Hi everyone.

Ive already posted this problem in the "Calculus" section like week ago, but nobody replied and i assumed that ive chosen a wrong section.

My problem is:

Find the local minima of the polynomial f(x)=ax^3+bx^2+cx+d, which is made by cubic 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)

f2:=f(x2)

f3:=f(x3)

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

,

Naena