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I'm trying to find a power series representation for $$\frac{-x}{x^2 + 3x + 2}$$ I think I've done it correctly save the last step. I create partial fractions and have to subtract two different power series. I skip the more simple steps as I have verified those are correct. If more steps are needed please let me know

Get the partial fractions from our original equation

$$ \frac{1}{x+1} - \frac{2}{x+2} $$

Turn them into their respective power series

$$\sum_{n=0}^\infty (-1)^n(x)^n - \sum_{n=0}^\infty (-1)^n(\frac{x}{2})^n$$

This is where my math fails me, subtracting these just by subtracting their terms gives me

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

which equals the following.

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

Now I know this is wrong, but I can't seem to make sense of my notes or any online resources to tell me where my mistake is. I'm very shaky on my power series so forgive me for simple mistakes. Thanks for reading and any help is appreciated!

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  • $\begingroup$ What makes you feel you can do that subtraction? You can see by putting in successive values for $n$ that it won't yield the same terms of the series $\endgroup$ – imranfat Nov 23 '17 at 17:55
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    $\begingroup$ Where did $(-1)^n$ go? $\endgroup$ – Math Lover Nov 23 '17 at 18:01
  • $\begingroup$ $(-1)^n - (-1)^n = 0$, or so I think $\endgroup$ – Howard P Nov 23 '17 at 18:19
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You seem to have lost the $(-1)^n$. You should have $$\sum_{n=0}^\infty (-1)^n\frac{2^nx^n- x^n}{2^n}= \sum_{n=0}^\infty (-1)^n\frac{2^n- 1}{2^n}x^n \; .$$

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  • $\begingroup$ Why does the $(-1)^n$ not become 0? when I subtract both sides they cancel out, don't they? $\endgroup$ – Howard P Nov 23 '17 at 18:21
  • $\begingroup$ Ah, wait, nevermind. I see my error now $\endgroup$ – Howard P Nov 23 '17 at 18:21

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