Motose, Kaoru

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Mathematical Journal of OkayamaUniversityVolume 47, Issue 1 2005 Article 5JANUARY 2005Ramanujan’s Sums and CyclotomicPolynomialsKaoru Motose∗∗Hirosaki UniversityCopyright c©2005 by the authors. Mathematical Journal of Okayama University is produced byThe Berkeley Electronic Press (bepress). http://escholarship.lib.okayama-u.ac.jp/mjouMath. J. Okayama Univ. 47 (2005), 65{74RAMANUJAN'S SUMS AND CYCLOTOMICPOLYNOMIALSKaoru MOTOSEThis paper consists of three parts. First, we shall show that some formulasabout Ramanujan's sum and cyclotomic polynomials are equivalent to thede¯nition of cyclotomic polynomials. Next, we shall show also that cyclo-tomic polynomials are strictly increasing for x ¸ 32 : The last part is to showthat values of some classical series are equivalent to ¯rst three coe±cientsof s in the inverse 1³(s+1) of the zeta function.Let ¡n be the cyclic group of n th roots of unity in the complex number¯eld C and let ¢n be the set of primitive n th roots of unity. Then we have¡n = f³kn j0 · k · n¡ 1g and ¢n = f³kn j (k; n) = 1; 1 · k · ngwhere ³n = e2¼in and (k; n) is the greatest common divisor of k and n:Classifying elements of ¡n by the orders, ¡n is the disjoint union of subsets¢d with djn; namely,¡n =[djn¢d:Let n be a natural number and let k be an integer. Then Ramanujan'ssum cn(k) is de¯ned as follows (see [1, p.55-56]):cn(k) =X´2¢n´k:x1. From the de¯nition of Ramanujan's sum, we can see immediatelysome statements in Lemma 1. For these statements, we use the followingnotation±kjn =½1 if kjn0 otherwise.From (1) in Lemma 1, it is natural to consider a polynomial with coe±-cients of Ramanujan's sums, namely,Rn(x) = cn(0) + cn(1)x+ cn(2)x2 + ¢ ¢ ¢+ cn(n¡ 1)xn¡1:Lemma 1. Let ' denote Euler's function, let ¹ be MÄobius function. Thenwe have(1) cn(k) = cn(`) if k ´ ` mod n:(2) Rn(0) = cn(0) = '(n):(3) cn(k`) = cn(`) if (k; n) = 1:This paper was ¯nancially supported by Fund for the Promotion of International Sci-enti¯c Research B-2, 2004, Aomori, Japan.651Motose: Ramanujan's Sums and Cyclotomic PolynomialsProduced by The Berkeley Electronic Press, 200566 K. MOTOSE(4) cn(k) = cn(¡k) = cn(n¡ k):(5) Rn(1) =Pn¡1k=0 cn(k) = 0:(6)Pdjn cd(k) = ±njkn, namely, cn(k) =Pdjn ¹(nd )d±djk (see [1, p.237]).(7) cn(k) = cn(`) if (k; n) = (`; n):(8) cn(k) = cn(1) = ¹(n) if (k; n) = 1:(9) cst(k) = cs(k)ct(k) if (s; t) = 1 (see [1, p.56]).(10) Rn(x)1¡xn =Pk¸0 cn(k)xk =Pdjn ¹(nd )d1¡xd :Proof. (1) » (4) are easy from the de¯nition.(5) follows from the next equations.n¡1Xk=0cn(k) =n¡1Xk=0X´2¢n´k =X´2¢nfn¡1Xk=0´kg = 0where ´ runs through the primitive nth roots of unity.