Determining Power Series Representations of Rational Functions

# Determining Power Series Representations of Rational Functions

We are now going to look at some examples of determining power series representations for rational functions. Before we do so, we must recall a very important power series representation that we've already looked at, namely:

(1)
\begin{align} \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n \quad \mathrm{for \: \mid x \mid ≤ 1} \end{align}

If we can get any rational function to appear in an analogous form to $f(x) = \frac{1}{1 - A}$, then the power series of $f(x)$ will be $\sum_{n=0}^{\infty} (A)^n$ and the interval of convergence will be $\mid A \mid < 1$. We will now demonstrate some examples of this useful technique below.

## Example 1

Determine a power series representation of the function $g(x) = \frac{1}{1 + 3x}$.

We note that the function $g$ could be rewritten as $g(x) = \frac{1}{1 - (-3x)}$, and so in this example $A = -3x$. Using the power series above we get that:

(2)
\begin{align} \frac{1}{1 + 3x} = \sum_{n=0}^{\infty} (-3x)^n = \sum_{n=0}^{\infty} (-3)^n x^n \quad \mathrm{for \: \mid -3x \mid < 1} \end{align}

It is always best to state the interval of convergence as neatly as possible. We note that $\mid -3x \mid = \mid -3 \mid \mid x \mid = 3 \mid x \mid < 1$ and so $\mid x \mid < \frac{1}{3}$.

## Example 2

Determine a power series representation of the function $p(x) = \frac{3}{1 + x^3}$.

In this example we want to rewrite $p(x)$ as follows: $p(x) = 3 \cdot \frac{1}{1 - (-x^3)}$. Now we have the form we are looking for, and so:

(3)
\begin{align} \quad p(x) = 3 \cdot \frac{1}{1 - (-x^3)} = 3 \sum_{n=0}^{\infty} (-x^3)^n = 3 \sum_{n=0}^{\infty} (-1)^n x^{3n} = \sum_{n=0}^{\infty} (-1)^n 3x^{3n} \quad \mathrm{for \: \mid -x^3 \mid < 1} \end{align}

Simplifying our interval of convergence we obtain that $\mid x \mid < 1$.

## Example 3

Determine a power series representation of the function $j(x) = \frac{2}{3 + \sin x}$.

In this example we want to rewrite $j(x)$ as follows: $j(x) = \frac{2}{3} \cdot \frac{1}{1 - \left (-\frac{\sin x}{3} \right)}$. Once again, we have the form we are looking for and so:

(4)
\begin{align} \quad \quad j(x) = \frac{2}{3} \cdot \frac{1}{1 - \left(-\frac{\sin x}{3} \right)} = \frac{2}{3} \sum_{n=0}^{\infty} \left (-\frac{\sin x}{3} \right)^n = \frac{2}{3} \sum_{n=0}^{\infty} (-\frac{1}{3})^n \sin ^n x \quad \mathrm{for \: \biggr \rvert \frac{- \sin x}{3} \biggr \rvert < 1} \end{align}

By simplifying our interval of convergence we obtain that $\mid \sin x \mid < 3$, which is true for all values of $x$, and so our interval of convergence is all of $\mathbb{R}$.

## Example 4

Determine a power series representation of the function $m(x) = \frac{\sin x}{x + 3x^3}$.

In this example we want to rewrite $m(x)$ as follows: $m(x) = \frac{\sin x}{x} \cdot \frac{1}{1 - (-3x^2)}$. Now we have the form we want, and so:

(5)
\begin{align} \quad \quad m(x) = \frac{\sin x}{x} \cdot \frac{1}{1 - 3x^2} = \frac{\sin x}{x} \sum_{n=0}^{\infty} (-3x^2)^n = \sin x \sum_{n=0}^{\infty} (-1)^n 3^n x^{2n-1} \: \mathrm{for \: \mid -3x^2 \mid < 1} \end{align}

By simplifying our interval of convergence we obtain that $\mid x \mid < \frac{1}{\sqrt{3}}$.