CHAPTE

Power Series
R4
Example 1:
Power Series ∞

∑ 2𝑛 𝑥 𝑛 = 1 + 2𝑥 + 4𝑥 2 + ⋯
𝑛 =0

Sequence a list of the numbers that are in order,

denoted by:
Curly brackets { } or braces {2,4,6,8, 10, …} or Starting from 0, substitute 0 to 𝑛 in the equation
briefly 2𝑛𝑥𝑛:
{ xn }
2(0)𝑥0 = 1
Series is when we add the numbers in a sequence:
Substitute 1 to 𝑛:
2+4+6+8+10+ …
2(1)𝑥1 = 2
Partial Sum is when sum up only a part in a
sequence: Substitute 2 to 𝑛:
Partial sum of the first 3 terms: 2+4+6 = 12
2(2)𝑥2 = 4

Infinite Series an infinite sequence is summed up And so on. Simply add “…” since the series is from
0 up to infinity (∞).
Power Series is the sum of all the terms in a
sequence which is a set of numbers in order.
Example 2:
A power series in powers of x – x0 is a series of the
form:

∞ 𝑛=0

Starting from 0, substitute 0 to 𝑛 in the equation

�𝑛 (𝑥 − 𝑐)𝑛 = �0 + �1(𝑥 − 𝑐) + �2(𝑥 − 𝑐)2
+⋯ 1 𝑛:
2𝑛+2 𝑥 2 01+2 𝑥0
Where 𝑥 is a complex variable, �0, �1, …
are complex (or real) constants called the
= 14 Substitute 1 to 𝑛:
coefficients of the series, and 𝑐 is a complex (or
real) constant, called the center of the series. If 𝑐 =
0, we obtain as a case of a power series in powers 2 11+2 𝑥1 = 18 𝑥 Substitute 2 to 𝑛:
of x:
∞

∑ �𝑛 𝑥𝑛 = �0 + �1𝑥 + �2𝑥2 + ⋯ And so on. Simply add “…” since the series is from
𝑛=0 0 up to infinity (∞).

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 Exercise Problems
Example:
∞ 𝑛
(𝑥−1)
∑ n! 1. When a = 1 and r = x
𝑛 =0

∞ 𝑛
(𝑥+2)
∑ (−1)𝑛 n! 2.
𝑛 =0
Wherein 1 is the initial value and x is the value you
need to multiply to get the 2nd term. Just
ANSWERS: remember, if the you change the sign before the x,
the geometric sequence also changes.
1. 1 + (𝑥 − 1) + + +
⋯
𝟐! 𝟑!
Example:
2. 1 − (𝑥 + 2) + + +
When a = 1 and r = - x
⋯
𝟐! 𝟑!

GeometricPowerSeries Other examples:

When a = x and r = 2x
Geometric series are among the simplest examples
of infinite series with finite sums, although not all of
them have this property. Historically, geometric
series played an important role in the early
development of calculus, and they continue to be When a = 2 and r = - 5x
central in the study of convergence of series.
Geometric series are used throughout mathematics,
and they have important applications in physics,
engineering, biology, economics, computer science,
queueing theory, and finance.

Geometric series has the formula, � 1−�. Which is

equivalent to: Exercise Problems

� 1− � = � + �� + ��2 + ��3 + ⋯ Find the Geometric Series of the following:

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It only works when a and rare functions of x, where 1.
a is the initial value and x is the value you need to 1−3𝑥

multiply to get the 2nd term. 2𝑥

2
 2. 𝑛
1−𝑥 �𝑛𝑑
2
𝑛
3. 1−𝑥2 �
𝑛
𝑥2
then
4.
1−2𝑥

4𝑥2
� 𝑛
5. 𝑛
1+4𝑥
3

MULTIPLICATION
6. 1−5𝑥2
𝑥2
With the same definitions for f(x) and g(x), the
7. power series of the product of the functions can
1+2𝑥 be obtained as follows:
1
8.
1+2𝑥 �
𝑥2
9.
1−4𝑥

10.
𝑥

Power Series Table

Operations inower
P Series
Here are some functions and its equivalent power
series:

