On some oscillating sums

Uniform Distribution Theory 3(2008), no.1, 35–72 uniform distribution theory ON SOME OSCILLATING SUMS J. Arias de Reyna∗— J. van de Lune ABSTRACT. This paper deals with various properties (theoretical as well as computational) of the sums Sα(n) = Pn j=1(−1)bjαcwhere αis any real number (mostly a positive real quadratic). Communicated by Michael Drmota Introduction This paper deals with the sums Sα(n) = n X j=1 (−1)bjαc where αis any real number. If the value of αis clear from the context we will simply write S(n) instead of Sα(n). Apparently the interest in these sums was initiated (in 1976) by an unsolved problem proposed by H. D. Ruderman [3]: Prove that the series ∞ X n=1 (−1)bn√2c n converges and estimate its value. Indeed, when applying Abel-summation (summation by parts) to this series our sum (with α=√2) emerges naturally. This fact is reflected in the 1978 issue of the Amer. Math. Monthly, where one finds a solution by D. Borwein [7] (with an editorial reference to 11 other solutions). Many generalizations were presented, one of them being: If αis any real quadratic irrational then the series P∞ n=1(−1)bnαc/nsconverges for every s > 0. Further solutions may be found in Bundschuh [6] and van de Lune [4]. 2000 Math em a t i c s S u b j e c t C l a s s i f i c a t i o n: 11K31, 11J70. Ke yw or ds: Exponential sum, continued fraction, quadratic irrational, algorithm. ∗Supported by grant MTM2006-05622. 35 J. ARIAS DE REYNA — J. VAN DE LUNE The last related publication seems to be Schoissengeier [17], where it is proved that if the regular continued fraction expansion of β=α/2 is {b0;b1, b2, . . . }with convergents pk qkthen the series P∞ n=1 (−1)bnαc/n converges if and only if the series ∞ X k= 0,2-qk (−1)klog bk+1 qk converges. Our sums S(n) are also interesting in themselves as may be seen from the following example: For α=√2 we compute S(n) for n= 1, 2, 3, . . . , and keep track of those nfor which S(n) assumes a value for the first time (i.e., is larger/smaller than ever before). Here (and in the sequel) we define S(0) = 0. The sequence of these n’s (the record-holders) will be denoted by t1,t2,t3, . . . . We found t→1 3 8 20 49 119 288 696 1681 4059 9800 S→ −11−2 2 −3 3 −4 4 −5 5 −6 In making our computations we observe that tk+1 = 2tk+tk−1+ 1 for all k≥1. Here (and in the sequel) we define t0= 0. Also, the sequence sign(S(tk)) appears to be purely periodic. In this paper we will show that similar results are true for all real quadratic irrationals. We will see that the behavior of our sums is intimately connected with the regular continued fraction expansions α={a0;a1, a2, a3, . . . }and β=α/2 = {b0;b1, b2, b3, . . . , }. In this vein Brouwer and van de Lune [5] have shown that S(n)≥0 for all nif and only if the partial quotients a2iare even for all i≥0. In Fokkink, Fokkink and van de Lune [9] we find a description of a fast algorithm for the computation of S(n) for (very) large nin terms of the regular continued fraction expansion of β=α/2. For an explicit program we refer to the Appendix. We just mention that by means of this program we easily found that S√2(101000) = −10, S√2(1010000) = 166, Sπ(1010000) = 11726. A closely related (but somewhat less general) algorithm is given in an interesting paper by O’Bryant, Reznick and Serbinowska [15]. It will turn out that the sequence (−1)bjαcexhibits various symmetries. We will determine these explicitly in terms of the convergents of α. We will also show that for any real irrational αthe sum S(n) is not bounded, so that the corresponding sequence of record-holders tkactually is an infinite sequence. 36 ON SOME OSCILLATING SUMS In addition we will prove that for every j≥1 there is an index ksuch that tj−tj−1=Qk, where Qkis the denominator of a certain regular continued fraction approximant of α, as defined in Section 2. We will translate this into an algorithm which (given sufficiently many initial convergents of α) computes the entire sequence of record-holders. We will also give explicit formulae for the numbers tjand the corresponding Qk. Finally we will study the function H(α) = Λ(1 −α), where Λ(α) = lim sup n→∞ Sα(n) log n. It will turn out that this function is finite for all real αwith bounded partial quotients, and, in a certain sense, is a modular function since H(α+ 2) = H(α), H³−1 α´=H(α). We will also present a fast algorithm for the computation of H(α). In Section 6 we solve the problem of the order of supn≤NSα(n) for almost all real α(see Theorem 29). This order is similar to that of supn≤NnD∗ n(jα), where D∗ nis the discrepancy of the sequence {jα}. We think that our proof of the lower bound of (41) might very well be the most straightforward. In an Appendix we present a fast implementation of the FFL-algorithm (already implicitly described in FFL [9]) to compute Sα(n) for any irrational real α. 1. Preliminary results We begin with some well known facts. Lemma 1 . We have x∈Rand n∈Z=⇒ bn+xc=n+bxc(1) x∈R r Z =⇒ bxc+b−xc=−1.