Representing Functions by Power Series

A power series

is said to represent a function f with a domain equal to the interval I of convergence of the series if the series converges to f(x) on that interval.

That’s if:

0n

nnxa

Ixxfxan

nn

;)(0

Example

)1,1(1

1

)1,1(1

1)(

0

xifx

toconvergesseriesthisbecause

onx

xffunctionthe

representsxseriespowerThen

n

Theorem

cn

cxadxxf

cxnaxf

havewercrcxThenreconvergencofrediusthehaving

cxaseriespowerthebydrepresentebexfLet

n

n

n

n

nn

n

nn

0

1

0

1

0

1

)()(.2

)()(.1

:),,(.0

)()(

Examples

Example(1)

?)1(

1)(

2xxg

functiontherepresentsseriespowerWhat

We notice that

And we know that:

)1

1(

)1(

1)(

2

xxxg

)1,1(;1

1)(

0

xxx

xfn

n

Solution

1

1

1322

0

32

0

32

)1,1(;4321)1(

1

,,

)1,1(;11

1)(

:'

)1,1(1

1)(

1

n

n

n

n

n

n

n

n

n

nx

xnxxxxx

getwetermbytermatingDifferenti

xxxxxxx

xf

sThat

onx

xffunctiontherepresents

xxxxxseriespowerThe

Example(2)

?)1ln()( xxg

functiontherepresentsseriespowerWhat

We notice that

And we know that:

)1,1(;1

1)(

0

xxx

xfn

n

cxx

dx

x

dx

)1ln(

11

Solution

1

1432

0

1

0

1

0

32

0

)1,1(;)1ln(

:

)1,1(;1432

1)1ln(

00)01ln(

)1,1(;1

)1ln(

1

1)1ln(

)1,1(;11

1)(

,

n

n

n

n

n

n

n

n

n

n

n

n

xn

xx

isstatementequivalentAn

xn

xxxxx

n

xx

cc

xcn

xdxxx

xx

and

xxxxxxx

xf

haveWe

Question

What about the convergence at the end points?

1. The function ln(x-1) is not defined at x = 1

2. We can show easily that the series is convergent if x = -1 (how?)

But does it converge to ln2?

The answer to this question has to

wait till we introduce Able’s Theorem

Approximating ln2

3ln

:

)(,sec

69115.02ln

,

69115.06

}21

(

5

}21

(

4

}21

(

3

}21

(

2

}21

(

2

1}

2

1ln(

,

}2

1ln(,

384

1

160

1

64

1

24

1

8

1

2

11

}21

(

6

}21

(

5

}21

(

4

}21

(

3

}21

(

2

}21

(

2

1)

2

11ln(}

2

1ln(

1,1(;1432

)1ln(

}2

1ln(2ln

,

65432

165432

1432

ofionapproximatanGive

Question

ionapproximatsuchin

accuracytheerrortheestimatetoablebewillwetionnexttheIn

soand

getweseriestheoftermsfourfirsttheonlyuseweIf

accuracyofdgreeanytoedapproximatbecanThus

n

xn

xxxxxx

haveWe

n

n

Example(3)

?arctan)( xxg

functiontherepresentsseriespowerWhat

We notice that

And we know that:

)1,1(;)1()(

)(1

1

1

1)(

0

2

0

2

22

xxx

xxxf

n

nn

n

n

cxx

dx

arctan1 2

)1,1(

111

1

)(1

)1(

1

1)(

2

2

2642

0

2

2

x

xx

xsatisfyingxallfor

xxxx

x

xxf

n

n

nn

Solution

)1,1(;12

)1(119753

)1,1(;12

)1(arctan

000arctan

)1,1(;12

)1(arctan

)1,1(;)1(1

)(1)1(1

1

;

12119753

0

12

0

12

0

22

26

0

4222

xn

xxxxxxx

xn

xx

cc

xcn

xx

xdxxx

dx

xxxxxx

haveWe

nn

n

nn

n

nn

n

nn

n

n

nn

Question

What about the convergence at the end points?

We can show easily that the series is convergent if x = 1or x = -1 (how?)But does it converge to arctan1 = π/4 & arctan(-1) = π/4 respectively ?

