New PDF release: Elementary Number Theory, Group Theory and Ramanujan Graphs

By Giuliana Davidoff, Peter Sarnak, Alain Valette

ISBN-10: 0521531438

ISBN-13: 9780521531436

This article is a self-contained examine of expander graphs, particularly, their particular building. Expander graphs are hugely hooked up yet sparse, and whereas being of curiosity inside of combinatorics and graph concept, they could even be utilized to laptop technological know-how and engineering. just a wisdom of effortless algebra, research and combinatorics is needed as the authors give you the invaluable historical past from graph conception, quantity concept, staff conception and illustration conception. hence the textual content can be utilized as a short advent to those matters and their synthesis in sleek arithmetic.

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Corollary. Let (X m )m≥1 be a family of connected, k-regular, finite graphs, with g(X m ) → ∞ as m → ∞. For every ε > 0, there exists a constant C => 0, √ such that the number of eigenvalues of X m in the interval [−k, (−2 + ε) k − 1] is at least C |X m |. Proof. 8. 10. Let f be the function k k which is 1 on − √k−1 , −2 , 0 on −2 + ε, √k−1 , and interpolates linearly between 1 and 0 on [−2, −2 + ε]. Then, for every m ≥ 1, k νm − √ , −2 + ε ≥ k−1 √k k−1 k − √k−1 f (x) dνm (x). 10, this gives lim inf νm − √ m→∞ k k−1 , (−2 + ε) ≥ 2 f (x) dν(x).

Let µ be the number of negative minimal residues of the sequence q, 2q, . . , p−1 · q. Then S( p, q) has the same parity as µ. 3. Quadratic Reciprocity 51 To see this, for k = 1, . . , p−1 , write kq = p kqp + u k , with u k ∈ 2 {1, . . , p − 1}; note that u k is nothing but the remainder in the Euclidean division of kq by p. If u k < 2p , then u k is the minimal residue of qk, so that u k = ri for exactly one i; if u k > 2p , then u k − p is the minimal residue of kq, so that u k − p = −r j for a unique j.

Z [i] is prime if and only if, whenever π divides a product αβ (α, β ∈ Z [i]), it divides either α or β. Proof. (⇒) If π divides αβ, we have αβ = π σ for some σ ∈ Z [i]. We may assume that π does not divide α and must then show that π divides β. Consider (π, α): since it divides π , which is prime, we must have (π, α) = 1. Then, by our previous result, 1 = π γ + αδ 42 Number Theory for some γ , δ ∈ Z [i]. Then β = πβγ + αβδ = πβγ + π σ δ = π(βγ + σ δ), showing that π divides β. (⇐) If π = αβ, in particular π divides αβ.

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Elementary Number Theory, Group Theory and Ramanujan Graphs by Giuliana Davidoff, Peter Sarnak, Alain Valette

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