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AN ANALYTICAL CALCULUS

VOLUME II

ANANALYTICAL CALCULUS

FOR SCHOOL AND UNIVERSITY

BY

E.A.MAXWELLFellow of Queens' College, Cambridge

VOLUME II

CAMBRIDGEAT THE UNIVERSITY PRESS

1966

CAMBRIDGE UNIVERSITY PRESSCambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi

Cambridge University PressThe Edinburgh Building, Cambridge CB2 8RU, UK

Published in the United States of America by Cambridge University Press, New York

www. Cambridge. orgInformation on this title: www.cambridge.org/9780521056977

© Cambridge University Press 1954

This publication is in copyright. Subject to statutory exceptionand to the provisions of relevant collective licensing agreements,no reproduction of any part may take place without the written

permission of Cambridge University Press.

First published 1954Reprinted 1966

This digitally printed version 2008

A catalogue record for this publication is available from the British Library

ISBN 978-0-521-05697-7 hardbackISBN 978-0-521-09036-0 paperback

CONTENTS

PREFACE vii

CHAPTER VII: THE LOGARITHMIC AND

EXPONENTIAL FUNCTIONS

PAGE PAGE1. The logarithm 1 6. The exponential function 182. First properties of the log- d ^ d

arithm 2 7. The relations / = - , / = y 203. The graph of log a; 4 dx xdx4. The use of logarithms in 8. The integration of ex 24

differentiation 10 9. The reconciliation of logea?5. The use of logarithms in integ- and log10a? 27

rating simple rational func-tions 11

REVISION EXAMPLES I I I , 'ADVANCED' LEVEL 28

CHAPTER VII I : TAYLOR'S SERIES ANDALLIED RESULTS

8. The series for since and cosy 509. The binomial series 51

10. The logarithmic series 5311. The exponential series 5412. Approximations 5513. Newton's approximation to

a root of an equation 5614. Leibniz's theorem 62

REVISION EXAMPLES IV, 'ADVANCED' LEVEL 66

CHAPTER IX: THE HYPERBOLIC FUNCTIONS

1. The hyperbolic cosine and sine 84 4. The graph y = sinho? 952. Other hyperbolic functions 87 5. The inverse hyperbolic cosine 963. The graph y = cosh a? 94 6. The inverse hyperbolic sine 98

CHAPTER X: CURVES

1. Parametric representation 100 6. The length of a curve in polar2. The sense of description of a coordinates 105

curve 101 7. The 'gradient angle' tp 1073. The 'length' postulate 102 8. The angle from the radius4. The length of a curve 103 vector to the tangent 1095. The length of a curve in 9. The perpendicular on the

Cartesian coordinates 104 tangent 110

1. A series giving since2. A series giving 1/(1 +x)3. Expansion in series4. The coefficients in an infinite

series6. Taylor's theorem6. Maclaurin's theorem7. Maclaurin's series

384041

43444949

VI C O N T E N T S

Chapter X: Curves—continuedPAGE PAGE

10. Other coordinate systems 111 14. The circle of curvature 11911. Curvature 113 15. Envelopes 12312. A parametric form for a 16. Evolutes 126

curve in terms of s 116 17. The area of a closed curve 12813. Newton's formula 117 18. Second theorem of Pappus 132

R E V I S I O N E X A M P L E S V, ' A D V A N C E D ' L E V E L 133

R E V I S I O N E X A M P L E S VI , ' S C H O L A R S H I P ' L E V E L 138

CHAPTER XI : COMPLEX NUMBERS

1. Introduction 158 11. The product of two complex2. Definitions 161 numbers 1723. Addition, subtraction, multi- 12. The product in an Argand

plication 163 diagram 1734. Division 164 13. De Moivre's theorem 1745. llqual complex numbers 165 14. The nth roots of unity 1776. The complex number as a 15. Complex powers 178

number-pair 166 16. Pure imaginary powers 1797. The Argand diagram 168 17. Multiplicity of values 1818. Modulus and argument 169 18. The logarithm of a complex9. The representation in an number, to the base e 183

Argand diagram of the 19. The sine and cosine 184sum of two numbers 171 20. The modulus of ez 186

10. The representation in an 21. The differentiation and integ-Argand diagram of the ration of complex numbers 186difference of two numbers 172

R E V I S I O N E X A M P L E S V I I , ' S C H O L A R S H I P ' L E V E L 192

CHAPTER X I I : SYSTEMATIC INTEGRATION

1. Polynomials 200 6. Trigonometric functions 2142. Rational functions 200 7. Exponential times polynomial3. Integrals involving *J(ax + b) 203 in a? 2154. Integrals involving 8. Exponential times polynomial

*J(ax2 + 2hx + b) 203 in sines and cosines of5. Integrals involving *J(ax2 + 2hz multiples of a? 215

+ b): alternative treatment 207

CHAPTER X I I I : INTEGRALS INVOLVING ' I N F I N I T Y '

1. 'Infinite'limits of integration 217 2. 'Infinite'integrand 222

R E V I S I O N E X A M P L E S V I I I , ' A D V A N C E D ' L E V E L 226

R E V I S I O N E X A M P L E S I X , ' S C H O L A R S H I P ' L E V E L 227

A P P E N D I X , FIRST STEPS IN PARTIAL DIFFERENTIATION 236

ANSWERS TO EXAMPLES 245

I N D E X 271

P R E F A C E

Appreciation for help received was expressed inthe Preface to Volume I, but I would recordhow much deeper my indebtedness becomes asthe work progresses.

E. A. M.

QUEENS' COLLEGE, CAMBRIDGE

June, 1953

CHAPTER VII

THE LOGARITHMIC AND E X P O N E N T I A L

FUNCTIONS

THE particular functions which we have used in the earlierchapters (Volume i) are the powers of x, the ordinary trigono-metric functions, and combinations of them such as polynomials.

We now introduce an entirely new function, the logarithm. Theneed for it arises, for example, when we seek to evaluate theintegral

\xndx

for n = — 1. The standard formula

/•xndx = -xn+l

71+1

becomes meaningless; the integral cannot be evaluated in terms ofthe functions at present at our disposal.

1. The logarithm. Consider the integral

Cdx

To make the discussion precise, we shall fix the lower limit, givingit the value unity; the effect of this is merely to remove ambiguityabout the arbitrary constant. The integral is a function of itsupper limit, which we denote by the letter x, replacing thevariable in the integration by the letter t. The function is thus

where (Vol. I, p. 87) f'(x) = - .x

2 LOGARITHMIC AND EXPONENTIAL FUNCTIONS

The function defined in this way is called the logarithm of x,usually written ,

logicor loge#,

the suffix e being inserted for reasons to be given later (p. 27).

2. First properties of the logarithm. We now prove someof the basic properties to which the logarithm owes its importance.The reader will note the very close connexion with logarithms tothe base 10', with which he is presumably familiar.

(i) log 1 = 0.

This follows immediately, since (Vol. i, p. 83)

(ii)

For log #£

log ay = log x + logy.(-r*Now use the substitution

in the latter integral.

t =

We have

dt =

f /Vr

= XU

the relation

= xdu

(remembering that x is CONSTANT here, the variable of integrationbeing t). Also the values xfxy of t correspond to the values \,yof 11. Hence ^ ^

Jx t J i XU J i U

We therefore have the required relation

log xy = log x + log y.

FIRST PROPERTIES OF THE LOGARITHM 3

COROLLARY. log {xjy) = log x — log y.

For logo; = log j P ] y\

= log(x/y) + \ogy.

(iii) log(#w) = nlogx.

Cx dtIn the relation logic = 1 —,

h t

make the substitution u = tn.We have the relation du — ntn~xdt.

Also the values 1, x of t correspond to the values 1, xn of u. There-fore, since _

11 ntn '

we have logo: = ———Ji nt

= — (applying the substitution)

so that log (xn) =

Note, n may have any real value, and is not necessarily apositive integer.

(iv) The value of log x increases indefinitely as x does.Suppose that N is aiiy large number and m the largest integer

such that 2m<N. Then

. .T r2dt r*dt r8dt c** dt cNdt\ogN= - + - + - + ...+ + 7 .

Now consider —.J2P-1 t

Throughout the interval (2P-1,2^), the variable t is less than 2*\

so that - exceeds —• Hence, by the basic definition of an integral

(Vol. i, p. 81),

r dt f2* dt __ 1 I""]2* _ 2JJ-2^-1 _ 1


Applying this inequality to the successive integrals in the formulaCN dt

for log N, and noting that — is positive, we obtain the relationJ2"1 t

logN> J + J + + ... + J (m terms),

so that logiV, which exceeds \m> increases without bound.

COROLLARY. The value of logx tends to MINUS infinity as x tendsto zero.

For lc

so that

or

As N tends to infinity, 1/iV tends to zero, and the result follows.

3. The graph of log JC.If y log a?,

then dx x*

Hence -^- is positive for all positive values of x, so that log x is aCiX

steadily increasing function of xfor positive x (Fig. 62). Also the

gradient - is large and positivex

when x is small and positive,decreases as x increases, takingthe value 1 when x = 1, andtends to the value zero as x in-creases indefinitely. Moreover,as above, y itself tends to 'minusinfinity* when x tends to zero,increases with x, taking thevalue 0 when x = 1, and 'tendsto infinity' as x increases in-definitely.

lg'

The general shape is therefore that shown in the diagram.

THE GRAPH OF LOG X 5

The value x = 0 imposes a downward barrier on the logarithm,and the function log a; is undefined for negative values of x.

Note. lixx = x2i being positive, then loga^ = logic2; and, what ismore important, the converse property holds, that, if log xx = log x2,then xx — x2. In fact, if xx > x2, then loga^ > log#2, since the loga-rithm is an increasing function; and if xx < x2, then log xx < log x2.

It is also clear from the graph that, if c is a given number, thenthe relation

logo? = c

defines x uniquely.

WARNING. The value of the integral

C~zdt

J-7is fully determinate, but we cannot use the argument:

J_2T- 3

since log (— 3) and log (— 2) are non-existent. We must proceed asfollows:

Substitute t = —u,

so that dt = — du.

Then-2 t J 2 -u J a u

UNDEE NO CIECUMSTAKCBS may we evaluate an integral such as

J-2 t

where the variable of integration t runs through the value zero atwhich (l/£) has no meaning.


We give two typical examples to show how logarithms arise inphysical applications.

ILLUSTRATION 1. An electric circuit contains a resistance R, acoil of self-inductance L, and a battery of electromotive force E,supposed constant. To find an equation to determine the currentt seconds after a switch in the circuit has been closed.

The equation for the current x is known to be

dxso that L -r- = B- Rx.

dt R

Hence the differentials dt9dx areconnected by the relation

dtE-Rx'

and so t ~ JE-Rx

[Note. We are changing from the conception of # as a functionof t to that of t as a function of x.]

The value of the integral may be written down at once, but thebeginner may prefer to use the substitution

so that — Rdx = du,

L Cdu L, n

giving t = - - I— = --logu + C,

assuming that we are dealing with a case in which u is positive.

Hence t = - -^ log {E - Rx) + C,

where C is an arbitrary constant. Now x = 0 when t = 0, and so

0 = - | lor C = | l o g J2.

THE GRAPH OF LOG#

Hence t = ^ {log E - log (E - Rx)}

L E

dxAn alternative method for dealing with the relation L —

atwill be given later (p. 22).

ILLUSTRATION 2. To find the work done when a given quantity ofa perfect gas expands from volume vx to volume v2 at constant absolutetemperature T.

It is known that the volume v and pressure p are connected bythe relation _

pv = RT,where R is constant. Also the work done is known to be

= pdv.

Hence W =

Flog v | "

The following illustrations are typical of integrals involvinglogarithms.

ILLUSTRATION 3. To find

C4xdx

We have



= \ t&nxdx.

Write u = cos#,so that du = — sin xdx.

Hence I = - — = —logu = log (l/u)J v>

= log sec x.


/== Lcwlogo;fe (?i4= — 1).

On integration by parts (Vol. I, p. 103), we haverxn+l l

logo;— -,-dxB Jn+1 x

xn+l-

n+lxn+l r xn

log x — r axn+l 8 Jn+lxn+l xn+l

l


coso;

Ccos xdx C cos xdxsm2a;

Let u = sin a;,

so that du = GO&xdx.

TTT , _ Ccos xdx C cos xWe have / = s—= hi r

J cos2 a; J l~sm

T h e n

, , 1 + sin a?flog-2 & l

THE GRAPH OF LOGiC

This may also be expressed in the form

in x\2 , /I + sin x

= log (sec x + tan x).

Another form for the answer isI X+

ILLUSTRATION 7.* To find

T —

Notice that the differential coefficient of the denominator is

2x + 4,

and express the numerator in the form

Then 7 = 2JX*+4:X+ I'd JX*-

dx

f dx

= 21og (a;2 + 4a; +13) - J tan"1 ^ - .\ 3 /

ILLUSTRATION 8.* To find

Notice that the differential coefficient of the denominator is

2a;-6,

and express the numerator in the form

f (2a?-6)+23.

/ = _ _L 1 j_ 23

* An important type.

Mil

10 LOGARITHMIC AND EXPONENTIAL* FUNCTIONS

EXAMPLES I

Find the following integrals:

dx1.

- dx. 5.

x+1'dx

Evaluate the following integrals:

7. I d-. 8. I —-. 9.Ja x J_3 aj+1 Jo

r~3 d*, „. r_^_. i2. rDifferentiate the following functions with respect to a;:

13. log(3a; + 2). 14. logtana;. 15. logcoseca;.

16. a;2loga;. 17. xnloex. 18.


19. flog xdx. 20. [l^^dx. 21. fJ J J

22. fa; log a;da. 23. [4^-. 2±. [ .J Jsma; J si

9^ I v"w ' v / ^ w Qfi j v / ^/ wW _ | (^a; + ojaa;

28. fJ-

4. The use of logarithms in differentiation. The differen-tiation of a fraction (in which the numerator and the denominatormay themselves be products of factors) is often made easier by themethod known as logarithmic differentiation, illustrated in thefollowing examples.

ILLUSTRATION 9. To differentiate the function

THE USE OF LOGARITHMS IN DIFFERENTIATION 11

Take logarithms. Then

log y = 3 log x + log (1 + a?2) — 4 log (1 — x) — 2 log (1 + 2x).

Differentiate. Then

1 dy 3 2x 4 42/ dx x \+x2 l—x l-h2#'

and the value of -~ follows at once.dx

With a little practice, the two steps may be taken together:

ILLUSTRATION 10. To differentiate the function

( 1 - 2x)2 sin3 x

Take logarithms and differentiate. Then

Idy —4 3 cos a; 16a;— fydx l — 2x

EXAMPLES IIUse the method of logarithmic differentiation to differentiate

the following functions:

cos2#T+2'

x2(l+x)2 ccsina: (1+a;2)2

• ( 1 + 4 ) 2 ' ( 1 + ) 3 ( 1 - ^ ) ' 2a; cos2 a;'

5 a ; ( l a ; ) 3 (1 + COSOJ)2

• • —I iTT: • 8 . -r-— • rr. 9 .

5. The use of logarithms in integrating simple rationalfunctions. A rational function of x is an expression of the form

u(x)v(x)'

where u{x), v(x) are polynomials in x. We shall later (p. 200) give adetailed treatment of the integration of such functions; here wegive a preliminary account of the simpler cases.


If the degree of u(x) is higher than that of v(x), we can divideu(x) by v(x), and obtain an expression of the form

p(x) + -7-Y,^v v(x)

where p(x) is a polynomial, and w(x) is a polynomial whose degreeis less than that of v(x).

The integration of the polynomial p(x) is immediate. We maytherefore confine our attention to the form

w(x)

where w(x), v(x) are polynomials in x9 the degree of w(x) being lessthan that of v(x). The method is to express this quotient inpartial fractions; details may be found in a text-book on algebra,but, for convenience, a brief account of the calculations involved isinserted for reference.

In order to explain what is required, we consider some typicalexamples. (Different mathematicians use varying methods. Thosewhich follow have the advantage of giving independent checks ofaccuracy in some of the more complicated cases.)

2x( i ) / ( )

The denominator consists of the two linear factors (x — 2), (xeach occurring to the first degree only. We seek to express f(x) inthe form . R

x-2x + 3'

TT7 , A B 2xWe have - + •

Multiply throughout by x — 2. Then

A Blx-2) 2xx+3 # + 3'

This holds for all values of a;; in particular, for x = 2. Then

2.2 4~~ 2 + 3 " 5"

Hence A = - .5

INTEGRATING SIMPLE RATIONAL FUNCTIONS 13

In practice, these steps are usually telescoped, as we nowillustrate in finding B. Multiply throughout by 3 + 3 and then put3 = - 3. Thus

D 2(-3) 6

Hence f(x) =

(-3-2) 5

4 65(3-2) 5(3 + 3)

3+1(ii) f(x)

The denominator consists of two linear factors (3 + 2), (3—2),of which (3—2) occurs to degree 3. We seek to express f(x) in theform

A B C D•(3-2)3+(3-2)2 + 3-2>

A B C D x+1SO that + ^ +

Multiply throughout by ru + 2 and then put a; = — 2. Thus

- 2 + 1 - 1 = 1"" (_2 -2 ) 3 " " - 6 4 " " 64*

Multiply throughout by (x — 2)3 and then put x = 2. Thus

2 + 2 4

In order to find C, D, we use these values of A, B:

C D x+l 1 3+

64(o; + 2) (a? -2)3

- 123+8-48^-9664(3 + 2) (3-2)3

3 3 -63 2 -43 + 2464(3 + 2) ( 3 -2 ) 3 '

At this point, we are able to check accuracy for the highestcommon factor of the denominators on the left-hand side is (x — 2)2.


Hence x + 2 and x — 2 MUST be factors of the numerator on theright. By division, we find that

„ G D x-6Hence -r + - = - —

Multiply by (x — 2)2 and then put x = 2. Thus

° 64~~16"

(a;-6) 1

64(a;-2)2 64(*-2)'

again checking accuracy by the cancelling of x — 2.

Finally, D=-^i'

Hence

Ax— 1

The denominator consists of the linear factor (#—1), repeated,and the quadratic factor (x2 + x + 1). We require to express f(x) int h e f o r m A B Cx + D

the numerator above the quadratic factor being of the form Cx + D.We thus have

A B Cx + D 4x-l+ +

Multiply throughout by (x— I)2, and then put x = 1. Thus

^ ~ 1 + 1 + 1 " 1#

INTEGRATING SIMPLE) RATIONAL FUNCTIONS 15

^ B Cx + D Ax-1 1Hence f- -

x—\

„ 5 Oz + D a:-2Hence - +

the cancelling of the factor x — 1 providing a check of accuracy.Multiply throughout by x — 1, and then put x = 1. Thus

1-2 1

Hence

1 + 1 + 1 3'

x-2 1

on cancelling the factor, a?— 1. Hence

and so

The following illustrations exhibit some further points aboutthe calculation of partial fractions, and also show how theintegration of rational functions is carried out.


1 =

The numerator is not of less degree than the denominator, sowe begin by dividing out:

xx _ 2a:2-1


Consider, then, the function

Factorize the denominator, so that

2a:2-1

We therefore have to find constants A,B9C9D such that

A B C D 2#2-l

{ (x — 1 )2 (x = 1

2 and then put | _ . ThusA =

(l + l)2 4 '

Hence5 i> 2a; 2 -1 1

+ 4:(x-l)2

Hence, by the usual process,

It follows that

l h L 1 x 3 x 1 3 . .II *^ At "I \ O I A I 1 \ • A I . 1 \ O A I 1 \ / ti'«* /

INTEGRATING SIMPLE RATIONAL FUNCTIONS 17


/ = f Udx

We must first factorize the denominator. It vanishes whenx = 2, so that x — 2 is a factor, and, after division, we find that itis (x - 2) (x2 + 2x + 5). We therefore seek to express

// N 1 3 1 3

m the form + -

. , , A Bx + C 13so that +x-2

Following a routine which should now be familiar, we have

13

Bx + C 13 1(x-2)(x2 + 2x + 5) x-2

%-2x-x2

""xHence


EXAMPLES III

Integrate the following rational functions:

1.

4.

7.

10.

13.

15.

18.

21.

23.

x2 - 9 '

a*s - 1 "

1

2.

5.

11.

1 3 ) - 1 6 - ;

19.

22.

4a;+ 7*

3.

- . 9.

12.

14.

17.

20.

x-2

x2+l '

2x

24.

6. The exponential function. Imagine the graph y = logx(Fig. 62) to be turned, as it were, through a right angle andviewed through a mirror, and ythe axes then renamed to givethe curve shown in the diagram(Fig. 64). Then y is a certainfunction of x with the propertythat

x = logy.

Thus y is an 'inverse' functionof log x in a sense similar to thatin which sin"1 a; is (Vol. I, p. 38)an inverse function of sin a:.

0

Fig. 64.

THE EXPONENTIAL FUNCTION 19

We write the relation x = logy

to give y in terms of x, in the form

y = exp x,

where exp x, whose properties we now study, is called the exponen-tial function. It is defined, as the graph implies, for all values of x,increasing steadily from zero to 'infinity' as x increases from 'minusinfinity' to 'infinity'.

The exponential function (of a real variable x) is necessarilyPOSITIVE.

From the relation logy = x,

we have, by differentiation with respect to x9

1 ^ = 1ydx 'dy

so that ~T~ = V*dx *

Hence the differential coefficient of exp x is exp x itself.It is convenient to have a name for the value of the function

when x = 1, and for this we use the letter c.Thus 1

exp 1 = e,

or, in equivalent form, logc = 1.

From the graph, we have the relation

[The value of e, to four significant figures, is 2-718.]

We now seek to identify the function expx in terms of theconstant e and the variable x. If

y = expo;,

then log {exp x) = log y= x.

Also the relation log(^) = nlog# leads, on replacing x,n by theletters e, x respectively, to the relation

log(ex)


since loge = 1. Hence

log{exp#} = log^*).

But we have proved (p. 5) that, if the logarithms of twonumbers are equal, then the numbers themselves are equal,and so _

exp x = ex.The exponential function exp a; is therefore identified as the

number e raised to the power x.

1. The relations -~ = - , ~ = v.dx x dx J

(i) THE LOGARITHM.It follows from the definition of a logarithm that the relation

dx x

yields for positive x the result

where C is an arbitrary constant.If x is negative, say __

X — *""" Uy

where u is positive, then

dy _dydx __ dydu dx du dx'

dy 1hence -=— = —

du x1

so that y = \ogu-\-G

We may therefore conclude that, whether x is positive or nega-tive, the equation ,

dx x

leads to the relation y — log | x \ + C,

where\x\is the numerical value of x.


In practice, it is customary to use the form

with the tacit assumption that x is positive; but this needs care.The relation y = log x + G may be put into an alternative form

by writing C = log a, where a is also an arbitrary constant. Then

y = logz + loga

= log ax (ax assumed positive).

By the definition of the exponential function, we then have

ax = ev,

or x = b ev,

where b is likewise an arbitrary constant, assumed to have thesame sign as x.

(ii) THE EXPONENTIAL FUNCTION.

We turn now to the equation

dy= y

dx 1Writing this in the form — = -

dy ywe see that interchange of x,y in the above relation x = bev leadsto the result , _

y = bex.dy

Hence the equation -~ = yCLX

leads to the relation y = bex,

where b is an arbitrary constant, assumed to have the same sign as y.More generally, the relation

leads to the relation y = bekx:


For the substitution x = ujk

gives dy = dydx=ldydu dxdu kdx9

so that — = r

du k

Hence y = beu

We give two typical examples to show how exponential func-tions arise in physical applications.

ILLUSTRATION 13. We return to Illustration 1 (p. 6) of acircuit with resistance R, self-inductance L and electromotiveforce E. The equation for the current x at time t is

dxHence £ — = E — Rx.

dt

Write E-Rx = u;

, du __ du dxdt dx dt

= (-BIL)u.Hence u = Ae~iB/L)t,

where A is an arbitrary constant, so that

If x = 0 when t = 0, then

E = Ae° = A,

and so E-Bx = Ee-{R'L)',

or


ILLUSTRATION 14. To find the variation of pressure with height inan atmosphere obeying the law

pv =* constant,

where p, v denote pressure and volume respectively.Consider a vertical filament of air whose cross-sections have

area SA (Fig. 65). Let the pressures at heights x9x + Sx bep,p + 8p. Then the element of volume (shaded in the diagram)of height Sx and base SA is in equilibrium under pressure roundits sides, which does not concern us, and alsounder the following vertical forces: P + Si>

(i) pSA upwards; j

(ii) (p + 8p)8A downwards;

(iii) pSASx downwards,

where p is the weight per unit volume atheight x. Hence

pSA -(p + Sp) SA - pSxSA = 0,

or 8p + pSx = 0,

Now let p09 p0, v0 be the values of p9 p, v atground level. Since p is the weight per unitvolume, the relation

pv = povo

qy rp

is equivalent to - = —,P Po

and so

In the limit, this is

Po

SAFig. 65.

dx Vn

and so p ,

where A is an arbitrary constant. But p = p0 when x = 0, so that

p0 = Ae° = A.

Hence the pressure at height x is given by the relation

p


8. The integration of e*.

To find I exdx9

we have merely to note that the relation (p. 19)

leads at once to the result

ex = exdxy

so that the value of exdx is ex itself.

COROLLARY. eaxdx = - e°*.


/ = eax sin bxdx.

On integration by parts, we have

1 fl/ = -eaxsinbx— \-eax.bcosbxdxa Ja

1 b C= - eax sin bx — eax cos bxdx.

a aj

Integrating again by parts, we have

/ = -eaxsinbx—^eaxcosbx-\—^ \eax( — bsia Ja a2

a a ; 6 —s/.a2

Hence (1 + - I / = - eax sin 6# — 5 eax cos 6a;,

\ a2/ a a2

so that / = -g—^2 (a sin te — 6 cos 6a;).

THE INTEGRATION OF ex

ILLUSTRATION 16. To prove that, if

then

We have

Hence

y = eax sin bx9

dx2 dx

-2L =- a eax s u l bx + b eax cos bxdx

= ay+ beax cos bx.

d2y dy . „ , ,„ „„ . .-=~ = a~- + abeax cosbx -b2eax smbxdx2 dx

25

so that

EXAMPLES IV

Find the differential coefficients of the following functions:

1 P2X

4. e^cosa.

7. ^(e^ + c-x)2.

10. (l + e^)si]ia;.

Find the following

13. \e2xdx.

16. \x2e-x*dx.

r19. j e ooB«<to.

• J

2. e*\

8. *e*Sin*.

11. e3iCcos4^.

integrals:

14. \e~5xdx.

17. \xexdx.

r20.JBin»ooB«6^-d

f23. e3a5cos4a;da;.

3.

6.

9.

12.

15.

18.

x. 21.

24.

( l+a:2)e-

I-*2'

enana;.

\2xex*dx.

\x2exdx.

\sec2xeta'nxd,

((l+x)e2xdx


ILLUSTRATION 17. To find a formula of reduction for

In=\xnexdx.

On integration by parts, we have the relation

In = ex.xnex.nxn~1dx

= xnex-nln_v

This is the required formula.

ILLUSTRATION 18. To find a formula of reduction for

In = ex$mnxdx.

On integration by parts, we have the relation

In = e^.sin71^ — ex.nsmn-1xcosxdx

= ex sin,71 x — n. ex sin71"1 x cos x

+ n \ex {(n — 1) sin71"2 x cos2 x — sin71 #} dx

— n ex sin*1"1 x cos x

+ n lex {(n — 1) sin™-2 o;( 1 — sin2 a:) — sinn x] dx

= ex sin71 a; — ex sin71"1 x cos a;

Hence

= e^ sinw OJ — w- e35 sin71"1 x cos a: + n(n — 1) Jw_2.

This is the required formula.

EXAMPLES V

Obtain formulse of reduction for the following integrals:

1. \xneaxdx. 2. le^sin^da:. 3. \eaxoosnbxdx.

Evaluate the following integrals:

4. x5exdx. 5. ex&in4:xdx. 6.Jo Jo Jo

THE RECONCILIATION OF LOGe# AKD LOG10# 27

9. The reconciliation of logejc and logio A:. The reader willrecall the elementary definition:

The logarithm of a number N to the base a is the index of the powerto which a must be raised to give N.

If N = ak,

then l°ga^ = b*

In particular, if y = ex,

then x = Iogey.

By this relationship the work which we have just done is reconciledto the more elementary approach, and our use of the word'logarithm' is justified.

Note. The relation -=- (logo;) = -

is true only for the base e.

For other bases we must proceed as follows:

Let y = loga#.

Then x = d».

Take logarithms of each side to the base e. Then

iogex = ylogea.

Differentiate. Then - = log,a-^,

so that

Thus, if

then

™-&

dx

y

dydx

xlogea*

= l°gloa;,

1

#loge103-2


REVISION EXAMPLES III

'Advanced9 Level

1. Differentiate (1 +#)nlog (1+x) (i) with respect to x9 (ii) withrespect to (l+x)n.

If y =

prove that 4x(l+x)<^+2(l + 2x)(^--y = 0.

2. Prove that, if y = u4sin2# +J3cos2#, then

dx2 dx

Given that z = Ae2x + Be~2x + C, find a differential equationsatisfied by z and not containing the constants A,B,C.

3. Find dy/dx in terms of t when

Prove that -^ = 1 when t = 0, and find a second value of t fordx

wnicn — = 1.

T-» rf2^/ 2 /I — 2A3

Prove that -z~ = — - 1 •dx2 3 \ 1 - ^ /

4. Differentiate with respect to x:

log sin2 a;,


sin2x cos a;, eax (cos aa; + sin ax),

6. Find the differential coefficient of <Jx from first principles.Differentiate with respect to x:

REVISION EXAMPLES III 29

7. Differentiate the following with respect to x:

csiTi 9 /y i ^an"~11 ———— I oQu/>» ' i 1 ^^ />t2 /

8, Differentiate the following with respect to x:

, sin2 3x, x

9. Differentiate the following with respect to x, expressing yourresults as simply as possible:

» /_.,, i . . . x . _ . / i . . . t x /,,,v . , . ._ - , /3 + 5 c o s o ; \+ i77)» (m) s i n ~ T5 + 3coso;/

10. Find, from first principles, the differential coefficient of\\x2 with respect to x.

Differentiate the following with respect to x, expressing theresults as simply as possible:

, tan*"1 (x2), loge tan 2x.(x+1)

11. Differentiate the following expressions with respect to x,giving your results as simply as possible:

. (

12. Differentiate the following functions of x with respect to x:

sin a;-cos a;/ 3 \ 2 l(l+x\ si\ x2) 9 /s/\I — x)' si

13. Prove from first principles that

-j- tan x = sec2 xax

and deduce the values of -7- tan"1 x and -7- cot"1 x.dx dx

Differentiate with respect to x:1+x2

=,1 —x


14. Differentiate the following expressions with respect to x>simplifying your results as much as you can:

x3

, sin2 x cos3 x, sin""1 (y]x).± -\~x

Prove that, if y = e~2x cos 4x, then

15. Find from first principles the differential coefficient ofwith respect to x.

Differentiate the following expressions with respect to x,simplifying your results as much as you can:

x

16. A particle moves along the #-axis so that its displacementx from 0 at time t is e'cos2£. Find its velocity and acceleration attime t = 77.

Prove that the values of t for which the particle is at rest formtwo arithmetic progressions, each with common difference TT, andthat the successive maximum displacements from O form ageometric progression

where a is an acute angle such that tan a = J.

17. Two circles, with centres 0 and P, radii a ft. and b ft.respectively, intersect at A and B; the chord AB subtends angles20 and 2<f> at 0 and P respectively; the area common to the twocircles is denoted by A and you may assume that

2A = a2(20-sin 260+ &2(2<5&--sin 2<£).

Prove that, if P moves towards 0 with a speed of u ft. per sec,t h e n dL ^ . a

at

18. Find the equations of the tangent and normal at any pointon the curve x = a cos3 t,y = a sin31, when t is a variable parameter.

Show that the axes intercept a length a on the tangent and alength 2a cot 2t on the normal.


19. A rod AB of length a is hinged at A to a horizontal tableand turns about A in a vertical plane with angular velocity o>. Aluminous point is situated vertically above A at a height h(>a).Find the length of the shadow when the rod makes an angle 6with the vertical, and prove that the length of the shadow isaltering at the rate haa)(hcosd — a)l(h — acos6)2.

20. Prove that, if a, 6 are positive and 96 > a, then

asinx + bsinSx

will have a maximum value for some value of x between 0 and |TT.Find this maximum value when a = 3,6 = 0-5, proving that it

is a maximum and not a minimum.

21. Given that y = x5-5x* + 5x2+l,

find the stationary values of y. Determine whether these valuesare maximum or minimum values or neither.

22. The perpendicular from the vertex A to the base BG of atriangular lamina cuts BG at D; CD = q,DB = p (where q<p)and AD = h. The lamina lies in the quadrant XOY with B onOX9 (7 on OF, and A on the side of BG remote from 0. It movesso that B, C slide on OX, OY. Prove that, if LOBG = 0, then OAis maximum (not minimum) when

tan 26 = .q-p

Prove also that OA = CA when

tan 6 = .q-p

23. On a fixed diameter of a circle of radius 6 in., and onopposite sides of the centre 0, points A,B are taken such thatAO = 3 in., OB = 2 in. The points A,B are joined to any pointP on the circle. Prove that, as P moves round the circle, AP + BPtakes minimum values 13 in. and 11 in., and takes a maximumvalue 5^7 in. twice.

24. Prove that, for real values of x, the function

3 sin a?

cannot have a value greater than /3 or a value less than — </3.Sketch the graph of this function for values of x from — IT to TT.


25. A particle P falls vertically from a point A, its depth belowA after t seconds being at2, where a is constant. B is a fixed pointat the same level as A, and at a distance b from A. Prove that therate of increase of Z.ABP at time t is

2abt

and show that this rate of increase is greatest when

LABP = 30°.

26. The angle between the bounding radii of a sector of a circleof radius r is 0. Both r and 0 vary, but the area of the sectorremains constant and equal to c2. Prove that the perimeter of thesector is a minimum when r = c and 0 = 2 radians.

27. If a variable rectangle has a diagonal of constant length10 inches, prove that its maximum area is 50 square inches.

28. A straight line with variable slope passes through the fixedpoint (a, 6), where a, b are positive, so as to meet the positive partof the o;-axis at A and the positive part of the y-axis at B. If 0 isthe origin, prove that the minimum area of the triangle OAB is 2ab.

Find also the minimum value of the sum of the lengths of OAand OB.

29. Prove that of all isosceles triangles with a given constantperimeter the triangle whose area is greatest is equilateral.

30. A variable line passes through the point (2,1) and meets thepositive axes 0X,0Y at A, B respectively. If 0 denotes the angleOAB (0<6< 77), express the area of the triangle OAB in termsof 9, and prove that the area is a minimum when tan 0 = J.

Find also the value of 0 if the hypotenuse of the triangle is aminimum.

31. Find the turning points on the graph of the function

2x

stating (with proof) which is a maximum and which is a minimum.Sketch the graph and find the equation of the tangent at the

point on the curve where x = J.


32. The slant height of a right circular cone is constant andequal to Z. Prove that the volume of the cone is a maximum whenthe radius of the base is

33. A lighthouse AB of height c ft. stands on the edge of avertical cliff OA of height b ft. above sea level. From a small boatat a variable distance x from 0 the angle subtended by AB is 0.Prove that

tan 0 = —r .

Prove also that, if a is the maximum value of 0, then

c

Integrate with respect to x:

2x—734 ® ^

4-x35. (i) COB«2*, (ii) Bx2_2x_v

36. (i) 5=i* (ii) a;tan-1 a;, (iii)w cos4* v ' » v /

3 7 ( i ) ( ) i 2 8

7r38. (i) 3 ^ 2 n a ; + 6> (ii) ^sina:, (iii) (a*-

39. Integrate with respect to x:

cos3 a;sin2 a;'

Evaluate — ^rr^-—TT dx

J2 (2a;-3)(2a;+1)

to three significant figures, given that log10e = 0-4343.

40. Interpret by means of a sketch the definite integralt-x2)dx9

JoJo

and evaluate this integral.


41. Integrate the following functions with respect to x:

(I-*2)* ' x+1 *

f2Evaluate xlogexdx.Ji

42* By a suitable change of variable, prove that

dxf»» dx r*Jo l + sin# Jo

and by means of the substitution t = tan^a;, or otherwise, evaluateone of these integrals.



[InEvaluate sin 3# cos 2x dx.

siiJo

45. Explain the method of integration by parts, and employ itto integrate ^(1 — x2) with respect to x.

Integrate with respect to x

sin*"1^ 1 . .171 1\> Ti 2T#> ex(cosx-smx).](l-X2) (1X2)$


. 5 x2+l

47. Integrate with respect to #:

1 1sin2 # + 2 cos2 #' a;3+l#

By integration by parts, or otherwise, integrate sin-1 x.



