Chapter 4 Arithmetic Functions and Dirichlet Series.2014

Chapter 4 Arithmetic Functions and Dirichlet Series. 2014 [4 lectures ]

Deﬁnition 4.1 An arithmetic function is any function f : N → C.

Examples 1) The divisor function d (n) (often denoted τ (n)) is the number of divisors of n, i.e. d (n) = 1. Xd|n Note that if d|n then n = d (n/d ) = ab say, so d (n) equals the number of pairs of integers ( a, b ) with product n, i.e.

d (n) = 1. ab n X=

This can be extended to give dk (n), the number of k-tuples of integers (a1, a 2, ..., a k) with product n, i.e.

dk (n) = 1. a1a2...a k=n X 2) The function σ (n) is the sum of divisors of n, i.e.

σ (n) = d. Xd|n This can be extended, so for any ν ∈ C deﬁne

ν σν (n) = d . Xd|n 3) The Euler Totient or phi function is the number of integers ≤ n which are co-prime to n, so φ (n) = 1. 1≤r≤n gcd(Xr,n )=1

Those students who did MATH10101 will have seen φ (n) as the number ∗ of invertible elements in Zn, or equivalently the cardinality of Zn.

1 4) The function ω (n) counts the number of distinct prime factors of n, so

ω (n) = 1. Xp|n Recall the notation k deﬁned by pa|| n if, and only if, a is the largest power of p that divides n. So pa|| n if, and only if, pa|n and pa+1 ∤ n. 5) Then deﬁne Ω ( n) to be the number of primes that divide n counted with multiplicity , i.e. Ω ( n) = a = 1. pakn p r≥1 X Xpr|Xn Some further important examples of arithmetic functions are

•) 1 ( n) = 1forall n, •) j (n) = n for all n, sometimes known as id,

1 if n = 1 •) δ (n) = sometimes known as e (n) . 0 otherwise, To an arithmetic function f : N → C we can associate a “generating” function.

Deﬁnition 4.2 Given f : N → C deﬁne the Dirichlet Series of f to be

∞ f (n) D (s) = , f ns n=1 X deﬁned for those s ∈ C at which the series converges.

In this Chapter we are not particularly interested in the question of where a given Dirichlet Series converges, though the answer is known in the following example:

Example 4.3 If f = 1 then

∞ 1( n) ∞ 1 D (s) = = = ζ (s) , 1 ns ns n=1 n=1 X X the Riemann zeta function, which is known to converge when Re s > 1 and converges absolutely only in this half plane.

2 ∞ ∞ Recall (or see Background Notes 0.2) that if n=1 an and m=1 bm are absolutely convergent with sums A and B respectively then the double sum ∞ ∞ P P n=1 m=1 anbm converges absolutely to AB. Being absolutely convergent means we can rearrange the double however we like. In particular we can P P group together anbm in diﬀerent ways and then sum over the groups. For example

∞ Deﬁnition 4.4 The Cauchy Product (or Convolution ) of n=1 an and ∞ m=1 bm is to collect all terms anbm for which the sum n + mPis constant, i.e. n + m = N. That is P ∞ ∞ ∞

an bm = anbm . n=1 ! m=1 ! N=1 mn +m=N ! X X X X This has applications to the products of power series. Example 4.5 If we deﬁne the exponential function by the power series ∞ xr ex = . r! r=0 X then it can be shown the series converges absolutely on R, In this case the Cauchy Product is the well known result ex+y = exey for all real x and y. Alternatively

∞ Deﬁnition 4.6 The Dirichlet Product (or Convolution ) of n=1 an and ∞ b is to collect all terms a b for which the product nm is con- m=1 m n m P stant, i.e. nm = N. That is P ∞ ∞ ∞

an bm = anbm . n=1 ! m=1 ! N=1 mnm =N ! X X X X Let f and g be two arithmetic functions such that their associated Dirich- let Series converge absolutely at some s ∈ C. Then, at this s we can look at the Dirichlet convolution of the two series,

∞ f (n) ∞ g (m) ∞ f (n) g (m) D (s) D (s) = = f g ns ms ns ms n=1 m=1 N=1 nm =N ! X X X X ∞ 1 = f (n) g (m) . (1) N s N=1 nm =N ! X X 3 So the Dirichlet product of Dirichlet series of arithmetic functions gives a Dirichlet Series of a new arithmetic function. This motivates the

Deﬁnition 4.7 If f and g are arithmetic functions then the Dirichlet Con- volution f ∗ g is deﬁned by

f ∗ g (n) = f (a) g (b) ab n X= n n = f (d) g = f g (d) d d Xd|n Xd|n for all n ∈ N.

