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Analytic Number Theory Notes on a course taught at the University of East Anglia by G R Everest in Spring 1999 July 20, 2020 Dr C Roettger Mathematics Department Iowa State University 463 Carver Hall, 50011 Ames, IA, USA [email protected] https://faculty.sites.iastate.edu/roettger/ 1 Contents 1 Elementary Number Theory and Easy Asymptotics 3 1.1 The Log of the Zeta Function . 7 1.2 Euler's Summation Formula . 11 1.3 Multiplicative arithmetical functions . 15 1.4 Dirichlet Convolution . 20 2 The Riemann Zeta Function 23 2.1 Euler Products . 23 2.2 Uniform Convergence . 25 2.3 The Zeta Function is Analytic . 28 2.4 Analytic continuation of the Zeta Function . 30 3 The Functional Equation 35 3.1 The Gamma Function . 35 3.2 Fourier Analysis . 37 3.3 The Theta Function . 41 3.4 The Gamma Function Revisited . 46 4 Primes in an Arithmetic Progression 54 4.1 Two Elementary Propositions . 54 4.2 A New Method of Proof . 55 4.3 Characters of Finite Abelian Groups . 61 4.4 Dirichlet characters and L-functions . 65 4.5 Analytic continuation of L-functions and Abel's Summation Formula . 69 A Proof of Theorem 3.7 73 B Abel's Limit Theorem 76 2 Figure 1: Level of Difficulty of the Course Riemann zeta function level Riemann hypothesis Easter vac 6 33 elementary lectures Analytic Number Theory 1 Elementary Number Theory and Easy Asymp- totics Recommended text: Tom APOSTOL, "Introduction to Analytic Number Theory", 5th edition, Springer. ISBN 0-387-90163-9. Course content: • Integers, especially prime integers • connections with complex functions How did the subject arise? E. g. Gauss and Legendre did extensive calcula- tions, calculated tables of primes. By looking at these tables, Gauss reckoned that the real function π(x) := #fp ≤ x : p is primeg grows quite regularly, namely π(x) log(x) lim = 1; (1) x!1 x 3 although the individual primes are scattered rather irregularly among the natural numbers. Equation (1) is known as the Great Prime Number The- orem - or just THE Prime Number Theorem. The first proof of it used the developing theory of complex functions. Definition 1.1 In this course, a prime is a positive integer p which admits no divisors except 1 and itself 1. By definition, 1 is not a prime. Theorem 1.2 (Fundamental Theorem of Arithmetic - FTA) Every integer n > 1 can be expressed as a finite product of primes, unique up to order. Example: 6 = 2 · 3 = 3 · 2. From the FTA follows the Corollary 1.3 There must be infinitely many primes. Proof 1 (Euclid): If there are only finitely many primes, we can list them p1; : : : ; pr. Define N := p1 · ::: · pr + 1: By FTA, N can be factorized, so it must be divisible by some prime pk of our list. Since pk also divides p1 · ::: · pr, it must divide 1 - an absurdity. 2 Proof 2 (Euler): If there are only finitely many primes p1; : : : pr, consider the product r −1 Y 1 X := 1 − : pk k=1 Note that the product is well defined, since 1 is not a prime and since, by hypothesis, there are only finitely many primes. Now expand each factor into a geometric series: 1 1 1 1 1 = 1 + + 2 + 3 + ::: 1 − p p p p 1In Ring theory, primes are defined in a different way, i. e. an element x of a commu- tative ring R, not a unit, is prime in R iff for all products in R divisible by x, at least one factor must be divisible by x. For positive integers, this amounts to the same as our definition. 4 Put this into the equation for X: 1 1 1 1 1 1 X = 1 + + + + ::: · 1 + + + + ::: 2 22 23 3 32 33 1 1 1 1 1 1 · 1 + + 2 + 3 + ::: ··· 1 + + 2 + 3 + ::: 5 5 5 pr pr pr 1 1 1 1 = 1 + + + + + ::: 2 3 4 5 X 1 = ; n n the harmonic series! But this series diverges, and again we have reached an absurdity. 