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Thread: Advanced cubic interpolation problem

  1. #1
    Oct 2008

    Lightbulb Advanced cubic interpolation problem

    Hi everyone.

    I have the following problem:
    Our task is to find the local minima of the polynomial f(x)=ax^3+bx^2+cx+d, which is made by interpolation from these values:

    Suppose we have x1, x2, x3
    and f(x1), f(x2), f(x3)
    and f ' (x1) -> f derived in x1,

    where x1,x2,x3 is in R
    and x1<x2<x3,
    and f(x1)!=f(x2)!=f(x3)
    and f ' (x1) < 0

    I should find a proof which showes, that the desired minima=x0 is always equal to:

    x0=x1 + (x3 - x1) * { [1/3(w+z) - f1'] / [w+f13-f1'] }

    where f1:=f(x1)

    f12:=[f(x2) - f(x1)] / [x2 - x1]
    f13:=[f(x3) - f(x1)] / [x3 - x1]
    f23:=[f(x3) - f(x2)] / [x3 - x2]

    f1' := f ' (x1)

    v:= {[x3-x1]/[x2-x1]} * {f1' +f23 - f13 - f12}

    z:= v - f13 + f1'

    w:= sqrt { (z^2) - (3f1' * v) } ,

    where sqrt(a) means a^(1/2) of course.

    If anyone could help me with this madness , or maybe post some useful link about this type of interpolation (the closest thing i found was discussion about Neville's algorithm at wikipedia), i would be really glad.

    Thanks in advance,

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