We see that f ∗ g = g ∗ f, i.e. Dirichlet Convolution, ∗, is commutative . Note that for any arithmetic functions f and g we have

(f ∗ g) (1) = f (1) g (1) and n (1 ∗ f) ( n) = 1 f (d) = f (d) . d Xd|n Xd|n

With this deﬁnition the right hand side of (1) is Df∗g (s) . By the assump- tion above on convergence of products of series, this Df∗g (s) will converge (absolutely) at all s ∈ C for which Df (s) and Dg (s) converge absolutely, and further satisﬁes Df (s) Dg (s) = Df∗g (s) . (2)

Example 4.8 For n ≥ 1 we have n 1 ∗ 1( n) = 1( d) 1 = 1 = d (n) , d Xd|n Xd|n thus d = 1 ∗ 1.

Example 4.9 For n ≥ 1 we have n 1 ∗ j (n) = 1 j (d) = d = σ (n) , d Xd|n Xd|n ν so σ = 1 ∗ j. Similarly σν = 1 ∗ j .

4 Exercise 4.10 on Problem Sheet. Recall δ (n) = 1 if n = 1 , 0 otherwise. Show that f ∗ δ = δ ∗ f = f for all arithmetic functions f.

Thus δ is the identity under ∗. Inverses When does an arithmetic function have an inverse under ∗? That is, for what f does there exist g such that f ∗ g = δ? In the appendix it is shown that f has an inverse under ∗ if, and only if f (1) =6 0 . Associativity Given three arithmetic functions f, g and h, note that

(f ∗ g) ∗ h = (f ∗ g) ( a) h (b) = f (c) g (d) h (b) ab n ab n cd a ! X= X= X= = f (c) g (d) h (b) cdb n X=

= f (a1) g (a2) h (a3) , a1a2a3=n X on relabeling. The symmetry of the ﬁnal sum shows that

(f ∗ g) ∗ h = f ∗ (g ∗ h) , i.e. ∗ is associative , and we can simply write f ∗ g ∗ h with no ambiguity.

Example 4.11 dk = 1 ∗ 1 ∗ ... ∗ 1, convolution k times.

The convolutions found above can be used within Dirichlet Series.

Example 4.12

∞ d (n) = D (s) = D (s) ns d 1∗1 n=1 X

= D1 (s) D1 (s) by (2)

= ζ2(s) .

Since the individual factors ζ(s) converge absolutely for Re s > 1, the ﬁnal result holds for Re s > 1.

5 Example 4.13 Similarly

∞ d (n) k = ζk(s) ns n=1 X for Re s > 1.

Example 4.14 The Dirichlet series for the sum of divisors function σ is

∞ σ (n) = D (s) = D (s) ns σ 1∗j n=1 X

= D1(s) Dj(s) by (2)

∞ j (n) ∞ n = ζ (s) = ζ (s) ns ns n=1 n=1 X X = ζ (s) ζ (s − 1) , valid for Re s > 1 and Re ( s − 1) > 1, i.e. Re s > 2.

Question given an arithmetic function can you write it as a convolution of ‘simpler’ functions?

This question is easier to answer if the function has additional properties.

Deﬁnition 4.15 An arithmetic function f is multiplicative iﬀ,

∀m, n ∈ N, gcd ( m, n ) = 1 ⇒ f (mn ) = f (m) f (n) , completely (or totally ) multiplicative iﬀ

∀m, n ∈ N, f (mn ) = f (m) f (n) , strongly multiplicative iﬀ multiplicative and

∀p, ∀r ≥ 1, f (pr) = f (p) .

6 a1 a2 ar Note Factoring n = p1 p2 ...p r into distinct primes p1, p 2, ...., p r, then f multiplicative means that

r a1 a2 ar ai a f (n) = f (p1 ) f (p2 ) · · · f (pr ) = f (pi ) = f (p ) . (3) i=1 a Y pYkn So we need only evaluate f on prime powers. If f is completely multiplicative then a1 a2 ar a f (n) = f (p1) f (p2) · · · f (pr) = f (p) , a pYkn while if f is strongly multiplicative then

f (n) = f (pa) = f (p) = f (p) . a a pYkn pYkn Yp|n So in the last two cases we need only evaluate f on primes.

