Analytic Number Theory Andrew Kobin Fall 2013 Contents Contents Contents 0 Introduction 1 1 Preliminaries 2 1.1 Basic Number Theory . .2 1.2 Basic Analysis . .3 1.3 Euler-Maclaurin Summation . .4 1.4 The Bernoulli Numbers . 13 2 Euler's Work 15 2.1 On the Sums of Series of Reciprocals . 15 2.2 Newton's Identities . 20 2.3 Euler's Product Form . 22 2.4 The Prime Number Theorem . 31 3 Complex Analysis 35 3.1 Arithmetic . 35 3.2 Functions and Limits . 37 3.3 Line Integrals . 44 3.4 Differentiability . 46 3.5 Integration in the Complex Plane . 52 3.6 Singularities and the Residue Theorem . 57 4 The Zeta Function 62 4.1 The Functional Equation . 62 4.2 Finding the Zeros . 68 4.3 Sketch of the Prime Number Theorem . 69 4.4 Generalized Riemann Hypothesis . 71 i 0 Introduction 0 Introduction These notes were compiled from a semester of lectures at Wake Forest University by Dr. 1 X 1 John Webb. The primary focus is the Riemann Zeta Function: ζ(s) = ns n=1 1 X 1 Example 0.0.1. ζ(1) = , the harmonic series, is a divergent series. n n=1 1 X 1 π2 Example 0.0.2. We know ζ(2) = converges to , but how? n2 6 n=1 Euler's Results π2 Proved that ζ(2) = 6 900+ papers over his lifetime { Some 15 of them dealt with the Zeta Function Riemann wrote one paper on the topic { Revolutionized analytic number theory x Prime Number Theorem: π(x) ≈ , where π(x) is the number of primes less than log(x) or equal to x. Riemann gave a map of how to prove this theorem. Modern Research Riemann Hypothesis L-functions Calculating zeroes of the Zeta Function 1 1 Preliminaries 1 Preliminaries 1.1 Basic Number Theory Z is the integers f0; ±1; ±2;:::g. Definition. For an integer n, we say d is a divisor of n, d j n, if there exists an integer c such that dc = n. Properties (1) n j n for all n 2 Z (2) If d j n and n j m then d j m (3) If d j n and d j m then d j (xn + ym) (4) If d j n then ad j an for all a 2 Z (5) If ad j an then d j n if a 6= 0 (6)1 j n for all n 2 Z (7) n j 0 for all n 2 Z (8) If 0 j n then n = 0 (9) If d j n and n 6= 0 then jdj ≤ jnj (10) If d j n and n j d then jdj = jnj Definition. Given n; m 2 Z, we say that d ≥ 1 is the greatest common divisor of n and m if d j n, d j m, and for all e that divides n and m, jej ≤ d. Notes: Given another divisor of n; m, say e, then e j gcd(n; m) Given gcd(n; m), there exist integers x and y such that xn + ym = gcd(n; m) f(xn + ym) j x; y 2 Zg = fk · gcd(n; m) j k 2 Zg Definition. An integer p ≥ 2 is prime if its only positive divisors are 1 and p. Definition. An integer n > 1 is composite if it is not prime. Note: 1 is not prime. Theorem 1.1.1 (Fundamental Theorem of Arithmetic). Any integer n, jnj > 1, can be uniquely written as the product of powers of distinct primes e1 e2 er n = p1 p2 ··· pr ; where pi is prime and ei 2 Z for all i. Example 1.1.2. 2013 = 3 · 11 · 61. 2 1.2 Basic Analysis 1 Preliminaries 1.2 Basic Analysis Definition. If f is a function, f(x) converges to L 2 , denoted lim f(x) = L, if for all R x!1 " > 0 there exists N > 0 such that for all x > N, jf(x) − Lj < ". Definition. A function f(x) diverges (to 1), denoted lim f(x) = 1, if for all M > 0 x!1 there exists some N > 0 such that for all x > N, f(x) > M. This can be adapted for −∞ as well. Definition. Given functions f(x) and g(x) defined on R (or Z) ≥ a, with g(x) > 0 and monotonic on [a; 1), we say that f(x) = O(g(x)) if for all x ≥ a there exists some constant M > 0 such that jf(x)j ≤ Mg(x), also denoted f(x) << g(x). Definition. Given functions f and g, f(x) >> g(x) if there exists some constant m > 0 such that