Problem of the Week #50 - March 11th, 2013

[Chris L T521]

Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Let $f(x)=\exp(-|x|)$ for $-\pi\leq x\leq \pi$ be a $2\pi$-periodic function.

Show that it's Fourier series is $\displaystyle\frac{e^{\pi}-1}{\pi e^{\pi}}+\frac{2}{\pi e^{\pi}}\sum_{n=1}^{\infty} \frac{1}{n^2+1}(e^{\pi}-(-1)^n) \cos(nx)$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Chris L T521]

This week's problem was correctly answered by BAdhi. You can find his solution below.

Spoiler

The exponential Fourier series of a periodic function $f(x)$ with a period of $2\pi$ is,

$$f(x)=\sum\limits_{n=-\infty}^{\infty} c_n e^{inx}$$
where,
$$c_n=\frac{1}{2\pi}\int \limits_{-\pi}^{\pi}f(x)e^{-inx}dx$$

when $f(x)=\exp(-|x|)$ for $\pi\leq x\leq \pi$, with a period of $2\pi$,

$$\begin{align*}
c_n&=\frac{1}{2\pi}\int \limits_{-\pi}^{\pi} e^{(-|x|)}e^{-inx}dx\\
&=\frac{1}{2\pi}\left[\int \limits_{-\pi}^{0}e^{x}e^{-inx}dx+\int \limits_{0}^{\pi}e^{-x}e^{-inx}dx\right]\\
&=\frac{1}{2\pi}\left[\int \limits_{-\pi}^{0}e^{x(1-in)}dx+\int \limits_{0}^{\pi}e^{-x(1+in)}dx\right]\\
&=\frac{1}{2\pi}\left( \left[\frac{e^{x(1-in)}}{(1-in)}\right]_{-\pi}^{0}- \left[ \frac{e^{-x(1+in)}}{(1+in)} \right]_{0}^{\pi} \right)\\
&=\frac{1}{2\pi}\left( \frac{1-e^{-\pi(1-in)}}{(1-in)}-\frac{e^{-\pi(1+in)}-1}{(1+in)}\right)\\
&=\frac{1}{2\pi}\left( \frac{1-e^{-\pi}(e^{i\pi})^n}{(1-in)}-\frac{e^{-\pi}(e^{i\pi})^{-n}-1}{(1+in)} \right)\\
&=\frac{1}{2\pi}\left( \frac{1-(-1)^n e^{-\pi}}{(1-in)}-\frac{(-1)^{-n}e^{-\pi}-1}{(1+in)} \right)\\
&=\frac{1}{2\pi}\left( \frac{[1-(-1)^n e^{-\pi}](1+in)-[(-1)^{n}e^{-\pi}-1](1-in)}{(1-in)(1+in)} \right)\\
&=\frac{1}{2\pi}\left( \frac{2[1-(-1)^ne^{-\pi}]}{(1+n^2)} \right)\\
&=\frac{[1-(-1)^ne^{-\pi}]}{\pi(1+n^2)}
\end{align*}$$

since,
$$c_n=\frac{a_n-ib_n}{2}\qquad n>0\\
c_{n}=\frac{a_{-n}+ib_{-n}}{2}\qquad n<0$$ where $a_n$ and $b_n$ are coefficients of trigonometric Fourier series and $b_0=0$

then,
$$\frac{[1-(-1)^ne^{-\pi}]}{\pi(1+n^2)}=\frac{a_n}{2}-i\frac{b_n}{2}$$
comparing imaginary and real components,
$$a_n=\frac{2[1-(-1)^ne^{-\pi}]}{\pi(1+n^2)}, \qquad b_n=0$$

Then the trigonometric fourier series will be,

$$\begin{align*}
f(x)&=\frac{a_0}{2}+ \sum \limits_{n=1}^\infty a_n \cos(nx)\\
&=\frac{[1-e^{-\pi}]}{\pi}+\sum \limits_{n=1}^\infty \frac{2[1-(-1)^ne^{-\pi}]}{\pi(1+n^2)} \cos(nx)\\
&=\frac{e^{\pi}-1}{\pi e^{\pi}}+\frac{2}{\pi e^\pi}\sum\limits_{n=1}^\infty \frac{1}{(1+n^2)} (e^{\pi}-(-1)^n)\cos(nx)
\end{align*}$$