ANALYTIC NUMBER THEORY (Spring 2019) Eero Saksman Chapter 1 Arithmetic functions 1.1 Basic properties of arithmetic functions We denote by P := f2; 3; 5; 7; 11;:::g the prime numbers. Any function f : N ! C is called and arithmetic function1. Some examples: ( 1 if n = 1; I(n) := 0 if n > 1: u(n) := 1; n ≥ 1: X τ(n) = 1 (divisor function). djn φ(n) := #fk 2 f1; : : : ; ng j (k; n))1g (Euler's φ -function). X σ(n) = (divisor sum). djn !(n) := #fp 2 P : pjng ` ` X Y αk Ω(n) := αk if n = pk : k=1 k=1 Q` αk Above n = k=1 pk was the prime decomposition of n, and we employed the notation X X f(n) := f(n): djn d2f1;:::;ng djn Definition 1.1. If f and g are arithmetic functions, their multiplicative (i.e. Dirichlet) convolution is the arithmetic function f ∗ g, where X f ∗ g(n) := f(d)g(n=d): djn 1In this course N := f1; 2; 3;:::g 1 Theorem 1.2. (i) f ∗ g = g ∗ f, (ii) f ∗ I = I ∗ f = f, (iii) f ∗ (g + h) = f ∗ g + f ∗ h, (iv) f ∗ (g ∗ h) = (f ∗ g) ∗ h. Proof. (i) follows from the symmetric representation X f ∗ g(n) = f(k)f(`) k`=n (note that we assume automatically that above k; l are positive integers). In a similar vain, (iv) follows by iterating this to write X (f ∗ g) ∗ h(n) = f(k1)g(k2)h(k3): k1k2k3=n Other claims are easy. Theorem 1.3. If f(1) 6= 0, the arithmetic function f has a unique inverse f −1 such that f ∗ f −1 = f −1 ∗ s = I: Proof. Assume that f(1) 6= 0. The condition f ∗ f −1(k) = I(k) (1.1) is satisfied for k = 1 if we set f −1(1) := 1=f(1): Assume then that n ≥ 2 and f −1(1); : : : ; f −1(n − 1) are chosen so that (1.1) holds true for k ≤ n − 1. Then it holds also for k = n if we choose −1 X f −1(n) := f(k=d)f −1(d) : f(1) djn d<n Definition 1.4. M¨obiusfunction µ : N ! R satisfies ( (−1)`; if n = q ··· q with distinct primes q ; : : : q ; µ(n) = 1 ` 1 ` 0; otherwise: In other words, µ(n) = (−1)!(n) if n is squarefree, otherwise µ(n) = 0. Especially, µ(1) = 1: Theorem 1.5. µ ∗ u = I: P Proof. We thus need to verify that djn µ(d) = 0 if n ≥ 2: Assume that p1; : : : ; p` are the distinct primes that divide n. As µ(d) = 0 if d has a square factor, we simply have X µ(d) = µ(1) + µ(p1) + ::: + µ(p`) + µ(p1p2) + : : : µ(p`−1p`) + ::: + µ(p1 : : : p`) djn ` ` ` = 1 − + − ::: + (−1)` = (1 − 1)` = 0: 1 2 ` 2 P Corollary 1.6. (M¨obiusinversion formula) If f(n) := djn g(d); 8n ≥ 1; then P g(n) = djn µ(d)f(n=d). Proof. According to the assumption f = u∗g. The claim follows by taking convolution on both sides with µ and recalling Thm 1.5. Definition 1.7. We denote the identity function by j. Thus j(n) = n for all n ≥ 1. α Moreover, given α 2 C we set jα(n) = n for n ≥ 1: P In the course NT it was shown that djn φ(d) = n, or in other words, φ ∗ u = j. It follows that φ = j ∗ µ. (1.2) Definition 1.8. Arithmetic function f is multiplicative if f 6= 0 and f(mn) = f(m)f(n) if (m; n) = 1: Moreover, f is completely multiplicative if the above condition holds even without the condition (m; n) = 1. Example 1.9. Function φ is multiplicative. Same is true for µ directly from the definition. Functions j and jα are completely multiplicative. Theorem 1.10. If f and g are multiplicative, then also f ∗ g is multiplicative. Proof. Let (m; n) = 1 If djnm, we may write uniquely d = d1d2, where d1jm and d2jn. Hence X X m n (f ∗ g)(mn) = f(d)g(mn=d) = f(d1d2)g d1 d2 djmn d1jm; d2jn X = f(d1)f(d2)g(m=d1)g(n=d2) d1jm; d2jn X X = f(d1)f(m=d1) f(d2)g(n=d2) d1jm d2jn = (f ∗ g)(m)(f ∗ g)(n): Lemma 1.11. If g (6≡ 0) and f ∗ g are multiplicative,then also f is. Proof. Clealy, unless f ≡ 0, we must have f(1) = 1: Of course also g(1) = 1. Assume then f(1) = 1 and that N ≥ 2 so that we have proven multiplicativity condition f(mn) = f(k) for all k = mn ≤ N − 