A rational function f(x) has the following power series representation for the interval $-3

Question

A rational function f(x) has the following power series representation for the interval $-3<x<3$. $f(x) = x - \frac{x^2}{3} + \frac{x^3}{3^2} +...$. Find a closed-form expression for f(x).

Now, I found that $f(x)=\sum_{n=1}^\infty (\frac{-1}{3})^{n-1}(x)^n$, but how do I continue from here.

Yes I recognised this but the 'x' is not to the same power as the $\frac{-1}{3}$ doesn't this pose a problem? — user2250537, Mar 31 '16 at 22:28
First term $x$, common ratio $-\frac{x}{3}$. Now can you find the sum? — André Nicolas, Mar 31 '16 at 22:29
Does that mean the closed form expression is $\frac{x(1-(\frac{-x}{3})^n}{1-(\frac{-x}{3})}$ — user2250537, Mar 31 '16 at 22:33
It is an infinite geometric series, sum $\frac{x}{1+x/3}$, which maybe looks better as $\frac{3x}{3+x}$. Note that for convergence we need $|x/3|\lt 1$, so $|x|\lt 3$. — André Nicolas, Mar 31 '16 at 22:43
You may recall that if $|r|\lt 1$ the infinite geometric series $a+ar+ar^2+\cdots$ has sum $\frac{a}{1-r}$. Here $a=x$ and $r=-x/3$. — André Nicolas, Mar 31 '16 at 22:48

score 1 · Accepted Answer · answered Mar 31 '16 at 22:46

If you need the terms to have the same exponent for everything, just make it happen:

$\begin{array}\\ f(x) &=\sum_{n=1}^\infty (\frac{-1}{3})^{n-1}(x)^n\\ &=x\sum_{n=1}^\infty (\frac{-1}{3})^{n-1}(x)^{n-1} \qquad\text{(pull x out so exponents the same)}\\ &=x\sum_{n=0}^\infty (\frac{-1}{3})^{n}(x)^{n} \qquad\text{(shift exponents)}\\ &=x\frac1{1-\frac{-x}{3}}\\ &=\frac{x}{1+\frac{x}{3}}\\ &=\frac{3x}{3+x} \qquad\text{(this is optional)}\\ \end{array} $

Perfect! Thank you so much ;) – user2250537 Mar 31 '16 at 22:48 — user2250537, Mar 31 '16 at 22:48

A rational function f(x) has the following power series representation for the interval $-3

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