-, (1 — cos2 x)2 sin x, xex.1-x'

Prove that ~r- ^ &x — s i n ~ x ^J3 -y(1 — X )

49. By integration by parts, show that

roc

Jois equal to a — sin a,

a3 1 fa

and also to 777—«rr xBsin(ot — x)dx.3! 3!J0

From graphical or other considerations prove that, if 0 < a < 77,then - a -

I xzsin(aL~x)dx< x3dx,Jo Jo

/•aand deduce that I x3 sin (oc — x)dx< Ja4,

Jo


Evaluate I xsinxdx.Jo


smSxcosSxdx.0


-, cos3 x sin2 xm

Evaluate x2sinxdx.x2 sirJo


/ 3 \ 2

53. Integrate 11 +-^l with respect to x, and evaluate

sJo54. Evaluate the definite integrals:

cos2 2a; da;, #Jo Ji

If ^w = I xnsinxdx9 and%>l, show thatJo

and evaluate w4.

55. (i) Prove that, if

In = seGnxdx,

then (n — 1) In = tan # secw"2 a; + (7i — 2) ln_2.

Use the formula to evaluate

Jo

(ii) Find the positive value of x for which the definite integral

rx i

Jo 7(dt

is greatest, and evaluate the integral for this value of x.

56. Prove that

— (sinm+1 x cos71"1 x)ax

= (m + n) sinm x cos71 x — (n — 1) sinm x cos71"2 xdx,

and deduce a formula connecting[In [in

smmXGO&nxdx, sin™ x cos71"2 xdx.Jo Jo

Evaluate

sin3 x cos5 x dx, sin3#cos5a;d#.J o J -j?r

KEVISION EXAMPLES III 37

57. Prove the rule for integration by parts and use it in finding

eaa; cos cxdx, I eax sin2 bxdx.

CaIf In denotes (a2 — x2)ndx, prove that, if n > 0,

Jo

58. (i) Find a reduction formula for

n(l+x2)n+*dx,

Jo

and evaluate the integral when n = 2.

(ii) By integration by parts, show that, if 0 < m < n, and

Deduce that / = 0.

r dn~1

xm~x \xn{\ - x)n}dx.ax

CHAPTER V I I I

TAYLOR'S S E R I E S AND ALLIED RESULTS

1. A series giving sin A;. By repeated application of themethod given in Volume i (p. 53), we may establish that, whenx is positive, sin a; lies between the following pairs of functions:

x3

(i)

(ii)

(iii)

(iv)

X —

X3

x-~{ +

X —

X3

3! +

X5

5!

X

X3

3!

ITx>71

and

and

and

and

o f '

Xs

xs

X3

* - 3 ! H

X5

X5

X5

x>7!'

x"1

~7! Ha;9

h 9 ! '

A simple inductive step completes the argument. Moreover, theresults as stated are equally true when x is negative, though thedirections of the inequalities must then be reversed; for example,if x is positive, then 3

X > OXXX si/ *r •*/ 7-7 9

O I

whereas, if x is negative,x*

x< smx<x — —.o!

In either case sin# is between x and x ——.

There is, however, a more significant form in which the state-ments may be cast:

\x I3

(i) sin# differs from x by not more than ^-A-, where | x \ stands

for the numerical value of x;

x3

(ii) sin a; differs from ^"""ol

\ x \by not more than —•j-;

A SERIES GIVING SIN X 39

#3 25

(iii) sin x differs from x — —r + —-.61 5!\x\by not more than —~-;

£ 3 /p5 z 7

(iv) sin a; differs from # - ^ 7 + - £ 7 - ^o I o I 71

la;!9

by not more than ^—j- ;

and so on.

We are therefore led to a series

/ ^ . 2 t l 1

whose nth term is ( — I)71-1 -— —-, with the property that sin#v - 1 ) 1

differs from the sum of the first n terms by less than

Let us examine this 'difference' term j - ~ - — — , writing it in the(2n-{-1)!

form I I I I I I I I I I I I

~F*~2~*~3~#~4~*"#"272

Suppose that x has some definite value, positive or negative. Ifwe 'watch' n increase a step at a time, there will come a pointwhen 2n+l exceeds \x\. Thereafter, the later factors in theproduct are less than 1; moreover the factor

tends to zero as n continues to increase. By taking n sufficientlylarge, we may thus ensure that the value of sin# differs from thatof the sum of the first n terms of the series

by as little as we please. In that case, sin x is called the 'sum toinfinity' of the series, and we describe the series as an expansion ofsin x in ascending powers of x.

40 TAYLOR'S S E R I E S AND ALLIED RESULTS

EXAMPLES I

1. Complete the inductive step in the argument to prove thatsin a; lies between

/y.3 zy.5 /y.2fl.— 1

^ 3 /^

a n d ' - s i + s i — • + ( - 1

2. Obtain the expansion for cos a; as a series of ascending powersof x in the form

x2 x* x« x2n~2

2. A series giving 1/(1 + JC). Consider the sum

Sn= l-x + x2-xs+... + (-l)n-1xn-\

consisting of n terms. By direct multiplication, we have

xSn= x- x2 + x*-... + ( - 1 ) " - 2 ^ " 1 + (-1)*-1**,

so that (1 + x)Sn = 1 + ( - I)*-1**,

or ^ n = , + )L—r •n 1+x l + x

Hence( — l)nxn

__'y3 4_ j _ / _ _ l \n-l ~n-l 4. v /—Us - r . . . - r i L) *" i ^ ! >

\ + xas is probably familiar.We have therefore obtained a series

1 — x + x2 — xs+...,

whose nth term is ( — I)n-Ixn""19 with the property that 1/(1+a?)

differs from the sum of the first n terms in this series by precisely theamount \ xn/(l+x) |.

Consider, then, the 'difference' term \xn/(l+x) |. When x liesbetween —1,1, so that — 1 < x < 1, this difference can be made assmall as we please by taking n sufficiently large, and so we mayensure that, when | x \ < 1, the value of 1/(1 +x) differs from thatof the sum of the n terms

by as little as we please.

A SEBIES GIVING 1/(1+#) 41

On the other hand, when |a?|>l, the difference \xn/(l+x) \becomes larger and larger with increasing n, and the value of

so far from approximating to 1/(1+#), oscillates wildly as thenumber of terms increases when x is positive, and increasesbeyond all bounds when x is negative.

We have therefore obtained a series

which represents the function 1/(1+ x) for a certain range ofvalues of x (namely —1<#<1), but whose sum has no valuewhen | x \ > 1; in contrast to the series

x3 x5 x7

which was shown (p. 39) to be an expansion for sin# for allvalues of x.

The intermediate values x= + 1 , x = — 1 require separateconsideration:

When x = + 1 , the series is

whose sum to n terms is oscillating, being 1 when n is odd and 0when n is even.

When x = — 1, the series is

1 + 1 + 1 + 1 + ...,

and the sum of the first n terms increases indefinitely as nincreases.

In neither of these cases can we assign a meaning to the sum'to infinity'.

3. Expansion in series. The two examples given in §§1,2illustrate the way in which a function f(x) can be expanded as aseries of ascending powers of x in the form

possibly for a restricted range of values oix. They suffer, however,by referring to very particular functions, and the treatment


given for sin a; and 1/(1+ x) leaves us with no idea of how toproceed in more general cases.

In the next paragraph we shall give a formula for the coefficientstto»ai>a2>--- i n terms of f(x) and its differential coefficients, andlater we proceed to a more detailed discussion. First, however, wemust say a few words about the meaning of the 'sum to infinity*in general; for a fuller treatment, a text-book on analysis shouldbe consulted.

Suppose that we have a series whose successive terms are, say,uvu2,uz, ...,un,.... It may happen that the sum of the first nterms ~

5

tends to a limit S as n tends to infinity.

I For example, if x=$= 1,

xn

1 _ / y 1 v>

Xn1 xso that, for this series, 8n = ,

X ~~ X JL "" X

and, if | x \ < 1, S = lim Sn = -^—.)

In this case we call S the sum to infinity of the series. We write

8

and say that the series converges to 8.On the other hand, if 8n does not tend to a limit as n tends to

infinity, the series has no 'sum to infinity', and the expression

has no arithmetical meaning.If the terms uvu2,... of the series depend on x (as in the

particular example just quoted) so also does the sum 8n of thefirst n terms, and the sum to infinity when there is one. We shallbe concerned exclusively with series of the form

...+anxn+...,

where the coefficients ao,al9... are constants, and we have seen(§§ 1, 2) that such a series may converge for all values of X or forsome values only.

EXPANSION IN SEEIES 43

Note. It is important to realize that the word 'converge'implies definite tending to a limit. A series such as

for which 8n = 1 when n is odd and 0 when n is even, has afinite sum for all values of n, but does not converge. The seriesis said to oscillate boundedly.

4. The coefficients in an infinite series. We first assumethat expansion in an infinite series is possible, and seek a formulato determine the coefficients:

To prove that, iff(x) is a given function which CAN be expanded in

ejorm f(x)==a + 2 + ...+anxn +...,

then an = -—r^>n n\

where f(n)(0) is the value offin)(x) when x = 0.We assume without proof that (in normal cases) we can

differentiate the sum of an infinite series by differentiating theterms separately and adding the results, as we should for a finitenumber of terms. Then

f'(x) =f"(x) =

and so on. Putting x = 0, we have successively

/(0) = a09

/'(0) = al9

/"(0) = 2a2,

/ '"(0) = 3.2a3,

/«v>(0) = 4.3.2.a4,and, generally,

/<*>(0) = n(n-l)...Z.2an = n\an

Hence

Note. This work gives no help about whether the function CANbe expanded; for that we must go to the next paragraph.

4-2

44 TAYLOB'S S E R I E S AND ALLIED RESULTS

ILLUSTRATION 1. A 'particle falls from rest under gravity in amedium whose resistance is proportional to the speed. To find anexpression for the distance fallen in time t.

If x is the distance fallen, then the acceleration is x downwards;the forces acting per unit mass are (i) gravity, of magnitude gdownwards, (ii) resistance of magnitude kx upwards. Hence

x = g — kx.

The formula just given, when adapted to this notation, is

t2 .. t3 ...X = XQ-JT tXQ + -jry XQ + — XQ + . . . ,

where xQ, x0, x09... are the values of x9 x, x,... when t = 0.From the initial conditions,

x0 = 0, xo = 0,

so that x0 = g — kx0 = g.

By successive differentiation of the equation of motion, we have

x o = -Jcxo = -Jcg,

XQ = KXQ = /c g9

and so on. Hence

The series converges for all values of t.

EXAMPLES II

Use the formula of § 4 to find expansions for the followingfunctions:

1. sin#. 2. cos#. 3. !/(!+«).

5. Taylor's theorem. We come to a somewhat difficulttheorem on which the validity of expansion in series can be based.

Suppose that f(x) is a given function of x, possessing as manydifferential coefficients as are required in the subsequent work.

TAYLOR'S THEOREM 45

To prove that, if a, b are two given values of x, then there exists anumber £ between a, b such that, for given n9

Write

( a ? ) ~(ft-*)*

where Rn is a number whose properties will be described asrequired, and k is a positive integer to be specified later. Wepropose to use Rolle's theorem (Vol. I, p. 60) for F(x) exactlyas we did (Vol. I, p. 61) for the mean value theorem, of whichthis is, indeed, a generalization; we therefore want the relationsF(a) = F(b) = 0.

By direct substitution of b for x in the expression for F(x), weh a v e F(b) = 0.

In order to obtain the relation F{a) = 0, we substitute a for xon the right-hand side and equate the result to zero; thus

0 = ftb)-f(a)- (b-a)f'(a)-^^f"(a) -...

We must therefore give to En the value

Rn-Ab) -/(a) - ( 6 - a ) / » J±

We have now ensured the relations

= F(a) = 0,

and so, by Rolle's theorem, there exists a value £ of x betweena, 6 at which F'{x) = 0.


The next step is to evaluate F'(x). This will involve a numberof terms of which

is typical, and the differential coefficient of this term is

Hence, remembering that Rn is a constant, we have

F'(x) = -f[x)-{{b-x)f"{x)-f(x)}

(n-2)\J (X)j+ (b-a)* "n

J {X~ (n-l)\J {X)+ {b-af Kn>

after cancelling like terms of opposite signs. But there is anumber £ between a, b for which F'(g) = 0, and so

M (6-q) (6- f l

Equating this to the value of Rn obtained above, we have

The value of k is still at our disposal. If we put k = n (as wemight have done from the start, of course, had we so desired) wehave

(6-q)»

TAYLOR'S THEOREM 47

and the result enunciated at the head of the paragraph follows atonce. This gives us the most usual form for Taylor's theorem,and we have adopted it in the formal statement; but there areadvantages in keeping k more general, as we see below.

The theorem is therefore established.The expression

is called the remainder after n terms. It is, in the first instance,what it says, namely a remainder, the difference between

Kb)

(h — n\% lh — n\n-l

and /

The theorem just proved enables us, however, to express thisremainder in the suggestive form (with k = n)

> ( 0

by choosing £ suitably. This expression is known as Lagrange'sform of the remainder.

By giving k other values, we obtain various forms for theremainder. In particular, when k = 1, we have

= (6-q)(6-g)n-i" ~ (n -1) ! J (*}

This may be expressed alternatively by writing f, which liesbetween a, b, as a + 6(b—a), where 6 lies between 0 and 1. Then

This is called Cauchy's form of the remainder. Though less 'insequence' than Lagrange's form, it enables us to deal with someseries for which Lagrange's form does not work.


ALTERNATIVE TREATMENT. There is an alternative treatmentof Taylor's theorem, based on integration by parts, which leads toyet another form of the remainder. The method is essentially acontinued application of the formula

Cb(b-t)

JO V^

the term /(fc) (a +1) vanishing for t = b when & > 0.

This formula may be expressed more concisely by writing

bk

so that uk = T*f{k) {<*>) +

Hence ux = bf (a) + u2

r6

Moreover, % = / ' (a +1) dt = f(a + 6) —/(a).Jo

Equating the two values of uv we obtain the 'Taylor' relation(with our original C6' now replaced by ea + 6')

w h e r e

TAYLOR'S THEOREM 49

EXAMPLE I I I

1. Prove that, if Rn(%) is the function otx defined by the relation

+(* ~ ^ V... + _

then i?» - ^ ^

and deduce that Rn{x) = f*^"^ f{n\t)dt.Jb (n—1)1

6. Maclaurin's theorem. If we write 6 = a + h, Taylor'stheorem (with the Lagrange remainder) becomes

where f is a certain number between a,a + h.

A convenient form is found by putting a = 0 and then renamingh to be the current variable x:

f(x) = | J ^ ^

where f is a certain number between 0,x. This important resultis known as Maclaurin's theorem. Compare p. 43.

With the Cauchy form of remainder, the corresponding result is

where ^ is a certain number between 0,1.

7* Maclaurin's series. The remainder Rn in Maclaurin'stheorem appears in the form

or

50 TAYLOR'S S E M E S AND ALLIED RESULTS

where £ lies between 09x and 6 between 0,1. Then

f(x) = /(0) + xf (0) +. . . + ~ ^ /<»-D (0) + Rn.

It may be possible to prove that, as n becomes larger and larger(x having a definite value for a particular problem) the remainderRn tends to the limit zero. When this happens, the sum of thefirst n terms of the series

tends to the limit /(#), and so the sum to infinity of the seriesexists, and is f(x).

The condition for the remainder to tend to zero may involve x9

so that it is fulfilled for some values of x but not for others.It is on this basis that the possibility of obtaining an expansion

rests. The succeeding paragraphs give the details for a numberof important functions.

8. The series for sinjc and cos*. Let

f(x) ss sin x.Then

f'{x) a= cos#, f"(x) = —sinx, / '"(#) = —cosx, ...,so that

Hence the Maclaurin series is

a:3 x5 x*or * _ _ + _ _ _ + . . . .

To see whether the series converges to sin x, we consider Rn, theremainder after n terms, where

ryfll

The numerical value |/(m)(£)l is certainly not greater than 1, sincefln) (£) is a s i n e o r a cosine. Hence

THE SEBIES FOR SIN X AND COS X 51

But (see p. 30) Urn ^ = 0,

for all values of xt and so the series converges to sin x for allvalues of x.

By similar argument, we obtain the expansion

x2 x4 x*1 + +

convergent for all values of x.

9. The binomial series. Let

where p may be positive or negative, and not necessarily an integer.

f'"(x)=P(P-l)(p-2)(l+z)P-\

and, generally,

f{n)(x) = p(p- l)...(p-n

Hence /(0) = 1,

f"(0)=p(p-l),

f"(0)=p(p-l)(p-2),

The Maclaurin series is thus

, ft(ffl)...(fftt+l)"1 ' 1 X "+"••••

n\If p is a positive integer, the series terminates, giving the

expansion, familiar from any text-book on algebra,

)v= l+c1x + c2x2+...+cpx

pt

w h e r e Cn . = _ ^


/ / p is NOT a positive integer, we obtain an infinite series, andthe conditions for convergence become important. We prove thatthe expansion

where p is not a positive integer, is valid for values of x in the interval

— 1 <x< 1.

The Cauchy form of remainder* gives

xn( 1 — 0)n

Now l - 0 < l + 0a;

whether x is positive or negative, since —

„ /1-0V*-1 _Hence I - — ^ <1 .

\1 + Ox)Also, if p > 1, (1 + ftu)»-i < (1 +1 x l)^-1,

' (l + Ox)1-*

1

For any given x in the interval — 1 < x < 1, the product

is therefore less than an ascertainable positive number A.Consider next the product

(n-l)I

which we denote by the symbol %w. Then

(n- l ) lun " n\

p — n= x.

n

* The proof which follows may be postponed, if desired.

THE BINOMIAL SERIES

Take n so large as to be greater than p. Then53

u,n+l n—pn

\x\.

Write 2/ = (l + |#|), so that 0<y< 1. Let n0 be the first positiveinteger such that (n—p) \x\jn<y, that is, such that

Then Uno+2

<y.y-y y- y,

so that |^w+i| < \un \yn~~n°+1

Since uno is a definite ascertainable number, and since \y\< I,it follows that

as 7&->oo.But \Rn\<A\un\,

when A is the positive number already defined. Hence

as w->oo, and the validity of the expansion is established.

10. The logarithmic series. Let

Then

and generally,

54 TAYLOR'S SEBIES AND ALLIED RESULTS

Hence /(0) = log 1 = 0,

so that the Maclaurin series is

or x-iz* + $a?-l&+ ...+{-l)n-1-71

This expansion for log (1 +x) is valid for all values of x in therange

When x = 1, we have the result

The case x = — 1 is reflected in the graph (Fig. 62, p. 4) wherey-> —oo as #->0.

EXAMPLE IV

I. Use the method given for the binomial series to prove thevalidity of the logarithmic series when — 1 < x < 1.

I I . The exponential series. Let

fix) =e\Then / ' (x) = ex,

f" (x) = e*,

and, generally^ f{n) (x) = ex.

Hence /(0) =/'(0) =/"(0) = ... = 1.

We therefore have the Maclaurin series

2 ry& rv>Yl

+ + + +

It may be proved that this expansion for ex is valid for all valuesofx. [See Examples V (1) below.]

THE EXPONENTIAL SERIES 55

COROLLARY. The work of this paragraph enables us to fill a gapin the discussion of the exponential function (pp. 18-20) byobtaining an expression for the number e. Putting x = 1, we have

EXAMPLES V

1. Use the method given for the sine series to prove the validityof the exponential series for all values of x*

By direct calculation of the differential coefficients and substitu-tion in Maclaurin's formula, obtain expansions for the followingfunctions as series of ascending powers of x:

2. e2x. 3. log (1 + 2x). 4. sin2#.

5. l/(l+a;). 6. 1/(1-a;)2. 7. cos4x.

8. e~2x. 9. yl{l + 2x). 10. log (l-3z).

12. Approximations. If the successive terms in the expansion

become rapidly smaller, a good approximation to the value of thefunction f(x) may be found by taking the first few terms.

ILLUSTRATION 2. To estimate 7(3-98).Writing the expression in the form

we may expand by the binomial series (p. 51) to obtain

= 2{1 - -0025 - K-000025) + ^ (-000000125)...},

and a good approximation is

2(1--0025)

= 1-9950.

A limit may be set to the error by means of Taylor's theorem,as follows:


Since, for a general function f(x),

f{x) = | J

the error cannot exceed the greatest value of

rt (-005)2 1 -00000625or, here, 2 x —2! ' 4(1 - -0050)* - (1 - -0050)*'

In the most unfavourable case, with 0 = 1 , this gives a value lessthan -0000063 for the error.

EXAMPLES VI

Estimate the values of the following expressions:

1. V(4'02). 2- V(8*97)-3. ^(8-02). 4. ^(26-98).

5. ^(31-97). 6. l/V(9-03).

13. Newton's approximation to a root of an equation.Suppose that we are given an equation

f(x) = 0

and know that there is a root somewhere near the value x = a.Newton's method, which we now describe, shows that, undersuitable circumstances, a better approximation to the root is

a 7WIf the correct root being sought is | == a + h, then

so that, by Taylor's theorem for n = 2, with Lagrange's form ofthe remainder,

NEWTON'S APPROXIMATION 57

for a value of 77 between a and a + h. If h is reasonably small, thenthe term involving h2 may be regarded as negligible for practicalpurposes. We therefore have

f(a) + hf'(a) = 0,

or

so that the root is approximately

a"7T«TThis crude statement, however, should be supplemented by

more careful analysis, and the following graphical treatment showsthe precautions which ought to be taken.

We begin with an examination of the curve

y=f(x)

near the point £, taking first the case where the gradient is positiveand the concavity 'upwards' near that point, so that f(x)J"(x)are both positive.

near

Fig. 66.

(i) Let X be the point (f, 0) on the curve, and let A be the pointfor which x — a, where a > £; draw AM perpendicular to Ox, andlet the tangent at A meet Ox in P.

58 TAYLOR'S SBBIES AND ALLIED RESULTS

Then OM = a,

AM=f(a),

AM-p^ = tan^=/(a),

so that PM = {jy\,

and OP = a-

But under our assumptions that f{x)J" (x) are positive near X(so that the gradient is positive and the concavity upwards) thetangent AP lies between AM and the curve, so that P lies betweenX and M. Hence, under these conditions, OP is a better approxima-tion than OM to OX. That is, a — V^ is a better approximation

/'(a) rr

than a to the root.It is assumed that /'(a) is not zero, and, indeed, that/'(#) is not

zero near the required root.(ii) Suppose next that, with the same diagram, B is the point

for which x = b, where b < £, so that B lies 'below' Ox; draw BNperpendicular to Ox, and let the tangent at B meet Ox in Q. Then

BN = — f(b) since f(b) is negative,

QH = tan fo =f'(b),

so that

and OQ^b-jj^-y

But now we cannot be sure that Q is nearer to X than JV, for Qmay be anywhere to the right of X according to the shape of the

curve. Hence we cannot be sure whether b — ~-L is, or is not, a

better approximation than b to the root.

N E W T O N ' S APPBOXIMATION 59

Similar argument applied to the accompanying diagram(Fig. 67) shows that the results (i), (ii) are also true if (the con-cavity still being 'upwards') the gradient is negative near X; P iscloser than M to X, whereas Q may or may not be closer thaniVtoX.

Fig. 67.

We have therefore proved, so far, that, iff" (x) is positive near £,

the approximation a — J±JL {S better than a itself when f (a) is positive.

It is easy to verify in the same way that, if /"(#) is negative

near £, the approximation a — JTT\ is better than a itself when f(a)

is negative. (The whole diagram is merely 'turned upside down'.)In other words, we can be sure of a better approximation if

f"{x) retains the same sign near x = a, that sign being also thesign off(a).

In more complicated cases, it is wise to draw sketches such aswe have shown in order to determine how the tangent at A cutsthe X-axis. It is, however, clear from the graphs that, for ordinary

functions, the approximation a — jrrk is ALWAYS better than a

itself once we come sufficiently close to the correct answer. Inother words, once it is ascertained that an approximation isreasonably good, it may confidently be expected that Newton'sapproximation will make it better still.

5-2

60 TAYLOR'S SERIES AND ALLIED RESULTS

ILLUSTRATION 3. An example where the answer is apparentfrom the outset may help to make the principle clear. Considerthe equation

fso that

It is obvious that x = 0 is one root, but let us attempt to reachthat value by approximation from (i) x = 1, (ii) x = — 1.

When x = 1,

/(!) = ¥ , /'(I) = H, /"(!) = ¥ ,so that /(I), /"(I) have the same sign; moreover/"(#) remainspositive near x = 1 (in fact, down to x = — f, where/(#) is negative,indicating a point on the other side of the root). Hence we expectNewton's formula to give a better approximation; and since

_ / 0 ^ = = 19_ 14/'(I) 33 33'

this is actually the case.On the other hand, when x = — 1,

Here, again,/( —1),/"(- 1) have the same sign; but f"(x) changessign from negative to positive at x = — f, which is near — 1; weare therefore in a doubtful region; and since

the approximation is actually worse.

ILLUSTRATION 4. To find an approximation to that root of the

equation ^-Zx+l = d

which lies between 0,1.

We have f(x) = x3 - 3x + 1,

f'(x) = 3x*-B,

f"(x) = 6x.

NEWTON'S APPROXIMATION 61

Since f"(x) is positive when x lies in the given interval 0,1, it isadvisable to begin with an approximation which makes f(x) alsopositive.

Now &

We should therefore like an approximation between \ and \ whichkeeps f(x) positive, and inspection shows that \ appears verysuitable. We then have

so that

and the corresponding approximation is

H TV = ft ='3472.

The correct root is '3472..., so we have already obtained fourcorrect figures.

ILLUSTRATION 5. / / 77 is a small positive number, to find anapproximation to that root of the equation

sin x = rjx

which lies near to x = 7r.(The intersection of the graphs y = sin x, y — rjx shows that

there is a root near to TT.)Since sin n = 0 and 77 is small, the approximation x = IT is

reasonably good.Write f(x) = sin x — -qx,

so that f'(x) = cos x-7],

f"(x) = —sin a;.

Then / ( T T ^ - T ^ , / ' (77H-I-77, / ' » = 0.

Although /"(TT) is actually zero, so that the curve (Fig. 68)y = sinx-rjx has an inflexion at x = TT, the concavity is 'down-wards' in the interval O,TT and /(IT) is negative; moreover thegradient /'(TT) is also negative. The accompanying sketch showsthat the tangent lies between the ordinate x = 77 and the curve,so that Newton's method will improve the approximation.


The corresponding solution is

— 7 7 y z r — 7 T — i •

(-1-7,) 1+r, l+7jTo the first order in -q this is, on expansion of (1 + ij)"

o

Fig. 68.

Note. This solution is less than 77, as we expected from thediagram.

14, Leibniz's theorem. The theorem which follows is usefulin calculating the higher differential coeflRcients necessary for aMaclaurin expansion.

To prove that, iff{x) is the product of two functions u, v, so that

f(x) = uv9

then

f(n){x)

where ncp is the binomial coefficient

nCp~n\

nCp~p\(n-p)V

We use the method of mathematical induction, assuming theresult to be true for a certain integer N9 so that

LEIBNIZ'S THEOREM 63

Now differentiate this expression to obtain f(N+1)(x). Thedifferential coefficient of a product such as u{N~p)vip) is

M(N-p+l) v(p) _|_ u(N-p) v(P+Dt

As we write down these terms for the series on the right, we putthe answer in two lines, the top line consisting of terms such asU(N-P+DV(P) a n ( j the lower of ^w-^dH-D; also we displace thelower line one place to the right, thus:

Now the coefficient of u{N~p+1)v{p) is

N\ N\p\(N-p)\

Hence

It follows that, if the theorem is true for any particular value N9

then it is true for N +1 , N + 2, and all subsequent values. But it iseasily established when N «= 1, being merely the result

It is therefore true generally.

Note. The expression is symmetrical when regarded from thetwo ends, and will equally well be written in the form

fln)(x) = uv{n) + ... + nf


EXAMPLES VII

Use Leibniz's theorem to find the following differentialcoefficients:

d4 . d4

!• -m (#8sina;). 2. -r—= i(XX iXjJU

3^ ifix QQg Qx), 4.

6. To ap2% Leibniz's theorem in finding aMaclaurin expansion for

We have

f'{x) =

on simplification. Hence

(ar«+1) {/(*)}• = 1 .Differentiate. Then

(o:2+ l).2/'(^)/"(^) + 2 4 / ' ( ^ = 0,

or (x2+l)f"(z) + xf'(x) = 0.

Differentiate n times, using Leibniz's theorem. Write theexpansion from (x2+ l)f"{x) on the first line, and from xf'(x) onthe second; note that the expansions terminate since (x2 +1)'" = 0,x" = 0. We obtain

(x2 + 1 )/<"+»> («) + n . 2a; ./««+!> (a;) + n ^ " j " ^ . 2 ./<»> (»)

n .1 ./(w) (x) = 0.

L E I B N I Z ' S THEOREM 65

We require values when x = 0, so that

= 0,

or /<n+2)(0) = -n2f{n)(0).

But /(0) = log 1 = 0,

and so /"(0) =/(iv)(0) = ... = /(2rl)(0) = 0.

Also /'(0) = 1,

so that /'"(0) = - 1 2 . 1 ,

/(v)(0)= + 32.1,

/(vii)(0) = ~5 2 .3 2 . l ,

and so on. Hence the Maclaurin expansion is

l x s 1.3 a;5 1.3.5 a;7

° r X 2 3 + 2 ^ " 5 ~ 2 X 6 7+~"

converging to log {x + <j(x2 +1)} for such values of x (not consideredhere) as make the series convergent.

EXAMPLES VIII

Use the method of Illustration 6 to obtain the followingexpansions:

1. f{x)= sin"1 a;.

2. f(x)== tan-1 a;.

3. Prove that, if f{x)== cos (m sin-1 x), then

( 1 - # 2 ) / < " + 2 > ( Z ) - ( 2 T I + 1 ^ ^ = 0.

4. Prove that, if/(a;) = (sin-1^)2, then

-n2f<n)(x) = 0.


REVISION EXAMPLES IV

'Advanced' Level1. Differentiate

(i) ^ ^ }If y = eaainbx, and y',y" are the first and second differential

coefficients of y with respect to x show that

Calculate the values of y', y" when x = 0.

2. Differentiate, ,1-2:*; . / / l -a? \tan"1 — , sin"1 / ,

2 + x *J\l+x)'

log (sec x + tan x),

Find the values of x for which

x2- 8 # -

is a maximum or minimum, distinguishing the maximum fromthe minimum.

3. If y = ext8Hixt prove that

Prove that y = uexta,nx also satisfies this equation when u is afunction of x such that

Verify that u = cot x is such a function.

4. Differentiate 2/ = secm#tanna\Deduce that the kth differential coefficient of secm# can be

expressed in the form secma;PA.(tana:), where Pfc(tana;) is a poly-nomial in tan x of degree k.

Evaluate J^(tan x) when both m and k are taken equal to 4.

REVISION EXAMPLES IV 67


- + 2log -, x — tan x 4- \ tan3 x>

simplifying your answers as much as you can.Prove that the first function has a maximum at x = 1 and that

the second function (whose differential coefficient vanishes atx = 0) has neither a maximum nor a minimum at x = 0.

6. Find the first four differential coefficients of sin4 x.Show that the function

has turning values at x = 0 and at x = JTT, and determine whetherthey are maxima or minima.

7. Prove that, if y = sin2(a;2), then

Prove that the result of changing the independent variable inthis equation from x to | , where £ = x2, is

where a, b are constants to be determined.Prove also that, if A9B are any constants, the function

y = | + A cos (2a;2) + B sin (2x2)

satisfies the first differential equation.

8. Differentiate/ l\w

(cos x + sec x)n

with respect to x, and find the nth differential coefficient of

K)'for all positive integral values of n, distinguishing the casesn < 2 and n > 2.


9. Show that the polynomial

4 16f(x) = x -(2ir-5)x* + —ATT-3)x5

77 77"

and its differential coefficient/'(a;) have the same values as sina?and its differential coefficient respectively, at the values x — 0,x = ± \n.

J %i7T

f(x)dx as an approximation for0

I sinxdx is less than 0*1 per cent.Jo

10. (i) Given that —- = ex} -j- = since,

ax axand that u = 1, v = 0 when x = 0, show that

d2(uv) _~dtf~ =

when # = 0.(ii) Given that y = axn + bx1"71, prove that

x-7- + (n-l)y = (2n-l)axn,

ax

( i n d t Jand form an equation in x,y9-~-9 -=-^ which does not contain a or 6.

ax ax


Prove that -r— 1 = , —r,

and find the nth differential coefficient of

x

12. Find the nth derivatives with respect to x of

W i and x>—v(ii) sin a; and x sin #.


13. (i) Find the derivative of y = sin2 a; with respect to z = cos xby evaluating the limit of 8y/8z.

(ii) Differentiate with respect to x:

sin#

(iii) Prove thatdy2 dx2l \dxt

14. Find from first principles the differential coefficient of \jxz

with respect to x.What is the differential coefficient of ljxB with respect to #2?Differentiate with respect to x:

logecosa;, xj{\-x% ( 1 ^ 2 ) 2 .

15. Find the nth differential coefficient of y with respect to x ineach of the following cases:

(i) y = l/x2, (ii) y = sin2#, (iii) y = e2*sin2#.

There are three values of k for which the function y = xke~x

satisfies the equation

Find these values.


l-2x3 tan2 X c i - n ~ l -

Prove that, when y = at2 + 2bt + c, t = ax2 + 2bx + c, and a,b,care constants,

17. Differentiate ctanaJ,

Show that, if y = sin 6, a; = cos 0, then

70 TAYLOE'S SEKIES AND ALLIED RESULTS

18. A particle moves along the axis of x so that its distancefrom the origin t seconds after starting is given by the formula

x = a cos ht + \dkt,

where a, k are positive constants. Find expressions for the velocityand acceleration in terms of t, and prove that the particle is notalways moving in the same direction along the axis.

Find the positions of the particle at the two times t = TT/6& andt = 13TT/6&; also find the position of the particle at the instantbetween these times at which it is momentarily at rest, anddeduce the total distance travelled between the two times.

19. A particle moves in a plane so that its coordinates at time tare given by x = el cos t9y = e'sintf. Find the magnitudes of thevelocity and of the acceleration* at time t and prove that theacceleration is always at right angles to the radius vector.

Draw a rough sketch of the path of the particle, from timet = 0 to t = 77, and indicate the direction of motion at times 0,| T T , JTT, f 77, 77.

20. A particle moves in a plane so that its position at time t isgiven by x = a cos pt,y — b sin pt. Find expressions for (i) themagnitude v of the velocity at time t; (ii) the magnitude / of theacceleration at time t; and prove that there is no value of t forwhich / = dvjdt.

Prove also that the resultant acceleration makes an angle 6with the normal to the path, where

2ab tan 0 = (a2- 62) sin 2pt.

21. A point moves in a straight line so that its distance at time tfrom a given point 0 of the line is x, where

x =

Find its velocity at time t, and prove that the acceleration is then

— t2 Bint —2t cost+ 2smt.

Determine the times (£>0) at which the acceleration has aturning value, distinguishing between maxima and minima.

* If v,f are the velocity and acceleration respectively, then v2 = x2+y2,

BEVISIOST EXAMPLES IV 71

22. A particle moves along the axis of x so that its distancefrom the origin t seconds after starting is given by the formulax = a cos pt. Prove that the velocity of the particle changes direc-tion once, and only once, between the times t = 0, t = 2TTJP andthat the change of direction occurs at the point x = — a.

The distance from the origin of a second particle is given by theformula . 1 n .

x = a cos pt + \a cos 2pt.Write down expressions for its velocity and acceleration at time t.Show that between t = 0 and t = 2TTJP the velocity of the particlechanges direction three times, and find the values of x at whichthese changes occur.

23. Prove that the equation of the tangent to the curve given by

at the point where t = tan a is

y — #tana+.tan3a + t a n a + l = 0.

Show that the curve lies on the positive side of the line x = 1and is symmetrical about the line y = — 1, and prove that the areabounded by the line x = 4 and the curve is ^ .

24. Prove that there are two distinct tangents to the curve

y = x*-x + 3

which pass through the origin. Find their equations, and theirpoints of contact with the curve.

Give a rough sketch of the curve.

25. A curve is defined by the parametric equations

where a is a positive constant. Prove that(i) the curve passes through the points A,0,B whose coordi-

nates are (0,6a), (0,0), ( - 3a, 0).(ii) the point t is in the first quadrant when - 1 < t < 1 and in

the third quadrant when 1 < t < 2.Make a rough sketch showing the part of the curve correspond-

ing to values of t between — 1 and + 2.Find the equation of the tangent to the curve at O, and prove

that the area bounded by the arc AB and the chord AB is 27a2/2.


