Unable to differentiate sum of broadcast

[ForceBru]

let
    @variables q[1:5]
    
    Symbolics.gradient(
        sum(q .* log.(q)), collect(q)
    )
end

This outputs a vector of zeros: [0, 0, 0, 0, 0], but the correct gradient consists of elements like 1 + log(q[k]):

import sympy as sp

q = sp.symbols("q1:5", real=True)
the_sum = sum(
    q_ * sp.log(q_) for q_ in q
)
sp.Matrix([
    [the_sum.diff(q_) for q_ in q]
]).T

Output:

┆Issue is synchronized with this Trello card by Unito

[Yanksi]

Seems that the symbolic array is not working properly when computing gradients, I have a similar issue here.

@variables t[1:3]
println(Symbolics.derivative(cos.(t)[1], t[1]));
println(Symbolics.derivative(Symbolics.scalarize(cos.(t)[1]), t[1]));
println(Symbolics.jacobian(cos.(t), t));

Output:

0
-sin(t[1])
Num[-sin(t[1]) 0 0; 0 -sin(t[2]) 0; 0 0 -sin(t[3])]

The derivative computed by the first line is clearly not correct, while the problem goes away when I forcibly convert the symbolic expression with symbolic arrays to a scalar expression.

[bmxam]

I do have a similar failing example with Symbolics v5.4.0. I have used the scalarize trick found here #580 (comment).

using Symbolics
using LinearAlgebra

@variables Ω[1:3] x[1:3] y[1:3] τ

# the below expressions should all have the same jacobian with respect to `x`
eqn1 = x / τ - cross(Ω, x) # note the absence of `y`
eqn2 = x / τ - cross(Ω, x - y)
eqn3 = x ./ τ .- cross(Ω, x .- y)
eqn4 = Symbolics.scalarize(x / τ - cross(Ω, x - y))

Symbolics.jacobian(eqn1, x) # OK
Symbolics.jacobian(eqn2, x) # KO
Symbolics.jacobian(eqn3, x) # KO
Symbolics.jacobian(eqn4, x) # OK

For me it is a dangerous bug since we start by building simple expressions to check everything is ok (and it is), and then we complexify things with confidence but the result is wrong.