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Need to show that:

$$\frac{1}{x^2+x+1}=\frac{2}{\sqrt3}\sum_{n=0}^{\infty}\sin\bigg(\frac{2\pi(n+1)}{3}\bigg)x^n$$

There is a hint given that $x^3-1=(x-1)(x^2+x+1)$ but I don't seem to get how I could use it. If I try to integrate the expression on the left I get:

$$\frac{2}{\sqrt3}\arctan\bigg(\frac{2x+1}{\sqrt3}\bigg)$$

Which has some resemblance to the expression I am trying to come up with.

I know the expansion for $\arctan(x)$ function, but it does not involve sines. So can I arrive at this expression explicitly of should make some useful observations allowing me to incorporate sines into solution?

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2 Answers 2

up vote 4 down vote accepted

The hint given to you is that $$\frac1{1+x+x^2}=(1-x)\frac1{1-x^3}=(1-x)\sum_{n\geqslant0}x^{3n}=\sum_{k\geqslant0}a_kx^k $$ with $a_{3n}=1$, $a_{3n+1}=-1$ and $a_{3n+2}=0$ for every $n\geqslant0$. The rest is a (not too useful, if you ask me) rewriting of the sequence $(a_k)_{k\geqslant0}$ based on the sine function.

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$$let : C_x = \sum_{n=0}^{\infty} \cos \left( \frac{2\pi(n+1)}{3} \right )x^n$$

$$and : S_x = \sum_{n=0}^{\infty} \sin \left( \frac{2\pi(n+1)}{3} \right )x^n$$

$$C_x + iS_x = \large{\sum_{n=0}^{\infty} e^{\frac{2\pi i(n+1)}{3}} x^n = e^{i\frac{2\pi}{3}} \sum_{n=0}^{\infty} (xe^{\frac{2\pi i}{3}} )^n}$$

$$= \frac{e^{i\frac{2\pi}{3}}}{1 - xe^{i\frac{2\pi}{3}}} \Rightarrow S_x = \Im \left( \frac{e^{i\frac{2\pi}{3}}}{1 - xe^{i\frac{2\pi}{3}}} \right ) \overset{\mathbf{do \ it \ by \ yourself}}{=} = \frac{\sqrt{3}}{2(x^2 + x + 1)}$$

$$\Rightarrow \frac{2}{\sqrt{3}}S_x = \frac{1}{x^2+x+1}$$

$$note : \Im = Im $$

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@Sarunas see my answer , please . –  what'sup Aug 11 '13 at 15:31

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