continued
1111 ___
___-
-j
1
+
-
1
+
6
-
+
1*
the cycle belonging to the given surd, and s/13
+
7
"
L JL _L
+
4
1
|
1
_L IT IT 6+
I'
EXERCISE LVI 1. Verify some of the following results, where the values of as far as the middle of the first cycle
6, r,
a are given
:
fb. 0.3. 1.2 (i)
s/13
I
,
(Hi)
1
.
4
.
3
.
3
a 3
.
1
.
1
.
1
r
-j
.
.
Cb. 0.4. 2. 3. 3 (ii)
I
0.5. 1.4. 5.
Cb
.
r
.
1
.
6
.
5
.
3
.
2
.
3
I
a
.
5
.
1
.
1
.
3
.
6
.
3
f
6
^31
v/19 \ r
.
(v) s/91 \ r
.
I
.
[ a
.
9
.
^43^
(iv)
[
9.1.8. 7.7 10
.
9
.
3
.
14
.
3
1.1.5. 1.5
8 7.5.9. 7 8 9.5. 12.7.4. 15.3. {ba. 10. 2.3. 1.2.4. 1.6. .
r
.
.
1.
10
.
.
.
.
3
.
5
.
2
.
5
.
2
.
1
.
3
.
1
T6.0. 6. 1.5. 4. 5.
5
j
.
I
.
a 4
.
.
10 3
6
5
r
.
1
.
7
.
6
.
3
.
9
.
2
.
9
a
.
6
.
1
.
1
.
3
.
1
.
5
.
1
GENERAL CASES r
N/^Tl
(vii)
__
VaM- 2
(ix)
J
6
.
r
.
.
o
.
a
.
1
.
1
1
541
O.a-l.a-1
("6. \l~tf~- 1 J
(viii)
r
1
.
2(a -
.
l.a.2rt.2a
[a.a-1.
fb.Q.a.a
f
J I
r
a
.
1
.
2
.
1
.
a
.
a
.
2a
Va 2 - 2
(x)
-|
I
b
.
r
.
a
.
1)
1
.
1
.
2(a-~l)
.a-l.a-2.a-2 I
a-
1
.
2a - 3
.
.
1
.
2 a -2
.
.
2a - 3 1
(a>2) 6
.
r
.
r
Va(a4 1H
(xi)
1
.
a
.
a
.
a
.
1
Va^Ht
(xii)
la.a.2.2a
(xiii)
*Ja(a
4 4)
r 6
.
r
.
.
1
.
la.a +
1.
^
p.O. r
-I
1
.
a
.a -2. 2
4
.
.
a
La.a.i(a-l). (a> 1 and odd)
a41 2a - 1
.
a-2 4
.
1
.
-l(a
-
1)
a
-1
a
.
a
.
.
a
.
.2
.
1
(a> 1 and odd)
4
(-!)
.
In certain cases, a solution of one of the equations x- can be written down at once thus
Ny 2 ~-
2,
1
or
i4
;
if
A = a~
if
A =a
a2 -
T
if if
Apply
3,
this to write
2
,
then
J- 2,
then
1
A7 = a (a 4- 1 A ^ a i 4, 2
down
the
then
),
2
(a
(2a
: j- 1 )
-f 1
2
2
-
)
N -
1
A
.
-
2
T
Aa
2
22
T1 1
;
-= 1
;
;
a --A.I ~ T 4. then -A 1 when integral solutions of x N = 5l, 47, 56. 2
2
T
2
2
?/
Find a solution of x 2 -109?/ 2
-4, and apply the H.C.F. process to show
that
+7 __. ^
v/109
*-
A
JL J_ JL _JL Y^ 2+ T+ 94 14
^-
4,
Obtain the following
* *
2
*
:
I4l4l4l42
;
^
1+
6
84 24 14 34
2
;
*
/17_ = "9 +
V3 5.
111111 24 F+ ff IT 24*4*
Find two positive integral solutions of 2 2 2 ^ 2 -7, (i) 5^ ~13t/ ^7, (ii) 5x -13i/
(iii)
3x 2 -
2
17i/
^7.
Find positive integral solutions of x 1 ~~ 621?/ 2 ~4 and of a; 2 621z/ 2 1. Referring to Art. 7, verify that if 623/25 and 7775/312 are substituted for v/621 in the fraction ( N /62l421)/18, the first gives two cycles and the second three 6.
cycles of the fraction.
SUNDRY THEOREMS
542 If A=:(2ra
7.
+ 3) -4, prove 2
-l
_
~~-
that
JYI
111__
__
-I
___ m-f 1+ 4(w-f-l)+
2+
* If a
8,
'
1
#
odd, find a positive integral solution of
is
x 2 -(a 2 + 4)y 2 = a;
(i)
(ii)
x 2 -(a 2 + 4)*/ 2 -^ -a.
9. Prove that r n ^2a lf and that if r n ~2a l9 then a w term of the reciprocal part of the a cycle. T Show also that if A is an odd prime and rn = 2a l9 then
= 2fl,,
therefore
6n
+ 6n
1
and a n
is
the mid-
A must be equal =a n rn = 2a so r
<], 1 b n ~b n+l Hence a n is the mid-term of the reciprocal part of the a cycle. 1. is an odd prime, we must have a n a.i or a t Hence a x = 1 or 2, and N = 3, 5, 7. On trial it is found that JV 3.] rw
[If
n <(N/-V-f'6J/rw
,
1
,
.
to 3.
that If
N
10. If r n = r nn = a 19 then a n = a w}1 = l, so that a n a n+1 are the mid- terms of the reciprocal part of the a cycle. ~ I(a + b n )la l ~\ or 2, according as b n or b n = a l If 6 n a x l [a n ~ = &n+i 2! -6 n ffj, so that 6 n & w ,i. This is impossible. Hence a n = l and ,
^a
l
.
,
= J(ffi + &nu)/i = l-]
4i
11. If aj
is
an odd prime, or a power of such a number, and
rn
a^ then
=!.
X-bn +l9
[For r n r nn of
a;
2
=JV(mod
= (b n + b n
therefore a n
r n -ir n since each
and
Gfj),
t
}Ir n
l
N-b n
two
and
irrationals z
p', q 9 q'
therefore b n and 6 nfl are solutions a l is an odd prime, b n ~\~b n+1 ~a l9
continued fractions which are equivalent
are identical, prove that
(pz+p')/(qz + q')
z'
where p,
z'
,
^aja^ --!.]
12. If, after a certain stage, the simple
to
2
Show
are integers.
where
pq'~p'q=^:l
also that the converse
is
true.
number (c) of elements in the cycle belonging to \/N is even, prove that Pc/zf Jci2 = (Pc-^^)l^c^ a^ c rding as c is or is not divisible by 4. = 21 illustrate the fact that equations (J) of Art. 6, (5), (iv), (ii) By taking = 19. is a prime, arid verify the equations for are not necessarily true unless 13. If the
(
(i)
N
9
N
14. If the (i)
Also, (ii)
and
N
by taking
A'
If
ra
verify T
[(i)
number
Prove that
A
is
=
J
(c
when
1
iV
N
of elements in the cycle belonging to *JN is odd sum of two squares which are prime to one another. 205, show that the converse of this is not generally true.
(c)
is
)
and n
= 29.
a factor of
:
the
p r 2 + L]
J
(c -f 1),
prove that
MISCELLANEOUS EXERCISES
(A)
L Show
that the product of any two numbers, as 19 and 35, can be found as In the left-hand column, each number is the jg quotient when the preceding number is divided by 2. Any 70 9 remainders are disregarded. In the right-hand column, 4 each number is twice the preceding. 2 1 560 The numbers 140 and 280 which are opposite even numbers ~ 665 on the left are crossed out, and the sum of the other numbers 1 9 x 35 on the right is 19 x 35. [The reckoning shows that
on the
right.*
2 19x35^(1 +2 + 2
2. Show that n 8 is the sum of n consecutive odd numbers, and find or the two middle numbers.
tjie
middle
3. Take any sum of money less than 9 8s. 7d. Reverse the order 12, say of pence and pounds, obtaining 7 8s. 9d. The difference between these sums is 1 19s. lOd. Reversing the order of pence and pounds, we obtain 10 19s. Id. Adding the last two sums, we have 12 18s. lid. Prove that the result will be the same whatever sum less than 12 is chosen.
4.
r t r2 ,
,
a/b is a proper the remainders
If ...
rn
show that
fraction and q l9 q 2 ... q n are b is divided by a, r 19 r 2 ,
quotients and
the
,
when
.
.
?"n~i>
respectively,
<^
ql
Rn =
where
(
1)
.
b
tfn (ii)
For some value of form
n,
Rn = 0,
h
I
p where p, (iii)
5.
and
q
r
9
s
are positive integers, their
q, ...
Express in this
can always be expressed in the
so that a/b
1111
way
Prove that a -f b
-f
2 (a 4
-4-
and
f-f
J-f
number being
finite.
.
+ d and a f 6 - c - d are factors of 6 4 + c 4 -f d*) - (a 2 + 6 2 + c 2 -f d 2 ) 2 4- Sabcd, c
find the remaining factors.
6.
Solve
7.
If x, y, z are real, prove that (x
cannot 8.
(i)
among (ii)
lie
between
1
and
2
-f
y
2
+ z 2 )/ (yz + zx + xy)
2.
Find the simplest equation with integral coefficients which has -Vf + s/f and -Vf-Vf
its roots.
What
Prove that 4(z
2
are the other roots of the equation ?
+ #4 l)*-21x*(x+ l) 2 ^(z-
*Known
as the Russian peasants'
method
2
l) (2a;
+
2
l) (z
+ 2) 2
of multiplication.
.
MISCELLANEOUS
544 9. If #-ft/
+ 2 = o;
10. If a~\~b
+c+d
2
3
-f ?/
4-2
2
and
(A)
-2, show that
a;
f
?/
|
2
t
|
= 0,
prove that the results of rationalis-
ing the equations *Jax
*Jby
+ *Jez 4- V^ ~- 0,
*Jay + \'dz
v/fa: 4-
0,
Find the equation whose roots arc given by a a a %=*<-*, + (ai + s* + S3 s ~ are the roots of x x -f 4 ~ 0. 2
(i=l,
),
where x l9 x 2
equation
being given that 13. If
^
x
15. If
it
has a pair of roots whose
------- f -----/ + x -me x f tno x + nm value of x is m(nc + bd)/(a + b).
-ac')
a< 6 < c <
r/,
and a -
are real for all values of
+m+n
0,
positive
}-d
0,
c>
^
/>
-
r/,
c)
prove that
&' 2
>a'c'.
the roots of
^k(x
-
b)(x-d)
k.
show that the expression b 2 m'z + c'2 /i 2 2brmn ~ 2canl - 2ablm
a 2/ 2 if
zero.
2 prove that 6 >ac and
z
(ca'
l
is
and oH-6-f c
_-f
(x a)(x
16. If
sum
nid
the only finite 14. If
2, 3),
2
x%
,
12. Solve the
is
-f >/cJ
State the two other equations which lead to equivalent results.
are equivalent. 11.
-f
-I-
a&c(a + & + c)
is
positive.
Show
that the expression (ex ~-az) z - (ay -bx)(bz -cy) is the product of two linear functions of x, y, z, and that, when a, b, c are real, the coefficients in the linear functions are real if 6 2 5? 4ac. 17.
abed
from the equation
18. If the cubic derived
-
+ --, -- -- -f ----x+b x+c x+d
-,
x-\-a
has a pair of equal roots, then either one of the numbers the numbers c, d or
a, b is
equal to one of
1111 abed
-+T=-+119.
20.
Express in factors - a 2 ) (ca - b 2 (be )
Show
that
if a, b, c
+ (ca- b 2
)
(ab
- c2)
-f
(ab
are real, the roots of
are real. 21.
Eliminate
x, y, z
from the equations x
y
z
y
z
x
__
x z
:
y x
y
z
y
~
c 2 ) (be
- a 2 ).
MISCELLANEOUS 22,
10 n - (5
Prove that
23. (a) Solve
If a:,
(6)
i/,
(xz are
3
l)
divisible
is
... -f
n+1 by 2
.
(x-n)^-O.
unequal and
show that each product
- 1 and that equal to ~ yz + zx -f xy 3 (a: -f y 4- 2) -f 9 =
is
xyz-2(x + y + z) + 9~0.
and 24.
+ */l7) n - (5 - \/17) n
+ (z-2) 3 + (x -3) 3 +
545
(A)
By the method of partial fractions, prove sum of n fractions of the type
that, if a, 6,
c, ...
are
ft
/&
numbers,
then the
is
(a-b)(a~~c)(a-d) ... (a-k) and is unity if
m is an integer less than n- I
zero if
24
21
25. If
n
is
m
n -
I
.
a positive integer,
+ 2\n-2 \n _ _ ,
26.
Under what conditions
27.
Show
9
2
+4
jn-f3|n-3 -
*
.
-
...
are the following equations consistent
that, if the roots of x*
n terms
to
'
?
- ax 1 -f 6x - care not in A.P., then there are form x y-\-\ such that the transformed
in general three transformations of the cubic in y has its roots in G.P.
28. (a)
Rrove algebraically that,
if a, 6, a,
j8
are real numbers, then
the positive root being indicated by the radical sign. (b) If a and b are positive and a-f-6=-l, show that \ 2
29.
Examine,
1
/
\2
for different positive or negative values of
x and p, the conver-
gence or divergence of "
x r\ r (1)27-;
rN/(^ +
r-\
1
--
)-Vw
(11)^^5
.
30. Show how to find the nth term and the sum of n terms of a recurring series of which the scale of relation is of the form ^n -f jm n _ 1 + n _ a = ^ anc* of which the first four terms are known. Apply the method to the series
^
(i) (ii)
31.
(a)
Show
(b)
Eliminate
1,2,5, 14..., 1,2,5, 12....
that 2Z'6 2 c 2 (6 2
--
x, y, z 2
z
= a3
,
q = 3)
b--2, ^=-1)
- 6 2c 2 ) - a 4 [(6 4 - c 2a 2 ) (a 2 - 6 2
c 2 ) (a 4
- 27 (a 26 2 4- a 2c 2 x*y 4-
(i>=-4,
)
)
4
-f (c
- a 26 2 ) (c 2 - a 2 )].
from
y
2
z 4-
2 2/
a:
= 68
,
z zx
+ z 2y = c 3
,
JM/Z
=
3
rf .
MISCELLANEOUS
546
Show
32. (a)
If #
(b)
Show
a, 6, c, r
is less
.
33.
number n can be expressed
that any root of the
where each of the quantities
than >/A #
1S
(A)
is
...
n, 1 or
n~ l
but greater than
,
nearer *to
Ns
*u than
AT
is
form
.
%JN
-^ either
'
in the
9
show that
-
-
or x.
that
a z n(ai
~
___
-a
y)
2
(a-a 2 )" 2 (a-a 3 )-
Write out the terms of the product in the numerator, and give the resulting expression its correct sign. 34. Discuss the reality of the roots of
x 4 + 4z 3 - 2x* - I2x + a = 0, for all real values of a.
[Denoting the equation
by/(#)0,
35. Solve the equation 46u; side as the sum of two cubes. 36. If
(
1
+ x) n
1
+ c^x
{-
3
c zx z
-f
72#
2
consider the roots of f'(x)~Q.] -f
+ ...+ c nx n show ,
[Denoting the expressions by ^ and 37.
Prove that there
18&- 11=0, by expressing the left-hand that
show that f-==^-, ttic ct^c
v,
only one set of real values of
is
x,
y
t
z
etc.]
which
satisfy the
equation
and
find them.
[Writing
a, 6, c, d, for
o + 6 a + c 2 4-d a = i 38. If
39.
oj is
a
;
fifth
Prove that,
(l-x), (x-y), (y-z)> 2 hence (a-b) = 0.]
Z
root of 2 and
if r 2
p 2^,
a;
= eo + w a
,
z,
we have a+b + c\-d = l and
prove that
the product of a pair of roots of the equation
x 1 ~\-px* 4- qx 2 4- rx + 5 is
equal to the product of the other pair. x* - # 3 - 16# 2 - 2x + 4 = 0. Solve 40. Obtain the roots of
in the
41.
form Prove
= (3 + and (4?i-f l)a = 27r.
that
positive integer
a;
4
"* 1 +
1
77r " 2
.
2
1 (a:
+ 2o:
cos r a -1),
where n
is
a
MISCELLANEOUS numbers
42. If three real
satisfy the equations
x+y+z=5
prove that none of them 43.
by
yz 4- zx
4-
xy = 8,
nor greater than 2^.
1
when divided 3 respectively, and show that 206 is the
for all the positive integers which,
leave remainders
1, 2,
them.
leant of
44.
and
than
is less
Find a general formula
5, 6, 7, will
547
(A)
Show
that a determinant of order n can be expressed as one of "order
n-l
as follows.
a l^l c l
If
^1
,
then Jrt 1 "- 2
[Multiply A by
=
at
-&!
...0 ...0
tfi
a 1 ...0
-c,
-^ 45.
where as r
Show that, if n is a positive integer,
^
and
when n
7? 2
is
is
1,
7 (a-f
n JL
is
~\~
where
number of
the
3 36 2 or
rr^-TT 4 44
solutions of the equation
N
N
or is equal to + 1 according denotes the quotient and r the remainder
N
3 3a 2
^3^3
Prove that
47. If
j
are positive integers or zero,
not equal to divided by 6.
or
is
46.
...
~rr 4 4
r>)
a b being positive,
>
>
a positive integer,
OIL
3.4 ~ n(n-\) ~ ~ *
~\~
1
.2
4.5 w(n-l)(n-2) 3 1.2
r ^r
~"7^
I
48. If a, 6,
d are
c,
real
and b2
- ac
ad -be
show that each
ratio
= (~
)
ad -be r 2 - bd
'
and that a*c + bd* + a*d*
0,
the real values of the
\a/
roots being taken in 49.
all cases.
Prove that the condition that
(ax + by -\-cz) (bx + cy + az)(cx -\-ay + bz) can be expressed in the form P* + Q* where P and Q are linear functions of x y> z is a 3 + 6 3 + c 3 ^3a6c; 9
and
find values of
50. If
Zx m ^n
x i9 x 2 if
,
P and
...
xn
Q.
are
positive
numbers such that
Zx=n,
prove that
MISCELLANEOUS
548 51. If r
to i(r-f
terms
1)
/ } 4,
is
# n
52. If (
equal to
-I
j
+
,
7,
-
prove that
>
}
|.
-l-a^-f a^-f tf 2 # (r
-
-
an odd integer and q r =^ ~>
is
(A)
1
2
-h...
r
+ #r
-f ...
a r+l - 2na r - (r -
)
1
show that
,
)
fl^ - 0.
Also show that
53.
Sum
54.
Examine the convergence of
27*
the series
ft
;
th
series
x n /(x* n for
any value of
55. If
56.
m
is
Prove
a positive integer, show that
that, as the real variable x changes steadily (x '
where a and 6 are real and r/ an interval of length 4(6 a) Prove that,
,/;,
if
a,
58.
ft,
-"V-6
is
f
'
al]
values except those in
locate this range of values precisely.
a
a3
6
1
fr
a4 b*
c3
^*
-
1
|-0,
1
1
'*
The graph of y~---
Prove that,
if
,
----.
l)(x
x^
then x^
100,
61. If f'(x)
is
Prove that
show that
positive,
1)}.
has a turning point at P(2, -1). is
a
maximum
-%((x+ f(x)
is
1 ft
i-
(x
at P.
-
1 ft] -=
increasing.
2x + x cos x - 3 sin x >0
if
0< x<
.
4
Prove that
T75 + 5. 11.31.3.5 ^ 44.64.6.8 ^
-f
-
-
4-
#,
Find
4)
and show that y
6,
i-
Sketch the curve. *x~
the approximation being correct to at least eight decimal places.
and that
-h oo
in the expansion of (1 \-x) n (I in ascending powers of x)' a positive integer, the sum of the first n -t- r coefficients is
the values of a and
62,
to
c are all different, then
(x
60.
oo
1
+ 3)+4r(r 59.
-
a)
Prove that,
where n
from
2
assumes twice over
and
;
-
r
r,
and
is
x.
the function
57.
whose wth term
+x n +1)
-
:
4-
. . .
to infinity =
1
.