Lemma 4.16 If f is a non-zero multiplicative function then f (1) = 1 .

Proof If f is non-zero then there exists n for which f (n) =6 0. Yet multiplicativity means f (n) = f (1 × n) = f (1) f (n) .

Divide by the non-zero f (n) to get 1 = f (1) .

Deﬁnition 4.17 An arithmetic function is additive if,

∀m, n ∈ N, gcd ( m, n ) = 1 ⇒ g (mn ) = g (m) + g (n) , completely (or totally ) additive if

∀m, n ∈ N, g (mn ) = g (m) + g (n) , strongly additive if additive and

∀p, ∀r ≥ 1, g (pr) = g (p) .

a1 a2 ar Note Factoring n = p1 p2 ...p r into distinct primes p1, p 2, ...., p r, then g additive means that r a1 a2 ar ai a g (n) = g (p1 ) + g (p2 ) + · · · + g (pr ) = g (pi ) = g (p ) . (4) i=1 a X pXkn 7 So we need only evaluate g on prime powers. If f is completely additive then

g (n) = a1g (p1) + a2g (p2) + · · · + arg (pr) = ag (p) , a pXkn while if it is strongly additive then

g (n) = g (pa) = g (p) = g (p) . a a pXkn pXkn Xp|n So in the last two cases we need only evaluate g on primes. For student to check that if g is a non-zero additive function then g (1) = 0.

Important So we see from (3) and (4) that to calculate multiplicative and additive functions it suﬃces to ﬁnd their values only on prime powers (and only on primes if either strongly multiplicative or strongly additive).

Example 4.18 The functions 1, j and jν are (completely) multiplicative while ω is (strongly) and Ω (completely) additive. Note that this means 2ω and 2Ω are multiplicative, as will be used later. Question how to identify a multiplicative function? If it is the convolution of two other arithmetic functions we can use Theorem 4.19 If f and g are multiplicative then f ∗ g is multiplicative.

Proof Assume gcd ( m, n ) = 1. Then d|mn if, and only if d = d1d2 with d1|m, d2|n (and thus gcd ( d1, d 2) = 1 and the decomposition is unique ). Therefore

= f ∗ g (m) f ∗ g (n)

8 Example 4.20 The divisor functions d = 1 ∗ 1, and dk = 1 ∗ 1 ∗ ... ∗ 1 ν convolution k times, along with σ = 1 ∗j and σν = 1 ∗j are all multiplicative.

Note The function 1 is completely multiplicative but d is not , so convolution does not preserve complete multiplicativity. Question , was it obvious from its deﬁnition that σ was multiplicative? I suggest not.

Example 4.21 For prime powers

pa+1 − 1 d (pa) = a + 1 and σ (pa) = . p − 1 Hence pa+1 − 1 d (n) = (a + 1) and σ (n) = . a a p − 1 pYkn pYkn Given multiplicative functions f and g then f ∗g is multiplicative. To describe the convolution, it suﬃces to evaluate it on powers of prime. First note that f ∗ g (1) = f (1) g (1) = 1, as required for any multiplicative function.

Important For a prime power

pr f ∗ g (pr) = f (d) g . r d Xd|p Yet the only divisors d of pr are of the form pk with 0 ≤ k ≤ r, thus

f ∗ g (pr) = f pk g pr−k 0≤k≤r X # # = g (pr) + f (p) g pr−1 + f p2 g pr−2 + · · ·

· · · + f p#r−2 g p2 #+ f p#r−1 g(p) + f (pr) .

We will use this often as we will# also use # #

f ∗ g (pr) = f (pa) g pb . a+b=r X #

9 Recall how in an earlier Chapter we showed that ζ (s) has a Euler Product, so for Re s > 1, 1 −1 ζ (s) = 1 − . (5) ps p Y The same method of proof can be used to prove

Theorem 4.22 If f is multiplicative and Df (s) is absolutely convergent at s0 ∈ C then f (p) f (p2) f (p3) D (s) = 1 + + + + ... f ps p2s p3s p Y f pk = 1 + pks p k≥1 # ! Y X for all s with Re s > Re s0. If further f is completely multiplicative then

f (p) −1 D (s) = 1 − f ps p Y

s for Re s > Re s0 and as long as |f (p) /p | < 1 for all primes p.

