Operations with Power Series: Learn It 3

Giselle Miller

Operations with Power Series: Learn It 3

Differentiating and Integrating Power Series

Once you have a power series that converges on some interval, two natural questions arise about the function it represents:

Is the function differentiable? If so, how do we find its derivative?
How do we find the indefinite integral of this function?

The answer turns out to be remarkably straightforward: you can differentiate and integrate power series term-by-term, just like polynomials.

Recall the general expression of the power rules for derivatives and integrals of polynomials.

Power Rules for Polynomials

Derivative: [latex]\frac{d}{dx}(x^n) = nx^{n-1}[/latex]
Integral: [latex]\int x^n dx = \frac{1}{n+1}x^{n+1} + C[/latex] (for [latex]n \neq -1[/latex])

These familiar rules extend naturally to power series. If you have: [latex]f(x) = \sum_{n=0}^{\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots[/latex]

Then you can find:

Derivative: [latex]f'(x) = c_1 + 2c_2 x + 3c_3 x^2 + 4c_4 x^3 + \cdots[/latex]

Integral: [latex]\int f(x) dx = C + c_0 x + c_1 \frac{x^2}{2} + c_2 \frac{x^3}{3} + c_3 \frac{x^4}{4} + \cdots[/latex]

Evaluating the derivative and indefinite integral in this way is called term-by-term differentiation and term-by-term integration of a power series, respectively.

The ability to differentiate and integrate power series term-by-term gives you a powerful tool for finding power series representations of new functions. You can start with a known power series and use calculus operations to build series for related functions.

For example, given the power series for [latex]f(x) = \frac{1}{1-x}[/latex], you can differentiate term-by-term to find the power series for [latex]f'(x) = \frac{1}{(1-x)^2}[/latex]. Similarly, using the power series for [latex]g(x) = \frac{1}{1+x}[/latex], you can integrate term-by-term to find the power series for [latex]G(x) = \ln(1+x)[/latex], an antiderivative of [latex]g[/latex].

We’ll show how to do this in the next examples. First, let’s state the main result regarding differentiation and integration of power series.

theorem: term-by-term differentiation and integration for power series

Suppose [latex]\sum_{n=0}^{\infty} c_n (x-a)^n[/latex] converges on [latex](a-R, a+R)[/latex] and defines function [latex]f(x)[/latex].

Then [latex]f[/latex] is differentiable on [latex](a-R, a+R)[/latex] and:

[latex]f'(x) = \sum_{n=1}^{\infty} nc_n (x-a)^{n-1} = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + \cdots[/latex]

For integration:

[latex]\int f(x) dx = C + \sum_{n=0}^{\infty} c_n \frac{(x-a)^{n+1}}{n+1} = C + c_0(x-a) + c_1\frac{(x-a)^2}{2} + c_2\frac{(x-a)^3}{3} + \cdots[/latex]

Both resulting series converge on the same interval [latex](a-R, a+R)[/latex].

The proof of this result is beyond the scope of this course.

While the derivative and integral series have the same radius of convergence as the original series, their behavior at the endpoints might differ. Always check endpoint convergence separately when needed.

Note that while term-by-term differentiation and integration preserve the radius of convergence, the behavior at the endpoints can change. The differentiated and integrated series might converge or diverge at the endpoints even when the original series behaves differently there. We’ll see examples of this endpoint behavior in the upcoming problems.

Use the power series representation

[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].

Watch the following video to see the worked solution to the above example.

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).

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]

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 the above example.

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 “6.2.4” here (opens in new window).