ASYMPTOTICS OF THE EULERIAN NUMBERS REVISITED: A LARGE DEVIATION ANALYSIS

Transcription

1 ASYMPTOTICS OF THE EULERIAN NUMBERS REVISITED: A LARGE DEVIATION ANALYSIS GUY LOUCHARD ABSTRACT Using the Saddle point method and multiseries expansions we obtain from the generating function of the Eulerian numbers A nk and Cauchy s integral formula asymptotic results in non-central region In the region k = n n α > α > /2 we analyze the dependence of A nk on α This paper fits within the framework of Analytic Combinatorics INTRODUCTION The Eulerian numbers A nk have been the object of renewed interest recently see for instance Janson [0] They are defined by the recurrence A n+k = n k + 2A nk + k A nk 0 k n The initial conditions vary in the litterature We choose here A 00 = 0 A 0 = This is used for instance in Bender [2] and Flajolet and Sedgewick [6] chv I I I The Eulerian numbers correspond for instance to runs in permutations Let us mention OEIS [] A which stores these numbers and gives many references to the litterature The bivariate generating function exponential in z ordinary in w is given by We have g z w := n=0 k=0 k=0 A nk A nk z n w k = = w w e w z w hence we can define a random variable RV J n this corresponds to the number of runs in a random permutation such that P[J n = k] = A nk From Flajolet and Sedgewick [6] chi X we know that the roots of the denominator are with h j w = f w + 2ij π w j Z f w = w lnw As w f w is close to whereas the other poles h j w with j 0 escape to infinity This fact is consistent with the limit form g z = z which has only one simple pole at If one restricts w to w 2 there is clearly at most one root of the denominator in z 2 given by f w Thus we have for w close enough to g z w = f w + Rz w z Date: June Mathematics Subject Classification 05A6 60C05 60F05 Key words and phrases Eulerian numbers Asymptotics Saddle point method Multiseries expansions Analytic Combinatorics

2 2 GUY LOUCHARD with Rz w analytic in z 2 and [z n ]g z w = f w n+ + O 2 n Notice that f w does not correspond to a discrete RV but if we set w = e t f e t n+ corresponds to a sum of n + independent RV uniformly distributed on [0] see for instance Tanny [4] In the rest of this paper we set m := n + µ := m The corresponding mean and variance are given by M = m 2 σ 2 = m 2 The generating function of the mean and second moment are given respectively by 2 z 2 z z 3 They are obtained either from M and σ 2 or more simply by setting w = e t in f w z and differentiating wrt t Note that the exact mean and second moment generating functions are derived from g z w as 2 2z + z 2 2 z 2 6 2z + 5z 2 7z 3 + z 4 6 z 3 Of course asymptotically by classical singularity analysis exact and asymptotic moments are the same In the central region k = M + xσ x = O J n is asymptotically normal This has first been proved by David an Barton [4] Without being exhaustive a very complete bibliography can be found in Janson [0] let us also mention Bender [2] Carlitz et al [3] Tanny [4] The first two terms of a correction were given in Sira zdinov [5] and Nicolas [3] A complete analysis is given in Gawronski and Neuschel [7]: There exists polynomials q ν ν such that for any l 0 as n uniformly for all k Z A nk = 6 /2 πm e x2 + q ν x = 2 ν H 2ν+2s x6 s x = k m 2 2 m l q ν x ν= ν j = m ν k j! + O m l 3/2 B 2j +2 j + 2j + 2! summing over all non-negative integers k k ν with k + 2k νk ν = ν and letting s = k + k k ν B k are the Bernoulli numbers and the Hermite polynomials satisfy H j x = j e x2 /2 d j d x j e x2 /2 A very simple proof is given in Janson [0] As far as the large deviation is concerned let us mention Bender [2] Hwang [9] Esseen [5] improves Bender s result as follows: k j

