Introduction To Infinite Series (1897)

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i! tt.

STAT.

UBRAIt

INTRODUCTION TO

INFINITE SERIES BY

WILLIAM

F.

OSGOOD,

Pir.D.

ASSISTANT PROFESSOR OF MATHEMATICS IN HARVARD UNIVERSITY

CAMBRIDGE b$ Ibarparfc

1897

"Ulniversitp

Copyright, 1897, by

HARVARD UNIVERSITY.

fft fl/ffti

PKEPACE. TN an introductory course on the

Differential

and Integral Calculus

the subject of Infinite Series forms an important topic.

presentation of this subject should have in view

first to

make

The the

beginner acquainted with the nature and use of infinite series and secondly to introduce him to the theory of these series in such a that he sees at each step precisely

what the question

never enters on the proof of a theorem actually requires proof.

till

at issue is

way and

he feels that the theorem

Aids to the attainment of these ends are

(a) a variety of illustrations,

:

taken from the cases that actually arise

in practice, of the application of series to computation both in pure

and applied mathematics (b) a meaning and scope of the more ;

full

diagrams and graphical illustrations

The pamphlet

that follows

the kind here indicated.

The

is

and careful exposition of the

difficult

theorems

;

(c)

the use of

in the proofs.

designed to give a presentation of

references are to Byerly s Differential

Calculus, Integral Calculus, and Problems in Differential Calculus ,

and to B. O. Peirce

Ginn

&

s

Short Table of Integrals;

all

published by

Co., Boston.

WM. CAMBRIDGE, April 1897.

F.

OSGOOD.

LNTKODUCTIOISr.

1.

Consider the successive values of the variable

Example.

=

sn

for

n

=

r

1 -f-

r2

-f-

-f~

rn

~1

Then

Let r have the value J.

1, 2, 3,

s2 *3

If the values

-|-

= =

+J - +4+i 1

i

be represented by points on a S,

=

line, it is

Sa

I

FIG.

1J if

easy to see the

S,

S4

2.

1.

law by which any s n can be obtained from its predecessor, lies half way between * H _ 1 and 2. namely Hence it appears that when n increases without limit, :

The same

result could

formula for the sum

a

-f-

s n of

,

ar

=

1, r

=

s

tl

=.

2.

have been obtained arithmetically from the the first n terms of the geometric series

-\-

ar 2

-\-

a (I -

8

When

l

/t

Lim

Here a

sn _

.s

-\-

ar n-1

,

r)

J,

w increases without

and hence as before Lim

SH

limit,

=. 2.

^ approaches

as

its

limit,

INTRODUCTION.

2

2.

any

an

Definition of

2.

Let M O w 1? w 2

Infinite Series.

,

+

U

Denote the sum of the

Allow n

1

.....

+M + 2

n terms by

first

sn

be

series

(1)

:

Then

to increase without limit.

..... ,

and form the

set of values, positive or negative or both,

either a) sn will

approach

U:

a limit

Lim n or b) s n approaches no limit. Infinite Series, because n

is

=

SH

=

U;

co

In either case we speak of (1) as an allowed to increase without limit. In

case a) the infinite series is said to be convergent and to have the In case b) the infinite or converge towards the value U.

U

value*

series is said to

be divergent. series above considered

The geometric vergent

is

an example of a con

series. 1 1

..... .....

+ 2 + 3+ + 1-1

are examples of divergent series.

,

Only convergent

series are of use

in practice.

The notation uo is

+u +

.....

often used for the limit

U *

d

\

C7,

u

inf. (or to infinity)

or simply

+u +

.....

v

often called the sum of the series. But the student must not forget not a sum, but is the limit of a sum. Similarly the expression "the sum of an infinite number of terms" means the limit of the sum of n of these terms,

that

as

7

is

7 is

n increases without

limit.

I.

a)

3.

ALL OF WHOSE TERMS ARE POSITIVE.

SERIES,

Let

Example.

CONVERGENCE.

it

be required to test the convergence of the

series

where n\ means 1-2 3

....... n

moment

Discarding for the

the

first

and

read

is

factorial

term, compare the

sum

n".

of the

next n terms

19ft IZIZO

1.9

9 l^O

l

with the corresponding

1

ft

sum

2

n ^L<2

(Cf.

Each term

of

term in

and hence the sum

S.n ,

a- n

or, inserting the

after the first

factors

1).

two

is

less

than the corresponding

discarded term and denoting the

terms of the given series by

1

sum

of the

first

n

SM ,

no matter how large ?i be taken. That is to say, s n is a variable that always increases as n increases, but that never attains so large a value as 3. To make these relations clear to the eye, plot the successive values of s n as points on a line.

CONVERGENCE.

3, 4.

= + = + + 1-

s2

*3

1

1

=2.

1

1

=2.5

l-fl

$4

*5=l

+ jL + JL

+ + l

~

*6

+

7

+ f7

111

^j-

-

s

j _i_

]_

_i

o

s,

I

2.667

=

2.708

-9717

I

-

I

2

718

=

H

When

i

I

1

~4T ~5l

~2~!~^3~!

=

1

n increases by 1, the point representing s H + 1 always moves but never advances so far as the point 3. Hence there

to the right,

must be some point (i.e. e


3),

which

e,

sn

3 or

either coinciding with

approaches as

its

lying to the

left

of 3

limit, but never reaches.

To

the value of e to 8 judge from the values computed for s 1? s 2 three places of decimals is 2.718, a fact that will be established <s

,

,

later.

4.

The reasoning by which we have e, although we do not as yet

FUNDAMENTAL PRINCIPLE.

here inferred the existence of a limit

know how to compute the numerical value of that limit, is of prime importance for the work that follows. Let us state it clearly in general form. If s n

is

a variable that

1)

always increases wlien n increases:

but that 2) always remains less than some definite fixed number, s tt

for

all

values of n, then

sn

<

A

approaches a

Lim

fs lt

=

limit,

U.

U:

A:

CONVERGENCE.

4, 5.

This limit, U,

is

A

not greater than

:

U
s3 s 4 U H+H

sa

s, 1

The value

A may be

A h

1

1

the limit itself or any value greater than the

limit.

Exercise. creasing, but

State the Principle for a variable that is always de is always greater than a certain fixed quantity, and

draw the corresponding

figure.

Direct Comparison. On the prin based the following test for the

First Test for Convergence. of the preceding paragraph ciple convergence of an infinite series. 5.

z

\

a

series

test.

If a

less

in the series to-be tested (a), then (a) is does not exceed that of the series ((3).

=

to

than the corresponding terms

a convergent

and

series,

its

value

.....

o

i


Lira n

=

it

a)

let

n

Then

(

of positive terms, the convergence of which it is desired series of positive terms already known to be convergent

can be found whose terms are never

For

.....

+u +u +

Let be

is

Sn

since

follows that

and hence by

<^

sn

=

A.

QO

A

and sn

4

Sn

<^

sn


Sn

,

A

approaches a limit and this limit

is

not greater

than A.

Remark. It is frequently convenient in studying the convergence of a series to discard a few terms at the beginning (m, say, when is & fixed number) and to consider the new series thus arising. That

m

the convergence of this series is necessary and sufficient for the convergence of the original series is evident, since

CONVERGENCE.

6

5, 6.

u is constant and hence s n will converge toward a limit and conversely. Examples. Prove the following series convergent.

.....

+ +r + + ++-^ ^ ^ 5 3 r

r

4

7

!

!

Write

r

,

<

does,

1 <

!

3-4

2-3

1-2 Solution.

r 16

9

if s n _ m

in the fo-in

s

+! then

Lim

2

j

4. 1^

I

,

T^

22

A New

6.

1

1

,

g2

I

Test-Series.

.

I

^2

.....

+ +

l+

1

o-6

8 4

1

=

sn

>

has just been seen that the series

It

converges when the constant quantity also converges whenever inequalities is at once evident

that

2

P

,

it

p

1

p

>

>

We

2.

The

.

will

:

I+1 2?

III

4;

I_~ I-~ P P

T~

5p

~r

QP

"^

1

3*

2 ~ ^ _ = J_ 2 <-

2*

!<-!^^ 4/ 7 ,

;>

^

now prove

truth of the following

i
/)

1

JL

^-i "-

1

(5,

CONVERGENCE.

7.

m

Hence, adding

7

of these inequalities together,

I+QP +

+ /O -.

OP

I

rt

+

1 1

< ^-*-

1

\/>

"

1 -I }V)

1

M

|

1

by

r

O^J

|

1 ^>

=

I

T

converges toward the limit

many terms

1

Ap

I

1

I

.

Qp

1

+

.

p

L- 41

1- H J

p

then, since

;

get

^ -f -JL +

+)p

_i_

Denote 1/2

we

0, r

r -f r 2 I

and the

1

series

<^

r 3 4-

-4I

Consequently ho matter how

.

of the series

2~P

+ 3; + 47 +

be taken, their sum will always be less than

,

and

this series is

4. therefore convergent, by the principle of Series (3) is useful as a test-series, for many series that could not

be shown to be convergent by the aid of the geometric For example, be so shown by reference to it. 1

7.

I

/

Q

I

f)

Divergent Series.

p

1

.

/

Q

A

I

I

A

I

The series (3) has been proved convergent Thus the series

^>

2^2

3

a convergent series, for

7 p

r 4

3

1.01.

J

7

4

Now

consider what the nu

merical values of these roots in the denominators are

In fact

^ 2 = 1.007, ^ 3 = 1.011, ^ 4 ^ 100 = 1.047 and ^ 1000 = 1.071;

thousand terms of the last

term

is

can

+_l_ + ^_+ + _J_ 9 OA/O 4:y4: zyz

for every value of

is

series,

:

1.014. that

is,

when a

have been taken, the denominator of the a number so slightly different from 1 that multiplied by series

significant figure of the decimal part appears only in the second place. And when one considers that these same relations will be still more strongly marked when p is set equal to 1.001 or 1.0001, one may well ask whether the series obtained by putting p 1,

the

first

1.

is

not also convergent.

1

1

CONVERGENCE.

8

This

is

11 I

r

n+l

since each of the

we can

For

however not the case.

?i

1

r

n terms, save the

last, is

111 ^V.

I

+2n

7, 8.

n

+ n^

f^\

2n~~2

greater than 1/2 n. Hence add a definite number of

strike in in the series anywhere,

terms together and thus get a sum greater than this as often as

we

,

and we can do

For example,

please.

i+\>\

j+i+t+t>j "=

Hence the sum

++

-

the

of

first

+>

n terms increases without

n

limit as

increases without limit,

Lim

or

n

The

series (4) is called the

How

=

sn

=

oo

oo

harmonic

series.

the apparently sudden change from convergence for p 1 to be accounted for? in series (3) to divergence when p is

^>

1

The

When p is only slightly greater than 1, explanation is simple. series (3) indeed converges still, but it converges towards a large value, and this value, which

is

of course a function of p, increases

without limit when p, decreasing, approaches limit exists,

8.

and the

Test for Divergence.

Exercise.

namely

When p

5 for convergence,

Let o

+w +

(

i

of positive terms that is to be tested for divergence. series of positive terms already known to be divergent be

a

no

1,

Establish the test for diver

gence of a series corresponding to the test of :

1.

series is divergent.

series

o

+

i

+

a)

If a

(ft)

can be found whose terms are never greater than the corresponding terms in the series to be tested (a), then (a) is a divergent series.

Examples. ,

.

J_4.

J

.

^

j_

+ ^

9

CONVERGENCE.

8, 9.

1

1

_

i

4_ i

_i_

i

i _L I

JL

i

_L_

This last series can be proved divergent by reference to the series

2

Let

""

4

""

"""

6

111 + + + 246

-

sn

1

+ v~ 2n

-

-

series in parenthesis is the harmonic series, and its sum in hence S H increases without limit creases without limit as n increases

The

;

and the

series is divergent.

The

Second Test for Convergence.

9.

Test- Ratio.

Let the series

to be tested be

+ Ul +

U and form the

When n definite

Then

test-ratio

increases without limit, this ratio will in general approach a fixed limit (or increase without limit). Call the limit r.

if r

the series is convergent,

1 <^

\ there is

r

+ Un+

no

test

"

=

n

+1

un

*>

=

r

1

,

<^

r

"

1

,

^>

r

"

First, let r

1. <^

to the values of

ifr^>

1,

it

is

divergent, if

:

Then

as

=

1,

Convergent ; Divergent;

No

Test.

n increases, the points corresponding about the point r, and hence if

u, + 1 /u n will cluster l

r

o

1

I

FIG.

y 4

i

1

4.

a fixed point y be chosen at pleasure between T and 1, the points u n + l /u n will, for sufficiently large values of n, i.e. for all values of n equal to or greater than a certain fixed number m, of y,

and we

shall have

lie

to the left

CONVERGENCE.

10

m

n =-

Adding p

<"

The sum

"H+3

"

I

>

-j- z,

(

7

<\

we

of these inequalities,

u

2 2 y 4- y 4- y

9

^w + 3

^ u m + iy

\

/

\

um y

O i

get

4- y p )

-I-

<^

i

w 1

y

beginning with w w + i never, therefore, V rises as high as the value u m Hence the ^-series converges.

