Limiting behaviour of large Frobenius numbers. V.I. Arnold

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1 Limiting behaviour of large Frobenius numbers by J. Bourgain and Ya. G. Sinai 2 Dedicated to V.I. Arnold on the occasion of his 70 th birthday School of Mathematics, Institute for Advanced Study, Princeton, ew Jersey, U.S.A. 2 Mathematics Department, Princeton University, Princeton, ew Jersey, U.S.A.

2 . Introduction Consider n-tuples a = a, a 2,..., ) of positive integers which are co-prime, i.e., the largest common divisor lcd) of all a j is. Frobenius number F a) of a is the smallest F such that any integer t F can be written in the form t = n x j a j j= with non-negative integers x j. V.I. Arnold in [] introduced the ensembles of a for which a + a = σ tend to infinity and studied the behaviour of F a) under the limit transition σ. In particular, he formulated the hypothesis according to which F a) grows for typical a as σ + n. Other hypotheses and results of Arnold can be found in [2]. In this paper we consider different ensembles of large a and study the same question of the growth of F a) in these ensembles. amely, take and denote by Ω the set of all a for which a j, j =, 2,..., n, and lcd a) =. Using elementary probability methods one can show that the limit lim Ω n exists and is positive see, for example, [3]). It gives the probability of a Ω in the ensemble of all possible n-tuples a with entries less than. Below P denotes the uniform probability distribution on Ω. We study in this paper the behaviour of F a) for typical a in the sense of P ) as. he case n = 2 follows easily from the famous result of Sylvester according to which F a) = a )a 2 ) see [5]). his implies that F a) has the limiting distribution as. 2 Below in Sections 2 and 3 we consider the next case n = 3. Based on some facts from the theory of continued fractions we prove heorem. As there exists the limiting distribution of 3/2 F a). he proofs of the needed facts will be a subject of another paper by C. Ulcigrai and one of us Ya. Sinai). It is hopeless to write down explicitly the limiting distribution in heorem. Probably the methods of this paper can be used for estimating its decay at infinity. Below we introduce another function F a) about which we prove in Lemma that in typical situations it behaves as F a). But F a) is much easier for the analysis of the problem because it is formulated as a max-min problem. In Section 2 we discuss the case n = 3 when lcda i, a j ) = for at lease one pair a i, a j. In Section 3 we discuss the general case of n = 3.

3 In Section 4 we consider n > 3. heorem 2 of this section shows that for slightly modified ensembles the distributions of are uniformly in ) bounded in the sense that F a) + n P,α { F a) + n } D ɛd) where Ω,α is the ensemble of a for which a j α,, j n and P,α is the uniform distribution in this ensemble, 0 < α < is the fixed number, ɛd) does not depend on and ɛd) 0 ad D. In Appendix we prove a general Lemma which was used in an earlier version of this paper and can have different applications. amely, we show that l { α : M m= e 2πimα DM ln M } ɛ D) where l is the Lebesgue measure on [0, ] and ɛ D) does not depend on M, ɛ D) 0 as D. he analysis of the behaviour of M M m= e 2πimα importance but we do not go further in this direction. as function of M for typical α is of some ow we give the defintion of the function F a) and prove Lemma which shows in what sense F a) and F a) are equivalent. For any a Ω introduce the arithmetic progression Π r = {r + m, m 0}, 0 r <. Consider the equality x a + x 2 a x n = r + mx,..., x n ) which shows that x a + x 2 a x n r mod ). ) Here x j 0 are integers. Put m r = min 0 x,x 2,...,x n < mx, x 2,..., x n ) and denote F a) = max r r + m r ). Lemma. F a) F a) F a) provided that F a) > 0. Proof. ake t F a). hen t Π r for some r, 0 r <, i.e., t = r + m. herefore t = r + m r + m m r ) = x a + + x n + m m r ) 2

4 for some 0 x j <, j =,..., n. Since t F a) we have x a + +x n F a) and m m r. his gives the needed representation of t and the inequality F a) F a). o prove the inequality from the other side, take r such that F a) = r + m r = max r r + m r ). We shall show that t = r + m r ) cannot be represented in the form t = y a + + y n with non-negative y j, j n. Indeed, if such representation is possible we would have y a + + y n = r + m for some m m r and by definition of t t = y a + + y n + y n = r + m + y n = r + m r ). herefore m + y n = m r proved.. Since m m r this is possible only if y n < 0. Lemma is Certainly, instead of we could take any other a j. In a typical situation we expect that x grow as n. herefore typically F a) F a). he financial support from SF, Grant o given to the second author is highly appreciated. 2. he case n = 3 and lcda i, a j ) = for some a i, a j Without any loss of generality we may assume that i =, j = 3. In this case the maxmin problem for F a) can be solved more or less explicitly. Write for positive integers x, x 2 or x a + x 2 a 2 = r + mx, x 2 )a 3 2) x a + x 2 a 2 r mod a 3 ) 3) 3

