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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 + r
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<1; and let (>/5 - 2)n =/'. Then, since 0< V5 -2< 1, we have also 0'< 1. n w -2V +/ -/' = (5 + 2) - (v/5 - 2) =an integer Again, since n is odd, and thus than and and are less since 1, / /', hence, / /' positive Ex.
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.
Whether a number is regarded small depends on the purpose to which numbers are applied. A
7.
large or error of 6 inches
i
would be large in measuring a
but small in
table,
settir
out a mile course. If for purposes of comparison we choose some positive number c whuih we regard as small, then any number x is said to be small if x <. Any number x is said to be large if x >N where N is some previously \
|
\
|
chosen positive number, which is regarded as large. We say that x is large or small compared with y when x/y is large or small. If x/y is neither large nor small, x and y are said to be large or small
numbers of
We
the
same
order. '
use the abbreviation
order as If
'
y
is
0(x)
to indicate that y
is
same
of the
x.
is
small,
numbers which are
same order
of the
2
as
e,
,
e
3
...
,
are
orders respectively. called small numbers of the first, second, third, 2 2V 3 , ... are called If JV is large, numbers of the same order as N, , . . .
N
large If
orders respectively. numbers of the first, second, third, x - y is small, we say that x and y are nearly equal. . . .
Meaning of
8.
'
that x varies in such a less
To say that x
Tends.'
way
that
its
small.
is
If
from If
expressed by writing x-> 0. zero, where a is constant,
we say
or/rom the x tends to a and
above,
is
left,
always
less
than
a,
we say that x tends
)
is
great that number may be. In this case we also say that
from above, l/x tends to
;
oo
Any
,
we may
- x tends to - 00 Thus, and - l/x tends to - oo
collection of
.
number
as x tends to zero
.
numbers
is
the numbers themselves are the elements of the the
have what
is
called
an aggregate or
set.
^
of
called
to a
x becomes and choose, however
elements exceeds any positive integer however choose, great, we say that their number is infinite^ & If
and
to say that
remains greater than any positive number that
set
to a,
and we write x-> a -0.
tends to infinity (x -> oo
Aggregate.
that x tends
is
below, or from the
9.
how
always greater than a, we say that x tends to a This is expressed by writing x -> a -f 0. right.
x tends to a and
from To say that x
to say
choose, no matter
may
x - a tends to we write x -> a. If
is
numerical value becomes and remains
than any positive number that we This
tends to zero
an
infinite set.
^^
APPROXIMATE VALUES 10.
System Everywhere Dense.
system of rationals
many
infinitely
For
if
is
An
important property of the that between any two rationals a and b there are
is
rationals.
a
This fact
15
k
is
any
positive rational,
it is
sometimes expressed by saying that
easily seen that
the system of rationals is
everywhere dense. 1 1
.
A
Sequence.
according. to
some
succession of
definite rule,
Such a
denoted by (u n ).
numbers u ly u 2 u 3 ,
called a sequence,
is
rule defines
,
...
un
...
,
which
is
u n as a function of
,
formed
generally
the positive
integral variable n.
The rule may be quite arbitrary, and it is unnecessary that we should be able to express u n in terms of n by an algebraical formula. For instance, u n may denote the nth prime number, or the integral part of Jn.
A
sequence in which each term
is
followed by another term
is
called
an
infinite sequence.
12. Approximate Values. Suppose that the object of an experiment is to determine a certain number represented by A. No matter what care may be taken to ensure accuracy, it is unlikely that the exact value of A can be found. All that can be done in most cases is to find two numbers a and a' between which A must lie. If we find that a
is an approximate value of A, A -a is called the and Error, (A a)/A the Relative Error, in taking a to
a positive real number, there and one only, whose n-th power is equal to a. 2.
//a
If a = z n where #
is
is
a positive rational, there
is
is
one positive real
only one such number
x,
for the nth If
powers of unequal positive rationals are unequal. no such x exists, divide the system of rationals into a lower
class
A
and an upper class A' according to the following rule The class A is to contain every negative number, zero, and every positive :
',
number x such that x n a. Then (i) no rational escapes classification for no rational x exists such ;
that x n =a. (ii)
Every #
(iii)
There
is
x',
for
sc
n
no greatest x and no
n .
least
x
f .
This follows from the last
theorem. Therefore the classification defines a real number Again, for every
x and every
.
x'',
xn
xn
and
.
Also every rational is an x or an x', therefore these may be chosen so x is as small as we like. Hence, as in Theorem 1, they may be where e is any positive number, however small. chosen so that x' n -xn
that x'
<,
n are defined
Thus both a and
by the same
classification of rationals,
a classification which satisfies the conditions of the theorem of Art. 4 and n
therefore
=a.
a positive real number, the principal n-th root of a defined as the positive real number whose nth power is equal to a. This Definitions.
is is
If
written J^a, and
a
is
is
generally called the n-th root of n
Thus
(;/a)
a..
=a.
It follows that
ya.y]8-y(aj8) If
n
is
odd,
we have (-
When n
is
WH^a^ y(.
and
an odd
simply the n-th
n==
(-
1 )n
)
w== -a.
we therefore define the -a as - ^a thus
integer,
root, of
(^a
principal n-th root, or
;
even, the 7^th power of every real number is positive, and therefore no real number exists which is the nth root of a negative number. If
n
is
SURDS
208 Indices
which are Rational Fractions.
If p, q are positive integers,
?
aq
is
p defined as *]a* or (!ja) This definition assigns no meaning to a x is
.
considered in Art.
Surds.
8.
If
a
when x
is
This case
irrational.
9.
is
not a perfect nth power, ^Ja
is
called a surd of the
nth order.
Two rational
surds of the same order are like surds
when
their quotient is
otherwise they are unlike surds.
:
The following theorems and examples depend on the number cannot be equal to an irrational.
Theorem
1.
If
x+Jy^a + jb
fact that a rational
x, y, a, b are rationals, then
where
x=a
and y = b,or else
y and b are the squares ofrationals. For suppose that x-^a and let x = a 4- z,. then z + Jy
Jb,
and by squaring,
Therefore ,Jy is rational, and from the given equation the other hand, if x = a, then y 6.
is
Jb
rational.
On
Ex.
1.
// a
+ b^/p + c*jq
where
a, 6, c are rationals
and Jp
t
^/q are unlike surds,
thena = Q, 6=0, c=0. By transposing and squaring we can show that 2 c*q ~a b~p. 2ab^/p
If
ab^Q, the left-hand side would be irrational and the right-hand side would
be rational ; which
is
impossible.
Therefore a6=0, and consequently a
or
6~0.
a=0, then b*Jp + CsJq~Q; and, if 6^0, then Vp/\A?~ -c/6; and which is not the case. */p *Jq would be like surds Therefore 6=0 and c = 0. //
N
x and first let Suppose that no rational x exists such that a = x x '. n>l. Then, by Art. 7, rationals x, x' exist such that a Divide the system of rationals into a lower class A and an upper class A'
Proof.
by the following rule The class A is to contain every rational x such that ax N. Then (i) no rational escapes classification for it is assumed that no for a x
;
x
This concludes the proof of the statement (i). r Again, s being any positive integer, by Ch. Ill, 16, the coefficient of x 2 s m in is the sum of all terms of the form (bx + ex + ... 4 kx )
EXPANSIONS where
. . .
y,
j8,
ic
have any positive integral or zero values such that jS
and
+ 2y + ...+w/c = r, K
y-K..+
/?+
from (A) that the coefficient the sum of all terms of the form It follows
is
n(n-l)...(n-* + T-7f
where the values of of s
by
Ex.
1.
Here
j8,
y,
JPt'nrf f/ie coefficient
n~
--J,
j8,
y, 8
x r in the expansion of f(x)
of
l)
fl
c/
A;
. . .
,
j
of X* in the expansion of (\
r
^4, and the
( v
~i)( -i--l)( v -i ~2) I _JLI *_ -_ _z
coefficient is the ...
(a
sum 1
.
2#(
3
8
terms of the form
- 4) v ( - 2) 6 ,
have positive integral or zero values such that
and the corresponding values of a are given by
~~
+ 2x - 4# 2 -
of all
+ 1)
i
-
where
(B)
S ............................... (C)
c; i
..............................
are given by (B) and the corresponding values = n-s. (ii) follows on writing a
...
The statement
(C).
355
/J
+ 2y + 33 -4,
a+/?+y + 8=~-|-.
The possible values of a, /S, y, 8 are shown in the margin. The term corresponding to the first set of values is
a
y
ft
2>
>
-3i,
2,
1,
-4,
4,
0,
-2j,
0,
2,
S
'
j
Similarly it will be found that the other terms are ^, 6, so that the required coefficient is
15,
-3 + 15+4f + 6=22-|.
EXERCISE XXXV 1.
By
n considering the expansion of (1 -a:)~
A ~r H H
n(n +
n(n +
1)
,
show that
l
r
r^,:
2
L 2.
Show
that, if
z
= ~^, and x+y
2 |
|
<1, then n-f2);
3. If n, r are positive integers (including zero), show that n f| - 1 2 . t he expansion of (1 2r)2 x) /(l -a;) is (n
n+r-i
^
+
+
n - 2 [Expand {2 - ( 1 x)} l ( 1 a;) .] 4. If w is a positive integer and (1
show that
+ c*i + a, +
+x) n .
+ an_! = $n (n 4- 2) (n + 7) 2n
-4 .
the coefficient of
COMPARISON OF COEFFICIENTS
356 5. If
n
a positive integer, show that
is
4
3.4n(n-l)
= coefficient of z n 6.
If n
Prove that
Show
n
if
!?L 8.
{1
p
in
{2
-
-
(1
n x)} l(l
(n
+ l)(n + 2)
- x)*=(n z + In 4- 8)2W ~ 3
"~~ ""
1)
(
"
f
-# (2 -x)}~ 1 = (I -x)~ 2 .]
'
JL??_
that, for all values of
p 4-^ = 1 and
n(tt-l)(n~2)
+" ~
'
*
m and n
9
coefficient of
x r in the expansion of
n, r are positive integers (n>r),
put 1 q for p, and show that the hand side when expanded is
(1
+x)
n~m
.
show that
p
r
,
coefficient of q k
^coefficient of z fc in (1 +3;) n x (1 +a?)-
+0^+
value, find the
sum
2 r==Q mc x
of the series,
r
IL^J; 14. If
'
n
[Divide by
any
.
a positive integer,
is
n n
= the 9. If
5 n(n - l)(n -2)
a positive integer, show that
is
[From the expansion of 7.
.
12.
r 9
when
on the
left-
- r - fc+1 >.]
n may have
m is
,-,!-=,;
13.
ur =Ao + A lr + A s r(r-I) + ...+A h r(r-l)...(r-h + l) and ~ h~1 h 4- A l nx + A 2n(n- l)xh 2 -f + A h n(n ~I) ...(n-h + l),
(x)-A Qx
. . .
prove that
15.
Prove that
CfCfJlJI
[Consider the product 16.
Show
(1
+ Cf^Cfilf H-(7J +1 C;Si; + ... T +^ x (1 - X )~^- T \] -x)~(
that
3 2n + a
1
to 40 terms -6' JJ.
-3
2714-3*
the summation being taken for n. 9 r such that p + q + r [Consider the expansion of
all
positive integral values, including zero, of
p q y
(a 4-6
+c)
2n+3
- (6 *
-f
c
-a)
2n + 3
- (c 4-a - ^>) 2n+3 - (a 4-6 -c) 2n + 3 .]
