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 ϕ. It counts the number of relatively prime positive integers less than n. For example, ϕ(15) = ϕ(3)ϕ(5) (multiplicative), but ϕ(9) = 6 6= ϕ(3)ϕ(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.
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Example 2.4 Let n ≥ 1 be an integer. Then X ϕ(d) = n. d|n
Proof. The idea is that we can use the property that if m, n are relatively prime positive integers, then ϕ(mn) = ϕ(m)ϕ(n) (this fact follows from the Chinese remainder theorem). e1 ek e1 ek Using this fact we can break up ϕ(n = p1 . . . pk ) = ϕ(p1 ) . . . ϕ(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 ϕ(d) = (ϕ(1) + ϕ(p1) + ··· + ϕ(p1 )) ... ϕ(1) + ϕ(pk) + ··· + ϕ(pk ) d|n because when we expand the product on the RHS, each term will be ϕ(d) for some d | n. Now we’re done because it’s easy to see that ϕ(1) + ϕ(p) + ··· + ϕ(pe) = pe when p is a e1 ek prime, and thus the right-hand side is p1 . . . pk = n. Exercise 2.5. Verify that ϕ, 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 ϕ(p ) = p − p for p a prime and k ∈ N. Then for n = p1 . . . pk , we get
e1 e1−1 e2 e2−1 ek ek−1 ϕ(n) = (p1 − p1 )(p2 − p2 ) ... (pk − pk ).
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). d|n de=n
Here are some properties of ∗:
• The identity of ∗ is the Dirichlet delta function δ. P The reason for this fact is that d|n 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.
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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) d|n X = 1 = τ(n). d|n Exercise 3.4. Verify that
• 1 ∗ id = σ.
• id ∗ id = n · τ.
• 1 ∗ ϕ = 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 (ϕ ∗ 1 = id) Let n ≥ 1 be an integer. Then X ϕ(d) = n. d|n
Proof. Rephrasing the problem, we wish to show that
ϕ ∗ 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 d|n 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.
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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). d|n d|n
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) = d|n f(d), then f is also multiplicative. Furthermore, P if f is multiplicative and g(n) = d|n 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 d|n 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).
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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 | n. So f ∗ 1 = s = δ. But µ ∗ 1 = δ, so f = µ by uniqueness of convolution, as desired.
Proposition 5.3 P∞ −s Let s be a complex number with <(s) > 1 and define ζ(s) = n=1 n . Then
∞ 1 X µ(n) = ζ(s) ns n=1
Definition 5.4 (Dirichlet series). Let {an} be any complex sequence and s be a complex number. The Dirichlet series for an is
∞ X an . ns n=1
P∞ 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
∞ ! Y X f(pm) 1 + . pms p∈P m=1 Remark 5.6. We are manipulating F (s) as a formal power series and thus assuming convergence. Observe that the above identity holds because when we expand the product out, we get exactly the sum of
f(pe1 ) f(pek ) f(pe1 ) ··· f(pek ) f(n) 1 ··· k = 1 k = e1s eks e1s eks s p1 pk p1 ··· pk n Proof of Proposition 5.3. Note that the Euler product of ζ(s) is simply
Y Y 1 (1 + p−s + p−2s + ... ) = 1 − p−s p∈P p∈P by the geometric series formula. So 1 = Q 1 − 1 , which is the Euler product ζ(s) p∈P ps P∞ µ(n) for n=1 ns , as desired.
5 Karen Ge (August 3, 2017) Smol Results on the M¨obiusFunction 6 Extra Problems
Problem 6.1. Prove that for any integer n ≥ 1,
2 X 3 X (τ(d)) = τ(d) . d|n d|n
Problem 6.2 (Bulgaria 1989). Let Ω(n) denote the number of prime factors of n, counted with multiplicity. Evaluate 1989 X 1989 (−1)Ω(n) . n n=1 Problem 6.3. Suppose that X n µ f(d) = n. d d|n Describe f(n).
Pn n Problem 6.4. Prove that if f is an arithmetic function and g(n) = k=1 f k , then
n X n f(n) = µ(j)g . j j=1 Q Problem 6.5. Let f and g be arithmetic functions such that f(n) = d|n g(d). Prove:
Y nµ(d) g(n) = f . d d|n
Problem 6.6. Let n ∈ N. Show that X (µ(d))2 n = . ϕ(d) ϕ(n) d|n
Problem 6.7. Let p be a prime. We say that g is a primitive root (mod p) if gp−1 = 1 and ga 6= 1 for 0 < a < p−1. (Why does g exist?) In other words, the set {ga : a = 1, . . . p−1} × generated by g is exactly Zp . Prove that the sum of the primitive roots (mod p) is µ(p − 1).
Problem 6.8. Get a bound on the number of square-free numbers less than some number √ x in terms of x. Can you get an error term better than O( x)?
n 1 X Problem 6.9. Consider µ (n) = µ(k), the average value of µ(n). Prove that a n k=1
lim µa(n) = 0. n→∞ This result is actually equivalent to the prime number theorem! However, this equiva- lence needs a lot of machinery to prove. See [6], [3] for more details.
Remark 6.10. More precise estimates about the growth rate of nµa(n) are related to the Riemann Hypothesis. Again, see [3].
6 Karen Ge (August 3, 2017) Smol Results on the M¨obiusFunction References
[1] Summation, Evan Chen, October 13th, 2016.
[2] The Sum of Primitive Roots of Unity, Yimin Ge, June 9th, 2009.
[3] Introduction to Analytic Number Theory Math 531 Lecture Notes, A.J. Hildebrand, Fall 2005.
[4] Math 105 notes, week 3, Carl Pomerance, Fall 2013.
[5] Euler Product, Sondow, Jonathan and Weisstein, Eric W.
[6] Average Values Continued: ϕ and µ, Pete L. Clark, Spring 2009.
[7] Number Theory for Mathematical Contests, David A. Santos, August 13, 2005.
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