Differentiation and Integration of Power Series

# Differentiation and Integration of Power Series

We will now look at differentiating and integrating power series term by term, a technique that will be very useful. We first note that power series have terms which are polynomials, and polynomials are relatively easily to differentiate and integrate. The following theorems will allow us to differentiate and integrate power series as we would have expected.

## Differentiation of Power Series

 Theorem 1 (Differentiation of Power Series): Let the power series $\sum_{n=0}^{\infty} a_n(x-c)^n = a_0 + a_1x + a_2x^2 + a_3x^3 + ...$ converge to $f(x)$ on the interval $(c-R , c+R)$. Then $f$ is differentiable on the same interval $(c-R, c+R)$ and $f'(x) = \sum_{n=1}^{\infty} na_n(x - c)^{n-1} = a_1 + 2a_2(x-c) + 3a_3(x-c)^2 + ...$.

We will note that if a power series is differentiated, then the differentiated series has the same interval of convergence as the original series EXCEPT possibly the loss of one or both end points of the interval of convergence if the original series was convergent at these point.

Of course, we could go further and derive the following formulas for second, third, etc… derivatives of power series as follows:

• $f''(x) = \sum_{n=2}^{\infty} n(n-1)a_n(x - c)^{n-2}$ over the interval $(c-R, c+R)$.
• $f'''(x) = \sum_{n=3}^{\infty} n(n-1)(n-2)a_n(x-c)^{n-3}$ over the interval $(c-R, c+R)$.

Let's look at an example of differentiation with regards to power series.

### Example 1

Find a power series representation of the function $f(x) = \frac{2}{(1 - x)^2}$.

We first note that $\frac{d}{dx} \frac{1}{1 - x} = \frac{1}{(1 - x)}^2$. We already know a power series for $\frac{1}{1 - x}$, namely $\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n$ for $-1 < x < 1$, and so if we multiply this equality by 2 and then differentiate both sides as follows, we will obtain our power series:

(1)
\begin{align} \frac{1}{1 - x} = \sum_{n=0}^{\infty} x_n \\ \frac{2}{1 - x} = 2\sum_{n=0}^{\infty} x_n \\ \frac{d}{dx} \frac{2}{1 - x} = \frac{d}{dx} 2 \sum_{n=0}^{\infty} x_n \\ \frac{2}{(1 - x)^2} = 2 \sum_{n=0}^{\infty} nx_n^{n-1} \\ \frac{2}{(1 - x)^2} = \sum_{n=0}^{\infty} 2 n x^{n-1} \\ \end{align}

We note that the power series we differentiated, $\frac{1}{1 - x}$ is not convergent at its endpoints, and so the derivative series is also not convergent at its endpoints and so our series represents $f(x)$ for $-1 < x < 1$. The graph below represents the function $f(x)$ in blue, and the series $\sum_{n=0}^{\infty} 2 n x^{n-1}$ in red as a representation of $f(x)$ on $(-1, 1)$: ## Integration of Power Series

 Theorem 2 (Integration of Power Series): Let the power series $\sum_{n=0}^{\infty} a_n(x-c)^n = a_0 + a_1x + a_2x^2 + a_3x^3 + ...$ converge to $f(x)$ on the interval $(c-R , c+R)$. Then $f$ is integrable over any subinterval of $(c-R, c+R)$ and $\int_{0}^{x} f(t) \: dt = \sum_{n=0}^{\infty} \frac{a_n}{n+1}(x - c)^{n+1} = a_0(x - c)+ \frac{a_1}{2}(x - c)^2 + \frac{a_2}{3}(x-c)^3 + ...$.

If a power series is integrated, then the integrated series has the same interval of convergent as the original series EXCEPT possibly the gain of one or both end points of the interval of convergence if the original series wasn't convergent at the end points already.

We will now look at an example of where integrating power series can be useful.

### Example 2

Find a power series representation of the function $g(x) = \ln (3 - x)$.

First consider the function $\frac{1}{3 - x}$. Note that if we integrate this function, we get that $\int \frac{1}{3 - x} \: dx = -\ln (3 - x)$. Let's first come up with a power series for the function $\frac{1}{3 - x} = \frac{1}{3} \cdot \frac{1}{1 - \frac{x}{3}}$.

(2)
\begin{align} \quad \frac{1}{3} \cdot \frac{1}{1 - \frac{x}{3}} = \frac{1}{3} \sum_{n=0}^{\infty} \left ( \frac{x}{3} \right )^n \end{align}

The power series above is a representation of $\frac{1}{3 - x}$ on the interval $-3 < x < 3$. Now we will apply theorem 2 of integration to determine a power series for $g(x)$:

(3)
\begin{align} \quad \frac{1}{3} \cdot \frac{1}{1 - \frac{x}{3}} = \frac{1}{3} \sum_{n=0}^{\infty} \left ( \frac{x}{3} \right )^n \\ \quad \int_0^x \frac{1}{3 - x} dt = \int_0^x \frac{1}{3} \sum_{n=0}^{\infty} \left ( \frac{t}{3} \right )^n \: dt \\ -\ln (3 - t) \biggr \rvert_0^x = \frac{1}{3} \sum_{n=0}^{\infty} \frac{t^{n+1}}{3^n(n+1)} \: \biggr \rvert_{0}^{x} \\ -\ln (3 - x) + \ln (3) = \frac{1}{3} \sum_{n=0}^{\infty} \frac{x^{n+1}}{3^n(n+1)} \\ \ln (3 - x) = \ln (3) - \sum_{n=0}^{\infty} \frac{x^{n+1}}{3^{n+1}(n+1)} \end{align}

We now have to check to see if we've gained convergence of the end points.

First check $x = -3$ which produces the series $\ln(3) - \sum_{n=0}^{\infty} \frac{(-1)^{n+1}3^{n+1}}{3^{n+1}(n+1)} = \ln(3) - \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}$ which converges by the alternating series test.

Now check $x = 3$, which produces the series $\ln(3) - \sum_{n=0}^{\infty} \frac{3^{n+1}}{3^{n+1}(n+1)} = \sum_{n=0}^{\infty} \frac{1}{n+1}$, which is the harmonic series and diverges.

Therefore $g(x) = \ln (3) - \sum_{n=0}^{\infty} \frac{x^{n+1}}{3^{n+1}(n+1)}$ for $-3 ≤ x < 3$, i.e., on the interval $[-3, 3)$.

The graph below depicts our function $g(x)$ is blue and our power series representation in red. 