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Mathematical Foundations 2: Unit 3/Chapter 2

18

Introduction A power series in x is a series of the form

a a x a x a xrr

0 1 22+ + + + +... ... ,

where the coefficients a a a ar0 1 2, , ,..., ... are constants. • The possibility of representing a given function of x by a power series, which may be

used to approximate the value of the function for a range of values of x, will be investigated.

Example 1 Consider the power series 112

13

2 3+ + + +x x x! !

... for which the general term

is ( )ur

x rrr=

−=−1

11 2 31

!, , ,..., .

Evaluate the sum of the first n terms, , when 0.5nS x = :

n un Sn 1 1 1 2 0.5 1.5 3 0.125 1.625 4 0.020833333 1.645833333 5 0.0026041666 1.6484375 6 0.00026041666 1.648697917 7 0.000021701388 1.648719617 8 0.0000015500 1.648721167

Observations • As n increases un appears to be approaching zero, and Sn appears to be approaching some fixed number. On this evidence, when x = 0 5. the series appears to be convergent. In fact to 4 decimal place accuracy it appears that ( )Sn 05 16487. .= , as S S S6 7 8 16487, , ,.. .= to 4 decimal place accuracy. • Comparing the graphs of

( )S x x1 1= + ,

( )S x x x212

21= + + ! ,

( )S x x x x312

2 13

31= + + +! ! and

( )S x x x x x412

2 13

3 14

41= + + + +! ! !


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with the graph of ex (see Appendix 1) suggests that, as n increases the graph of Sn gets closer to the graph of ex. The closest agreement is obtained for values of x close to x = 0 . • Since e0 5 16487. .= , and Sn(0.5)→1.6487, it appears that this power series could be used to represent ex. • The graphs suggest that the series converges to ex, and theoretically it could be used to evaluate ex for any value of x It remains to consider how the above power series representing ex is derived. • A similar consideration of the power series x x x x− + − +1

22 1

33 1

44 ... , indicates that

the series appears to converge to ( )loge x1+ , but only when − < < +1 1x . (For the relevant graphs see Appendix 2.)

2.1 Maclaurin’s Series A Maclaurin’s Series uses information about a function and its derivatives at x = 0 to represent the function by a power series close to x = 0 . Example 1 Assume ex may be represented by the power series a a x a x a x0 1 2

23

3+ + + +... i.e. e a a x a x a x xx = + + + =0 1 2

23

3 0+... close to Then: • 0

0 0 1e a a= � =

• Differentiating: 2 3

1 2 3 4

01 1

2 3 4 ...

i.e. 1

xe a a x a x a x

e a a

= + + + +

= � =

• and again: 2

2 3 4

0 12 2 2

2 3 2 4 3 ...

i.e. 2

xe a a x a x

e a a

= + × + × +

= � =

• and again: e a a x

e a a

x = × + × × +

= × � = •×

3 2 4 3 2

3 23 4

03 3

13 2

...

i.e.

Repeating this process enables as many coefficients as required to be determined.


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Based on the above information the first four terms in the power series for ex are:

2 3 2 31 1 1 1

2 3 2 2! 3!1 1 ... 1 ...x x x x x x×+ × + × + + = + + + + • For a general function, ( )f x , the Maclaurin’s Series is given by:

( ) ( ) ( ) ( ) ( ) ( )f f x f x f xr

f xr r0 012

013

01

02 3+ ′ + ′′ + ′′′ + + +! !

...!

...

where ( ) ( ) ( ){ }0 evaluated at 0r

rr

df f x x

dx≡ =

• To determine the Maclaurin’s series for a given function ( )f x , up to and including

the term in xn, determine the first n derivatives of ( )f x and evaluate them at x = 0 . The coefficients of the Maclaurin’s series for ( )f x may then be found.

