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HIGHER ALGEBRA BY S.
BARNARD, M.A.
FORMERLY ASSISTANT MASTER AT RUGBY SCHOOL, LATE FELLOW AND LECTURER AT EMMANUEL COLLEGE, CAMBRIDGE
AND J.
M. CHILD,
B.A., B.Sc.
FORMERLY LECTURER IN MATHEMATICS IN THE UNIVERSITY OF' MANCHESTER LATE HEAD OF MATHEMATICAL DEPARTMENT, TECHNICAL COLLEGE, DERBY FORMERLY SCHOLAR AT JESUS COLLEGE, CAMBRIDGE
LON-DON
MACMILLAN
fcf'CO
LTD
*v
NEW YORK
ST MARTIN *S PRESS
1959
This book
is copyright in all countries which are signatories to the Berne Convention
First Edition 1936
Reprinted 1947, ^949> I952> *955, 1959
MACMILLAN AND COMPANY LIMITED London Bombay Calcutta Madras Melbourne
THE MACMILLAN COMPANY OF CANADA LIMITED Toronto ST MARTIN'S PRESS INC
New York
PRINTED IN GREAT BRITAIN BY
LOWE AND BRYDONE (PRINTERS) LIMITED, LONDON, N.W.IO
CONTENTS
ix
IjHAPTER
EXEKCISE
XV
(128).
Minors, Expansion in Terms of Second Minors (132, 133). Product of Two Iteterminants (134). Rectangular Arrays (135). Reciprocal Two Methods of Expansion (136, 137). Use of Double Deteyrrtlilnts,
Symmetric and Skew-symmetric Determinants, Pfaffian (138-
Suffix,
143),
ExERtad XVI
(143)
X. SYSTEMS OF EQUATIONS. Systems (149, 150). Linear Equations in Line at Infinity (150-152). Linear Equations in Three Unknowns, Equation to a Plane, Plane at Infinity (153-157). Definitions, Equivalent
Two Unknowns, EXEKCISE XVII
(158).
Systems of Equations of any Degree, Methods of Solution for Special Types (160-164).
EXERCISE XVIII
XL
(164).
RECIPROCAL AND BINOMIAL EQUATIONS. The Equation Reduction of Reciprocal Equations (168-170). xn - 1=0, Special Roots (170, 171). The Equation xn -A =0 (172). The Equation a 17 - 1 ==0, Regular 17-sided Polygon (173-176). EXERCISE
XIX
(177).
AND BIQUADRATIC EQUATIONS. The Cubic Equation
(roots a, jS, y), Equation whose Roots are Value of J, Character of Roots (179, 180). Cardan's Solution, Trigonometrical Solution, the Functions a -f eo/? -f-\>V> a-f a> 2 4- a>y (180, 181). Cubic as Sum of Two Cubes, the Hessftfh (182, 183). Tschirnhausen's Transformation (186). (
-y)
2
,
etc.,
EXERCISE XX (184). The Biquadratic Equation
A=y + aS,
(roots a,
,
y, 8) (186).
The Functions
the Functions /, J, J, Reducing Cubic, Character of Roots (187-189). Ferrari's Solution and Deductions (189-191). Descartes' Solution (191). Conditions for Four Real Roots (192-ty). Transformation into Reciprocal Form (194). Tschirnhausen's Transetc.,
formation (195).
EXERCISE
XXI
(197).
OP IRRATIONALS. Sections of the System of Rationals, Dedekind's Definition (200, 201). Equality and Inequality (202). Use of Sequences in defining
a Real Number, Endless Decimals (203, 204). The Fundamental Operations of Arithmetic, Powers, Roots and Surds (204-209). Irrational Indices, Logarithms (209, 210). Definitions, Interval, Steadily Increasing Functions (210). Sections of the System of^Real Numbers, the Continuum (211, 212). Ratio and Proportion, Euclid's Definition (212, 213).
EXERCISE XXII
(214).
CONTENTS
x CHAPTER
XIV/INEQUALITIES. Elementary Methods (210, 217)
Weierstrass' Inequalities (216).
For n Numbers a l9 a 2 a
n (a* -!)/*
(a" -I)/*,,
xa x ~ l (a-b)$a x -b x
(l+x)
n
^l+nx,
\*JACJJ
>
n
n
(219).
^ xb x
~l
(a
- 6),
(219).
(220).
Arithmetic and Geometric Means (221, 222). -
-
^
V
n and Extension
(223, 224).
(224).
XV. SEQUENCES AND Definitions,
Maxima and Minima
(223).
EXERCISE XXIII
LIMITS.
Monotone
Theorems,
Sequences
(228-232).
E*
ponential Inequalities and Limits, /
l\m
/
i\n
and
1) >(!+-) m/ n/ \ /
1 \
n
lim (1-fnj n_ >00 V
EXERCISE
XXIV
l\-m /
/
=e,
nj
(1)
if
m>n,
(232,233).
(233).
General Principle of Convergence (235-237). Limits of Inde termination (237-240).
Theorems
~n
l\"w
=lim(l--) \
1 \
<(l--) nj \
(1--) mj \
Bounds of a Sequent
:
Increasing Sequence (u n ), where
u n -u n ^ l
^
1
u n >0 and u n+l lu n -*l, then u nn -*L u n l, then lim If lim (Ui+u 2 + ...
(2) If (3)
n (4) If
lim
+u n )jn
a n ~a, and lim
n-~>oo
lim (a n 6 1 n >ao
fe
w
= 6,
then
n~>oo
+ a w _ 1 6 1 + ...H-a 1 6 n )/n=o6,
(240-243).
Complex Sequences, General Principle of Convergence EXERCISE
XXV
I.
n->oo
>oo
(244).
