11.

9 Exercises

8. Find a power series representation for the function and determine the interval of
convergence:
x
f ( x )=
2 x2 +1
∞
x n
Answer: 2
=∑ (−2 ) . x 2 n+1
2 x +1 n=0
Steps: Expressing the given function in a form that allows for determining the initial term
and common ratio:
x x
=
2 x +1 1−(−2 x 2)
2

From this expression, we can determine the following:

Initial term ( a )=x∧common ratio ( r ) =−2 x 2

Therefore, we can now express the given function as a power series, using the formula for
sum of a geometric series:
∞
a
S=∑ a r n=
n=0 1−r
∞ ∞
n
f ( x )=∑ x(−2 x 2)n =∑ (−2 ) . x 2 n+1
n =0 n=0

This is power series representation of the given function.

 11. Express the function as the sum of a power series by first using partial fractions.
Find the interval of convergence.
3
f ( x )= 2
x −x−2
Answer:
∞ n +1
1
f ( x )=∑ (−1 ) +
n =0
[ () ]n

2
. xn

Radius of convergence=1
Steps: To find the partial fraction form of the given function, we rewrite it as:
3 A B
= +
( x+1 ) ( x−2) x+1 x −2
By using suitable manipulation and simplifying:
3 A x−2 B x +1
= . + .
(x +1)(x −2) x+ 1 x−2 x−2 x +1
3 Ax−2 A+ Bx+ B
=
(x +1)(x −2) ( x +1)(x−2)
Comparing coefficients on both sides, we get the following two linear equations:
A+ B=0∧−2 A+ B=3
Solving for A and B, we get A = -1 and B = 1.
3 A B −1 1
f ( x )= = + = +
x −x−2 x+1 x−2 x +1 x−2
2

This is partial fraction form of the given function. Now we find the power series for each
individual term.
−1
To find the power series for , we rewrite the function in a form that allows for
x+1
determining the initial term and common ratio:
−1 −1
=
x+1 1−(−x )

From this expression, we can determine the following:

Initial term ( a )=−1∧common ratio ( r )=−x
Therefore, we can now express the given function as a power series, using the formula for
sum of a geometric series:
∞
a
S=∑ a r n=
n=0 1−r
 ∞ ∞
n n+1
1 st Term=∑ (−1)(−x) =∑ (−1 ) . xn
n=0 n=0

1
To find the power series for , we rewrite the function in a form that allows for
x−2
determining the initial term and common ratio. Divide numerator and denominator by 2:
1 1
−
1 2 2
= =
x−2 x x
−1 1−( )
2 2
From this expression, we can determine the following:
−1 x
Initial term ( a )= ∧common ratio ( r )=
2 2

Therefore, we can now express the given function as a power series, using the formula for
sum of a geometric series:
∞
a
S=∑ a r n=
n=0 1−r
∞ ∞
−1 x n n+1
1
2 nd Term=∑ (
n=0
)( ) =∑ (−1 )
2 2 n=0 2 () . xn

Combining both these results:

∞ ∞ n+1
1
f ( x )=1 st Term+2 nd Term=∑ (−1 )
n=0
n+1
. x n +∑ (−1 )
n=0
()
2
. xn

Combining these two terms and simplifying, we get:

∞ n+1
1
f ( x )=∑ (−1 )
n =0
n+1
. x n + (−1 ) ()
2
. xn

∞ n +1
1
f ( x )=∑ (−1 )
n =0
[ n+1
+(−1) ()2 ] . xn

To determine interval of convergence, we can apply the ratio test. Applying ratio test, the
series will converge if the following is satisfied:
 an +1
lim
n→∞ | | an
<1

For the sake of simplicity, we will write these terms separately, simplify them, making it
convenient and avoiding lengthy equations.

n+1
1
[
a n= (−1 )
n +1
+(−1) ()
2 ] . xn

n +1+1
1
[
a n+1= (−1 )
n++11
+(−1) ()2 ] . x n +1

n+ 2
1
[ ( ) ]. x
|[ (−1 )n+2 +(−1)

|
n+1

an +1 2
lim
n→∞ | | an
=lim
n →∞
(−1 )
n +1 1
+(−1) ( ) ] . x
n+1
n
2

n
1
We know that lim
n→∞ |( ) |
2
=0

[ (−1 )n+2 +(−1)( 0) ] . x =¿−x∨¿

lim
n→∞ | [ (−1 )n+1 +(−1)( 0) ] |
For the series to be convergent:
|−x|<1

The interval of convergence is (-1, 1) and radius of convergence is 1.