The case and may be If r

=

the terms w,

of

that T ^>

1 (

or T

=

treated in a similar manner,

oo ) is

an exercise for the student. there is no test. For consider series

left as

1

(3).

The

test-

ratio is -

UH

and hence T

=

1,

(n-f- 1)

may have. But when p

no matter what value p

^>

1,

Thus it appears 1, (3) diverges. converges; and when p that T can equal 1 both for convergent and for divergent series. Remark. The student will observe that the theorem does not say (3)

<^

that the series will converge

than

1

when n

u u + i/ u n snall b e series this ratio

if

u + l /u lt

increases, but that I

Thus

GSS than 1.

n/(n

-f- 1) is

less

lt

it

becomes and remains less demands that the limit of

in the case of the

than

1

But the limit is not less than 1. Are the following series convergent or divergent?

the series diverges.

Examples.

harmonic

for all values of w, and yet

2

i.?.

+ 5^3 + 3^3 i.

i

-

2

.

11 100

3 7

28

2-2

A + Jl + 2ioo

5

i

3100

2*

2S

4. i

4100

+ J*! + j*L + 100 n 100 2

8

+

.....

CONVERGENCE.

10, 11.

10.

A

The following

Further Test-Ratio Test.

gence and divergence If omitted

is

11

sometimes useful

test for

conver

the proof of the rule is

;

.

approaches a

limit, let this limit

be denoted by

+ i)/-

(

and

o-=2;

a-.

Then

i

+

the series

!

the series converges.

Test the following series 1

1^5

1^3 2-4

2

:

2-4-6 n

l-3\ 3 2 1

22

f\-Z 3\

,

4 y

V

6

+^ - + - -+1

L

a

4

2

r

022 3

i

a

la2 4

a

Apply any of the foregoing tests to determine the convergence or the divergence of the series on pp. 45, 46 of Ityerly s Problems in Differential Calculus. Z)

11.

SERIES

Alternating

series 1)

2)

let

3)

let

WITH BOTH POSITIVE AND NEGATIVE TERMS. Theorem.

Series.

be alternately positive

each u be

less

than (or equal

Lim u n fi

Then

the series is convergent.

Let the terms of the given

and negative

=

CO

to}

its

:

predecessor

:

CONVERGENCE.

12

The following

series

11

serve as an example.

may

+ ~ + Let

Proof.

and plot the points points s l5 s 3

,

s5 ,

st , s2 ,

s3 ,

.....

*2/

-

..... Then we shall show that the ..... always move to the left, +u

i-

1

FIG.

but never advance so far to the

5.

left as s 2 , for

4 they approach a limit,

the principle of

Lira s 2m +

m = QO

1

=

t/i

Hence by

example.

:

t/i.

.....

.....

s 2/)i , Similarly, the points s 2 , s 4 , ,s 6 always move to the right, but never advance so far to the right as s 1? for example ,

;

hence by the same principle they also approach a

Lim

.j

m.:

oo

=

Finally, since 2w

lim WI

but lim

1(2,,,

=n

;

,s

=

2ra

+1

-

O

"T~

s 2m

-

.

m a

I

*II

z

2ra

5

+ lim m

?6 2m

;

<x>

here the third hypothesis of the theorem comes

hence

;

__ -

Ul

Thus

2

m=oo

QO

TT

over

U

=:

+ 1 z= lim

into play for the first time

or simply U.

2//t

U

limit,

U2

approaches a

J

- --

s rt

its limit.

TT

limit,

o

1

1

llll FIG.

I

C7,

continually springing

vHi^ 1

<

1

6.

It remains to supply the the reasoning of the proof. this reasoning depends. on which establishment of the facts analytical

Such

is

=

2

-2

2m _

and the parentheses are

2

all

2w

_l

positive (or null)

Next

s zm

The proof

13

CONVERGENCE.

11, 12.

is

now

+l

>

and

s2

s 2w

sx .

<

complete.

Examples.

log 3

log 2

Iog4

12. The Limit of Error in the Alternating Series. Suppose it be required to find the value of series (5) correct to &, say, to 3 places of decimals.

For

this

purpose

it

is

not enough to

know merely

that the series

converges, and hence that enough terms can be taken so that their sum s n will differ from the limit by less than .001, for n might be

U

so great, say greater than 10,000, that it would be out of the question to compute s n And in any case one must know when it is safe to .

stop adding terms. The rule here is extremely simple. The sum of the first n terms of series (5), sn differs from the value of the series, U, by less than ,

the numerical value of the (n -\- l)st term. stop adding terms as soon as we come to a

In other words, we may term which is numerically

smaller than the proposed limit of error. The transition from For, consider Fig. (6).

s n to s n + 1 consists in the addition to sn of a quantity numerically greater than the distance from s n to U. This quantity is precisely the (n -{- l)st term of the

series.

Hence the

For example,

rule.

let it

be required to compute the value of the series

I-I.I+ .I- 4 .1+ 1

3

2

32

3

1

(7 )

34

33

correct to three places of decimals.

= = = = (i)* ()

(W

.3333

i (|) 3 5 I (^)

.0123

i

.0000

\ 4 \ (i) 6 | (|)

.0008

-

0556

.0031

=

.0002

.3464 .3464 or, to 3 places, the value of series (7) *

The 4th place is retained throughout the place in the final result.

is

.0589

=

.2875

.288.*

work

to insure

accuracy in the third

14

CONVERGENCE.

Show

1.

Examples.

12, 13.

that the value of the series _ ~"~

4 2*

5

2

to three places of decimals is .405. 2. How many terms of the series

-4+i-J would have

to be taken that the

sum might represent

the series correct to 3 places of decimals

13.

A

any

convergent series

General Theorem.

the value of

?

Let .

be

of positive,

and negative

Lim w n

=

n

More

.

oo

generally,

Lim [X 71

where

Then

terms.

p

is

=

any

The proof

+

+!

+

.....

+ W-i] =

.

CO

integer, either constant or varying ivith n.

of this theorem flows directly out of the conception of a

Let

limit.

and plot the points

we say

"s

which the

n

.....

s 1? s 2 , $,,

approaches a limit

U"

is

Then what we mean when

that there

is

a point I/ about

as n increases.

This does not necessarily re quire that (as in the series hitherto considered) s n should always come s n s cluster,

-

Thus s 3 may lie further away steadily nearer to U, as n increases. than s 2 does. But it does mean that ultimately the s s will

from

U

-- U -a

1

s4

U-(? 1

sa U+rf

Mill

1

cease to deviate from

1

/t

s,

1

1

U by more

than any arbitrarily assigned quan In other words, let 8 be taken at pleasure tity, 8, an interval extending to a dis and off (= 1/1,000,000, say) lay then for the tance 8 from U in each direction, (U 8, U -f- 8); n all of values of more values for n, larger greater than precisely,

however small.

a certain fixed

number m,

s n will

lie

within this interval.

can be stated algebraically in the following form

U

8 <

sn

<

(7+8,

:

when n

>

m.

This

CONVERGENCE.

13, 14.

Having thus stated what

we now

If

n ^>

The

is

meant by

turn to the proof of the theorem.

m, both

sn

and s n+p

15

sn s

approaching a limit U,

The sum

wi\l lie in the interval

distance between these points

is

(U

8,

therefore less than 2

U-{-

8).

Hence

8.

-28 no matter what value p may have. But if a quantity depends on n and can be made to remain numerically as small as is desired by Thus as its limit, when n increasing ?i, then it approaches

=

the proposition is established. It is to be noticed that while the condition if

the series

to converge,

is

is

it

in

no wise

Lim u n

==.

is

<x>

.

necessary,

sufficient for the conver

series (4) the general term approaches The harmonic series the series diverges. however does not satisfy the more general condition of the theorem

gence.

Thus

harmonic

in the

as its limit, but

still

;

for if

we put p

=

n,

and does not converge toward

new proof It

may

as its limit.

This fact affords a

of the divergence of the harmonic series. be remarked that the more general condition

Lim [u n where

p may

vary with n in any

ivise

u e choose,

dition for the convergence of the series.

14.

Convergence.

The General Case. ^o

be any series and

+u +

is

a sufficient con

See Appendix.

Let

.....

i

( a)

let

%+

..... *>!+

denote the series of positive terms,

the series of negative terms, taken respectively in the order in which For example, if the it-series is they occur in (a).

2

2

CONVERGENCE.

16 then the v-series

is i

and the

14.

w-series

+p+p+

is

~_

1

_

9

1

1

03

T5

~

Let

Then, whatever value n may have, Sn

Here

m

etc.

When n

limit,

~

sn

can be written in the form

Vm

TP

denotes the number of positive terms in S B m their sum, increases without limit, both increase without andp
,

m

and two cases can arise. I. Both m and rp approach

Case


Lim tn

=

a-

=

m

limits

:

Lim

T7 ",

p

QO

=

rp

=

TF;

cc

so that both the ^-series and the ^y-series are convergent.

Hence

the ^-series will also converge,

Lim n

and

=

s

=

n

U,

oo

U=

V

W.

The above example comes under

this case.

Case

I will

be of

principal interest to us.

Case

II.

At

least

one of the variables

For example, suppose the 3, 1

As it

-r

w-series

_. 3

g

1

1

y


m , TP

,

approaches no

limit.

were

_. -5 1

these examples show, the u- series

33

T^

1

may

7 I

then be convergent and

may be

divergent. Exercise. Show that

if

the w-series converges and one of the

w-series diverges, the other must also diverge. Let us now form the series of absolute values* of the terms of the

v-,

* By the absolute value of a real number is meant the numerical value of that 3 is 3; of 2^ is 2. number. Thus the absolute value of Graphically it means the distance of the point representing that number from the point 0.

CONVERGENCE.

14, 15.

and write

^-series

uH If

will

we

it is

this series as

be a certain

un

v, if

is

positive

a certain w,

;

if

un

is

negative.

set

clear that

= *. +

<

From is

17

this relation

a convergent

we deduce

at

*,-

once that in Case I the w -series

series.

Conversely, if the u -series converges, then both the v-series and

and ice have Case I. For both the v-series and the w-series are series of positive terms, and no matter how many terms be added in either series, the sum Hence by cannot exceed the limit U toward which s n converges.

the w-series converge,

4 each of these series converges. whose absolute value series

the principle of

are convergent said to be u are -series converge) whose absolutely or are series said to be other convergent unconditionally convergent Definition. (i.e. w-series

Series

;

The reason not absolutely convergent or conditionally convergent. and for the terminology unconditionally conditionally convergent will appear in

Since the it-series surely converges and then absolutely convergent

Test for Convergence.

15. if

34.

the

tt

-series

is

it

converges

is a series made up exclusively of positive terms, the tests for convergence obtained in I. a) can be applied to the tt -series and from its convergence the convergence of the it-series can

since the ^ -series

The series that occur most frequently in elementary analysis either come under this head and can be proved convergent in the manner just indicated, or they belong to the class of alternat

thus be inferred.

ing series considered in

The

test of

9 can

11.

be thrown into simpler form whenever the

test-

the rule being that when t is approaches a limit, numerically less than 1, the series converges absolutely; when t is numerically greater than 1, the series diverges ; when t is numerically ratio u n + \/u n

equal

;

to 1, there is

no

test

1

( *

:

1

t

1

t <^

or

<^

<^

^>

t

Convergence;

,

1

t

1

or

t

=

1

,

Divergence

,

No

Test.

:

CONVERGENCE.

18

15, 16.

For, the test-ratio u n + l /u n is numerically equal to the test-ratio of the series of absolute values, u n + i/u n Now if a variable f(n) a n when its numerical oo, limit, H, approaches value, being the .

=

distance of the point representing /(n) from the point 0, approaches a limit too, namely the numerical value of from 0) (distance of

H

H

.

Hence lim

=

n

*>

u n+1 un

where T equals the numerical value of t. If then follows that r 1 and the u -series converges. <^

1

t <^

The

<^

1, it

^-series is

then absolutely convergent. The second part of the rule will be proven in the next paragraph. Example. Consider the series ~

Xs

X2 2"""

X*

~~

""

4

"3

n n

+

Hence the e. when

=

.

JU

v

u

oo

converge when x

series will

1

__ _ --

Lim n

1

xn

1

is

numerically less than

1

,

i.

-iO
When

x =.

\

or

1

,

this test fails to give

any information concern

ing the convergence of the series. But it is then seen directly that in the first case the series is convergent, in the second case, divergent. 16.

To

Divergence.

establish the divergence of a series

^o

with positive and negative terms,

divergence of the

^

-series, as the

~

1

2

shows.

It will

however

.....

+%+ 1

it

is

not enough to establish the

example of the

series

1

+ 3 ~~ 4

suffice to

show that the terms do not approach

as their limit, and this can frequently be done most conveniently by showing that the terms of the ^ -series do not approach 0.

Thus and

if

or

>1

uw+1 .

>

1,

*<--!,

when

then

f

n

1 >

>

m.

1(5,

CONVERGENCE.

17.

u

Hence

"

i

that

the

is, all

=

quantity p limit,

if

B

s

<

>

uu

or

^w

>

=

>

;

u n and u n cannot approach

as their

oo.

