Representation of functions by Power Series
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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.
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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
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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.
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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
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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
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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
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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
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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!
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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
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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
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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
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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
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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
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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