Möbius Inversion Formula. Multiplicative Functions Zvezdelina Stankova-Frenkel UC Berkeley

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1 BERKELEY MATH CIRCLE Möbus Inverson Forula. Multplcatve Functons Zvezdelna Stanova-Frenel UC Bereley Contents 1. Multplcatve Functons 1 2. Drchlet Product and Möbus Inverson 3 3. War-up Probles 5 4. The Euler Functon ϕ(n) 6 5. War-up Probles 7 6. Applcatons to Probles 7 7. Probles wth [x] and Multplcatve Functons 8 8. Hnts and Solutons to Selected Probles 9 1. Multplcatve Functons Defnton 1. A functon f : N C s sad to be arthetc. In ths secton we dscuss the set M of ultplcatve functons, whch s a subset of the set A of arthetc functons. Why ths subset s so specal can be explaned by the fact that t s usually easer to calculate explct forulas for ultplcatve functons, and not so easy to do ths for arbtrary arthetc functons. 1 Defnton 2. An arthetc functon f(n) : N C s ultplcatve f for any relatvely pre n, N: f(n) f()f(n). Exaples. Let n N. Defne functons τ, σ, π : N N as follows: τ(n) the nuber of all natural dvsors of n #{d > 0 }; σ(n) the su of all natural dvsors of n d; π(n) the product of all natural dvsors of n d. As we shall see below, τ and σ are ultplcatve functons, whle π s not. Fro now on we shall wrte the pre decoposton of n N as n p α 1 1 pα 2 2 pα r r for dstnct pres p and α > 1. It s easy to verfy the followng propertes of ultplcatve functons: 1 But on a deeper level, M s specal partly because t s closed under the Drchlet product n A. For ths one has to wat untl after Secton 2.

2 Lea 1. If f s ultplcatve, then ether f(1) 1 or f 0. Further, f f 1, f 2,..., f are ultplcatve, then the usual product f 1 f 2 f n s also ultplcatve. Usng the pre decoposton of n N, derve the followng representatons of τ, σ and π. Lea 2. The functons τ(n), σ(n) and π(n) are gven by the forulas: r r p α +1 1 τ(n) (α + 1), σ(n) p 1, π(n) n 1 2 τ(n). Conclude that τ and σ are ultplcatve, whle π s not. Exaples. The followng are further exaples of well-nown ultplcatve functons. µ(n), the Möbus functon; e(n) δ 1,n, the Drchlet dentty n A; I(n) 1 for all n N; d(n) n for all n N. Tang the su-functons of these, we obtan the relatons: S µ e, S e I, S I τ, and S d σ. These exaples suggest that the su-functon s ultplcatve, provded the orgnal functon s too. In fact, Theore 1. f(n) s ultplcatve ff ts su-functon S f (n) s ultplcatve. Proof: Let f(n) be ultplcatve, and let x, y N such that (x, y) 1. Further, let x 1, x 2,..., x and y 1, y 2,..., y be all dvsors of x and y, respectvely. Then (x, y j ) 1, and {x y j },j are all dvsors of xy. S f (x) S f (y) f(x ) Hence S f (n) s ultplcatve. j1 f(y j ),j f(x )f(y j ),j f(x y j ) S f (xy). Conversely, f S f (n) s ultplcatve, then let n 1, n 2 N such that (n 1, n 2 ) 1 and n n 1 n 2. We wll prove by nducton on n that f(n 1 n 2 ) f(n 1 )f(n 2 ). The stateent s trval for n 1: f(1) S f (1)( 1 or 0.) Assue that t s true for all 1 2 < n. Then for our n 1, n 2 we have: S f (n 1 n 2 ) d n f(d 1 d 2 ) d n f(d 1 d 2 ) + f(n 1 n 2 ) d n f(d 1 )f(d 2 ) + f(n 1 n 2 ). d 1 d 2 <n On the other hand, S f (n 1 )S f (n 2 ) f(d 1 ) f(d 2 ) d 1 n 1 d 2 n 2 d n d 1 d 2 <n d 1 d 2 <n f(d 1 )f(d 2 ) + f(n 1 )f(n 2 ). Snce S f (n 1 n 2 ) S f (n 1 )S f (n 2 ), equatng the above two expressons and cancelng approprately, we obtan f(n 1 n 2 ) f(n 1 )f(n 2 ). Ths copletes the nducton step, and shows that f(n) s ndeed ultplcatve. 2

