Combining and Multiplying Power Series

Learning Outcomes

Combine power series by addition or subtraction
Create a new power series by multiplication by a power of the variable or a constant, or by substitution
Multiply two power series together

Combining Power Series

If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new power series, also with the same interval of convergence. Similarly, we can multiply a power series by a power of x or evaluate a power series at ${x}^{m}$ for a positive integer m to create a new power series. Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for $f\left(x\right)=\frac{1}{1-x}$ , we can find power series representations for related functions, such as

y = \frac{3 x}{1 - x^{2}} and y = \frac{1}{(x - 1) (x - 3)}

$y=\frac{3x}{1-{x}^{2}}\text{ and }y=\frac{1}{\left(x - 1\right)\left(x - 3\right)}$ .

In Combining Power Series we state results regarding addition or subtraction of power series, composition of a power series, and multiplication of a power series by a power of the variable. For simplicity, we state the theorem for power series centered at $x=0$ . Similar results hold for power series centered at $x=a$ .

theorem: Combining Power Series

Suppose that the two power series $\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}$ converge to the functions f and g, respectively, on a common interval I.

The power series $\displaystyle\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}\pm {d}_{n}{x}^{n}\right)$ converges to $f\pm g$ on I.
For any integer $m\ge 0$ and any real number b, the power series $\displaystyle\sum _{n=0}^{\infty }b{x}^{m}{c}_{n}{x}^{n}$ converges to $b{x}^{m}f\left(x\right)$ on I.
For any integer $m\ge 0$ and any real number b, the series $\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(b{x}^{m}\right)}^{n}$ converges to $f\left(b{x}^{m}\right)$ for all x such that $b{x}^{m}$ is in I.

Proof

We prove i. in the case of the series $\displaystyle\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)$ . Suppose that $\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}$ converge to the functions f and g, respectively, on the interval I. Let x be a point in I and let ${S}_{N}\left(x\right)$ and ${T}_{N}\left(x\right)$ denote the Nth partial sums of the series $\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}$ , respectively. Then the sequence $\left\{{S}_{N}\left(x\right)\right\}$ converges to $f\left(x\right)$ and the sequence $\left\{{T}_{N}\left(x\right)\right\}$ converges to $g\left(x\right)$ . Furthermore, the Nth partial sum of $\displaystyle\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)$ is

\begin{array}{cc} \sum_{n = 0}^{N} (c_{n} x^{n} + d_{n} x^{n}) & = \sum_{n = 0}^{N} c_{n} x^{n} + \sum_{n = 0}^{N} d_{n} x^{n} \\ = S_{N} (x) + T_{N} (x) . \end{array}

$\begin{array}{cc}\hfill {\displaystyle\sum _{n=0}^{N}} \left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)& ={\displaystyle\sum _{n=0}^{N}} {c}_{n}{x}^{n}+{\displaystyle\sum _{n=0}^{N}}{d}_{n}{x}^{n}\hfill \\ & ={S}_{N}\left(x\right)+{T}_{N}\left(x\right).\hfill \end{array}$

Because

\begin{array}{cc} lim_{N \to \infty} (S_{N} (x) + T_{N} (x)) & = lim_{N \to \infty} S_{N} (x) + lim_{N \to \infty} T_{N} (x) \\ = f (x) + g (x), \end{array}

$\begin{array}{cc} \hfill {\underset{N\to\infty}\lim} \left({S}_{N}\left(x\right)+{T}_{N}\left(x\right)\right) & ={\underset{N\to\infty}\lim} {S}_{N}\left(x\right)+{\underset{N\to\infty}\lim} {T}_{N}\left(x\right)\hfill \\ & =f\left(x\right)+g\left(x\right),\hfill \end{array}$

we conclude that the series $\displaystyle\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)$ converges to $f\left(x\right)+g\left(x\right)$ .

$_\blacksquare$

We examine products of power series in a later theorem. First, we show several applications of Combining Power Series and how to find the interval of convergence of a power series given the interval of convergence of a related power series.

Example: Combining Power Series

Suppose that $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ is a power series whose interval of convergence is $\left(-1,1\right)$ , and suppose that $\displaystyle\sum _{n=0}^{\infty }{b}_{n}{x}^{n}$ is a power series whose interval of convergence is $\left(-2,2\right)$ .