(2) α∈Rand n≥0 =⇒Sα+2(n) = Sα(n) (3) α∈R r Z =⇒S−α(n) = −Sα(n).(4) In the sequel we will always assume that αis real, and that all continued fractions are regular. This also applies to our programs (which have been designed primarily for real quadratic irrationals). In case α=p q, with (p, q) = 1, is rational it is easily seen that the sequence S(n) is (1) periodic (and hence bounded) with period 2qif pis odd and (2) 37 J. ARIAS DE REYNA — J. VAN DE LUNE unbounded (of true order n) if pis even. Therefore, we will restrict ourselves from now on to irrational real α. In view of Lemma 1 we may, without loss of generality, even restrict ourselves to 0 < α < 1. It will turn out that the properties of S(n) heavily depend on the regular continued fraction expansion {a0;a1, a2, . . . }of α, in particular on its convergents P−2 Q−2 =0 1,P−1 Q−1 =1 0,Pj Qj ={a0;a1, . . . , aj},(j≥0) (5) which satisfy the relations Pj+1 =aj+1Pj+Pj−1, Qj+1 =aj+1Qj+Qj−1,(j≥ −1) (6) and (Pj, Qj) = 1 (j≥ −2), PjQj−1−QjPj−1= (−1)j−1,(j≥ −1).(7) If an+1 ≥2 we will also make use of mediants. These are the irreducible fractions P/Q defined by P Q=hPn+Pn−1 hQn+Qn−1 ,(n≥0, h = 1,2, . . . , an+1 −1). For these fractions we have PQn−PnQ= (−1)n, and we say that Pn/Qnis the convergent next to the mediant P/Q. Note that αalways lies strictly between P/Q and the convergent next to P/Q. Just for the sake of easy reference we state the following known lemma (see [11, Chapter VII, Exercise 13]). Lemma 2 . If a b<c dwith band d > 0and cb −ad = 1, then every fraction m n with n > 0and a b<m n<c d satisfies n≥b+d. Lemma 3 . Let P Q=hPn+1+Pn hQn+1+Qn, with 0≤h < an+2, be a mediant or a convergent of α. If 1≤m < Q +Qn+1, with Q-m, then bmαc=jmP Qk.(8) In particular this is true for 1≤m < Q. P r o o f. Suppose the assertion of the Lemma is false. Then we can find an integer kbetween mα and mP/Q, so that k/m lies strictly between αand P/Q, and is not equal to these extremes. (Indeed k/m 6=αsince αis irrational and k/m 6=P/Q since P/Q is in lowest terms and by hypothesis Q-m.) Since α 38 ON SOME OSCILLATING SUMS lies between Pn+1/Qn+1 and P/Q, it follows that k/m is lying strictly between P/Q and Pn+1/Qn+1, Now observe that P Q−Pn+1 Qn+1 =PnQn+1 −QnPn+1 QQn+1 =−(−1)n QQn+1 (9) so that, by Lemma 2, we would have m>Q+Qn+1, which contradicts our hypothesis. ¤ Lemma 4 . Let P Q=hPn+1+Pn hQn+1+Qn, with 0≤h < an+2, be a mediant or a convergent of α. Then (−1)bQαc= (−1)n+P.(10) P r o o f. As in the proof of the previous Lemma, αlies between the fractions P/Q and Pn+1/Qn+1. By a computation as in (9) we have ¯¯¯ P Q−α¯¯¯<¯¯¯ P Q−Pn+1 Qn+1 ¯¯¯=1 QQn+1 . For n≥0 the two numbers Qand Qn+1 are ≥1, so that |Qα −P|< Q−1 n+1 ≤1. It follows that bQαc=Pif Qα > P, and that bQαc=P−1 if Qα < P. But, by the theory of continued fractions the first case happens when nis even, and the second when nis odd. Hence, (10) is true in both cases. ¤ The following theorem shows the fundamental connection between the sums Sα(n) and the convergents of α. Proposition 5 . Let αbe any irrational real number, and Pn Qnwith n≥0one of its convergents. Let Qn≤m<Qn+Qn+1 and put m=hQn+rwith 0≤r < Qn. Then bmαc=     hPn+brαcif r6= 0, hPnif r= 0 and nis even, hPn−1if r= 0 and nis odd. (11) P r o o f. If r6= 0 then Qn-mand Qn-r. By Lemma 3 we have bmαc=jmPn Qnk=jhPn+rPn Qnk=hPn+jrPn Qnk=hPn+brαc. If r= 0 then m=hQn. Also, mα lies between the two numbers mPn/Qn and mPn+1/Qn+1, and the distance between these two numbers is ¯¯¯mPn Qn−mPn+1 Qn+1 ¯¯¯=m QnQn+1 ≤1. 39 J. ARIAS DE REYNA — J. VAN DE LUNE (For the last inequality observe that always 1 ≤Qn≤Qn+1. So QnQn+1 ≥ Qn+Qn+1 > m, unless Qn= 1. In this case Qn= 1 ≤m < 1 + Qn+1 and the inequality follows.) When nis even we have mPn Qn =hPn< mα < m Pn+1 Qn+1 ≤1 + hPn and it follows that bmαc=hPn. When nis odd we have mPn Qn =hPn> mα > m Pn+1 Qn+1 ≥hPn−1. Thus, in this case we get bmαc=hPn−1, and the proof of (11) is complete. ¤ Corollary 6 . Let αbe any irrational real number, and Pn Qnwith n≥0one of its convergents. Let Qn≤m<Qn+Qn+1 and put m=hQn+rwith 0≤r < Qn. Then (a)Pneven =⇒S(m) = (−1)nh+S(r).(12) (b)Pnodd =⇒S(m) = (S(Qn)−S(r)if his odd S(r)if his even. (13) P r o o f. We compute the sum S(m) = m X j=1 (−1)bjαc= = h−1 X k=0 Qn−1 X s=1 (−1)b(kQn+s)αc+ h X k=1 (−1)bkQnαc+ r X s=1 (−1)b(kQn+s)αc. Applying (11) we get S(m) = h−1 X k=0 Qn−1 X s=1 (−1)kPn+bsαc+ h X k=1 (−1)kPn−[[ nis odd ]] + r X s=1 (−1)hPn+bsαc where, following Iverson’s notation, [[ X]] = 1 if the proposition Xis true, and [[ X]] = 0 if Xis false. Simplifying we get S(m) = h−1 X k=0 (−1)kPnS(Qn−1) + h X k=1 (−1)kPn−[[ nis odd ]] + (−1)hPnS(r).