In the series of

Example.

m

n

,

= m on are greater than a certain positive

from

u m and hence

when n

19

=

15,

a;;

hence

verges for all values of x numerically greater than may be represented graphically as follows

1

this series di

These results

.

:

1

1

Divergent

Divergent

Convergent

For what values of x are the following series conver what values divergent? Indicate these values by a diagram

Exercise.

gent, for similar to the one above.

x

x*

"

. |

I

H

V~2

V3

Ans. a;

1


10 x

3

17.

-\-

,

x5

x

X*

X

10 2 a 2

+ + x

THEOREM.

x

1 <^

>.2

1

H

[

2

!

-fa;

2

,

Conv.

x

;

^>

1,

x

1,

<^

Div.

1 ,

10 3 x 3

+

3

!

-|-

a;

3

+

Ze^

absolutely convergent series; p

,

pt

..... ,

/o 2

5

Then

quantities not increasing numerically indefinitely.

a ^2/

se ^

/

the series

converges absolutely.

For,

let

an p ,

n

be the absolute values of a >t

,

/a rt

respectively, // a and form the M,

positive quantity greater than any of the quantities p series

20

CONVERGENCE.

The terms

17, 18.

of this series are less respectively than the terms of the

convergent series

Ha and each

.....

+ Ha\ + //a + 2

made up exclusively of positive terms. converges and the series

Hence

series is

first series

the

converges absolutely.

Examples.

1

The

.

series

since

sin 3

T

~~^~

,

converges absolutely for

all

I

2

32

converges absolutely and sin If

-f-

!

+

-f- a-!

and

cos x

6 X sin

2

nx

never exceeds unity numerically. an( l

^1

+

+

^2

.....

are

series, the series

+a

-f- 6 2

a?

series

.....

+ I-T 5

any two absolutely convergent a

For the

x.

.....

+

tf

~~5*~

values of

I. -1

2.

sin5#

a;

"

2

2

cos 2 x

SU1 2a; -f-

..... .....

-f-

converge absolutely. 3.

Show

that the series e"*

cos x

-f-

e~ 2;r cos 2x

.....

-\-

converges absolutely for all positive values of x. 4. What can you say about the convergence of the series 1

+

r

+r

cos

2

cos 2

.....

-f

18. Convergence and Divergence of Power Series. ascending integral powers of a;,

where the called a aj,

but

others.

coefficients

power

it

series.

a

will in general

In the

,

a l5 a 2

Such a

,

series

+

series of

j

may converge

for

all

of #,

is

values of

converge for some values and diverge for

latter case the interval

distances in each direction Divergent

A

..... ..... are independent

a o 4~ a i x H~ a x2 <2

?

from

of convergence extends equal x =. 0, and the series con-

the point

r

r

Direr gent

Convergent

verges absolutely for every point x lying within this interval, but not necessarily for the extremities of the interval.

CONVERGENCE.

18.

21

Let x be any value of x for which the a terms of the power series n x^ do not increase without limit a n a? Then a n aj is less than the absolute values respectively of a M x

The proof

as follows.

is

;

,

,

"

.

,

some fixed positive quantity (7, independent of n, a, the power series may converge and For #

=

all

it

may

,

Let h be any value of x numerically less than x value.

Then

where r

=

7i

the power

/a;

1

values of n.

for

;

h

diverge.

its

numerical

=

h.

For

Hence the terms of the absolute value

series

series converges absolutely for x


+ a\h + a

a

2

*

h

+

.....

are less respectively than the terms of the convergent geometric series

C + Or and the

+

.....

+

Or 2

series

a

+

!

h

-\-

a2 h 2

converges absolutely. From this theorem it follows that

..... -\-

if

the power series converges for

converges absolutely for all values of x within an interval from to x and reaching out to the same distance on the stretching and if diverges for x other side of the point x x^ it diverges

x =.

a*

,

it

=

;

x lt If for all values of x lying outside of the interval from x t to now the series ever diverges, consider the positive values of x for

which it diverges. They x r, where r in general

=

a region extending down to a point and such that the series greater than

fill

is

converges absolutely for all values of x numerically less than r this is what was to be proved.

A

;

and

simpler proof of this theorem can be given for the special case For then oo that a n + 1 /a n approaches a limit, L, when n

=

Lim n or

t

=. Lx.

lutely for

all

absolutely is

=

GO

u ni __ a

Hence when values of x

when x

is

lim n

=

L=

.

a a xn

co

0, the

while

power if

L ~\-

series

converges abso

0, the series

converges 15) ( numerically less than 1/Z/, and diverges when x ;

numerically greater than 1/L.

This proves the proposition.

II.

SERIES AS

a)

19.

One

A MEANS OF COMPUTATION.

THE LOGARITHMIC SERIES.

most important applications of

of the

infinite

series in

analysis, and the one that chiefly concerns us in this course, is that of computing the numerical value of a complicated analytic expres sion, for

example, of a definite integral

like

when

the indefinite integral cannot be found. In fact, the values of the elementary transcendental functions, the logarithm, sine, cosine, Let us see how a table etc., are computed most simply in this way.

of logarithms can be computed from an infinite series. series for the function log e (1 -f- li) can be obtained as follows.

A

Begin with the formula

1

The function

(1 -|-

can be represented by the geometric series

a?)"

Integrate each side of this equation between the limits fly.

/h

1

+

_ 1

a-

f*h

/Vt /

l

.

clx

f*h

xdx

I

Jo

Jo

+ Jo

x*dx

/

and h

log

(l+/

t

)

=

7i

In deducing the above formula

h + it

h

3

-=---

:

.....

Evaluating these integrals we are led to the desired formula 2

:

:

.....

(8)

9

has been assumed that the theo

integral of a sum of terms is equal to the sum of the Now integrals of the terms can be extended to an infinite series. an infinite series is not a sum, but the limit of a sum, and hence the

rem that the

extension of this theorem requires justification,

v.

39, 40.

SERIES AS A MEANS OF COMPUTATION.

19, 20.

23

Obtain the formula

Exercise.

1 tan"

h

=

,

)

Hence evaluate

.....

+ ---

h

the series

-J+4- ..... 12 the value of series (8) was computed J and h J and it thus appears

20. In the examples of

to three places of decimals for h

that

log 1J

To

find looO 2

=

.287 (5),

we could

=

log 2

But

=

=

;

U=

log

.405 (5).

the value h substitute in (8) \ /

1

J

this series is not well

+k

+

i

=

1

:

.....

adapted to numerical computation.*

In

fact to get the value of log 2 correct to the third place of decimals, it would be necessary to take 1000 terms. simple device however

A

Write

makes the computation easy. 2

= 14

and then take the logarithm of each side

:

= log | + log = .287 (5) + .405 Hence, to three places, log 2 = .693. log 2

Next, to find log different

way, for

5.

=

.693 (0).

series must be applied in still a h be set equal to 5, h =. 4, and the series therefore set

Here the

if 1 -f-

We

does not converge.

(5)

= 4 + l=4(l.+ J), log 5 = 2 log 2 + log 1J = 1.386 (0) + .223 (2) = 5

where log 1J

From

is

computed

directly

from formula

1.609 (2), (8).

the values of log 2 and log 5, log 10 can at once be found.

log 10

=

log 2

+ log 5 =

.693 (0)

+

1.609 (2)

=

2.302 (2)

or to 3 places, log 10

of great importance, for its value must be compute the denary logarithm from the natural

This latter logarithm

known *

in order to

The formula

We

is

2.302.

is

nevertheless useful as showing the value of a familiar

could not find by direct computation the value of this series say, seven places, because the work would be too long. series, (6).

to,

SERIES AS A MEANS OF COMPUTATION.

24

20, 21.

logarithm. By the formula for the transformation of logarithms from the base c to the base 6, "

Iog c 6

we have Hence

logio^

for

=

log,

A

,

log, 10

example

Compute

Examples.

log 20,

Iog 10 20,

log 9,

Iog 10 9,

log 13,

Iog 10 13.

pu pose well when only a Suppose however we wished to know log 2 correct to 7 places of decimals. Series (8) would then In fact, it would require 16 terms of give less satisfactory results. Series (8) is thus seen to serve its

21.

few places of decimals are needed.

the series to yield log 1 J to 7 places. From (8) a new series can be deduced as follows.

Then

(8)

Let h

x.

becomes log (1

Next replace h

x)

in (8)

log(l

-

by x

+*)=:

x

x-1

xs

2

3

-

:

+ *_| +| -

Subtracting the former of these series from the latter and combining the logarithms we get the desired formula :

We have subtracted on the We have not we have limits ;

in

right hand side as if we had sums. of sums. This step will be justified

35.

We

now apply series (9) to the determination of log 2 to seven x must be so chosen that

will

places,

1

x

zn 2,

x

i.e.

11_

=$

lill _ + __ +

and

\

.

j

J

SERIES AS A MEANS OF COMPUTATION.

21.

25

The advantage of this series over (8) is twofold first, it suffices to J, and compute the value of the series for one value of &, x second, the series converges more rapidly than (8) for a given value of JB, since only the odd powers of x enter. :

=

(-i)

Q)

3 5

(I)

= = =

7

(|) 9

(|) 11

(I) 13

(|) 15

(|)

= = =

.333

333

33

.037

037

04

.004

115

23

.000

457

25

.000

050 005

81

000 000

63

.000 .000

.000

The term TV

=

(|) 8

|

(|) (i) 7

823

05 32

.000

065

.000

005

65

11

.000

000

51

.000

000

05

.000

000

00

.346

573

59

(i) -

(i-)

= =

13

TV T^

07

68

.000

33

9

(i)

| TV

65

333

345

= =

5

| i

.333

:= .012

(|) 15

(i)

16

has no effect on the eighth decimal place. But not enough to justify us in stopping here. must show that the remainder of the series from this point on cannot influence (-J)

We

this is

Now

this place either.

the remainder

A ~ JL

J_

.

15 3 15

The value nor

is

_!_

-1

19

3 19

--

~

3 15

15

+

_! ~

17 3 17

is

17 3 2

19

the series

~ 34

of the series in brackets cannot be readily determined it is obviously less than the value of the ;

that necessary, for

from it by discarding the coefficients if, ^|, than the geometric series

series obtained i.

e.

1,1, + 3*

32

and hence the remainder

9

1 =

~*

in question is less

r^^i

~~ s

than

119 15

3"

8

and so does not

We

affect the eighth place. obtain then finally for log 2 the value

2

X

.346

573

or to seven places log 2

5(9)

=

=

.693

.693

147

147

2.

1(8)

etc.,

SERIES AS A MEANS OF COMPUTATION.

26

Show

Examples.

that

= =

log 1J log 5

Compute

log 2

by

log 2

.223

143

(4)

1.609

437

(8).

aid of the formula

log 2

Knowing

21, 22.

=

log

|

= -- log |

and log 5 we can log 10

=

log J.

find log 10

2.302

:

585.

Compute

Example.

2

log

Iog 10 9

,

to six places.

Series (9) is thus seen to be well adapted to the computation of If y denote any positive number and x be so deter logarithms.

mined that

1+3 = r-

y

.

-

e

"

-

i

*=FFT

1 and -\- I and series (9) converges then x always lies between towards the value log y. For values of y numerically large the convergence will be less rapid and devices similar to those above

explained must be used to get the required result. In the actual computation of a table, not all the values tabulated

A

few values are computed are computed directly from the series. in this way and the others are found by ingenious devices. b)

THE BINOMIAL SERIES.

(a

22. In elementary algebra the Binomial Theorem exponent l m( m l a m-* b * am ma m -^b

is

established.

integral

+ b)

+

for a positive

~

:

+

_|_

(to

m

-\- 1

terms)

Consider the series

If

/A

is

a positive integer, this series breaks off with ^ this point on, each numerator contains

then, from

Thus

if /A

1 -\-

=

2,

-^-

terms, for

as a factor.

we have

-"2x

-\- 1

1*2

x2

-)

x8 1

2

8

-\- etc.

(subsequent terms

all

0),

SERIES AS A MEAXS OF COMPUTATION.

22. or simply

1

2x

-J-

-f-

x2

In this case the series

.

son with the binomial formula (a value (1 a;)**:

1

,

b

=

is

m

x,

27

seen by compari =.

to

/A)

have the

+

any number not a positive integer (negative num never breaks off, i.e. it becomes an infinite series. Let us see for what values of x it converges, for The general term of only for such values will it have a meaning. however

If

/x

is

ber, fraction, etc.) the series

the series

is

2

1

?T~

3

~.

Hence /ji

UH + I

1)

(/A

u

n

(/A

(n -\- 1 Ti-l-1)L

(M

1)_

J

!

n

2

1

n

/u,

1)

-j-

n i**(i*< 1

_

n

(/x,

2

1

1

)

1

x*

fji/n

and

Lim 71

=

00

all values of x numerically less For x the values 1 special investiga 1, 15.) ( which we will not into here. necessary, go

Consequently the series converges for than unity. tion

_

is

We may

Divergent

__

1

1

Convergent

note in passing that when series, a fact that

_ _ Dirergenl

x

1

<^

becomes an alternating is

=

is

<^

the series finally when the series

useful

used for computation.

