A Weak First Digit Law for a Class of Sequences

Transcription

1 International Mathematical Forum, Vol. 11, 2016, no. 15, HIKARI Lt, A Weak First Digit Law for a Class of Sequences M. A. Nyblom School of Mathematics an Geospatial Sciences RMIT University, Melbourne, Victoria 3001, Australia Copyright c 2016 M. A. Nyblom. This article is istribute uner the Creative Commons Attribution License, which permits unrestricte use, istribution, an reprouction in any meium, provie the original work is properly cite. Abstract A non-ergoic approach is employe to establish Benfor s law for the leaing igit = 1 for the sequence of powers of two. For the sequence of powers {α n }, 1 < α 10/, this metho is extene to obtain a weak first igit law by establishing Benfor like lower an upper bouns to the asymptotic relative frequency of terms with a given leaing igit. Mathematics Subject Classification: 11A, 11K36 Keywors: Frequency istribution, Benfor s law 1 Introuction When examining numerical ata which is erive from large naturally occurring ata sets such as population, lengths of rivers or stock prices, often (but not always, a curious phenomena known as Benfor s law is observe. This phenomenological law concerns the frequency istribution of the leaing igits of numerical ata, which typically spans many orers of magnitue. Such a set of numerical ata will satisfy Benfor s law if the proportion of ata values having a leaing igit of {1, 2,..., } is approximately log 10 ( Thus with numerical ata sets which obey this law, the number 1 woul appear

2 68 M. A. Nyblom as a leaing igit with a frequency of approximately 30%, while larger igits such as woul appear as a leaing igit with a frequency of approximately 4.5%. Benfor s law was first empirically observe in 1881 by the American astronomer Simon Newcomb [2], but remaine largely forgotten until 138, when Frank Benfor [1], a physicist, re-teste the phenomenon against many sources of ata as well as on some infinite integer sequences. He foun in both case, that the frequency istribution of the leaing igit exhibite a close aherence to the logarithmic rule. For a sequence of positive reals {a n }, if one efines N (n = #{0 k n : a k has leaing igit }, N then {a n } will satisfy Benfor s law if lim (n n = log n+1 10(1+ 1. To establish that an integer sequence such as {2 n } satisfies Benfor s law, one can appeal to an Ergoic property of the irrational numbers prove by Weyl [3]. Namely, if x is a positive irrational number, n a positive integer an nx the integer part of nx, then the sequence of fractional part of nx, given by c(n = nx nx, is homogeneously istribute over the interval [0, 1]. That is, given any arbitrary reals 0 a < b 1 then #{0 k n : c(k (a, b} lim n = b a. (1 Now if 2 n has a leaing igit of, then 10 k < 2 n < ( k for some integer k. Upon taking logarithms one fins log 10 ( + k < n log 10 (2 < log 10 ( k, an as log 10 ( + 1 [0, 1], one euces k = n log 10 (2, an so log 10 ( < c(n < log 10 ( + 1, where c(n = n log 10 (2 n log 10 (2. Thus when a k = 2 k with a = log 10 ( an b = log 10 ( + 1, we have N (n = #{0 k n : c(k (a, b}, from which the require frequency istribution follows via (1, as log 10 (2 is irrational. In this article an entirely elementary proof will first be presente, which establishes the logarithmic rule for the frequency istribution of terms in the sequence {2 n } having leaing igit = 1. This is possible, as one can show for a n = 2 n that N 1 (n = n log 10 (2 + 1, via an argument base on counting the number of integers in an interval. By applying a variation to this argument, one can then obtain a weak first igit law for the sequence a n = α n, for a fixe 1 < α 10, where in particular it is shown that for all n L n N (n U n, with L n log 10 (1 + 1 log 10(α an U n log 10 ( log 10(α as n. Although this result is only suggestive of the logarithmic rule of Benfor s law, the avantage of the approach taken here rests on the non-ergoic nature of the argument employe.

