Basic Fourier Series Question

Question

Let $f$ be a $2π$ periodic function where $$f(x) = \frac{π - x}2$$ over $[0, π]$.

It is known that the Fourier series of $f$ is $$\sum_{n=1}^{\infty}\frac{\sin nx}n$$

At which points in $[-π, π]$ does $$f(x)=\sum_{n=1}^\infty\frac{\sin nx}n$$ holds?

Answer 1

The $2\pi$-periodic function $f$ is odd (this is clear from the Fourier series expansion); on $[0,\pi]$ its expression is given by

$$f(x)=\frac{\pi-x}{2}.$$

In summary (this follows from plotting $f$), due to symmetry and $2\pi$-periodicity $f$ is discontinuous at $x=0$, and continuous on $[-\pi,\pi]-\{0\}$. Then the Fourier sum

$$F_k(x):=\sum_{n=1}^k\frac{\sin nx}{n} $$

converges pointwise to $\tilde{f}$ on $[-\pi,\pi]-\{0\}$, where $\tilde{f}(x)=\frac{\pi-x}{2}$ if $x\in(0,\pi]$ and $\tilde{f}(x)=\frac{x-\pi}{2}$ if $x\in[-\pi,0)$.

At $x=0$ we have

$$\lim_{k\rightarrow \infty} F_k(0)=\frac{f_+(0)+f_0(0)}{2}=\frac{\pi}{2}, $$

with $f_+(0):=\lim_{x\rightarrow 0^+}f(x)$ and $f_+(0):=\lim_{x\rightarrow 0^-}f(x)$.