(243, 244).
CONTENTS XVI. \CONVERGENCE OP SERIES
xi
(1).
Definitions, Elementary Theorems, Geometric Series (247, 248). Introduction and Removal of Brackets, Series of Positive Terms. p D'Alembert's of Order Terms, Comparison Tests, 271 /w Changing ,
and Cauchy's Tests
XXVI
EXERCISE
(248-254).
(254).
Terms alternately Positive and Negative (256). Absolute Convergence, with Terms Positive or Negative.
Series with Series
General Condition for Convergence, Pringsheim's Theorem, Introduction and Removal of Brackets, Rearrangement of Terms, Approximate Sum, Rapidity of Convergence or Divergence (256-261). Series of Complex Terms. Condition of Convergence, Absolute Conn n vergence, Geometric Series, Zr cos nd, Sr sin n6. If u n /u n+l = l+a n /n, where a n ->a>0, then u n -*Q. Convergence of Binomial Series (261-263).
XXVII
EXERCISE
(264).
^
XVII^CoNTiNuous VARIABLE.
<*> Theorems Meaning of Continuous Variation, Limit, Tending to on Limits and Polynomials (266-268) Continuous and Discontinuous Functions (269, 270). Continuity of Sums, Products, etc., Function Fundaof a Function, lirn {f(x)}, Rational Functions, xn (271). ,
.
mental Theorems
Derivatives, Tangent to a Curve, Notation (272). of the Calculus, Rules of Differentiation (273-277). Continuity of of {f(x)} and xn (278, 279). Meaning of Sign {f(x)}, Derivatives of J'(x) (279). Complex Functions, Higher Derivatives (2*9, 280). Maxima and Minima, Points of Inflexion (280-282).
EXERCISE XXVIII
(282).
Inverse Functions, Bounds of a Function, Rolle's Theorem, MeanValue Theorem (284-288). Integration (289). Taylor's Theorem, Lagrange's Form of Remainder (290, 291). Function of a Complex Variable, Continuity (291, 292).
XXIX
EXEBCISE
(293).
XVIIL, THEORY OF EQUATIONS TIONS
(2),
POLYNOMIALS
RATIONAL FRAC-
(2),
(1).
Multiple Roots, Rolle's Theorem, Position of Real Roots of/(&)=0 Newton's Theorem on Sums of Powers of the Roots of f(x) =0 (297). Order and Weight of Symmetric Functions (298, 299). (296, 296).
Partial Derivatives, Taylor's x, y,
...
Euler's
.
(299-302).
Theorem
for Polynomials in
for Polynomials, J
x
A Theorem on Partial Fractions (302). lf
EXERCISE
Theorem
XXX
(304).
+ ...=0,
(303).
dx
+y
x and in
+ ...~nu dy
The Equation
CONTENTS
xii
CHAPTER
XIX. EXPONENTIAL AND LOGARITHMIC FUNCTIONS AND
SERIES.
The Exponential Continuity, Inequalities and Limits (306, 307). x Theorem, Series for a (307, 308). Meaning of an Irrational Index, Derivatives of a x log x and x n (309). Inequalities and Limits, the x way in which e and log x tend to oo Euler's Constant y, Series for 2 The Exponential Function E(z), Complex Index (310-312). log ,
,
(312, 313).
Series for sinx, cos x arid Exponential Values (313). in Summing Series (314).
Use of Exponential Theorem
EXERCISE
XXXI
(315).
Logarithmic Series and their Use in Summation of Series, Calculation of Logarithms (315-319). The Hyperbolic Functions (319-321).
EXERCISE
XXXII
XX. CONVERGENCE
(321).
(2).
Series of Positive Terms.
2
-
'
Cauchy's Condensation Test, Test Series
Rummer's, Raabc's and Gauss's Tests
(325).
Binomial and Hyper-geometric Series (328, 329). %
(326-328).
De Morgan and
Bertrand's Tests (330).
Terms Positive or Negative. Theorem, Abel's Inequality, and Abel's Tests (330, 331). Power Series, Interval and Radius of Convergence, Criterion for Identity of Power Series (332Binomial Series l-f?iz4-... when z is complex (334). Multi334). plication of Series, Merten's and Abel's Theorems (335-338). Series with
Dirichlct's
EXERCISE XXXIII
(338).
XXI. JBlNOMIAL AND MULTINOMIAL THEOREMS. Statement,
Vandermonde's Theorem
Binomial Theorem.
(340).
Euler's Proof, Second Proof, Particular Ins tancesj 34 1-345).
Num-
";
1 + x (345-348). erically Greatest Term, Approximate Values of EXERCISE XXXIV (349). Use of Binomial Theorem in Summing Series, ^Multinomia^Theorem
(351-355).
EXERCISE
XXXV
(355).
XXII. RATIONAL FRACTIONS
(2),
PR,ECURRING)SERIES
AND DIFFERENCE
EQUATIONS. Expansion of a Rational Fraction (357-359).
EXERCISE
XXXVI
(359).
Expansions of cos nd and sin nOjsin 9 in Powers of cos 6 (360). Recurring Series, Scale of Relation, Convergence, Generating FuncLinear Difference Equations with Constant tion, Sum (360-363). Coefficients (363-365).
EXERCISE
XXXVII
(365).
Difference Equations, General
EXERCISE
XXXVIII
(370).
and Particular Solutions
(367-370).
CONTENTS
xiii
CHAPTER
XXIII. THE OPERATORS 4> E, D. The Operators J, E, Terms of u l9 Au^ A 2 u l9
INTERPOLATION. A ru x u x+r
Series for ...
