MULTIPLICATIVE FUNCTIONS: A REWRITE OF ANDREWS CHAPTER 6

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1 MULTIPLICATIVE FUNCTIONS: A REWRITE OF ANDREWS CHAPTER 6 In these notes we offer a rewrte of Andrews Chapter 6. Our am s to replace some of the messer arguments n Andrews. To acheve ths, we need to change the order of presentaton of the materal. The man am s to study the three mportant functons: Euler s ph functon, ϕ(n); the number of dvsors of n, d(n); and the sum of the dvsors of n, σ(n). Each functon s multplcatve, n the sense defned below. Defnton A functon f : N R s completely multplcatve f f(mn) = f(m)f(n) (m, n N). An easy nductve argument shows that we have and so f(m 1 m 2... m s ) = f(m 1 )f(m 2 )... f(m s ) f(p α 1 1 p α p αs s ) = f(p α 1 1 )... f(p αs s ) = (f(p 1 )) α 1... (f(p s )) αs ( ) Thus, a completely multplcatve functon s determned by ts values on the prme numbers. The smplest natural examples are gven by f(n) = n c. But the nterestng functons of number theory have only the weaker property that f(mn) = f(m)f(n) (m, n N, gcd(m, n) = 1). For such functons, only the frst equaton n ( ) holds true. Equvalently, a multplcatve functon s known when we know ts values on every power of each prme number. Recall that U k s the set of nvertble elements n Z k ; equvalently t s the set of all those elements x Z k whch are coprme to k. Euler s ph-functon s just the number of elements n U k. THEOREM 1. Euler s ph-functon s multplcatve. Proof. Let m, n be coprme. We are gong to show that the set U mn s n one-one correspondence wth the Cartesan product U m U n. Ths then gves mmedately that ϕ(mn) = ϕ(m)ϕ(n), as requred. Gven x U mn, reduce x modulo m and also modulo n. Thus we get a Z m, b Z n such that x a (mod m) and y b (mod n). Suppose that p a and p m. It 1

2 follows that p x, and also p mn. Snce x U mn, we must have p = 1 and so a U m. Smlarly, b U n. So we get a functon F : U mn U m U n defned by F (x) = (a, b). It remans to show that F s one-one and onto. To see that F s one-one, suppose that F (x) = F (y) wth x, y U mn. Then x a y (mod m), x b y (mod n). Ths gves m (x y) and n (x y). But m, n are coprme, and so mn (x y), that s, x = y. To see that F s onto, take any a U m and any b U n. By the Chnese Remander Theorem, there exsts x Z mn such that x a (mod m) and x b (mod n). To see that x U mn, suppose that a prme p dvdes both x and mn. But then p dvdes ether m or n, whch contradcts the fact that a U m and b U n. Thus x U mn and we have F (x) = (a, b). So F s onto, and the proof s complete. We can now get the formula for ϕ(n). For n = p α, we see that the only elements not coprme to n are 0, p, 2p,... p 2, p 2 + p,... p α p and so Snce ϕ s multplcatve, we get ( ) ϕ(n) = ϕ p α = ϕ(p α ) = p α p α 1. (p α p α 1 ) = n (1 1p ). The last product above gves the fracton of n for whch the numbers are coprme to n. When n s a large prme, ths fracton s very nearly 1. When n s the product of the frst k prmes, then the fracton goes to zero as k goes to nfnty. Ths last asserton s by no means obvous, but t corresponds to the fact (to be proved later) that the nfnte seres (1/p ) dverges. 2

3 We are now gong to see how to buld a new multplcatve functon from an old one. As an mmedate corollary we shall see that our other two functons d(n) and σ(n) are also multplcatve. THEOREM 2. Let f be multplcatve and let F (n) = k n f(k). Then F s multplcatve. Proof. Let m, n be coprme, and let k mn. By the Fundamental Theorem of Arthmetc, k = ab where a m, b n and a, b are coprme. Conversely, each such choce of a, b gves a product ab whch s a factor of mn. Hence, F (mn) = f(k) = f(ab) = f(a)f(b) = f(a) f(b) = F (m)f (n). k mn a m,b n a m,b n a m b n Take f(n) = 1 (whch s completely multplcatve) and we see that d(n) s multplcatve. Take f(n) = n (whch s also completely multplcatve) and we see that σ(n) s also multplcatve. To get the formulas for d(n) and σ(n), we just have to calculate that and d(p α ) = α + 1 σ(p α ) = 1 + p + p p α = pα+1) 1 p 1. As another applcaton of Theorem 2, take f = ϕ and we clam that the assocated functon F s gven by F (n) = n. To get ths, we just have to show that F (p α ) = p α. But F (p α ) = ϕ(1)+ϕ(p)+ϕ(p 2 )+ +ϕ(p α ) = 1+(p 1)+(p 2 p)+ +(p α p α 1 = p α as requred. We fnd below another way to get the same result. Our fnal specal functon s a very curous one, called the Möbus functon, denoted by µ(n). We let µ(1) = 1, µ(p 1 p 2... p s ) = ( 1) s, and µ(n) = 0 otherwse, that s, f n has at least one repeated prme factor. It s routne by 3

4 case-checkng to see that µ(n) s also multplcatve. Apply Theorem 2 wth f = µ to get assocated multplcatve functon M(n). Clearly M(1) = 1 and M(p α ) = µ(1) + µ(p) + µ(p 2 ) + + µ(p α ) = = 0. It follows that M(n) = 0 whenever n > 1, n other words, µ(k) = 0 (n > 1). k n The next theorem s called the Möbus Inverson Theorem. Snce summaton corresponds to ntegraton and dfferences (here wth spacng h = 1) correspond to dervatves, you may thnk of the theorem as a dscrete verson of the Fundamental Theorem of Calculus. THEOREM 3. Let f, g be any (real) functons on N. Then f(n) = k n g(k) (n N) ( ) f and only f g(n) = k n µ(k)f(n/k) (n N) ( ). Proof. Suppose that ( ) holds. Then µ(k)f(n/k) = µ(k) g(e) = µ(k)g(e) = g(e) e k j e k n kk =n keh=n ee =n µ(j). From above, the nner summaton n the last equaton s zero except when e = 1 and so e = n. Thus the sum s f(n), as requred. The other half of the proof s smlar. A par of functons (f, g) satsfyng ether of the equvalent condtons ( ) or ( ) s called a Möbus par. Thus (n, ϕ), (d, 1), (σ, n) are all Möbus pars. So also s (µ, δ), where δ s the Drac delta functon whch s 1 at 1 and zero elsewhere. Our fnal theorem s about the multplcatve propertes of a Möbus par. We have already proved one half of t. The other half s smlar and s left to the reader. 4

5 THEOREM 4. Let (f, g) be a Möbus par. Then f s multplcatve f and only f g s multplcatve. When we nvert the Möbus par (n, ϕ), we get the formula ϕ(n) = k n µ(k) n k. Of course ths s easy to check when n = p α. It s temptng to thnk that we now have a quck proof that ϕ s multplcatve. Unfortunately ths s not the case, snce we would need to check that one of the Möbus par equatons holds for all n. Remark. In combnatorcs, the Möbus Inverson Theorem has been generalzed by replacng N by certan partally ordered spaces. 5