Check the answer by putting
nl.
CHAPTER XXII RATIONAL FRACTIONS
RECURRING SERIES AND DIFFERENCE
(2),
EQUATIONS 1.
Expansion of a Rational Fraction.
fraction in x,
Any
rational proper
whose denominator does not contain # as a
factor,
may
be
written in the form
P
+a _ - - -+p
+ a 1 a; + a 2z2 + ... 2 l+piX+p^x + ...
a _
Q
mx rx
r
Various ways of expanding such a fraction in a series of the form
Zu n x n
are given below.
can be shown that the
obtained by any process is convergent for all values of x in any interval including zero, then by Ch. XX, 20, any other method of expansion yields identically the same series. If it
2.
Theorem
.
The
series
rational fraction
P/Q can
be
expanded in a convergent
series of the form
U + u l x + u 2x2 + if
and only
if
For, by the
x \
\
<
A |
method
of several terms of the
|,
where A
+u n xn + ...
...
is the root
of partial fractions,
of
,
$=
with the least modulus.
expressed as the sum a constant, r a positive
P/Q can be
form A(x-a,)~- r where ,
A
is
integer and a is any root, real or imaginary, of the equation Q Q. All of these terms can be expanded in a convergent series of ascending
powers of x if, and only the least modulus.
When
this
the various Ex.
1.
is
if,
x |
\
<
A |
|,
where A
the case, the expansion of
is
the root of
the fraction
be expanded in a series of ascending powers of x? Equating the denominator to zero, the roots are
-iV2
and t(i
and their moduli are -s/2il, >/5, ^5. Hence the expansion is possible if and only
with
P/Q can be obtained by adding
series.
For what values of x can
$=
if |
x \<*J~ -
1.
RECURRING SERIES
358
Methods of Expansion.
3.
where the Art.
1.
(1)
Suppose that
assumed to be convergent and P,
series is
are as described in
Q
Multiplying by Q,
Expanding the right-hand side and equating uQi u v u2 ... are determined by the equations
coefficients, tjie values of
,
Hence, after a certain stage, any r successive coefficients are connected by a linear relation. A series having this property is called a Recurring Series. (2)
The process
equivalent to synthetic division carried Ch. Ill, 3.) (See
- 9, + 12 6 - 3 + 1 =4, the next term in (d). diagonally in (c), (6), (a) do this example by the method in Art. 3, (1), he will see that the reckoning is essentially the same as that just described.
put
6,
;
If the reader will
(3)
The method of Partial Fractions.
Ex.2. If We have
where
Hence
Now hence
a,
j8
1/(1
-2ar+5x 2 )=w
a;
1
1
are the roots of i
+w1 + ...+Mwarn + ...
x*
find the value of un
_
= _1 /__?_ _ _J )~a-8 \1 -a* 1 -j -j8*)~a-j8
- 2x + 5 =0, so that we
may
take a = 1 + 2*,
i
l a=v/5(cos0 + isin0), 0=V5(cos0 -tsin0), where 0=tann +1 n ^ 1 =5*(n + 1 /. a ).2tsin(n + l)0; -/3
w n =i.5*(n fl )sin
.
-
(n
+ l)0, where 0=tan"-
!
2.
2,
ft
= 1 - 2i.
HOMOGENEOUS PRODUCTS
A
(4)
fraction can also
expanded by the Multinomial method and that -of Partial Fractions and equating
rational
Theorem.
359
Using this we can obtain
be
many important
coefficients,
results, as in
the following
exercise.
EXERCISE XXXVI Use synthetic division to expand
1.
w (l-a;)
v
2
'
1 (l-2*)' as far as the terms containing a; 6 Find also the .
Show
2.
that, if
p and
coefficient of xn in
each expansion.
2 q are real numbers, the condition that 1/(1 4- px + qx ) series of ascending powers of a; is as follows
can be expanded in a convergent 2
p >4#, then
(i) if
<
x
|
\
:
- Vp 2
\
(p'
*q)/2q
where p'-\p\\
|,
(ii)ifp*<4q, then \x\
x lies between a and show that the fraction l/(x -
|
|
|
|
. . .
[Let If
The
|a|
,
= (a- ]8){l/(a;- a) -!/(*- J8)}.
fraction
|a|<||<|j8|, prove that
and
H
[Hn i.e.
is
sum
of the homogeneous products of n dimensions which can show that dn +*(b-c) + b n +*(c-a)+cn +*(a-b)=-Hn (b--c)(c-a)(a--b). the coefficient of x n in the product
4. If n is the be formed with a,
6, c,
The
in the expansion of 1/(1 -ax)(l -bx)(l -ex). fractions.]
result
obtained by the
is
method of partial 5.
Prove that, for
sufficiently small values of x, 1
1
a n =p n
where
n
-(n-
""
2
l
-px + qx*~~
~
+
l)p ~*q
n
-p
n
~*q*
-
. . .
,
I
the (r+ l)th term being 6.
(
T
l)
r
C^- p
n - zr r q
.
use the identity
If a-f/?=jp, aj9=5 , f
/
to prove that
an + p" =pn - pr
n~2 j3
Li the (r + l)th term being
[Expand both
sides,
(
I)f
g
+
1
-
__ *>\ ,
p
f-
w~ 4
g
1
-
2
-
1
fix
. . .
-px + qx*
,
!j_
^~>-l)^-^-2) - (n-2r + l) yn _ ifgr
equate coefficients and use Ex.
5.]
SCALES OF RELATION
360 7.
+ p=p, 4=9,
If a
to prove that
-
a
p
the (r + l)th term being
[Expand both 8. If
%
is
sides
(
(-1?
^n
*~ r "
and use Ex.
~^
(r
(ii)
+ 1) th term
552* sin
-*V-- y
.
5.]
n 0)
-
~ (2 cos B) n ~
2
+
r
Li
the
"' (n
a positive integer, prove that
2 cos nB = (2 cos
(i)
-
use the identity
being
(2 cos
-
r
0r- -
-
the (r+ l)th term being
(
n n~r~ ^
^
2
I
n (2 cos 0)
~4 -
...
,
_
l)
(
^r~
(2 cos
1
(n
*) "' (*>-
^
~ T ~ l)(n
(
(2 CO8
"- 3
4)
y-
"'
(
-
(2 cos *)
12
~*~ 2)
)n -sr.
"~ 2r)
1)'
(2 cos
...
,
0)-->.
7"
[In Exx. 6, 7 put a 9.
(ii)
Show The
that
(i)
= 2, p = z~
l
where
z~ cos
0-fi sin
0.]
the coefficient of x m in the expansion of -~
coefficient of x* n in the
expansion of
-- -
--
-
(1
-x)(\
-x
;
2
)(l
- x4 )
1
+#
is
(n f
2 Recurring Series. (1) Let u Q + u 1 x + u 2x + ... be a series which any r + 1 successive coefficients are connected by the equation
4.
u n +PiU n -i+P2U n _ 2 +...+p r u n _ r = where r
is
a fixed number and
p l9 p 2
,
...
pr
)
is
-, ''
1 )
2 .
in
.................... (A)
are constants.
Such a
series is
called a Recurring Series of the r-ih order.
Some authors
equation (A) the Scale of Relation of the series 1 +p 1 x+p 2 x 2 + ... + p r x r others use this term to denote the polynomial We shall take it to mean either of these things. call
u l9
;
.
u r _^ are known, the subsequent coefficients can be found in succession by equation (A). Thus a recurring series of the rth order depends on 2r constants. Hence if the first 2r coefficients are given, in general the series can be continued as a recurring series of the rth order, and in one way only. Also If the coefficients
it
u
,
...
can be continued as a recurring series of the (r + l)th order in a doubly number of ways. For to do this we can give ur and ur+l any values.
infinite
2r terms belong to a recurring series of necessary that this may be so, and then the series can be continued as a recurring series of the (r - l)th order.
may happen that the first order r-1. Two conditions are It
CONVERGENCE OF RECURRING SERIES Ex.
361
Discuss the question of continuing 2un x n as a recurring series when the
1.
first
six terms are known.
un +pu n _ l + qu n _ 2 + run _ 3
Let the scale of relation be
Then
are given
q, r
p,
u3 4- pu2
4-
qu
0.
by
4- r u
w4 4- pu9 4- #1*2 4- r^j
0,
Now Sunx n
can be continued as a recurring definite values and rr^O. This will be the case
w e 4- pw 4 4- g?/ 3
0,
4-
ru 2
order
series of the third
if
0.
have
p, q, r
if
and
(A)
"3
In case either determinant then
is
zero,
suppose that the scale
%
u & +pu^ 4- gw 3 =0.
That these equations may be
one of the
u n ^-pu n _ l
W 4 4-^ 3 4-
first
+ p W2
4-
^ Wj = 0,
consistent, each of the determinants in (A)
vanish, so that two conditions are necessary that the second order. If it is
is
u2 4- p^ + qu = 0,
order, the scale being
Eun xn may
must
be a recurring series of
un +pu n ^ 1 =0, then
Ut+pUQ Q, +pu 1 = 0, ... uB +pu4 0, 2 conditions u l -u^u 2 = Q u 2 2 -u u 3 =0 w 3 2 -w 2 w 4 u2
so that the four
must be
(2)
9
9
1
u4 z
0,
-u 3 u5 =Q
satisfied.
Any
u n xn
recurring series
is
convergent for sufficiently small values
of x.
For
let
the scale be
u n +p l u n _ l +p 2 u n __ 2 +...+p r u n _ r = 0,
and let U Q Then by (A),
|,
|
|,
... |
pl
|,
p2
|
|,
be denoted by UQ ',
...
u,
p, p%',
...
(A) ....
'
the greater of the numbers Pi +p% + +p r an d 1, and un ^ the same way the greatest of the set u' ... In __ n 2 l9 UQ.
where # is
u |
..................
is
u^
where
w^
is
,
the greatest of u^_ s _ l) u'n _ 8 _ 2
Continuing thus, by multiplication
where
A
is
the greatest of Wn-i> Wn-2>
Thus |
,
...
u$.
we can show that |
...
2u n x n
\
u \
u nx n \^\gx\ n .A,
n convergent if x | for then w nx (w4-l)th term of a convergent series of positive terms.
and
un
is less
is
|
|
|
than the
GENERATING FUNCTION
362
n
when convergent, of any recurring series Zu n x a rational proper fraction, which is called the generating function of The sum
(3) is
to infinity
,
the series.
Let the scale be
U n +PlU>n-l+P2U n _ 2 +...+pr Un __ r = Q, and
..................
(A)
let S
= w + w 1 x + w 2 x2 -f ...
to QO ......................... (B)
p^, p2^> p rxr and adding, we find that + p& 4- p 2x2 4 4 p rx?) = a 4- djX 4 a 2x2 4 + a r _ 1 xr - 1
Multiplying (B) by s(1
where
. . .
a
=w
al
,
=w
. . .
4r
,
......
(C)
a 2 = t/ 2 4 jo^
^WG,
the coefficients of powers of x higher than x r
~l
vanishing on account
of equation (A).
Thus
s is
It follows
A
ti;Aer6
i
equal to a rational fraction, of which the denominator
from Art. 2 that
// w n
(4)
i
a recurring
is
Q=
iA6 root of
To prove
a
of which the scale
is
(1
-x)
n of r+l
is
r,
then
Zun xn
. . .
