The Euler Totient, the Möbius and the Divisor Functions


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1 The Euler Totiet, the Möbius ad the Divisor Fuctios Rosica Dieva July 29, 2005 Mout Holyoke College South Hadley, MA
2 Ackowledgemets This work was supported by the Mout Holyoke College fellowship for summer research ad the Mout Holyoke College Departmet of Mathematics ad Statistics. I would like to thak Professor Margaret Robiso for all the help ad guidace she gave me this summer. I would also like to thak Professor Jessica Sidema ad all the other studets participatig i the Mout Holyoke College Summer REU 2005 for their support ad time. 2
3 1 Itroductio The theory of umbers is a area of mathematics which deals with the properties of whole ad ratioal umbers. Aalytic umber theory is oe of its braches, which ivolves study of arithmetical fuctios, their properties ad the iterrelatioships that exist amog these fuctios. I this paper I will itroduce some of the three very importat examples of arithmetical fuctios, as well as a cocept of the possible operatios we ca use with them. There are four propositios which are metioed i this paper ad I have used the defiitios of these arithmetical fuctios ad some Lemmas which reflect their properties, i order to prove them. 2 Defiitios Here are some defiitios to illustrate how the fuctios work ad describe some of their most useful properties. 2.1 Arithmetical fuctio A real or complex valued fuctio with domai the positive itegers is called a arithmetical or a umbertheoretic fuctio. 2.2 Multiplicative fuctios A arithmetical fuctio f is called multiplicative if f is ot idetically zero ad if f(m) = f(m)f() wheever (m, ) = 1. A multiplicative fuctio f is called completely multiplicative if f(m) = f(m)f() for all m,. 3
4 2.3 The Möbius fuctio The Möbius fuctio is a arithmetical fuctio, which takes the followig values: µ(1) = 1 ad for = p a 1 1 p a p a m m, where > 1, we defie µ() to be: µ() = ( 1) m if a 1 = a 2 =... = a m = 1, µ() = 0 otherwise. This defiitio implies that the Möbius fuctio will be zero if ad oly if has a square factor larger tha oe. Let us look at a short table of the values of µ() for some positive itegers: µ() The Möbius fuctio is a example of a multiplicative, but ot completely multiplicative fuctio, sice φ(4) = 0 but φ(2)φ(2) = 1. Oe if its most 4
5 importat applicatios is i the formulas for the Euler totiet, which is the ext fuctio I will defie. 2.4 The Euler totiet The Euler totiet fuctio is defied to be the umber of positive itegers which are less or equal to a iteger ad are relatively prime to that iteger: for 1, the Euler totiet φ() is: φ() = 1, where the idicates that the sum is oly over the itegers relatively prime to. Below is a table of the values of φ() for some small positive itegers: φ() There is a formula for the divisor sum which is oe of the most useful properties of the Euler totiet: 5
6 Lemma 1: for 1 we have φ(d) =. d Sice the Euler totiet is the umber of positive itegers relatively prime to we ca calculate φ() as a product over the prime divisors of, where 1: Lemma 2: φ() = p (1 1 p ). The followig formula gives a relatio betwee the Euler totiet ad the Möbius fuctio: Lemma 3: for 1 we have: φ() = d µ(d) d. The Euler totiet is aother multiplicative fuctio which is ot completely multiplicative because φ(4) = 2 but φ(2)φ(2) = The divisor fuctios For a real or a complex umber α ad a iteger 1 we defie σ α () = d d α to be the sum of the αth powers of the divisors of, called the divisor fuctio σ α (). These fuctios are also multiplicative. 6