26. Prove that the slope of the curve whose equation is

is always positive.Show that the curve has a point of inflexion where x = — 1, the

slope there being \%

Prove also that the tangent at the point (0,1) meets the curveagain at the point (— 3, — 2).

Sketch the curve, indicating clearly the point of inflexion andthe tangents at the points (—1,4) and (0,1).

27. A,0,B are three fixed points in order on a straight line,and AO = p, OB = q. A fixed circle has centre 0 and radius agreater than p or q, and P is a point on this circle. Show that theperimeter of the triangle APB is a maximum when OP bisects theangle APB9 and find the corresponding magnitude of the anglePOA.

28. Prove that the maximum and minimum values of thefunction y = x cos 3x occur when 3 tan 3x — Ijx, and discuss thebehaviour of the function when x = 0.

By considering the curves y = tan 3x, y = \jx, show thatmaximum values occur near the values x = § kir, and minimumvalues near x = J(2&+ 1)TT, the approximation becoming moreexact as x becomes larger.

29. Given that a2#4 + 62?/4 = c6, where a, b, c are constants, showthat xy has a stationary value c3/>/(2a6). Is this value a maximumor a minimum ?

30. (i) A right circular cylinder is inscribed in a given rightcircular cone. Prove that its volume is a maximum when itsaltitude is one-third that of the cone.

(ii) A right circular cone of height h stands on a base ofradius h tan a. A cylinder of height h — x is inscribed in the cone.Prove that 8, the total surface of the cylinder, is equal to

2TT{X2 (tan2 a — tan a) + xh tan ex},

and prove that, when tan a > ^, S increases steadily as x increases.

31. P is a variable point in the circumference of a fixed circle ofwhich AB is a fixed diameter and 0 is the centre. Prove that,


when OP is perpendicular to AB, then the function AP + PB hasa maximum value and the function AP3 + PBZ has a minimumvalue.

32. If/(#) is a function otx and/'(#) is its differential coefficient,show that, when h is small, f(a + h) is approximately equal tof(a) + hf'(a).

Without using trigonometrical tables, find to three significantfigures (i) the value of cos 31°, and (ii) the positive acute anglewhose sine is 0*503. Give your answer to (ii) in degrees and tenthsof a degree.

33. Calculate

(i) <$/8-05 to 4 significant figures,

(ii) cos 59° to 3 significant figures.

[Take TT to be 3-142.]

34. Determine to two places of decimals that root of theequation ^

-^— = 3-2104x + 2

whose value is nearly equal to 8.

35. Determine to 3 places of decimals the value of that root ofthe equation xs

which lies between 1-5 and 1-6.

36. Prove, graphically or otherwise, that, if n is a large positiveinteger, there is a root of the equation xmnx = 1 nearly equal to2n7T. Show that a better approximation is 2rnr-\~ (1/2/iTr).

37. Prove that, if TJ is small, the equation

0 + sin0cos0 = 2rjcos0

has a small root approximately equal to

v-h3-[You may assume the power-series expansions for sin 9 and

cos 0.]6 M II


38. Prove that the result of differentiating the equation

ax

n+ 1 times (w > 1) with respect to x is

Hence verify that, if h is a positive integer and

then z is a solution of the equation

(1 + a;2) | + 2(i + l)a; J- + fc(* + 1) z = 0.

39. By using Maclaurin's theorem, or otherwise, obtain theexpansion of log (1 -f sin x) in ascending powers of x as far as theterm in x*.

40. Prove that, if y = loge cos x, then

== a

dxHence, or otherwise, obtain the Maclaurin expansion of logc cos xas far as the term in #4.

Deduce the approximate relation

41. Prove that, if y = etanrc, then

where t = tano;.Prove that the expansion of y as far as the term in xz is

y= l+x + ix* + ±x*.

42. Prove that, if y = sin (log a;), where x > 0, then

EEVISION EXAMPLES IV 75

By induction, or otherwise, prove that

where n is a positive integer.

43. If y = e^loga;, show that

Find the equation obtained by differentiating this equation ntimes.

44. Prove that, when y = eax sin bx,

and that -p = t/(a + 6 cot bx).CbX

00 £

Prove that, if eaxsmbx = T. —%xn,

then cn+2 - 2acn+1 + (a2 + b2)cn = 0,

and find the values of cl9 c2, c3.

45. If cos y = cos oc cos #, where #,^, a lie between 0 and \TTdii du

radians and a is constant, find the values oty,-^-, -~ when x = 0.

Taking a; to be so small that xz and higher powers of x arenegligible, use Maclaurin's theorem to show that

y as a + ^#2 cot a.

Hence calculate y in degrees, correct to 0-001°, if a = 45° anda; = 1° 48'. [Take IT = 3-142.]

46. By using Taylor's theorem obtain the expansion of

tan (x + \IT)

in powers of x up to the term in #3.Hence calculate the value of tan 44° 48' correct to four places of

decimals. [Take TT = 3-142.]6-2

76 TAYLOB'S S E R I E S AND ALLIED RESULTS

47. Prove that, if y = loge I : — I , then

(i) | -

and (ii) the expansion of y in a series of powers of x as far as theterm in #4 is 2x + \xz.

Find, correct to four significant figures, the value of y whenx = 1° 48', taking TT = 3-142.

48. Prove that, if y = (sin"1 #)2, then

The Maclaurin expansion of 2/ in powers of x is taken to be

Given that y = 0 when a; = 0, prove that a0 = 0 and ax = 0. Provealso that aw+2 = w2an when n > 0, and hence show that

49. If y = w"dM, prove that

3-<>•*•>£•Find the values of the first four differential coefficients of y

when # = 1, and, by using Taylor's expansion in the form

f1'1deduce that the value of uudu is approximately 0«1053.

h

50. If y = tana:, prove that

and find the third, fourth, and fifth derivatives of y.

BEVISION EXAMPLES IV 77

Hence find the expansion of tana; in a series of powers of xup to x5.

51. Prove that, if y = sin^ttsin^a;), then

Show that the first two terms in the expansion of the principalvalue of y in ascending powers of x are

mx + Jm( 1 — m2)x3.

52. Find the indefinite integrals:

x e~x* dx, - 2~ fa* x* e* dx-

If u, v are functions of x, and dashes denote differentiation withrespect to x, show that

(uv"f + u'"v)dx — uv" — u'v' + u"v + constant.

53. Show that \f(x)dx*= \f(a-x)dx.Jo Jo

Deduce that I- :——)dx= ta,n2xdx,Jo \l+sin2aj/ Jo

and evaluate the integral.


f i * C dx f x* J\xlogxdx, \——, \- ~.dx.

Prove that, when the expression e^sina; is integrated n times,the result is

2-tn ex s i n (£ _ in wj

where Pn_ x is a polynomial of degree w — 1 in a;.

78 TAYLOR'S SERIES AND ALLIED RESULTS

55. Find the indefinite integrals of


32cos4#, 27x2(logx)2, ~-^j-> a*.JL — X


4

Evaluate the definite integrals:

C\ f*» dx) * * * * * ' Jo 1 + BOOBV



f dx r 2 O _ f_ , 1 , f cosJ ( l - 3 x ) 3 J J J2~cos2o;

60. Find the values of

s i n 5 ^ ^ cos4a;^, ex sin2 x dx.Jo Jo Jo

61. (i) Find the indefinite integrals of

x{x+l)2'

(ii) By substitution, or otherwise, prove that

f1/V2 f1

i r s i n - 1 ^ ^ = | , x*J{x*+ l)dx =Jo Jo


cos3 3x, -— , Xs sin x.3 —2cosa;


Use the method of integration by parts to integrate

twice with respect to x.

63. By means of the substitution x = oc cos2 0 -f j8 sin2 0, or other-wise, prove that

Cfi dxJa^{{x-oc){p-x)}^7Tt

Prove also that

[ - a) (j8 -

64. P r o v e t h a t

| c o s 2 Odd = JTT, | C O S 4

Jo Jo§77.

Find the area bounded by the curve r = a(l + cos0) anddetermine the position of the centre of gravity of the area.

65. Find the equation of the normal at (the point (f, sin £) tothe curve whose equation is y = sin x.

Prove that, if £ lies between 0, TT, the normal at P divides thearea bounded by the #-axis and that arc of the curve for which0 x 77 in the ratio

(2 - cos £ - cos3 £) : (2 + cos £ + cos3 £).

66. The ellipse - 2 + v2= l

r a2 b2

is rotated through two right angles about the #-axis. Prove thatthe volume generated is ^Trab2.

(i) Prove that, if a and 6 are varied subject to the conditiona + b = £, then the greatest volume generated is 2TT/81.

(ii) The volume is cut in two by the plane generated by therotation of the y-axis. Prove that the centre of gravity of eitherpart of the volume is at a distance fa from the plane of separation.

67. The complete curve x2/a2 + y2\b2 = 1 is rotated round the2/-axis through two right angles. Find the volume generated bythe area enclosed by the curve.


The semi-axes a and b of the curve are each increased by €.Prove that, if e is small, the increase in volume is approximately

68. OA is a straight rod of length a in which the density at apoint distant x from 0 is b + ex, where b and c are constants. Findthe distance of the centre of gravity of the rod from 0.

69. The gradient at any point (x, y) of a curve is given by

dx

and the curve passes through the point (2,0). Find its equationand sketch the graph, indicating the turning points.

Find the distance from the y-axis of the centre of gravity of auniform lamina bounded by the curve and the positive halves ofthe x and y axes.

70. A lamina in the shape of the parabola y2 — 4ax bounded bythe chord x = a is rotated (i) about the axis of y, (ii) about theline x = a. Prove that the volumes generated in the two cases are


sin2#, sin3 x9 x2 cos x.

The portion of the curve y = sin a; from x = 0 to x = \n revolvesround the axis of y. Prove that the volume contained betweenthe surface so formed and the plane y = 1 is JT^TT2 — 8).

72. The coordinates of a point on a curve referred to rectangularaxes are {at2,2at), where t is a variable parameter which liesbetween 0 and 1. Make a rough graph of the curve.

Calculate (i) the area enclosed by the curve and the linesx = a, y = 0; and (ii) the area of the surface obtained by revolvingthis part of the curve about the #-axis.

73. Find the area contained between the #-axis and that partof the curve x = 2£2 -f 1, y = t2 — 2t which corresponds to values of tlying between 0 and 2.

Find also the coordinates of the centroid of this area.

REVISION EXAMPLES IV 8J

74. Prove that the tangent to the curve

Jx + Jy = Jaat the point (x,y) makes intercepts ^{ax)^(ay) on the axes.

A solid is generated by rotating about the #-axis the area whosecomplete boundary is formed by (i) the arc of this curve joiningthe points (a, 0), (0,a), and (ii) the straight lines joining the originto these two points. Prove that, when this solid is of uniformdensity p, its mass M is x^Trpa3.

Prove also that the moment of inertia of the solid about Ox is

75. Find the area bounded by the curve x = 2a(t3 — l),y = Sat2

and the straight lines x = 0, y = 0.Prove that the tangent to the curve at the point t = 1 meets

the curve again at the point t = — | , and find the area bounded bythe parts of the tangent and of the curve that lie between thesepoints.

76. Prove that the parabola y2 = 2ax divides the area of theellipse 4#2 + 3y2 = 4a2 into two parts whose areas are in the ratio

77. The portion of the curve y2 = 4ax from (a, 2a) to (4a, 4a)revolves round the tangent at the origin. Prove that the volumebounded by the curved surface so formed and plane ends per-pendicular to the axis of revolution is ^-Tra3, and find the squareof the radius of gyration of this volume about the axis of revolution.

78. Find the coordinates of the centre of gravity of the areaenclosed by the loop of the curve whose equation is r = acos20,which lies in the sector bounded by the lines 6 = ± \TT.

Find also the volume obtained by rotating this loop about theline 0 — 77.

79. Find the coordinates of the centre of gravity of the loop ofthe curve traced out by the point x = 1 — t2, y = t — ts.

Find also the volume obtained by rotating this loop about theline x = y.

80. A plane uniform lamina is bounded by the curve y2 = 4a#and the straight lines y = 0, x = a. Find the area and the centreof gravity of the lamina.


The lamina is rotated about the axis Ox to form a (uniform)solid of revolution. Find the centre of gravity of the solid and,assuming the density of the solid to be p, find its moment ofinertia about the axis Ox.

81. A lamina in the shape of the parabola y2 = 4=ax, bounded bythe chord x = a, is rotated (i) about the axis of y, (ii) about theline x = a. Prove that the two volumes thus generated are in theratio 3 : 2.

82. Prove by integration that the moment of inertia of auniform circular disc, of mass m and radius a, about a line throughits centre perpendicular to its plane is |ma2.

The mass of a uniform solid right circular cone is M, and theradius of its base is a. Prove that its moment of inertia about itsaxis is

2.

Jo83. Evaluate cosn0d0 when n = 1,2,3,4.

Jo

The area bounded by the axis of #, the line x = a, and the curve

x = asin0, y = a(l — cos 0),from 0 = 0 to 0 = \n, revolves round the axis of x. Prove thatthe volume generated is ^7ra3(10 — 3TT).

84. The portion of the curve x2 = 4:a(a — y) from x = — 2a tox = 2a revolves round the axis of x. Prove that the volume con-tained by the surface so formed is ffrra3, and find its radius ofgyration about the axis of revolution.

85. Sketch the curve r = a(l + cos#), and find the area itencloses and the volume of the surface formed by revolving itabout the line 0 = 0.

86. Find the three pairs of consecutive integers (positive,negative, or zero) between which the roots of the equation

lie, and evaluate the largest root correct to two places of decimals.

87. Show that the equation

x*-3x-7 = 0

has one real root, and find it correct to three places of decimals.

BEVISION EXAMPLES IV 83


has one real root, and find it correct to three places of decimals.

89. For a function / having an Nth differential coefficient,Taylor's theorem expresses f(a + x) as a polynomial of degreeJV— 1 in #, together with a remainder. State the form taken bythis polynomial, and one form of the remainder.

Prove by induction, or otherwise, that

dn

-=— (e* sin 3^3) =dX

Hence find the coefficients an in the Maclaurin series Hanxn of the

function e^sina;^.By means of Taylor's theorem, show that when x>0 the

J V - 1

difference between exsinx^j3 and 2 anxU *s n o ^ greater thann - 0

(2x)Nex

[A proof showing that the difference is not greater than

(2x)Nex

is acceptable if the form of remainder which you have quotedleads to the result.]

90. If y

show that (l + x)2-^+(l+x)y= 1.dX

Deduce the first four terms in the Maclaurin series for y inpowers of x.

91. Show that y = {x + ^ ( l + x2)}k

satisfies the relations y'y](l+x2) = ley,

Deduce the expansion

Verify that this agrees with the series derived from the binomialseries when h = 1.

CHAPTER IX

THE HYPERBOLIC FUNCTIONS

1, The hyperbolic cosine and sine. There are two functionswith properties closely analogous to those of cos x and sin x. Theyare called the hyperbolic cosine of x, and the hyperbolic sine of x,and are written as cosh x and sinh#. We define them in terms ofthe exponential function as follows:

cosh# = \{ex-\-e~x),

sinhz = \{ex-e~x).

We establish a succession of properties similar to those of thecosine and sine:

(i) To prove that cosh2 a; — sinh2# = 1.

The left-hand side is

l{{e2x + 2 + e~2x) - (e2x - 2 + e~2x)}

= 1.

(ii) To prove that

cosh (x + y) — cosh x cosh y 4- sinh x sinh y

It is easier to start with the right-hand side:

(e* + e~x) (& + e~y) + {ex - e~x) (e^ - e~y)}

== i{(ex+y + ex~y + e-x+y + e~x-y) + (ex+y - ex~y -

= cosh (x + y).

THE HYPERBOLIC COSINE AND SINE 85

(iii) Similarly

sinh (x + y) = sinh x cosh y + cosh x sinh y.

(iv) As particular cases of (ii), (iii),

cosh 2x = cosh2 x + sinh2 x,

sinh 2# = 2 sinh a; cosh a\

COROLLARY (i). cosh2x = |(cosh 2x + 1),

sinh2 a; = ^(cosh2# — 1).

COROLLARY (ii). Since

cosh a;—1 = 2 sinh2-

and sinh2 ~ is positive, it follows that cosh x is greater than unity for

all (real) values ofx, i.e. cosh x 1.

Note. coshO = 1,

sinh 0 = 0.

(v) To prove that cosh a; = cosh (— x).

The right-hand side is

= cosh a;.

(vi) To prove that sinh x = — sinh (— x).

For sinh ( - x) = l{e~x - e-<-*>) = i(e-* - ex)

= —sinh a:.

iV'o e. Sinh a; is positive when a; is positive, and negative when xis negative. For example, if x is positive, then ex is greater thane~x, since e is greater than 1. Hence \(ex — e~x) is positive.

(vii) To prove that

-r~ (cosh a;) = sinh a:,CLX

•j- (sinh x) = cosh a;.ax

86 THE HYPERBOLIC FUNCTIONS

We have -=- (cosh x) = - -=- (ex + e~x)dxK ' 2dx

and

= cosh a;,

(viii) To prove that

coshrfa; = sinha;,

= cosh#.

These results follow at once from (vii).

(ix) By theformulce of (vii), we have the relations

-=-s (cosh a;) = cosh#,cLx

-r-g (sinh#) = sinha;.dx

Thus cosha:, sinha; both satisfy the relation

dx* y

(x) The following expansions in power series are immediateconsequences of Maclaurin's theorem:

x2 x* xB

1 + + +

The series converge to the functions for all values of x.

OTHER HYPERBOLIC FUNCTIONS 87

2. Other hyperbolic functions. The following functions aredefined by analogy with the corresponding functions of elementarytrigonometry:

sinh xKJCVXIXX U/

cotha;

cosecha;

seen a;

cosh a;'

cosh a?sinh a;'

1sinh x9

1~~ cosh a;"

The relations sech2# + tanh2# = 1,

are found by dividing the equation

cosh2 x — sinh2 x = 1

by cosh2#, sinh2 a; respectively.Note the implications

sech2 x<l9 tanh2 x<l.

The differential coefficients are easily obtained from thedefinitions:

(i) If

y = tanh a; = *(sinha;) (cosh a?)-1,

then -j- — (cosh x) (cosh x)"1 — (sinh a;) (cosh a;)~2 sinh x&x

= l - tanh 2 a;

= sech2 a;.

(ii) if

y = cotha; = (cosh a:) (sinha:)"1,

then -J- = (sinh x) (sinh x)"1 — (cosh a:) (sinh a:)~2 cosh xCLX

= l-coth2a;

= — cosech2as.

88

(iii)

then

(iv)

then

If

If

THE HYPERBOLIC FUNCTIONS

y =dy _dx

y

dydx

• cosecha; = (sinha;)"1,

= — (sinh#)-2cosh#

• — cosecho;cotha?.

= sech a; = (cosh a;)*"1,

= — (cosha;)-2sinha?

= — sech x tanh x.

ILLUSTRATION 1. A body of mass m falls from rest under gravityin a medium whose resistance to motion is gv2/k2 per unit mass whenthe speed is v. To prove that the speed after t seconds is k tanh(<#/&).

Let x be the distance dropped in time t. Then the accelerationdownward is x and the forces are

(i) mg downwards due to gravity,

(ii) mgx2/k2 upwards due to the resistance.

Hence mx = mg — mgx2jk2.

(it)Write x = v. Then — = g-gv2jk2,

ator k2t

Substitute v = &tanh 8.

(This substitution is possible so long as v is less than k, since(p. 87) tanh20<l.)

Then k2. k sech2 6-^ = gk2(l- tanh2 6)

= gk2 sech2 6.

Hence ft = Vso that 0 = |- + C,

where C is an arbitrary constant. It follows that

OTHER HYPERBOLIC FUNCTIONS

Now we are given that v = 0 when t = 0, so that

0 = tanhC.

Hence C = 0,

and so v = &tanh(<7£/&).

Note that v = /ctanh (gt/k)

egt/k __ e-gt/k= k

89

1 _ e-2flr</fc

= JCl+e-2gl/kt

Now as ^ increases, e2gt/k tends to infinity (see, for example, thediagram (Fig. 64) for ex on p. 18), so that e~2gt/k->0. Hence

as t increases. In other words, v tends to a terminal value k as thetime of falling increases.

The example which follows deserves close attention. It bringsout very clearly the points of similarity between the trigonometricand the hyperbolic functions.*

ILLUSTRATION 2. Suppose that aparticle P (Fig. 69), of mass m, is freeto move on a fixed smooth circularwire, of radius a, whose plane is verti-cal. A light string, of natural lengtha and modulus of elasticity 2kmgjoins P to the highest point B ofthe wire. We wish to examine whathappens if P receives a slight dis-placement from the lowest point ofthe wire.

Let AB be the diameter throughB, and 0 the centre of the circle.Denote by 6 the angle LAOP.

Fig. 69.

* It may be postponed or omitted by a reader who finds the mechanicsdifficult.


If x denotes the horizontal distance of P to the right of thevertical diameter AB, and if y denotes the depth of P below 0,then

x = a sin 0,

y = a cos 9.

Hence, differentiating with respect to time,

x = adcosd,

y == — adsind,

and x = aScos 9 — a02 sin 6,

y = — ad sin 0 — ad2 cos 9.

The acceleration/ of P in the direction of the (upward) tangent is

thus / = x cos 9 — y sin 9

= a§.

[This is a standard formula of applied mathematics.]

Now the forces on the particle are(i) the reaction R along PO, which has no component along the

tangent;

(ii) gravity mg, whose component along the (upward) tangent is

— mg sin 9;

(iii) the tension T9 where, by definition of modulus,m _ (modulus) (extension)

~~ natural length

2kmg{2a cos -|0 — a]a

= 2kmg{2 cos 19-1);

the component of T along the (upward) tangent is thus

T sin | 0

OTHER HYPERBOLIC FUNCTIONS 91

Equating the component of the acceleration to the componentof the forces, we have the equation of motion

ad = -gsin9 + 2kg(2 cos J0 -1 ) sin \9.

Now suppose that 0 is small. By the work given on p. 38, thevalue of sin0 is nearly 6 itself, while cos0 is nearly equal to 1.Hence the equation is approximately

Hence we reach the equation

a8 = g(k-l)6,

valid during the time while 0 is small.The argument now divides, according as k is less than or greater

than unity; that is, according as the string is 'fairly slack' or'fairly tight'.

(i) Suppose that k<l.

Then 0 = - ^ -

where an2 = (1 — k) g.

It may be proved that, when 9 = — n2 9, then 9 MUST be of theform

9 — A cosnt + Bsmnt,

where A,B are constants. In the meantime, the reader may easilyverify the converse result, that this value of 9 does satisfy therelation.

If we suppose that the particle is initially drawn aside so that 9has the small value ex, then 9 = a when t = 0, so that

If also the particle is released from rest, then 0 = 0 when t = 0,so that

-0 = 0.

Hence 0 = a cos nt.7-2


Thus if 0 is initially small, it remains small, and its valueoscillates between ±a. The equilibrium is stable at the lowestpoint.

(ii) Suppose that k>\.

Then *-<*zl>l!*

where ap2 = (k—l)g.

This equation is not satisfied by sines and cosines, but we canexpress the relation between 6 and t in the form

where, again, the reader may verify the converse result that thisvalue of 9 does satisfy the equation.

With the same initial conditions as before, we have

so that A = B

Hence 6 = | a

= a cosh pt.

As t increases, cosh#£ increases steadily to 'infinity' (since epl

does), so that 6 ceases to be small. The equilibrium is unstableat the lowest point.

The differential equation, in fact, ceases to be accurate once 6ceases to be small.

It is instructive to consider the same problem under thealternative initial conditions that the particle is projected fromthe lowest point with speed v. Thus 6 = 0,ad = v when t = 0. Wetake the two cases in succession:

(i) 6 = Acosnt + Bsinnt,

where 0 = A,

vja = nB.

Hence 6 = (v/an) sin nt.

(ii)

where

so that

OTHER HYPERBOLIC

O-Aefi + Be-*

0 = A + B,

vja = pA —pB,

A-7?-. *=2ap

FUNCTIONSt

V

~2ap%

93

Hence 6 = -— (ept - (

= (v jap) sinh pt.

The analogy between the pairs of solutions

oc cos nt, otcoshpt

and (vjan) sin nt, (vjap) sinh pt

affords striking confirmation of the analogy between the twoclasses of functions.

EXAMPLES I

Differentiate the following functions:

1. sinh 3a;. 2. cosh2 2a;. 3. artanha?.

4. sinh2 (2a;+1). 5. sinh # cos #. 6. secha:sin2a:.

7. (1 +a;)3 cosh3 3a;. 8. a;2tanh24x. 9. log sinh a:.

10. log (sinh x + coshx). 11. e^x. 12. xe~tl


»•/•1 O I a-! YI y% A.*y* fit* 14. I Qi in Ti * v /If 1 i

J J16. j a; sinh a;da;. 17. \ex cosh xdx. 18. | sinh 3a; cosh xdx.

C C19. a;sinh2#c?a\ 20. cosh3 xdx. 21.

J J22. \x2 cosh xdx. 23. e2*sinh5a;da;. 24. I tanha;sech2a;da;.

•I


Establish the following formulae:

25. sinh A + sinh B = 2 sinh —-— cosh —-—•

26. sinh A — sinh B = 2 cosh —-— sinh —-—.2 2

A + B A —B27. cosh A + cosh B = 2 cosh —-— cosh —-—.

2 2

A + B A — B28. cosh A — cosh B = 2 sinh —-— sinh —-—.

2 2

29. Prove that, for all values of u, the point (a cosh u% b sinh w)lies on the hyperbola whose equation is

and that the tangent at that point is

= 1.a o

(But note that that point is restricted to the part of the hyperbolafor which x is positive.)

3. The graph y = coshx. The two relations

y = cosh a;, >

dydx

= sinh

give us sufficient informationto indicate the general shape ofthe curve:

Since

cosh ( - x) = cosh (x),

the curve is symmetrical about the i/-axis; and since

1,

the curve lies entirely above the line y = 1. The value of yincreases rapidly with x.

THE GRAPH 95

Taking x to be positive (when it is negative the correspondingpart of the curve is obtained simply by reflexion in the t/-axis) wehave sinh# positive, so that the gradient is positive. Moreover,

= cosh a;,dx*

which is positive, and so (Vol. I, p. 54) the concavity is 'upwards'.The general shape of the curve is therefore that indicated in the

diagram (Fig. 70).

4. The graph y = sinhx. We have the relations

y = sinha;,

O

Fig. 71.

-¥~ — cosh#.dx

dinSince ~- is positive, the gradient is

dxalways positive, and y is an increas-ing function of x, running from — ooto +00 as x increases from — ooto +00.

At the origin, -— = 1, so that theax

curve crosses the #-axis there at an angle of |TT. Also

ffiy_dx2

which is positive for positive x and negative for negative. Hencethe curve lies entirely in the first and third quadrants, with(Vol. i, p. 54) concavity 'upwards* in the first and 'downwards' in the

third. At the origin, -j~ = 0, so that (Vol. I, p. 55) the curve

has an inflexion there.

Since sinh (— x) = — sinh (x),

the curve is symmetrical about the origin.The general shape of the curve is therefore that indicated in the

diagram (Fig. 71).


5. The inverse hyperbolic cosine. The problem arises inpractice to determine a function whose hyperbolic cosine has agiven value x. If the function is y9 then

x = cosh y,

and we use the notation

y = cosh"1 a;

to denote the inverse hyperbolic cosine.The graph (Fig. 72) y = cosh"1 x is

found by 'turning the graph y = cosh#through a right angle' and then re-naming the axes, as shown in thediagram. The graph exhibits two pro-perties at first glance:

(i) / / x < 1, the value of cosh"1 a; does not exist]

(ii) If x>l, there are TWO values of cosh"1 a;, equal in magnitudebut opposite in sign.

We can express cosh"1 a; in terms of logarithms as follows:

If y = cosh"1 a;,

then x = cosh y

Fig. 72.

so that e*v-2xev+l = 0.

Solving this equation as a quadratic in ev, we have

and so, by definition of the exponential function,

y = log{x±J(x2-!)}.

These are the two values of y. Moreover their sum is

= 0.

Hence the two roots are equal and opposite. The positive root islog {x + <](x2 -1)} and the negative log {x - <J(x2 - 1)}.

THE INVERSE HYPERBOLIC COSINE 97

3oefficient o:

cosh y = x

To find the diflFerential coefficient of cosh"1 a;, we differentiatethe relation

with respect to x. Then

dy 1or dx sinh y ± /(cosh2 y — 1)

1±

The gradient is positive when y is positive and negative when yis negative. In particular, if we take the POSITIVE value of

" 1 ^ , thendy 1dx

We can also obtain the result from the formula

taking the positive value. For if

then

-u-r dy dy duHence T~~T~~

1

Note the corresponding integral

17~2—v\ ~ c08^"1 x (positive value)J <J(x — I)


6. The inverse hyperbolic sine. On 'turning the graph ofsinh a; through a right angle and taking a mirror image', and thenrenaming the axes, we obtain the graph (Fig. 73)

y = sinh""1 /

of the inverse hyperbolic sine of x> that is, of the number whosehyperbolic sine is x. The function sinh"1 a; is a SINGLE-VALUED

function, uniquely determined for all values of x.To express sinh"1 a; in terms of

logarithms, we write

sinh""1 a; = y,so that

x = sinh y = \{ey — e~y)9

giving e2y - 2xey - 1 = 0.

Solving this equation as a quad-ratic in ey, we have

O

Fig. 73.

But (p. 19) the exponential function is positive, so that wemust take the positive sign for the square root. Hence

orwithout ambiguity.

To find the differential coefficient of sinh"1 x, we differentiatethe relation

sinh ywith respect to x. Then

ordy= 1 1dx cosh y ± (sinh2 y + I)

1

But, from the graph (Fig. 73), the gradient is always positive,and so dy __ 1

dx~~*J{x2+lY

THE INVERSE HYPERBOLIC SINE

We can also obtain this result from the formula

For if

then

Hence

dudx "

1

s/(*2-fn

dy__dx

x + (x2 + l)*

YT2 4- 1 \ 4. ^1

1 w

1

Note the corresponding integral

—z—rr = sinh*1 x

EXAMPLES II

Find the differential coefficients of the following functions:

1. x cosh"1 x. 2. sinh-1(l + ^2). 3. tanh-1^.4. sech"1^. 5. cosech""1^. 6. log (cosh*1 #).

7. ajcosh-1^2^!). 8. (cosh"1 a;)2. 9. l/(sinh-1o;)-


2 rJ

dx 11 r dx 12 r ***

J V ( 2 + 2 + 5 ) ' —4ar-l«)'

CHAPTER X

CURVES

1. Parametric representation. Hitherto we have regarded acurve as defined by an equation of the form

y =/(*)•

For many purposes it is more convenient to adopt a parametricrepresentation whereby the coordinates x,y of a point T of thecurve are expressed as functions of a parameter t in the form

(Of course, there is no reason why the parameter should not bex itself.) For economy of notation, however, we often write

x = x(t)9 y = y(t).

Familiar examples from elementary coordinate geometry arethe representations

x = at2, y = 2at

for the parabola y2 = 4ax, and

x = ct, y — c/t

for the rectangular hyperbola xy = c2.We confine ourselves to the simplest case, in which x(t),y(t)

are single-valued functions of t with as many continuous differen-tial coefficients as the argument may require.

The 'dot' notation

x,y,x,y,...

will be used to denote the differential coefficients

dx dy d2x d2y

The tangent at the point ef of the curve

PABAMETEIO EEPEESENTATION 101

may easily be found; it is the line through that point with gradient

dx x

= g'(t)f'(t)'

For the reader familiar with determinants, an alternative formof equation may be given:

The equation of any straight line is

Ix + my + n = 0.

If this line passes through the point x = f(t), y = g(t), then

lf(t) + mg(t) + n = 0;

if it passes through the point x =f(t + St), y = g(t+St), then

lf(t + St) + mg{t + 8t) + n = Q.

Subtracting, we have the relation

l{f(t + St) -/(*)} + m{g(t + St) - g(t)} = 0,

or, on division by St,

st - ° -For the tangent, we must take the limiting form of this relation

as 8 ^ 0 , namely lf(t) + mg>{t) = 0.

Hence, on eliminating the ratios I :m :n, we obtain theequation of the tangent in the form

x y 1

f{t) g(t) 1 = 0 .

/'(*) g'(t) o

2. The sense of description of a curve. We regard thatsense of description of a curve as positive which is followed by avariable point for increasing values of the defining parameter.For example, if the points A9P,Q (Fig. 74) correspond to thevalues a,p, q respectively, and if a <p < q, then the sense is APQ.

It is important to realize that sense is not an inherent property;it is a man-made convention. Thus different parametric repre-sentations may give rise to different senses along the curve.

102 CTJEVES

For example, the positive quadrant of the circle x2 + y2 = 1 maybe expressed with x as parameter in the form

yB

O X

T

A x

Fig. 74. Fig. 75.

As x increases from 0 to 1, the arc is described in the sense BA ofthe diagram (Fig. 75). On the other hand, if the polar angle 6 istaken as parameter, we have

x = cos 9, y = sin 9

As 9 increases from 0 to |TT, the arc is described in the sense AB.

3, The'length* postulate. If we confine our attention to thesimplest case, where the curve has a continuously turning tangent,as in the diagram (Fig. 76), then our in-stinctive ideas will be satisfied if we ensurethat the length of a curved arc PQ is nearlythe same as that of the chord PQ when thepoints P,Q are very close together. Forthis purpose, we shall base our treatment oflength on the postulate

chord QP= 1.

Fig. 76.

Our aim is to let the derivation of all the standard formulaeof the geometry of curves rest on this single assumption, togetherwith the normal manipulations of algebra, trigonometry and thecalculus.

THE ' L E N G T H ' POSTULATE 103

The warning ought perhaps to be added that in a more advancedtreatment it would be necessary to examine whether length'exists at all and to proceed somewhat differently. The presenttreatment suffices when dxjdt, dyjdt are continuous; this is true for'ordinary' cases such as we shall be considering.

4. The length of a curve. Suppose that U PtP' are threepoints on the curve (Fig. 77)

* P'/

x = x(t), y = y(t) u

given by the values u,p,p + 8p of theparameter. For convenience, we assumethat

u<p<p + 8p,so that the curve is described in the

sense UPP'.If P,P' are the points (x, y), (x + Sx,y + Sy) respectively, then

the length of the chord PPr is

whether Sx, 8y are positive or negative.Now the length of the arc UP is a function of p, which we may

call s(p). Thus, since arc PP' = arc UP1 - arc UP, we have

arc PP' = s(p+ Sp)-s(p).

s(p + 8p)- s{p) _ s(p + Sp) - s(p) chord PP'Sp ~~ chord PP' * Sp

arc PP'

But

chord

If we proceed to the limit, as Sp -> 0 so that P' -> P, then

lim

arc PP' _1

'~ 'chord PP'

Sx dx 8y dy$P dp sP-+oSp dp

104 CTJBVES

Hence s'(p) •

where the POSITIVE square root must be taken since 5 increaseswith p.

Replacing p by the current letter t, we have the relation

Integrating, we obtain the formula

measured from the point U with parameter u.

5. The length of a curve in Cartesian coordinates. If thecoordinate x is taken as the parameter t, then the formula of § 4becomes

so that * W

measured from the point where x = a, where the positive sense ofthe curve is determined by x increasing.

In terms of the coordinate y, we have similarly

measured from the point where y = 6, where the positive sense ofthe curve is determined by y increasing.

ILLUSTBATION 1. To find the length of the arc of the parabolay2 = Aax from the origin to the point (x, y), where y is taken to bepositive.

The parametric representation is

x = at2, y = 2at,

so that x = 2at, y = 2a.

Hence [l2aM2

Jo

LENGTH OF CXJBVB IN CARTESIAN COORDINATES 105

To evaluate the integral, write

=U(tI=U(t*+l)dt

on integration by parts. Hence

Hence / = |*V(*2 + l) + ilog{t + j(t*+ 1)},

so that s = a* V(Z2 + 1) + a log # + j(t2 + 1)}.

6. The length of a curve in polar coordinates. Let theequation of the curve in polar coordinates be

If 8 is taken as the parameter, then the formula of § 4 becomes

Now x = r cos 8, y — r sin 8,

where r is a function of 8. Hence

dx dr n . n dy dr .j n 7 / j V VyiO V / Oi.ll l/j J f\ jh aX±X V ~J~ T COS Uj

cLU (LU ctu du

To) +1^1 = TZ + r •

It follows that 8'(0) = 7 ( ( ^ +A,

so that 5(0) =

measured from the point where 8 = a, where the positive sense ofthe curve is determined by 8 increasing.

106 CURVES

ILLUSTRATION 2. To find the length of the curve (a CARDIOID)

given by the equation r = a ( i + COS 0).

The shape of the curve is shown in the diagram (Fig. 78).

We have — = — a sin 0, so thatdo

= 2a2 (1 + cos 0) = 4a2 cos2 J0.

