Pochhammer symbol of a differential, and hypergeometric polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T16:22:47Z http://mathoverflow.net/feeds/question/107159 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107159/pochhammer-symbol-of-a-differential-and-hypergeometric-polynomials Pochhammer symbol of a differential, and hypergeometric polynomials Emilio Pisanty 2012-09-14T09:36:06Z 2012-09-19T02:23:40Z <p>I have a minor result which I'm sure has come up somewhere before but I can't seem to find it.</p> <p>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,</p> <p>$$\ff(b+k;b;z) =\sum_{n=0}^\infty\frac{(b+k)_n}{(b)_n}\frac{z^n}{n!}$$</p> <p>where $(b)_k=b(b+1)\cdots(b+k-1)$ is the Pochhammer symbol. By exploiting the identities $$\frac{(b+n)_k}{(b)_k}=\frac{\Gamma(b+k+n)\Gamma(b)}{\Gamma(b+k)\Gamma(b+n)}=\frac{(b+n)_k}{(b)_k} \textrm{ and }nz^n=z\frac{d}{dz}z^n,$$ one can easily prove that $$\ff(b+k;b;z)=\frac{\left(b+z\frac{d}{dz}\right)_k}{(b)_k}e^z,$$</p> <p>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.</p> <p>Even better, if you put this together with <a href="http://dlmf.nist.gov/13.2.E39" rel="nofollow">Kummer's first transformation</a>, $$\ff\left(a;b;z\right)=e^{z}\ff\left(b-a;b;-z\right),$$ and the <a href="http://dlmf.nist.gov/18.11.E2" rel="nofollow">expression for Laguerre polynomials in terms of hypergeometric functions</a>, $L^{(\alpha)}<em>{n}\left(x\right)=\frac{\left(\alpha+1\right)</em>{n}}{n!}\ff\left(-n,\alpha+1,x\right)$, you get an analogous result for Laguerre polynomials,</p> <p>$$L^{(b-1)}_{k}\left(x\right)=\frac{1+k/b}{k!}e^z\left(b+z\frac{d}{dz}\right)_ke^{-z}.$$</p> <p>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.</p> http://mathoverflow.net/questions/107159/pochhammer-symbol-of-a-differential-and-hypergeometric-polynomials/107191#107191 Answer by Tom Copeland for Pochhammer symbol of a differential, and hypergeometric polynomials Tom Copeland 2012-09-14T15:40:57Z 2012-09-19T02:23:40Z <p>Formally using the inverse Mellin transform for x>0:</p> <p>$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$</p> <p>$$=e^x \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$ </p> <p>Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$</p> <p>then</p> <p>$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x)$$</p> <p>where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.</p> <p>For an elaboration, see the notes <a href="http://tcjpn.wordpress.com/2011/11/16/a-generalized-dobinski-relation-and-the-confluent-hypergeometric-fcts/" rel="nofollow">The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions</a>.</p> <p>See also Rodriguez-like formula on pg. 59 of Bateman's (et al.) <a href="http://apps.nrbook.com/bateman/Vol1.pdf" rel="nofollow">Higher Transcendental Functions Vol. I</a>:</p> <p>$$(x\tfrac{d}{dx}+\alpha)_{n} h(x) = x^{1-\alpha}D^{n}[x^{n+\alpha-1}h(x)],$$</p> <p>with $D=\tfrac{d}{dx}$, leading to</p> <p>$$e^x \binom{x\tfrac{d}{dx}+\alpha+n}{n}e^{-x}=e^x x^{-\alpha}\tfrac{D^{n}}{n!}[x^{n+\alpha}e^{-x}]=L_{n}^{\alpha}(x).$$</p> <p>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 <a href="http://people.math.sfu.ca/~cbm/aands/page_505.htm" rel="nofollow">Abramowitz and Stegun</a>):</p> <p>$$K(a,b,x)= e^x \tfrac{(b-1)!}{x^{b-1}}\int_{0}^{x} e^{-t}\tfrac{(x-t)^{a-1}}{(a-1)!} \tfrac{t^{b-a-1}}{(b-a-1)!} dt=e^x \tfrac{(b-1)!}{x^{b-1}}D^{-a}[e^{-x}\tfrac{x^{b-a-1}}{(b-a-1)!}],$$</p> <p>leading to</p> <p>$$e^x {x^{-\alpha}}\tfrac{D^{\beta}}{\beta!}[x^{\beta+\alpha}e^{-x}]=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x).$$</p>