Power Series for Functions in Powers Other than X

Power Series for Functions in Powers Other than X

Sometimes we want to represent a function as a power series where the center of convergence is not zero. We will now look at some examples of obtaining power series for functions in powers other than $x$.

Example 1

Find a power series representation of the function $\frac{1}{4 + x}$ in powers of $x - 2$, and determine the radius of convergence for this power series

Let $t = x - 3$. Then $x = t + 3$, and so:

(1)
\begin{align} \quad \frac{1}{4 + x} = \frac{1}{7 + t} \end{align}

Therefore we have that:

(2)
\begin{align} \quad \frac{1}{7 + t} = \frac{1}{7} \frac{1}{1 - \left ( - \frac{t}{7} \right )} = \frac{1}{7} \sum_{n=0}^{\infty} \left ( -\frac{t}{7}\right )^n = \sum_{n=0}^{\infty} \frac{(-1)^n t^n}{7^{n+1}} \end{align}

Substituting $t = x - 3$ back and we get that:

(3)
\begin{align} \quad \frac{1}{4 + x} = \sum_{n=0}^{\infty} \frac{(-1)^n (x - 3)^n}{7^{n+1}} \end{align}

We must have that $\biggr \rvert \frac{-t}{7} \biggr \rvert < 1$ or equivalently, $\mid t \mid = \mid x - 3 \mid < 7$ which implies that $-4 < x < 10$. Therefore the radius of convergence is $7$.

Example 2

Find a power series representation of $\frac{1}{x^2}$ in powers of $x + 2$.

Let $t = x + 2$. Then $x = t - 2$, and so $\frac{1}{x^2} = \frac{1}{(t - 2)^2}$. We have that:

(4)
\begin{align} \quad \frac{1}{t - 2} = \frac{1}{-2 + t} = \frac{1}{-2} \frac{1}{1 - \left ( \frac{t}{2} \right )} = \frac{1}{-2} \sum_{n=0}^{\infty}\left ( \frac{t}{2} \right )^n = - \sum_{n=0}^{\infty} \frac{t^n}{2^{n+1}} \end{align}

Now we differentiate the power series above to get that:

(5)
\begin{align} \quad \frac{d}{dx} \frac{1}{t - 2} = \frac{-1}{(t - 2)^2} = \frac{d}{dx} \left ( - \sum_{n=0}^{\infty} \frac{t^n}{2^{n+1}} \right ) = - \sum_{n=0}^{\infty} \frac{nt^{n-1}}{2^{n+1}} \end{align}

Therefore:

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

Substituting back $x = t - 2$ and:

(7)
\begin{align} \quad \frac{1}{x^2} = \sum_{n=0}^{\infty} \frac{n(x + 2)^{n-1}}{2^{n+1}} \end{align}