5 where 0 r < a 3. Since a, a 3 are co-prime there exists a, a < a 3 for which a a mod a 3 ). It follows easily from the estimates of Kloosterman sums that for any fixed 0 < α < α 2 < and α a α 2 the inverse a is asymptotically uniformly distributed on [,..., a 3 ]. Presumably this is also true in our ensemble. Rewrite 3) as follows: where r ra mod a 3 ), a 2 = a a 2 mod a 3 ) and x + a 2 x 2 r mod a 3 ) 4) a 2 x 2 r x ) mod a 3 ) 5) he equation 5) has atural geometric interpretation. Consider S = [0,,..., a 3 ] as a discrete circle. he shift R by a 2 mod a 3 ) is the rotation of the circle S and {a 2 x 2 } is the orbit under the action of R of the point zero. hen 5) means that r x belongs to this orbit. From Lemma F a) = max r min x a +x 2 a 2 r mod a 3 ) 0 x, x 2 < a 3 x a + x 2 a 2 ) = = 3/2 max r = 3/2 max r min x a +x 2 a 2 r mod a 3 ) min x +x 2 a 2 r mod a 3 ) x a + x 2 a 2 x a + x 2 a 2 ) ). 6) First we localize our ensemble. Choose α = α, α 2, α 3 ), 0 < α j < for j =, 2, 3 and ɛ > 0 and define Ω,α,ɛ Ω as a subset of those a = a, a 2, a 3 ) for which a j α j ɛ. hen P,α,ɛ is the notation for the uniform probability distribution on Ω,α,ɛ. heorem will follow if we prove heorem wrt to the distribution P,α,ɛ see also the end of this section). In the ensemble Ω,α,ɛ the ratios a, a 2 are ɛ-close to α, α 2. We shall use some facts from the theory of continued fractions and from the theory of rotations of the circle see [5]). ake ρ = a 2 a 3 and expand it into continued fraction: 4

6 ρ = h + h 2 + h h s0 7) where h j are integers. Let ρ s = h + h = p s q s + h s be the s-approximant of ρ. One can find odd intervals 2p ) = {0,,..., m 2p } and even intervals 2p {a 3 m 2p,..., a 3 }, p, such that if 2p ) j = R j 2p ), 2p) j = R j 2p) then the intervals 2p ) j, 0 j < q 2p and 2p) j, 0 j < q 2p are pair-wise disjoint and their union gives the whole circle S. his means that 2p ) j, 2p) j, constitute some partition of S which we denote by η p). he partitions η p) increase, η p+) η p). heir exact structure depends on the elements of the continued fraction 7). We shall show that in 6) it is enough to consider x D, x2 D where D is sufficiently large depending on ρ see below). ake s such that q s < q s. If x 2 > D choose k so that qs +k x 2 < q s +k +. Clearly, k increases of D increases. Put x = x + a 2 q s +k p s +k a 3 ) = x + a 3 ρ q s +k p s +k ), x 2 = x 2 q s +k. It is easy to see that and x + a 2 x 2 x + a 2 x 2 mod a 3 ) x a + x 2 a2 = x a + x 2 a2 + + a 3ρ q s + k p s + k ) a q s +k a2. 8) 5

