Support Math 4 Wisdom Study Groups Featured Investigations Featured Projects Contact Andrius Kulikauskas m a t h 4 w i s d o m @ g m a i l . c o m +370 607 27 665 Eičiūnų km, Alytaus raj, Lithuania Thank you, Participants! Thank you, Veterans! Dave Gray Francis Atta Howard Jinan KB Christer Nylander Kirby Urner Thank you, Commoners! Free software Open access content Expert social networks Patreon supporters Jere Northrop Daniel Friedman John Harland Bill Pahl Anonymous supporters! Support through Patreon! 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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