(14) 40 ON SOME OSCILLATING SUMS By Lemma 3 we have Sα(Qn−1) = SPn/Qn(Qn−1). Therefore, if Pnis even we have Sα(Qn−1) = 0 (see [5, Lemma 5.1]), so that in this case S(m) = (−1)nh+S(r). If Pnis odd and heven, then the two sums in (14) are equal to 0 and we get S(m) = S(r). Finally, when Pnand hare odd, the first sum in (14) is equal to S(Qn−1), the second is (−1)n+1, and we get S(m) = S(Qn−1) + (−1)n+1 −S(r) = S(Qn)−S(r) since, by Lemma 4, we have (−1)bQnαc= (−1)n+Pn= (−1)n+1.¤ 2. Symmetries of the sequence of signs We consider four types of symmetries of the sequence of signs (−1)bjαc. Each of them leads to a useful transformation of the sums S(n). In this section we define these symmetries and present some theorems in order to obtain these symmetries from the sequence of convergents of α. It should be noted that our definitions differ slightly from those in [9]. Definition 7 . The integer n≥1will be called a point of repetition (a REP, for short) if 1≤k≤n=⇒(−1)bkαc= (−1)b(n+k)αc or, equivalently, if bkαcand b(n+k)αchave the same parity for 1≤k≤n. Note. There seem to be α’s which do not yield any REP’s. For example, α=√2 seems to be such a number. The first few REP’s for α=πare: n= 2, 7, 14, 21, 28, 35, 42, 49, 56, 226, 452, 678, 904, 1130, 1356, 1582, 1808, 2034, 2260, 2486, 2712, 2938. Lemma 8 . If nis a REP for α, then 0≤k≤n=⇒Sα(n+k) = Sα(n) + Sα(k).(15) P r o o f. For k≥1 we have Sα(n+k) = n+k X j=1 (−1)bjαc= n X j=1 (−1)bjαc+ k X j=1 (−1)b(n+j)αc= =Sα(n) + Sα(k). As usual we define Sα(0) = 0, and for k= 0 the result is trivial. ¤ 41 J. ARIAS DE REYNA — J. VAN DE LUNE Definition 9 . The integer n≥1will be called a point of contra-repetition (a CREP, for short) if 1≤k≤n=⇒(−1)bkαc=−(−1)b(n+k)αc or, equivalently, if bkαcand b(n+k)αchave different parities for 1≤k≤n. The first few CREP’s for α=πare: n= 1, 3, 106, 113, 339, 565, 791, 1017, 1243, 1469, 1695, 1921, 2147, 2373, 2599, 2825, 3051, 3277, 3503, 3729. In the same way as in Lemma 8 we prove the following Lemma 10 . If nis a CREP for α, then Sα(n+k) = Sα(n)−Sα(k),0≤k≤n. (16) The next Theorem provides REP’s and CREP’s for α. See note at the end of this section regarding the scope of this theorem. Proposition 11 . Let αbe any irrational real number. If h∈N,n≥0and 2hQn< Qn+Qn+1, then b(j+hQn)αc=hPn+bjαc,1≤j≤hQn.(17) Thus hQnis a REP for αif hPnis even, and a CREP if hPnis odd. P r o o f. We may apply (11) with m=jand m=j+hQn. If j=kQn+rwith 0≤r < Qn, then j+hQn= (h+k)Qn+r. Thus, by (11), bjαc=kPn+A(n, r),b(j+hQn)αc= (h+k)Pn+A(n, r) where A(n, r) depends only on nand rand is the same in both cases. This proves (17). ¤ Definition 12 . The integer n≥2will be called an end-point of reflection (an EREF, for short) if 1≤k≤n/2 =⇒(−1)bkαc= (−1)b(n+1−k)αc or, equivalently, if bkαcand b(n+1−k)αchave the same parity for 1≤k≤n/2. The first few EREF’s for α=πare given by: n= 3, 5, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 331, 557, 783, 1009, 1235, 1461. Lemma 13 . If nis an EREF for α, then Sα(n+ 1 −k) = Sα(n)−Sα(k−1),(1 ≤k≤n/2).(18) 42 ON SOME OSCILLATING SUMS P r o o f. For 2 ≤k≤n/2 we have S(n) = n X j=1 (−1)bjαc= n+1−k X j=1 (−1)bjαc+ n+1−1 X j=n+1−(k−1) (−1)bjαc= =S(n+ 1 −k) + k−1 X r=1 (−1)b(n+1−r)αc=S(n+ 1 −k) + k−1 X r=1 (−1)brαc= =S(n+ 1 −k) + S(k−1). For k= 1 the result is clear. ¤ Definition 14 . The integer n≥2will be called an end-point of contrareflection (an ECREF, for short) if 1≤k≤n/2 =⇒(−1)bkαc=−(−1)b(n+1−k)αc or, equivalently, if bkαcand b(n+1−k)αchave different parities for 1≤k≤n/2. The first few ECREF’s for α=πare given by: n= 2, 4, 6, 13, 211, 218, 225, 444, 670, 896, 1122, 1348, 1574, 1800, 2026, 2252, 2478, 2704, 2930. Lemma 15 . If nis an ECREF for the real irrational α, then Sα(n−k) = Sα(n) + Sα(k),0≤k≤n/2.(19) It follows that Sα(n) = 0 for neven, and Sα(n) = (−1)bn+1 2αcfor nodd. The proof is similar to that of Lemma 13. Proposition 16 . Let αbe any real irrational. If n≥ −1and 1≤h≤an+2 then b(Qn+hQn+1 −j)αc+bjαc=Pn+hPn+1 −1,1≤j < Qn+hQn+1.