Proof Left to student, but see appendix if stuck.

Recall that when convergent, an inﬁnite product is non-zero. Hence (5) gives us

Corollary 4.23 ζ (s) =6 0 for Re s > 1.

Thus for Re s > 1 the inverse 1 /ζ (s) is well-deﬁned and writing it as an Euler Product gives

1 1 ∞ µ (n) = 1 − = , ζ (s) ps ns p n=1 Y X where the function µ is found by multiplying out the inﬁnite product.

10 a1 a2 ar Deﬁnition 4.24 M¨obius Function µ (1) = 1 and for n = p1 p2 ...p r > 1, written as a product of distinct primes , then

r (−1) if a1 = a2 = ... = ar = 1 , µ (n) = ( 0 if some ai ≥ 2. Note ∞ µ (n) 1 6 = = , n2 ζ (2) π2 n=1 X from the result for ζ (2) seen in Chapter 1. It is easily seen that µ is multiplicative, but do convince yourself.

Recall the deﬁnition of δ as δ (n) = 1 if n = 1, 0 otherwise. Thus Dδ (s) = 1 for all s ∈ C. We also have 1 1 = ζ (s) = D (s) D (s) = D (s) ζ (s) 1 µ 1∗µ

for Re s > 1. Hence Dδ (s) = D1∗µ (s), i.e.

∞ δ (n) ∞ (1 ∗ µ) ( n) = ns ns n=1 n=1 X X for such s. We would like to conclude that δ = 1 ∗ µ but to do so we would have to prove that if DF (s) = DG (s) for some domain of s then F = G. We do not do so in this course. All we can deduce from

Dh (s) = Df (s) Dg (s) = Df∗g (s) is that it “suggests” h = f ∗ g, which we then have to prove in an alternative manner. If f and g are multiplicative then so is h and we only have to show equality on prime powers, i.e. h (pr) = ( f ∗ g) ( pr) for all primes p and r ≥ 1. For if we do, then

h (n) = h (pr) = (f ∗ g) ( pr) = ( f ∗ g) ( n) . p p Y Y

So we have to ﬁnd an alternative justiﬁcation for the fundamental result

11 Theorem 4.25 M¨obius Inversion 1 ∗ µ = δ, i.e. 1 if n = 1 µ (d) = 0 otherwise. Xd|n In other words, since δ is the identity under ∗ then µ is the inverse of 1 under ∗. Proof The two functions 1 and µ are both multiplicative and thus, so is 1 ∗µ. Since multiplicative functions are given by their values on prime powers it suﬃces to show that 1 ∗ µ (pr) = δ (pr) for any prime p and all r ≥ 1. If r = 0 then p0 = 1 and both 1 ∗ µ (1) = 1 and δ (1) = 1. If r ≥ 1 then (1 ∗ µ) ( pr) = µ (d) = µ pk d|pr 0≤k≤r X X # = µ pk since µ pk = 0 for k ≥ 2, 0≤k≤1 X # # = µ (1) + µ (p)

= 1 − 1

= 0 = δ (pr) . The following is also often called M¨obius Inversion and looks more general than the previous result - but it is not! Corollary 4.26 For arithmetic functions f and g we have f = 1 ∗ g if, and only if, g = µ ∗ f. Proof (⇒) If f = 1 ∗ g then µ ∗ f = µ ∗ (1 ∗ g) = ( µ ∗ 1) ∗ g since ∗ is associative = δ ∗ g = g

12 (⇐) I leave the implication in the other direction to the student.

To give us arithmetic functions with which to give examples illustrating future results:

Deﬁnition 4.27 An integer n > 1 is square-free if no square divides n, a1 a2 ar alternatively, if n = p1 p2 ...p r , distinct primes, then no ai ≥ 2. Deﬁne Q2 (n) = 1 if n square-free and 0 otherwise.

a1 a2 ar An integer n = p1 p2 ...p r , distinct primes, is k-free if no ai ≥ k. Deﬁne Qk (n) = 1 if n is k-free (short for k-power free) and 0 otherwise.

