I'm 18 years old and this is my first time writing a question asking for help. So I hope people out there can really help me by not just giving me tips only. My question is how to solve this fourier series ?
Given that
$$f(t) = \begin{cases}50, &\text{for }t\in(0,2) ;\\ 100 , &\text{for }t=2.\end{cases}$$
Compute the fourier series.
I like to solve Maths question but I end up I can't resolve it because I don't have proper reference book on this topic and my teacher doesn't teach this topic in detail. So guys at least give some work done so that I able to understand clearly.
The Fourier series of a function $f(t)$ defined on the interval $0<t\leq 2$ is an expression $${1\over 2}a_0+\sum_{n=1}^\infty \bigl(a_n\cos(n\pi t)+b_n\sin(n\pi t)\bigr)$$ where the coefficients $a_n$ and $b_n$ are given by $$a_n=\int_0^2f(t)\cos(n\pi t)\,{\rm d}t,\qquad n=0,1,2,\dots$$ $$b_n=\int_0^2f(t)\sin(n\pi t)\,{\rm d}t,\qquad n=1,2,3,\dots$$ When you substitute your function $f(t)$ into the expressions above and compute the coefficients, you will find that the value $f(2)=100$ does not matter, since changing a function at only one point will not change the value of the integral. ("The area under a point is zero".) The integrals are therefore $$a_0 =\int_0^2 50\,{\rm d}t=100$$ $$a_n=\int_0^2 50\cos(n\pi t)\,{\rm d}t={1\over 50n\pi}\sin(n\pi t)\Bigr|_0^2 =0, \qquad n=1,2,3,\dots$$ $$b_n=\int_0^2 50\sin(n\pi t)\,{\rm d}t=-{1\over 50n\pi}\cos(n\pi t)\Bigr|_0^2 =0, \qquad n=1,2,3,\dots$$
This means you have a rather "boring" Fourier series, since all terms vanish with the exception of one term, and that's why I wondered if the function $f(t)$ was given correctly.
You then have the Fourier series $${1\over 2}a_0=50$$ which will "converge" to $50$ for all $t$. In particular, it will not converge to $f(t)$ when $t=2$.