(6): Classifying elements by orders in the cyclic group ¡n =Sdjn¢d; wehave Xdjncd(k) =XdjnXµ2¢dµk =Xµ2¡nµk =n¡1X`=0(³kn)` = ±njkn:(7) and (8) are easy from (6).(9): It is clear that (k; st) = (k; s)(k; t) and for every divisor d of st, thereexist d1; d2 such that d = d1d2; d1 j s and d2 j t; and conversely. Thus (9)follows from (6) and the following equations.cst(k) =Xdj(k;st)¹(std)d = (Xd1j(k;s)¹(sd1)d1)(Xd2j(k;t)¹(td2)d2) = cs(k)ct(k):(10) follows from the following equations.Rn(x) =n¡1Xk=0(X´2¢n´k)xk =X´2¢nn¡1Xk=0(´x)k=X´2¢n1¡ (´x)n1¡ ´x = (1¡ xn)X´2¢nXk¸0(´x)k= (1¡ xn)Xk¸0cn(k)xk:2Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 5http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/5RAMANUJAN'S SUMS AND CYCLOTOMIC POLYNOMIALS 67Moreover, we have from (6).Rn(x) =n¡1Xk=0cn(k)xk =n¡1Xk=0Xdjn¹(nd)d±djkxk =Xdjn¹(nd)dn¡1Xk=0±djkxk=Xdjn¹(nd)dnd¡1X`=0xd` =Xdjn¹(nd)dxn ¡ 1xd ¡ 1 :¤Remarks. 1). The formula in Lemma 1 (6) is very useful.cn(k) =Xdjn¹(nd)d±djk:This formula shows that cn(k) is an integer and cn(k) represents importantarithmetic functions, in fact, about MÄobius ¹ function, we can see ¹(n) =cn(1) from this and we have the well known important property:Xdjn¹(d) =Xdjncd(1) = n±nj1 =½n for n = 10 for n > 1:About Euler ' function, we have '(n) = cn(0) from the de¯nition of cn(k)and we have usual formula'(n) = cn(0) =Xdjn¹(nd)d and n = n±nj0 =Xdjncd(0) =Xdjn'(d):2). From Lemma 1 (6) and (9), we can compute cn(k) by the primefactorizations of n and k as follows: In case n = pr is a power of a prime pand k = psk0 with (k0; p) = 1; we havecpr(k) = pr±prjk ¡ pr¡1±pr¡1jk =8<: pr ¡ pr¡1 for s ¸ r¡pr¡1 for s = r ¡ 10 for s < r ¡ 1:In general, for prime factorizations of n =Qsi=1 prii ; using Lemma 1 (9), wehave cn(k) =Qsi=1 cprii(k): Thus we have the next formula (see [1, p. 238])from this consideration,cn(k) =¹( n(k;n))'(n)'( n(k;n)):This means that Ramanujan's sums cn(m) are equivalent to Euler's function'(n) + MÄobius function ¹(n).3Motose: Ramanujan's Sums and Cyclotomic PolynomialsProduced by The Berkeley Electronic Press, 200568 K. MOTOSE3). Moreover, from Lemma 1 (6), we have (see [1, p. 250-251]).³(s)Xn¸1cn(k)ns=Xn¸1Pdjn cd(k)ns=Xn¸1±njknns=½¾s¡1(k)ks¡1 for k > 0 and s > 1³(s¡ 1) for k = 0 and s > 2where ³(s) is the Riemann zeta function and ¾`(m) =Pdjm d`.In particular,Xn¸1cn(k)n2=6¼2¾1(k)kfor k > 0 andXn¸1'(n)ns=³(s¡ 1)³(s)for s > 2:4). In Lemma 1, (6) and (10) are equivalent. We can see already that(10) follows from (6). On the other hand, we have (6) from (10) and nextequations.Xk¸0cn(k)xk =Xdjn¹(nd)d11¡ xd =Xdjn¹(nd)dX`¸0xd` =X`¸0(Xdjn¹(nd)d)xd`:In the next theorem we shall show some formulas are equivalent to thede¯nition of cyclotomic polynomials.Theorem 1. The next formulas are all equivalent for monic polynomials©n(x) over C.