Chapter 2 Arithmetic Functions and Dirichlet Series.

Transcription

1 Chater 2 Arithmetic Functions and Dirichlet Series. [4 lectures] Definition 2.1 An arithmetic function is any function f : N C. Examles 1) The divisor function d (n) (often denoted τ (n)) is the number of divisors of n, i.e. d (n) = 1. d n Note that if d n then n = d (n/d) = ab say, so d (n) equals the number of airs of integers (a, b) with roduct n, i.e. d (n) = ab=n 1. This can be extended to give d k (n), the number of k-tules of integers (a 1, a 2,...,a k ) with roduct n, i.e. d k (n) = 1. a 1 a 2...a k =n 2) The function σ (n) is the sum of divisors of n, i.e. σ (n) = d n d. This can be extended, so for any ν C define σ ν (n) = d n d ν. 3) The Euler Totient or hi function is the number of integers n which are co-rime to n, so φ (n) = 1. 1 r n gcd(r,n)=1 Those students who did MATH10101 will have seen φ (n) as the number of invertible elements in Z n, or equivalently the cardinality of Z n. 1

2 so 4) The function ω (n) counts the number of distinct rime factors of n, ω (n) = n 1. Define the notation by a n if, and only if, a is the largest ower of that divides n. So a n if, and only if, a n and a+1 n. 5) Then define Ω (n) to be the number of rimes that divide n counted with multilicity, i.e. Ω (n) = a = 1. a n Some further examles are ) 1 (n) = 1 for all n, r 1 r n ) j (n) = n for all n, sometimes known as id, ) δ (n) = { 1 if n = 1 0 otherwise, sometimes known as e (n). To an arithmetic function f : N C we can associate a generating function. Definition 2.2 Given f : N C define the Dirichlet Series of f to be D f (s) = f (n), defined for those s C at which the series converges. In this Chater we are not articularly interested in the question of where a given Dirichlet Series converges, though the answer is known in the following examle: Examle 2.3 If f = 1 then D 1 (s) = 1 (n) = 1 = ζ (s), the Riemann zeta function, which is known to converge when Res > 1 and converges absolutely only in this half lane. 2

3 Recall (or see Background Notes 0.2) that if a n and m=1 b m are absolutely convergent with sums A and B resectively then, however the terms a n b m are arranged as a single series, the single series converges absolutely to AB. Definition 2.4 The Cauchy Product (or Convolution) of n=0 a n and m=0 b m is to collect all terms a n b m for which the sum n + m is constant, i.e. n + m = N. Then the result above says that if n=0 a n and m=0 b m are absolutely convergent then ( ) a n b m = a n b m. N=0 n+m=n n=0 m=0 This is most often used with ower series, when ( a n z n b m z m = a n b m )z N. n=0 m=0 n=0 n+m=n Definition 2.5 The Dirichlet Product (or Convolution) of a n and m=1 b m is to collect all terms a n b m for which the roduct nm is constant, i.e. nm = N. Then the result above says that if a n and m=1 b m are absolutely convergent then ( ) a n b m = a n b m. N=1 nm=n Let f and g be two arithmetic functions such that their associated Dirichlet Series converge absolutely at some s C. Then, at this s we can look at the Dirichlet convolution D f (s)d g (s) = = N=1 f (n) m=1 g (m) m s = m=1 ( N=1 nm=n f (n) ) g (m) m s ( ) 1 f (n)g (m). (1) N s nm=n So the roduct of Dirichlet series of Arithmetic functions gives a Dirichlet Series of a new arithmetic function. This motivates the 3

4 Definition 2.6 If f and g are arithmetic functions then the Dirichlet Convolution f g is defined by f g (n) = ab=nf (a)g (b) = d n ( n f (d)g = d) d n f ( n d) g (d) for all n N. It should be obvious (but see the Problem Sheet) that f g = g f, i.e. Dirichlet Convolution,, is commutative. With this definition (1) can be written as D f (s)d g (s) = D f g (s) (2) at all s C for which D f (s) and D g (s) converge absolutely. Examle 2.7 For n 1 we have 1 1 (n) = ( n 1 (d) 1 = d) d n d n 1 = d (n), thus d = 1 1. Examle 2.8 For n 1 we have 1 j (n) = ( n 1 j (d) = d) d = σ (n), d n d n so σ = 1 j. Similarly σ ν = 1 j ν. Exercise 2.9 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. This begs the question of when does an arithmetic function have an inverse under? That is, for what f does there exist g such that f g = δ? On the roblem sheet you are asked to show that f has an inverse under if, and only if f (1) 0. 4