Find the interval of convergence of the series $\displaystyle\sum _{n=0}^{\infty }\left({a}_{n}{x}^{n}+{b}_{n}{x}^{n}\right)$ .
Find the interval of convergence of the series $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{3}^{n}{x}^{n}$ .

Show Solution

Since the interval $\left(-1,1\right)$ is a common interval of convergence of the series $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ and $\displaystyle\sum _{n=0}^{\infty }{b}_{n}{x}^{n}$ , the interval of convergence of the series $\displaystyle\sum _{n=0}^{\infty }\left({a}_{n}{x}^{n}+{b}_{n}{x}^{n}\right)$ is $\left(-1,1\right)$ .
Since $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ is a power series centered at zero with radius of convergence 1, it converges for all x in the interval $\left(-1,1\right)$ . By Combining Power Series, the series

\sum_{n = 0}^{\infty} a_{n} 3^{n} x^{n} = \sum_{n = 0}^{\infty} a_{n} {(3 x)}^{n}

$\displaystyle\sum _{n=0}^{\infty }{a}_{n}{3}^{n}{x}^{n}=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(3x\right)}^{n}$

converges if 3x is in the interval $\left(-1,1\right)$ . Therefore, the series converges for all x in the interval $\left(-\frac{1}{3},\frac{1}{3}\right)$ .

Watch the following video to see the worked solution to Example: Combining Power Series.

You can view the transcript for “6.2.1” here (opens in new window).

try it

Suppose that $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ has an interval of convergence of $\left(-1,1\right)$ . Find the interval of convergence of $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(\frac{x}{2}\right)}^{n}$ .

Hint

Find the values of x such that $\frac{x}{2}$ is in the interval $\left(-1,1\right)$ .

Show Solution

Interval of convergence is $\left(-2,2\right)$ .

In the next example, we show how to use Combining Power Series and the power series for a function f to construct power series for functions related to f. Specifically, we consider functions related to the function $f\left(x\right)=\frac{1}{1-x}$ and we use the fact that

\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n} = 1 + x + x^{2} + x^{3} + \dots

$\frac{1}{1-x}=\displaystyle\sum _{n=0}^{\infty }{x}^{n}=1+x+{x}^{2}+{x}^{3}+\cdots$

for $|x|<1$ .

Example: Constructing Power Series from Known Power Series

Use the power series representation for $f\left(x\right)=\frac{1}{1-x}$ combined with Combining Power Series to construct a power series for each of the following functions. Find the interval of convergence of the power series.

$f\left(x\right)=\frac{3x}{1+{x}^{2}}$
$f\left(x\right)=\frac{1}{\left(x - 1\right)\left(x - 3\right)}$

Show Solution

First write $f\left(x\right)$ as

f (x) = 3 x (\frac{1}{1 - (- x^{2})})

$f\left(x\right)=3x\left(\frac{1}{1-\left(\text{-}{x}^{2}\right)}\right)$ .

Using the power series representation for $f\left(x\right)=\frac{1}{1-x}$ and parts ii. and iii. of Combining Power Series, we find that a power series representation for f is given by

\sum_{n = 0}^{\infty} 3 x {(- x^{2})}^{n} = \sum_{n = 0}^{\infty} 3 {(- 1)}^{n} x^{2 n + 1}

$\displaystyle\sum _{n=0}^{\infty }3x{\left(\text{-}{x}^{2}\right)}^{n}=\displaystyle\sum _{n=0}^{\infty }3{\left(-1\right)}^{n}{x}^{2n+1}$ .

Since the interval of convergence of the series for $\frac{1}{1-x}$ is $\left(-1,1\right)$ , the interval of convergence for this new series is the set of real numbers x such that $|{x}^{2}|<1$ . Therefore, the interval of convergence is $\left(-1,1\right)$ .

