Find fixed point of multivariable function in Julia - julia

Question

I need to find the fixed point of a multivariable function in Julia.
Consider the following minimal example:
function example(p::Array{Float64,1})
q = -p
return q
end
Ideally I'd use a package like Roots.jl and call find_zeros(p -> p - example(p)), but I can't find the analogous package for multivariable functions. I found one called IntervalRootFinding, but it oddly requires unicode characters and is sparsely documented, so I can't figure out how to use it.

Answer 1

There are many options. The choice of the best one depends on the nature of example function (you have to understand the nature of your example function and check against a documentation of a specific package if it would support it).
Eg. you can use fixedpoint from NLsolve.jl:
julia> using NLsolve
julia> function example!(q, p::Array{Float64,1})
q .= -p
end
example! (generic function with 1 method)
julia> fixedpoint(example!, ones(1))
Results of Nonlinear Solver Algorithm
* Algorithm: Anderson m=1 beta=1 aa_start=1 droptol=0
* Starting Point: [1.0]
* Zero: [0.0]
* Inf-norm of residuals: 0.000000
* Iterations: 3
* Convergence: true
* |x - x'| < 0.0e+00: true
* |f(x)| < 1.0e-08: true
* Function Calls (f): 3
* Jacobian Calls (df/dx): 0
julia> fixedpoint(example!, ones(3))
Results of Nonlinear Solver Algorithm
* Algorithm: Anderson m=3 beta=1 aa_start=1 droptol=0
* Starting Point: [1.0, 1.0, 1.0]
* Zero: [-2.220446049250313e-16, -2.220446049250313e-16, -2.220446049250313e-16]
* Inf-norm of residuals: 0.000000
* Iterations: 3
* Convergence: true
* |x - x'| < 0.0e+00: false
* |f(x)| < 1.0e-08: true
* Function Calls (f): 3
* Jacobian Calls (df/dx): 0
julia> fixedpoint(example!, ones(5))
Results of Nonlinear Solver Algorithm
* Algorithm: Anderson m=5 beta=1 aa_start=1 droptol=0
* Starting Point: [1.0, 1.0, 1.0, 1.0, 1.0]
* Zero: [0.0, 0.0, 0.0, 0.0, 0.0]
* Inf-norm of residuals: 0.000000
* Iterations: 3
* Convergence: true
* |x - x'| < 0.0e+00: true
* |f(x)| < 1.0e-08: true
* Function Calls (f): 3
* Jacobian Calls (df/dx): 0
If your function would require a global optimization tools to find a fixed point then you can e.g. use BlackBoxOptim.jl with norm(f(x) .-x) as an objective:
julia> using LinearAlgebra
julia> using BlackBoxOptim
julia> function example(p::Array{Float64,1})
q = -p
return q
end
example (generic function with 1 method)
julia> f(x) = norm(example(x) .- x)
f (generic function with 1 method)
julia> bboptimize(f; SearchRange = (-5.0, 5.0), NumDimensions = 1)
Starting optimization with optimizer DiffEvoOpt{FitPopulation{Float64},RadiusLimitedSelector,BlackBoxOptim.AdaptiveDiffEvoRandBin{3},RandomBound{ContinuousRectSearchSpace}}
0.00 secs, 0 evals, 0 steps
Optimization stopped after 10001 steps and 0.15 seconds
Termination reason: Max number of steps (10000) reached
Steps per second = 68972.31
Function evals per second = 69717.14
Improvements/step = 0.35090
Total function evaluations = 10109
Best candidate found: [-8.76093e-40]
Fitness: 0.000000000
julia> bboptimize(f; SearchRange = (-5.0, 5.0), NumDimensions = 3);
Starting optimization with optimizer DiffEvoOpt{FitPopulation{Float64},RadiusLimitedSelector,BlackBoxOptim.AdaptiveDiffEvoRandBin{3},RandomBound{ContinuousRectSearchSpace}}
0.00 secs, 0 evals, 0 steps
Optimization stopped after 10001 steps and 0.02 seconds
Termination reason: Max number of steps (10000) reached
Steps per second = 625061.23
Function evals per second = 631498.72
Improvements/step = 0.32330
Total function evaluations = 10104
Best candidate found: [-3.00106e-12, -5.33545e-12, 5.39072e-13]
Fitness: 0.000000000
julia> bboptimize(f; SearchRange = (-5.0, 5.0), NumDimensions = 5);
Starting optimization with optimizer DiffEvoOpt{FitPopulation{Float64},RadiusLimitedSelector,BlackBoxOptim.AdaptiveDiffEvoRandBin{3},RandomBound{ContinuousRectSearchSpace}}
0.00 secs, 0 evals, 0 steps
Optimization stopped after 10001 steps and 0.02 seconds
Termination reason: Max number of steps (10000) reached
Steps per second = 526366.94
Function evals per second = 530945.88
Improvements/step = 0.29900
Total function evaluations = 10088
Best candidate found: [-9.23635e-8, -2.6889e-8, -2.93044e-8, -1.62639e-7, 3.99672e-8]
Fitness: 0.000000391

Answer 2

I'm an author of IntervalRootFinding.jl. I'm happy to say that the documentation has been improved a lot recently, and no unicode characters are necessary. I suggest using the master branch.
Here's how to solve your example with the package. Note that this package should be able to find all of the roots within a box, and guarantee that it has found all of them. Yours only has 1:
julia> using IntervalArithmetic, IntervalRootFinding
julia> function example(p)
q = -p
return q
end
example (generic function with 2 methods)
julia> X = IntervalBox(-2..2, 2)
[-2, 2] × [-2, 2]
julia> roots(x -> example(x) - x, X, Krawczyk)
1-element Array{Root{IntervalBox{2,Float64}},1}:
Root([0, 0] × [0, 0], :unique)
For more information, I suggest looking at https://discourse.julialang.org/t/ann-intervalrootfinding-jl-for-finding-all-roots-of-a-multivariate-function/9515.