Power series representation.

**Kuma** · April 9th 2011, 12:28 PM

Hi.

Find the power series representation of:

f(x) = x/9-x^2

factor out x/9 and we get:

1/(1-x^2/9) or 1/(1-(x/3)^2)

Now my question is. The solution shows that we use the geometric series representation, 1/1-r. In this case, shouldn't r be x^2/9. The solution uses x/3.

**TheCoffeeMachine** · April 9th 2011, 12:47 PM

Originally Posted by Kuma

Now my question is. The solution shows that we use the geometric series representation, 1/1-r. In this case, shouldn't r be x^2/9. The solution uses x/3.

It doesn't matter! You could let $r = \frac{x}{3}$ or $r = x^2/9$ and both would give you the answer:

$\displaystyle \frac{1}{1-\frac{x^2}{9}} = \sum_{k = 0}^{\infty}\left(\frac{x^2}{9}\right)^k = \sum_{k=0}^{\infty}\left(\frac{x^2}{3^2}\right)^k = \sum_{k=0}^{\infty}\left(\frac{x}{3}\right)^{2k}$ and $\displaystyle \frac{1}{1-\left(\frac{x}{3}\right)^2} = \sum_{k=0}^{\infty}\left(\frac{x}{3}\right)^{2k}$

This is because in general $\displaystyle \frac{1}{1-x^n} = \sum_{k=0}^{\infty}x^{nk}$ (you can see why by simply letting $x^n = r$ ).

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