## Differentiating and Integrating Power Series

### Learning Outcomes

• Differentiate and integrate power series term-by-term

Consider a power series $\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots$ that converges on some interval I, and let $f$ be the function defined by this series. Here we address two questions about $f$.

• Is $f$ differentiable, and if so, how do we determine the derivative ${f}^{\prime }?$
• How do we evaluate the indefinite integral $\displaystyle\int f\left(x\right)dx?$

We know that, for a polynomial with a finite number of terms, we can evaluate the derivative by differentiating each term separately. Similarly, we can evaluate the indefinite integral by integrating each term separately. Recall the general expression of the power rules for derivatives and integrals of polynomials.

### Recall: Power rule for Derivatives and Integrals

1. $\frac{d}{dx} \left (x^n \right) = nx^{n-1}$
2. $\int{x^n\: dx} = \frac{1}{n+1} x^{n+1} + C \:(\text{for}\: n \ne -1$)

In this section we show that we can take advantage of the simplicity of integrating and differentiating polynomials to do the same thing for convergent power series. That is, if

$f\left(x\right)={c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots$

converges on some interval I, then

${f}^{\prime }\left(x\right)={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+\cdots$

and

$\displaystyle\int f\left(x\right)dx=C+{c}_{0}x+{c}_{1}\frac{{x}^{2}}{2}+{c}_{2}\frac{{x}^{3}}{3}+\cdots$.

Evaluating the derivative and indefinite integral in this way is called term-by-term differentiation of a power series and term-by-term integration of a power series, respectively. The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for $f\left(x\right)=\frac{1}{1-x}$, we can differentiate term-by-term to find the power series for ${f}^{\prime }\left(x\right)=\frac{1}{{\left(1-x\right)}^{2}}$. Similarly, using the power series for $g\left(x\right)=\frac{1}{1+x}$, we can integrate term-by-term to find the power series for $G\left(x\right)=\text{ln}\left(1+x\right)$, an antiderivative of g. We show how to do this in the next two examples. First, we state Term-by-Term Differentiation and Integration for Power Series, which provides the main result regarding differentiation and integration of power series.

### theorem: Term-by-Term Differentiation and Integration for Power Series

Suppose that the power series $\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}$ converges on the interval $\left(a-R,a+R\right)$ for some $R>0$. Let f be the function defined by the series

$\begin{array}{cc}\hfill f\left(x\right)& ={\displaystyle\sum _{n=0}^{\infty}}{c}_{n}{\left(x-a\right)}^{n}\hfill \\ & ={c}_{0}+{c}_{1}\left(x-a\right)+{c}_{2}{\left(x-a\right)}^{2}+{c}_{3}{\left(x-a\right)}^{3}+\cdots \hfill \end{array}$

for $|x-a|<R$. Then f is differentiable on the interval $\left(a-R,a+R\right)$ and we can find ${f}^{\prime }$ by differentiating the series term-by-term:

$\begin{array}{cc}\hfill {f}^{\prime }\left(x\right)& ={\displaystyle\sum _{n=1}^{\infty}}n{c}_{n}{\left(x-a\right)}^{n - 1}\hfill \\ & ={c}_{1}+2{c}_{2}\left(x-a\right)+3{c}_{3}{\left(x-a\right)}^{2}+\cdots \hfill \end{array}$

for $|x-a|<R$. Also, to find $\displaystyle\int f\left(x\right)dx$, we can integrate the series term-by-term. The resulting series converges on $\left(a-R,a+R\right)$, and we have

$\begin{array}{cc}\hfill {\displaystyle\int f\left(x\right)dx}& =C+{\displaystyle\sum _{n=0}^{\infty }{c}_{n}\frac{{\left(x-a\right)}^{n+1}}{n+1}}\hfill \\ & =C+{c}_{0}\left(x-a\right)+{c}_{1}\frac{{\left(x-a\right)}^{2}}{2}+{c}_{2}\frac{{\left(x-a\right)}^{3}}{3}+\cdots \hfill \end{array}$

for $|x-a|<R$.

The proof of this result is beyond the scope of the text and is omitted. Note that although Term-by-Term Differentiation and Integration for Power Series guarantees the same radius of convergence when a power series is differentiated or integrated term-by-term, it says nothing about what happens at the endpoints. It is possible that the differentiated and integrated power series have different behavior at the endpoints than does the original series. We see this behavior in the next examples.