5 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 = f (c)g (d)h(b) = cdb=n a 1 a 2 a 3 =n f (a 1 )g (a 2 )h(a 3 ) on relabelling. The symmetry of the final sum shows that (f g) h = f (g h), i.e. is associative, and we caimly write f g h with no ambiguity. Examle 2.10 d k = , convolution k times. The convolutions found above can be used within Dirichlet Series. Examle 2.11 d (n) = D d (s) = D 1 1 (s) = D 1 (s)d 1 (s) by (2) = ζ 2 (s). Since the individual factors ζ (s) converge absolutely for Res > 1, the final result holds for Res > 1. Examle 2.12 Similarly for Res > 1. d k (n) = ζ k (s) Examle 2.13 The Dirichlet series for the sum of divisors function σ is σ (n) = D σ (s) = D 1 j (s) = D 1 (s)d j (s) by (2) = ζ (s) j (n) = ζ (s)ζ (s 1) 5 = ζ (s) n

6 valid for Res > 1 and Re (s 1) > 1, i.e. Res > 2. Question given an arithmetic function can you write it as a convolution of simler functions? This question is easier to answer if the function has additional roerties. Definition 2.14 An arithmetic function f is multilicative if, m, n N, gcd (m, n) = 1 f (mn) = f (m)f (n), comletely (or totally) multilicative if m, n N, f (mn) = f (m)f (n), strongly multilicative if multilicative and, r 1, f ( r ) = f (). Note Factoring n = a 1 1 a 2... ar r multilicative means that into distinct rimes 1, 2,..., r, then f f (n) = f ( a 1 1 )f ( a 2 2 ) f ( ar r ) = a nf ( a ). If f is comletely multilicative then f (n) = f ( 1 ) a 1 f ( 2 ) a2 f ( r ) ar = a nf () a, while if it is strongly multilicative then f (n) = f ( a ) = f () = f (). a n a n n If f is multilicative and f 0 then f (1) = 1. To see this, f 0 means there exists n 1 for which f (n) 0. But then f (n) = f (n 1) = f (n)f (1) since f is multilicative, and we can cancel the non-zero f (n) to get 1 = f (1) as desired. Definition 2.15 An arithmetic function is additive if, m, n N, gcd (m, n) = 1 g (mn) = g (m) + g (n), comletely (or totally) additive if strongly additive if additive and m, n N, g (mn) = g (m) + g (n),, r 1, g ( r ) = g (). 6

7 Note Factoring n = a 1 1 a 2... ar r additive means that into distinct rimes 1, 2,..., r, then g g (n) = g ( a 1 1 ) + g ( a 2 2 ) + + g ( ar r ) = a ng ( a ). If f is comletely additive then g (n) = a 1 g ( 1 ) + a 2 g ( 2 ) + + a r g ( r ) = a nag (), while if it is strongly additive then g (n) = a n g ( a ) = a n g () = n g (). So we see that to calculate multilicative and additive functions it suffices to find their values only on rime owers (and only on rimes if strongly multilicative or additive). Examle 2.16 The functions 1, j and j ν are multilicative while ω and Ω are additive, in which case 2 ω and 2 Ω are multilicative. Question how to identify a multilicative function? If it is the convolution of two other arithmetic functions we can use Theorem 2.17 If f and g are multilicative then f g is multilicative. Proof Assume gcd (m, n) = 1. Then d mn if, and only if d = d 1 d 2 with d 1 m, d 2 n (and thus gcd (d 1, d 2 ) = 1). Therefore f g (mn) = d mnf ( mn ) (d)g d = d 1 m d 2 n ( ) m n f (d 1 d 2 )g d 1 d 2 = ( ) m ( ) n f (d 1 )g f (d 2 )g d 1 d 2 d 1 m d 2 n since f and g are multilicative = f g (m)f g (n) 7