Let ,
1
so that v n is a polynomial in
,
r steps of this kind,
where &
degree
n 2 - x) = w 4 v^ 4 1 2^ 4 ... 4 v nx 4
s (1
where vn = u n -u n _ l
|,
.
we can proceed as in Ch. VIII, 7. s = u 4 ux 4 u%x 2 -f 4 u nx n . . .
then
if\x\<\ A
wjAicA Aas $Ae Zeas^ modulus.
positive integral function of
series
this,
2Ae recurring series is convergent
is
we
. . .
,
n of degree r-1.
By
find that
independent of
x,
and therefore
s(I-x)
r+ l
~u +(k-u Q )x,
which proves the theorem, Ex.
2.
Find
the
sum l
This
by
is
to infinity 2
of
42 2*43 2a: 2 44V4..., when |*|<1.
a convergent recurring
series,
and the
scale
is
(1 -a;)
8,
= =
-3
.
l*x
-3
.
2 2 *2 - 3
.
3.
and
S--
8 .
Denoting the
sum
FINITE DIFFERENCE EQUATIONS The sum
(5)
n terms of a recurring
to
similar to that in Ex.
Find
3.
the
to
can be found by a process
or in the last example.
(3),
sum
series
363
n terms
+ u^x 4- u 2x2
of w
+
. . .
,
where u n
+pu n _ t + qu n , z =
n>l.
for
=u + u lx+ u2 x 2 + ... + u n _ xn ~ 2 + pu n _ 2 xn - + pu n _^xn pas n = 2?tt # -f >Wi# +
Let
l
6n
l
1
then
. . .
Since w r +7>w r _ 1
+ ^w r _ 2 =0
for
::
*n (1
+ JP^ +
^
2 )
t
r>l, by addition,
~ uo + w i +P ?/o) x ~ u n xn (
which determines the value of sn unless a; is a root a of the equation 1 In this case the sum may be deduced from the general result by supposing that #,
Given the
(6)
uv
w n -i-^ 1 w n _ 1 +
scale
ur _ v we can
...
find the value of
4-j? r w n _ r
= 0, and
the values of w
,
u n by
(i) finding the generating function (ii) expressing this as the sum of partial fractions of the form ~ ~ m where a is a root of xr A(l-ocx)" +p 1 x r l + p 2xr 2 + ... -f 7? r = 0; and n (iii) expanding the partial fractions and collecting the coefficients of x ...
;
}
.
Examples
Examples of Finite Difference Equations.
5.
any
are given in the next article.
r
+1
consecutive terms of the sequence (u n )
u n +PiU n ~i+P2U n _ 2 +
equation
where p l9 p2 ...pr are constants. ,
is
From
...
Suppose that are connected by the
+;p r w n _ r = 0,
................. (A)
this point of view, equation (A)
called a Linear Finite Difference Equation with constant coefficients. So far as equation (A) is concerned, the first r terms of the sequence
may have any
values whatever, that
is
to say they are arbitrary constants.
Thus the process described in the last section enables us to find the general value of u n which satisfies equation (A), and shows that this value involves r arbitrary constants. The general value of u n is called the general solution, and the process of finding it is called solving the equation. The process leads to the following results. The general
(i)
(A
-f
wJ5)a
n
where a,
,
u n + pu n _ l + qu n _ 2 ^=Q is wn = ^4a n -f J3/?" or are the roots of x 2 +px + q = Q and A, B are arbitrary
solution of j8
constants, the first or second form being taken according as (ii)
The general
where a,
j3,
y
solution of
//a^^y,
then
2 +px + qx + r = Q and A,
no two of
the roots are equal. n a n -f
u n = (A+ nB) u n = (A+nB + n*C)oL n
If a = P 7^ y, the solution 2A
u n +pu n ^ l + qu n _ 2 + ru n _3 = Q
are the roots of x 3
constants, provided that
a^/?
is
Cy
B,
or
a = j8.
is
C
are arbitrary
.
.
B C A -
-
-
GENERAL SOLUTIONS
364 For any
(iii)
other value of r the general solution of equation (A) has
similar form.
We If
give the proof of
w
,
(ii).
%, u 2 have any values, then
+ u 2*x2* + + u^x l
UQ
where
a, 6, c
where
a,
j3,
. . .
+ u n xn +
are functions of ^
y are the roots
. . .
=
-a + bx + cx2
_
u v w2
,
~ 2
==
3
P
a
.
J 7: (say),
Q
Also
.
of
# 3 + px* 4- qx 4- r = 0. If
(i)
no two of
where A, B,
a,
j8,
are equal,
y
C
are arbitrary constants. the fractions on the right and equating coefficients, Expanding
(ii)
If
a = /? 9^ y, then
P/4 (iii)
If
w
JJ
u n = (4
and
f^
JD
^L
-f
nfi)a
n
+ Cy w
.
a = j3 = y, then
B
P__A__ ~ + T - ocx
Q
(^
+
(l^axp
(1-a;
u n = ^4o
and
where
JB',
C' are arbitrary constants.
The first term of a sequence is 1, the second is 2, of the two preceding tertrw. Find the n-th term.
j&r. 1.
sum
Denoting the sequence by u l9 w 2 , ... we have u -u and n -2^Q
Hence
Eliminating A,
,
B from these (a
-
equations,
a
-a
we
0K = (2 - )3)a" -'
Now a+0 = l, a/3=-l and 2
1,
other term is the
U 2 ^2.
2-Ao^ + B^, l=.4a + ^, where
1
Un^AoP+Bp
^=
n~i-u
n
and every
a,
j8
are the roots of
find that
-
(2
- )"-'.
a>/8, a- 8=v'5, Therefore -1=0), and similarly 2-a=/?
hence 2-y8
if
J
2
.
I
-d
-
= l+a=a 2
(for
GENERAL TERMS Ex.
Find
2.
a recurring
the n-th term of
which
series of
Find also the sum of the first n terms when x ~ 1 and Denote the series by w -f u^x + u 2 x 2 + . . .
,
365 the first
.
let
the scale be
7+2p+# = and 20 + 7^4-2^=0, giving p= -2, un -2u n ^ l -3u n _ z -Q. The roots of x*-2x-3^Q are 3, - 1, and therefore Then
is
un =A.3n + B(-l) n
A+B^\,
where
Thus the nth
-^2.
3.4
n
term-w^a;"- -i-[3 + ( 3
.*.
sum to n terms = -
1
l-3#
4
---h
--
1
Now the
sum
to
Ex.
3.
The
solution
n terms
are arbitrary constants.
The values
of a,
ft
Hence the values
~iin
a,
==
;
...
n-1 + u n _ 2 =0 ^w a rea/ /orm. are the roots of x*-x + \
and
j8
Changing the constants,
this
^3
.77
may
(7,
be written
are
an
of
,,!.. solution
;
.
1
-w
scale
n ~l
-2 + 7- 20 +
1
of un + />^3n where
1
and the
v\n
of the series
Can
un
]x
1+s
J'iwrf the general solution is
l)
!-(-*)" _
/
and the
-3,
,
iZL-^^ limLJL
lim
q=
B=^\. n ~l
--
l-3w z n
4
/.
A-%
Hence 1
four terms are
,
j3
n are cos
t
sin
-^~, o
o
AA
un
is
7T
r + B am -
cos u
and so
~u
EXERCISE XXXVII 1.
Find the wth term and the sum to n terms of the recurring (i) 1
2.
is
Show
+2+5+
Show
vergent
.
;
1+2 + 5 + 12 +
(ii)
x |
|
-
<
sum
to infinity
..
.
series
:
.
is 7J-.
un -2un _ 1 + 4un _ 2 -3un _ 3 = Q
that, if
if
infinity is
+ ..
that the recurring series
convergent, and that the 3.
14
;
and
if
the
first
the series
9
three terras are
1
2unxn
+ 2x + 3#
2 ,
** ( i
^ ~\
9
_1_ 1*2^i ~^~ fjjL,
//Ix
; i
_
f?l* 4Utfe
I
n^
4.'/*^ xC/ <
-
^f*^ O*<
!
3 2 [The moduli of the roots of x -2# + 4x-3=0 are
1, 1/^/3, l/>/3.]
the
is
con-
sum
to
EXAMPLES OF RECURRING SERIES
366
3un -lun_ l -{-5un _ 2 -~u n _ 3 =0 and
4. If
%=
u 3 = 17,
u 2 = 8,
!,
the
find
un and u l
values of 5.
Find the nth term of the recurring
6.
If u n
series
l-f2-f34-54-74-94-...
.
the nth term of the recurring series 14-24-34-84-134- 30 4-... , find the sum to infinity of the
is
show that un ={(3n-4)( -l) n -f 2n + 2 }, and 8eries u If w n
7.
-^~w _ n
un ain6~u 2 q
+ gw n _ 2 =:0, where ^ 2 <4g, and -^7r<6<^rr show
1
9
2
2
?L:J
sin (n
- 1)0
2
-u^
sin
tan#=
where
(n- 2)0,
that
\/(4-7>
2 ) .
J
Zu n x n Er n xn
If
8.
and
1
,
+p'x + q'x
2
then
,
are recurring series of which the scales are 1+px + qx 2 2(un + vn )xn is a recurring series whose scale is (
ZW n ZV n
9. If
1 -\-px 4-
2 flo:
) (
1 4- 2>'z 4- q'x*)-
are recurring series of which the scales are and vn + p'u n 4- q'u n__ 2 = 0,
,
un + pu n _^ 4- qu n _i ^
^
7^ 4^', then Sun vn is a recurring series whose ^ 4<^ and un ~PP' un-i + (PV +P' 2<1 - 2^0^-2 -pp'OU'lln-z + '^4-g ~0 n n /n n /n n " n then tively, 4- ^?a /3 4- 6V + Dai' p Un r n ^Aoc where ^4, B, C, I) are constants. Hence the scale of, Zun vn xn is 1 - OLOL'X) ( 1 - *P'X) 1 - affix) I ~ OL'fi'x).] (
where
2
>'
^>
2
7
r/
scale
is
respec-
'
,
(
10. If 1
2un
- 2kx 4- A: 2x 2
,
xn
Evn
are recurring series whose scales are then Eun r n xn is a recurring series whose scale ,
Hence show that Ununxn
Eun
(
xn
xn
is
a recurring
series
whose
scale
l+px + qx 2 and is
is
(1
a recurring series whose scale is 1+px + qx 2 whose scale is [1 - (p 2 ~ 2q)x 4- g 2 # 2 ] ( 1 - qx). are the roots of z 2 4-pz4-?=-0, then u n * = A(ot*) n [If a, fi where A, B, C are constants. Hence the scale of Eu n 2 x n is 11. If
is
a recurring
12. If
-
is
-
l+px + qx
13.
,
then
Uun
2
xn
series
+ B(fi*) n + Cq n
,
(l-a 2^)(l-^)(l-^).] , 2
= ^r u xn n==0 n
Prove that each of the COS a sin
use Ex. 11 to show that
series
4- X COS
oc + x
,
sin
(0
4-
a)
4-
X 2 COS (20 4- a)
4- ...
,
2
(04-a)+# sin (204-a)4-... 2 being 1 2x cos 4- x in each ,
Show
also
and that the sum to n terms of the second series may be obtained from expression by changing cos to sin in the numerator.
this
a recurring series, the scale that the sum to n terms of the is
case.
first series is
cos a - x cos (a - 0) - x n cos (a -f nO) + xn + l cos (a _____._____
+n -
Id) 9
FORMATION OF DIFFERENCE EQUATIONS Finite Difference
6.