7 If we look at the trivial case whe α = 0 we say that σ 0 () is the umber of divisors of. I the case that α = 1 we defie σ 1 () as the sum of the divisors of. Sice the fuctio is multiplicative we kow that for = p a 1 1 p a p am m the σ α () = σ α (p a 1 1 σ α (p a 2 2 )...σ α (p am m ).There is a formula for the divisor fuctio of a iteger power of a prime: Lemma 3: σ α (p a ) = 1 α + p α + p 2α p aα = pα(a+1) 1 p α 1 if α 0 σ 0 (p a ) = a + 1 if α = 0 The ext defiitio I will itroduce is the Dirichlet product of arithmetical fuctios, which is represeted by a sum, occurrig very ofte i umber theory. 2.6 Dirichlet product of arithmetical fuctios The Dirichlet product of two arithmetical fuctios f ad g is defied to be a arithmetical fuctio h() such that: (f g)() = h() = d f(d)g( d ). If we look at the formula for the relatio betwee the Euler totiet ad the Möbius fuctio, we will see that for a fuctio N, such that N() = the φ = µ N. 7
8 3 Propositios ad their proofs 3.1 Propositio 1 For a positive iteger we have that: φ() = d µ 2 (d) φ(d) where the sum is over all the divisors of. Proof: We kow by Lemma 2 that φ() = p (1 1 p ) so if we let = p a 1 1 p a p a m m, we ca express φ() i the followig φ() = (1 1 p 1 ) (1 1 p 2 )... (1 1 p m ) Takig a commo deomiator for each of the terms i the paretheses we see that: Thus we have that φ() = (p 1 1) (p 2 1)... (p m 1) p 1 p 2... p m. φ() = (p 1 1) (p 2 1)... (p m = 1) p 1 p 2... p m p 1 p 2... p m (p 1 1) (p 2 1)... (p m 1) This equatio is our result for the left had side of the idetity we have to prove. We will deote φ() with the iitials LHS for the rest of the proof. 8
9 Now we look at the right had side of the idetity above. From the defiitio of the Möbius fuctio we kow that for = p a 1 1 p a p a m m µ() = ( 1) m ad µ() = 0, whe has a square term. Therefore µ 2 (d) = 1 if d has o square term ad µ 2 (d) = 0 if d has a square term. Thus our sum will be over oly the square free divisors of sice if a divisor is ot square free we will have a zero term. For the rest of this paper d 1 will represet a divisor of which is square free: d 1 µ 2 (d 1 ) φ(d 1 ) = 1 φ(d 1 ) d 1 Sice d 1 is square free d 1 will be ay product of the prime factors of, where each prime could be used oly oce i the prime factorizatio of each divisor d 1. The last statemet meas that d 1 takes o each of the values 1, p 1,..., p m, p 1 p 2, p 1 p 3,..., p m 1 p m, p 1 p 2 p 3,..., p 1 p 2... p m. The RHS will thus become d 1 1 φ(d 1 ) = 1 φ(1) + 1 φ(p 1 ) φ(p 1 p 2 ) φ(p 1 p 2... p m ) d 1 1 φ(d 1 ) = 1+ 1 p (p 1 1) (p 2 1) (p 1 1)... (p m 1) The commo deomiator of this sum will be (p 1 1) (p 2 1)...(p m 1), so after we get the sum over a commo deomiator the right had side becomes: RHS = (p 1 1)... (p m 1) + (p 2 1)... (p m 1) (p m 1) (p 1 1) (p 2 1)... (p m 1) 9
10 We ca ow rearrage the terms i the umerator, startig with the last oe ad for the umerator of the right had side, RHS N, we get the followig: RHS N = 1+(p 1 1)+...+(p m 1)+(p 1 1) (p 2 1)+...+(p 1 1) (p 2 1)... (p m 1) Whe we look carefully at each of the terms i this equatio we ca see that each term is actually the φ fuctio of some prime or of some product of primes. We ca therefore rewrite the umerator as: RHS N = φ(1) + φ(p 1 ) φ(p m ) + φ(p 1 p 2 ) φ(p 1 p 2... p m ), Thus for the RHS N we fid that: RHS N = By Lemma 1 we kow that: l p 1 p 2... p m φ(l) φ(d) =, d thus whe = p 1 p 2...p m we have the RHS N = p 1 p 2...p m. The right had side of the idetity we wat to prove becomes RHS = sice this equals the LHS, our idetity is prove: p 1... p m (p 1 1)... (p m 1) ad φ() = d µ 2 (d) φ(d). 10
