Krawczyk fails with infinite interval

[dpsanders]

julia> rts = roots(x->x^2 - 2, -∞..∞, Krawczyk)
0-element Array{Root{Interval{Float64}},1}

This is presumably related to the same issue about what happens when the derivative is zero at the midpoint. We should just explicitly check that and move the midpoint if necessary.

[dpsanders]

Why does it work for Newton?

[ys1999]

Sir, Krawczyk is failed in this case because in the code contractors.jl contract function for Krawczyk and Newton for both is

function newtonlike_contract(op, C, X, tol)
    # use Bisection contractor for this:
    if !(contains_zero(C.f(X)))
        return :empty, X
    end

    # given that have the Jacobian, can also do mean value form

    NX = op(C.f, C.f′, X) ∩ X

    isempty(NX) && return :empty, X
    isinf(X) && return :unknown, NX  # force bisection

    if NX ⪽ X  # isinterior; know there's a unique root inside
        NX =  refine(X -> op(C.f, C.f′, X), NX, tol)
        return :unique, NX
    end

    return :unknown, NX
end

As we can see from code that if NX is empty than it will return empty status. 'op' function for Krawczyk ( single variable) is

function 𝒦(f, f′, X::Interval{T}) where {T}
    m = Interval(mid(X))
    Y = 1 / f′(m)

    m - Y*f(m) + (1 - Y*f′(X)) * (X - m)
end

so for the function f(x)=x^2-2 and interval X=-Inf..Inf will give m=[0,0] and if f′(m) become zero and due to this Y will become ∅ ( this below case ) and it will result in 'empty' status.

julia> X=0..0
[0, 0]

julia> p=1/X
∅

julia> isempty(p)
true

[ys1999]

This problem can be solved by using this function for Krawczyk ( single variable) -

function 𝒦(f, f′, X::Interval{T}) where {T}
    m = Interval(mid(X , where_bisect))
    Y = 1 / f′(m)

    m - Y*f(m) + (1 - Y*f′(X)) * (X - m)
end

where_bisect is not exactly equal to 0.5 , so it will not give m=[0,0]. where_bisect =0.49609375(approx 0.5)

[dpsanders]

@ys1999 Yes, you are right. I believe this is fixed in #93.

[Kolaru]

Fixed by #93 using the proposition of @ys1999.