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.

  • $\begingroup$ Hint: 3=2+1. Yes, that's a hint. $\endgroup$ – user65203 Mar 18 '16 at 18:06
  • 1
    $\begingroup$ 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

Let's try the following. For $\;|x|<2\iff \left|\frac x2\right|<1\;$ :

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

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

  • $\begingroup$ 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

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 $f(0)=-1/2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.