DDNS Discrete Dynamics in Nature and Society 1607-887X 1026-0226 Hindawi 10.1155/2017/9092515 9092515 Research Article An Identity in Commutative Rings with Unity with Applications to Various Sums of Powers Andjić Miomir 1 http://orcid.org/0000-0001-9722-9044 Meštrović Romeo 2 Tempesta Piergiulio 1 Faculty for Information Technology University “Mediterranean” Vaka Djurovića BB Podgorica Montenegro unimediteran.net 2 Maritime Faculty Kotor University of Montenegro Dobrota 36 85330 Kotor Montenegro ucg.ac.me 2017 27 02 2017 2017 19 12 2016 08 02 2017 27 02 2017 2017 Copyright © 2017 Miomir Andjić and Romeo Meštrović. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Let R=(R,+,·) be a commutative ring of characteristic m>0 (m may be equal to +) with unity e and zero 0. Given a positive integer n<m and the so-called n-symmetric set A=a1,a2,,a2l-1,a2l such that al+i=ne-ai for each i=1,,l, define the rth power sum Sr(A) as Sr(A)=i=12lair, for r=0,1,2,. We prove that for each positive integer k there holds i=02k-1(-1)i2k-1i22k-1-iniS2k-1-i(A)=0. As an application, we obtain two new Pascal-like identities for the sums of powers of the first n-1 positive integers.

1. Introduction and Statements Results 1.1. Recurrence Formulas for <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M18"><mml:mrow><mml:mi>n</mml:mi></mml:mrow></mml:math></inline-formula>-Symmetric Sets of Commutative Rings with Unity

Here, as always in the sequel, R=(R,+,·) (briefly R) will denote a commutative ring of characteristic m>0 (m may be equal to +) with unity e and zero 0. Throughout this paper N, Z, and Zn (n=2,3,) will, respectively, denote the set of positive integers, the ring of integers, and the ring of residues modulo n.

Definition 1.

Given a positive integer n such that n<m, we say that a subset A of R is n-symmetric if it is satisfied: aR belongs to A if and only if ne-a also belongs to A.

Similarly, (a1,a2,,an)Rn is called n-symmetric tuple if it satisfies ne-ai=an+1-i for all i=1,2,,n.

It is easy to see that if 2e is not invertible in R, then a finite subset A of R is n-symmetric if and only if (1)A=a1,,al,ne-a1,,ne-al,where {a1,,al} is a finite subset of R such that ne-aiaj for all i and j with 1i<jn. Clearly, in this case B=(a1,,al,ne-al,,ne-a1) is a n-symmetric 2l-tuple of R.

If 2e is invertible element in R with the inverse (2e)-1, then a finite n-symmetric set AR is of the above form or of the form (2)A=a1,,al,ne·2e-1,ne-a1,,ne-al,where {a1,,al} is a finite subset of R such that ne-aiaj for all 1i<jn and ai(ne)·(2e)-1 for all 1in (in particular, {(ne)·(2e)-1} is a finite n-symmetric set). Clearly, in this case, B=(a1,,al,(ne)·(2e)-1,ne-al,,ne-a1) is a n-symmetric (2l+1)-tuple of R.

Our main result is as follows.

Theorem 2.

Let R=(R,+,·) be a commutative ring of characteristic m>0 with unity e and zero 0. For a nonnegative integer r and a n-symmetric set A={a1,,as}R with 0<n<m, define the rth power sum Sr(A) as (3)SrA=i=1sair.Then for each positive integer k it holds (4)i=02k-1-1i2k-1i22k-1-iniS2k-1-iA=0.

Using the obvious facts that (ie)r=ire, {e,2e,,(n-1)e} is n-symmetric set of ring R and {0,e,2e,,(n-1)e} is n-1-symmetric set of R (with S0(n)=ne) for each 1<n<m; Theorem 2 immediately yields the following result.

Corollary 3.

Let R=(R,+,·) be a commutative ring of characteristic m>0 with unity e and zero 0. For positive integers r and n>1 and a set A={e,2e,,(n-1)e}R let (5)Srn=i=1n-1ire(we also define S0(n)=(n-1)e). Then for each positive integer k it holds (6)i=02k-1-1i2k-1i22k-1-iniS2k-1-in=0,i=02k-2-1i2k-1i22k-1-in-1iS2k-1-in=nn-12k-1e.