To find the power series representation, use partial fractions to write $f\left(x\right)=\frac{1}{\left(1-x\right)\left(x - 3\right)}$ as the sum of two fractions. We have

\begin{array}{cc} \frac{1}{(x - 1) (x - 3)} & = \frac{- \frac{1}{2}}{x - 1} + \frac{\frac{1}{2}}{x - 3} \\ = \frac{\frac{1}{2}}{1 - x} - \frac{\frac{1}{2}}{3 - x} \\ = \frac{\frac{1}{2}}{1 - x} - \frac{\frac{1}{6}}{1 - \frac{x}{3}} . \end{array}

$\begin{array}{cc}\hfill \frac{1}{\left(x - 1\right)\left(x - 3\right)}& =\frac{\text{-}\frac{1}{2}}{x - 1}+\frac{\frac{1}{2}}{x - 3}\hfill \\ & =\frac{\frac{1}{2}}{1-x}-\frac{\frac{1}{2}}{3-x}\hfill \\ & =\frac{\frac{1}{2}}{1-x}-\frac{\frac{1}{6}}{1-\frac{x}{3}}.\hfill \end{array}$

First, using part ii. of Combining Power Series, we obtain

\frac{\frac{1}{2}}{1 - x} = \sum_{n = 0}^{\infty} \frac{1}{2} x^{n} for | x | < 1

$\frac{\frac{1}{2}}{1-x}=\displaystyle\sum _{n=0}^{\infty }\frac{1}{2}{x}^{n}\text{for}|x|<1$ .

Then, using parts ii. and iii. of Combining Power Series, we have

\frac{\frac{1}{6}}{1 - \frac{x}{3}} = \sum_{n = 0}^{\infty} \frac{1}{6} {(\frac{x}{3})}^{n} for | x | < 3

$\frac{\frac{1}{6}}{1-\frac{x}{3}}=\displaystyle\sum _{n=0}^{\infty }\frac{1}{6}{\left(\frac{x}{3}\right)}^{n}\text{for}|x|<3$ .

Since we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact and part i. of Combining Power Series, we have

\frac{1}{(x - 1) (x - 3)} = \sum_{n = 0}^{\infty} (\frac{1}{2} - \frac{1}{6 \cdot 3^{n}}) x^{n}

$\frac{1}{\left(x - 1\right)\left(x - 3\right)}=\displaystyle\sum _{n=0}^{\infty }\left(\frac{1}{2}-\frac{1}{6\cdot {3}^{n}}\right){x}^{n}$

where the interval of convergence is $\left(-1,1\right)$ .

try it

Use the series for $f\left(x\right)=\frac{1}{1-x}$ on $|x|<1$ to construct a series for $\frac{1}{\left(1-x\right)\left(x - 2\right)}$ . Determine the interval of convergence.

Hint

Use partial fractions to rewrite $\frac{1}{\left(1-x\right)\left(x - 2\right)}$ as the difference of two fractions.

Show Solution

$\displaystyle\sum _{n=0}^{\infty }\left(-1+\frac{1}{{2}^{n+1}}\right){x}^{n}$ . The interval of convergence is $\left(-1,1\right)$ .

In the previous example, we showed how to find power series for certain functions. In the next example we show how to do the opposite: given a power series, determine which function it represents.

Example: Finding the Function Represented by a Given Power Series

Consider the power series $\displaystyle\sum _{n=0}^{\infty }{2}^{n}{x}^{n}$ . Find the function f represented by this series. Determine the interval of convergence of the series.

Show Solution

Writing the given series as

\sum_{n = 0}^{\infty} 2^{n} x^{n} = \sum_{n = 0}^{\infty} {(2 x)}^{n}

$\displaystyle\sum _{n=0}^{\infty }{2}^{n}{x}^{n}=\displaystyle\sum _{n=0}^{\infty }{\left(2x\right)}^{n}$ ,

we can recognize this series as the power series for

f (x) = \frac{1}{1 - 2 x}

$f\left(x\right)=\frac{1}{1 - 2x}$ .

Since this is a geometric series, the series converges if and only if $|2x|<1$ . Therefore, the interval of convergence is $\left(-\frac{1}{2},\frac{1}{2}\right)$ .

try it

Find the function represented by the power series $\displaystyle\sum _{n=0}^{\infty }\frac{1}{{3}^{n}}{x}^{n}$ . Determine its interval of convergence.

