Automatic differentiation through spherical harmonic transforms
This example finds a positive value of in:
for some such that . We do this by using derivative information through:
using FastTransforms, LinearAlgebra
The colatitudinal grid (mod ):
N = 15
θ = (0.5:N-0.5)/N
0.03333333333333333:0.06666666666666667:0.9666666666666667
The longitudinal grid (mod ):
M = 2*N-1
φ = (0:M-1)*2/M
0.0:0.06896551724137931:1.9310344827586206
We precompute a spherical harmonic—Fourier plan:
P = plan_sph2fourier(Float64, N)
FastTransforms Spherical harmonic--Fourier plan for 15×29-element array of Float64
And an FFTW Fourier analysis plan on :
PA = plan_sph_analysis(Float64, N, M)
FastTransforms plan for FFTW Fourier analysis on the sphere for 15×29-element array of Float64
Our choice of and angular parametrization of :
k = [2/7, 3/7, 6/7]
r = (θ,φ) -> [sinpi(θ)*cospi(φ), sinpi(θ)*sinpi(φ), cospi(θ)]
#1 (generic function with 1 method)
Our initial guess for :
λ = 1.0
1.0
Then we run Newton iteration and grab an espresso:
for _ in 1:7
F = [sin(λ*(k⋅r(θ,φ))) for θ in θ, φ in φ]
Fλ = [(k⋅r(θ,φ))*cos(λ*(k⋅r(θ,φ))) for θ in θ, φ in φ]
U = P\(PA*F)
Uλ = P\(PA*Fλ)
global λ = λ - (norm(U)^2-1)/(2*sum(U.*Uλ))
println("λ: $(rpad(λ, 18)) and the 2-norm: $(rpad(norm(U), 18))")
end
λ: 0.5565017029393282 and the 2-norm: 1.8510924318185522
λ: 0.5031571262839712 and the 2-norm: 1.104184591487036
λ: 0.5010418434316171 and the 2-norm: 1.0040147091934828
λ: 0.5010383094266805 and the 2-norm: 1.0000066984258862
λ: 0.5010383094167955 and the 2-norm: 1.0000000000187361
λ: 0.5010383094167953 and the 2-norm: 1.0000000000000004
λ: 0.5010383094167954 and the 2-norm: 0.9999999999999999
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