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Thread: Interpolation

  1. #1
    Junior Member
    Joined
    Jan 2010
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    30

    Interpolation

    Show that the polynomial interpolating the following data has degree three.

    $\displaystyle \begin{tabular}{|r|r|}
    \hline
    $x$&$f(x)$\\
    \hline
    -2&1\\
    -1&4\\
    0&11\\
    1&16\\
    2&13\\
    3&-4\\
    \hline
    \end{tabular}
    $

    If I set up a lagrange polynomial, I get $\displaystyle -x^3-x^2+7x+11$. Is this sufficient to show it is of degree 3? The official solution is $\displaystyle \Delta^3f(x_{0}) = -6, \Delta^4f(x_{0}) = 0, \Delta^5f(x_{0}) = 0$ using Newton's forward-difference formula but I don't know what delta is supposed to be.

    Thanks.
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Thanks
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    Hello, Makall!

    Show that the polynomial interpolating the following data has degree three.

    $\displaystyle \begin{tabular}{|r|r|} \hline $x$&$f(x)$\\ \hline -2&1\\ -1&4\\ 0&11\\ 1&16\\ 2&13\\ 3&-4\\ \hline \end{tabular} $

    I am not familiar with "Newton's forward-difference formula",
    . . but I'm sure it is similar to the following.


    Take the differences of consecutive terms,
    . . then take the differences of the differences, and so on.


    $\displaystyle \begin{array}{c|ccccccccccc}
    \text{Function: }f(x) & 1 && 4 && 11 && 16 && 13 && \text{-}4 \\
    \text{1st diff: }\Delta f && 3 && 7 && 5 && \text{-}3 && \text{-}17 \\
    \text{2nd diff: }\Delta^2f &&& 4 && \text{-}2 && \text{-}8 && \text{-}14 \\
    \text{3rd diff: }\Delta^3f &&&& \text{-}6 && \text{-}6 && \text{-}6
    \end{array}$


    Since the third differences are constant,
    . . the generating function is of the third degree (a cubic).


    Your function is correct . . . Good work!

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