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I have to find a power series representation for $f(x) = \frac{x-1}{x+2}$. In rearranging the function so as to attain a form suitable for representation as a power series I get $$(x-1) * \frac{1}{2}\left(\frac{1}{1-(-\frac{x}{2})}\right) $$

Which yields

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

So I know I can express my first sum as $\sum_{n=1}^\infty(-1)^n\frac{x^n}{2^n}$, and after browsing a few similar questions to mine I found that the second can be expressed as $\frac{1}{2}(1+\sum_{n=1}^\infty(-1)^n\frac{x^n}{2^n})$, so that ultimately I'd have

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

Which differs from the book's answer

$$-\frac{1}{2} - \sum_{n=1}^\infty(-1)^n\frac{3x^n}{2^{n+1}}$$

This one's got me stumped. Where'd the 3 in the book's answer come from? I'm not sure what I'm doing wrong.

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  • $\begingroup$ Hint: 3=2+1. Yes, that's a hint. $\endgroup$ – user65203 Mar 18 '16 at 18:06
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    $\begingroup$ You know you can just straight perform polynomial long division to get the power series? $\endgroup$ – Simply Beautiful Art Mar 18 '16 at 18:17
  • $\begingroup$ Thanks for that! Using long division hadn't occurred to me. $\endgroup$ – enharmonics Mar 18 '16 at 18:40
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Let's try the following. For $\;|x|<2\iff \left|\frac x2\right|<1\;$ :

$$\frac{x-1}{x+2}=1-\frac3{x+2}=1-\frac32\cdot\frac1{1+\frac x2}=$$

$$1-\frac32\sum_{n=0}^\infty(-1)^n\left(\frac x2\right)^n=1-\frac32-\frac32\sum_{n=1}^\infty(-1)^n\frac{x^n}{2^n}=-\frac12-\sum_{n=1}^\infty(-1)^n\frac{3x^n}{2^{n+1}}$$

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  • $\begingroup$ Just out of curiosity, is there any particular reason why the book shifts the beginning of the index of summation to $n =1$? Would writing the answer as $1 - \frac{3}{2} \sum_{n=0}^\infty(-1)^n(\frac{x}{2})^n$ be okay? $\endgroup$ – enharmonics Mar 18 '16 at 18:41
  • $\begingroup$ @enharmonics I really don't know. I can't see any standard, important reason. Perhaps it is just tje author's personal taste, perhaps it is related to some other problem. $\endgroup$ – DonAntonio Mar 18 '16 at 18:43
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Your series gives $$\frac12+\frac{-x}{2 (2 + x)}=\frac{1}{x+2}$$

You can also see that your solution is wrong by checking $x=0$ which should give you $f(0)=-1/2$.

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