Representation of functions by Power Series

We will use the familiar converging geometric series form to obtain the power series representations of some elementary functions.

If ırı < 1, then the power series:

The sum of such a converging geometric series is:

0

converges.n

n

ar

0 1n

n

aar

r

Conversely, if an elementary function is of the form:

Its power series representation is:0

n

n

ar

1

a

r

By identifying a and r, such functions can be represented by an appropriate power series.

Find a geometric power series for the function:

0 1n

n

aar

r

We write in the form and identify 3

and 4 1

aa r

x r

3( )

4f x

x

43

344 4

4 xx

Make this 1 Divide numerator and denominator by 4

3/ 4

1 / 4x

a = 3/4 r = x/4

Use a and r to write the power series.

0

( ) n

n

f x ar

0

3

4 4

n

n

x

10

3

4

n

nn

x

2

1

3 3 3 3.... .....

4 16 64 4

n

n

xx x

Find a geometric power series for the function:3

( )4

f xx

Let us obtain the interval of convergence for this power series.

1 1

1

2

3 4

4 3

n n

n

n n

n

a x

a x

4

x

10

3( )

4

n

nn

xf x

2

1

3 3 3 3.... .....

4 16 64 4

n

n

xx x

The series converges for 14

x

-4 < x < 4

f(x)

naWatch the graph of f(x) and the graph of the first four terms of the power series.

The convergence of the two on (-4, 4) is obvious.

Find a geometric power series centered at c = -2 for the function:

0

( )1 ( )

n

n

aa r c

r c

We write in the form3

4 1 (

)

a

x r c

3( )

4f x

x

31/ 2

6 ( 2) 1 ( 2) / 66

6 6x x

Make this 1Divide numerator and denominator by 6

a = ½ r – c = (x + 2)/6

0

( ) ( )nn

f x a r c

0

1 2

2 6

n

n

x

0

( 2)

2 6

n

nn

x

The power series centered at c is:

( 2

4 [ )]

3

2x

3

26

( )x

2

2

1 2 ( 2) ( 2)....... ....

2 2 6 2 6 2 6

n

n

x x x

x = -2

f(x)

na

Find a geometric power series centered at c = 2 for the function:

3( )

2 1f x

x

We write in the standard form 3

2 1 1

a

x r

3 3

2 1 1 2x x

Since c = 2, this x has to become x – 2

3

1 2( 42)x

Add 4 to compensate

for subtracting 4 3

3 2( 2)x

Divide by 3 to make

this 1 3/

3 2(

3

2)3 3

x

3

12

1 ( 2)x

Make this – to bring it to standard form

3

12

1 ( 2)x

Find a geometric power series centered at c = 2 for the function:

3( )

2 1f x

x

3

( )2 1

f xx

3

12

1 ( 2)x

1a 2( 2)

3r c x

0

( ) ( )nn

f x a r c

0

21 ( 2)

3

n

n

x

0

2 ( 2)( 1)

3

n nn

nn

x

2 2 3 3

2 3

2( 2) 2 ( 2) 2 ( 2) 2 ( 2)1 ........ ( 1) ......

3 3 3 3

n nn

n

x x x x

This series converges for:2

( 2) 13x

3( 2)

2x

3 3( 2)

2 2x

1 7

2 2x

x = 2

Find a geometric power series centered at c = 0 for the function:

4 4/44 4

4

4xx

11

14x

11

14x

Make this – to bring it to standard form

2

4( )

4f x

x

We first obtain the power series for 4/(4 + x) and then replace x by x2

1a 1

4r x

0

n

n

ar

0

11

4

n

n

x

0

( 1)4

nn

nn

x

2

4( )

4f x

x

0

2

( 1)4

n

n

nn

x

Replace x by x2

2

0

( 1)4

nn

nn

x

2 4 6 2

2 4 61 ..... ( 1) .....

4 4 4 4

nn

n

x x x x

2The series converges for 4x 2 2x

f(x)

na

Find a geometric power series centered at c = 0 for the function:

We obtain power series for 1/(1 + x) and 1/(1 – x) and combine them.

