On the distribution of the leading statistics for the bounded deviated permutations

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(1)國立臺灣師範大學數學系碩士班碩士論文. 指導教授：林延輯博士. On the distribution of the leading statistic for the bounded deviated permutations. 研究生：簡維良. 中華民國一 ○ 三年六月.

(2) 致謝本篇論文能夠完成，我要特別感謝我的指導教授林延輯老師。平常在討論問題時，老師總是有耐心為我解答問題，同時也會在錯誤的觀念予以指正；在寫論文遇到困難時，老師往往都能給予我極具突破性的想法，和需要的技術指導。在跟老師學習的這兩年中，我不僅僅是得到了知識，也學到了如合作研究的方法和態度，也期許自己能持續進步，將正確的數學知識和研究精神向學生延續下去。林延輯老師，謝謝您！另外，我要感謝師大數學系的游森棚老師和大同大學資訊工程學系的張薰文老師，感謝兩位老師擔任我的口試委員，並對我論文後續的修正上提供非常具有建設性的建議。也感謝游森棚老師在平日組合學課程中的指導，使我能更了解組合學的深刻內涵和豐富的思想，並有機會進一步充實和進步。我還要感謝張毓麟老師在複分析上的課程指導，加強了我對複分析領域上的認識，更熟稔各項內容和主題，這對我完成論文有很大的幫助。最後，我要謝謝在師大數學系碩士班中 M410、M407a 和 M407 的所有同學，謝謝你們這兩年的陪伴，不管是課業上，還是教學助理的工作上，都能適時給予我幫助，和諧和愉悅的研究室氣氛，更是我在精神上很大的支持力量。謹將這份論文獻給所有協助我、關心我的師長、同學和朋友！當然，還有我的家人！. 維良 2014.7.

(3) 摘要本篇論文的研究主題是要在均勻分佈的假設下，探索有界偏差排列的第一位置統計量的分佈情形，而我們猜測其將為常態分佈。定義好與第一位置統計量相關的隨機變數後，藉由考慮雙變數生成函數，我們可以計算此隨機變數的平均數和變異數，本篇論文將把這方法應用在三個特殊的情形上。因為這個雙變數生成函數的係數並沒有 closed form，在計算過程中，我們會使用 Hayman's 公式求其漸進式。最後，使用電腦計算，而這三個特殊的情形確實收斂到常態分佈，證實了我們的猜測。. 關鍵字：有界偏差排列，第一位置統計量，常態分佈，雙變數生成函數， quasi-powers 定理，Hayman's 公式.

(4) Abstract The purpose of the thesis is to investigate the distribution of the leading statistic in the bounded deviated permutations distribution in. Snl,1r .. Define the random variable.    1 2. Snl,1r , assuming the uniform. Xn. to take the value k if. 1  k  1.  n1  Snl,1r . By considering the bivariate generating function A(z,u),. we could calculate the expected value and the standard deviation for method is then applied to three specific cases, coefficients. for. n ,k. Sn1,21 , Sn1,31. and. X n . The. Sn2,2 1 . Since the. of the bivariate generating function do not have a closed form, we. will apply the Hayman’s method to get its asymptotic formula. Finally, by running computer programs, the convergence of the normal distribution on these three cases are verified.. Key word: bounded deviated permutation, leading statistic, bivariate generating function, normal distribution, quasi-powers theorem, Hayman’s method.

(5) On the distribution of the leading statistic for the bounded deviated permutations. Author Wei-Liang Chien. Advisor Dr. Yen-Chi Lin. Department of Mathematics, National Taiwan Normal University Taipei, Taiwan July, 2014.

(6) Content ………………………………………… 1. 1. Introduction. 2. Bounded deviated permutations 3. Random variable. …………………… 2. …………………………………… 4. 4. Convergence to normal distributions. ……………… 5. 5. Main results 5.1 Case 1: (ℓ,r)=(1,2). ………………………… 6. 5.2 Case 2: (ℓ,r)=(1,3). ………………………… 10. 5.3 Case 3: (ℓ,r)=(2,2). ………………………… 14. 6. Conclusion and further discussion References. ………………… 18. …………………………………………… 18.

