fourier issue arising from input miscommunication

Question

xdomain = Table[i , {i, -10, 10, .1}];
ListPlot[InverseFourier[
  Fourier[E^-#^2 & /@ xdomain]*E^#^2 & /@ xdomain]]

So I want to numerically fourier transform a function, multiply that fourier'd function by a second function, and then inverse fourier transform it.

The code above doesn't work. I get a blank graph, I think, because it doesn't properly multiply the two vectors that I commanded to multiply. I'm not sure why I can't get {a,b,c}*{1,2,3} to multiply inside my fourier function, but I have tried putting evaluate functions in different spots inside to see if that would fix it.

Answer 1

1. You need brackets around E^#^2 & /@ xdomain
2. ListPlot plot only real lists so you need to apply Re, Im or Abs

3. E^-#^2 & /@ xdomain is overkill. Just Exp[-xdomain^2]

   xdomain = Range[-10, 10, 0.1];
   
   ListPlot[Re@InverseFourier[Fourier@Exp[-xdomain^2] Exp[xdomain^2]]]
   

P.S. It seems that you want to apply a deconvolution. The method with the inverse Fourier transform is unstable. For this purpose there is a stable build-in function ListDeconvolve.

Answer 2

Oneliners can be elegant, but sometimes they burdens Mathematica a bit too much. Let's do it one step at the time:

xdomain = Table[i, {i, -10, 10, .1}];
F1= E^-#^2 & /@ xdomain; (* no problem here *)
F2 = E^+#^2 & /@ xdomain; (* no problem here *)
ftot=InverseFourier[Fourier[F1]*F2]; (* no problem here *)

The problem, then, is not in the multiplication. So, where is it? Well, the problem is that ftot is a complex function. You can plot its magnitude, its phase or its real and imaginary part. But if you try to listplot it as it is, you will get a blank graph. Perhaps you want:

ListPlot[Abs[ftot]]

or

ListPlot[Re[ftot]]