Power Series for Functions in Powers of x

Power Series for Functions in Powers of x

We will now look at some examples of evaluating power series for functions in powers of $x$ by using the following power series for the geometric series:

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

Example 1

Determine a power series representation of $\frac{1}{2-x}$ in powers of $x$.

We have that:

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

This series is a representation of the function $\frac{1}{2 - x}$ if $\biggr \rvert \frac{x}{2} \biggr \rvert < 1$ or rather, if $-2 < x < 2$.

Example 2

Determine a power series representation of $\frac{1}{(2 - x)^2}$ in powers of $x$.

First notice that $\frac{d}{dx} \frac{1}{2 - x} = \frac{1}{(2 - x)^2}$. Therefore since $\frac{1}{2 - x} = \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} \cdot x^n$ if $-2 < x < 2$ (which we saw from example 1), then it follows that:

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

This series is a representation of the function $\frac{1}{(2 - x)^2}$ if $-2 < x < 2$.

Example 3

Determine a power series representation of $\ln (2 - x)$ in powers of $x$.

First notice that $\int \frac{1}{2 - x} \: dx = -\ln (2 - x) + C$ and so it follows that since $\frac{1}{2 - t} = \sum_{n=0}^{\infty} \frac{t^n}{2^{n+1}}$:

(4)
\begin{align} \quad \int_0^x \frac{1}{2 - t} \: dt = \sum_{n=0}^{\infty} \left ( \int_0^x \frac{t^n}{2^{n+1}} \: dt \right ) \\ - \ln (2 - t) \biggr \rvert_0^x = \sum_{n=0}^{\infty} \frac{t^{n+1}}{2^{n+1}(n+1)} \biggr \rvert_0^x \\ - \ln (2 - x) + \ln (2) = \sum_{n=0}^{\infty} \left ( \frac{x^{n+1}}{2^{n+1}(n+1)} - \frac{0^{n+1}}{2^{n+1}(n+1)} \right ) \\ -\ln ( 2 - x) + \ln (2) = \sum_{n=0}^{\infty} \frac{x^{n+1}}{2^{n+1}(n+1)} \\ -\ln (2 - x) = -\ln (2) + \sum_{n=0}^{\infty} \frac{x^{n+1}}{2^{n+1}(n+1)} \\ \ln (2 - x) = \ln (2) - \sum_{n=0}^{\infty} \frac{x^{n+1}}{2^{n+1}(n+1)} \\ \end{align}

This series is a representation of the function $\ln (2 - x)$ if $- 2 ≤ x < 2$.

Another way to solve this problem is by integrating with indefinite integrals, and then solving to find the value of the constant $C$ as follows:

(5)
\begin{align} \frac{1}{2 - x} = \sum_{n=0}^{\infty} \frac{x^n}{2^{n+1}} \\ \int \frac{1}{2 - x} \: dx = \sum_{n=0}^{\infty} \int \frac{x^n}{2^{n+1}} \: dx \\ -\ln ( 2 - x) = \sum_{n=0}^{\infty} \frac{x^{n+1}}{2^{n+1}(n+1)} + C \ln ( 2 - x) = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{2^{n+1}(n+1)} + C \end{align}

Now notice that if $x = 0$ then $\ln (2) = C$, and so we have that:

(6)
\begin{align} \ln (2 - x) = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{2^{n+1}(n+1)} + \ln (2) \\ \ln (2 - x) = \ln (2) -\sum_{n=0}^{\infty} \frac{x^{n+1}}{2^{n+1}(n+1)} \end{align}