7 he expression a 3ρ q s + k p s + k ) a = a 3 ρ q s + k p s + k ) a decreases as k takes values increases because ρ q s +k p s +k ) behaves as q s. On the other hand, q s +k +k O) and increases as k increases. herefore, the sum of the last two terms in 8) becomes negative if k is large enough. Since in 6) we are interested in the minimal values of x 2, x 2 = x 2 q s +k give smaller values for the expression 6). hus it is enough to consider x 2 D for sufficiently large D depending on ρ. Let us show that x 2 > D2 for another sufficiently large D2. Indeed, take k 2 so that q s k 2 < D2 qs k 2 + and consider the partition η s k 2 ). ake any element = [y, y 2 ] of this partition and for r = y 2 the value of x are y 2 y ), y 2 y )+l, y 2 y ) + l + l 2,... where l, l 2,... are the lengths of the elements of η s k 2 ) which follow. If x 2 D2 x qs k 2 + then it is clear that min of a + x 2 a 2 is attained at x = y 2 y. On the other hand, for r consider an element of the partition η s ) containing r. ake x =. It is clear that x 2 q s + and x a + x 2 a 2 is much smaller than in the previous case. hus D2 x2 D. In the above mentioned paper by C. Ulcigrai and the second author [6], in preparation) the following problem was considered. ake large R and some fixed number k. For any irrational ρ consider q s such that q s R < q s and h s k,..., h s,..., h s+k. In [7] it is proven that with respect to the Gauss density there exists the joint limiting distribution of ln 2+ρ) q s /R, q s /R, h s k,..., h s,..., h s+k. Presumably, the same limiting distribution appears for any probability distribution P,α,ɛ but we do not consider this question in more detail. For any s and k, consider the elements, of η s k) which contain 0 for the point 0 is the right end-point while for it is the left end-point). he partitions η s k+), η s k+2),..., η s+k) generate a finite partition of which we denote by ν. he structure of this partition is determined by h s k, h s k+,..., h s+k. Denote by the finite set of endpoints of elements of ν. ake b = a 3 a 2 and = a 2 mod a 3 ). Denote f k = max y min x +a 2 x 2 y mod a 3 ) a 2 x 2 x a + x ) 2 a 2. he same arguments as before show with P,α,ɛ -probability tending to as k the solution of the main max-min problem for F is given by f k. In this sense it is a function of q s R, qs R and h s k,..., h s+k and has a limiting distribution as. 6

8 3. he case n = 3 and arbitrary b ij = lcda i, a j ) Since all a j have no common divisors, b 3 and b 23 are co-prime. Again for given r, 0 r < a 3, we consider the equation x a + x 2 a 2 = r + mx, x 2 )a 3. We write a = b 3 a, a 2 = b 23 a 2, a 3 = b 3 b 23 a 3. Clearly a and a 3, a 2 and a 3 are co-prime. Let First we consider the equation r = r b 3 b 23 + r, 0 r < b 3 b 23, x = b 23 x + x, 0 x < b 23, x 2 = b 3 x 2 + x 2, 0 x 2 < b 3, a = b 23 a + a, 0 a < b 23, a 2 = b 3 a 2 + a 2, 0 a 2 < b 3. x b 3 a + x 2b 23 a 2 r mod b 3 b 23 ). 9) We can find unique solution which we denote by x, x 2 such that x a + x 2a 2 = r + tb 3 b 23 where t can take values 0 or. After that we consider the equation which remains after dividing both sides of 8) by b 3 b 23 : x a + a 2a 2 = r x a x 2a 2 t + ma 3. 0) Denote r = r x a x 2a 2 + t. Clearly x a x 2a 2 + t can take finitely many values depending only on {b ij }. It is easy to see that the limits of probabilities of these values exist as. 7

9 he equation 9) is similar to 6) because a and a 3 are co-prime. We can write x + x 2a 2a ) = r a ) + m a 3 and use the same arguments as in Section 2. In particular, we consider the expansion of a 2a ) into continued fraction, take s for which q s, q s < and find the value of s for which s) + qs number gives the limiting distribution of 3/2 F a). takes its minimum. he limiting distribution of the last 4. he case n > 3 For n > 3 our result is weaker. Again we consider the equation or x a + x 2 a x n = r + m ) x a + x 2 a 2 + x n r mod ). 2) he left-hand side is the orbit of the abelian group generated by n ) commuting rotations R j where R j is the shift mod of S = {0,,..., } by a j, j n. We shall prove the following theorem. heorem 2. Consider the ensemble Ω,α Ω such that α < a j, j n, where 0 < α < is a fixed number. ake the set D Ω,α of a, a 2,..., ) Ω,α such that for any r S the equation 2) has a solution with 0 x j D n. hen P,α ) ɛd) D where ɛd) 0 as D. Here P,α is the uniform probability distribution on Ω,α. Proof. It is easy to see that a n m=0 { exp 2πim r } { exp 2πi n j= ma j x j } = if 2) holds 0 if 2) fails 8

