Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

"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

share|cite|improve this question
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
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 $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.