Pochhammer symbol of a differential, and hypergeometric polynomials

Question

I have a minor result which I'm sure has come up somewhere before but I can't seem to find it.

Consider a confluent hypergeometric function of the form $$\newcommand{\ff}{{}_1F_1} \ff(b+k;b;z)\textrm{, for }k\in\mathbb{N}.$$ Numerical tests suggest that this is always a polynomial of degree $k$ multiplied by an exponential. One can prove this in a dull fashion by using $\ff(b;b;z)=e^z$ and then applying recurrence relations, but I found a cleaner way using the series definition,

$$ \ff(b+k;b;z) =\sum_{n=0}^\infty\frac{(b+k)_n}{(b)_n}\frac{z^n}{n!} $$

where $(b)_k=b(b+1)\cdots(b+k-1)$ is the Pochhammer symbol. By exploiting the identities $$ \frac{(b+k)_n}{(b)_n}=\frac{\Gamma(b+k+n)\Gamma(b)}{\Gamma(b+k)\Gamma(b+n)}=\frac{(b+n)_k}{(b)_k} \qquad\textrm{and}\qquad nz^n=z\frac{d}{dz}z^n, $$ one can easily prove that $$ \ff(b+k;b;z)=\frac{\left(b+z\frac{d}{dz}\right)_k}{(b)_k}e^z,$$

by being somewhat liberal with the meaning of the Pochhammer symbol. This is clearly the desired polynomial-times-exponential, and provides an explicit expression for the polynomial that looks kind of like a Rodrigues formula.

Even better, if you put this together with Kummer's first transformation, $$\ff\left(a;b;z\right)=e^{z}\ff\left(b-a;b;-z\right),$$ and the expression for Laguerre polynomials in terms of hypergeometric functions, $L^{(\alpha)}_{n}\left(x\right)=\frac{\left(\alpha+1\right)_{n}}{n!}\ff\left(-n,\alpha+1,x\right)$, you get an analogous result for Laguerre polynomials,

$$ L^{(b-1)}_{k}\left(x\right)=\frac{1+k/b}{k!}e^z\left(b+z\frac{d}{dz}\right)_ke^{-z}. $$

Are these results familiar to anyone? Do they fit inside a larger framework? They are not the best thing since sliced bread but they do have a nice simplicity to them, and particularly I would like to cite the appropriate reference if they have appeared before.

Answer 1

Formally using the inverse Mellin transform for x>0:

$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$

$$=e^x \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$

Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$

then

$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x)$$

where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.

For an elaboration, see the notes The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.

See also Rodrigues-like formula (attributed to Poole, 1936) on pg. 59 of Bateman's (et al.) Higher Transcendental Functions Vol. I:

$$(x\tfrac{d}{dx}+\alpha)_{n} h(x) = x^{1-\alpha}D^{n}[x^{n+\alpha-1}h(x)],$$

with $D=\tfrac{d}{dx}$, leading to

$$e^x \binom{x\tfrac{d}{dx}+\alpha+n}{n}e^{-x}=e^x x^{-\alpha}\tfrac{D^{n}}{n!}[x^{n+\alpha}e^{-x}]=L_{n}^{\alpha}(x).$$

This can be generalized by using the fractional integro-derivative representation of $K(a,b,x)$ (see Eqn. 13.2.1 on pg. 505 of Abramowitz and Stegun):

$$K(a,b,x)= e^x \tfrac{(b-1)!}{x^{b-1}}\int_{0}^{x} e^{-t}\tfrac{(x-t)^{a-1}}{(a-1)!} \tfrac{t^{b-a-1}}{(b-a-1)!} dt=e^x \tfrac{(b-1)!}{x^{b-1}}D^{-a}[e^{-x}\tfrac{x^{b-a-1}}{(b-a-1)!}],$$

leading to

$$e^x {x^{-\alpha}}\tfrac{D^{\beta}}{\beta!}[x^{\beta+\alpha}e^{-x}]=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x).$$

Edit (Sept. 2019):

Abstracting, (as Jeans said) "leaving the operators hungry for something to differentiate":

$$\binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}={x^{-\alpha}}\frac{:Dx:^{\beta}}{\beta!}x^{\alpha}=L_{\beta}^{\alpha}(-:xD:)$$

$$=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,-:xD:),$$

where, by definition, for any two operators $:AB:^{\beta}=A^{\beta}B^{\beta}$.