(7) On the distribution of the leading statistics for the bounded deviated permutations Wei-Liang Chien July, 2014. 1. Introduction. The study originated from this question: In how many ways can one list the numbers 1; 2; : : : ; n such that apart from the leading element the number k can be placed only if either k 1 or k + 1 already appears?(see[1],Exercise1.7) The answer is 2n 1 : And then we’d like to ask a question: when giving a permutation = 1 2 2, the value k can be assigned to only if at least one n+1 2 Sn+1 , for i of the values from k ` to k + r has appeared among 1 ; 2 ; ; n+1 ; how many ways will be? These special permutations will be called bounded deviated permutation within `;r (`; r), denoted by Sn+1 . Calculate some cases of these permutations, we …nd that the distribution of the leading statistic has one peak. n We conjecture thatoit will converge to `;r a Gaussian distribution owing to the number 2 Sn+1 : 1 = k + 1 being a natural. `;r combinatoric object. To verify our conjucture, on the Sn+1 we de…ne a random variable `;r Xn = k if 1 = k + 1 for = 1 2 n 2 Sn+1 . In the mean time, probability function `;r : 1 =k+1gj jf 2Sn+1 is given, too. of random variable P (Xn ) = `;r jSn+1 j In the section 2, wen…rst de…ne the bounded o deviated permutation within [`; r], and `;r 2 Sn+1 : 1 = k + 1 , its exponential generating function will then by setting n;k = n o n o P n z z2 zl zr z2 be S `;r (z) = = exp z + + + + + exp z + . In the section n;k n! 2 l 2 r. `;r 3, we assume the permutation Sn+1 are uniformly distributed and de…ne the random `;r variable Xn on the Sn+1 : We’d like to …nd the mean value and standard deviation of the random variable by the generating function S `;r (z), then by general quasi-power theorem –stated in the section 4 –we con…rm that the random variable converges to a Gaussian distribution. In the section 5, we calculate some cases, including the expected value and variance of Sn1;2 , Sn1;3 , Sn2;2 , and these cases are our main results. In the process, since the coe¢ cient n;k has no closed form, we replaced the true value by its asymptotic, in which Heyman formula is the main tool we use. Di¤erent techniques are used when computing these three cases, respectively. In the end, we verify our cases by computer.. 1.

(8) 2. Bounded Deviated Permutations. At …rst, we give a de…nition of bounded deviated permutations: De…nition 1 A permutation = 1 2 n 2 Sn is bounded deviated within (`; r) if, for i 2; the value k can be assigned to i only if at least one of the values in [k `; k + r] has appeared among 1 ; 2 ; ; i 1 ; or equivalently, min f 1 ; for all i. 2;. ;. i 1g. `. max f 1 ;. i. 2;. ;. i 1g. +r. `;r 2: Denote by Sn+1 the set of (`; r)-bounded deviated permutations.. For example, 4356217 is bounded deviated within (1; 1) ; but 4365217 is not. In this section we investigate the combinatorics of these permutations. The key result is that a bounded deviated permutation within (`; r) can be decomposed into a pair of permutations, such that the …rst of which has cycle length bounded by ` `;r and the other bounded by r after standardization. From that we can enumerate Sn+1 immediately. We …rst describe the basic sequences that bounded deviated permutations, and there are be decomposed into. De…nition 2 For r; k 2 N; a plus-r word in Sn is a permutation = 1 2 n 2 Sn such that 1 r; and i max f 1 ; 2 ; ; i 1 g + r for all i 1: The set of all such words is denoted by rn : The exponential generating function of the numbers of permutations, all of whose cycles have lengths at most r is known to be Sr (z) = exp z +. z2 + 2. +. zr r. ;. hence it is also the exponential generating function for j rk j : Now we turn to the set of bounded deviated permutations. fundamental decomposition of permutations.. (1). We …rst introduce the. De…nition 3 Let = 1 2 n 2 Sn : The upward subsequence (resp. downward subsequence) of is the subsequence + (resp. ) of which consists of all numbers that are larger (resp. smaller) than 1 : Starting with a permutation 1 2 n 2 Sn ; the length of the upward (resp. downward) subsequence of is n (resp. 1). De…ne the reduced word of + to be the 1 1 word red( + ) obtained by subtracting 1 from each number of + ; whereas the reduced word of to be the word red( ) obtained by subtracting each number from 1 : The following observation is then the direct translation of the de…nition of bounded deviated permutations. 2.

(9) Proposition 4 Let `; r; n 2 N: r ` n 1: 1 1. Then. `;r 2 Sn+1 if and only if (red(. `;r The enumeration of Sn+1 is now clear: after. 1. is chosen, we assign (. ); red(. +. )) 2. 1) positions. 1. the remaining will be for + : Then chose a word in ` 1 1 for the redeced word for ; another word in rn 1 for the redeced word for + : From this procedure it is clear that n X n `;r ` r Sn+1 = j 1 n j : j 1 j=1 for. 1;. `;r With the aid (1), we can write down the exponential generating function of Sn+1 to be. Theorem 5 (Eu-Lin-Lo , [2] ) S. `;r. (z) =. X. `;r Sn+1. n 1. = exp z +. zn = S` (z) Sr (z) n! z2 + 2. +. 3. z` `. exp z +. z2 + 2. +. zr r. :.

