# Symbolics.jl simplify exp(0) to 1
**URL:** https://discourse.julialang.org/t/symbolics-jl-simplify-exp-0-to-1/135970
**Category:** New to Julia
**Tags:** symbolics
**Created:** [March 2, 2026, 6:32pm UTC](https://discourse.julialang.org/t/symbolics-jl-simplify-exp-0-to-1/135970 "2026-03-02T18:32:41Z")
**Posts on this page:** 2
**Page:** 1
### Author:  [@jhai](https://discourse.julialang.org/u/jhai)
#### Post date: [March 2, 2026, 6:32pm UTC](https://discourse.julialang.org/t/symbolics-jl-simplify-exp-0-to-1/135970/1 "2026-03-02T18:32:41Z")
I’m trying to check a system of linear differential equations using symbolic variables:
X'(t)=R(\alpha)X(t)
This is my code
```julia-auto
using Symbolics
using SymbolicUtils
using LinearAlgebra
@variables α t
R = [cos(α) -sin(α); sin(α) cos(α)]
X = [im*exp(exp(α*im)*t) -im*exp(exp(-α*im)*t);
exp(exp(α*im)*t) exp(exp(-α*im)*t)]
# Check the fundamental matrix
Dt = Differential(t)
dX = Dt.(X)
dX = expand_derivatives.(dX)
dX = expand.(dX)
H = simplify.(dX.-R*X)
display(H)
H = substitute.(H, (Dict(exp(0) => 1),))
display(H)
```
However the terms in the final output always contain things like “exp(0)”, eventhough the solution should be correct:
```julia-auto
2×2 Matrix{Complex{Num}}:
0 cos(-α)*sin(t*exp(0)*sin(-α))*exp(t*exp(0)*cos(-α)) - cos(α)*sin(t*exp(0)*sin(-α))*exp(t*exp(0)*cos(-α)) + (sin(-α) + sin(α))*exp(t*exp(0)*cos(-α))*cos(t*exp(0)*sin(-α)) + im*(-cos(-α)*exp(t*exp(0)*cos(-α))*cos(t*exp(0)*sin(-α)) + cos(α)*exp(t*exp(0)*cos(-α))*cos(t*exp(0)*sin(-α)) + sin(t*exp(0)*sin(-α))*(sin(-α) + sin(α))*exp(t*exp(0)*cos(-α)))
0 cos(-α)*exp(t*exp(0)*cos(-α))*cos(t*exp(0)*sin(-α)) - cos(α)*exp(t*exp(0)*cos(-α))*cos(t*exp(0)*sin(-α)) - sin(t*exp(0)*sin(-α))*(sin(-α) + sin(α))*exp(t*exp(0)*cos(-α)) + im*(cos(-α)*sin(t*exp(0)*sin(-α))*exp(t*exp(0)*cos(-α)) - cos(α)*sin(t*exp(0)*sin(-α))*exp(t*exp(0)*cos(-α)) + (sin(-α) + sin(α))*exp(t*exp(0)*cos(-α))*cos(t*exp(0)*sin(-α)))
```
For the past hour I’ve tried every combination of “substitute.”, “expand.”, “simplify.” , but it doesn’t even get rid of the exp(0).
It would also be nice to have things like \cos(-\alpha) = \cos(\alpha) and \sin(-\alpha)=-\sin(\alpha).
Thanks for any help!
---
### Author:  [@langestefan](https://discourse.julialang.org/u/langestefan)
#### Post date: [March 2, 2026, 6:47pm UTC](https://discourse.julialang.org/t/symbolics-jl-simplify-exp-0-to-1/135970/2 "2026-03-02T18:47:55Z")
It would be nice to have such simple rules available by default. You could define the rules yourself.
For example:
```julia
rules = RuleSet([
@rule(exp(0) => 1),
@rule(sin(-(~x)) => -(sin(~x))),
@rule(cos(-(~x)) => cos(~x))
])
H2 = map(H) do h
re = simplify(real(h), rewriter=rules)
im_part = simplify(imag(h), rewriter=rules)
re + im * im_part
end
```
Will yield
```julia
2×2 Matrix{Complex{Num}}:
0 cos(-α)*sin(t*sin(-α))*exp(t*cos(-α)) - cos(α)*sin(t*sin(-α))*exp(t*cos(-α)) + (sin(-α) + sin(α))*cos(t*sin(-α))*exp(t*cos(-α)) + im*(-cos(-α)*cos(t*sin(-α))*exp(t*cos(-α)) + cos(α)*cos(t*sin(-α))*exp(t*cos(-α)) + (sin(-α) + sin(α))*sin(t*sin(-α))*exp(t*cos(-α)))
0 cos(-α)*cos(t*sin(-α))*exp(t*cos(-α)) - cos(α)*cos(t*sin(-α))*exp(t*cos(-α)) - (sin(-α) + sin(α))*sin(t*sin(-α))*exp(t*cos(-α)) + im*(cos(-α)*sin(t*sin(-α))*exp(t*cos(-α)) - cos(α)*sin(t*sin(-α))*exp(t*cos(-α)) + (sin(-α) + sin(α))*cos(t*sin(-α))*exp(t*cos(-α)))
```
I’m not sure why the `cos(-α)` is not taking…
edit:
This works:
```julia
rules = RuleSet([
@rule(exp(0) => 1),
@rule(sin((-1) * (~x)) => -(sin(~x))),
@rule(cos((-1) * (~x)) => cos(~x)),
])
```
that yields
```julia
2×2 Matrix{Complex{Num}}:
0 0
0 0
```