3 Corollary 1. The su-functon S f (n) of a ultplcatve functon f(n) s gven by the forula: r ( S f (n) 1 + f(p ) + f(p 2 ) + + f(p α ) ). 2. Drchlet Product and Möbus Inverson Consder the set A of all arthetc functons, and defne the Drchlet product of f, g A as: f g(n) f(d 1 )f(d 2 ). d 1 d 2 n Note that f g s also arthetc, and that the product s coutatve, and assocatve: (f g) h(n) f (g h)(n) f g h(n) f(d 1 )f(d 2 )f(d 3 ). d 1 d 2 d 3 n Wth respect to the Drchlet product, the dentty eleent e A s easy to fnd: { 1 f n 1 e(n) 0 f n > 1. Indeed, chec that e f f e f for any f A. If we were worng wth the usual product of functons f g(n) f(n)g(n), then the dentty eleent would have been I(n) 1 for all n N, because I f f for all functons f. In the set A, however, I s certanly not the dentty eleent, but has the nce property of transforng each functon f nto ts so-called sufuncton S f. Defne S f (n) f(d), to be the su-functon of f A, and note that S f s also arthetc. Chec that: I f f I S f for all f A. It s nterestng to fnd the Drchlet nverse g A of ths sple functon I n our set A,.e. such that g I I g e. Ths naturally leads to the ntroducton of the so-called Möbus functon µ: Defnton 2. The Möbus functon µ : N C s defned by 1 f n 1 µ(n) 0 f n s not square-free ( 1) r f n p 1 p 2 p r, p j dstnct pres. Lea 3. The Drchlet nverse of I s the Möbus functon µ A. Proof: The lea eans that µ I e,.e. { 1 f n 1 (1) µ(d) 0 f n > 1. 3

4 Ths follows easly fro the defnton of µ. Indeed, for n p α 1 1 pα 2 2 pαr r > 1 we have r µ(d) µ(d) µ(1) + µ(p 1 p ).,d sq.free 1 1 < < r Here the last su runs over all square-free dvsors of n. For cobnatoral reasons, r ( ) r µ(d) ( 1) (1 1) r 0. 0 To suarze, we have shown that the Drchlet nverse of the functon I(n) s the Möbus functon µ(n): µ I I µ e. Unfortunately, not all arthetc functons have Drchlet nverses n A. In fact, show that Lea 4. An arthetc functon f has a Drchlet nverse n M ff f(1) 0. 2 The rght noton to replace nverses n A turns out to be su-functons, whch s the dea of the Möbus nverson theore. Theore 2 (Möbus nverson theore). Any arthetc functon f(n) can be expressed n ters of ts su-functon S f (n) f(d) as f(n) µ(d)s f ( n d ). Proof: The stateent s nothng else but the Drchlet product f µ S f n A: µ S f µ (I f) (µ I) f e f f. Note: Here s a ore tradtonal proof of the Möbus Inverson Forula: µ(d)s f ( n d ) µ( n d )S f (d) µ( n d ) f(d 1 ) d 1 d f(d 1 ) µ( n d ) f(d 1 ) µ( ), d 2 d 1 n d 1 d 1 n d 2 where n/d 1, d 2 d/d 1. By property (1), the second su s non-zero only when 1,.e. d 1 n, and hence the whole expresson equals f(n). Ths proof, however, hdes the product structure of A under the Drchlet product. It s natural to ass the opposte queston: gven the Möbus relaton f(n) µ(d)g( n d ) for two arthetc functons f and g, can we deduce that g s the su-functon S f of f? The answer should be obvous fro the product structure of A: f µ g I f I (µ g) S f (I µ)g e g g, and ndeed, g s the su-functon of f. If you prefer ore tradtonal proofs, you can recover one fro the above by recallng that ultplyng f by I s the sae as tang the su-functon of f. 2 For those who care about group theory nterpretatons, ths ples that M s not a group wth the Drchlet product, but the subset M of t consstng of all arthetc functons f wth f(1) 0 s a group. 4

