I think I've the graph wrong, just struggling with the Fourier series.
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1What are you exactly struggling with? – Tom Himler Jun 29 '18 at 14:06
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Can you upload what you have tried? – Alex Vong Jun 30 '18 at 01:19
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Your function is even on $[-\pi,\pi]$. So you end up with a Fourier cosine series only. For $n=1,2,3,\cdots$, $$ a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(n x)dx \\= \frac{1}{\pi}\int_{-\pi}^{0}(\pi +x)\cos(n x)dx + \frac{1}{\pi}\int_{0}^{\pi}(\pi-x)\cos(n x)dx $$ For $n=0$, $$ a_0 = \frac{1}{2\pi}\int_{-\pi}^{0}(\pi +x)dx+\frac{1}{2\pi}\int_{0}^{\pi}(\pi-x)dx. $$
The Fourier series is
$$ f \sim \sum_{n=0}^{\infty}a_n \cos(n\pi x). $$

Disintegrating By Parts
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