The answer to this question has to

wait until after we introduce Able’s Theorem!

Approximating arctan(0.5)

?)(565.26

46365.013

)2

1(

11

)2

1(

9

)2

1(

7

)2

1(

5

)2

1(

3

)2

1(

)2

1()5.0arctan(

,13

)2

1(

11

)2

1(

9

)2

1(

7

)2

1(

5

)2

1(

3

)2

1(

)2

1()5.0arctan(

,

)1,1(;12

)1(119753

arctan

,

13119753

13119753

12119753

thatatarrivewedidHowelyapproximattocorresponsThis

onlytermsseventhfirstthegConsiderin

soand

xn

xxxxxxxx

haveWen

n

everywhereabsolutelyconverges

n

x

seriesThe

Example

n

n

0 !

:

)4(

Rxn

x

nxnx

foorP

xn

n

x

;10

1

1lim

!

)!1(lim

:

1

Showing that this series converges to e

x

n

n

n

n

n

n

eytoleadwillthatshowwillWe

y

y

Thus

y

xxx

xxxxfy

getwetermbytermatingdifferenti

n

xxxxx

n

xxfyLet

n

xxxxx

n

x

seriesThe

1

,

!3!21

4!4

13

!3

12

!2

110)(

,

!!4!3!21

!)(

!!4!3!21

!

:

32

32

432

0

432

0

x

n

n

x

cc

xc

xccx

en

xThus

ey

Thus

ceee

getweWhyfBecause

eey

getwesignvalueabsolutetheDiscarding

eeey

cxy

cxy

dy

dxdxy

y

y

yequationaldifferentitheSolving

0

0

!

,

01

:?),(1)0(

,,,

ln

1:

Approximating e

7.2008.0041,0151.0500.02120

1

24

1

6

1

2

111

,

65.215.05.026

1

2

111

,

!

12

!

1!

1

!4

1

!3

1

!2

111

!!4!3!21

,

20

1

432

e

ionapproximattheatarrivewetermsixththeatStopping

e

ionapproximattheatarrivewetermfourththeatStopping

nn

nee

n

xxxxxe

haveWe

nn

nx

Question

Approximate 3√e

Able’s Theorem

],[

).(

)(

&.2

)()(

&)(.1

;

),(;)(

0

0

0

0

0

0

rronfunctiontherepresentsseriesthethenrandratconverges

whichrronxabydrepresenteandrandrbothatcontinuousisfIf

Corollary

rfraThen

rxatcontinuousisfconvergesraseriestheIf

rfraThen

rxatcontinuousisfconvergesraseriestheIf

Then

rrxxaxf

Let

n

nn

n

nn

n

nn

n

nn

n

nn

n

nn

toconvergesthatseriesafindtocresulttheUsed

ffunctiontherepresentsseriesthewhichonervalexacttheeertotheorem

sAbleandbandaresultsthewithtogotherexampletheofresulttheUsec

xand

xatcontinuousisitthatshowandfGraphb

xatandxatconvergesseriesthethatshowa

onxxf

functiontherepresentsn

xseriesthethatshownhaveWe

ecxampletobackGo

toconvergesthatseriesafindtocresulttheUsed

ffunctiontherepresentsseriesthewhichonervalexacttheeertotheorem

sAbleandbandaresultsthewithtogotherexampletheofresulttheUsec

xatcontinuousisitthatshowandfGraphb

xatconvergesseriesthethatshowa

onxxffunctiontherepresentsn

xseriesthethatshownhaveWe

ecxampletobackGo

n

nn

n

n

..

.intmindet

'...

1

1.

11.

)1,1(arctan)(

12)1(

)3(.1

2ln..

.intmindet

'...

1.

1.

)1,1()1ln()(

)2(.1

0

12

1

Home Quiz (2)

Homework

2

4

3

2

2

4

)1()()()9(

)arctan()()8(

3arctan)()7(91

1)()6(

)23ln()()5(

)3(

1)()()4(

3)()3(

23

5)()2(

3

1)()1(

,

x

xxfxf

xf

xxfx

xf

xxf

xxfxf

x

xxf

xxf

xxf

ffortionrepresentaseriespowerafindseiesgeometricafromStarting

x