The following examples illustrate this procedure. Example 2 Determine the Maclaurin’s series for sin x , up to and including the term involving x5, and deduce the general term. Solution Here ( )f x x= sin ( )∴ = =f 0 0 0sin ( )′ =f x xcos ( )∴ ′ = =f 0 0 1cos ( )′′ = −f x xsin ( )∴ ′′ = − =f 0 0 0sin ( ) ( )f x x4 = − cos ( ) ( )∴ = − = −f 4 0 0 1cos ( ) ( ) ( )f x x x5 = − − =sin sin ( ) ( )∴ = =f 5 0 0 0sin Note: ( ) ( ) ( )f x f x5 = , therefore the cycle 0 1 0 1, , ,− is just repeated!

( ) ( ) ( ) ( ) ( ) ( )∴ = = = −f f f6 7 80 1 0 0 0 1, , , etc .

Also the coefficient of 5x involves ( ) ( )f 5 0 , therefore derivatives of ( )f x of order higher than 5 are not really required, although they can be used to deduce the general term. Thus:

( )2 3 4 51 1 1 12! 3! 4! 5!

3 51 13! 5!

sin 0 1 0 1 0 1 ...

...

x x x x x x

x x x

= + × + × × + × − × + × × + × × +

= − + −


21

Observe that the terms alternate in sign and involve only odd powers of x. It can be shown that this series converges for all values of x.

Example 3 Determine the Maclaurin’s series for 1

1 2+ x, up to and including the

term involving x3, and deduce the general term. Solution The coefficient of x3 involves ( )′′′f x , therefore derivatives of order up to and including 3 are required:

Here ( ) ( )f xx

x=+

= + −11 2

1 2 1 ( )0 1f∴ =

( ) ( ) { }

( )

′ = − × + × +

= − +

−

−

f x xddx

x

x

1 1 2 1 2

2 1 2

2

2 ( )0 2f ′∴ = −

( ) ( ) { }

( )

′′ = − × − × + × +

= +

−

−

f x xddx

x

x

2 2 1 2 1 2

8 1 2

3

3 ( )0 8f ′′∴ =

( ) ( ) { }

( )

′′′ = − × × + × +

= − +

−

−

f x xddx

x

x

3 8 1 2 1 2

48 1 2

4

4 ( )0 48f ′′′∴ = −

Thus:

( ) 2 3

2 3

1 8 481 2 ...

1 2 2! 3!1 2 4 8 ...

x x xx

x x x

−= + − × + × + × ++

= − + − +

It can be proved that the power series for 1

1 2x+ only converges when 1

2 .x <

Thus, the Maclaurin’s series may be used to approximate 1

1 2+ x only when x

lies between − +12

12 and .

The interval − < < +12

12x is called the interval of convergence of the

power series. Example 4 Determine the Maclaurin’s series for xe x− up to and including the term involving x3.


22

Solution As in Example 3, derivatives of order up to and including 3 are required: Here ( )f x xe x= − ( )0 0f∴ =

( ) { } { }

( ) ( )

′ = × + ×

= + − = −

− −

− − −

f xddx

x e xddx

e

e x e x e

x x

x x x1 ( )0 1f ′∴ =

( ) { } ( ) { }( ) ( )

( )

′′ = − × + − ×

= − × + − × −

= −

− −

− −

−

f xddx

x e xddx

e

e x e

x e

x x

x x

x

1 1

1 1

2

( )0 2f ′′∴ = −

( ) { } ( ) { }( ) ( )

( )

′′′ = − × + − ×

= × + − × −

= −

− −

− −

−

f xddx

x e xddx

e

e x e

x e

x x

x x

x

2 2

1 2

3

( )0 3f ′′′∴ =

Thus:

2 3 2 312

2 30 1 ... ...

2! 3!xxe x x x x x x− −= + × + + + = − + +


23

Exercises 2.1 1 Determine the Maclaurin series for cos 2 ,x up to and including the term involving 6.x 2 Determine the Maclaurin series for ( )ln 1 2 ,x− up to and including the term involving

4.x 3 Determine the first four non-zero terms in the power series for 1 3 .x+ 4 Determine the Maclaurin series for sin ,x x up to and including the term involving 6.x


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2.2 Standard Maclaurin’s Series A list of standard power series, i.e. Maclaurin’s series for some standard functions, is given in the Facts and Formulas leaflet. The interval of convergence is stated alongside each power series. • Standard power series may be used to determine the powers series for functions which

involve combinations of standard functions, within the common interval of convergence.

