6.2 Properties of power series (Page 5/10)

Calculus volume 2 Page 5 / 10

Term-by-term differentiation and integration for power series

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

\begin{matrix} f (x) & = \sum_{n = 0}^{\infty} c_{n} {(x - a)}^{n} \\ = c_{0} + c_{1} (x - a) + c_{2} {(x - a)}^{2} + c_{3} {(x - a)}^{3} + ⋯ \end{matrix}

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

\begin{matrix} f^{'} (x) & = \sum_{n = 1}^{\infty} n c_{n} {(x - a)}^{n - 1} \\ = c_{1} + 2 c_{2} (x - a) + 3 c_{3} {(x - a)}^{2} + ⋯ \end{matrix}

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

\begin{matrix} \int f (x) d x & = 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} + ⋯ \end{matrix}

for $| x - a | < R .$

The proof of this result is beyond the scope of the text and is omitted. Note that although [link] 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.

Differentiating power series

Use the power series representation

$\begin{matrix} f (x) & = \frac{1}{1 - x} \\ = \sum_{n = 0}^{\infty} x^{n} \\ = 1 + x + x^{2} + x^{3} + ⋯ \end{matrix}$

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

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

on the interval $(−1, 1) .$ Determine whether the resulting series converges at the endpoints.
Use the result of part a. to evaluate the sum of the series $\sum_{n = 0}^{\infty} \frac{n + 1}{4^{n}} .$

Since $g (x) = \frac{1}{{(1 - x)}^{2}}$ is the derivative of $f (x) = \frac{1}{1 - x},$ we can find a power series representation for g by differentiating the power series for f term-by-term. The result is

$\begin{matrix} g (x) & = \frac{1}{{(1 - x)}^{2}} \\ = \frac{d}{d x} (\frac{1}{1 - x}) \\ = \sum_{n = 0}^{\infty} \frac{d}{d x} (x^{n}) \\ = \frac{d}{d x} (1 + x + x^{2} + x^{3} + ⋯) \\ = 0 + 1 + 2 x + 3 x^{2} + 4 x^{3} + ⋯ \\ = \sum_{n = 0}^{\infty} (n + 1) x^{n} \end{matrix}$

for $| x | < 1 .$ [link] does not guarantee anything about the behavior of this series at the endpoints. Testing the endpoints by using the divergence test, we find that the series diverges at both endpoints $x = ± 1 .$ Note that this is the same result found in [link] .
From part a. we know that

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

Therefore,

$\begin{matrix} \sum_{n = 0}^{\infty} \frac{n + 1}{4^{n}} & = \sum_{n = 0}^{\infty} (n + 1) {(\frac{1}{4})}^{n} \\ = \frac{1}{{(1 - \frac{1}{4})}^{2}} \\ = \frac{1}{{(\frac{3}{4})}^{2}} \\ = \frac{16}{9} . \end{matrix}$

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

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

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Integrating power series

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

$f (x) = ln (1 + x)$
$f (x) = {tan}^{−1} x$

For $f (x) = ln (1 + x),$ the derivative is $f^{'} (x) = \frac{1}{1 + x} .$ We know that

$\begin{matrix} \frac{1}{1 + x} & = \frac{1}{1 - (− x)} \\ = \sum_{n = 0}^{\infty} {(− x)}^{n} \\ = 1 - x + x^{2} - x^{3} + ⋯ \end{matrix}$

for $| x | < 1 .$ To find a power series for $f (x) = ln (1 + x),$ we integrate the series term-by-term.

$\begin{matrix} \int f^{'} (x) d x & = \int (1 - x + x^{2} - x^{3} + ⋯) d x \\ = C + x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + ⋯ \end{matrix}$

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

$\begin{matrix} ln (1 + x) & = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + ⋯ \\ = \sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{x^{n}}{n} \end{matrix}$

for $| x | < 1 .$ [link] does not guarantee anything about the behavior of this power series at the endpoints. However, checking the endpoints, we find that at $x = 1$ the series is the alternating harmonic series, which converges. Also, at $x = −1,$ the series is the harmonic series, which diverges. It is important to note that, even though this series converges at $x = 1,$ [link] does not guarantee that the series actually converges to $ln (2) .$ In fact, the series does converge to $ln (2),$ 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 $(−1, 1] .$
The derivative of $f (x) = {tan}^{−1} x$ is $f^{'} (x) = \frac{1}{1 + x^{2}} .$ We know that

$\begin{matrix} \frac{1}{1 + x^{2}} & = \frac{1}{1 - (− x^{2})} \\ = \sum_{n = 0}^{\infty} {(− x^{2})}^{n} \\ = 1 - x^{2} + x^{4} - x^{6} + ⋯ \end{matrix}$

for $| x | < 1 .$ To find a power series for $f (x) = {tan}^{−1} x,$ we integrate this series term-by-term.

$\begin{matrix} \int f^{'} (x) d x & = \int (1 - x^{2} + x^{4} - x^{6} + ⋯) d x \\ = C + x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + ⋯ \end{matrix}$

Since ${tan}^{−1} (0) = 0,$ we have $C = 0 .$ Therefore, a power series representation for $f (x) = {tan}^{−1} x$ is

$\begin{matrix} {tan}^{−1} x & = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + ⋯ \\ = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{2 n + 1}}{2 n + 1} \end{matrix}$

for $| x | < 1 .$ Again, [link] 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 $x = 1$ and $x = −1 .$ As discussed in part a., using Abel’s theorem, it can be shown that the series actually converges to ${tan}^{−1} (1)$ and ${tan}^{−1} (−1)$ at $x = 1$ and $x = −1,$ respectively. Thus, the interval of convergence is $[−1, 1] .$

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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