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  {\large \textbf{Mathematical Tripos Part IB: Easter 2001}}

  {\Large \textbf{Numerical Analysis -- Detailed Example
      9\footnote{Corrections and suggestions to these notes should
        be emailed to \texttt{a.shadrin@damtp.cam.ac.uk}. All handouts
        are available on the WWW at the URL 
        \texttt{http://www.damtp.cam.ac.uk/user/na/PartIB/Handouts.html}.}}}

  \vspace{8pt}
  \textbf{\textit{\LARGE Interpolation: the Newton formula}}
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Let $x_0=3$, $x_1=8$, $x_2=5$, $x_3=1$, $x_4=7$; $f_0=-459$, $f_1=
-18864$, $f_2=-3105$, $f_3=-13$, $f_4=-11263$. We wish to find
$p\in\BB{P}_4[x]$ s.t.\ $p(x_k)=f_k$, $k=0,\ldots,4$ and commence by
forming the table of divided differences:
\begin{displaymath}
  \begin{array}{ccccccccc}
    {\bf -459}\\
    & \searrow\\
    & & {\bf -3681} \\ 
    & \nearrow & & \searrow\\
    -18864 & & & & {\bf -786}\\
    & \searrow & & \nearrow & & \searrow\\
    & & -5253 & & & & {\bf -73}\\
    & \nearrow & & \searrow & & \nearrow & & \searrow\\
    -3105 & & & & -640 & & & & {\bf -4}\\
    & \searrow & & \nearrow & & \searrow & & \nearrow\\
    & & -773 & & & & -89\\
    & \nearrow & & \searrow & & \nearrow\\
    -13 & & & & -551\\
    & \searrow & & \nearrow\\
    & & -1875\\
    & \nearrow\\
    -11263
  \end{array}
\end{displaymath}
Therefore
\begin{Eqnarray*}
  p(x)&=&f[x_0]+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)\\
  &&\mbox{}+f[x_0,x_1,x_2,x_3] (x-x_0)(x-x_1)(x-x_2)\\
  &&\mbox{}+f[x_0,x_2,x_3,x_4]
  (x-x_0)(x-x_1)(x-x_2)(x-x_3)\\
  &=&-459 -3681(x-3)-786*(x^2-11x+24)-73(x^3-16x^2+79*x-120)\\
  &&\mbox{}-4(x^4-17x^3+95*x^2-199*x+120)\\
  &=&-4x^4-5x^3+2x^2-6x.
\end{Eqnarray*}

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