# Interpolation

• Mar 21st 2010, 03:50 AM
Makall
Interpolation
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}
$

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

Quote:

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.

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

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

Your function is correct . . . Good work!