I am struggling with the Fourier coefficients for a piecewise constant function:
Let
\begin{equation} f(\xi)=\begin{cases} 2, \ \ \ \ -3\le \xi<-\frac{3}{2} \\ 1, \ \ \ \ -\frac{3}{2}\le \xi<\frac{1}{5} \\ 3, \ \ \ \ 0.2\le \xi<1 \\ 6, \ \ \ \ 1\le \xi\le3 \end{cases} \end{equation}
be piecewise constant function on the symmetric interval $[-3,3]$.
Then I get, with $h_j=(\xi_j-\xi_{j-1})$ and have verified that this is correct with Mathematica:
\begin{equation} \begin{split} &\alpha_0=\frac{1}{2a}\int_{-a}^af(\xi)\text{d}\xi=\frac{1}{6}\sum_{j=1}^4\eta_j(h_j)={}\\ &\frac{1}{6}\bigg[\bigg(2(-\frac{3}{2}+3)\bigg)+(\frac{1}{5}+\frac{3}{2})\bigg)+3(1-\frac{1}{5})+6(3-1)\bigg)\bigg] =\frac{191}{60} \end{split} \end{equation}
But when calculate the $\alpha_k$ coefficient, which should be the same structure, only with $\cos\omega k\xi$ multiplied to it, with $\omega=\frac{\pi k}{n}=\frac{\pi k}{4}$ (n must be 4, since there are 4 intervals in the piecewise function):
I get the incorrect result:
\begin{equation} \begin{split} &\alpha_k=\frac{1}{a}\int_{-a}^af(\xi)\cos(\omega k \xi)\text{d}\xi=\frac{2}{\pi k}\sum_{j=1}^4\eta_j\sin\bigg(\omega k \frac{h_j}{2}\bigg)\cos\bigg(\omega k\frac{\xi_j+\xi_{j-1}}{2}\bigg)={}\\ &\frac{2}{\pi k}\bigg[2\sin\bigg(\frac{\pi k}{4} k \frac{(-\frac{3}{2}+3)}{2}\bigg)\cos\bigg(\frac{\pi k}{4} k\frac{(-\frac{3}{2}-3)}{2}\bigg)+\sin\bigg(\frac{\pi k}{4} k \frac{(\frac{1}{5}+\frac{3}{2})}{2}\bigg)\cos\bigg(\frac{\pi k}{4} k\frac{(\frac{1}{5}-\frac{3}{2})}{2}\bigg)+{}\\ &3\sin\bigg(\frac{\pi k}{4} k \frac{(1-\frac{1}{5})}{2}\bigg)\cos\bigg(\frac{\pi k}{4} k\frac{(1+\frac{1}{5})}{2}\bigg)+6\sin\bigg(\frac{\pi k}{4} k \frac{(3-1)}{2}\bigg)\cos\bigg(\frac{\pi k}{4} k\frac{(3+1)}{2}\bigg)\bigg] ={}\\ &\frac{2}{\pi k}\bigg[6 \sin \left(\frac{\pi k^2}{4}\right) \cos \left(\frac{\pi k^2}{2}\right)+3 \sin \left(\frac{\pi k^2}{10}\right) \cos \left(\frac{3 \pi k^2}{20}\right)+2 \sin \left(\frac{3 \pi k^2}{16}\right) \cos \left(\frac{9 \pi k^2}{16}\right)+{}\\ &\sin \left(\frac{17 \pi k^2}{80}\right) \cos \left(\frac{13 \pi k^2}{80}\right)\bigg] \end{split} \end{equation}
But how can it be incorrect? The formula is correct indeed.
Or have I misunderstood something here?
What is the right set of calculation steps?
Thanks
The sine-cosine form of the Fourier series for
$$f(\xi)=\left\{ \begin{array}{cc} 2 & -3\leq \xi \leq -\frac{3}{2} \\ 1 & -\frac{3}{2}\leq \xi <\frac{1}{5} \\ 3 & \frac{1}{5}\leq \xi <1 \\ 6 & 1\leq \xi \leq 3 \\ \end{array}\right.\tag{1}$$
is given by
$$f(\xi)=\frac{a_0}{2}+\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \left(a_n \cos\left(\frac{n \pi \xi}{3}\right)+b_n \sin\left(\frac{n \pi \xi}{3}\right)\right)\right)\tag{2}$$
where
$$a_0=\frac{1}{6} \int_{-3}^3 f(\xi) \, d\xi=\frac{191}{60}\tag{3}$$
$$a_n=\frac{1}{3} \int\limits_{-3}^3 f(\xi) \cos\left(\frac{\pi n \xi}{3}\right) \, d\xi=-\frac{\sin\left(\frac{\pi n}{2}\right)+3 \sin\left(\frac{\pi n}{3}\right)+2 \sin\left(\frac{\pi n}{15}\right)-8 \sin(\pi n)}{\pi n}\tag{4}$$
$$b_n=\frac{1}{3} \int_\limits{-3}^3 f(\xi) \sin \left(\frac{\pi n \xi}{3}\right) \, d\xi=\frac{-\cos \left(\frac{\pi n}{2}\right)+3 \cos \left(\frac{\pi n}{3}\right)+2 \cos \left(\frac{\pi n}{15}\right)-4 \cos (\pi n)}{\pi n}\tag{5}$$
Figure (1) below illustrates the Fourier series for $f(\xi)$ defined in formula (2) above in orange overlaid on the blue reference function $f(\xi)$ defined in formula (1) above where formula (2) above is evaluated at $K=50$.
Figure (1): Illustration of formula (3) for $f(\xi)$ (orange) overlaid on formula (1) for $f(\xi)$ (blue)