On the discrepancy of circular sequences of reals

On the discrepancy of circular sequences of reals Fan Chung Ron Graham Abstract In this paper we study a refined measure of the discrepancy of sequences of real numbers in [0, ] on a circle C of circumference. Specifically, for a sequence x = (x, x,...) in [0, ], define the discrepancy D(x) of x by D(x) = inf inf n xm xm+n n m where x i x j = min{ x i x j, x i x j } is the distance between x i and x j on C. We show that sup x D(x) 3 and that this bound is achieved, strengthening a conjecture of D. J. Newman. Introduction A basic question in the study of the distribution of real sequences is the quantitative estimation of the extent by which an arbitrary sequence must deviate from some measure of regularity. This topic has an extensive literature, much of which is surveyed in [], [], [4], [9], and [0]. As an example, de Bruijn and Erdős in [6] considered the following measure of irregularity for sequences on a circle C of circumference (see also [], [3], and [4]). For x, y C, define the distance d(x, y) between x and y to be x y = min{ x i x j, x i x j }. Thus, x y is just the length of the shorter arc in C joining x and y. For an infinite sequence x = (x, x,...) in C, define ω(x) = lim inf n Theorem [6]. For any sequence x in C, inf n x i x j. i<j n ω(x) = 0.73.... () ln 4 Furthermore, the bound in () is best possible, as shown by taking x n = {log (n )} where {z} denotes the fractional part of z. While this sequence x is optimal with respect to ω, it is certainly not as spread out as one would think that a well-distributed sequence should be. In University of California, San Diego

particular, consecutive points x n and x n+ are very close together. In this note, we consider a related but much more sensitive measure of discrepancy, suggested by the following conjecture of D. J. Newman (see [7]). For an infinite sequence x = (x, x,...) in [0, ], define Conjecture (D. J. Newman). δ(x) = inf lim inf n x m x m+n. n m sup δ(x) = 0.447.... x This conjecture was motivated (in part) by the fact that occurs frequently in extremal problems involving discrepancy. For example, it follows from standard results in diophantine approximation (e.g., see []), that for any real θ 0, lim inf n{nθ}. () n Furthermore, the constant cannot be replaced by any smaller constant since () holds with equality for θ = +, for example. For an infinite sequence x = (x 0, x, x,...), we define the discrepancy D(x) of x by D(x) = inf n inf m n x m x m+n. For this measure of discrepancy, any run of n consecutive terms of x must be just as well dispersed as the first n terms of x. In this sense, it is a more sensitive measure of irregularity of distribution. A natural question concerning this definition is about the choice of inf versus liminf. As it turns out, the same bounds hold for both versions and therefore we here adapt the stronger notion. More details are discussed in Section 3. In this paper, we will prove that Newman s conjecture is valid. In particular, we determine the best possible constant which turns out to be somewhat smaller than. For the discrepancy of a sequence x of reals on a circle C of circumference, we will prove the following. Theorem. sup x D(x) = α 0 = 3. Note that α 0 = 3 = 0.38966... < = 0.447.... The analogous problem of bounding the discrepancy of sequences x which lie in the interval [0, ] (as opposed to the circle C) was considered by the authors in []. For this case, we define the discrepancy D (x) = inf n inf m n x m x m+n. As it turns out, the extremal value of D (x) is slightly larger than it is for D(x). In particular, the following result [] was established.

Theorem. For any sequence of reals x = (x, x,...) with x i [0, ], we have ( D (x) = inf inf n x m+n x n + ) n m F k k = β = 0.3944967... where F k denotes the k th Fibonacci number. Furthermore, this upper bound is sharp. For example, it is achieved by the sequence x defined by x n = β i ɛ i (n) F i (3) where the ɛ i (n) {0,, } are determined as follows. Every integer n has a unique expansion as n = i ɛ i (n)f i where ɛ i (n) {0,, } and if ɛ i (n) = = ɛ j (n) for i < j then there is an index k with i < k < j such that ɛ k (n) = 0. Before proceeding we first state several facts concerning Fibonacci numbers that will be needed. The Fibonacci numbers F n are given by the recurrence F 0 = 0, F = and F n+ = F n+ + F n for n 0. (i) F n = (τ n σ n ) where τ = ( + )/ and σ = ( )/. Thus τ + σ = and τσ =. (ii) n F k = F n+. k= (iii) For all s t, F s+ < t i=s+ +. F i F t F t+ F s F s+ In the remainder of the paper, we will use other various well-known properties of the Fibonacci numbers, all of which are standard and can be found in [8]. 3