367
Let (u n )
be a sequence in u n ^ r are connected by
Equations.
all of the terms u n u n __ v u n _%, ... an equation which holds for all values of n^r, where r is a fixed number. Such an equation is called a Finite Difference Equation.
which some or
An
,
important class of these equations has been considered in Art.
5.
Here we give methods which apply to various cases which occur in ordinary algebra. The general theory belongs to the Calculus of Finite Differences. To solve a difference equation is to express u n as a function of n in the most general form. The result is called the general in Art. 5
it is
clear that this
and from what has been said solution must involve one or more arbitrary solution,
constants.
A
particular solution
obtained by giving these constants special
is
values.
un
a given function of n containing one or more arbitrary constants, a difference equation may be obtained as in the following examples. If
Ex.
is
// u n
1.
~-Gn-2
where
C
an arbitrary
is
constant, obtain the corresponding
difference equation.
The required equation
is
found by eliminating
un ~Cn-"l
- l)w (n n
giving
C'
from
^n-\~ V(n -
and
1)
-2,
-nw n- 1 = 2. .
n n where A, B are arbitrary constants and // un ~AoL +Bf$ numbers neither which is zero, prove that unequal of u - (* + $)u -=0. fl _
Ex.
2.
The
result is
n _i +aj8t*
n
found by eliminating A
Solution of
7.
,
a,
ft
are given
8
B from
some Elementary Types.
n // u n = au n _ l9 then u n = Ca where C is an arbitrary constant. For w n /w n _ 1 ==M n _. 1 /w w ._ 2 ==... =w 2 /w 1 = a, and u v may have any value. (1)
// u n = a n u n-i where a n
(2)
where
C
is
For *or
an
Wi/a i
n> then
arbitrary constant.
Un ~ l ~Un un-\ u n-Z *
un =
therefore
and
a given function of
is
mav have any
value
...
al
we
^-aa --a na n -i ul
a^
. . .
choose.
a ny
a
<*2i
w n = Ca xa2
...
an
SOLUTION OF ELEMENTARY TYPES
368 (3)
= bn
// u n a n u n _i
where a n and b n are given functions of n, the
general solution is
where
C
is
an
arbitrary constant.
Let vn = a 1 a 2 ...a n
The
result
,
and divide each
side of the given equation
by v n
.
is
Un/Vn~Un-lK-I = b n/ v n' In
this,
writing n
Whence by
n~
1,
2 in succession for w,
...
we have
addition,
wK This
2,
the
gives
^1/^1
= &iM + 6 2 /v 2 +
fe
the value of
question, for
in
revsult
+ & nK - iKu l lv l -bl jv l
is
arbitrary.
In examples of this type, it is better to apply the rather than to use the actual result. -l)u n - nu n _ 1 ~2. we have Dividing by n(n-l), u n \n - u n ^l(n - 1) =2/ Ex.
1.
1,
nn
-2,
...
2 in succession for
U.2
I
Hence un ~ Cn-2 where C. (See Art. 6, Ex. 1.) Ex.
2.
Find
(4)
we
2.3
C~ 2 + u v
the general solution of
(n
-
1).
n and adding,
(n-l)n) and
since u t
u n - nu n _ { =
I
is
an arbitrary constant, so
also
M.
and proceeding the equation becomes un j\n--u n _ l l\n-\ l, find that wn + (7) where C is an arbitrary constant.
Dividing by as before,
just described
Solve the equation (n
Writing n
is
method
n,
|
The general
~[n(n
solution of
^a n + c according as a and
?/
n
n
~aw n _ 1 = cjSn
+Y03--a)
is
or
are unequal or equal.
j3
For dividing by an we have ,
Writing n-1,
n-2,
...
2 in succession for n and adding,
a
LINEAR DIFFERENCE EQUATIONS u n = coc n
Therefore
=u
where B( Hence,
l
#
-
gn r
a
if
+ -^2 + [a a -
{
.
.
.
n
369
l
+ r-n \ a
j
an arbitrary constant.
is
/ot-cft/oi)
B
(B
2
_ an
-
n
n
,
/3-a which-
may
be written
where A(^B-cft/(ft~oc)) If a = ]8, then i/ n = cna n
The general
(5)
is
an arbitrary constant.
-
w n ~(a-f j3)w n _ 1
solution of n
u n ^AoL + Bft according as a and
/?
n
-faj3w n _ 2
=
u n ^(A + nB)oc.
or
are unequal or equal, where A,
i
n ,
B are arbitrary constants.
For the equation may be written
^-atV-i^K^-a^o). Hence by where C
is
u n ~ocu n
(1),
^^C^
t
an arbitrary constant, and therefore, by (4), w n ==^a w + Cj8 n+1 /()8-a) or (4+Cw)a n ,
according as a 7^)8 or
Cft/ (ft -of)
is
a = /J.
This establishes the result in question, for an arbitrary constant if a^/3. (See also Art. 5.)
Linear Difference Equations.
8.
au n + bu n _ l + cu n __<>+ is
With regard (i)
it
An
equation of the form
ku n , r = l
r is a fixed
n or constants.
to equation (A),
The general
... 4-
Here
called a linear difference equation.
k, I are given functions of are of this type.
(1)
.....................
number, and
(A)
a, 6,
...
All the equations in Art. 7
should be noticed that
u r+l9 wr4 2 >---
solution involves r arbitrary constants, for
can be found in succession in terms of u l9 u^
...
u r) which
,
may
have any
values. (ii)
// v n
is (he
general solution of
au n -f 6w n _ x -f cw n _ 2 +
and
wn
is
solution
any particular For it of (A).
. .
.
+ ku n , r = 0,
solution of equation (A), then v n is obviously a solution, also
+ wn it
is
is the
general
the general
solution, because v n involves r arbitrary constants. (iii)
The method
of dealing with such equations as
au n -f bu n ^ where
a,
fc,
...
k, r
-f
. . .
are constants and
trated in the next example.
-f
ln
ku n _ r = /, is
a given function of n,
is illus-
GENERAL TERM OF A SEQUENCE
370 Ex.
-Su^j-f 6w n _ 2 =n
Solve the equation u n
1.
The general
-5%^-f 6w n _ 2 =0
un
solution of
A
*^tl
Next search
Assume
w n =a + fru + en 2
a + bn+cn -5{a + b(n-l)+c(n-l)
powers of
coefficients of
6
-J-,
+ 2n
.
is
^
a
J,
-^
n,
we
then for
,
2
}
A
/
all
values of n,
+ 6{a
find that
2a
26-14c=0,
2c^l, c
5n
+/J.O
z
giving
-f
for a particular solution of
as a trial solution
Equating
2
~ n d so a solution
is
15) ...................................... (B)
Again, take the equation
and assume as a
a(5
Thus a solution
n
a 5n
un
trial solution
-5.5n " 1
.
,
.
then
+6.5n - a
)=5
n ,
? giving a = 6^.
is tS.
Finally, take the equation
Since 2
un
is
a root of x 2
an. 2 n ,
-5x + 6=0,
it is clear
we have a{n
- 2. giving a =
.
2n
-5(w -
Thus a solution
l)2
n -x
that a
.
2W
is
not a solution.
+ 6(n -2)2n ~ 2 } =2n
Trying
,
is
u n = -n.2n + l .......................................... (D)
The general
solution of the given equation
sides of equations (A)-(D),
and
is
obtained by adding the right-hand
is
EXERCISE XXXVIII The first term of a sequence is 1, and every other term the preceding terms. Find the nth term. 1.
2.
If
3.
The
2ttn
-tt n _ 1
first
the arithmetic
is
the
sum
of
all
=na, show that
is a, the second is 6, and every other term of the two preceding terms. Show that the nth terra is
term of a sequence
mean
thus proving that the nth term tends to J (a
+ 26)
as a limit as
n ->
oo
is
.
4. The first term of a sequence is 1, the second is a, and every other term the geometric mean of the two preceding terms. ~ Show that the nth term is ax where x ^i:{2 + ( - i) n 2 }.
is
DIFFERENCE EQUATIONS
371
5. If ww = />_! gw n _ a and un tends to a limit other than zero as w-> oo are 1 and -~q; also show show that |g|
^i
,
x
6.
i-
Show
+ qui
Ui
hm unn
that
.
,
1+q
that the equation
un - (a 4 ft + y) U n _i 4 (fty
4-
ya 4 aj8)
tt n _2
~
can be written in the form
Hence prove that
Un-oMn-i^Bp + Cy" fi^y or j9 y, where J5, C
OT
are arbitrary constants. according as as of the in Art. 5, (ii). solution equation given general 7.
If
un
nu n _ l + (-l) n and w ~l, show that un j\n-*e~* as ^->oo.
K/|n_= Wn_ /ln_-J-K-D 1
8.
If un = ?i(u n _ l
+ un _ 2
[Write the equation 9.
If
wn
(^
t*
n
)y
n
/[n..]
un
find
- (w 4-
in terms of
w and ^, and show that
u n _ l ~ - (^n-i ~ nu n-2)-]
1)
- l)(^ n ~i + ^n-2) an(i
V 3
j
l
M2 = l
= 0,
"
1
"
(
prove that
1)n 1
,
li
/'
L?
[Write the equation in the form
un - nun ^ = - [u^.! - (n 10.
Obtain a particular solution of
by assuming un =an + b. Hence find the general solution of the equation. Find the general solution of un -5u n _ l -i-Qu n _ 2 n [Assume un ~a .l as a trial solution.] 11.
12.
Find the general solution of
[Take as a trial solution 13.
wn -wW -i-Wn_2=ft 2 un ~ an 2 + bn + c.]
.
Find a particular solution of
un + u n-i sinna by assuming un a cos no, -f b sin noc. Hence show that the general solution is
~~ 2(1 14. If
Deduce the
u n un _i
=-
[nu n = (n l)?/n_ 2
un
in terms of
-f
cos
)
,
find
;
no^e that there are two forms.}
u^
ln
.
GENERAL SOLUTIONS
372 15.
Show
that the general solution of the equation
Vn-l + aun + bun-l + C = can be written in the form
where
A
an arbitrary constant and a, /3 are the roots of 2 - ab =0 a: - (a - b)x 4- c
is
;
unless
a.
=
fi,
in
which case
[Write the equation as ^(^ n _i4-a)4-&w n _ 1 4-c^0. v The equation then reduces to so that u n 4 a /
.
v n-l
vn
unless vn _j 16. If
4-
(6
- a)v n _ 1
4- (c
- ab)v n _ 2
= 0.]
Vn-i + 3ww - 4
w_ 1
-2=0, prove
that
>l.2 n 4-5 n ^
/I
17. If
<
wn w w _ 1
4- 5itw 4-
WTJ_I 4-
9
0,
2w
_,, ""
1
4-5 n
~1
<
then 2
,
18. If
tt
n+1
=-~
-'
2n-wn
and w l = a, prove that
a4-w(l-a)
and [Put
wn ^nvw .]
0,
Put
CHAPTER XXIII THE OPERATORS 1.
The Operator
variable n, (i)
J, E, D.
INTERPOLATION
Let u n be a function of the positive integral
A.
and consider the sequence u v u 2 u3 ,
The meaning
symbol A
of the
is
,
...,
un
...
,
.
defined by
^n^tVrt-f^, and A
regarded as denoting the operation which when applied to u n the Here n may have any value, so that difference u n+l - u n produces is
.
Jw 1 = w2 -^^ 1
Au 2 = u3 -u 2
,
etc.
