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?

  • $\begingroup$ What is wrong with $(1)$? (Just be sure it converges i.e. $|x-2|<1$) $\endgroup$ – Simply Beautiful Art Sep 17 '16 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, $x=2.5$:


There is nothing wrong here.

  • $\begingroup$ 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. $\endgroup$ – Perturbative Sep 17 '16 at 14:34
  • $\begingroup$ @Perturbative Cheers and good sleep. $\endgroup$ – Simply Beautiful Art Sep 17 '16 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.

If we look at the series expansion of the first method \begin{align*} f(x)=\sum_{n=0}^\infty2(x-2)^n\qquad\qquad\qquad |x-2|<1\tag{1} \end{align*} and the second method gives \begin{align*} f(x)=\sum_{n=0}^\infty\left(\frac{2}{3^{n+1}}\right)x^{n}=\frac{\frac{2}{3}}{1-\frac{x}{3}}\qquad\qquad \left|\frac{x}{3}\right|<1\tag{2} \end{align*}

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 $x=3$.

                  enter image description here

  • 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:

\begin{align*} \frac{2}{3-x}&=\frac{2}{(3-x_0)-(x-x_0)}\\ &=\frac{2}{3-x_0}\cdot\frac{1}{1-\frac{x-x_0}{3-x_0}}\\ &=\frac{2}{3-x_0}\sum_{n=0}^\infty\left(\frac{x-x_0}{3-x_0}\right)^n\\ &=\sum_{n=0}^\infty\frac{2}{(3-x_0)^{n+1}}(x-x_0)^n\qquad\qquad \left|\frac{x-x_0}{3-x_0}\right|<1 \end{align*}

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.

  • $\begingroup$ just wanted to thank you for this awesome answer to this question! $\endgroup$ – Perturbative Sep 17 '16 at 17:00
  • $\begingroup$ @Perturbative: My pleasure. Your question indicated you could already look somewhat behind the scene! :-) $\endgroup$ – Markus Scheuer Sep 17 '16 at 17:01

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