11 3.2 Propositio 2 For all with at most 8 distict prime factors we have that φ() > 6. Proof: We will first prove the propositio for a with 8 distict prime factors. Let = p a 1 1 p a p a 8 8 so usig Lemma 2 for the Euler totiet we see that: φ() = p a 1 1 p a p a 8 8 (1 1 p 1 )(1 1 p 2 )...(1 1 p 8 ) Sice p 1, p 2, p 3, p 4, p 5, p 6, p 7, p 8 are distict prime factors we kow that p 1 2, p 2 3, p 3 5, p 4 7, p 5 11, p 6 13, p 7 17, p 8 19, because these are the first 8 distict primes. Therefore 1 p 1 1 2,..., 1 p , 1 p ad the (1 1 p 1 ) (1 1),...(1 1 2 p 8 ) (1 1 ). We ca ow substitute i 19 the equatio for φ() ad sice we will substitute each term i paretheses with a term which is less or equal to the iitial oe we will get: therefore φ() (1 1 2 )...( ), φ() , ad the φ() , so φ() 0, 171. But 6 that φ() > 6 for = pa p a , 167, which meas Each of the factors (1 1 p 1 ), (1 1 p 2 ),..., (1 1 p 8 ) is less that oe sice the smallest possible prime is two, so each of the terms i paretheses will be less tha oe, which meas that whe we multiply the product by it, we will decrease its value. So if our iteger has less tha 8 distict prime factors 11
12 the value for its Euler totiet will be greater tha the value of the Euler totiet of a iteger with 8 distict prime factors. Thus, we have proved that for all itegers with 8 or less distict prime factors φ() > Propositio 3 Let f(x) be defied for all ratioal x i 0 x 1 ad let F () = f( k ) The F () = (k,)=1 f( k ) A)F = µ F, the Dirichlet product of µ ad F. Proof: Let us look at the Dirichlet product of the two fuctios. µ F = d µ(d)f ( d ) = d µ(d) d f( kd ). Agai, sice we have the Möbius fuctio all divisors d, which are ot square free will give us zero for the sum. We will deote the square free divisors by d 1 ad let = p a 1 1 p a p a m m. The the divisors d 1 will be all the primes p 1,..., p m ad all the possible products of these primes. The Möbius fuctio will take the values 1 ad 1, whe we have odd ad eve umber of primes i our divisors, respectively. The our Dirichlet product will become: 12
13 µ F = d 1 µ(d 1 ) d 1 f( kd 1 ) = A A = F () F ( )... F ( )+F ( )+...+F ( )...+( 1) m F (1), p 1 p m p 1 p 2 p m 1 p m ad whe we substitute with the formula we have for F () we get: A = f( k ) p 1 f( kp 1 )... pm f( kp m ) + p 1 p 2 f( kp 1p 2 ) p m 1 pm We kow that F () = f( kp m 1p m )... + ( 1) m (k,)=1 p 1...pm f( k ) = f( k ) (k,) 1 f( kp 1...p m ). f( k ) Let us look at the secod term i the last differece. We have that (k, ) 1 which meas that k will have at least oe of the prime factors of. Thus, k {p i b i i [1, m], b i Z } ad sice we kow that k, the p i b i ad therefore for each value of i, b i p i. We ca the rewrite the secod term i our differece as the sum of the sums with each of the b i s as a variable. But sice we are lettig b i be aythig less tha p i each of the sums will cout the factors with more tha oe distict prime factor of i the umerator of f( k ) twice. This observatio will be true for all the subsequet sums, too  the sums which start with two distict prime factors of i the variable b i will cout all the other sums with three prime distict factors of 13
14 twice ad so o. This way whe we are writig out the sum ad start with the first group of sums  the oes for which b i has at least oe of the distict prime factors of  we will have to subtract the secod group of sums  with 2 or more of the distict prime factors of, which o its ow will subtract all the sums with 3 or more of the distict prime factors of, thus we will have to add those sums ad the we would have added twice the ext sums ad so o. Therefore for the secod term we will get a alteratig series of sums, so that we ca accout for all the itegers less tha, which have commo factors with ad get rid of all the terms which repeat. Thus, (k,) 1 f( k ) = p 1 p m 1 pm f( kp 1 ) pm f( kp m 1p m ) ( 1) m f( kp m ) p 1...pm p 1 p 2 f( kp 1p 2 )... f( kp 1...p m ) We ca ow substitute i the formula we derived for F () ad it will become: F () = (k,)=1 f( k ) = p m 1 pm f( k ) [ p 1 f( kp 1 )+...+ f( kp m 1p m ) ( 1) m pm p 1...pm f( kp m ) f( kp 1...p m )] p 1 p 2 f( kp 1p 2 )... Opeig the parethesis ad applyig the egative sig i frot of them we will get: 14