,
;
U1 + u2 + u3 +
...
in
and
Interpolation, Lagrange's
(373-379).
d \n (uv) ( j- ) Vcte/ /
Bessel's
The Operator
Formulae (379-382).
f) 9
Value of
(382, 383).
EXERCISE
XXXIX
(384).
XXIV. CONTINUED FRACTIONS
(1).
Definitions, ForniationjoConvergents, Infinite Continued Fractions
(388-391).
EXERCISE
Simple and ^eeuTrirSg; Continued Fractions (391-394).
XL
(394).
Simple Continued Fractions, Properties of the Convergents, an Irrational as a Simple Continued Fraction (396-401). Approximations, Miscellaneous Theorems (402-406). Symmetric Continued Fractions, Application to Theory of Numbers (406-409). Recurring Con tinned Fractions (409-411).
EXERCISE XLI
(411).
XXV. INDETERMINATE EQUATIONS
OF THE FIRST DEGREE.
axby axbycz...=k
Solutions of the Equation
EXERCISE XLII
Simple
c (414-417).
Two
Equations in
x, y, z
;
(417-419).
(419).
XXVI. THEORY OF NUMBERS
(2).
Congruence, Numbers less than and prime to n, Value of q)(n) E
'
(424-427).
EXERCISE XL1II
(427).
Roots of a Congruence, the Linear Congruence, Simultaneous Congruences (428-431). Theorem on Fractions (431). The General Congruence, Division (mod n) 9 Congruences to a Prime Modulus (432-434).
EXERCISE
XLIV
(434).
XXVII. RESIDUES OF POWERS OF A NUMBER, RECURRING DECIMALS. Residues of a, ag ag* 9
Decimal Equivalent of m/n Number of Figures in the Period, Short Methods of Reckoning, Prime Factors of 10* - 1, 9
(
EXERCISE
XLV
= 1,2,...
10), (439-444).
(442).
The Congruence xn -sa (mod p) Methods of Solution (444 EXERCISE XLVI (445). 9
45).
CONTENTS
xiv CHAPTER
EXERCISE XLVII
(447).
Separation of the Roots, Sturm's Theorem, Fourier's Theorem (447-456).
EXERCISE XLVIII (456). Newton's Method of Approximating to a Root, Fourier's Rule, Nearly Equal Roots (456-460). Horner's Method (461-465). EXERCISE XLIX (466).
FUNCTIONS, CURVE TRACING. Implicit Functions, Rule for Approximations (467-469). for Roots of a Cubic Equation (when all Real) (470).
EXERCISE
L
Series
(471).
Tangents, Asymptotes, Intersection of a; Straight Line and Curve (473). Curve Tracing, Newton's Parallelogram (474-480).
EXERCISE LI
XXX.
(480).
INFINITE PRODUCTS. Convergence, Absolute Convergence, Derangement of Factors, Convergence discussed by Use of
Expansion as a Series (485-488). Logarithms (490). EXERCISE LII (491).
XXXI.yPERMUTATIONS, COMBINATIONS AND DISTRIBUTIONS. Combinations with Repetitions, Things not all Different (493). Arrangement in Groups, Distribution in Parcels (494-497). Derangements, General Theorem (498, 499). Partition of Numbers, Table of p Partitions of n, Euler's Use of Series (500Distributions,
504).
XXXII.
EXERCISE LI1I
(504).
EXERCISE TJV
(506).
^PROBABILITY. First Principles, Exclusive Events, Independent and Interdependent Events (508-513). Probability estimated by Frequency, Expectation, Successive Events (513-518). Probability of Causes, Value of Testimony, Appli cation of Geometry (518-522).
EXERCISE
LV
(523).
XXXIII. CONTINUED FRACTIONS
(2).
Expression of a Quadratic Surd as a Simple Continued Fraction, the Form The form V~AjB, Cycle of l )lr l (527-530). 2 2 Quotients, the 6 and r Cycles, Solution in Integers of Bx -Ay 2= z x The Form of Solutions 9 ^/N ~Ny (530-534). Integral (534-538). The Cycle belonging to (^/N + 6)/r found by the G.C.M. Process (538).
(VNb
=M
M
EXERCISE LVI
(540).
MISCELLANEOUS EXERCISES
A
(543),
B
(555).
HIGHER ALGEBRA CHAPTER
I
THEORY OF NUMBERS In *
'
this section
number
we
discuss properties peculiar to whole numbers.
'
fs
(1)
taken to
mean
l
The word
whole number,' and, unless otherwise stated^
positive whole number*
Division.
1.
(1)
Let b be any positive whole number, and consider
the sequence ...-36,
-26, -6,
0, 6, 26, 36,...
continued indefinitely both ways. Any whole number a (positive, negative or zero) is either a term of this sequence, or it lies between two consecutive
Thus two numbers q and
terms.
r
can be determined uniquely so that
and
a-=bq + r
To
divide a
ditions
is
6
is
to find the
called the quotient
numbers
and
r
q
and
r
which satisfy these con-
the remainder.
= 0, we
say that a is divisible by 6 or is a multiple of 6, and that a divisor or a factor of a. Among the divisors of a number we count
If r
6 is
q
;
by
0^r<6 ........................... (A)
number itself and 1. Whatever 6 may be,
the
= 6.0 4-0; hence
(2) If r
= 6~/, we have a = 6-fl-r'
if
and
R
then /
r
>|6
Hence
and
it is
and Ex.
0,
1.
For
if
1,
2,
must be regarded as
by every whole number.
divisible
also
zero
Every number
is
any number 5-2, 5-1.
is
always possible to find numbers
Q
JR<6.
of one of the forms 5n, 5n
1,
5n
divided by 5, the remainder
2. is
one of the numbers
DIVISION
2 Ex.
Every square number
2.
