Symbolic derivation of complex valued functions with Symbolics.jl

New to Julia
,
1

when there is im (the imaginery number i) in the function I get zero for the derivative which is obviously wrong. We can replace im with a symbol like I and then differentiate, but I was wondering
maybe there is a better way to do it.

using Symbolics
@variables t
D = Differential(t)
z = im*t 
D(z)
expand_derivatives(D(z))  #gives 0!
3 Likes
2

@shashi

3

I played a bit with this

using Symbolics
@variables x::Float64
D = Differential(x)
@show D
@show D(x) 
@show expand_derivatives(D(x))

@variables z::Complex{Float64}
D = Differential(z)
@show D
@show D(z)
@show expand_derivatives(D(z))

yielding

D = Differential(x)
D(x) = Differential(x)(x)
expand_derivatives(D(x)) = 1
D = Differential(z)
D(z) = Differential(z)(z)
expand_derivatives(D(z)) = 0

Bug?

2 Likes
4

That definitely looks like one.

5

@Vahid_Hosseinzadeh: are you going to report it? Else I’ll do.

6

This is a more nice way to see it. Thank you.

7

You mean issue on Github? Anyway be free to report it. Thanks @goerch

8

Done.

1 Like
9

This happens because we sneakily represent @variables z::Complex as Complex{Num}(term(real, z), term(imag, z)) in order to be able to call functions restricted to Complex{<:Real}…

10

It’s also unclear to me what form the answer should take in this case. @YingboMa pointed out it is possible to convert C into R^2 and differentiate, but the result is not a complex number…

11

You are not alone in this. Here is a different view of the same problem.

1 Like
12

Check Complex numbers · ChainRules

If we use ChainRules in Symbolics, then we can propagate Union{Jacobian2x2, Complex}, and if we are propagating Jacobian2x2 then convert it to Complex if du_dx == du_dy and -du_dy == dv_dx. We can have a list of holomorphic functions so that we don’t compute Jacobian2x2 unnecessarily. The return type of derivative would be Union{Jacobian2x2, Complex}. Maybe we want to error if not all the elements are Complex when computing Jacobian and gradient.

1 Like
13

Thanks, that makes sense. Maybe @stevengj has a good idea for this… It would be nice to have differentiation with respect to complex numbers work in a usable way.