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Smol Results on the Möbius Function 1 Introduction 2 Multiplicative Functions

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Smol Results on the Möbius Function 1 Introduction 2 Multiplicative Functions Smol Results on the M¨obiusFunction Karen Ge August 3, 2017 1 Introduction We will address how M¨obiusfunction relates to other arithmetic functions, multiplicative number theory, the primitive complex roots of unity, and the Riemann zeta function. 2 Multiplicative Functions The M¨obiusfunction is an example of a special class of functions, called multiplicative functions. We will see why these functions are nice. Definition 2.1. An arithmetic function is a function f : N ! C. It is • completely multiplicative if f(mn) = f(m)f(n) for any m and n. • multiplicative if f(mn) = f(m)f(n) for relatively prime m and n. Example 2.2 (Completely multiplicative functions) The following are examples of completely multiplicative functions. • The identity function id: id(n) = n ( 1 n = 1 • The Dirichlet delta function, δ(n) = 0 n ≥ 2: • The constant function 1, given by 1(n) = 1. Example 2.3 (Multiplicative functions) The following are examples of multiplicative functions which are not completely multiplicative. • Euler's '. It counts the number of relatively prime positive integers less than n. For example, '(15) = '(3)'(5) (multiplicative), but '(9) = 6 6= '(3)'(3) = 4 (so it is not completely multiplicative). • The M¨obiusfunction µ, defined by ( (−1)m if n has m prime factors, all distinct µ(n) = 0 n is not square-free: For example, µ(10) = 1 = µ(2)µ(5), µ(105) = −1 = µ(3)µ(5)µ(7), but µ(12) = 0 6= µ(2)µ(6) = −1. • σ, the sum of divisors function. • τ, the number of divisors function. 1 Karen Ge (August 3, 2017) Smol Results on the M¨obiusFunction Example 2.4 Let n ≥ 1 be an integer. Then X '(d) = n: djn Proof. The idea is that we can use the property that if m; n are relatively prime positive integers, then '(mn) = '(m)'(n) (this fact follows from the Chinese remainder theorem). e1 ek e1 ek Using this fact we can break up '(n = p1 : : : pk ) = '(p1 ) :::'(pk ) and compute the e1 ek sum by focusing on one prime at a time. More explicitly, let's write n = p1 : : : pk . Then we have X e1 ek '(d) = ('(1) + '(p1) + ··· + '(p1 )) ::: '(1) + '(pk) + ··· + '(pk ) djn because when we expand the product on the RHS, each term will be '(d) for some d j n. Now we're done because it's easy to see that '(1) + '(p) + ··· + '(pe) = pe when p is a e1 ek prime, and thus the right-hand side is p1 : : : pk = n. Exercise 2.5. Verify that '; µ, σ are indeed examples of multiplicative functions. Of course, the product of two multiplicative functions is also multiplicative. As we saw above, the nice property of multiplicative functions is that it is sufficient to determine their values on prime powers. For example, we can easily check that k k k−1 e1 ek '(p ) = p − p for p a prime and k 2 N. Then for n = p1 : : : pk , we get e1 e1−1 e2 e2−1 ek ek−1 '(n) = (p1 − p1 )(p2 − p2 ) ::: (pk − pk ): 3 Dirichlet Convolution Given two arithmetic functions f and g, we define their Dirichlet convolution as X X (f ∗ g)(n) = f(d)g(n=d) = f(d)g(e): djn de=n Here are some properties of ∗: • The identity of ∗ is the Dirichlet delta function δ. P The reason for this fact is that djn δ(d)f(n=d) = 0 · f(1) + ··· + 1 · f(n=1) = f(n) as all of the terms for d 6= 1 are multiplied by 0. • The operation ∗ is commutative. • The operation ∗ is associative because X ((f ∗ g) ∗ h)(n) = (f ∗ (g ∗ h)) (n) = f(d1)g(d2)h(d3): d1d2d3=n • It distributes over addition: f ∗ (g + h) = f ∗ g + f ∗ h. • Most important: the convolution of two multiplicative functions is also multiplicative. 