My Question: My Goal is to determine the Fourier series for $f(x)=\max \{0, \frac{\pi}{2}-\lvert x\rvert \} \quad$ for $x \in [-\pi, \pi ]$ This function is $2\pi$-periodic.
My Approach: i found out, $f(x)$ is an even function, because its graph is symmetric to the y-axis. The Expression: $\frac{\pi}{2}-\lvert x \rvert$ has to "zeros":
$\frac{\pi}{2}-\lvert x \rvert = 0$
$\frac{\pi}{2}= \lvert x \rvert $
$x_{0}=-\frac{\pi}{2}$
$x_{1}=+\frac{\pi}{2}$
the graph must look somehow like that.. so the function is linear all over the Intervall:
But now Comes the hard stuff. I'm stuck in building the Fourier series for $f(x)$ The integral should be builded maybe this way, but i am not sure if if am right:
$\int (\frac{\pi}{2}-x)\cdot \cos (nx)\ dx = \frac{\pi}{2n}\sin (nx) - \frac{x}{n}\sin (nx) - \frac{1}{n^2}\cos (nx)$
What do you think, is that right? And what would be the next step? My textbook doesn't describes the further calculation.
p.s. edits for improving language and latex
Since the function is even, the coefficients of $\sin nx$ are all zero. The constant term is the average value of the function on $[-\pi,\pi]$: $$ \frac{1}{2\pi}\int_{-\pi}^\pi f(x)\,dx\;=\; \frac{1}{2\pi}\left(\frac{\pi^2}{4}\right) \;=\; \frac{\pi}{8} $$ (The integral here was evaluated using the formula for the area of a triangle.)
The coefficient of $\cos nx$ is given by the formula $$ a_n \;=\; \frac{1}{\pi}\int_{-\pi}^\pi f(x)\,\cos nx\;dx $$ Since $f(x)$ is nonzero only on $[-\pi/2,\pi/2]$, this is the same as $$ a_n \;=\; \frac{1}{\pi}\int_{-\pi/2}^{\pi/2} f(x)\,\cos nx\;dx $$ Moreover, since $f(x) \cos nx$ is an even function, we can restrict to $[0,\pi/2]$ and double the integral: $$ a_n \;=\; \frac{2}{\pi}\int_0^{\pi/2} f(x)\,\cos nx\;dx \;=\; \frac{2}{\pi}\int_0^{\pi/2} \left(\frac{\pi}{2} - x\right)\,\cos nx\;dx $$ The integral on the right can be evaluated using integration by parts. The result is: $$ a_n \;=\; \begin{cases}\dfrac{2}{\pi n^2} & \text{if }n\equiv 1,3\pmod{4}, \\[6pt] \dfrac{4}{\pi n^2} & \text{if }n\equiv 2\pmod 4, \\[6pt] 0 & \text{if }n\equiv 0 \pmod{4}.\end{cases} $$ Thus $$ f(x) \;=\; \frac{\pi}{8} + \frac{2}{\pi}\cos x + \frac{1}{\pi}\cos 2x + \frac{2}{9\pi}\cos 3x + \frac{2}{25\pi}\cos 5x + \frac{1}{9\pi}\cos 6x + \cdots. $$ In summation form, $$ f(x) \;=\; \frac{\pi}{8} \,+\, \sum_{k=0}^\infty \frac{2}{\pi(2k+1)^2} \cos\bigl((2k+1)x\bigr) \,+\, \sum_{k=0}^\infty \frac{1}{\pi(2k+1)^2} \cos\bigl((4k+2)x\bigr). $$ By the way, the following animation shows thee convergence of this Fourier series for the first twelve terms: