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Math Help - 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.

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

    If I set up a lagrange polynomial, I get -x^3-x^2+7x+11. Is this sufficient to show it is of degree 3? The official solution is \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)
    Posts
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    Thanks
    742
    Hello, Makall!

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

    \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.


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