# Interpolation

• Mar 21st 2010, 03:50 AM
Makall
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.
• Mar 21st 2010, 05:43 AM
Soroban
Hello, Makall!

Quote:

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!