Careful, this notation is not universal. Also, Q2 (n) is often written as 2 |µ (n)| or µ (n), there being no such alternatives for Qk (n) when k ≥ 3. It is easily seen that Q2 and Qk are multiplicative, but convince yourself. The deﬁnition can be extended to k = 1, for the only integer that is 1-free, i.e. divisible by no ﬁrst power of a prime, is n = 1. Thus Q1 (n) = 1 if n = 1, 0 otherwise, i.e. Q1 = δ.

Aside The deﬁnition of the M¨obius function can be written as (−1) ω(n) if n is square-free, µ (n) = ( 0 otherwise.

The form ( −1) ω(n) suggests deﬁning a further arithmetic function.

Deﬁnition 4.28 The Liouville Function λ is deﬁned by λ (1) = 1 and λ (n) = ( −1) Ω( n) for all n ≥ 2.

Convince yourself that this is multiplicative and λ (pr) = ( −1) r for all powers of primes. This function is used a lot in the Problem Sheet to the section.

Question If Dirichlet convolution “combines” arithmetic functions can we factor a given function into “simpler” functions?

First Method of factorization is based on the idea that the properties of a multiplicative function are strongly dictated by the values on the primes. So given G look through known functions to ﬁnd, if it exists, a function g satisfying g (p) = G (p) for all primes. Then we might think of G as a ‘pertubation’ of g and look for a function f such that G = g ∗ f. Note that on primes

G (p) = ( g ∗ f) ( p) = g (p) f (1) + g (1) f (p) = G (p) + f(p)

13 since g (p) = G (p). Hence f (p) = 0 for all primes. So we can think of f being a ‘pertubation’ of the zero function, 0( n) = 0 for all n and thus f a ‘small’ function.

Because Q2 (p) = 1 and 1( p) = 1 for all primes p that we may think that Q2 = 1 ∗ f for some ‘small’ function f. Important If given F which you suspect can be written as 1 ∗ f for some ‘simpler’ f then, by M¨obius inversion, f = µ∗F . If further F is multiplicative then f will be also and you need only calculate the values of f on prime powers, pr f (pr) = µ (d) F = µ pk F pr−k d d|pr 0≤k≤r X X # # = µ pk F pr−k since µ pk = 0 for k ≥ 2, 0≤k≤1 X # # # = F (pr) − F pr−1 , for r ≥ 1. We will use this# often so needs to be remembered . Since f is multiplicative we will have f (1) = 1 and this need not be calculated. And when r = 1, f (p0) = f (1) = 1 and so f (p) = F (p) − 1.

Example 4.29 If Q2 = 1 ∗ f2 describe f2.

Solution The function Q2 is multiplicative so, by the above, for r ≥ 1 we have

r r r−1 f2 (p ) = Q2 (p ) − Q2 p # 0 − 0 if r − 1 ≥ 2 = 0 − 1 if r − 1 = 1   1 − 1 if r − 1 = 0

 −1 if r = 2 , = 0 if r =6 2 . a1 a2 ar Thus if n = p1 p2 ...p r then f2 (n) = 0 if any ai =6 2. If all ai = 2 then 2 2 r n = ( p1p2...p r) = m say, and f2 (n) = ( −1) = µ (m) since m is square-free. Therefore µ (m) if n = m2, i.e. n is a square, f (n) = 2 0 otherwise. 14

We may think of f2 as ‘small’ since f2 (p) = 0 for all primes p and is thus a pertubation of the 0 function, 0( n) = 0 for all n.

Notation In fact we will write µ2 in place of f2, since this better represents the deﬁnition of the function. Thus

Q2 = 1 ∗ µ2.

Note The previous result Q2 = 1 ∗ µ2 is equivalent to

Q2 (n) = µ2 (m) = µ (d) . 2 Xm|n Xd |n Example 4.30 for student . The function 2ω is multiplicative and satisﬁes 2ω(p) = 2 for all primes. Similarly the divisor function d (p) = 2 for all primes. Thus we may think of 2ω as a ‘pertubation’ of d. Therefore, describe the function f such that 2ω = d ∗ f.

We can use the decomposition of Q2 to factor the Dirichlet series for this function.

Example 4.31 ∞ Q (n) ζ (s) 2 = , ns ζ (2 s) n=1 X for Re s > 1.