2 Karen Ge (August 3, 2017) Smol Results on the M¨obiusFunction Exercise 3.1. Check that convolutions of multiplicative functions are multiplicative. Exercise 3.2. Show that if f is a multiplicative arithmetic function, then an inverse g of f under convolution (i.e. f ∗ g = δ) exists if and only if f(1) 6= 0. Also show that this inverse is unique. (Hint: compute the values of g inductively.) Example 3.3 We have 1 ∗ 1 = τ. Proof. X (1 ∗ 1)(n) = 1(d)1(n=d) djn X = 1 = τ(n): djn Exercise 3.4. Verify that • 1 ∗ id = σ. • id ∗ id = n · τ. • 1 ∗ ' = id (first example). • δ ∗ f = f for any f. Now we can use our theory to simplify the proof of our earlier example. Example 3.5 (' ∗ 1 = id) Let n ≥ 1 be an integer. Then X '(d) = n: djn Proof. Rephrasing the problem, we wish to show that ' ∗ 1 = id: The above is true for prime powers n = pe, because in that case the LHS is 1 + (p − 1) + (p2 − p) + ··· + (pe − pe−1) = pe. But since both the LHS and RHS are multiplicative, we are done. 4 M¨obiusInversion We now know that given a multiplicative function f, we have a good handle on g(n) = P djn f(d) because g is the Dirichlet convolution f ∗ 1. However, occasionally we instead have g and want to recover f. The way to do this is through M¨obiusinversion. Lemma 4.1 (M¨obiusis inverse of 1) We have µ ∗ 1 = δ. 3 Karen Ge (August 3, 2017) Smol Results on the M¨obiusFunction Theorem 4.2 (M¨obiusinversion formula) Let f and g be any arithmetic functions (possibly not multiplicative). Then X X g(n) = f(d) () f(n) = µ(d)g(n=d): djn djn In other words, if g = f ∗ 1 then f = g ∗ µ. Proof. Assume g = f ∗ 1. Then g ∗ µ = (f ∗ 1) ∗ µ = f ∗ (1 ∗ µ) = f ∗ δ = f: The reverse direction is similar. Corollary 4.3 P If g is multiplicative and g(n) = djn f(d), then f is also multiplicative. Furthermore, P if f is multiplicative and g(n) = djn f(d), then g is also multiplicative. P k For example, σk(n), the sum of the kth powers of the divisors of n, is just djn d . k Since f(d) = d is completely multiplicative, it follows that σk(n) is multiplicative. 5 Cute Facts Recall Lemma 4.1, which states that µ ∗ 1 = δ. If the proof below has too many words to understand, consider an example, say n = 23 · 34 · 5 · 7 · 11 · 132 · 17. Proof of Lemma 4.1. First note that the identity holds for n = 1. For n ≥ 2, we observe that there is a bijection between the divisors d of n for which µ(d) 6= 0 and the subsets of the prime factors of n. Indeed, whenever µ(d) 6= 0, the factorization of d consists of distinct prime numbers which thus form a subset of the prime factors of n. So the identity reduces to proving that the number of odd-sized subsets is equal to the number of even-sized subsets for any non-empty set (in this case, the set is that of the distinct prime factors of n). This fact is true by the binomial theorem. Say n has k distinct prime factors. The negative terms in (1 − 1)k = 0 are of the form n − 2a+1 ; 0 ≤ a ≤ b(n − 1)=2c, and so count the number of subsets of size 2a + 1. The n positive terms are of the form 2a ; 0 ≤ a ≤ bn=2c ; and so count the number of subsets of size 2a. The sum of the negative terms is negation of the total number of odd-sized subsets and the sum of the positive terms is the total number of even-sized subsets. Since the sum of these two totals is (1 − 1)k = 0, they must cancel exactly. Remark 5.1. An alternate proof of the last fact involves a fun combinatorial operation called \toggling." Fact 5.2 (M¨obiusand primitive roots of unity). Let X k f(n) = ζn: 1≤k≤n gcd(k;n)=1 Then f(n) = µ(n): 4 Karen Ge (August 3, 2017) Smol Results on the M¨obiusFunction Proof. We recall another function that looks an awful lot like the Dirichlet delta function. Define n X k s(n) = ζn; k=1 the sum of the complex nth roots of unity. Clearly s(1) = 1: By definition, ζn satisfies n n−1 Pn k ζn − 1 = 0 = (ζn − 1)(ζn + ··· + ζn + 1). Since ζ 6= 1, it follows that k=1 ζn = 0 for n > 1 So we in fact have ( 1 n = 1 s(n) = 0 n ≥ 2: So s(n) is exactly δ(n). Now we observe that every nth root of unity is a primitive dth root of unity for some d j n. So f ∗ 1 = s = δ. But µ ∗ 1 = δ; so f = µ by uniqueness of convolution, as desired. Proposition 5.3 P1 −s Let s be a complex number with <(s) > 1 and define ζ(s) = n=1 n : Then 1 1 X µ(n) = ζ(s) ns n=1 Definition 5.4 (Dirichlet series). Let fang be any complex sequence and s be a complex number. The Dirichlet series for an is 1 X an : ns n=1 P1 f(n) Definition 5.5 (Euler product). Let F (s) = n=1 ns be the Dirichlet series for a multiplicative function f(n): Then the Euler product for F (s) is the same function 1 ! Y X f(pm) 1 + : pms p2P m=1 Remark 5.6.
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