” Superior Mathematics from an Elementary point of view ” course notes Jacopo d’Aurizio To cite this version: Jacopo d’Aurizio. ” Superior Mathematics from an Elementary point of view ” course notes. Master. Italy. 2017. cel-01813978 HAL Id: cel-01813978 https://cel.archives-ouvertes.fr/cel-01813978 Submitted on 12 Jun 2018 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. \Superior Mathematics from an Elementary point of view" course notes Undergraduate course, 2017-2018, University of Pisa Jack D'Aurizio Contents 0 Introduction 2 1 Creative Telescoping and DFT 3 2 Convolutions and ballot problems 15 3 Chebyshev and Legendre polynomials 30 4 The glory of Fourier, Laplace, Feynman and Frullani 40 5 The Basel problem 60 6 Special functions and special products 70 7 The Cauchy-Schwarz inequality and beyond 97 8 Remarkable results in Linear Algebra 121 9 The Fundamental Theorem of Algebra 125 10 Quantitative forms of the Weierstrass approximation Theorem 133 11 Elliptic integrals and the AGM 137 12 Dilworth, Erdos-Szekeres, Brouwer and Borsuk-Ulam's Theorems 147 13 Continued fractions and elements of Diophantine Approximation 158 14 Symmetric functions and elements of Analytic Combinatorics 173 15 Spherical Trigonometry 183 0 Introduction This course has been designed to serve University students of the first and second year of Mathematics. The purpose of these notes is to give elements of both strategy and tactics in problem solving, by explaining ideas and techniques willing to be elementary and powerful at the same time. We will not focus on a single subject among Calculus, Algebra, Combinatorics or Geometry: we will just try to enlarge the \toolbox" of any professional mathematician wannabe, by starting from humble requirements: • knowledge of number sets (N; Z; Q; R; C) and their properties; • knowledge of mathematical terminology and notation; • knowledge of the main mathematical functions; • knowledge of the following concepts: limits, convergence, derivatives, Riemann integral; 1 CREATIVE TELESCOPING AND DFT • knowledge of basic Combinatorics and Arithmetics. A collection of problems in Analysis and Advanced integration techniques, kindly provided by Tolaso J. Kos and Zaid Alyafeai, are excellent sources of exercises to match with the study of these notes. These notes are distributed under a Creative Commons Share-Alike (CCSA) license. Personal use is allowed, distribution is allowed with the only contraint of making a proper mention of the author (Jack D'Aurizio). Commercial use, modification or inclusion in other works are not allowed. 1 Creative Telescoping and DFT It is soon evident, during the study of Mathematics, that the bijectivity of some function f does not grant that the explicit computation of f −1(y) is \just as easy" as the explicit computation of f(x). Some examples are related with the ease of multiplication, against the hardness of factorization; the possibility of computing derivatives in a algorithmic fashion, against the lack of a completely algorithmic way to find indefinite integrals; the determination of a Galois group of an irreducible polynomial over Q, against the difficult task of finding a polynomial having a given Galois group. In the present section we will outline two interesting techniques for solving (or getting arbitrarily close to an actual \solution") a peculiar inverse problem, that is the computation of series. Definition 1. We give the adjective telescopic to objects of the form a1 + a2 + ::: + an, where each ai can be written as bi − bi+1 for some sequence b1; b2; : : : ; bn; bn+1. With such assumption we have: a1 + a2 + ::: + an = (b1 − b2) + (b2 − b3) + ::: + (bn − bn+1) = b1 − bn+1: Essentially, every telescopic sum is simple to compute, just as any convergent series with terms of the form bi − bi+1. A peculiar example is provided by Mengoli series: the identity N N X 1 X 1 1 1 = − = 1 − n(n + 1) n n + 1 N + 1 n=1 n=1 grants we have X 1 = 1: n(n + 1) n≥1 The following generalization is also straightforward: Lemma 2. X 1 1 8k 2 +; = : N n(n + 1) · ::: · (n + k) k · k! n≥1 In a forthcoming section we will also see a proof not relying on telescoping, but on properties of Euler's Beta function. The first technical issue we meet in this framework is related to the fact that recognizing a contribution of the form bi − bi+1 in the general term of a series is not always easy, just like in the continuous analogue: if the task is to R b 0 find a f(x) dx, it is not always easy to devise a function g such that f(x) = g (x). Here there are some non-trivial examples: Lemma 3. X 1 π arctan = : 1 + n + n2 4 n≥1 Page 3 / 195 Proof. If we use \backwards" the sum/subtraction formulas for the tangent function, we have that tan(x) ± tan(y) tan(x ± y) = 1 ∓ tan(x) tan(y) implies: 1 1 ! 1 1 n − n+1 1 arctan − arctan = arctan 1 = arctan 2 n n + 1 1 + n(n+1) 1 + n + n π so the given series is telescopic and it converges to arctan(1) = 4 . Exercise 4. The sequence of Fibonacci numbers fFngn≥0 is defined through F0 = 0;F1 = 1 and Fn+2 = Fn+1 + Fn for any n ≥ 0. Show that: X (−1)n+1 p arctan = arctan( 5 − 2): Fn+1(Fn + Fn+2) n≥1 Exercise 5. Show that: X sinh 1 π X 1 π π arctan = ; arctan = − arctan tanh : cosh(2n) 2 8n2 4 4 n2Z n≥1 Exercise 6. Prove that: N X 1 2n N + 1 2N + 2 = ; 4n n 22N+1 N + 1 n=0 for instance by considering that by De Moivre's formula we have Z π 1 2n 1 2n An = n = cos (x) dx 4 n 2π −π 2n 1 P2n 2n (2n−j)ix −jix R π kix since cos(x) = 4n j=0 j e e and −π e dx = 2π · δ(k), so the only non-vanishing contribution is related to the j = n term, and N X 1 2n 1 Z π 1 − cos2N+2(x) N + 1 2N + 2 = dx = (2N + 2) A = 4n n 2π 2 N+1 22N+1 N + 1 n=0 −π sin (x) R dx follows from the integration by parts formula ( sin2 x = − cot x). Prove also that X 2k 1 = 1 k (k + 1)4k k≥0 by recognizing in the main term a telescopic contribution. Give a probabilistic interpretation to the proved identities, by considering random paths on a infinite grid (Z × Z) where only unit movements towards North or East are allowed. We now outline the first (really) interesting idea, namely creative telescoping: even if we are not able to write the main term of a series in the bi − bi+1 form, it is not unlikely there is an accurate approximation of the main term that can be represented in such a telescopic form. By subtracting the accurate telescopic approximation from the main term, the original problem boils down to computing/approximating a series that is likely to converge faster than the Page 4 / 195 1 CREATIVE TELESCOPING AND DFT original one, and the same approximation-by-telescopic-series trick can be performed again. For instance, we might employ creative telescoping for producing very accurate approximations of the series X 1 ζ(2) = n2 n≥1 that will be the main character of a forthcoming section. 1 1 In particular, for any n > 1 the term n2 is quite close to the telescopic term 2 1 : n − 4 1 4 1 1 = = 2 − 2 1 n − 4 (2n − 1)(2n + 1) 2n − 1 2n + 1 1 1 1 and we have n2 − 2 1 = − (2n−1)n2(2n+1) , so: n − 4 X 1 X 1 X 1 ζ(2) = 1 + = 1 + − n2 n2 − 1 (2n − 1)n2(2n + 1) n≥2 n≥2 4 n≥2 2 X 1 = 1 + − 3 (2n − 1)n2(2n + 1) n≥2 5 2 gives us ζ(2) < 3 (that we will prove to be equivalent to π < 10), and the magenta \residual series" can be manipu- lated in the same fashion (by extracting the first term, approximating the main term with a telescopic contribution, considering the residual series) or simply bounded above by: X 1 X 1 3 22 < = ζ(2) − (2n − 1)n2(2n + 1) (2n − 1) n − 1 n + 1 (2n + 1) 2 9 n≥2 n≥2 2 2 74 74 5 from which the lower bound ζ(2) > 45 follows. We may notice that the difference between 45 and 3 is already pretty small. In this framework the iteration of creative telescoping leads to two interesting consequences: the identity X 1 X 3 ζ(2) = = ; n2 n22n n≥1 n≥1 n providing a remarkable acceleration of the series defining ζ(2), and Stirling's inequality: nn p 1 nn p 1 2πn exp ≤ n! ≤ 2πn exp e 12n + 1 e 12n further details will be disclosed soon. It is also possible to employ creative telescoping for proving that: X 1 5 X (−1)n+1 ζ(3) = = n3 2 n32n n≥1 n≥1 n a key identity in Apery's proof of ζ(3) 62 Q.