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Find the Fourier series of f(x) = cos μx in (-π,π), where μ is not an integer.
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$\begin{align*} f(-x) &= \cos \left(-\mathrm{\mu}x\right)\\[2ex] &= \cos \mathrm{\mu}x\\[2ex] & = f(x)\\[2ex] \end{align*}\\[2ex] \therefore{}\ \cos \mathrm{\mu}x \ is \ an \ even \ function.\\[2ex] \therefore{} b_n = 0$

$\begin{align*} a_o&=\frac{1}{\pi{}}\int_0^{\pi{}}f\left(x\right)\ \ dx\\[2ex] &= \frac{1}{\pi{}}\int_0^{\pi{}}cos\ \mathrm{\mu}x\ \ \ dx\\[2ex] &=\frac{1}{\pi{}}\ {\left[\frac{\sin{\mathrm{\mu}x}}{\mathrm{\mu}}\right]}_0^{\pi{}}\\[2ex] &=\frac{1}{\mathrm{\mu}\pi{}}\sin{\mathrm{\mu}\pi{}}\\[2ex] &=\frac{\sin{\mathrm{\mu}\pi{}}}{\mathrm{\mu}\pi{}} \end{align*}\\ $

$\begin{align*} \ \ \ a_n&=\frac{2}{\pi{}}\int_0^{\pi{}}f\left(x\right)\text{ cosnx}\ dx\\[2ex] &=\frac{2}{\pi{}}\int_0^{\pi{}}\cos\ \mathrm{\mu}x\text{ cosnx}\ dx\\[2ex] &=\frac{1}{\pi{}}\ \int_0^{\pi{}}\left[\cos\left(\mathrm{\mu}+n\right)x+\cos\left(\mathrm{\mu}-n\right)x\ \right]dx\ \\[2ex] &=\frac{1}{\pi{}}\ {\left[\ \frac{\sin\left(\mathrm{\mu}+n\right)x}{\left(\mathrm{\mu}+n\right)}+\ \frac{\sin\left(\mathrm{\mu}-n\right)x}{\left(\mathrm{\mu}-n\right)}\right]}_0^{\pi{}}\\[2ex] &=\frac{1}{\pi{}}\ \left[\ \frac{\sin\left(\mathrm{\mu}+n\right)\pi{}}{\left(\mathrm{\mu}+n\right)}+ \frac{\sin\left(\mathrm{\mu}-n\right)\pi{}}{\left(\mathrm{\mu}-n\right)}\right]\\[2ex] &=\frac{1}{\pi{}}\ \left[\ \frac{\sin\left(\mathrm{\mu}\pi{}+n\pi{}\right)}{\left(\mathrm{\mu}+n\right)}+\ \frac{\sin\left(\mathrm{\mu}\pi{}-n\pi{}\right)}{\left(\mathrm{\mu}-n\right)}\right]\\[2ex] &=\frac{1}{\pi{}}\ \left[\ \frac{{(-1)}^n \sin\left(\mathrm{\mu}\pi{}\right)}{\left(\mathrm{\mu}+n\right)}+\ \frac{{(-1)}^n \sin\left(\mathrm{\mu}\pi{}\right)}{\left(\mathrm{\mu}-n\right)}\right]\\[2ex] &=\frac{{(-1)}^n \sin\left(\mathrm{\mu}\pi{}\right)}{\pi{}}\left[\frac{\mathrm{\mu}-n+\mathrm{\mu}+n}{{\mathrm{\mu}}^2-n^2}\right]\\[2ex] &=\frac{2\mathrm{\mu}\ \sin\left(\mathrm{\mu}\pi{}\right)}{\pi{}}\left[\frac{{(-1)}^n}{{\mathrm{\mu}}^2-n^2}\right]\\[2ex] \end{align*}\\ \displaystyle \therefore{}\cos{\mathrm{\mu}x} = \frac{\sin{\mathrm{\mu}\pi{}}}{\mathrm{\mu}\pi{}}+\ \frac{2\mathrm{\mu}\ sin\left(\mathrm{\mu}\pi{}\right)}{\pi{}}\sum_{n=1}^{\infty{}}\left(\frac{{(-1)}^n}{{\mathrm{\mu}}^2-n^2}\cos{nx}\right)\\[2ex] \ \ \ \ \ \displaystyle \cos{\mathrm{\mu}x}= \frac{2\mathrm{\mu}\ sin\left(\mathrm{\mu}\pi{}\right)}{\pi{}}\ \left[\frac{1}{2{\mathrm{\mu}}^2}-\ \frac{cosx}{{\mathrm{\mu}}^2-1^2}+\ \frac{cosx}{{\mathrm{\mu}}^2-2^2}-\ \frac{cosx}{{\mathrm{\mu}}^2-3^2}+…\right]$

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