(20) So Qn+hQn+1 −1(when ≥2) is an EREF for odd Pn+hPn+1, and an ECREF when Pn+hPn+1 is even. P r o o f. Let P=Pn+hPn+1 and Q=Qn+hQn+1. The fraction P Qis a mediant for 1 ≤h < an+2, whereas P Q=Qn+2 Pn+2 if h=an+2. In both cases we may apply Lemma 3 to get bjαc=jjP Qk,1≤j < Q. Observe that (20) is equivalent to jP−jP Qk+jjP Qk=P−1,1≤j < Q. 43 J. ARIAS DE REYNA — J. VAN DE LUNE (3) When k= 0, we have J−1= [0,1), J0= [1,1 + Q1) with P0=a0. Thus, P0even implies a06= 1 and J−1∩J0=∅. So, assertion (b) is (vacuously) true for k= 0. Thus, in what follows we may assume k≥1. Assume k≥1, Pkeven and Jk−1∩Jk6=∅. We want to prove that ak= 1. Since Pk−1is odd there is only one record-holder t∈Jk−1and t=t0+Qk−1 with t0< Qk−1. Since t∈Jkwe get Qk≤t0+Qk−1. Therefore (ak−1)Qk−1+ Qk−2≤t0. Since k≥1 we have Qk−1≥1, and ak>1 leads to t0< Qk−1≤ (ak−1)Qk−1≤t0which is a contradiction. Conversely, if k≥1, Pkis even and ak= 1, then since (Pk−1, Pk) = 1 the number Pk−1is odd and Pk=Pk−1+Pk−2implies that also Pk−2is odd. Then, by case (1) considered above, the only record-holder t∈Jk−1and the record-holder t0∈Jk−2satisfy Qk−2< t0≤Qk−1≤Qk−2+Qk−1< t, and t=t0+Qk−1. It follows that Qk−2+Qk−1=Qk≤tand t∈Jk. We only need to show how many record-holders are contained in Jkwhen Pk is even. They are all the numbers t+hQk< Qk+Qk+1 = (ak+1 + 1)Qk+Qk−1 with h≥1 and t < Qka particular record-holder. In case ak>1 this recordholder is t∈Jk−1and t /∈Jk. Thus Qk−1≤t < Qk. We see that the allowed values of hare 1 ≤h≤ak+1. In case ak= 1 the only record-holder in Jk−1is contained in Jkso that t < Qk−1and it is easily seen that the allowed values of hare given by 1 ≤h≤ ak+1 + 1. Formula (28) is an easy consequence of the above results. ¤ The above theorems justify the following procedure (written in Mathematica Version 5.2) in order to obtain the sequence of “all” record-holders. We assume that we have previously defined the numbers P[n] and Q[n] for n < kMax (kMax being an appropriate limit). Program to compute the record-holders for α T = {0}; (* T will contain the sequence of record-holders *) t = 0 ; (* The last obtained record - holder *) For[n = 0, n <= kMax, n++, If[OddQ[P[n]], (* then *) If[t < Q[n], t = t + Q[n]; T = Append[T, t]], (* else *) While[t + Q[n] < Q[n] + Q[n + 1], t + = Q[n]; T = Append[T, t]] ] ]; Print[T] 50 ON SOME OSCILLATING SUMS 4. The case of a quadratic irrationality In the case of a real quadratic irrational αthe numbers Pkand Qkcan be given explicitly. We did not find the formulas in Proposition 24 in the standard textbooks dealing with continued fractions. Proposition 24 . Let α∈Q(√d)be a real quadratic irrationality, and let kbe the length of the period of the regular continued fraction of α. Then Pnk+j=Ajωn 1+Bjωn 2, Qnk+j=Cjωn 1+Djωn 2,0≤j < k, n ≥n0(29) where ω1and ω2are certain conjugate units in the ring of algebraic integers in Q(√d). P r o o f. Let the continued fraction expansion of αbe α={a0;a1, a2, . . . , ah, b1, b2, . . . , bk}(30) and Pn/Qnthe corresponding convergents. We consider the number βwith continued fraction {0; b1, b2, . . . , bk}. Let pn/qnbe the convergents of β. Note that p0= 0, p1= 1, q0= 1 and q1=b1. It is well known that the two quadratic irrationals αand βgenerate the same field Q(α) = Q(β) = Q(√d). For n≥0 and 1 ≤j≤kwe have Qh+nk+j=bjQh+nk+j−1+Qh+nk+j−2. In particular Qh+nk+1 =b1Qh+nk +Qh+nk−1=q1Qh+nk +p1Qh+nk−1. We claim that in general for n≥0 and 1 ≤j≤k Qh+nk+j=qjQh+nk +pjQh+nk−1. We will prove this by induction on the number j. Assuming that we have proved the result for all numbers less than j+ 1 we have Qh+nk+j+1 =bj+1Qh+nk+j+Qh+nk+j−1 =bj+1(qjQh+nk +pjQh+nk−1) + qj−1Qh+nk +pj−1Qh+nk−1 = (bj+1qj+qj−1)Qh+nk + (bj+1pj+pj−1)Qh+nk−1 =qj+1Qh+nk +pj+1Qh+nk−1. 51 J. ARIAS DE REYNA — J. VAN DE LUNE We can write this equation in matrix form Qn:=       Qh+kn+k Qh+kn+k−1 Qh+kn+k−2 . . . Qh+kn+1       =Ω      Qh+kn Qh+kn−1 Qh+kn−2 . . . Qh+kn−k+1       =Ω Qn−1, n ≥0 where Ωis defined by Ω=    qkpk0. . . 0 qk−1pk−10. . . 0 ....................... q1p10. . . 0     . The linear transformation Ωhas k−2 eigenvalues equal to 0 whereas the other two are the solutions of the equation ¯¯¯¯ qk−ω pk qk−1pk−1−ω¯¯¯¯ =ω2−(qk+pk−1)ω+ (−1)k= 0. Its two solutions are algebraic integers. We call them ω1and ω2. They are quadratic irrationals in the same field Q(α). In order to see this we prove that the discriminant of ωis the same as that of β. In fact, βis the solution of the quadratic equation pk+βpk−1 qk+βqk−1 =βor qk−1β2+ (qk−pk−1)β−pk= 0. Therefore, the discriminant of βis ∆ = (qk−pk−1)2+ 4qk−1pk= (qk−pk−1)2+ 4(qkpk−1+ (−1)k−1) = (qk+pk−1)2−4(−1)k which coincides with the discriminant of ω. Since ω1ω2= (−1)kand ω1and ω2are not rational, exactly one of them is in absolute value larger than 1. This one we call ω1. Let u1and u2be the corresponding eigenvectors. Every vector in the image of Ωis a linear combination of these two vectors. In particular there exist constants Aand Bsuch that Q0=Au1+Bu2. Then we have Qn=Ωn(Au1+Bu2) = Aωn 1u1+Bωn 2u2. That is, for n≥0 and 1 ≤j≤k, Qh+kn+j=AUjωn 1+BVjωn 2 52 ON SOME OSCILLATING SUMS where Ujand Vjare the coordinates of the two eigenvectors. So, Ujand Vjare conjugate numbers in the field Q(α). ¤ Proposition 25 . Let α∈Q(√d)be a real quadratic irrationality, and kthe length of the (pure) period of the continued fraction of α. Then there are natural numbers b,L,Kan integer multiple of k, and a function φsuch that the sequence of record-holders for αsatisfies tb+nL+j−tb+nL+j−1=QnK+φ(j),0≤j < L, n ≥0.