Example 1 Determine the power series for e xx sin , up to and including the term

involving x5. Solution Using standard power series:

e x x x x x

x x x x

x = + + + + + +

= − + −

1 12

2 13

3 14

4 15

5

13

3 15

5

! ! ! !

! !

...

sin ...

, and

where only terms up to and including x5 have been considered. Therefore, retaining only those terms which contribute:

( ) ( )( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

2 3 4 5 3 51 1 1 1 1 12! 3! 4! 5! 3! 5!

3 5 3 2 31 1 1 1 13! 5! 3! 2! 3!

3 41 13! 4!

3 5 2 4 3 5 4 51 1 1 1 1 1 16 120 6 2 12 6 24

2 3 41 1 1 1 1 1 16 2 6 6 120 12 24

sin 1 ... ...

1

...

...

xe x x x x x x x x x

x x x x x x x x x

x x x x

x x x x x x x x x

x x x x

= + + + + + + × − + −

= × − + + × − + × −

+ × + × +

= − + + − + − + + +

= + + − + + − + + − + 5

2 3 51 13 30

...

...

x

x x x x

+

= + + − +

The power series for ex and sinx are both convergent for all values of x, therefore the power series for exsinx is also convergent for all values of x.


25

Exercises 2.2 1 Using standard power series from the Facts and Formulas leaflet write down power

series for 1 x+ and sin ,x up to and including the term involving 4.x Hence, determine the power series for 1 sin ,x x+ up to and including the term

involving 4.x 2 Using standard power series from the Facts and Formulas leaflet write down power

series for ( ) 11 x −+ and ,xe up to and including the term involving 4.x

Hence, determine the power series for ,1

xex+

up to and including the term involving

4.x 3 Using standard power series from the Facts and Formulas leaflet write down power

series for ( ) 12 x

−+ and ( )ln 1 2 ,x− up to and including the term involving 4.x

Hence, determine the power series for ( )ln 1 2

,2

x

x

−+

up to and including the term

involving 4.x


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2.3 Approximations using Power Series The following examples indicate how power series may be used to obtain approximations. Example 1 Using the power series for 1+ x obtain the value of 105. to 4 decimal place accuracy.

Solution ( ) ( )105 1 0 05 1 112

12. .= + + = +, and x x

Therefore, the power series for ( )1 0 05

12+ =x x with . will give an

approximation to 105. , provided a sufficient number of terms to give 4 decimal place accuracy are taken into account.

Note: In order to obtain an answer accurate to 4 decimal places it is necessary to carry out intermediate calculations to at least 6 decimal places; this will ensure that the 4th rounded decimal place is accurate.

Using the power series for ( )1 nx+ with n = 1

2 :

( )

12

3 3 51 1 1 1 1 12 3 42 2 2 2 2 2 2 2 21

2

2 3 41 1 1 12 8 16 128

1 1 ...2! 3! 4!

1 ...

x x x x x

x x x x

× − × − × − × − × − × −+ = + × + + + +

= + − + − +

Setting x = 0 05. :

2 3 41 1 1 12 8 16 1281.05 1 0.05 0.05 0.05 0.05 ...

1 0.025 0.0003125 0.0000078125+terms which do not contribute to 4dp accuracy1.024695312 1.0247 to 4 decimal place accuracy

= + × − × + × − += + − +

= ≈

Example 2 Evaluate

x xdxcos0

1

� to 3 decimal place accuracy.

Solution This definite integral cannot be evaluated exactly using standard integration techniques. However, an approximate answer may be obtained by approximating the integrand by a power series. Using the standard power series for cos x : cos ...! ! ! !x x x x x= − + − + −1 1

22 1

44 1

66 1

88


27

Therefore:

x x x x x x x x xcos cos ...! ! ! != = − + − + −12

12

52

92

132

1721

214

16

18

and:

[ ]

0

1

0

1

x xdx x x x x x dx

x x x x x

cos

...