(nearly),
MISCELLANEOUS 63. Solve the equations
:
y*
+ z*-x(y + z)^a,-} )~ b
[Show that 64.
6
+ c - a = 2 (x 2 - yz),
By expanding + (n -
2
.
3
ax + by + cz =
65. If
that this
is
>
.
etc.]
2
1
in
)~
two
different ways, prove that
2-' +
.
1
which - H x y
x, y, z, for
Show
1)
-2x-3x
(1
549
(A)
h
and
a, 6, c
-
stationary, are given
is
z
maximum
a true
are
or
show that the values of
positive,
minimum,
by
if
xyz>Q>
tan x~ I + x 66. Show that the equation has an infinite number of real roots, and find graphically the approximate value of the smallest positive root. 67.
Assuming that
+ a;)}~ 1 can be expanded
a;{log(l
in
ascending powers of
a:,
f~\d the first four terms of the expansion. By help of this result prove that a capital
interest at r per cent, per
annum
/23026
\
h
(
sum accumulating at compound be increased ten-fold after
will
1-15
years approximately.
j
68. If
x
y~y
3
2
-\-2y
-y*
and
9
it
is
assumed that y can be expanded
in
ascending powers of x show that 9
69.
Draw graphs
of 2 _2(z ~6z)
70.
in
2(x*-$x)
2
By expanding
two ways, prove that
71.
Show
maximum difference
that
(a
--- #)(4-3o:
value and one
between them
minimum / is
J
(
a
V
for various values of a 72.
Prove that
has four real roots,
where a
),
value.
1\ }
is
a positive constant, has one
Find these values and show that the
3 .
What
is
a/
?
x* if
+
2
- ISx 2 + 4dx + 9 = d*
^ 1728.
the least value of this difference
MISCELLANEOUS
550
r*x^aR + bW,
73. If
and the quantities denoted by
all
is
Prove that,
if
1
approximately
where
the letters are real and positive, prove that jc,
74.
(A)
^ 2^ab.
small, one root of x* + x 2 -(I+2c*)x-I^O
is
f ie 2
and
,
find corresponding approximations to the other
roots.
75. If
,
/?,
y are the roots of x 3 - ax 2 4- bx
c
= 0,
the area of the triangle, of which the lengths of the sides are a,
If the triangle
is
right- angled,
a(4ab 76. If
n
is
show that
-a 3 - 8c)(a
2
- 26) ^
_<3
71
(ii)
71
P.2
2
.3 3
... rc
n
when n
<(^-
/?/
78.
y
y, is
a positive integer, prove that nn 2n n - 1 when __ <
(i\
77. If
/?,
is
positive,
Prove that
all
show that
log
y
between 2
lies
f
-
1\ -
y
and
2 -~
j
the roots of the equation
8< k<
are real if
11.
7
f
79. Find a function (ax H- 2bx + c)/(a x 2 -f 2b x + c') which has tm ning values 2 respectively, and has the value 6 when x = 0. 3 and 4 when x = 2 and 2
80. If a, 6 are
81.
Show
i
unequal positive quantities, show that
that two pencils of straight lines in one plane, containing p and q no two lines being coincident or parallel, divide the plane into
lines respectively,
pq + 2p + 2q-l
parts.
Show
that the number 30 may be divided into three unequal parts in 61 ways, while if equal parts are allowed, there are 75 ways. 82.
83. In
how many ways can a batsman make
more than 4 84.
off
any
Prove that the equation
has no real roots 85.
14 runs in 6 balls, not scoring
ball ?
if r
4
.
Find the minimum values of a m + a.m /a + x\ I
_.
2
_
1
AT/
for positive values of x,
where
,
CITlH
-
3 a, 6,
m are positive and
~~3 m>l.
MISCELLANEOUS 86.
551
(A)
Find the condition that ax + b/x can take any
of x.
real value for real values
- (x - a] (x ~
b)/ (x -c)(x-d) in terms of y where y~(x~ d)f (x-c); and hence, or otherwise, show that f can take all real values if (c-a)(c~b)(d~a)(d-b)
Express f
is
negative. 87.
maximum and minimum
Find the
values of
and draw a rough graph of the curve. 7 2n - 48n - 1 is divisible by 2304. 88. Prove that 89. If
4
then
I
#,_
yl
3
2/3
I
-c).
that an approximate solution of x log x 4- x -
90.
Show
91.
Find the
--y3 ~ j3
limits of
greater than
Prove that
f/^a;-a;
(ii)
,
ar =
1 -f r~ -
+ o ai +
94.
Show
95. If
sum
that a
farthings in (n
+
2
I)
of
e is small,
y-^x-x*.
(iii)
,
2
+ 6 2 + c 2 -6c-ca-o6
is
-- (a-c), where a>b>c. &
+
Li 1
where
v/3
or than
if
then
3
prove that square root of a
|(2a-6~c)
= e,
as x, y tend to zero along the curves
,
2
92. If a, 6, c are real,
1
I
x ~y
y=x-x
(i)
93.
-x
(x l
b""2
Z
-^
-f ... -f
y-
,
l
a 2+--- to
oo
=2^6.
n pence can be made up of
pennies, halfpennies
and
different ways.
x>2, prove that n n(n + l)
^^
" t
"'""
n *~*~
n(n-l)
+"
and deduce that
n+ n(n-l)
(r-l)
^
96.
_
n(n-l)(n-2) (r-l)(r-2)
^+
j
_n(n +..._
+ l)
_
...(n
+ r-1) .
Find approximations to the roots of
where a 97. If
is
small,
3wn -7w n _ 1 + 5w n _ 2
~u n _ 3
and w ~l>
^i
8,
w2
17,
prove that
MISCELLANEOUS
552 98.
Expand
approximation for
Deduce that
+ V(9 + 3z 2 )}/(3 _#)
{2a;
the fifth power of
ei
and show that
x, e
i
(A)
n ascending powers of x, as far as x it leads to a good
for small values of
x .
= 1-2840
....
99 If a, 6 are positive, and p and q are positive rationals such that 1 /p + prove that ab ^ a p !p + b //qt .
I,
1 /q
<
By means
100.
of the expansions of
ex
and
log (1 +x), prove that,
when n
is large,
1 H
Show
that e
is
)
given approximately by
with an error of about 0-46 per cent. 101. If
a time,
p
Show
102.
is
r>p>q, show that the number of combinations of p+q things r at things being of one sort and q of another, is p + q -r + 1. that one root of the equation
approximately 1-0103, and find the other roots correct to two places of decimals.
Prove that,
103.
if
n
is
a positive integer, the
coefficient of
xn
in
(l+x + x z
n )
is
1
104.
The number of ways of selecting
105.
Having given that
r pairs
from n different things
is
prove that 106. If a,
then a 2
,
fc
2
107. If
and x 2 ^ 3.
,
6, c
are real
and
c 2 are all greater
than
1
or
all less
than
x + y + z~xyz and x 2 =yz, then y and
1.
z
may have any
real values
108. Solve 109.
Show
that the result of eliminating x and y from the three equations
1.11
x-a y-a
a
1 9
x-b
11
y-b
9
b
and is
110.
Prove that
log e {log, ( 1
+ *)*} = - \x + ^-a: 2 - |a;3 -
....
Find, without using tables, the value of log e (log e 1-01) correct to five places of decimals, having given log e 10 2-302585.
MISCELLANEOUS 111.
--
~ in a -that the coefficient of a371 * 1 in the expansion of (x + 4)(x +&) # of is ascending powers
Show
series of
Find to
112.
553
(A)
# ~^
113. If
equation
five places of
3
is
a
3
given by
+ x + 1 = 0.
decimals the real root of x3
where ^ is small, show that an approximate root of the x a{l+%p log e a + ^n 2 log e a ( I + log e a)}. -J-
i
114.
Prove that
(
1
+
x^ =
1 -f
x + x* + f* 3 + V* 4 +
115. The difference between the A.M. and the G.M. of a set of nearly equal numbers is to a first approximation equal to the quotient of half the difference between the square of their A.M. and the A.M. of their squares by any of the
numbers. 116.
Find the sum of the terms
after the
nth in the expansion of
(1
+ #)/(! ~x) 2
powers of x. Prove that the ratio of this sum to the sum of the corresponding terms in the 1 expansion of (l-x)" can be made equal to any given number A, which is than suitable choice of x. Explain clearly why the restriction 2n, by greater
in ascending
upon A
is
necessary.
m
marks are 117. An examination consists of three papers, to each of which assigned as a maximum. One candidate obtains a total of 2m marks on the three papers. Show that there are f (m + l)(w + 2) ways in which this may occur. 118.
Show
that the 2n-th convergent of
3434 ------
-
43+4-3+ n n
...
differs
from the
12 + 4( -
value of the continued fraction by 13/{9 l) }. with a like [Prove that (i) p 2 n 1 Ijp2n-z + 12# 2n _ 4 - l) n and ? 2 n-i/4 = 2n 2n + ( (ii) ? 2 (<7 -#an-i)/5 .
,
result
for
the
q'a;
n^
119. If the infinite product '
2/ is
expressed in the
form
a
then 120.
(2
Find the
l+xz
_
1
- z - xz z 1
r +*
. . .
- l)a i r+
~f
2V
V
a rx r +
. . .
,
= 4(ar + ar _j).
expansion in ascending powers of z of
,
show that
f/l + V4a?+l\) H
n+ a
j/ ,
I
+ wo; H
v
/l-Vto-fl\ n+2
-(2;
..-x M ..-2) o: j^r
2
l
/
(n-2)(n-3)(7t-4)^ rx3 +
H
. . .
.
Prove that
where m and ?i are positive integers, where
2V
V
2 c^x + a 2 x -f
coefficient of z n in the
2
121.
-f
p r = m(m + n -r-
1)
is
divisible
(r^m),
by >
r
(x
-
2
I)
,
and that the quotient is
~(r-f-l)n
(r
MISCELLANEOUS
554 122.
among
123.
(A)
Prove that the number of ways in which n prizes may have one at least is
may
Prove that when x
is
sufficiently small
1
Jog (1 +#)~~
Also, if n
>1
1
1
#
2
11 24*
12*
,
"
#
(n
L__ |
+ 2)
"
'
\n 'L
A man
owes a sum of a, and repays b at the end of each year, partly interest at 100?* per cent, and partly to repay capital. Show that the will be repaid in A years, where A is the integer equal to or next greater than
124. to
be distributed
q people so that everybody
pay the
sum
125.
Show
which n are
that the number of combinations taken n axe b, and the rest unlike is (
and that
n together of 3n
this
sum
is
%
4-
26'Jt-
i
w equal to 2
+ 3CJU + (n + 2).
.
. .
~1
of
+ nC? + n + l,
126. If a, b are positive integers, the probability that ^(a 2 f 6 2 )
integer
letters
a,
is
a positive
gV
is
x + (b+c)y + bcz = b~c, x + (c + + caz - c ~ a,
127. If
a)y
x + (^
-
&2
4- ft)y 4-
- a - 6,
where a^b^c, prove that 2;o;-2/ 2 = 3. 128. Ladders 15 ft. and 10 ft. long are placed in a passage, each with its top against a wall and its foot against the opposite wall. If the point where the ladders cross is 5 ft. from the ground, find the width of the passage. (Sunday Times. ) [If the width of the passage b ft. from the ground, then -
a
Hence
if
6
5(t/-|-l),
i/
ft.
+ ! = ?> and b
and the tops of the ladders are a a2
-& 2 -125.
5
show that 0
giving
x
is
4
4-2?/
3
+ 5i/2 --2?/-1^0,
= 0-5761287, 6^7-880644, ^ = ^(100-
6 2 )^ 6- 155929.]
ft.
and
MISCELLANEOUS EXERCISES
(B)
N
A
of party consisting of 7 robbers ami a monke}' had stolon a number In the night a nuts, which were to be divided equally among the robbers. robbor woke and decided to take his share. He found that, if he gave one to the monkey, the remaining nuts could be divided into 7 equal lots. He gave one to the monkey and took his share. The other robbers woke in succession, found that the same thing was possible, and each gave one to the monkey and took one-seventh of the remaining nuts. In the morning they divided what v,ere left, and they found that, if they gave one to the monkey, the nuts could be divided 8 is 7 - 6 = 5764795. into 7 equal lots. Prove that the least value of - 7, - 6 of i\ satisfies the numerical conditions 6 fur value [The ~ - 6. Thus - 6 is unaltered - 7 ~-7 = -7- the set of 1.
N
I
;
1,
by
1)
(
Hence the values of
corresponding to what each robber did.
2,
Let a/b be a proper fraction and
c l9 r 2 , c 3 ,
...
an
N
infinite
operations
are given by
sequence of
r n the If q l9 g 2 positive integers. remainders when ac L9 r L c z nc 3 ... r n _^r n respectively, are divided by 6, show that ,
,
,
--
......
,
c^ qr^
CL
b
,
qca...^
E n = --
r ~
1
where
Ci<- a
(iiiJV-ar^ (iv) If a
is
...c n
-
;
b
...c
(mod6);
a prime to b and a<6, then
a
^? +
_??
CL
c^'o
1
6
-
(h .j..
|
c,c 2 c 3
where the of
n
series terminates if c x c 2 ... c n is divisible by otherwise the series is convergent.
:
Express the following fractions in the form
3.
ft
3
3
8
+ J'- 4--? + 3.5 3.5.7 (i)
In the 4.
two
last
Show
~
6.
Use Ex.
Sum
(ii)
;
i
;
some value
:
JLn
4.
3.5... (2n +
cases, find the value of q n
for
(iii)
'
l)
J.
.
that "
"
32
8 5.
J
4--
6,
2
3"
.
2, (iv), to
show that
e is
irrational.
the series
n 2 + 2(w-l) 2 -f3(rc-2) 2 -f..., where n
is
a positive
Prove that
integer.
$I
+ TS+-"+ 12
where
2N
Sn
is
the
sum
-
n
n
+
-
of the squares of the
to infinity ' first
,
6
n positive
integers. B.C.A.
MISCELLANEOUS
556 7.
Prove
(B)
and x + y + z = l, then
that, if x> y, z are positive
\
\y
From
8.
a cask containing x apples, a
man
half the contents of the cask
sells
and half an apple to one customer, half the remainder and half an apple to a second, and so on, always selling half of what are left and half an apple. In no case is an apple divided. After n sellings, a apples are left. Show that z = 2 n (af 1)-1.
Show
9.
mon
line
m 10.
if,
n
TA-
II
/?
if,
= --<
.,
y~"n
d
Cp
have a com-
y
where ad -
3 d
-f-
d are connected by the equation
a, 6, c,
mx~ly~v
.
COL -f
and
ny-mz = \, h-nxp, ZA + mfji + nv =
that the planes
and only
a2 +
+6 + cy
=-ay
.
prove that
.
,
e/
Show
also that a,
y are the roots of
/?,
x
_
.
ex
|2r-2
where
r
A:
-ri
r
,
^), ^,
13. If
-dx + b _____-ex
-a
-
Ca - a
,
k r =kjc r
that
^+
-
-da + -6
aa +6 ---COL f a
=a
.
+
...
a positive integer greater than unity, and
is
10 ou .u . 12. Show that
where
s show
-
11. If
.
+d
-
a P + 6 p + cP + a+& m---> ~ c
r are positive quantities
a-j-^-t .
m
such that
x r = x(x-l)(x-2) ...(x-r+l), prove that (2n
-
!)_!
.
(2n-3) w _3
.
(2n-5) n _ 5 ...>nn
.
nw _ 3 nn _ B .
.
7i n
_7
each product being continued as long as the suffixes are greater than zero. 14. If
L
TH,
Z',
w', I"
and
X/
7W
are integers,
and
if
/)9
is
not rational, and
show that / = Z and m m'. Also show that no two of the numbers r
,
(2Z
+ 5m)a + /^,
X
-f
(2Z
5m' -f
/X
1
)a
-f
Z^,
-f
(2Z
5m") a -f (Z
/x
-f 1) jS
can be equal. 15. is
One
root of the equation
3 - 33 - 33"
;
x 6 - 9x 5 + 18x*
f
9o;
3
+ 27 x 2 - 54x -36-0
find all the roots.
+ c ^V, x) + c 2 = O
a 1 yz + b l (y + z)
16. If
a&x
4-
6 a (z
4-
are true for an infinite series of values of x, y, c 2 --
26 2 6 3
,
agCj, 4-
1
t
z,
a x c 3 = 26 1 6 3
,
prove that, in general, a 1 c 2 + a 2 c 1 ~26 1 6 2 .
if
MISCELLANEOUS
557
(B)
show that
17. If a, b, c are unequal,
ax
b~ c
4-
+ cxa ~ b ^a + b + c.
~ bxc a
18. If
a (z 4- x + 6)
where y and
z are unequal,
bzx (z
4-
= 0,
x)
+ ?/ 2z 2 + fy/z (y + z) = 0.
Prove that 2
a4-4-y4-5
a + j3 + y
is
4-
prove that
a (y 4- 2 + 6 ) 19.
z*x 2
4-
+8
a(y 4- 8) 4- y8(a
2(a4-]8)(y4-8)
equal to zero. 20.
Prove that rt
~(L-x where
A r ~(
according as n 21.
-
7""' 1
I)
cos n
sin 2
~
4
is
and
-
r&
TIT
,
X"
r
,
4-
tan-
5
= 1,
n ...
2, 3,
^(n
~
or
1)
-^
odd or even.
Determine the condition that
(x
n
shall be divisible
by
1).
22.
Show
that the
(
are identical
four terms in the expansions of
first
1 4- x) n 4-
a
1 4-
f
-
and j
\
if
a
b
c
p
q
r
'
and
find the values of a, 6, c in terms of p, q,
23.
a
b
pz
q
2
c r2
'
r.
Find the limiting values of i
(cos x)
i
.1
x
(cos x)
,
xz
(cos x)
,
x3 ,
as x tends to zero through positive or negative values. 24.
Show
that, if
I
l9
m
l9
n^
;
/2 ,
m
2,
n2
-f
w
a
;
13 ,
m
3,
n 3 are
real quantities satisfying
the six relations, 2
!
-f
m^ 4- ^! = 2
/2
2
a
-h
n 22 = /
=l 3 -f n 2 r? 3
4-
then
4-
and
n 3?3
ll
m
l
4-
^
w^a
0,
MISCELLANEOUS
558 25. If z 3
-j-3/fcz
3
+ <7~
[/x(z-hi/)
-v(z-f
3 /x) ]
A = g 2 + 4th 3
where
Show 26. If
that the cubic 4z 3
u
(i)
If
u
(ii)
If
u
for all values of
z,
show that
.
has two equal roots.
27a 2 (z-f a)
- c) n + (c- a) n + (a- 6) n where n is a positive integer, prove that 2 divisible by 27a Z!bc, then n is of the form 3k 1, where k is an
(b is
(B)
,
integer,
divisible
is
2 2 by (Ha - Hbc) then u ,
is
of the form 3& +
1.
27. If all the roots of the equation
are rational
28.
and negative, show that
Prove that,
if
n
is
a positive integer, and 1
m -2xn
x'
29. Investigate the
r
~(a-f 2r7r)/n, then
^tt-xsina-tfsin (oc~6 r ) r "=0 x 2 ~-2x cos O r + l
_
cosa + l~~nsina
maximum and minimum
5
l) /(x
+ 1), and
real roots if
0
values of (x+
5
trace its graph.
Prove that the equation (x+ I)*~m(x 5 + 1) has three and only one real root ( - 1) if m<0 or m>16. 30. If
divisible 31.
n and n + 2 are both prime numbers, prove that by n (n + 2)
(n -2) in
-1-2
is
.
Prove that,
if e is small,
has a root nearly equal to
1
the equation
-f
4e 2
,
and
find approximations to the other
two
roots.
32. If a, b,
show that
JJL
c, ... k, I
lies
are positive
numbers arranged -Z /xa
between the greatest and a/6, 6/c, c/d,
33. Discuss the
least of the ...
,
in descending order,
and
numbers
k/l.
convergence of the series whose nth terms are (J*)!*-.
(-!)
w+1
|^, .
34.
where If z
Find the seventh roots of unity and show their positions in the z plane z x + iy. is any one of the imaginary roots, find the equation whose roots are z2
and
MISCELLANEOUS 35. If
and
559
(B)
d denotes the determinant ax
a2
a3
a4
a
a6
!
#2
aa
a
a4
a5
ax
aa
a
D denotes the determinant obtained by A = a + 2a 2
x
!
