Anne Wacker

2022-01-01

Find the complex form of the Fourier Series expansion of

Charles Benedict

Expand $\mathrm{cos}\left(ax\right)=\frac{1}{2}\left({e}^{iax}+{e}^{-iax}\right)$
Then
${f}_{n}=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\mathrm{cos}\left(ax\right){e}^{-\in x}dx$
$=\frac{1}{2\pi }\frac{1}{2}{\int }_{-\pi }^{\pi }\left({e}^{iax}+{e}^{-iax}\right){e}^{-\in x}dx$
$=\frac{1}{4\pi }{\int }_{-\pi }\left({e}^{i\left(a-n\right)x}+{e}^{-a\left(a+n\right)x}\right)dx$
This is a straightforward integration. Take care when a is an integer (in which case the Fourier series is trivially obtained from the above expansion).
If $a\in \mathbb{Z}$, then the formula for $\mathrm{cos}$ gives ${f}_{n}=\frac{1}{2}{\sigma }_{|n||a|}$
${f}_{n}=\frac{1}{4\pi i}{\left(-1\right)}^{n}\left(\frac{{e}^{ia\pi }+{e}^{-ia\pi }}{a-n}-\frac{{e}^{-ia\pi }-{e}^{ia\pi }}{a+n}\right)$
$=\frac{1}{4\pi i}{\left(-1\right)}^{n}\left({e}^{ia\pi }-{e}^{-a\pi }\right)\left(\frac{1}{a-n}+\frac{1}{a+n}\right)$
$=\frac{1}{2\pi }{\left(-1\right)}^{n}\mathrm{sin}\left(a\pi \right)\frac{2a}{{a}^{2}-{n}^{2}}$
$=\frac{1}{\pi }{\left(-1\right)}^{n+1}\mathrm{sin}\left(a\pi \right)\frac{a}{{n}^{2}-{a}^{2}}$

Jacob Homer

${c}_{n}=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\mathrm{cos}ax{e}^{-\in x}dx$
We make integration by parts twice to
$I\phantom{\rule{0.222em}{0ex}}={\int }_{-\pi }^{\pi }\mathrm{cos}ax{e}^{-\in x}dx=\frac{i}{n}\mathrm{cos}ax{e}^{\in x}{\mid }_{-\pi }^{\pi }+\frac{ai}{n}{\int }_{-\pi }^{\pi }\mathrm{sin}ax{e}^{-\in x}dx$
$=-\frac{a}{{n}^{2}}\mathrm{sin}ax{e}^{-\in x}+\frac{{a}^{2}}{{n}^{2}}I$
I think you can handle it from here.

karton

Complex form of Fourier series is given by
$f\left(x\right)\sim \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{n}\left(f\right){e}^{inx}$
You need to find Fourier coefficients
${c}_{n}\left(f\right)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left(x\right){e}^{-inx}dx$
Thus, simply integrate by parts (twice)
${c}_{n}\left(f\right)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left(x\right){e}^{-inx}dx$
$=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\mathrm{cos}\left(ax\right){e}^{-inx}dx.$

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