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Problem of the Week #27 - October 1st, 2012

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Find the Fourier series representation for the $2p$-periodic function
\[f(x) = a\left(1-\left(\tfrac{x}{p}\right)^2\right),\quad -p\leq x\leq p,\, a\neq 0.\]

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's problem was correctly answered by BAdhi. You can find his solution below.

The angular frequency - $\omega$ of this periodic function is $2\pi \frac{1}{2p}=\frac{\pi}{p}$Let the Fourier series of the function $f(x)$ be,


$$f(x)=\frac{a_0}{2}+ \sum \limits_{n=1}^{\infty} [a_n \cos(n\omega x)+b_n\sin(n\omega x)]$$


the function $f(x)=a\left( 1-\left( \frac{x}{p}\right) ^2 \right)$ is symmetrical over the y-axis, so function is an even function which makes the sin terms disappear ($b_n=0$) from the fourier series. Then the fourier series will be,


$$f(x)=\frac{a_0}{2}+ \sum \limits_{n=1}^{\infty} a_n \cos(n\omega x)$$


before continuing with finding coefficients of the fourier series, Lets find results of the following integrals.


$when\; \omega =\frac{\pi}{p} \implies \omega p=\pi$


$$\int \limits_{-p}^{p} \cos(n\omega x)\, dx =\left[ \frac{\sin(n\omega x)}{n\omega}\right]_{-p}^{p}=\frac{1}{n\omega}(\sin(n\pi) +\sin(n\pi))=0 \qquad \qquad ...(1)$$




$$\begin{align}
\int \limits_{-p}^{p} x^2\cos(n\omega x)\, dx&= \left[x^2\underbrace{\frac{\sin(n\omega x)}{n\omega}}_{=0\; from\; (1)}\right]_{-p}^{p}-\int \limits_{-p}^{p} 2x\frac{\sin(n\omega x)}{n\omega}\, dx\\
&=-2\left[ \left[-x \frac{\cos(n\omega x)}{(n\omega )^2}\right]_{-p}^{p}-\int \limits_{-p}^{p} \frac{-\cos(n\omega x)}{(n\omega )^2}\, dx \right]\\
&=2\left[ \frac{p\cos(n\pi )-(-p)\cos(-n\pi )}{(n\omega )^2}+\underbrace{ \left[ \frac{sin(n\omega x)}{(n\omega )^3} \right]_{-p}^{p}}_{=0\; from\; (1)} \right] \\
&=2\left[\frac{p\cos(n\pi)+p\cos(n\pi)}{(n\omega)^2}\right]\\
&=\frac{4p\cos(n\pi)}{(n\omega)^2}\\
&=\frac{4p(-1)^n}{(n\omega)^2} \qquad \qquad ...(2)
\end{align}$$


Now let's find the coefficients of the fourier series of the function $f(x)$,


$$\begin{align}
a_0&=\frac{2}{2p} \int \limits_{-p}^{p} f(x)\, dx\\
&=\frac{1}{p} \int \limits_{-p}^{p} a\left(1-\left(\frac{x}{p}\right)^2 \right) \, dx\\
&=\frac{a}{p}\left[ x-\frac{x^3}{3p^2} \right]_{-p}^{p}\\
&=\frac{a}{p}\left[ \left( p-\frac{p^3}{3p^2}\right) - \left(-p-\frac{(-p)^3}{3p^2}\right) \right]\\
&=\frac{a}{p}\left[ p-\frac{p}{3}+p-\frac{p}{3}\right]\\
&=\frac{4a}{3}
\end{align}$$


$$\begin{align}
a_n&=\frac{2}{2p} \int \limits_{-p}^{p} f(x)\cos(n\omega x)\,dx\\
&=\frac{1}{p} \int \limits_{-p}^{p} a\left( 1-\left( \frac{x}{p}\right) ^2 \right) \cos(n\omega x)\, dx\\
&=\frac{1}{p} \int \limits_{-p}^{p} a\cos(n\omega x) -\frac{a}{p^2}x^2\cos(n\omega x)\, dx\\
&=\frac{a}{p} \underbrace{\left[ \int \limits_{-p}^{p} \cos(n\omega x)\, dx\right]}_{=0\; from\; (1)} -\frac{a}{p^3} \underbrace{\left[ \int \limits_{-p}^{p} x^2\cos(n\omega x)\, dx\right]}_{=\frac{4p(-1)^n}{(n\omega)^2}\; from \; (2)} \\
&=-\frac{4a(-1)^n}{(np\omega)^2}\\
&=-\frac{4a(-1)^n}{(n\pi)^2}
\end{align}$$


Enventually, the fourier series of the function f(x) can be stated as,


$$f(x)=\frac{2a}{3}- \sum \limits_{n=1}^{\infty} \frac{4a(-1)^n}{(n\pi)^2}\cos \left(\frac{n\pi x}{p}\right)$$
 
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