8 Examle 2.18 The divisor functions d = 1 1, and d k = convolution k times, along with σ = 1 j and σ ν = 1 j ν are all multilicative. Question, was it obvious from its definition that σ was multilicative? I susect not. Examle 2.19 For rime owers d ( a ) = a + 1 and σ ( a ) = a Hence d (n) = a n (a + 1) and σ (n) = a n ( a ). Recall how in Theorem 1.8 we showed that ζ (s) has a Euler Product, so for Res > 1, ζ (s) 1 1 ) 1. (3) s The same method of roof can be used to rove Theorem 2.20 If f is multilicative and D f (s) is absolutely convergent at s 0 C then D f (s) 1 + f () + f (2 ) + f ) (3 ) +... s 2s 3s 1 + f ( k) ) ks k 1 for all s with Res > Res 0. If further f is comletely multilicative then D f (s) 1 f () ) 1 s for Res > Res 0 and as long as f ()/ s < 1 for all rimes. Proof Left to student, but see aendix if stuck. When convergent, an infinite roduct is non-zero. Hence (3) gives us Corollary 2.21 ζ (s) > 0 for Res > 1. 8

9 Thus for Re s > 1 the inverse 1/ζ (s) is well-defined and the Euler Product gives 1 ζ (s) 1 1 ) µ (n) =, s where the function µ is found by multilying out the infinite roduct. Definition 2.22 Möbius Function µ (1) = 1 and for n > 1, written as a roduct of distinct rimes n = a 1 1 a ar r, then { ( 1) r if a 1 = a 2 =... = a r = 1, µ (n) = 0 if some a i 2. Note µ (n) n 2 = 1 ζ (2) = 6 π 2, from the result for ζ (2) seen in Chater 1. It is easily seen that µ is multilicative. Recall the definition of δ as δ (n) = 1 if n = 1, 0 otherwise. Thus D δ (s) = 1 for all s C. We also have 1 = ζ (s) for Res > 1. Hence D δ (s) = D 1 µ (s), i.e. 1 ζ (s) = D 1 (s)d µ (s) = D 1 µ (s) δ (n) = (1 µ) (n) for such s. We would like to conclude that δ = 1 µ but to do so we would have to rove that if two Dirichlet Series are equal iome region of C then their coefficients are equal. Such a result is true but we don t rove it in this course. So we have to find an alternative justification for Theorem 2.23 Möbius Inversion 1 µ = δ. In other words, since δ is the identity under then µ is the inverse of 1 under. 9

10 Proof The two functions 1 and µ are both multilicative and thus, so is 1 µ. Since multilicative functions are given by their values on rime owers it suffices to show that 1 µ ( r ) = δ ( r ) for any rime. If r = 0 then 0 = 1 and both 1 µ (1) = 1 and δ (1) = 1. If r 1 then (1 µ) ( r ) = d r µ (d) = = 0 k 1 = µ (1) + µ () = 1 1 = 0 = δ ( r ). 0 k r µ ( k) µ ( k) since µ ( k) = 0 for k 2, The following is also often called Möbius Inversion and looks more general than the revious result - but it is not. Corollary 2.24 For arithmetic functions f and g we have Proof If f = 1 g then f = 1 g if, and only if, g = µ f. µ f = µ (1 g) = (µ 1) g since is associative = δ g = g I leave the imlication in the other direction to the student. There there is also yet another version of the Möbius Inversion found in the Problem Sheet. Question If Dirichlet convolution combines arithmetic functions can we factor a given function into simler functions? Given f that you susect can be written as 1 g for some simler g then, by Möbius inversion, g = µ f. 10

