# Chapter : Power Series

1

**CHAPTER:POWER SERIES**

Contents

0Introduction

1Maclaurin’s Theorem

2Standard expansions

0 Introduction

A few minutes play with a scientific calculator will show that for most values of the variable x, the values of the functions sin x, cos x, ex, ln x, and so on, are not rational numbers or recognizable irrational numbers expressible as roots. If the internal logic of a calculator or computer is limited to the basic arithmetic operations of addition, subtraction, multiplication, and division, how does it calculate the value of these and other such functions? Power series is a way of representing these functions by polynomial series which will allow us to find their respective approximate values to any desired degree of accuracy.

Definition: A power series is a series of the form a0 + a1x + a2x2 +a3x3 + . . . + anxn + . . . to , where the coefficients a0, a1, a2, a3, . . . , are constants.

If Sn = a0 + a1x + a2x2 +a3x3 + . . . + anxn , and if Sn S as n then the power series is said to be convergent and have a sum S.

1 Maclaurin’s Theorem

Under certain conditions, a function f(x) can be expressed as an infinite series of ascending powers of x. Assuming that such a series exists, we may write f(x) = a0 + a1x + a2x2 +a3x3 + . . . + arxr + . . . where ai are constants. Differentiating with respect to x, we have,

f’(x)= a1 + 2a2x + 3a3x2 + 4a4x3 + . . . + rarxr-1 + . . .

f’’(x)= 2a2 + 3.2a3x + 4.3a4x2 + . . . .+ r(r-1)arxr-2 + . . .

f’’’(x)= 3.2a3 + 4.3.2a4x+ . . . .+ r(r-1)(r-2)arxr-3 + . . .

......

f r(x)= r(r-1)(r-2) . . .3.2.ar + . . .

Substituting x = 0, we find that

f (0) = a0a0 = f (0)

f ’(0) = a1a1 = f ’(0)

f ’’(0) = 2a2a2 =

f ’’’(0) = 3.2a3a3 =

......

f r(0) = r(r-1)(r-2) . . . 3.2arar =

Substituting these expressions for ai back into the original f(x), we get

f(x) = f(0) + f ’(0)x + x2 + x3 + . . . + xr + . . .

This result is called**Maclaurin’s Theorem** and the series obtained is known as the **Maclaurin’s Series** for f(x).

It is possible to find a Maclaurin’s series for any function f(x) whose derivatives f ’(0), f ’’(0), f ’’’(0), . . . can be determined.

The series must converge to the sum f(x) in order to be useful. Hence, for many functions, Maclaurin’s

Theorem holds only within a restricted range of values of x.

Example 1.1

Use Maclaurin’s theorem to obtain the expansion for f(x) = (1+x)n where n is a real number.

[1 + nx + x2 + ... for - 1 < x < 1 ]

Solution:

Example 1.2

Use Maclaurin’s theorem to obtain the expansion of f(x) = ex.

[1 + x + + + ... + + ... for all value of x]

Solution:

Example 1.3

Use Maclaurin’s theorem to obtain the expansion for f(x) = ln (1+x). [x - + - + ... for -1< x 1]

Solution:

Think! Can you find the expansion of f(x) = ln x using Maclaurin’s theorem?

Example 1.4

Use Maclaurin’s theorem to obtain the expansion for f(x) = sin x. [x - + - + ... for all x in radian]

Solution:

2 Standard Expansions

a)(1 + x)n= 1 + nx + x2 + ... for - 1 < x < 1

b)(1 + x)-1= 1 - x + x2 - x3 + ... + ( - 1)r xr + ...for - 1 < x < 1

c)(1 - x)-1= 1 + x + x2 + x3 + ... + xr + ...for - 1 < x < 1

d)ex = 1 + x + + + ... + + ...for all values of x

e)ln (1 + x)= x - + - + ...+ (- 1)r + 1 + ...for - 1 < x 1

f)sin x= x - + - + ... + (- 1)r + ...for all values of x in radians

g)cos x= 1 - + - + ... + (- 1)r + ...for all values of x in radians

Trivia : This series is named after Colin Maclaurin (1698 - 1746), the Scottish mathematician who studied it. The series actually appeared earlier in the work of the Swiss physician and mathematician Johann Bernoulli (1667 - 1748), but his book on differential calculus, which was written in 1691, was not published until 1922, long after Maclaurin’s name had become associated with the series. (Taken from Modern Technical Mathematics with Calculus by Robert A Carman and Hal M Saunders, 1985)

Example 2.1

Expand, using Maclaurin’s Theorem, (8 + 3x)1/3 in ascending powers of x as far as the term in x3, stating

the values of x for which the expansion is valid. Hence, obtain an approximate value for . [2.05829]

Solution:

Example 2.2

Assuming that x is so small that x3 and higher powers may be neglected, obtain a quadratic approximation

to the function . [ 2 - x + x2 + ...for - 1 < x < 1]

Solution:

Example 2.3

Show that, if n > 1, ln = 2 ( + + + . . . ).

Hence, calculate the value of ln 2 correct to 4 decimal places. [0.6930]

Solution:

Example 2.4

Given that y = e cosx, show that = y cos x sin x. By further differentiation, or otherwise, show that the first three non-zero terms in the Maclaurin expansion of e cosx are e ex2 + ex4.

Solution:

Example 2.5

The variables x and y are related by = 2xy 1, and y = 1 at x = 0. Show that, at x = 0, = 4 and

= 12. Find Maclaurin’s series for y up to and including the term in x4,and hence find an approximation to the value of y at x = 0.1, giving your answer to 3 decimal places.

[y = 1 x + x2x3 + x4 + ... , y 0.909 ]

Solution:

Example 2.6

Given that y = exsin x, express in the form kex sin (x + ), stating clearly suitable values of k and . Express the next 2 derivatives in similar form. With the aid of Maclaurin’s theorem, or otherwise, express y as a series of ascending powers of x as far as the term in x3. Use this series to find an estimate of the value of

e1/2 sin . [k = , = ; y” = 2 ex sin (x + ); y ‘’’ = 2 ex sin (x + );y =x + x2 + ; ]

## Solution

Example 2.7 (J84/1/21)

Given that y = e , show that 4(1 + x) + 2 = y. By further differentiation of this result, or otherwise, show that the series expansion for y in ascending powers of x up to the term in x4 is

y = e (1 + x + x3 + kx4 ) where k is a constant to be determined. [k = ]

## Solution

Summary (Power Series)

**Maclaurin’s Theorem**

f(x) = f(0) + f ’(0)x + x2 + x3 + . . . + xr + . . .

Standard Expansions

- (1 + x)n= 1 + nx + x2 + ... for - 1 < x < 1

- (1 + x)-1= 1 - x + x2 - x3 + ... + ( - 1)r xr + ...for - 1 < x < 1

- (1 - x)-1= 1 + x + x2 + x3 + ... + xr + ...for - 1 < x < 1

- ex = 1 + x + + + ... + + ...for all values of x

- ln (1 + x) = x - + - + ...+ (- 1)r + 1 + ...for - 1 < x 1

- sin x= x - + - + ... + (- 1)r + ...for all values of x in radians

- cos x= 1 - + - + ... + (- 1)r + ...for all values of x in radians

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