Hence the length of the curve is

I 2a cos \6dQ = 4a | sin \B 1 *

= 4a[sin ( 77) — sin (— |TT)]

= 8a.Note. If we had taken the limits of integration as 0, 2TT, we

should apparently have had the result that the length isf2ir r -\2n

2a cos 100*0 = 4a sin £0

= 4a [sin TT — sin 0]= 0.

It is instructive to trace the source of error. This lies in ourassumption that ^ cog2 id) = 2a cos \0.

When 0 lies between — TT, TT, this is true, since cos|0 is thenpositive. But in the interval TT, 2TT, the value of cos | 0 is negative,so that V(4a2 cos2 ¥) = - 2a cos id-Hence we must use the argument:

fJo

Jo JnCn P

= 2acos^0a0 —Jo Jn

r v r I2*= 4a sin 10 — 4a s i n | 0

L Jo L in8a.

THE 'GRADIENT ANGLE* \ft 107

1. The 'gradient angle* 4>, If the tangent at a point P of the

curve

makes an angle if; with the #-axis, then

The diagrams (Fig. 79) represent the four ways (indicated bythe arrows) in which a curve may leave' a point P on it, theparameter being such that the positive sense along the curve isthat of the arrow.

o

(i) FIRST QUADRANT

dx + ; dy + ; cos y> + ; sin xp +.

(ii) SECOND QUADRANT

dx —; dy 4-: cos y) —; sin ip + .

(iii) THIRD QUADRANT

— ;dy — ; cosy — ; siny> —.

(iv) FOURTH QUADRANT

; dy —; cosy + ; s iny^ .

Fig. 79.

dxi

dxThus -=- is positive for (i), (iv) and negative for (ii), (iii); while

fit]~- is positive for (i), (ii) and negative for (iii), (iv).ds

8-2

108 CURVES

But\axj

-r-l = cos2*/*;

ds)

( dy\2

-pi = sin2^.

Hence -=-, -p have the numerical values of cos ip, simp respectively;

and, if we define the angle \p to be the angle I whose tangent is -pi

from the positive direction of the #-axis round to the positivedirection along the curve, in the usual counter-clockwise sense ofrotation, then the relations

dx , dy . ,-y- = cos w, ~- = sm ibds ds T

hold IN MAGNITUDE AND IN SIGN for each of the four quadrants,as the diagram implies.

We therefore have the relations

dx dy

true for every choice of parameter by correct selection of theangle ip.

EXAMPLES I

1. If the parameter is x, then ip lies in the first or fourthquadrant.

2. If the parameter is y, then ip lies in the first or secondquadrant.

3. If the parameter is 6, then ip lies between 8 and 8 + IT(reduced by 2n if necessary).

4. What modifications are required in the treatment given in§ 7 if the curve is parallel to the y-axis ? Prove that the formulaedx dy— = cos «/r, — = sm w are still true.ds T ds r

ANGLE FROM RADIUS VECTOR TO TANGENT 109

8. The angle from the radius vector to the tangent.The direction of the tangent to the curve

r=f(0)may be described in terms of the angle </> 'behind' the radiusvector. For precision, we define <£ as follows:

T

Fig. 80. Fig. 81.

A radius vector, centred on the point P (Figs. 80, 81) of thecurve, starts in the direction (and sense) of the initial line Ox;after counter-clockwise rotation through an angle 0, it lies alongthe radius OP produced. A further counter-clockwise rotation <j>brings it to the tangent to the curve, in the POSITIVE sense; thisdefines <f>.

If we assume, as usual, that 0 is the parameter, then the positivesense along the curve is that in which the length increases with 0.Hence the angle <f> lies between 0 and 77, as the diagrams (Figs. 80, 81)indicate.

By definition of ifs (p. 108) we have the relation

with possible subtraction of 2TT if desired.It is important to remember when using this formula that 0 is

the parameter used in defining ifs.

To find expressions for sin<£, cos<£, tan <f>:From the relations

x r cos 0, y = r sin 0,

we havedx dr n d0— = — cos a - r - rds ds ds

dy dr—- = -=-ds ds

d0ds

cos 0,

110 CURVES

so that (p. 108), when 6 is the parameter,

-=- cos 6 — r -j- sin 8 = cos I}J = cos (5 + <j>) = cos </> cos 9 — sin <£ sin

si -r- = sini/r = si = cos <f> sin 0 + sin ^ cos 0.

Solving for sin ^, cos , we have the formulae

. . dd

dr

so thatdr

9. The perpendicular on the tangent. Let the line ON(Fig. 82) be drawn from the pole 0 on to the tangent at the pointP of the curve

r =Then, by elementary trigonometry,

.ddds

This formula may be cast intoan alternative useful form:

1 = 1^1p* r^Xdd}

'dr

I I IA i ~T7» I •

If we write

so that

then

dude

1 _ /^YP* ~ \de)

Fig. 82.

THE PERPENDICULAR ON THE TANGENT 111

Note. Since <f> lies between 0, TT, the value of sin <f> is necessarilypositive. We may retain the formula

p = rsin<f>

generally if we allow p to take the sign of r. In any case, thenumerical value of p is that of rsin<£.

The expression for p in terms of Cartesian coordinates follows atonce: for . ,

p = r sin <p

= rsin(ifj— 6)

= r sin ip cos 6 — r cos iff sin 0

= xsimfjwhere (x9 y) are the coordinates of P.

It should be noticed that the step <f> = ifi—6 depends on theconventions adopted when 0 is the parameter. If another para-meter (for example, x) is used, the sign attached to p may requireseparate checking.

10. Other coordinate systems. The Cartesian coordinates(x, y) and the polar coordinates (r, 6) are by no means the onlycoordinates available for defining the position of a point of acurve. Others in use are the intrinsic coordinates (s9ift) and thepedal coordinates (p, r). The passage between Cartesian and polarcoordinates is familiar. To pass from Cartesian to intrinsiccoordinates, we have the relations

dx dy

or the equivalent

To pass from polar to pedal coordinates, we have

p* r*\d0) r2'

We do not propose to develop the theory of these newcoordinate systems any further. The following illustrationdemonstrates the use of some of the formulae.

112 CURVES

ILLUSTRATION 3. The PEDAL CURVE of a given curve with respectto a given pole.

The locus of the foot of the perpendicular from a given point 0on to the tangent at a variable point P of a given curve is calledthe pedal curve of 0 with respect to the given curve.

Referring to the diagram (Fig. 82) on p. 110, we see that N isthe point of the pedal curve which corresponds to P . The polarcoordinates of N are (rv d^, where

Let us find the pedal coordinates (pl9 r±) of N in terms of the pedalcoordinates (p} r) of P . We have at once

Also, if <f>± is the angle 'behind' the radius vector for the locus of N,

ddt d0 dSdp dp dp'

Now p — if

so that rsin<£ rsirdp

dOdrs i n © . / ~z ~ ~j~ \

dr dp

drdp

fdra n ? > W S m

t a n cf>-j-(r sin <j>

.dd>dp

^dp

Ydp

roosd)d<f>\rCOS^dp)

) = t a n ^ |

OTHER COORDINATE SYSTEMS 113

Hence

since each angle Kes between 0, TT and 9 is being used as parameter.It follows that

^ 1 = 7-1sin^1

= p sin cf>

and so the pedal coordinates (pl9 r±) of N are (p2/r,p).

COROLLARY. Since t an^ 1 =^ -^ - , and <£ = <f>l9 we have the

relation 7 ,

true for any curve.

11. Curvature. The instinctive idea of curvature, or bending,might be expressed in some such phrase as 'change of angle withdistance', and it is just this conception which we use for ourformal definition.

DEFINITION. (See Fig. 83.) The

CURVATURE K at a point P of a curve is

defined by the relation

dif;ds

The value of K may be positive ornegative, according as ifs increases ordecreases with s.

For many purposes, the calculation ofK is best effected by the direct use of thedefinition. It is, however, useful to be able to obtain the formulaein the various systems of coordinates which we have described.

(i) CARTESIAN PARAMETRIC FORM.

If x, y are functions x(t),y(t) of a parameter t, the relationsdx . dy . , .-=- = cos ib. — = sin 0 assume the formds T ds T

x = s cos ifj, y = s sin if/.

O

Fig. 83.

114 CURVES

Differentiating, x = s cos ift — sijj sin iff

't — s2Ksinif/,

since

similarly y = s sin I/J + S2K COS 0.

Hence y cos I/J— x simp = S2K,

or ~ ™~" ~ ——• o /c«

, i i yx — xy

so that #c = y .o y .

where s

with (p. 104) POSITIVE square root.

Note. We must choose our parameter to avoid the possibiKtythat x = 0, y = 0 simultaneously. This can be done, for example,by identifying t with x, when x = 1.

(ii) CARTESIAN FORM.

In the particular case when x is the parameter, we havedx

x = — = 1, and x = 0. Hence, by (i),dx

dx2

Tx

nwrthe denominator is POSITIVE since -7- is positive when x is the

CuX

parameter.It follows that, with our conventions, the sign of K is the same as

d2ythe sign of -r-|. Thus (Vol. 1, p. 54) the sign of K is positive when

CuX

the concavity of the curve is 'upwards', and negative when theconcavity is 'downwards'.

CURVATURE 115

(iii) POLAR FORM.

When 8 is the parameter, we have the relation (p. 109)

sothat K = dAfds ddds

Now (p. 110) cotd> = ~ ,

so that

or

Hence f&

/dry

\de)d?r

the positive sign being taken since 0 is the parameter.

Hence

(iv) PEDAL FORM.

We use the Corollary (p. 113)

tan^ = ^ ,

; drgiving • ' •

116 CURVES

But -=- = cos <£, so that T- = sec 6. Henceds T dr T

tan<£ = drdp'

so thatI dpr dr

The sign conventions used in this proof imply that 0 wasoriginally taken as parameter on the curve; otherwise, the signof K may require independent examination.

12. A parametric form for a curve in terms of s.Suppose that 0 is a given point on a curve (Fig. 84). Choose thetangent at 0 as #-axis and the normal at 0 as i/-axis. We seek toexpress x, y in terms of s as parameter, assuming that the condi-tions are such that a Maclaurin expan-sion is possible.

The Maclaurin formulae (p. 49) are

9Z

.' /A\ I nitf

Now

O

Fig. 84.

dx

so that

and

so that

dsi

&y • i

-£ = sm r,

d2y

4 ds

-Sinj/r - f

PARAMETRIC FORM FOR A CURVE IN TERMS OF S 117

dKIf we write K0, KQ to denote the values of /c, -=- at the origin, then,

CIS

since iff = 0 there,

*'((>) = 1; *"(<)) = 0; ^ " (0 ) = ^ ^

y'(0) = 0; */"(0) = *0; </"'(<)) = *£.

We therefore have the expansions

13. Newton's formula. To prove that the curvature at theorigin of a curve passing simply through it, and having y = 0 astangent there, is o

/co= l im- | .X-+Q**'

We use a method like that of the preceding paragraph to obtainthe expansion, assumed possible, of y as a series of ascendingpowers of x. We have that, if

y=f(x),

then

Thus

so that

/(0)

y

*

r» =

==

no)0) + xf

dydx

sec;

sec

sec

K sSv

= tan?/r,

dx

ds dx2\fs.K sec

2^Hence -~ = K0 + (terms involving x),

x

and so, assuming conditions to be such that the remainder tendsto zero with x9 2vi

118 CUKVES

ILLUSTRATION 4. To find the curvature of an ellipse, of semi-axesa,b, at an end of the minor axis.

Take the required end of the minor axis as origin (Fig. 85) andthe tangent there as the #-axis. The equation of the ellipse is

x* (y-bf+ l

Hence

o

Fig. 85.

For values of y near the origin, the negative square root must betaken, giving

6

so that

Hence

a;2 ~ a

K = Km

i 2 *

THE CIRCLE OF CURVATURE 119

14. The circle of curvature. The formulae

dy

dx*

show that, if the two curves

y=/(*),

both pass through a point P(g9rj), so that

and if, further, /'(£) = gr'(f)

and /"(£) = / ' ( | ) ,

then the two curves have the same gradient and curvature at P.In particular, the circle which passes through P, touches the

given curve y = f(x) at P, and has the same curvature as the givencurve at P, is called the circle of curvature of the given curve at P.

Denote by , „

the values of y,y',y" for the given curve at the point P(xP,yP).Suppose that the equation of the circle of curvature at P is

where G (a, jS) is the centre of the circle of curvature and p itsradius. By differentiation of this equation with respect to x, weobtain the relations

Since x, y9 y\ y" are the same for the circle of curvature as for thegiven curve,

120 CURVES

TT rt (1 "I" V P )

xience p = yp + - -f^—yp

XP —n >2/p

Vp

where /cP is the curvature of the given curve at P.We therefore have the formulae

a - xj

VP

I—• i 9

Kp

the sign being selected to make p positive.If the coordinates x,y of a point on the curve are given as

functions of a parameter t, then

dy

d*ydx2

dy Idxdt 1 dt

dxd2y

idx

dyd2xdt dt2 dt\2 dx'

THE CIRCLE OF CURVATURE

so that, if dots denote differentiations with respect to t,

dy ^ydx x9

d2y xy — yx

Hence

121

OC = XT> — -

_ _ _ i

~Kp

The point C (oc, /?) is called the centre of curvature of the given curveat P, and p its radius of curvature. We are adopting a conventionof signs in which the radius of curvature is essentially positive.

ILLUSTRATION 5. The curvature of a circle, and the sign of thecurvature.

Too much emphasis may easily be given to questions about thesign of curvature. Usually common sense and a diagram will settleall that is wanted. We give, however, an exposition for the casewhen the given curve is itself a circle, so that the reader may, if hewishes, be enabled to examine moreelaborate examples.

The difficulties about sign arise,with our conventions, as a result ofvarying choices of the parameterused to determine the curve. Webegin with the simplest case, in whichthe parameter is selected so that thecircle is described completely inone definite sense as the parameterincreases.

Let A (u,v) be the centre of the given circle (Fig. 86), and a itsradius. Take a variable point P (x, y) of the circle, and denote by t

O

Fig. 86.

122 CURVES

the angle which the radius vector AP makes with the positivedirection of the #-axis. As t (the parameter) increases from 0 to 2TT,the point P describes the circle completely in the counter-clock-wise sense. Then

x = u + acost, y = v + asint9

so that x = — a sin t, y = a cos t9

x = — acost, y = — asin£.

Hence x2 + y2 = a29

xij — yx = a2 (sin21 + cos21) = a2.

We therefore have the relations

, (a cos t) (a2)a = (^ + a cos £) — ^ — - = w,

= v9

a3

Hence for all points on, a given circle, the centre of curvature is atthe centre of the circle, and the radius of curvature is equal to theradius of the circle.

The curvature at any point istherefore ± I/a. With the presentchoice of parameter, the formulaof p. 114 shows at once thatK = + I/a, but other parametersmay give different signs.

For example, if # is the para-meter, the sense of description of

the curve is LPM in the upper .part of the diagram (Fig. 87) andLQM in the lower, these being Fig. 87.the directions taken by P9Q re-spectively as x increases. We know (p. 114) that, when x is theparameter, K is positive when the concavity is 'upwards' andnegative where it is 'downwards'. Thus K = — I/a in the arc

o

THE CIBCLE OF CURVATURE 123

LPM, and + I/a in the arc LQM. In fact, if P(x, y) is in the arcLPM, then

2/ '{a?-{x-u)%

where the positive sign is attached to the square root. Hence

-(x-u)y {a2-(x-uyf9

•*2

so that

, „ - 1 (x-u)2

and y = {a2-(x-u)2f {a2 -(x-\a2

Applying the formula (p. 114) for K, we have

a2

K = —{a2 - (a - uffl {a2 - (a? -

a"

Similarly we may prove that K = + I/a in the arc LQM, where

15. Envelopes. We have been considering a curve as the pathtraced out by a point whose coordinates are expressed in terms ofa parameter. An analogous (dual) problem is the study of asystem of straight lines

Ix + my + n = 0

when the coefficients I, m, n, instead of being constants, are givento be functions of a parameter t, say

/=/(*), m = g(t), n = h(t).

A familiar example is the system

x — yt + at2 = 0

consisting of the tangents to the parabola

y2 = 4ax.9-2

124 CURVES

If we take a number of values of t, we obtain correspondingly anumber of lines, which lie in some such way as that indicated inthe diagram (Fig. 88). They look, in fact, as if a curve could bedetermined to which they are all tangents. More precisely, thediagram assumes that the individual lines are numbered in an

Fig. 88.

order corresponding to increasing values of the parameter, and thepoints of intersection of consecutive pairs (1,2), (2,3), (3,4),...have been emphasized by dots. These dots appear to lie on acurve, and it is easy to conceive of the lines as becoming tangentsto that curve as their number increases indefinitely. In that case,the lines are said to envelop the curve, and we make the followingformal definition:

DEFINITION. Given a system of lines

Ix + my + n = 0,

v)hose coefficients Z, m, n are functions of a parameter t, the locus ofthat point on a typical line of the system, which is the limiting posi-tion of the intersection of a neighbouring line tending to coincidencewith it, is called the envelope of the system.

Consider, for example, the system of which a typical line is

x-yt + at2 = 0.

Another line of the system is

# — yu + au2 = 0.

They meet where x = aut, y = a(u +1).

ENVELOPES 125

That point on the typical line to which the intersection tends isfound by putting u = t in the expressions for the coordinates,g i v i l l g x = at2, y = 2at.

This is the 'parametric representation of the envelope, whoseequation is therefore 2

We may now find the rule for determining parametrically theenvelope of the system of which a typical line is

Another line of the system is

xf(u) + yg(u) + h(u) = 0.

Where these lines meet, it is also true that

4/M -/(*)}+vW) - 0(0}+{* W - *«} = o,or, on division by u — t, that

JM-ftt) { g(u)-g(t) {Mn)-h(t) = a

To emphasize the limiting approach of u to tf write u = t + St.Then the point of intersection of the two lines V , H' also lies onthe line

J(t+8t)-f(t) , ^g(t + 8t)-g(t) , h(t+8t)-h(t) _X U +y 8t + ft " 0>

In the limit, as S£->0, this is the line

zf'(t)+yg'(t)+h'(t) = O.

Hence the envelope is the locus, as t varies, of the point of intersectionof the lines

ILLUSTRATION 6. To find the envelope of the system

zcoat + ysint + a = 0.

The envelope is the locus of the point of intersection of this linewith the line . ± x

— x sin t + y cos t = 0.

126 CURVES

That point is x = — acost, y = — asintf,

and so the envelope is the circle

Note. Envelopes exist for many families of curves as well as forfamilies of straight lines; but we are not in a position to give atreatment of the more general case.

EXAMPLES II

Find the envelopes of the following families of straight Knes:

2. xsect — yta,nt — a = 0.

3. #eosh£ — ysinht — a = 0.

4. 2tx+(l-t2)y + (l + t2)a = 0.

5. tx + y-a(tz + 2t) = Q.

6. tzx-ty-c(t*-l) = 0.

16. Evolutes. Particular interest attaches to the envelope ofthose lines which are the normals of a given curve. Using thenotation of § 14, denote by

yp> yp> yp"

the values of y, y\ y" (where dashes denote differentiations withrespect to x) for the given curve at the point P{xF,yP). Theequation of the normal at P is

or x + yP'y = xP + yP'yP.

Taking xP as the parameter, the envelope of this line is thelocus of its intersection with the line

so that

X = Xpye"

EVOLUTES 127

Comparison with the results in § 14 establishes the theorem:The centre of curvature at a point P of a given curve is that point

on the normal at P which corresponds to P on the envelope of thenormals.

DEFINITION. The envelope of the normals, which is also thelocus of the centres of curvature, is called the evolute of the givencurve.

ILLUSTRATION 7. To find the evolute of the rectangular hyperbola

xy = c2.

Parametrically, the hyperbola is

x = ct, y = c/t,

and the gradient at this point is — 1/t2. Hence the normal is

or tzx —

For the envelope, we have

The centre of curvature, being the point of intersection of thesetwo lines, is given by

= 3c*4+ c,

so that x =

v = —u —l9t 2

The evolute is the curve given parametrically by these twoequations.

EXAMPLES III

Find the evolutes of the following curves:

1. The parabola (at2, 2at).

2. The ellipse (a cos t, b sin t).

3. The rectangular hyperbola (a sect, atan£).

128 CUBVES

17. The area of a closed curve. Consider the closed curvePUQV shown in the diagram (Fig. 89). For simplicity, we supposeit to be oval in shape, and also, to begin with, to lie entirely in thefirst quadrant.

The coordinates of the pointsof the curve being expressed inthe parametric form

x = x(t), y = y(t),

we suppose that the positivesense, namely that of t increas-ing, is COUNTER-CLOCKWISEround the curve, as implied by Fig. 89.the arrows in the diagram. [Ifthis is not so, replacement of t by —t will reverse the sense.] Thusthe curve is described once by the point (x9 y) as t increases from avalue t0 to a value tv Moreover, since the curve is closed, the two

values t0, t± give rise to the SAME point, so that

x(t0) = xfa); y{t0) = y(tx).

A simple example is the circle

x = 5 -f 3 cos t9 y = 4 + 3 sin t,

described once in the counter-clockwise sense as t increases from0 to 2?r. The values t = 0,1 = 2TT both give the point (8,4).

In order to calculate the area, draw the ordinates AP9BQwhich just contain the curve, touching it at two points P, Q whoseparameters we write as tP,tQ9 respectively. For reference, let Ube a point in the lower arc PQ and V in the upper. Then the areaenclosed by the curve can be expressed in the form

area AP VQ - area APUQ.

The area APVQ is given by the formula

area APF# =Xp

integrated over points (x, y) on the arc PVQ. Let us suppose firstthat the 'junction' point given by t0 or tl9 does not lie on this arc.

THE AREA OF A CLOSED CURVE 129

Then the parameter t varies steadily along the arc, and thearea is

dx T.

On the other hand, the 'junction' point J must then lie on the lowerarc PUQ, so we write the formula for the area APUQ in the form

CXQ

area APUQ = ydx

integrated over points (x, y) on the arc PUQ, giving

rj rxQ

area APUQ = ydx + ydxJxP JJ

£ dx , ClQ dxd t + )

where the value tx or tQ is taken for J according to the segment ofthe arc PUQ over which the respective integral is calculated, t± forPJ and t0 for JQ. In all, the area enclosed by the curve is thus

r_ydx - fa dx

dt+jtyTtdx 7 ft* dx 1 /**» dxdt+hdt+L

Similarly, if J lies on the arc PVQ, we have the formulae

rj rxQ

area APVQ = ydx+ ydxJxp JJ

dx ,. Ctq dx .

CXQarea APUQ = ydx

JXp

I 'Q dx .

130 CURVES

so that the area enclosed by the curve is

--[V^dt

Hence, in both cases, the area of the closed curve is

** dx ,.

In the same way, if we draw thelines CB,DS parallel to Ox (Fig. 90),just containing the curve, to touch itat B, S, and if L, M are points on theleft and right arcs B8 respectively,then the area of the closed curve is

area CB3IS - area CBLS.

yD

c

0

y(V R

S

/

X

Fig. 90.

Now area CB31S = I xdyJVB

integrated over points {x,y) on the arc BM8. If the 'junction'point J is not on this arc, we have

For the area CEL8, the 'junction' point J must then lie on the arcBLS, so we have

In all, the area of the closed curve is (in brief notation)

£-{£•£}

THE AREA OF A CLOSED CURVE 131

Similarly for the 'junction' point on the arc BM8, we have

IMHD - r + r + rJt0 ha JtRT1 dy ,]f dt

Hence the area of the closed curve may be expressed in either of the

-rt.yTtdt>

where the closed curve is described completely, in the counter-clockwisesense, as t increases from t0 to tv

A useful alternative form is found by taking half from each ofthese:

The area of the closed curve is

1 /V dy dx\ .

ILLUSTRATION 8. To find the area of the ellipse

The ellipse is traced out by the point

x = a cos t9 y — bsint

as t moves from 0 to 2TT. Then

dx = — a sin tdt, dy — b cos tdtf

and -^Uxdy-ydx)

1 \.Vn

•• r r a b .

132 CUEVES

The restriction for the curve to lie in the first quadrant is notessential. For, if it does not, a transformation of the type

for suitable values of a, b, can always be employed to bring it intothe first quadrant of a fresh set of axes. But this shows that thearea enclosed is

dy

since t0, tx give the SAME point of the curve. Hence the area is

dy

18.* Second theorem of Pappus. Suppose that a surface ofrevolution is obtained by rotating an arc PQ (Fig. 91) about the#-axis (assumed not to meet it). We proved (Vol. I, p. 130) that,if PQ is the curve = ^ x

the area S so generated is given by the formularb

8= 2<7ryds.Ja

Now it is easy to prove that the^/-coordinate of the centre of gravity ofthe arc is given by the formula

I yds ,Ja

o A

Fig. 91.

B

Iwhere I is the length of the arc PQ. (Compare the similar work inChapter vi.) Hence 8 = 2rrrjl

* This paragraph may be postponed, if desired.

SECOND THEOREM OF PAPPUS 133

Thus if a given curve, lying on one side of a given line, is rotatedabout that line as axis to form a surface of revolution, then the area ofthe surface so generated is equal to the product of the length of thecurve and the distance rotated by its centre of gravity.

ILLUSTRATION 9. To find the centre of gravity of a semicircular arc.Suppose that the circle is of radius a (Fig. 92), and that the

centre of gravity, lying on the axis of symmetry, is at a distance 77from the centre.

On rotating the semicircle about itsbounding diameter, we obtain the sur-face of a sphere, whose area is known tobe

Hence

so that 7] =2a

Fig. 92.

REVISION EXAMPLES V'Advanced9 Level

1. Find the equations of the tangent and normal at any pointof the cycloid given by the equations

x = a(2ip + sin 2\jj), y = a(l — cos 2ip).

Prove that ifs is the angle which the tangent makes with the axisof x; and verify that, if p and q are the lengths of the perpendicularsdrawn from the origin to the tangent and normal respectively, thenq and dpjdijj are numerically equal.

2. Define the length of an arc of a curve and obtain an expres-sion for it.

A curve is given in the form

x = — t, y = cosh£ + t.

Express t in terms of the length s of the arc of the curve measuredfrom the point (1,1).

The coordinates of any point of the curve are expressed interms of a and are then expanded in series of ascending powers of s.

134 CURVES

Prove that the first few terms of the expansions are

3. Find the radius of curvature of the parabola y2 = 4A# at apoint for which x = c, and deduce that, if A varies (c remainingconstant), this radius of curvature is a minimum when A = \c.

4. Prove that the radius of curvature of the curve

y = \x2-\\ogx (x>0)

at the point (x, y) is (l+o;2)2/4a;.Find the point at which this curve is parallel to the #-axis, and

prove that the circle of curvature at this point touches the y-a>xm.

5. Prove that, if iff is the inclination to the a;-axis of the tangentat a point on the curve y = alogsec(#/a), then the radius ofcurvature at this point is a sec ip.

6. A particle moves in a plane so that, at time t, its coordinatesreferred to rectangular axes are given by

x = a cos 2t + 2a cos t, y = asin2£ + 2asin£.

Find the components of the velocity parallel to the axes and theresultant speed of the particle.

Show that />, the radius of curvature at a point of the path, isproportional to the speed of the particle at that point.

7. The coordinates of the points of the curve 4yz = 27#2 areexpressed parametrically in the form (223,322). By using thisparametric representation, or otherwise, prove that the length ofthe curve between the origin and the point P with parameter tx is

2(1 + <?)*- 2,

and that the radius of curvature at P is numerically equal to

8. The tangent to a curve at the point (x, y) makes an angle \JJwith the #-axis. Prove that the centre of curvature at {x, y) is

dy nt dx\

BEVISION EXAMPLES V 135

A curve is given parametrically by the equations

x = 2a cos £ +a cos 2£, y = 2a sin t — a sin 2t.

Prove that ifs = —|£.P is a variable point on the curve, and Q is the centre of curva-

ture at P; the point R divides PQ internally so that PR = \PQ.Prove that the locus of R is the circle x2 + y2 = 9a2.

Prove also that no point of the curve lies outside this circle.Draw a rough sketch of the curve.

9. The coordinates of a point of a curve are given in terms of aparameter t by the equations x = te1, y = t2eK Find dyjdx in termsof t, and prove that

d2ydx2 ""

Prove also that the radii of the circles of curvature at the twopoints at which the curve is parallel to the #-axis are in the ratioe 2 : l .

10. Express cosh4 9 in the form

a0+ax cosh 0+a2 cosh 20 + a3 cosh 3d+a4 cosh 40.

11. The area lying between the curve

y = cosh (x/2\),

the ordinates x = + A2, and the #-axis is rotated about this axis.Prove that the volume of the solid of revolution so formed is

7rA(A + sinh A).

Show that the volumes given by A = 1 , A = 1 + S differ byapproximately TT(2 + e) S when S is small.

12. Prove that the two curves

y1 = ^ cosh 2x, y2 = tanh 2x

touch at one point and have no other point in common.Sketch the two curves in the same diagram and, with them,

the curves , . ,yz = cosh a;, j / 4 = smh#.

Prove that the four curves intersect in pairs for two values of xand indicate clearly in the diagram the relative positions of thefour curves for all values of x.

136 CURVES

13. If s be the length of the arc of the catenary y = a cosh (xja)from the point (0,a) to the point (x, y), show that

s2 — y 2 — a 2 .

Find the area of the surface generated when this arc is rotatedabout the y-axis.

14. Sketch the curve whose coordinates are given parametricallyby the relations x = a{t + sint),y = a(l + cost) for values of tbetween — TT9 TT, and find the length of the curve between thesetwo points.

15. Find the radius of curvature of the parabola y2 = x at thepoint (2,^/2).

16. Find the radius of curvature and the coordinates of thecentre of curvature at the point (0,1) on the curve y = cosh a;,

17. For what value of A does the parabola

2 / 2 _

have the same circle of curvature as the ellipse

( x-~a\2 y2

at the origin ?

18. Find the radius of curvature of the curve

y = sin ax2

at the origin.

19. Draw a rough graph of the function cosh a;.The tangent at a point P(x, y) of the curve y = c cosh (x/c)

makes an angle I/J with the #-axis. Show that y — c sec 0,If the tangent at P cuts the ?/-axis at Q, show that

PQ = xylc.

20. Find the radius of curvature of the curve

at the point (a, a).ay2

BEVISION EXAMPLES V 137

21. Find the radius of curvature at the point t = \TT on thecurve . x n.

x = a sin t, y = a cos 2t.22. Find the radius of curvature of the curve

y

at the point (x, y) and find a point on the curve for which the centreof curvature is on the y-axis.

23. Show that the radius of curvature of the epicycloid

x = 3cos£ — cos 3 , y = 3sin£ — sin3£

is given by 3 sin t.

24. Sketch the curve whose equation is

r = asin30 (a>0).

Find the area of the loop in the first quadrant and show thatthe radius of curvature at the point (r, 0) is

(5+ 4 cos 60)** a (7+ 2 cos 60)'

25. Show that the two functions

sinh"1 (tan a;), tanh""1 (sin x)

have equal derivatives, and hence prove that the functions them-selves are equal when — \TT < x < \TT.

26. Find the point of maximum curvature on the curve

y = logo;,

and the curvature at this point.

27. If y is the function of x given by the relation

sinhy = tana),

where —\TT<X<\n> prove that

cosh y = sec x, y = log tan (\x + \TT).

Show that y has an inflexion at x = 0, and draw a rough sketchof the function in the given range.

IO Mil

138 CURVES

REVISION EXAMPLES VI

'Scholarship* Level1. Transform the equation

d2u 2z du u

to one in which y is the dependent and x the independent variable,W h e r e u(l-z2)* = y and (l + z2)/(l-z2) = x,

obtaining the result in the form

2. If f(x) is a polynomial which increases as x increases, showthat, when x > 0,

is also a polynomial which increases as x increases.Show that, when x > 0, the expression

(x2-2x + 2)ex 2

x xis an increasing function of x.

3. Prove that

Prove also that the function

satisfies the equation

4. If j / = sin (msiii"1^), show that

REVISION EXAMPLES VI 139

Hence or otherwise show that the expansion of sinm0 as apower series in sin 9 is

. J (m 2 - l ) . 9Q (m2-l)(ra2-32) . Anmsin0 l-^———-sm2fl + - ~- - s in 4 0- . . .

5. By induction, or otherwise, prove that, if y = cot"1 a?, then

dny—*- = (—l)n(n—l)\sinnysmny.(AJX

Hence, or otherwise, prove that, if

( • »asm ex \

1+X COS OLj

then -— = (— 1 ) n ~ 1 —. — sin n (oc — z) sin™ (a — z).dxn v y sm n a

6. The function / n (re) is defined by the relation

where y = e^2. Prove tha t

dy

and deduce that

and that /w(a;) is a polynomial in # of degree n.

Prove that / ^ ( a ) = fn'(x) + 2^(a;),

and hence express /n+2(^) in terms of fn{x), fn'(x) and fn"(x).D e d u c e t h a t » n. N , n * ,, x o i ? / \ ^

fn"(x) + 2xfn'(x)-2nfn(x) = 0.7. If = sin-1 a;, prove that

xdy-X

Determine the values of y and its successive derivatives whenx = 0, and hence expand y in a series of ascending powers of x.

140 CUBVES

8. Prove that, if y = tan"1^, then

dnyU^-T-~ = (n— 1) \cosny cos {ny + l(n— 1)TT}

for every positive integer n.Deduce, or prove otherwise, that u satisfies the differential

equation

9- K

prove the relations

^ t l = 2(n+ l)x^ + 2(n+ lfy

10. If yn(x) = J ^

prove that yn = x - ^

n dx dx

Hence show that the polynomial yn satisfies a certain linear differ-

ential equation of the second order, i.e. linear in -r^r> -r^> yn.I dx2 dx *n \

11. Show that, if the substitution u = ta+1 ,~ is made inat

the equation du~dt

then

Deduce that, if a is not a negative integer, then the value of

-~ when t = 0 is Afan

where A is the value of y when t = 0.

EEVISION EXAMPLES VI 141

12. Prove that the polynomial y = (x2 — 2x)n satisfies the equation

13. A function f(x) is defined for 0 < x <f as follows:

l, f(x) = -

Discuss the continuity of f(x) and its successive derivatives.Sketch the graphs of/(#) and its derivatives throughout the wholeinterval (0,f).

Prove that the function». , 3 . TTXf(x)—sm-

has a stationary value at x = 1, and determine whether it is amaximum or a minimum,

14. Prove that, if y = e~x2/2 and n is a positive integer, then

The functions /w (#) are defined by the formula

Prove that (i) f^+1 = (n + l)/w,

(fi)/•**== */»-/»»

(Hi) fn+2-tfn+1 + (n+ l ) / n = 0,

(iv) /w is a polynomial in a: of degree w.

15. Obtain the equations

for a curve C.Prove that the equation of the general conic having 4-point

contact (i.e. 4 'coincident' intersections) with C at the origin 0 is

where p = l//c and A is an arbitrary constant.Deduce that the length of the latus rectum of a parabola having

4-point contact with a circle of radius a is 2a.

142 CTJEVES

16. Obtain the equations

The point Q on the curve is such that P is the middle point ofthe arc OQ; the tangent at the origin 0 cuts the chord PQ at T.Find the limiting value of the ratio TP : TQ as Q tends to 0.

Prove that, correct to the second order in s, the angle between

the tangents to the curve at P and Q is KSI l + — s\.

17. Show that, if f(x) has a derivative throughout the intervala^x^b, then

/(&)-/(*) = (b-a)f'{a+8(b-a)}9

where 0 lies between 0 and 1.Find the value of 6 if/(#) = tan#, a = JTT, 6 = |TT.Draw an accurate graph of the function y = tana; between the

limits \TT, JTT, taking 1 in. to represent ^rr as the #-axis and 1 in.to represent O2 on the y-axis.

Illustrate the theorem by means of your graph.

18. Given that eay = cos a;

and that yn denotes dny/dxn9 prove that

and express yn+2 in terms ofyvy2, ...,2The expansion of y as a power series in x is

Prove that bn is zero when n is odd and that, if a is positive, bn isnegative when n is even.

19. Prove that the nth. differential coefficient of eaV3cos# is

2neajV3cos|•K)-Deduce, or obtain otherwise, the expansion of the function as a

series of ascending powers of x giving the terms up to x% and thegeneral term.

Find the most general function whose nth differential coefficientis exV3cos#.

KEVISION EXAMPLES VI 143

20. (i) Prove by induction that, if x = cot 6, y = sin2 0, then

dny = , r

dxn K

(ii) Prove that -=— -^— = , Z^L.*

where fn{x) is a polynomial of degree n in as.Prove further that

21. State, without proof, Rolle's theorem, and deduce thatthere is a number £ between a, 6 such that

explaining what conditions must be satisfied by the function f(x)in order that the theorem may be valid.