(31) Also, there exists a finite sequence of signs (εj)L j=1 such that εjS(tb+nL+j)>0, n ≥0.(32) Let Mbe the number of jsuch that εj= 1, and mthe number of jwith εj=−1. Then m+M=L, and S(tb+nL+j) = (nM +ajif εj= 1 −nm +ajif εj=−1.(33) P r o o f. Assume that αhas the continued fraction (30). For n≥1 the pair (Pn, Pn+1) modulo 2 has only three possible values (1,0), (0,1) and (1,1). Therefore, there exists a multiple Kof k(Kwill be k, 2kor 3k) and c > h + 2 such that Pc−2≡Pc+K−2and Pc−1≡Pc+K−1. Given the periodicity of the partial quotients of αand the recursive formulas (6), we will have Pc+m≡Pc+K+m,(mod 2), m ≥ −2.(34) Without loss of generality we may assume that Kis even (if necessary take 2K instead of K). By Proposition 23 the number of record-holders in the interval Jndepends only on the parity of Pnand the numbers anand an+1. Since Kis a multiple of the period kit follows from (34) that for n≥cthe intervals Jnand Jn+Kcontain the same number of record-holders. By Propositions 20 and 21 the characters maximum/minimum of these record-holders will be the same since Khas been taken even. Therefore, the number, characters and relative positions of the record-holders in the union of intervals SK j=0 Jc+nK+jdo not depend on n. Let tb,tb+1, . . . , tb+L−1be the record-holders contained in SK j=0 Jc+j. Then the record-holders contained in SK j=0 Jc+nK+jwill be the numbers tb+nL+jwith 0≤j < L. The numbers tb+nL+jwith a fixed jare either all maxima or all minima. Put εj= 1 for a maximum and εj=−1 for a minimum. Then, obviously, we will have εjS(tb+nL+j)>0. 53 J. ARIAS DE REYNA — J. VAN DE LUNE If n≥band if the record-holder tn∈Jvthen the record-holder tn+L∈Jv+K. This record-holder will be the only one in the given interval or will have the same position between the record holders in the respective intervals Jvand Jv+K. It follows that if tn−tn−1=Qrthen tn+L−tn+L−1=Qr+K. Thus we can define a function φsuch that tb+j+nL −tb+j+nL−1=Qφ(j)+nK,0≤j < L. Finally, let Mbe equal to the number of maxima in the period of recordholders, and mthe number of minima. If tb+nL+jis a maximum, then tb+nL+L+j is also a maximum and S(tb+nL+L+j) = M+S(tb+nL+j) since there are M maxima on the period of record-holders. This fact, together with reasoning similar to that for the case of minimum establishes our formula for S(tb+nL+j). ¤ Corollary 26 . With the same notations as in Theorem 24 and Proposition 25 tb+nL+j=Ej+Fjωκn 1+Gjωκn 2,0≤j < L, n > n1(35) where K=κk,κbeing a positive integer. P r o o f. Summing equations 31 for 0 ≤j < L we get tb+(n+1)L−1−tb+nL−1= L−1 X j=0 QnK+φ(j). For every fixed jlet φ(j) = ujk+rwith 0 ≤r < k. Then nK +φ(j) = (nκ +uj)k+rand by (29), for n≥n1, we will have QnK+φ(j)=Cjωnκ+uj 1+Djωnκ+uj 2. Thus there exist constants n1,C0 jand D0 jsuch that tb+(n+1)L−1−tb+nL−1=C0 jωnκ 1+D0 jωnκ 2, n ≥n1. Summing this for n1≤n≤N−1 we get tb+NL−1=tb+n1L−1+ N−1 X n=n1 C0 jωnκ 1+D0 jωnκ 2. Summing the geometric series we see that there are constants E−1,F−1and G−1such that tb+NL−1=E−1+F−1ωNκ 1+G−1ωNκ 2, N > n1. 54 ON SOME OSCILLATING SUMS From this equation, by induction, we obtain (35). For example: tb+NL =E−1+F−1ωNκ 1+G−1ωNκ 2+QNK+φ(0) = =E−1+F−1ωNκ 1+G−1ωNκ 2+C0ωNκ 1+D0ωNκ 2=E0+F0ωNκ 1+G0ωNκ 2. ¤ 5. Connection with modular functions The functions Λ(α) = lim sup n→∞ Sα(n) log n, λ(α) = lim inf n→∞ Sα(n) log n are finite at the points αfor which Sα(n) = O(log n), in particular at real quadratic irrationals. We may restrict ourselves to one of them since S−α(n) = −Sα(n) (by (2)) so that λ(α) = −Λ(−α).(36) Defining H(α) = Λ(1 −α) we have Theorem 27 . For every real irrational α H(α+ 2) = H(α), H³−1 α´=H(α).(37) P r o o f. By (3) in Lemma 1 we have Λ(α+2) = Λ(α), so that H(α+2) = H(α). It is convenient to extend the definition of Sα(n). For any real number xwe put Sα(x) = X 1≤n≤x (−1)bnαc. It is easy to show that with this definition we have Λ(α) = lim sup x→∞ Sα(x) log x, λ(α) = lim inf x→∞ Sα(x) log x. Now we prove that α∈I, α > 1,1 α+1 β= 1 ⇒Λ(α) = −λ(β); λ(α) = −Λ(β).(38) (Here Istands for the set of all irrational real numbers.) Assume that α > 1 is irrational and that 1 α+1 β= 1. A well known theorem by Beatty says that bnαc 55 J. ARIAS DE REYNA — J. VAN DE LUNE and bnβcthen form a partition of the natural numbers. This can be translated into a property of the sums Sα(n). In [15] we found the proof of Sα(x/α) + Sβ(x/β) = O(1) based on Beatty’s theorem. Dividing by log xwe get Λ(α) = lim sup x→+∞ Sα(x/α) log(x/α)= lim sup x→+∞ Sα(x/α) log x= = lim sup x→+∞ O(1) −Sβ(x/β) log x=−lim inf x→+∞ Sβ(x/β) log(x/β)=−λ(β). In the same way we get λ(α) = −Λ(β). We can write the main equation in (38) in the form Λ(α) = −λ³α α−1´,for α∈I, α > 1.