 13. (a) Use differentiation to find a power series representation for
1
f ( x )=
( 1+ x )2
What is the radius of convergence?
(b) Use part (a) to find a power series for
1
f ( x )=
( 1+ x )3
(c) Use part (b) to find a power series for

x2
f ( x )=
( 1+ x )3
Answer: (a)
∞
n
f ( x )=∑ (−1 ) ( n+1 ) x n
n =0

Radius of convergence is 1.
∞
1 n
(b) f ( x )= ∑ (−1 ) ( n+1 ) ( n+2 ) x n
2 n=0

Radius of convergence is 1.
∞
1 n
(c) f ( x )= ∑ (−1 ) ( n ) ( n−1 ) x n
2 n=2

Radius of convergence is 1.

Steps:

(a) We start by integrating the given function. Substitute u = 1 + x and dx = du:

1 1 −1
∫ ( 1+ x )2 dx=∫ u2 . du= u
+C

Substituting back for ‘u’, we get:

−1 −1
+C= +C
u 1+ x
Converting the equation to a form that can be represented as power series:
−1 −1
+C= +C
1+ x 1−(−x )
 This is a geometric series with:

Initial term ( a )=−1∧common ratio ( r )=−x

∞ ∞
−1 n n +1
+C=∑ (−1 )(−x ) +C ∑ (−1 ) . x n+ C
1−(−x ) n=0 n=0

Differentiate this term to reach where we want:

∞ ∞
d
dx ( n=0
)
∑ (−1 )n+1 . x n +C =∑ (−1 )n+1 . n. x n−1
n=0

Expanding the terms we get the power series expression for the given function:
∞ ∞
n+1 n
f ( x )=∑ (−1 ) .n . x n−1=0+1−2 x +3 x 2 … ..=∑ (−1 ) . ¿ ¿
n =0 n=0

Applying the ratio test to check for convergence and determine radius of convergence:

a n=¿

Applying the ratio test rule:

an+1
lim
n→∞ |( )| an
<1

lim ¿ ¿
n→∞

Simplifying, and later dividing the numerator and denominator with ‘n’, we get:
lim ¿ ¿
n→∞

lim ¿ ¿
n→∞
 2
lim
n→∞
(−1).(n+ 2). x
|( (n+1)
=lim
n→∞ )| |( (−1).(1+ ) . x
1
(1+ )
n
n
=
)| |
(−1 ) (1+ 0 ) x
1 |
=¿−x∨¿

The series converges when:

|x|<1

Therefore the radius of convergence is 1.

(b) We can start differentiating directly from part (a), since we know that the extra term
−1
that needs to be included is , which is a result of the first integration.
2

∞ ∞
d −1
f ( x )= ( ∑
dx 2 n=0 )
(−1 )n . ( n+ 1 ) . x n+C =
−1
∑
2 n=0
(−1 )n . n . ( n+1 ) . x n−1

Same as done previously, we can expand the term and rewrite the series as:

∞ ∞
−1 n 1 n
∑ (−1 ) . n. ( n+1 ) . xn −1 = ∑ (−1 ) .( n+2) . ( n+1 ) . x n
2 n=0 2 n =0

The radius of convergence for a differentiated series will be the same as that of the
original series. The radius of convergence is 1 for the new series as well.

(c) With the inclusion of the term in the numerator, we know that the function is x 2 times
the series in (b).
 ∞
1 n
f ( x )= ∑ (−1 ) . ( n+2 ) . ( n+ 1 ) . x n+2
2 n=0

Applying the ratio test to check for convergence and determine radius of convergence:

a n=(−1 )n . ( n+ 2 ) . ( n+1 ) . x n +2∧an +1=(−1 )n+ 1 . ( n+3 ) . ( n+2 ) . x n+3

Applying the ratio test rule:

an+1
lim
n→∞ |( )| an
<1

n +1
(−1 ) . ( n+3 ) . ( n+2 ) . x n +3
lim
n→∞ |( )|
(−1 )n . ( n+2 ) . ( n+1 ) . xn +2
<1

Simplifying, and later dividing the numerator and denominator with ‘n’, we get:

lim
n→∞ |( (−1 ) . ( n+3 ) . x
( n+1 ) )| =lim
n→∞
|( )| ( 3n ) . x = (−1) .( 1+0 ) . x =¿ x∨¿
(−1 ) . 1+

( n)
1+
1 | 1+ 0 |

The series converges when:

|x|<1

Therefore the radius of convergence is 1.