2
a 6 + 2a 3 a 4 , a52
At
4-
^4
= a 4 + 2a 2
2
2ia 4 -f 2a 2 a 3
x
^4 5
,
changing a r into A r where ^ 3 = a 2 2 -f 2a 1 a 3
a 2 + 2a 3 a 5
,
= a 3 2 -f 2^(15 4- 2a
2
a4
,
show that d*~D. 36. If
n, p, g are positive integers
m,
and
if x, y, z all
^(yn^g^ + ytn^n^^^gm^^yn) a;(y-^)+yP(2-a^) + p (a^-y)
wre (
tend to
m,n
prove that
a,
n
)
__
2:
Prove that,
37.
in the expansion of
where - !<#
Show
38.
zero)
l)x
2 .
that the number of distinct sets of three positive integers (none is the odd integer 2n -f 1, is given by the least integer containing
whose sum
39.
Show that
whose sum 40.
is
the
number of distinct
an even integer 2n
is
positive for positive values of x, 41.
which
sets of three positive integers (none zero) the integral part of \ri*.
Prove that
I-x is
m(m
+ ix*-(l+x)e~ x
z
and increases
as x increases.
Find to 6 places of decimals the root of lies
between 5 and
Prove that the in powers of * is 42.
6.
coefficient of
xm
in the
expansion of
1
-
(1
-
v
7
,
*) (1
- *)( 1 *)
43. If the volume of a right circular cone is 20 cubic feet, prove that, when the cone is such that the curved surface is a minimum, the radius of the base is approximately 2-381 feet. 44. Arrange the following numbers in order, .so that as x -> a each number to the preceding may tend to infinity :
ar
2 ,
2X ,
xx
,
ex ,
a;lo8
r ,
(log
2 a;) ,
^
the ratio of
MISCELLANEOUS
560 45.
---
(B)
Prove that
n
/
-r
'
v
|r)2[n-r
and deduce, or otherwise prove, that ln
+1
lrc
1
46.
Prove that,
+2
a, 6, c are
+3
ln-f-4 "
n-3"
|3
|w^4~~'"
[3 [4
if
a
where
|2
[n-2
[2
|n
"
b
c
a
b
not zero and no two of
bm
al
c
-
-
m x y j+ m + n^o. ,
,
I
en
~ are equal, then
n
I
47. If a 1 a 2 ^3 an = the a's is greater than -
An
prove that the
,
48. Solve the equations
sum
of the products r together of
:
x + y + z + w = l,
+ dw~ A, + ^ 2t^ = A 2 a*x + 6 3 y -f c 3 ^ + d 3 w = A 3 c 22
/ 4-
Prove that, if a, 6, c, d, A are all than three of x, y, z, w are positive. 49.
Prove that 2 t/
-f
yz
+
real
,
.
and unequal, at
least
two and not more
if
z2
a2
z2
,
+ 2#-f # 2 ~6 2
,
x z + xy + y z
y2 -f zx 4- ^2/ = 0,
c 2,
a&c-0.
then 172 50.
Prove that 7^=7-, -
51.
Prove that
if
1 is divisible
by
73.
m is prime and p
8f
52. If a 2 -fj8 2 = *a0, j3 2 are all different, then
53.
-1 =
+ y 2 = /c)5y, y 2 + 8 2 -fcy8,
82
+
2
^AcSe,
and
Prove that
1 1
ttfo"
1
)
n(n-l)(n-2)(n-3)
_
2n-f2r
54. If
ar
prove that
and that
1
+3a t 4- 6a 2 +
... -f
a 2n_r
;
(4w + l)a,n = (2n
+ 1)(2 +f
n )
,
if a,
)3,
y,
MISCELLANEOUS
561
(B)
55. If of 3n letters there are n a's, n 6's and n c's, the number of combinations of these r together is equal to the number of combinations 3n ~ r together. Also, if n>r>2n + 1, the number of combinations is
|(n + l)(n + 2) -f (r n)(2n number of combinations is
and the maximum according as 56.
n
is
|{3(n-f even or odd. 1 4- -
Assuming that
+~
-f
2
-f
l)
--
H-
. . .
or
1}
r)
-J(n-f I)
- log n tends
2 ,
a limit v as
to
n -*
oo
,
prove that (i)
lim
^
57.
Expand
lim
n v j' -a:
l/-vf* /"I 1 V
2
obtain an expression for
1
^^~3
\
,
,
g
n~
7
-/
*
# 4 and ,
if
'
>
-2
t^
and prove that
n,
-
r hm
- (2x ~ x 2 )
_ ^03
--(j*
f
'
)*
/yi
v
lim
'""
as far as the term containing f%\ Ulj H 2-
Findi
Find
1
+
)
T-"! 1
59.
^
2
"~
lo g
KQ 58.
h
n ~
(
log(l -a:
^ n ~ "2
^
2
(n
1 (li)
,
where
,
,_>o c^-rf*
a, 6, c,
are positive
c
* 2 + y a +2 2 = f a + i7 2 -f
60. If
and show that
a;
+ y + 2 = f 4-^-f 3a:
2
2
^=(i7-{)
#
and
2
and
= 1,
+
2
4-f
2
^!.
61. If n points are taken in space so that not more than 3 of them lie in any plane and not more than 2 in any straight line, show that the points define \n(n - 1) straight lines and \ \n 1 polygons of n sides.
------
- in which a is x - 1 and is repeated any number of times, must have one of three equal to that if x satisfies the equation 2# 3 + 3# 2 - 3x - 2 ~ 0, the fraction and values, 62.
a
Prove that the continued fraction
a~ a
...
a
satisfies this equation.
63. If (1
+ x) n ~Co -cx+ -\
P
...c n x n ,
where n
a positive integer, prove that
.
'
38
2*
is
(n
+
w+lV
a
l)
2
3
n+1/'
64. If a, 6, c are real and I, ra, n are integers, find the values of l ~ b) m (x - c) n has maxima or minima. (x ~a) (x
I,
x
Determine which of these values give maxima and which minima, m, n are all even, (ii) when I, m, n are all odd. 65. Solve the equations
y + z = a(l -yz),
^
66.
o Sum
i
-
the series
which
(i)
when
:
+ x~b(I
-zx),
2
8
z
,
for
3
13
x + y + z -xyz~c(\ 30
55
+ 7^ + 77; + "l+7T1 + T + T^H~r7 4 |5 3 o |2 1
~xy~ yz-zx).
to
-
MISCELLANEOUS
562
(B)
convergence of the series
67. Discuss the
I2n
E.
,n
Z
!
w) 2 (l for all values of z, real or complex, distinguishing cases of absolute
and conditional
convergence.
The equations of four
68.
(1) (3)
lines are
y = a, z = a'; x = c, y = c';
= 6, * = &';
(2) 3
x=y=z.
(4)
Find the equations of the line drawn from the point (k, k, k) to cut the lines (1) and (2). Prove that two lines can be drawn to cut all the four given lines, and that these k where lines cut the line (4) in two points determined by x = y = z 9
(k-a)(k~b)(k-c)^(k-a')(k-h')(k-c'). 69. Discuss the reality
and equality of the roots of x* - 2A(3* 2 - 4x + 3) + 3A 2 =
for different real positive values of A.
A
70. bag contains 21 balls of G different colours, there being 6 of the first Show that the number of colour, 5 of the second, 4 of the third, and so on. different selections of 6 balls which can be made from the bag is 259.
Af(a)=f(a + w)-f(a)
71. If
f(a + xw) =f(a)
prove that
Use
A 2f(a) = AJ(a + w) -
9
+ xAf(a) + -^r^-
1)
J/(o), etc.
4 /() -f 2
- - .
.
given that
this to find log 7-063,
J 7
= 0-845098
log 7 -2
= 0-857332
log
--
x(x
J2
log
J3
log
log
12234
-334 11900
log 7-4
16
= 0-869232
-318 11582
log 7-6 = 0-880814 72.
A
series is
such that the
sum
of the rth term and the
(r
+ l)th
is
4 always r
.
Prove that (i)
(ii)
73. If
the rth term the
n
is
sum
is
%r(r
l)(r
2
r(r 2
of r terms
is
-r1
->
- -
-
1)
W^r
2
(
r l) c
7^ '-
;).oO
;
r
+ ^{l-(~ 2
r l)
}.
a positive integer,
f
7)
74.
-
Prove that,
if -
%
,
- are fractions s
nator of any fraction whose value
lies
such that qr-ps
between
p q
and
\,
then the denomi-
r
- is
s
not
less
than q -f s.
Prove that there are two and only two fractions with denominators which lie between jf and -J.
less
than 19
MISCELLANEOUS
563
(B)
Determine ranges of x for which the function (log x)jx (i) increases and decreases as x increases. Hence prove the following theorems, wherein n is a given positive number and only positive values of x are considered 75.
(ii)
:
The equation n x
(i)
x has two
roots,
one root or none according to the value
of n.
The inequality n x >x n
(ii)
a consequence of either
is
x>n
x
or
accord-
ing to the value of n. State the critical values of n.
From
a bag containing 9 red and 9 blue balls, 9 are drawn at random, the Show that the probability that 4 balls of each colour will be included is a little less than -J. 76.
balls being replaced.
For
77.
-
continued fraction
the
(I
altered if a l9 a 29
An
78.
...
approximate value of
provided that <^< error
is
a n are permuted
OTT.
approximately
------
prove that
...,
^ -f-
cyclically.
measured
<^,
radians,
in.
when
Establish this result
<^
is
3 sin
(f>
<^)
/180. fc
i
nearly equal to
cos
r>
<
Hence express approximately the acute angles of a right-angled 2 terms of the sides and deduce that, if c 2 ~a'2 then
is
/ (2 -f
small and show that the
is
values of a and
\TT for all positive
triangle in
,
h.
79. Given a l things of one kind, a 2 things of another kind, etc., a r things of an rth kind, show that the number of different groups which can be formed from one or more of the above things is (a 1
and show that n members is |-
80.
and
if
verify
if
[(n
When n^2,
r~4 and -M) (n
-f
+
l)(a t
+
l)...(a r +l)-l,
a l
2) (n
+ 3) - (n - ax
)
tl
(n
,
- Oj +
1)
(w
- a x + 2)].
Zn^Vn-i + ^n-z*
if
xQ ~I,
xl
a^, yQ by induction that xn+l yn -xn yn+l = ( ~
Q, l)
1,
2/i
n+1 ,
and deduce that
2/n_i2/n
2/22/3
81.
number of groups having
the
Find the equation determining the values of x for which
stationary.
Hence, or otherwise, exceeds
m
show
that,
if
m
is
an
integer,
sin mx/sin.
sin 2 war/sin 2
x
is
x never
2 .
82. A quadratic function of x takes the values y l9 y zt y%> corresponding to three equidistant values of #. Prove that if 2/i + ,V 3 >2?/ 2 , the minimum value of the function is 2 _ ,
83.
Show
that
(n
.
+ l) n ~ l (n 4-2) n >3 n ( ]n ) 2
MISCELLANEOUS
564 84. The any order. by 11. 85. If
a number are 1, 2, 3, 4, 5, 6, 7, 8, 9, written at random in that the odds are 115 to 11 against the number being divisible
digits of
Show
w, n are
show that
positive,
\
86.
(B)
The
w-f
1
x 2n in the expansion of
coefficient of
87. If
-f
(
1 -f x*)
n /
(
1
- x) 3
=
cz 2 2
2
is
= d,
&z + by 2 prove that
88.
Z2
2/3
23
Prove that
sin a
sin
cos a
cos
sin 2a
sin 2/J
sin
sin 3 a
sur
sin j
sin a
sin
sin
cos a
cos
89.
2/2
-4
^
sin
p
cos y
t
y
J
()3
y) sui
^(y
~ a)
sin v(a
~
.
j8)
j3).
2y --
y
- sin
(p -y) sin (y a) sin (a
/?)
sin (a
-f-
/?
+ y).
y
Denoting the number of combinations of n
unlike, the
Show from 3n
by
W 6' r
,
show that
number of combinations
letters taken r together, all the of the letters^are alike, the rest being
if r
is
that the number of combinations in groups of n which can be formed letters, of which n are a and n are b and the rest unlike, is
VV
and show that
this
Prove that 1
if
n
is
.
+ Vn-s + - + ~ 2 n (w + 2). 1
a positive integer,
n n(n-l) x+
t(w -!)(*_- 2) 3
1T2
TT2i"
1.
.
91. Investigate the convergence of the stories
whose nth term
is
92. Prove that
^__
"* ,/V
and
93. If
A,
nn ;
_ 2n(2n
n(w-fl)
where
2 sin (a +
COS y
,
letters being unlike,
90.
sin
/Lt,
t
2
+ at~b both have
v are integers
[For instance,
^
2
and
+ 5^-f6 and
jtx
is
prime to
rational factors, then
v.
MISCELLANEOUS For the fraction
94.
p
2
2
3
565
(B)
2 . . .
prove that
,
95. Discuss the reality of the roots of
16* 4 + 24* a +16^+9 = 0, for all real values of k, 96. If ax*
for
all real
and solve the equation when k = ^2.
+ 2hx + b>0
show that
for all real values of x,
values of x, y,
z,
the coefficients being real.
97. 2 n players of equal skill enter for a tournament. They are drawn in pairs, and the winners of each round are drawn again for the next. Find the chance that two given competitors will play against each other in the course of the tournament. Also show that if n -7 the chance that a given competitor either wins or is beaten by the actual winner is tV
98. If thore are four relations,
for i = 1, 2 j= 1, 2 A fy + B& + Cfy + Djdj = (B,C)_(C A)_(A,B)_(A D)_(B,D)_(C,D) ~ ~ ~ ;
show that
(a~d) 99. If sin
""(ft, d)
(c,
d)
(b, c)
x~xy 2 where x 2 and y - I
(ii)
;
9
9
(c,
a)
(a,
are both small,
'
ft)
show that
x*= -12(t/~l)+(y-l) 2 approx. c
of Craps, each player in turn acts as banker.' The banker throws with two dice numbered 1 to 6. If he throws 7 or 11 he wins. If he throws If he throws any other number, he throws again and con2, 3 or 12, he loses. tinues to throw until either the number he threw first or 7 turns up. In the first case he wins, in the second he loses. Show that the odds against the bank are 251 to 244. 100, In the
101
.
where
If
a,
/?
game
un u n _i + aun + bu n
_!
+ c = 0, show
that
are the roots of
unless a = jS, in which case 1
Un - a Hence obtain the general results so obtained agree
102. If
solution of the first equation,
with those in Exercise
rv
cosx
1
COS
XXXVIII,
X
1
COS a
COS
COS a
1
COS
cos 2 2 2 prove that sin y 8in z = cos a
-f
ft
oos 2 /3
cos y
-2
oos a oos
ft
y
1
ft
oos y.
and show that the 15, p. 372.
MISCELLANEOUS
566 103. If o
*-.**-?* =*, x x d +
+y
(B)
!=,<) or z = a-c. *++ o a
then either
y
a 2x* + b*y* + cW = 0,
104. If
aW + b y z
l-
and
2
+c
a z
3
z*^Q,
2
-f
6
3
4 ?/
-f c
4 3
z
a e z 3 + & 6 */ 3 4- cV = a 4 ^ 2
and
1 ,
2
y 4
2
=--& = - -c
a;
prove that
3
= 0,
&y + c z 4
-f-
2 .
Y are two places on a road, 4 miles apart. A man A starting from X at any time between one and four o'clock, walks to Y at 4 miles an hour. A man B, starting from Y at any time between one and four o'clock, walks to X at 105.
and
J
y
4 miles an hour. Show that the odds in favour of the men meeting on the way are as 5 4, all times of starting being equally likely. [Suppose that A and B start at x hrs. and y hrs. past one, respectively. Taking axes Ox, Oy at right angles, set off OL~3 units along Ox and draw a square on OL. The total number of cases is represented by the area of the square, and the number of favourable cases by the part of the square between the lines :
*-y= 106.
1.]
that, in a game of whist, the chance that a hand is void of a suit is Also the chance that in 25 deals a certain player has a hand void of
Show
about -$. a suit at
least twice is
about 0-358.
[Chance that a hand 107. If
is
void of a suit
4
.
u~ax* + 4bx* + 6cx + 4:dx + e and
show that the
z
roots of
w=
are given
by
Gy3J t
-rC'JJ.]
is
any root of
ANSWEES EXERCISE
I
PAGE
10.
7.
28 2 -27 2
82
;
11. 23. 6,28,496.
-3 2
10. 212, 213,
.
27. 120.
28. 360.
...
,
22. 20, 4836.
219, 220.
30. (i)**'"**; -
(ii)
(p-1)
EXERCISE 22. 19.
(?)
5-394, 0-006
(ii)
;
2-594
;
II
0-002
;
EXERCISE 31.
1.
5* 3 -20z 2 + 80a;-314; 1258.
3.
ia-V*' + tfa-tt
5.
3z 2 + a;-3;
8.
2.
(iii)
11. 13.
15.
2-276
0-002.
;
III
z 4 - 3* 3 - 6* 2 - &x - 10
;
-21.
fc^-i^ + fcc'-fta + A; ~^A-3z~7. 6. 3z 3 + o;-3; 5a + l. 7.13. 3 2 3 (i)(x + 5) -13(z + 5) -f56(*4-5); (ii) (x + I) ~(x+ l) + 80 4.
;
;
+ 5^2.9. (ii)jf-75y + 7; (iii) -n + 7n(n-hl)-67i(7i + l)(n4-2)4
9. (1)2/3
10.
T(p-1)
~^;
0,^.
3
-
i/
2
15?/ -f
257.
12.
(x-3y)(2x + y)(x* + xy + y 7z 2 - lla; - 6.
z ).
16.
+ 4x 2 - 6x - 2. 3* - Ixy + 4y 2
23.
x 2 - 3y 2 + 3z*
14. 3a; 3 2
.
18.
20. a( 21.
a=3, 6=-3,
c
= l.
32. 22. p* - 3q z + 3r 2 - s 2 = 0. 24.
26.
a=5, 6- -3, c = l. a = 8, 6=0.
25. a
-w = 0. 2
= l, 6-2.
27.
29. a 2 + 2)(a? + 2y-3). 31. (i) (x-l)*(x + 5); x + 3)*(x-3). A = l,(2a? + y-2)(aj-y + l); A = - 7/8, (* - 7y + 4) (x - y - 4)/8. A=-l,(y-l)(y-2); A- -2, -(
28. (3a;-4y 30. 33. 34.
(
EXERCISE IV .
16. 120.
21. 23.
(i)
17. 60.
#i"("+i)
;
(n)|tt
18. 15.
20.
(i)
2 2 ".
+ l.
(-l)Ca ifn=2m; (-l)(n + l)05, if n=2m + l. f
HIGHER ALGEBRA
568
EXERCISE V PAGE
43.
1.
5.
a
(s-l) 2 (s-1)
8. JC:=:r 3
10.
.
2.
2*4-
1.
3.
.
6.
3z~4.
7.
+ l, 7=3.
x*
+ x-3.
4.
= i(x + 2), y = J(a;-2). Z^l-fz 2 7^1 JC
'
9.
,
A=-2 = l; 4= -*,=-!, 0=1, !>=-*. ft
EXERCISE VI 50.
3.
18.
51.
-2x* + 3Zx
2
p^a+f-g-fi, 25.
24. 6.
-2.
26.
31.
-4(6-c)(c-a)(a-6)(a + 6 + c). l
33.
l
2 ~(6-c)(c-a)(a-6)(Z a + 3Z 6c). 2 2 2 (l-a&c)(l-a ~& -c -f2a&c).
36. (16)(25)(34)
(12)(24)(36)(65)
;
(16)(65)(53)(24).
;
EXERCISE 60.
1. 2, 7T/6;
4. (i) (5
61.
5.
-7T/6; 2,
2,
^O
(x
+ 1/ 2
5 F 2 )
(a:- iy)(x*
(i)1625;
/
cos
/
*>
2
;
(iii)
(iii)
+1
)
\
+ y*)
(ii)
y\ --
tan"" 1
7i
L_
(x -2/ 7.
13/125.
2
7.
3.
i( 1
+ 0/25.
*
(
-
(18
+ cot-
sin n( tan~" x \
y i*
1 ._
l~2t; i(ltV3).
13.
~7r/3.
EXERCISE 67.