ADDITION AND SUBTRACTION �

𝐬� �
When two functions f and g are decomposed into � � �
� 𝟑 �
power series around the same center c, the power
�
series of the sum or difference of the functions can
𝑥 𝑥 �
be obtained by term wise addition and
� �
subtraction.
�
That is, if 𝐬
�
� 𝑛

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 � (
𝐭� � 𝟏
� � +
� � �
� 𝑥 )
� �
�𝐭
� 𝑥
Exercise Problems
𝐬
�
�
� 1. Find the
�𝑥 power
𝐬� � series
� representation for tan−1 𝑥
−𝟏
2. Find the power series representation for
�
ln (1 + 𝑥)
� 𝑥
�
ANSWERS:
� 𝑥
𝐬− 1. 𝑡�𝑛 𝑑𝑥
𝟏
𝑛
�
𝑛
𝐭�
𝑥� 𝑛
𝑑𝑥
� 𝑥 �
�
−𝟏
�
� = (−1)𝑛 (( 𝑥2𝑛+1))

𝑥
� 𝑥
� Starting from 0, substitute 0 to 𝑛 in the
𝑛
𝑥 𝑥 ((𝑥2𝑛+1)):
𝐥� equation (−1)
𝑥

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 Substitute 3 to 𝑛:
0
((𝑥2(0)+1)) = 𝑥 (𝑥3+1) 𝑥4
(−1) (−1)3 ( )=−
4

Substitute 1 to 𝑛:

(𝑥2(1)+1) 𝑥3 𝑥2 𝑥3 𝑥4
ln(1 + 𝑥) = 𝑥 − + − +⋯
(−1)1 ( )=− 2 3 4
3

Substitute 2 to 𝑛:

(𝑥2(2)+1) 𝑥5
Theorems of Power Series
= (−1)2 ( )=−
5

Substitute 3 to 𝑛: Theorem 1
3 (𝑥 2(3)+1 ) 𝑥7
= (−1) ( ) =−
2(3) + 1 7 The power series

𝑥3 𝑥5 𝑥7 � �𝑛𝑥𝑛
𝑡�𝑛 −1 𝑥 = 𝑥 − + − +⋯ 𝑛
3 5 7
is convergent for
|x| < R,
2. 𝑑𝑥 where R is the radius of convergence. Moreover,

𝑛 𝑥𝑛𝑑𝑥
� �𝑛𝑥𝑛
𝑛

is continuous and infinitely differentiable within

Starting from 0, substitute 0 to 𝑛 in the the interval of convergence,
𝑛+1 |x| < R

equation (−1)𝑛 ( ): Note/Proof:

𝑛
The convergence properties of the power series
(𝑥0+1) are a consequence/result of the ratio test.
(−1)0 ( )=𝑥

Substitute 1 to 𝑛: Theorem 2
(𝑥1+1) 𝑥2
(−1)1 ( )=− If the power series
2
� �𝑛𝑥𝑛
Substitute 2 to 𝑛: 𝑛
is convergent at x = R, then it is a continuous
(𝑥2+1) 𝑥3
function within the interval of convergence
(−1)2 () =
3

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including the endpoint at x = R. Look at this Eq. 2
example:
Setting x = 1 above yields the divergent series
Example 1: 1−1+1−1+1−1+· · ·. Hence, the conditions of Abel’s
theorem are not satisfied, in which case we cannot
conclude that
� �𝑛𝑥𝑛
𝑥→
𝑛 𝑛𝑥 𝑛
𝑛=0

