I read this at many places such as wiki or elsewhere quote :
It is possible to give explicit examples of a continuous function whose Fourier series diverges at $0$.
For instance, the even and $2π$-periodic function f defined for all $x$ in $[0,π]$ by
$$ {\displaystyle f(x)=\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}\sin \left[\left(2^{n^{3}}+1\right){\frac {x}{2}}\right].}$$
I am confused by this.
If I plug in $x=0$ for $f(x)$ then I get
$$ {\displaystyle f(0)=\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}\sin \left[\left(2^{n^{3}}+1\right){\frac {0}{2}}\right]} = 0$$
Because the sine of zero is zero.
Also if I use the formula to compute the Fourier coefficients of $f(x)$ I get exactly
$$ {\displaystyle f(x)=\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}\sin \left[\left(2^{n^{3}}+1\right){\frac {x}{2}}\right].}$$
So no contradiction there.
Notice the text said continuous, not differentiable.
Sure $f'(x)$ equals
$$ {\displaystyle f'(x)=\sum _{n=1}^{\infty }{\frac {2^{n^{3}}+1}{n^{2}}}\sin [(2^{n^{3}}+1){\frac {x}{2}}]}$$
And that diverges.. euh for $x>0$. But at $x=0$ it gives zero, again because sine of zero is zero. So it seem to be not differentiable at zero.
But the cube root of $x$ is also not a differentiable at zero. (although it does not jump from zero to infinity when going from zero to positive )
So I am very confused about this " counterexample "
And why this cubic term ?
Why not just any $f'(x)$ that diverges at zero such as
$$ {\displaystyle f(x)=\sum _{n=1}^{\infty }{\frac {3}{n^{2}}}\sin \left[\left(2^{n}+15\right){\frac {x}{2}}\right].}$$
??
I do not see how the series diverges at $0$ or the function is not equal to its Fourier series or anything like that.
I just see a non-differentiable function that is defined for all real $x$ because
$$\sum _{n=1}^{\infty}\frac {1}{n^2}$$
converges and sine is always between $-1$ and $1$.
Im new to Fourier series as you might have guessed, so sorry for the dumb question maybe.