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! Charlier 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 {$g>0, g+h>0, d, f\in\mathbb{R}, d^2-4g\neq 0$} The Charlier polynomials arise when {$d\neq 0, g=0$}. Set {$f=\frac{h}{d}$}. In my notation, set {$cf=\frac{\alpha\beta c}{-\alpha -\beta}$}, which means {$f=\frac{\alpha\beta}{-\alpha -\beta}$} {$P_{n+1}(x)=[x - d(n + \frac{h}{d^2})]P_n(x) - nhP_{n-1}(x)$} Set {$a=\frac{h}{d^2}$}. In my notation, this means {$a=\frac{\alpha\beta c}{(-\alpha-\beta)^2}$}. {$P_{n+1}(x)=[x - d(n + a)]P_n(x) - nhP_{n-1}(x)$} Define {$P_n(x)=d^nC_n^{(a)}(x/d)$}, which is {$C_n^{(a)}(x)=\frac{1}{d^n}P_n(dx)$}. {$\frac{1}{d^{n+1}}P_{n+1}(dx)=\frac{1}{d^{n+1}}[dx - d(n + a)]P_n(dx) - \frac{1}{d^{n+1}}nhP_{n-1}(dx)$} {$C_{n+1}^{(a)}(x)=\frac{1}{d}[dx - d(n + a)]C_n^{(a)}(x) - \frac{1}{d^{2}}nhC_n^{(a)}(x)$} {$C_{n+1}^{(a)}(x)=[x - (n + a)]C_n^{(a)}(x) - anC_n^{(a)}(x)$} Kim and Zeng Set {$f= -\beta$}, {$-\alpha = 1$}, {$c=\frac{a}{-\beta}$}. We get {$L(x^n)=\sum_{\sigma\in S_n}(-\beta)^{\textrm{fix}\;\sigma + \textrm{dec}\;\sigma - \textrm{cyc}\;\sigma}a^{\textrm{cyc}\;\sigma}$} and then we set {$\beta = 0$}. Then we are summing over permutations where each cycle is a singleton or has a single decedence, thus can be identified with a part in a partition of {$[n]$}. Consequently, {$L(x^n)=\sum_{\pi\in\Pi_n}a^{\textrm{bloc}\;\pi}$} where {$\Pi_n$} is the set of partitions of {$[n]$} and {$\textrm{bloc}\;\pi$} is the number of parts in the partition {$\pi$}. The recurrence relation becomes {$P_{n+1}(x)=[x-((-\alpha)+1)n+a)]P_n(x)-[(-\alpha)n(n-1) + n\frac{a}{-\alpha}]P_{n-1}(x)$} My notation Start here: {$P_{n+1}(x)=[x-(((-\alpha)+(-\beta))n+cf)]P_n(x)-(-\alpha)(-\beta)[n(n-1) + nc]P_{n-1}(x)$} Set {$f= -\beta$}, {$-\alpha = 1$}, {$c=\frac{a}{-\beta}$}. We get {$P_{n+1}(x)=[x-((1 - \beta)n+a)]P_n(x)-[n(n-1)\beta - na]P_{n-1}(x)$} Let {$\beta=0$}, then we have, as desired: {$P_{n+1}(x)=[x-(n+a)]P_n(x) - naP_{n-1}(x)$} Alternatively... Set {$f=-\beta$} and set {$cf=a$}. We have: {$P_{n+1}(x)=[x-(((-\alpha)+(-\beta))n+a)]P_n(x)-(-\alpha)[(-\beta)n(n-1) + an]P_{n-1}(x)$} Let {$\beta=0$}. We have: {$P_{n+1}(x)=[x-((-\alpha)n+a)]P_n(x)-(-\alpha)anP_{n-1}(x)$} What happens if we simply start with setting {$\beta=0$}? Then the last term will go to zero. {$P_{n+1}(x)=[x-(-\alpha n+cf)]P_n(x)$} When {$\alpha = 1$}, this generates a falling factorial {$(x-cf+n-1)_n=\frac{(x-cf+n-1)!}{(x-cf-1)!}$} So we need {$c$} to grow inversely to {$\beta$}. Thus we set {$c=\frac{a}{-\beta}$} {$P_{n+1}(x)=[x-(-\alpha n+\frac{fa}{-\beta})]P_n(x)-(-\alpha)(-\beta)[n(n-1) + n\frac{a}{\beta}]P_{n-1}(x)$} But now we have the problem that the term {$\frac{fa}{-\beta}$} will go to infinity if {$\beta$} goes to zero. So we need to set {$f=-\beta$}. So we need all of these conditions. This all suggests to me that the real physics is in the combinatorics. As the quantum narrative proceeds, the recurrence relations break down - they cannot include all of the details - such as {$\alpha$}. I am curious how this manifests in the combinatorics.
This page was last changed on January 12, 2024, at 08:01 PM