VII
-. D
2,
;
-0/13; (ii)(- 2 + 2)/8; 2
(i)
(ii)
6.
I2xyz.
-
30.
+ a + 26)(a + 6 + 2c).
32. 3(6H-c-f-2a)(c
62& 2 y-
-f
27.
-(6-fc-2a)(c + a-26)(a + 6-2c).
35.
3
q 1.
29.
34.
2z
4.
y-12xyz.
VIII
2-i, -4-3i.
EXERCISE IX 78.
3.
KX + 4r = (l+* + iy)/(l-*-iy), then X = and
2 a (l -a: -i/ )/[(l
-
x 2 + t/ 2 = l.
Z^O, y-2?//[(l-a;) + 2/ ]; Geometrically, take any line ^4OJ5, so that BO OA, and draw AL, 9 parallel to Ox to meet a line through 1, parallel to BOA, in L and con2
2
if
BM
M
struct
OMN
similar to
opposite directions.
1+3,
1-zand
hence L 5.
If
OM
is
If
OL1, with L* ION,
A
represents (l+z)/(l-z). If |z| =
a right angle
a=p + iq, px + qy=p + q 2
pendicular to qx =py.
2 ,
;
and
z,
l,
LOM
:
described in
N
then L, M, represent O^lLl, J5O13f arerhombi;
N lies on Oy.
which passes through
(p, q)
and
is
per-
ANSWERS EXERCISE X
569
PAGE
86.
1.
1 = 0; 4z - llz + 9z - 2=0
4x- 3
(i)
4
2x*-3x-
3
(iii)
(i)4z
3.
t(3
-8z 2 + 8;r-3=0;
3
N/6)
-
2, 9/2,
t(
;
-
1
(iv)
6.
4.
-
2, 1/3,
-2,
9.
8. 4, 2, 1, i.
+ 9x 2 +
llx-f 3=0
2z - 9x*+llx
;
-3=0;
x3 -4
(ii)
N/5).
1/2.
2x* 3
;
2.
5.
(ii)
2
3/2, 1(1
\/5).
7.
1/2.
5, 3, 1,
10. 3,
3, 6.
-2,
-
1.
1/2,
-
1/2.
2 4 x - Sx - 15 (ii):r -23x 4-59x-52=0. 3 3 2 14z + lla; -75 = 0; (ii) a; 4 25* 3 + 375z 2 - 1260*- 11700=0. z 3
11.
(i)
12.
(i)
;
EXERCISE XI 94.
3.
4,
7.
V,
-2. -1.
4.
2.
11,
-3. 15, 3,6, -4. 8.
14. 6. 18. 2, 2, 2/3,
-
3-1,
5.
4,
9.
7,
-
2,0.
<>.
6.
-0.
13.
4.
17.
3/2,2/3.
16. 2/3. 19. 3/2, 2/3.
1/2.
EXERCISE XII 99. 17.
iv3a 2 + 2a& + 36 2 ).
-a, -6, -i(a4~&
EXERCISE i
3
'
i
i
2.
^ i; J5___^___? T x
9 ^
1 i
*-3 x-2
^T
1
\2
(x-2)
^_
3
(x-2)
4
'
3
'
J-+-
x-l
i
x
2
* 5.
XIII
ft
5.
x-2
x-3
X4-2
|
-1) Q
(a:
_ _____
j^
_i_
______ 4.
3
x + 2) 3 *
!
10.
"' 13
.
x"2
-fl" (x 2 -fl) 2
_1
x 2 +~2
x
" L(x+l)
2
x 2 + x-fl'(x 2 -fx+l) 2
1,1,1,1
-I)
14.
JLr
1 3
2(x-fl)
3
HIGHER ALGEBRA
570 PAGE
104.
15.
*_
-l + -JL_ + 1 -ax
-ox
I
;
'
'x-a'
(a-b)(a-c) 18.
Put x^din Ex.
19.
x + a + b + c + Z. r. 7-7-7 (a~b)(a-c)
1.
\
2.
j
+ l)(2n- I).
l-n(2n
7.
Tzn(n + l)(?i-f 2)(3n + 13). 10. 2926. ^(47i 2 -f7i-l). J Miw 1). !)(/<- f2)(3/* iV/t^j
14.
pui>x~d.
;
19.
*
2
a
(i)
(n
})
i-/i(6n
+
8.
(ii)
iw(n
23. 715.
(B)
2.^.
1.^IV\UIV
A
I
I
13. 4300.
- l)(9w a -9w -2), where
21. 344.
EXERCISE XIV 117.
12. 50336.
%n(n + l)(n + 2,).
15.
'2).
-3-l);
20. 220.
(A)
11. 286.
t-
-f-
16. -iVw("
'
4.
3.
9.
x-a
1
x-a
EXERCISE XIV 116.
-b)(a-c)
17.
a*
105.
'
(a
3.
_
f
6.
n(29n 2 +
5
138r*4- 157) *
"
36(ft4-l)(rZ'4-2)(tt4-3)
10
*
2.5....(3n + 2j"
2
llf
'
1 _ Li?_i -Jil^^tl) 2 2.4.6 .... (2w + 2)"
~
*
1~,
17.
^
^y/2.
.
18.
-
19.
140
4(2n + 7)(2ra
i
Q|
i __.
r
i
i I
i I
si
I
n(n-f 2)
ANSWERS
571
PAGE '
(l-.r)
n)'
~
26.
-
,
fc
where k^=2 n
(I-*) 2
2(1-.*:)
-
------
27.
.
3
^
,
where & = 2 r
28. 1
EXERCISE XV 129. 8. 131.20.
a3 + & 3
(i)
- (a 3 + 6 3
(ii)
;
2
(0-y)(y-a)(y-|8);
(i)
36.
(iii)
;
)
-
(ii)
(jB
y)(y
- a)(a0)(a +
+
y)
EXERCISE XVI 143. 146.
1.
19.
2
V|7
0,
39
2
+ &2 + c2
).
= 30 2 + 20 + 14 2 + 5 2
2 .
EXERCISE XVII 158.
1.
^ ii
-Tir;
-2, -1,3.
(ii) 1,
-2:
3:
2.
10.
3.
-3/2.
EXERCISE XVIII 165.
3.
4. (1
*N/2,
5. (0, 0),
V,
IT J-N/2),
-f
9. (1,1,
(
(
\/3, q= N/3),
(
7. 2, 1, i,
in
any
(
- iTt'), i>/57~, t,
(
-2
-2
iN/2,
T 0-
order.
6. 2, 3, 4,
in
8. (3, 5),
-f,
(
any
tD
2
x4:
or -4^36
2
;
y, a) y,
(W54^, W^T ii^S), or
a;
= -y
wy
*v2).
-|),
order.
-^).
(f,
2 or -*
(a-iVSdbl^, i^ v
a fourth root of f
.
4 2 2 2 2 2 4 [Use the identity, # + x y + y = (# + xy + y ) (x
12.
T
- i=F iV57).
1),(-|, -f,7).
10. x, MX, 11.
-1
(2, 0), (0, 2),
The only
solutions are (0, a), (a, 0).
xy0 or a
2 ,
and that the
[Remove
-xy + y
fractions
rangement of
= i{l
with ftny ar _
signs.
+ 26 + 2c - 2a)}, etc., with any arrangement of signs. b 2 + a 2c 2 -b 2 c*)labc, etc., the upper or lower sign to be taken
14.
a;
15.
x
16.
&*=a a (& a + c 2 -a 2 ), etc., where
(a
).]
and show that
latter alternative is impossible.] _ t
13. (1, 1),
2
s/(l
z
throughout. 17. (a,6,c)
2o
or
a
a Jfc
2 2 2 a 2 2 2 2 a =(6 + c -a )(c + a -6 )(a + 6 -c ).
a;=(a6-a6 c~l)/(a6
2
c2
-6 2c + 6),
etc.
[Put
s=
HIGHER ALGEBRA
572 PAGE
165.
= (a + 6 + c)/(a-&-c),
or
18. (1,1,1)
etc.
(-3, -3, -3), (>J2, dW2, 1TV2), in any order.
19. (1,1,1),
20. (i,|, -i).
J^V(&
21. If
2
+ c 2 ~a 2 ),
B=^/(c + a*-b 2
(Voc,
),
2z~A+B,
2z=J3-hC', 2y=C+A, 2 2 [Show that 2x(y - 2) = 6 - c .]
then
22.
2
N/(
giving eight solutions.
(i(a-6+c)VQ, J(6 + c-a) Q = Za*~2Ebc.
N/6c),
C=
;
F
VG) where
Zbc^Q, the only solutions are (0, 0, 0), (a, 0, 0), (0, 6, 0), (0, 2bc=Q, the equations are not independent. [Show that
23. If
0, c).
If
x(y-z)/bc =
n>i
,
24.
0, 0,
rt ,
--
x=
or
-
.
..
-
z
= ....]
-
-
a(ca-b )(ub-c
25.
(-!,-!,
26.
The values of
27.
*=y=a: =y =a.
/
etc.
(
-2,
3, |, f).
28. (f,V).
166. 29. (2,1), (-1, -2), (-5,4). 31. a 3 -f-6 3 + c 3 ^a 2 (6 + c)-f6 2 (c + a) (i)
,
)
-1), (i,i,i), (if,0). x, y, u, v are (3, -2, |, |) or
/
32.
-
2
+ c 2 (a + 6).
Subtract the second equation from the first, and divide by x-y; 4 3 5 then, from the fact that prove (x + y*)(l -xy}~ 2a -a ; 2 z are roots of x, y (2a-l)t -\-a(2a-l)t + a (a-l)=Q, obtain (ii)
(2a~l)(o;
from
(i)
2
+ 2/ 2 )=a 2 (2a- l)(z 3 -ft/ 3 )=a 3 (a-2), and z xy = 2a* -a + l =a (a l)/(2a 1) ,
for
;
(iii),
(iv),
(v),
multiply the result
- 2a -
obtain 3a 4
39.
40.
a 4 + 6 4 =c 2 (a 2 + 6 2 ).
167. 42. 16:5: -21.
46.
EXERCISE XIX 1.
i(3dbv/5),
3.
i(-lW3), i(-lW15). 1, i(-3V15), i(-lW15).
5. 7.
-1, -1,
8.
-1,
9.
-1, -1,
t,
2.
i(5iV21).
*,
i(3
(-
4. i( 6.
-1,
+ K/7).
i[3-f N/33
>/(6 N/33-42)J,
K-1 + W3).
16.
i[3
k=-3.
EXERCISE XX 184.
(ii)
1, and the results follow. 2 = ...^...-Za + 2i;bc. 2x(b + c-a) 2 3 3 4 2 2 4 2c 3a& 3a 6 4ac 3 ) + 16d 3 (a 2 - 6 2 ) 2 -0. -f -f 36 (a ) (a
34.
177.
thus
2-68061
0-80125
4-22524.
1.
(i)
2.
2 cos 40 or 1-53209, 0-34730, -1-87939. (ii) 0-69073, 0-11479, -6-30553. (i)
;
(ii)
;
(ii)
by a 4-1, and
ANSWERS
573
EXERCISE XXI PAGE
198.
9.
10.
-lN/2, l2i. i(-l
11. t(l
13. 14.
12. |(1
*s/3).
-2>/2, -4 2*. a 2 3x + 3), i(# -'# + 1)(# 2
\
199.
1(1
N/13)",
{x
2
s/5),
1(3
N/17).
- x(2 -
i>/3) 4-
i
-
{a;
- V5
v2 + 2\^5,
18. s/5
v 2 - 2^5
the other substitutions are given
;
by g + 2# + 5^0. 2
20.
Roots as in Ex. 10
the other substitutions are given by
;
-3=0.
EXERCISE XXII 214.
7.
8.
= t(y -9y); z*-10a; 2 + 1^0; 8
s/2
- i(i/ 3 - lly) -18i/- 110 = 0;
-s/3= 3 7/
N/
;
z9
- 15z 6 - 117z 3 - 125^0.
EXERCISE XXIV 233.
1.
Oscillates
(i)
between
- 2 and + 2
;
(iv)
1
and 3
(ii)
;
converges to 2
diverges
;
;
(iii)
oscillates
between
(v) diverges.
EXERCISE XXVI 254. 255.
1.
(i)
7.
Divergent.
Convergent
Convergent.
15.
Convergent.
Divergent
if
19. Convergent.
23. a?
24.
oscillates finitely.
(ii)
8.
11.
18.
;
12.
Convergent.
16. Divergent.
p-q+
1
^ 0, 25.
10. Divergent.
Convergent.
14. Divergent.
17.
Convergent.
convergent
20. Convergent.
x^l.
Divergent.
13.
9.
Divergent.
if
2>--fl<0. 22. Convergent.
21. Divergent.
x^l.
26.
x
29.
x>a.
30.
x^l.
31.
39.
.
40.
&.
41. -h.
27.
x>I. 38.
28.
x
A-
EXERCISE XXVII 264.
10.
265.
13.
(i)
Between i(5-Vil) and
(ii)if
or between 2
i,
and
|a:|
1-t.
EXERCISE XXVHI 282. 283.
1.
15.
i, i.
2. 0,
-oo.
f, 0.
3.
4.
1.
5.
7/3.
6.
GO.
(
(ii)
-(l-&r(l+aO
(v)
-n
"" cos n 1 x sin
x
2 ;
;
(iii)
(vi)
(a
-2z 2 )/(a 2 + z 2 );
n tann-1 x
sec 2 x.
(iv)
7.
fa;*.
8.
1.
HIGHER ALGEBRA
574 PAGE
283.
-(hx + by+f)l(ax + hy + g);
16.
(i)
21.
y^ax.
-ay)l(ax-y*).
(x*
(ii)
when z^l/3.
24. 4/27,
the graph has a point of inflexion at (1, 0), where the gradient is zero; another at (3/2, -1/16), where the gradient is -J; a minimum at (7/4, -27/256).
25. 1,7/4;
284.
27.
Between x~ -2 and x 0, and for x>3 at ( - 2, - 65) min. at (3, - 190).
;
max. at
(0,
-1); min.
;
29. 18 cubic inches.
EXERCISE XXX a
= 2,
s=
304. 305.
18.
315.
2.
2e
9.
+z+ -[(z~l)e* 3?
3.
l,
Zx^
Use
1,
or a = if, * = 3/5, 3/5, 5/6. substitute the values of 27s 3 , 2J^ 2 , 2k.
5/2;
3 and -272:(a; + 3) ,
EXERCISE XXXI 3.
c-1.
4.
e.
5.
6.
3e/2.
15e.
7.
1-2/e.
8.
-1/c.
l].
EXERCISE XXXII 322.
5. 9.
(ii)
0-84510.
6.
-3/4m; + l/(4w +
1-04139.
(ii)
13
12.2-log,2.
o^o
323.
,r
&1
c--,.-;
17.
1-010299956?
19.
^
20.
-4
(x
t
+ 2) +
6
6
C
15.
1
7.
r
1
-
1-11394.
s
1
1
,
(ii)
21og e 2-l.
11.
10. log, 2.
l).
-^-^.^l.c^-;a^. 0-3010299957.
;
^
r
+ _L_ (5 _ a;2)lo g i_ a;) (
+ 3^ + 2^) +
%x
x2
1
+
-^
(8
.
- x 3 log )
(1
- *).
EXERCISE XXXIII 339.
10.
Convergent for
13.
(i)
If
all
values of
y>a + ]8+l;
(ii) if
(see Art. 11,
y>a +
Ex.
1).
]8.
EXERCISE XXXIV 349.
1.
8th term.
5.
6th and 7th terms.
7.
f
<
23. 1-5860098; 5
.
10~
5th term.
3.
8th term.
6.
5th and 6th terms.
4.
llth term.
8.
.
350. 21. 10-0033322284; 5 25. 1-2431626;
llth term.
2.
10~ 13
.
8
4.10~ 8
.
5
22. 1-414214;
.
24. 1-70998;
.
26. 1-319508;
3
.
3
.
10~ 7
10~ 6 .
.
.
10~ 7
.
ANSWERS
575
EXERCISE XXXV PAGE
356.
10.
(n-f 12.
P-
13.
P-g,
where
_
+ 2+
+
Q--^ 2
^
~
1.
57i
n+ ^9
2
-?i~9 ~~
II* 2 + 26^ + 57a: 4 + I20x 5 2 n 2 ~ - 3. - 1 + 3 2* - 3 W + 2o; + 2x 2 - 4.T3 - 34z 4 - I48,r 6
(i) 1 4- 4o; 4-
1 (ii)
f
7?
;
.
;
EXERCISE XXXVII 365.
1.
366.
4.
1 (iJKl + S"- ),
(20/i
- 27
-f
i(2?i
~ 3 3 n ), J (20w 8
5.
K(23-3n)2^ + (-l)].
370.
1.
2n
371.
8.
n
- 34n -f 27 - 3 3 ~ n ). 6.
-J + l +
EXERCISE XXXVIII
10.
4>
~2 .
=
A .2 n -f-.3 n + i(4n-f7).
W2
2n-l 2n-3 "~'
1
11.
a
2n
EXERCISE XXXIX 384.
1.
s 3 + 2a?-l.
2.
(i)J(4n-l);
(ii)
and
and
EXERCISE XXXVI 359.
*
15
.
+
131 ""'
13n
HIGHER ALGEBRA
576
EXERCISE XL PAGE
394.
1.
+2
H,
|,
(i) J,
-ffs
A
Q
1
,
J.
+3
+4 -3' -"12
'
5 D
T+r+BTsTi* +
1
'
l
A
J.
111
*
6.
(ii) 1,
;
a
1
a-2+
'
^~2^3~4-
-I
1
-^-^-
3
2a 2
a
5*
-a~l *
'a-2'a-l'
2a 2
-
3a
EXERCISE XLI 412.
8.
10.
-f& = 0-242253 12
yards
. . .
^=
,
0-242268
....
3 inches.
;
EXERCISE XLII 419.
127, 55
(i)
3.
7
5.
29 + 41*, 12+17*.
7.
155
+ 8*,
13. 3,
8
;
4
(ii)
2.
+ 27*.
+ 225*,
10. 39, 8
420.
30, 13.
1.
137
+ 199*. 11.
18.
17. 67.
20. 14, 2, 3
22. 2, 1, 3
20 + 29*.
4.
9+
6.
24 + 7*, 11 +
13*,
12*. 9.
1,
20;
8, 9.
A-A; #-
12.
3, 2,
2
;
x = 205 -17*;
13,
12,
x = 201 -
17*.
21. 11, 3, 7. 23.
4, 3, 1.
2.3.7.
1149 + 1540*.
24. 1149;
26.
25.
27.
+ 8*.
f + A.
1, 13, 8.
;
;
1
14. 57, 43,
13, 10.
;
+ 27*,
8. 6, 1.
10, 20.
;
19
One
12 12s. 8d.
;
EXERCISE XLIII 428.
12. 2,
9
;
3,
6
;
4, 13
;
5,
7
;
8,
15
;
10, 12
;
11, 14.
EXERCISE XLIV 434.
1.
(i)9+13*
3.
435.
8.
11.
;
(ii)75
+ 179* (i)78 + 1217*; + f + A-117 + 77 k.
2. (i)
A
149
;
+ 77*;
(iii)
379 + 770*.
(ii)
30+179*.
(ii)
1800+ 1861*
;
9.
(iii)
f+
12. 3
540+ 1009* + H-2.
;
A
+ 23*, -7 + 23*.
(iv)
1683 + 1901*. 10.
ANSWERS
577
PAGE
435.
15. 2,
14. 1735 + 3465*, + 66*, 9 + 66*. 2 x + l=s(inod 7) has no solution. 4
16.
2,
etc.
13
13.
;
-3, -4.
17. 3.
EXERCISE XLV 442.
1.
those of 19 in reverse order
1, 10, 5, 12, etc., i.e.
since 7
number
the 6th
is
2 1 2 5 , 2 7 2 11 2 13 , 2 17 , or ,
2.
1, 3, 9, 10, 13, 5, 15,
number or
444.
12.
,
,
is
divisible
2, 20, 36, 32, 37,
14. Tf
16.
is
2n-fl
since
preceding,
3 2n +*,
;
is
i.e.
every other
prime
to
17-1,
6.
41, hence recurring periods contain 5 digits the remainders in the division of 2 by 41.
by
= 271; w = 123, ^' ^903342366757. p-1 occurs as a remainder.