Toward what value does the series converge when x lies between 1 and 1 ? The answer to this question is as follows For all

+

:

values of x for ivhich (1

+

the

binomial series converges,

its

value

is

*)":

(1

+

The proof III).

Let us

a.-)"

==

1

+ M* +

f

^- + **

(10)

of this theorem will not be considered here (v. Chap. first see whether the series is of any value for the pur

poses of computation. Example I. Let it be required to compute places.

.....

^J

35 correct to

five

SERIES AS A MEANS OF COMPUTATION.

28

We by the

must throw the radicand into a form adapted to computation series. We do this as follows. Since 2 5 32 we write

^ The second

(i

22.

35

factor can be

i

=

+

off,

^

series.

+ f^^ (A) + 703 + .000 040 .000

(A)

.000

3

2

003

1.018 08(4)

$

breaking

+ ^)s.

(1

.018 750

and

Show we

Exercise.

2

computed by aid of the

+ *)* == + i A + 1

=

=

35

2.036 17.

that in the above computation did, with the fifth term.

as

we

are justified in

ffi

would be com

Example II. Find -\J 15 to five places. Here we have a choice between the expressions 15

and

15

In the

first

case (1

8+ 7=2 = 27 12 = 3

(1+S)

zz

in the

-(- J)fc,

3

(1

second

-*)

(1

puted by aid of the series. In practice however there is no question as to which expression to use, for the second series converges more rapidly than the first. 1.

Examples. 2.

Show

that

Complete the computation of ^ 15. 2.080 09 and $9 ^/ 2000

=

2

3.

Compute then by writing

first

by

->/

=

2

letting

f

^ =.

.

4.

Find

5.

Obtain from (10) the following formulas

^

i,

by any method.

2 to five places

:

l+x 1

=

1

2x 1

2

X2

3.T

1

2

3

X

,

, "

2^1 2-4

~l~

1-3-5 2-4-6

2-4-6

x

zz=

=

2.961 94.

i ;

SERIES AS A MEANS OF COMPUTATION.

23, 24.

29

The Computation of 23. Series for sin~ l li and tan~ l h. The method set forth in 19 is applicable to the representation 1 1 ^ (v. Exercise, and 19) by series. sin"

dx

T^ .tan- 1

=

/*

l

+

h

This

these series the value of set

h

1,

we

is

h*

3

1

TT

.

,

2

3

2

If in series

:

not well adapted to computation.

=

obtained by putting h

2

(12)

can be computed.

i in

series (11)

2

-4

5

:

+

~~ 6

+

5

.....

4 o

get the equation

series, like series (6), is

better series

h8

1

+ 23+^4

=/i

o

From we

of

tan"

/*-

(12)

TT.

2

This series yields readily three or four places of decimals

;

but

if

more elaborate methods are necessary. 262 (v. Jordan, Cours d Analyse, Vol. I, 1893). Exercise. If the radius of the Earth were exactly 4000 miles, to how many places of decimals would you need to know ir in order to is

greater accuracy

desired,

;

compute the circumference correct to one inch? number of places by Jordan s method.

24.

The Length of

Arc of an

the

the ellipse be given in the form

x

Then will

=

a

sin

Determine

TT

to this

Let the equation of

Ellipse.

:

y

<

,

=

6 cos

<

.

the length of the arc, measured from the end of the minor axis,

be

/* a

I

Jo where (a 2

b 2)/a 2

=

V

& si

1

The integral that here presents itself 1. as an Elliptic Integral and its value cannot be found in the usual way, since the indefinite integral cannot be expressed in terms

is

e

2

<

known

of the elementary functions. last

V

e

2

sin 2

can however be obtained by for x in the

substitution of

esin<

22 gives the formula

example of 1

Its value

The

the aid of infinite series.

<

=

1

~

e

2

sin 2

<

-i-

e

4

sin 4

<

SERIES AS A MEANS OF COMPUTATION.

30 Hence

40)

(v.

=

s

a

24, 25.

r

i

-

<

i

/"0

2

e

sin 2

I

4

e <

2

/*0 I

J

4

2

Jo

"i

sin 4

d>(7

J

These integrals can be evaluated by the aid of the formulas of IV of Short Table of Integrals. In particular, the length of a will S be found quadrant by putting ^ TT and using the formula 240 of the (No. Tables)

Peirce

s

=

<

?

.

sm"

1-3-5 2-4-6

Jf The

elliptic integral

1)

(?i

z= -

o

_ n

TT

-

97,

,

2

an even integer,

then becomes the integral known as the Complete Second Kind it is denoted by

Elliptic Integral of the

E

;

ft

E

I

Jo

(No. 248 of the Tables).

V

e

1

2

sm 2

:

.

<

d

Hence

S

=

a#.

=

the ellipse reduces to a circle and S ^-n-a. Examples. 1. Compute the perimeter of an ellipse whose major axis is twice as long as the minor axis, correct to one tenth of If e :=

one percent. 2. A tomato can from which the top and bottom have been removed is bent into the shape of an elliptic cylinder, one axis of which is

Find what size to make the new top and twice as long as the other. If the original can held a quart, how much will the new bottom. can hold ?

25. The Period of Oscillation of a Pendulum. It is shown in Mechanics (v. Byerly s Int. Cal., Chap. XVI) that the time of a complete oscillation of a pendulum of length

=4*

is

given by the formula

I

\

where a denotes the

K

I

V

1

initial inclination of the

pendulum

to the vertical.

known

as the Complete Elliptic Integral of the First Kind and The substitution of fcsin< force its value is computed as follows. is

in the series for (1

2 o?

)~5 gives the formula (v. Exs.,

2

2

4

22).

31

SERIES AS A MEANS OF COMPUTATION.

25, 26.

Integrating and reducing as in

we

24,

obtain the formula

If the angle through which the pendulum oscillates is small, an sufficiently accurate for most purposes will be approximation for

T

obtained by putting k

the usual

Then

0.

K z=

ITT

and

pendulum formula.

Exercise. less

=

Show

that

if

a <

5,

this

approximation

is

correct to

than one tenth of one percent.

APPROXIMATE FORMULAS IN APPLIED MATHEMATICS.

c)

26.

It is often possible to replace a complicated

mathematics by a simpler one which

is

still

formula in applied

correct within the limits

of error of the observations.*

The

of Expansion.

Coefficient

sion of a solid

is

meant the

By

the coefficient of linear

expan

ratio

V I is the length of a piece of the substance at temperature The coefficient of cubical expansion the length at temperature t is defined similarly as

where

,

.

n__V ~

P where F,

V stand for the volumes

-V -

,

at temperature

,

t

respectively.

Then

V

V

as

is

we

at once clear if

of an edge being is as follows.

I

at t.

I

8

13

I

V

I

/

3

consider a cube of the substance, the length The accurate expression for a in terms of ft

_

.

5

%

^/

1

+

(3

1

=

/3

7

}

P+

See Kohlrausch, Physical Measurements,

1-6.

SERIES AS A MEANS OF COMPUTATION.

32

26, 27.

the error made by Since (3 is small, usually less than .0001, neglecting all terms of the series subsequent to the first is less than the errors of observation and hence we may assume without any loss

of accuracy that a =:

j^ /:?

/?

,

:= 3 a

.

Double Weighing. Show that if the apparent weight of a body in one scale pan is p lt when placed in the other scale

when placed

pan, p 2 (the difference being due to a slight inequality in the lengths of the arms of the balance) the true weight p is given with sufficient accuracy by the formula ,

:

P

iCPi +P*)In an experimental determination of

27. Errors of Observation.

a physical magnitude it is important to know what effect an error in an observed value will have on the final result. For example, let it

be required to determine the radius of a capillary tube by measuring the length of a column of mercury contained in the tube, and weigh From the formula ing the mercury.

W where

w

of the

column

TTl^lp

,

denotes the weight of the mercury in grammes, in centimetres, p the density of the

I

the length

mercury (== 13.6),

and r the radius of the tube, we get I

Jw TT

^

pi

to I

Now

the principal error in determining r arises from the error in I Let I be the true value, I observing /. -f- e the observed value

=

of the length of the column puted value of the radius.

;

=

E

from the error of observation

assumed

negligible.

e,

the error in observing

1

Since

e

is

first

small

term

;

iv

being

Hence x

only the

E

r -\r the true value, r the com is the error in the result arising Then

we

e

.

3

e

i

2

get a result sufficiently accurate

and hence, approximately,

by taking

SERIES AS A MEANS OF COMPUTATION.

27, 28.

33

Thus

for a given error in observing Z, the error in the computed value inversely proportional to the length of the column of mercury a result not a priori obvious, for r itself is inversely propor used,

of r

is

V

tioned only to An Exercise.

1"

surveys a

engineer

field,

using a chain that

is

of one percent of its length. Show that the error thus arising in the determination of the area of the field will be two tenths of one percent of the area.

by one tenth

incorrect

A

clock regulated by a pendulum is 28. Pendulum Problems. If it is carried to a located at a point (A) on the earth s surface. h feet above the level of point neighboring (-B), (A), show that it will lose

^T

seconds a day,

li

i.

e.

one second for every 244 feet of

elevation.

The number

of seconds

N that

the clock registers in 24 hours is T of the oscillation of the pen

inversely proportional to the period

dulum.

Hence

25)

(cf.

N where the unprimed to (B).

T

letters refer to the location (-4), the

If the clock

was keeping

g

where

R

(+

117.)

If h does not

letters

86,400.

/t )a>

(Cf Byerly s .

Hence Z

first

N=

denotes the length of the radius of the earth.

Diff. Cal.,

the

primed

true time at (A), then

exceed 5 miles, li/E

<

NR

.001. h 2 /fi 2

+h

<

.000 001, and

N correct to seconds term of the series gives N N N = th h

:

-

Examples. the sea level.

1.

The summit

How many

of Mt.

Washington

is

6226 feet above

seconds a day will a clock lose that keeps

accurate time in Boston Harbor,

if

carried to the

summit of the

Mountain ? 2.

A

pendulum that beats seconds on the surface of the earth

is

observed to gain one second an hour when carried to the bottom of a How deep is the mine? mine.

SERIES AS A MEANS OF COMPUTATION.

34 29.

Exercises.

1.

Show

29.

that the correction for expansion and is given by the formula

contraction due to heat and cold

=

n

where a denotes the

43, 200 a

coefficient of

t,

linear expansion,

t

the rise in

temperature, and n the number of seconds lost in a day. For brass, a .000 019, t being measured in degrees centigrade. Thus for a brass pendulum n .82 and a rise in temperature of

=

=

causes the clock to lose a

,

over 4 seconds a day. man is standing on the deck of a ship and his eyes are h ft. 2. above the sea level. If denotes the shortest distance of a ship

5

little

A

D

away whose masts and rigging he can measured

to him, 7^ the height,

the water,

show

If h

=

1.23

=

(V

whose

hull is invisible

which the hull

rises out of

can be neglected,

that, if refraction

D

see, but

in feet, to

+

h

V

miles

7

-

>i)

D

16 ft,, 10 miles (nearly). that an arc of a great circle of the earth, 2J miles long, recedes 1 foot from its chord. h,

3.

Show

4.

Assuming that

the sun s parallax

tance of the sun from the earth 5.

earth 6.

Show is

is 76, prove that the dis about 94 million miles.

is

8".

that in levelling the correction for the curvature of the How much is it for two miles ?

8 in. for one mile.

The weights

of an astronomical clock exert, through faulty when the clock

construction of the clock, a greater propelling force

has just been wound up than increase the amplitude of the

when

it

has nearly run down, and thus 2 to 2 4 on each side

pendulum from

Show that if the clock keeps correct time when it has nearly run down, it will lose at the rate of about .4 of a second a day when it has just been wound up.

of the vertical.

7. Two nearly equal, but unknown resistances, A and jB, form two arms of a Wheatstone s Bridge. A standard box of coils and a resistance x to be measured form the other two arms. A balance is obtained when the standard rheostat has a resistance of r ohms. When however A and B are interchanged, a balance is obtained

when

the resistance of the rheostat

proximately,

x 8.

The

focal length

/of

=

b(r

a lens

is

is

r

+r

).

PI

ohms.

Show

given by the formula

--- + /

1

P.

that,

ap

SERIES AS A MEANS OF COMPUTATION.

29.

p and p 2

where

35

denote two conjugate focal distances.

t

simpler approximate formula for

/

that will answer

Obtain a

when p and p z

are nearly equal.

A

ranchman 6 feet 7 inches tall, standing on a level plain, How much must agrees to buy at $7 an acre all the land in sight. he pay? Given 640 acres make a square mile." Admission Exam, "

9.

in Sol.

Geom., June, 1895.

Show

assumed the altitude of the zone ranchman s eyes above the ground and had made no other error in his solution, his answer would have been 4 cents too small. that

the candidate had

if

in sight to be equal to the height of the

Show

10.

that for small values of

approximately correct (h

li the following equations are be either positive or negative)

may

+ h) m = = + 2h (1

Hence

(1

i

+ h) +

2

1

;

h

(1

7i,

&,

+ 70

Z,

p,

(1

.....

+

fc)

(1

are

+

1

all

mh 1

.

+h= +

h

1

l>

5

+

+h numerically small, then, approximately,

.....

