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Math Help - Hermite-Interpolation Addition

  1. #1
    guuy
    Guest

    Red face Hermite-Interpolation Addition

    Hello,
    it's me again, the guy with the Hermite-interpolation.
    I now decided to post a concrete task so maybe it's easier to answer.

    The following points and the accompanying gradients are given:

    x0=0 f(x)=3 f'(x)=0
    x1=3 f(x)=3 f'(x)=27
    x2=-2 f(x)=-57 f'(x)=72

    I now would like to find out a polynomial function with goes through all those mentioned points and the same gradient in those points.

    Thanks a lot in advance.

    guy.
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by guuy View Post
    Hello,
    it's me again, the guy with the Hermite-interpolation.
    I now decided to post a concrete task so maybe it's easier to answer.

    The following points and the accompanying gradients are given:

    x0=0 f(x)=3 f'(x)=0
    x1=3 f(x)=3 f'(x)=27
    x2=-2 f(x)=-57 f'(x)=72

    I now would like to find out a polynomial function with goes through all those mentioned points and the same gradient in those points.

    Thanks a lot in advance.

    guy.
    I don't know anything about that Hermite-something. But I played with your question using what little I know about points, gradients and characteristics of some known equations/functions. And I found an equation/function that you're looking for.

    Graphing those ordered pairs on the same x,y plane, I saw that the graph of the unknown function is rising steeply at point (-2,-57), then it level off at (0,3), then goes down after (0,3), and then rises steeply again at (3,3).
    That looks like a vertical S-curve. A cubic curve in x.
    So I played with:
    f(x) = Ax^3 +Bx^2 +Cx +D -----(1)
    So,
    f'(x) = 3Ax^2 +2Bx +C -----------(2)

    At (x=0, f'(x)=0), in (2):
    0 = 3A(0) +2B(0) +C
    C = 0 --------------------------***
    So, (2) becomes
    f'(x) = 3Ax^2 +2Bx ------------(2a)

    At (x=3, f'(x)=27), in (2a):
    27 = 3A(3^2) +2B(3)
    9 = 9A +2B --------------------(3)

    At (x=-2, f'(x)=72), in (2a):
    72 = 3A((-2)^2) +2B(-2)
    72 = 12A -4B
    18 = 3A -B
    B = 3A -18
    Plug that into (3),
    9 = 9A +2(3A -18)
    9 = 9A +6A -36
    A = 45/15 = 3 ---------------------------***
    So, (2a) becomes f'(x) = 9x^2 +2Bx -----------(2b)

    Back to (x=3, f'(x)=27), in (2b):
    27 = 9(3^2) +2B(3)
    27 = 81 +6B
    B = (27-81)/6 = -9 ---------------------***

    So (1) becomes f(x) = 3x^3 -9x^2 +D -----(1a)

    At (x=0, f(x)=3), in (1a):
    3 = 3(0) -9(0) +D
    D = 3 ---------------------------------***
    So (1a) becomes f(x) = 3x^3 -9x^2 +3 .

    And, f(x) = 3x^3 -9x^2 +3 is the function you are looking for.

    -----------
    I checked that on all the given ordered pairs and it checked allright.
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