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Subsections

1 Introduction

1.1 Notations

1.2 About the Gamma function

Define, for $ n\in \mathbb{N}\setminus 0 $ and $ a>0 $, $ \Gamma \left( n+a\right) =\left( a+n-1\right) \, \cdots \, \left( a+1\right) \, a\, \Gamma \left( a\right) $, and you obtain :

$\displaystyle \Gamma \left( n+\frac{1}{2}\right) =\frac{1}{2^{2n}}\Gamma \left(...
...ac{1}{2}\right) \frac{\Gamma \left( 2\, n+1\right) }{\Gamma \left( n+1\right) }$ (1)

Define $ \Gamma $ in a more complicated way, and you obtain : $ \Gamma \left( \frac{1}{2}\right) =\sqrt{\pi } $, the validity of EQ. 1 for all $ n>0 $ (integer or not) and therefore ($ n=-1/4 $) $ \Gamma \left( \frac{3}{4}\right) \Gamma \left( \frac{1}{4}\right) =\pi \, \sqrt{2} $

1.3 Revisiting the hypergeom function

$ \mathrm{hypergeom}\left( [1,\, b,\, a,\, -2],\, [3,\, -2,\, 1],\, z\right) =\mathrm{hypergeom}\left( [-2,\, a,\, b],\, [-2,\, 3],\, z\right) $

$ \mathrm{hypergeomV}6\left( [1,\, b,\, a,\, -2],\, [3,\, -2,\, 1],\, z\right) =\mathrm{hypergeomV}6\left( [a,\, b],\, [3],\, z\right) $


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Previous: About Orthogonal Polynomials Up: About Orthogonal Polynomials Next: 2 Orthogonal polynomials over   Contents


douillet@ensait.fr
2002-01-30