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Derivation of the Classification of Orthogonal Sheffer Polynomials

I am writing out Meixner's proof from his paper

  • Meixner, Josef. Orthogonale Polynomsysteme mit einer besonderen Gestalt der erzeugenden Funktion, J. London Math. Soc. 9 (1934), 6–13; JFM 60.0293.01; Zbl 0008.16205.

as presented in this paper

and this explanation





{$A(s)=\sum_{n=0}^{\infty}\frac{P_n(0)}{n!}s^n$} upon setting {$x=0$}

Recurrence relation

{$P_{n+1}(x)=(x+l_{n+1})P_n(x) +k_{n+1}P_{n-1}(x)$}

See Tom Koornwinder's note.

{$l_{n+1}=l_1+nl$}, {$k_{n+1}=n(k_2+(n-1)k)$}, {$(l\in\mathbb{R},k_2<0,k\leq 0)$}


{$P_{n+1}=(x+l_1 + nl)P_n(x)+n(k_2+(n-1)k)P_{n-1}(x)$}

Differential equation

  • How to algebraically derive a differential equation that works for the case when {$f\neq 0$}?
  • How to combinatorially derive an interpretation that satisfies the case when {$f\neq 0$}?
  • Contact Dongsu Kim and Jiang Zeng regarding this.

{$$A(s)=\sum_{n=1}^{\infty}\frac{P_n(0)}{n!}s^n$$}Given formula for {$A(s)$}
{$$A'(s)=\sum_{n=1}^{\infty}\frac{P_n(0)}{(n-1)!}s^{n-1}$$}Derive formula for {$A'(s)$}
{$$A'(s) = \sum_{m=0}^{\infty}\frac{P_{m+1}(0)}{m!}s^m$$}Set {$m=n-1$}
{$$P_{n+1}(0)=nlP_n(0)+n(k_2+(n-1)k)P_{n-1}(0)$$}Given recurrence relation for orthogonal Sheffer polynomials
{$$\frac{1}{n!}P_{n+1}(0)=\frac{l}{(n-1)!}P_n(0) + \frac{k_2}{(n-1)!}P_{n-1}(0) + \frac{k}{(n-2)!}P_{n-1}(0)$$}divide by {$n!$}
{$$\frac{1}{n!}P_{n+1}(0)=\frac{l}{(n-1)!}P_n(0)s^{n-1} + \frac{k_2}{(n-1)!}P_{n-1}(0)s^{n-1} + \frac{k}{(n-2)!}P_{n-1}(0)s^{n-1}$$}multiply by {$s^{n-1}$}
{$$\sum_{n=2}^{\infty}\frac{P_{n+1}(0)}{n!}s^{n-1} = l\sum_{n=2}^{\infty}\frac{P_n(0)}{(n-1)!}s^{n-1} + k_2\sum_{n=2}^{\infty}\frac{P_{n-1}(0)}{(n-1)!}s^{n-1} + k\sum_{n=2}^{\infty}\frac{P_{n-1}(0)}{(n-2)!}s^{n-1}$$}sum from n=2 to {$\infty$}
{$$\frac{1}{s}\sum_{n=2}^{\infty}\frac{P_{n+1}(0)}{n!}s^{n} = l\sum_{n=2}^{\infty}\frac{P_n(0)}{(n-1)!}s^{n-1} + k_2\sum_{n=2}^{\infty}\frac{P_{n-1}(0)}{(n-1)!}s^{n-1} + k\sum_{n=2}^{\infty}\frac{P_{n-1}(0)}{(n-2)!}s^{n-1}$$}in the first sum, pull out {$\frac{1}{s}$}
{$$\frac{1}{s}[A'(s)-P_2(0)s - P_1(0)] = l[A'(s) - P_1(0)] + k_2[A(s) - P_0(0)] + k[sA'(s)]$$}reexpress sums (minus initial terms for n=1, n=0) as functions (minus these initial terms)
{$$A'(s)[\frac{1}{s}-l - ks] + [- P_2(0) - \frac{P_1(0)}{s} + lP_1(0) + k_2P_0(0)] = k_2A(s)$$}collect terms
{$$A'(s)[\frac{1}{s}-l - ks] + [- P_2(0) - \frac{P_1(0)}{s} + lP_1(0) + k_2] = k_2A(s)$$}by convention {$P_0(x)=1$}
{$$A'(s)[\frac{1}{s}-l - ks] + [- l_1P_1(0) - \frac{P_1(0)}{s}] = k_2A(s)$$}{$P_2(0)= -l_2P_1(0) + k_2P_0(0)$} and {$l_2=l_1+l$} so {$-P_2(0) + lP_1(0) + k_2=-l_1P_1(0)$}
{$$A'(s)[\frac{1}{s}-l - ks] = k_2A(s)$$}by convention {$l_1=0$} which means {$P_1(x)=x$} so {$P_1(0)=0$}
{$$\frac{A'(s)}{s}[1-ls - ks^2] = k_2A(s)$$}factor out {$\frac{1}{s}$}
{$$\frac{A'(s)}{sA(s)}=\frac{k_2}{1-ls - ks^2}$$}divide on either side

Explicit formula for {$t(D)$}


When {$\beta\rightarrow\alpha$} we have {$t(D)=\frac{D}{1+\alpha D}$}.

In the general case we have {$(e^{(\alpha - \beta)D}-1)g(x)=g(x+\alpha - \beta)-g(x)$}.

Moments and Distributions

From moments to distribution

Thanks to John Harland!