(1) xn ¡ 1 =Qdjn©d(x), namely, ©n(x) =Qdjn(xd ¡ 1)¹(nd ):(2) ©n(x) =Q´2¢n(x¡ ´):(3)Pm¸0cn(m)xm = x©0n(x)©n(x); where ©0n(x) =d©n(x)dx :(4) Rn(x)1¡xn =1x©0n(1x)©n(1x):(5) x©0n(x)©n(x)=Pdjn ¹(nd )dxdxd¡1 , namely,nxnxn¡1 =Pdjn x©0d(x)©d(x):Proof. (1) ) (2): Let ® be of order n. Then there exists a divisor d of nwith ©d(®) = 0 and so ®d ¡ 1 = 0 , namely, d = n: Conversely, let d be theorder of ¯ and assume ©n(¯) = 0: Then d is a divisor of n: If d < n, thenwe obtain(xd)nd¡1 + ¢ ¢ ¢+ xd + 1 = xn ¡ 1xd ¡ 1 =Qhjn©h(x)Qhjd©h(x)= ©n(x)g(x)where g(x) is 1 or a product of cyclotomic polynomials. Thus we have (2)from the next contradiction.nd= (¯d)nd¡1 + ¢ ¢ ¢+ ¯d + 1 = ©n(¯)g(¯) = 0:4Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 5http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/5RAMANUJAN'S SUMS AND CYCLOTOMIC POLYNOMIALS 69(2) ) (3): The next equations show (3) from (2)x©0n(x)©n(x)=X´2¢nxx¡ ´ =X´2¢n11¡ ´x=X´2¢nXm¸0´mxm=Xm¸0X´2¢n´mxm=Xm¸0cn(m)xm:(3) ) (4): From (3) and Lemma 1 (1) (or Lemma 1 (10)), we obtainRn(x)1¡ xn =Xk¸0Rn(x)xnk =Xk¸0cn(k)xk =1x©0n(1x)©n( 1x):(4) ) (5): It follows from (4) and Lemma 1 (6) thatx©0n(x)©n(x)=Rn( 1x)1¡ 1xn=Xk¸0cn(k)xk=Xk¸0Xdjn¹(nd)d±djkxk=Xdjn¹(nd)dXk¸0±djkxk=Xdjn¹(nd)dX`¸01xd`=Xdjn¹(nd)d11¡ 1xd=Xdjn¹(nd)dxdxd ¡ 1 :(5) ) (1): We set S(x) =Qdjn©d(x): It follows from (5) thatS(x)nxn¡1xn ¡ 1 =XdjnS(x)©d(x)0©d(x)= S(x)0 and so S(x)0(xn ¡ 1) = S(x)nxn¡1:Thus we have µS(x)xn ¡ 1¶0= 0 and hence S(x) = c(xn ¡ 1):Since S(x) is monic, we have our assertion. ¤Remark. We shall show again the formulas Lemma 1 (6) from Theorem 1.We can see the next equations from Theorem 1 (5).x©0n(x)©n(x)=Xdjn¹(nd)d11¡ 1xd=Xdjn¹(nd)dXk¸01xkd=Xk¸0Xdjn¹(nd )dxkd:Thus we obtain from Theorem 1 (3)Xm¸0cn(m)xm=Xk¸0Xdjn¹(nd )dxkd:5Motose: Ramanujan's Sums and Cyclotomic PolynomialsProduced by The Berkeley Electronic Press, 200570 K. MOTOSEThe formula in Lemma 1 (6) also follows from the above equation.cn(m) =Xdjm; djn¹(nd)d =Xdjn¹(nd)d±djm:x2. We stated in [5, Corollary] that cyclotomic polynomials are strictlyincreasing for x > 1: However, it was incorrect and we restated in [6, Theo-rem 3] that this hold for x ¸ 2: We ¯rmly believe that it is valid for x > 1:However we cannot prove now it. It is equivalent to Rn(x) > 0 for 1 > x > 0by Theorem 1 (4), and it is also equivalent to Rn(x)0 · 0 for 1 > x > 0because Rn(0) = '(n) and Rn(1) = 0: In this section, we shall prove that itis true for x ¸ 32 :Proposition 1. Let n > 1 be a natural number and let r be the number ofdistinct odd prime factors in n: Then we have ©n(x) is strictly increasingfor x ¸ 1 + 12¢r! : In particular, ©n(x) is strictly increasing for x ¸ 32 :Proof. Let t be the product of distinct primes dividing