3 ASYMPTOTICS OF THE EULERIAN NUMBERS REVISITED: A LARGE DEVIATION ANALYSIS 3 Let a := k/m uniformly in all 0 < k < m Let ta be the solution of Set a = e ta e ta ta δa = e ta tae ata σ 2 a = ta 2 e ta e ta 2 Then A nk = δa m 2πmσa + O m As noted by Esseen further terms can be obtained in this asymptotic All these papers use the solution ρ of 2 mw f w k f w = 0 which actually corresponds to the Saddle point of the Saddle point method see Sec2 In this paper we are interested in the extreme large deviation case k = m m α /2 < α < the choice of this range is justified in Sec3 This range was already the object of our analysis of Stirling numbers of first and second kind see Louchard [] [2] Let us summarize the motivation of this paper: Previous papers simply use ρ as the solution of 2 They don t compute the detailed dependence of ρ on α neither the precise behaviour of functions of ρ they use We will use multiseries expansions: multiseries are in effect power series in which the powers may be non-integral but must tend to infinity and the variables are elements of a scale The scale is a set of variables of increasing order The series is computed in terms of the variable of maximum order the coefficients of which are given in terms of the next-tomaximum order etc This is more precise than mixing different terms Our work fits within the framework of Analytic Combinatorics In Sec2 we revisit the asymptotic expansion in the central region and in Sec3 we analyze the non-central region k = m m α α > /2 α is chosen such that m α is integer We use Cauchy s integral formula and the Saddle point method Sec4 provides a justification of the Saddle point technique we use here 2 CENTRAL REGION In this section as a warm-up we rederive the first terms of the asymptotics We use the Saddle point technique for a good introduction to this method see Flajolet and Sedgewick [6]

4 4 GUY LOUCHARD chv I I I Let ρ be the Saddle point and Ω the circle ρe iθ By Cauchy s theorem A nk f z m 2πi Ω z k+ d z = π ρ k f ρe iθ m e kiθ dθ 2π π = π ρ k e m lnf ρeiθ kiθ dθ 2π π = f ρ m π [ { ρ k exp m 2π π 2 κ 2θ 2 i }] 3 6 κ 3θ 3 + dθ i 4 κ i ρ = lnf ρe u u u=0 See Good [8] for a neat description of this technique Let us now turn to the Saddle point computation We let ρ denote the root of smallest modulus of 2 with k = M + xσ This amounts to [ w lnw w lnw 2 w w ] 2lnw The solution is given by the asymptotic expansion µ µ 2 x3/2 w µ = 0 6lnw ρ := + 2x3/2 + 6x2 µ µ x3 3 /2 5µ x4 5µ 4 + O µ 5 In the sequel we will only give the first terms of our expansions of course Maple knows much more This leads to T = k lnρ = x3 /2 µ x 2 x3 3 /2 5µ x4 5µ 2 + O T 2 = µ 2 lnf ρ = x3 /2 µ + x2 2 + x3 3 /2 5µ + 3x4 20µ 2 + O T + T 2 = x2 2 x4 20µ 2 x6 T 3 = exp T + T 2 + x2 2 µ 3 µ 3 050µ 4 + O µ 6 = x4 20µ 2 + /050x6 + x 8 /800 µ 4 + O Now we turn to the integral in 3 The first κ i are given by κ[2] = 2 x2 20µ 2 + 2x4 525µ 4 + 2x6 κ[3] = x3/2 60µ + 79x3 3 /2 6300µ 3 + O κ[4] = 20 + x2 42µ 2 + O 2625µ 6 + O µ 8 and similar expressions for the next κ i that we don t detail here We proceed as in Flajolet and Sedgewick [6] chv I I I Let us choose a splitting value θ 0 such that mκ 2 θ0 2 mκ 3θ0 3 0n For instance we can use θ 0 = µ /2 We must prove that the integral K mk = 2π θ0 here and in the sequel an b n means a n b n n θ 0 µ 4 µ 5 e m lnf ρeiθ kiθ dθ µ 6

5 ASYMPTOTICS OF THE EULERIAN NUMBERS REVISITED: A LARGE DEVIATION ANALYSIS 5 is such that K mk is exponentially small This is done in the Appendix Now we use the classical trick of setting [ ] m κ 2 θ 2 /2! + κ l iθ l /l! = u 2 /2 Computing θ as a series in u this gives by Lagrange s inversion 6 θ = u i a[i ]/µ = 3 [u 2 /2 + 3x2 2ix 5µ 2 + O µ 4 + u 2 5µ ix3 525µ 4 + O This expansion is valid in the dominant integration domain T 4 := 3/2 2 /2 π /2 µ l=3 u µ a θ 0 = µ /2 µ 6 ]/ + O u 3 µ Setting dθ = dθ du du we integrate on u = [ ]: this extension of the range is justified as in Flajolet and Sedgewick [6] chv I I I This gives [ / + µ 2 + Now it remains to compute 3x T 5 := T 3 T 4 = 3/2 2 /2 57x x / ] µ 4 + O 20 µ 6 [ + x x π /2 µ x x x x / µ 2 / µ 4 + O Note that the coefficient of the exponential term is asymptotically equivalent to the dominant term of as expected The first three terms correspond to Note that our derivation is simpler 2πσ then Nicolas computation in [3] 3 LARGE DEVIATION k = m m α > α > /2 m α INTEGER m We have k = m m α m α integer We set 2 ε := m α ε = m α µ exp/ε µ = m The multiseries scale is here {m α µexp/ε} Set τ = exp /ε Our result can be summarized in the following local limit theorem: Theorem 3 With /2 < α < the rate of growth of A nk / is the following: µ 6 ] 5 6 A nk = e m ε m T 5 + T 6 τ + T 7 τ 2 + O τ 3 with T 5 = 2π εµ 2εµ 3 T 6 = µ 2 ε + 2ε 2 + T 7 = µ4 2 + ε 2 + ε 2 / µ 2 + O 8ε ε 3 ε 2 µ 2 + ε 4 0 3ε ε 2 ε + O µ 4 µ 2 2 here and in the sequel an b n means a n = ob n