1.2. The Application of Theorem <xref ref-type="statement" rid="thm1.2">2</xref> for the Sums of Powers on a Finite Arithmetic Progression in <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M110"><mml:mrow><mml:mi mathvariant="double-struck">Z</mml:mi></mml:mrow></mml:math></inline-formula>

Using the obvious fact that for 1<n<m and a,dZ with d0, {a,a+d,,a+(n-1)d} is a (2a+(n-1)d)-symmetric set of the ring Z of integers, as a consequence of Theorem 2 we immediately obtain the following recurrence formula for the sums of powers on a finite arithmetic progression in Z.

Corollary 4.

Let a, d0, r1, and n2 be integers, and let (7)Srn;a,d=j=0n-1a+jdr(we also define S0(n;a,d)=n). Then for each positive integer k it holds that(8)i=02k-1-1i2k-1i22k-1-i2a+n-1diS2k-1-in;a,d=0.

If a and d0 are arbitrary real numbers, then, expanding via binomial formula every term (a+jd)r (j=0,,n-1) of the power sum Sr(n;a,d)=j=0n-1(a+jd)r (with a fixed r{1,2,}), it follows by Bernoulli’s formula given in Remark 8 that Sr(n;a,d) can be expressed as a polynomial in variables n,a, and d. Hence, this is also true for the sum on the left hand side of (8) which vanishes for all integers a and d0. This yields the following extension of Corollary 4.

Corollary 5.

For complex numbers a and d0, and positive integers r1 and n2 set (9)Srn;a,d=j=0n-1a+jdr.Then formula (8) of Corollary 4 is satisfied for each positive integer k.

Remark 6.

A calculation of sums Sr(n;a,d) can be traced back to Faulhaber  in 1631 and Bernoulli  in 1713. The recurrence formula (8) is in fact a summation formula which expresses S2k-1(n;a,d) as a sum involving power sums S2k-1-i(n;a,d)(i=1,2,,2k-1) with the corresponding “coefficients” (-1)i-12k-1in/2i.

In particular, since the set {1,2,,n-1} is n-symmetric (the case when a=d=1 with n-1 instead of n in Corollary 4) and the set {0,1,2,,n-1} is n-1-symmetric (the case a=0, d=1 in Corollary 4), the formula (8) immediately yields the following two Pascal-like identities for the sums of powers of the first n-1 positive integers (cf. (6) with R=Z and e=1).

Corollary 7.

For nonnegative integers r and n2 define (10)SrnSrn;0,1=j=1n-1jr.Then for each positive integer k it holds that(11)i=02k-1-1i2k-1i22k-1-iniS2k-1-in=0,(12)i=02k-2-1i2k-1i22k-1-in-1iS2k-1-in=nn-12k-1.

Remark 8.

Finding formulas for sums of powers Sk(n) has interested mathematicians for more than 300 years since the early 17th-century mathematical publications of Faulhaber (1580–1635) . Bernoulli (1654–1705) had given a comprehensive account of these sums in his famous work Ars Conjectandi , published posthumously in 1713. The second section of Ars Conjectandi (also see [3, pp. 269-270]) contains the fundamental Bernoulli’s formula which expresses the sum Sr(n)=i=1n-1ir(r=0,1,2,) as a (r+1)th-degree polynomial function on n whose coefficients involve Bernoulli numbers. Namely, the celebrated Bernoulli’s formula (sometimes called Faulhaber’s formula) gives the sum Sk(n)(k1) explicitly given as a (k+1)th-degree polynomial of n (see, e.g., [3, Section  6.5, pp. 269-270], where it is given an induction proof on k). (13)Skn=1k+1i=0kk+1ink+1-iBi,where Bi(i=0,1,2,) are Bernoulli numbers defined by the generating function (see, e.g., ) (14)i=0Bixii!=xex-1.It is easy to find the values B0=1, B1=-1/2, B2=1/6, B4=-1/30, and Bi=0 for odd i3. Furthermore, (-1)i-1B2i>0 for all i1. It is well known that Bn=Bn(0), where Bn(x) is the classical Bernoulli polynomial (see, e.g., [4, 710]). These and many other properties can be found, for instance, in [11, Section  5.3. pp. 525–538] or . In particular (see, e.g., , where a similar formula was established), the usual recurrence is well known(15)Bk=-1k+1i=0k-1k+1iBi,B0=1,k=1,2,.Notice that Bernoulli numbers and polynomials from a more general point of view were studied by many authors (e.g., in [14, 15] the method of generating function was applied to introduce new forms of Bernoulli numbers and polynomials; also see ). Finding formulas for Sk(n) has interested mathematicians for more than 300 years since the time of Bernoulli (see, e.g., [17, 18]). Recall that, by the well-known Pascal’s identity proven by Pascal (1623–1662) in 1654 , (16)i=0kk+1iSin=nk+1-1.Recently, q-analogues of the sums of powers of consecutive integers have been investigated extensively (see, e.g., ).