7i=41, w
Because
and 6
1,
2, 13, 14, 15, 3, 10.
11, 16, 14, 8, 7, 4, 12, 2, 6
10,5, 11, 14,7, 12,
3,
90999
the
in
after
a period of 3, a factor of (19-1);
1, 7, 11,
;
7
EXERCISE XLVI 445.
4. 25.
1.
1, 7, 11.
2.
15.
3.
5.
5.
8, 33.
6.
2, 10, 19.
7.
19fc9, 19&2
8.
(25, 214), (54, 2157), (67, 4120).
18& + 4.
EXERCISE XLVII 447.
1.
10.
2.
15.
3.
5. 5.
4. 4.
4.
EXERCISE XLVIII 456.
2.
(
-
2,
4.
(
6,
6.
(
- 3,
8.
Two
10.
(
- 1), (I, 2), (3, 4). - 7) and two in (0, - 4) and two in (5,
(
1).
5.
(
6).
7.
(2, 3), (3, 4).
9.
(-2, -
in (1,2).
- 6, -
5),
-
1, 0), (4, 5).
(
- 8, - 9), (13, 14), (14, 15). - 9, - 10) and two in (3, 4).
3.
11.
One
1), (0,
root in
l)and two in
(1, 2).
EXERCISE XLIX 466.
2-6306166.
1.
1-3569, 1-6920.
2.
1-7838.
4.
16-0428539.
5.
0-3472964, 1-5320862, -1-8793826.
6.
0-1147994, 0-6907309.
7.
13-8440609, 14-2895592.
5-2100150,5-2973245.
9.
17-7459667. -1-9216606.
8.
10. 1-2783089, 1-5511616.
3.
11. 1-4007219, 1-5823564.
12. 1-0156820, 1-52858586.
13. 3-4334634, 3-6617081.
14. 1-7320508, 1-8666161.
15. 4-4707625, 4-7814825.
16. 2-59861938, 2-76190631.
17. 2-1010205, 2-1084952.
18. 7-3355540.
19.
(i)
3-3548487
;
(ii)
4-4641016
;
(in)
0-6386058.
(1, 2).
;
HIGHER ALGEBRA
578
EXERCISE
LIII
PAGE
c+
504.
1.
(i)6720;
9.
1024.
10.
116280.
505.
13.
(i)
20.
(i)44; (ii)20. n
17.
27.
(i)
506.
480.
(ii)
2.
22.
.
\\n.
57
47
(ii)
;
1771
(i)
(ii)
;
a~ l
.C+*>-i c m
1895040; (ii) 145680. 969 (iv) 165 (iii) 885 ;
;
28. 23,
15.
(iii)
;
7.
\n-lj\r.
29. 42.
(v) 552.
;
30. 15.
EXERCISE LIV 506.
1.
C%-('$.
EXERCISE LV 523.
1.
6.
;
(ii)
5/12
(iii)
5/9.
1/12;
(ii)
125/1296;
(iii)
1/36
(i)
8. (i)
10.
15.
(ii)
23.
12.
18.
(ii)(ft-l)/2;
(n-
(iii)
1/8.
9.
25.
;. (ii)
2/7.
(m + l)(2n-m -2)/n(n-
(iii)
(i)
7.
20/27,
(ii)
21. (n
l)2w,
1).
496/729.
+ l)(3n + 2) pence.
2257/54145.
(ii)
;
1/7
(i)
2(n-7w-l)/n(fl-l);
2162/54145
(i)
4. 5/33.
14/33.
155/648.
l/2w.
19. (i)l//*;
22.
;
236/270725.
13. (i)2//i;
524.
3.
(i)6/55; (ii)4/ll.
(i)
Zp
(H)
24. 73/648.
525.
25. (0-55) 7
OC Ab.
/
;
\
(i)
.
A p-,
O
I
]6
.
13 5i
3-0
2197
13
13 .
.
4U~ 2-0-82.I
/' 1_2 P = 4 tf
/;; \ 11 ) I
i
13
I
I
'
13
49
5"6"
^
13
~4.*=P
.
>
|
(iii)
97 Z/.
Tsp +
J ~ U/ ?71--^
(\\
i$p'-"-p. 2fi
'
2
-^01-30
~ 833
2_^__3_26..
49
/::\ 11 /
>
V
P /_ O. a
24 50
_2_R
51
.
.
26 49
_2 _5 .
__
.
48'-/?,
-
526.
29.
(i)6/w(w~l); (ii)6(-3)/?i(w-l);
33.
(i)
35.
1/2.
36.
(i)
1/6
;
(ii)
pp'p" to
34.
7/9.
(1
-p)(l
-j>')(l
-XO
(i)
;
(iii)
1/4
(n-3)(n-4)/n(w-
;
(ii)
^'(1 -X')
(ii)
to (1 -jp)(l
EXERCISE LVI 541.
542.
5.
(i) (2, 1),
6.
(i)
8.
(i)[|a(a-l)
(50, 31)
;
(ii) (3,
(25, 1), (623, 25),
...
;
2), (29, 18)
(7775, 312),
+ l, i(a-l)];
(ii)
[Ja(a
;
(iii) (5, ...
2), (12, 5).
.
+ 1) +
1,
1).
2/3.
J(a
+ 1)].
-^)X'.
ANSWERS
579
MISCELLANEOUS EXERCISES (A) or
n -! and n 2 + l. 2
4.
2.
ri*
5.
a + c-6-d, a + d-6-c.
i-i + i-A + iis;
(iii)
*=-i
6.
2
-
8.
10.
1L
^. V _,, + 1 o=o.
21. a6 26.
29.
.
>/o
= c + l.
a?=i(n +
23.
or
l)
(a~6) + c a ~c 3 =:2a~6. Convergent if a? <1 or if a;= 1.
a-4-6^0 and (i)
c1
|
the series
is
Convergent
(ii)
30.
12.
1(14-3^);
(i)
p> i,
if
Divergent divergent if
|
p> 1,
if
convergent
divergent
if
a? |
|>
1
.
When
a?=
p^%.
if
(ii)
31. d 6 (
>9, none
34. If 35.
46z= -24 + 15. -.
3.
y
4,
/
- 7 sea <0, four
real,
io
4* A;
+ 10
5, 2
49.
Equate the factors to 4
59.
X
a=
l,
6
or
= l,
A:
43.
i(5N/17).
P 4- Q,
4
- x)
r- log (1
where
or
o>
o>
real. 2 .
4.
x=~2*J2 ic
2* A; 2
- 1 two
!
39.
53.
.
a<
real,
P-fcuQ,
Convergent unless x =
54.
.
X
206 + 2 lOn.
= 0. [y0, ~l,x= 4, are asymptotes; a;
-
66
etc -
-
y is a minimum at (
2
(i)
(ii)
-1+e-l 4
/ 76.
Since
1
\m
+
2 .'.
6 * 05
74.
71. 32/9.
56.
1.
_l- -i
.
.
4
/4\ 3
/3\ 2
r
2 ,
/
n
" X*1 1
...()
<8.
(l (|) (|) L) Consider the arithmetic and geometric means of the factors of P.2 2 .3 3 ..... nn
when
written at full length thus
1.2.2.3.3.3.... 83.31.
85.
Minimum
87.
H
is
a
values
maximum
when
(i)
value
;
x~a
9
(ii)
a;=^(a-
and - 4^f
are
minimum
values.
2,
-i).]
HIGHER ALGEBRA
580 91.0,
2,
96.
QO.
Put p^llm, g = l/7i.
99.
l+l
+
+
?-^
108. 4*, 6*,
-0-68233.
112.
-4-61015.
110.
-1047.
102. 946,
3
2-a,
^>
123.
MISCELLANEOUS EXERCISES 3
ul+^+^i+ir^
-
(iii)
& = 0,
g 4w+a
= 3 4- 6n, 2
(
;
=n
* )q
(B)
'>
g 4w ^3 = 5 + 6n,
+ 2w, for* = 0,
1, 2, ....
6.
15. 3
-3?k-3*k 2 and of the form 6k
21.
n
22.
The values of - a,
23.
31.
is
1,
3^fc,
+
where
has the values
A;
-
1.
6, c
are
pz
~-~
2 .
and similar expressions.
:
(p-q)(p-r)
and
16
29.
e~i, 0.
2+8e-f32e 2
1,
0.
--3-7c.
and
e
33.
(i)
(ii)
x < 4, divergent Convergent Oscillates between finite limits. if
|
35.
D = IJ(a
36.
Put
l
+ a 2 o> + a a w 2
-i-
64.
_
a 4 o> 8 + a 5 a> 4
x=a + h, y=a + k, z=a +
x ^ 4,
l,
2 )
,
where
where
# -= -
oscillates if
4.
-1.
5
=
1
.
h, k and Z->0.
44. (logs) 2 , 2lg*,
41. 5-674619.
58.
if
\
a:
2 ,
X x^e*, 2 9
e
x ,
x*.
59. (Ioga-log6)/(logc-logd;).
Let a, j9, (a<)3) be the roots of Zl(x~b)(x-c)=Q. (i) If I, m, n are odd, x a gives a max. and x=j8 a min. value, (ii) If Z, m, n are even, cc=a and x = j3 give max. values, and x=a, 6, c give min. values, each zero.
c-a Put
c-6
=
67. Converges absolutely if
|l
verges, but not absolutely.
diverges
if
||>J.
If
||=t,
it
con-
ANSWERS
581
'
(a-k)(b-k)
(a'~ k) (b'-k)
(a'
> 2, the roots are all real if A < 0, all are imaginary
69. If A
-k)(b-k)'
< A < 2, two are real and two imaginary if A = 1 or 2.
if
; ;
equal roots
71. 0-8489892. 73.
For the
first part,
use Chap.
show that the second part, r
sum
15, p. 223, putting
of the series <
6 to oo
to e
and
+l
w= -1.
H
=
4n-t-l decreases as
3?t
increases as x increases from
75. (logo;)/a;
from
XIV,
<
(ii)
One root if < n < 1 two roots if nx > xn if x < n < e, or if e < n < x. ;
A rough graph of
(log x)/x should
84. Consider the number of sum is 17 or 28.
85. This
is
so if
n
/ I
-
n
ways
\wM-i
1
rr
(
in
)
P<4,
all
;
< n < ee
x increases 1
;
no root
digits
the roots
* = N/2, -
if
n > ee
.
can be chosen so that their
then use Chap. XIV, 11.
and comparison test (iii), p. 250. are imaginary; if A; 2 ^4 two are real and two
imaginary. If
.
3
be drawn.
which 4
>1
1
91. Convergent; use Euler's constant, p. 311, 95. If
-
.
1 (i)
For the
=-oo.-8in tcos- + Binwhere Bin-j
or
;
INDEX The numbers Abel's inequality, 330 test for convergence, 331 theorem on multiplication of series, 338 Absolute error, 15 least residues, 421
value, 13
residues
of
terms, 8
Approximations, "M, 15-17 to roots of equations, 456-461 to sums of series, 260 Approximate value as a fraction ?'(l +x) (Newton), 347, 348 of log e (l+a;), 323 of log e (a/6) (Napier), 323 Archimedes' axiom (Eudoxus), 13 Associated residues, 425
of
Asymptote, 473 Bernoulli's numbers, 113, 114, 378
theorem, 115 Bessel's interpolation formula, 381 Bicycle -gear as revolution counter, 435 Bilinear substitution, 80
Binomial equations, 170, 171 solution of
a"-l,
series, 263,
reciprocal
form,
Tschirnhauseri's transformation, 105
Bordered determinant, 138 Bounds of a function, 286 of a sequence, 237 of,
Cardan's solution of cubic, 180 Cauchy's condensation test, 325 test for convergence, 253 Change of order of terms, 240, 258 Cofactors of determinant, 123 Combinations and permutations, 403
Commutative law, 13 Comparison tests for convergence, 250 Complex functions of a real variable, 270 Complex numbers, 52 et seq. modulus and amplitude, 56 product and quotient, 60 use as operator, 76
Complex sequences, 243 Complex variable, 201, 334 in binomial series, 328 limits and continuity, 202
Associative law, 13
Binomial
into
258
60, 70
Arithmetic mean, 221 Arithmetical progression,
transformation 104
Brackets, introduction and removal
Absolutely convergent series, 256, 261 Aggregate, 14, 211 Alternating functions, 45 Amplitude of complex number, 56
Argand diagram, 56, 77 product and quotient,
refer to pages.
173 328
for complex variable, 334 Binomial theorem, 34, 204 general statement, 340 elementary proof, 342 Euler's proof, 341 Biquadratic equation, 186-190
condition for four real roots, 102, 453 Euler's solution, 100 Ferrari's and Descartes' solutions, 189-
102 functions /, J, and the discriminant A, 187, 188 reducing cubic, 187-101
roots of equations, 66 Composite numbers, 4 Conjugate numbers, 60 partitions, 402 Congruences, linear, 428 reduction of, 433 x a ~b(modp), 445 Continued fractions, 380 approximations by, 402 equivalent surds, 401, 527 calculation of convergents, 535 Continued product, 33 Continuous functions, 260-271 variable, 266 Continuum, 212 Convergence, general principle of, 235, 240, 244, 257 Convergence of infinite products, 486 Convergence of sequences, 231, 235, 244, 257 Convergence of series, 247 et seq. binomial series, 263, 328, 334
INDEX comparison test, 250 hypergeometric series, 328 (for other tests see under the respective proper names) Cube roots of unity, 65 Cubic as sum of two cubes, 182 Cubic equation, 85 auxiliary quadratic, 182, 184
Cardan's solution, 180 character of roots, 180 functions 0, H, and the discriminant J, 179 irreducible case, 180 roots as power series, 469, 472 Curve tracing, 474 et seq. Cusp, 474 Cyclic expressions, 46
D'Alembert's test for convergence, 252 Dedekind, definition of irrational, 201 theorem on section of system of real numbers, 211 De Gua's rule, 90 Do Moivre's theorem, 59, 63
De Morgan's and
Bertrand's test for con-
vergence, 330 Derivatives of polynomials, 273 x n 309 of a x , log x, Descartes' rule of signs, 88 ,
Determinants, 119 bordered, 138
Expansion off(x + k), 35, 290 Exponential curve, 309 function and series, 19, 306, 312 inequalities and limits, 232, 306, 310 values of sine and cosine, 313
Exponential theorem, 307 Expectation, 514 Eudoxus, theorem of, 13 Euler's constant, 311 solution of biquadratic, 199
theorem on polynomials, 301 Factorial n, powers of primes in, 9 Factors of cos mj> - cos nO t 68 of x n l, 64 of x 2n -2x n coaO + l, 68 Fer mat's theorem, 424, 433 Euler's extension, 425, 437 Ferrari's solution of biquadratic, 189 Finite difference equations, 363, 367 Fourier's theorem on position of roots of
an equation, 454 in connection with Newton's method, 457 Function of a function, 271, 278
rule
Fundamental laws of arithmetic,
13, 53,
of order, 12
et seq.
Gauss's test of convergence, 327 General principle of convergence, 235,
expansion, 132, 137
reciprocal, 136
symmetric, 138-140 use of remainder theorem, 126 Differences, method of, 106 Differentiation, 273, 276 Dirichlet's test for convergence, 331 Discontinuous functions, 269 Discriminant of biquadratic, 188 of cubic, 179 of quadratic in x, y, 30
Displacements and vectors, 70 Distributions, 494, 496
of,
240, 244, 257 Generating function, 362 Geometric series, 248, 262 Gradient of tangent, 274 Greatest term in binomial expansion, 345
Hessian of cubic, 183 Higher derivatives, 280 Highest common factor, 40 ^Homer's method, 461 319 Hyperbolic functions, Hypergeometric series, 328
Imaginary roots of equations, number of, 89, 90
Distributive law, 13 Division (mod n), 432 Division transformation, 24
number and sum
Errors in excess and defect, 15 Exclusive events, 509
54
minors and cofactors, 193 product of two, 134
Divisors,
583
6
Double points on curves, 473 Double roots of equations, 42 Elements of
set, 14 Endless decimals, 16, 203 Equation of a plane, 154
Equations, cubic, biquadratic, 85 equivalent systems, 149 numerical, 447 rational roots, 92 symmetric functions of the roots, 95
65, 82
Implicit functions, 467 Independent events, 511 Indeterminate equations, 415 Indetermination, limits of, 239 Induction, method of, 9 Inequalities, fundamental, 216, 219
A.M.^Q.M., 221 exponential, 239 logarithmic, 306, 310 Infinite continued fractions, Infinite products, 485
390
derangement of factors, 488 expansion as series, 488
INDEX
584
sums of powers of roots of an equation,
Infinite sequences, 15
Inflexion, points of, 282, 473 line of inflexions, 484 Integration, 289 Interdependent events, 512
297 upper limits to roots, 446 Node, 473
Intermediate convergents, 402 Interpolation, Lagrange's formula, 379 Interval, 210 of convergence, 333 Inverse functions in general, 83, 284 circular functions, 285 hyperbolic functions, 321 Inverse probability, 518, 528 Inversions, 48 Irrational numbers, 12, 17 Dedeldnd's definition, 201 Irrational indices, 209
Operators A, E, D, 373, 378, 382 cos 6 + 1 sin 6, or E(id), 75 Order and weight of functions, 298
Rummer's
test for convergence, 306
Lagrange's interpolation formula, 379 theorem on divisibility, 426, 433 Least positive residues, 421 Leibniz'
theorem on differentiation of
product, 383 Limits, exponential and logarithmic, 232, 306, 310 of a function, 228, 229 of indeterrnination, 239 of sequence, 15, 16 to roots of equation, 90 Line at infinity, 152 Linear congruences, 429 Linear difference equations, 369
Logarithmic curve, 309 function and series, 306, 310-316 Logarithms, 210, 317
Maclaurin's theorem, 291 Magnification, 80 Matrices, 135
Maxima and minima, by
five sides,
178
of seventeen sides, 175 Power-series, 332 criterion for identity, 334 Prime and composite functions, 42 Primitive roots, 438, 444
Principal value of amplitude, 56 of inverse circular functions, 285 of inverse hyperbolic functions, 321
Principal nth root, 207 Pringsheim's theorem on convergence, 257 Probability, 508 of causes, 518
geometrical methods, 522, 526 Product of consecutive integers, 7 Lagrange's theorem, 426, 433
Quadratic function of
.r,
y,
29
Raabe's test for convergence, 326 Radius of convergence, 333 Ratio and proportion, 212 Rational fractions, 100, 357 integral functions, 23 roots of equations, 92
differentiation,
inequalities, 223
Mean
value theorem, 288 Minors of determinants, 123, 132 Modulus of complex number, 56-59 Monotone sequences, 230 Multinomial theorem, positive integral index, 35-37 real index,
Pfaffian, 141
Polygon of
Real numbers,
280
by
Partial derivatives, 299 Partial fractions, 100, 302 Partition of numbers, 500 Perfect numbers, 11
354
Multiple roots of equations, 82 Multiplication of series, 335-338 Abel's theorem, 338
E(x)xE(y)=E(x + y), 336 Merten's theorem, 337
Newton's approximation to >/l 4- x, 348 approximation to roots of an equation, 456 method of divisors, 92 parallelogram, 479
18,
201
Reciprocal equations, 168 determinants, 136 Rectangular arrays, 135 Recurring continued fractions, 392, 409 decimals, 439 series,
360
Relative error, 15 Remainder theorem, 26 use in determinants, 126 Residues, associated, 425 of terms of A. P., 8 of terms of G.P., 436 Reversion of series, 467 Rolle's theorem, 287
deductions from, 295 Roots of equations, number position
and character
of,
81
of, 87,
Roots, of complex numbers, 62 of i 1,64 of congruences, 428 Rule of Sarrus, 128
448, 454
INDEX Scalar quantities, 71 Scale of relation, 360 Section of rational system, 200 of real system, 211 Sequence, definition of, 15
bounds, upper and lower limits, 237, 239 general principle of convergence, 235,
244 monotone, 230 Series, approximation to sum, 260 of complex terms, 261
summation
of,
106
Taylor's, Maclaurin's, 290 Series for cos 6, sin 0, 313 for cosw0, sinn#, 61, 360 for cos n 0, sin n 6, 62 for Jog c 2, 312 for tan- 1 .*-, 294 Series summablc by the binomial theorem,
351 exponential series, 3 1 4 logarithmic series, 317 n r in terms of
,
...