10 ake M > 0 and consider the weight on Z cx) = x M M, 0 x 2M 0 otherwise. o show that 2) has a solution for any r S, it will suffice to show that for any r Z a r) = x,x 2,...,x n Z cx ) cx 2 ) cx n ) a n exp { 2πim n j= a j x j } m=0 { exp 2πimr } 0. 3) We write Z a r) = a n m=0 { exp 2πi mr } n { cx) exp 2πi ma } j x. a j= n x Z It is easy to check that x Z cx) e 2πiθx = e 2πiMθ 2M + { } 2 sin π2m + )θ 4) 2 sin πθ for any θ. Separating in 3) the contribution of m = 0 and m 0 we can write Z a r) = M n + Z a ) r) = M n + a n m= mr 2πi e an n j= x Z cx) e 2πi ) ma j an x. We shall consider M = A n. herefore M n view of 4) for Z a ) r) we have the estimate = An. In our case α an. In Z ) a a n m= n j= sin π2m + )θj 2 sin πθ j ) 2 2M + 5) 9

11 where θ j = ma j. his estimate does not depend on r. herefore, if we show that the expectation of the rhs of 5) is bounded by some constant then the Chebyshev inequality gives the statement of the theorem. It is easy to check that ) 2 sin π2m + )θj 2 sin πθ j sin πmaj C )) 2 + M 2 for some absolute constant C. herefore Z ) a C 2 a n m= n j= M ) 2 sin πm a j + M 2 = C 2 a n m= n j= M ma. 6) j 2 + M for another absolute constant C 2. In the last formula is the distance till the nearest integer number. Let us average the rhs of 6) wrt P,α, i.e., consider Z 2) = n+ α a n m= α a,..., lcda,...,an) = n j= M ma. 7) j 2 + M Assume that m as = b/q for some b q and b, q) =. hen the rhs of 7) can be written n q= q b q b,q)= α a,...,an lcda,...,,q) = n j= M b a. 8) q j 2 + M If lcda,...,, q) = then certainly a k is not a multiple of q for some k =,..., n. Hence 0

12 α a,..., n j= M b q a j 2 + M n k= α a,..., a k / qz n j= M b q a j 2 + M = = n ) α a a qz M ba q 2 + M α a M b q a 2 + M ) n 2 9) ow we shall estimate both sums in 9). Lemma 2. i) If q > M then α a M b q a 2 + M ; ii) If q M then α a M b q a 2 + M M q ; iii) If q M then α a a / qz M ba q 2 + M q M. Proof. Partition [α, ] onto approximately α) intervals I q γ of the length q. Denote by π q the quotient map Z Z/qZ. hen for each I γ we have {π q ba) a I γ } = Z q and herefore the sum {π q ba) a I γ, π q a) 0} = Z q {0}. α a M b q a 2 + M

13 behaves as q q z=0 q ) M z q 2 + M and for another absolute constant C 3 if q M and q z=0 M z q 2 + M C 3q if q < M. q z=0 M z q 2 + M C 3M his gives the statements i) and ii) of the lemma. For q < M the sum q z= M z q 2 + M q2 M. Lemma is proven. Return back to 9). For q M Lemma 2 shows that it is not more than n while ) n 2 ) n 3 for q < M it is not more than q M n M q = M q n. Substituting these estimates into 7) we get assuming n > 3 Z 2) n+ M q q b q b,q) = n + + n+ q M q b q b,q) = 2 ) n 3 M n < q

14 M n 3 < C 4 + C 4 q M q n 3 < C 4 + M ) n 2 for another constant C 4. hus Z 2) is bounded. his implies the statement of the theorem. heorem 2 shows that in any ensemble Ω,α the family of probability distributions of F a)/ + n is weakly compact. However, it does not imply the existence of the limiting distribution of F a)/ + n but gives only the existence of limiting points. References [] V.I. Arnold, Weak asymptotics for the number of solutions of Diophantine problems, Funct. Anal. Appl., 33:4 999), [2] V.I. Arnold, Arithmetical turbulence of self-similar fluctuations statistics of large Frobenius numbers of additive semi-groups of integers, Preprint IC/2006/037, ICP, Miramare, Funct. Anal., Appl., 24:3 990), -8. [3] L.B. Koralov and Ya. G. Sinai, Probability heory and Random Processes, Springer- Verlag, in press). [4] J.J. Sylvester, Mathematical questions with their solutions, Educational imes, 4, 884), 2. [5] Ya. G. Sinai, opics in Ergodic heory, Princeton University Press, 994. [6] Ya. G. Sinai and C. Ulcigrai, Renewal-type Limit heorem for the Gauss Map and Continued Fractions, in preparation). 3

