Anne Wacker

2022-01-01

Find the complex form of the Fourier Series expansion of

$f\left(x\right)=\mathrm{cos}ax,\text{}-\pi x\pi$

Charles Benedict

Beginner2022-01-02Added 32 answers

Expand $\mathrm{cos}\left(ax\right)=\frac{1}{2}({e}^{iax}+{e}^{-iax})$

Then

${f}_{n}=\frac{1}{2\pi}{\int}_{-\pi}^{\pi}\mathrm{cos}\left(ax\right){e}^{-\in x}dx$

$=\frac{1}{2\pi}\frac{1}{2}{\int}_{-\pi}^{\pi}({e}^{iax}+{e}^{-iax}){e}^{-\in x}dx$

$=\frac{1}{4\pi}{\int}_{-\pi}({e}^{i(a-n)x}+{e}^{-a(a+n)x})dx$

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 $f}_{n}=\frac{1}{2}{\sigma}_{\left|n\right|\left|a\right|$

${f}_{n}=\frac{1}{4\pi i}{(-1)}^{n}(\frac{{e}^{ia\pi}+{e}^{-ia\pi}}{a-n}-\frac{{e}^{-ia\pi}-{e}^{ia\pi}}{a+n})$

$=\frac{1}{4\pi i}{(-1)}^{n}({e}^{ia\pi}-{e}^{-a\pi})(\frac{1}{a-n}+\frac{1}{a+n})$

$=\frac{1}{2\pi}{(-1)}^{n}\mathrm{sin}\left(a\pi \right)\frac{2a}{{a}^{2}-{n}^{2}}$

$=\frac{1}{\pi}{(-1)}^{n+1}\mathrm{sin}\left(a\pi \right)\frac{a}{{n}^{2}-{a}^{2}}$

Then

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

Jacob Homer

Beginner2022-01-03Added 41 answers

We make integration by parts twice to

I think you can handle it from here.

karton

Expert2022-01-09Added 613 answers

Complex form of Fourier series is given by

You need to find Fourier coefficients

Thus, simply integrate by parts (twice)

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