(39) For every irrational y < 0 we have H(y) = Λ(1 −y) = −λ³1−y −y´= =−λ³1−1 y´= Λ³1 y−1´= Λ³1 + 1 y´=H³−1 y´. (The first equality is the definition of H, the second an application of (39) with 1−y > 1, the third an algebraic identity, the fourth an application of (36), the fifth an application of the first equation in (37), and the last one also an application of the definition of H.) But then, by the symmetry of this equation, it is true for all y∈I.¤ Theorem 28 . For αa quadratic irrational and with the same notations used in Theorem 24 and Proposition 25 we have Λ(α) = M κlog ω1 and λ(α) = −m κlog ω1 .(40) P r o o f. If M= 0, then there is at most a finite number of maximum-record- holders and a constant Csuch that S(n)≤Cfor all n∈N. So Λ(α) = 0 and equation (40) is true. When M > 0 there are infinitely many maximum-record-holders. So, given x, there is a record-holder satisfying tb+(n−1)L+j< x ≤tb+nL+j. By Corollary 35 56 ON SOME OSCILLATING SUMS there is a constant C(of the order of |ωκ 1|) such that tb+nL+j≤Ctb+(n−1)L+j. It follows that log x∼log tb+nL+jand we have Λ(α) = lim sup x→+∞ Sα(x) log x≤lim sup n→∞ S(tb+nL+j) log tb+nL+j = = lim sup n→∞ nM +aj log(Ej+Fjωnκ 1+Gjωnκ 2)=M κlog ω1 . Since this is the limit of Sα(n)/log nfor a particular sequence, it is also less than or equal to the lim sup = Λ(α), proving (40). ¤ 6. The order of the sums O’Bryant, Reznick and Serbinowska wonder in [15] whether Sα(n) = O(log n) is the correct type of growth of Sα(n) for a quadratic irrational α. They say that it seems unlikely that Sα(n) = O(log n) for almost all α, but a proof of this is elusive. They also ask for necessary and sufficient conditions on α (in terms of its continued fraction expansion) in order to have Sα(n) = O(log n). In this section we will answer the first and second of these questions. Applying Theorem 28 to a real quadratic irrational α, we have lim sup n Sα(n)/log n > lim inf nSα(n)/log n and both are finite real numbers. Hence Sα(n) = O(log n) and Sα(n) = Ω(log n). Thus, O(log n) is the correct rate of growth of Sα(n) for any fixed real quadratic irrational α. Our Theorem 18 shows that for any irrational αthe sum Sα(n) is not bounded. This is about all that can be said in general. In [15] it is proved that for any positive function ψ(n)≥1 that increases to infinity, we can find an αwith |Sα(n)| ≤ ψ(n) for all n. There is an important connection between our sums and the concept of discrepancy. Given a sequence of real numbers (xj) in the interval [0,1], its discrepancy D∗ n(xj) is defined by D∗ n(xj) = sup t∈[0,1)¯¯¯ #{j≤n: 0 ≤xj< t} n−t¯¯¯. It is easy to show that (see, for example, [15]) Sα(n) = 2n³#{j≤n:{jα/2}<1 2} n−1 2´. 57 J. ARIAS DE REYNA — J. VAN DE LUNE Here, as usual, {x}denotes the fractional part of x. So we have |Sα(n)| ≤ 2nD∗ n({jα/2}). There is a substantial literature on the discrepancy of the sequence {nα} (a good survey can be found in [10]). For example, the following result is due to Khintchine:Let ϕ(n)be a positive increasing function. Then sup n≤N nD∗ n({jα}) = O(log N·ϕ(log log N)) for almost all α∈Rif and only if ∞ X n=1 1 ϕ(n)<∞. It follows that for almost all α∈(0,1) we have sup n≤N|Sα(n)| ≤ sup n≤N nD∗ n({jα/2}) = O(log N·ϕ(log log N)) when ϕ(n) is an increasing positive function with P∞ n=1 1 ϕ(n)<∞. We are going to extend this result to the following Theorem 29 . Let ψ(x)and ϕ(x)be positive increasing functions such that Z∞ 1 dx ψ(x)= +∞and Z∞ 1 dx ϕ(x)<+∞. Then for almost all α∈(0,1) we have Ω(log N·ψ(log log N)) ≤sup n≤N|Sα(n)| ≤ ≤sup n≤N nD∗ n({jα/2}) = O(log N·ϕ(log log N)).(41) P r o o f. We only need to prove the first inequality. By Proposition 23 the number of record-holders < Qn+Qn+1 is larger than or equal to the sum X 0≤k≤n Pkeven ak+1. It follows that the maximum of |Sα(k)|for k < Qn+Qn+1 is, when Pnis even, at least 1 2an+1. Thus, with m=n+ 1 we have (once more using Iverson’s notation) sup n≤Qm+Qm−1|S(n)| ≥ 1 2[[ Pm−1even ]] am.(42) Now we need some measure theory. Let E⊂[0,1] be the set of those irrational numbers α∈[0,1] for which the partial quotients a1=k1,a2=k2, . . . , ar=kr 58 ON SOME OSCILLATING SUMS take definite values, and let E(k) be the subset of these numbers for which the additional partial quotient ar+1 is equal to k. Then we have (see [1, p. 60]) |E| 3k2<|E(k)| ≤ 2|E| k2.