...

! ! ! !

! ! ! !

! ! ! !

� �= − + − + −

= − × + × − × + × −

= − × + × − × + × −

12

32

92

132

172

32

72

112

152

192

12

14

16

18

23

12

27

14

211

16

215

18

219

0

1

23

12

27

14

211

16

215

18

219

�

0.66666 0.14285 0.01190 0.00018 0.000000.53553 0.536 to 3 decimal place accuracy.

= − + − += ≈


28

Exercises 2.3 1 Use standard power series to show that if powers of x greater than the third are

neglected, then

( ) 31 1 191 2 sin 3 ln 1 2 .

3 2 6xx x x e x−− + − + − ≈ −

2 Use the power series for ( )ln 1 2x− to obtain ln 0.98 accurate to 5 decimal places.

Check your answer using your calculator.

3 By first expressing the integrand as a power series (using standard power series)

obtain approximations to each of the following integrals, accurate to 3 decimal places:

(i) 12

3

0

1 x dx+� (ii) 12

1sin x

dxx�


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2.4 Taylor’s Series Taylor’s series are a generalization of Maclaurin’s series, in that they use information about a function and its derivatives at any point to obtain a power series which approximates the function close to that point.. • The Taylor’s series which approximates the function ( )f x close to x a= is:

( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( )f a f a x a

f ax a

f ax a

f a

rx a

rr+ ′ − +

′′− +

′′′− + + − +

2 32 3

! !...

!...

where ( ) ( ) ( ){ } evaluated at .r

rr

df a f x x a

dx= =

• Setting a = 0 in the Taylor’s series gives the Maclaurin’s series. • Setting x a h= + gives an alternative form of the Taylor’s series which is often used:

( ) ( ) ( ) ( ) ( ) ( ) ( )2 3

... ...2! 3! !

rrh h h

f a h f a hf a f a f a f ar

′ ′′ ′′′+ = + + + + + +

Example 1 Determine the Taylor’s series for loge x x about = 3 up to and including the

term involving ( )x − 3 5 . Given that log .e 3 1098612= determine log .e 31 to 4 decimal place accuracy. Solution The coefficient of ( )x − 3 5 involves ( ) ( )f 5 3 therefore derivatives of order up

to and including 5 are required. Here ( )f x xe= log ( )3 log 3ef∴ =

( )′ =f xx1

( ) 13

3f ′∴ =

( )′′ = −f xx1

2 ( ) 2

13

3f ′′∴ = −

( )′′′ =f xx2

3 ( ) 3

23

3f ′′′∴ =

( ) ( )f xx

44

6= − ( ) ( )44

63

3f∴ = −


30

( ) ( )f xx

55

24= ( ) ( )55

243

3f∴ =

Thus:

( ) ( )

( ) ( )

( )

( ) ( ) ( ) ( )( )

2

2

3 4

3 4

5

5

2 3 41 1 1 13 18 81 324

511215

1 1 1log log 3 3 3

3 3 2!

2 1 6 13 3

3 3! 3 4!24 1

3 ...3 5!

log 3 3 3 3 3

3 ...

e e

e

x x x

x x

x

x x x x

x

� �= + × − + − × × −� ��

� �+ × × − + − × × −� ��

+ × × − +

= + − − − + − − −

+ − +

Setting 3.1:x =

2 3 41 1 1 13 18 162 324

511215

log 3.1 log 3 0.1 0.1 0.1 0.1

0.1 ...

1.098612 0.033333 0.000556 0.000006 0.0000003 ...1.431395 1.4314 to 4 decimal place accuracy

e e= + × − × + × − ×

+ × += + − + − += ≈


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Exercises 2.4 1 Determine the first 6 terms in the Taylor’s series for e x about x = 2 . Use this series to approximate the value of e2 1. accurate to 4 decimal places, given that e2 7 389056= . .

2 Determine the first 4 terms in the Taylor’s series for 2

1+ x about x = 2 .

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