6 6 GUY LOUCHARD Proof We have The Saddle point equation 2 becomes k = m m α = m ε 7 w + + lnw + lnwεw lnwε = 0 To first order we have lnw /ε So we set we now have to the next order ρ = e ξ ξ = + η ε ερ η + ρ + ρ ε = 0 or hence This gives ηερ + ηε + ρ = 0 η ρ ε ε ρ = e ξ = e /ε η/ε τ η ε hence η τ η ε τ ε ε + τ ε 2 = τ ε τ2 ε 3 We derive by bootstrapping from 7 again we only provide a few terms here we use more terms in our expansions η = ε τ + ε 3 + ε 2 τ ε 2ε ε 4 4 ε ε 2 τ 3 + O τ 4 ε and successively ρ = expη/ε ie τ ρ = τ ε 2 + 2ε 4 + ε 3 ε 2 τ + 2 2ε ε 5 3 ε ε 3 ε 2 τ 2 + O τ 3 lnρ = + η ie ε lnρ = ε ε 2 τ + ε 4 + ε 3 τ 2 ε 2 + f ρ = τ + ε ε + 2ε 3 + ε 2 ε lnf ρ = lnε + ε + ε 2 + ε τ + O τ 2 3 2ε ε 5 4 ε ε 3 ε 2 τ + O τ 2 τ 3 + O τ 4 For the first part of the Cauchy s integral we have T = lnf ρ εlnρ = lnε + τ + now we extract the dominant part lnε + T 2 = expµ 2 T lnε + = µ 2 τ + 2ε 2 τ 2 + O τ ε2 2ε 2 µ2 + 2 µ4 τ 2 + O τ 3

7 ASYMPTOTICS OF THE EULERIAN NUMBERS REVISITED: A LARGE DEVIATION ANALYSIS 7 Also κ[2] = ε ετ + O τ 2 κ[3] = 2ε 3 + 6ε 2 τ + O τ 2 Again we must choose a splitting value θ 0 = m β β < 0 such that mκ 2 θ0 2 mκ 3θ0 3 0n This leads to β > 2 αβ < 2 3 α that is why we restrict the range to /2 < α < We can then use θ 0 = m /2 α We must prove that the integral K mk = 2π θ0 θ 0 e m lnf ρeiθ kiθ dθ is such that K mk is exponentially small This is done in the Appendix Proceeding further we derive θ = u i a[i ]/µ = u εµ + iu2 3εµ 2 u3 36εµ 3 + iu4 270εµ 4 + O τ This expansion is valid in the dominant integration domain u θ 0µ a = m Setting dθ = dθ du du we integrate on u = [ ] This gives successively [ T 3 = 2π εµ 2ε 2εµ ε ε 5 µ T 4 = T 3 T 2 = T 5 + T 6 τ + T 7 τ 2 + O τ 3 with T 5 = 2π εµ 2εµ 3 T 6 = µ 2 ε + 2ε 2 + T 7 = µ4 2 + ε 2 + ε 2 / µ 2 + O 8ε ε 3 ε 2 µ 2 + ε 4 0 3ε ε 2 ε + O ε 4 24ε 3 2ε 2 µ 3 µ 4 µ 2 τ ] + O τ 2 This concludes the proof Given some desired precision how many terms must we use in our expansions? It depends on α For instance in T 7 we encounter terms like ε 2 µ 2 and ε 4 So we must compare α + with 4 α The critical value is α = 2/3 To check the quality of our asymptotic we have chosen m [90500] and α = 2/3 Figure shows lna nk circle and the ln of expression 5 line Figure 2 shows the quotient of lna nk by the ln of expression 5 without the τ term in 5 line and with this term circle Of course the good influence of the τ term is less effective for large m Another way is to fix m to 00 for instance n = 000 The maximum value for k is m m /2 = 969 We must set k larger than the central domain for instance larger than m m/2 = 58 But notice that the term T 6 starts with two negative terms µ 2 so k must be large so that τ is ε small enough to compensate these negative terms It appears that k = 860 is large enough in our case So our α range is [/207] Figure 3 shows lna nk circle and the ln of expression 5 line