2

2 1 1( )

1 1 1

xh x

x x x

1 1

1 1 ( )x x

0

1( )nn

x

0

( 1)n n

n

x

0

1

1n

n

xx

2

2 1 1( )

1 1 1

xh x

x x x

0 0

( 1)n n n

n n

x x

0

( 1) 1n n

n

x

3 5 2 1

0

2 2 2 ..... 2 n

n

x x x x

2 1

0

2 n

n

x

f(x)

na

STOPHERE!

Find a geometric power series centered at c = 0 for the function:

We obtain power series for 1/(1 + x) and obtain the second derivative.

1 1

1 1 ( )x x

0

1( )nn

x

0

( 1)n n

n

x

1

1

d

dx x

2

3 2

2 1( )

( 1) 1

df x

x dx x

0

( 1)n n

n

dx

dx

1

1

( 1)n n

n

nx

2

2

1

1

d

dx x

1

1

( 1)n n

n

dnx

dx

2

2

( 1) ( 1)n n

n

n n x

Replace n by n + 2 to make index of

summation n = 0

2

0

22 2( 1) ( )( 1)n n

n

xn n

0

( 1) ( 2)( 1)n n

n

n n x

Find a geometric power series centered at c = 0 for the function:

We obtain power series for 1/(1 + x) and integrate it

1 1

1 1 ( )x x

0

1( )nn

x

0

( 1)n n

n

x

1

1dxx

0

( 1)n n

n

x dx

1

10

( 1)1

nn

n

xc

n

2 1 1( ) ln(1 )

1 1f x x dx dx

x x

Then integrate the power series for 1/(1 – x ) and combine both.

0

1

1n

n

xx

1

1dxx

0

n

n

x dx

1

20 1

n

n

xc

n

Find a geometric power series centered at c = 0 for the function:

2 1 1( ) ln(1 )

1 1f x x dx dx

x x

1

1dxx

1

20 1

n

n

xc

n

1

1dxx

1

10

( 1)1

nn

n

xc

n

2( ) ln(1 )f x x 1

10

( 1)1

nn

n

xc

n

1

20 1

n

n

xc

n

1

0

( 1) 11

nn

n

xc

n

2 4 62 2 2

.......2 4 6

x x xc

2 2

0

2

2 2

n

n

xc

n

2 2

0

( 1)

1

n

n

xc

n

2 2

2

0

( 1)ln(1 )

1

n

n

xx c

n

When x = 0, we have

0 = 0 + c c = 02 2

2

0

ln(1 )1

n

n

xx

n

Find a geometric power series centered at c = 0 for f(x) = ln(x2 + 1)

2

2

2ln( 1)

1

d xx

dx x

We obtain the power series for this and integrate.

2 2

2 2

1 1 ( )

xx

x x

2

0

2( 1)n n

n

x x

2 1

0

( 1) 2n n

n

x

2 2 1

0

ln( 1) ( 1) 2n n

n

x x dx

2 2

0

2( 1)

2 2

nn

n

xc

n

2 2

0

2

2( 1)

( 1)

nn

n

xc

n

2 2

0

( 1)1

nn

n

xc

n

When x = 0, we have 0 = 0 + c c = 0

2 22

0

ln( 1) ( 1)1

nn

n

xx

n

-1 < x < 1

f(x)

na

Find a power series for the function f(x) = arctan 2x centered at c = 0

2

2arctan 2

1 4

dx

dx x

We obtain the power series for this and integrate.

1 1

1 1 ( )x x

0

( 1)n n

n

x

2

2

1 4x

Replace x by 4x2

2

0

2 ( 1) 4nn

n

x

2

2arctan 2

1 4x dx

x

2

0

2 ( 1) 4nn

n

x dx

2

0

2 ( 1) 4n n n

n

x dx

2 1

0

2 ( 1) 42 1

nn n

n

xc

n

Find a power series for the function f(x) = arctan 2x centered at c = 0

2 1

0

arctan 2 2 ( 1) 42 1

nn n

n

xx c

n

2 1

2

0

2 ( 1) 22 1

nn n

n

xc

n

2 12 1

0

( 1) 22 1

nn n

n

xc

n

When x = 0, arctan 2x = 0 0 = 0 + c c = 02 1

2 1

0

arctan 2 ( 1) 22 1

nn n

n

xx

n

1 1

2 2x

3 5 78 32 1282 ....

3 5 7

x x xx

f(x) = arctan 2x

-0.5 -0.5

na