ORMAL UMBERS AD UIFORM DISTRIBUTIO WEEKS -3 OPE PROBLEMS I UMBER THEORY SPRIG 28, TEL AVIV UIVERSITY Contents.. ormal numbers.2. Un-natural examples 2.3. ormality and uniform distribution 2.4. Weyl s criterion 96 3.5. Proof of Weyl s criterion 5.6. Polynomial sequences 7.7. Metric Theory.8. Quantitative aspects of uniform distribution: Discrepancy 2 References 5.. ormal numbers. Definition. A real number α, is simply normal to base, if in the decimal expansion α =.a a 2..., each digit occurs with the same frequency namely /: For any digit t {,,...,, 9}, lim # {n : a n = t} = We say that α is normal to base if for every k, every string t... t k of k digits t j {,,..., 9} appears with the same frequency, namely / k : lim # {n : a n+ = t, a n+2 = t 2,..., a n+k = t k } = k Date: March 25, 28.
2 We can naturally change base to any other base b 2, with the same definitions. Exercise. Show that a rational number is not normal to any base. The natural questions are: Is there a normal number?.2. Un-natural examples. Champernowne s number 933, []: Concatenate all natural numbers into a decimal expansion.23456789234... Copeland Erdős number 946 [2], conjectured by Champernowne: Concatenate all primes into a decimal expansion..2357372329... Davenport and Erdős 952 [3], conjectured by Copeland and Erdős: Let fx be any polynomial in x, all of whose values for x =, 2,... are positive integers. Then the decimal.ff2f3..., where fn is written in the scale of, is normal in base. Open Problem. Find a single example of a naturally occurring number which is normal to some base. For instance, is the golden ration ϕ = + 5/2 normal? It is conjectured that any real algebraic number is normal to all bases..3. ormality and uniform distribution. Let X = {x n } R/Z [, be a sequence of numbers on the unit circle. We say that X is uniformly distributed mod if for every subinterval I [,, the proportion of elements of X contained in I equals the length of I: lim # {n : x n I} = I. Likewise, if we take a sequence x n of real numbers, we use the fractional parts {x mod }. ote that a uniformly distributed sequence X [, is necessarily dense. The converse is not true: A standard example is the sequence of fractional parts {log n : n =,... } Exercise 2. Show that the sequence of fractional parts {{log n} : n =, 2,... } is dense in [,.
UIFORM DISTRIBUTIO 3 As we shall later see, if α R\Q is irrational, then the sequence {αn : n =, 2,... } is uniformly distributed mod. The relation to our previous discussion of normal numbers lies in the following observation: Proposition.. α is normal to base resp., base b if and only if the sequence {α n : n =, 2,... } is uniformly distributed modulo resp. αb n. Proof. It suffices to consider intervals I = [a, b with endpoints being finite decimals, i.e. of the form a =.a...a r, in fact when we only change the last digit exercise: why?. For simplicity, take I = [.23,, 24. ow note that if α =.a a 2..., then the fractional part of n α =.a n+ a n+2 a n+3..., so that { n α} I = [.23,, 24 a n+ =, a n+2 = 2, a n+3 = 3. Thus the number of n for which the fractional parts {α n } I is exactly equal to the number of occurrences of the string 23 in the first + 3 digits of α. ormality of α means that this number is asymptotically / 3. oting that the length of our interval I is / 3, we see that we get the number of n such that { n α} I is asymptotically I, which is what is required in the definition of uniform distribution. This observation shifts the focus from normality to the general context of uniform distribution of sequences. So far we have not proved that any sequence is uniformly distributed. Before doing so, we develop Weyl s breakthrough method for establishing this..4. Weyl s criterion 96. Theorem.2. Let X = {x n : n =, 2,... } R. Then X is uniformly distributed modulo if and only if for all integers k, lim e 2πikxn =. n= In what follows, we shall abbreviate ez := e 2πiz Example: We shall show that for any irrational α, the fractional parts of αn are u.d.: By Weyl s criterion, it suffices to show cancellation in the Weyl sums n= e2πikαn. But for this sequence, these are just