(43) For every mput Mα(m) = supn≤Qm+Qm−1|Sα(n)|. We are going to show that, given the set E(with k≥2), we have |{α∈E:Mα(k+ 3) ≥L}| ≥ |E| 200L. All numbers in Ehave regular continued fraction expansions starting with {0; k1, k2, . . . , kr, . . . }. Thus, they share the same convergents P0/Q0, . . . , Pr−1/Qr−1,Pr/Qr. Since (Pr−1, Pr) = 1 modulo 2, these two numbers can be: both odd (1,1), the first even and the second odd (0,1), or the first odd and the second even (1,0) (for all numbers in the set E). Now we decompose Ein four disjoint subsets E1,1,E1,0,E0,1, and E0,0defined by: E1,1={α∈E:ar+1 ≡1 (mod 2), ar+2 ≡1 (mod 2)} with similar definitions for E1,0,E0,1,E0,0. We have to consider three cases. Assume first that (Pr−1, Pr)≡(1,1) modulo 2. In this case all the numbers α∈E0,1satisfy Pr+1 =ar+1Pr+Pr−1≡Pr−1≡1 (mod 2) and Pr+2 =ar+2Pr+1 +Pr≡Pr+1 +Pr≡0 (mod 2). Therefore, by (42), for α∈E0,1we have Mα(r+ 3) = sup n≤Qr+3+Qr+2 |Sα(n)| ≥ ar+3 2 so that we have {α∈E:Mα(r+ 3) ≥L} ⊃ [ k≥2L E0,1(k). Here E0,1(k) denotes the set of those α∈Efor which ar+1 is even, ar+2 is odd and ar+3 =k. 59 J. ARIAS DE REYNA — J. VAN DE LUNE For[j = 1, j <= m2, j++, nonperiodic = Append[nonperiodic, {minima[[j]], "min"}]]; nonperiodic = Sort[nonperiodic]; nonperiodic = Table[nonperiodic[[j]][[2]], {j, 1, Length[nonperiodic]}]; period = Append[nonperiodic, Lperiod]; (* =================================================================== *) (* 8. We simplify the period *) (* Now we simplify the period {max, {max, min, max}} -> {{max, max, min}} *) u = Length[Last[period]]; v = Length[period] - 1; While[(v > 0) && (period[[v]] == Last[Last[period]]), period = Insert[period, RotateRight[Last[period], 1], -1]; period = Delete[period, -2]; period = Delete[period, -2]; u = Length[Last[period]]; v = Length[period] - 1]; (* We simplify the pure period {max, min, max, min} -> {max, min} *) lp = Length[Last[period]]; div = Divisors[lp]; pureperiod = Last[period]; v = 1; CheckValue = False; While[CheckValue == False, CheckValue = True; d = div[[v]]; For[j = 1, j <= d, j++, For[k = 0, k < lp/d, k++, If[pureperiod[[j]] != pureperiod[[d*k + j]], CheckValue = False]]]; v++]; period = Append[Drop[period, -1], Take[Last[period], d]]; (* 9. Print the results of the above analysis *) Print["* α= ", α]; Print["* Type of S-extremes in period = ", period]; Print["* Length of this period = ", Length[period[[1]]]]; PeriodRecordHolders = Table[S[α, T[[n]]], {n, 2, 1 + 2*Length[period[[1]]]}]; (* ! We only observed pure periods ! *) Print["* Regular CF(α) = ", RegularContinuedFraction[α]]; Print["* Sign sequence of S(t_n) -> ", Sign[PeriodRecordHolders]]; Print["* Units : ω1 = ", ω1," ω2 = ", ω2]; Print["* t_n of Maxima -> ", MM]; Print["* t_n of minima -> ", mm]; T = Delete[T, 1]; Print["* ’All’ Record-Holders t_n -> ", T]; (* The next line requires the loading of the FFL routine of the Appendix *) Print["* S(t_n) -> ", Table[S[α, T[[j]]], {j, 1, Length[period[[1]]]}]]; Print["* κ= ", κ]; Print["* Λ(α) = ", Λ= FullSimplify[ M/(κ*Log[ω1])], " ≈", N[Λ]]; Print["* λ(α) = ", λ= FullSimplify[ - m/(κ*Log[ω1])], " ≈", N[λ]]; "Done"]; One may check the record-holders for αby the following simple program 66 ON SOME OSCILLATING SUMS Mathematica Code for Generating the Record-Holders. α=√2(* For example *) n = 0; s = 0; sMax = 0; sMin = 0; While[0 == 0, n += 1; If[EvenQ[Floor[n*α]], s += 1, s -= 1]; If[s > sMax, sMax = s; Print[" n= ", n, " s= ", s]; Goto[A]]; If[s < sMin, sMin = s; Print[" n= ", n, " s= ", s]]; Label[A]] 8. Some remaining open problems 1. In all of our computations, the sequence of the signs of Sα(tn) always turned out to be purely periodic. We were unable to prove the consistency of this surprising observation. 2. It seems that there is always a system of recurrence relations for the recordholders tn. For example, for α=√3 we find that t4n= 2 t4n−1+t4n−4+ 1 t4n+1 =t4n+t4n−1+ 1 t4n+2 =t4n+1 + 2 t4n+ 1 t4n+3 =t4n+2 + 2 t4n+ 1. We have not pursued this subject any further. 3. It seems that the inhomogeneous sums Pn j=1(−1)bjα+βcexhibit certain characteristics very similar to those of the homogeneous sums dealt with in this paper. For example for the sums Pn j=1(−1)bj√2 + 1 2cwe find the recurrences: t2n= 2 t2n−1−t2n−4 t2n+1 = 3 t2n+ 1. 4. The probabilistic distribution of the values of Sα(n) for n= 1, 2, 3, . . . appears to be very regular and stable. Is there a Gaussian distribution lurking in the background? 5. Finally there is the problem of the distribution of Sα(n) over the residue classes mod m(with m > 2). 6. Although we did not study general irrational α’s, we observed various regularities of Sα(n) for α= a simple form composed with the number e. 67 J. ARIAS DE REYNA — J. VAN DE LUNE For example, α=e,e1/m ,e−1/m ,e1/m−1 e1/m+1 . It seems that for the last of these the record-holders are given by tk=     kif 1 ≤k≤2m 2(4mr −m+ 1)tk−1−tk−2if k= 4mr2−2mr + 1, (r > 1) 2tk−1−tk−2if k6= 4mr2−2mr + 1, (r > 1). 