= +H+ + + i

= +h+k 1

(1+p)

V (i

V If

1 -f-

lc

l

..... ,

TAYLOR S THEOREM.

III.

30.

It is not the object of this chapter to

prove Taylor s Theorem, done satisfactorily in any good treatise on the Differ ential Calculus but to indicate its bearing on the subject under con sideration and to point out a few of its most important applications. It is remarkable that this fundamental theorem in infinite series since this is

;

admits a simple and rigorous proof of an entirely elementary nature. Rolle

Theorem, on which Taylor

s

of the

Mean

From

Rolle

lie

Theorem follows

s

s

Theorem depends, and

the

Law

at the very foundation of the differential calculus.

at once the theorem contained in the

equation J

(13)

This latter theorem with the

is

frequently referred to as Taylor

R =/

Remainder

lt

-

(

(x

s

Theorem

It includes the

-\-OJi)

Law

Mean

of the

/(a-o

+ h)

/(.TO)

=

hf

(x

+

Oh}

(14)

as a special case and thus affords a proof of that Law. If in (13), when n increases indefinitely, n converges towards as its limit,

R

the series on the right hand side of (13) becomes an infinite power series, representing the function f(x -\- Ji) throughout a certain

region about the point XQ

f(x

+ h) =

This formula

is

/(**>)

known

:

+ f (x )h

as Taylor s

h2 +f"(x )

Theorem and

+

.....

(15)

the series as Taylor s

Series.

The value X Q is an arbitrary value of x which, once chosen, The variable x is then written as x -\- h. The object fast. is

as follows.

is

held

of this

It is desired to obtain a simple representation of the

terms of known elements, for the purpose of com One of of value the function or studying its properties. the puting with known series coefficients. is a forms such of the simplest power function f(x)

in

TAYLOR

30, 31.

Now

it is

by one and the same and even when this is possible, the

a:,

converge rapidly enough for large values of the argu Consequently we confine our

series will not

ment

37

usually impossible to represent f(x)

series for all values of

power

THEOREM.

S

to be of use in computation.

attention to a limited

domain of values, choose an x

in the midst of

domain, and replace the independent variable x by

this

X

The values

XQ ~\-

of x for the

values of h will be, h

chosen that /(o),/

f

(o?

ll

7i,

=

X

XQ.

in question may not be small, but the XQ If x is so corresponding to x ad inf. are all finite, then ),/"(ie )

=

.

,

i.

e.

for values of h

numerically small, will usually* be given by Taylor

example

will aid in

making

Let f(x)

Then it is at once Theorem for x

=

s

Theorem.

above general statements.

clear the

log x.

clear that f(x) cannot be developed by Taylor s oo It is just at this log for/(0)

=

=

0,

we have

point that the freedom that good stead; for if we take x will all

/"(XQ),

where

domain

the value of f(x) for values of x near to x0l

An

7i,

be

.

in the choice of

greater than

finite

and/(a;

-f-

x stands us

in

f

x 0, then f(x ), ( o), can be developed by h)

Theorem, the series converging for all values of h lying Taylor x The proof is given for x between x and 1 in the Diff. s

=

.

Thus we have a second proof

130.

CaL, log (1

+

31.

7i),

Two

(formula (8) of

(13).

Any

series.

19).

Applications of Taylor

This theorem,

it

will

of the development of

s

Theorem with

be observed,

is

the

Remainder,

not a theorem in infinite

function whose

expressed in the

first n derivatives are continuous can be form (13), while the expression in the form (15)

requires the proof of the possibility of passing to the limit

n

=

qo

when

.

Thus

(13) is a more general theorem than (15) and it avoids the It is because of the applica necessity of a proof of convergence.! tions that (13) and (15) have in common, that it seemed desirable to treat *

some applications of (13)

Exceptions to

this

rule,

here.

though possible, are extremely rare in ordinary

practice. f It is desirable that (13) should be applied much more freely th;m has hitherto been the custom in works on the Infinitesimal Calculus, both because

affords a simple means of proof in a vast variety of cases and because many proofs usually given by the aid of (15) can be simplified or rendered rigorous by the aid of (13). The applications given in this section are cases in point. it

TAYLOR

38

THEOREM.

S

31.

First Application: Maxima, Minima and Points of Inflection; Curvature. Let it be required to study the function f(x) in the

=X +/

neighborhood of the point x

+ h) = f(x

f(x

)

Plot the function as a curve

Q.

+

h

(*o)

J/"(a>

+

Oh)

li\

* :

and plot the curve

The

latter

curve

nates, y l and

?/ 2

is

Consider the difference of the ordi-

a right line.

:

.

A

y*

=

+

i/"(a

2

0ft)

ft

.

Hence it appears that ?/! ?/ 2 is an infinitesimal of the second order. This property characterizes the line in question as the tangent to the curve in the point oj and thus we get a new proof that the equation ,

of the tangent

is

)+f (x )(x

y=f(x

x

).

Next, suppose

Then

f(x

The equation

/

(2

it

"

== f(x

of the tangent

will in general

}

(cc)

is

+ h)

positive

at

this

)

is

+ f^

(x,

+ Oh) -^-^

now

=

# and be continuous near the point # it will therefore be positive in the

point

;

neighborhood of this point and hence ya

Ik

>

both for positive and for negative values of 7i, i. e. the curve lies x above its tangent and has therefore a minimum at the point x

=

it

can be shown that

Similarly vatives vanishing, f(x) has a

Lastly,

f

(xo) *

if

/

(2

"

)

(a; )

maximum

<

0, all

x

in the point

.

let

=0, The student should

/ (2 illustrate

(^o)

=

0,

each case in

/( this

2

l

.

the earlier deri

+ 1)

(*o) =j= 0.

by a figure.

TAYLOR

31.

Then

/

{2

y,

+ 1)

"

will

(aj)

- = y,

THEOREM.

S

+ (.*

/<*

39 fc2n+l

+

.

ft)

(2n+1)

,

=

be continuous near a? x and it will same sign for small values of 7i, positive or changes sign with h. Hence the curve lies on tangent on opposite sides of the point x and

in general

therefore preserve the +1 but

negative; opposite sides of its this is then a point of inflection. 7r"

Exercises.

1.

Show

that the condition for a point of inflection not

parallel to the o;-axis is /"(%)=0,

/

2

"

+ 1)

)=b

/->(aR D

being continuous near

(#)

35

=

a?

a 1

"

/<

+1 >(o

b )=|=

0,

.

Show

that a perpendicular drawn to the tangent from a point is an infinitesimal of infinitely near to a point of inflection

2.

P

.....

P

higher order than the second. The osculating circle was defined (Diff. Col. 90) as Curvature. a circle tangent to the given curve at and having its centre on the

P

We

inner normal at a distance p (the radius of curvature) from P. and a be taken infinitely near to will now show that if a point

P

P

perpendicular

FM be dropped from P P

the osculating circle at P", then of the third order referred to the arc

Let

on the tangent at P, cutting in general an infinitesimal

is

P"

PF

as principal infinitesimal.

P

P

be taken as the origin of coordinates, the tangent at being and let the ordinate the axis of x and the inner normal the axis of y Here y be represented by the aid of (13). ;

*

=

0,

x=h, y =

and

The radius

/(0)

$f"(0)x*

of curvature at

P is

and the equation of the osculating *

Hence the

*

=

+

circle is 2

(y

p)

=

P-

lesser ordinate y of this circle is given

y r= p

n

2

V

2

p

*?

-

=

P

P (1

I

* by the formula :

I

Instead of the infinite series, formula (13) might have been used here, with But we happen to know in this case that the function can be developed

4.

by Taylor

s

Theorem

(15).

TAYLOR

40

and

y

y

S

THEOREM.

31, 32.

= )

K*^*>:-*?From

this

result

follows

B

that

y )/x approaches in general a finite limit different from 0, and hence that ?/ y is an infini tesimal of the third order, referred to x as principal infini

tesimal.

PM

But

proposition. Exercise.

Show

(y

PM

and

PP

are of

the

same

order.

that for any other tangent circle y

.

Hence the y

is

an

infinitesimal of the second order.

Second Application : Error of Observation Let x denote the magni tude to be observed, y the f (x) magnitude to be computed from the observation. Then if .r be the true value of the observed magni .

=

tude, x =. x -\- h the value determined by the observation, h will be the error in the observation, and the error // caused thereby in the result will be (cf. (14))

H = f(x + h)

f(x

)

= f (x

+0h)h.

In general / (#), will be a continuous function of x and thus the value of f(x Oh is Oh) will be but slightly changed if a?

+

replaced by x.

+

Hence, approximately,

H = f (x)h and

this is the

formula that gives the error in the result due to the

error in the observation.

The Principal Applications of Taylor s Theorem without the Remainder, i. e. Taylor s Series (15) consist in showing that the 1 fundamental elementary functions e x sin a, cos a-, logo;, o^, SB, 32.

sin"

:

,

1 tan"

a;

can be represented by a Taylor

s

explicitly the coefficients in these series.

Series, It is

and

in

determining shown in Ch. IX of

the Diff. Cal. that these developments are as follows.*

2!

1

xs

z5

,

5!

3! COS

99=1

1

These developments hold for *

The developments

x4

x*

all

values of x.

for sin" a; and tank s are to be sure obtained by in tegration; but the student will have no difficulty in obtaining them directly

from Taylor

s

Theorem.

1

TAYLOR

32, 33.

log*

-i

l tan~"

T

a?

=

log(l

Z

* 4.

"

=

+

1

+ h) =

XS

S

-

h

3

^

2^4

5

1

+

THEOREM.

1-

+

,

;

-J-

o

o

These developments hold for all values of 7i last two formulas, of a?) numerically less than Exercise.

Taylor

s

-

-4-

-I-

2 3

+

41

Show

(or, in the

case of the

1.

that sin a; can be developed about any point X Q by that the series will converge for all values of h.

Theorem and

Hence compute

sin 46

correct to seconds.

33. As soon however as we pass beyond the simple functions and Theorem, we encounter a difficulty that is In order namely to show that f(x) can be usually insurmountable. expanded by Taylor s Theorem it is necessary to investigate the general expression for the n-ih derivative, and this expression is try to apply Taylor s

To avoid this difficulty recourse usually extremely complicated. methods of obtaining the expansion. more or less indirect is had to For example,

let it

be required to evaluate

ra

/ The

indefinite integral

do;.

cannot be obtained and thus we are driven to

develop the integrand into a series and integrate term by term. Now x e~ r )/x^ if we try to apply Taylor s Theorem to the function (e the successive derivatives soon

ever proceed as follows

become complicated.

We

:

= -x+

X1

1

and hence, dividing through by

+

.T,

X - 3T + s

we have

3l

+

5l^

.....

can how

TAYLOR

42

^

dx=2

Examples.

Do

(l

S

THEOREM.

+ 3 37+5^7+

33.

)

made up

functions, into a

502-

the examples on p. 50 of the Problems.

General Method for the Expansion of a Function. function /(a?),

= *.H4

power

To

develop a

manner out of the elementary the general method is the following.

in a simple series,

The fundamental elementary functions having been developed by 32, we proceed to study some of the simplest Taylor s Theorem, operations that can be performed on series and thus, starting with the developments already obtained, pass to the developments de sired.

ALGEBRAIC TRANSFORMATIONS

IV.

OF SERIES. has been pointed out repeatedly ( 19, 21, 24) that since not a sum, but a limit of a sum, processes appli if applicable, this cable to a sum need not be applicable to a series

34.

an

It

infinite series is

;

fact requires proof. For example, the value of a

which the terms are added. terms be extended to series?

value

Its

less

is

terms as follows i

The

than

1

-)-

i

=

(12)

f

+ * - i + i + i- * -M + A -J +

general formula for three successive terms

if

the result it is

is

its

Rearrange

3

4 A;

1

.....

(i

2k

+ I) - i + (i + A)

an alternating

(-8)

is

each pair of positive terms be enclosed in parentheses

+ i) - i +

(i

For

J

:

4k and

sum is independent of the order in Can this interchange in the order of the Let us see. Take the series

*

:

+

(y)

series of the kind considered in

(

11).

easy to verify the inequalities

Hence the

converges toward a value greater than (1 -)- ^) of the first n terms of (/?) differs from a properly chosen sum of terms of (y) at most by the first term of a parenthesis, series (y)

The sum

J z= J.

a quantity that approaches as its limit when n =. Hence the series (/?) and (y) have the same value and the rearrangement of <x>

terms in

from

(a)

has thus led to a series

(/?)

.

having a different value

(a).

In fact

it is

possible to rearrange the terms in (a) so that the

have an arbitrarily preassigned value, C. For, if positive, say 10 000, begin by adding from the positive terms series will

new

C

is

ALGEBRAIC TRANSFORMATIONS OF SERIES.

44

enough have been taken so that

till

their

sum

34, 35.

will just

exceed C.

This will always be possible, since this series of positive terms Then begin with the negative terms diverges.

and add just enough

to reduce the

sum below

As

C.

soon as this

has been done, begin again with the positive terms and add just enough to bring the sum above (7; and so on. The series thus obtained value

is

is

the result of a rearrangement of the terms of (a) and its

C.

