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Laguerre polynomials

Chihara starts with the general recurrence relation

{$P_{n+1}(x)=[x - (dn + f)]P_n(x) - [n(gn + h)]P_{n-1}(x)$} where {$g>0, g+h>0, d, f\in\mathbb{R}$}

The Laguerre polynomials arise when {$d\neq 0, \rho=d^2-4g =0$}.

He sets {$f=(h+g)\frac{1}{\sqrt{g}}$}

{$P_{n+1}(x)=[x-(dn + (h+g)\frac{1}{\sqrt{g}})]P_n(x) - [n(gn + h)]P_{n-1}(x)$}

Factor out the {$\sqrt{g}$} and the {$g$}.

{$P_{n+1}(x)=[x-(n\frac{d}{\sqrt{g}} + \frac{h}{g}+1)\sqrt{g}]P_n(x) - g[n(n + \frac{h}{g})]P_{n-1}(x)$}

In my notation

In my variables, {$h=c(-\alpha)(-\beta)$}, {$g=(-\alpha)(-\beta)$} and {$f_C=c_Af_A$}. Thus

{$f_C=(h+g)\frac{1}{\sqrt{g}}=(c+1)\sqrt{\alpha\beta}$}

Furthermore, {$\alpha=\beta$}. Thus, in my notation {$f=\frac{c+1}{c}|\alpha|$} and assuming {$\alpha < 0$}, so that {$f=\frac{-c-1}{c}\alpha$}, and recalling {$d=-\alpha -\beta$}

{$P_{n+1}(x)=[x-((-\alpha)+(-\beta))n+cf)]P_n(x)-[(-\alpha)(-\beta)n(n-1) + nc]P_{n-1}(x)$}

{$P_{n+1}(x)=[x-(-2n-c-1)\alpha)]P_n(x)-[\alpha^2 n(n-1) + nc]P_{n-1}(x)$}

Apparently, Chihara sets {$\alpha=-1$}, thus {$g=\sqrt{\alpha^2}=1$}, yielding in my notation

{$P_{n+1}(x)=[x-(2n + c + 1)]P_n(x)-[n(n-1) + nc]P_{n-1}(x)$}

Kim and Zeng write

{$L_{n+1}(x)=(x-(2n+\beta))L_n(x)-n((n-1) + \beta)L_{n-1}(x)$}

and so given that {$c_A=\beta_{KZ}$} it is not clear why we end up with the clash {$c_A+1\neq \beta_{KZ}$}

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This page was last changed on January 11, 2024, at 05:30 PM