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Optimum term in quadratic expansion #109
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Assuming fexact(α, β, x) = ((2*β*x + α)^2 - α^2)/(4*β) this way julia> y = Interval(-1, 1)
[-1, 1]
julia> fprod(0.0, 1.0, y), fsum(0.0, 1.0, y), fexact(0.0, 1.0, y)
([-1, 1], [0, 1], [-0, 1])
julia> y = Interval(-3, -2)
[-3, -2]
julia> fprod(1.0, 2.0, y), fsum(1.0, 2.0, y), fexact(1.0, 2.0, y)
([6, 15], [5, 16], [6, 15])
julia> y = Interval(-3, 3)
[-3, 3]
julia> fprod(1.0, 2.0, y), fsum(1.0, 2.0, y), fexact(1.0, 2.0, y)
([-21, 21], [-3, 21], [-0.125, 21]) |
well the formula above doesn't work if |
where should this go? I can open a PR for this. Also does my assumption (alpha and beta are not intervals) hold? If it doesn't, it may make dependency problem worse |
i no longer remember the context in which this method was needed. (seems to be the scalar version of this one). then i suggest to define the method on quadratic_expansion(x::Interval, α::Real, β::Real) (NB this library doesn't depend on LazySets so there is no risk of confusion between LazySets.Interval and IntervalArithmetic.Interval)
yes, that assumption holds
go for it 👍 |
Next that function should replace this line ? With a quick glance, the rest of the code should not suffer from dependency issues? |
Doing
(α + βx) x
looses the dependency between the arguments, so it may be less precise using interval arithmetic. If there is a general rule, one has to find it by taking the derivative of the expression as in the paper, which is a particular case with alpha, beta being something like t and 1/2t^2.Originally posted by @mforets in #91 (comment)
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