The square
these are divided by
and 5n -
l,
Ex.
For (2k +
by
2
l)
is of
one of the forms 5n,
odd number
form Sn + 1. and either k or k + l must be even, so that k(k +
-4k(k + l) + l,
(1
// both a and b are divisible by
(2)If
of the
l) is
r is the
Division.
remainder when a
so also is
c,
divided by
is
ma 6,
nb.
then cr is the remainder
ca is divided by cb.
For
a = 6
if
then
ca
(3)
(cb)q
and
+ cr
and
// both a and b are divisible by
divided by b, then is divided by b/c.
For also
is
2.
Theorems on
when
.
,
2. )
5wL
is
1.
Tlte square of every
3.
divisible
number
of one of the forms (5w) 2 (Sw^l) 2 , (5m 2) 2 If 5, the remainders are 0, 1, 4 ; and, since 4 = 5-1, the forms are
of every
if
is r,
(i)
r is divisible
c, and r is the remainder when a is and (ii) r/c is the remainder when a/c
a~bq + r and 0^r<&, then which
is
equal to a a/c
3.
by
c
-
bq.
since a and Thus we have
= (b/c)q + r/c where
6 are divisible
by
c,
so
O^r/c <6/c.
Theorems on the Greatest Common
Divisor.
is divided by b, the common divisors of (1) If a and b are the same as those of b and r. For, since a = bq + r, every common divisor of b and r is a divisor of a
r is the
remainder when a
;
and, since r
= a-bq,
common
divisor of a
every This proves the statement in question.
and
b
is
a divisor of
r.
(2) If a and b are any two numbers, there exists a number g, and one only, such that the common divisors of a and b are the same as the divisors of g.
For numbers (q v rx ), and uniquely, so that
(q 2 , r 2 ), etc.,
a = 6y 1 -f-r1
,
can always be found in succession,
6=r1 ? 2 + r2
,
6>r1 >r2 >r3
where
rx ...
= r2
3
+ r3
,
etc., ............... (A)
^0.
number of positive integers less than b remainder must occur and, supposing that rn ^ =0, the Since the
;
r -2
= r n-l?n + rn>
rn-l
=
is
limited, a zero
process terminates
GREATEST COMMON DIVISOR divisors
numbers have the same common
(A), the following pairs of
Hence, by
3
:
(a, 6)
(6,
rj,
(r
v
r2 ) y
...
(rn ^ v r n )
,
;
^
e common divisors of r n _ 1 and r n are the divisors and, since fn-i VZn+i* Hence the number g exists, and its value is r n of r n .
.
Since g
common
is
divisible
numbers and
divisor of these
as the greatest (3)
by any common
common measure
divisor of a
known
is
(G.C.M.) of a
and
the greatest in elementary arithmetic
and
it is
ft,
ft.*
If a and b are each multiplied by any number m, or are divided by a divisor m, then g, the greatest common divisor of a and ft, is multiplied
common
or divided by m.
For each (4) If a,
of the r's in the proof of (2) ...
c,
ft,
and
divisor of a
ft,
,
is
multiplied or divided by m.
and g l
are several numbers,
g 2 that of g 1 and
c,
ft,
d,
c,
. . .
),
(gv
ft,
c,
d,
. . .
common (# 2 ,
),
c, d,
Hence we arrive eventually at a number g, whose divisors of a,
the greatest 4.
ft,
and
d,
divisors . . .
),
common
so on, then :
etc.
divisors are the
common
uniquely determined. This number g is divisor , or the greatest common measure of a, 6, c, ...
c, ...
common
the greatest
g3 that of g 2 and
the following sets of numbers have the same (a,
is
;
also g
is
.
Numbers Prime
Two numbers,
to each other.
a and
6,
are
said to be prime to each other when their greatest common divisor is 1, so that they have no common divisor except 1. This is often expressed
by saying that a is prime to theorems are fundamental.
ft,
or that
ft
is
prime to
// the product ab is divisible by a number m, and factor a, then m is a divisor of the other factor ft. (1)
a.
m
The following
is
to
prime
one
For, if a is prime to m, the greatest common divisor of a and m is 1 hence the greatest common divisor of aft and mb is ft. But, by hypothesis, m is a divisor of aft, and is therefore a common divisor of aft and mb ;
;
hence
m is equal to,
// a is prime a divisor of N.
(2) is
or
to 6,
is
a divisor
of,
ft.
and each of these numbers
is
a divisor of
2V,
then ab
Suppose that N = aq, then ft is a divisor of aq, and since ft is prime to one factor a, it must be a divisor of the other factor q. Let q = wft, then
N = waft, and *
The
aft is
a divisor of N.
process, here stated algebraically,
is
that used in Elementary Arithmetic.
PRIME NUMBERS
4 If a
(3)
is
prime
to b, positive integers x,
ax-by For
it
y can be found such
that
1.
follows from equations (A) of Art. 3, (2), that TI
= a - bq v
r2 -
-
aq z + b (1
+ qfa),
Continuing thus, every remainder can be expressed in the form where x, y are positive integers. If
a
prime to
is
6,
the last remainder
and the theorem
is 1,
Theorems on Prime Numbers.
5.
A number
(ax-~by),
follows.
which has no
divisors except itself and 1 is called a prime number, or simply a prime. Numbers which are not prime are said to be composite.
A
(1)
prime number, p,
is
prime
to
every
number which
not a multiple
is
For, if a is any such number, q and r can be found such that a~pq + r where 0
therefore to a. (2)
least
If a prime p is a divisor of a product abed one of the factors a, 6, ... k.
