By Benz W.

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4 Continuous Hahn Definition. pn (x; a, b, c, d) = in (a + c)n (a + d)n 3 F2 n! −n, n + a + b + c + d − 1, a + ix 1 . 1) Orthogonality. If Re(a, b, c, d) > 0, c = a ¯ and d = ¯b, then ∞ 1 2π Γ(a + ix)Γ(b + ix)Γ(c − ix)Γ(d − ix)pm (x; a, b, c, d)pn (x; a, b, c, d)dx −∞ Γ(n + a + c)Γ(n + a + d)Γ(n + b + c)Γ(n + b + d) = δmn . (2n + a + b + c + d − 1)Γ(n + a + b + c + d − 1)n! 2) Recurrence relation. (a + ix)˜ pn (x) = An p˜n+1 (x) − (An + Cn ) p˜n (x) + Cn p˜n−1 (x), where p˜n (x) := p˜n (x; a, b, c, d) = and n!

N }. β+N −x N −x References. [13], [31], [32], [39], [43], [50], [64], [67], [69], [123], [127], [130], [136], [142], [143], [181], [183], [212], [215], [251], [271], [274], [286], [287], [290], [294], [295], [296], [298], [301], [307], [323], [336], [338], [339], [344], [366], [385], [386], [399], [402], [407]. 6 Dual Hahn Definition. Rn (λ(x); γ, δ, N ) = 3 F2 −n, −x, x + γ + δ + 1 1 , n = 0, 1, 2, . . 1) where λ(x) = x(x + γ + δ + 1). Orthogonality. For γ > −1 and δ > −1 or for γ < −N and δ < −N we have N (2x + γ + δ + 1)(γ + 1)x (−N )x N !

14) Remark. 1) : Rn (λ(−a + ix); a + b − 1, c + d − 1, a + d − 1, a − d) Wn (x2 ; a, b, c, d) ˜ n (x2 ; a, b, c, d) = . =W (a + b)n (a + c)n (a + d)n References. [43], [62], [64], [67], [69], [145], [274], [297], [301], [323], [331], [341], [343], [399]. 3 Continuous dual Hahn Definition. Sn (x2 ; a, b, c) = 3 F2 (a + b)n (a + c)n −n, a + ix, a − ix 1 . 1) Orthogonality. If a,b and c are positive except possibly for a pair of complex conjugates with positive real parts, then ∞ 1 2π Γ(a + ix)Γ(b + ix)Γ(c + ix) Γ(2ix) 2 Sm (x2 ; a, b, c)Sn (x2 ; a, b, c)dx 0 = Γ(n + a + b)Γ(n + a + c)Γ(n + b + c)n!

### A Beckman-Quarles type theorem for finite desarguesian planes by Benz W.

by Paul

4.5

Categories: Geometry And Topology