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

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September 4

Heliosphere

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

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:

$f_{m}(\theta _{j})=\int _{0}^{2\pi }f(\theta _{j},\phi )e^{im\phi }~d\phi ~~~for~~~m=0,1...$ (1)

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

Forward Transform:

$X(n)={\frac {1}{N}}\sum _{k=0}^{N-1}x(k)\exp ^{-jk2\pi n/N}~~~for~~~n=0,....N-1$ (2)

Inverse Transform:

$x(n)=\sum _{k=0}^{N-1}X(k)\exp ^{+jk2\pi n/N}~~~for~~~n=0,...N-1$ (3)

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.

${\displaystyle }$

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

$\sum _{k=0}^{n-1}{1^{jk \over n} \over {\sqrt {n}}}a_{k}^{*}$ for $j=0\cdots n-1$

where $a_{k}^{*}$ is the complex conjugate of $a_{k}$, and $1^{x}=e^{2\pi ix}$ has nothing to do with the transcendental numbers $e$ and $\pi$. Note that $(1^{x})^{*}=1^{-x}$. This transform is its own inverse, because

$\sum _{j=0}^{n-1}{1^{kj \over n} \over {\sqrt {n}}}\left(\sum _{k=0}^{n-1}{1^{jk \over n} \over {\sqrt {n}}}a_{k}^{*}\right)^{*}=a_{k}$ 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)