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Christoffel-Darboux Formula

For three consecutive Orthogonal Polynomials

\begin{displaymath}
p_n(x)=(A_nx+B_n)p_{n-1}x-C_np_{n-2}(x)
\end{displaymath} (1)

for $n=2$, 3, ..., where $A_n>0$, $B_n$, and $C_n>0$ are constants. Denoting the highest Coefficient of $p_n(x)$ by $k_n$,
$\displaystyle A_n$ $\textstyle =$ $\displaystyle {k_n\over k_{n-1}}$ (2)
$\displaystyle C_n$ $\textstyle =$ $\displaystyle {A_n\over A_{n-1}}={k_nk_{n-2}\over{k_{n-1}}^2}.$ (3)

Then


\begin{displaymath}
p_0(x)p_0(y)+\ldots+p_n(x)p_n(y) = {k_n\over k_{n+1}} {p_{n+1}(x)p_n(y)-p_n(x)p_{n+1}(y)\over x-y}.
\end{displaymath} (4)

In the special case of $x=y$, (4) gives


\begin{displaymath}[p_0(x)]^2+\ldots+[p_n(x)]^2={k_n\over k_{n+1}} [p_{n+1}'(x)p_n(x)-p_n'(x)p_{n+1}(x)].
\end{displaymath} (5)


References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 785, 1972.

Szegö, G. Orthogonal Polynomials, 4th ed. Providence, RI: Amer. Math. Soc., pp. 42-44, 1975.




© 1996-9 Eric W. Weisstein
1999-05-26