By Roger B. Nelsen
ISBN-10: 0883857219
ISBN-13: 9780883857212
Like its predecessor, Proofs with no phrases, this ebook is a set of images or diagrams that aid the reader see why a specific mathematical assertion should be real, and the way you'll be able to start to cross approximately proving it. whereas in a few proofs with no phrases an equation or might sound to assist advisor that technique, the emphasis is obviously on offering visible clues to stimulate mathematical idea. The proofs during this assortment are prepared through subject into 5 chapters: geometry and algebra; trigonometry, calculus and analytic geometry; inequalities; integer sums; and sequences and sequence. academics will locate that a number of the proofs during this assortment are well matched for lecture room dialogue and for assisting scholars to imagine visually in arithmetic.
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Extra resources for Proofs without words 2. More exercises in visual thinking
Sample text
The survey article [141] provides an excellent introduction to this chain of ideas, and we recommend this to the reader. But the origins of the chain go back to the 1916 paper of Ramanujan which acted as a catalyst for this development. In his epic paper of 1916, Ramanujan [162] considered the function ∞ 1 − qn (z) = q 24 , q = e2πiz . n=1 As we have seen, expanding the right-hand side as a power series in q defines the celebrated Ramanujan τ -function: ∞ 1 − qn q n=1 24 ∞ = τ (n)q n . n=1 In his paper, Ramanujan made three conjectures concerning τ (n): (1) τ (mn) = τ (m)τ (n), for (m, n) = 1, (2) for p prime and a ≥ 1, τ (p a+1 ) = τ (p)τ (p a ) − p 11 τ (p a−1 ), (3) |τ (p)| ≤ 2p 11/2 .
One can renormalize and show that √ an (f ) yKir 2π |n|y e2πinx f (x, y) = a0 (f )y s + a0 (f )y 1−s + n=0 where Kir (y) = 1 2 ∞ −∞ e−y cosh t−irt dt 3 Upper Bound for Fourier Coefficients and Eigenvalue Estimates 47 with λ = 1/4 + r 2 . Maass [117] proved that the series n=0 an (f ) |n|s extends to a meromorphic function for all s ∈ C analytic everywhere except possibly at s = 0 and s = 1, and satisfies a functional equation. The analog of the Ramanujan conjecture in this context is that for any > 0, an (f ) = O(n ).
We now show that this is a contradiction. We do this by showing that some Fourier coefficient of E(z, (1 + it0 )/2) is non-zero. That is, we need to check ∞ e−πry(u+u −1 ) 0 du = 0. u1+it0 If we set u = eθ , we have to show that ∞ −∞ e−πry(e θ +e−θ )−it 0θ dθ = 0. In other words, it suffices to show that ∞ e−πry(e θ +e−θ ) cos t0 θ dθ = 0. 0 This integral is of the form ∞ 0 e−y(a θ +a −θ ) cos θ dθ, a > 1. 54 4 The Ramanujan Conjecture from GL(2) to GL(n) We would like to determine its behaviour as y tends to infinity.
Proofs without words 2. More exercises in visual thinking by Roger B. Nelsen
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