n: Then we have©n(x) = ©t(xnt ): Thus we may assume n is square free. The conditionx ¸ 1 + 12¢r! is equivalent to2 ¢ r! ¸ 1x¡ 1 :We set f(x) = xd ¡ 1¡ dx+ d: Then f(1) = 0 and f(x) is not decreasingfor x ¸ 1: Thus we have the next inequality1x¡ 1 ¸dxd ¡ 1 >dxd + 1for a natural number d:We set n = p0m and m = p1p2 ¢ ¢ ¢ pr where p0 = 1 or 2, andp1; p2; : : : ; pr are distinct odd primes. Then we have'(n) = '(m) = (p1 ¡ 1)(p2 ¡ 1) ¢ ¢ ¢ (pr ¡ 1) ¸ 2 ¢ 4 ¢ 6 ¢ ¢ ¢ 2r = 2r ¢ r!:In case p0 = 1; we have the next inequality from the above and©n(x) =Qdjn(xd ¡ 1)¹(nd ):x©0n(x)©n(x)=Xdjn¹(nd)dxdxd ¡ 1 =Xdjn¹(nd)d(1 +1xd ¡ 1)= '(n) +Xdjn¹(nd)dxd ¡ 1> '(n)¡X¹(nd)=¡11x¡ 1 = '(n)¡2r¡1x¡ 1¸ '(n)¡ 2r ¢ r! ¸ 0:6Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 5http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/5RAMANUJAN'S SUMS AND CYCLOTOMIC POLYNOMIALS 71In case p0 = 2; we have the next inequality from the above and©n(x) = ©2m(x) =Qdjm(xd + 1)¹(md):x©0n(x)©n(x)=Xdjm¹(md)dxdxd + 1=Xdjm¹(md)d(1¡ 1xd + 1)= '(m)¡Xdjm¹(md)dxd + 1> '(m)¡X¹(md)=11x¡ 1 = '(n)¡2r¡1x¡ 1¸ '(n)¡ 2r ¢ r! ¸ 0:¤x3. In this section, we shall show that values of next series u(k) fork = 0; 1; 2 are equal to ¯rst three coe±cients of s in the inverse 1³(s+1) ofthe zeta function. These results are closely related to the prime numbertheorem (see [3]).We de¯ne the next series and a sum. Since u(k) is convergent (see [3,pp.594-595, x158]), r`(k) is convergent (see Lemma 2 (2)).r`(k) :=Xn¸1cn(`)(logn)kn:u(k) := r1(k) =Xn¸1¹(n)(logn)kn:v`(k) :=Xdj`(log d)k:Concerning three values u(k) for k = 0; 1; 2; u(0) = 0 was proved byMangoldt 1898 [4], u(1) = ¡1 by Landau 1899 [2, 3] and these general,but equivalent, results were proved by Ramanujan 1918 [7], in addition tou(2) = ¡2° where ° is Euler's constant. However, we can see from Theorem3 that these values were already found by Mangoldt 1898 ([4, p.440]). Theseare combined by the next lemma.Lemma 2. We obtain assertions.(1)1³(s+ 1)=mXk=0(¡1)ku(k)skk!+O(sm+1) as s! 0:(2) r`(k) =kXt=0µkt¶u(t)v`(k ¡ t):7Motose: Ramanujan's Sums and Cyclotomic PolynomialsProduced by The Berkeley Electronic Press, 200572 K. MOTOSEProof. (1) We set f(s) = 1³(s+1) : It is well known thatf (k)(t) = (¡1)kXn¸1¹(n)(logn)knt+1for t > 0:Since u(k) is convergent from [3, pp.594-595, x158] and f (k)(t) is right con-tinuous at a point 0 from [3, p.107, x30], we havelimt!+0f (k)(t) = (¡1)ku(k):On the other hand, for s > 0 in the neighbourhood of t > 0;f(s) =mXk=0f (k)(t)(s¡ t)kk!+O((s¡ t)m+1):Hence, making t! +0; we havef(s) =mXk=0(¡1)ku(k)skk!