...
hk,
it is
a divisor of at
the prime p is not a divisor of a, by Theorem (1) it is prime to a, hence it is a divisor of bed ... k. If, in addition, p is not a divisor of 6, it must be a divisor of cd ... k. Continuing thus, it can be shown that if p is
For
if
not a divisor of any of the numbers
a, b, c,
...
h, it
must be a
6.
Theorems on Numbers, Prime or Composite.
(1)
Every composite number
N
divisor of k.
N has at least one prime divisor. N
not a prime, it has a divisor, n, different from and from 1, which is not greater than any other divisor. Further, this divisor must be a prime for, otherwise, it would have a divisor less than itself and greater than 1 and this latter would be a divisor of N. This contraFor, since
is
;
,
dicts the hypothesis that It follows that every
n
not greater than any ot,her divisor. composite number, N, can be expressed as the is
product of prime factors. has at least one prime factor, p, we have N~pa, For, since where l<.p
N
factor, q
Thus,
;
and a = qb, where l
N =*papqb
;
COMPOSITE NUMBERS But the numbers
N
the set
pressed in the form,
N are limited
than
must
...
a, b,
y
less
finally
N=pqr
...
5
and N>a>b>...
;
end in a prime. Hence, where p, q, r, ... u are
u,
,
,
therefore
N
can be ex-
all
primes, not
all different.
necessarily That is to say,
any composite number, N, can be expressed
N=p where p,
q, r, ...
u are
,
a
b
.q .r
all different
c
u
as
s ,
primes.
(2) A composite number can be expressed as the product of prime factors in one way only. a b c = P A Q B R C where p, g, r, For suppose that P, Q, R, ^=p q r a b c are primes then, since the prime P is a divisor of the product p q r ...
N
. .
. . .
. . .
.
. . .
,
,
,
;
a divisor of one of the factors p, q,r, ... and is therefore equal to one of them. In the same way each of the set P, Q, R, ... is equal to one of it is
,
and no prime factor can occur in one of the expressions q, r, ... which does not occur in the other. Suppose then that N^p a qr c ...^p A q Br c ...
the set p, for
N
,
.
a --A one of them must be the greater. Let A>a> and suppose that + e, then q b r c .... t m ^p c qB r .... t M but this is impossible,
If
9
Aa
.
.
.
;
as the left-hand side of the equality right-hand side is divisible by p.
Hence, a must be equal to thus the two expressions for
A
;
a number prime to p, and the
is
and
similarly, 6
= B,
. . .
m~M
,
and
;
N are identical.
The' above theorem
is
one of the most important in the Theory of
Numbers, and the following propositions are immediately deducible. (3)
//
m
product ab (4)
// a
is ...
is
A
each of the numbers, a,
i,
...
,
k, it is
prime
to the
then a n is prime to b n , where
n
is
any
integer
;
and
*-*""*'
number
greater than 1
For, if 2V
to
to b,
prime
conversely. (5)
prime k.
N
and
is
a prime, if
less than, or
it
is
equal
to,
= a&, where b^a, then
not divisible by
any prime number
v^V.
N^a
2 ;
that
The sequence of primes is endless. For, if p is any prime, the number p 4-
is,
a^.*jN.
(6)
1 is
j
divisible
or by any smaller prime.
by p must have a prime
than
p
exists.
divisor greater than p,
If
greater than
then
and in
p+1
is
p and
is
not
not a prime,
either case a
it
prime greater
NUMBER
DIVISORS OF A
6
which are
...
N-
,
than
less
problem of discovering whether a large number one of great difficulty. Ex.
If n
1.
is
any number, prove
that
n(n + I)(n + 2)
Of the two consecutive numbers, n and n +
be obtained in order by
The process consists in writing 1 and erasing all multiples
a process called tlie(Sieve of Eratosthenes). down in order all the numbers from 1 to of the primes 2, 3, 5, 7,
N can
number
All the primes less than a given
1,
one
%/JV.
Nevertheless, the
prime or composite
is
is divisible
by
6.
by 2
divisible
is
is
and one of
;
the three consecutive numbers, n, n + 1, n + 2, is divisible by 3. Hence the product and, since 2 is prime to 3, n(/i-f-l)(n + 2) is n(tt + l)(tt-f-2) is divisible by 2 and by 3 ;
divisible
Ex.
We
G.
by
Prove that 3 2n+1
2.
have
3
+1
2r2
=3
.
+ 2"+ 2
is
divisible
7.
by
= 3 (7 + 2) n = Ik + 3
9W
.
2 n,
by the Binomial theorem,
and
7.
The
where p,
q,
the divisors
Then,
Divisors of a Given Number N. Let N=p a are primes ; and let n be the number, and s r, ... N, including 1 and N. of
.
,
n = (a + _1
(1)
pa+l
l)(6 yft+l
p-l For the divisors of
+ l)(c + l) _ l r c+l _
...
For,
if
N
the
rc
.
...
,
sum, of
;
r-l
q-l
N are the terms in the expansion of
N
and
1, is
%(n +
the product of
l) or %n, according as
a perfect square, each of the numbers, a,
is
.
I
and the expressions for n and s follow immediately. (2) The number of ways in which N can be expressed as two factors, including not, a perfect square.
b
q
N
6, c, ...
,
is,
is
or is
even,
and therefore n is odd but if N is not a perfect square, at least one of these numbers is odd, and therefore n is even. Further, if ^( = 1), d2 rf 3 ... d n _ 2 d n _ v d n (=N), are the divisors of A' '
;
,
,
,
in ascending order, the different
two
and
factors are
did n
,
d^dn
,
N
ways
of expressing
N
as the product of
:
d 2d n _ v
c^ n _
d 2dn , l9
dsd n _ 2
is,
or
is
...
2>
Hence, the number of ways according as
,
,
d xd x
d yd y+l
,
is
either
when
,
x or
not, a perfect square.
,
when
y,-i.e.
n=*2a?-l,
= 2y.
either |(w
+ l)
or \n,
PRODUCT OF CONSECUTIVE INTEGERS
7
The number of ways in which N can be expressed as the product of two m ~ 1 where m is the number mctors, which are prime to one another, is 2 of N. Different prime factors of (3)
,
For, such factors are the terms in the expansion of
~~ and their number is 2 W Hence the number of pairs is 2 m 1 For example, if # = 2 2 3 3 5 = 540, n = (2 + l)(3 + l)(l + 1) = 24, .
.
.
.
8.
The Symbol
l[x/y].
If
a
is
m==3
and
'
a fraction or an irrational number, the
symbol / (a) will be used to denote the integral part of where 0
// n v n 2 n 3 ,
,
...
are
,
any
>
integers,
and
Thus
a.
s is their
if
x = qy + r
sum and a
is
any
number, then
= aq^ + r l9 n 2 = aq% + r2 w 3 = ay3 -f r3
Let n
,
Hence,
etc.
;
then
/[/a]
and, since Ji^/f^i/a],
The
(2)
,
highest
r
52
==
^[ n 2/a ]>
>
^e resu
power of a prime p which
is
lt
follows.
n
contained in
is
\
For, of the numbers from 1 to n inclusive, there are I[n/p] which are 2 2 and so on hence the by p of these I[n/p ] are divisible by p
divisible
;
;
;
result follows.
9.
by
Theorems.
(1)
The product of any n consecutive
integers is divisible
\-n.
For (m + l)(w + 2) last expression is
occurs in
[m
n^
...
(m-f n)/\ n =
\
m + nj\ m
w,
and to show that the
it is sufficient to show that any prime p which Thus we occurs to at least as high a power in -f n.
an integer
m
j
have to show that /[ (m
+ n)/p] -f /[ (m + n)/p*] + 1[ (m -f n)/p 3] +
Now
/[(m + n)/p]>/[w/p]-f /[n/p], and the same is true by p p*, ... in succession hence the result in question. 2
ft
. . .
,
,
:
if
we
replace
NUMBERS IN ARITHMETICAL PROGRESSION
8 (2)
is
Ifn
a prime,
is divisible
C?
by n.
n(n - l)(n 2)
For by the preceding since n is a prime and r r is
Hence,
...
-r+
(n
1) is divisible
supposed to be less than n, r a divisor of (rc-l)(n-2) ... (n-r + 1) is
is
by
[/% aijid
prime to
;n.
|
|
and Thus
a prime,
t*
ft
t/
except the first
and
last,
text-books about
'
divisible
is
l)/[r
by
n.
in the expansion of (l+x)
all the coefficients
n ,
are divisible by n.
supposed to be acquainted with what is said in elementary for permutations and combinations and the binomial theorem
The reader
NOTE.
(n-r +
...
^(ft-l)
is
*
'
a positive integral index.' In what follows, F$ denotes the number of permutations, and C the number of combinations, of n things taken r at a time. Ex.
Find
1.
We have
the highest
power of 5 contained in
7[158/5]-31,
158. |
2
/[158/5 ]=/[31/5J-:6,
therefore the required power has an index
1
/|158/f>J=y|<$/5J
:
= 31 +6 + 1 =38.
an odd prime, the integral part of (x/5 +2) n -2 W+1 is divisible by 20n. Let (V5 + 2) #+/ where 0/5 - 2)n =/'. Then, since 0< V5 -2< 1, we have also 0
If n
2.
is
n=
;
N =2(Cf Moreover, since n
by 20w
;
which
is
.
is
2
Sty- 1 ) + C%
a prime,
C^
is
.
23
.
5*( n
divisible
~3
)
by
+ ... + C n,
.
2n
~z .
and therefore
5
+ 2n ).
N -2 n +
1
is
divisible
the required result.
10.
Numbers
(1)
Let a be prime
in to
Arithmetical Progression. n, then if the first n terms of the arithmetical
+ a, x + 2a, ... are divided by n, ... n - 1, taken in a certain order.
sion x, x 0, 1, 2,
.
,
the
progres-
remainders are the numbers
,
we suppose
that two of the terms as x + ma, x
+ m'a leave equal remainders, then their difference (m~m')a would be divisible by n. This is impossible, for a is prime to n and m-m' \
|
,
//
the progression is continued
beyond the n-th term the remainders recur in >
same order. For the terms x + ma, x + m'a leave the same remainder
the
If a and n are not prime to one another and g is divisor, the remainders recur in a cycle of n/g numbers. (2)
For
let
a~ga', n=gri, so that
a' is
prime to
n'.
if
m = m' + qn\.
their greatest
The terms
common
METHOD OF INDUCTION
9
if, and only if, a(w-w') is divisible by a! (m m') is divisible by ri. Since o! is prime to w', this can when m - m' is divisible by n'. Thus the first ri terms leave
x f m'a leave the same remainder that
is, if
o ly happen different remainders and, after that, they recur in order.
Method of Induction.
11.
Many theorems
to whole
relating
numbers can be proved by a process known as mathematical induction. In some cases this is the only method available. The method may be described as follows.
-
x
'
Let/(w) be a function of an integral variable n. Suppose that a certain statement S, relating to/(w), is true when na. Further, suppose we can prove that, true when n = + 1.
if
8
is
true
when n = m,
it is
also
m
Then
S
since
in succession, that
Ex.
is,
- 3w -
that 2 2n
Show
1.
2n -3n-l, Let/(tt) = 2 for n = l.