{$\hat{\omega}(\xi)=\sum_{n=0}^{\infty}\frac{(-2\pi i)^n\mu_n}{n!}\xi^n$}

{$\omega(x)=(\check{\hat{\omega}})(x)=\int\hat{\omega}(\xi)e^{2\pi i x\xi}\textrm{d}\xi$}

Orthogonality measure


polynomialsmomentsobjects countedgenerating function {$g(t)$}weight function {$w(x)$}{$u(t)$}
Meixner{$L(x^n)=a(n)$}ordered Bell numbers (weak orderings of a set){$\sum_{n=0}^{\infty}a(n)\frac{t^n}{n!} = \frac{1}{1-(e^t-1)}$}step function {$\frac{a^k(\beta)_k}{k!}$} on interval {$[k,k+1)$}{$\frac{1}{\beta - \alpha}\textrm{ln}(1-\alpha t)+\frac{1}{\alpha - \beta}\textrm{ln}(1-\beta t)$}
Charlier{$L(x^n)=B_n$}Bell numbers (partitions of a set){$\sum_{n=0}^{\infty}B_n\frac{t^n}{n!} = e^{e^t-1}$}step function {$\frac{a^k}{k!}$} on interval {$[k,k+1)$}{$-\frac{1}{\alpha}\textrm{ln}(1-\alpha t)$}
Laguerre{$L(x^n)=n!$}permutations{$\sum_{n=0}^{\infty}n!\frac{t^n}{n!} = \frac{1}{1-t}$}{$x^\alpha e^{-x}$}{$\frac{t}{(1 - \alpha t)}$}
Hermite{$L(x^n)=j(n)$}involutions without fixed points{$\sum_{n=0}^{\infty}j(n)\frac{t^n}{n!} = e^{\frac{t^2}{2}}$}{$e^{-x^2}$}{$t$}
Meixner-Pollaczek{$L(x^n)=A_n$}zigzag permutations{$\sum_{n=0}^{\infty}A_n\frac{t^n}{n!} = \textrm{sec}\;t$}{$\frac{|\Gamma(\frac{n+ix}{2})|^2}{|\Gamma(\frac{n}{2})|^2}e^{-x\;\textrm{tan}^{-1}\delta} $}{$\frac{1}{\overline{\alpha} - \alpha}\textrm{ln}(1-\alpha t)+\frac{1}{\alpha - \overline{\alpha}}\textrm{ln}(1-\overline{\alpha} t)$}

Note the progression. In terms of the generating functions, we start by assembling collections of nontrivial terms from the exponential, yielding partitioned orderings. We then switch to the exponential function, yielding partitions of unordered sets. We then collapse further so that n! in the numerator and denominator cancel out, yielding permutations. We then lose the odd terms and have building blocks that give only even terms, transpositions. We then seem to link together transpositions into zigzag permutations, and they manifest steps in a circular order.

Think of {$N=\frac{-\gamma}{\alpha\beta}$} as a positive integer. Then the Meixner polynomials are the same as the Kravchuk polynomials.

The weight function for the Meixner polynomials is {$w((\alpha-\beta)n)=(\frac{\alpha}{\beta})^n\binom{N}{n}$}.

If {$\alpha - \beta \neq 0$} then the entire system is moving with a mean {$pN$} and a variance {$p(p-1)N$}.

For Meixner-Pollaczek polynomials the unit of space-time is {$\alpha - \bar{\alpha}=2bi$}. The weight function is {$w((2bi)n)=w(\frac{\alpha}{\bar{\alpha}}|\alpha|n) = (\textrm {cos}\;2n\theta + i\; \textrm{sin}\;2n\theta)\binom{N}{n}$}. This is in terms of {$\theta$} where {$\frac{\alpha}{\bar{\alpha}}=\textrm{cos}2\theta + i\textrm{sin}2\theta$}.


Math Stack Exchange. Generating function for fixed-point free involutions.


Consider the space {$\mathcal{P}$} of polynomials with differentiation operator {$D$}.

Define the formal power series {$t(s)=s + a_2s^2 + a_3s^3 + \dots$}

We have {$t(D)=D+a_2D^2+a_3D^3+\dots$}

What can we say about a linear operator {$\Lambda$} for which {$D\Lambda=\Lambda t(D)$} ?

Applying to polynomial {$P(x)=p_nx^n+p_{n-1}x^{n-1}+\cdots +p_0$}, we have

{$D(\lambda(P(x))=\lambda(t(D)P(x))=\lambda[np_nx^{n-1}+(n-1)p_{n-1}x^{n-2}+\cdots + p_1 + a_2n(n-1)p_nx^{n-2} + a_2(n-1)(n-2)p_{n-1}x^{n-3}+\cdots + a_22p_2 + \cdots + n(n-1)(n-2)\cdots 1 \cdot a_np_n]$}

{$=np_nx^{n-1} + [(n-1)p_{n-1}+a_2n(n-1)p_n]x^{n-2}+[(n-2)p_{n-2}+a_2(n-1)(n-2)p_{n-1}+a_3n(n-1)(n-2)(n-3)p_n]x^{n-3}+\cdots$}

{$\cdots + [2p_2+a_2\cdot 3\cdot 2p_3 + \cdots + a_{n-1}n!p_{n-1}]x + [p_1 + a_22p_2 + a_33\cdot 2p_3 + \cdots + a_nn!p_n]$}

Thus the derivative of {$\lambda (P(x))$} has degree {$n-1$}, which means that {$\lambda(P(x))$} has degree {$n$}, just like {$P(x)$}.


Assume that {$\mu=0$} and thus {$\Lambda P_n(x)=x^n$}.

Then {$D\Lambda = \Lambda t(D)$} implies

The differential equation

{$t(u)$} satisfies the differential equation

{$t'(u)=1-\lambda u - \kappa t(u)^2$}

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This page was last changed on January 04, 2024, at 08:33 PM