If f(x)=zsinx, find all the values of f when a = 0,6 = §77.Illustrate the result with reference to the graph of sin a;.

22. If a is small, the equation sin x = OLX has a root nearly equalt o 7T. S h o w t h a t f 1 , 2 /1 2 • i \ <n

TT{1 - a + a2 - (f772 + 1 )a3}is a better approximation, if a is sufficiently small.

23. A plane curve is such that the tangent at any point P isinclined at an angle (k +1)0 to a fixed line Ox, where k is a positiveconstant and 0 is the angle xOP. The greatest length of OP is a.Find a polar equation for the curve.

Sketch the curve for the cases k = 2, k = | .

24. Prove that

d*d* [d*) dy2dy* \dy*)

dx) \dy)

25. Obtain the expansion of sin re in ascending powers of x. Forwhat values of x is this series convergent.

A small arc PQ of a circle of radius 1 is of length x. The arcPQ is bisected at Qx and the arc PQt is bisected at Q2. The chordsPQ,PQ1,PQ2 are of lengths c,\cx,\c% respectively. Prove that

5-(c — 20cx + 64c2) differs from x by a quantity of order x1.

144 CURVES

26. Sketch the graph of the function

where a, b are both positive. Prove that there are always at leasttwo points of inflexion.

Find the abscissae of the points of inflexion when a = | ° , 6 = 5.

(J11 (JOT (I 77 fi 0T

s ' * • & v ta temls o t

(dxVcPy)

dt9 dt9 dt*9

Prove that

dy) dx* + \dx) dy* + dx* dy*

28. The functions </>(X),I/J(X) are difFerentiable in the intervala^x^bf and if*'(x) >0 for a^a;^6. Prove that there is at leastone number £ between a, b such that

If ^(x) = a:2, if*(x) = a;, find a value of | in terms of a and 6.

29. Prove by differentiation, or otherwise, that

for all real x and positive y.When does the sign of equality hold ?

30. (i) Prove that, for positive values of x,

(ii) Find whether e'^'sec2^ has a maximum or a minimumvalue for x = 0.

31. Show that, if /(0) = 0 and iff'(x) is an increasing functionfix)

of x, then y = ^ -^ is an increasing function of a; for x > 0.a?

32. Prove that, if a; > 0,

(1 - -|-a;2 + a:4) sin a; > ( # - ia^ + rioar5) cos a;.

BEVISION EXAMPLES VI 145

33. If x> 0, prove that (x-l)2 is not less than #(Discuss the general behaviour of the function

for positive values of x and with special reference to x = 1.Sketch the graph of the function.

34. Prove that, if x is positive,

2x

Prove also that, if a, h are positive, then

log (a + Oh) — log a — 0{log (a + h) — log a},

considered as a function of 9, has a maximum for a value of 0between 0 and | .

35. If the sum of the lengths of the hypotenuse and one otherside of a right-angled triangle is given, find the angle between thehypotenuse and that side when the area of the triangle has itsmaximum value.

36. A pyramid consists of a square base and four equaltriangular faces meeting at its vertex. If the total surface area iskept fixed, show that the volume of the pyramid is greatest wheneach of the angles at its vertex is 36° 52'.

37. Find the least volume of a right circular cone in which asphere of unit radius can be placed.

38. Find the numerical values of

y = sin(o; +J7r) + |sin4a;

at its stationary values in the range — 7r<#<7r. Distinguishbetween maxima, minima, and points of inflexion, and give arough sketch of the curve.

39. Prove that, if

0=00^% (O<0<TT),

dn9then j - ^ = ( - l)n (n - 1)! sin718 sinnd$

when n is any positive integer.

146 CURVES

Show that the absolute value of dn6jdxn never exceeds (n— 1)!

if n is odd, or (n — 1)! cosn+11 1 if n is even.\2n + 2j

40. Find the two nearest points on the curves

y*-±x = 0, x2 + y2-6y + 8 = 0,

and evaluate their distance.

(i) find the maximum and minimum values of y;

(ii) find the points of inflexion of the curve;

(iii) sketch a graph showing clearly the points determined in(i), (ii), and also the position of the curve relative to the line y = x.

42. Prove that the maxima of the curve y = e~kxsinpx, whereJc,p are positive constants, all lie on a curve whose equation isy = Ae~kx, and find A in terms of k,p.

Draw in the same diagram rough sketches of the curvesy _ e-kx^ y _. _e-kx a n ( j y _ e-kx s i n ^# for positive values of x.

43. The tangent and normal to a curve at a point 0 are takenas axes. P is a point on the curve at an arcual distance s from 0.Prove that the coordinates of P are approximately

x = s - i*2s3, y = ^ 2 + $fcV,

where K is the curvature at 0, and AC' the value of dKJds at O.The tangents at 0 and P meet at 2\ A circle is drawn through

0 to touch PT at 2\ Prove that the limiting value of its radiusas P tends to 0 is l/(4/c).

44. Obtain the coordinates of the centre of curvature at a point(x, y) of a curve in the form (# — p sin if/, y + p cos 0), where p is theradius of curvature and tan 0 the slope of the tangent at the point.Prove also that, when s is the length of arc between (x, y) and afixed point on the curve, dx/ds = COSI/J, dyfds — sin^r.

The centres of curvature of a certain curve lie on a fixed circle.

Prove that p ~- is constant.rda


45. Obtain the formula p = rdrjdp for the radius of curvature ofa curve in terms of the radius vector and the perpendicular fromthe origin on the tangent.

Prove that a curve for which p = p satisfies the equationr2 = p2 + a2, where a is constant, and deduce that the polarequation of the curve is

<J(r2 -a2) = aO + a cos"1 (a/r),

where the coordinate system is chosen so that the point (a, 0) is onthe curve.

46. (i) Prove that the (p, r) equation of the cardioid

r = a(l + cos#)

is 2ap2 = r3.

Hence, or otherwise, prove that the radius of curvature is

fa cos \ 0.

(ii) Prove that the equation of the circle of curvature at theorigin of the curve

is 3(x2 + y2)

47. If y is a function of x, and

x = g cos oc — 7] sin a, y = | sin a + rj cos a,

where a is constant, express ^ and ^-f in terms of ~ and ^~dx dx2 dt; dj;2

d2y d2t]

Deduce that

and interpret this result.

48. A curve is such that its arc length s, measured from acertain point, and ordinate y are related by

y2 = 52 + C2,

where c is a constant. Show that referred to suitably chosenrectangular axes the curve has the equation

y = c cosh (x/c).

148 CTJBVES

If C is the centre of curvature at a point P of the curve and G isthe point in CP produced such that CP = PG, prove that thelocus of G is a straight line.

49. Find the radius of curvature at the origin of the curve

y = 2x + 3x2-2xy + y2 + 2y*9

and show that the circle of curvature at the origin has equation

50. Find the values of x for which y = x2(x — 2)3 has maximumand minimum values, and evaluate for these values of x thecurvature of the curve given by the equation above.

51. Sketch the curve defined by the parametric equations

x = a cos31, y = a sin31.

Show that the intercept made on a variable tangent by thecoordinate axes is of constant length.

Find the radius of curvature at the point ct\

52. Find the points on the curve y(x—l) = x at which theradius of curvature is least.

53. Sketch the curves

y = x 2 — x z , y2 = x 2 — #3,

and find their radii of curvature at the origin.

54. Prove that all the curves represented by the equation

yn+l _ (_ab_\n

a b

for different positive values of n9 touch each other at the point

/ ab ab \b9 a + b)'

Prove that the radius of curvature at the point of contact isequal to (


55. Establish the (p, r) formula for the radius of curvature of aplane curve. For a certain curve it is known that the radius ofcurvature is anlrn"1

9 where n 4= — 1 and a is positive, and also thatp = a/(n+1) when r = a. Show that it is possible to express thecurve by the polar equation rn = (n+ l)anGosn0.

56. Show that the curvature of the curve a/r = coshw0 has astationary value provided that Sn2 is not less than 1. Determinewhether this value is a maximum or a minimum.

57. 0 is the middle point of a straight line AB of length 2a, anda point P moves so that AP. BP — c2. Show that the radius ofcurvature at P of the locus is 2c2/*3/(3r4 + a4 —c4), where r = OP.

58. Find the integrals:

dx> jeP*ooabxdx (a #0,6 4=0), J -

59. Prove that the area enclosed by the curve traced by thefoot of the perpendicular from the centre to a variable tangent ofthe ellipse b2x2 + a2y* = a2ft2 is ±7r(

60. (a) Prove that, if

F(x) = e2x [Xe~2lf{t)dt-e* f V

then (i) F(0) = 0,

(ii) J"(0) = 0,

(iii) I"'{z)-3F'(x) + 2F(x) =/(*).(6) Find the most general function <f>{x) which is such that

I t<f>(t)dt = x2cf>(x).

61. Sketch the curve whose polar equation is r2 = a2(l + 3 cos 6),and find the area it encloses.

*o /-\ T> j-.ce x- J.' j.1. • w#cos# + sina;62. (I) By differentiating the expression ^zi > o r

s in QO

otherwise, find f xdxJsin6#'

(ii) Find f

150 CURVES

63. Sketch the general shape of the curve

x = atcost, y = at&mt

for positive values of the parameter t.Find an expression for the length of the arc of the curve

measured from the origin to the point t.

64. (i) Prove that, if n is a positive integer,

[ex cos nxdx = —^—- {Ae** — 1},

where A has one of the values ±1 , ±n; and classify the cases.(ii) Find the area bounded by the parabola y2 = ax and the circle

#2 + t/2 = 2a2.

65. Find the indefinite integral

log (1 + x2) tan*1 xdx./••

66. Show that the four figures bounded by the circle r = 3a cos 8and the cardioid r = a(l + cos 6) have areas

67. The polar equation of a curve is

r 2 -2rcos0 + sin20 = 0.

Sketch the curve and prove that the area enclosed by it is TT/^2.

68. Determine the function </>(x) such that

X(j)(t)e

tdt=(l+x)2ex,\XJo

Prove also that it is possible to find a pair of quadratic functionsf(x), g(x) such that

s l + f*g(t)dt,Jo

and determine these functions.


69. A plane curvilinear figure is bounded by the parabolax2 = -y from the origin to the point (13,5); by the hyperbolax2 — y2 = 144 from the point (13,5) to the point (15, 9); and by theparabola y2 = -x from the point (15,9) to the origin. Prove thatthe area of the figure is 33^+ 72 log f.

70. Prove that

Verify the identity

- xf = (1 + x2) (4 - 4x2 + 5#4 - 4x5 + x6) - 4,

and, by using this identity in the second of these integrals, provethat

22 1 22 1_ _ ^ fj ^ _ _

7 630 7 1260'

71. Determine constants A,B9C,D such that

)+tf+l-

Hence or otherwise prove that

° ' 5 0 2 < I 7~2—7u^<0'503-Jo (% +1)

72. Determine A9B,C,D such that

x2 _ d_ (Ax5 + Bx* + Cx\ D2+lf~~ dx ) +

and show that

73. If yr(x) satisfies the equation

show that ym(x)yn(x)dx = 0 (m#=7i).J-i

152 CUBVES

74. Polynomials fQ{x)if1(x)>f2(x)f... are defined by the relation

Prove that fm(x)fn(x)^x = ° (m + n)>

Show that , if <f>(x) is any polynomial of degree m,

0

where aw = ^ ^

75. If m, n are positive integers greater than unity, prove that

7 m w = cosmxGO&nxdxJo

m m j

m-n m~^n+1 m + n

Hence show, if p,q are positive integers, that

TT( p + a)!

76. Find a reduction formula for

(1 — x2)n cosh axdx.

Evaluate (1— a;2)3 cosh a; & .Jo

77. If y = sinp a; cos2 0: /(1 — &2sin2#) and p,q,k are constants,

axHence, by taking suitable values for p, q, express

m = Jo VU-&2sin2z)in terms of / m _ 2 ,7 m _ 4 .


78. State and prove the formula for integration by parts, andshow that p

xn( 1 - x)mdx =Jo79. Find the volume and area of the surface of the solid obtained

by rotating the portion of the cycloid

x = a(0 + sin0), y = a(l + cos#)

between two consecutive cusps about the axis of x.[Consider the range — TT 6 < TT.]

80. Prove that the envelope of the system of lines

is the curve y2(x + y— 1) = xz.

81. Find the envelope of the line

x sin 3t — y cos 3t = 3a sin t,

expressing the coordinates of a point of the envelope in terms ofa and t.

82. Through a variable point P (at2,2at) of the parabola y2 = 4axa line is drawn perpendicular to SP, when S is the focus (a, 0).Prove that the envelope of the line is the curve 21 ay2 = x(x — 9a)2.

83. Sketch the locus (the cycloid) given by

x = a(t + aint), y = a(l + cost)

for values of the parameter t between 0, 4TT.Prove that the normals to this curve all touch an equal cycloid,

and draw the second curve in your diagram.

84. Find the envelope, as t varies, of the straight line whoseequation is xcoaH + ysinH = a.

Sketch this envelope, and find the radius of curvature at oneof the points of the curve nearest to the origin.

85. As t varies, the line

x-t2y+2aP = 0

envelops a curve F. Show that for each value of t, other thant = 0, the line cuts F at a point P distinct from the point at whichthe line touches F.

154 CUBVES

Find the equation of the normal to F at P, and deduce that thecentre of curvature at P is given by

x = - 20atz - (3a/4J), y = 48a£5 - 3a*.

Discuss the case t = 0.

86. Find the equation of the normal and the centre and radiusof curvature of the curve ay2 = x3 at the point (at2, at3).

Show that the length of the arc of the evolute between thepoints corresponding to t = 0, t = 1 is (13*/6)a.

87. Show how to find the envelope of the line

y = tx+f(t).

Show that the length of an arc of the envelope is given by

Hence obtain a formula for the radius of curvature in terms of t.

88. Prove that the equation of the normal to the curve

may be written in the form

x sin t — y cos 14- a cos 2t = 0,

and find parametrically the envelope of the normal.

89. Find the equations of the tangent and normal at any pointof the curve

x = 3 sin t — 2 sin31, y = 3 cos t — 2 cos31,

Prove that the evolute is

= 2*.

90. The coordinates of any point on the curve xs -f xy2 = y2 canbe expressed in the form

x = t2l(l+t2), y=P/(l + t2).

Find the equations of the tangent and normal at any point, anddeduce that the coordinates of the corresponding centre ofcurvature are . 2 1M 4 .


91. Prove that the complete area of the curve traced out by the

^ (2a cos t + a cos 2t, 2a sin t—a sin 2t)is 2ira2.

92. Find the area inside the curve given by

x = a cos t + c sin let, y = b sin t + d cos Jet (0 < t < 2TT),

where & is an integer and a, 6, c, d are positive.[It may be assumed that c,d are so small in comparison with

a, b that the curve has no double points.]

93. Sketch the curve

x = a sin 2t, y = b cos31,

where a > 0, b > 0, and find the area it encloses.

94. Show that the curve

x = (£-l)e- ' , y = te

has a loop, and find its area.

95. Trace the curve x = cos 2t, y = sin 3t for real values of t, andfind the area of the loop.

96. Show that, for the range — a^x^a, the area between thecurve

and the #-axis is f

97. Sketch the curve given by

x = 2a(sin31 + cos3 £), y = 26(sin3 £ — cos31),

and prove that its area is Z-nab.

98. Find the area of the loop (— 1 < £ < 1) of the curve

_ l-t2 _t(l-t2)X y

99. Prove that the area of the curve

x = a cos t + b sin t + c, y = a' cos £ + 6' sin t + c'

is n(ab' — a'b).

156 CURVES

100. Explain the reasons for using the formula

277-2/ ds

to calculate the area of a surface of revolution.Apply the formula to show that the area of the surface obtained

by revolving the curver = a(l + cos0)

about the line 0 = 0 is equal to ^-TTO,2.

101. The curve traced out by the point

x = a log (sec t + tan t) — a sin t>

y — a cos t,

as t increases from — \n to + \n, is rotated about the axis of x.Prove that the whole surface generated is equal to the surface ofa sphere of radius a, and that the whole volume generated is halfthe volume of a sphere of radius a.

102. Find the length of the arc of the catenary

y = c cosh (xjc)

between the points given by x = + a.Find also the area of the curved surface generated by rotating

this arc about the #-axis.

103. Evaluate the area of the surface generated by the revolu-tion of the cycloid

x = a(t — sin t)> y = a(l — cos t)

about the line y = 0.

104. Two points A(0, c), P(£, rj) lie on the curve whose equation

y = ccosh(#/c),

and s is the length of the arc AP. If the curve makes a completerevolution about the #-axis, prove that the area 8 of the curvedsurface, bounded by planes through A and P perpendicular to the#-axis, and the corresponding volume V are given by

c£ = 2V -


105. A torus is the figure formed by rotating a circle of radius aabout a line in its own plane at a distance h( > a) from its centre.Find the volume and surface area of the torus.

106. AB, CD are two perpendicular diameters of a circle. Findthe mean value of the distance of A from points on the semicircleBCD, and also the mean value of the reciprocal of that distance.

Prove that the product of these means is

87r-2>/2log(l+>/2).

107. Prove that the mean distance of points on a sphere ofradius a from a point distant / ( ^ a) from the centre is

a+(f/3a).

What is the mean distance i f />a?

108. Calculate the average value, over the surface of a sphereof centre 0 and radius a, of the function (1/r3), where r is thedistance of the point on the surface of the sphere from a fixedpoint C not on the surface and such that OC = / . Distinguishbetween the two cases f> a, f<a.

109. The centre of a disc of radius r is O, and P is a point on theline through O perpendicular to the plane of the disc. Prove that,if OP = p, the mean distance (with respect to area) of points ofthe disc from P is

Find the mean distance (with respect to volume) of the interiorpoints of a sphere of radius a from a fixed point of its surface.

110. Prove that the mean value with respect to area over thesurface of a sphere, of centre O and radius a, of the reciprocal ofthe distance from a fixed point C is equal to the reciprocal of OCif C is outside the sphere, but equal to the reciprocal of the radiusa if C is inside the sphere.

111. A point P is taken on an ellipse whose foci are 8,H. Thedistance SP is denoted by r, and the angle HSP by 6. Show thatthe mean value of r with respect to arc is the semi-major axis a,and that the mean value of r with respect to 8 is the semi-minoraxis 6.

CHAPTER XI

COMPLEX NUMBERS

1. Introduction. It may be helpful if we begin this chapterby reminding the reader of the types of elements which he isalready accustomed to use in arithmetic. These are

(i) the integer (positive, negative or zero);

(ii) the rational number, that is, the ratio of two integers;

(iii) the irrational number (such as ^2, TT, e) which cannot beexpressed as the ratio of integers.

It is now necessary to extend our scope and to devise a systemof numbers not included in any of these three classes.

In order to exhibit the need for the new numbers, we solve insuccession a series of quadratic equations, similar to look at, butessentially distinct.

Following the usual 'completing the square' process, we have thesolution

x2-6x = - 5 ,

x2-6x + 9 = - 5 + 9,

= 4,

(x-3)2 = 4.

The important point at this stage is the existence of two integers,namely + 2 and —2, whose square is equal to 4. Hence we canfind integral values of x to satisfy the equation:

# - 3 = +2 or z - 3 = - 2 ,

so that x = 5 or x — 1.

The equation x2 — 6x -f 5 = 0 can therefore be solved by taking xas one or other of the rational numbers (integers, in fact) 5 or 1.

INTRODUCTION 159

II. X2-6X+7 = O.

Proceeding as before, we have the solution

x2-6x = - 7 ,

x2-6x + 9 = - 7 + 9,

= 2,

(x-S)2 = 2.

Now comes a break; for there is no rational number whosesquare is 2, and so we cannot find a rational number x to satisfythe given equation. We must therefore extend our idea of numberbeyond the elementary realm of integers and rational numbers.This is an advance which the reader absorbed, doubtless uncon-sciously, many years ago, but it represents a step of fundamentalimportance. The theory of the irrational numbers will be found intext-books of analysis; for our purpose, knowledge of its existenceis sufficient. In particular, we regard as familiar the concept of theirrational number, written as ^2, whose value is 1*414..., with theproperty that (yj2)2 = 2.

Once the irrational numbers are admitted, the solution ofthe equation follows:

# - 3 = +>/2 or x-3 = -<]2t

so that x = 3 + sj2 or £=3-^/2.

The equation x2 — 6x + 7 = 0 can therefore be solved by takingx as one or other of the irrational numbers 3 + <J2 or 3 — 2.

III.As before,

X2

X2

X2

- 6 # + 1 0 = 0 .

_6# =—10,

-6x + 9 = - 1 0 + 9,

(£-3)2 =_1 .

Again comes the break; for there is no number, rational orirrational, whose square is the NEGATIVE number —1. We havetherefore two choices, to accept defeat and say that the equationhas no solution, or to invent a new set of numbers from which a

160 COMPLEX NUMBERS

value of x can be selected. The second alternative is the purposeof this chapter.

To be strictly logical, we should now proceed to define these newnumbers and then show how to frame the solution from amongthem. But their definition is, naturally, somewhat complicated,and it seems better to begin by merely postulating their 'existence';we can then see what properties they will have to possess, and thereasons for the definition will become more apparent.

Just as, in earlier days, we learned to write the symbol '^2' fora number whose square is 2, so now we use the symbol 'J( — 1)' fora 'number* (in some sense of the word) whose square is — 1; but,in practice, it is more convenient to have a single mark instead ofthe five marks J, (, —, 1,), and so we write

as a convenient abbreviation.We shall subject this symbol i to all the usual laws of algebra,

but first endow it further with the property that

This will be its sole distinguishing feature—though, of course, thatfeature is itself so overwhelming as to introduce us into an entirelynew number-world.

It may be noticed at once that the 'number' — i also has — 1 forits square, since, in accordance with the normal rules,

= - 1 .

We round off this introduction by displaying the solution of theabove equation:

x — 3 = t or x — 3 = — t,

so that x = 3 + i or x=3 — i.

By selecting x from this extended range of 'numbers', we aretherefore able to find two solutions of the equation.

It may, perhaps, make this statement clearer if we verify how3 + i, say, does exactly suffice. By saying that 3 -f i is a solution ofthe equation x2 — 6x + 10 = 0, we mean that

+10 = 0.

INTRODUCTION 161

The left-hand side is

1+t2

= 0

= right-hand side,

so that the solution is verified.

We now investigate the elementary properties of our extendednumber-system, and then, when the ideas are a little morefamiliar, return to put the basis on a surer foundation.

EXAMPLES I

Solve after the manner of the text the following sets of equations:

1. x2-4:X + 3 = 0, x2-±x+l = 0, x2-4x + o =0 .

2. x2 + 8x+15 = 0, #2 + 8a;+ll = 0, z2 + 8a; + 20 = 0.

3. x2-2x-3 =0 , z 2 - 2 z - 4 =0 , x2-2x+10 = 0.

2. Definitions. The numbers of ordinary arithmetic (positiveor negative integers, rational numbers and irrational numbers) arecalled real numbers.

If a, b are two real numbers, then numbers of the form

a + ib (i2 = - l )

are called complex numbers. When 6 = 0, such a number is real.When a = 0, the number

ib (b real)

is called a pure imaginary number.Two numbers such as

a + ib, a — ib (a, b real)

are called conjugate complex numbers.A single symbol, such as c, is often used for a complex number.

162 COMPLEX NUMBERS

then a is called the real part of c and b the imaginary part. Thenotations

a = Me, b = Jcare often used.

The conjugate of a complex number c is often denoted by c, sothat, if c = a 4- ib, then c = a — ib. Clearly the conjugate of c is citself.

If c is real, then c = c.The magnitude

+ V(2 + &2) (a, 6 real)

is called the modulus of the complex number c = a + i6; it is oftendenoted by the notations

\c\, \a + ib\.

It follows that | c | = | c |.

Moreover, we have the relation

cc = | c |2,

for cc = (a + i6) (a — i&)

Finally, since c = a + ib, c = a — ib,

we have the relations

Jrc = b = -.(c-c).

Note. All the numbers now to be considered are actuallycomplex, of the form a + ib, even when 6 = 0. Correctly speakingwe should use the phrase 'real complex number' for a numbersuch as 2, and 'pure imaginary complex number' for 2i. But thisbecomes tedious once it has been firmly grasped that all numbersare complex anyway.

DEFINITIONS 163

EXAMPLES II

The following examples are designed to make the readerfamiliar with the use of the symbol i. Use normal algebra, exceptthat i2 = - l .

Express the following complex numbers in the form a + ib,where a,b are both real:

1. (3 + 5i) + (7-2i). 2.

3. (4 + 5i)(6-2i). 4.

5. 3 + 2i + i(5 + i). 6.

7. (1+i)2. 8.

9. ( 2 - i ) 2 - ( 3 + 2i)2. 10.

Prove the following identities:

n.U-t . 12. ^ « 1(1--).

13 * l

i6' 3^4ip + iq

3. Addition, subtraction, multiplication. Let

c = a + ib9 w~u-\-iv (a, b, u, v real)

be two given complex numbers. In virtue of the meaning whichwe have given to i, we obtain expressions for their sum, differenceand product as follows:

(i) SUM C + W =

(ii) DIFFERENCE C — IV = (a — u) + i(b — v).

(iii) PRODUCT CW = (a + ib)(u + iv)

= au + ibu + iav + i2bv

= (au — bv) + i(bu + av).

The awkward one of these is the product. At present it is bestnot to remember the formula, but to be able to derive it whenrequired.

164 COMPLEX NUMBERS

EXAMPLES III

Express the following products in the form a + ib (a, 6 real):

1. (2 + 3i)(4 + 5i). 2. (l + 4i)(2-6i).3. ( _ i _ ; ) ( _ 3 + 4i). 4.

5. i(4 + 3i). 6.

7. (3-f)(3 + t). 8. (2 + 3i)3.

4. Division. As before, let

c = a + ib, w = u + iv (a, b, u, v real).

Then the QUOTIENT is, by definition, the fraction

c _a + ib

It is customary to express such fractions in a form in which thedenominator is real. For this, multiply numerator and denomi-nator by w = u — iv. Then

ib)(u — iv)w (u + iv) (u — iv)

(au + bv) + i(bu — av)= u2 + v*

au + bv .bu — av= u2 + v2+l u2 + v2'

It is implicit that u, v are not both zero.

For example, (2 + i) ~ (3 - i)

2 + i

9-i2 " 9+1

10

DIVISION 165

EXAMPLES IV

Express the following quotients in the form a + ib (a,b real):

1. -—-.. 2. -—-^. 3. „ . .l - i 4 + 3^ 1%

A 2 — 3i 3 + 5i ,, cos 20 + i sin 295. —.. 6. jr1 —6i# " cos^ + i s in^

5. Equal complex numbers. To verify that, if two complexnumbers are equal, then their real parts are equal and their imaginaryparts are equal.

Let c = a + ib, r==p + iq

be two given complex numbers {a,b,p,q all real), such that c = r.Then .,

a + %b = p + iq,or a— p =i(q — b).

Squaring each side, we have

But a—p,q — b are real, so that their squares are positive or zero;hence this relation cannot hold unless each side is zero. That is,

a = p, b = q,as required.

COROLLABY. / / a, 6 are real and

a + ib = 0,

then a = 0, 6 = 0.

EXAMPLES V

Find the sum, difference, product and quotient of each of thefollowing pairs of complex numbers:

3-5i. 2. -

3. 4, 2i. 4. 3 + 2i, - i .

5. 2 + 3i, 2-3i . 6. - 3 + 4i, - 3 - 4 ^ .

166 COMPLEX NUMBERS

Find, in the form a + ib (a, b real) the reciprocal of each of thefollowing complex numbers:

7. 3 + 4i. 8. -5+12^.

9. 6i. 10. - 6 - 8 ^ .

Solve the following quadratic equations, expressing youranswers in the form a±ib:

11. #2 + 4#+13 = 0. 12. # 2 -

13. x2 + 6x+l0 = 0. 14. 4x2 + 9 = 0.

15. x2-8x + 25 = 0. 16. x2 + 4x + 5 = 0.

6. The complex number as a number-pair. A complexnumber

consists essentially of the pair of real numbers a, b linked by thesymbol i to which we have given a specific property (i2 negative)not enjoyed by any real number. Hence c is of the nature of anordered pair of real numbers, the ordering being an importantfeature since, for example, the two complex numbers 4 + 5i and5 + 4i are quite distinct.

The concept of number-pair forms the foundation for the morelogical development promised at the end of § 1, for it is preciselythis concept which enables us to extend the range of numbersrequired for the solution of the third quadratic equation(x2 — 6x + 10 = 0). We therefore make a fresh start, with a seriesof definitions designed to exhibit the properties hitherto obtainedby the use of the fictitious 'number' i.

We define a complex number to be an ordered number-pair (ofreal numbers), denoted by the symbol [a, 6], subject to the follow-ing rules of operation (chosen, of course, to fit in to the treatmentgiven at the start of this chapter):

(i) The symbols [a, 6], [u,v] are equal if, and only if, the tworelations a = u,b = v are satisfied;

(ii) The sum[

is the number-pair [a + u,b + v];

THE COMPLEX NUMBER AS A NUMBER-PAIR 167

(iii) The product r ... r n

is the number-pair [au — bvtbu + av\.Compare the formulae in § 3.A number [a, 0], whose second component is zero, is called a

real complex number) a number [0,6], whose first component iszero, is called a pure imaginary complex number. These terms areusually abbreviated to real and pure imaginary numbers. (Comparep. 162.)

In order to achieve correlation with the normal notation for realnumbers, and with the customary notation based on the letter ifor pure imaginaries, we make the following conventions:

When we are working in a system of algebra requiring the useof complex numbers,

(i) the symbol a will be used as an abbreviation for the number-pair [a, 0],

(ii) the symbol ib will be used as an abbreviation for thenumber-pair [0,6].

It follows that the symbol a + ib may be used for the number-pair [a, 0] + [0,6], or [a, 6].

In particular, we write 1 for the unit [1,0] of real numbers, andi x 1 for the unit [0,1] of pure imaginary numbers; but no con-fusion arises if we abbreviate i x 1 to the symbol i itself.

We do not propose to go into much greater detail, but thereader may easily check that the complex numbers defined in thisway by number-pairs have all the properties tentatively proposedfor them in the earlier paragraphs of this chapter. We ought,however, to verify explicitly that the square of the number-pair iis the real number — 1. For this, we appeal directly to the defini-tion of multiplication:

[a, 6] x [u} v] = [au — bv, bu + av].

Write a = u = 0, 6 = v = 1. Then

or, in terms of the abbreviated symbols,

i x i = — 1.

168 COMPLEX NUMBERS

Finally, we remark that, although the ordinary language of realnumbers is retained without apparent change, the symbols incomplex algebra carry an entirely different meaning. For example,the statement

2x2 = 4

remains true; but what we really mean is that

[2,0] x [2,0] = [2 x 2 - 0 x 0,0 x 2 + 2 x 0]

= [4,0].

From now on, however, we shall revert to the normal symbolismwithout square brackets, writing a complex number in the forma + ib as before. It must always be remembered that number-pairs are intended.

7. The Argand diagram. Abstractly viewed, a complexnumber a + ib is a 'number-pair' [a, b] in which the first com-ponent of the pair corresponds to the real part and the secondcomponent to the imaginary. A number-pair is, however, alsocapable of a familiar geometrical interpretation, when the twonumbers a,b in assigned order are taken to be the Cartesiancoordinates of a point in a plane. We therefore seek next to linkthese two conceptions of number-pairso that we can incorporate the ideas ofanalytical geometry into the develop-ment of complex algebra.

With a change of notation, denote acomplex number by the letter z, where •??

•P(x,y)

We do this to strengthen the implicationof the real and imaginary parts x and yas the Cartesian coordinates of a point Fig- 93«P(x,y) (Fig. 93).

There is then an exact correspondence between the complexnumbers z = x + iy and the points P(x, y) of the plane:

(i) If z is given, then its real and imaginary parts are determinedand so P is known;

(ii) If P is given, then its coordinates are determined, and so zis known.

THE ARGAND DIAGRAM 169

We say that P represents the complex number z. The diagramin which the complex numbers are represented is called an Arganddiagram, and the plane in which it is drawn is called the complexplane, or, when precision is required, the z-plane.

EXAMPLES VI

Mark on an Argand diagram the points which represent thefollowing complex numbers:

1. 3 + 4i. 2. 5-i. 3. 6i.

4. - 3 . 5. 1+i. 6. - 2 - 3 i .

7. cos 0 + isin 0 for 0 = 0°, 30°, 60°,..., 330°, 360°.

8. 2cos2|0 + i8iu0 for 0 = 0,45°, 90°, ...,315°, 360°.

8, Modulus and argument. Given the point P(x, y) repre-senting the complex number

we may describe its position alternatively by means of polarcoordinates r, 0 (Fig. 94), where r is the distance OP, ESSENTIALLY

POSITIVE and 0 is the angle LxOP which OP makes with thex-axis, measured in the counter-clockwise sense from Ox. Thus ris defined uniquely, but 0 is ambiguous by multiples of 2TT.

We know from elementary trigo-nometry that y

x = r cos 0, y — r sin 0,

whatever the quadrant in which ^\pP lies.

Squaring and adding these rela-tions, we have

so that, since r is positive,

Dividing the relations, we have

tan 0 = yjx,

or 0 = tan-%/#).

Fig. 94.

170 COMPLEX NUMBERS

The choice of quadrant for 9 depends on the signs of BOTH yAND x; if x, y are + , + , the quadrant is the first; if x, y are —, + ,the second; if x, y are —, —, the third; if x, y are 4-, —, the fourth.

The two numbers r = J{x2 + y2)y

9 = tan-1^/*),

with appropriate choice of the quadrant for 9, are called themodulus and argument respectively of the complex number z. Ifr, 9 are given, then

z = r(cos 9 + i sin 8).

We have already (p. 162) explained the notation \z\ = r, andproved the formula zz = | z j 2 .

ILLUSTRATION 1. To find the modulus and argument of the complexnumber —l+i<j3.

If the modulus is r, then

r 2= (_l )2 + (^3)2 = 1 + 3 = 4,

so that r = 2. The number is therefore

so that, if 9 is the argument,

cos 9 = — J, sin 0 = ^ - .

Hence 0 *= f TT, and so the number is

2{cos |TT + i sin f TT}.

EXAMPLES VII

1. Plot in an Argand diagram the points which represent thenumbers 4 -f 3i, — 3 + 4i, — 4 — 3i, 3 — 4i, and verify that the pointsare at the vertices of a square.

2. Plot in an Argand diagram the points which represent thenumbers 7 + 3i, 6i, — 3 — i, 4 — 4i, and verify that the points are atthe vertices of a square.

MODULUS AND AKGUMENT 171

3. Find the modulus and argument in degrees and minutesof each of the complex numbers

^3 -i, 3 + 4i, -5+12 i , 3, 6-8i , - 2 i .

4. Prove that, if the point z in the complex plane lies on thecircle whose centre is the (complex) point a and whose radius isthe real number Jc, then

\z — a\ = k.

5. Find the locus in an Argand diagram of the point representingthe complex number z subject to the relation

\z + 2\ = \z-5i\.

9. The representation in an Argand diagram of the sumof two numbers. Suppose that

are two complex numbers represented in an Argand diagram bythe points

respectively (Fig. 95),Their sum is the number

Orepresented by the point P(x, y), where Fig. 95.

By elementary analytical geometry, the lines Px P2 and OP have

the same middle point ( ^ ± ^ , ^ ± * ! ? ) f so that OPXPP2 is a\ 2 2 /

parallelogram. Hence P (representing zx + z2) is the fourth vertexof the parallelogram of which 0Pv OP2 are adjacent sides.

In particular, OP is the modulus \z1 + z2\ of the sum of the twonumbers zlt z2.

172 COMPLEX NUMBERS

10, The representation in an Argand diagram of thedifference of two numbers. To find the point representing thedifference

z==z2-zl9

we proceed as follows:The relation is equivalent to

so that OP2 (Fig. 96) is the diagonal of the parallelogram of whichOPX, OP (where P represents z) are adjacent sides.

Hence OP is the line through 0 parallel to P±P2y the sense alongthe lines being indicated by the arrows, where OP, P±P2 are equal inmagnitude.

COROLLARY. If P1(x1, yj, P2(x2, y2)represent the two complex numbersz1 = x1 + iy1,z2 = x2 + iy2, then the linePiP2 represents the difference z2 — zly

in the sense that the length PiP2 is themodulus |z2~""zil ana* the angle which ^ ' ^^ l ]* 1

makes with the x-axis is theargument of z2 — zv

This corollary is very important Fig. 96in geometrical applications.

[The reader familiar with the theory of vectors should comparethe results in the addition and subtraction of two vectors.]

11. The product of two complex numbers. In dealingwith the multiplication of complex numbers, it is often moreconvenient to express them in modulus-argument form. Wetherefore write

zx = rx(cos 9X + i sin 0J, z2 == r2(cos 92 + i sin 92).

