Differentiating and Integrating Power Series

Learning Outcomes

Differentiate and integrate power series term-by-term

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

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

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

[latex] \frac{d}{dx} \left (x^n \right) = nx^{n-1} [/latex]
[latex] \int{x^n\: dx} = \frac{1}{n+1} x^{n+1} + C \:(\text{for}\: n \ne -1 [/latex])

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

[latex]f\left(x\right)={c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots [/latex]

converges on some interval I, then

[latex]{f}^{\prime }\left(x\right)={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+\cdots [/latex]

and

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

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 [latex]f\left(x\right)=\frac{1}{1-x}[/latex], we can differentiate term-by-term to find the power series for [latex]{f}^{\prime }\left(x\right)=\frac{1}{{\left(1-x\right)}^{2}}[/latex]. Similarly, using the power series for [latex]g\left(x\right)=\frac{1}{1+x}[/latex], we can integrate term-by-term to find the power series for [latex]G\left(x\right)=\text{ln}\left(1+x\right)[/latex], 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 [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}[/latex] converges on the interval [latex]\left(a-R,a+R\right)[/latex] for some [latex]R>0[/latex]. Let f be the function defined by the series

[latex]\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}[/latex]

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

[latex]\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}[/latex]

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

[latex]\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}[/latex]

for [latex]|x-a|<R[/latex].

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

[latex]\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}[/latex]

for [latex]|x|<1[/latex] to find a power series representation for

[latex]g\left(x\right)=\frac{1}{{\left(1-x\right)}^{2}}[/latex]

on the interval [latex]\left(-1,1\right)[/latex]. Determine whether the resulting series converges at the endpoints.

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

Show Solution

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.

You can view the transcript for this segmented clip of “6.2.4” here (opens in new window).

try it

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

Hint

Show Solution

Example: Integrating Power Series

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

[latex]f\left(x\right)=\text{ln}\left(1+x\right)[/latex]
[latex]f\left(x\right)={\tan}^{-1}x[/latex]

Show Solution

For [latex]f\left(x\right)=\text{ln}\left(1+x\right)[/latex], the derivative is [latex]{f}^{\prime }\left(x\right)=\frac{1}{1+x}[/latex]. We know that

[latex]\begin{array}{cc}\hfill \frac{1}{1+x}& =\frac{1}{1-\left(\text{-}x\right)}\hfill \\ & ={\displaystyle\sum _{n=0}^{\infty}}{\left(\text{-}x\right)}^{n}\hfill \\ & =1-x+{x}^{2}-{x}^{3}+\cdots \hfill \end{array}[/latex]

for [latex]|x|<1[/latex]. To find a power series for [latex]f\left(x\right)=\text{ln}\left(1+x\right)[/latex], we integrate the series term-by-term.

[latex]\begin{array}{cc}\hfill {\displaystyle\int {f}^{\prime }\left(x\right)dx}& ={\displaystyle\int \left(1-x+{x}^{2}-{x}^{3}+\cdots \right)dx}\hfill \\ & =C+x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+\cdots \hfill \end{array}[/latex]

Since [latex]f\left(x\right)=\text{ln}\left(1+x\right)[/latex] is an antiderivative of [latex]\frac{1}{1+x}[/latex], it remains to solve for the constant C. Since [latex]\text{ln}\left(1+0\right)=0[/latex], we have [latex]C=0[/latex]. Therefore, a power series representation for [latex]f\left(x\right)=\text{ln}\left(1+x\right)[/latex] is

[latex]\begin{array}{cc}\hfill \text{ln}\left(1+x\right)& =x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+\cdots \hfill \\ & ={\displaystyle\sum _{n=1}^{\infty}}{\left(-1\right)}^{n+1}\frac{{x}^{n}}{n}\hfill \end{array}[/latex]

for [latex]|x|<1[/latex]. Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this power series at the endpoints. However, checking the endpoints, we find that at [latex]x=1[/latex] the series is the alternating harmonic series, which converges. Also, at [latex]x=-1[/latex], the series is the harmonic series, which diverges. It is important to note that, even though this series converges at [latex]x=1[/latex], Term-by-Term Differentiation and Integration for Power Series does not guarantee that the series actually converges to [latex]\text{ln}\left(2\right)[/latex]. In fact, the series does converge to [latex]\text{ln}\left(2\right)[/latex], but showing this fact requires more advanced techniques. (Abel’s theorem, covered in more advanced texts, deals with this more technical point.) The interval of convergence is [latex]\left(-1,1\right][/latex].

The derivative of [latex]f\left(x\right)={\tan}^{-1}x[/latex] is [latex]{f}^{\prime }\left(x\right)=\frac{1}{1+{x}^{2}}[/latex]. We know that

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

for [latex]|x|<1[/latex]. To find a power series for [latex]f\left(x\right)={\tan}^{-1}x[/latex], we integrate this series term-by-term.

[latex]\begin{array}{cc}\hfill {\displaystyle\int {f}^{\prime }\left(x\right)dx}& ={\displaystyle\int \left(1-{x}^{2}+{x}^{4}-{x}^{6}+\cdots \right)dx}\hfill \\ & =C+x-\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5}-\frac{{x}^{7}}{7}+\cdots \hfill \end{array}[/latex]

Since [latex]{\tan}^{-1}\left(0\right)=0[/latex], we have [latex]C=0[/latex]. Therefore, a power series representation for [latex]f\left(x\right)={\tan}^{-1}x[/latex] is

[latex]\begin{array}{cc}\hfill {\tan}^{-1}x& =x-\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5}-\frac{{x}^{7}}{7}+\cdots \hfill \\ & ={\displaystyle\sum _{n=0}^{\infty}}{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{2n+1}\hfill \end{array}[/latex]

for [latex]|x|<1[/latex]. Again, Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the convergence of this series at the endpoints. However, checking the endpoints and using the alternating series test, we find that the series converges at [latex]x=1[/latex] and [latex]x=-1[/latex]. As discussed in part a., using Abel’s theorem, it can be shown that the series actually converges to [latex]{\tan}^{-1}\left(1\right)[/latex] and [latex]{\tan}^{-1}\left(-1\right)[/latex] at [latex]x=1[/latex] and [latex]x=-1[/latex], respectively. Thus, the interval of convergence is [latex]\left[-1,1\right][/latex].

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

You can view the transcript for this segmented clip “6.2.4” here (opens in new window).

try it

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

Hint

Show Solution

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

[latex]{c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots ={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\cdots [/latex]

for all values x in some open interval I about zero, then the coefficients c_n should equal d_n for [latex]n\ge 0[/latex]. We now state this result formally in Uniqueness of Power Series.

theorem: Uniqueness of Power Series

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

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

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

Proof

Let

[latex]\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}[/latex]

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

[latex]\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}[/latex]

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

[latex]\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}[/latex]

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

[latex]_\blacksquare[/latex]

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.

Module 6: Power Series