Numerical Jacobian

New to Julia
1

Hello!

I am looking for the best package to compute a numerical Jacobian.

Best,
Juan

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2

For π‘Œ = g(𝑋), CoupledFields.gradvecfield([a b], X, Y, kernelpars) returns 𝑛 gradient matrices, for 𝑛 random points in 𝑋.
For parameters [π‘Ž 𝑏]: π‘Ž is a smoothness parameter, and 𝑏 is a ridge parameter

using CoupledFields
g(x,y,z) = x .* exp.(-x.^2 - y.^2 - z.^2)
X = -2 .+ 4*rand(100, 3)
Y = g.(X[:,1], X[:,2], X[:,3])

 kernelp.ars = GaussianKP(X)
 βˆ‡g = gradvecfield([0.5 -7], X, Y[:,1:1], kernelpars)

Also CoupledFields doesn’t require a closed-form function, it can be used if you only have the observed fields 𝑋 and π‘Œ.

3

Your best bet is to use ForwardDiff.jl which uses automatic differentiation

julia> using ForwardDiff

julia> ForwardDiff.jacobian(x->exp.(x), rand(2))
2Γ—2 Array{Float64,2}:
 2.33583  0.0
 0.0      1.59114
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4

Yeah that’s what I’ve been trying to use. However it seems to me that if I have a function that does not have a closed form ForwardDiff does not worj. Or at least it is not working for me.

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5

You could also use DiffEqDiffTools.jl to calculate the jacobian via finite differencing.

Also if you post your function and the error message we could figure out why ForwardDiff isn’t working.

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6

FiniteDIfferences.jl to calculate via fiinite differencing.

https://www.juliadiff.org/FiniteDifferences.jl/latest/pages/api/#FiniteDifferences.jacobian

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7

@jmcastro2109: when you asked via e-mail you also mentioned that you are using these for indirect inference. The problem with that is that for discrete-choice problems, derivatives may not be easy to approximate using (finite) samples, so I would consider using a derivative-free optimization method.

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