8 8 GUY LOUCHARD FIGURE α = 2/3 lna nk circle and the ln of expression 5 line as function of m FIGURE 2 α = 2/3 quotient of lna nk by the ln of expression 5 as function of m without the τ term in 5 line and with this term circle Figure 4 shows the quotient of lna nk by the ln of expression 5 without the τ term in 5 line and with this term circle

9 ASYMPTOTICS OF THE EULERIAN NUMBERS REVISITED: A LARGE DEVIATION ANALYSIS FIGURE 3 n = 000 lna nk circle and the ln of expression 5 line as function of k FIGURE 4 n = 000 quotient of lna nk by the ln of expression 5 as function of k without the τ term in 5 line and with this term circle 4 APPENDIX JUSTIFICATION OF THE INTEGRATION PROCEDURE 4 The central region We must analyze Rm lnf ρe iθ kiθ

10 0 GUY LOUCHARD Let us first notice that kiθ does not contribute to the analysis Next we have ρ = + 2x3/2 + O µ µ 2 Hence this leads to analyze ρe iθ R ln ln ρe iθ ρe iθ = ln ln ρe iθ = 2 ln 2 cosθ θ 2 The first term has a dominant peak at 0 42 The non-central region Now we must analyze Set δ = τ = e /ε Hence lnδ = ε We have ρe iθ R ln ln ρe iθ = 2 ln + x3/2 µ + O Rm lnf ρe iθ kiθ ρ = δ lnδ 2 + O δ µ 2 = 2 2ρ cosθ + ρ 2 ln lnρ 2 + θ 2 lnδ 2 + θ 2 + lnδ + O δ = lnδ lnlnδ θ2 2lnδ 2 + O = ε + lnε ε2 θ O ε4 The θ contribution has a dominant peak at 0 Acknowledgments lnδ 4 We thank MD Ward for a first reading of this paper preliminary version The insightful comments of the referees are gratefully acknowledged REFERENCES [] The On-line Encyclopedia of Integer sequences OEISorg [2] EA Bender Central and local limit theorems applied to asymptotic enumeration Journal of Combinatorial Theory Series A 5:9 973 [3] L Carlitz DC Kurtz R Scoville and OP Stackelberg Asymptotic properties of Eulerian numbers Zeitschrift für Wahrscheinlichkeitstheorie und verwandte Gebiete 23: [4] FN David and DE Barton Combinatorial Chance Charles Griffin and co London 962 [5] C G Esseen On the application of the theory of probability to two combinatorial problems invoving permutations In Proceedings of the Seventh Conference on Probability Theory Braşov pages [6] P Flajolet and R Sedgewick Analytic Combinatorics Cambridge University Press 2009 [7] W Gawronski and T Neuschel Euler-frobenius numbers Integral Transforms and Special Functions 240: [8] I J Good Saddle-point methods for the multinomial distribution Annals of Mathematical Statistics 284: [9] HK Hwang Large deviations for combinatorial distributions I: Central limit theorems Advances in Applied Probability 6: [0] S Janson Euler-Frobenius numbers and rounding Online Journal of Analytic Combinatorics [] G Louchard Asymptotics of the stirling numbers of the first kind revisited: A saddle point approach Discrete Mathematics and Theoretical Computer Science 22:

11 ASYMPTOTICS OF THE EULERIAN NUMBERS REVISITED: A LARGE DEVIATION ANALYSIS [2] G Louchard Asymptotics of the stirling numbers of the second kind revisited: A saddle point approach Applicable Analysis and Discrete Mathematics 72: [3] JL Nicolas An integral representation for Eulerian numbers In Sets graphs and numbers Budapest 99 volume 60 pages [4] S Tanny A probabilistic interpretation of Eulerian numbers Duke Mathematical Journal 40: [5] SH Sira zdinov Asymptotic expression for Euler numbers in Russian Izv Akad Nauk UZSSR Ser Fiz-Mat Nauk 6: UNIVERSITÉ LIBRE DE BRUXELLES BELGIUM DÉPARTEMENT D INFORMATIQUE CP 22 BOULEVARD DU TRIOM- PHE B-050 BRUXELLES BELGIUM address: louchard@ulbacbe