Then zxz2 = rx r2(cos 9± + i sin 0X) (cos 92 + i sin 92)

= /•^{(cos 9X cos 92 — sin 9X sin 92)

+ i(sin 9X cos 92 + cos 0± sin 92)}

= ^^{cos {0t + 92) + i sin (91 + 92)}9

which is a complex number of modulus rxr2 and argument 0±+ 92.

THE PRODUCT OF TWO COMPLEX NUMBERS 173

Hence(i) the modulus of a product is the product of the moduli',

(ii) the argument of a product is the sum of the arguments.

We may obtain similarly the results for division:(iii) the modulus of a quotient is the quotient of the moduli',

(iv) the argument of a quotient is the difference of the arguments.

12. The product in an Argand diagram. In the diagram(Fig. 97), let PVP2,P represent the two complex numbers zvz2

and their product z = zxz2 respectively, and let U be the 'unit'point 2 = 1 .

Then (§ 10)

LxOP = 91+92,

where 0l9 92 are the arguments of zl9 z2,and so

» LxOP-LxOP9

O u

Fig. 97.

= LU0P±.

Moreover, if rv r2 are the moduli of zv z2,

OP\OP2 = (rxr2)/r2 = rjl= OPJOU.

Now in measuring the angle LP20P or LUOPV we proceed, byconvention, from the radius vector OP2 or OU, through an angle 9V

in the counter-clockwise sense, to the vector OP or OPV It may bethat 9X itself does not lie between 0, TT, but nevertheless the factthat the angles LP20P, LU0P1 so described are equal ensures thatthe Euclidean angles LP2OP9 LUOP1 of the triangles AP20P,hU0Px are also equal. Moreover, the sides about these equalangles have been proved proportional, and so the triangles aresimilar.

Hence follows the construction for P:Let Pl9P2 represent the two given numbers zl9z2; let 0 be the

origin and U the unit point 2 = 1 . Describe the triangle P0P2

similar to the triangle PxOU, in such a sense that LP20P = LUOPV

Then the point P represents the product zxz2.

174 COMPLEX NUMBERS

COKOLLABY. The effect of multiplying a complex number z by iis to rotate the vector OP representing z through an angle \TT in thecounter-clockwise sense. This follows at once if we put zx = i in thepreceding work, so that Px in the diagram becomes the point (0,1)or, in polar coordinates, (1, £TT).

Thus i acts in the Argand diagram like an operator, turning theradius vector through a right angle.

EXAMPLES VIII

1. Mark in an Argand diagram the points which represent(a) 3 + 2i, (6) 2 + i, (c) their product. Verify from your diagramthe theorem of the text.

2. Repeat Ex. 1 for the points

(a) 1, (6) 3-5i, (c) their product;

(a) 2 - i , (6) 2 + i, (c) their product.

13. De Moivre's theorem.(i) To prove that, if n is a positive integer, then

(cos 6 + i sin 0)n = cos nd + i sin nO.

We use a proof by induction. Suppose that the theorem is truefor a particular number N, a positive integer, so that

(cos 0 + i sin 0)N = cos JV^ + i sin NO.Then

(cos 9 -f i sin 0)*+* = (cos N9 + i sin N6) (cos 6 + i sin 6)

= (cos Nd cos 6 — sin NO sin 6)

+ i(cos NO sin 0 + sin NO cos 6)

Hence if the theorem is true for any particular value N, it is truefor N + 1, N + 2,... and so on. But it is clearly true for N = 0, andso the theorem is established.

(ii) To prove that, if n is a negative integer, then

(cos 0 + i sin 0)n = cos nO + i sin nO.

Write w = - w .

DE M O I V R E ' S T H E O R E M 175

Then m is a positive integer, and so

(cos 9 + i sin 9)n = (cos 9 + i sin 0)~m

1(cos d + i sin

1(as above)

cosmd — isinmd(cos m9 + i sin m#) (cos md — i sinmO)

cos mO — isinmOcos2m0 —i2sin2ra0

cosra0 —isinrafl

= cosrafl —isinmfl

= cos (— n0) — i sin (— nO)

= cos nd + i sin nd,as required.

(iii) To prove that, if n is a rational fraction, then

(cos 9 + i sin 9)n = cos n(9 + 2^TT) + i sin w

i5 an integer, positive or negative.Suppose that

n =

where p, q are integers (q positive) without a common factor.Consider the expression

(cos0 +isinfl)1/*,

and suppose that it CAN be expressed in the form

c o s cf> + i s i n <f>.

If so, then cos 6 + i sin 9 = (cos <f> + i sin </>)*

= cos g<£ + i sin #<£ [by (i)].

Equating real and imaginary parts, we have

cos 9 = cos q<f>, sin 0 = sin q<f>.

176 COMPLEX NUMBERS

These equalities hold simultaneously if, and only if,

where k is any integer, positive or negative. Hence

A n

Q qso that

(1 2k \ /I 2k \~9-\ 7r|+isin(-0H TT).9 9/ \9 ? /

Raising each side to the power jp, by (i) or (ii) above, and thenwriting p\q = n, we obtain the relation

(cos 9 + i sin 0)n = cos n(9 + 2&7r) + i sin w(0 + 2kir).

The expression on the right-hand side takes various valuesaccording to the choice of k. That is to say, there are severalvalues for (cos 9 + i sin 9)n when n is a rational fraction; the casen = \ is familiar, leading to two distinct square roots. We see byelementary trigonometry that distinct values are obtained when kassumes in succession the values 0,1,2, ...,g—l, where q is thedenominator in the expression of n as a proper fraction in itslowest terms. Thereafter they repeat themselves, any block ofq consecutive values of k giving the whole set.

SUMMARY. The formula

(cos 6 + i sin 6)n = cos n(9 + 2&TT) + i sin n(9 + 2k-rr)

is true for all values of n, where n may be a positive or negativeinteger or rational fraction. When n is an integer, k may be takenas zero; when n is fractional, distinct values are obtained on theright-hand side for k = 0, l ,2, . . . ,g— 1, where q is the denominatorin the expression of n as a proper fraction in its lowest terms,

EXAMPLES IX1 J3

1. Express ^ + i~ in modulus-argument form r(cos 6 + i sin 9)and hence, with the help of your tables, find (i) its square, (ii) itssquare roots, (iii) its cube roots, (iv) its fourth roots.

2. Repeat Ex. 1 for the number 4 + 3i.

3. Repeat Ex. 1 for the number 12 — 5i.

COS

THE WTH ROOTS OF UNITY 177

14. The nth roots of unity. In virtue of the formula

1 = cos 0 -f i sin 0,

it follows from De Moivre's theorem that, if n is any integer(assumed positive here) then an nth root of unity, that is

can be expressed in the form

i (-.2kn) + isin ( - . 2kir\\n ) \n )

for & = 0,l,2,...,7&—1. Hence for any given positive integer n,there are n distinct roots of unity, namely

2k-n . . 2kircos h % sin ,

n n

where k = 0,1,2, ...,w— 1.

For example:When n = 2, there are two square roots, namely

2&7T . . 2klT

cos — + t s i n — ,

where k = 0,1. When k = 0, this gives cosO, or 1; when k = 1, itgives cos 77, or — 1.

When n = 3, there are three cube roots, namely

2kir . . 2k7Tcos — + i s i n — ,

where £ = 0,1,2. When k — 0, this gives 1 itself. The valuesk = 1,2 give

2TT . . 2TT , ,cos — +1 sin —-==!(-

6 6

4TT . . 4TT „cos —- 4- * sin —- = J( — 1 — i ^3) .

o o

These complex cube roots of unity are usually denoted by thesymbols o>, o>2, either being the square of the other. They are alsoconnected by the relation

= 0.

178 COMPLEX NUMBERS

EXAMPLES X

Use De Moivre's theorem to express the following powers in'modulus-argument' form r(cos 8 + isin 6):

1.(1+0*- 2< ( 1 3 ^ - 3. (l + iV3)6.

4. (V3-»)*. 5. (l-»)*. 6. (V3 + 0*.

7. Find expressions for

(i) 1*, (ii) 1*

analogous to the roots evaluated in the text.

15. Complex powers. The reader will remember that, whenseeking an interpretation for symbols such as

a0, a~n, a1/n, a~1/n

for real positive a and positive integral n, he was guided by theprinciple that the formula of multiplication

should hold for all values of p and q. This led to the interpretations

a0 = 1, a~n = \\an, a1/n = tya, a~1/n =

We now consider the meanings which should be given to an whena,n are complex numbers, still retaining the validity for theformula of multiplication.

(i) When n is real and rational, the treatment is comparativelysimple, for we know that any complex number a can be expressedin the form , » . . / »

a = r(cos O + t&in 6),

and so, by De Moivre's theorem,

an = rn{cos n(9 + 2lc7T) + i sin n(9 +2 JCTT)}where the values h = 0,1,. . . , q— 1 (q being the denominator in theexpression of n as a proper fraction in its lowest terms) givedistinct values for an. Thus an is, in general, a g-valued function.(For example, if n = | , there are two square roots.)

COMPLEX POWERS 179

(ii) When n is complex, say

n = x + iy, (x, y real)

then we must interpret an in such a way that

ax+iy _ a

We have already dealt with the factor ax for real and rational valuesof x, so that we are left to consider what meaning may be given tothe expression aiy when the power iy is pure imaginary.

16. Pure imaginary powers. In the preceding paragraph wereduced the interpretation of an to the case, which we now con-sider, when n is pure imaginary. With a slight change of notation,we write

n = ix,where x is real.

(i) When a is real and positive.We note first that any real positive number a (other than zero)

can be expressed in the form

so that, if the laws of indices are to be preserved,

an = (el0K«a)n = enloe*a.

Writing n logea = ix logea = ix' (x' real),

we obtain the expression, as an imaginary power of e, in the form

which is found to be more amenable to treatment. Dropping thedash, then, we consider what interpretation should be given to then u m b e r ** (a; real).

We take the hint from the expansion of p. 54, which, if validfor pure imaginary numbers, would yield the relation

/ X2 X*

180 COMPLEX NUMBERS

Now (p. 50) we have the relations, valid for all real x,

X5

and so we are led to propose for consideration the relation

eix = cos x + i since.

Before we can accept this as a valid interpretation for eix%

however, we must satisfy ourselves that it obeys the index law

eix x eiy = eUx+y) fa y r e aJ)#

Under the proposed interpretation, the left-hand side is

(cos x + i sin x) x (cos y + i sin y)

= (cos x cos y — sin x sin y) + i(sin x cos y -f cos x sin y)

= cos (x + y) + i sin (x + y),

and this is precisely the interpretation to be given to the right-hand side also.

The interpretation is therefore consistent with the seriesexpansions and with the index law, and so we DEFINE theinterpretation of eix (x real) to be given by the relation

eix = cos x + i sin x.

Note. The mere writing of ix in the expansion in series doesnot of itself allow us to use the notation eix under the index laws.The step that the product of the expressions proposed for eix, eiv

is indeed eiix+v) is vital to the interpretation.

(ii) When a is any complex number.We are now in a position to give an interpretation to the

expression aix for any complex number a. We have seen that amay be written in the form

a == r(cos 9 + i sin 6) (r positive)

and, by what we have just done, we have the two relations

r = = e iog ,r

cos 8 4- i sin 0 = eid.

PURE IMAGINARY POWERS 181

Hence a = eloe<rxeid

= eu x eid,

say, where u} or loger, is a real number, positive or negative. Forinterpretations consistent with the laws of indices, we must there-fore take aix = {eU)ix x {eid)ix

= eiux x e~6x

= e~0x (cos ux + i sin ux)

= e~^{cos (x loge r) + i sin (# log,, r)}.

COROLLARY. Any (non-zero) complex number can be expressed inthe form eu+iv, where uyv are real.

17. Multiplicity of values. There is one difficulty which wehave slurred over, as it should not be over-emphasized at thepresent stage. An example will illustrate the point.

Let us consider what interpretation we are to give to the symbol

Our first task is to express the number to be raised to the power i(in this case, i itself) in the form reid, and this we do easily bynoticing that . = o

for any integer Jc, positive or negative. Hence the interpretation(i s fe(2k+i)7riy

-s e(2k+*)7TiXi

gives an infinite succession of values, all of which, oddly enough,are real.

In a full discussion, we should therefore be on our guard toinclude many-valued powers. A safe way of doing this is to includethe factor e2kni (which is just cos 2&77 + i sin 2&TT, or 1, for anyinteger k) in the expression in exponential form of the numberwhose power we seek. That is to say, in order to evaluate an, wewrite a in the form

and allow k to take integral values.

182 COMPLEX NUMBEES

Ifc should be noticed that we established the interpretation

eix — ii

under the conditions of De Moivre's theorem, namely that a; is areal number, being a positive or negative integer or rationalfraction. When x is not integral, there is really an ambiguity ofinterpretation, since

eix = (ct)z

__ eUl+2k7r)x

= cos (1 + 2kir) x + i sin (1 -f 21CTT) X,

where distinct values are obtained for k = 0,1, ...,# — 1 as usual,q being the denominator when the fraction x is expressed in itslowest terms. Our interpretation is therefore subject to theconvention k = 0.

ILLUSTRATION 2. To find an expression for

(l+iV3

We first write 1 + i 3 in the form

= 2ei^7T+2kn) (k integral).

We next use the fact that

Now (1 + i V3)4

AISO (1 + i V3)** = 2^ X

= 2*f x e~

Finally, 2 = el0«*2,

giving 2** = e*il0«*

MULTIPLICITY OF VALUES 183

so that, in all,

= l6e-<*-+^> {cos (f 77 + 1 log, 2) + i sin (§77 + J log, 2)},

where & is any integer, positive, negative, or zero.

EXAMPLES XI

Find expressions for

1. (l+iy. 2. (I-*)2"*. 3. (V3 + »)H.

4. (l-iV3)3+2*. 5- (^

18, The logarithm of a complex number, to the base e.We have seen (p. 181) that any (non-zero) complex number canbe expressed in the form

a==eu+ivm

We DEFINE the logarithm of a to be the complex number

u + iv (u, v real).

This is consistent with the treatment already given for realnumbers (v = 0) and, in virtue of the interpretation already givento indices, it retains the property

We have now, of course, released the restriction (p. 5) that amust be positive in order to have a logarithm. It need not evenbe real.

The ambiguous determination of v9 as seen by the equivalentformula

a == eu+uv+2kn) (j integral)

means that the logarithm is ambiguous to the extent of additivemultiples of 2iri.

Note. The relation (pp. 170, 180)

z = reid,

where r is the modulus and 9 the argument of z, gives the formula

logz = log I z I + i*argz.

184 COMPLEX NUMBERS

NOTATION. The notation

Logea

is often used to denote a value of the logarithm when ambiguitymay be present. Then the notation

denotes that determination of the logarithm whose imaginarypart lies between — IT and TT, and logea is called the principal valueof Loge a.

This distinction, which is often of importance, will not, in fact,be used much in this book, and we shall usually take the principalvalue without further comment.

19. The sine and cosine. We have made no statements sofar about the formula

eix — cosa;-fisin#

except when x is real; naturally, because cos a;, sin a; are otherwiseundefined. Our next task is to define the trigonometric functionsof a complex variable z = z + iy (x,y real), and this we do by whatis in some ways a reversal of the process hitherto adopted. Wefirst observe that, for real x,

eix = cos x + i sin x,

e-ix = c o g x __ i s m x^

so that cos a;

sin a; = —,(eix — e~ix).

We now adopt these formulae, and make them the basis of thefollowing more general definitions:

/ / z is a complex number, then

sin z = —. (eiz - e~iz).

It follows at once that these definitions give the normal functionswhen z is real, and also that

eiz = cos z + i sin z.

THE SINE AND COSINE 185

We must, however, now make sure that they retain the NORMAL

properties of the trigonometric functions.Firstly, we have

4(cos2 z + sin2 z) = (eiz + e~iz)2 - (eiz - e~iz)2

= (e2iz + 2 + e~2iz) - (e2iz - 2 + e~2iz)

= 4,

so that cos2 z + sin2 z = 1.

Again,

4(cos z1 cos z2 — sin z± sin z2)

= (e^1 + e-i2!i) (eiz*

on reduction, so that

cos zx cos z2 — sin zx sin z2 = cos (zx 4- z2).

Also

4i(sin zx cos z2 4- cos zx sin z2)

so that sin zx cos z2 + cos zx sin z2 = sin (zx + z2).

These are the formulae on which the theory of real angles isbased, and so the structure will stand for complex 'angles' also.

Further details are left to the reader.

Note. The functions |cosz|, |sinz| are not now subject to therestriction of being less than unity. For example, the equation

cos z = 2

is satisfied if eiz + e~iz = 4,

or 62te_4C<*+i = 0,

so that eiz = 2 + ^/3.

Thus iz = log"(2 + ,/3)

or z = - i log (2 + /3).

186 COMPLEX NUMBERS

For reference we record the following formulae:

(i) cosh iz = \ (eiz + e~iz)

= cos z;

(ii) sinh iz = \{eiz - e~iz)

= isinz;

(iii) e2k7Ti = cos 21CTT + isin 2TCTT = 1 (k integral);

(iv) e(2A:+1)^ = cos(2fe+l)7r + i s i n ( 2 i + l ) 7 7 = - l ;

(V) e(2k+»ni = c o s (2fc + 1) 77 + i sin (2Jfe + | ) 77 = i.

20. The modulus of e*. If z is complex, so that

z=x + iy (x, y real),

then ez = ex+^ = exeiy

Hence the modulus of ez is ex and its argument is y, the argumentbeing undetermined to the extent of multiples of 2TT.

COROLLARY. An important corollary, found by putting x = 0,is that , . ,

1^1 = 1,when iy is a pure imaginary.

21. The differentiation and integration of complexnumbers. We confine our attention to the only case which weshall use, the complex functions of a REAL variable x. The generaltheory for a complex variable is much beyond the scope of thisbook.

By a complex function of a real variable x, we shall mean afunction w(x) which is either given in the form

w(x) = u(x) + iv(x),

where u(x), v(x) are real functions of x, or can be reduced to thatform. For example, the function eix can be reduced to

cos x + % sin x.

DIFFERENTIATION AND INTEGRATION 187

We then use the following DEFINITIONS:

,.. dw _du .dvW dx~ = dx' + idx>

(ii) wdx = udx + i vdx.

Particular interest is attached to the function ecx, where c may becomplex of the form a + ib. We have

= — {eax (cos bx + i sin bx)}

= 7- [{e<lx c o s bx} + {{e™ sin bx}].

Hence, by definition,

— (e0*) = (a eax cos bx — be™ sin bx)CiX

+ i(a e°* sin bx + 6 eax cos 6a:)

a e°* (cos 6a; + i sin bx) + i6 e ^ (cos 6a; + i sin 6a;)

(a + ib)eax.eibx

ce™.

Hence the rule — (eM) =ao;

holds whether c is real or complex.

Similarly we may prove that the formula

\ecxdx = - e«

holds whether c is real or complex.13-2

188 COMPLEX NUMBERS

Because of its importance, we ought perhaps to mention alsothe differentiation and integration of #c, where x is assumed to bereal but where c may be complex. Since

xc = ecio«x9

we have -j- (xc) = ec]ogx.c.-j- (loga;) (as above)ax ax

in accordance with the similar rule for real functions.It follows that

\xcdx = -^-jXc+1 (c*- l )

whether c is real or complex.

ILLUSTRATION 3. To prove the rule

whenf(x),g(x) are complex functions of the real variable x.

Write f(x) = u(x) + iv(x) = u + iv,

g(x) =p(x) + iq(x) ==p + iq.

Then f(x) g(x) = (up — vq) + i(uq + vp),so that

sOTrt*»= [(u'p - v'q) + (up' - vq')] + i[{u'q + v'p) + (uq' + vp')]

Reversal of this formula leads to the rule for integration by parts.Hence this method is also at our disposal for these functions.

The two illustrations which follow show how complex numbersmay be used to sum series and to evaluate integrals.


ILLUSTRATION 4. To find the sum of the first n terms of the series

1 + cos 0. cos 0 + cos2 9. cos 2 9 + ... + cos71"19. cos (n - 1 ) 9,

where 9 is a real angle.Write

S= cos0.sin0 + cos20.sin20+...+ cos71"10. sin (71 - 1)0.Then

C + iS = 1 + cos 0 eie + cos2 0 e2ie + ... + cos"-10 e<n~1)ie.

This is a geometric progression, of TI terms, with first term 1 andratio cos9ei0. It may be proved, exactly as for real numbers, thatthe sum for first term a and ratio r is

a{l-rn)1-r

so that C + iS = —

In order to find 'real and imaginary parts', multiply numeratorand denominator by 1 — cos 0 e~ie. The new denominator is

(1 - cos 0 eie) (1 - cos 0 e-*°) = 1 - cos 9(eie + e-^) + cos2 0= 1-cos 0(2 cos 0) +cos2 0 = l-cos 20= sin20.

Thus

C + ttf = -J^ {(1 - cos* 0 en<*) (1 - cos 0 e-'*)}

sm20 l

Equating real parts, we have

C = -r-oTT {l-COSn0COS7l0-COS20 + COSn+10COS(n- 1)0}sm20 ^ \ / >

= . 2 {sin2 0 - cosn 0(cos n9 - cos 0 cos (n - 1) 0)}

= — - ^ {sin2 0 - cosn 0( - sin 0 sin (n - 1) 0)}sin (/

_ cosn0sin(ri-l)0"~ sin 0

It is implicit in the working that sin 0 is not zero. That case caneasily be given separate treatment.

190 COMPLEX NUMBERS

ILLUSTRATION 5. To find the real integral

xexcosxdx.

Write C=\xex cos x dx,

8= xexmxxdx9

so that C + iS = \x ex eix dx

= (xea+i)xdx.

Integrating by parts (p. 188), we have

C + iS = —!-; eu+o*.x- —U [e^*A.dx1+^ 1+iJ

Now

Hence

= [\x — \ix + \i\ ex (cos x + i sin #).

Equating real parts, we have

C = \

EXAMPLES XII

Use the method of the two illustrations to find:

2. sin# — #sin20+ #2sin30 — o;3sin40+... (to n terms).Find the integrals:

3. zcoszote. 4. e^sinzda;. 5. [e^smZxdx.

6. I xe%*sin xdx. 7. xsin Zxdx. 8. I xer**cos 3xdx.


Finally, we give an illustration to show how the use of complexnumbers can help in dealing with the geometry of a plane curve.We use for the curve the notation of the preceding chapter.

ILLUSTRATION 6. Let P(x9 y) be a point of a plane curve, andlet s9 if* denote as usual the length of the arc measured from afixed point and the angle between the tangent at P and the #-axis.(It is assumed that x9 y are real.)

Writez = x + iy.

m i dz dx ,dyThen - = - - + 4/

ds ds ds

= cos ip -f i sin if* (p. 108)

= e** (p. 180).In particular, -7- = 1 .

We also have the relation

<Pz_ .ds2 ds

(p. 113).This is equivalent to

d2z .ds2 ds2

so that, taking moduli on each side,

A . dx Ay . . . .Again, -=— % -~ = cos \p — i sin \p

so that(dx .dy\(d2x .d2y\ .\ds ds) \ds2 ds2)

Equating imaginary parts, we obtain the formula

_ dxd2y dyd2xds ds2 ds ds2'

192 COMPLEX NUMBERS

REVISION EXAMPLES VII'Scholarship' Level

1. Two complex numbers z, z' are connected by the relationz' = (2 + z)/(2 —z). Show that, as the point which represents z onthe Argand diagram describes the axis of y, from the negative endto the positive end, the point which represents z' describes com-pletely the circle x2 + y2 = 1 in the counter-clockwise sense.

2. Explain what is meant by the principal value of the logarithmof a complex number x + iy as distinct from the general value.

Show that, considering only principal values, the real part of

is 2*log«2e-*7rl cos (JTT loge 2).

3. Express tan (a-fit) in the form A + iB, where A,B are realwhen a,b are real.

Show that, if x + iy = tan J d + fq), then

i# = e-v sin | — e~2ri sin 2 | -1- e~3^ sin 3£ —...,

when 7] is positive; and that there is a like expansion with thesign of 7] changed, valid when TJ is negative.

4. Prove that, if

sin (x + iy) = cosec (u + iv),

where x, y, u, v are real, then

cosh2 v tanh2 y = cos2w, cosh2 y tanh2 v = cos2#.

5. Solve the equation

i)n = 0,

giving the roots as trigonometrical functions of angles between0 and 77.

6. Prove that, if x + iy = c cot (u + iv), then

x -y csin 2u sinh 2v cosh 2v — cos 2u'

REVISION EXAMPLES VII 193

and show that, if x, y are coordinates of a point in a plane, thenfor a given value of v the point lies on the circle

x2 + y2 + 2cy coth 2v + c2 = 0.

Also verify that, if alta2 denote the radii of the circles for thevalues vl9 v2 oiv, and d denotes the distance between their centres,then 2 *> 72

* 2 = cosh 2(vt — Vo).2axa2

7. Obtain an expression for | o,o&{x-\-iy) |2 in terms of trigono-metric and hyperbolic functions of the real variables x, y.

Show that | cos (x + ix) \ increases with x for all positive valuesof x.

8. Express -

in the form X + iY, where z1 = x1-i-iy1, z2 = x2 + iy2.Deduce that the points which represent three complex numbers

zl9 z2, z3 in the Argand diagram cannot all lie on the same side ofthe real axis if

9. Prove that the roots of the equation

(z-l)» =

where n is a positive integer, are

(4=0,

Deduce, or prove otherwise, that

S cot- —-—— = 0, 2 cosec2- —-—— = n2.

10. Write down the complex numbers conjugate to x + iy andto r(cos# + isin 6).

Prove that, if a quadratic equation with real coefficients hasone complex root, the other root is the conjugate complex number.Deduce a similar result for a cubic equation with real coefficients.

Given that x = 1 + 3i is one solution of the equation

a4+16z2+100 = 0,

find all the solutions.

194 COMPLEX NUMBERS

11. Express each of the complex numbers

in the form r(cos 0 + isin 0), where r is positive.Prove that z\ = z2, and find the other cube roots of z2 in the

form r(cos 9 -f i sin 9).

12. The complex number z = x + iy = r(cos 5 + i sin 0) is repre-sented in the Argand diagram by the point (x,y). Prove that, ifthree variable points zvz29zz are such that z3 = Az2+(1 — X)z19

where A is a complex constant, then the triangle whose verticesare zvz2>zz is similar to the triangle with vertices at the points0,1, A.

ABC is a triangle. On the sides BC,CA,AB triangles BCA\CAB\ ABC are described similar to a given triangle DEF. Provethat the median points of the triangles ABC, A'B'C are coincident.

13. The complex numbers a,b, (1 — k)a + kb are represented inthe Argand diagram by^the points A}B,C. Prove that the vectorAB represents b — a, and that

(i) when k is real, C lies on AB and divides AB in the ratiok: 1-k;

(ii) when k is not real, ABC is a triangle in which the sidesAB, AC are in the ratio 1 : | k \ and the angle of turn from AB toAC is 9, where k = | k \ (cos 9 + i sin 9).

The vertices of two triangles ABC and XYZ represent thecomplex numbers a,b,c and x,y,z. Prove that a necessary andsufficient condition for the triangles to be similar and similarlysituated is

i l l

= 0.

1

a

X

1

b

y

1

c

z

14. [In this problem small letters stand for complex numbersand capital letters for the corresponding points in the Arganddiagram.]

The circumcentre, centroid, and orthocentre of a triangle ABCare S, G, H respectively. Prove that

g = £(a-f 6 + c), 2$ + h = a + b + c.


ABC is a triangle with its circumcentre at the origin. Theinternal bisectors of the angles A,B,C of the triangle meet thecircumcircle again at L,M,N respectively. Prove (by puregeometry if you wish) that the incentre P of ABC is the ortho-centre of LMN, and deduce that

p =Prove also that the excentres Pv P2, P3 of the triangle ABO are

given by

px = l — m — n, p% = —l + m — n, p3 = —l — m + n,

and that the circumcentre K of P1P2P3 is given by

k = — (Z-f m + n).

Prove, finally, that KP1 is perpendicular to BO.

15. Points A,B in the Argand diagram represent complexnumbers a,b respectively; 0 is the origin, and P represents one ofthe values o{yj(ab). Prove that, if OA = OB = r, then also OP = r,and OP is perpendicular to AB.

A,B,0 lie on a circle with centre 0, and represent complexnumbers a, b, c respectively. Prove that the point D which repre-sents — bcja also lies on the circle, and that AD is perpendicularto BO.

The perpendiculars from B, O to CA,AB meet the circle againat E, F respectively. Prove that OA is perpendicular to EF.

16. Prove by means of an Argand diagram, that

Find all the points in the complex plane which representnumbers satisfying the equations

(i) z2-2z = 3, (ii) | z - 2 | = 3, (iii) | z- 11 + | z-2 \ = 3.

17. Find the fifth roots of unity, and hence solve the equation

(2#- l ) 5 = 32a5.

18. Prove that, if a, /? are the roots of the equation

= 0,n sinnd

a-j8

where cot 9 = x+ 1.

196 COMPLEX NUMBERS

19. A point z = x + iy in the Argand diagram is such that|z | = 2 , # = l , and y>Q. Determine the point and find thedistance between the point and \z2.

Show also that the points 2, z, -z2, |z3, Jz4, ^-z5 are the verticesof a regular hexagon.

20. The points zi,z2,z3 form an equilateral triangle in theArgand diagram, and zx = 4 + 6i, z2 = (1 — i)zv Show that z3 musthave one of two values, and determine these values.

What are the vertices of the regular hexagon of which zx is thecentre, and z2 is one vertex ?

21. A regular pentagon ABODE is inscribed in the circlex2 + y2 = 1, the vertex A being the point (1,0). Obtain the complexnumbers x + iy of which the points A,B9C,D,E form a representa-tion in an Argand diagram.

22. Use De Moivre's theorem to find all the roots of the equation

(2x- I)5 = (z-2) 5

in the form a + ib, where a, b are real numbers.

23. Mark on an Argand diagram the points ^3 + i, 2 + 2^3 , andtheir product. What is the general relation between the positionsof the points a + ib and ( /3 + i) (a + ib) ?

A triangle ABO has its vertices at the points 0,2 + 2^3 ,— l+i^/3 respectively. A similar triangle A'B'C has its verticesA', B' at the points 0, Si respectively. Find the position of C".

24. Find all values of (5 — 12i)*, (2i — 2)*, expressing the answersin the form a + ib, where a, b are fractions or surds.

25. Prove that the triangle formed by the three points repre-senting the complex numbers f,g,h is similar to the triangleformed by the points representing the numbers u, v, w if

fv + gw + hu =fw + gu + hv.

26. By expressing 3 + 4i, l + 2i in the form r(cos 9 -i- i sin 6), orotherwise, evaluate

in the form a + ib, obtaining each of the real numbers a, b correctto two significant figures.


27. Identify all the points in the z plane which satisfy thefollowing relations:

z-2(i) 2* + 2 = 23f (ii)

(iii)

z + 2

2 - 1

Z+l( iv) | Z -

28. Prove that three points z1,z2)zz in the complex plane arecollinear if, and only if, the ratio

(28-21)/(22-21)

is real.

29. Find the real and imaginary parts and the modulus foreach of the expressions

1 .... a + ib 1 + cosa + isina(1) TT2i ' (U) ~b+Ta9 {m)

30. Find the logarithms of - 1, 2 - i ,

31. The three points in an Argand diagram which correspondto the roots of the equation

-r = 0

are the vertices of a triangle ABC. Prove that the centroid of thetriangle is the point corresponding to p.

If the triangle ABC is equilateral, prove that p2 = q.

32. Find the cube roots of 4-3i .

33. If the vertices A, B, C of an equilateral triangle represent thenumbers zl9z29zz respectively in an Argand diagram, state thenumber represented by the mid-point M of AB, and show that

J3) ± * ( 2 - « )

If 21 = 2 + 2£, 22 = 4

and if G is on the same side of AB as the origin, find the number z3.

34. If 2 = cos 0 + i sin 6, express cos 9, sin 9, cos n09 sin nO in termsof 2, where n is an integer.

Hence express sin7 9 in the form

198 COMPLEX NUMBERS

35. Prove that

. + sin0 —icos0\6icosey

a) = -&cos0/

cos 60-asm 60.

36. Prove that


|» — 1 — 2i | = 3

represents a circle in an Argand diagram.If 2 lies on | z — 1 — 2i | = 3, what is the locus of the point

Find the greatest and least values of | z — 4 — 6i \ if z is subjectto the inequality 12 — 1 — 2% \ ^ 3.

38. Find the modulus of

Sz + i

when | z | = 1.

39. Two complex numbers z, zx are connected by the relation

__ 2 + zZl"2^Tz

Prove that, if z = iq, where q is real, then the locus of z1 in anArgand diagram is a circle. Describe the variation in the ampli-tude of zx as q increases from — oo to + oo.

40. If a, b are real and n is an integer, prove that

has n real values, and find those of

41. If x,y are real, separate

sec (x + iy)

into its real and imaginary parts.


42. If aib,c,d are real, express each of a2 + b2 and c2 + d2 as aproduct of linear factors.

Deduce that the product of two factors, each of which is a sumof two squares, is itself a sum of two squares.

43. The intrinsic coordinates of a point on a plane curve ares, xfs, and the cartesian coordinates are x, y. The complex coordinatex -f iy is denoted by z and the curvature by K. Prove that

dz . d2zds ds2

Hence, or otherwise, prove that

K=(z"* + y"*)*=(z'y"-x"y'),

where dashes denote differentiations with respect to s.Prove further that

= K(2K2-I

44. Any point P is taken on a given curve. The tangent at Pis drawn in the direction of increasing s, and a point Q is taken at aconstant distance I along this tangent. In this way Q describes acurve specified by X, F, 8>Z analogous to the specification x,y,s,zfor P. [Notation of previous problem.] Show that

x'

x"

x'"

y'

y"

y"'

1

K

K2

(ii) the curvature K of the derived curve at Q is given by theformula

where K = dfc/ds.

CHAPTER X I I

SYSTEMATIC INTEGRATION

AT first sight the integration of functions seems to depend asmuch upon luck as upon skill. This is largely because the teacheror author must, in the early stages, select examples which areknown to 'come out'. Nor is it easy to be sure, even with years ofexperience, that any particular integral is capable of evaluation;for example, #sin# can be integrated easily, whereas sinz/a;cannot be integrated at all in finite terms by means of functionsstudied hitherto.

The purpose of this chapter is to explain how to set about theprocesses of integration in an orderly way. This naturally involvesthe recognition of a number of 'types', followed by a set of rulesfor each of them. But first we make two general remarks.

(i) The rules will ensure that an integral of given type MUST

come out; but it is always wise to examine any particular examplecarefully to make sure that an easier method (such as substitution)cannot be used instead.

(ii) It is probably true to say that more integrals remainunsolved through faulty manipulation of algebra and trigonometrythan through difficulties inherent in the integration itself. Thereader is urged to acquire facility in the normal technique of thesesubjects. For details a text-book should be consulted.

1. Polynomials. The first type presents no difficulty. If f(x)is the polynomial

f(x) = aoxn + axx

n-1 +. . . + an,

then (f(x)dx = ^ ^ + ^ * ! + ... +an*.J tb ~T~ J- 7h

2. Rational functions. (Compare also p. 11.) A rationalfunction f(x) of the variable x is defined to be the ratio of twopolynomials, so that

W)

RATIONAL FUNCTIONS 201

for polynomials P{x),Q(x). If the degree of P(x) is not less thanthat of Q(x), we can divide out, getting a polynomial (easilyintegrated) together with a rational fraction in which the degreeof the numerator is less than that of the denominator.

We therefore confine our attention to the case in which thedegree of P(x) is less than that of Q(x). It is assumed, too, that allcoefficients are real.

It is a theorem of algebra that any (real) polynomial, and, inparticular, Q(x), can be expressed as a product of factors, of whichtypical terms are

where oc, jS, a, h, b are real constants, but where

h2<ab

so that the quadratic ax2 + 2hx + b cannot be further resolved intoreal factors.

[We ought to add that, for a given polynomial, the difficulty offactorization may be very great indeed.]

It is a further theorem of algebra that the rational functionmay then be expressed in the form

yf Al , ^2 . • Am ]

Bnx + Cn \a

where the first summation extends over all linear factors ocxand the second over all quadratic factors ax2 + 2hx + b.

