LinkBack URL
About LinkBacksHi, I think something is wrong with my working because I have a 2 where it should only be a 1.
Q: Show
Working:
I have got the fourier transform ofwhich is
.
So from here I use:
Thus:
 + \sum_{n=1}^\infty F(2 \pi n))
![= \frac{\pi}{a} [2 \sum_{n=1}^\infty e^{-4n\frac{\pi}{a}} -1]](http://latex.codecogs.com/png.latex?= \frac{\pi}{a} [2 \sum_{n=1}^\infty e^{-4n\frac{\pi}{a}} -1])
![= \frac{\pi}{a}[\frac{2}{1-e^{-4\frac{\pi}{a}}} -1]](http://latex.codecogs.com/png.latex?= \frac{\pi}{a}[\frac{2}{1-e^{-4\frac{\pi}{a}}} -1])
So where did I go wrong to get the 2 instead of a 1?
Thanks
Hi, I think something is wrong with my working because I have a 2 where it should only be a 1.
Q: Show
Working:
I have got the fourier transform of
So from here I use:
Thus:
So where did I go wrong to get the 2 instead of a 1?
Thanks
  |