T 6(6
+
(a
1). ..(&
+ r-
1)
+ r-l)
,
16),
106
110
Simple continued fractions, 391 Smallness, orders of, 14 Skew- symmetric determinants, 138, 140 n 1, 171 Special roots of x n 515 Stirling's approximation to |
Sturm's theorem, 448, 452 Substitutions, 47 Successive events, 515 Sums of powers of roots, 297 of two squares, 407, 408
585
Surds, 208 expressed as continued fractions, 401, 427, 435 JP, 7, 8 Symbols, I(r/y), ,
[n,
0(x), 14 t as an operator, 75 l /- (*),83
Symmetric continued
fractions,
406
determinants, 138
Symmetric functions, 44 of roots cf equations, 95 order and weight of, 298
Synthetic division, 24, 84 Systems of equations, linear, 149 of any degree, 160
Tangent to a curve, 274 Tangents and asymptotes, 473 Taylor's theorem and series, 290 for polynomials, 300 Testimony, value of, 521 Test-series for convergence S\jn'p 251 Z'l/n(lo g n), 325 Tetrahedron, volume of, 159 Three planes, 156 Transformation, Z = (az + b)f(cz+d), 80 Transformation of equations, 82-84 Transpositions, 44 ,
;
Tschirnhausen's transformation for cubic, 186 for biquadratic, 195
Undetermined
coefficients,
28
Vandermoride's theorem, 340 Vectors, 70-76
Weierstrass' inequalities, 216 Whit worth's theorem, 499 Wilson's theorem, 425, 433
<&
= 0,
INCONSISTENT EQUATIONS (ii)
//
(a 1 62 )
= 0,
if
(ai& 2 )
= 0, we
For
151
the equations are inconsistent unless also
have w x 62 - u2 b
(b^) =0 and
s
Hence the equations cannot have a common solution unless (b^) and c a ( i %) are
both
zero.
If (a 1 62 ), (ft^), (c^) are
(iii)
same, and there are infinitely For in this case
all zero, the
many
two equations are one and the
solutions.
and
u^a^u^av Hence any solution of one equation is a solution of the other, or else an a is zero and a 6 is zero. The latter alternative cannot occur. For example, suppose that a^O; then by hypothesis fc^O and a 2 61 = a 1 62 = 0, therefore a2 = and b2 ^0. Hence the equations are one and the same. uj) 2 =zuj)i
The case in which
(iv)
the equations are inconsistent
may
be regarded from
another point of view.
Suppose the (6^2)
-
bja
We
&2/a
finite
;
y to vary in such a way that (a^-^O while then #->
equality.
= say that in the limiting case when (a^) 0, the equations are - fe a l or - 62 a2 x by infinite values of x, y which are in the ratio
shall
satisfied
(3)
coefficients of x,
and (c^) remain l9 2 tend to
:
:
Ifa^x + b l y =
and a 2x + b<$ = 0, then
either
x = 0,y = Qor
else
.
(a^) = 0.
Conversely, if (afi^^Q these equations have solutions other than (0,0), they are consistent.
i.e.
This
is
a particular case of the preceding.
The following Ex.
1.
=0
=0 or
and (a x c2 ) ^=0, then (6 x c 2 ) (o^) satisfy the equations a x 6 2 aj} { =0,
//
For a lf a 2
results are required later. else
a x =0, a a =0.
a^ - afa ~0.
are not all zero and Ex. 2. // (a 1 6 8 )=0, (6 1 cs )=0 and (c l a 2 )=0, where a lf b lt #2 ^2> c a are not all zero, then of the pairs (a l9 a 2 ), (b l9 6 2 ), (c lf c2 ) there is at least one in
q
which neither quantity is zerb. For suppose that fl^O, then 6 t =0,
if a a =0 we should have o^ =0, a^ =0 which the contradicts =0, Cg hypothesis. Hence, if a^O, then
;
Ex. 3. Let A =(0^3) and l& A\> B oe the cofactors of a lt 6j ..... A =0 and CB = Q, then either Cl =0, C2 =0 or else A B =0, B3 =0. For we have a l C l 4 a2 C 2 = 0, &!
and therefore
Prove that if
^=0, C2 =0, or else (6 1 c 2 )=0, (c 1 a 8 )=0, i.e. ^4 3 =0, B^-0. Or thus: Using identities of the type (^^3)= Ja t it will be seen that (OjBa), (C 2 A 9 ), (C 2 B3 ) are all zero. Whence the result follows. therefore either
,
(Ci-4 t ),
LINE AT INFINITY
152 4.
Three Linear Equations.
Consider the equations 0,
Let ^^(a^Cg), and
let
.4^
B
...
l9
be the cofactors of a x b l9 ,
(1) //"
For on account of the
.
J = 0.
identities
+ a 2 C 2 -f o3 C3 = 0,
a l Cl
C^ +
we have
...
Hence, if the equations are must have
(7 2
w2
satisfied
by the same values
of x, y,
we
AQ.
= If A
Ae equations are consistent and have an unique solution except when every C is zero ; in which case each of the equations (assumed to be distinct) is inconsistent with either of the others. (2)
For by the preceding, u l C l + u 2 C 2 + w3 <73 = 0. (i)
// no
C
is zero,
the equations are not independent.
= and w 2 = 0, w3 = are (ii) // CjL-0, C 2 7^0, C 3 =Q, then u 2 C 2 4- u 3 C 3 the same equation. In both of these cases, since C3 ^0, the equations w 1 = 0, w 2 = have an unique solution, which is the solution of the system.
= 0, C2 = 0, C3 9^0, then w 3 ==0. This case is excluded. (iv) // every C is zero and the equations are distinct, each is inconsistent with either of the others. In this case we may say that the equations are (iii)
7/C'1
satisfied
5.
by infinite values of
The Line
say that the
x,
y which are in
the ratio
For the sake
at Infinity.
of
- 6X a r :
complete generality we
'
equation/
Ox + 0y-hc = represents the line at infinity. w = meets the line at infinity
The is
where
c^O,
ideal point in
which a straight line on that line.
called the point at infinity
Parallel straight lines are to be regarded as meeting on the line at infinity.
For the
lines
w = 0, u + k = Q
intersect
on the
line
Ox-f Oy
+ k = 0.
These
conventions enable us to include various special cases under one heading. Illustration.
// u 1 =0, w 2
and A =0.
then the necessary at infinity) is
0,
u3 =0
(as above) are the equations to three straight lines,
sufficient condition that they
may
meet at a point (which
may
be
SYMMETRICAL FORM 6.
Two
Equations
in
Three Unknowns.
Consider the equations
^O,
........................... (A)
= 0,
........................... (B )
excluding the case in which a l9 b l9 (1)
Of the
three
153
(0^), (^2)5
(
or a 2 , 6 2 , Cg are all zero,
Cj
c i a a)>
that at least one,
oppose
say (a l b 2 )
>
is
not zero, then the equations are equivalent to
- %a 2 + u2a l = 0,
Wj&jj
~ w^i ^
which are the same as
-(aA)y + (Cia2 )z=(a ld2 ), -(b l c 2 )z +
........................... (C)
(a l b2 )x^(b l d 2 ) ............................ (D)
z, there are definite values of x and y, and (c^) are not both zero. If (6 1 c 2 = and (c 1 a 2 ) = 0, then x and y have definite values and z may have any value. (In fact, we have ^ = 0, c2 = 0.) Thus in every case, by Art. 3, Ex. 1, there are infinitely many solutions.
Corresponding to any value of
provided that
c (fc 1 2 )
)
(2)
(a v
Suppose 2 ),
(b ly
quantity
is
that
(a^), (b^), (c^) are
all
(See Art. 3, Ex. 2.)
zero.
Also
Then
zero.
6 2 )j (Cj, c 2 ) there is at least one, say (a l9
of the pairs
a2 ), in which neither
we have
t^aj-wg^ss -((*&) ............................... (E) Thus the equations are of the form 1^ = 0, + ^ = 0, where k~(a l d2)/a 2 and they are inconsistent unless k = 0.
^
(3)
In particular
it
follows that the equations
a i x + b^j are always consistent are equivalent to
y
(i.e.
-f
cz = 0,
a 2x
+ 6gt/ 4- c^ = 0,
they have solutions other than
(0, 0, 0)).
They
^/(b^) =y/(cl a 2 )=z/(a 1 b 2 ),
provided that no denominator
is
zero.
=0, they are equivalent to x = 0, y/(Cia 2 )=z/(a l b2 ). = 0, then x = 0, y 0, and z may have any value. (6^2) =0 and (0^2)
If (6^2)
If
(4) If (a,
j3,
y)
is
any solution
of (A), (B), these
may
be written
which are equivalent to
(x-a)/(V2 ) = 2/-^)/(c1 a2 ) = (z-y)/(aA) (
provided that no denominator
is
=0, they are equivalent to
If
=0 and
(0^2)
= 0,
................. (F)
zero.
If (6^2)
(6^)
)
then x=oc,
y==)8,
and
z
may have any
value.
INTERSECTION OF TWO PLANES
154 NOTE.
Much
Equations
labour can be saved by the following considerations. unchanged by the simultaneous cyclic substitutions (Oi6 1 c 1 ), If then we make these changes in any equation derived from (A) and
(A), (B) are
(a*6 2 c2 ), (xyz). (B),
the resulting equation will also hold.
from
For example, equation (D) can be derived
(C) in this way.
The Equation
7.
to a Plane.
The equation
(1)
represents a plane unless a, 6, c are all zero. For the sake of complete generality we say that the
Ox + 0y + 0z + d = represents an (2)
to
Two
'
*
equation
d^O
where
ideal plane called the plane at infinity.
= 0, w2 = be the equations Eeferring to Art. 6, let In general these intersect in a line, and we say that the
^
Planes.
two
planes. equations to the line are 1^=0, u2 Q. The equations may be written in one of the forms described in Art. 6, (4), where (a, /S, y) is any point on the line. The equation to any plane through the line u 1 ==0, w2 = is of the form
u l + ku 2
t
where k
Parallel Planes.
is
a constant.
(a^), (b^), (c^) are
If
zero, the planes
all
^ = 0,
u2 =
do not intersect, that is to say they are parallel. Hence the equation to any plane parallel to the plane ^==0 form ul + k^0, where A; is a constant. All this follows from Art.
We
is
of the
6, (2).
therefore say that parallel planes intersect in a line on the plane at
infinity.
Ex.
1.
The condition
whose equations are
that the line
(x-a)ll = (y-fllm = (z-y)ln
may
be parallel to the plane
is la
+ mb + nc == 0.
If the line
// also au + bp+cy + d=Q, then the line is in the plane. meets the plane at (x, y, z) and each fraction in (A) is denoted by
where
2.
The condition
~~
may and If
meet in a point
if
(x,
y
lll
-a ) = 0, /
m /
y
then
~~
n
z)
= m/m and
which
satisfies
value.
the last equation unless
Hence the
results follow.
whose equations are
~
'
m'
I'
- a') (mn' - m'ri)
(
the lines are parallel if
the lines meet at
r exists
may have any
thai the lines
I
r,
+ mb + nc)-faa
r(la
If /a-fra6 + wc=0, no value of o + bp -f cy + d = 0, in which case r
Ex.
............................... (A)
+(f$- fi') (nl'
~ n'
'
- n'l) + (y y') (lm' I'm) =0,
/
n/n'. f
(x -a)/J=r, (x-
mr~wV
/
/
4-(j8~^ )=0,
,
we have
nr-nV-f (y -/)=0.
CRITERION FOR UNIQUE SOLUTION These equations are
by the same
satisfied
a -a'
P~p' y-y'
(finite)
J
I'
m
m'
n
ri
values of
155
r, r' if
=0,
provided that (mri), (nl') (lm') are not all zero : but if these three are all zero, the values of r, r' are infinite. In the latter case equations (A) are satisfied by infinite values of x, y, z and the lines are parallel. 9
Three
8.
Unknowns.
Three
in
Equations
equations i2/
~*~
ciz
Consider the
+ d-i = .(A)
Let A==(a 1 b 2c3 ) and Also
A
A l ^(d 1 b2c,)
let
On
let
B
19
ly
...
be the cofactors of a l9 b l9
A 2 ^(a 1 d2c3
y
)
)
...
in A.
J 8 -(
account of the identities
.
we have and and
Suppose
(1) it is
u l B l -i-u 2 B2 + u^B3 ^ Ay + A 2y
similarly,
that
given by
We
........................ (C)
A^O. The last identities show that if a solution exists 3= -4/J, y= -'A^/A, z=-AJA .................... (D)
,
can show by actual substitution that these values of #, y, z satisfy Or we may use the following method, in which every step
the equations.
is reversible, so that verification is unnecessary.
Because
Suppose that
zero.
Because
that
2 (A^C^) =A ^0,
is
A^^Q,
to
A^gC^^O,
so that no one of
U2 = o,
w3 = 0,
u^ = 0,
* It
L
2>
not
Ax - - A r
^ = 0, Jy=~J
essential to show that suffixes p, q r exist, This step has been omitted in the text-books. is
19
is
zero.*
B2 ^0 and C3 ^ 0, the equations are equivalent
Hence the equations have the unique zero.
(A^B^C^
A B C3 is
the equations are equivalent to
In the same way, because to each of the sets,
therefore at least one term of
t
2
;
%==0, w2 = 0,
solution,
(
Az= - J 3
- A-JA, -
such that none of A$ t J5a See Chrystal, Algebra, Vol. I, p. 360.
all different,
.
,
Cr
B.CJU
is
THREE PLANES
156 (2)
= and A v A 2 A 3
If A
,
are not all zero, the equations are inconsistent.
For example, suppose that J 1 ^0, then
since
UiAi + UiAz + UsA^Av it
(E)
follows that the equations have no common solution. If we suppose the coefficients to vary so that J->0,
A v J2 J3
and at
least
one of
tends to infinity. d in with the d. d large terms, we see that in Neglecting l9 2 3 comparison general the following ratios tend to equality ,
remains
then at least one of
finite,
x, y, z
,
:
A :B :C
x:y:z,
1
l
A2
lJ
:
J5 2
:
C2
,
A^
:
B3 C3 :
.
In the limiting case when A=0, we say that (he equations have a solution (x, y, z) such that at least one of x, y, z is infinite and
x:y: z = A l B l
C^A
:
:
:
2
2
:
C a = 4 3 #3 C8 :
:
.
A l9 J 2 A 3
are all zero, the equations are not all independent and there are infinitely many solutions ; or else each of the equations (assumed to (3)
If A,
,
be distinct) is inconsistent with either of the others.
u^A^ + u 2 A 2 + u^A 3 =
For we have
A
If no
(i)
A =
//
(ii)
l
9
(F)
the equations are not independent,
is zero,
A^O, ^
3
7^0, then u 2 A 2 + u B A 3
= Q; and w 2 = 0, w3 =
same equation.
are the
the equations M 1 = 0, w 2 = solutions which are solutions of the system.
In both cases, since
many
A.^Q,
have
infinitely
A = 0, A 2 =-Q, A^Q, we have w 3 sO, which case is excluded. If, in the above, every A is replaced by the corresponding B or If
(iii)
(iv)
L
C,
Therefore the only remaining case is that in which minor is A zero, and then every two of the equations are inconevery of or they are one and the same. sistent similar results follow.
;
Ex.
We
1.
// a,
6,
c
are unequal, solve the equations
have 1
1
1
1
1
1
a
b
c
d
b
c
a2
62
c2
b*
c
x
and the values
9.
of y, z are obtained
Three Planes.
by the
(i)
(ii)
which
If
A^O
(d-b)(d-c) (a-6)(a-c)'
2
cyclic substitutions (a6c), (xyz).
Referring to 8, let
equations to three planes, then
x+y+z~I,
^=0,
w 2 = 0, Wg=0 be the
:
they meet in the point given by the equations in Art.
// A =0 and A l9 A 2 J 3 ,
8, (1).
are not all zero, two of the planes meet in a line
is parallel to the third plane.
FOUR PLANES (iii)
// A,
J x J 2 ^s >
,
157
are a M zero the planes have a >
common
line of inter-
section or they are all parallel.
Proof of
Suppose that
(ii).
Jx ^
We
0.
have
J 1 ........................... (A) A l ............................. (B) ,
On account of (A), at least one A is not zero. Let A l = w2 0, w 3 = are not parallel and, on account of (B), section does not meet the plane % = 0. Observe that
which
all the
planes are parallel
is parallel to the line,
The truth
of
(iii)
follows
Hence we conclude
0,
then the planes
their line of inter-
to the line
w 2 = 0, w3 = 0. from Art.
that, if
or else they hare a
infinity,
^
8, (3).
J=0,
common
a point at which may be
the planes either meet in line of intersection,
at infinity.
Four
10.
Equations
in
Three
Unknowns.
Consider the
equations
u^a^x + b^j + c x z -f d x =0, = 0, = 0. Let
A = (^i^a^) an ^
^ ^i>
BI>
^ e ^he cof actors
If the equations are consistent, then For, as in Art. 4, (1), we have
(1)
of a l9 b
...
.
A =0.
u l D 1 + w 2 D2 + w 3 J53 -f ^4^4 = d (2)
//
A =0,
^Ae equations are consistent
when every D is
and have an unique
solution except
zero.
This follows from the identity == 0.
D
is zero, the equations are inconsistent, but they are // every as having a common infinite solution, unless every minor of regarded
(3)
to
be
A
is
which case there are infinitely many solutions. explained with reference to four planes in Exercise XVII, 17. The methods already described apply to systems of linear equations con-
zero, in
This
is
necting more than three variables,
and lead to
similar results.
SIMULTANEOUS EQUATIONS
158
EXERCISE XVII
SIMULTANEOUS LINEAR EQUATIONS 1.
Solve by means of determinants
:
x-y + z~Q, 2x 4- 3y - 5z = 7, 3x - 4y - 2z = - 1.
(i)
(ii)
y+z ~
x
-
w
1,
3y- 22 4-3^ 2.
With regard
to the equations
4x 4- 7y - 142 = 10,
show that the
first
l(2x
and
find
3.
I
:
m
:
n.
2# + 32/-4z= -4,
can be expressed in the form 4-
3y - 4z) + m (x + y + z)
= n,
Hence show that the equations are
inconsistent.
With regard to the equations 4x + ly- I4z = -24, 2x + 3y-4z~ -4, first can be put in the form
x + y + z6,
show that the and
find
I
:
m.
4. If x, y, z
Hence show that the equations are not independent. are not
ax
all
-f-
zero
where
-f-
63
+ c 3 H- 3a&c ==
Having given the equations x cy + bz, y x, y, z are
not
az + cx,
-a 2 )^i/ 2 /(l -b 2 )^z 2 /(l -c 2
x*j(l
Having given the equations a = b cos C c cos J5, 6 c cos -f-
deduce the following
A -f-cos 26c cos A 6
and 8.
2
Having given that =1 a? 4- y 4-
prove that
.4
+a
cos
C
f
c
,
).
=a
cos
b cos
jE? -f-
:
cos 2
7.
bx + ay
z
prove that
all zero,
and 6.
bx -f cy -f- az ~cx -}~ ay + bz = 0,
a3
prove that 5.
and
by + cz
2
jftH-cos -he 2
2
- a2
(74-2 cos ,
^4
a/sin J.
cos
B cos (7
=6/sin
J5
1,
=c/sin 6
ax + by + cz=d, a zx 4- 6 2t/ 4- C 2z = cZ 2 a 3 x 4- 6 3 ?/ 4- c*zd 3 -(d-a)(d- b) (d - c) a*x + b*y + c*z= d* ~(d -a)(d -b)(d -c)( ,
If the equations
cy
4-
bz
6~c
__
az c
-f
ex _ bx + ay _ ax
a
4-
6y 4-
a -b
are consistent, prove that either
a + b + c~Q .or (b -c)(c -a)(a -b)=abc. [Put each fraction equal to k and eliminate x, y, z, &.]
f
,
.
^4,
APPLICATION TO GEOMETRY 9. If the three
that
equations of Art. 8 are distinct and
J 2 =0 and J 3
[Since therefore
unless b 1 :
are consistent
if
Cj
=6 2
:
c2
=63
:
A =0 and Ji=0, show
if
ca .
d=Q, A l A^: A 3 =B l Bz B3 and since d l A l +d< J 2 = d l Bl + dpB2 + d3 Bz =Q unless ^4 2 ^ 3 are all zero.] :
:
10. If the equations
Also
:
.159
^
,
xby + cz + du,
t
,
y~cz + ax + du,
uax + by + cz z=ax-\-bydu, and no one of a, 6, c, d is equal to - 1, then
a= -
1,
then one of 6,
c,
d
is
equal to
1.