15 F () = + f( k ) p 1 p m 1 pm f( kp 1 )... pm f( kp m 1p m )... + ( 1) m f( kp m ) + p 1...pm which is exactly our result for d 1 µ(d 1) d 1 proved the propositio. f(kd 1 B)µ() is the sum of the primitive th roots of uity: µ() = (k,)=1 e 2πik/ p 1 p 2 f( kp 1...p m ), f( kp 1p 2 ) +... ) = µ F. We have Proof: Let f(x) = e 2πix  this is a valid fuctio for f(x) because the expoetial fuctio is defied for all ratioal x satisfyig the coditio o x. The for = p a 1 1 p a p am m : F () = e 2πi k F () = (k,)=1 e 2πi k Let us look at the right had side of the idetity we wat to prove. We will deote the square free divisors of d 1 ad the divisors with squares i them d 2, where d 1 ad d 2. We ca use part A of Propositio 3 ad substitute i the formula for the Dirichlet product. Thus, sice F () = (µ F )(): 15
16 F () = d µ(d)f ( d ) = µ()f (1) + d 2 µ(d 2 )F ( d 2 ) + d 1 µ(d 1 )F ( d 1 ) Whe the argumet of the Möbius fuctio has a square its value is 0. Therefore, the sum over the square divisors is zero. Let us look at the other terms i F.For = 1 F (1) = e 2πik = e 2πi = cos(2π) + isi(2π) = = 1, thus F () = µ() + d 1 µ(d 1 )F ( d 1 ). We will ow solve the secod term i our equatio. F ( d 1 ) = d 1 e 2πi d 1 k = e 2πi d 1 + e 2πi 2d e 2πi d 1 ( 2) d1 + e 2πi d 1 ( 1) d1 + e 2πi We wat to prove that µ() = F (), where f(x) = e 2πix, which meas that we eed to prove that F ( d 1 ) = 0. I order to do this we will multiply F ( d 1 ) by a fuctio which is ot of value 1 but will still retur the same fuctio. I our case e 2πi d 1 1, sice d 1 by the defiitio of d 1. Thus, e 2πi d 1 F ( ) = e 2πi d 1 d 1 ad summig over we get d 1 d 1 e 2πi d 1 k = e 2πi (k+1)d 1 = A, A = e 2πi 2d 1 + e 2πi 3d e 2πi d 1 ( 2+1) d1 + e 2πi d 1 ( 1+1) d1 + e 2πi d 1 ( +1) d1 16
17 If we look carefully at the terms of both products we will see that if we rearrage the terms ad let the last term of the secod product become the first the two products will be completely the same. The, e 2πi d 1 F ( d 1 ) = F ( d 1 ), which meas that e 2πi d 1 F ( d 1 ) F ( d 1 ) = 0 ad we kow that e 2πi d 1 1 by the defiitio of d 1 ad sice the oly solutios to F ( d 1 )(e 2πi d 1 1) = 0 are e 2πi d 1 = 1 or F ( d 1 ) = 0, therefore F ( d 1 ) = 0. So the the right had side of the idetity we wat to prove becomes F () = µ() + d 2 0.F ( d 2 ) + d 1 µ(d 1).0 = µ() ad by the idetity we have from part A of Propositio 3 we have thus prove 3.4 Propositio 4 For 1 we have µ() = (k,)=1 e 2πik/. σ 1 () = d φ(d)σ 0 ( d ). Proof: Let = p a 1 1 p a p am m. By the defiitio of the divisor fuctio σ α ad the Euler totiet we kow that they are both multiplicative which meas that, for example, σ 1 () = σ 1 (p a 1 1 p a p a m m ) = σ 1 (p a 1 1 ) σ 1 (p a 2 2 )... σ 1 (p a m m ). From the properties of the divisor fuctio it follows that σ α (p a ) = pα(a+1) 1 p α 1 whe α 0 ad σ α (p a ) = a+1 whe α = 0. Therefore i our case whe α = 1 we solve the equatio: σ 1 () = σ 1 (p a 1 1 p a p am m ) = σ 1 (p a 1 1 ) σ 1 (p a 2 2 )... σ 1 (p am m ) = 17