� �𝑛𝑥𝑛 is continuous at x = 1. For x = 1, the left-hand side

𝑛 of eq. (2) yields 1/2. Although one can make a case
for assigning 1 2 to the series 1 − 1 + 1 − 1 + 1 − 1 +
� 𝑛 �𝑛�𝑛
· · ·, the latter series is clearly not convergent
𝑛=0 according to the standard mathematical definition
of convergence.
where lim 𝑥→−�+ means that x approaches R
from the right, i.e. from inside the interval of
convergence, |x|<R.
Theorem 3
Application:
Consider a power series
∞ 𝑥𝑛
ln(1 + 𝑥) = ∑ (−1)𝑛+1 ,𝑥<1 � �𝑛𝑥𝑛
𝑛=0
𝑛
𝑛=1
with radius of convergence R. Then, term-by-term
Eq.1 differentiation and integration of the power series
is permitted and does not change the radius of
In this case, the radius of convergence is R =
1. Moreover, if we set x = 1 in the convergence. That is,
equation, the resulting series is
∞ ∞
conditionally convergent (because of
the alternating series test). Thus, the 𝑑� = 𝑑 𝑛
= ∑ 𝑑 �𝑛𝑥𝑛
power series for ln(1+x) is continuous ∑ �𝑛𝑥
at x = 1, which allows us to conclude 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑛=0 𝑛=0
that:
ln(2) = ∑
𝑛 𝑛�𝑛𝑥𝑛−1 , |𝑥| < �
𝑛=1
𝑛=0 Eq. 3

The converse of Abel’s theorem is sometimes

false. As an example, look on the infinite
geometric series, ���� �=�

∑𝑛𝑥𝑛
1+𝑥 𝑑𝑥�𝑛𝑥𝑛
𝑛=1 𝑛=0

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 ∞ 𝑛
�𝑛𝑥𝑛+1
=∑
𝑛+1 � ���𝑛 �𝑥𝑛 , |𝑥| < �
𝑛
𝑛=0

Eq. 4 Eq. 8

For values of x within the interval of where the radius of convergence of the sum and of
convergence, |x| < R, it is permissible to the product is at least as large as the minimum of
interchange the order of the infinite R1 and R2, i.e. R ≥ min {R1, R2}. The subtraction of
summation and the differentiation or two series is then defined simply by changing the
integration. This feature is one of the reasons signs of all the bn above before adding the two
that power series are so nice—they behave for series. The division of the two series, f1(x)/f2(x),
the most part-like ordinary polynomials. can be performed if and only if b0 6= 0. If this
condition holds,
Note/Proof:
The convergence properties of the power
series are a consequence of the ratio test. (See
note on
(
Theorem 1) 𝑐𝑛𝑥𝑛 , |𝑥| < � ′
𝑛

Theorem 4 Eq. 9
Given two power series with radii of where the radius of convergence satisfies R′ ≥ min
convergence {R1, R2, x0}, with x0 identified as the zero of f 2(x)
R1 and R2, respectively, nearest to x = 0. The coefficients c n in eq. (9) are
determined recursively using:
∞
𝑛
�1(𝑥) = ∑ �𝑛𝑥𝑛 , |𝑥| < �1
𝑛=0 �0

Eq. 5 𝑐0 = , 𝑐𝑛 ��𝑐𝑛
�0
,𝑛 …
� �𝑛𝑥𝑛 , |𝑥| < �2
𝑛

Eq. 6 Theorem 5

then the sum and product of the two-power The power series representation of a function,
series are given respectively by:
� �𝑛𝑥𝑛
𝑛
𝑛
� 𝑥 , |𝑥| < �
𝑛 with nonzero radius of convergence |x|< R is
unique.
Eq. 7
Note/Proof:

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This is a consequence of Taylor’s theorem in For this example, we need to notice that,
calculus, which provides an explicit formula for ∞ 𝑥𝑛
the coefficients of a power series, �𝑥 = ∑
𝑛!
𝑑𝑛� 𝑛=0

�𝑛 = 𝑛! 𝑑𝑥𝑛 |𝑥=0
Therefore,
∞ 2

�−𝑥2 = ∑ ( −𝑥 )𝑛
Power Series
𝑛!
Representation 𝑛=0

Multiplicatio = 1 − 𝑥2+ 𝑥4 − 𝑥5 − ⋯
n 2 6
We use Multiplication as another
technique in getting new power series We already know from the table that 𝑐𝑜𝑠𝑥 = 1 −
𝑥2 𝑥4 𝑥6
representation. Suppose that
so,
f (x) =
�−𝑥2 • cosx
(�𝑛𝑥𝑛)
� = ( 1 − 𝑥2 𝑥4
) • ( 1 − 𝑥2 + 𝑥4 )
� + 2 2! 4!
𝑥2 4

= 1 − −+𝑥2 + 𝑥4 + 𝑥4 + ⋯
g (x) =
𝑥

2 24 2 2
(�𝑛𝑥𝑛)
𝑛
= 1+(
3 25
=1− 2
𝑥 2 + 24 𝑥 4 + ⋯
Then we can get the power series for f (x)
• g (x) by multiplying these equations.