5 Corollary 2. For two arthetc functons f and g we have the followng equvalence: g(n) f(d) f(n) µ(d)g( n d ). Ths ncdentally shows that every arthetc functon g s the su-functon of another arthetc functon f: sply defne f by the second forula as f µ g. Notce that Theore 1 does not use the Drchlet product at all, but nstead t hdes a ore general fact. For an arthetc functon f, ts su-functon s S f I f, and by Möbus nverson, f µ S f. Thus, Theore 1 sply proves that f f M, then the product of the two ultplcatve functons I and f s also ultplcatve, and that f S f M, then the product of the two ultplcatve functons S f and µ s also ultplcatve. Ths naturally suggest the ore general Theore 3. The set M of ultplcatve functons s closed under the Drchlet product: f, g M f g M. Proof: Let f, g M, (a, b) 1, and h f g. Then h(a)h(b) [ f g(a) ] [f g(b) ] d 1 a d ab d 1 a,d 2 b f(d 1 )f(d 2 )g( a d 1 )g( b d 2 ) f(d 1 )g( a d 1 ) d 2 b f(d)g( ab ) f g(ab) h(ab). d Thus, h s also ultplcatve, so that f g M. d 1 a,d 2 b f(d 2 )g( b d 2 ) f(d 1 d 2 )g( ab d 1 d 2 ) For the group theory fans: fnd the explct Drchlet nverse of any ultplcatve functon f 0, and conclude Lea 5. The set M {f 0} s a group under the Drchlet product. 3. War-up Probles Proble 1. Fnd, n N such that they have no pre dvsors other than 2 and 3, (, n) 18, τ() 21, and τ(n) 10. Proble 2. Fnd n N such that one of the followng s satsfed: π(n) , π(n) , π(n) 13 31, or τ(n) Proble 3. Defne σ (n) d. Thus, σ 0 (n) τ(n) and σ 1 (n) σ(n). Prove that σ (n) s ultplcatve for all N, and fnd a forula for t. Proble 4. Show that τ(n) s odd ff n s a perfect square, and that σ(n) s odd ff n s a perfect square or twce a perfect square. Proble 5. If f(n) s ultplcatve, f 0, then show r ( µ(d)f(d) 1 f(p ) ). 5

6 Proble 6. If f(n) s ultplcatve, then show that h(n) µ( n d )f(d) s also ultplcatve. Conclude that every ultplcatve functon s the su-functon of another ultplcatve functon. 4. The Euler Functon ϕ(n) Defnton 4. The Euler functon ϕ(n) assgns to each natural nuber n the nuber of the ntegers d between 1 and n whch are relatvely pre to n (set ϕ(1) 1.) Lea 6. ϕ(n) s ultplcatve. Proof: Consder the table of all ntegers between 1 and ab, where a, b N, (a, b) a 1 a a + 1 a + 2 a + 2a 1 2a 2a + 1 2a + 2 2a + 3a 1 3a ja + 1 ja + 2 ja + (j + 1)a 1 (j + 1)a (b 2)a + 1 (b 2)a + 2 (b 2)a + (b 1)a 1 (b 1)a (b 1)a + 1 (b 1)a + 2 (b 1)a + ba 1 ba Note that f ja + s relatvely pre wth a, then all nubers n ts (-th) colun wll be relatvely pre wth a. In the frst row there are exactly ϕ(a) nubers relatvely pre wth a, so ther ϕ(a) coluns wll be all nubers n the table whch are relatvely pre wth a. As for b, each colun s a syste of reanders odulo b because (a, b) 1 (chec that n each colun the b ntegers are dstnct odulo b.) Thus, n each colun we have precsely ϕ(b) relatvely pre ntegers to b. Hence, the total nuber of eleents n ths table relatvely pre to ab s exactly ϕ(ab) ϕ(a)ϕ(b), and ϕ(n) s ultplcatve. We can use ultplcatvty of the Euler functon to derve a forula for ϕ(n). For a pre p, all nubers d [1, p ] such that (d, p ) 1 are exactly those d dvsble by p: d p c wth c 1, 2,..., p 1,.e. p 1 n nuber. Hence ϕ(p ) p p 1. Therefore, r r r ϕ(n) ϕ(p α ) (p α p α 1 ) p α (1 1 p ) n(1 1 p 1 ) (1 1 p r ). Lea 7. The su-functon S ϕ (n) of the Euler functon ϕ(n) satsfes: ϕ(n) n. Proof: The ultplcatvty of the Euler functon ϕ(n) ples that S ϕ (n) s ultplcatve, so that S ϕ (n) S ϕ (p α 1 1 ) S ϕ(p αr r ). Ths reduces the proble to a pre power n p a, whch together wth ϕ(p j ) p j p j 1 easly ples S ϕ (p a ) p a. Rear: We can prove the lea also drectly by consderng the set of fractons { 1 n, 2 n,..., n 1 n, n } { a1, a 2,..., a n 1, a } n n b 1 b 2 b n 1 b n 6

