The Fourier series of $1/2-x$

Question

Let $u(x) = 1/2-x, 0<x<1$ and extend it periodically over $\mathbb{R}$. Calculate its Fourier series.

The solution is claimed to be $\hat f_n = 1/(2\pi in),n\neq 0$ and $\hat f_0 = 0$. But I did not manage to reach this answer.

Anyone help?

Answer 1

Sure. The coefficients of the Fourier series are given by

$$c_n = \frac{1}{{2l}}\int_0^{2l} {\left( {\frac{1}{2} - x} \right)} e^{ - in\pi x/l} dx = \int_0^1 {\left( {\frac{1}{2} - x} \right)} e^{ - 2in\pi x} dx$$

When $n=0$, this clearly gives $c_0=0$. When $n\ne0$, we have, from integration by parts,

$$\begin{array}{l} c_n = \frac{1}{{4\pi ni}}\left( {e^{ - 2\pi in} - 1} \right) + \frac{{e^{ - 2\pi in} }}{{2\pi in}} - \frac{1}{{2\pi ni}}\int_0^1 {e^{ - 2in\pi x} } dx \\ = \frac{1}{{4\pi ni}}\left( {e^{ - 2\pi in} - 1} \right) + \frac{{e^{ - 2\pi in} }}{{2\pi in}} + \frac{1}{{4\pi^2 n^2 }}\left( {e^{ - 2\pi in} - 1} \right) \\ \end{array}$$ The result then follows since $e^{ - 2\pi in} = 1$ for all integers $n$.