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Find all roots of a function in a guaranteed way with Julia

Results 73 IntervalRootFinding.jl issues
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I new issue found while writing the documentation. The current situation is awkward. I think we could go with the following scheme: **Positional arguments** 1. The function. 2. The search...

The example in `test/smiley_examples.jl` has 41 roots, however, only 20 are found on some setups. On other setups all 41 are identified correctly. This inconsistency was first discovered and discussed...

Remove old code: - [x] bisection.jl - [x] move out definitions of `N` used in `branch_and_prune` - [ ] krawczyk.jl - [ ] newton.jl - [ ] find_roots stuff

Problem taken from https://discourse.julialang.org/t/solvers-fail-on-nonlinear-problem-which-has-solutions/20051: ``` function f(X, p) x1,x2,x3,x4,x5,x6,x7,x8,x9,x10 = X return SVector( -p[17] * x1 + -2 * p[1] * (x1 ^ 2 / 2) + 2 * p[2]...

Currently, the functions passed to `roots` must return `SVector` which has several drawbacks: - Require one more explicit import from the user. - `IntervalBox` and `SVector` work in a quite...

For Krawczyk and Newton method, the fact that a contracted interval is subset of the original interval is a sufficient condition for the presence of a solution. The proposition here...

``` julia> f(p) = abs(1 / ( (1+p)^30 ) * 10_000 - 100) f (generic function with 1 method) julia> roots(f, -1000..1000, Newton) 2-element Array{Root{Interval{Float64}},1}: Root(∅, :unique) Root(∅, :unique) ```

Add mean value forms for scalar and vector functions, and a 3rd-order Taylor form for scalar functions.