11 If further f is multilicative then g will be too and you need only calculate the values of g on rime owers, where it satisfies g ( r ) = ( ) r µ (d)f = µ ( k) f ( r k) d d r for r 1. = 0 k 1 0 k r µ ( k) f ( r k) = f ( r ) f ( r 1), To give us an arithmetic function with which to give examles illustrating future results: Definition 2.25 An integer n > 1 is square-free if no square divides n, alternatively, if n = a 1 1 a 2... ar r, distinct rimes, then no a i 2. Define Q 2 (n) = 1 if quare-free and 0 otherwise. An integer n = a 1 1 a ar r, distinct rimes, is k-free if no a i k. Define Q k (n) = 1 if n is k-free and 0 otherwise. Careful, this notation is not universal. Often Q 2 (n) is written as µ (n) or µ 2 (n), there being no such alternatives for Q k (n) when k 3. It is easily seen that Q 2 and Q k are multilicative. The definition can be extended to k = 1, for the only integer that is 1-free is n = 1. Thus Q 1 (n) = 1 if n = 1, 0 otherwise, i.e. Q 1 = δ. The definition of the Möbius 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 2.26 The Liouville Function λ is defined by λ (n) = ( 1) Ω(n). Can we factorize Q 2 in any meaningful way. Note that Q 2 () = 1 for all rimes, exactly the same as 1 () = for all rimes. So a starting oint is to see if Q = 1 f for any meaningful function f. Examle 2.27 If Q 2 = 1 f describe f. 11

12 Solution The function Q 2 is multilicative so, by the above, for r 1 we have f ( r ) = Q 2 ( r ) Q 2 ( r 1 ) = 0 0 if r if r 1 = if r 1 = 0 = { 1 if r = 2, 0 if r 2. Thus { µ (m) if n = m f (n) = 2, i.e. n is a square, 0 otherwise. Notation In fact we will write µ 2 in lace of f, since this better reresents the definition of the function. Note The revious result Q 2 = 1 µ 2 is equivalent to Q 2 (n) = m n µ 2 (m) = d 2 n µ (d). But Q 2 = 1 µ 2 also leads to Q 2 (n) = D Q2 (s) = D 1 µ2 (s) = D 1 (s)d µ2 (s) by (2) = ζ (s) µ 2 (n) = ζ (s) m=1 µ (m) m 2s = ζ (s) ζ (2s). And this is valid wherever the final Riemann zeta functions are both 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 multilicative one method is the use of Dirichlet Series and Euler Products. 12

13 Method Assuming f is multilicative we have f (n) 1 + f ( l) ) = ls l 1 ( f ( l) ), (4) l 0 ls recalling that f (1) = 1. In all our examles f ( l) will not deend on, only l, so we can write a l = f ( l). Then we need sum l 0 a l ls. Write y = 1/ s and the series becomes a ower series a l y l. (5) l 0 The aim is to write this series as roduct and quotient of terms of the form 1 y m for various integers m 1. For if we have a factor of the form (1 y m ) 1, relacing y by 1/ s we find a factor of the right hand side of (4) of ( 1 1 ) 1 = ζ (ms). ms Further, if the final exression for the sum of (5) contains a factor of 1 y k for some k 1, then on relacing y by 1/ s we find a factor of the right hand side of (4) of ( 1 1 ) = 1 ks ζ (ks). See Background Notes 0.3 for methods for summing various series. As a way of illustrating this method: Recall that Q k is the characteristic function of the k-free integers. Examle 2.28 Show that Q k (n) = ζ (s) ζ (ks) for Res > 1. 13

14 Solution The function Q k is multilicative so, without yet considering the regions of convergence for the Dirichlet Series, we have the Euler Product Q k (n) 1 + ( Q ) ) k l 1 + ( Q ) ) k l ls ls l 1 k 1 l s ). 2s (k 1)s Write y = 1/ s when each bracket is a finite geometric series of the form 1 + y + y y k 1 = 1 yk 1 y. Therefore Q k (n) = ( ) 1 1/ ks 1 1/ s = ( 1 1 s ) 1 ( ( 1 1 ) ) 1 1 ks = ζ (s) ζ (ks), (6) having used (3). We can now consider convergence. Since the ζ-functions on the right hand side are absolutely convergent in Res > 1, the final result is valid in this half lane. Note that where µ k is defined by 1 ζ (ks) = µ k (n) = The Möbius Function is µ 1. m=1 µ (m) m ks = µ k (n) { µ (m) if n = m k, 0 otherwise. The right hand side of (6) can be written as ζ (s) ζ (ks) = D 1 (s)d µk (s) = D 1 µk (s) = 14 1 µ k (n)