4 geometric progressions, so as long as kα / Z which is guaranteed by irrationality of α if k, so that Hence n= e 2πikαn = n= e 2πikαn 2 ekα e + kα ekα. ekα, as, so that Weyl s criterion is satisfied. Example: The sequence of fractional parts of log n is not uniformly distributed mod earlier we saw that it is dense mod. To see this, we need to see that Weyl s criterion does not hold. We will show that elog n n Indeed, we use summation by parts to evaluate the Weyl sum. Recall that for a differentiable function ft, and any sequence {a n }, we denote by At := n t the partial sums of the sequence, then the weighted sum n a nfn can be written as x a n fn = Axfx Atf tdt n x and more generally, if z < x then a n fn = Axfx Azfz z<n x a n x Exercise 3. Using summation by parts, show that log n = x log x x + Olog x n x for some constant C. n= n = log + C O z Atf tdt
UIFORM DISTRIBUTIO 5 We apply this with ft = e 2πi log t = t 2πi, and a n =, so that At = t to obtain n elog n = 2πi which is clearly not o. t 2πit 2πi dt = +2πi 2πi t 2πi dt + 2πi = +2πi 2πi +2πi + 2πi + O = +2πi + O log, + 2πi t 2πi {t}dt t dt.5. Proof of Weyl s criterion. The argument is all about approximating various functions by step functions and by trigonometric polynomials. For a periodic function f, we ask whether the following holds * lim fx n = n fxdx Lemma.3. Let X R/Z. The following are equivalent: The sequence X is uniformly distributed. 2 * holds for all continuous f. 3 * holds for all Riemann integrable f. Proof. Clearly 3 implies. For the converse, note that uniform distribution implies that * holds for all step functions, that is linear combinations of the indicator functions of intervals. But for any Riemann integrable function, by definition, given any ε >, there are step functions s f s + with s + s < ε. Choose ε > so that for all ε,. s + x n n= s + xdx < ε
6 and then fx n n= fxdx = + + fx n s + x n n= n= s + x n s + xdx s + xdx fxdx ow the first expression is non-positive, since f s +, so that the LHS is bounded above by the sum of the last two expressions. By., the second expression lies in ε, ε, and since s + s < ε, the third expression is in [, ε. Hence fx n n= Arguing with s + replaced by s will give and hence fx n n= fx n n= fxdx < 2ε. fxdx > 2ε. fxdx < 2ε. for all ε. Hence 3 holds. Since continuous functions are Riemann integrable, we have 3 2. For the implication 2, argue similarly, approximating the indicator function of an interval above and below by continuous functions. Proof of Weyl s criterion: Since the exponential functions e k x = eks are continuous, by the above Lemma uniform distribution implies that * holds for them, that is one direction follows. We need to show that assuming that * for fx = ekx, for all k k = is obvious, implies uniform distribution, and by the Lemma, it suffices to show * holds for continuous functions. ote that * for all ekx for all integer k implies * holds for all trigonometric polynomials, by linearity. By the Weierstrass approximation theorem, given a continuous function f, for any ε > there
UIFORM DISTRIBUTIO 7 is a trigonometric polynomial T x so that f T < ε. One then argues as in the Lemma that fx n fxdx fx n T x n n= + + n= n= T x n T x fx dx T xdx The first and third expressions are at most ε, since f T < ε, and the second expression is at most ε for all ε. Thus we have uniform distribution..6. Polynomial sequences. Weyl s criterion allows one to get a first step up on a number of problems. As an example, we present one of Weyl s original breakthroughs, namely that the sequence αn 2 is uniformly distributed mod for α irrational. However, unlike the linear case αn, this does not come for free, and needed some significant extra ideas. We start with Dirichlet s theorem on rational approximations. Lemma.4. Let x R be a real number. Then for any integer Q, there are integers a Z, q, with q Q so that x a q < qq In particular for the fractional part {qx} we get {qx} < /Q. If x is irrational, then we must have q as Q. Proof. We consider the Q + numbers {nx} [,, n Q. Dividing the unit interval [, into Q subintervals of length /Q, we use Dirichlet s box principle to deduce that at least one of these subintervals contains two points of the sequence, that is there are m n Q so that {nx} {mx} < Q Writing {mx} = mx M, {nx} = nx with M, Z we obtain nx mx M = n mx M < Q Taking q = n m [, Q] and a = ± M we obtain qx a < Q