Au n = v n *Aen Vj + Va + Va + .-.+Vn^Wn^-ti!. = w a - tij, = v n = w w+1 - ti w Vj *?_! w n !*__!, ...
(ii)//
,
For
,
whence the (iii)
,
result follows
by
addition.
d(u n + v n )=Au n + dv n A (u n + v n ) = (w w+1 + v w+1 ) - (u n + w n ) ~ *n) + (V i - V n =A
// u w v n are functions of n, then
For
(
In this connection
then As n
is
it
n-fl
w+
not equal to
.
)
should be noted that
.
if
Au v +Au% + ...+Au n
unless
w 1 = 0, for
and
Following the index notation of Algebra, the symbol that the operation A is to be repeated, and we write (iv)
,
and so on
;
thus
J2
denotes
TABLE OF DIFFERENCES
374
This process may be continued to any extent and, from the way in which the successive differences are formed, it is obvious that the numerical 3 coefficients in the expressions for J 2w n , J w n ... are the same as those ,
in the expansions of (1
-x)
2 ,
(1
-x)
3 ,
Hence we conclude that
....
4 ru n = u n+r -Clun+r _ 1 + C'2 un+r
(A) 2 -...+(-I)nu n in Art. 4. more concisely proved For the sequence uQ u l9 w2 ... a table of differences is conveniently
a result which
,
is
,
written thus:
,
UQ
u2
Ul
2 Thus, for the sequence O ,
u
I2,
2 2 , 32 ,
Au
...
W3
,
4
1
1
A*u A*u
(v) If 'u x is
_.
where u n 9
222
a function of any variable
n2 we write ,
16...
5
3
...
7
x, the
meaning
A applied to u x is defined by Aux ^u x ^-u x All that has been said about the symbol A still
of the'operation
.
successive differences of II u x>
I/
holds, for in finding the
u x we are merely concerned with the sequence
07 x+V ux+2>
Illustrations,
(i)
u n = n(n~ l)(n-2), then
If
J rw n -0
and (ii)
If
w n = n3
2. Special
,
since
for
r>3.
w 3 = n(n~l)(w-2)-f 3w(w-l)-hw, by the preceding,
Cases,
(i)
The following notation
xr =x(x-l)(x-2)...(a;-r-f 1) and 3_ r = If
x
that
is
is
variable
and J
is
-
often used
-^-^-
x(x + l)(x-f 2)
...
;
we
write
-
applies to x, then
Jx r = rx r _
1
........................................
(A)
THE OPERATOR & Similarly
we can show that Ax^ r
-
=
ra( f +D,
375 ...........................
(B)
being assumed that no denominator vanisJies. These formulae may be compared with
Putting n =2, we find that (7 = 0. Hence the result. Similarly for the second Of course, the rule in Ch. VIII, 3, might have been applied.
A
(ii)
by
Polynomial.
a?
x,
dent of
1,
#,
a? -2, ... such that
a polynomial in x of degree r, dividing in succession, we can find a a^ ... ar indepenIf
ux
is
,
ux = ao + a i^i + a 2 x2 +
J w^ = a l + 2a2 x x + 3a 3#2 -f
and then
and so
on.
series.
Finally
A ru x = a r
+ a rx r
,
. . .
J r ^wx = 0.
and
r
.
. . .
,
|
2.
?x.
For
//
A mu
x
w
as a factor.
\
The Operator
3.
.
when applied
to w n ,
is
E.
The
the operation denoted by E by unity. Thus for all values of n,
effect of
to increase
n
Eu n ^u n+l a+
.
a constant, the following equations define the meaning of E, Ea and a$, regarded as operators.
If
a
is
(E + a) u n = w n ^! + a?/ n = (a + J0) w n Further,
and so on
;
we
write
thus,
if
r is
y
a positive integer,
,
E + a,
LAWS OF OPERATIONS
376 Hence,
p, q are positive integers,
if
E*E*u n = E* (Eu n ) = Eu n ^ =
E*E*u n = E*>+
so that
Again,
.
a, 6 are constants,
if
(E + a)(E + b)u n = (E + a) (u n+l + bu n )
+ abu n Thus, so far as addition (subtraction) and multiplication are concerned, the operator E combines with itself and with constants according to the laws of Algebra.
The same
for A, for
is true
4* n = ti w+1 -w n = (^-l)ti n More defined
4.
x
9
generally,
if
ux
is
a function of any variable
x,
the operation
E
is
by
Fundamental Theorems.
and
.
r is
a positive
// u x
is
a function of the variable
integer, then
(i)^ux ^ux+r -C[ux+r ^^C 2 ux+r ^-...+(-lYu X) ......... (A) = ux + C[Aux + ClA 2 ux +...+A r u x ....................... (B) (ii) ux+r r
The first theorem Proofs. on writing x for n. More easily thus
follows from the reasoning in Art.
1, (iv),
:
(i)
J'.-(-l)X = {Er - CJ^- 1 + Cj'- -...+(- l) r}ux
Again,
(iii)
(ii)
If u x
u x+r =Eru x = (l +A) ru x and therefore ,
is
a polynomial in x of degree
For the left-hand (iv)
//
s n is the
side
sum
=4^ =
to
n(n-l)
l,
then
(Art. 2,(ii)).
n terms of the A
r
series
n(n-l)(n-2) '
u
-f
u 2 -f u3 -h
. . .
,
then
METHOD OF DIFFERENCES For w
=
n
l+^) ~X>
377
therefore
Similarly
Adding and using the Ex.
4
3,
Find a
1.
results of Art. 2,
n which has
cubic function of
Ex.
the result follows at once.
1,
the values
-3, -
1,
1,
13 when n
= \,
2,
respectively.
Denoting the function by u n
and A r un =
,
we have
for.r>3, since u n
Also
tt
is
a cubic function.
= ^ w -X =
-
_
+6
j&a;.
Sum
2.
the series
2.3+3.6 + 4. Here un ~ (n + l)(^ 2 +2). Writing down the as
on the
first
right,
This
four terms,
that
is
ll
+ ...+(
a cubic function of
we
un
find,
^=6, 4^ = 1
?i,
A r un
for
44
18
Aun A 2u
A^u^U, J 3 u 1 = 6.
so that
6
26
12
n>3. 90
46 20
14 6
Hence by
(iv),
i9 12
.
Ex.
3.
//
positive integers,
For by Art.
if
2,
A
(Cf.
Ch. VIII,
5 = *n -C7J(a: + l = (-l) n \n then 8 if r>n and S
applies
(ii).
,n(n-l)(n-2)(n-3)
1A 14
-
to
x,
S=^(l
4,
Ex.
1.)*
r,
if r
- E) rxn ^(-l) r A rx n f
are
= n. and
the
result
follows
BERNOULLI'S NUMBERS
378 Ex. 4.*
(Bernoulli's Numbers.)
For nr =tf n O r =(l+J)n O r and
If
Br
is
defined as in Ch. VIII, 8, (5), then
-l+$-...
+ (-ir
m>r;
if
i
}r
1)
...
......................... (A)
...
Li
II.
#r = l f + 2r + ... +nr we have f(n + l)n (n + l)n(*-l)
Hence
if
,
l)n(n-
(n-f
(*~r +
l
'
*'~\
LT+JL
[3
[2_
Now, by
definition,
Br
coefficient of
n
in
Sr
hence
,
Again, equating the coefficients of w in (B), A3
J "T + {AZ Hence
~- +( ~ 1)f "
Ar^
^ =i
and (-l) rJ5r =J5r Hence equation (A) holds for r>0.
5.
5.
J0r =l,
Apply equation
in
Operators '
'
following
"
r>l,
if
Moreover,
Ex.
r=
r}
proof
(A) of
for
r>l
Ex. 4 to show that B^
a Fractional Form.
of the
theorem in Art.
4, (iv)
(Ch. VIII, 8,
(5)).
^.
Many
writers give the
:
En -l n(n-l)
.
n(n-l)(n-2)
i
Here the operator is written in a fractional form in order to transform a polynomial in E into a polynomial in J. This result was obtained his collected works).
by Cayley by a much more
difficult process (Vol.
IX, pp. 259-262 of
LAGRANGE'S INTERPOLATION FORMULA The
when
result,
obtained, can be verified by multiplication
379 and addition ;
this fact justifies the process.
But
Thus the
at every stage the operator must be regarded as a whole.
E -l n
equation Sn
(E -
does not imply that
In fact the last statement
is
~~E^T Ul 1) s n
= (En - 1) %.
not true, for
(E-l)s n = s n+1 -s n = u n+l
(E -l)u l ^un ^ l -u l n
and
.
We
conclude that, in such transformations, the operator can be put in the form /(J)/^(J), where f (A) and (d) are polynomials in J, provided that
f(A)
is divisible
6.
by <(J), andf(A),
(d)
are not regardedas separate operators.
Suppose that y
Interpolation.
a function of x whose value
is
has been determined by experiment, or otherwise, for x = a,
The problem
of interpolation intermediate values of x.
The graphical method is to plot the points draw a smooth curve through the points.
(x, y) for
We
'
*
6, c, ...
.
to find approximate values of y for
is
x = a,
6, c, ...
assume that
,
and
this curve
represents as nearly as possible the graph of the function y. Let us suppose that the equation to the curve so drawn is y=f(x). It is natural to take for f(x) the simplest function of x which satisfies the conditions.
If
y
is
known
for
n values
we assume that
of x,
where the n constants p p^ ... are to be determined by making the curve go through the n points. If only two values of y are known, the curve is the straight line joining ,
the corresponding points, and we have the rule of proportional parts as used in working with tables of logarithms and trigonometric functions. 7. ively,
Lagrange's Formula. we assume that b)(x
(x
y=syi
For y
is
when x = a, Similarly
c)
(a-b)(a-c)
If
+y*
(x
y=y
y2 y$ for x=a,
-a)(x
6, c
,
l9
(x
c)
(b-a)(b-c)*
y3
a)(x
respect-
b) *
(c-a)(c-6)
a quadratic function of x of which the values are yl9 y& y3 6, c. if
y = j/1
y2
,
y s- y
>
-
^3
when x = a,
-^4
(x-b)(x-c)(x-d) \
'
_L^
'
b,
.
+ three
c, .
d we take ..
similar terms
;
*(a-b)(a-c)(a-~d)
with similar formulae when more than four values of y are given. B.O.A. 2B
INTERPOLATION
380
8
.
If
the values of y are
known
of applying Lagrange's formulae,
we
For example, suppose that y
Assume that y is
a positive integer,
at equal intervals, instead
use the operators
y&
y\,
a cubic function of x
is
x
for values of
;
y%,
then
E
for
j/3
J ry =
and A. x v x2
X
XQ
for
r^4, and
,
x$.
,
if
n
= (l+A) n y0)
n
y n ^f! y therefore
Hence the equation
represents a cubic curve which passes through the four points (# ?/ ) (x l9 T/J), ... and gives an approximate value of y corresponding to any value ,
of x within the specified range. Ex.
1.
Given
tJiat
sin 45 =0-7071,
sin 50 =0-7660,
in 55 =0-8192,
A-rn
60 =0-8660,
n62. Let
1*3
= 10 4 sin (45+5x).
Construct a table of differences as below.
Au
u
A 2u
A*u
589
-57
Mj-7660 532 "
7
-64 468
J^ = 589,
/.
Assuming that
r
J w
J 2 w = - 57,
-=
A92
7.