15 Proof ∞ Q (n) 2 = D (s) = D (s) ns Q2 1∗µ2 n=1 X

= D1 (s) Dµ2 (s) by (2)

∞ µ (n) = ζ (s) 2 ns n=1 X ∞ µ (m) = ζ (s) ns n=1 nX=m2

∞ µ (m) = ζ (s) m2s m=1 X ζ (s) = . ζ (2 s) And this is valid wherever the ﬁnal Riemann zeta functions are all absolutely convergent, i.e. Re s > 1.

Question Given an arithmetic function f how can we get an indication that it can be factored? If f is multiplicative then one method is the use of Dirichlet Series and Euler Products.

Alternative method of factorization If Df (s) factors as Dg (s) Dh (s) then Df (s) = Dg (s) Dh (s) = Df∗g (s) which, as seen earlier, suggests f = g∗h. But how to factor the Dirichlet series? Idea, write it as an Euler product and factor each term in the product. So, assuming f is multiplicative we have, by Theorem 4.22

∞ f (n) f (p) f (p2) f (p3) = 1 + + + + · · · ns ps p2s p3s n=1 p X Y f pℓ = , (6) pℓs p ℓ≥0 # ! Y X having used f (p0) = f (1) = 1.

16 In all our examples f pℓ will not depend on p, only ℓ, so we can write a = f pℓ for ℓ ≥ 0. Write y = 1 /p s and the series within (6) becomes ℓ # # ℓ aℓy . (7) ℓ X≥0

The aim of this method is to write this series as product and quotient of terms of the form 1 − ym for various integers m ≥ 1. For if we have a factor of the form (1 − ym)−1, replacing y by 1 /p s we ﬁnd a factor of the right hand side of (6) of 1 −1 1 − = ζ (ms ) . pms p Y Further, if the sum of (7) contains a factor of 1 − yk for some k ≥ 1, then on replacing y by 1 /p s we ﬁnd a factor of the right hand side of (6) of

1 1 1 − = . pks ζ (ks ) p Y

As a way of illustrating this method:

Recall that Qk is the characteristic function of the k-free integers.

Example 4.32 Show that

∞ Q (n) ζ (s) k = ns ζ (ks ) n=1 X for Re s > 1.

Solution The function Qk is multiplicative so, without yet considering the regions of convergence for the Dirichlet Series, we have the Euler Product

∞ Q (n) Q (p) Q (p2) Q (p3) k = 1 + k + k + k + · · · ns ps p2s p3s n=1 p X Y 1 1 1 = 1 + + + · · · + , ps p2s p(k−1) s p Y

17 ℓ s since Qk p = 0 for all ℓ ≥ k, and = 1 elsewhere. Write y = 1 /p when each bracket is a ﬁnite geometric sum of the form # 1 − yk 1 + y + y2 + · · · + yk−1 = . 1 − y Therefore ∞ Q (n) 1 − 1/p ks k = ns 1 − 1/p s n=1 p X Y −1 1 −1 1 −1 = 1 − 1 − ps pks p p ! Y Y ζ (s) = , (8) ζ (ks ) having used (5). We can now consider convergence. Since the ζ-functions on the right hand side are absolutely convergent in Re s > 1, the ﬁnal result is valid in this half plane.

Note that 1 ∞ µ (m) ∞ µ (m) ∞ µ (m) ∞ µ (n) = = = = k ζ (ks ) mks (mk)s ns ns m=1 m=1 n=1 n=1 X X nX=mk X where µk is deﬁned by

µ (m) if n = mk, µk (n) = ( 0 otherwise.

The M¨obius Function is µ1. Example 4.32 shows that 1 D (s) = ζ (s) = D (s) D (s) = D (s) . Qk ζ (ks ) 1 µk 1∗µk This result, written as

∞ Q (n) ∞ 1 ∗ µ (n) k = k ns ns n=1 n=1 X X 18 for Re s > 1 ‘suggests’ a decomposition of the function Qk as Qk = 1 ∗ µk, but you need to prove this directly, perhaps by showing equality on prime powers. (See Problem Sheet)

Note that when k = 1 we have seen that Q1 = δ, while µ1 = µ and so Qk = 1 ∗ µk reduces down to the M¨obius inversion δ = 1 ∗ µ.