[Ex.
XVI,
10.]
Using the notation of Art. 8, show that if t* l =0, u 2 =0, w 8 =0 are the equations to three straight lines, the coordinates of the vertices of the triangle formed by 11.
them
are
triangle 12.
(4, |),
if
13.
(# 4 ,
2/4,
and that the area of the
|), (^, |),
i JV^A^-
is
Prove that
if
c^O,
=
valent to
and
,
(
^=0,
the equations
w 2 = 0, u^=0 of Art. 8 are equi-
X
A^
deduce the solution in the form already given.
Show
that the condition that the points (x l9 y l9 2 X ), (x 2 , y29 z z ) be in the same plane is
24 )
9
,(x^ 9
t/ s ,
may
i
y\
~ "i
9
2/3
23
1
4
2/4
Z4
1
.
r
n ^*
i
*
14. Using the notation of Art. 10, consider the lines whose equations are U!=Q, 1*2=0 and w a =0, w 4 =0. It is required to find the equation to a plane which passes through either of these and is parallel to the other. therefore the required planes are [We have u 1 D 1 +u 2 D 2 -\-u 3 D 3 -^-u t ~\=0 and
D^A
15. Consider the lines
whose equations are
X-aS
S-ql/-j8Sm n Show
9
m
that the equation to the plane which contains the
to the second
first line
and
is
parallel
is
Deduce the condition that the [The required equation and aV + bm' + en' =0.]
is
may meet in a point or be parallel. + b (y - ft) + c (z - y ) = where al + bm + en =
lines
a (x - a)
16. Using the notation of Art. 10, show that if ^=0, % 8 =0, t* 8 =0, w4 =0 are the equations to four planes, the vertices of the tetrahedron bounded by them /'A
are the points
B
f^r> y^
the tetrahedron
C
\
jr)
>
is **-A
/A
B
(jr*
ff
z
ID
>
C
\
77
)
>
e ^c
>
an(^ * na ^ tn
volume of
DDD
shown in works on Coordinate Qeom&try that the volume of the tetrahedron whose vertices are (x l9 y l9 Zj), etc., is one-sixth of the determinant in Ex. 13.] [It is
EQUATIONS OTHER THAN LINEAR
160 17.
be the equaUsing the notation of Art. ]0, let t^ 0, w 2 0, w 3 0, w 4 and suppose that J = 0, then If at least one Z>, say D 19 is not zero, the planes meet at the point
tions to four planes, (i)
(ii)
Suppose that every
D is zero,
then
(a) If all the fninors of A are not zero, the four planes are parallel to a certain line, that is to say, they have a common point at infinity. (/?) If every minor of A is zero, the planes have a common line of intersection or they are all parallel.
For example, if and we consider the planes u l9 u 2 i/ 3 Since D 4 =0, two of these, say u l9 u 29 meet in a line which is parallel to the third. Also, considering the planes, u l9 u. u^, since /) 3 =0 the plane w 4 contains or is parallel to the line u l ~0, M 2 =0. Hence all the planes are parallel to the line
A^O
.
,
9
x\ (6^2)
Given the equations a 2 c 3 + a 3 c 2 - 26 2 6 3 = 0,
18.
a 3c l
= yj (Cjda) - zf (a^). -f
iC 3
-
26^3
= 0,
a tc 2
+ a^ - 2bJ) 2 = 0,
prove that
gl
[Prove that
L lf
20. If
Jf ,,
a;,
~ 2&1
^ 2 (i
Cl
=
c i-^i 2 )
-.
X= N
prove that P
where
=
etc.,
are the cofactors of
y are connected with J,
1'
N X-N Y=L -M
given by
t
l
1
l lt
m
lt
etc., in
the determinant
by the equations of Ex. #,
considered in detail in another volume.
The question of Here we illustrate
various methods of dealing with systems of special types. (1)
Systems which are symmetrical with regard
Ex.
1.
Find
to
x
9
y, ....
the rational solutions of
x *+ y * +558=50,
(y+ 2 )(z +*)(*+#) = 14,
3 ir'+^+z - 32^2 = 146.
If
x=p, Syz~q x\jzr t
2 Z>
y
the given equations are equivalent to
-2g = 50,
are
^2 .
Systems of Equations of any Degree. is
Y
y are given by
= m v -f
elimination
and X,
t,
prove that the corresponding values of
11.
19,
pq-r = U,
3
p -3^-146;
SPECIAL TYPES OP EQUATIONS whence we If
-
- 23,
r
find that
p =2, then
Thus
3
x, y, z
q
~
p
I50p -f 292 =0, - 60 and x,
~
2
p
giving
-
or
161 1
^147.
the roots of
y, z are
0*- 20 2 -230 + 60=0. may have the values 3, 4, - 5 taken in any
order,
and there
is
no other
rational solution.
(2)
A
solution
sometimes be discovered by considering the properties of
may
a particular determinant. Ex.
Solve
2.
x 2 - yz = a,
y
2
zx = b,
z2
- xy
c.
Consider the determinant
x
z
z
y x
y
The
cofactors of x 9 y,
if (x, y, z) is
be
.
z
are yz - x", zx ~ y 2 , xy - z 2
in the top line,
z,
/.
Hence
3 - 3 - 3 z 3xyz - x y
y x
2
2
(zx~y )(xy-z )-(yz-x
z 2 )
;
^Ax.
a solution,
-a 2
Ax, and similarly, ca
-b 2 =Ay
a, b, c, and adding, 3abc - a 3 - 6 s - c8 = A (ax
+ by+cz)
f
ab
-c 2 ~Az.
Multiplying these by
- A2
.
Therefore the only possible solutions are given by ~~
a 2 -be ~~b 2 -ca "c 2 -ab It is easily verified
in
which there
(3)
A
Ex. 3.
is
no solution unless
+ 6s -f c3 - 3a6c)
*
are solutions unless a 3
by substitution that these
a, b, c are all zero.
change of the variables. Solve
x(x-a)=yz, Let x
3
*J (a
l/X, y
1/7, z
= l/Z,
X -YZ 2
a
z(z-e)=xy.
y(y-b)=zx,
then the equations are equivalent to
Y*-2 o
c
Therefore, as in Ex. 2,
where the
A
first
a 2 -be,
B=b 2 -ca, 0=c 2 -ab,
and k
is
to be determined.
Substituting in
equation \ k_(k__ a
&_
A\A
k =0, giving the solution
therefore, either
or
k (BC
~A
with similar values for
2
8 )= a ABC, giving k (a
y, z.
(0, 0, 0),
+ 63 -f c 8 - 3a6c) = - A EC
;
SOLUTION BY SUBSTITUTION
162
In
(4)
In equations involving
useful.
Ex.
the case of two variables x,
4.
x, y, z,
we may
solution
t'x.
f
is
+ bt + c2) -. x {i + mj),
x^Q,
if
the solutions are given
x
x 2 (a'+b't + c'P) -x(l' + m't).
by
+ mt)/(a + bt + d 2 ),
(l
2 (a + bt + ct
where
)
(V
y
tx,
2 m't) = (' + 6'e + c'J (Z + mi). = become cy* my, c'y 2 m'y,
-f
)
If x~Q and y^O, the equations - y m/c. In this one value of < is infinite.
solution
7/a
(5)
sometimes
+ bxy -f cy2 = lx + my, a'x2 + b xy + c'ya = I'x -f m'y. = tx the equations become (0, 0), and if y
x *( a
Hence
write
is
Solve
ax*
One
y = tx = y tz, z
the substitution
y
x = x 1 ,y = y l
,
/.
cm'-c /m =
and
can be guessed, then a suitable substitution
...
is "V
x ~~~ Ju\ *&r. 5.
One
i
-A-
"V
y ~~ y\
)
*
i
j
/S'o/ve
solution
is (6 -t-c, c -fa, ar
a
+ 6), and
if
we
write
= 6+c + JT, y=c+o-l-7,
=
2
the equations become
Hence
if
none of the
three, 1
F
X, F, Z,
+ |__1 Z""
a
I 9
Z
i# zero,
+ !___!
X~
Special cases.
If
This
JT=0,
gi ^es
7=0
I
L_^I
1
X+Y~~
9
;
c
1
2_1_1 X~a b
'
with similar equations.
we have
c'
the only solution, beside the obvious one.
and
Z^O,
then a=0,
6=0 and Z may
have any
value. If
JC=0, 7^,0,
Z^Q,
then 6=0,
c=Q and
7,
^ may
have any values such that
7Z + a(7+Z)=0. Ex.
6.
The
eight solutions of
x*
+ 2yz=a,
y*
+ 2zx=b,
z*+2xy=c,
.......................... (A)
are given by
where
-X"
and
By
=2a-6~c2v/(a2 + 6 2 -fc2 -bc-ca-ab)
k=>J(a + b+c). addition,
From where
2
(+i/ + z) 2 =a + 6+c,
/.
x + y + z-k.
the second and third of the given equations, by subtraction,
X~k~Zx.
SIMULTANEOUS QUADRATICS Adding the second and third
163
and subtracting the
of equations (A)
first,
we
easily
find that
^V
A/ -
3
and solving
for
X
z ,
we obtain the values
Another solution.
.
=6+c-a;
stated in the question.
Equations (A) are equivalent to
-A
(say),
2 2 2 (x + w y + ojz) ~ a + o>6 + co c = B,
(x+uy +o> 2z) 2 =a + oj 26 +o>c = (7, where
o> is
an imaginary cube root
of
1.
Therefore the eight solutions are given by
x+w zy+a)Z=*/B, root of A and so for *JB,
x + y + z=JA,
where *JA
From
is
either square
z
x+a)y+a) z=,JC
9
............... (B)
*JC.
equations (B), by addition, 3a?= N /^+Z,
where X=*JB+>JC, and therefore
Also,
from the second and third of equations
(B),
%c
and
(
These Ex.
7.
^^
results agree
Show
)
with the preceding
that, if the
if
we change the sign of X, which is permissible.
equations
(A) are consistent, then
abc
From
+ 2fgh-af*-bg*-ch*=Q ................................ (B)
equations (A), by multiplication,
ax *(b*y* + c 2 z*) + ...+...+2abcxzy*z*=Sfghx*y*z* ..................... (C)
6V + c z = (4/ 2 4
Also
2
- 26c) y*z2
,
with two similar equations. Substituting in (C) and dividing by x*y*z* (assumed not to be zero), the result (B) follows. Special cases. (i)
If
Considering the original equations
z=0, y^O, 3^0,
then
C2 a =0,
it will
be seen that
and c=0;
hence the equations become, by* = 2fyz, &y*=0, and
we have 6=0, and/=0.
=0, y=0 and 2^0, the equations reduce to the single equation c 8 =0; (ii) and thus do not constitute a system of equations. Thus, if the equations (A) form a system of equations with a solution other than (0, 0, 0), then the condition (B) holds. If
COMMON ROOT OF TWO QUADRATICS
164 Ex.
Prove
8.
that, if
x 2 + y 2 + z 2 - 2yz - 2zx - 2xy =. 0,
and no two of x
then, if
(b-c)
a, b, c are equal, z
y
2
xyz
"
(c- a)
2
(a~b
2
+ b 2y 2 -f c 2z 2 - 2bcyz - 2cazx - 2abxy
a^x 2
)
'
The given equations can be written
-ax + by + cz)
x(
+y-
Multiplying these by (x find that
-by + cz)
-\-y(ax
+ by - cz)
(ax
z),
-f
+ by -cz)=Q,
z(ax
respectively
and subtracting, wo
xy{y(a-b)+z(c-a)}=xy{z(b-c)+x(a-b)}; similar equations can be obtained. is zero, we may write
and two z, y, z,
y(a-b)+z(c Hence,
if
-a)k,
no two of a,
+x(a-b)=k,
z(b -c)
x~
are equal,
b, c
Hence, assuming
-
-~-
-
a) (a
A
~ 2
2
-a(b~c) +b(c-a) +c(a-b)* :.
none of
the three,
x(
k(b -c)
(c
whence also
that
j~
,
0)
ax -by - cz
(b+c)(c-a)(a~b)'
2 2 2 Zax(ax-by-cz)^X (b-c)(c-a)(a-b) 2a(b -c .
)
-A 2 (6-c)2(c-a) 2 (a-6 and the result in question follows. The consideration of special cases
is left
to the reader.
EXERCISE XVIII 1.
If (a,
ft), (a', j8')
are the roots of
u prove that
a^^^-^^ and consequently
common
root is
the necessary
E=0,
and
R = (ac'- a c) f
2.
Eliminate
x,
sufficient condition that the equations 2
- (ab' - a'b) (be' - b'c).
y from the equations
and show that the (a, 0), (a', j3')
roots of the resulting equation in z are aa', are the roots of
ax 2 + bx + c-0,
Show
have a
where
that this equation
a'x 2
a/3', /?
where
+ b'x + c'Q.
is
f a 2a' zz* - aa'bb'z* + (b z a'c + b /2ac - 2ocaV) z 2 - bcb'c'z + c 2c /2 = 0.
Prove also that the equation whose roots are a/a', from the last equation by interchanging a' and c'.
x
a//T,
/a
,
//T is
obtained
SYSTEMS OF EQUATIONS
165
Solve the equations in Exx. 3-28. 3.
* 3 + 3^ = 8,
4.
X 2y + xy 2 + x + y^3,
5.
x(l-y*)=2y,
6.
x + y + z = $,
7. (y
*2 8
'
xy+(x + y)=2.
+y
l)
y(I-x*)=:2x.
x 2 + y 2 + z 2 ^ 29,
+ z)(z + x)(x + y)=*, 9
[Put
2(x - l)(y -
(x
a;- 1
+
25 8
"
'
9.
10.
-=l6-xy,
^--^Q-xy. x
y
12 l^
f
^ ---
15. [(
a>
2
a z --y~
+
4.
i y -. _ -a z -x- a ?
l-p, and prove that
2 2 ==a
+ 2/"f z,
2 2:
2
4-a;
6-i-
- x 2 = 6 s - 7/ 2 yz + zx {-xy= a 2
+ y)(a? + 2)=o
c
2
1
X
,
[Put 18, ayz
V
y
b
-X(a-b) + a-b
z-f re,
- 22
.
etc.]
1
1
1_ d
A2
a
16.
1
B
+ (l-2/) 2 -(l-o:)(l-2/)(l-a: y),
a;l-A, y
[Put
i/
,
/,
1
13. (l~rr) 2
14.
x
-----4. y2 a y a a?
V
,
1 -
O
<2
,
z
11 C
x
c
~~
.
a
x= + by + cz= bzx + cz -f ax ~
19.
20.
_
= -_ = ..2
+ 0;
X4-7/-1
21.
22.
x(bc-xy)=y(xy~ac), xy(ay+bx - xy) ~abc(x + y - c).
-x)=by (z-x)(z-y)=cz. 2 xy(c-z)cz 2 zx(b-y)=by
23. (x-i/)(2:-2)=aa;,
24.
yz(a-x)=ax
2 ,
(y ~-z)(y
t
.
,
25. 26. 27.
28.
'
'
x+y=x' + y'=r,
112 112 = --.
-
x
y'
-,
-
y
y
\^e + 1 + *Jy~-2 = *Jx -1 + N/jMh2 = 3.
x'
x
and
ELIMINATIONS
166 29. If
3
5a?
5t/
-2
pr0ve
,
8
y+\=x,
that
and
solve
the
equations. 30. If
a
-}-^ o
=~+ x
=a + b, show
that either
+ y=a + b.
(X+Y + I)(aX + bY -a-b)=Q.]
[Puta^aJf, y=bY, zndshov? 31. Eliminate #, y, z
or x
-4-^4-1=0 b
a
y
--2)
from
---
-
and ax + by + cz-Q.
C
CL
x3 -y*~ y*-x*=a* and aM-y= -a, and a:, y are unequal, prove that 2 8 2 2 2 ~ (iii) 3a (s 4-t/ )= (ii) 3a -3a + 3a- 1 =0, xy = a*(a-l)l(2a-l), 4 4 2 3 3 = and (v) 3a(a; + t/ 4- l) = 2+4a 3a(z 4-y ) 2-a,
32. If (i)
(iv)
1,
.
yV + (y + z)
33. If
2
2
z
+
2
2 2 a: !/ 4-
then will
z
(z-fir)
(x
Ayz -
+ +
2
+ i/) 2
1
+ A,
A2^ = l4-A and Aa:y
-f-
x^y,
= 1 -f A.
111111111 __ __ __
34. If x, y, z satisfy the equations
__
__
i
x
y-\~z
__ y
_,
9
a
_._
i
z
_
__
_
i
9
+x
b
z
x-\-y
9
c
+ c a)=y(c-\-a-b)z(a + b c): and hence solve the equations. + by* -f cz 2 = 0, ayz bzx + cxy = and x3 + y3 -f z 3 -f Xxyz = 0, prove that
prove that x(b 35. If ax*
36. If a
-f-
+ 6-fc=0 and x + y + z = Q, prove
that
2 2 [Use the identities (6y + cz ax) = (bz + cy) ,
37.
Given that
a 2o: 2
and Sh
etc.]
-f-
+ c zz 2 - 2bcyz - 2cazx - 2a6a:y = 0,
b 2y 2
W 38. Eliminate x, y, z
from (
showing that the result
(x
+ y-z)(y+z~x)~bzx,
(y
+ z-x)(z+x-y)=;cxy,
is
abc~(a + b + c-4) 39. Eliminate x, y, z
41. If
a;
2
2
a?
4-t/ -f2
.
from
x + y-z=a, 40. Eliminate
2
a:
2
+
a t/
-2 2 ==6 2
,
x8 + y8 -2 3 =c 3
,
xyzd*.
and y from 2
= l,
+ 2y 4-32=3, a;
3o;
+ y4-2 = 3,
-8y-f52=
3.
prove that
CONSISTENT EQUATIONS Show
42.
167
that the equations
are consistent,
and
x y
find
:
:
z.
and x 2 - ayz ~y 2 - bzx = z 2 - on/ f (a; 2 + 1/ 2 + z 2 ), a 2 -f & 2 4-c 2 = 2a&c + l and z a /(l -a 2 )^2/ 2 /(l -6 2 )=z 2 /(l -c 2 ).
43. If x, y, z are not all zero
prove that
Show
44.
that
z=c is a solution of 3 3 px* + #y -f rz = &M/Z,
#
if
a, ?/=&,
then
xa(qb* ~rc 3 ), is
i/
= &(rc 3 -pa 3 ), z=c(pa 3 -gfc 8
)
another solution. 45. If
3/
7V
-(y + z-x)
*-
2
+ x-y) = -
(z
(x
+ y-z) and J-fw + n=0, show
46. Eliminate x, y from (fc~#)(c-t/)=a 2 , (c-a;)(a-i/)=& 2 , (a -x)(b
that
-y)=c
2 .
47. If the four expressions
u2 +
xz+zu), are equal, prove that each 48. If x lt x 2 ,
... Xf,
2 2 z 2 equal to %(x + y + z + u ).
is
are all different
and 2
prove that 49.
(i)
2
2
2
_
,
x1 2 + xs 2 ~(k*- 4k 2 + 2) a:^. from the equations x ax + by + cz + eft = 0, a a; 6'y -I- c'z -f rf^ =0,
Eliminate
t
x, y, z,
-I-
a
-
x showing that the result
&
-
-f-
c -f
-
d
y
- + 6' - + c- d' T =0, x y z
+ -=0,
z
x
a'
.
.
-f-
t
t
is
L MN + Lmn + Mnl + Nlm = 0,
m = (ca
f
where
l-(bc
)9
L = (ad'), (ii)
Use
this to prove Exercise
n = (ab'),
/
),
N=(cd').
M=^(bd'),
XVI,
25,
(iii).
L
I
[(i) LV }
Prove that
...
...
2
2
(x*-t )yz the unwritten fractions being obtained
(LMN). Now put x\
2 t/
,
z\
t
2
for x, y, z 9
...
.
v. ,
(y -z*)xt
by the 1
cyclic substitutions (xyz), (Imn), in Ex. XVI, 25, (i).
