Consider $$ f(t)= \begin{cases} 1 \mbox{ ; } 0<t<1\\ 2-t \mbox{ ; } 1<t<2 \end{cases}$$
Let $f_1(t)$ be the Fourier sine series and $f_2(t)$ be the Fourier cosine series of $f$, $f_1(t)=f_2(t), 0<t<2$. Write the form of the series (without computing the coefficients) and graph $f_1$ and $f_2$ on [-4,4] (including the endpoints $\pm 4$) using *'s to identify the value of the series at points of discontinuity.
I think we have:
$f_1(t)=\sum \limits_{n=1}^{\infty} b_n \sin \frac{n \pi t}{2}$
$f_2(t)=\frac{a_0}{2}+\sum \limits_{n=1}^{\infty} a_n \cos \frac{n \pi t}{2}$
I think we have $f_2=1$ and for $0<t<2, f_1=f_2=1$
Can we do anything else? Can someone help me with the end?
Thank you
Ok at first we gonna plot our function
We know that on jump discontinuities it will converge to the arithmetic mean of them, so the first approximation is just taking $\frac{1}{2}$. This gonna look like
The Cos terms gonna look like
The Sin terms are looking like