8 as claimed. ow if x / Q is irrational, then {qx} and hence we cannot have {qx} < /Q for infinitely many Q s. Hence q as Q. Theorem.5. For irrational α, the sequence {αn 2 : n =, 2... } is uniformly distributed modulo. By Weyl s criterion, it suffices to show that for any nonzero integer k, the Weyl sums S := ekαn 2 = o k,α n= the implied constants are allowed to depend on α and k. We square out the sum S 2 = + 2R ekαn 2 m 2 m<n Writing n = m + h, with m + h, so that we obtain m<n n 2 m 2 = h2m + h = h 2 + 2hm, ekαn 2 m 2 = h= h= h ekαh 2 h m= m= eγ h m eα 2k h m where we have set γ := 2kα, which is irrational if and only if α is irrational. ow we sum the geometric progression, which is the inner sum: h m= h, eγ h m = min ehγ ehγ h+ ehγ, hγ where x := distx, Z. Thus we obtain S 2 + Hence it suffices to show that h= min, 2, else 2 sinπhγ. hγ h γ Z
Proposition.6. For γ / Q, h= UIFORM DISTRIBUTIO 9 min, = o 2 hγ Proof. We use Dirichlet s lemma to obtain coprime a, q, with q so that γ a q < q or γ = a q + θ, θ <. q Divide the range of summation [, ] into consecutive intervals of length q, plus perhaps an extra leftover interval. So if M q < Mq then writing h = qk + j, k < M, j =,..., q, we have γh = ah q + θh = aj q q where θ <. Thus h= min, hγ q M max θ < θh + ak + = aj q q + θ q j= q q max θ < i= min, min, aj q + θ q i q + θ q where we have changed variables aj mod q i mod q, as we may since a is coprime to q, so that as we vary over all residues j mod q, the set aj mod q varies over all residues i mod q recall aj + z only depends q on aj mod q. For at most one value of i mod q, we have i q + θ q < 2q and for that value we replace the corresponding term in the sum by. For the remaining i s, we have Hence we find q max θ < i= min, i q + θ q i q + θ q i 2q + q i= q/i + q log q
Therefore h= min, + q log q 2 hγ q q + log Recall that since γ is irrational, we have q as, and hence the above is o 2. An elaboration of the above argument gives Theorem.7. Let P x = a d x d + a d x d + + a x + a be a polynomial of degree d 2, with at least one of the ceofficients a d,..., a irrational the constant term a does not matter. Then the sequence {P n : n =, 2,... } is uniformly distributed mod..7. Metric Theory. Theorem.8. Let {an} be a sequence of distinct integers: an am if n m. Then for almost all α R, the sequence {αan} is uniformly distributed mod. ote that the proof does not provide a single example of such an α. As an example, we may take an = n, and find that almost all α are normal to base. Likewise, given any integer b 2, almost all α are normal to base b. Intersection this union of subsets having full measure shows that in fact almost all α are normal to any case b. Proof. By Weyl s criterion, it suffices to show that for almost all α, and any integer k, the normalized Weyl sums S kα; := ekαan tend to zero. In fact it suffices to show that given k, this holds for almost all α apriori depending on k. Replacing an by kan, it suffices to take k =, and we write S α; = Sα;. We may also restrict to α [, by periodicity. From now on we take = M 2, and first show that for almost all α, S α; M 2. ow one way to show this is to show that for almost all α, the series S α; M 2 2 < M= converges, since this forces the individual terms of the series to tend to zero. n=
UIFORM DISTRIBUTIO ext, note that by what we learnt in measure theory about the Lebesgue integral, it suffices to show that S α; M 2 2 dα < M= since for a function to be integrable, a necessary condition is that it is finite almost everywhere. By Fatou s Lemma, S α; M 2 2 dα S α; M 2 2 dα M= M= so it suffices to show that the sum of the integrals converges. squaring out the Weyl sum gives S α; M 2 2 dα = M 4 = M 4 M 2 M 2 m= n= M 2 M 2 m= n= e αam an dα δ am, an But since the an s are integers. ow and only now! we use the assumption that an are distinct, to deduce that only the diagonal terms survive, giving S α; M 2 2 dα = M 4 M 2 m= = M 2 and so we deduce that S α; M 2 2 dα < so that for almost all α, M= S α, M 2. Finally, we move from M 2 to general : Given, we can find M so that M 2 < M + 2, and then we claim that S α, = S α, M 2 + O and hence will deduce that for almost all α, Sα,.
