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I was reading this question, and made a wrong contribution which I deleted. Now I would like to understand things. Here is the problem:

Consider $f(x)=\cos 2x$ on $[0,\pi]$:

enter image description here

$f(x)$ is not even on $[0,\pi]$.

$$a_m=\frac{2}{\pi}\int_0^{\pi}\cos 2x \cos mx =\frac{2}{\pi}\frac{m \sin m \pi}{m^2-4}=0,m=0,1,3,4,5...$$

and $a_2=1$. But for $m$ odd we have

$$b_m=\frac{2}{\pi}\int_0^{\pi}\cos 2x \sin mx =\frac{4}{\pi}\frac{2m-1}{4m^2-4m-3}$$

and for $m$ even we have $b_m=0$. Now I define the following function: \begin{align} F(x)&=\frac{2}{\pi}\sum_{m=1}^{\infty}\frac{m (1-\cos m \pi)}{m^2-4}\sin [mx]\\ &=\frac{4}{\pi}\sum_{m=1}^{\infty}\frac{2m-1}{4m^2-4m-3}\sin [(2m-1)x] \end{align}

Now define $\displaystyle F_n(x)=\frac{4}{\pi}\sum_{m=1}^{n}\frac{2m-1}{4m^2-4m-3}\sin [(2m-1)x]$ and lets plot $F_1(x)$ against $f(x)$:

enter image description here

Now lets plot $F_{20}(x)$ against $f(x)$:

enter image description here

And this is what happens for $n=100$:

enter image description here

So can I say then say $F_n(x)$ converge to $\cos 2x$ on $(0,\pi)$ this sounds puzzling. Maybe I made a mistake. Could anyone please explain me what is going on here?

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1 Answers1

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The series you got is the odd extension of this function to $[-\pi,\pi]$and periodic with period $2\pi$.

You are seeing in the partial sums the Gibbs phenomenon.

If you were to take the Cesaro sums of the series you would see it converge to $0$ at $x=0$ and similar points, see Fejer's theorem.

  • Many thanks for the links and answer. I did not know Gibbs phenomenon. So you mean that $F_n(x)$ converges to zero at all points? – Math-fun Feb 24 '15 at 13:57
  • No, only at $x=k\pi$ the Cesaro sum of the series converges to $0$. In the interval $(0,\pi)$ it (the Cesaro sum of the series) converges to $\cos(2x)$. But in $(-\pi,0)$ it converges to $-\cos(2x)$. That is why it is not the Fourier of the even extension of the function $\cos(2x)$ in $[0,\pi]$, but the odd extension of it. Try making those plots but in $[-\pi,\pi]$ and you will see it happen. –  Feb 24 '15 at 14:01
  • This I see: at $x=k\pi$ I have that $\sin[(2m-1)k\pi]=0$, hence the sum is zero. But I thought I should figure out convergence for other points. Plotting on $(-\pi,0)$ I get $-\cos 2x$. – Math-fun Feb 24 '15 at 14:04
  • At the points in $(0,\pi)$ the series converges (the usual convergence) to $\cos(2x)$. just because $\cos(2x)$ is differentiable there. –  Feb 24 '15 at 14:08