For given positive integer k, the recurrence (11) presents a formula for expressing S2k-1(n) as a sum involving power sums S2k-1-i(n)(i=1,2,,2k-1) with the corresponding “coefficients” (-1)i-12k-1in/2i. For example, taking k=1,2,3,4 into (11), we, respectively, obtain (17)S1n=n2S0n,(18)S3n=3n2S2n-3n24S1n+n38S0n,(19)S5n=5n2S4n-5n22S3n+5n34S2n-5n416S1n+n532S0n,(20)S7n=7n2S6n-21n24S5n+35n38S4n-35n416S3n+21n532S2n-7n664S1n+n7128S0n,where S0(n)=n-1.

Substituting (17) into (18), we get (21)S3n=3n2S2n-n34S0n.Substituting the above formula and (17) into (19), we obtain (22)S5n=5n2S4n-5n32S2n+n52S0n.Substituting the above two formulas and (17) into (20), we find that (23)S7n=7n2S6n-35n34S4n+21n52S2n-17n78S0n.More generally, for any fixed k1, iterating the formula (11) we can express the sum S2k-1(n) as a linear combination of the polynomials n2k-1-2iS2i(n) in n with i=0,1,,k-1. This is given by the following result.

Corollary 9.

Let (akl), k=1,2,, l=1,2,,k, be real numbers recursively defined as (24)ak+1,l+1=122k+12k+12l22l-j=1k-l2k+12j22k+1-2jak+1-j,l+1with a11=1/2, k=1,2,, and 1lk. Then for each k1 it holds that(25)S2k-1n=i=0k-1ak,i+1n2k-1-2iS2i.Furthermore, for each k1 (25) is a unique representation of the power sum S2k-1(n) as a linear combination of the functions n2k-1S0(n),,nS2k-2(n).

Remark 10.

Unfortunately, the analogous formula to (11) with 2k instead of 2k-1 does not exist. Namely, suppose that for some even m2 there exist real numbers a1,a2,,am such that for every n(26)Smn=i=1mainiSm-in.Considering the sum Sm(n) as a polynomial in n, then it is well known that Sm(n) is divisible by n for each m1, and Sm(n) is divisible by n2 if and only if m is odd (see, e.g., ; this also immediately follows by Bernoulli’s formula). However, the first of these facts and the equality (26) show that S2k(n) is divisible by n2 for each k1, a contradiction.

Although formula (11) cannot be directly used for recursive determination of the expressions for Sk(n), it can be useful for establishing various congruence involving these sums .

1.3. The Application of Theorem <xref ref-type="statement" rid="thm1.2">2</xref> to the Sums <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M234"><mml:msub><mml:mrow><mml:mi>φ</mml:mi></mml:mrow><mml:mrow><mml:mi>k</mml:mi></mml:mrow></mml:msub><mml:mo stretchy="false">(</mml:mo><mml:mi>n</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>

The Euler totient function φ(n) is defined to be equal to the number of positive integers less than n which are relatively prime to n. Each of these φ(n) integers is called a totative (or “totitive”) of n (see [11, Section  3.4, p. 242], where this notion is attributed to J. J. Sylvester). Let t(n) denote the set of all totatives of n; that is, t(n)={jN:1j<n,gcd(j,n)=1}. Given any fixed nonnegative integer k, in 1850 A. Thacker (see [11, p. 242]) introduced the function φk(n) defined as (27)φkn=ttntk,where the summation ranges over all totatives t of n (in addition, we define φk(1)=0 for all k). Notice that φ0(n)=φ(n) and there holds φk(n)=Sk(n) if and only if n=1 or n is a prime number.

