Simplifying Symbolics.jl expression

General Usage
1

I’m adding to the list of posts about simplifying a Symbolics.jl expression, but I wanted to check if I did something wrong or if it is an internal limitation. I want to check if two (lengthy) expressions are identical, and am trying to simplify their difference. However, simplify(expr) doesn’t simplify things such as this:
Result of simplify(expr) |> simplify: -2 * G1 * cos(w1°w2) * G2 * rt2 * r2 + 2 * G1 * cos(w1°w2) * G2 * rt2 * r2

(I ran simplify twice, and a third pass didn’t change anything).

Edit: I found a simplified example.

image of b1 coming from substition computations, and b2 defined by hand. b1 doesn't simplify to 0, but b2 does.
image of b1 coming from substition computations, and b2 defined by hand. b1 doesn't simplify to 0, but b2 does.713×566 35.6 KB

Maybe the fact that the computed expression’s (b1) terms are not reordered is the problem ?

Edit 2: a MWE

julia> using Symbolics
[ Info: Precompiling Symbolics [0c5d862f-8b57-4792-8d23-62f2024744c7]

julia> @variables a b c d e f
6-element Vector{Num}:
 a
 b
 c
 d
 e
 f

julia> (a^2 + b^2) * cos(c + d) * e + e * cos(c + d) * (-a^2 - b^2)
(a^2 + b^2)*e*cos(c + d) + (-(a^2) - (b^2))*e*cos(c + d)

julia> (a^2 + b^2) * cos(c + d) * e + e * cos(c + d) * (-a^2 - b^2) |> simplify
0

julia> (a^2 + b^2) * cos(c + d) * e * f + e * cos(c + d) * (-a^2 - b^2) * f |> simplify
(a^2 + b^2)*e*f*cos(c + d) + (-(a^2) - (b^2))*e*f*cos(c + d)

I suspect the problem here is the terms in the computed expression (b1) are not reordered.

2

Just change simplify to expand, and the output becomes 0.
It’ll be great to have a smarter simplify, but in this case, expansion is clearly the solution.

1 Like
3

This was indeed the solution for the two last examples, thank you!

For the first one (quoted below), it didn’t work: differences of identical terms do not cancel out.

I got it working with the same expression on another Pluto notebook though, it might just be an error I didn’t spot.