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Math Help - Advanced cubic interpolation problem

  1. #1
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    Oct 2008
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    Lightbulb Advanced cubic interpolation problem

    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
    Attached Thumbnails Attached Thumbnails Advanced cubic interpolation problem-graph.jpg  
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  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
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    Quote Originally Posted by Naena View Post
    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
    I think the main reason you got no response last time was that your explanation of the problem is very difficult to follow.

    CB
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  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    7

    Trying to simplify it

    Ok, i will try to put it in a simpler way:

    Lets say we have x1, x2, x3 from R and also their functional values f1, f2, f3 from R. That means we have three points in R^2.
    For example if x1=2, x2 = 5 , x2 = 12 and f1= 15, f2 = 7, f3 = 13, then these are our three points:
    (2,15) (5,7) (12,13)
    Now lets say we also know the value of the f1' , which is f derived in x1.
    For example, if f= -7x^3 + 2x^2 + x + 8 , then f1' = f' (x1) = -21(x1)^2 + 4(x1) + 1 = -75
    we also know that this derivation f1' is lower than zero.
    Now, lets say that we want to find a function f, which looks like: ax^3 + bx^2 + cx + d , where a,b,c,d are constants from R.
    This function should be made from our points (x1,f1), (x2,f2), (x3,f3) and it should be the "best" function , that we can make from them.
    This method of aproximation is called interpolation: Interpolation - Wikipedia, the free encyclopedia
    Normally, if we had 4 points, it would be very easy to perform a standard polynomial interpolation of third degree: Polynomial interpolation - Wikipedia, the free encyclopedia
    On this wiki site Polynomial interpolation - Wikipedia, the free encyclopedia
    is algorithm "Constructing the interpolation polynomial". But we cant use it, because it needs n+1 points for the polynomial of n-th degree, and we have only 3 points for 3-th degree.
    So we have only 3 rows in our matrix, but in my opinion, we could add the last one as the 3x1^2 + 2x1 + x1 + 0 = f1' .
    So now we have a 4x4 matrix which we should be able to solve:
    (x1^3 x1^2 x1 1) (a) = ( f1 )
    (x2^3 x2^2 x2 1) (b) = ( f2 )
    (x3^3 x3^2 x3 1) (c) = ( f3 )
    (3x1^2 2x1 1 0) (d) = ( f1')
    or written different:
    x1^3 + x1^2 + x1 + 1 = f1
    x2^3 + x2^2 + x2 + 1 = f2
    x3^3 + x3^2 + x3 + 1 = f3
    3x1^2 + 2x1 + 1 + 0 = f1'

    so we solve it ( of which i wasnt capable ) and find a,b,c,d

    now we have our desired polynomial f(x) = ax^3 + bx^2 + cx + d.
    But our task is to find its minimum, so we make its first derivation, and find when its equal to zero:
    3ax^2 +2bx + c = 0
    which means
    xa=[-2b+sqrt(4b^2-12ac)] / 6a
    or
    xa=[-b+sqrt(b^2-3ac)]/3a

    therefore:
    xb=[-b-sqrt(b^2-3ac)]/3a

    but our minimum is xb, not xa, because first derivation of f in x1 -> f1' is lower than zero.
    but we already know a, b, and c, so we find xb and we are done.
    BUT!

    It should be in exactly this shape:
    xb=x1 + (x3 - x1) * { [1/3(w+z) - f1'] / [w+f13-f1'] }
    where:
    f12:=[f(x2) - f(x1)] / [x2 - x1]
    f13:=[f(x3) - f(x1)] / [x3 - x1]
    f23:=[f(x3) - f(x2)] / [x3 - x2]
    z:= v - f13 + f1'
    w:= sqrt { (z^2) - (3f1' * v) } ,

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

    Which is kind of problem for me.

    If u guys can solve it and prove that it really looks like this... respect...

    I tried to put it in a simpler way, hope it helped, but if you still find something confusing or unclear, please - dont hesitate and reply, everything is helpful.

    Thanks,
    Naena
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