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I feel like I'm on the right track, but I don't know if I need to do something else to finish it off...

\begin{align*} f(3x)&=\frac{2}{3-3x}\\ &=\frac{2}{3(1-x)}\\ &=\frac{2}{3}\cdot\frac{1}{1-x}\\ &=\frac{2}{3}\cdot\sum_{n=0}^{\infty}x^{n} \end{align*}

I feel like - OK, so I substituted $x$ with $3x$ so that I could make the fraction look like how I needed it to - but do I need to do something else to reverse the substitution? Or do I have to do this completely differently?

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This is correct, now you can substitute $x$ by $x/3$ and you will get the final answer. – zuggg Jul 6 '13 at 6:49
up vote 2 down vote accepted

You method looks great, but you aren't quite finished! The only thing I would change in the beginning is to let $x=3u$ rather than letting $x=3x$. It's a notational thing, but letting $x=3x$ is not a true substitution.

Using my method, at the end you would have $$ f(3u)=\frac{2}{3}\cdot\sum_{n=0}^\infty u^n. $$

To put the function back in terms of $x$, we can make use of our substitution by plugging back in that $3u=x$ and that $u=\frac{x}{3}$ resulting in

$$ f(x)=\frac{2}{3}\cdot\sum_{n=0}^\infty \left(\frac{x}{3}\right)^n. $$

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Ah.. I like that. Just like u-substitution in integration, it is quite easy to see how to back-substitute. Thanks for the insight. – agent154 Jul 7 '13 at 2:37

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

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