(�𝑛𝑥𝑛) (�𝑛𝑥𝑛) Example 2:

𝑛 𝑛 �𝑥
Find the power series representation of
1−𝑥
=( • (�0 + �1𝑥
+ �2𝑥3 + = �𝑥 • 1

1−𝑥

= �0 �0 + ( �0 �1 + �1 �0 )𝑥
We know that,
+ ( �0 �2 + �1 �1 + �2 �0 )𝑥 2 + ⋯
𝑥 𝑥2 𝑥3
�
and 1
=(1+
Example 1: 1−𝑥
𝑥 + 𝑥2 + 𝑥3 + ⋯ )
Find the power series representation of �−𝑥2 •
cosx Therefore,

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 �𝑥 So,
1
1 −𝑥
2𝑥
1+𝑥
= ( 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ ) • ( 1 + 𝑥 + 𝑥2 + 𝑥3
+ = 2𝑥 [1 − 𝑥 + 𝑥2 − 𝑥3 + ⋯ ]
2 6
= [2𝑥(1) − 2𝑥(𝑥) + 2𝑥(𝑥2) − 2𝑥(𝑥3) + ⋯ ]
⋯)
= 2𝑥 − 2𝑥2 + 2𝑥3 − 2𝑥4 + ⋯
1 + 1) 𝑥3 + ⋯
5 8
= 1 + 2𝑥 + 𝑥 2 + 𝑥 3 + ⋯
2 3

Substitution Alternatively, in shorthand:

We derive the series for a given
function using another function for which we �
already have a power series representation. 𝑛 𝑛
Then, we do the following: 𝑥

1. We need to figure out which substitution can

= ∑ 2𝑥(−1)𝑛𝑥𝑛
be applied to transform the function we know
the power series representation to transform
the function we want to have a series = ∑(−1)𝑛2𝑥𝑛+1
representation. 𝑛=0

2. We need to apply the same substitution to the

known series. Doing so will give us the series
representation we wanted. Example 2: Find the series representation for
=
3. The domain of the new function can be �(𝑥) 1

obtained by applying the same substitution of

4−𝑥
the domain of the known series
representation. We know that the geometric series is

𝑥𝑛
Example 1:
1−𝑥
Find a power series representation for �(𝑥) = 1 1 1
2𝑥
�(𝑥) = 4 − 𝑥 = 4 [ 1 − 𝑥 ]
1+𝑥 4
2𝑥 1 1
Since = 2𝑥 , we have With 𝑥 = 𝑥 , 𝑤� 1−
𝑥 ��𝑡 = 1 + 𝑥 + ( 𝑥)
2
+(
𝑥 3
1+𝑥 1+𝑥 ) +
44 4 4
4

⋯
Therefore,
= 1 − 𝑥 + 𝑥 2 − 𝑥3 + ⋯ 1 𝑥 𝑥2 𝑥3

9
 = 1 ++ +
+⋯
𝑥 𝑥2 𝑥3
=1+ + +
4 16 64
+⋯

Exercises:
𝑥2
1. Find a power representation of by
2+𝑥
substitution. 2. Power series
=
representation for �(𝑥) 1

8+3𝑥

𝑥3 𝑥3 𝑥3
3. Since 𝑠𝑖𝑛𝑥 , find the
𝑥
power series representation of
𝑠𝑖𝑛𝑥
𝑥
4. Power series representation of 1+𝑥2

5. Using substitution, find the power series

representation for �𝑥2

Answers:
𝑛
𝑥𝑛+2

1. (−1) 2𝑛+1

4. 𝑥 − 𝑥3 + 𝑥5 − 𝑥7 + ⋯
𝑥2𝑛
5.
𝑛!

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