15 Appendix. Below we prove some estimate which was not used in the previous proofs but is of some independent interest. A similar statement was proven by A. Kochergin private communication). Lemma Let for 0 < α < S α) = t= exp{2πitα} and A D) = {α : S α) D ln }. hen la D)) ɛ D) where ɛ D) 0 as D, where l is the Lebesgue measure. Proof. ake { two positive numbers } C, C 2, < C < 2C < C 2, introduce the intervals k) = α : α Ck+, k = 0,,..., K. Without any loss of generality we may C k assume that C K+ =. Clearly, K ln ln C. Consider B,k C, C 2 ) = {α : ν,k α) C 2 C k } where ν M,k α) is the number of m, m M, such that {mα} M k), and B C, C 2 ) = K k=0 B,k C, C 2 ). hen for α B C, C 2 ) S α) = t= exp{2πitα} K k=0 2πC k ν,k α) C 2 2π K + ) C 2 ln 2π ln C. 4

16 his is the needed inequality with D = C 2 2π ln C. hus we have to estimate the measure of the complement of B C, C 2 ). Clearly l B C, C 2 )) K l B,k C, C 2 )) k=0 where B is the complement to B. Let χ k α) be the indicator of B,k C, C 2 ). hen ν,k C, C 2 ) = χ k tα) t= and by Chebyshev inequality = l l{α : ν,k α) C 2 C k } = l { α : = t= { α : } χ k tα) C 2 C k t= ) χ k tα) Ck C ) [ ) ] 2 E χ k tα) Ck C ) t= C 2 C + ) 2 C 2k E j= j)eχ k α)χ k jα) C 2 C + ) 2 C 2k C 2 C + )C k ) C k C ) ) he expectation is taken with respect to the Lebesgue measure. We shall estimate the last sum. It will be done separately in four steps.. } Step : j < C. Here Eχ k α)χ kjα) Eχ k α) = Ck C ) and j<c M j) since C 2 > 2C. [ Eχ k α)χ k jα) C 2 C + ) 2 C 2k C k C ) ) 2 ] Ck C ))C C 2k C 2 C + ) 2 C k+ 5

17 Step 2: C j < C k+. In this case Eχ k α)χ k jα) = 0 and therefore there is nothing to estimate. Indeed, χ k α) is the indicator of the set k). [ he function χ k jα)] is C the indicator of the arithmetic progression of the intervals k j + s, C k+ j j + s, j s 0. From our condition on j it follows that k) can intersect only with the interval for which t = 0 this intersection is empty. Step 3: C k+ j 3 C k C ). he number 3 does not play any essential role and can be replaced by any bigger number. Here k) intersects with not more than three intervals from the above mentioned arithmetic progression. herefore Eχ k α)χ k jα) 3Ck and j [ j) Eχ k α)χ k jα) j C 2 C + ) 2 C 2k C k C ) ) 2 ] 3C k C 2 C + ) 2 C 2k C k+ 3 j C kc ) j 3 C k+ 2 C k C 2 C + ) 2 C k C ) = C k C 2 his is the estimate which we need. Step 4: j 3. In this case Eχ C kc ) kα)χ k jα) is close to Eχ k α)) 2 C k = C ) Indeed, we can increase k) by adding an interval near each end-point so that the [ ] new set k) will consist of an integer number of intervals s, s+. herefore j j Eχ k α)χ k jα) l k)) Ck C ) = l 2 k)) + 2Ck C ) j l k)) + 2 ) C k C ) j ) 2. and 3 C k C ) j [ j) Eχ k α)χ k jα) 6 C 2 C + ) 2 C 2k C k C ) ) 2 ]

18 3 C k C ) j j)2c k C ) jc 2 C + ) 2 C 2k 2C k C ) C 2 C + ) 2 C k ln Ck C ) 3 = 4C )k ln C C 2 C 2 + ) 2 C k. ow we can finish the proof of the Lemma. From all our estimates it follows that l { α : ν,k α) C 2 C k } 4 ln C C k C k and therefore his implies the statement of the lemma. k l{α : ν,k α) C 2 C k } const ln C C. March 20, 2007:gpp 7