The main results In the circle C of circumference, we will assume the point 0 C is located at the rightmost point of C and the positive direction on C proceeds in a counterclockwise direction (so that 4 is at the top of C). For an infinite sequence x = (x 0, x, x,...), we will assume without of loss of generality that x 0 = 0, and x. Given a finite sequence x n = (x 0 = 0, x, x,..., x n ), we will let π(x n ) denote the sequence specifying the order in which the x i appear as we move around C in a counterclockwise direction starting from 0. For example, if x 7 = ( { 3 8 k} : 0 k 7) = (0, 3 8, 6 8, 8, 4 8, 7 8, 8, 8 ) then π(x 7 ) = 0, 3, 6,, 4, 7,, (see Figure ) where use the convention that the sequence in the angle brackets denotes the order in which the x i are encountered as we go around C in the counterclockwise direction. That is, starting at x 0 = 0, we see x 3 = 8, then x 6 = 8, then x = 3 8, etc. As mentioned previously, {x} denotes the fractional part of x. {6 3 8} = 8 { 3 8} = 3 8 {3 3 8 } = 8 {4 3 8} = 4 8 {0 3 8 } = 0 {7 3 8 } = 8 { 3 8 } = 7 8 { 3 8 } = 6 8 Figure : Circle C with the points { 3k 8 }, 0 k 7. Now for a positive integer N, define the restricted discrepancy D N (x) of x by D N (x) = inf inf n x m x m+n. n N m With this restricted discrepancy we will establish the following theorem which will be a step in proving Theorem. Theorem 3. sup x D N (x) = Fn F n+ for F n N < F n+. 4

First, we will show that the bound in Theorem can be achieved. Define ȳ = (y 0, y, y,...) with y k = {k α 0 }, k 0 where we recall that α 0 = 3. Claim. D(ȳ) = α 0. Proof. To begin, observe that we only need to check that inf n y n α 0. (D(ȳ) n is certainly at most α 0 since y = α 0.) Furthermore, we only have to verify the claim for those y n which have the property that y n < y s, for all s < n. A quick calculation shows that the order of the first 6 terms of ȳ is 0, 3,, 4,,. Thus, extending the order to the next three terms, since y is between y 0 and y, then y 6 must be between y and y 3, and y 7 must be between y and y 4. Consequently, the distances y 6 and y 7 cannot be smaller than y. However, y 8 lies between y 3 and y, and since this interval also contains 0, then y 8 might be a new champion small distance (in fact, it is). A check shows that it is in the interval between y 0 and y 3. Continuing this argument, we see that y 9 must be between y and y 4, y 0 must be between y and y, y must be between y 3 and y 6, and y must be between y 4 and y 7. Thus, none of these 4 points can be closer to 0 than y 8 is. However, y 3 will lie in the interval defined by y and y 8, and since this interval also contains 0, then y 3 could be a new champion. (Again, it is, and in fact, it lies between y 0 and y.) Notice that the indices at which new champions occur are exactly the Fibonacci numbers. This fact is well known and follows from the following calculations. In particular, F t α 0 F t = ( τ (τ t σ t ) (τ t σ t ) = (σ t σ t+ ) = ( σ 4 ) σ t = σ t. F r α 0 = F r + σ r > F r, F r+ α 0 = F r + σ r+ < F r. This implies that when the time comes for y Ft to be located, it lies between 0 and y Ft (and not between 0 and y Ft ). Therefore, for t, F t F t α 0 = F t σ t = (τ t σ t σ t ) = (( ) t σ t ) ( σ 4 ) = α 0. )

