Determining Power Series Representations of Functions Examples 1

# Determining Power Series Representations of Functions Examples 1

Recall from the Determining Power Series Representations Of Functions page that:

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.

We will now be looking at some more examples of determining power series representations of functions using this technique.

## Example 1

Determine a power series representation of the function $t(x) = \frac{x^3 - \log x^2 + 2^x}{5 + \cos x}$.

This example may look incredibly difficult, but it isn't harder than anything we've look at so far. We first want to rewrite $t(x)$ as follows: $t(x) = \frac{x^3 - \log x^2 + 2^x}{5} \cdot \frac{1}{1 - \left ( - \frac{\cos x}{5} \right)}$. Therefore we have that:

(1)
\begin{align} \quad \quad t(x) = \frac{x^3 - \log x^2 + 2^x}{5} \cdot \frac{1}{1 - \left ( -\frac{\cos x}{5} \right)} = \frac{x^3 - \log x^2 + 2^x}{5} \sum_{n=0}^{\infty} (-\frac{1}{5})^n \cos^n x \quad \mathrm{for \: \biggr \rvert -\frac{\cos x}{5} \biggr \rvert < 1} \end{align}

Upon simplifying the interval of convergence we get that $\mid \cos x \mid < 5$ which is true for all $x$, so our power series represents $t(x)$ for all of $\mathbb{R}$.

## Example 2

Determine a power series representation of the function $w(\varphi) = \frac{\varphi^{2} - \ln ( (\sin (\varphi))^2)}{2^{\varphi} + e^{\varphi}}$.

Once again this example may look complicated, but it really is not. We first want to rewrite $w(\varphi)$ as follow: $w(\varphi) = \frac{\varphi^2 - \ln ( (\sin (\varphi))^2)}{2^{\varphi}} \cdot \frac{1}{1 - \left ( - \left (\frac{e}{2} \right)^{\varphi} \right )}$ and therefore:

(2)
\begin{align} \quad \quad w(\varphi) = \frac{\varphi^2 - \ln ( (\sin (\varphi))^2)}{2^{\varphi}} \cdot \frac{1}{1 - \left ( - \left (\frac{e}{2} \right)^{\varphi} \right )} = \frac{\varphi^2 - \ln ( (\sin (\varphi))^2)}{2^{\varphi}} \sum_{n=0}^{\infty} (-1)^n \left ( \frac{e}{2} \right )^{\varphi n} \quad \mathrm{for \: \biggr \rvert -\left (\frac{e}{2} \right)^\varphi \biggr \rvert < 1} \end{align}

We note now that $\biggr \rvert -\left (\frac{e}{2} \right)^\varphi \biggr \rvert < 1$ implies that $\biggr \rvert \left (\frac{e}{2} \right)^\varphi \biggr \rvert < 1$, which is true if and only if $\varphi < 0$ and so therefore our interval of convergence is $\varphi < 0$. The diagram below illustrates our power series representation of $w(\varphi)$.