Differentiation and Integration of Power Series Examples 1

Recall from the Differentiation and Integration of Power Series page that:

• If $f(x) = \sum_{n=0}^{\infty} a_n (x - c)^n$ on the interval $(c - R, c + R)$ then $f$ is differentiable on $(c - R, c + R)$ and:

(1)

\begin{align} \quad f'(x) = \sum_{n=1}^{\infty} na_n(x - c)^{n-1} = a_1 + 2a_2(x - c) + 3a_3(x - c)^2 + ... \end{align}

• If $f(x) = \sum_{n=0}^{\infty} a_n (x - c)^n$ on the interval $(c - R, c + R)$ then $f$ is integrable over any subinterval of $(c - R, c + R)$ and:

(2)

\begin{align} \quad \int f(x) \: dx = \sum_{n=0}^{\infty} \frac{a_n}{n+1} (x - c)^{n+1} = a_0(x - c) + \frac{a_1}{2} (x - c)^2 + \frac{a_2}{3} (x - c)^3 + ... \end{align}

Furthermore, if the original series converges at either of the endpoints $c - R$ or $c + R$, then the differentiated series may not converge at these endpoints.

If the original series does not converge at either of the endpoints $c - R$ or $c + R$, then the integrated series may converge at either of these endpoints.

We will now look at some examples of differentiating and integrating power series.

Example 1

Find a power series representation for the function $f(x) = \ln \mid a - x\mid$ where $a > 0$, and determine the center of convergence $c$ and the radius of convergence $R$ for this series. Does the power series representation for $f(x) = \ln \mid a - x \mid$ converge at either of the endpoints $c - R$ or $c + R$? Determine the interval of convergence for this power series

We first note that $\ln \mid a - x \mid = -\int \frac{1}{a - x} \: dx$.

We already have that:

(3)

\begin{align} \quad \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n \end{align}

Therefore we have that:

(4)

\begin{align} \quad -\frac{1}{a - x} = -\frac{1}{a} \frac{1}{1 - \frac{x}{a}} = -\frac{1}{a} \sum_{n=0}^{\infty} \left ( \frac{x}{a} \right )^n = -\sum_{n=0}^{\infty} \frac{x^n}{a^{n+1}} \end{align}

We now integrate the series above to get that:

(5)

\begin{align} \quad \int_0^x - \frac{1}{a - t} \: dt = \int_0^x -\sum_{n=0}^{\infty} \frac{t^n}{a^{n+1}} \: dt \\ \quad \left [ \ln (a - t) \right ]_{t=0}^{t=x} = \left [ - \sum_{n=0}^{\infty} \frac{x^{n+1}}{a^{n+1} (n + 1)} \right ]_{t=0}^{t=x} \\ \quad \ln \mid a - x \mid - \ln \mid a \mid = - \sum_{n=0}^{\infty} \frac{x^{n+1}}{a^{n+1}(n + 1)} \\ \quad \ln \mid a - x \mid = \ln \mid a \mid - \sum_{n=0}^{\infty} \frac{x^{n+1}}{a^{n+1}(n + 1)} \\ \end{align}

The center of convergence is $0$. Furthermore, we note that from the original series, $\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n$ for $\mid x \mid < 1$. Therefore, $\frac{1}{1 - \frac{x}{a}} = \sum_{n=0}^{\infty} \frac{x^n}{a^n}$ for $\biggr \rvert \frac{x}{a} \biggr \rvert < 1$ or equivalently, for $\mid x \mid < a$. Therefore the radius of convergence is $a$.

We will now test to see if our power series representation converges at either of the endpoints. First let's check $-a$. We have that:

(6)

\begin{align} \quad \ln \mid -a \mid - \sum_{n=0}^{\infty} \frac{(-a)^{n+1}}{a^{n+1}(n + 1)} = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}a^{n+1}}{a^{n+1}(n+1)} = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n+1} \end{align}

The series above converges as the alternating harmonic series. Now let's check the endpoint $a$. We have that:

(7)

\begin{align} \quad \ln \mid a \mid - \sum_{n=0}^{\infty} \frac{a^{n+1}}{a^{n+1}(n+1)} = \sum_{n=0}^{\infty} \frac{1}{n+1} \end{align}

The series above diverges as the harmonic series.

Thus we have that the interval of convergence is $[a, a)$.

Example 2

Determine a power series representation for $f(x) = \tan^{-1} x$, and find the center or convergence and radius of convergence.

Recall that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1 + x^2}$ and so $\tan^{-1} x = \int \frac{1}{1 + x^2} \: dx$. We have that:

(8)

\begin{align} \quad \frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n} \end{align}

We now integrate this series to get that:

(9)

\begin{align} \quad \int_0^x \frac{1}{1 + t^2} \: dt = \int_0^x \sum_{n=0}^{\infty} (-1)^n t^{2n} \: dt \\ \quad \left [ \tan^{-1} (t) \right ]_{t=0}^{t=x} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n + 1} \\ \quad \tan^{-1} x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n + 1} \\ \end{align}

This series has center of convergence $0$ and converges for $\mid -x^2 \mid < 1$, or equivalently, $\mid x \mid < 1$.