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$$f(x)=\begin{cases}0&-5\leq x<0\\3&0\leq x<5\end{cases}\\\text{with } f(x+10)=f(x)$$

is the function I need to make a Fourier series on, I know where to start, but I get confused along the way and I don't know in general how to "finish up" a Fourier series.

My own solution so far so you can follow my thoughts, used variables etc: page 1 page 2

Hope my handwriting is readable and thank you in advance!

Since the quality of the images aren't good enough, I'll just write it out on here: $\tilde{f}_k = \int_{-5}^{5} f(x)e^{-ikx}dx$

$\tilde{f}_k = \int_{-5}^{0} 0e^{-ikx}dx + \int_{0}^{5} 3e^{-ikx}dx$

Integrating this, I got:

$\tilde{f}_k = \frac{1}{ik}(3-3e^{-5ik})$

for k = 0 I got $\tilde{f_{0}} = \frac{3}{2}$

After this it gets fuzzy for me for what exactly I need to do, I filled in the formula but the steps after this become weird and unclear. It's kinda different in each example I've seen.

The solution to this problem should be:

$f(x) = \frac{3}{2} + \frac{6}{\pi}\sum_{n=0}^{+\infty}\frac{1}{2n+1}\sin(\frac{(2n+1)\pi x}{5})$

Thomas Andrews
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Monkey D. Ruffy
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    You should type it all out. No one wants to click on arbitrary hyperlinks, even if the page location is well known. – Matthew Cassell Jan 06 '22 at 19:43
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    Please, since you are able to write in LaTeX, type all in math mode. No links, images or photos. – Enrico M. Jan 06 '22 at 19:44
  • If the interval is really $[-5,5],$ then you shouldn’t be using $e^{ikx},$ but some $e^{i\alpha kx}$ for some $\alpha\neq1.$ You also have to divide by a constant, I think. – Thomas Andrews Jan 06 '22 at 19:44
  • Seems to be a duplicate of https://math.stackexchange.com/questions/4350458/cant-figure-out-how-to-get-to-the-solution-of-the-fourier-series-of-this-functi. – Steven Clark Jan 07 '22 at 03:42
  • Yea I reposted it 3 times to get the question right and fully typed out with LaTeX until people were happy – Monkey D. Ruffy Jan 07 '22 at 09:28

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