Testwiki:Reference desk/Archives/Mathematics/2017 September 4

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September 4

(Moved to Science Desk: Wikipedia:Reference_desk/Science#Heliosphere_Moved_from_Math_Desk..)

Finding a relation between two expressions

I have a function f(θ,ϕ) defined on a 2-sphere. Next I fix a certain angle θj. This angle is of course the angular distance from the North Pole. I need to consider only a slice of the function along one of the parallels. I need a Fourier transform expression for this function:

Template:NumBlk

I also need to consider the numerical approximation of the integral above (Discrete Fourier Transform):

Forward Transform:

Template:NumBlk

Inverse Transform:

Template:NumBlk

I am mired in computations. This is a very small part of them, but this part affects other results. There are many FFT's and DFT's on the web and every time I compute individual n members they are different from method to method but if I use them for the Inverse transform using corresponding methods, of course, I get a perfectly restored original function no matter which method I use.

In fact I need the result of the first m-related expression fm(θj) but computing integral (1) is computationally prohibitive. I wonder if a coefficient could be found analytically connecting the complex numbers fm(θj) and X(n)? It could look like this:

fm(θj)=AnX(n)

I use the Inverse transform for controls only and in the final variant I will not need it.

Thanks. --AboutFace 22 (talk) 17:32, 4 September 2017 (UTC)

I do not think that such an expression exists. Moreover any relation between fm and X(n) is likely to be non-linear and to depend on function f itself. For example, if f(φ)=exp(iaφ), then fm=(exp(i2πa)1)/i/(am) but X(n)=(exp(i2πa)1)/(exp(i2πa/Ni2πn/N)1)/N. Ruslik_Zero 18:43, 4 September 2017 (UTC)
I use this simple discrete fourier transform formula
k=0n11jknnak* for j=0n1
where ak* is the complex conjugate of ak, and 1x=e2πix has nothing to do with the transcendental numbers e and π. Note that (1x)*=1x. This transform is its own inverse, because
j=0n11kjnn(k=0n11jknnak*)*=ak for k=0n1.
Bo Jacoby (talk) 18:54, 4 September 2017 (UTC).

Thank you @Ruslik0 and @Bo Jacopy. Intuitively I was ready for it. --AboutFace 22 (talk) 18:58, 4 September 2017 (UTC)