Also 2 2n
-
Hence
/(n)
if
=(3 + l)
divisible
is
-f(n)
1)
=2 2 w +1 (
= 3fc, where 9, so
by
fc
by
1)
;
so the theorem
9,
- 2 2W - 3 = 3(2^ -
an
is
is/(n +
divisible
is
)
a-f 3,
...
9.
by
=0 and
when n = a + l, a + 2,
true
it is
1 is divisible
then /(I)
f(n + n-1
Again, 1
when n = a, when
true
is
is
true
1).
integer,
and, since /(I) is so divisible, it follows that is, the theorem holds for ;
in succession that /(2), /(3), /(4), etc., are so divisible all values of n.
Or more Ex.
2.
using the method of (9) Ex. 2,
easily,
If
n
is
a
1
i
i
~Zi
n+2
n+1
1,111 __ 2n-l 23
111
i
i
j
i
T
i
7^
i*
^|*
1
1
.
1_
_
n Again,
= (3 + l) n - 3n - 1,
etc.
positive integer, prove that
11 ___T__
n
f(ri)
vn~^
if
t^en
Vti ^ f
1 -_
2n
= i>l + l
" ...+
23
1
1 I
i
~T~
i
4
2n -
,
1
1 -J_
2n + 1
*
"*
'
~~
I
"""rki"**
2n-l
2n
2n + l
1
^n+i-^wHence
if
succession,
un -vn then ,
it is
true
t*
n+ i=t;n+l
forn=2,
3, 4,
Now
.
...
,
that
the statement holds for n = l is,
for
any value of
n.
;
hence, in
ELEMENTARY THEORY OF NUMBERS
10
EXERCISE q
is
I
the quotient and r the remainder when a is divided by the quotient when a is divided by b + 1> provided that r^q.
1.
If q
2. If
is
a and b are prime to each other, show that - 6 have no common factor other than 2 (i) a + b and a 2 a* ab a +6 have no common factor other than b and + (ii)
6,
show
th*|
id
;
3.
and b are prime to each
If a
(a
have no 4. If 5.
If
common a
is
a+b
factor, unless
prime to b and
y,
and n is a prime, prove + )l(a b) and a + 6
that
other,
n + bn
and
is
b
is
3.
a multiple of n.
prime to
x,
then ax + by
ab' -a'b = l,
X=ax + by and 7 a'z-H&'t/, where X and y is the same as that of x and y.
prime to
is
ab.
common
the greatest
divisor of
If 2>
is
7.
a prime, and j9=a 2 -6 2 then a=%(p + 2 b 2 in two ways. Express 55 in the form a
8.
If
takes the values
6.
a;
,
(i)
1, 2, 3, ...
z 2 + x+
17,
(ii)
,
l) 9 b
\(p-
1).
in the expressions
2z 2 + 29,
(iii)
x*
+ a; + 41,
the resulting values of the expressions are primes, provided that in
(i)
#<16,
in
x<29, and in (iii) #<40. Verify for x = 15, 28, 39 respectively.
(ii)
9. Iff(x) is a polynomial, it cannot represent primes only. [Let uf(x) and v~f(x + ku), where k is any integer. Prove that u factor of v.]
is
a
10. For the values 2, 3, 4, ... 10 of x, the number 2 3 5 7 + x is composite. Hence write down nine consecutive numbers none of which is a prime. 11. If n is any odd number, then n(n 2 - 1) is divisible by 24 and if n is an odd prime greater than 3, then n 2 - 1 is divisible by 24. .
.
.
'
;
12.
that 2 n
Show
13. If
n
is
of the form
+1
prime to
or 2 n -
5,
1 is divisible
then n
2
+1
or
n z
by
3,
according as
1 is divisible
5m -f 1.
by
5,
n
is
and
odd or even.
therefore
n4
is
14. If n is prime to 5, then n 6 - n is divisible by 30 hence the fifth power of any number has the same right-hand digit as the number itself. 15. Show that 2 -f 1 or 2^ - 1 is divisible by 5, according as n is odd or even. 2n 2n 16. Show that 5 1 is divisible by 13, according as n is odd or -f 1 or 5 ;
even.
- 1 is divided by 13, show that the remainder is either 0, 2, or 8, as n is of the form 3m, 3m -f 1, or 3m - 1 ; hence prove that 3 W - 1 or according 3 2n -f 3 n + 1 is divisible by 13, according as n is, or is not, a multiple of 3. 18. If 2n -f 1 is a prime, then n must be a power of 2. 17v If 3 n
19.
Show
that 7 2n - 48n -
1 is divisible by 2304. 20. Show that 7*" + 16n 1 is divisible by 64. 21. Prove that 2 2W + 1 - 9?i 2 -f 3n - 2 is divisible by 54. 22. Find the number of divisors of 2000, and their sum.
1*
DIVISORS Let
23. is
s be the
called
a
perfect Show that, if 2 n *nd the three least '
,
24. If
<
1
;
8 the products, of the divisors of L and M are prime to one another,
numbers, and F, R,
a are the
r,
ft,
M
N
sum of the divisors of N, excluding itself; if s=N, then number. n~1 n - 1 is a and 1) is a perfect number (2 prime, then 2 numbers given by this formula.
NLM, and
L, respectively, where rove that (i)n=r.0, (ii)P=R s
''J7,
n
$ 25. If
the number, and
is
11
P
.Sr
.
the product, of the divisors of
N
prove that
9
=lfn.
N
26. If the product of the divisors of is
N
N
9 excluding the product of two primes or the cube of a prime.
itself, is
equal to
N
9
then
27. If N has 16 divisors, it cannot have more than 4 prime factors, a, 6, c, d, and it must be of one of the forms abed, a 36 3 a 36c, 7 6, a 15 Hence find the smallest number having 16 divisors. .