20. Find a power series representation for the function and determine the radius of
convergence.
 x 2+ x
f ( x )=
( 1−x )3
Answer: The power series representation of the given series is:
∞
x2 + x
= n2 x n
3 ∑
( 1−x ) n=1
The series is convergent when |x| < 1
Steps: We first find the power series for the denominator term, by integrating it twice.
We substitute u=1−x∧dx=−du:

1 −1
∫ ( 1−x )3 . dx=∫ ( u )3 . du=(−1 ) (
−1
2 )( u1 )+C 1
2

Substituting back for u, we get:

(−1 ) ( −12 )( u1 )+C 1= 12 . ( 1−x1 ) +C 1

2 2

Integrating again, using substitution, we get:

1 1 1
∫ ( .
2 ( 1−x ) 2 )
+C 1 . dx=
2(1−x)
+C 1 x+C 2

1
We can rewrite in a form of a power series, having
2(1−x )
1
Initial term ( a )= ∧common ratio ( r )=x
2
∞
1 1
+C 1 x+C 2= ∑ x n +C 1 x+C 2
2(1−x ) 2 n=0

To reach back to what was required, we differentiate twice, the above expression:

∞ ∞
d 1
(∑ x n +C 1 x+ C 2 = 12 ∑ nx n−1+ C 1
dx 2 n=0 )
n=0

Differentiating again:
∞ ∞
d 1
(∑ nx n−1 +C 1 = 12 ∑ n. ( n−1 ) . x n−2
dx 2 n=0 )
n=0
Now bringing back the numerator term which we earlier neglected:
∞
x 2+ x 1 n−2
f ( x )= 3
= ∑ n . ( n−1 ) . ( x 2+ x ) . x
( 1−x ) 2 n=0
Simplifying:
∞ ∞
1 n −2 1
∑ n . ( n−1 ) . ( x 2+ x ) . x = ∑ n . ( n−1 ) . ( x n + x n−1 )
2 n=0 2 n =0

Separating the terms, we finally get:

∞ ∞ ∞
1 n n−1 n. ( n−1 ) . ( x n ) n. ( n−1 ) . ( x n−1 )
∑ n . ( n−1 ) . ( x + x ) =∑
2 n=0 2
+∑
2
n=2 n=2

We know that:
∞ ∞
n . ( n−1 ) . ( x n−1) n . ( n+1 ) . ( xn )
∑ 2
=∑
2
n =2 n=1

Using this, we can rewrite the sum as:

∞ ∞ ∞ ∞
n . ( n−1 ) . ( x n ) n . ( n−1 ) . ( x n−1) n . ( n−1 ) . ( x n ) n . ( n+1 ) . ( xn )
∑ 2
+∑
2
=∑
2
+∑
2
n =2 n =2 n=2 n=1

Since the term for n = 1 is zero, we can further rewrite the sum as:
∞ ∞ ∞ ∞
n . ( n−1 ) . ( x n ) n . ( n−1 ) . ( x n ) n. ( n−1 ) . ( x n ) n. ( n+1 ) . ( x n )
∑ 2
+ ∑ 2
=∑ 2
+∑ 2
n =2 n =1 n=1 n=1

Simplifying this further, we get:

∞ ∞ ∞
n . ( n−1 ) . ( x n ) n . ( n−1 ) . ( x n ) n. ( n−1 ) . ( x n ) n . ( n+1 ) . ( x n)
∑ 2
+ ∑ 2
=∑ 2
+
2
n =1 n =1 n=1

∞ ∞ ∞
n . ( n−1 ) . ( x n ) n . ( n+1 ) . ( x n ) n . ( xn )
∑ + = ∑ . ( 2n )=∑ n2 x n
n =1 2 2 n=1 2 n=1

This is the final power series representation of the given series.

Applying the ratio test to check for convergence and determine radius of convergence:

a n=n2 x n∧an +1=( n+1 )2 . x n +1

Applying the ratio test rule:

an+1
lim
n→∞ |( )| an
<1

( n+ 1 )2 . x n+1
lim
n→∞ |( )| n2 xn
<1

1
Simplifying, and dividing the numerator and denominator with we get:
n2

1 2
lim
n→∞ |(
2
( n+ 1 ) . x n+1
2 n
n x )| |(
= lim
n→ ∞
2

)|
( n+1 ) . x
n2
= lim
n →∞
|(( ) )|
1+
n
1
.x

1 2
lim
n→∞
|( ( ) )|
1+
n
1
.x
=1.|x|=¿ x∨¿

The series is convergent if:

|x|<1

Therefore the radius of convergence is 1.