Example 4.33 Show, by looking at Euler Product of the Dirichlet Series on the left, that ∞ 2ω(n) ζ2 (s) = ns ζ (2 s) n=1 X for Re s > 1.

Solution Left to the student but as a hint the left hand side has an Euler product 2 2 2 2 1 + + + + + . ps p2s p3s p4s p Y Thus you will be required to sum 1 + 2 y + 2 y2 + 2 y3 + ....

The result on 2 ω gives

2 1 D ω = ζ (s) = D (s) D (s) = D (s) , 2 ζ (2 s) d µ2 d∗µ2 where d is the divisor function. This result ‘suggests’ that

ω 2 = d ∗ µ2, which again needs to be proved. (See Problem Sheet.)

ω Once we know 2 = d∗µ2 it can be combined with the deﬁnition d = 1 ∗1 and the result Q2 = 1 ∗ µ2 to give

ω 2 = (1 ∗ 1) ∗ µ2 = 1 ∗ (1 ∗ µ2) = 1 ∗ Q2. (9) There are many such connections between Arithmetic functions, some of which are the content of questions on the Problem Sheet and all are collected on a page on the Course web site.

Euler’s phi function Recall the deﬁnition of Euler’s phi function as

φ (n) = {1 ≤ r ≤ n, gcd ( r, n ) = 1 } .

19 We ‘pick out’ the condition gcd( r, n ) = 1 using the δ function, for which δ (n) = 1 if n = 1, zero otherwise. For then

1 if gcd( r, n ) = 1 δ (gcd ( r, n )) = (10) ( 0 otherwise.

We can then use M¨obius inversion, in the form δ (m) = d|m µ (d) to get

Example 4.34 Show that Euler’s phi function satisﬁes φ =Pµ ∗ j, i.e. n φ (n) = µ (d) . d Xd|n Solution

n n φ (n) = 1 = δ (gcd ( r, n )) by(10) , r=1 r=1 gcd(Xr,n )=1 X

n = µ (d) by M¨obius inversion δ = 1 ∗ µ. r=1 X d| gcd(Xr,n ) Yet d| gcd ( r, n ) if, and only if, d|r and d|n. Continuing

n n φ (n) = µ (d) = µ (d) 1. r=1 d|r d|n r=1 X Xd|n X Xd|r on interchanging the summations. In this double summation we have that d|n, so n = ℓd say, and we also have d|r, so r = kd say. Thus in the inner sum we are counting the number of k ≥ 1 for which r ≤ n, i.e. kd ≤ ℓd, that is, k ≤ ℓ. There are ℓ = n/d such values. Therefore this inner summation equals n/d and thus n φ (n) = µ (d) . d Xd|n (Make sure you understand why this inner sum is exactly n/d ).

Since µ and j are multiplicative we conclude that φ is multiplicative.

20 Looking at φ on prime powers pa φ (pa) = µ (d) = µ pk pa−k d d|pa 0≤k≤a X X # = µ pk pa−k = pa − pa−1. 0≤k≤1 X # This actually should have been obvious from the definition, the only natural numbers ≤ pa, not coprime to pa are the multiples of p of which there are pa−1 in number. So the number of natural numbers ≤ pa, coprime to pa is the difference pa − pa−1. Thus, by multiplicativity, 1 φ (n) = φ (pa) = pa − pa−1 = n 1 − . p pakn pakn p|n Y Y # Y Finally , we saw an important arithmetic function earlier in the course, namely von Mangoldt’s function Λ( n) defined to be log p when n is a power of the prime p, zero otherwise. We have not studied it here because it is not multiplicative. The important result of Λ was

Λ ( d) = log n, (11) Xd|n which was proved without motivation - though it was fundamental to proving Chebyshev’s inequalities and Merten’s results. But where did it come from?

If we write ℓ (n) = log n we can see that the result is the convolution Λ ∗ 1 = ℓ. Then formally we can consider

∞ Λ ( n) ζ′ (s) D (s) = D (s) D (s) = ζ (s) = − ζ (s) Λ∗1 Λ 1 ns ζ (s) n=1 X ∞ log n = −ζ′ (s) = ns n=1 X

= Dℓ (s) .

This is the motivation for Λ ∗ 1 = ℓ, i.e. (11).

21 Note that M¨obius inversion applied to (11) gives Λ = µ ∗ ℓ, i.e. n Λ ( n) = µ (d) log . d Xd|n