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"Compute the Fourier series of the periodic function $f(x)$ that is defined in $\mathbb R$ as follows:

$$f(x) = |x-2n \pi| $$ for all $x$ s.t. $(2n-1)\pi < x <(2n + 1)\pi.$

Give the definition of a tempered regular distribution, and explain why $f(x)$ is such a distribution. Finally compute the 2nd distributional derivative of $f(x)$."

I am stuck on the first one in finding the Fourier series.

I have problem finding the coefficient in front of the cosine, (the integral), assuming n is real I end up with:

integral from (2n-1)*pi to (2n+1)*pi of |x-2n*pi| cos(kx) dx

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Hints: try to plot your function. Its definition is a bit tricky. After that, you could recognize some symmetry of $f$. This simplifies the computation of the Fourier coefficients. Once this is more clear, it would be good if you could add some computations to your question. –  Avitus Jun 25 '13 at 11:18

1 Answer 1

up vote 0 down vote accepted

The function is even with period $\tau=2\pi$, and in the first half-period $f(x)=x$ and the cosine coefficients are

$\displaystyle b_n=\frac{4}{\tau}\int_0^{\frac{\tau}2} x \cos (nx)\mathrm{d}x=\frac{4}{\tau} \left[\frac{x}{n}\sin(nx)+\frac{1}{n^2}\cos(nx)\right]_0^{\frac{\tau}2}=\Big\{ ^{-\frac{8}{n^2\tau} \ \ n \text{ odd}}_{0 \ \ n \text{ even}}$

for $n\not=0$ and $\frac{\tau}2$ for $n=0$.

[edit] Misunderstood the question. Set $X=x-2n\pi$ and the above is the fourier series of $f(X)=|x|$, then use $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$.

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but what if the number n is not an integer in f(x)=|x-2n*pi|? –  Jessie Jun 26 '13 at 6:25

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