Finding f(x) from F^{-1}(√F(f)) = x·f(x)

[enfield]

$$\dfrac{\mathcal{F}^{-1} \Big( \sqrt{\mathcal{F}(f(x))} \Big)}{f(x)} = g(x)<br />$$

g(x) is known, and for an example let's say g(x) is something simple like $$g(x) = x$$

so we have $$\mathcal{F}^{-1} \Big( \sqrt{\mathcal{F}(f(x))} \Big) = x \cdot f(x)$$

my question is, how do i find f(x)?

it's basically like you Fourier transform f(x), do something to it (in this case take the root), then inverse Fourier transform back to the original variable, and you want the new function to have a specified relationship to the old function (in this case have it be the old one multiplied by x).

one specific example of that kind of thing is Fourier transforming the function f(x), multiplying by another function, and then inverse Fourier transforming it back. There the new function is the convolution of f(x) and the function you multiplied the Fourier transform of f(x) by.

Any ideas? thanks.

 

[chiro]

Hey enfield.

Have you tried looking at the integro-differential equation that describes your function f(x)?

One suggestion is try and expand your LHS with the inverse Fourier transform as the integral involving the inverse tranform integral with the F of square root of x and then on the RHS you have f(x)g(x).

You can then differentiate both sides and the idea is to do it in such a way that gets rid of the integral and allows you to solve for F(f(x)) (i.e. the Fourier transform of f(x)).