% m211oh69.tex - The Lagrange-Sylvester Interpolation Formula 
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\begin{center}\tcr{\rmH The Lagrange-Sylvester \vspace{5pt}\\
Interpolation Formula} \end{center} \vspace{-25pt}
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\huge

\begin{description}  
\iffalse
\item[\tcr{Theorem 1:}] \tcg{Lagrange Interpolation Formula)} The 
\tcr{unique} polynomial \mc{f(x)} of degree \mc{\le n -1} which passes through \mc{(x_{\scr1}, y_{\scr1}), \ldots, (x_{\scr n}, y_{\scr n})}
is given by the formula\vspace{-5pt}
$$
\tcb{f(t)}  \tcb{=}
  \tcb{=}  
\tcb{\sum_{\scr i=1}^{\scr n} \textstyle{ y_{\scr k} \frac{g_{\scr k}(t)}{g_{\scr k}(x_k)}}}, \vspace{-5pt}
$$
where
\beqn
\tcb{ g_k(x)} 
& \tcb{=} & \tcb{(x - x_{\scr1})\cdots(x-x_{\scr k-1})(x - x_{\scr k+1})\cdots(x-x_{\scr n})}\rule[-15pt]{0pt}{30pt} \\
& \tcb{=} & \tcb{\frac{(x - x_{\scr1})(x - x_{\scr2}) \cdots 
(x - x_{\scr n})}{(x - x_k)}}. 
\eeqn
\fi

\item[\tcr{Theorem 9}]\tcg{(Lagrange-Sylvester Interpolation Formula)}
Let:\,
\begin{minipage}[t]{16cm}
\mc{\lm_{\scr1}, \ldots, \lm_{\scr s} \in \C} \mbox{\tcg{(distinct)}},\, 
\mc{m_{\scr1},\ldots, m_{\scr s} \in \N},\\
\mc{c_{ik} \in \C,\; 1\le i \le s,\, 0\le k \le m_{\scr i} - 1}
\end{minipage}\vspace{5pt}
be given, and put \vspace{-20pt}
$$
\tcb{h(t) \; = \;\sum_{\scr i=1}^{\scr s} \sum_{\scr k=0}^{\scr m_i - 1} c_{\scr ik}e_{ik}(t),}\vspace{-10pt}
$$ 
where the \mc{e_{\scr ik}(t)} are the \tcg{``constituent polynomials''} of \mc{g(t) := (t - \lm_{\scr1})^{m_{\scr1}}\cdots(t - \lm_{\scr s})^{m_{\scr s}}.}
Then \mc{\deg(h) < n:= \deg(g)},
% =  m_{\scr1} + \ldots + m_{\scr s}}, 
and \mc{h(t)} satisfies the relations
\beq
\hspace*{30pt}\tcb{h^{\scr(k)}(\lm_{\scr i})\; = \; c_{\scr ik}, \hspace{15pt} 1\le i \le s,\; 0\le k \le m_{\scr i} - 1.}
\eeq
Moreover, \mc{h(t)} is the \tcr{unique} polynomial of degree \mc{\le n-1}
satisfying these relations.
%\fi

\item[\tcr{Remarks:}] \tcr{1)} Clearly, \tcg{Theorem 9} \tcr{$\Rightarrow$}
\tcg{Theorem 8}.\\
 (Take \mc{c_{\scr ik} = f^{(\scr k)}(\lm_i)}.)

\tcr{2)} It can be shown that the polynomials \mc{e_{\scr ik}(t)} satisfy the relations\vspace{-10pt} 
$$
\tcb{e^{\scr (l)}_{\scr ik}(\lm_{\scr j}) = \left\{\begin{array}{cl}
1 & \mbox{if \mc{i = j} and \mc{k = l}} \\
0 & \mbox{otherwise}
\end{array}\right..}\vspace{-10pt}
$$
This can be used to prove \tcg{explicit formulae} for the \mc{e_{\scr ik}(t)}'s.

 
\end{description}
\end{document}                          