A question about Fourier series

Question

I have recently discovered what the "Fourier series" of a function is. So the Fourier series of $f(x)=x^2$ in $[0,2\pi]$ is $f(x)=\dfrac{4\pi^2}{3}+\displaystyle\sum_{n=1}^\infty\left(\frac{4}{n^2}\cos(nx)-\frac{4\pi}{n}\sin(nx)\right).$ I plot $x=0$ and so $0=\dfrac{4\pi^2}{3}+4\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^2}\Leftrightarrow \displaystyle\sum_{n=1}^\infty\dfrac{1}{n^2}=-\dfrac{\pi^2}{3}$ which is obviously wrong... Where is my fault (btw the Fourier series of $x^2$ is correct) ?

Answer 1

Read Fourier series.

Your function $f(x)$ is simply not periodic. Your function contains a jump at $x=0,2\pi$.

Edit:

I recommend the questionier and the previous answers to use the nomenclature:

• $f$ is not periodic: $f(x)=x^2$ where $x \in [0,2 \pi]$
• $g \neq f$ is a periodic extension: $g(x)=x^2$ with $g(x)=g(x + P)$ where $x\in [0,2 \pi]$ and $P=2\pi$.
• $h \neq g$ (not necessarily) and $h \neq f$ is the periodic Fourier series of $g$ defined here with $h(x) = \sum_{n=-N}^N c_n \cdot e^{ i \tfrac{2\pi nx}{P}}$ and $x \in [0,2 \pi]$.

Otherwise you are mixing different stuff.

Since $h$ is not the Fourier series of $f$ you can not except a uniform/absolute convergence especially at the point of interest $f(0),h(0)$ or $f(2\pi), h(2\pi)$.

Regards

Answer 2

Here is the representation of your series truncated after the 10th terms

For this partial sum, $0$ is a point of continuity located with a value close to the midvalue between the values of $f(x)$ at its endpoin. But, clearly, the final state there (with the infinite series development) will be a discontinuity.

A classical result in the theory of Fourier series is that the convergence of the series in this point of discontinuity is indeed towards the midvalue between $f(0)=0^2$ and $f(2\pi)=(2\pi)^2$, i.e., towards the value :

$$\tilde{f}(0)=2 \pi^2$$

with this value of $\tilde{f}(0)$, one gets :

$$2 \pi^2=\dfrac{4\pi^2}{3}+4\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^2}$$

giving :

$$\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^2}=\dfrac{\pi^2}{2}-\dfrac{\pi^2}{3}=\dfrac{\pi^2}{6}$$

as awaited.

Answer 3

The periodization of $f(x)=x^2$ for $0\leq x<2\pi$ is discontinuous at $x=0$. For piecewise differentiable functions (as yours), there is a well known result that states that

$S_nf(x)\rightarrow \frac{f(x-)+f(x+)}{2}$, where $S_n$ is the $n$-th partial sum $S_nf(x)=\sum_{|k|\leq n} c_ne^{-ikx}$, $f(x-)$ is the left limit of $f$ and $f(x+)$ is the right limit of $f$.

In your case

$$\frac{(2\pi)^2+0^2}{2}=\frac{4\pi^2}{3} + 4\sum_{n\geq1}\frac{1}{n^2}$$