This proves the Claim.. Proof of Theorem Suppose x = (0, x, x,...) has D(x) = α α 0 = 0.38966.... Thus, we must always have x m x m+n α n. First, consider (0, x, x ) (where we have assumed that x ). If 0 < x < x then x = x + x x α( + ) so that α 3, which is a contradiction. Hence, x > x. In other words, the order of these three points is 0,,. In this case we have so that x + x + x x x + α( + ) x 3 α 3 α 0 < 0.47. Of course, the same argument shows that x m x m+ < 0.47 for all m. The next issue is to decide where x 3 can go. If we had the order 0,,, 3 (i.e., if x 3 were between x and 0), then we would have so that = x + x x + x x 3 + x 3 α( + + + 0 3 ) = α( 3 ) α 3 0 < α 0 which is a contradiction. Hence, we must have the order 0, 3,,. In fact, this must be the relative order of any four consecutive terms of x. In particular, since 3 is between 0 and, then 4 must be between and. Now, where can go? If were between 0 and 3, so that we had the order 0,, 3,, 4,, then = x + x x 3 + x 3 x 4 + x 4 x + x α( + ++ + 7 ) = α( 0 ) which implies α 0 7 < 0.3704 < α 0, a contradiction. Hence, we must have the order 0, 3,, 4,,, i.e., is between and 0. Thus, we can conclude that if D(x) α 0 then any six consecutive terms of x must be in the same order 0, 3,, 4,, as we go around C in the positive direction. Here, we will begin to abuse our notation and identify the point x i with its index i (for typographical convenience). In this case, consider the sequence 0,,, 3, 4,, 0 (which goes around C exactly twice). We then have which implies α 0 6 = 3. α( + + + + + ) = 6 α 6

Let us do two more steps of this process before moving to the general case. Since lies between and 0, then 6 lies between 3 and, and 7 lies between 4 and. Since none of these intervals contains 0, then no new smaller distance to 0 can occur. However, 8 lies between and 3, and this interval does contain 0. So we need to decide on which side of 0 that 8 could lie. Suppose that 8 were in between and 0. Then consider the sequence 0, 8,, 4, 3, 0. Since this goes around C just once, then by our assumption that D 8 (x) D(x) = α α 0 we have, α( 8 + 3 + + + 3 ) = 67 4 α so that we have α 4 67 = 0.38... < α 0 = 0.380966..., a contradiction. Thus, if D(x) α 0 then 8 must lie between 3 and 0. Further, 9 must lie between 4 and, 0 must lie between and, must lie between 6 and 3, and must lie between 7 and 4. None of these intervals contain the point 0. However, 3 must lie between 8 and, and this interval does contain 0. Again, there are two possibilities for the location of 3. If 3 were between 0 and 8, then consider the sequence 0, 3, 8, 3, 4,, 0. This sequence goes around C once, so that we have (since D 3 (x) D(x)), α( 3 + + + + + ) = 74 6 α. This implies that α 6 74 = 0.3736... < α 0, a contradiction. Hence, 3 must go between and 0 (and the same relative order must hold for any 4 consecutive terms of x). In general, we must distinguish two cases when a new term is placed into the interval containing the point 0. If follows by induction that this point must be of the form F t for some t. We have just seen this for the points F 3 =, F 4 = 3, F =, F 6 = 8 and F 7 = 3. Note that the Fibonacci numbers with odd indices occur before 0 and the Fibonacci numbers with even indices occur after 0. The first case is when the new point has the form F n+, where we can assume that n 3. Then the interval containing 0 is bounded by F n from below and F n from above. Suppose that F n+ were between 0 and F n. Then consider the sequence 0, F n+, F n, F n, F n +, F n +, F n + 3,..., F n, F n, 0. This path goes around C exactly F n times. Hence, since D Fn+ (x) α, we have ( α + + + F n 3 + ) F n. F n+ F n F n F n 7