7 where (a, b ) 1. The set D of all denonators {b } s precsely the set of all dvsors of n. Gven a dvsor d of n, d appears n the set D exactly ϕ(d) tes: n a a a b d d n, where (a, d) 1 and 1 a d. Countng the eleents n D n two dfferent ways ples n ϕ(d). We can use ths Rear to show agan ultplcatvty of the Euler functon: ndeed, ths follows fro the fact that ts su-functon S ϕ (n) n s tself obvously ultplcatve. 5. War-up Probles Proble 7. Show that ϕ(n ) n 1 ϕ(n) for all n, N. Proble 8. Solve the followng equatons: ϕ(2 x 5 y ) 80. ϕ(n) 12. ϕ(n) 2n/3. ϕ(n) n/2. ϕ(ϕ(n)) Proble 9. Show that ϕ(n)ϕ() ϕ((n, ))ϕ([n, ]); ϕ(n)ϕ((n, )) (n, )ϕ(n)ϕ(). 6. Applcatons to Probles Proble 10. For two sequences of coplex nubers {a 0, a 1,..., a n,...} and {b 0, b 1,..., b n,...} show that the followng relatons are equvalent: a n b for all n b n 0 Proble 11. Solve the equaton ϕ(σ(2 n )) 2 n. ( 1) +n a for all n. Proble 12. Let f(x) Z[x] and let ψ(n) be the nuber of values f(j), j 1, 2,..., n, such that (f(j), n) 1. Show that ψ(n) s ultplcatve and that ψ(p t ) p t 1 ψ(p). Conclude that 0 ψ(n) p n ψ(p)/p. 7

8 Proble 13. Fnd closed expressons for the followng sus: µ(d)τ(d) µ(d)ϕ(d) µ(d)σ(d) µ 2 (d)ϕ 2 (d) µ(d)τ( n d ) µ(d) ϕ(d) µ(d)σ( n d ) µ( n d )ln d µ(d) d (t,n)1 1 t<n t 1 Proble 14. Consder the functon ζ(s), the so-called Reann zeta-functon. ns n1 It converges for s > 1. In fact, one can extend t to an analytc functon over the whole coplex plane except s 1. The faous Reann Conjecture clas that all zeros of ζ(s) n the strp 0 Re s 1 le on the lne Re s 1/2. For ζ(s) show the followng foral denttes: ζ(s) p ζ(s) p s ζ(s) 2 τ(n) n s n1 µ(n) n s ζ(s)ζ(s 1) n1 Proble 15. [Please, do not dscuss ths proble - t s on Monthly Contest 8.] Suppose that we are gven nfntely any tcets, each wth one natural nuber on t. For any n N, the nuber of tcets on whch dvsors of n are wrtten s exactly n. For exaple, the dvsors of 6, {1, 2, 3, 6}, are wrtten n soe varaton on 6 tcets, and no other tcet has these nubers wrtten on t. Prove that any nuber n N s wrtten on at least one tcet. n1 σ(n) n s Proble 16. Let f(n) : N N be ultplcatve and strctly ncreasng. If f(2) 2, then f(n) n for all n. 7. Probles wth [x] and Multplcatve Functons [ n ] Proble 17. Prove that τ() and σ() [ n ]. 8