3 A weak first igit law for a class of sequences 6 2 First igit law for powers of two We begin with a technical lemma concerning the counting of integers in an interval of the form [a, b. In what follows, efine for a real number x, the floor an ceiling of x as x = max{n Z : x n} an x = min{n Z : n x} respectively. Furthermore, we shall make use of the fact that x+n = x +n for all n Z an x R. Lemma 2.1 Suppose 0 a < b with b a = L, then the number of integers containe in the interval [a, b is either L or L + 1. Proof: Let I(a, b enote the number of integers containe in the interval [a, b. We note that for two integers 0 m n, the number of integers in the interval [m, n] is equal to n m + 1. Case 1: b N. In this instance I(a, b = b a + 1 since [ a, b ] [a, b. As b b an a a one fins I(a, b b a+1, but as b a+1 is the largest integer less than or equal to b a + 1, we euce that I(a, b b a + 1 = b a + 1 = L + 1. Now as a < a + 1, then b a > b a 1 b a 1 an as b a = b a is the largest integer less than or equal to b a, we further euce b a b a 1 = b a 1, that is I(a, b L. Hence I(a, b is either equal to L or L + 1. Case 2: b N. If a N, then I(a, b = (b 1 a + 1 = b a = L, since [a, b 1] [a, b. Alternatively if a N, then I(a, b = (b 1 ( a = b a 1, since [ a, b 1 [a, b an a = a + 1. But as a < a < a + 1 we fin b a > b a > b a 1, that is I(a, b + 1 > b a > I(a, b an so I(a, b = b a = L. Theorem 2.1 The frequency istribution of the terms in the sequence {2 n } having leaing igit = 1, is log 10 (2. Proof: Recall N 1 (n = #{0 m n : 2 m has leaing igit 1}. If 2 m has k igits an leaing igit 1, then 10 k 1 2 m < 2 10 k 1. Upon taking logarithm to base ten yiels k 1 m log 10 (2 < log 10 (2 + k 1. (2 Now consier the continuous linear function f(x = log 10 (2x, an suppose the interval of values on which f(x satisfies (2 is [a, b. As f(b f(a = log 10 (2, we euce that the interval [a, b must have unit length. Thus from Lemma 2.1, the interval [a, b can contain either one or two positive integers. To

4 700 M. A. Nyblom show there is exactly one such integer in [a, b, suppose there are two integers m 1 < m 2 such that both 2 m 1 an 2 m 2 have k igits with leaing igit 1. As m 2 m 1 + 1, observe 2 m 2 2 m 1 2 m 1 (2 1 1 = 2 m 1 an so 2 m 2 2 m 1 will have at least k igits, but as both 2 m 2 an 2 m 1 have k igits with a leaing igit 1, the ifference 2 m 2 2 m 1 can have at most k 1 igits, a contraiction. Hence there is exactly one power of two having k igits with leaing igit 1. Consequently N 1 (n must be equal to the positive integer k such that 2 n [10 k 1, 2 10 k 1 [2 10 k 1, 10 k or equivalently k 1 n log 10 (2 < k, that is k 1 = n log 10 (2 an so N 1 (n = n log 10 ( Finally as x 1 < x x one fins that n log 10 (2 < N 1(n n log 10(2 + 1 from which the esire limiting value of log 10 (2 follows upon taking n., 3 A weak first igit law for α n In Section 2 we were able to etermine the correct frequency istribution for the powers of two having a leaing igit of = 1, since an explicit formula for N 1 (n coul be erive. By varying the argument use to prove Theorem 2.1, we can obtain upper an lower estimates of N (n, for the sequence {α n }, where 1 < α 10. These estimates can then be use to obtain upper an lower bouns on the frequency istribution for the terms of the sequence {α n } having leaing igit, which are suggestive of the logarithmic rule preicte via Benfor s law. Theorem 3.1 Given the sequence {α n }, where 1 < α 10, then for all n L n N (n U n, with L n log 10 (1 + 1 log 10(α an U n log 10 ( log 10(α as n. Proof: Recall N (n = #{0 m n : α m has leaing igit }, moreover from assumption observe log 10 (α log 10 ( 10 log 10( +1. If αm has k igits an leaing igit, then 10 k 1 α m < ( k 1. Upon taking logarithm to base ten yiels log 10 ( + (k 1 m log 10 (α < log 10 ( (k 1. (3 Now consier the continuous linear function f(x = log 10 (αx, an suppose the interval of values on which f(x satisfies (3 is [a, b. As f(b f(a =

