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

= September 4 =

Heliosphere
(Moved to Science Desk: Reference_desk/Science.)

Finding a relation between two expressions
I have a function $$f(\theta,\phi)$$ defined on a 2-sphere. Next I fix a certain angle $$\theta_j$$. 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:

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

Forward Transform:

Inverse Transform:

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 $$f_m(\theta_j)$$ but computing integral (1) is computationally prohibitive. I wonder if a coefficient could be found analytically connecting the complex numbers $$f_m(\theta_j)$$ and $$X(n)$$? It could look like this:

$$f_m(\theta_j) = A_n X(n)$$

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 $$f_m$$ and $$X(n)$$ is likely to be non-linear and to depend on function $$f$$ itself. For example, if $$f(\varphi)=\exp(ia\varphi)$$, then $$f_m=(\exp(i2\pi a)-1)/i/(a-m)$$ but $$X(n) = (\exp(i2\pi a)-1)/(\exp(i2\pi a/N-i2\pi n/N)-1)/N$$. Ruslik_ Zero 18:43, 4 September 2017 (UTC)


 * I use this simple discrete fourier transform formula
 * $$\sum_{k=0}^{n-1} {1^{j k\over n}\over\sqrt n}a_k^*$$ for $$j=0\cdots n-1$$
 * where $$a_k^*$$ is the complex conjugate of $$a_k$$, and $$1^x=e^{2\pi i x}$$ has nothing to do with the transcendental numbers $$e$$ and $$\pi$$. Note that $$(1^x)^*=1^{-x}$$. This transform is its own inverse, because
 * $$\sum_{j=0}^{n-1} {1^{k j\over n}\over\sqrt n}\left(\sum_{k=0}^{n-1} {1^{j k\over n}\over\sqrt n}a_k^*\right)^*=a_k$$ for $$k=0\cdots n-1$$.
 * 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)