For the Fourier series in this question: Math Question
How do I find out where the Fourier series converges, I initially thought I just apply this equation: $ f\left(x\right)=\frac{1}{2}\left\{f\left(x+\right)+f\left(x-\right)\right\}$ at points of discontinuity
From the graph I thought the only point of discontinuity is at $x=2$ and the Fourier series converges to 3/2 at $x=2$.
But the answer says it also converges to -1/2 at $x=4$. I'm not sure why it does this at $x=4$
The function f(x) can be evaluated as defined in formula (1) below.
(1) $\quad f(x)= \left\{ \begin{array}{cc} x & 0\leq x<2 \\ 3-x & 2\leq x<4 \\ f((x \bmod 4)) & \text{True} \\ \end{array}\right.$
The following figure illustrates the function $f(x)$ defined in formula (1) above has discontinuities at $(x \bmod 4)=0$ as well as $(x \bmod 4)=2$. The jump at $(x \bmod 4)=0$ is from $-1$ to $0$ so the half-step at $(x \bmod 4)=0$ evaluates to $-\frac{1}{2}$ which is illustrated by the red dots in Figure (1) below. The jump at $(x \bmod 4)=2$ is from $2$ to $1$ so the half-step at $(x \bmod 4)=2$ evaluates to $\frac{3}{2}$ which is illustrated by the green dots in Figure (1) below.
Figure (1): Illustration of $f(x)$ defined in formula (1)