most recent 30 from http://mathoverflow.net 2013-05-18T12:46:13Z http://mathoverflow.net/feeds/question/99379 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99379/product-of-hypergeometric-functions Joe Lucke 2012-06-12T17:48:24Z 2012-06-14T14:45:04Z

<p>I am looking for the product of Gaussian hypergeometric functions of the form $_2F_1(a,b;c;\lambda z) _2F_1(d,e;f;\phi z)$. I would like, but do not expect, an answer of the form $_pF_q(\mathbf{a},\mathbf{b};(\lambda*\phi) z)$, where $\mathbf{a}$ is the numerator $p$-vector and \mathbf{b} is the denominator $q$-vector for the $(p,q)$-hypergeometric function. I have not found relevant answers in the typical references, e.g., Slater. </p>

http://mathoverflow.net/questions/99379/product-of-hypergeometric-functions/99383#99383 Igor Rivin 2012-06-12T19:01:07Z 2012-06-12T19:01:07Z

<p>For results of this sort, see Driver and Love, Products of hypergeometric functions and the zeros of 4F3 polynomials, and references therein (Numerical algorithms, 2001)</p>

http://mathoverflow.net/questions/99379/product-of-hypergeometric-functions/99612#99612 Joe Lucke 2012-06-14T14:45:04Z 2012-06-14T14:45:04Z

<p>I slightly misrepresented my problem. I wanted the product of ${}_2F_1(a,b;c;λz)\times {}_2F_1(d,e;c;ϕz)$. Note that the third argument $c$ is identical in the two functions and that the two final arguments are proportional $λz$ and $\phi z$. There appears to be no general solution to this product, although there are many restricted solutions. I had hoped my restrictions might have a solution, but apparently after reading Bailey's, Watson's, Preece's and Driver and Love's manuscripts, this is not to be. Srivastava provides a solution in terms of a double hypergeometric function, but I was hoping for something simpler. My posting contained a dreamed-up form I would have liked. Ignore it. Sorry. Thanks for the Driver and Love reference.</p>