This implies that ( α + 3 ) F n αf n 3 F n+ F n Thus, α F n α 0 F n 3 = ( τ n σ n ) τ (τ n 3 σ n 3 ) = ( σ n σ n ) = (σ 4 )σ n = σ n 3. F n+ + 3 = F n τ n 3 ( F n+ + 3 F n ). σ n 3 To get a contradiction, we would like to show i.e., Now (4) is equivalent to τ n 3 ( F n+ + 3 F n ) < α 0 = τ, + 3 >. (4) F n+ F n τ n τ n (F n + 3F n+ ) > F n+ F n, or τ n (τ n σ n + 3τ n+ 3σ n+ ) (τ n+ σ n+ )(τ n σ n ) = τ 4n + σ 4n (τ + σ ) = τ 4n + σ 4n 3. () However, () holds provided we have (3τ 4n 4 + τ 4n 6 σ 4 3σ 6 ) > τ 4n + σ 4n 3 or (3 τ + τ 6 )τ 4n 6 > (σ 4 + 3σ 6 ) + σ 4n 3. (6) 8

However, this is immediate since (3 τ + τ 6 ) = 3 ( ) =.840... > 0 whereas the right-hand side of (6) is less than 0. This shows that F n+ must go in between F n and 0. In this case, consider the sequence 0,,, 3,... F n+, F n, 0. This path goes around C exactly F n times and so we have the inequality In this case we deduce that D Fn+ (x) α(f n + F n+ ) F n. F n F n+ + = F n F n+ F n+ Fn+ + = F n F n+ F n F n+3 = F n+ F n+3. The other case is when the new point in question is F n, a Fibonacci number with an even index. The argument is this case is similar to the preceding case. Here, the interval containing the point 0 is bounded by F n from above and by F n from below. (In other words, as we traverse C in the positive direction, we see the points F n, 0, F n in that order.) Suppose that F n is between F n and 0. Then consider the sequence 0, F n, F n, F n, F n, F n 3,..., F n +, F n, 0. This covers C exactly F n times. Thus, we have ( α + ) + F n 3 F n. F n F n Therefore, ( α + ) F n αf n 3 F n F n F n α 0 F n 3 = F n τ F n 3 = ( τ n σ n τ (τ n 3 σ n 3 ) = ( σ n σ n ) = (σ 4 )σ n = σ n 3 = τ n 3. ) 9

Thus, α ( τ n 3 F n + So, we will obtain a contradiction to (7) if we can show in other words, In fact, it is true that τ n 3 ( This follows from the fact that F n + F n ) < α 0 = + > F n F n τ n. > F n τ n. F n τ n = τ n σ n τ n = (τ 3 + σ 4n 7 ) F n ). (7) τ, < τ 3 = + <. Thus, if D(x) α 0 then F n must go in between 0 and F n. Since we have shown that D(ȳ) = α 0 for the sequence ȳ = (y, y, y 3,...) with y k = {kα 0 }, then the order of the y k around C must also be in the same as we have deduced here. Finally, for N = F n+, we have already noted that D(x) = inf n inf m n x m x m+n α 0 implies D N (x) = inf n N inf m n x m x m+n F n+ F n+3. Since D(x) D N (x) for all N then sup x This proves Theorem. D(x) = lim D N(x) = lim N n F n+ = F n+3 τ = α 0. 0

To complete the proof of Theorem 3, it will suffice to exhibit a sequence x such that D N (x) = F n F n+ for F n N < F n+. To do this, we define We need to verify that x k = { k Fn F n+ }, k 0. inf inf k x m x m+k F n. (8) k<f n+ m F n+ As before, it will suffice to take m = 0. Also, it will be enough to show that (8) holds when k = F m for some m. So we need to prove that { F m Fn F n+ } F n F n+, m < n +. (9) Scaling our problem up by a factor of F n+, we have the equivalent condition F m (F m F n (mod F n+ )) F n, m < n +, (0) where r(mod s) denotes min{r(mod s), s r(mod s)}. Note that r(mod s) = (s r)mod s. For example, for n = 4, F 7 = 3, F 9 = 34, we have m F m (F m 3(mod 34)) 3 = 3 3 3 = 3 3 3 8 = 6 3 4 3 = 3 3 = 3 6 8 = 6 3 7 3 = 3 3 8 = 3 Note that in this example, F m 3(mod 34) = F 9 m (mod 34), m 8. Of course, this is no accident! We recall a general Fibonacci identity: Changing variables, this becomes: F r+ F s F r F s+ = ( ) s+ F r s. F n+ F m F n F m = ( ) m F n+ m. () However, F n (mod F n+ ) = F n (mod F n+ ).

