Finding a Power Series representation for the function $f(x) = \frac{2}{3-x}$

Let's say I want to find a Power Series representation of the function $f(x) = \frac{2}{3-x}$

Now I know we can write this as a geometric series

$$\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}$$

But I see two possible ways two write it as a geometric series. $(1)$ and $(2)$ below

Method $(1)$:

We let $a=2$, and $r = x-2$, thus we transform $\frac{2}{3-x}$ into the form $\frac{a}{1-r}$ and we write the Power Series for $f$ as follows:

$$f(x) = \sum_{n=0}^{\infty}\ 2(x-2)^n$$

Method $(2)$:

We divide both numerator and denominator by a factor of $3$ to go from $\frac{2}{3-x}$ to $$\frac{\frac{2}{3}}{1-\frac{x}{3}}$$

and we can then write $f$ as follows:

\begin{aligned}f(x) &= \sum_{n=0}^{\infty}\ \frac{2}{3}\left(\frac{x}{3}\right)^n \\ &= \sum_{n=0}^{\infty}\left(\frac{2}{3^{n+1}}\right)x^n \end{aligned}

But only $(2)$ is correct, and $(1)$ is incorrect, but I can't seem to see why. What I did in $(1)$ seemed like perfectly valid algebraic manipulations, so why does $(1)$ result in an erroneous answer?

• What is wrong with $(1)$? (Just be sure it converges i.e. $|x-2|<1$) Commented Sep 17, 2016 at 14:28

I don't see anything wrong with $(1)$, but just remember one little thing:

$$\sum_{n=0}^\infty ar^n=\frac a{1-r}$$

if $|r|<1$.

For yours, this translates to $|x-2|<1$. As far as I see, method $(1)$ is perfectly valid with those restrictions. For example, $x=2.5$:

$$\sum_{n=0}^\infty2(1/2)^n=\frac2{1-1/2}=\frac2{1/2}=\frac2{3-(2.5)}=f(2.5)$$

There is nothing wrong here.

• SimpleArt, You're correct, there's nothing wrong with $(1)$, I just realized that now, this is what little sleep and copious amounts of caffeine does. Thanks for the answer, I'll accept it as soon as possible. Commented Sep 17, 2016 at 14:34
• @Perturbative Cheers and good sleep. Commented Sep 17, 2016 at 14:35

Both methods are quite ok and valid and there are even many more possibilities to expand the function into a power series.

In order to better see what's going on, we should at first look somewhat closer at the function $f$. We consider the full specification of $f$ and choose as domain and codomain

\begin{align*} &f:\mathbb{R}\setminus\{3\}\longrightarrow\mathbb{R}\\ &f(x)=\frac{2}{3-x} \end{align*} Note that we have a singularity, a simple pole at $x=3$. This is crucial when expanding a function as power series.

As important as domain and codomain is the radius of convergence when expanding a function as power series. \begin{align*} \sum_{n=0}^\infty a x^n=\frac{a}{1-x}\qquad\qquad |x|<1 \end{align*}

It is the radius of convergence which determines where the representation of $f$ as power series is valid.

Attention: We have to explicitely state in (1) and (2) the range of validity since the power series is not defined outside this range. On the other hand $f$ can be defined on a much larger domain $\mathbb{R}\setminus\{3\}$.

Here's a graphic which illustrates both methods. We see the graph of $f$ with the asymptote at $x=3$.

• Method 1: The point $A=(2,2)$ is the center of an interval with length $2$ showing the validity of the power expansion. We clearly see the interval is bounded by the asymptote.

• Method 2: The point $B=\left(0,\frac{2}{3}\right)$ is the center of an interval with length $6$ showing the validity of the power expansion.

More expansions:

The graphic indicates what's going on, when we expand $f$ in a power series at a point $x=x_0$. The point $x_0$ is the center of an interval and the length is determined by the distance from $x_0$ to the asymptote $x=3$.

We can now expand $f$ at any point $x_0\in \mathbb{R}\setminus \{3\}$ as follows:

Of course, setting $x_0=2$ we obtain $\sum_{n=0}2(x-2)^n$ and setting $x_0=0$ we obtain $\sum_{n=0}^\infty\frac{2}{3^{n+1}}x^n$ and get back both methods as special cases.