9
$\begingroup$

I am trying to solve exercises for the coming exam, and I am stuck on this exercise:

Evaluate the integral $$\int_{-\pi}^\pi \Big|\sum^\infty_{n=1} \frac{1}{2^n} \mathrm{e}^{inx}\,\Big|^2 \operatorname d\!x$$

A page before it intoduced the Parseval's identity, so I guess it is related to it.

I tried to solve it, but whan ever it try is bad.

Can you please give me some hints? Thanks!

$\endgroup$
  • $\begingroup$ I didn't know that double dollar $$_$$ works in the title too. It gets beautiful. $\endgroup$ – Billy Rubina Feb 1 '14 at 23:06
  • 3
    $\begingroup$ You have to apply Parseval's identity to the function $g$ defined by : $$ \displaystyle \forall x \in \mathbb{R}, \; g(x) = \sum_{n=1}^{+\infty} \frac{1}{2^{n}} e^{inx} $$ which means that you (almost) only have to compute the Fourier coefficients of $g$, which shouldn't be too hard, I think. $\endgroup$ – jibounet Feb 1 '14 at 23:09
  • $\begingroup$ Duplicate $\endgroup$ – Mhenni Benghorbal Feb 2 '14 at 2:23
  • $\begingroup$ @PristineKavalostka This is explicitly discouraged, though. $\endgroup$ – AlexR Feb 24 '14 at 23:26
9
$\begingroup$

Define $$g(x) = \sum\limits_{n = 1}^{\infty} \frac{e^{inx}}{2^n}$$

This is absolutely convergent everywhere, and defines a continuous function (by, for example, the Weierstrass $M$-test). The Fourier coefficients of $g$ are easy to compute: Define $$c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} g(x) e^{-inx} dx$$

By orthogonality of the functions $\{e^{inx}\}$, we find that

$$c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} \sum_{k = 1}^{\infty} \frac{e^{i(k - n)x}}{2^k} dx = \frac{1}{2\pi} \frac{2\pi}{2^n} = \frac{1}{2^n}$$ Then Parseval's identity shows that

$$\int_{-\pi}^{\pi} |g(x)|^2 dx = 2\pi \sum\limits_{n = 1}^{\infty} |c_n|^2 = 2\pi \sum_{n = 1}^{\infty} 4^{-n} = \frac{2\pi}{3}$$

$\endgroup$
  • $\begingroup$ Is there something you don't know? You are a genius!!! Thank you!!! $\endgroup$ – Billie Feb 1 '14 at 23:26
  • $\begingroup$ Can you please explain me how did you get: $c_n = \frac{1}{2\pi} \int_{- \\pi}^{\pi} \sum_{k = 1}^{\infty} \frac{e^{i(k - n)x}}{2^k} dx = \frac{1}{2\pi} \frac{2\pi}{2^k} = \frac{1}{2^k}$ $\endgroup$ – Billie Feb 1 '14 at 23:32
  • 1
    $\begingroup$ It should be $\frac{1}{2\pi}\frac{2\pi}{2^n} = \frac{1}{2^n}$. $\endgroup$ – TonyK Feb 1 '14 at 23:34
  • $\begingroup$ @TonyK Still don't get how he got to it $\endgroup$ – Billie Feb 1 '14 at 23:34
  • $\begingroup$ $\int_{-\pi}^{\pi} e^{i m x} dx = 2\pi$ if $m = 0$, and $0$ otherwise. $\endgroup$ – user61527 Feb 1 '14 at 23:34
8
$\begingroup$

\begin{align} \int_{-\pi}^\pi \Big|\sum^\infty_{n=1} \frac{1}{2^n} \mathrm{e}^{inx}\Big|^{\,2} \operatorname d\!x &= \int_{-\pi}^\pi \sum^\infty_{n=1} \frac{1}{2^n} \mathrm{e}^{inx} \overline{\sum^\infty_{m=1} \frac{1}{2^m} \mathrm{e}^{imx}} \operatorname d\!x= \sum_{m,n=1}^\infty \int_{-\pi}^\pi\frac{1}{2^{m+n}} \mathrm{e}^{i(m-n)x}\,dx \\ &=\sum_{n=1}^\infty\frac{2\pi}{2^{2n}}=\frac{\frac{2\pi}{4}}{1-\frac{1}{4}}=\frac{2\pi}{3}, \end{align} since $$ \int_{-\pi}^\pi\frac{1}{2^{m+n}} \mathrm{e}^{i(m-n)x}\,dx=\frac{2\pi}{2^{m+n}}\,\delta_{m,n}. $$

$\endgroup$
  • $\begingroup$ your second equation misses the constant $\frac{2\pi}{2^{m+n}}\delta_{m,n}$ ($\int e^{i(m-n)x} dx = 2\pi\delta_{m,n}$). $\endgroup$ – AlexR Feb 24 '14 at 23:27
4
$\begingroup$

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#00f}{\large\int_{-\pi}^{\pi}\verts{\sum^{\infty}_{n=1}{1 \over 2^{n}}\,\expo{inx}}^{2}\,\dd x} =\int_{-\pi}^{\pi}\verts{\expo{\ic x}/2 \over 1 - \expo{\ic x}/2}^{2}\,\dd x =\int_{-\pi}^{\pi}{\dd x \over \bracks{2 - \cos\pars{x}}^{2} + \sin^{2}\pars{x}} \\[3mm]&=2\int_{0}^{\pi}{\dd x \over 5 - 4\cos\pars{x}} =2\int_{0}^{\infty}{1 \over 5 - 4\pars{1 - t^{2}}/\pars{1 + t^{2}}}\,{2\,\dd t \over 1 + t^{2}} =4\int_{0}^{\infty}{1 \over 9t^{2} + 1}\,\dd t \\[3mm]&={4 \over 3}\int_{0}^{\infty}{1 \over t^{2} + 1}\,\dd t= {4 \over 3}\lim_{t \to \infty}\arctan\pars{t} = {4 \over 3}\,{\pi \over 2} =\color{#00f}{\large{2 \over 3}\,\pi} \end{align}

$\endgroup$
4
$\begingroup$

Hint:

$$\sum^\infty_{n=1} 2^{-n} e^{inx} = \frac{e^{ix}/2}{1-e^{ix}/2}$$

After squaring and simplifying, contour integration around $|z|=1$ should do the trick.

$\endgroup$

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.