,
28.
Find the smallest number with 24
29.
Prove by induction that w(n + l)(w + 2)
30.
Find the highest power of the prime p (i)p
31. If less
N
is
r
divisors.
-l
...
in
(u)p
+r -
(n JV,
r
1) is divisible
by
\r.
when
-p
.
expressed as a polynomial in a prime p, with each of the coefficients s is the sum of these coefficients, prove that the power of p con-
than p, and
tained in |
32.
when 33.
N
is
(N - s)i(p -
1)
Show
N
is
that the index of the highest power of 2 contained in f - r when a power of 2, and is equal to 2 - 1.
N
Show
that
N
|2tt-l/(|n|w-l)is odd or even according
as
n
N
is
N-1
or
is
not,
|
is,
a
power of 2. 34.
Prove that
2n
is
divisible
[
35.
Prove that,
is divisible
by
m
by n
if g is
the greatest
1.
|
common divisor of m and n -f 1, then g
m 4- n
. \
n -f 1.
. \
|
Prove that the power of 2 in 2 in n + 2. n+1 [n 36.
.
n -f
.
|
3n
I
is
greater than or equal to the power Jf
.
|
37.
Prove that, ifn
38. If a, 6,
is
an integer 39.
c, ...
,
are
divisible
41 4f 2.
1
greater than 2, then
numbers whose sum
by
,
The
-f
1
is
by|n.|n + l.|n-t-2.
a prime, p then 9
that
wn
is
integer next greater than (3
integer next greater than (s/7
If 2"
3nis divisible |
p.
If tt n = (3 + v^5) n + (3 - V5) n show
Hence prove that the 40.
is
=&2/ prove that x
1
+ \^3) an
and y
-*
1
an
integer,
+ */5) n
is
is
divisible
and that
divisible 2n by 2
are divisible
n by 2
.
.
by the same power B.C.A.
CHAPTER
II
RATIONALS AND IRRATIONALS 1. Rational Numbers. number system is enlarged by
the introduction of fractions,
(i)
possible
in.
order that division
may be
always
;
the introduction of negative numbers and zero, so that subtraction
(ii)
may
Starting with the natural numbers, the
be always possible.
We
thus obtain the system of rational numbers or rationale, consisting of positive and negative integers and fractions, with the number zero. This system can be arranged in a definite order so as to form the rational scale.
With c
reference to
equal to
'
and
any two
'
less
than
'
rationals x
and
y,
the terms
*
greater than/
are defined as referring to their relative
positions on the rational scale. '
To say that x>y or y
To say
The Fundamental Laws of Order are as follows (l)Ifa=6 = 6 and 6 = c, then a = c. (3) If a^b and 6>c, or if (2) If a a>6 and 6^c, then a>c, (4) If a>b then -a< - 6. 2.
:
then 6 = a.
Hence we deduce the following rules for where it is assumed that zero is not used as a (5) If
b
+ x,
a-x = b-x
(6) If
a = b and x = y, then
(7) If
a>b
inequalities,
:
)
ax
bx
9
a/x
b/x.
then
and
ax^bx and a/x>b/x
(8) If
divisor
and
a= a+x
Also
equalities
a>b and x>y,
according as x
then a + x>b
both positive, then ax>by.
a-x>b-x. is
positive or negative.
+ y, and
if
a and y or b and x
ai!
REPRESENTATION OF NUMBERS BY POINTS Fundamental Laws of Arithmetic.
3.
Any two
13
rationals can
of addition, subtraction, multiplication and the result in each case /jvision, being a definite rational number, excepting lat zero cannot be used as a divisor. This is what is meant when it is
combined by the operations
j8
iid that the
system of rationals
The fundamental laws
;
f
(1)
(3)
4.
an
and multiplication are
+ 6 = 6 + a, ab = ba,
+ 6) + c = a + (6 + c), - ac + be, (a + b)c (a
(2)
(4)
(5)
The
closed for these operations.
a
a nd
fifth
is
of addition
= a(bc). (ab)c
and third constitute the Commutative Law, the second and
first
the Associative Law, the fourth the Distributive Law.
Theorem
integer
n
of Eudoxus.*
If a and b are any two positive rationals, nb>a. This simply amounts to saying that an greater than a/6.
exists such that
integer exists which
is
Representation of Numbers by Points on a Line. Take a
5.
OX
take a point 1 so that the segment 01 X'OX as axis in contains the unit of length. To find the point a which is to represent any Divide the segment 01 into n equal parts positive rational a, let a = m/n.
straight line
and
;
set off a length Oa, along
which represents - a
is
OX, equal
OX', at the
in
m
i
FIG.
The point
of these parts.
Ola
i
-a
X'
to
same distance from
as the point a.
I
i
X
1.
Points constructed thus represent the rational numbers in the following and one only. respects (i) For every number there is one point :
(ii)
The point point a
X
is generally taken to the right of to the right of the point 6.
Absolute Values.
6.
- a,
is
The points occur in the order in which the corresponding numbers stand on the rational scale.
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
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
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 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.
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
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)
= (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
(x)
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
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
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? (x)
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. (q + I)
<(#)
=
and
has a root between q,
so that
<(0),
(q)
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
Taking 9 as a trial figure, we find that this is too large, for decreasing the roots of (C) this shows that (0) and <^ (9) have opposite by 9 changes the sign of the last term ;
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: .
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
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.
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 (y), the (r-2)th approximation in x 3 ifj(y)
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
= 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 \ a limit different from
(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,
.* we have
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 6 n therefore the integral part of (^i + b n )jr n and so if r n = 2 fl
.
,
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
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.
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
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
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
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.
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
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
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
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
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
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,
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
