i. $f(x) = \operatorname{sgn}(x)$ for $-\pi < x < \pi$ where
$$\operatorname{sgn}(x) = \begin{cases} 1, & x>0, \\ 0, & x=0, \\ -1, & x<0. \end{cases} $$
ii. $f(x) = \displaystyle\frac{\pi - x}{2}$ for $-\pi < x < 2\pi$
I want to find the Fourier series of the functions given above. Its been a long time since I have done these, so I was wondering if anyone could help set them up for me, but not go through the computation as I want to do that myself, and then if it is not to much to ask provide a solution. I would say that is okay since solutions don't mean much if one does not know how to get to it, so I wanted to see if the work I go through gets to the right solution. If thats to much to ask then just help with setting up would be appreciated!!!!
"Fourier series" could mean $$ a_0 + a_1\cos x+b_1\sin x + a_2\cos(2x)+b_2\sin (2x)+\cdots\tag1 $$ or it could mean $$ \cdots+c_{-2}e^{-2ix}+c_{-1}e^{-ix}+c_0 + c_1e^{ix} +c_2 e^{2ix} + \cdots.\tag2 $$ Either way, the basic idea is this: \begin{align} & \int_{-\pi}^\pi f(x)\cdot((\sin\text{ or }\cos)(nx)\text{ or }e^{inx})\,dx \tag 3 \\[8pt] = {} & \int_{-\pi}^\pi (\text{infinite series (1) or (2) above})\cdot((\sin\text{ or }\cos)(nx)\text{ or }e^{inx})\,dx. \tag 4 \end{align} If you evaluate the integral in $(4)$, all but one of the terms will be $0$. Which one depends on $n$, and one whether you're looking at $\sin(nx)$ or $\cos(nx)$ or $e^{inx}$. What you get will be an expression involving the coefficient $a_n$ or $b_n$ or $c_n$. That expression will be equal to whatever you get when you evalutate the integral in $(4)$. So you get \begin{align} & \text{expression from evaluating }(3) \\[8pt] = {} & (c_n\cdot\text{expression from evaluating }(4)) \end{align} and thus you find $c_n$ (or $a_n$ or $b_n$, as the case may be).