2 Indeed assuming M 2, S α, S α, M 2 = n=m 2 + eαan M M 2 eαan 2 so that S α, S α, M 2 M 2 + M M 2 2 on trivially bounding ex as claimed. 2 M 2.8. Quantitative aspects of uniform distribution: Discrepancy. If a sequence X is uniformly distributed mod, it is in particular dense in the unit interval. This means that every subinterval of fixed length contains at least one point of the sequence, in fact the proportion of points falling into it is roughly the length of the interval. The next question is: What about shrinking intervals? That is, we ask that every interval of length at least /m contains at least one element from the first elements of the sequence {{x n } : n }, where m as. Definition. The discrepancy of the sequence X is D = sup #{n : x n I} I I [, Clearly if D then X is uniformly distributed; the converse also holds. Exercise 4. Show that / D. Recall that Weyl s criterion is that X is uniformly distributed if and only if all the normalized Weyl sums S k, := ekx n n tend to zero k. What is important is that we can use Weyl sums to estimate the discrepancy D. Theorem.9 Erdős-Turan. For any K and, D K K + + 3 k S k,. k= n=
UIFORM DISTRIBUTIO 3.8.. Example: The linear case x n = αn mod. We want to use the Erdős-Turan inequality. We use the bound for the geometric series for irrational α S k, = ekαn kα n= and so by the Erdős-Turan inequality, D K + K k= so we need to estimate the sum K k= k S k, K + K k= k kα. Here the Diophantine type k kα of α plays a special role. We say that α is of bounded type, or badly approximable if there is some constant c = cα > so that qα > c q. This is a measure zero condition, and is equivalent to the continued fraction expansion of α to have bounded partial quotients: α = [a ; a, a 2,... ], with a k M. The main example are quadratic irrationalities: Lemma.. Let α = D, D > an integer which is not a perfect square. Then for any integers p, q p q D > c q 2, c = /3 D Proof. We may assume that p/q D < /, otherwise there is nothing to prove. Let fx = x 2 D which is the minimal polynomial of D. Then p q D = fp q p + D > fp/q 3 D q since < p/q + D = 2 D + p/q D < 2 D + / < 3 D. Moreover, fp/q = p2 Dq 2 q 2 q 2 because p 2 Dq 2 is an integer, which is not zero because D is irrational if D, hence is at least in absolute value. Altogether we obtain as claimed. p q D > 3 D q 2
4 ote: The argument extends to give Liouville s theorem, that for a real algebraic number of degree d, we have p/q α > c/q d with c = cα effectively computable. Roth s theorem improves that to p/q α > c/q 2+ε, for all ε >, with c = cα, ε > some constant not effective. Open Problem 2. A real algebraic number of degree d 3 is not of bounded type, equivalently the partial quotients in the continued fraction expansion are not bounded. Proposition.. Suppose that α is irrational of bounded type. Then D{αn}, log 2. ote: This can be improved to Olog /. Corollary.2. If α is of bounded type, then every interval of length α log 2 / contains an element of the form αn, n. Lemma.3. Suppose that α is irrational of bounded type. Then and At := k t GK := K k= kα k kα t log t. log K2 Proof. The key is that if α is of bounded type, then the points kα, ] are well spaced in the sense that 2 mα nα j, m n j Indeed, mα = mα M for M mα for some integer M; hence for a suitable L Z, mα nα = ±m ± nα L so that if m n j then mα nα m ± n α m + n 2j as claimed. Hence, an interval of length < /2K will contain at most one element of the form kα, k K.
UIFORM DISTRIBUTIO 5 We use this to bound the sum At by dividing the range of summation into intervals of length δ = /2t At = /δ j= k t kα jδ,j+δ] kα each interval contains at most one summand, and the summand is bounded above by /jδ, so that At /δ j= jδ δ log δ t log t. To bound GK, we use the bound on At and get rid of the factor /k by using summation by parts: Exercise 5. Suppose that α is irrational of bounded type, and set x = distx, Z. Let At := k t kα, We saw that At t log t. Show that GK := K GK log K 2. We can now bound the discrepancy: D K + K k= k S k, K + Taking K = gives D log 2 /. References K k= k= k kα k kα log K2 + K [] Champernowne, D. G. The Construction of Decimals ormal in the Scale of Ten. J. London Math. Soc. 8, 933. [2] Copeland, Arthur H.; Erdős, Paul. ote on normal numbers. Bull. Amer. Math. Soc. 52, 946. 857 86. [3] Davenport, H.; Erdős, P. ote on normal decimals. Canadian J. Math. 4, 952. 58 63.