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1.2.1 NGF Interpolation

Let us first construct a polynomial of order $m$ by using only points to the right of $x_{k}$ (and $x_{k}$ itself). Written in terms of forward differences, we have

Newton-Gregory Forward interpolation:
$\displaystyle F_{m}(x)$ $\textstyle =$ $\displaystyle f_{k} + {u \choose 1} \Delta f_{k} + {u \choose 2}
\Delta^{2} f_{k} + \dots$  
  $\textstyle =$ $\displaystyle f_{k} + \sum_{l=1}^{m} {u \choose l}
\Delta^{l} f_{k} + O[(\Delta x)^{m+1}]$  

where

\begin{displaymath}
{u \choose l} \equiv \frac{u (u-1) \dots (u-l+1)}{l!}
\end{displaymath}

By $O[(\Delta x)^{m+1}]$ we denote the neglected remainder, stressing its gross dependence on the step size $\Delta x$. Close analysis would show that the remainder term is actually

\begin{displaymath}
R = O \left[ f^{(m+1)}(x') \frac{(x-x')^{m+1}}{(m+1)!} \right]
\end{displaymath}

where $x=x'$ is the position of the maximum of $\vert f^{(m+1)}(x)\vert$ in the interval $[x_{k}, x_{k+m}]$. Putting

\begin{displaymath}
x - x' \equiv \xi   \Delta x
\end{displaymath}

we have

\begin{displaymath}
R = O \left[ f^{(m+1)}(x') \frac{\xi^{m+1}}{(m+1)!} (\Delta x)^{m+1} \right]
\end{displaymath}



EXAMPLE: Taking $m=2$ in the general NGF formula we obtain the parabolic approximation

\begin{displaymath}
F_{2}(x) = \mbox{$f_{k}$}+ \frac{\Delta \mbox{$f_{k}$}}{\Del...
...}{(\Delta x)^{2}}
(x-x_{k}) (x-x_{k+1})
+ O[(\Delta x)^{3}]
\end{displaymath}




Franz J. Vesely Oct 2005
See also:
"Computational Physics - An Introduction," Kluwer-Plenum 2001