-
M x(x-l)(x-2) _j
x = 1-4 and
so that
u
=7071 D
= 1-4x589 MA
=
2
=
824-6
=
- 0-7 x 0-4 x 57
|
ILl-JpZJ
A 3 u Q ^ 1 -4
x 0-4 x 0-7
=
0-392
7895-992
I
15-96
u x = 7880-032 .'.
The
-
r>3, approximate values of w^are given by
for
x(x-l) -L_J
52=45 + 5z,
J 3w
sin 52 =: 0-7880 to four places.
correct seven-place value
is
0-7880108.
- 15-96
BESSEL'S INTERPOLATION If
wo only
FORMULA
381
use the values of sin 50 and sin 55, the rule of proportional parts would
give sin 52
=0-7660+f
x 0-0532 =0-7873.
If UQ and several other terms of the sequence U Q u v u%, ... are known and approximate values of the missing terms are required, we may proceed ,
as follows.
= 4-7046, w 3 = 5-6713, w 5 = 7-1154, find approximate w = 4-3315, and w 4 Four points on the graph of u x are known. We therefore assume that it can be 4 represented by a cubic function of x. This is the same as assuming that J wa = 0. Ex.
2.
^
//
values of u
.
.
Hence we have approximately 0,
/.
+ ^i) ^11-4619. = w 4 6-3199 (approx.).
w 2 ^5-1420,
NOTE. In this example w ~tan77, w 1 -tan78, w 3 tan 80, W 5 ~tan82, we have found that tan 79 ^5-1420, tan 81 ^6-3199 (approx.). The correct values are 5-1446 and 6-3138. It should be noticed that the
approximations just found are
given by )
9.
=5-1879,
much
closer
so
than those
w 4 = \ (u 3 + u 5 )= 6-3933.
Bessel's Formula.
second differences
In using Mathematical Tables, the effect of be easily and accurately allowed for by using a
may
formula due to Bessel.
Let u_ l9 u
,
u lt u
be four consecutive values of a function u XJ as
and suppose that values and 1.
given in the tables, of
x between
of
u x are required for values u -i
Denoting ences by a
9
on the
first 6, c
a
and second differand d e, as shown
,
b
9
right, Bessel's
formula
u*
is
f e
c
\
x(x-l) d + e which
is the
same as a) ............................ (B)
The values little
labour
is
of
uQ9
a, 6, c
can be taken from the tables, so that very
required in using the formula.
THE DIFFERENTIAL OPERATOR
382
if
To show that this gives a good approximation, and to estimate the error, we neglect differences of the fourth and higher orders, we have
Now
e-d^f; and
w -w_;i=a, b-a=^d,
x(x-l).
+ l)x(x-l),
(x
'
V
e-d\
x(x~l) fd + e so that
if
?'
x is
Hence the
(x
+ l)x(x-
1)
the approximation given by (A),
error
C/5^ to6/65
Ifo. 1.
therefore
is
approximately izx(x- l)(2x~
of square roots
to
I)/.
find
Using Barlow's tables, we take
045020= a -:
11 1-085553
045001=6 M1
= N /12350 = ............... 044983 -c.
^2-^/12360 = ................ (The differences
a, 6, c are
z^-4,
given in the tables,)
For >/12344, we have
c-a-^ -
-000037. -J*(3?-l)=--06, w = 111-085553
Ja?(a?
-
1
)
(c
-
6= a) -
-0180004
-00000224
~
N /12344=!ll-1035o7>
The
result is correct to the last figure.
10.
The Operator D. du
the symbol
-r-
is
When we d
dn
write
u/ d
\
n
regarded as an operator, which
is
sometimes denoted by
118
If
a+
a
is
a constant, the following equations define the meaning of
A Da and aD, regarded as operators.
D + a,
THEOREM
LEIBNIZ' Hence
(D + a) (D + 6) u =
a, b are constants,
if
383 -I-
(Z)
a) (Dw
-f
6w),
= Z)2 w -f 6Z)w -f aDw -f afrw = {D2 + (a + b)D + ab}u. (/) 4- a) (D -f 6) u
and
Thus, so far as addition, subtraction and multiplication are concerned, the operator D combines with itself and with constants according to the laws of Algebra.
The n-th Derivative of a Product.
11.
d
du
dv
dx
dx
dx'
x
of
D D
Let the symbols
l9
If w, v are functions
denote differentiation with regard to
2
x,
D
suppose that only operates on u and its differential coefficients, that 2 operates only on v and its differential coefficients, so that
and and
D
Dn
D
As
/
l
\
(uv)
=
ax
and addition and / ^
&u
t
D
dv
D 2 (uv) = u jr\
v,
i
\
ax
\dz/^ that
WV '"~'
is,
-j
1+
*'
(uv)
=^
This result
Ex.
is
(
1+11 -
uv >~\
known
If ij~u
+v
and
-
(uv)
n -f D n = (D l 2
.
)
uv.
where u m
=x + (xz - ^)
vm
,
=x -
-f
uv n
,
x.
- k) ",
/^
W = 0,
1, 2, ...
.
we have
Differentiating with regard to x,
_ i +x Wi
'
2
)>
often written in the
is
(
2)( u
2
with regard to
=
tn-i
-
. . .
suffixes indicate differentiation
ww
ax
21
2
as Leibniz' Theorem,
=
1.
~=
are independent of each other, they obey the laws of multiplication, therefore
form where the
d
,
2
n \
,
and hence
;
_ jt)"^ = tt w (a; 2 - Jt)"" (x z
;
hence (# a - k)^u l =u/m.
Differentiating again,
.'.
Similarly
~*
*-
*
(x
-
2
(#
2
k)
-
wa
!
-f-
arM!
k) v^-^-xvl
= (x* - kf*
- w/m 2
0.
- v/m?
0,
.
ujm*
;
- y/m 2 =0. whence by addition, k) y 2 -f ar^ (x Differentiating w times by Leibniz' theorem, 7 2 ^-i (* k) yn +* + n.2x. t/w+1 + 2
-
-
This example leads to a remarkable theorem on cubic equations, given
in
Ch.
XXIX.
USK OF OPERATOR A
384
EXERCISE XXXIX 1.
Find the polynomial of the third degree in n which has the values when w = l, 2, 3, 4 respectively. (Use Art. 4.)
2, 11,
32, 71, 2.
Use the method of Art.
P+3 +5 + 3
(i)
3
u n = xn and A
3.
If
4.
Show
(ii) I
;
apph'es to
/-l, 2 or
S be = u r (* n/) r
the
(ii)Lct
.
show that
0,
according as n
of the series .
(
r
~y)
n
n~2
or ?z->2.
n(n +
l),
I,
2
sum
n~ 3
if
n>2
of the series.
S=
then
',
?/
.
;t,
2
[In each case let
Also J
+ 22
(w-l) C?4-(n-3) r^ + (w-5) Ct4-...^2
(iii)
r?
.
xn -C'i(x-y) n
2
Let
22
2
that
(i)
(i)
sum to n terms 3 2 + 3 a 4 2 + ...
Ex. 2 to find the
4, ...
(
1
=
- E) n ii
-
1 )
(
nJ w
.
etc.
n, |
w r = (a:-r)(y-r), then S = ( -
l)
nA n
Also
u^.
w if n>2 Aut = (x-l)(y-\)-xy=-x-y+\ J M = 2, J w r = n w = r then S = i-{(^+ l) ~(E I) )w (iii) Let w r n n Also w = 0, Jw = l, ^ w = 2, and JX = 0if /. S-=l-{(2 + A) -A }u 9 8
9
2
,
;
2
.
5.
^x(x-l)(x~2)...(xr + l), show that - 1 )a:r_ 2 -h (x -h n) r = a- r + ( ?rar r _ l + CJr (r
If x r
the last term being r(r-
Verify that,
6.
If
if
n
1)
...
(r
3 and r
xr = x r -f o^^!
-}-
~n-\- l)x r _ n if
. . .
a^l-r(r-l),
[Equate to 7.
Show
,
,
if
prove that
6=A-(r-l)(r-2)(3r-5).
the coefficients of x r
JV+^fc+JrJ'rj-JL
(ii)
8.
.
~1
" and x r 2 on the
right.]
that
(i)
[Use Ex.
.
the identity becomes
2,
6a:r _2 -f
.
r>n and C^jr
J rx r + 2 =
J
[x
2
-f 7-a:
4-
;
^r (3r -f 1)]
. |
rjh
2.
6.]
Prove that xn -Ci(x+l) n + Cl(x + 2) n -...+(-\Y(x\-r) n
+\
= - Yl {& + r-r + iV(3r -f 1)}
and [Use Ex.
(
7.1
1
|
when
n~r-\
when n = r 4- 2. r+_2
1,
EXAMPLES OF IDENTITIES 9. If
385
none of the denominators vanish, prove that 1
x
(-ir\n
= __
Deduce that
4- C? ^-fC? ^-"'+ 1
+
(
1)n
1
1 -
n(n-l)
1
n
-
-
(x+l)(x + 2)
2
n(n-l)(n-2)
I
3 10.
Prove that 2
1 *
~r 71
r-
22
1.2
m+2
m-f-1
n.2.3 n ~ l
n
^3
m(w-f-l) 11. If u n
2W
yt(n-l) T
nt/^x/.^-.um(m + l)(m + 2)
,
'
/
i\n.
m(m-f 1)
...
(m-fn)
~\l(an + b), show that
(i)
A ru ~( -
r
l)
r
111
(an
+ b)(a .n 1
n ' n(n-l)
a '
n(n-l)(n~2) "
a2 '
"3
= l/a;(a: + l)(a; + 2)...(a: + r-l), show that + n)_ T = a?_ r -6 ?ra- r+1) +(7?r(r+l)a;_(r +8)-...
12. If a:_ r
I
(
n+ 1
to
(
13. If n is a positive integer vanishes, prove that l
+
n
n n ~l) (
and y
is
any number such that no denominator
n(n-l)(n-2)
n. ^n(n-l)
terms.
_
y+
l
n(n-l)(n-2)
/.x (in)
each
series being
continued to n 4-
[These results follow
1
terms.
from Ex. 12 by putting successively
r
= l, x=
-(t/4
1)
;
USE OF THE OPERATOR J
386 14.
Show
tfcat
ln + :
1
+ 2)... 1
V
/
,*//*.
i
I\/,M
i
* //
o\
IW/-M
i
-...to n +
-.
_
15. If
>
...
'
16. If
(a-n-f
a(a-
a-b a~6-fl
a-6-fr-l
tti=7
^2=7-77 6(6
b
_
^ rWn=
.
1)
...
(a-n-f 1)
a(a-
...
1)
(a-n-f
1)
e tc., as in Ex. 15, prove that
1)
a-6 a-6-fl a-6-f2 ^ T 6^1 6^2~
[Putting n = l in Ex. 15, find
17.
T;>
+2
prove that
66-16
1
ln '
1)
a-6 a-6 + 1
w
'
o\
..
___
__a~b ~~ a(a- 1) U==
i
terms=
1
--...-
^^
~w
o\/ M
i
Au 0t A*u
a-6-fr-l *
6-r-fl
etc., in
,
succession from the equations
Prove that fI
_? )
__
"*
a
M n
a 6
+
a
)(1
6
6/\
\
j
...
?L 6
Ti-
n(n - l)(n-2) a(a- l)(a-2)
n(n-l)a(a-l)
JT~
(1
6-2/"\
1/\
6(6-1)
6(6-l)(6-2)"
[3
to n-fl terms.
[Using Ex. 16, the left-hand side = ( 18.
