Question

how to prove that the fourier series expansion of this function
\[f(x)=1, -\pi 11 years ago

Answer 1

i have already found the fourier series expansion

this is a odd function so
\[a_0=a_n=0\]
\[b_n=\frac{ 2 }{ \pi }\int\limits_{0}^{\pi}-\sin(nx)dx=\frac{ 2 }{ n*\pi }[(-1)^n-1]\]
\[b_n=0 ,n=even\]
\[b_n=\frac{ -4 }{ n \pi },n=odd\]
so \[f(x)=-\frac{ 4 }{ \pi }[sinx+\sin3x/3+\sin5x/5+.......]\]

Answer 2

now very much confused what to do next
@hartnn
@ganeshie8
can u show something

Answer 3

So the Gibbs phenomenon has to do with the overshoot of the approximation as you approach the function's discontinuity. To show a function's Fourier representation adheres to the phenomenon, you have to show that the \(n\)th partial sum of the series evaluated at \(\dfrac{\text{period}}{2n}\) approaches
\[\frac{\pi}{2}\int_0^1\text{sinc}x~dx\]
There's an excellent example on the Wikipedia page that takes you through the computation:
http://en.wikipedia.org/wiki/Gibbs_phenomenon#The_square_wave_example

Answer 4

yeah thank u very much now it strikes me
if i break them like this
\[S_1=\frac{ -4 }{ \pi }sinx\]
\[S_2=\frac{ -4 }{ \pi }[sinx+\sin3x/3]\]
\[S_3=\frac{ -4 }{ \pi }(sinx+\sin3x/3+\sin5x/5)\]
and i also plotted them like

so theres a osillatory behavior
and it is obeying gibbs
so thanks again
@SithsAndGiggles