In the same

way

it

can be shown generally that

+

"o

"i

+w + 2

if

.....

any convergent series that is not absolutely convergent, its terms can be so rearranged that the new series will converge toward the pre-

is

Because of

assigned value C.

such series are often called

this fact

35 justifying the denoting conditionally convergent, Theorem 1 of of absolutely convergent series as unconditionally convergent. There is nothing paradoxical in this fact, if a correct view of the

For a rearrangement of nature of an infinite series is entertained. terms means a replacement of the original variable s n by a new vari able s n in general unequal to S H and there is no a priori reason why ,

,

these two variables should approach the same limit. The above example illustrates the impossibility of extending a Most of such priori to infinite series processes applicable to sums.

processes are however capable of such extension under proper restric and it is the object of this chapter to study such extension for some of the most fundamental processes.

tions,

35. THEOREM

In an absolutely convergent series the terms can

1.

be rearranged at pleasure without altering the value of the series.

suppose

First, sn

all

the terms to be positive and let

=u + u +

.....

1

+

_!

n

After the rearrangement

U.

lim s n

;

=

co

let

=

For s n always in approaches the limit U when n creases as n increases but no matter how large n be taken (and then held fast), n can (subsequently) be taken so large that s n will include

Then

sn

,

;

all

the terms of

s

>

and more too

;

therefore

<x>

.

.

ALGEBRAIC TRANSFORMATIONS OF SERIES.

35.

45

no matter how large n be taken,

or,

U.

, <

Hence

We ated

s

n,

U

approaches a limit

may now

^

U.

turn things about and regard the i/-series as gener of the u -series, and the above

by a rearrangement of the terms

reasoning shows that

U

7

<^

;

U =

hence

U.

q. e. d.

The second step in the above proof was abbreviated Exercise. by an ingenious device. Replace this device by a direct line of reasoning.

Secondly,

let

the series

.....

+U +U + 2

l

be any absolutely convergent series and

u

Let

let

.==.*-* u v w. + u\ + u .....

~

S n

V

W

V and

1 25

22

28

THEOREM

W;

-

hence

+ 12

let

J

.

U

U.

series

I

7

TX

w

v

1

Find the value of the

III 21

"

M/

u Exercise.

14)

z

be the series after the rearrangement and

But

(Cf.

I+ 29

2*

J_

l

2 11

2^

.

7

2.

F=

V

+

V,

+

.....

are any two convergent series, they can be added term by term, or

If they are absolutely convergent, the third series will also be abso lutely convergent and hence its terms can be rearranged at pleasure.

Let

Then S

n+

t

n

=

(

U <

ALGEBRAIC TRANSFORMATIONS OF SERIES.

46

When

n =.

oo

,

hand side converges toward

the left

U+ V = (u + It

v

)

+

(

Ul

+ vJ +

V\ hence

.....

may be

remains to sliow that the parentheses

U -\-

35.

dropped.

shown in the same way as in the case which arose in 34. The proof of the second part of the theorem presents no and may be left to the student. Exercise.

Show

that

U is

any convergent

THEOREM

3.

This

is

difficulty

if

+

u

series, c

?*!

.....

+u + z

any number,

If

U= V=

u v

+n + +v + l

l

?<

2

..... .....

+

r 2 -f

are any two absolutely convergent series, they can be multiplied together like sums; i.e. if each term in the first series be multiplied into each

term in the second and the series of these products formed, this series toward the limit UV. For example

will converge absolutely

U Vl

+

"M^o

+U

VZ

+

%!?!

+U

2

VQ

+

.....

This theorem does not hold for series that are not absolutely con vergent.

Let

s

=

u

.....

?

then

lim

n

=

s

t it

=

H

n_

l

UV.

oo

The terms

of the product s n t n are advantageously displayed in the scheme. following They are those terms contained in a square n terms on a side, cut out of the upper left hand corner of the scheme.

+.*& I

The theorem asserts that if any series be formed by adding the terms of this scheme, each term appearing in this series once and

ALGEBRAIC TRANSFORMATIONS OF SERIES.

35.

47

for example, the terms that lie on the oblique lines, the only once, successive lines being followed from top to bottom :

Vo + Vi +

+V +

MI^O

.....

2

(a)

,

converge absolutely toward the limit UV. show that one series formed in the prescribed way from the terms of the scheme, for example the series formed by fol lowing the successive boundaries of the squares from top to bottom this series will

It is sufficient to

and then from right to

namely the

left,

series

converges absolutely toward the limit UV. For any other series can then be generated by a rearrangement of the terms of this series.

Let

Sy denote

First suppose

the

Then,

positive.*

sum

of the first

N terms in

(/?).

the terms of the w-series and the v-series to be

all

if

n2

Hence

<

lim

N=

+

(n

JV<

v

=

2

I)

,

UV.

oo

the ^-series and the v-series are any absolutely con vergent series, form the series of absolute values

Secondly,

if

v

The product

But

of these series

is

the convergent series

this series is precisely the series of absolute values of (a),

therefore (a) converges

value toward which

it

UV.

is

converges

and

remains to show that the

It

absolutely.

Since

$v

approaches a

increasing, passes through all integral values, Sx will continue to approach a limit, and this will be the same limit, if

limit

when N,

N

2 passes only through the values n

lim

7V= But

S

tlt

=

Sy

:

= n

co

lim

=

S

and

s n tn

.

tl2

oo

lim n

=

/S, l2

=

UV.

oo

This proves the theorem. *

The

case that

the double sign

(

some of the terms

^

)

in

are

must not however be excluded

the inequality below:

S.y

fC s n +

it, l

+ i.

;

hence

ALGEBRAIC TRANSFORMATIONS OF SERIES.

48

For example,

35, 36.

let

+ a^x +

a

f(x)

be two convergent power

a.2

x2

.....

-\-

,

x any point lying at once within the

series,

Then

region of convergence of both series. series is given by the formula

the product of these

This formula can be used to give the square, or by repeated appli cation,

any power of a power

Thus

series.

it

gives as the square of

the geometric series

+ x + x* + + 2x + 3x* + l

the series l

..... .,

a result agreeing with the binomial expansion of (1 -}- #)~ 2 Exercise. Find the first four terms in the expansion of

e~ x s\ux COBX

x

-

V Square the expansion of

e

2

a

1

x series for e

los (I

and 2

1

and show that the

.

x)

-\-

+v

result agrees with the

*.

36. One more theorem is extremely useful in practice. would carry us beyond the bounds of this chapter.

Its proof

Let

be any polynomial in y and series in x :

=

y

Then

the powers of y

a

+ c^x + a 2

:

y be given by the convergent power

let

?/

,

y

2

x2

.....

8 ,

+

.....

n y can be obtained at once as

x by repeated multiplications of the ^-series by itself, power then formed by multiplying these the terms of the polynomial (?/) the cofficients 6, and the polynomial series power respectively by series in

<

/;



n (//)

thus represented as a power series in x by the addition of these

terms.

Suppose however that instead of the polynomial infinite series

Under what <(//)

power

as a

we had an (?/)

:

restrictions series in

power x?

series in

can

the

above process of representing

x be extended to representing

as a (y)

ALGEBRAIC TRANSFORMATIONS OF SERIES.

06.

One

restriction

immediately obvious.

is

49

Since a power series

represents a continuous function (v. 38) the values of y corre small values x will to of lie near to and thus the point a sponding must surely lie within the interval of convergence ( r y r) of

=

Suppose a now our theorem

the series (y).

And

satisfied.

condition

is

<^

<^

;

is

then this condition

is

always

precisely this, that no further

necessary.

THEOREM

If a

4.

the above process

=

0,

no further restriction is necessary; i.e. (y) as a power series in x is always

of representing



applicable.*

Remark. The point of the theorem just quoted is this. We know from 35 that each term in the y series can be expressed as a power series in x bn y

:

=f (x) = a w + Oi

n

and hence that

<

(?/)

#(30

+ ajt* +

o

n

.....

can be expressed in the form

=/o (*).+/i ()+/(*)+ .....

remains to prove (and it is precisely this fact that the theorem as a fact not true in general of a convergent series of the form serts, It

+ /i(a)-h/t(*) +

/o(*)

where

fn (x)

series all the

denotes a power series) that if we collect from these terms of common degree in x and then rearrange them

form of a single power

in the

and secondly, Examples.

its

=

series, first, this series will converge,

value will be <j>(y).

1.

Let

Let y

powers of x.f

*(y)

.....

e>

it

=

be required to develop e* sinx according to x sin.r. Then

=l+y + \tf + ^

=

*

But the theorem is the one that usually arises in practice. The case an r holds provided only that r, the only difference being that the Cf. coefficients in the final series will then be infinite series instead of sums.

still

<^

<^

25. Stolz, Allgemeine Arithmetik, Vol. I, Ch. X, f Even when it is known that a function can

be developed by Taylor s Ch. Ill; Diff. Cal., Ch. IX; Int. CaL, Ch. XVII) it is usually simpler to determine the coefficients in the series by the method here set forth than by performing the successive differentiations requisite in the application of

Theorem

Taylor

s

(v.

formula.

The example

in

hand

illustrates the truth of this statement.

ALGEBRAIC TRANSFORMATIONS OF SERIES.

50

36.

Find the first 4 terms in the expansion of sin Obtain a few terms in the development of each of the following functions according to powers of x. 2.

3.

log cos ft.

Let cos x

Suggestion.

and

log cos x

V

=

2x

1

=

1 -|-

x2

cos

(9

y

-^ x

-f

a;

then

;

4

iV ^

2

6

^/ i

+

_

A:

2

sin 2 ft

Theorem 4

gives no exact information concerning the extent of the region of convergence of the final series. It merely asserts that This deficiency is supplied by an elementary there is such a region.

theorem in the Theory of Functions.*

But for many applications it is not necessary to know the exact For example, let it be required to determine region of convergence. the following limit. log cos a? 4-

== *

1

V

Mm

1

+

2

+^

Both numerator and denominator can be developed according powers of x. The fraction then takes on the form

to

s J x -\- higher powers of x x 8 -j- higher powers of x .1.

Cancel x s from numerator and denominator and then as its limit.

The

limit of the fraction is then seen to

let

x approach

be

3.

The

usual method for dealing with the limit 0/0 is applicable here, but the method of series gives a briefer solution, as the student can readily verify.

Example.

Determine the

Va

lim x

limit

=

2

--ft 2 l

An

Va*

+

ft

is

to the proof of the

2

cos x

important application of Theorem 4 following theorem.

* Cf. Int. Cal., 220; Higher Mathematics, Ch.VI, Functions of a Variable, by Thos. S. Fiske John Wiley & Sons. ;

Complex

ALGEBRAIC TRANSFORMATIONS OF SERIES.

36.

51

THEOREM. The quotient of two power series can be represented as a power series, provided the constant term in the denominator series is

not

:

2

j

r

CQ

+az +

a^

j

Cj

X

C% X

4

r

2

if

show

It is sufficient to

a

0.

_|z

that 1

a{ x

a

+

>

2 .t

can be so represented, for then the power can be multiplied into the numerator series

Let

y

~

1 o

_ 1

+y

=

al x

1

a

a2 x2

1

~~

+4

1

-\-

series that represents it

-\-

y

y

.

2

y

s

3

4

o


o

provided y/a is numerically less than unity, i. e. y numerically less Thus the conditions of Theorem 4 are fulfilled and the func than a .

4~ y) can be expressed as a power series in x by develop n n ing each term ( l) y /a n + l into such a series and collecting from these series the terms of like degree in x.

tion l/(a

COROLLARY. denominator 7i

i

OQ-\-

/y

j (>

L

If the

of the first m powers of x in the can be expressed in the form

coefficients

series vanish, the quotient

X

jt

i

n 0_

i

/y.2

U.2

X-\-

-f-

I*n

nj

^

co

r ^ ^ ^1

1

r Xo ^2

1

i

t-

"T~

-\1

i l

Ml

ri

+ l_l

i

-z=r ~r

2

-f-

-^-1

|

-r

i

T"

For -|- 6 t /7 Cv

I "

/>.

/

i

a?

x2

-|- bz

/7 ^ m

/vim -4-

b l -|- 5j

1

-\-

+1

/yiW

I

r,

1

=

1

m

~Tn (C

JU

and

it

only remains to set

Show

Examples.

tana;

c

n

=

^

tt

-m

,,

.

+

C1

0,

+C

an(^ divide

=x+1

.^

3

+

a* 1

1

-a;

-

xs

+

2

X2 1 "

a?

+

)

into each term.

be given in

37.

,

-4-

and develop sec x and esc x to three terms. A more convenient mode of determining the expansions

("

that

ctna^ :=

will

I

-4-1

I

coefficients in these

CONTINUITY, INTEGRATION AND DIFFERENTIATION OF SERIES.

V.

AVe have had numerous examples in the fore Continuity. Is the functions of continuous represented by power series. going converse true, namely, that every power series represents, within its

37.

That this question interval of convergence, a continuous function? is by no means trivial is shown by the fact that while the continuous functions of ordinary analysis can be represented (within certain limits) by trigonometric series, i. e. by series of the form

a trigonometric series does not necessarily, conversely, represent a continuous function throughout its interval of convergence.