1 and (m; n) = 1. Let then N = mn with (m; n) = 1: We may compute as in the proof of Thm 1.10 X h(mn) = f(d1)f(d2)g(m=d1)g(n=d2) + f(mn) d1jm; d2jn d1d2<mn X X = f(d1)f(m=d1) f(d2)g(n=d2) + f(mn) − f(m)f(n) d1jm d2jn = h(m)h(n) + f(mn) − f(m)f(n): 3 Since h(m) = h(m)h(n); it follows that f(mn) = f(m)f(n): The claim follows now by induction on N. Corollary 1.12. If f is multiplicative and f 6≡ 0, also f −1 is multiplicative. Proof. This follows from Lemma 1.11 since now f and f∗f −1 = I are multiplicative. Corollary 1.13. Arithmetic functions equipped with addition and convolution (as the product operation) form a ring. Element f is invertible iff f(1) 6= 0. Multiplicative functions form a subgroup (with respect the convolution) of the set of invertible ele- ments. Theorem 1.14. (i) If f is completely multiplicative, then f −1(n) = µ(n)f(n) for all n ≥ 1. (ii) Conversely, if f is multiplicative and f −1(n) = µ(n)f(n) for all n ≥ 1, then f is completely multiplicative. Proof. Exercise. In NT it was shown that Y φ(n) = n (1 − 1=p): pjn Q` αk If the prime decomposition of n is n = k=1 pk , we also have (Exercise) ` Y τ(n) = (αk + 1); k=1 ` Y pαk+1 − 1 σ(n) = k ; pk − 1 k=1 ` Y (αk + 1)(αk + 2) u ∗ u ∗ u(n) = : 2 k=1 1.2 Estimation of summatory functions Many arithmetic functions behave fairly irregularly. However, it makes perfect sense and is mathematically useful to try to understand their average behaviour. This is most easily done by studying the corresponding summatory functions, of which also the prime counting function is an important example: X X π(x) := 1 = χP(n): p≤x n≤x 4 This function describes the average density of primes, and the large size irregularities of the prime distribution, so it is especially important in number theory. We shall study it later on, as well as the M¨obissum function X M(x) := µ(n): n≤x At this stage we make a closer study of the divisor sum X D(x) := µ(n)τ(n) n≤x and the following sum that counts the number of squarefree numbers X Q(x) := jµ(n)j: n≤x For that purpose (and for the sake of the whole course) let us revisit the standard Landau symbols. Let F and G be functions that depend on x 2 [a; 1). We say that F (x) = O(G(x)) (" big O of ") if there is finite constant C > 0 such that jF (x)j ≤ CG(x) for large enough x. Similarly F (x) = o(G(x)) (" small o of ") if limx!1 F (x)=G(x) = 0. If F (x) lim = 1; x!1 G(x) we denote F (x) ∼ G(x) (as x ! 1). The less stringent requirement that F and G grow (or decrease) about the same rate means that F (x) C−1 ≤ ≤ C as x large enough; G(x) and this is denoted by F (x) G(x): The above definitions are modified in a natural way when one considers e.g. the situations where x ! x0 or n ! 1: Example 1.15. ex = o(1) as x ! 1 −2x2 + x − 5 = O(x2) as x ! 1, 1 1 ∼ as x ! 1, x2 + x x2 2 1 as x ! 0. x2 + sin2(2x) x2 5 Definition 1.16. Let f be an arithmetic function. The corresponding summatory function ( or sum function ) is F : R ! C, where F (x) = 0 for x < 1 and X F (x) = f(n); for x ≥ 1: n≤x Example 1.17. The sum function that obtained from u is bxc if x > 0, and 0 otherwise. Lemma 1.18. Let f and g be arithmetic functions with F and G the corresponding summatory functions. Then X X X (f ∗ g)(n) = F (x=m)g(m) = G(x=`)f(`): m≤x m≤x `≤x Proof. Left hand side is X X X f(`)g(m) = f(`)g(m) n≤x m`=n `m≤x X X X = g(m) f(`) = F (x=m)g(m): m≤x `≤x=m m≤x The second claim follows then by symmetry. Example 1.19. As τ(n) = u ∗ u(n), and the sum function corresponding to u is bxc; we obtain by the above lemma X X D(x) = τ(n) = bx=mc: (1.3) n≤x m≤x This can also be deduced directly by noting that bx=mc is the number of positive integers less than x that are divisible by m.