The integrals from the first summation are of the type

Adx

which give

202 SYSTEMATIC INTEGRATION

The integrals from the second summation are of the type

f Bx + O

and require more detailed consideration.Note first that

/<

ax + h 7-dx

and so, by writing

Bx + C = (J5/a) (ax + A) + (a(7 - hB)ja,

we can reduce our problem to the evaluation of integrals such as

dxhWrite a(ax2 + 2hx + b) = (ax + h)2 + ab- h2,

and make the substitution (remembering that ab — h2 is positiveby assumption) , . ,, , ,9N

J r ' ax + h = tj(ab-h2).C dx _ r qP^dt^ab-h2)

1 J (ax2 + 2hx + b)v~] {t2 (ab - h2) + (ab - h2)}*

aP~x C dt

Our final problem is therefore to evaluate

dt

and for this we need a formula of reduction. On integration byparts, we have r ., (

•'•in / . n . - i \ / « I / J O . T

RATIONAL FUNCTIONS 203

Hence 2plp+1 -(2p-l)Ip = ^

or, replacing p by p — 1,

By applying this formula successively, we make the evaluation ofIp depend on that of /p_1? Ip-2> • ••> &nd, ultimately, on Iv But

= tan"1 *,

and so the whole integration is effected.

3. Integrals involving \/(ax + b). Rational functions oix and^(ax + b) may be integrated readily by means of the substitution

or s = -(J2-&).a

The result is the integration of a rational function of t.

Note. The reader is unlikely to remember all the details to begiven in § 4 following. The methods should be thoroughly under-stood, but it may well be found necessary to refer to the book fordetails in actual examples. If desired, §§ 5, 6 and 7, which will beof more immediate practical value, may be read next.

4. Integrals involving y(ax2 + 2hx + b). We first take stepsto simplify the quadratic expression under the square root sign,

(i) Suppose that a is positive. Then

a(ax2 + 2hx + b) = (ax + h)2 + ab- h2.

Write ax + h = x';

xf2+p2 (ab>h2)then a(ax2 + 2hx + b) =

[x'2-q2 (ab<h2)

where p2 = ab — h2, q2 = h2 — ab respectively.(ii) Suppose that a is negative. Then — a is positive, so we write

- a(ax2 + 2hx + b) = h2 - ab - (ax + h)2.14-2


If h2 — ab is negative, the right side is necessarily negative, so thatax2 + 2hx + b is also negative and (in real algebra) has no squareroot. We therefore take h2 — ab to be positive, and write

thus, ifas before, we have

ax + h = x',

- a{ax2 + 2hx + b) = r2- x'2.

If, then, we have to evaluate a rational function of x andy](ax2 + 2hx-\-b), we may first apply the transformation

ax + h = xr.

The integrand becomes a rational function of x' and of a surdwhich may assume one or other of the three forms

(i) yj{x'2 +p2) a> 0, ab-h2> 0,

(ii) J(x'2 -q2) a> 0, ab-h2< 0,

(iii) <]{r2-x'2) a<0,ab-h2<0.

We may now drop the dashes and treat x as the variable.

(i) The surd <J(x2+p2).

Consider the transformation

x =2pt

The graph (Fig. 98) indicates (whatcan also be proved algebraically)that all values of x are obtained byallowing t to vary continuouslyfrom — 1 to + 1 . We thereforeimpose the restriction

O

Fig. 98.on the values of t which we select.

We have(l-<2)2

INTEGRALS INVOLVING J(ax2+2JlX+b) 205

Since £2<1, we have, without ambiguity (assuming, as we may,that p is positive),

Hence a rational function of x,J(x2+p2) is transformed into arational function of t, and may be integrated accordingly.

(ii) The surd J(x2-q2).Consider the transformation

t2+l

The graph (Fig. 99) indicates (what can also be proved algebrai-cally) that all values of x for which x > q are obtained by allowing tto vary continuously from + 1 to +00.

1

~ ^ J

-1

„ J

1

x

0

— —

/

1

1

q " H

1

Fig. 99.

[We can restrict ourselves to positive values of x, since a rangeof integration which involved the two signs for x would have topass through the region —q<x<q where <J(x2 — q2) is undefined.We could, of course, equally restrict ourselves to negative valuesof x if necessary.]

We may therefore impose the restriction

Now dx —-4qtdt


Since t>l, we have, without ambiguity (assuming, as we may,that q is positive),

Hence a rational function of x, ^{x2 — q2) is transformed into arational function of t, and may be integrated accordingly.

(iii) The surd <](r2-x2).

Consider the transformation

x =t2-!t2+l r.

Fig. 100.

The graph (Fig. 100) indicates(what can also be proved alge-braically) that all values of x forwhich x2<r2 are obtained byallowing t to vary continuously from 0 to oo. We therefore imposethe restriction . ~

on the values of t which we select.

4rtdtWe have dx =

r2-x2 =(t2+l)2'

Since t>0f we have, without ambiguity (assuming, as we may,that r is positive), ~

Hence a rational function of x9 ^{r2 — x2) is transformed into arational function of t, and may be integrated accordingly.

Note. The work just completed proves that the integrations arePOSSIBLE; it does not necessarily give the easiest method. Forexample, the quadratic surds may also be subjected to thefollowing substitutions:

(i) For y](x2+p2), let x = #tan 9, or x = p sinh 9;

(ii) For sl{x2 — q2), let x = gsec 9, or x = gcosh 9;

(iii) For <j(r2-x2), let x = r sin 9.

ALTERNATIVE TREATMENT OF INTEGRALS 207

5. Integrals involving V(ax2+2hx + b): alternative treatment.Integrals of rational functions of #, <J(ax2 + 2hx + b) may also be

treated by methods which, by leaving transformation until the laststages, retajn the identity of the surd. We begin by reducing sucha rational function to a more amenable form.

Since even powers of yj(ax2 + 2hx + b) are polynomials and oddpowers are polynomials multiplying the square root, we mayexpress any polynomial in x, ^](ax2 + 2hx + 6) in the form

where P,Q are polynomials in x. Hence any rational function,being by definition the quotient of two polynomials, is

where P,Q}R,S are polynomials in x. Multiply numerator anddenominator by R — S yj(ax2 + 2hx + 6), so that the new denominatoris the polynomial R2 — S2(ax2 + 2hx + b) and the numeratorPR - QS(ax2 + 2hx + b) + (QR - PS) <J(ax2 + 2hx + b), and we obtainthe form

0 '

where A,B,C are polynomials in x. We already know how to dealwith the rational function AjC, so our problem reduces to theintegration of

0 '

It is found (surprisingly, perhaps) more convenient to have thesurd on the denominator, so we multiply numerator and denomi-nator by yj(ax2 + 2hx + b), obtaining

U

For -77—

where F is a rational function of x.

20S SYSTEMATIC INTEGRATION

By § 2 above, the problems to which F gives rise involve astypical termsJ xm (ra^O),

1

Ax + B

We have therefore to consider the three types of integral:

xmdxJ

... I*U J(iii)J (Px2 + 2g# + r)n ^(ax2 + 2hx + 6)

(i) The evaluation of the integral

C xmdx

Preparing the ground for an integration by parts, we observethe identity

- J= \

Now performing the integration, we have, on the right-hand side,X f

-m+lj

Equating the two sides, we have the recurrence relation

dm + 2) alm+i + (2m + 3) hlm+1 + (m+l)blm

= a;m+1 /(aa;2 + 21tx + b).


This enables us, once 70, Ix have been determined, to calculate72, /3, /4 , . . . successively, and so to evaluate Im for any positiveintegral value of m.

For 70, we must reduce the surd to one or other of the formsenumerated earlier in this section, giving

f—

For Il9 we have to consider the integral

1 = J/ i + hIQ=

JNOW

so that 7j = - {yl(ax2 + 2hx + b) -ex

(ii) The evaluation of the integral

dx

We can reduce this type to the form of type (i) and evaluateit at once by the substitution.

dx = ~72^«

Then ax2 + 2hx + b = alk + 7) + 2^(& + 7) + b

where a = aJe2 + 2hJc 4-6, /? = ak + h, y = a.


The integral is thereforedt

tn t

which is of the first form

xmdxk (ra 0).

(iii) The evaluation of the integral

(Ax + B)dxpx2 + 2qx + r)n J(ax2 + 2hx + 6)'

where n 1, g2 <^r.The general case is very difficult. It is, of course, possible to

avoid it by expressing the rational function as a sum of complexpartial fractions of the type \j(x — h)n

y where h is complex. Thisreduces the integration to type (ii)

-, 7i—77—o—^ rr (& complex)I \X — 1c\n' (nor -4- 2hor -4- h)

already discussed. The final return to real form is an added pointof difficulty.

We begin with an algebraic lemma, designed to reduce twoquadratic expressions simultaneously to simpler form.

LEMMA. TO establish the existence of a (real) transformation ofthe type

which reduces the quadratic expressions

ax2 + 2hx + b,

px2 + 2qx + r

ut2 + vto the forms

simultaneously.

u'P + v'


Since (*+l)2 (ax2 + 2hx + b)

under the transformation, we see that the coefficient of t vanisheson the right-hand side if a, j8 are chosen so that

aa/8 + % + j3) + 6 = 0.

Similarly the coefficient of t from px2 + 2qx + r vanishes if

poLp + q(aL + P) + r = 0.

Also a, j8 necessarily satisfy the equation in 9

On eliminating the ratios a/2: a + /3:1 between these three rela-tions, we obtain the equation

i h b

p q ri -e e2

o,

which is a quadratic in 9 with (by definition) roots a, /?. Hencea, j3 may be found and the transformation determined. (If, ex-ceptionally, a/h = pjq, the two quadratics differ only by a constant.The substitution ax + h = at reduces the integral In to one or otherof the types discussed on p. 213.)

Moreover a, p are real. If not, they must be conjugate complexnumbers, since all coefficients are real. Suppose that

a = A + i/x, /? = A — ijj, (A, jix r e a l ) .

Since p«.p + g(a + j8) + r = 0,we have the relation

Multiply by _p (which is not zero, by the nature of the problem).T h e n ^2A2 + 2 gA +pr +p22 = 0,

or (pX + q)2 + (pr - q2) + (p/x)a = 0.

Since p, g, A, JU, are all real, the two squares are positive; and, byhypothesis, pr — q2 > 0. Hence the left-hand side is positive. Weare therefore led to a contradiction, and so the supposition thata, j8 are not real is untenable.

Moreover, the condition pr — q2 > 0 shows that a, j3 are DIFFERENT

numbers.


SUMMARY. We can find two distinct real numbers, a,j8, theroots of the quadratic equation

a h b

P = 0,

1 -6 62

which enable us, by means of the transformation

x =ctt + p

t + l '

to reduce the two quadratic forms

ax2 + 2hx + b = 0,

px2 + 2qx + r = 0 (q2 < pr)

, ,, , ut2 + vto the forms

n't2 + v'

simultaneously.Let us now return to the integral. On substituting for x in the

expression Ax + B, we obtain an expression of the form

Ct + Dt+l '

where C = Aoc + B, D = AfS + B; also

Hence

Ct + D a-ftt+l '(t+l)2 dt

(urt2 + v')n

(t+l)2n ' t+l

__ [(ot-P)(Ct + D)(t+l)2n-2dt=J!

The numerator, which is a polynomial in t, can be expressed inthe form P + Qt, where P,Q are polynomials in t2; and so, writingu' s, or

u'


we obtain P and Q as polynomials in s. Hence the numerator canbe expressed as a polynomial in s = (u't2 + v'), together with t timesanother polynomial in (u't2 + v')9 the order of the whole numeratorbeing 2n — 1 in t, which is less than that of the denominator.Hence In is found as a sum of integrals of the form

rv)' k") (u't2 + v'

For the latter, put y2 = ut2 + v,

so that ydy = utdt;

hence K = -K u

ydylu' o— (y2

[u

if

— v) + v'

dy

)k

y

{ury2+{uv'-u'v)}k*

which reduces the problem to the integration of a rational function.Finally, consider

Write uft2 + v' = - ,

2u'tdt = —-9dz9z*

so that t= I—-.— I,

u — (uv' — u'v) zUt2 + V =

uz

Then UkJ[-^2 / (A-)U- /(—j4* ]K I I V'J/ y * KI \ 1 — t i c IX A / 111 i'iltr 1/0)1 71

j ^ AIL 6 /y yjL^c/ «t-y j / \ ^M/ — yu/U — U/ vj /if

_ __ 1 r zk~1dz~~2J yl[(l-v'z){u-(uv'-u'v)z}Y

But the integral is of the form

whose evaluation we considered in the preceding section.The whole integration may therefore be effected.


6. Trigonometric functions. Since

1 — tan2 Ix . 2 tan \xcos a; = -—•—r-f-. sin a; —1 + t a n 2 ^ 1 + t a n 2 ^ '

trigonometric integrals may be reduced by means of thesubstitution M A 1

t = tan \x.

Then cos x = , . ,o, sin a; =

2dt

Hence a rational function of since and cos# is transformed into arational function of t, and may be integrated by the methods of § 2.

For reduction formulae for

I&ii\mxcosnxdx

and similar integrals, see Vol. I, pp. 106-10.The substitution t — tan \x should not be applied blindly. For

example, if the integrand has period TT, the alternative

t = tana;may be better.

Thus, consider the integral

dxrC

Since{sec (x + TT) + cos (x + TT)}2 (sec x + cos a;)2'

we may use the substitution

t = tana;.

(sec2 x +1) 2

dt

Now

This is the integral of a rational function, and may be evaluatedaccording to the usual rules. Alternatively, the substitutiont = -y/2 tan 6 makes the integral trigonometric again, but leads toan easy solution.

EXPONENTIAL TIMES POLYNOMIAL IN X 215

7. Exponential times polynomial in JC. A polynomial is asum of multiples of powers of x, and so, in order to evaluate theintegral ^

\ePxf{x)dx,

where f(x) is a polynomial, we need only consider integrals of thetype r

un = epx xn dx (n ^ 0).

Integrating by parts, we have the relation

1 n Cun = - evxxn evxxn-\ dx

n P PJleading to the recurrence formula

1 nun = - ePxxn — un_xp p

which enables us to express un in terms of

\eJ

= \epxdx = -epx.J P

8. Exponential times polynomial in sines and cosines ofmultiples of JC. Consider the integration of a sum of terms of thetype

epxsinaxxsina2x ... sin amx cos bxxcosb2x... cosbnx,

involving m + n factors in sines and cosines. The use of theformulae such as

2 sin ux cos vx = sin (u + v) x + sin (u — v) x,

2sinw#sin vx = cos (u — v)x — cos (u + v)x,

and so on, enables us to reduce the number of terms in a typicalproduct to m + n— l,m + n — 2,m + n — 3, . . . successively, whileretaining the same type of expression. Ultimately we reach oneor other of the forms

C = \epx cos qxdx,

S= \epx sin qxdx,


whose solution should now be familiar. We find

°p2^2 - ? C 0 S 9*)-

WARNING. The work of this chapter will enable the reader toevaluate (ultimately) any integral that comes within its scope.But there are many functions which cannot be integrated in termsyet known to him. Some of these are very innocent to look at;for example, „ . „

fsm x j C , ,dx, \exdx,

MC 1 --dx

Thorough familiarity with the FORMS of the integrals that can beevaluated should help in the avoidance of these pitfalls.

[For examples on the work of this chapter, see RevisionExamples VIII and IX, pp. 226, 227.]

CHAPTER XIII

INTEGRALS INVOLVING 'INFINITY'

1. 'Infinite' limits of integration. It is often necessary toevaluate an integral such as

!"f(x)dxJa

under conditions where b tends to infinity; we then speak about'the integral from a to infinity'

fJa

f(x)dx.

Similarly we meet the integral

f(x)dx,

or, combining both possibilities,

f(x)dx.rJ-ccOur problem is to discuss what is meant by such integrals, and

to show how to evaluate them. The general theory is difficult, sowe confine ourselves to the simplest cases. We also assume thatthe function f(x) is continuous throughout the range of integration.

We define the integral to infinity

fJa

f(x)dxJa

by means of the relation

f(x)dx= lim f(x)dx.Ja N->coJa

In the cases with which we shall be concerned, the indefiniteintegral F(x) off(x) is considered to be known, so that

rN

= F(N)-F(a).fJa15

218 INTEGRALS INVOLVING 'INFINITY*

We can therefore evaluate the integral to infinity in the form

*f(x)dx = Hm F(N)-F(a).

It may be added that it is often possible to evaluate the definite

integral f(x)dx even when the indefinite integral /(#)cfo

cannot be found. A well-known example is dx whose value

is

ILLUSTRATION 1. To evaluate

rcoxndx.

Consider the integral

xndx

which, by elementary theory, is

[n+l

or n+l n+l

If n + 1 is POSITIVE, then Nn+1 increases indefinitely with N, sothat

hm { -}N^ooin+l n+l)

does not exist.If n+l is NEGATIVE, then Nn+1 tends to zero as N increases

indefinitely, so that

hm

Hence

^ 1 1 } 1-} = - .

l n+l) n+l

r°° ixndx = =- (n< - 1).

h n+l

' INFINITE' LIMITS OF INTEGRATION


/•oo

I xe~xdx.Jo

We know that, on integrating by parts,

xe~xdx = —xe-x+\ l.e~xdx

219

so that

Now

rN r IN

xe~xdx= -(x+l)e-x\

WW-Z£ ^_ _ + . . .

and, for positive values of N, the denominator is certainly greaterthan %N2, so that

Thus (N+l)e-N-+O

as N increases indefinitely. Hence

r xe~xdx = Km {- (N+ l)e~N+ 1}iV->oo

= 0+1

= 1.


dx


where M, N are large positive num-bers. The value of this integral is

tan"1 a;L \-M

or tan*1 N — tan"1 (— M). Fig. 101.

15-2


Care must be exercised in selecting the correct angles from themany-valued inverse tangents. The graph

y = tan"1 a;,

shown in the diagram (Fig. 101), illustrates the 'parallel' curvesalong which y must run, and the values selected must be confinedto one of them. The most natural choice is to work with the curvethrough the origin 0. Then for large negative values of x, thevalue tan~1( — M) just exceeds — \n\ as x increases, tan"1 x risescontinuously through, say, — \TT, — JTT, — £77,0, JTT, £TT and so on,approaching the value \TT for the limit of tan"1 N.

Hence/•oo ,7™

-f~ = lim {tan-1 i f}- lim {tan-1 (-

= 7T.

ILLUSTRATION 4. (Change of variable.) To evaluate

dx

hN dx

rJo

rConsider the integral uN= ., oxo.

Jo (1+x2)2

Make the substitution a; = tan 0,dx = sec20d0.Then sec4

between appropriate limits.When x = 0, we may conveniently take 0 = 0. As a; increases,

0 also increases, assuming a value very near to \TT for large valuesof N. In the limit, we obtain the value ^n. Hence

= Jir.

ILLUSTRATION 5. (Formula of reduction.) To evaluate/•oo

Jm= xme-xdx>Jo

where m is a positive integer, greater than zero.

' I N F I N I T E ' L IMITS OF I N T E G R A T I O N 221

Let Im = xme~xdxJor IN [N

= —xme-x\ +m\ xTn~1e~xdx.L Jo Jo

Nm

Now N™e~N = ~ =

and the denominator, being certainly greater than

Nm+1l(m+l)\9

greatly exceeds the numerator for large values of N, whatever mmay be, so that

Also xme~x = 0

when x = 0, since m > 0.

Hence lim — a^c-* = 0,2V-»ooL JO

and so, as N-+00, Jm-

The formula of reduction enables us to make the value of Jm

depend on that of Jo; for

Jm = mJm_x

= m(m-l)Jm_2

ra(ra-l)...2.U0

m\J0.

Moreover, Jo = I erxdxJo

r o = f °Jo

= lim - e~x

N-*-ao[_ Jo

= lim \-e~*

= 1.


/•oo

Hence xm erx dx = ml/•oo

Jo

EXAMPLES IEvaluate:

1

6. e~x8inxdx.Jo

. r^. 2. f j k 3. fJl ^2 J2 W * Jo

4. r. 5. r. 6.Ji 1 + a:2 J-ool+sc2 Jo

/•oo

7. eax, stating for what values of a the integration is possible.Jo

2, 'Infinite' integrand. It may happen that, in evaluatingthe function

\f(x)dx,Ja

we find that the integrand f(x) tends to infinity—or, indeed, hassome other discontinuity—in the range of integration. Suppose,for example, that f(x) increases without bound near x = c. If c isinside the interval (that is, not equal to a or 6) we 'cut it out' byconsidering the sum

J*c-e fb

f(x)dx+ f(x)dx,a Jc+17

where €,77 are small positive constants, over ranges which justmiss c on either side. We then define the value of the integral to be

rc-e rblim f(x)dx+]im f(x)dx,

supposing that these limits exist.When c is at a or 6 the modification is obvious; only one limiting

value is then required.

ILLUSTRATION 6. To determine the values of n for which theintegral

exists.

' I N F I N I T E ' INTEGRAND 223


I — n

As € tends to zero, the term c1"71 also tends to zero if the exponent1— n is positive; otherwise e1~n increases indefinitely. Hence theintegral exists provided that

n< 1,

and its value is then {1 — 0}\ — nK J

1

l - n

The case n = 1 requires separate treatment. We then have

j ~^= | 1 O gH =1°g1-1°g€>so that (compare p. 4) the integral does not exist.

The required condition is therefore

n< 1.


dx

This integral involves two infinities; the upper limit of integra-tion is infinite, and the integrand is infinite when x = 1. The twophenomena must be kept separate.

We consider the integral

r =Jl+e

dx


Write x-l = t*,

dx = 2tdt.

2tdtThen

- / •

between appropriate limits.Now we have cut out the value x = l,t = 0 by making the

lower limit 1 + e. Thus t is never zero, so that the factor t may becancelled from numerator and denominator of the integrand.Hence

between appropriate limits.Consider the range of variation of t as x increases continuously

from 1 -f e to N. We have

having committed ourselves to the positive square root by puttingyl(x—l) = t in the integrand during the substitution. Hence tincreases continuously from the small value <Je to the large valueyl(N— 1); and as it does so, tan~1(^) increases continuously fromjust above zero to just short of \n. Thus, in the limit,

/ = ITT-0


dxfJaThe denominator in the integrand becomes zero at x = a and at

x = 6, and so we must consider the integral

. rb-* dx

' I N F I N I T E ' INTEGRAND 225

Make the substitution

x = a cos21 + 6 sin21,

dx = 2(6 — a) sin t cos £d£.

Then a; — a = {b — a) sin2 £,

6 — x = (b — a) cos2 £.

Consider the range of integration in the variable t. When x = a,we have asin2£ = 6sin2£, so that sint = 0; and when x = 6, wehave 6 cos2 2 = a cos21, so that cos t = 0. We must select asstarting-point for £ a value for which sin t = 0 (corresponding to# = a), and the obvious value is t = 0. The relation

dx— = 2(6 — a) sin £ cos £U/t

shows that, at any rate for small values of t, the variables x, tincrease together; and this process continues until t = %TT, atwhich point x has the value 6. The range is therefore 0, \TT. Butwe have had to exclude these points themselves because of troublewith the integrand, and so the range must run from just abovezero to just short of \TT\ say from e' to \TT — rj'. Thus

, • 2(6 — a) sin t cos tdtV{(6- a)2 sin21 cos21}'

Moreover the POSITIVE value of ^{(6 — a)2 sin21 cos21) in the interval0, \TT is (6 — a) sin t cos t, so that

^ n*-i 2(6 - a) sin £ cos tdt~ Je' (b — a) sint cost

We have excluded the end-points, so that sin t and cos t are notzero in the range of integration; we may therefore cancel factorsin the numerator and denominator of the integrand, giving

=/; 2dt

Proceeding to the limit as e',7?' tend to zero independently, we

77.


REVISION EXAMPLES VIII

'Advanced* Level

1. Find the following integrals:

r dx

J )Ydx r x2dx r dxl)(a._3)' J (a:-l)(a;-3)' J J{(x-l)(3-x)

O T t + (^2+l) X

2. Integrate

Show that j — . — = -j—z—rjr cos"1 -1 (6 < a).Jo a + bsmx <J(a2-b2) \a/v '

3. Evaluate the integral

Cb___xdx___Ja<j{(x-<*)(b-x)}

by means of the substitution x = a cos2 0 + b sin2 0.Prove that the integrals

fV(l-a;)8dz, fz8(l -z)7dteJo Jo

are equal, and show that their common value is

7!8!

4. Integrate

a

5. Integrate

X

(1 + X2)(

Prove that, when

r

,3

)(X-

a,b

16! '

-2) '

X

:)' <J(x2 + 4:X + l

are positive,

cos2a;^

Xs

77

a2 cos2a; + b2 sin2a; a(a + 6)#

6. Integrate

(2x + 3) 1 x e x

REVISION EXAMPLES VIII 227

7. Integrate

Jltl , flog*)2-8. Find a reduction formula for

and evaluate the integral when n = 2.

9. By integration by parts, show that, if 0 < m < n, and

then / = - m\ x™-1 n_1 {xn( 1 - x)n) dx.

Deduce that 7 = 0.

10. Taking y =

verify that the result of changing the variable from y to x in theintegral #

J V<s/2— :i)

is

f2 rfccDeduce that ^ — ^ ^ — - ^ = loge2.

REVISION EXAMPLES IX

'Scholarship9 Level

cos nxdx1. Show that, if un

then, provided that tt# 1,

Hence, or otherwise, show that, if n is a positive integer,

«„ =77/(3.2").


2. Prove that

Jo a W J + 6 W g = 2 6' Jo l 0 g t a n

sin201ogtan0d0 = JTT.Jo

3. (i) Prove that values of a,b can be chosen so that thesubstitution , ,

at + bX = H T

yields the resultdx f (t+l)dtf dx _ f

J (2x2 - 6x + 5) V(5z2~ 12a + 8) " J

(ii) Find the indefinite integral of

t+l

4. Evaluate J^sin-x)^,, J ^ ^ ) - -

5. Prove that

6. Evaluate

7. Evaluate

J(cos 2x - cos 4a) da = i J6 - J V2 log (2 + V3)-o

f1 da f1 ^4fcr- - , - y - ^ .

r^atan-1^ C*n

Jo (1+a2)2 ' Jo a*oos2 9d0

Find

r°° xdx8. Evaluate

Jo

Find

Jo

-r T C2ncos(n — l)x — cosnx ,9. I i in== aa,11 Jo 1 -cosa

show that In is independent of w, where n is a positive integer.

REVISION EXAMPLES IX 229

Hence evaluate In and prove that

C2n ($m\nx\2 1

Jo \sin^/

10. Evaluate

Ca xdx' CLX I %j i ** \j\jij %\J -

~"2 2\ > T^~- ^ aX'Q> —X ) Jo

11. Evaluate the integrals:

J o p T ^ ^ 1 Jodx

2 cos2 # + 2 cos x sin a; + sin2 x*

dx

x2 dx12. Evaluate

-z2)(l+a;2)2#

00

13. Evaluate

where a, 6, c are positive.

14. Prove that

f00 rf^ aJo #2 + 2a; cos a + 1 sin a

Evaluater°° dx r«I /V«4 _l O/y«2 pr%Q /v -4— 1 I *v

,/0 "^ i •«•*/ vUij LX T^ x /o

15. Prove that, when b > a > 0,

(0<a<7r).

Cn sin 0 ccJo <J{a2 + b2 —

-2ab cos 6) b'

0 cos 0 d# 2a

Make it clear at what points of your proofs you use the condition&>a>0.

16. Evaluate the definite integrals:

230 INTEGKALS INVOLVING 'INFINITY'

17. Evaluate

7 _ f Ipl^ j i -cw (o<8<w)

when (i) 0 < r < 1, (ii) r > 1.Prove that, S being fixed, / tends to one limit as r-> 1 through

values less than 1, and to a different limit as r-> 1 through valuesgreater than 1.

Show, also, that neither limit is equal to the value of I when

r = 1.

18. Show that, by proper choice of a new variable, the integra-tion of any rational function of sin x and cos x can be reduced tothe integration of a rational algebraic function of that variable.

Integratesin a; 1

19. Integrate

Prove that

in(# —a)' sin a; cos a; + sin a; + cos a;—1#

(x2— 1)

dx a.(0 < OL < TT),

p_Ji (iE + cosa)^ 2 — 1) sin ex

rrocr0 l4-cosasin# sina(0 < OL < \TT).

20. Integrate —^— , 2 — ^ (b<a).

Evaluate

Cn x sin xdxl\l a s i n x)

Prove that

I /(sin 2x) sin xdx = y]2\ /(cos 2x) cos xdx.Jo Jo

21. Evaluate the integrals:


22. Find the integrals:

Evaluate '

23. Prove that, if a is positive, the value of the integral

1—acos0rJoJo 1 — 2a cos 0 + a2

is 77 —cot"1 a or —cot"1 a, according as a<\ or a > l , where thevalue of cot"1 is taken between 0, 77.

What is the value of the integral when a = 1 ?

24. Show that, if P, Q are polynomials in s, c (where s = sin 0,c = cos 0), of which Q contains only even powers of both s and c,then

can be expressed as a sum of integrals of the form 12 (5) <fo,

Z?2(c)dc, |2?3(<)(Z£, where £ = tan0 and each of the functions

Rx, R2, R3 is a rational function of its argument.Apply this method to obtain

... f(l + sin 8)(1 + cos 0) f 1(l) 0/> ^0, (u) :—p: dd.w J l + cos20 ' v y J l + sinfl

In case (ii) obtain the integral also by the substitution

x = tan^0,

and reconcile the results obtained by the two methods.

25. Show that the substitutionb( 1\2 d( 1\2

a\ tj c\ t)reduces the integral

\F{x9 J(ax + 6), J{cx + d)} dx,

where F(x, y, z) is a rational function of x> y, z, to the integral of arational function of t.

232 INTEGRALS INVOLVING 'INFINITY'

Hence, or otherwise, find

J (x +-,dx.


dxn J (5 + 4cos#)n

in terms of In_l9 In_2 {n 2), and use it to show that

f1** da;;oJo


dx

Evaluate the integral when n = 1 and when w = 2.

28. If Ip q = siJo

show that (p + q)Iv a =

and evaluate /a_x 5 where a is any positive real number.

29. If In = xne-axcosbxdx, Jn= xne-ax&inbx,Jo Jo

where w is a positive integer and a, 6 are positive, prove that

Show that (a2 + fe2)**^1) In = n! cos (n + 1 )a,

(a2 + 6«)*(»+D Jn = w! sin (n + l)a,

where tan a = bja and 0 < a < \ir.

30. (i) Prove that, if

dx

then (a2 + 62) Mn - 2a«n_1 + «n_2


(ii) Prove that, if n is an integer and m > 1, then

sinm x cos 2nx dx = m(m — 1) sinm~2 x cos 2nx dx.

o Jo

31. Find the reduction formula for

dx

Prove that, if a and ab — h2 are positive and n is a positiveinteger, then

J _oo (ax2 + 2te + 6)w+x " 2n 'ab- h2] _« (ax2

2.4. . .32. Show that

J (aa2 + 2hx + 6)w+* (A2 - ah)**1 J Vi/ ff ; ^'

where g2 = a^)2 + 2hp + 6,

Deduce that, when a,b,h + <J(ab) are positive,

dX l

\a*

Jo (aa:2 + 2te + 6)*

33. Find a reduction formula for the integral

f dx

Prove that


34. (i) Integrate \

(ii) Find a formula of reduction for

j (l+x2)neaxdx.

35. (i) Evaluate

f1 dxI (Q

(ii) If

prove that

?mn — (2w — 1) un_x cos a + (n — 1) wn_2 = 2 sin | a

when n > 2, and evaluate w0, i !.

36. Obtain reduction formulae for the integrals

r»(log^ ( V s i n ^ ,Ji *2 Jo

and evaluate the first integral for any positive integer n.

pi

prove that(2p - g + 2) 7p + (2p -


Jo

38. If /, Ja

verify by differentiation that, when n is a positive integer or zero,

2(n +1) k(k -1) In+2 - (2n + 1) (2* -1) In+1 + 2nln

Show further that I-^ can be integrated by the substitution


Hence, or otherwise, find

dx

39. Show that, if

In

i i*°°

then 2Jn = /n_2, 2


cos


r**_d^_Jo eosna;'

and evaluate the integral for the cases n = 1,2.

41. Prove that, ifdx

then

Hence evaluate the integral


xndx4 =

and use it to evaluate

J xz dx

16-2

APPENDIX

First Steps in Partial Differentiation

The functions which we have considered in this volume havealways involved a single, variable. The work on functions ofseveral variables belongs to a later stage, but it may be convenientto set down one or two of the most elementary properties—mainlydefinitions and first applications.

Consider, as an illustration, the expression

As x, y, z take various values, so also does u. For example,

if

then

if

then

if

then

3=1, y = -2, 2=3,

w = - 7 2 ;

3 = - 1, y = 0, 2 = 3 ,

w = 0;

3 = 2, 2 /= - 2 , 2 = 1 ,

t* = - 1 2 8 ;

and so on.We say that u is then & function of the three independent variables

x,y,z. To denote this functional dependence, we may use then o t a t i o n u(xyz)

The function u no longer has a unique differential coefficient.Each of the variables x,y,z is capable of its own independentvariation, and each of these variations produces a differentialcoefficient of its own. More precisely, we use the notation

— = the differential coefficient of u with respect to x only,

calculated on the assumption that y, z are constant;

= the differential coefficient of u with respect to y only

calculated on the assumption that z9x are constant;

= the differential coefficient of u with respect to z only

calculated on the assumption that x, y are constant.

APPENDIX 237

Thus, if u = x*y*z2,

2 to 3*22 tothen f = 4xyz, = 3xyzt

ox cy oz

As another illustration, suppose that

u = cos (ax + by2)

is a function of the two variables x, y. Then

du . . . o. du _. . . , ov

— = — a sin (ax + by2), — = — 26^/ sm (aa; + 02/2).mi r. .. Bu du duThe functions ^r-, -^-, -7-

&? ^ dzare called the partial differential coefficients of w with respect tox,y,z respectively.

These partial differential coefficients are, in their turn, alsofunctions of the three variables x, y, z, and have their own partialdifferential coefficients. We write

d

d (du\_d2uBz\dz)~ dz2'

The 'mixed* coefficients are a little more awkward. We write

d2udx \dy) ~ dxdy9 dx \dz) ~~ dxdz9

du\ d2u d [du\ d2u3 (du\ _(du\

3 (du\ __Tz\dx]

dydz' dy\fa)~dydx'

d2u d (du\ 82udzdx' ~j

In practice, however, it may be proved that for 'ordinary'functions (a term which we do not attempt to make more precise)interchange of the order of partial differentiation leaves the result

238 APPENDIX

unaltered. Thus 00

d2udy dz dz dy9

d2u _ d2udzdx ~ dxdz'

d2u _ d2udxdy ~ dydx

For example, returning to our function

we have the relations

dxdy ex

dydx

dzdy dz

7)

dzdx dz

dxdz dx

EXAMPLES I

-. . du du d2u d2n d2u d2u +_ __ .Find —9 - - , ^ -o , T - ^ - , a— -> XT for e a c h o f t l i e followingco; dy dx2 dxdy dydx dy2 &

functions:

1. x*yz. 2. xy2. 3. a;2 + ^2.

4. excosy. 5. Iog(#-f2/). 6.

7. x* sin2 y. 8. e^sinz. 9.

10. secx + secy. 11. exsin22y. 12. a e .

13. Prove that, if f(x,y) is any jpolynomial in #,?/, then

a2/ _ a 8 / .

APPENDIX 239

14. Prove that, if f(x,y,z) is any polynomial in x9y,z, then

d2f _ d2fdydz " dzdy'

ILLUSTRATION 1. The 'homogeneous quadratic form*.

Let u = ax2 + by2 + cz2 + 2/?/z + 2gzx + 2hxy.

du~dx~~

du

These expressions are probably familiar from analyticalgeometry.

Then

QUy.o)

R{x,o,o)

Fig. 102.

To show how the partial differential ~ \x^z)coefficients are linked with the idea ofgradient, we use an illustrative example.

Let OXtOY be the axes for a systemof rectangular coordinates in a hori-zontal plane. This is illustrated in thediagram (Fig. 102), where the readermay regard himself as looking 'down7 -upon axes drawn in the usual position. /The straight line OZ is drawn verticallyupwards.

Given a point P in space, let the vertical line through it meetthe plane XOY in Q; draw QR perpendicular to OX. Denote by

x,y,z the lengths OR,RQ,QP respectively; then the triplet x,y,zmay be used as coordinates for the point P in space, just as thepair x, y is used for a point in a plane. If P is the point (x, y, z),then Q is the point (x, y, 0) in the horizontal plane; the coordinatez gives the height of P referred to the plane XO Y as zero level.(Of course, P may be below the plane, in which case z is negative.)

In particular, if x} y, z are connected by the relation

240 APPENDIX

where we assume f(x, y) to be a single-valued function defined foreach pair of values of x, y, then, as x, y (and consequently z) vary,the point Q moves about the plane XOY, while P describes thesurface whose height at any point is equal to the correspondingvalue of the function. We say that the surface represents thefunction f(x, y).

For instance, it is an easy example on the theorem of Pythagorasto show that the function

is represented by the hemisphere of centre 0 and unit radiuslying above the plane XOY.

