From: vashkevich on
>Hi all,
>I am helping a friend on this. We would like to evaluate the
>convolution of
>f(x)=exp(3 * x) / (1 + exp(x) ^ 5).
>However, in Maple and Mathematica, using symbolic calculations, the
>convolution results in numerical oscillation (lots of spikeness). We
>don't know why. We just couldn't get rid of the spikeness. And we
>thought at the end of day, our end-goal is to numerically evaluate
>some functions involving the convolution of f(x). So it might be
>better to directly handle the convolution in discrete domain and using
>Matlab's conv function.
>So we wrote the following paragraph; but we just couldn't get the
>scaling and positioning right. For example, let's call the convolution
>g(x)=f(x) ** f(x), g(0) should correspond to the 20000th element in
>the resultant z, but z(20000) doesn't give a correct number, compared
>to the theoretical result of the convolution.
>Could anybody please help us? We need a way to figure out the scaling
>factor on the convolution result and how does it map into the
>continuous convolution result. And hopefully with lots of samples, we
>would be able to approximate the continuous convolution using discrete
>convolution. We have to take this approach because the closed-form
>expression for g(x) gives a lot spikeness and we just couldn't get a
>stable convolution result via evaluating the closed-form expression at
>all points.
>Thanks a lot!
>x= [-10:deltat:10-deltat];
>y= exp(3 * x) ./ (1 + exp(x) .^ 5);
>z=conv(y, y)*deltat;

If understand convolution as g(x) = Integral( f(y)*f(x-y) dy), then
convolution in your case may be calculated

g(x) = f**f = x*exp(3*x)/(exp(5*x) - 1)

Now it is clear, why you obtained oscillation.
Analytical result has uncertainty at zero, i.e. with x=0, but
it can be deleted in first order of Taylor series

g~x*(1+3x)/(1+5x -1) = x(1+3x)/5x = (1+3x)/5

If I not to mistaken in calculation, it's true.

Good luck