Let # = cos a-f-tsina, y = cos -f t sin j5, etc., where a, j8, y, 8 are the angles which the sides a, 6, c, d make with a fixed line. By projecting along and perpendicular to this line, prove that the four given equations hold.] (ii)
CHAPTER XI RECIPROCAL AND BINOMIAL EQUATIONS
A
Reciprocal Equations.
1.
reciprocal equation
which possesses the following property a root. Let
a,
a xn
then
+ a^"-1 4-
a nx n
. . .
+ art^x*" 1 +
a reciprocal equation
is
(A)
any
root, then I/a is also
+ a n _yX + a n = 0,
.....................
(A)
are the roots of
...
I/a, l/)3,
Hence
is
an equation
be the roots of
...
/J,
a.
If
:
is
Denoting each
of these fractions
+ a x x + a = 0.
. .
.
if
by
and only
if
we have
k,
an a so that
We
&=
1.
say that Reciprocal Equations are of the first or second types +1 or k= -1, /&aZ i5 according as a r = a n _ r or a r = ~a n _ r according as k shall
,
= /or r 0, 1, 2, Theorem
1
...
n.
TAe
.
reciprocal equation of the first
Let f(x)
=a
x n + ajX**" 1
of the first type,
xnf (
\3//J
-f
.
.
Therefore/(z)
(z is
2m + 1
.
-f
an =
be a reciprocal equation.
that of
// this
2m + 1. Then
+ 1) 4-a 1 x(a;
divisible
by x -h
2m ~ 1 1,
+ 1) +
and
if
may be m +a mx (x + 1) = 0.
the equation ...
(x)
+x =
is
is
is
written
the quotient,
x+
X hence $(x)
a
=/()
Let n be odd and equal to
= f(x) a
any reciprocal equation depends on type and of even degree.
solution of
a reciprocal equation of the
first
type of degree 2m.
TYPES OF RECIPROCAL EQUATIONS Iff(x)
=
Let n = 2w +
xn
of the second type,
is
-
/( x/
)
=
169
-/(#)
Grouping the terms as before, we can show that /(x)
1.
2m (-} = by x 1, and that, if <(x) is the quotient, then x with the same result as before. = ~a m Again, if n = 2m, since a r ~ -a 2 m-r> ^ follows that a m therefore a m = Q. Hence the equation may be written divisible
<
/(x)
=a
-
2
(x
Therefore /(x)
1)
+ a 1 x(x 2 - 2 -
divisible
is
equation
We the
of the first
if
,
and
+a m _ 1 m -1 (x 2 - 1) = 0. tf
<^(x) is
the quotient,
~
in every case the solution of /(x)
= (x)
...
2 by x -!, and
/
Thus
+
1)
is
type and
say that a reciprocal equation type and of even degree.
=
depends on that of a reciprocal even degree.
of
of the standard
is
form when
it is of
first
Theorem
The
2.
solution of a reciprocal equation of the first type
and of
2m
degree depends on that of an equation of degree m. Let the equation be -1 a x2m 4- fljX 2 4 ax + a = 0. 4. . .
Dividing by
xm
,
this
may
be written
1 \
Let x f
-^=z
x
and x r +
-
xr
/
_
=t/ r so that ,
1\2
1
,
u*=z
+ -) -2 =
~
\
2
;
, A .(A) .
then
-2.
x/
l
x+
Also
r
= 3,
and so on. Thus equation z 1 is
= ^ + -r ) + (x*~* -f -i-A + -^ ^xv V x 1 /) f\ xr 2 / 7
"-
wr = 2w
therefore
Putting
l
} (x'~ x/\
4,
...
in succession,
(A) can be expressed as an equation of degree
one of its roots, the corresponding values of xare given by
a?
2
m in z, and
if
- z^x + 1=0.
THE BINOMIAL EQUATION
170
The reader may
verify that
#*.
Solve 6s5
1.
One
root
-
is
w 8 = z8 - 8z6 + 20s4 - 16z2 + 2.
- 7z,
14z 3
1*4 - 33** - 33z 2
4- 1
Dividing by x +
1.
+ 1 1* + 6 = 0. we have 6x4 + 5z3 - 38x2 + 5x + 6 =0.
1,
2 Dividing by # and grouping the terms,
If 2
= 3; + -,
this
x
2
.".
Whence we
becomes
=
find
a;
2"
6(z
2
an d x
or ~^sr
= 2,
|,
-3
that
-2)+5z-38=0, i8
gi y ^ n
hy
x-\
The Binomial Equation x n -1=0.
(1)
If a
(2)
//
m
common For
a
is
root of
xn -
prime
to n,
1
ar
so aZso is
=0,
=|- or
Thus the roots are -1,
or ~J.
2.
is
is
,
i^Aere r is
2, |,
any
-3,
integer.
m -l=0 and xn -l=0 have no
the equations
root except 1.
if
a
a
is
common
positive integers.
can be found so that If n n roots are (3)
then a pm = l and afln = l, where p, q are any pm ~qn = 1, and since is prime to n, j9 and 5
root,
Therefore
m
(x.
pm-qn=
(See Ch.
1.
a prime number and a
is
1, a,
a
2 ,
...
1
a*
is
I, 4, (3).)
any imaginary
Hence a = l.
root of
xn -
1
=0, $e
"" 1 .
For every one of this set is a root and no two of them are equal. For a a~ = l, and the suppose that the two a and of (a>b) are equal. Then a n b = =0 and x 1 o^~ l have a common This is root. equations imaginary - 6. is a for and n is thus n to a l
;
If
(4)
of
xn
-
n=pqr
1 =
...
are the
wAere p, q,r,
...
are primes or powers of primes, the roots
n terms of the product
a root o/x p -l=0, jS o/a?-l=0, yo/xr -l=0, ete. Take the case of three factors p q, r similar reasoning applying in
where a
is
9
;
cases.
Any term
of the product, for instance
(a
and
6
similarly
(/J )
n
=* 1
)
and (y )*
1
= 1.
5
/?}/ , is
= (a*) a *r = l,
a n 5
a
a root.
For
all
SPECIAL ROOTS any two terms are equal,
If
w hich
-
=0
1
and
;
since
aj8 y
--htsin -
where
,
n
xn -
of
.
.
n
prime to
is
p
have no common root except 1. (5) As explained in Ch. V, 17, the roots cos
if
instance,
c
=a
&/ '^9
c/
y
impossible, for jg*-*y-' is a root of z
is
^ft-fty-c'^a'-a^ and oca '~ a is a root of x p
for
171 6
= 0, /x
r
1
qr,
=0
>
r
-l=0
these equations
are
i
w-1. -i
1, 2, ...
The imaginary roots are therefore
2r7T
cos
2riT
.
4
.
sin
= l, .,
where
,
r
according as n is odd or even. If x n - 1 is divided by x - 1 or x* the quotient, then
is
or
...
2,
according as n is odd or even and a reciprocal equation, and if this is
1
is
transformed by the substitution z = x + -, the values of x o 2 cos
r = l, 2,
If r is less (1) v '
= 0,
1
same
but
to than n and prime r i
/
.-
I or, 1
a
is not
type.
TH
-.
&
&
-n
,,
.
h
i
sm
-
2rn\ m
n
)
=
cos
Therefore divisible
by
rm
is
n.
This
where
1,
/
2rm7T 2rm?r ---sm ---- = 1 h
Definition.
is
by
and since n
n,
of
xn -
1
The
roots of special r
prime to
number
#w -
1=0
is
n
and of
degree,
the
m
.
is
prime to
r,
m
must be
is
not a root of an equation of
called a special root of the equation.
are cos
--hi sin n
n
,
where
r
-l=0 has
n Thus, x
is
the
of
r
2f7T
cos
n
.
db
i
sin
2r7T
n
are special roots. These can be arranged in pairs, fa, -j, so that they are the roots of a reciprocal equation.
M
a root of
m
=0 which
the same type and of lower degree
2rir -is .
sin
n
impossible, for
Any root
t
,
i
n
divisible
n
.
h
,
,
.
then
-2rrr
then cos
n,
any equation of lower
root of
2r7r
cos
if
\
If
or
-
-
...
z are
n Special Roots of x -1 -0.
3.
xn -
10
v where
n
M-
71-1
rt
fj8,
^J, ^
etc.,
B.C.A.
SOLUTION OF BINOMIAL EQUATION
172
a special root of x n - 1 =0, the n roots are 1, a, a 2 ... a 71 1 For every one of the set is a root. And if two of them are equal, for ~ a b then a a b = l, so that a is a root of x~ b = l. This is instance, if a =a ""
If QL
(2)
is
.
,
>
impossible, for the last equation
a
6
,
c, ...
,
,
1
=0.
xn -
I =0, the complete set of special roots is are the numbers less than n and prime to it,
root of
//a is any special a ac ... where a, 6,
(3)
a
than xn -
of lower degree
is
including unity.
a is any number less than n arid prime to it, the remainders when 3a ... (n-l)a are divided a, 2a, by n are the numbers 1,2,3, ... w-1, taken in some order or other (Oh. I, 10). And since a n 1, it follows ~ that every number in the set oc a a 2a a 3a ... a (n 1)a occurs somewhere in the set a, a 2 a 3 ... a n-1 Therefore a a is a special root.
For
if
,
,
,
4.
The Equation x n -A^O.
(1)
Let
A =a(cos a + ( \
.
are
^/a
.
i
For by De its n values are
,
a
.
4
2r7rl
-h
_
,
where
J-,
n
then the roots of x n -
|,
|
-ft sin
w
I
,
where a =
sin a),
a + 2r7i cos
,
.
r
= 0,
_ 1, 2, ...
^
o,.
it
cos
1.
J
----
.
4- i
a
=
,
------
-
sin
cos
n
n
n values of
^A may
a\
.
-
-f i
n
\
sm -
cos
j
n/
\
-n
r?r
.
sin
h t
he found by multiplying
n
roots of xn
+ 1 =0
are
This
is
-
---------i-ivsm
Solve x 5 -
Ex.
1.
real root is 1.
1
~0.
Deduce
where
,
n
obtained by putting a^=l, a
The
= A(),
,
.
cos
The values
of
then
a;
22
-fz-1^0,
are
#2
is,
-
54
TT
.'.
cos -
=coa 0JO 36
s/5 ^-- +
.
1.
,+ai-f
-f
t
--
+ 1 =0.
a:
2-|(-ls/5).
giving
cos - o
Therefore the values of z are 2 cos
n ~
30 and cos 72.
or-
2=3 + -
1, 2, ...
1,
x*+x*+x* + z + 1 ~0, that Let
,
in (J).
77
the values of co#
Dividing by x
r
sin
-
(
r
=1
,
2)
.
3
--
5 l
,
and 2 cos ,
,
and
Now
55
-
54 .
5
2?r
cos --
72 =cos ,,^0
cos
=
V "-! = ^-
\
),
/
any value of
by the n-th roots of unity.
The
and
r,
all different.
follows that the
(2)
=
t
n
Moivre's theorem, this expression is a root for every
bmce
^4
- cos -;
5
.
SOLUTION BY QUADRATICS Ex.
2.
the roots.
173
What are the special roots of a; 15 -1=0. Find the equation of which these are Show that this can be reduced to an equation of the fourth degree. What are its
roots?
The numbers
< 15 and prime to 1,
The
2, 4, 7,
special roots are
cos
15 are
15-7, 15-4, 15-2, 15-1.
-~
i
sin
The non -special roots are the The L.C.M. of x3 - 1 and x* - 1
(r
ID
15
roots of z 3 is
2
4-
(x
x+
1
= 1,
1 5
)
(x
-
2, 4, 7).
and xb 1) and
Therefore the equation whose roots are the special roots
x - x1 + ar - x* 8
1
+#
- a; +
3
1
1
-0.
is
=0,-
thatis, If
ZX + -, this becomes
that
z*
is,
of which the roots are 2 cos
2f7T (r
-r-^-
15
5.
- z3 - 4z 2 + 4z + 1 =0,
= l,
2, 4, 7).
The Equation x 17 - 1 =0.
this equation
Gauss discovered that
the solution of
depends on that of four quadratic equations.
This can be
proved as follows.
The imaginary roots
of
2f7T
cos
-=y
x17 .
L
1
=0
2f77
sm -y=-,
are
= 1, ^
,
where
r
rk
o
o
2, 3,
...
8.
Let yr = cos then we have --y ,
yn-r = yr Also the
sum
of the
and
2yrys - y f+J 4- y r .s ...................... (A)
imaginary roots
is
-
1,
therefore
Equations (A) lead to an arrangement in pairs of y l9 y 2 ... yg which has a remarkable cyclic property. Starting with the pair y^y^ we have ,
On
trial, it will
like result.
Let then
appear that no y except y4 taken with ,
t/j,
,
will lead to a
EXPRESSION IN SURD FORM
174 It follows that
relation connecting a,
any
j8,
y, 8 continues to hold after
the cyclic substitution (a/3yS).
Such
By
relations are the following.
a + j3 + y +
(B)
8= -i ................................... (E)
Again, using equations (A),
ay -i
to* 2/1^8+
2/42/2
+w^^ ay = - -^ ...................................... (F)
therefore
Consequently
j88
we can show
Similarly
=
-^
...................................... (G)
that
16aj8=-l+4a-4j9, .............................. (H) 4a 2 =l+2 8 + y ................................... (I) j
Then by
+ S = 2/x ............................. (J)
a + y = 2A,
Finally let
A+^= ~i
(E)
Also
4A/x,
=
Hence by (H) and three equations derived from stitution (a/?yS), we find that
2/2>2/8
+ j3 = y ~2fy + y -0 2 + 8 -0 2/ ~2yy
2/6,2/7
y
^/*
2/
2 y l9 y^ are the roots of y
Thus
this
by the
cyclic sub-
2ocy
2
2/3.2/5
Since cos quadratic
2
2
(K)
+ 82/-iV =
decreases as 6 increases from
is that
which
to
77,
the greater root of each
is stated first.
Thus, starting with the last equation and working upwards, the values of y l9 y 2 ... ys can be found, and then the roots of x 17 - 1 =0 are given by ,
a
-2^3 + 1=0, Ex.
1.
etc.
Express cos
as a surd.
and
= Also
and by
(I),
~
(34 -2^17),
g^
(34 +2^/1 7).
REGULAR therefor.
cos --
=
f
{
-
17-GON
175
,
1
J^{17 +3^17 ^(34 -2^17) -2^(34 + 2^17)}.
NOTE,
The
(i)
cosines of
~,
9
...
-^=-
can be expressed in a similar form; the
surd values of the sines of these angles involve one more radical sign. (ii)
It
is
(Ex.
XIX,
22.)
length determined by a quadratic equation can be found geometrically. therefore possible, by the use of ruler and compasses only, to effect the following
Any
construction.
To
inscribe a regular 17-sided Polygon in a Circle. be the centre. Take the radius of the circle as unit of length. Let Draw x'Ox, y'Oy at right angles, cutting the circle in A, A' and JB, B'. 6.
Let the vertices of a regular inscribed 17-gon be marked 0, 1, 2, ... 17, the The ordinates y l9 y 2 ... ys of the points point coinciding with B. ,
To solve these geomet8 are given by the equations of Art. 5. 1, 2, 2 = 6 are the ordinates of the observe that of + the roots 2ay rically, y 2 2 = cuts the circle x + y 2ay + 6 = points where the axis x Thus to find the points (0, t/j), (0, y 2 ), ... (0, j/8 ) (two of which are ...
marked
The
t/ 3 ,
y5 in Fig.
27),
we have
centres of the circles are
to
draw the
marked
o>,
circles in the following table.
A,
/z,
The To draw the
a, etc., in Fig. 27.
p are easily drawn, giving the points a, last four circles notice that circles co, A,
j9,
y, 8.
:
(i)
The
circles a,
j3,
y, S meet
y=0
at the
as diameters, respectively. For, since the equation to the circle on
By,
same points as
the circles
on
J5jS,
J58, J5a
_ 2
which
is satisfied
by
x=
*J
- f$,
J5j8
as diameter
is
~ /
.y=0,
2
we can draw
the circles
j8
und
y.
REGULAR
176 (ii)
The diameters of the
to the circles
j8,
circles a,
/3,
17-GON y, S are equal to the tangents
from
B
y, 8, a, respectively.
For (diameter of tangent from JB to
- j3) = (2jS + y + 1) - 4j8, and the square Hence we can draw the circles a, 2/2 -f y.
Q a) = 4 (a Q = 2
j8
1
2
of 8.
14,
Construction.
r
Along Ox take OJ^^OA. Along Oy take and radius oo7 draw a circle cutting yOy in A, f
With centre co With p. centre A and radius A J draw a circle cutting yOy in a, y. With centre ft and radius /i7 draw a circle cutting yOy' in jS, 8. Draw the circle on By as diameter cutting Ox in G. With centre jS and radius $6 draw a circle f
cutting yOy' in
t/3 , j/ 5 .
Through y 3 y 5 draw parallels to x'Ox. These lines meet the given circle and 5, 12 of the required 17-gon, which can be comthe arc 35 at 4 and setting off arcs equal to 34. pleted by bisecting Or we may draw the circles a, y, 8 as explained above. ,
in the vertices 3, 14
RECIPROCAL AND BINOMIAL EQUATIONS
177
EXERCISE XIX
RECIPROCAL AND BINOMIAL EQUATIONS Solve the equations in Exx. 1-9
:
1.
x'-Sx* + Ilx*-Sx+ 1=0.
2. x*
3.
2# 4 + 3z 3 + 5* 2 + 3a; + 2=0.
4.
2x* + x* -I7x*
5.
2x* +
5x*-5x-2=0.
6.
3x 5 - 10* 4 - 3* 3 - 3s a - 10* + 3=0.
7.
2z 6
8.
2z 7 -a;-32; 4 - 3^-0; + 2=0.
9.
*
8
-z 5 -2z3 ~a; + 2=0.
-f
l+(s + l) =2(* + :r + 2
8
1 [Write this x* + x-* + (x + x~
+ Qx*-5x* + 6x + l=Q. + x + 2^0.
4 .
I)
+ 2)*~2(x + x~ l + 1) 4 and show ,
that this reduces
to 10. If sn
~l+u + u + ...+un
where u n
2
1
un+l -Un
j
and
*n=-T-9-2 *j
xn +x- n and
z=x + x~
l
prove that
,
*n= wn-i- 5 n-.
Hence show that
+sn = 1
[1
:
A
11. If a is 2
n= xn
+ a5
,
as
12. If 13.
15 the
+
C
C
JL
-10 depends.
l
and
(1 -ar)(l
-ar')=2-z.]
an imaginary root of # 7 - 1 =0, the equation whose roots are a -fa 6 , 8 a is *,=3 + z -22J - 1 =0.
+ a4
n
is
The
a prime, the special roots of x 2n
special roots of 2T7T
The
~
i
sin
special roots of
values are
d-
(
x9
-1=0
2f7T
values are cos 14.
n+ l equation on which the solution of x*
x~ n
1
cos -
b
i
sin -
where
,
x 12 )
,
r
by 12 [To find the equation, divide a; 6 L.C.M. of a; the x* i.e. by 1, L] \
15. If a,
/?,
are the roots of x n
are the roots of x*
~ 1,
-1=0 that
-1^0
1
+ x* + 1~0, and
=0. their
2, 4.
are the roots of
is,
-f 1
g-(N/3
by the
# 4 -a; a -f
=0, and their
1
t).
L.C.M. of
x2 -
1,
#3 -
1,
x*
-
1,
x9 -
1,
3 2 y are the roots of o# -f bx + cx + d=0, prove that the equation
1 l l * whose roots are a-f-, 5 + -, y-f a ^ y
is
+ (ac + bd)z* + (ab + bc + cd- Bad) z + (a - c) 2 + (6 - d) 2 = 0. 2 3 2 [Prove that adIJ{x -a;(a-fa- + l} = {(aa; + bx* + cx + d) (dx* + ex + bx + a)}, and write (x* + l)fx=z.] adz*
1
)
that there is a value of k for which x* - I5x 3 -&r 2 + 2 find this value. and 1, that x*-15x*-Sx* + 2=0 has two roots a, p such that [Show 16.
x*
4-
Show
is
divisible
by
kx -f
a/?
= l. Then
CONSTRUCTION FOR REGULAR PENTAGON
178 17.
For the equation x*+px* + qx* + j34-l=0, prove that (i)
and only if, 2 4(g-2)