We now assume, for convenience of language, that OX is due

east and 0 Y due north. We regard the surface

as a hill, and P as the position of a climber on it.

OR R'

Fig. 103.

Suppose that the climber is at the point P (Fig. 103) defined bythe values x, y of the easterly and northerly coordinates, and thathe wishes to climb to the point P' defined by x + h, y + k. The cruxof the difference between functions of one variable and functions

APPENDIX 241

of two lies in the fact that, whereas for one variable motion alongthe CURVE representing the function is defined all the way, fortwo variables the SURFACE may be traversed by an innumerablechoice of paths. Moreover, each way of leaving P will demand agradient all of its own. The partial differential coefficients are thebases of the mathematical expressions for such gradients corres-ponding to the various paths.

From the mathematical point of view, the obvious way to passfrom P to P' is firstly to move the distance h easterly, to B, andthen to move the distance k northerly. The climber thus describesin succession the two arcs PB, BP' shown in the diagram.

Now suppose that P' is very close to P, so that the arcs PB, BP'are very small. The arc PB may be regarded as almost straight,so that the 'rise' between P and B is proportional to the length h,say

Sz(easterly) = ah.

Similarly BP' is almost straight, so that the 'rise' between B and P'is proportional to k, say

Sz(northerly) = j8&.

If 8z is the total 'rise' between P, P' , then

Sz = Sz(easterly) + Sz(northerly)

= ah + fSJc.

If the climber had gone first northerly and then easterly,following the course PD9 DP' in the diagram, then, for distancesso small that the paths may be regarded as straight, PBP'D isapproximately a parallelogram, and so, once again,

Sz = ah + fik

for the SAME values of a, j8.Two simple observations complete the illustration. Geometri-

cally, a, j8 are the gradients of those curves which are the sectionsof the hill in the easterly and northerly directions respectively.Analytically, we see by putting k = 0 that a is the ratio Sz H- hcalculated on the assumption that y is constant; thus

242 APPENDIX

evaluated at P. Similarly we see by putting h = 0 that jS is theratio 8z-7-k calculated on the assumption that x is constant; thus

R=8zP By

calculated at P.

Hence the partial differential coefficients —, — are identified as theex cy

gradients of the surface z = f(x9 y)

in the x- and y-directions respectively.

EXAMPLES II

1. Show that the function

with the z-axis measured vertically upwards, is represented by ahemisphere.

Prove also that the gradients in the x- and ^/-directions at thepoint (x,y,z) are in the ratio(# — 1) : y, and that these gradientsare equal only for points on a certain vertical diametral plane.

2. Prove that the function

with the z-axis measured vertically upwards, is represented by a'bowl-shaped' surface whose horizontal sections are ellipses ofeccentricity £^/3.

Prove that the gradient in the ^-direction at the point (1,2,17)is 2, and that the gradient in the ^-direction at the point (3,1,13)is 8.

du dû dû3. Find —, -^9 7—- for each of the following functions:

(i) eax sin (by + cz), (ii) xyze~x\ (iii) (y2 + z2)log(a

4. Prove that, if

u = ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy,

d2u d2u d2u ând if "i~2+^~2+"^2" = °>

dx2 dy2 dz2

then a + b + c = 0.

5. Prove that, if

then

Deduce that

Prove also that

APPENDIX

r = yj(x2 + y2 + z

dr x d2r 1

j j —

2 ) .

X2

2r

243

Finally, there is a point of notation which the reader maymeet in physical applications. Consider, as an illustration, thetransformation n . n

x = r cos 9, y = r sin 8between the Cartesian and the polar coordinates of a point. Fourvariables are involved, of which two are independent—forexample, r and 6, When we form the partial differential coeffi-

cient —, we naturally have in mind that 6 is the other independent

variable, and the relation

x = r cos 6

., 8x Q xthus gives — = cos 0 = - .

But it is possible to express x in terms of r and y, in the form

x = J(r2-y2),

, . , 8x r rand then — = -r—^ rr = - .

dr J(r 2-y2) xThese two values are quite different; they are, indeed, calculatedunder the quite different hypotheses 6 = constant, y = constantrespectively.

To make sure what is intended, we often use the notation

to denote the partial differential coefficient of x with respect to rwhen 6 is the other independent variable (being kept constant

244 APPENDIX

during differentiation). With such notation, the two formulaegiven above may be expressed in the form

cx

The following examples should serve to make the notation clear.

EXAMPLES IIIGiven the relations x = r cos 9,y = r sin 9, establish the following

formulae:

1 fk\ =?.dr)0 r

2 d~y\ =¥.' dr)e r

dx

5. | r

9. ^ )=- r . 10.

.2. «. i .11. «- 1.

ANSWERS TO EXAMPLES

CHAPTER VII

Examples I:

+ l). 2. ilog(2z+l). 3. - J log(2-3s) .

4. lx* + logx. 5. | l o g ~ . 6.

X ~T i.

7. log 2. 8. log 2. 9. Jlogf.

10. JlogS. 11. Jlogf. 12. JlogJ^.13. . 14. 2cosec2x. 15. —coto;.

3 + 216. x + 2x\ogx. 17. xn~1-\-nxn-1\ogx. 18. 2xj(l+x2).

19. a; log a;-a;. 20. ^(loga:)2. 21. log sin x.

22. ia-logx-lx-. 23.

(1 — oos ^c\^ — 1 . 25. log (x2 + 5a; +12).

1 + COS X)

26. log {x2 -3x + 7).

27. | tan-1 ^ ^ - ^ W log (a;2-2a;+17).

28. - 12 tan-1 (x + 3) + log (a;2 + 6a; +10).

29. 9tan-1 fcp)+ flog{x2-8a; + 25).

30. ~

Examples II:

ydx 1+a; 1—a;* ydx 1+a;2

3. \*-l + iaAx+«*. 4. ! £ - ?1 2 3 d2/aa; a; 1 —2a;3 2/da; a; 1+a ; 1+a;4

246 ANSWERS TO EXAMPLES

^ Idy I L 3 1 n Idy 4x5. --f- = - + cotx-- + - . 6. --/- =

d l+ 1 df = + c o t x + . 6. / = , sydx x l+z 1—z ydx l + x* z

m Idy 4 37. --r- = — 8 cosec 4#.

ydx x l—x

1 dy 2 sin a; 1 + 2xydx 1 + cos x

\dy _ 1 , 4 t 9 16y dx 1 — X 1 + 2X

Examples III:

3. l o g ^ ^ . 4.

(or 4-1)2

5. |tan-1 x + 1 logf^-fr. 6.

*g(^)2(^Ty 8- -

9. A A

10. l o g C x - ^ - ^ - ^ . 11.

12. ^ t a n - H ^ - ^ l o g ) ^ T ^ +

13. ilog

(a;2+ 4) 25(x-l)

(«-!)(»-3) '(z-2)4

4-14.

15.

16.

17. ^ - ^ -


19. a:-

20.

21.

24. ^ -

r t a n " 1

Examples IV:

1. 2e2x.

4. ex cos a; — ex sin ic.

7. e2x — e~2x.

g^fl + 2x x2)9 ^ ^

2.

5. e s i n x cos#

8.

9 - 7 i 9 ^ - 1 0 -

(1—a;2)2

11. e3x(3cos4a; — 4 sin 4a;). 12.

13. %e2x. 14. -%e-5x.

16. -\e~x\ 17. (a;-l)ex.

19. e8111*. 20. -|es in2x.

22.

23.

24.

3.

6. — e~x(l—a:)2.

15. c^.

18. 6x(a;2

21. etanx-

Examples V:

1. aln = a^c^-n/^j.

2. (a2 + n2)/„ = eaxsin71"1 x(asinx-ncosx) + n(n-l)In_2.

3. (a2 + 62^2)7n

= eax cos71"16a;(a cos bx + nb sin 6z) + n(n — l)b2ln_2.

4. 120-44e. 5. A(41c*»-24). 6. - f ( l + e»).


REVISION EXAMPLES III

1. (i) (l + x^Q + nlogll + x)}. (ii) - + log(l+s).7b

2. ^ _ 4 ^ = 0 .dx3 dx

4. 2cot#,— 2x

5. 2 sin x — 3 sin3 x, 2aeaxcosaa;,

„ 1 . - * -2x

7. - I ,

)i' {]i)secx' { i i i

1-a; 2x

— 2

12/ 3\

tana; —cot a;.

1

*' 1 + sin2s#

16. Velocity en9 acceleration —en.

18. Tangent: xt&nt + y — asint = 0.

Normal: xco8t — ysint — acos2t = 0.

2a;2


aft sin 0 3V3

h — a cos 0 2

21. a; = 0, y = 1, minimum; 2 8 . (<Ja + y

x = I,?/ = 2, neither;

a; = — 2,2/ = 29, maximum.

30. Area: 2 + | co t 0 + 2tan0; angle: tan-1

31. ( — 1 , - 1 ) minimum; (1,1) maximum; 24a; —25?/+ 8 = 0.

34. (i) 2log(2z + 3 ) - l o g ( x - 1).

(ii) Jcos3# — cos a;.

(iii) a;2 sin a; + 2a; cos a; — 2 sin a;.

35. (i) £a;

(ii)

(iii)

36. (i) J sec3 x — sec x.

(ii) ^(l+aj^tan-^-la;.

(iii) log (4a; - 1) - log (a; + 1).

37. (i) 2 log (2a; - 1 ) - 2 log {x + 2).

(ii) J sin3 a; — J sin5 a;.

38. (i) 3 log (x - 3 ) - f log (3a; - 2 ) .

(ii) sin x — x cos a;.

(iii) fa4 8 + Ja4 sin 2 6 + ^-a4 sin 40, where a; = a sin 0.

39. — sin a; — coseca;, £ex(sina; — cos a;), 2-24.

40. ^ .3 2

21og2-f.

42. 1.

43. a;2 sin a; + 2a; cos x — 2 sin x, | log (1 + 3a;) - log (1 - 3a;).17


44. |(1 +x*)l, ixJ(l+

or45. Ja;^(l—rc^ + Jsin-1^, ^sin^a;)2 , -T— gr, e*cosa;.

46. 21og(l+^»), — cos a; + § cos3 x — J cos5 x%

x + 37 log (x - 6) - 26 log (a; - 5).

— 1

a: sin"1 a; + ^(1 — a:2).

48. x — log(l— x)9 — cosa; + f cos3# — |cos5a;, ex(x— 1).

50. tan"1 a;, J tan3 a; — tan x + #, 1.x

51. a; —log (a:+2), a; tan x + log cos x, J.

52. log (2a:2 -x - 3 ) , J sin3 a; - £ sin5 a;, 7r~2.

53. « - ~ ^ , A, Ilog2.

54. JTT, 16 log 2 — ^ , ^773-127T + 24.

55. (i) ^ + | l o g ( l + V 2 ) - (2) 1, K4V2-5) .

56.

57.

fin r\n. (m + n) sinm a: cosn a: rfa; = (n— 1) sinm a: cosn~2 x dx, ^ , 0.

Jo Jo

1Ya ^

58. (i) P ( l +x*)«+*dx = 2n+*+(2n+l) F{1 + z*)n-*<lx9Jo Jo67

24 J\


CHAPTER VIII

Examples II:

v.2 /y.4 /y.2n

3. l -

Examples V:

o 1 ( ^ ^

3. 2o;-i(2a;)2 + i(2i

K "I __ /y» _l 1*2 __ sy3 1

[/uXf ( i

2!

XV. "~~ OJU — ^yOJUJ —— • ,

Examples VI:

1. 2-005.

3. 2-0017.

5. 1-9996.

to)3

c)3+...-

.. + ( - 1

+ (^+l]

(3x)n

I ^ ' (<)~\n i1 n (-x) I •

nxn(2n+l)l

i l 3 5 ^ _ 3 )

2. 2-995.

4. 2-999.

6. 0-3328.17-2


Examples VII:

1. 1680a;4 sin x + 1344a;5 cos x - 336a;6 sin x - 32a:7 cos x + x8 sin x.

2. a;2sina; — 8a;cosa; — 12sin#.

3. e2*(122cos3a; + 597siii3a;).

4. 34e3*(9a;3 + 54a:2 + 90a; + 40).

5. x3 cos {\UTT + x} + 3na;2 cos {(n — 1) |TT + a;}

+ 3w(7i — 1) a; cos {(TI — 2) \TT + x}

+ n(n — 1) (TI - 2) cos {(w - 3) JT7 + a;}.

6. 2n~3e2a;{8a;3+127ia;2 + 6n (n - l ) a ; + n ( n - l ) ( n - 2 ) } .

7. 25.10.9.8(1 - 2a;)5 ( - 132a;2 + 55a; - 5 ) .

2 36 12'8. -—r.—:(3a;4-l)4(819a;2 + 312a; + 28).

o!

Examples VIII:

1 a;3 1 3 a:5 1 3 5 a;71- a? + 2 ' "3 + 2 - i ' 5 " + 2 ' I -6-T + """ '

2. x-\x* + \x*-\x'J + ....

REVISION EXAMPLES IV

1. (i) ax(\ ^

ab, a262.

- 1 - 1 2(-a;2-q;+2)

x = 1 maximum, x = — 2 minimum.

4. secmo;tann~1a;{7i+(m4-W')tan2a;}.

840 tan4 a; + 640 tan2 x + 56.

ANSWEBS TO EXAMPLES 253

6. 4s in3xcosx, 12sin2s—16sin4s,

24 sin x cos x — 64 sin3 x cos x, 256 sin4 x — 240 sin2 x + 24,

x = 0 minimum, x — \TT maximum.

# + -1 11 — - 1 , nsinx tan2 o;(cos x + sec a?)*1"1,

li when «> 2, 2 + \ when » - 2,

22a:—r when TI = 1.

a;3

10. (ii)

1 2nn\ n\777 9 t a n x, — — 1 — — -

12 m

(ii) sin(a; + |n7r), a; sin (a; + ^ T T ) + TI sin {a;

13. (i) - 2 cos a;.

.... cos3a; —sin3a; 1— x+2x2 ex — e(22)

(cosx + sinx)2'

3<l(\—x2y (1+s2)3

15. (i) ( - 1 ) * 1 " ' V ' . (ii) 2nsin(2s + 4nw).

(iii) 2n e2x I sin 2a; + n sin (2s + fa)

k= 0,1,2.

~~ », 2tanssec2a;et a n t a ; , ""

Q

17. sec2setajaa;, ^ 7 -

254 ANSWEBS TO EXAMPLES

18. Velocity, \ak{\ — 2sinAtf); acceleration, — ak2 cos Jet.

At rest when t = —r9 x = — (5TT —On) J.J!

Total distance, \ (n + 6 ^/3).o

19. Velocity, e'^2; acceleration, 2e'.

20. (i) v2 = j92(a2si

(ii) /2 = p4(a2co

21. Velocity, £2 cos t — U sin t — 6 cos £;

Minimum at J = ( 4 i + 1)|TT; Maximum at t = (4i

22. Velocity, — a p ( s i n ^ + sin2_pf);

Acceleration, — ap2(cos _p + 2 cos 2pt);

3a a 3a

24. 3x-y = Oat (1,3); 5a; + y =» 0 at (-1,5) .

25. a?-2y = 0.

27. ooB

29. Maximum. 32. (i) 0-857. (ii) 30-2.

33. (i) 2-004. (ii) 0-515. 34. 8-03.

O*J» X <J*J<u» o c < •€/ 2 ^^ 6 ^ ^ 1 2 ™T" • • •

r r U . "~~ "2*1/ — 1 2 * ^ •

43 Xt/^"^"2^ "~~ ^2iIC ~~ 1 ~" 7i^ i2/^"^"A^ - j - (% —" 1 — 2?2<} | / ^ ) J - 77,1/^—1) rrs 0

44. cx = 6, c2 = 2a6, c3 = 3a2 6 - 6 3 .

45. y = a, j / ' = 0, y" = cot a, 45-028°.

46. l + 2a;+2a;2 + fa3 + ..., 0-9930.

47. 0-06285.


49. y' = 1, y" = 1, y"' = 2, y" = 3.

50. / " = 2 + 8/ + 6/,

y* = 16y + 40/ + 24/,

2/v = 16+ 136/+ 240/+ 120/,

52. --|e-*f,

53. I - JTT .

54. Jo;2 log x — Jo;2, log tan \x, \ log I 1 — \ tan"1 x.

55. sin a;-£ sin3 a;, lo

x {(log a;)3 - 3(log a;)2 + 6 log x - 6}, -j- tan"1 (^5 tan %x).

56. 12a; + 8 sin 2a; + sin 4a;, a;3 {9(log a;)2 - 6 log x + 2},

log ( ~ W 2 tan-ia; -2a;, =-^-.& \ l - a ; / loga

57. 4 log a; - 2 log (1 + x)- log (1+ #2) - 2 tan"1 x9

21og2-l , i t an -H-

x tan x + log cos x9 % tan3 x — tan x + x.

59.-———-rs, ^tan2o; — xt (1 + x2) tan"1 a; — x, tan-1(sina;).6(1 — ox)

60. A, frr, | (e2"-l).

61. (i) - J - + log (^-A, log (a;2 + 2a; + 2) + tan"1 (x + 1).A + # \ •»• + xj

262. ^ sin 3a; - ^ sin3 3a;, -y- tan"1 ( /5 tan Jo;),

— a;3 cos x + 3a;2 sin a; + 6a; cos x — 6 sin a;,


64. $ 7ra2, (fa, 0). 65. x sec g + y - sin £ - £ sec £ = 0.

69. 2/ = 2 + 3 x - ^ ; (1,4), ( - 1 , 0 ) ; ft.

71. £a; — J sin 2x, ^ cos3 x — cos #, a;2 sin a; + 2a; cos a; — 2sina;.

72. fa2, f77a2(2V2-l). 73. V, (¥,-§)•

75. ±M j ^ 2 . 77.

i - >)••

80. |o2, ( K f a ) ; (|a,0),

83. /x = 1 , /2 = Jw, J3 = | , 74 =

84- V(^r)- 85>

86. - 1 and 0, 0 and 1, 2 and 3, 2-88.

87. 2-426. 88. -1-844.

89. ^sin^WTT. 90. a;-|a;2 + ^ x 3 -I

CHAPTER IX

Examples I:

1. 3cosh3a\ 2. 4 cosh 2a; sinh 2a;.

3. tanh x + a; sech2 x. 4. 4 cosh (2a; + 1) sinh (2a; + 1).

5. cosh x cos a; — sinh x sin a;.

6. 2 sech a; sin a; cos x — sech a; tanh x sin2 a;.

7. 3(1 + a;)2 cosh3 3a; + 9(1 + a;)3 cosh2 3a; sinh 3a;.

8. 2a; tanh2 4a; + 8a;2 tanh 4a; sech2 4a;.

9. cotha;. 10. 1.

11. cosha;e8inhaj. 12. e"tanha;(l-a;sech2a;).

• H ) 4 - -fan)*-

'(!<») zfi-

= s^-3^ #I

• 0 I 'feV L i . ^ • o „ . g

. 8

*Q • — = ^ '^

1+ I -

'f *8T+ I

Z '——£-ii- + a;I_qsoo 'iX

• II

— x '\z

-XZqso°x-XZ W ^ | *6I

'XT + XZ^\ 'LI

sauivxa 01


Examples III:

2. x = - (a2-b2) cos31, y = -ha2-b2)sin31.a o

3. x = 2asec3£, 2/ = —2atan3£.

REVISION EXAMPLES V

1. Tangent, xsinifj — ycosI/J — 2a\\ssini/r = 0;

normal, # cos ip + ysimfj— 2ai/j cos i/r — 2a sin if; = 0.

4. ( l , i ) .

6. ^ = — 2asin 2t — 2a sin f, y = 2a cos 2i -f 2a cos ;

speed, 4a cos |^ (numerical value).

13. 2Tra(a;sinh — a cosh — + a I. 14. 8a.\ a a

15. ¥• 16. 1, (0,2).

17. (6«/2a). 18. 1.

20. 1*5. 21. 1??.6 4

22. V2(l + 2a: + 2^2)^, (-1,-1).

24. ^ira2. 26. -1^/3 at a: =

REVISION EXAMPLES VI

6. fM(x) = K(z) + 4fl/;(aj) + 2(1 + 2x2)fn(x).


13. f(x), f'(x)f f'{x) continuous everywhere in the range;f'"(x) has a discontinuity at x = 1;Maximum.

16. £. 17. 0-55.

19.

2-nexv3cos I a; —-^j + arbitrary polynomial of degree n— 1.

21. 1-8, 4-5 (radians).

23. rfc =

Of? __ iM- 27 ±~777^

dx d2y d2x dy

TdxV '

flti flu \H'y* fl^ii rlt fff^ i

'dx = ~dildl' dx1^ TobY\dt)

d2x dy dx d2ydx dx Idy d2x dt2 dt dt dt2

dy dtj dt' dy2 idy

28. %{b + J(3a2 + b2)}. 29. Equality when ex~x = y.

30. (ii) Minimum. 35. \IT.

37. §77.

38. x = — Y|TT, y = —0-65, maximum;

x = — IQ-TT, y = — 1*19, minimum;

x = — |TT, y = 0, inflexion;

x = -23b-7r, ?/ = 1-19, maximum;

x = J^TT, y = 0*65, minimum;

x = ^77, i/ = 0-73, maximum;

x = il-tr, y = 0-44, minimum.18-2


41. (i) Maximum, y = 2—1); minimum, y = —1(^2+ 1).

(ii) Inflexions, (1,0), {-2 + ^/3, -i(3-,/3)},

a n d { - 2 - V 3 , -

drj d2rj

d£_ d2y __ d^' dx dv^ ' dx* / drj . \3#

1 — -7^ tan a cos a — -~ sm a

50. a; = 0, maximum, /c = —16;

x = f, minimum, * = V^-

51. 3a cos £ sin t (numerical value of).

52. (0,0), (2,2).

53. l, 2^2.

56. Minimum.

(a cos bx + b sin bx),

\x2 — \x <J(x2 — 1) + \ cosh"1 x.

60. (b) —, A an arbitrary constant.x

61. a2(a + 3sina), where cos a = —% ( TT < a < TT).

62. (i) Y5 log sinx — YsX cos #(3 cosec5 x + 4 cosec3 a; + 8 cosec #)

— - o cosec4 X — TS cosec2 x.

... 65 /a + bx\ 64 63 62 6 1(ii) -^v a6a6 6 \ x I a5x 2a*x2 3a3 x3 4a2a;4 5az5'

63. 4a«

i.NSWEES TO EXAMPLES 261

64. (i) n = 4k,X=l;n = 4:k+l,X = n;

n= 4&+2.A = - 1 ; n = 4& + 3, A = -n.

(ii) (l + inW.

65. (tan-1 x)2 - 2x tan- ' x + x tan-1 x log (1 + xz)+ log(l+a;2)-£{log(l + a;2)}2.

68. <£(*) = (l+x)(3 + x).

71. ^ = | , B = f, C = f, D = f.

72. ^ = ^,J5=iC = -^,i)

76.

3

77. cos3"1 x{psinP-1 x-(p + q + k2+pk2)sin^+1 x

79. Volume 5?r2a3, area

81. x = a(cos ^ cos 3£ + 3 sin t sin 3^),y = a(cos ^ sin 3£ — 3 sin t cos 3^).

85. P ( -8a£ 3 , - 6 a O ;Normal, 4 ^ + 2/+6a« + 32a^5 = 0;Inflexion at origin.

86. Normal, 2x + Sty - Sat* - 2at2 = 0;Centre of curvature, (— f at* — at2,4at3 + %at);Radius

87. Envelope, x = -f'(t), y =f(t)-tf'(t);

P=f"(t)(l + t2)K

88. Envelope, x = a sin 2(3 - 2 sin21),

y — a cos t(S — 2 cos21).

89. Tangent, x sin t — y cos £ + cos 2t = 0;Normal, a; cos £ + y sin t — 2 sin 2£ = 0.


90. Tangent, t(3 + t2)x-2y-t* = 0;Normal, 2x + t(2 + t2)y-t2(2 + t2) = 0.

92. 7r(ab-cdk). 93.

94. ± . 95.

98. 2-\<n. 102. 2csinh-; 7rc(2a + csinh —c \ c

103. ^?ra2. 105. Volume, 2>rr2a2h; area

4-/2 2 «2

106. -*-a, — log(l+V2). 107. / + "7T TTd O/

1 O 9 ' | a -

CHAPTER XI

Examples I:

1. 1 and 3; 2 + ^/3; 2±i.

2. - 5 and ~ 3 ; - 4 + ^/5; - 4 ± 2 i .

3. - 1 and 3; l±^\ 1 ± 3i.

Examples II:

1. 10 + 3*. 2. - 8 + 8J. 3. 34 + 22£. 4. - 1 6 - 3 i . 5. 2 + 7i.

6. 0 + 0i. 7. 0 + 2i. 8. 2+ Hi. 9. - 2 - 1 6 i . 10. -10 + 0i.

Examples III:

1. - 7 + 22i. 2.

3. 1-L 4.

5. - 3 + 4i. 6.

7. 10. 8. -46 + 9i

Examples IV:

1. i. 2. | | + -2V. 3. f-fi.

4. iV-fK 5. -ff + ffi. 6. oosO + imnO.


Examples V:

1. 5-2i, -

2. - 7 + 9i, - l - 5 i , - 2 - 3 4 i , H + i K

3. 4 + 2i, 4 -2 i , 8f, -2i. 4. 3 + i, 3 + 3i, 2 -3 i , - 2 + 3£.

5. 4, «, 13, - A + ift. 6. - 6 , 8i, 25,-A- —if f.

9. -Jf. 10. -

11. - 2 ± 3 i . 12. l±i.

13. -3±i . 14. ±|i.

15. 4±3i. 16. -2±t.

Examples VII:

3. 2 , -30°; 5, 53° 7'; 13,112° 36'; 3,0°; 10,-53° 7'; 2 ,-90°.

5. Straight line, 4x+ 10^-21 = 0.

Examples VIII:

1. (a) (3,2); (6) (2,1); (c) (4,7).

2. (i) (a) (1,0); (b) (3 , -5) ; (c) (3 , -5) .

(ii) (a) (2 , -1) ; (6) (2,1); (c) (5,0).

Examples IX:

1. (i) -0-5 + 0-866i.

(ii) ± (0-866 + 0-5i).

(iii) 0-940+ 0-342i, - 0-766 + 0-643i, - 0-174 -0-985i.

(iv) ± (0-966+ 0-259i), ± (0-259-0-966i).


2. (i) 7 + 24i.

(ii) ±(2-121+ 0-707*).

(iii) 1-671 + 0-364i, -1-151 + 1-265*", -0-520-l-629i.

(iv) ± (1-476+ 0-239i), ± (0-239-1-476*).

3. (i) 119-120i.

(ii) ± (3-535 + 0-708*).

(iii) 2-331+ 0-308i, -1-432+1-865*, -0-899-2-173*.

(iv) ±(1-890 + 0-187*), ±(0-187-1-890*).

Examples X:

5TT

3.

4. 2(cos^(12fcl) + isin^( lo 18

5. (

6.

„ ... , 2TT . . 2TT 4TT . . 4TT7. (I) 1, cos —±*sin—, cos —±*sin-—.5 5 5 5

(ii) ±1, ±i(l±t'V3)-

Examples XI:1. e-(in+2kn) {cog (1 l o g 2) + { gin (J log 2)}.

2. - 2e(2**-*ir) {sin (J log 2) + i cos (J log 2)}.

3# e-(&n+kn) | c o s ( | iOg 2) + i sin ( | log 2)}.

4. - 8e-(4A;7r-*;r) {cos (log 4) + i sin (log 4)}.

5. 26-(A:7r-^) {cos (£ log 2 - JTT) + i sin ( | log 2 - |TT)}.

6. - 4e-(/"r+**) {cos (J log 2) + i sin (J log 2)}.


Examples XII:

1 — cos 0 + cos 7i0 — cos (n + 1) 9' 2 (1 -cos 0) #

gin 0 4 - f -

1 + 2x cos 6 + x2

3. cosx + x sinx.

4. ^e^sina; — cos a;).

6. ^xe2x(2 sin x — cos #) + e2 ; r(4 cos a: — 3"sin x).

7. | sin 3# — \x cos 3a;.

8. -£sxe-*x(4: cos 3a: - 3 sin 3a;) --^>se~*x0 c o s %x- 2 4 s i n 3^)«

REVISION EXAMPLES VII

sin a cos a — sinh6cosh6~~ cos2a + sinh26' cos2 a + sinh2 b *

5. — tan-—-(4&— 1), fc = 1, . . . ,n .4n

7. cos2 a; + sinh2 y.

= (a:2 + yj) a;2 + (a:2 + yj)x1- (xt + a:2)(4 + 2/?) (4 + y\) ~ 2(^x^2 " 2 ) + 1#

(a:2 + y\) {x% + y\) - 2(Xlx2-

10. x — iy9r(cos6 — isin6); ±l±3i.

11. zx = 2(c

16. (i) ( - 1,0), (3,0);

(ii) circle, centre (2,0), radius 3;

(ill) ellipse, foci (1,0), (2,0), eccentricity | .

17. 1, co, co2, o>3, co4 where o) = cosfTr + isinf n-;

x = J(l + i cot ££77), k = 1,2, 3,4.

19. (1,^/3), distance 2.


20. «8 = 7 + 273 + 1(4 + 373) or 7 -273 + i(4-373).Vertices (4 + 4 sin \1CTT + 6 cos \kiry 6 + 6 sin \JCTT — 4 cos \1CTT),

21. cc

23. (<j3 + i)(a + ib) and a + ib subtend an angle ^TT at the origin,and the distance of (<j3 + i)(a + ib) from the origin istwice that of a + ib from the origin. C is at ±273 + 2%.

24. ±(3 — 2i); l + i , 72cosY|TT + i72sinJ^TT,72 cos jf 77 + i 72 sin yf TT.

26. -2300-2100*.

27. (i) (1,1), (1 , -1) ,

(ii) all points with x^ 0,

(iii) interior of circle, centre ( — f, 0), radius f,

(iv) interior of ellipse, foci (± 1,0), eccentricity £.

29. ( i ) i f - I ; i ,

" ; 1,

cos|cxcos|(a —j3) cos|cxsin^(a —j8)V / _ _ _ 1 Q > ^ _ „ 1 O >

COS i

COS i* ^

30. tV, ^ e

32. l-67-0-36i, -0-52+l-63i, -1-15-1-27*.

33. i

—i (sin 19 - 7 sin 59 + 21 sin 30 - 35 sin 9).

37. Circle, centre (5, - 1 ) , radius 3; 8, 2.

V(10 + 6y)3 8 ' "| 10-62/1*


39. Amplitude increases from — n to IT.

40. 2*cos|7T, — 2*COS|T7, — 2* cos f jr.

cos x cosh y + i sin a; sinh ycos2 a; + sinh2 y

42.

CHAPTER XIII

Examples I:

1. 1. 2. ^/2. 3. 2. 4. JTT.

5. |TT. 6. -|. 7. - I / a (a<0) .

REVISION EXAMPLES VIII

1. | l o g ^ | ; r + J logi^-9 ; coS-i(2-x).*C — x JO — X

2. l ^ " " 3 ^X —

3.

4.| (x + 6)s + | ( a — 6) (x + 6)*; § cos3# — cosx — \ cos5x.

N/(z2 + 4a: + 5) - 2 log {# + 2 + V(z2 + 4a; + 5)};

£(a2 + a;2) - a2 log (a2 + x2) - ^a4/(a2

a;{(log x)2 - 2 log x + 2}.

REVISION EXAMPLES IX

_ .... 1 (3 <"> V a I


JLi^2 8

8. ITT; -X/V(X 2 -1) ; ^4-|(2x3-3x)sin2x-^-(2a;2-I)cos2a;9. 4 = 2w.

11. 1 - J T T ; fw;

12. | ( | - l o g 3 ) .

10 y Z2a(a2-62)(a2-c2)'

14. J77 sec \a, \n sec |a.

16. fc|

17. 1

2tan-1 ( ( | ^ ) tan|] + S, r< 1,

- 2 tan-1 {£t | ) tan|) + 8, r > 1.

Limits iT + S , r < l ; —7r + S , r > l ; 7 = S when r = 1.

18. a; cos a + sin a log sin (x — a);

1 1 * 1 J. , 2 t + l , , x .\log-—-—j-tan-1—j—-, where t = tan|x.

19. lo

20

21.

1 Jx- ji=W) Jog(

ANSWEKS TO EXAMPLES 269

23. ITT.

2tan0 — sec#= — - — ^ , differing by a constant.

M _ 2 = - ( 5 +

27. (7i-2)^n_2~(72,-l)^n = 0, n>2;

31. If the given integral is un+1, then

33.

un-2

a

35. 77/ (a2 — 1); u0 = log (1 + cosec | a ) , % = u0 cos a — 1 + 2 sin Jo

36. un = nun_l9 un = n\\ un + un_2 =

38. - - ^ t a n - 1 ! ^ - ) .

40. (n-\

41.3(2a;2+l)3

42. Formula: see p. 208.

4) - i log {x + 1 + J(x* + 2x + 2)} +1 (2X* - 5x + 7) 4(x* + 2x + 2).


APPENDIX

Examples I:

1. 4:X3ys, 3x*y2, \2x2yz, 12xzy2, I2xzy2, 6x*y.

2. y2, 2xyy 0, 2y, 2y, 2x.

3. 2a:, 2y, 2, 0, 0, 2.

4. excosyf —exsinyy excosy, —exsiny, —exsiny, —excosy.

I l l 1 1 15.

7. 3a;2 sin21/, 2#3 sin i/ cos y, 6xsin22/, 6a:2 sin y cos y,6a;2 sin y cos i/, 2o;3(cos2i/ —sin2i/).

8. yexysmx + exycosx,xexv8inxf

y2 exy sin a; + 2yexy cos a; — exy sin a;,eX2/ sin x + #2/eXI/ sin a; + xexy cos a:,eX2/ sin x + a^e^ sin x + a:eX2/ cos a:, a?2 e^ sin a:.

9. tan y, j ^ , 0, r ^ , y - ^ , - ( 1 + y a ) s -

10. sec a; tan a;, sec y tan y, sec x tan2 a; + sec3 xf 0, 0,sec i/ tan2 y 4- sec31/.

11. exsin22y9 4ex sin 2y cos 2y, exsin22y, 4:exsin2ycos2y,

4e* sin 2y cos 2?/, Sex cos2 2y - 8e^ sin2 2y.

12. c^, a:e , 0, ey, ey, xey.

Examples II:

3. (i) aeax sin (by + cz), — b2eax sin (by+ cz), aceax cos (by+ cz);

(ii) yze~x%-2x2yze~x\ 0, ye~x*

INDEX

Approximations, 55-62Area of closed curve, 128Argand diagram, 168, 171-4

Cauchy's form of remainder, 47Centre of gravity of semicircular

arc, 133Complex numbers, 158—91

Argand diagram, 168, 171-4argument, 169conjugate complex numbers, 161De Moivre's theorem, 174differentiation and integration,

186logarithm, 183modulus, 162, 169number-pair, 166powers, 178pure imaginary, 161quotient, 164roots of unity, 177sine and cosine, 184sum, difference, product, 163

Convergence of series, 42Coordinates, intrinsic, 111

pedal, 111Cosh a;, 84Curvature, 113

circle of, 119Newton's formula, 117

Curves, 100-33angle 'behind' radius vector, 109gradient angle, 107length, 103parametric form, 116sense of description, 101

De Moivre's theorem, 174Differentiation, logarithmic, 10

partial, 236

Envelopes, 123Evolutes, 126Expansion in series, 41-3Exponential function, 18-20, 21-7

Formula of reduction, 200-16for infinite integrals, 220

Gradient angle \jjt 107

Hyperbolic functions, 84-99inverse, 96-9

Integrals involving 'infinity', 217-25

Integration, systematic, 200-16involving surds, 203, 207polynomials, 200rational functions, 201

Intrinsic coordinates, 111Irrational number, 158

Lagrange's form of remainder, 47Leibniz's theorem, 62Length of curve, 102-6Logarithm, 1-18, 27

complex numbers, 183in differentiation, 10series for, 53

Maclaurin's series, 49Modulus, 162, 169

of e\ 186

Newton's approximation, 56Newton's formula, 117

Pappus, second theorem of, 132Partial differentiation, 236Pedal coordinates, 111Pedal curve, 112Powers, complex, 178

Rational functions, 11integration of, 11, 200number, 158

RemainderCauchy's form, 47Lagrange's form, 47

Roots of unity, 177

272 INDEX

Series, binomial, 51 Sine and cosine of complex num-convergence, 42 bers, 184exponential, 54 Sinh#, 84logarithmic, 53Maclaurin's, 49oscillating, 43 Taylor's series, 38-62for since, 38, 50 Taylor's theorem, 44for sinha;, cosh#, 86 Trigonometric functions, integra-sum to infinity, 42 tion of, 214Taylor's, 38-62for (1 +x)~\ 40

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