Find fixed point of multivariable function in 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.