5
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Expand $\frac{1}{1+X}$ and replace $X$ by $x+x^2$. $\endgroup$ – hamam_Abdallah Oct 29 '16 at 20:29
  • 6
    $\begingroup$ Note that $f(x) = \frac{1-x}{1-x^3}$. Can you find a power series for $g(x) = \frac{1}{1-x^3}$? $\endgroup$ – Hans Engler Oct 29 '16 at 20:29
  • 1
    $\begingroup$ Note: You can use partial fraction decomposition with complex roots. This is sometimes interesting and/or useful as a technique, and might be of interest here. Have you computed explicitly the first few terms of the given power series? They will come out more simply than is apparent from the formula. $\endgroup$ – Mark Bennet Oct 29 '16 at 20:48
7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thanks for the reply! I find $$\frac{1}{1-x^3}=\sum_{0}^{\infty} (-x^3)^n \,\,\,\, |x|<1$$ But then I don't know how to deal with $1-x$.. And in the result there is a $\mathrm{sin}$ which I have no idea where it comes from if I use this method.. $\endgroup$ – Gianolepo Oct 29 '16 at 20:36
  • 1
    $\begingroup$ Compute the first few values of $\sin(2\pi(n+1)/3)$ to see what it is. This will also explain that $2/\sqrt{3}$ in your formula. $\endgroup$ – GEdgar Oct 29 '16 at 20:46
  • $\begingroup$ @Gianolepo Well, $(1-x)\frac{1}{1-x^3}$ equals $1-x$ times $\frac{1}{1-x^3}$, so you multiply them. You know how to multiply polynomials, and multiplying power series is no different. $\endgroup$ – arctic tern Oct 30 '16 at 3:56
  • $\begingroup$ rest is GEdgar's comment. The period is $3$. (0,1,2) (3,4,5) ... $\endgroup$ – student forever Oct 30 '16 at 16:12
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Note that $\omega^2=\omega^{-1}$ which may help to see where the sine function comes from. $\endgroup$ – Mark Bennet Oct 29 '16 at 20:51
2
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

$\endgroup$

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.