(10) 3. Random Variables. Starting from this section, we will investigate the distribution of the leading statistic of the bounded deviated permutation. `;r Assume the permutation Sn+1 are uniformly distributed. We de…ne the random `;r variable Xn on the set of all (`; r)-bounded deviated permutations Sn+1 by Xn = k if `;r = 1 2 1 = k + 1 for n 2 Sn+1 : We set o n `;r : = k + 1 2 S = 1 n;k n+1. `;r Enumeration on Sn+1 yields the following counting formula:. n;k. n ak b n k ; k. =. 0. where (an ), (bn ) are the counting sequences for `n ; EGF for an is 1 X zn z2 an = exp z + + n! 2 n=0. k r n;. n; respectively. We know that the z` + `. ;. likewise for the sequence (bn ) :. Now we de…ne a bivariate generating function A(z; u) = =. 1 X n X. n;k u. n=0 k=0 1 X n X. n. n!. n zn ak b n k u k n! k. n=0 k=0. =. kz. 1 X X. n=0 i+j=n. (. ai. ui z i z j bj i! j! 2. (zu) + = exp (zu) + 2. (zu) + `. `. ). z2 exp z + + 2. zr + r. (2) ;. and we will use (2) to compute the expected value and the variance for the random variable Xn : The expected value and variance of Xn can be computed as n. @ [z n ] @u A(z; u)ju=1 = [z n ]A(z; 1). and 2. 2 n. @ @ [z n ] @u A(z; u)ju=1 2 A(z; u)ju=1 @u = + [z n ]A(z; 1) [z n ]A(z; 1). 4. @ [z n ] @u A(z; u)ju=1 [z n ]A(z; 1). !2.

(11) 4. Convergence to a Normal Distribution. Due to the number. n. `;r : 2 Sn+1. 1. o = k + 1 is a natural combinatoric object, we guess. the the random variable Xn = k if 1 = k + 1 for = 1 2 Gaussian distribution. And we need the following theorem:. n. `;r 2 Sn+1 converges to the. Theorem 6 (Generalized quasi-powers, [3], p.690) Assume that, for u in a …xed neighbourhood of 1, the generating function pn (u) of a non-negative discrete random variable (supported by Z 0 ) Xn admits a representation of the form pn (u) = exp (hn (u)) (1 + o (1)) ; uniformly with respect to u; where each hn (u) is analytic in : Assume also the conditions, h0n (1) + h00n (1) ! 1. and. h000 n (u) 3. (h0n (1) + h00n (1)) 2. ! 0;. uniformly for u 2 : Then, the random variable Xn. Xn =. h0n (1) 1. (h0n (1) + h00n (1)) 2. converges in distribution to a Gaussian with mean 0 and variance 1. In generally, considering the exact form pn (u) = exp (hn (u)) ; we have p0n (u) = h0n (u) exp (hn (u)) ; 2 p00n (u) = h00n (u) exp (hn (u)) + (h0n (u)) exp (hn (u)) ; 3 000 0 00 0 p000 n (u) = hn (u) exp (hn (u)) + 3hn (u) hn (u) exp (hn (u)) + (hn (u)) exp (hn (u)) ; hence 2 n. = h0n (1) + h00n (1) =. and h000 n (u) = We will show that section.. n;k. =. p000 n (u) pn (u) n k. ak b n. 3 k. p0n (1) p00n (1) + pn (1) pn (1) p0n (u)p00n (u) (pn (u))2. +2. 2. p0n (1) pn (1) p0n (u) pn (u). 3. satisfy the desire form of the theorem on the next. 5.

(12) 5. Main Results. In this section, we will show three cases:(`; r) = (1; 2); (1; 3) ; and (2; 2) . As a motivation, let us look at the case (`; r) = (1; 1) …rst. Given a permutation = 1 2 n+1 2 Sn+1 ; if we assign the value k + 1 to the position 1 ; then we can pick k positions in the remainnig to put the numbers from 1 to k: The way is nk and each arrangement is unique. So we have n;k = nk . When n tends to in…nity, the random variable Xn = k really converges to Gaussian distribution. In our next three cases, the combinatoric number n;k = nk ak bn k has no closed form, so we need to calculate its asymptotic. And the generating functions are all admissible in the sense of Hayman’s theorem, because they are of the form f (z) = eF (z) ; where F (z) is a polynomial whose coe¢ cient are all positive integers. The Hayman’s theorem is stated as following: Theorem 7 (Hayman formula, [4], [5], p.183) P Let f (z) = an z n be an admissible function. Let rn be the positive real root of the 0 (r) . Then equation a (rn ) = n, for each n = 1; 2; ; where a (rn ) is given by a (r) = r ff (r) f (r ) p n as n ! 1; rnn 2 b (rn ). an. where b (rn ) is given by b (r) = ra0 (r) :. 5.1. Case 1 (`; r) = (1; 2):. We set A(z; u) = exp((1 + u)z + the expected value. n. z2 ); 2. can be computed as 2. n. 2. @ [z n ]z exp(2z + z2 ) [z n 1 ] exp(2z + z2 ) [z n ] @u A(z; u)ju=1 = = : = 2 2 [z n ]A(z; 1) [z n ] exp(2z + z2 ) [z n ] exp(2z + z2 ). Now we use the Hayman formula to compute the asymptotic formula for the coe¢ cients 2 of the formula exp(2z + z2 ) : let f (z) = exp 2z +. z2 2. then f 0 (z) = (2 + z) exp 2z + then. ; z2 2. f 0 (r) a (r) = r = 2r + r2 ; f (r) 6. ;.