2 Why exactly does the harmonic series diverge??? Proof 1 (typical year One proof): 1 1 1 1 1 1 1 1 + + + + + + + + ::: 2 3 4 5 6 7 8 1 1 1 1 1 1 1 > 1 + + + + + + + + ::: 2 4 4 8 8 8 8 1 2 4 = 1 + + + + ::: 2 4 8 1 1 1 = 1 + + + + ::: 2 2 2 which cleary diverges. 2 Exercise: Try to prove that P 1=n2 diverges using the same technique. Of course, this will not work, since this series converges, but you will see some- thing "mildly" interesting. PN 1 R N 1 Proof 2: Compare n=1 n with the integral 1 x dx = log(N). The result is: N X 1 log(N) ≤ ≤ log(N) + 1: (2) n n=1 The first inequality in (2) is enough to show the divergence of the harmonic series. But together with the second, we are given information about the 5 Figure 2: The harmonic series as integral of a step function speed of the divergence! Proof 2 is a typical example of a proof leading for- ward. 2 6 26.1.99 1.1 The Log of the Zeta Function Definition 1.4 Given two functions f : R ! C and g : R ! R+, we write f = O(g) as n ! 1 if there exist constants C and x0 such that jf(x)j ≤ Cg(x) for all x ≥ x0. This is used to isolate the dominant term in a complicated expression. Examples: • x3 + 501x = O(x3) • Any bounded function is O(1), e. g. sin(x) = O(1) • Can have complex f: eix = O(1) for real x. PN 1 Thus we can write 1 n = log(N) + O(1). You may also see f = o(g). This jfj means g ! 0 as x tends to infinity. The Prime Number Theorem can be written x x π(x) = + o or log(x) log(x) π(x) log(x) = 1 + o(1): x Theorem 1.5 P 1 The series p p diverges. Proof 1 (see Apostol p. 18): By absurdity. Assume that the series converges, i. e. X 1 < 1: p p P 1 1 Q So there is some N such that p>N p < 2 . Let Q := p≤N p. The numbers 1 + nQ, n 2 N, are never divisible by primes less than N (because those divide Q). Now consider 1 !t X X 1 X 1 < = 1: p 2t t=1 p>N t 7 We claim that 1 1 !t X 1 X X 1 ≤ (3) 1 + nQ p n=1 t=1 p>N because every term of the l.h.s. appears on the right at least once (convince yourself of this claim by assuming e. g. N = 11 and find some terms in the r.h.s.!) But the series on the l.h.s. of (3) diverges! This follows from the 'limit comparison test': an P If for two real sequences an and bn holds ! L 6= 0, then an P bn n converges iff n bn does. 1 1 Apply this test with an = 1+nQ , bn = n and L = 1=Q. This absurdity proves the theorem. 2 Proof 2: We will show that X 1 > log log(N) − 1: (4) p p≤N Remark 1.6 In these notes, we will always define N = f1; 2; 3;:::g (excluding 0). If 0 is to be included, we will write N0. Let N := fn 2 N : all prime factors of n are less than Ng Then X 1 Y 1 1 1 = 1 + + + + ::: n p p2 p3 n2N p≤N ! Y 1 = (5) 1 − 1 p≤N p If n ≤ N, then n 2 N , therefore X 1 X 1 ≤ : n n n≤N n2N But log(N) is less than the l.h.s., so −1 X 1 Y 1 log(N) ≤ = 1 − : (6) n p n2N p≤N 8 Lemma 1.7 For all v 2 [0; 1=2] holds 1 2 ≤ ev+v : 1 − v 1 Apply this with v = p (note primes are at least 2, so the lemma does apply): −1 Y 1 Y 1 1 1 − ≤ exp + : (7) p p p2 p≤N p≤N Now combine this with equation (6) and take logs: X 1 1 log log(N) ≤ + : (8) p p2 p≤N Finally, we observe that 1 X 1 X 1 π2 < = − 1 < 1 p2 n2 6 p n=2 Proof of lemma 1.7: Put f(v) := (1 − v) exp(v + v2). We claim 1 ≤ f(v) for all v 2 [0; 1=2]. To prove this, we check f(0) = 1 and f0(v) = v(1 − 2v) exp(v + v2) which is nonnegative for v 2 [0; 1=2]. 2 This completes Proof 2 of theorem 1.5. 2 We will prove later X 1 1 = log log(N) + A + O ; (9) p log(N) p≤N where A is a constant. Question: Is it possible to prove (9) with O(1) in place of A + O(:::) using only methods of Proof 2? 9 P 1 29.1.99 Third proof that p p diverges: The following identity holds for all σ > 1: X 1 log(ζ(σ)) = − log 1 − (10) pσ p 1 1 X X −1 X 1 X X 1 = − = + mpmσ pσ mpmσ p m=1 p p m=2 We will prove the identity (10) later. Note however, that the series involved converge absolutely. We claim: The last double sum on the right-hand side is bounded.
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