Using the obvious fact that t(n) is a n-symmetric set in the ring Z, Theorem 2 immediately yields the following recurrence relation involving the functions φk(n).

Corollary 11.

Let k1 and n2 be positive integers. Then (28)i=02k-1-1i2k-1i22k-1-iniφ2k-1-in=0.

Remark 12.

The following recurrence relation for the functions φk(n) was established in 1857 by J. Liouville (cf. [11, p. 243]): (29)dnndkφkd=Skn+11k+2k++nkwhich for k=0 reduces to Gauss’ formula dnφ(d)=n. Furthermore, in 1985 Bruckman and Lossers  established an explicit Bernoulli’s-like formula for the Dirichlet series of φk(n) defined as fk(s)=k=1φk(n)/ns (there φk(n) is called generalized Euler function).

For given integers a, d0, k1, and n2 define the function φk(n;a,d) as (30)φkn;a,d=ttna+tdk.Notice that φk(n;0,1)=φk(n) with the function φk(n) defined above. From the obvious fact that tt(n) if and only if n-tt(n), it follows immediately that the set {a+td:tt(n)} is a (2a+nd)-symmetric set in the ring Z. Therefore, by Theorem 2 we immediately obtain the following recurrence formula for the functions φk(n;a,d).

Corollary 13.

Let k be a positive integer. Then (31)i=02k-1-1i2k-1i22k-1-i2a+ndiφ2k-1-in;a,d=0.

2. Proofs of Theorem <xref ref-type="statement" rid="thm1.2">2</xref> and Corollary <xref ref-type="statement" rid="coro1.9">9</xref> Proof of Theorem <xref ref-type="statement" rid="thm1.2">2</xref>.

Suppose first that 2e is not invertible element in R. Then as noticed above, the set A has the form(32)A=a1,a2,,as=b1,,bl,ne-b1,,ne-bl,where {b1,,bl} (s=2l) is a finite subset of R such that ne-bibj for all i,j with 1i<jn.

Since the binomial formula holds in any commutative ring with unity, we have (33)2aj-ne2k-1=i=02k-1-1i2k-1i22k-1-iniaj2k-1-ifor all positive integers n and j with 1jl. After summation in (33) over j=1,,2l we obtain (34)j=1s2aje-ne2k-1=j=1si=02k-1-1i2k-1i22k-1-iniaj2k-1-i=i=02k-1-1i2k-1i22k-1-inij=1saj2k-1-i=i=02k-1-1i2k-1i22k-1-iniS2k-1-iA.On the other hand, from (32) we see that s=2l and we can assume that ai=bi for each i=1,,l and ai=ne-bi for each i=l+1,,2l. Using this and observing that -(2bje-ne)=2(ne-bje)-ne for each j=1,,l, we find that(35)j=1s2aje-ne2k-1=j=1l2bj-ne2k-1+j=l+12l2ne-bj-ne2k-1=j=1l2bje-ne2k-1+2ne-bj-ne2k-1=j=1l2bj-ne2k-1+-12k-12bj-ne2k-1=0.Comparing (34) and (35) immediately gives the desired identity (4).

Similarly, if 2e is invertible element in R, then as noticed in the previous section, the set A may be of the form (32) or of the form (36)A=a1,,al,ne·2e-1,ne-a1,,ne-al.Then since the central element (ne)·(2e)-1 of A satisfies the equality 2(ne)·(2e)-1-ne=0, in the same manner as in the first case, we arrive to (32). This completes the proof.

Proof of Corollary 9 is based on Corollary 7 and the following lemma.

Lemma 14.

Let S~k(x), k=0,1,2, be a sequence of polynomials of the real variable x defined as S~0(x)=x-1 and (37)S~kx=1k+1i=0kk+1ixk+1-iBi,k=1,2,,where Bi(i=0,1,2,) are Bernoulli numbers. Then for each nonnegative integer k the polynomials x2k+1S~0(x),x2k-1S~2(x),,x3S~2k-2(x),xS~2k(x) are linearly independent over the field of real numbers R.

