Power Series for Functions in Powers Other than x Examples 1

# Power Series for Functions in Powers Other than x Examples 1

On the Power Series for Functions in Powers Other than x page, we looked at some examples of obtaining power series representations of functions in powers other than $x$. We will now look at some more examples like that.

## Example 1

Find a power series representation of $\ln \mid 2 - x \mid$ in powers of $x - 1$. Determine the center of convergence, radius of convergence, and interval of convergence for this power series.

Let $t = x - 1$. Then $x = t + 1$, and so $\ln \mid 2 - x \mid = \mid 1 - t \mid$. Note that $\ln \mid 1 - t \mid = -\int \frac{1}{1 - t} \: dt$. We have that:

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

Therefore by integrating this series we get:

(2)
\begin{align} \quad -\int_0^m \frac{1}{1 - m} \: dm = - \int_0^m \sum_{n=0}^{\infty} m^n \: dm \\ \quad \left [ \ln \mid 1 - m \mid \right ]_{m=0}^{m=t} = - \sum_{n=0}^{\infty} \frac{t^{n+1}}{n+1} \\ \quad \ln \mid 1 - t \mid = - \sum_{n=0}^{\infty} \frac{t^{n+1}}{n+1} \\ \end{align}

Substituting back in $t = x - 1$ and we have that:

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

The center of convergence is $1$. We must have that $\mid t \mid < 1$ which is equivalent to $\mid x - 1 \mid < 1$ or $0 < x < 2$. Therefore the radius of convergence is also $1$. We now test the endpoints, $0$ and $2$ of our series.

For $x = 0$ we have:

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

This series converges as an alternating harmonic series. For $x = 2$, we have:

(5)
\begin{align} \quad - \sum_{n=0}^{\infty} \frac{(1)^{n+1}}{n+1} \end{align}

This series diverges as a regular harmonic series. Therefore the interval of convergence is $[0, 2)$.