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Mathematical Foundations of Software Development: by Hartmut Ehrig (auth.), Hartmut Ehrig, Christiane Floyd,

By Hartmut Ehrig (auth.), Hartmut Ehrig, Christiane Floyd, Maurice Nivat, James Thatcher (eds.)

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Extra info for Mathematical Foundations of Software Development: Proceedings of the International Joint Conference on Theory and Practice of Software Development (TAPSOFT) Berlin, March 25–29, 1985 Volume 1: Colloquium on Trees in Algebra and Programming (CAAP' 85)

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D|n where |f (a)| ≤ cF ak/2 , where ζ (s) is the Riemann zeta function. [Hint: Show first that the function F(z) − f(0)(2ζ (k))−1 Ek (z) is a cusp form. 35. 20. 36. Show that the function ∆ (z) = ((2ζ (4))−1 E4 (z))3 − ((2ζ (6))−1 E6 (z))2 is a nonzero cusp form of the space M12 (Γ1 ) and that ∆ (z) and E12 (z) span the space. ] Petersson Scalar Product. Every space Nk (K, χ ) of cusp forms of an integral weight k and a congruence character χ for a congruence subgroup K of the symplectic group can be endowed with the structure of a Hilbert space by means of the scalar product.

By substituting F(i/qy) = q(iy)k F ∗ (iy) in the first integral, we can rewrite the last relation in the form Φ(s; F) = ik q1−s ∞ q−1/2 F ∗ (iy)yk−s−1 dy + ∞ q−1/2 F(iy)ys−1 dy. 24). 23) both converge absolutely and uniformly on any compact subset of the s-plane. This proves part (1). 21). Finally, if, say, F ∈ N− (q, χ ), then it satisfies (−i)k q−k/2+1 F ∗ = −F, and so the functional equation turns into the equation Φ(k − s; F) = ik qs−k+1 (−1)ik qk/2−1 Φ(s; F) = −(−1)k qs−k/2 Φ(s; F). 2. Let F ∈ Mk (q, χ ) be an arbitrary modular form with Fourier expansion ∞ F(z) = f(0) + ∑ f(m)e2π imz .

Holomorphic on Hn−1 , and is bounded on each subset Hn−1 ε The next lemma gives useful characterizations of cusp forms in terms of Siegel operators. 29. Let K be a congruence subgroup of Sn and χ a congruence character of K and let F ∈ Mk (K, χ ). Then the following three conditions are equivalent: (1) The function F is a cusp form; (2) F satisfies (F|k M)|Φ = 0 for all M ∈ Gn (Q) = Gn where Φ is the Siegel operator; (3) F satisfies (F|k γ )|Φ = 0 Q2n 2n , for all γ ∈ Γn . Proof. 54) that condition 1 implies condition 2.

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