5 A weak first igit law for a class of sequences 701 log 10 ( +1, we euce that the interval [a, b must have length L = log 10 ( +1/ log 10(α 1. Thus from Lemma 2.1, the interval [a, b contains either L or L + 1 positive integers. Consequently, the number of terms of the sequence {α m } for which α n has k igits an leaing igit, is either L or L +1, moreover these upper an lower bouns are inepenent of k. If α n contains s igits, then 10 s 1 α n < 10 s, from which one euces that s 1 = n log 10 (α, an so {α 0, α 1,..., α n } n log 10 (α +1 k=1 [10 k 1, 10 k. Now as α n [10 n log 10 (α, 10 n log 10 (α [ 10 n log 10 (α, (+1 10 n log 10 (α [( n log 10 (α, 10 n log 10 (α +1, the leaing igit of α n may or may not be, an so the number of sequence elements in [10 n log 10 (α, 10 n log 10 (α +1 having a leaing igit, can only be boune between 0 an L + 1. While from above, the number of sequence elements having a leaing igit of in [10 k 1, 10 k, for k = 1,..., n log 10 (α, must be boune between L an L + 1. Thus from efinition, we have for all n 0 [L]( n log 10 (α N (n ([L] + 1( n log 10(α + 1. (4 Finally, using again the inequality x 1 < x x, one fins that the upper an lower bouns of (4 are respectively boune above an below by ( log10 ( +1 U n = ( (n log (α + 1 log10 ( +1 an L n = (n log 1 10 (α 1 log 10 (α log 10 (α, with L n log 10 (1 + 1 log 10(α an U n log 10 ( log 10(α as n. To conclue, we note the result of Theorem 3.1 can be strengthene for the terms of the sequence {( 10 n }, having a leaing igit of as follows. If we set α = 10 an =, then the length of the interval [a, b on which the linear function f(x satisfies (3 is L = 1, an so there can only be one or two terms of the sequence {( 10 n } having k igits with a leaing igit of. To show there is exactly one, assume there are two integers m 1 < m 2 such that both ( 10 m 1 an ( 10 m 2 have k igits with a leaing igit. As m 2 m 1 + 1, observe ( 10 m 2 ( 10 m 1 ( 10 m 1+1 ( 10 m 1 = ( 10 m 1 ( 1, but ( 10 m 1 ( 1 has k igits with a leaing igit of 1. However as both ( 10 m 2 an ( 10 m 1 have k igits with a leaing igit, the ifference ( 10 m 2 ( 10 m 1 can have at most k 1 igits, a contraiction. Now { 10 0, 10 1,..., 10 n } n log 10 ( k=1 [10 k 1, 10 k, an as the number of sequence elements having a leaing igit of in [10 k 1, 10 k, for k = 1,..., n log 10 ( 10, is precisely one, we can conclue as ( 10 n may or may not have a leaing igit of, that n log 10 ( 10 1 n log 10( 10 N (n n log 10( n log 10( ,

6 702 M. A. Nyblom from which the esire limiting value of log 10 ( 10 follows upon taking n. References [1] F. Benfor, The law of anomalous numbers, Proceeings of the American Philosophical Society, 78 (138, [2] S. Newcomb, Note on the frequency of use of the ifferent igits in natural numbers, American Journal of Mathematics, 4 (1881, [3] H. Weyl, Uber ie Gleichverteilung von Zahlen mo Eins, Mathematische Annalen, 77 (116, Receive: June 7, 2016; Publishe: July 25, 2016