Thus, Therefore, by () we have F m F n (mod F n+ ) = F m F n (mod F n+ ). F m F n (mod F n+ ) = F n+ m where the ( ) m factor has been absorbed into the mod function. Thus, (0) becomes F m F n+ m F n, m < n +. () If fact, we will prove the more general inequality: F m F n m F n, m < n, n. (3) (3) clearly holds for n = and n = 3. We prove by induction that it holds for all n 4. Since F m F n m = F m (F n m + F n m ) = F m F n m + F m F n m F n + F n 3 by induction = F n then (3) follows, as does (). This completes the proof of (9) and Theorem 3 is proved. 3 Concluding remarks. While we have defined our discrepancy in the form D(x) = inf n inf m n x m x m+n, we could have just as well have used the modified version D(x) = inf n lim inf x m x m+n. m It is not hard to show that the same bound holds for this modified version as well. The sequence x in (3) which achieves equality in Theorem has curious similarity to the well-known van der Corput sequence z which arises in the study of the uniform distribution of real sequences (see []). Here for each n 0, we expand n = i ɛ i(n) i into its binary expansion, and then define z n = i ɛ i(n) i. That is, we flip the powers of from the numerator to the denominator. This is just what did to form the x n, namely, we flipped the Fibonacci numbers in the expansion of n from the numerators to the denominators. Is this hinting at some general principal for generating well-distributed sequences?

It would be very interesting to understand the analogous problems for two (or more) dimensions. These questions are well known to be difficult. Even for the case of two-dimensional diophantine approximation (with the sup norm), the value of γ = sup θ,φ lim inf q max( qθ, qφ ) q is not known. It is known that γ 7 (see [3]). This implies that the analogue D (x) of D(x) for a sequence x [0, ] [0, ] with the sup norm d, namely D (x) = inf lim inf n d(xm+n, x m ), n m can remain above 7 ɛ for any ɛ > 0 for a suitable sequence x. This is probably not the best possible value, however. It would be very interesting to know the truth in this case. The authors wish to express their thanks to a diligent referee who made many excellent comments. References [] J. Beck and W. Chen, Irregularities of Distribution, Cambridge Tracts in Mathematics 89, Cambridge Univ. Pr., Cambridge, (987) xiv + 94 pp. [] J. Beck and V. T. Sós, Discrepancy Theory, in Handbook of Combinatorics, Vol II, R. L. Graham, M. Grötschel and L. Lovász, eds., Elsevier Science B. V, Amsterdam, (99), 40 446. [3] J. W. S. Cassels, Simultaneous diophantine approximation, J. London Math. Soc., 30, (9), 9. [4] A Panorama of Discrepancy Theory, W. Chen, A. Srivastav and G. Travaglini, eds., Lecture Notes in Mathematics 07, Springer, Heidelberg, (04), xii + 69 pp. [] F. R. K. Chung and R. L. Graham, On irregularities of distribution, Finite and Infinite Sets, Colloquia Mathematica Societatis János Bolyai 37, (98), 8. [6] N. de Bruijn and P. Erdős, Sequences of Points on a Circle, Indagationes Math.,, 46-49. [7] P. Erdős and R. L. Graham, Old and New Problems and Results in Combinatorial Number Theory, Monographies de L Enseignement Mathématique 8, Université de Genève, Geneva, (980), 8 pp. 3

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