Convert 2F1 to polynomial - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T11:48:08Z http://mathoverflow.net/feeds/question/90608 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90608/convert-2f1-to-polynomial Convert 2F1 to polynomial Remy 2012-03-08T18:42:46Z 2012-03-09T11:38:21Z <p>Is there any transformation to convert each of the following versions of ${}_2F_1$ to a polynomial? </p> <p>The first one is $${}_2F_1\left(\frac{1-a}{2}, -\frac{a}{2}; b;\frac{4z}{(1+z)^2} \right), \quad a\in\mathbb{R},\quad b\in\mathbb{Z},\quad b\ge 0,\quad z\in\mathbb{R}$$</p> <p>The second one is</p> <p>$${}_2F_1\left(\frac{b-a-1}{2}, \frac{b-a}{2}; b+1;\frac{4z}{(1+z)^2} \right)$$</p> <p>I checked the transformations reported in Mizan Rahman's paper (Quadratic Transformation Formulas for Basic Hypergeometric Series), but couldn't find a method. </p> <p>Further Explanation: The type of polynomial I am looking for is not an orthogonal polynomial. Instead I am looking for transformations such as</p> <p>$${}_2F_1\left(\frac{c}{2},\frac{c+1}{2};c;\frac{4z}{(1+z)^2}\right)=(1-z)^{-1}(1+z)^c{}_2F_1\left(0,1;c;\frac{z}{z-1}\right)$$ $$\quad\quad\quad\quad \quad=(1-z)^{-1}(1+z)^c$$</p>