Proof.

The proof easily follows by induction on k0. The base of induction (k=0) is satisfied because S~0(x)0. Suppose that the polynomials x2k+1S~0(x),x2k-1S~2(x),,x3S~2k-2(x),xS~2k(x) are linearly independent for some k0. Now consider the k+1 polynomials (38)x2k+3S~0x,x2k+1S~2x,,x3S~2kx,xS~2k+2xand assume that (39)i=0k+1aix2k+3-2iS~2ix0,where a0,a1,,ak+1 are real numbers. Note that the equality (39) can be written as (40)ak+1xS~2k+2x+x2i=0kaix2k+1-2iS~2ix0.From (37) we see that the linear term of the polynomial S~2k+2(x) is B2k+2x. This together with (40) yields that the linear term of the polynomial on the left hand side of (40) is ak+1B2k+2x. Accordingly, in view of the fact that B2k+20 for each k0, the identical equality (40) yields ak+1=0, which substituting in (40) gives (41)i=0kaix2k+1-2iS~2ix0.It follows from (41) that a0=a1==ak=0 because the polynomials x2k+1S~0(x),x2k-1S~2(x),,xS~2k(x) are linearly independent by induction hypothesis. Therefore, (39) yields a0=a1==ak=ak+1=0. This completes the induction proof.

Proof of Corollary <xref ref-type="statement" rid="coro1.9">9</xref>.

By using the identity (11) of Corollary 7 with 2k+1 instead of 2k-1, we find that(42)22k+1S2k+1n=i=12k+1-1i-12k+1i22k+1-iniS2k+1-in=i  odd·+i  even·=j=0k2k+12j+122k-2jn2j+1S2k-2jn-j=1k2k+12j22k+1-2jn2jS2k+1-2jn.As noticed above, iterating the formula (11), for any k1 we can express the power sum S2k-1(n) as a linear combination of the polynomials n2k-1-2iS2i(n) of n with i=0,1,,k-1. This means that for each k1 there exist real numbers bk1,bk2,,bkk such that (43)S2k-1n=i=0k-1bk,i+1n2k-1-2iS2i.Replacing k by k-j+1 into formula (43) yields (44)S2k+1-2jn=i=0k-jbk-j+1,i+1n2k+1-2j-2iS2i.Substituting (44) into (42) we obtain(45)22k+1S2k+1n=j=0k2k+12j+122k-2jn2j+1S2k-2jn-j=1k2k+12j22k+1-2jn2ji=0k-jbk+1-j,i+1n2k+1-2j-2iS2i=i=0k2k+12k+1-2i22in2k+1-2iS2in-j=1k2k+12j22k+1-2ji=0k-jn2k+1-2iS2inbk+1-j,i+1=i=0k2k+12i22in2k+1-2iS2in-i=0k-1j=1k-i2k+12j22k+1-2jn2k+1-2iS2inbk+1-j,i+1.Notice that, for each k1 by Bernoulli’s formula (see the first formula of Remark 8), (46)Skn=1k+1i=0kk+1ink+1-iBi.From (37) and (46) we see that S~k(n)=Sk(n) for each n=1,2,, and thus the polynomial identity (45) with respect to the positive integer variable n can be extended to the real variable x instead of n; that is, (45) is identically satisfied (on R) if we replace Sl(n) by S~l(x) (with l=2k+1 and l=2i). Accordingly, in view of Lemma 14, by comparing in (46) the coefficients with n2k+1-2iS2i(n), for i=0,1,,k, we immediately obtain (47)bk+1,l+1=122k+12k+12l22l-j=1k-l2k+12j22k+1-2jbk+1-j,l+1.Finally, since a11=1/2=S1(n)/(nS0(n))=b11, from the recurrence relations (24) and (47) we conclude that akl=bkl for all k=1,2, and l=1,2,,k.

Notice also that, by Lemma 14, (25) is a unique representation of the power sum S2k-1(n) as a linear combination of the functions n2k-1S0(n),,nS2k-2(n). This completes proof of Corollary 9.

Competing Interests

The authors declare that they have no conflict of interests.

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