jf(x)j ≥ mg(x) for all x > a. Definition. If f(x) >> g(x) and f(x) << g(x) then f and g are said to have the same order, denoted f(x) g(x). X 1 Example 1.2.1. = O(log(x)) p p prime p≤x Proof. What if we sum over all integers?: x X 1 n n=1 x x 1 X 1 X 1 X Z 1 We know ≥ . Recall the integral test: f(n) and f(x) dx both converge n p n=1 p=2 n=1 1 or both diverge (if f > 0 and f is monotone on [1; 1)). So we have x X 1 Z x 1 ≤ dt = log(x): n t n=2 1 In fact, x X 1 log(x) < < 2 log(x): n n=1 x x x X 1 X 1 X 1 Therefore ≤ < 2 log(x) for all x > 2. Hence = O(log(x)). p n p p=2 n=1 p=2 x x X 1 X 1 But is way bigger than so this is a bad approximation tool. n p n=1 p=2 3 1.3 Euler-Maclaurin Summation 1 Preliminaries Example 1.2.2. sin(x) << 1 Let g(x) = 1, then j sin(x)j ≤ g(x) for all x 2 R. So sin(x) << 1. But is sin(x) 1? No, since sin(x) = 0 at infinitely many points. Example 1.2.3. f(x) = sin(x) + x First, f(x) << x because if M = 2, jf(x)j ≤ 2x for x > 0. And f(x) >> x because if m = 1=2, jf(x)j ≥ 1=2x for x > 0. Thus f(x) x. Definition. Two functions f and g are asymptotic to each other, denoted f(x) ∼ g(x), if f(x) lim = 1: x!1 g(x) Example 1.2.4. f(x) = sin(x) + x, g(x) = x f(x) sin(x) + x sin(x) lim = lim = lim + 1 = 0 + 1 = 1 x!1 g(x) x!1 x x!1 x thus f(x) ∼ g(x). Proposition 1.2.5. If f(x) ∼ g(x) then f(x) g(x). Proof omitted. Note that the converse is not true in general. x X 1 Z x 1 Example 1.2.6. The integral test actually states that ∼ dt = log(x). n t n=1 1 1.3 Euler-Maclaurin Summation n−1 X Z n Let f(x) > 0 and strictly decreasing on [1; 1). Examine dn = f(k) − f(x) dx. k=1 1 f(x) 4 3 2 dn 1 1 2 4 6 n8 4 1.3 Euler-Maclaurin Summation 1 Preliminaries Proposition 1.3.1. dn < f(1) for any n > 1. n−1 X Z k+1 Proof. Rewrite dn = (f(k) − f(x)) dx. Then k=1 k n−1 X dn < (f(k) − f(k + 1)) = f(1) − f(n) k=1 by telescoping series. And since f(n) > 0, f(1) − f(n) < f(1). Hence dn < f(1). Let C(f) = lim dn. We know C(f) exists because dn is increasing but bounded. Then n!1 we can write n X Z k+1 C(f) = lim [f(k) − f(x)] dx: n!1 k=1 k 1 X Z k+1 Let Ef (n) = f(n) + dn − C(f). Then Ef (n) > 0 since dn − C(f) = [f(k) − f(x)] dx. k=n k Together, this gives us 1 X Z 1 f(k) = f(x) dx + C(f) + Ef (n): k=1 1 1 X 1 Goal: Approximate to at least 3 decimal places. n2 n=1 1 Definition. γ = C x is called the Euler constant. 1 Example 1.3.2. Let f(x) = x . Then n X Z n f(k) = f(x) dx + γ + Ef (n) k=1 1 Z n 1 We know that dx = log(n). To approximate the remaining terms, we will first prove 1 x the following theorem for the general case. n X Z n Z n Theorem 1.3.3. f(k) = f(x) dx + (x − bxc) f 0(x) dx + f(1), where f has a con- k=1 1 1 tinuous first derivative on [1; n]. Proof. We begin with n X Z n f(k) = f(x) dx + dn: k=1 1 5 1.3 Euler-Maclaurin Summation 1 Preliminaries To find dn, n−1 X Z k+1 dn = [f(k) − f(x)] dx k=1 k which we will integrate by parts. Let u = f(k) − f(x) dv = dx du = −f 0(x) dx v = x − (k + 1) we get to choose a constant and integrate: Z k+1 k+1 Z k+1 0 [f(k) − f(x)] dx = [f(k) − f(x)](x − (k + 1)) + (x − (k + 1))f (x) dx k k k Z k+1 = [(f(k) − f(k + 1)) · 0 − 0(−1)] + (x − (k + 1))f 0(x) dx k Z k+1 Z k+1 = 0 + (x − (k + 1))f 0(x) dx = (x − (k + 1))f 0(x) dx: k k Thus n−1 n−1 X Z k+1 X Z k+1 [f(k) − f(x)] dx = (x − (k + 1))f 0(x) dx k=1 k k=1 k n−1 X Z k+1 = (x − bxc − 1)f 0(x) dx k=1 k n−1 n−1 X Z k+1 X Z k+1 = (x − bxc)f 0(x) dx − f 0(x) dx k=1 k k=1 k n−1 X Z k+1 Z n = (x − bxc)f 0(x) dx − f 0(x) dx k=1 k 1 n−1 X Z k+1 = (x − bxc)f 0(x) dx − (f(n) − f(1)): k=1 k This is a formula we can work with.