### Example: Differentiating Power Series

1. Use the power series representation
$\begin{array}{cc}\hfill f\left(x\right)& =\frac{1}{1-x}\hfill \\ & ={\displaystyle\sum _{n=0}^{\infty}}{x}^{n}\hfill \\ & =1+x+{x}^{2}+{x}^{3}+\cdots \hfill \end{array}$

for $|x|<1$ to find a power series representation for

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

on the interval $\left(-1,1\right)$. Determine whether the resulting series converges at the endpoints.

Use the result of part a. to evaluate the sum of the series $\displaystyle\sum _{n=0}^{\infty }\frac{n+1}{{4}^{n}}$.

Watch the following video to see the worked solution to Example: Differentiating 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.

### try it

Differentiate the series $\frac{1}{{\left(1-x\right)}^{2}}=\displaystyle\sum _{n=0}^{\infty }\left(n+1\right){x}^{n}$ term-by-term to find a power series representation for $\frac{2}{{\left(1-x\right)}^{3}}$ on the interval $\left(-1,1\right)$.

### Example: Integrating Power Series

For each of the following functions f, find a power series representation for f by integrating the power series for ${f}^{\prime }$ and find its interval of convergence.

1. $f\left(x\right)=\text{ln}\left(1+x\right)$
2. $f\left(x\right)={\tan}^{-1}x$

Watch the following video to see the worked solution to Example: Integrating 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.

### try it

Integrate the power series $\text{ln}\left(1+x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{{x}^{n}}{n}$ term-by-term to evaluate $\displaystyle\int \text{ln}\left(1+x\right)dx$.

Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique? That is, given a function f and a power series for f at a, is it possible that there is a different power series for f at a that we could have found if we had used a different technique? The answer to this question is no. This fact should not seem surprising if we think of power series as polynomials with an infinite number of terms. Intuitively, if

${c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots ={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\cdots$

for all values x in some open interval I about zero, then the coefficients cn should equal dn for $n\ge 0$. We now state this result formally in Uniqueness of Power Series.

### theorem: Uniqueness of Power Series

Let $\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}$ and $\displaystyle\sum _{n=0}^{\infty }{d}_{n}{\left(x-a\right)}^{n}$ be two convergent power series such that

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

for all x in an open interval containing a. Then ${c}_{n}={d}_{n}$ for all $n\ge 0$.

#### Proof

Let

$\begin{array}{cc}\hfill f\left(x\right)& ={c}_{0}+{c}_{1}\left(x-a\right)+{c}_{2}{\left(x-a\right)}^{2}+{c}_{3}{\left(x-a\right)}^{3}+\cdots \hfill \\ & ={d}_{0}+{d}_{1}\left(x-a\right)+{d}_{2}{\left(x-a\right)}^{2}+{d}_{3}{\left(x-a\right)}^{3}+\cdots .\hfill \end{array}$

Then $f\left(a\right)={c}_{0}={d}_{0}$. By Term-by-Term Differentiation and Integration for Power Series, we can differentiate both series term-by-term. Therefore,

$\begin{array}{cc}\hfill {f}^{\prime }\left(x\right)& ={c}_{1}+2{c}_{2}\left(x-a\right)+3{c}_{3}{\left(x-a\right)}^{2}+\cdots \hfill \\ & ={d}_{1}+2{d}_{2}\left(x-a\right)+3{d}_{3}{\left(x-a\right)}^{2}+\cdots ,\hfill \end{array}$

and thus, ${f}^{\prime }\left(a\right)={c}_{1}={d}_{1}$. Similarly,

$\begin{array}{cc}\hfill f^{\prime\prime}\left(x\right)& =2{c}_{2}+3\cdot 2{c}_{3}\left(x-a\right)+\cdots \hfill \\ & =2{d}_{2}+3\cdot 2{d}_{3}\left(x-a\right)+\cdots \hfill \end{array}$

implies that $f^{\prime\prime} \left(a\right)=2{c}_{2}=2{d}_{2}$, and therefore, ${c}_{2}={d}_{2}$. More generally, for any integer $n\ge 0,{f}^{\left(n\right)}\left(a\right)=n\text{!}{c}_{n}=n\text{!}{d}_{n}$, and consequently, ${c}_{n}={d}_{n}$ for all $n\ge 0$.

$_\blacksquare$

### Try It

In this section we have shown how to find power series representations for certain functions using various algebraic operations, differentiation, or integration. At this point, however, we are still limited as to the functions for which we can find power series representations. Next, we show how to find power series representations for many more functions by introducing Taylor series.