Find the power series of $f(x)=\frac{1}{x^2+x+1}$

Question

I want to find the power series of $$f(x)=\frac{1}{x^2+x+1}$$ How can I prove the following? $$f(x)=\frac{2}{\sqrt{3}} \sum_{n=0}^{\infty} \mathrm{sin}\frac{2\pi(n+1)}{3} x^n \,\,\,\, |x|<1$$

In particular I would like to know how to proceed in this case. The polinomial $x^2+x+1$ has no roots so here I cannot use partial fraction decomposition: what method should I use?

Answer 1

$$\frac1{x^2+x+1}=\frac{1-x}{1-x^3}=(1-x)\frac{1}{1-x^3}$$ when $x\ne 1$.

Answer 2

The polynomial $x^2+x+1=\Phi_3(x)$ has no real roots, but it vanishes at $x=e^{\pm\frac{2\pi i}{3}}$.
In particular, by setting $\omega=\exp\left(\frac{2\pi i}{3}\right)$ and $\overline{\omega}=\omega^2=\exp\left(\frac{4\pi i}{3}\right)$, $$ \frac{1}{x^2+x+1} = \frac{1}{(x-\omega)(x-\omega^2)} = \frac{i\omega^2}{\sqrt{3}}\cdot\frac{1}{1-\omega^2 x}-\frac{i\omega}{\sqrt{3}}\cdot\frac{1}{1-\omega x} $$ where the RHS, expanded as the difference between two geometric series, equals $$ \frac{i}{\sqrt{3}}\sum_{n\geq 0}\left(\omega^{2n+2}-\omega^{n+1}\right)x^n =\frac{2}{\sqrt{3}}\sum_{n\geq 0}\sin\left(\frac{2\pi(n+1)}{3}\right)x^n$$ as wanted. That clearly simplifies, since $$ \frac{1}{1+x+x^2}=\frac{1-x}{1-x^3} = \sum_{m\geq 0}\left(x^{3m}-x^{3m+1}\right),$$ too, and the Taylor series at the origin is unique.

Answer 3

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Lets $\ds{r \equiv -\,{1 \over 2} + {\root{3} \over 2}\,\ic = \expo{2\pi\ic/3}.\quad r\ \mbox{and}\ \bar{r}\quad \mbox{are the roots of}\quad x^{2} + x + 1 = 0}$.

\begin{align} {1 \over x^{2} + x + 1} & = {1 \over \pars{x - r}\pars{x - \bar{r}}} = \pars{{1 \over x - r} - {1 \over x - \bar{r}}}{1 \over r - \bar{r}} = \bracks{2\ic\Im\pars{1 \over x - r}}{1 \over 2\ic\Im\pars{r}} \\[5mm] & = -\,{2\root{3} \over 3}\,\Im\pars{\bar{r}\bracks{1 \over 1 - \bar{r}x}} = -\,{2\root{3} \over 3}\,\Im\pars{\bar{r}\sum_{n = 0}^{\infty} \bracks{\bar{r}x}^{n}} \\[5mm] & = -\,{2\root{3} \over 3}\,\sum_{n = 0}^{\infty} x^{n}\,\Im\pars{\bar{r}^{\, n + 1}} = -\,{2\root{3} \over 3}\,\sum_{n = 0}^{\infty} x^{n}\,\Im\pars{\exp\pars{-\,{2\bracks{n + 1}\pi \over 3}\,\ic}} \\[5mm] & =\ \bbox[15px,#ffe,border:2px dashed navy]{\ds{% {2\root{3} \over 3}\,\sum_{n = 0}^{\infty} \sin\pars{2\bracks{n + 1}\pi \over 3}x^{n}}}\qquad\qquad\verts{x} < 1 \end{align}