Show
that the
sum
n J nw
n =(l -E) u
.]
of the products two together of
1
x is
l)
9
JL JL x + l'
i
x+2
equal to
n(n-l)
(
n-2
"--
I
~+ (n-2)(n-3)
1
"-
[2
[Equate the
coefficients of a* in
Ex. 17 and put
#=
-6.]
to
n ~ 1 term8
1
r
INTERPOLATIONS 19.
In Art.
4,
Ex.
387
has been shown that
4, it
Prove also that
and use each of these formulae to show that B^ = [Equation (D) of Ex. 4 can be written
^n /e=2, 5, 15, 52, 203, 877, 4140. we have SW =(J + l) n and Sn ^ = (E - l)n S on
23. If
+
L_
Sn = -Sn_ 1 SB = 5 ^ 1 n = l,
[(iii)
....
d
I
or~i (ii)
prove that
,
I
i
= Tn.!
Using
(ii),
show that
for
n=l,
rn .e=0,
2,
3, 4, 5,
-1, -1,
2, 9.
to
.]
<, prove that
CHAPTER XXIV CONTINUED FRACTIONS 1
Definitions.
.
Any
(1)
expression of the form .................................
i-: called a continued fraction,
is
The
and
-
Oj
may be supposed to be attached fraction may be expressed in the form
- Ui ~ F-a + JU
where the
and
a's
ai
^2/a 2
by
stopping at ,
b n /a n
.,
.
6's ...
&3
&2
~
"
is
,
convergents are
av +
,
1
is
.............. (B)
and so any
6's,
........... .......
called a convergent.
flu
. . .
~
..
........
(C) VI \
be positive or negative numbers. The quantities are called the elements of jF, and the fraction obtained
,
second, third,
3
may
any particular stage
,
bn
"...
1
--
to the
signs
continued
It
a
written
is
obvious that the fraction
is
6<> "
a:
,
H
6<>6q----
. . .
,
#2
unaltered
Thus the
first,
.
2^~ ^3
if
6n
,
an
,
b n+l
(u
= 2,
3, ...)
by any number k or if the signs of 6 n a n and 6n fl are Thus any continued fraction may be written in the form
are all multiplied all
(C)
changed. where a 2 a 3 ,
Ex.
1.
Prove
,
are
...
,
that, if
all positive.
JLJLA
2.
fraction
on the
left is
*
elements,
*
l
1 .
4+ 4+
4-f
The
n
each fraction contains
2+ 4+
to s equal ^
4+
2-h
1122 ---- -
-
2+4+4+44-
Formation of Convergents.
(1)
...
""
=x
1111 -
~
x ^ 2+4+2+4 +
Let u n denote the nth con-
vergent of the continued fraction, _.
62
60
a 2 + a3
We
shall write
+
bn
.
w=^) 1 /
a,
J
an + 1,
,
so that
w=^22
fi,=r 1 1
where
J a2
a2
THE RECURRENCE FORMULAE Observing that u3
389
be obtained from w 2 by changing a 2 into
may
a2 4- 63/03, we have u3 =
(
and so u$ = psjq& where y3 - a3j> 2 4- 63^ and Proceeding thus we can show that, if p n and values of n by the equations
? =
?-! + &nPn-*
Vn
?3
==
q n are defined for successive
= anqn -i + b n qn _ 2t
.................
(A)
the nth convergent of jP. p n/q n Equations (A) are called the recurrence formulae. It will be found convenient to write = 0, ............................. i and
then
is
yo==
and
it
will
Pz^^Pi^^Po an d
be seen that
important to state
For
Definition.
#Ae quantities
?2 ==a 29 i r
rather more precisely
all this
with the initial values
If F. convergent of
pQ = l 9
p n and
qo
=
Pi
9
= ^i>
q\
:
q n are defined in this way, then
,. .
.1
II
'
I
J.
I JL
I
............... (A)
,
=1
Assume that this holds for the values 2, 3, Proof. the nth convergent by u n , then
Now p
so
the continued fraction
p n and q n are defined by the equations = a n?n-l + 6wyn _2 Pn = ttfiPn-l + ^n-2 ?n
Theorem.
+ ^25 o> r
n^2.
equations (A) hold for (2) It is
(B)
?0
...
p n /qn r of n,
is the n-th
and denote
f,
also u r may q r _ z are independent of a r and b r be transformed into u r+1 by writing ar + br+1 / a r+i for ar therefore r
_v
gv_
j,
pr _z
,
;
,
(
. Ur+1
r
+
Hence
br+l
^ \
w^
,
,
Thus the theorem holds therefore for
w = 3,
4, 5,
TPr ~ Z
P '~ l+
...
=
-
_ 5^i (r^i
-
r +-'
for
n = r-fl;
in succession.
but
it
holds for
n = 2,
and
INFINITE CONTINUED FRACTIONS
390 (3)
For the fraction
p n and
q n arc defined
b
by equations
Again, the fraction
b2
ft
is
bn
b
(A) with the initial values
b3
obtained from the fraction in
by changing the
(2)
signs of the 6's, so
that the recurrence formulae are
Pn = fln^n-l
~ &nPn-2>
?n
= a n?n-l - 6 n?n-2'
In the following articles we shall call the reader's attention to certain types of continued fractions which will be considered in detail later. Ex.1.
Prove that (I-
x)-^l+~ |j 3^ y
Calculating the convergents of the continued fraction,
we have
2.1-3&.1 6-8o;~
which provea the statement in question.
Notice that no fractions are reduced
to
lower
terms, until the final stage.
3.
Infinite
Continued Fractions. ,_ If
JL'
/I t*i
I
T"
694
"
With regard
__6q?_
to the fraction
6n )
-
if
the
number
of elements is finite,
F
is
called a terminating continued
fraction.
We
may, however, suppose the b's and a's to be determined by a rule In this case F stands for an is no last element.
of such a kind that there infinite
continued fraction, which
regarded as defining the sequence of
is to be
convergents Pl/
?2/?2>
>
Pn/
>
and we say that
the continued fraction is convergent, divergent or oscillatory this as according sequence converges, diverges or oscillates. is defined as If the sequence converges, the value of the fraction
F
lim Pn/q n
,
and we write jF
= a 1! +
-
--
a2 +a3 +
...
SIMPLE CONTINUED FRACTIONS Let x stand for ^ 2
Illustrations.
after the first is
Again,
meaningless.
,
2
may
x=lJ2i,
leading to
where every element
be treated as a number,
a result which
is
clearly
x stands for
if
333
t000
2T2T2T"* and we assume that x
to oo
...
-=
^ 2
// we assume that x
-3/2.
we have # = 3/(2-x)
333
391
is
we have
a definite number,
x = 3/(2+se);
z2 + 2z-3 = 0;
/.
We
have thus proved that unity, but we have not shown
.'.
that
a value
z=*l
or
has a value,
the fraction
if
'
its
-3. value must be
exists.
In neither of these examples can the first step be justified without knowing that the continued fraction is convergent. 4.
An
Simple Continued Fractions.
_
1
a i T^___ i
_ _ 1
1
a 2 + a3
expression of the form
..."
+
>
an +
where every a is a positive integer, except that a l a simple continued fraction.
may
be zero,
is
called
Theorem. Any rational number can be expressed as a simple terminating continued fraction which can be arranged so as to have either an odd or an even number of quotients. Let
A/B
be the number,
of finding the G.C.M. of r2 ,
...
A
and B being positive integers. In the process and 5, let a v a2t ... be the quotients and rl9
A
the remainders, so that
A^a^B + r^ and therefore
IB
A= #1 + B l ^prrBr
-^
JS==r 1a 2
,
r
Again,
if
>!, we can
r1
,
1
=flo"f -
This process terminates, and
A B
+ r2
g,^
if
an
r,
7-
rr is
,
r
111
-^
write 1
=O+ r 3
etc.,
,
1
7-, etC.
the last quotient,
1
a n ~an -l+
andifa n
= r2a3 + r3
1
RECURRING CONTINUED FRACTIONS
392
Thus matters can always be arranged so that the continued either an odd, or an even, number of quotients, as required. Ex. 1. Express ^g$- as a simple continued fraction with add number of quotients. al
157
.
i.e.
(i)
fraction has
an even and
an
I
JL
=2 +
(ii)
-gg-
a4 =5
i
i
i
number of quotients can be made odd or even as required For instance, the penultimate convergents to 157/68 in the two forms above are 30/13 and 127/55; and it will be found that (127, 55) and (30, 13) are The
NOTE.
is
fact that the
of importance.
the least solutions in positive integers of
68* - 157y = + (See Art. 9
respectively.
Ex.
Use
2.
The
z -x)~ as a continued fraction.
is
giving
*
- x) 2
(1
^ l/2a?
-
1+
"^4/3+"
1/3
-
L
^9/2*+
1/3
2x
2s 3*
+
1
-3/2
1- 2+ -9/2* + l/3 x 2x
2x 3x
~ 1+ last three steps, the
+ - 4/3 + -9/2* +
??.
""
In the
1
3.)
the H.C.F. process to express (1
H.C.F. process
68* - 157y = -
and
1
and Ch. XXV,
ri2 + 3^T*
statement at the end of Art.
1
has been used.
Recurring Continued Fractions. If after a certain stage, the elements recur in the same order, we have a recurring continued fraction. 5.
,
'
The recurring elements form the the non-recurring elements,
'
'
*
recurring period
such exist, form the
if
or the
*
9
cycle,
*
'
acyclic part
and
of the
fraction.
The
cycle is usually denoted
last of the recurring elements, _____ _____ _ _ ~~~ _
.-
-
-
_____ _
-.
~.
1+2+3+4+3+4+ A
by putting
asterisks
under the
* . .
1Q AD
/l ATir'I'f'.Afl VA.\3JLlv Uv7vA
MY J
r4"\7
-
._,....
_.
.
_
1+2+3+4 *
Here 3-f
-
4+
is
first
thus
and the cycle J
-
1+2s
^e acyclic part. J *
*
.
and
SPECIAL TYPES Ex.
is
1.
equal
to
(a
n+1
are the roots of x* -00;
ft
-^n+l )l(
1
Hence show
(ii)
is the
// a,
(i)
(l
+ l/n)a
- ax + 6 of x 2
according as ot^fi or a=j5.
a 2 ^ 46, and that
value
Here pn and qn are given by
together with
p = l, p^^a, qQ
where A, B, A',
E
f
= aq
4n
^ = 1.
0,
a^jS, then by Ch. XXII,
First suppose that
p
(i),
B^a=a+^
=A +B
gvng
5,
Therefore
are constants.
p 1 =Aoc +
=1,
^l/= yw =
thus leading to the
Next
let
n+1
-]8
(
w+1
and
)/(-]8)
fc
n (
n -)8
)/(-^
= a=)S, so that a 2oc and 6=^2, then by Ch. XXII,
A~B~\
,
If
=
first result.
Proceeding as before, we find that n -1 n yn = na ;>n -(n + l)a a 2 >46, then
a,
j3
>
^n /gn = (l + l/n)a.
,
l-W
^
5, (i),
A' = Q, B' = l/a., hence
n
Let |a|>|/?|, then
are real.
qn
j8
/a
n ->0 as n->oo and
'
2
a 46, then a~/? and pn /qn (l+I/n)a-+oc. Hence, in both cases, the continued fraction is convergent and
If
Ex.
2.
//
pn lqn
is the
11 prove that
and
n
is
its
value
is
a.
n-th convergent of the recurring fraction 1
aT 6TaT
If
its
= 0.