Let us

first

function.

<

put into precise form what is meant by a continuous be continuous at the point x if

(x) is said to

lim X i.

e. if,

and y

a belt being

=



e,

(#<,)

<

=X

marked where

off

e is

(a?)

=



(>

)

;

e bounded by the lines y ) ~h an arbitrarily small positive quantity, <(#

FIG.

an interval

8,

(.*

when x

.T

O -(-

8), 8

^>

0,

can then always be found such

within this interval, will lie within this belt. These conditions can be expressed in the following form that,

lies

(x)

:

*(*o)

<



(x)

<

4>Oo)

+

e,

^

8

<

x

* <

+ 8;

CONTINUITY, INTEGRATION, DIFFERENTIATION.

37.

53

or

A simple sufficient condition that

+ U, O) +

U (X)

represent a continuous function

THEOREM

1.

V (x)

a

is

series

the series of continuous functions

.....

given by the following theorem.

is

If

+u

l

.....

+

(x)

a

,

<

x

<

ft

,

of continuous functions convergent throughout the interval

f (x) represented by this series will be con (a, /3), tinuous throughout this interval, if a set of positive numbers, Q 2 independent of x, can be found such that then the function

M

M

.....

,

,

M^

,

1)

|

u n (x)

=0,1, i, .....

;

.....

+ M, + M, +

3/

2)

it

a<x
,

\<Mn

a convergent series. We have to show that, x being any point of the interval, if a posi tive quantity e be chosen at pleasure, then a second positive quantity 8 can be so determined that is

()

/)

\f(x) Let s n (x)

=

u (x)

f(x)

s \

if

,

+ U! (x)

will

n( x)

/(ar

)

=

s

(x)

{

\x


+;+

/()=:.(*).+

Then

We

e <

|

n

(x

|<8.

_!

(aj),

*.(*) }

}

+

rn (x ).

rn (x)

show that the absolute value of each of the sn (x o)\,

chosen and

x |

*

M (

:tl

XQ

)j

r n ( xo) is

less

From

8. \

<^

.

tlian

this

8

if

quantities

is

F properly follows that the absolute *,

value of f(x) f(%o) is less than e; hence the proposition. Let the remainder in the Jtf-series be denoted by n

R

and

let

R

n be so chosen that

n

<^

^e,

:

and then held

fast.

Then,

since

UW it

^n,

follows that I

*

< I

The

(*)

absolute value of a quantity

A

<

shall

R.

from now on be denoted by

|

A

. \

54 for

CONTINUITY, INTEGRATION, DIFFERENTIATION. all

37,38.

values of x at once,* or I

rn (x)

Kit-,

a<x
sn (x) is the sum of a fixed number of continuous functions, a continuous function and hence 8 can be so chosen that

Since it is

I

*()

Kit,

*(so)

a>-3

|<s.

|

Hence

and the theorem

is

proved.

Show

Exercise.

that the series

since

sin3#

"I*"

~~32

sin5cc I

52

converges and represents a continuous function.

38. The general test for continuity power series

just obtained

can be applied

at once to

A

THEOREM

2. power series represents a continuous function interval of convergence. The function may however be discontinuous on the boundary of the interval.

within

come

its

Let the

series

be

convergent when

+a

a

f(x)

x

r

1

x

r

<^

<^

a 2 x2

-\-

and

;

+

..... ,

let (a, /?)

be any interval con

tained in the interval of convergence, neither extremity coinciding with an extremity of that interval. Let be chosen greater than either of the quantities a less but than r. Then (3

X

|

|

V

I

and the

<*

|<

I

n

,

,

\

X-,

|

a<x<(3;

series

a |

converges.

|

+

Hence

if

|

a,

we

|

X+

Theorem

I

1 will

continuous throughout the interval

By

a2

X + ..... 2

|

set

J.= the conditions of

|

.!*, be satisfied and therefore f(x)

is

(a, /?).

the aid of this theorem the following theorem can be readily

proved. * It is just at this point that the restriction on a convergent series of con tinuous functions, which the theorem imposes, comes into play. Without this restriction this proof would be impossible and in fact, as has already been

pointed out, the theorem

is

not always true.

55

CONTINUITY, INTEGRATION, DIFFERENTIATION.

38.

THEOREM. If a power series vanishes for a certain interval about the point x

=

+ a^x + a

2

a2

values of x lying in

all

:

.....

+

,

x

I

I <

<

,

then each coefficient vanishes:

=

a

=

First put x

0,

0,

=

then a

;

=

!

.....

and the above equation can be

written in the form

= From

this equation

x (a l

satisfied

=j=

a2 x

al

-{-

.....

)

.....

but it does not follow that this last equation is =. 0, and therefore a l cannot be shown to vanish by here as in the previous case. Theorem 2 furnishes a ;

when x

=

/!() since /j (x)

is

=a +a 1

lim /! (x)

But

lim x

=

/i

Let

difficulty. 2

..... a;+

by that theorem a continuous function of x x

By

+

az x

putting x convenient means of meeting this

Then

-j-

follows that

it

= provided x

-\-

(x)

=Q

=

./I

(0)

=

a,

.

a x =:

.-.

;

.

repeating this reasoning each of the subsequent coefficients can 0, and thus the theorem is established.

be shown to be

If two power series have the same value for

COROLLARY.

values of x in an interval about the point x are respectively equal:

a

=&

,

i

=

&i,

=

all

0, their coefficients

etc.

Transpose one sei-ies to the other side of the equation and the proof is at once obvious.

The Determination of the Coefficients c. It was shown in quotient of two power series can be represented

that the

power

By

36 as a

series.

the aid of the theorems of this paragraph a more convenient of determining the coefficients c can be established.

mode

each side of the equation by the denominator series

Multiply

:

56

CONTINUITY, INTEGRATION, DIFFERENTIATION.

=. a

c -f- (^i c -f-

Hence

6

a

Ci)

=

a

=

ac

x -\- (a 2 c

-)-

%c

x

-|-

a

38

c<

c

a

c2

A

simple mode of solving these equations for the successive c s is furnished by the rule of elementary algebra for dividing one poly nomial by another. Quotient:

a 3 ^8

a

fl

+

(&o-

a2 c

(& 2

a

ajCi

c2

2

)a?

+(6

a3

3

a2 cl

c-

a^)x^-\-

etc.

The equations determining the

first

the c s are precisely the condition that

term in the remainder shall vanish each time.

For example,

to develop tan

aj,

divide the series for sin x

by the

series for cos x. Quotient:

I

|

/y

_i_ 1 /^3 _j_ J2_ 1 3 I

15

5

I I

* X

T5^

I

etc.

Hence tan x

This method lary,

is

=

x

-\-

s $x

-\-

5 2 T ^- x

-\-

applicable even to the case treated in the corol

36.

Exercise.

Develop

3

+x+x

1

csc

aj

.

CONTINUITY, INTEGRATION, DIFFERENTIATION.

39.

57

39. The Integration of Series Term by Term. Let the continuous function f(x) be represented by an infinite series of continuous func tions convergent throughout the interval (a, ft) :

=

u 9 (x)

The problem

is this

f(x)

given by

+ Ul

(x)

+

i.

;

e. to

a

,

<

X

<

(3.

(A)

when

the integral of f(x) will be of the terms on the right of integrals

to determine

:

the series of the

equation (A)

.....

determine when f

C *u

1

(x )dx

(B)

i) a

The right hand a true equation. the term by term integral of the ?<-series.

will be

member

of (J5)

is

called

Let

Then

r&

r&

(*B I

f(x)dx=

sn

I

I

(x)dx-\-

tja

tJ a.

f(x)dx= Pu,(x)dx+ Pu,(x)dx+ e/a

rn

(x)dx

tJ a.

.....

t/a

+e/aPu

n

_ l (x)dx

(*B

+t/a Hence isthat

the necessary

and

r (x)dx. lt

sufficient condition that

^^^ Jaf

is

a true equation

r.OOdtes-O.

To

obtain a test for determining plot the curve

y

=

rn

FIG.

The

(B)

lim

when

this condition is satisfied,

(

9.

area under this curve will represent geometrically r n (x)dx.

58

CONTINUITY, INTEGRATION, DIFFERENTIATION.

Draw lines

through the highest and lowest points of the curve parallel The distance p n of the more remote of these lines

to the ie-axis.

from the

maximum

aj-axis is the

value that

r n (x)

attains in the

=

=

a belt bounded by the lines y p a and y p the curve lies wholly within this belt and the absolute value of

interval.-

Then

Lay

off

.

tl

the area under the curve cannot exceed the area of the rectangle bounded by the line y This area will converge p n or (ft a) p n as its limit if * toward

=

.

,

lim

n

and thus we if

(B)

we

have a

shall

p

tl

=

0,

QO

sufficient condition for the truth of

establish a sufficient condition that the

maximum

=

equation value p n

in the interval (a, ft) approaches rn (x) when n Now GO in the proof of Theorem 1 that if the series (A) satisfies the conditions of that theorem,

of

|

.

|

we saw

r n (x)

B

< |

a

HJ

<

x

<

n

Hence any such

Let the

arise in ordinary practice.

f(x)dx

=

c$P /

u

ya

t/a

test be

most of the cases that

formulated as follows.

>

/

Ul

(x)dx+

t/a

a true equation, if a set ofpositive numbers independent of x, can be found such that |

u n (x)

< |

n

3/

2) is

M

i. e.

CP

(x)dx+

will be

1)

0.

n

oo

Series (A) can always be integrated term by term,

3.

r$P I

=

can be integrated term by term and we have

series

in this result a test sufficiently general for

THEOREM

R =

lim

ft,

a

,

<

x

<

ft,

n

M

=

,

M\,

M

z

,

0, 1, 2,

.,

;

+ M + Mt+

a convergent

series.

The form

which the

l

been deduced is restricted to real But the theorem itself is equally appli complex variables and functions. It is desirable therefore in

test has

functions of a real variable. cable to

to give a proof that applies at once to both cases. *

This condition

restrictions

is

not satisfied by

hitherto imposed on (A).

all

series that are subject

Not every

series

merely

to the

of this sort can be

See an article by the author A Geometrical Method Treatment of Uniform Convergence and Certain Double Limits, Bul letin of the Amer. Math. Soc., 2d ser., vol. iii, Nov. 1896, where examples of series that cannot be integrated term by term are given and the nature of such series is discussed by the aid of graphical methods. integrated term by term.

for

the

:

Keeping the notation used above, the

CP

CP I

Ja

59

CONTINUITY, INTEGRATION, DIFFERENTIATION.

39, 40.

I

f(x)dx

u

CP

(x)dx+

I

relation

+JaCP u _

u l (x)dx-\-

1

Jo.

Jo.

+ JCP r I

n

l

(x)dx

(x)dx

n

a

holds and the proof of the theorem turns on showing that the hypotheses are sufficient to enable us to infer that

still

Then

it

Rn

37 by

Let the remainder of the Jf-series be denoted as in

:

follows, as in that paragraph, that

|

r n (x)

R

< \

.

n

Now /fl

/fl

rn

I

Ja

(x)dx

I

<

~

|

Ja

dx

rn (x)

<

\

R

n l

,

the second integral being extended along the same path as the first, and I denoting the length of the path. But lim (R n l) hence

=

when n

(x)dx converges toward

rn

I

oo

;

and the proof

is

i/a

complete.

of

40. We proceed now to apply the above some of the more common forms of series.

test to the integration

A power series

First Application: Power Series. term by term throughout any interval

of convergence and not reaching out

Let the

series

let

a |

,

a

|

|/8 I

and

if

+

a^x

X be chosen greater than ,

we

can be integrated

contained in the interval

to the extremities

of this interval:

be written in the form

f(x)

and

(a, /3)

but less than

V

\

<

\a n

set |

+a

2

^2

+

the greater of the two quantities

Then

r.

X",

a<x
\

an

X =M 1

n

the conditions of the test will be satisfied.

,

CONTINUITY, INTEGRATION, DIFFERENTIATION.

60 Ill

particular

C* h

+a

a h

I

Joo when x

If

40.

then h

r or

may be

h2

+a

x

h3

-

2

.....

+

,

r

li |

\

.

<

the series for f(x) converges absolutely, r. If however the series for

?*,

taken equal to r or

r or r, it f(x) does not converge absolutely or diverges when x may nevertheless happen that the integral series converges when r or r. In this case the value of the integral series will still h

Thus

be the integral of f(x).

+ x~ diverges when x =

the series

1

1

holds

still

when h

The proof

but the equation

;

1

:

of this theorem will be omitted.

Let be Second Application : Series of Poicers of a Function. a continuous function of x and let its maximum and minimum values <(#)

between

lie

r

converge when

and

r

r <^

\x\

when a

y

Then

r.

the series

<^

can be integrated term by term from a

C

Let the power series

ft.

to ft

+

ft

t/a

f(x)-d=a9 f^dx+ai e/a t/a

For

P(x)dx

if

value of

Y be so taken that

greater than the numerically greatest <

x

tw + |i ^+

i)

|

2)

converges

it is

(x) in the interval a

<

|

;

and

if

we


but less than

)8,

r-,

<

;

|"

|

:

K

!