+O(sm+1) as s! 0:(2) Using Lemma 1 (6), we obtain (2) from the next equations.r`(k) =Xn¸1cn(`)(logn)kn=Xn¸1fXhjn;hj`¹(nh)hg(log n)kn=Xh0¸1Xhj`¹(h0)(log h0 + log h)kh0=Xh0¸1Xhj`¹(h0)h0kXt=0µkt¶(log h0)t(log h)k¡t=kXt=0µkt¶fXh0¸1¹(h0)h0(log h0)tgfXhj`(log h)k¡tg=kXt=0µkt¶u(t)v`(k ¡ t):¤The next was stated in [4, p. 440].Theorem 2 (Mangoldt).1³(s+ 1)= s¡ °s2 +O(s3) as s! 0:8Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 5http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/5RAMANUJAN'S SUMS AND CYCLOTOMIC POLYNOMIALS 73Proof. It is well known from [4, p. 440](see also [3, p.164]) that³(s+ 1) =1s+ ° +O(s) as s! 0:Thus we have1³(s+ 1)=s1 + °s+O(s2)= s¡ °s2 +O(s3) as s! 0:¤The following theorem is our purpose in this section.Theorem 3. Let d(`) be the number of positive divisors of ` and let ° beEuler's constant. Then we obtain equivalent results.(a)1³(s+ 1)= s¡ °s2 +O(s3) as s! 0:(b) u(0) = 0; u(1) = ¡1 and u(2) = ¡2°:(c) r`(0) = 0; r`(1) = ¡d(`) and r`(2) = ¡d(`)(2° + log `):Proof. (a) , (b): It is clear from Lemma 2(1).(b) , (c): (b) is a special case of (c) because u(k) = r1(k): The ¯rsttwo equations in (c) are easy from (b) and Lemma 2(2). Using (Qtj` t)2 =Qtj` t`t = `d(`), the last equation in (c) follows from (b), Lemma 2(2) andthe next equations.r`(2) = u(2)v`(0) + 2u(1)v`(1) + u(0)v`(2) = ¡2°d(`)¡ 2Xtj`log t= ¡2°d(`)¡ 2 log(Ytj`t) = ¡2°d(`)¡ d(`) log `:¤Acknowledgement. I would like to thank the referee for useful advicesand corrections. Moreover I wish to thank Professor Noriaki Nagase for hisvaluable comments.References[1] G. H. Hardy and E.M. Wright, An Introduction to the Theory of Numbers,Clarendon Press, 1979.[2] E. Landau, Contribution µa la th¶eorie de la fonction ³(s) de Riemann, Comptes rendus,129(1899), 812-815.[3] E. Landau, Handbuch der Lehre von der Verteilung der Primezahlen , Chelsea, 1909.[4] H. Von Mangoldt, D¶emonstration de l'¶equationP1k=1¹(k)k= 0, Annalesscienti¯ques de l'¶Ecole Normale sup¶erieure, 15(1898), 431-454.[5] K. Motose, On values of cyclotomic polynomials, Math. J. Okayama Univ., 35(1993),35-40.9Motose: Ramanujan's Sums and Cyclotomic PolynomialsProduced by The Berkeley Electronic Press, 200574 K. MOTOSE[6] K. Motose, On values of cyclotomic polynomials. VII, Bull. Fac. Sci. Tech. HirosakiUniv., 7(2004), 1-7.[7] S. Ramanujan, On certain trigonometrical sums and their application in the theoryof numbers, Collected Papers, Chelsea, 1927, pp. 179-199.Kaoru MotoseDepartment of Mathematical System ScienceFaculty of Science and TechnologyHirosaki UniversityHirosaki 036-8561, Japane-mail address: skm@cc.hirosaki-u.ac.jp(Received October 18, 2004 )(Revised January 5, 2005 )10Mathematical Journal of Okayama University, Vol. 47 [2005], Iss. 1, Art. 5http://escholarship.lib.okayama-u.ac.jp/mjou/vol47/iss1/5

Ramanujan's Sums and Cyclotomic Polynomials

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