Hint

Write $\frac{1}{{3}^{n}}{x}^{n}={\left(\frac{x}{3}\right)}^{n}$ .

Show Solution

$f\left(x\right)=\frac{3}{3-x}$ . The interval of convergence is $\left(-3,3\right)$ .

Try It

Multiplication of Power Series

We can also create new power series by multiplying power series. Being able to multiply two power series provides another way of finding power series representations for functions.

The way we multiply them is similar to how we multiply polynomials. For example, suppose we want to multiply

\sum_{n = 0}^{\infty} c_{n} x^{n} = c_{0} + c_{1} x + c_{2} x^{2} + \dots

$\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots$

and

\sum_{n = 0}^{\infty} d_{n} x^{n} = d_{0} + d_{1} x + d_{2} x^{2} + \dots

$\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\cdots$ .

It appears that the product should satisfy

\begin{array}{cc} (\sum_{n = 0}^{\infty} c_{n} x^{n}) (\sum_{n = - 0}^{\infty} d_{n} x^{n}) & = (c_{0} + c_{1} x + c_{2} x^{2} + \dots) \cdot (d_{0} + d_{1} x + d_{2} x^{2} + \dots) \\ = c_{0} d_{0} + (c_{1} d_{0} + c_{0} d_{1}) x + (c_{2} d_{0} + c_{1} d_{1} + c_{0} d_{2}) x^{2} + \dots . \end{array}

$\begin{array}{cc}\hfill \left(\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}\right)\left(\displaystyle\sum _{n=-0}^{\infty }{d}_{n}{x}^{n}\right)& =\left({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots \right)\cdot \left({d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\cdots \right)\hfill \\ & ={c}_{0}{d}_{0}+\left({c}_{1}{d}_{0}+{c}_{0}{d}_{1}\right)x+\left({c}_{2}{d}_{0}+{c}_{1}{d}_{1}+{c}_{0}{d}_{2}\right){x}^{2}+\cdots .\hfill \end{array}$

In Multiplying Power Series, we state the main result regarding multiplying power series, showing that if $\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}$ converge on a common interval I, then we can multiply the series in this way, and the resulting series also converges on the interval I.

theorem: Multiplying Power Series

Suppose that the power series $\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}$ converge to f and g, respectively, on a common interval I. Let

\begin{array}{cc} e_{n} & = c_{0} d_{n} + c_{1} d_{n - 1} + c_{2} d_{n - 2} + \dots + c_{n - 1} d_{1} + c_{n} d_{0} \\ = \sum_{k = 0}^{n} c_{k} d_{n - k} . \end{array}

$\begin{array}{cc}\hfill {e}_{n}& ={c}_{0}{d}_{n}+{c}_{1}{d}_{n - 1}+{c}_{2}{d}_{n - 2}+\cdots +{c}_{n - 1}{d}_{1}+{c}_{n}{d}_{0}\hfill \\ & ={\displaystyle\sum _{k=0}^{n}{c}_{k}{d}_{n-k}}.\hfill \end{array}$

Then

(\sum_{n = 0}^{\infty} c_{n} x^{n}) (\sum_{n = 0}^{\infty} d_{n} x^{n}) = \sum_{n = 0}^{\infty} e_{n} x^{n}

$\left({\displaystyle\sum _{n=0}^{\infty}}{c}_{n}{x}^{n}\right)\left(\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}\right)=\displaystyle\sum _{n=0}^{\infty }{e}_{n}{x}^{n}$

and

\sum_{n = 0}^{\infty} e_{n} x^{n} converges to f (x) \cdot g (x) on I

$\displaystyle\sum _{n=0}^{\infty }{e}_{n}{x}^{n}\text{converges to}f\left(x\right)\cdot g\left(x\right)\text{ on }I$ .

The series $\displaystyle\sum _{n=0}^{\infty }{e}_{n}{x}^{n}$ is known as the Cauchy product of the series $\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}$ .

We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course. We now provide an example of this theorem by finding the power series representation for

f (x) = \frac{1}{(1 - x) (1 - x^{2})}

$f\left(x\right)=\frac{1}{\left(1-x\right)\left(1-{x}^{2}\right)}$

using the power series representations for

y = \frac{1}{1 - x} and y = \frac{1}{1 - x^{2}}

$y=\frac{1}{1-x}\text{and}y=\frac{1}{1-{x}^{2}}$ .