9 q 1 [ ] p Proble 18. Prove that for (p, q) 1: q p 1 [ ] q. p Proble 19. Prove that for n N: 1 <n ϕ( + 1) n 1 { } ( 1)! + 1. Proble 20. Let S(, n) { Z (od ) + n (od ) }. Fnd 8. Hnts and Solutons to Selected Probles Hnt 10. Defne a functon a : N C by a 0 f n 1 a(n) a r f n p 1 p 2 p r 0 f n not sq.free. S(,n) ϕ(). Soluton 11. We have ϕ(2 n+1 1) 2 n. If 2 n+1 1 1, then n 0 and ϕ(1) 1. Otherwse, 2 n+1 1 p α 1 1 pα 2 2 pαr r > 1 wth all p s odd. Fro a prevous forula, ϕ(p α 1 1 pα 2 2 pαr r ) 2 n p α p α p αr 1 r (p 1 1)(p 2 1) (p r 1). Therefore, all α 1, 2 n (p 1 1)(p 2 1) (p r 1), and all p 2 s + 1 for soe s 1. It s easy to see that f s has an odd dvsor > 1, then 2 s + 1 wll factor, ang p non-pre. Hence p 2 2q + 1 for soe q 0. We have reduced the proble to fndng all sets of pres p of the above type such that p 1 p 2 p r n+1 2(p 1 1)(p 2 1) (p r 1). Order the pres p 1 < p 2 < < p r. Chec consecutvely va odulo approprate powers of 2, that, f they exst, the sallest pres ust be: p 1 3, p 2 5, p 3 17, p However, f p 5 also exsts, then p , whch s not pre. Hence, r 4, and n 1, 2, 3, 4. Hnt 12. Modfy the table proof for ultplcatvty of the Euler functon ϕ(n). Hnt 13, lnd. Note that lnd s not ultplcatve. Consder the functon (n) defned by: (n) ln p f n s a power of a pre p, and (n) 0 otherwse, and use Möbus nverson. Soluton 16. We have f(1) 1 and f(2) 2. Let f(3) 3 + for N {0}. f(6) f(2) f(3) f(5) f(10) f(9) f(18) f(15) But f(15) f(3) f(5) (3 + ) (5 + ) Hence, 2 0,.e. 0 and f(3) 3. Now assue by nducton on that f(s) s for s 1, 2,..., 2 1, where 2. Then f(4 2) 2f(2 1) 4 2. Snce f(2 1) 2 1 and the functon s 9

10 strctly ncreasng, f(s) s for all s [2 1, 4 2]. In partcular, f(2) 2 and f(2 + 1) (4 2 > 2 + 1). Ths copletes the nducton step. [ n ] [ ] { n 1 1 when n, Soluton 17a. Set B n, for 1, 2,..., n. Then B n, 0 when n, and τ(n) B n,. Sung up, we obtan τ(),1 1 [ ] B, 1 n 1 n 1 1 [ ] Soluton 17b. Fro the prevous proble, B n, σ(),1 1 [ ] [ p Hnt 18. Use the dentty q (p 1)(q 1) both sus equal. 2 B, ] 1 n 1 n 1 1 [ ] ([ ] 1 n n [ ]) 1 [ ]. { when n, 0 when n, ( [ ] Hnt 19. Both sdes count the nuber of pres p n. Hnt 20. The gven su equals ([ ] + n [ [ ] l ϕ(). Further, where Σ(l) 1 n n and σ(n) [ ]) 1 [ ]. B n,. [ ] (q )p p 1 for 1, 2,..., q 1, and show that q ] [ n ] ) ϕ() Σ( + n) Σ() Σ(n), Σ(l) 1 q l d q ( l + 1 ϕ(d) 2 ). 10