9. Appendix. The FFL algorithm This algorithm is implicitly contained in [9]. It computes the value of Sα(n) for any irrational α. Here we present its implementation in Mathematica Version 5.2 and present a proof of it. In the first part of the algorithm, applying Lemma 1, we determine an irrational β∈(0,1) and a sign σsuch that for every natural number nwe have Sα(n) = σSβ(n). S[α_, M_] := Module[{j, R, m, a, b, k}, (* ========================================================= *) (* 1. Compute βand σsuch that Sα(n) = σ·Sβ(n)*) (* ========================================================= *) σ= 1; β=α; If[β< 0, β= -β;σ= -σ]; (* Now β> 0 *) β- = 2 Floor[β/2]; (* Now 0 < β< 2 *) If[β> 1, β=2-β;σ= -σ]; (* Now 0 < β< 1 *) In order to compute the value of Sβ(n) the FFL algorithm uses the denominators qkof the convergents of the number γ=β/2. We will have to compute these qkfor the indices ksatisfying qk−1≤n<qk. By induction we find that qk≥Fk+1, a Fibonacci number. So, we compute qkfor all ksuch that Fk−1≤³1+√5 2´k≤n. (* ========================================================= *) (* 2. Compute the necessary q[k] for M *) (* ========================================================= *) β=β/2; kMax = 1 + Ceiling[Log[2, M + 1]/Log[(1+Sqrt[5])/2]]; CF=ContinuedFraction[ β, kMax ]; 68 ON SOME OSCILLATING SUMS q[-2]=1; q[-1]=0; For[ k = 0, k < kMax, k++, q[k] = CF[[k+1]]* q[k-1]+ q[k-2] ]; Finally, the algorithm depends on two principles: (A) If qk−1≤m < qkand qk/2< m then Sβ(m) = Sβ(qk−m−1) + ((−1)k−1if qkis even 0 if qkis odd. (B) If qk−1≤m < qk,m≤qk/2, and m=hqk−1+rthen Sβ(m) = Sβ(r) + h(0 if qk−1is even (−1)k−1if qk−1is odd. Now we can supply the code for the function Sα(n). 9(* ========================================================= *) 10 (* 3. The FFL-routine proper *) 11 (* ========================================================= *) 12 m=M; R=0; 13 (* Throughout the program we will have S(M)= S(m)+ R *) 14 j = kMax - 1; 15 While[ m > 0, While[q[j] > m, j--]; 16 (* We have located j with q[j] <= m < q[j+1] *) 17 a = q[j]; b = q[j+1]; 18 If[ b - m < b/2, (* Then we apply principle A *) 19 R += If[ EvenQ[b], (-1)^j, 0]; 20 m=b-m-1, 21 (* Else we apply principle B *) 22 R += Floor[m/a] * If[EvenQ[a], 0, (-1)^j ]; 23 m = Mod[m,a] ]]; 24 (* end of While. *) 25 (* Output *) sigma * R ] 26 (* END of the FFL-routine and S[α,M] *) We prove the two principles of the algorithm. (A) By Theorem 17 the number 2qk−1 is an ECREF for γso that b(2qk−j)γc+bjγc= 2pk−1,1≤j < qk.(46) Put j= 2n. Since γ=β/2 we have b(qk−n)βc+bnβc= 2pk−1,1≤n < qk/2. 69 J. ARIAS DE REYNA — J. VAN DE LUNE Therefore 1≤n < qk/2 =⇒(−1)bnβc=−(−1)b(qk−n)βc. Thus qk−1 is an ECREF for βand we have Sβ(qk−1−n) = Sβ(qk−1) + Sβ(n),0≤n < qk/2. Moreover, we have Sβ(qk−1) = qk−1 X j=1 (−1)bjβc= qk−1 X j=1 (−1)b2jγc. By (46) we have (−1)b2jγc=−(−1)b2(qk−j)γc. If qkis odd we get Sβ(qk−1) = 0, and for qkeven the entire sum is equal to the central term (−1)bqkγc. By Lemma 4 this term is equal to (−1)k+pk. Since qkis even, pkis odd, and we get Sβ(qk−1) = (−1)k−1. Writing m=qk−1−nwe get Sβ(m) = (−1)k−1+Sβ(qk−1−m), qk/2< m < qk completing the proof of (A). (B) Now assume that qk−1≤m≤qk/2. Let hand 0 ≤r < qk−1be such that m=hqk−1+r. Let jbe such that qk−1≤j < j +qk−1≤m<qk/2 and let 2j=dqk−1+u. By applying Proposition 5 twice, first for 2j+ 2qk−1and then for 2j, we obtain (−1)b(j+qk−1)βc= (−1)b(2j+2qk−1)γc= (−1)(d+2)pk−1+buγc= = (−1)dpk−1+buγc= (−1)b2jγc= (−1)bjβc. (This for u6= 0 and by a similar reasoning for u= 0.) It follows that Sβ(m) = hqk−1+r X j=1 (−1)bjβc= r X j=1 (−1)bjβc+h qk−1 X j=1 (−1)bjβc. As observed in the proof of (A), the terms of the sum Pqk−1 j=1 (−1)bjβccancel (the term for j= 1 with that for j=qk−1−1, the term with j= 2 with that for j=qk−1−2, . . . ). If qk−1is odd the sum is equal to (−1)b2qk−1γc, and if qk−1 is even it is equal to (−1)bqk−1γc+ (−1)b2qk−1γc. By Proposition 5 we see that b2qk−1γcis equal to 2pk−1when k−1 is even, and equal to 2pk−1−1 when k−1 is odd. Thus in each case (−1)b2qk−1γc= (−1)k−1. So, when qk−1is odd we have Pqk−1 j=1 (−1)bjβc= (−1)k−1. 70 ON SOME OSCILLATING SUMS When qk−1is even we have qk−1 X j=1 (−1)bjβc= (−1)k−1+ (−1)bqk−1γc. Also we have (−1)bqk−1γc= (−1)pk−1+k−1, and qk−1being even, pk−1is odd. Therefore (−1)bqk−1γc= (−1)kand (−1)bqk−1γc+ (−1)b2qk−1γc= 0. This completes the proof of (B). Acknowledgment. The authors would like to thank Foster Dieckhoff ( Kansas City, MO ) for his linguistic assistance in preparing this paper, and his interest in our results. 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