15 and so the result (6) suggests that Q k = 1 µ k. Unfortunately you have to rove this directly by showing that Q k ( r ) = 1 µ k ( r ) for all rimes and r 1 (See Problem Sheet) Note that when k = 1 we have seen that Q 1 = δ, while µ 1 = µ and so Q k = 1 µ k reduces down to the Möbius inversion δ = 1 µ. Recall that the divisor function was defined as d = 1 1. From this it was shown that on rime owers d ( a ) = a + 1. What if we had defined d by what its values on rime owers? Could we factor such a function? Examle 2.29 Let d be the multilicative function defined on rime owers by d ( a ) = a + 1. Show that d = 1 1. Solution Since we are told that d is multilicative we have the Euler Product d (n) 1 + d () + d (2 ) + d (3 ) + d ) (4 ) +... s 2s 3s 4s s + 4 2s + 5 ) s 4s Thus we are led to the ower series 1 + 2y + 3y 2 + 4y 3 + 5y We now attemt to formally sum this, i.e. we do not worry about the validity of our methods, the justification comes from the result. Thus, with no justification, we integrate term-by-term 1 + 2y + 3y 2 + 4y 3 + 5y = d ( y + y 2 + y 3 + y ) dy Substituting back in, d (n) = = ζ 2 (s). = d dy = 1 (1 1/ s ) 2 = 15 ( ) 1 1 y 1 1 (1 y) 2. ( (1 1 ) ) 1 2 s

16 This is valid for Res > 1. Thus we have factorized the Dirichlet Series. To factorise the function d we note that for Res > 1, D d (s) = d (n) = ζ 2 (s) = D 1 (s)d 1 (s) = D 1 1 (s). This suggests that d = 1 1, but you have to rove this directly. Since all functions are multilicative this means showing that 1 1 ( a ) = d ( a ) = a+1 for all rimes and a 1. Yet 1 1 ( a ) = 1 = 1 = a + 1, d a d= r,0 r a as required. Hence d = 1 1. Recall ω (n), the number of distinct rime divisors of n, is an additive function of n. Therefore 2 ω is a multilicative function because, if gcd (m, n) = 1 then Examle 2.30 Write 2 ω(mn) = 2 ω(m)+ω(n) = 2 ω(m) 2 ω(n). 2 ω(n) as a roduct and quotient of Riemann zeta functions. Solution On Problem Sheet, but the solutiotarts as 2 ω(n) ) ) ) ) 1 + 2ω() + 2ω(2 + 2ω(3 + 2ω( s 2s 3s 4s s + 2 2s + 2 ) s 4s Thus we are led to the ower series 1 + 2y + 2y 2 + 2y 3 + 2y = 1 + 2y 1 y = 1 + y 1 y. This can not be the end of the matter, since as described above, we are looking for factors of the for 1 y k for k 1, not 1 + y. I would suggest continuing by multilying to and bottom by 1 y. 16

17 Examle 2.31 Show that Euler s hi functioatisfies φ = µ j, i.e. φ (n) = d n µ (d) n d. Solution φ (n) = = n r=1 gcd(r,n)=1 n 1 = r=1 d gcd(r,n) n δ (gcd (r,n)) by definition of δ, r=1 µ (d) by Möbius inversion δ = 1 µ, = d n = d n µ (d) n 1 on interchanging the summations, r=1 d r µ (d) n d. (Make sure you understand why this inner sum is exactly n/d). In articular, for a ower of a rime φ ( a ) = d a µ (d) a d = = 0 k 1 0 k a µ ( k) a k µ ( k) a k = a a 1. This actually should have been obvious from the definition, the only natural numbers a, not corime to a are the multiles of of which there are a 1 in number. So the number of natural numbers a, corime to a is a a 1. Since µ and j are multilicative we conclude that φ is multilicative. Thus φ (n) = φ ( a ) a a 1) = n ( 1 1 ). a n a n n 17