Add info about the function in the README

[David P. Sanders]

The function in the README has precisely 80 roots in (-5..5)^2
(assuming IntervalRootFinding is working correctly...):

julia> using IntervalArithmetic, IntervalRootFinding, StaticArrays

julia> f(x, y) = SVector( (x+3)*(y^3-7)+18, sin(y*exp(x)-1) )
f (generic function with 1 method)

julia> f(xx) = f(xx...)
f (generic function with 2 methods)

julia> rts = roots(f, IntervalBox(-5..5, 2), Newton, 1e-20)
80-element Array{IntervalRootFinding.Root{IntervalArithmetic.IntervalBox{2,Float64}},1}:
 Root([4.99884, 4.99885] × [1.68094, 1.68095], :unique)
 Root([4.98639, 4.9864] × [1.68053, 1.68054], :unique)
 Root([4.97379, 4.9738] × [1.68011, 1.68012], :unique)
 Root([4.96103, 4.96104] × [1.67968, 1.67969], :unique)
 Root([4.9481, 4.94811] × [1.67925, 1.67926], :unique)
 Root([4.935, 4.93501] × [1.67881, 1.67882], :unique)
 Root([4.92172, 4.92173] × [1.67836, 1.67837], :unique)
 Root([4.90826, 4.90827] × [1.6779, 1.67791], :unique)
 Root([4.89461, 4.89462] × [1.67743, 1.67744], :unique)
 Root([4.88077, 4.88078] × [1.67696, 1.67697], :unique)
 ⋮
 Root([2.91809, 2.9181] × [1.58188, 1.58189], :unique)
 Root([2.80939, 2.8094] × [1.57427, 1.57428], :unique)
 Root([2.68705, 2.68706] × [1.56525, 1.56526], :unique)
 Root([2.54714, 2.54715] × [1.55431, 1.55432], :unique)
 Root([2.3837, 2.38371] × [1.5406, 1.54061], :unique)
 Root([2.18717, 2.18718] × [1.5226, 1.52261], :unique)
 Root([1.94053, 1.94054] × [1.49727, 1.49728], :unique)
 Root([1.60901, 1.60902] × [1.45725, 1.45726], :unique)
 Root([1.10116, 1.10117] × [1.377, 1.37701], :unique)
 Root([0, 0] × [1, 1], :unique)

julia> all([root.status for root in rts] .== :unique)
true

[David P. Sanders]

(This is with the latest PR to IntervalRootFinding, changing bisection to not bisect exactly in the centre of an interval.)

[Patrick Kofod Mogensen]

Thanks! I'll try to write something intelligent about the problem :)

[David P. Sanders]

By the way, where did that function come from? It's pretty nasty ;)

[David P. Sanders]

Do you know any other software with which we can check if that number of roots is correct? Mathematica doesn't seem to be able to do it...

[Patrick Kofod Mogensen]

It's pretty nasty ;)

It's very nasty indeed! And above all, solving it is a very bad example of what this package claims it can do :) It's old... You'd have to ask svillemot, he added it in Dec 24, 2013 :) c434bd8#diff-04c6e90faac2675aa89e2176d2eec7d8

[David P. Sanders]

cc @svillemot ^

[David P. Sanders]

I found this:

https://www.sciencedirect.com/science/article/pii/S0377042700007111

which effectively seems to do what IntervalRootFinding.jl does, only without intervals.

It has some nice looking hard test cases.

[Sébastien Villemot]

I actually don't remember why I picked this particular example, sorry. But clearly I was not aware that it had 80 roots! I guess it should be replaced by a better-behaving one…

[Patrick Kofod Mogensen]

I actually don't remember why I picked this particular example, sorry. But clearly I was not aware that it had 80 roots! I guess it should be replaced by a better-behaving one…

It does serve as a warning though, we might just want to make it explicit :)