Example: Multiplying Power Series

Multiply the power series representation

\begin{array}{cc} \frac{1}{1 - x} & = \sum_{n = 0}^{\infty} x^{n} \\ = 1 + x + x^{2} + x^{3} + \dots \end{array}

$\begin{array}{cc}\hfill \frac{1}{1-x}& ={\displaystyle\sum _{n=0}^{\infty}}{x}^{n}\hfill \\ & =1+x+{x}^{2}+{x}^{3}+\cdots \hfill \end{array}$

for $|x|<1$ with the power series representation

\begin{array}{cc} \frac{1}{1 - x^{2}} & = \sum_{n = 0}^{\infty} {(x^{2})}^{n} \\ = 1 + x^{2} + x^{4} + x^{6} + \dots \end{array}

$\begin{array}{cc}\hfill \frac{1}{1-{x}^{2}}& ={\displaystyle\sum _{n=0}^{\infty}}{\left({x}^{2}\right)}^{n}\hfill \\ & =1+{x}^{2}+{x}^{4}+{x}^{6}+\cdots \hfill \end{array}$

for $|x|<1$ to construct a power series for $f\left(x\right)=\frac{1}{\left(1-x\right)\left(1-{x}^{2}\right)}$ on the interval $\left(-1,1\right)$ .

Show Solution

We need to multiply

(1 + x + x^{2} + x^{3} + \dots) (1 + x^{2} + x^{4} + x^{6} + \dots)

$\left(1+x+{x}^{2}+{x}^{3}+\cdots \right)\left(1+{x}^{2}+{x}^{4}+{x}^{6}+\cdots \right)$ .

Writing out the first several terms, we see that the product is given by

\begin{matrix} (1 + x^{2} + x^{4} + x^{6} + \dots) + (x + x^{3} + x^{5} + x^{7} + \dots) + (x^{2} + x^{4} + x^{6} + x^{8} + \dots) + (x^{3} + x^{5} + x^{7} + x^{9} + \dots) \\ = 1 + x + (1 + 1) x^{2} + (1 + 1) x^{3} + (1 + 1 + 1) x^{4} + (1 + 1 + 1) x^{5} + \dots \\ = 1 + x + 2 x^{2} + 2 x^{3} + 3 x^{4} + 3 x^{5} + \dots . \end{matrix}

$\begin{array}{c}\left(1+{x}^{2}+{x}^{4}+{x}^{6}+\cdots \right)+\left(x+{x}^{3}+{x}^{5}+{x}^{7}+\cdots \right)+\left({x}^{2}+{x}^{4}+{x}^{6}+{x}^{8}+\cdots \right)+\left({x}^{3}+{x}^{5}+{x}^{7}+{x}^{9}+\cdots \right)\hfill \\ =1+x+\left(1+1\right){x}^{2}+\left(1+1\right){x}^{3}+\left(1+1+1\right){x}^{4}+\left(1+1+1\right){x}^{5}+\cdots \hfill \\ =1+x+2{x}^{2}+2{x}^{3}+3{x}^{4}+3{x}^{5}+\cdots .\hfill \end{array}$

Since the series for $y=\frac{1}{1-x}$ and $y=\frac{1}{1-{x}^{2}}$ both converge on the interval $\left(-1,1\right)$ , the series for the product also converges on the interval $\left(-1,1\right)$ .

Watch the following video to see the worked solution to Example: Multiplying Power Series.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “6.2.3” here (opens in new window).

try it

Multiply the series $\frac{1}{1-x}=\displaystyle\sum _{n=0}^{\infty }{x}^{n}$ by itself to construct a series for $\frac{1}{\left(1-x\right)\left(1-x\right)}$ .

Hint

Multiply the first few terms of $\left(1+x+{x}^{2}+{x}^{3}+\cdots \right)\left(1+x+{x}^{2}+{x}^{3}+\cdots \right)$ .

Show Solution

$1+2x+3{x}^{2}+4{x}^{3}+\cdots$

Module 6: Power Series