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I tried to solve it in a way. The solution did not match. Please tell me where i went wrong.

$\cfrac {1} {1+x+x^2} = \cfrac 4 {4+4x+ 4x^2} = \cfrac 4{ 3+(2x+1)^2} = \cfrac 1{\sqrt 3}\cdot\cfrac 4{ 1+ \left(\frac {2x+1}{\sqrt 3} \right)^2 }$ And now, taking the $\cfrac 4 {\sqrt 3}$ as constant, and the rest part in the form $\cfrac 1{1+t^2} $, where $t= \frac {2x+1}{\sqrt 3}$ $$\frac 1{1+t^2} = 1 - t^2 + t^4 - \dots$$ Now, substituting t’s value, and multiplying with $\cfrac 4{\sqrt 3}$, the answer is something other than $\sum x^{3n}$. The answer given is $\sum x^{3n}$.

KeyC0de
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Upamanyu Ray
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  • For the sum you are given as answer, you can sum the geometric progression and see that it is not the answer. But it is related to the answer because $1-x^3=(1-x)(1+x+x^2)$ – Mark Bennet Aug 28 '15 at 20:09
  • Please use LaTeX to format your question, right now it is not clear what you mean and it is also quite difficult to try to edit it for you as there is no clear way to interpret what you might have thought of as fractions (e.g. your 1/ 1+x+x^2 would be strictly speaking equal to $\frac{1}{1}+x+x^2=1+x+x^2$). See this guide: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Hirshy Aug 28 '15 at 20:10
  • I have edited your post to try to get the layout right. You might want to check that I have done it faithfully to the original. If you look carefully you will see that in the first line you need to take out a factor of $3$ and not $\sqrt 3$, though your value for $t$ is correct. – Mark Bennet Aug 28 '15 at 20:19
  • Your approach yields a series in $t= \frac {2x+1}{\sqrt 3}$ while you are asked a series in $x$. The answer is the series $\sum\limits_na_nx^n$ where $a_{3n}=1$, $a_{3n+1}=-1$ and $a_{3n+2}=0$ for every $n\geqslant0$. – Did Aug 28 '15 at 20:23
  • A Taylor series needs a "base point". The Taylor series for $\mathrm{f}$, close to $x=a$ has the formula

    $$\mathrm{f}(x) \sim \sum_{n=0}^{\infty} \frac{\mathrm{f}^{(n)}(a)}{n!}(x-a)^n$$

    For some functions, e.g. $\mathrm{e}^x$, the base point is merely cosmetic. But for others, e.g. $(1-x)^{-1}$, it is very important and gives non-equivalent expressions.

     – Fly by Night Aug 28 '15 at 20:46
  • @MarkBennet Thank You for editing the post. And yes, you have done it faithfully to the original. But im still blank at where my approach went wrong. All im saying is that this might be a trivial matter, but since im new, i need help. – Upamanyu Ray Aug 29 '15 at 05:52
  • Possible duplicate of Find the power series of $f(x)=\frac{1}{x^2+x+1}$ –  May 20 '18 at 09:21

2 Answers2

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Hint: $$\frac 1{1+x+x^2}=\frac {1-x}{1-x^3}$$

Mark Bennet
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$$\frac{1}{1+x+x^2}=(1-x)\sum_{n\geq 0}x^{3n} = \sum_{n\geq 0}\left(x^{3n}-x^{3n+1}\right).$$

Jack D'Aurizio
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