i have the following:
$P \in \Pi_3$ is the interpolation polynom with $P(x_i)=f_i$, for $$x_i \quad -1 \quad 0 \quad 1 \quad 3$$ $$f_i \quad 5 \quad -6 \quad -9 \quad 33$$ (table)
I want to calculate $P(2)$ with the Neville scheme. But i dont get it how to do this.
Does anyone out here, know how to use the Neville scheme? I tried to get it here: http://en.wikipedia.org/wiki/Neville%27s_algorithm
Thanks!
I am certain there are many ways to go about this but I shall do it by using this similarly to the Neville scheme by using the finite difference table to get the interpolating polynomial. Notice they give us $4$ points, so we are looking for a polynomial of degree $3$. Start by making your finite difference table: $$ \begin{array}{c|c|cccc} x & y & & & & \\ \hline -1 & 5 & & & \\ 0 & -6 & & & \\ 1 & -3& & & \\ 3 & 33 & & & & \end{array} $$ and then fill it all out to get $$ \begin{array}{c|c|cccc} x & y & & & & \\ \hline -1 & 5 & -11 & 4 & 1 \\ 0 & -6 & -3 &8 & \\ 1 & -3& 21 & & \\ 3 & 33 & & & & \end{array} $$ You should know how I obtained this. For example, the first row. I get the $-11$ by taking the first difference: $$ \frac{-6-5}{0- (-1)}=\frac{-11}{1}=-11 $$ Then for the next column, you do the second difference: $$ \frac{-3-(-11)}{1-(-1)}=\frac{8}{2}=4 $$ Following these shows you enough to create the whole table. The difference table gives us the interpolating polynomail: $$ 5+-11(x+1)+4(x+1)(x-0)+1(x+1)(x-0)(x-1) $$ Exapnding this (though you don't have to), yields: $$ p_3(x)=x^3+4x^2-8x-6 $$ Then $p_3(2)=2$, which is our approximation.