Automorphic Forms, Representations, and L-Functions by W Casselman, Armand Borel, W. Casselman

By W Casselman, Armand Borel, W. Casselman

This was once the convention on $L$-functions and automorphic kinds. the 2 volumes are actually classics.

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The Appell functions generalize the hypergeometric function to two variables. , 1953b) ∞ F1 (a; b, b ; c; x, y) = (a)n+m (b)m (b )n m n x y , (c)m+n m! n! 36) ∞ F2 (a; b, b ; c, c ; x, y) = (a)n+m (b)m (b )n m n x y , (c)m (c )n m! n! 37) ∞ F3 (a, a ; b, b ; c; x, y) = (a)m (a )n (b)m (b )n m n x y , (c)m+n m! n! 38) ∞ F4 (a, b; c, c ; x, y) = (a)n+m (b)m+n m n x y . (c) m (c )n m! n! 41) respectively. Indeed π 2 , 2 F1 1/2, 1/2; 1; k 2 π E(k) = 2 F1 −1/2, 1/2; 1; k2 . 43) We refer to k as the modulus, while the complementary modulus k is k = 1 − k2 1/2 .

Moreover, β(x) ≥ 0, β(x) ≡ 0, on I1 , but β(x) ≤ 0, β(x) ≡ 0 on IN +1 . On Ij , 1 < j ≤ N , β either has a constant sign or changes sign from negative to positive at some point within 30 Orthogonal Polynomials the interval. 8). Thus, [a, b] can be subdivided into at most 2N subintervals where β(x) has a constant sign on each subinterval. 9) to have degree at most 2n − 2 such that p (x)β(x) ≥ 0 on [a, b], which gives a contradiction. Thus we must have at least 2N intervals where β(x) keeps a constant sign.

Clearly, β(x) is nondecreasing on Ij , Vj . Moreover, β(x) ≥ 0, β(x) ≡ 0, on I1 , but β(x) ≤ 0, β(x) ≡ 0 on IN +1 . On Ij , 1 < j ≤ N , β either has a constant sign or changes sign from negative to positive at some point within 30 Orthogonal Polynomials the interval. 8). Thus, [a, b] can be subdivided into at most 2N subintervals where β(x) has a constant sign on each subinterval. 9) to have degree at most 2n − 2 such that p (x)β(x) ≥ 0 on [a, b], which gives a contradiction. Thus we must have at least 2N intervals where β(x) keeps a constant sign.

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