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

  {\Large \textbf{Numerical Analysis -- Detailed Example
      8\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 Lagrange formula}}
\end{center}

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.\ $p(x_k)=f_k$, $k=0,\ldots,4$. We evaluate the
cardinal polynomials
\begin{Eqnarray*}
  L_0(x)&=&\prod_{l=1}^4 \frac{x-x_l}{x_0-x_l}=-\Frac{1}{80} x^4
  +\Frac{21}{80} x^3-\Frac{151}{80}x^2+\Frac{411}{80}x-\Frac72 ,\\
  L_1(x)&=&\prod_{\stackrel{\scriptstyle l=0}{l\neq 1}}^4
  \frac{x-x_l}{x_1-x_l}= \Frac{1}{105}x^4-\Frac{16}{105}x^3
  +\Frac{86}{105} x^2-\Frac{176}{105}x+1,\\
  L_2(x)&=&\prod_{\stackrel{\scriptstyle l=0}{l\neq 2}}^4
  \frac{x-x_l}{x_2-x_l}= \Frac{1}{48}x^4-\Frac{19}{48}x^3
  +\Frac{119}{48}x^2 -\Frac{269}{48}x+\Frac72,\\
  L_3(x)&=&\prod_{\stackrel{\scriptstyle l=0}{l\neq 3}}^4
  \frac{x-x_l}{x_3-x_l}= \Frac{1}{336}x^4-\Frac{23}{336} x^3
  +\Frac{191}{336} x^2 -\Frac{673}{336} x+\Frac52, \\
  L_4(x)&=&\prod_{\stackrel{\scriptstyle l=0}{l\neq 4}}^4
  \frac{x-x_l}{x_4-x_l}=-\Frac{1}{48}x^4+\Frac{17}{48}x^3
  -\Frac{95}{48} x^2+\Frac{199}{48}x-\Frac52.
\end{Eqnarray*}
Therefore
\begin{displaymath}
  p(x)=\sum_{k=0}^4 f_k L_k(x)=-4x^4-5x^3+2x^2-6x.
\end{displaymath}

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