18 p a p pam+1... m 1 p 2 1 p m 1 pa2+1 We will call this side the left had side, LHS. Let us look at the right had side, RHS, of the equatio. We see that our variable i this case is d, which represets the divisors of. These divisors will have the form d = p i 1 1 p i p i m m where 0 i 1, i 2,..., i r a 1, a 2,..., a r, respectively. This form of each of the divisors will allow us to represet them all sice this way we will be able to cout for the divisors with differet primes ad the differet powers these primes could have. The whe we substitute for d the RHS will become RHS = d φ(d)σ 0 ( a 1 d ) = a 2 i 1 =0 i 2 =0... a m i m =0 φ(p i 1 1 p i p i m m )σ o ( p i 1 1 p i ) p i m for which be the defiitio of ad sice the Euler totiet is a multiplicative fuctio we get the followig result: a 1 RHS =... i 1 =0 a m i m =0 φ(p i 1 1 )... φ(p i 2 2 )σ 0 ( pa 1 1 p a p a m m p i 1 1 p i ) p i m ad whe we divide i the argumet of the divisor fuctio we geerate a 1 RHS =... i 1 =0 a m i m =0 φ(p i 1 1 )... φ(p i 2 2 )σ 0 (p a 1 i 1 1 p a 2 i p a m i m m ) ad agai because the divisor fuctio is also multiplicative we ca write the equatio ad rearrage its terms so that a 1 RHS =... i 1 =0 a m i m=0 φ(p i 1 1 )σ 0 (p a 1 i 1 1 ) φ(p i 2 1 )σ 0 (p a 2 i 2 2 )... φ(p im m )σ 0 (p am im m ) 18
19 We have msums ad for each oe of them oly two terms of the equatios are variable  the oes whose variables are the respective i j, where j = 1,..., r. We ca the rearrage the terms so that each couple of terms will be summed over i the appropriate sum. We are allowed to do that sice for each sum oly terms chage ad the rest m 1 couples of terms are costats, which we ca get i frot of the sum. Applyig this rule m times we rearrage the RHS: a 1 RHS = i 1 =0 a 2 1 ) φ(p i 1 1 )σ 0 (p a 1 i 1 i 2 =0 φ(p i 2 2 )σ 0 (p a 2 i 2 2 )... a m i m =0 φ(p i m m )σ 0 (p a m i m m ) The fuctios for each sum are the same, the oly differece beig that they deped o a differet prime umber. The variables for these fuctios assume the same values, so it will be eough to solve for oe of these sums ad the this result will apply to all the other sums takig ito accout their respective primes. Let us take the first sum ad solve for it: We will deote the first sum with A, therefore ad the a 1 i 1 =0 φ(p i 1 1 )σ 0 (p a 1 i 1 1 ) = A A = φ(1)σ 0 (p a 1 1 )+φ(p 1 )σ 0 (p a )+...+φ(p a )σ 0 (p 2 1)+φ(p a )σ 0 (p 1 )+φ(p a 1 1 )σ 0 (1) whe we apply the formulas we kow for the fuctios i questio we get A = a 1 +1+(p 1 1)(a 1 1+1)+p 1 (p 1 1)(a 1 2+1)+p 2 1(p 1 1)(a 1 3+1)+ 19
20 +p 3 1(p 1 1)(a ) p a (p 1 1)(a 1 (a 1 1) + 1) + p a (p 1 1) we ca the factor out their commo factor so A = a (p 1 1)(a 1 + p 1 (a 1 1) + p 2 1(a 1 2) p a p a ) ad whe we ope the brackets iside A = a 1 +1+(p 1 1)(a 1 +p 1 a 1 p 1 +p 2 1a 1 2p 2 1+p 3 1a 1 3p p a p a ) we ow multiply the two terms i the brackets ad so A = a 1 +1+(a 1 p 1 a 1 +a 1 p 2 1 a 1 p 1 p 2 1 +p 1 +a 1 p 3 1 a 1 p 2 1 2p p 2 1 +a 1 p 4 1 a 1 p 3 1 3p p p a p a p a p a p a 1 1 p a 1 1 ) we ca see that some terms repeat but with opposite sigs, so these terms will give 0. Some other terms ca be combied together so whe we apply all operatios possible we will ed up with A = a ( a 1 + p 1 + p p p p a p a p a p a 1 1 ) so whe we add the first term the sum comes out to be A = 1 + p 1 + p p p p a p a p a p a 1 1 ad we ca see that this is a geometric progressio so whe we use the formula for a geometric progressio the first sum equals a 1 A = φ(p i 1 1 )σ 0 (p a 1 i 1 1 ) = 1 pa = pa p 1 p 1 1 i 1 =0 Sice this result will apply to all m sums we solve the right had side to be: a 1 a 2 a m RHS = φ(p i 1 1 )σ 0 (p a 1 i 1 1 ) φ(p i 2 2 )σ 0 (p a 2 i 2 2 )... φ(p im m )σ 0 (p am im m ) i 1 =0 i 2 =0 20 i m=0
21 so whe we substitute the result we got the product becomes RHS = pa p 1 1 which meas that 2 1 par p 2 1 p r 1 1 pa2+1 r 1 p r 1 par+1 so we proved the equality LHS = RHS σ 1 () = d φ(d)σ 0 ( d ). 21
22 Refereces [1] Tom M. Apostol, Itroductio to Aalytic Number Theory, Spriger Verlag, New York, Ic, USA,
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