Pn = <*Pn-
(ii)
prove that the n-th convergent of
that the continued fraction is convergent if
numerically greater root
(i)
or
)
+ 6=0,
393
even, the recurrence
1
'"
'
6-f
p n - (ab + 2)p n _ 2 qn -(ab + 2) qn formulae give
Pn
=t>Pn
Pn-2 = pn
whence eliminating p n -i and pn -3> we have n +2)pn -2
p
(ab
+Pn-t =0.
If n is odd, we have the same equations as before, except that a and b are interchanged, leading to the same result. And so for the g's.
BROUNCKER'S FRACTION
394 Ex.
3.
//
pn lqn
n4h
ta the
convergent of
3a
52
1+ 2+ 2+
2+
1
I
(2w-3)
1)
<
Hence show
this holds
and
that the fraction is convergent
If vn denotes either
if
n
Replacing the
we
n-
write
ft
1,
- 2,
-(2- lK_i=(- l) t;'s
by
p's
n ~1
'
2 ^Ti
that its value is w/4.
we have
or qn ,
pn
"
2+
Pn l3n = l ~ 3+5-- + ~
prove that
and
"
2 in succession for
...
n -1
(2-3)(2n-5)
and observing that
=
jt)
3
...
and
n, therefore
.
^=
!(,-,). 1, we have
^n-i
2n-l'
,
" where
^4 is
"" 4
independent of
Next, replacing the
v'a
n.
by
find that qn
Again, the series fraction converges
-
1
and
=1 1/3
its
n~ 1, we find
and observing that ?n
whence we
1
3
1
Putting g's
1/5
value
A =0.
that
an(^ 5 i
l
(fo
F
we
== l
-(2n- l)7n-i-0,
.3.5... (2n -f-
""'"" ,/. "*
.
,
1.3. ..(2-l)""
- ...
is
1),
leading to the
convergent and
its
first result.
sum
is Tr/4
;
therefore the
is Tr/4,
EXERCISE XL 1.
Find the
first
four convergent (unreduced) of
w JLJLJL* 2+3+4+5 r\
2.
W JLAJL! l-2-3-4' /-\
f
Without calculating the convergent^, prove that 2
3
2+ 3T
5_ 1 1 23 T 5^1+ 2T 3T 4
4
'
3. Express fff as a simple continued fraction having odd number of quotients.
4.
The reckoning on the 85
5.
3- 4-
52 57 5
an
85 104 19
I
this.
Express
(ii)
20
5*
^
in the
form a l
greater than unity. 6. Express (2a -a- l)/(2a*-3a) calculating the convergents. 2
by
an even and
right shows that
1_J_1
52""
Explain
(i)
a * ~ a* in the
...
,
where the
"""
form
-
1+^ /i + /a +
a's are integers
...
,
and verify
INFINITE CONTINUED FRACTIONS Show
7.
that the nth convergent of
JL_1__!
2-2-2- -
2 1 is
+ l)/n, and
(n 8.
Show
395
that the value of the infinite fraction
is
unity.
that the nth convergent of 1
1
1
1
rr^r^r^r is
n,
and that the
9.
infinite fraction is divergent.
There being (n +
1)
elements in
2~2- 210. If
show that
all,
1
'"
2^"
l_(n + l)a?-n nx-n + T
*
a?""
a z >46, the numerically smaller root of x 2 -ax + 6=0
A JL J A a
a
[This follows from Art. 5, Ex. 11.
For the fraction
b
is
equal to
a
#
1.]
b
b
b
a+ a+ a+
"
*
*
a
+
"
'
pn = - a -)/(- 0), gn -( n+1 - j3n+1 )/(a - ]3) where a, ^ are -a# -6-~-0. Hence show that if a, 6 are positive, the fraction is convergent and equal to the positive root of x~bj(a + x).
prove that
(
an
the roots of # 2
12. If
p n jqn
IL
the (unreduced) nth convergent of
and the number of terms tively,
then u n \vn
is
in
>
n,
qn
respectively are denoted
by un
,
v n respec-
the nth convergent of
Hence show that
13.
Prove that
,
x
l
fi\ v
14.
;
_z_ z = _ 1
(
_o/JLJL-i-
1^4^1-4^'
^
\2~2^2-"V"
Hence show the nth convergent of the
first
fraction reduces to 2n/(n
+ 1).
Prove that 1
showing that the 2o
9fy.
Prove that, each fraction containing n elements,
J_ J_ JL JL
15.
_
1- 1+ 1- 3+ 2-
I2
infinite
22
2
J3_
(^-1]
_1
continued fraction
is
11
1
divergent.
B.C.A.
SIMPLE CONTINUED FRACTIONS
396 Prove that
16.
~
' ' '
2a + 2n -I+5= k n a (a + 1) (a + 2) (a + n)
......
'
2a
if
showing incidentally that,
Use
show that the
this to
fraction in Ex. 15
Putting
a~ -m,
where
in Ex. 16,
m J-itJli
a
O
iftl
m is
~~ ju'tTl
~~
9
then
divergent. l)
a positive integer, show that 2
I
~~
x
(w--2)
jw-l) i.
is
!){?! -(a 4- w -
[Here gn -(a + H)g n _ 1 ^=(a + n17.
. .
.
2
Also prove this result by induction. 18.
For the fraction
-
+
1
Hence show that the n
...
prove that r
,
n+
^1_1_
fraction
12
is
13
~ convergent and equal to l/(e
1).
a positive integer, prove
is
n
by induction that 2 11 ~ n 1 "" ?T-~ "2+ w+~2 2"+ 1 + 2 n - -2
[
w"+ w - 1*+
-t-
'
For the fraction I
2
22
1_
r+ 7)
n
-
prove that
1+1+
1"*-
^
1
3a
11 + -
-
fraction
is
. . .
(
where the
,
...
l
_
,
...
3/3
>
.
2.
l *
*
'
~j
except that a x
1
1
then x2) ^3
"1 1 -
are called the (partial) quotients
X2 =
'
are of the form
+ ~~"~" r~T
a's are positive integers,
Here a l9 a 2
71
I)
""
convergent and equal to log
1
at
2
r+
+ -
Simple Continued Fractions xl
(?i-l)
"*
t>
Hence show that the 6.
...
\n-i I
n
20.
-
+
?n
^j
J>n_
in -hi
19. If
~^- _ 2+ 3
Gt
3
+
-
-
a4
be zero.
may and
if
11 ~ --
+ a5 +
are called the complete quotients.
Thus we have Xl
that
is
"~
JL ai f "
a2
_L_JL.
+ *"a n ^+ x n
to say, the continued fraction x l
is
l
obtained
its
from
by substituting the complete quotient x n for the quotient a n
Pn-\ and q n to q n _ v For if the numbers belonging to any (q
7n-i)
have a
would divide 1 1
.
3)
>
From
common
pn ,
the recurrence formulae (A), -
n
prime
to
of the pairs (p n q n ), (p n p n ^) factor g, then on account of equation (B), g ,
1.
Pn-l
is
- --
Pn-l/Pn-2
we have
and
-
=
n
?n-l
The first of these holds if we write n - ] n ~ 2, The second of these holds if we write n - 1, n - 2, ,
---;
-
2 for n.
3 for
n.
.
9
ODD AND EVEN CONVERGENTS
398
1111
and
Also PI/PQ^^I
therefore
JaAh^^J
/n
-,
(E)
(F)
11,
It follows that if
pr/qr pr '/([r ,
a*2
are respectively the r-th convergent of
and
+
an
-r
an +
i*
11
'
a-, n-l +
a,l
where in the second the quotients occur in the reverse order, then
p n '/q n ~p n lpn -i and pn ^ 1/?n-1 = ? n/?n-i '
For by the preceding,
by
and
Art. 10 these fractions are in their lowest terms.
12. It has been shown that
if
p'q', p/q are consecutive convergents
of a simple continued fraction, then pq' ~-p'q =
1.
where q'^q, and suppose that p/q is Conversely, let as a fraction continued with an even or an odd number expressed simple
pq-p'q=l
of quotients according as the arbitrary sign
that p'jq'
For
is the
p" jq"
if
is
-f
or
-
.
We
can prove
convergent immediately preceding p/q.
is
the last convergent but one,
by hypothesis and equation
(B) of Art. 9, f
pq
-p q = pq" -p"q, and f
r
therefore
p(q
-q")q(p' -p")> .
Now p else
is
prime to
q,
for
pq -p'q=
1,
therefore q divides
q'
-q"
or
= q"q'
Also
5'<<7
sequently
13.
and q"^q, therefore q cannot divide q'-q", and con-
= q" and p'=p", which q'
proves the theorem.
Sequences of Odd and Even Convergents. Let w n
then, from equations (D)
of Art. 9, it follows that
Wn-w-i
and
u n -u n _ 2
Therefore u n lies between w n _rl and w n _2 between the two preceding convergents. say, every convergent
are of opposite sign.
>
that
is
to
lies
Now W!
consequently W1
on, therefore
Thus
the
odd convergents form an increasing sequence, the even convergents
form a decreasing sequence, and every odd convergent convergent.
is less
than any even
CLOSENESS OF APPROXIMATION
399
Consequently, as n tends to infinity, both w 2n _! and u2n tend to limits, and these limits are the same, for by equation (D) of Art. 9,
finite
W2 n 7 U*n-l == l/?2n?2n-l-> 0.
Hence un tends to a
limit x such that for every n,
In other words,
and
(ii) its
continued fraction is convergent, (i) every infinite simple value is greater than any odd convergent and (iii) less than any
even convergent.
4. Each convergent is a closer approximation fraction than the preceding. 1
For
let
so that
z'
is
the (n + l)th complete quotient, then
= a n+lPn + ^n-l)/( a n+l?n + ?n-i) z = (z'p n +p n -.i)/(z q n + ?n-])
Consequently
2;'
and
(zg^
and
>? n _i
- j> n ) = "
2
r
'
(
f
and therefore
gn
and 2'=a
...
Pn+l/
Now
value of the continued
1
1
1
= %-{-
z
to the
(2;<7
- y n _!)
n _1
=
2;'^!, therefore
z-that
is
to say,
is
nearer to
^an to
p n/
pn -\l
5. ulwy convergent p/q of a simple continued fraction is a closer approxito the value of the fraction than any rational fraction r/s with a denominator less than q. 1
mation
For
let
z
al
preceding p/j. to z than p'/q'. lies
-\
--
By
...
and
,
Art. 14,
if
let
Moreover, between y/g and p'jq', and
that
?
|
ry'
nearer to z than ^/j, it is also nearer between p/q and p'/gr' therefore r/s ;
1
P'i
r_ Therefore
be the convergent immediately
r/s is
z lies
\~8
p'jq'
i.e.
9
q'
- ^jp' <^ |
;
9'
and, since
ry'
- sp'
is
an integer>
it
follows
16. If
is the
numerical value of the error, in representing the value z
of the continued fraction, by
and within wider For
then
,
limits
u n =p n /q n
if
p n/q n
then by Art. 13,
,
w n> is
ERROR
LIMITS OF
400
U n+l9
^>
^n+2>
an increasing or a decreasing sequence, therefore ~ ~ t*n Un+2 <\ W n *
I
I
Also by equations (D), I
W n - W n +2
whence we have the
I
=nW?n7nf2 and
I
w n ~ w n-h
inequalities (H).
Again, g n + a = a n +29Wi
+ ?n>
therefore
^ ?n +2/a n+2 Also qn+l >q nJ and therefore l/9 n ?n Hence the inequalities (I) follow from (H).
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