- #1
enfield
- 21
- 0
[tex] \dfrac{\mathcal{F}^{-1} \Big( \sqrt{\mathcal{F}(f(x))} \Big)}{f(x)} = g(x)
[/tex]
g(x) is known, and for an example let's say g(x) is something simple like [tex] g(x) = x [/tex]
so we have [tex] \mathcal{F}^{-1} \Big( \sqrt{\mathcal{F}(f(x))} \Big) = x \cdot f(x) [/tex]
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.
[/tex]
g(x) is known, and for an example let's say g(x) is something simple like [tex] g(x) = x [/tex]
so we have [tex] \mathcal{F}^{-1} \Big( \sqrt{\mathcal{F}(f(x))} \Big) = x \cdot f(x) [/tex]
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.
Last edited: