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2D via Lanczos to Tensor product basis? #4
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Hmm, for the disk what about something like P_k(ρ(x)) * T_j(y/ρ(x)), x*P_k(ρ(x)) * T_j(y/ρ(x)) whee For x^j * P_k(ρ(x)) * T_j(y/ρ(x)) for j = 0:3 so the expansion part is not so clear... |
Yeah I think it works on the disk since it's quite a lot like spherical harmonics without the z = cos(theta) variable transformation. Don't know about the quartic weight, but perhaps instead of Givens rotations the isométries can be factored in terms of short Householder reflectors. |
Works on the disk from what basis? I’m not talking about the radial coordinates basis |
I think the Dunkl-Xu will work. Just needs implementing |
It works like this. I assume we have no problem in the even k case. You write:
as polynomials in x times For odd k, you leave one power of |
@MikaelSlevinsky's transform work by first expanding in a tensor product basis, which contains non-polynomial terms, before transforming to a polynomial basis. For example, to compute OPs on the triangle we expand in the basis
(I'm using OPs on
0..1
here) which includes non-polynomial termsy/(1-x)^j
, before converting to the polynomial basisThe point is polynomials in
x
andy
are a subspace of the non-polynomial basis.Does this technique translate to OPs on the disk (a la Dunkl & Xu, not Zernike)
where
ρ(x) = sqrt(1-x^2)
? A tensor basis likedoes not seem to work here... seems like we'd need a sum-space frame that also includes
ρ(x)*T_k(x)*T_j(y/ρ(x))
....What about
ρ(x) = (1-x^4)^(1/4)
? Now we wouldn't know the OP basis but perhaps we can represent it by the conversion to a bigger basis a laLanczosPolynomial
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