^

2

I

+

set

|

an

Y |

1

M

n

,

the conditions of the test will be satisfied. Thus the integrations of 24, 25 are justified.

r,

then

CONTINUITY, INTEGRATION, DIFFERENTIATION.

40, 41.

Third Application.

If the function

+ satisfy the

+

!?/

same conditions as

2

2?/

+ a^(x)

a ^(x)

the series

(x)

+

.....

in the preceding theorem

any continuous function of x, then f(x)

and


61

and

if

\ff(x)

is

the series

+ a^(x)

+

O(V>P

.....

can be integrated term by term. The method of proof has been so fully illustrated in the two pre ceding applications that the detailed construction of the proof be left as an exercise to the student.

This theorem

Cauchy

needed in the deduction of Taylor

is

s

may

Theorem from

s Integral. I.

Examples.

Compute /

x*e~ x dx,

I

2.

Show

-v/

sin a? da?.

t/O

J"!

that

X5 .

Hitherto the limits of integration have always been the limits of the interval considered, a and j3. It becomes evident on a little reflection that if

the interval

(a, (3)

a

For,

all

any other

<

limits of integration,

are taken,

X

<

Theorem

still

a

X

ft,

<

aj

a;,

,

hold

<

lying within

:

ft.

the conditions of the test will hold for the interval

they hold for the interval

throughout the interval

any point of

(a?

,

x)

(a, ft).

Then

Term

by Term.

the derivative

the interval by the series

Let the function

f (x)

ivill

of the derivatives

be given

:

provided the series of the derivatives

u satisfies the conditions

(x)

if

(a, ft).

41. The Differentiation of Series f(x) be represented by the series :

at

3 will

+u

l

(x)+

.....

of Theorem 1 throughout the interval

(a, ft)

CONTINUITY, INTEGRATION, DIFFERENTIATION.

62 Let the

We

be denoted by

latter series

<

41.

:

(a;)

wish to prove that

^(x)=f By Theorem

the function

1

Theorem

<j>(x)dx

2 the series

+ Jo.

u (x)dx

I

ul(

I

t/a

.....

+ W + = /(*)-/().

MO()}

{<>(*)

:

r*x

f*x

=

f*x \

Jo.

represented by the ^i -series is con can be integrated term by term

(x)



tinuous and by

(x).

!

{

1

<X)

}

Hence, differentiating,

Show

Exercise.

q. e. d.

(x),

4>(x)=f

that the series

cos a?

cos 3

cos 5 a;

a;

~^

|

~1~

1

can be differentiated term by term. By the aid of this general theorem we can

at once

prove the follow

ing theorem.

THEOREM. A power series can be any point within (but not necessarily its

term by term at

a point on the boundary of)

interval of convergence. Let the power series be

+ax+

f(x)

x |

|

al

Then we want

f

-\-

x |

<^

2

2

x

to prove that

(x

=

)

x

X

?*,

<^

+

a,

|

r <^

\

<^

out the interval

r,

2

+

3 a 3 x2

-\-

+

a:

;

3a 8 a;

converges.

a

series of the derivatives if

X

con

be so chosen that

the conditions of the test will be fulfilled through can prove this as follows. Let x X, X).

We

(

|

\

<

2

r,

( <

X" \

a, 1

+2

I

a,

I

I

as-

I

+8

I

<^

<^

?%

series

.....

+ a.

x

r

:

r)

The

r.

It will serve as a test-series for the I

:

.....

+

for in that case,

X +| % \X +\a

|

series of the derivatives

?*,

<^

\

|

2tt g

,

.....

-\-

x

if

.....

+

^3

and form the

be any value of x within the interval x be so chosen that let

X

3

show that the

It will be sufficient to

verges when

2^

i

convergent when

|

differentiated,

at

convergence of 2

*

I |

|

and

41. if it

CONTINUITY, INTEGRATION, DIFFERENTIATION.

63

can be shown that

n from some

an

"~ I

a;

definite point,

J <

|

|

n

This

m, on. x <

|

|

\a n

\

X

will

,

tn

be the case

m

n >

if

.

=

But the expression on the left approaches when n oo for x n is of and less than 1 the limit can independent / therefore be obtained by the usual method for evaluating the limit Hence the condition that the former series may serve as oo 0. test- series is fulfilled and the proof is complete.

X

|

,

;

|

From

Exercise.

the formula

x

1

obtain by differentiation the developments for 1

1 2

(1

a)

3

(1

z)

(1

z)

and show that they agree with the corresponding developments given by the binomial theorem.

APPENDIX. A FUNDAMENTAL THEOREM REGARDING THE EXISTENCE OF A LIMIT. It was shown when n increases

13 that

in

when p

approaches the limit

And

stated that

is

was when n

will

approach a

limit.

n.

it

the variable s n approaches a limit

if

indefinitely, then

oo

constant or varies in any wise with

is

conversely, the limit of this expression no matter how we allow p to vary with n, then sn

,

if,

This latter theorem

is

important in the theory

however only a special case of a theorem regarding the existence of a limit, which is of fundamental impor of infinite series.

It

is

tance in higher analysis.

THEOREM.

Let f(x) be any function of x such that Urn [/(

)/(")]

=

when x and regarded as independent variables, Then f(x) approaches a limit when x GO nite. 05",

=

We will begin f(x )

/(#")

by

stating precisely approaches the limit

both become infi

.

what we mean by saying that when x and regarded as a?",

X

is independent variables, both become infinite. We mean that if taken as an independent variable that is allowed to increase without

limit

and then, corresponding to any given value of X, the pair of

values (V, both x and

chosen arbitrarily subject only to the condition that

is aj")

x"

are greater than

X (or at least as great),

the quantity In other

as its limit. then converge towards /(# ) /(#") an small let e denote words, positive quantity. arbitrarily can be so chosen that * will

/(a;

We

)

/(")

proceed

values that

c

now

|

<

on any

creasing and approaching the 1, J,

,

,

X by Xu X X 2,

e,.

=

l/i.

set

limit

*

X

and

x"

>

X.

e 1? e 2 , e 3 ,

for

steadily de the values

example Denote the corresponding values of

Then

s

>

X

Let us choose for the successive

to the proof.

to take

is

x

if

,

Then

For the notation cf

;

in general these latter values will .

foot-note, p. 53.

APPENDIX. steadily increase, and do always increase.

we can

=

Begin by putting e /f(^ Jf(i\ (x ) (X

I

)

i

X

Assign to x the value

any case choose them so that they

in

:

l

\

^

c

a/ />

ei

<^

,

/(XO

e.

V

^>

^_ ^\i

^

x

v"

,

^>

7?

^Y!.

Then

le

/(XO-/W i.

65

Cl


X

for all values of x greater than

x

|<e l5


+

Cl

The meaning

.

tion can be illustrated graphically as follows.

of this last rela

Plot the point /(Xj)

Then on a line and mark the points /(Xj) ^ and f(Xt ) -f- t the inequalities assert that the point which represents f(x) always lies within this interval, whose length is 2e^ provided x lt .

X

^>

/j )-i

/PL

t

/CEjH-4*

/+*

i

in

i

FIG.

10.

e of this interval by a 1? Denote the left hand boundary f(X\) the right hand boundary /(Xj) -|- ^ by ^^ Then, to restate con

cisely the foregoing results, i

Now


A

<

/(X

(a)

/(X )

e2

2

both of these points

in this case, let a 2

other point

/(X ) 2

denoted by fa (c)

val

;

the value

e2

=

ai

/(X

2

2)

+

/(X ) 2

.

+ lie

/3 2

04

e2

.

Then

:

X

2

Plot

.

Mark

three cases can arise

in the interval (a x , fa)

;

let

the

:

them be

;

that/(X2 )

e2

falls outside the inter

be taken coincident with

04

:

will lie in the interval (a 1? yS^

so near to

lies

denoted by a 2

.

a2

=

and

ttl

;

shall

the

be

;

in this case, let

other point

e2

Cl


and /(X2 ) ,

;

ft

;

;

denoted respectively by a 2 near to (6) /(X2 ) lies so val

e

Xi

denotes any value of the variable x greater than /(X2 ) this point lies in the interval (a l5 /5J.

the point

points

>

e2

2)

a;

a?

repeat this step, choosing for

i.e.

where

if

/? 2 e2

/? t

that/(X2 )

+

e2

falls outside the inter

be taken coincident with will lie in

p

l

:

the interval (a,, fa)

fa

=

and

fa

;

the

shall be

APPENDIX.

66

In each one of these three cases a2


<

X

if

ft

>

X

2

a2

ft

;

2e 2

<

.

The remainder of the proof is extremely simple. The step just described at length can be repeated again and again, and we shall have as the result in the general case the following

*^

if

Now

:

a.<2e.

consider the set of points that represent a 1? a 2 a,, in general toward the right as i increases, .

,

They advance

never recede toward the

.

.

.

.

.

they but no one of them ever advances

left,

so far to the right as ft. Hence, by the principle* of 4, they opproach a limit A. Similar reasoning shows that the points repre And since senting ft, ft, ... ft, ... approach a limit B. <

ft

a,

<

2

e,

,

A=

B. these limits must be equal From this it follows that f(x) converges toward the :

same

limit.

For ft,

,
and

if

when x

nitely at the

*>*;;

increases indefinitely,

we

allow

same time, but not so rapidly as

to increase indefi

i

to invalidate these in

we see that/(#) is shut in between two variables, a and ft, each of which approaches the same limit. Hence /(a;) approaches that limit also, and the theorem is proved. equalities,

t

In the theorem

above quoted n

in infinite series

is

the independent

the expression s n+p variable x, S H the function /(V) ^corre l thus that theorem is seen to be a x an( to sponds f( /(# ) ;

")

5

The domain of values for special case of the theorem just proved. the variable x is in this case the positive integers, 1, 2, 3, Another application of the present theorem is to the convergence

.....

of a definite integral

when

the upper limit becomes infinite.

f(x)= Then

1

/(x )

/(")

= f% Ja

(a?)

Let

P(x)dx.

dx

C*

Ja

V (x) dx

$ (x) dx.

\

Jxii

Xx $ (x)dx = /

This principle was stated, to be sure, in the form S,,t but it obviously continues to hold if we assume merely that *

n

^>

n.

S ,/

if >

n 5,,

1 ^>

n;

when

67

APPENDIX. as independent variables, both

when x and x \ regarded finite, the integral

become

in

/oo (x)dx

s. The domain of values for the variable x is in this is convergent. case all the real quantities greater than a. In the foregoing theorem it has been assumed that the independent

The theorem can however be

variable x increases without limit.

indefi readily extended to the case that x decreases algebraically sides. nitely or approaches a limit a from either side or from both

In the

case, let

first

x

=

x

=

y;

in the second, let

a H-y

if

x

is

always greater than

its limit

a

let

;

= a -1

x

y if

x

is

than a.

less

always

Then f(x)

if

=

we <

set

(;y)

theorem when will a and limit. hence approach Finally, if x (y) f(x) in approaching a assumes values sometimes greater than a and some times less, we may restrict x first to approaching a from above,

and the function y

-|- oo

,

<

(y) satisfies the conditions of the



secondly from below.

In each of these cases

that f(x) approaches a limit,

it

has just been seen

and since

lim

may now be taken the one above, the other below a, these two limits must be equal. are thus led to the following more general form of statement of the theorem. where x and

x"

We

THEOREM.

Let f(x) be such a function of x

that,

when x and regarded as independent variables, approach the limit a from above or from below or from both sides, or become positively or Then f(x) approaches a limit when x approaches negatively infinite. 1

x",

the limit

a from above or from or negatively infinite.

belotv or

from

both sides, or becomes

68

APPENDIX.

A TABLE OF THE MORE IMPORTANT FORMULAS. The heavy

line indicates the region of convergence.

- = +x+ + + 1

1

JL>

1,5^x H

1

a

101

-

bx

a

.

-\

x*

x*

a2

b* =

a3

_

xz

W

+a .

4

.....

X B, r

r

1

101

1-2

i

(1+x)

_ 2x + 3x

2

1-2-3

= a numerically. .

-

,,

APPENDIX.

1

Vi

11-31T +

= +oX + 1

tf

69

X*

1

1

.

/v.2 ~ *

5

1

X4

2^4

2-4-6

o

i

= o

sin

a?

=x

a;

3 ,

x*

x

1

-j-

o

2!

-3

/y.4

"

2

e*

3- 5

101 o

4! o

6!

rt6 *

APPENDIX.

70

sec x

= +

J X*

1

101 xs

1

2 3

ic

+ ^a

8

1

3 x5

^ 2^~4

1

-

3

2-4-6 7

101 x5

a^

7

________f_

(a;

=

r-V

Jo

1

5 x1

""

5~

APPENDIX.

71

For small values of x the following equations are approximately correct.

/(a

+

)==./(>

+ / (a)

v

= Vi + If

a;,

y, 2, 10,

+ x) (1 + x)

(1

are

all

cos (a

-\-

log (a

A a;

1

= +x+y+z+

+ y) (1 + z) (1 + y) _

+

=

numerically small, then, approximately, 1

(1

sin (a

2*

1

a;)

x)

+ x)

= cos a x sin a cos a = log a = sin

a

a?

/y*

RETURN

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