(13) thus a0 (r) = 2 + 2r; we …rst solve the equation 2rn + rn2 = n; we get 2+. rn =. p. 4 + 4n. 2. =. p. n 1+. =. p. n. =. 1+. p. 1+n=. 1 1 + 2n 8n2 1 1 1+ p + 2 n 8n 23. 1+. p. n. r. 1+. 1 n. 1 ;. hence p. n. 1 1 1+ p + 2 n 8n 32 n p n 1 1 n 1 p + = n 2n n 1 1 = (n) 2 exp n log 1 p + n 2n ( n 1 1 1 p + = (n) 2 exp n 2 n 2n p n (n) 2 exp n :. rnn =. n. 2. 1 1 p + n n. +O n. Also b (r) = ra0 (r) = 2r + 2r2 ; then b (rn ) = 2rn + 2rn2. 2rn2. 2n (n ! 1). and f (rn ) = exp 2rn +. rn2 2. n + rn 2 1 1 1+ p + 2 n 8n 32. = exp. p n exp n 2 n p exp + n 1+O n 2. = exp. 1 2. :. Finally, we have. an. f (r ) p n rnn 2 b (rn ) p e n2 exp (2 n p = n 4n 7. 1). 1 + O(n. 1 2. ) :. 3 2. !).

(14) 2. The asymptotic formula for the coe¢ cients of the formula exp(2z+ z2 ) can be computed :. p. e n e2 n 1 5 ( )2 p (1 + p + O(n 1 )); n 6 n 4n. z2 [z ] exp(2z + ) 2 n. thus [z n 1 ] exp(2z +. =. n. z2 ) 2 2. [z n ] exp(2z + z2 ) p 1 = n 1+O n 2. and 2. @ @ A(z; u)ju=1 [z n ] @u 2 A(z; u)ju=1 @u = + [z n ]A(z; 1) [z n ]A(z; 1). 2 n. = =. [z n 2 ] exp(2z +. z2 ) 2. [z n ] exp(2z p. + 3 +O n 2. n. z2 ) 2. +. 1 2. ;. hence. 3 n 4. 1. n. !2. 2 n). (. n. @ A(z; u)ju=1 [z n ] @u [z n ]A(z; 1). n4. 1 4. +O n. 3 4. :. Hence h0n (1). n. and h0n (1) + h00n (1). 2 n. Here pn (u) =. n X k=0. n! k n;k u = 2 i. ! 1 as n ! 1:. Z exp (1 + u) z +. z2 2. z n+1. C. dz;. where C is the counterclockwise oriented contour within center 0 and radius R. Note that jpn (u)j =. n! 2 i. Z exp (1 + u) z + C. Z exp n! 2 C. z2 2. z n+1. (1 + u) z + jzjn+1. n! exp (1 + u) R + 2 Rn+1 = n!g (u; R) ;. R2 2. dz. z2 2. n! 2. dz. Z. exp (1 + u) z + z n+1. C. Z exp (1 + u) jzj + n! 2 jzjn+1 C n! exp (1 + u) R +. 2 R=. Rn. bounded with respect to n and R, so does p0n (u); p00n (u); and p000 n (u):. 8. z2 2. R2 2. dz jzj2 2. dz.

(15) Since h000 n. p000 (u) (u) = n pn (u). 3. p0n (u)p00n (u) (pn (u))2. h000 n (u) (h0n (1) + h00n (1)). 3 2. =O n. p0n (u) pn (u). +2 5 4. 3. = G (u; R) ;. ! 0 as n ! 1:. Hence by generalized quasi-powers theorem, we have 1;2 Theorem 8 On Sn+1 ; the leading statistics. has the mean law.. n. p. n. Xn =. 1. 1 and the variance. 2 n. 1 p. n. 3 ; 2. and it admits a limit Gaussian. We verify above consequences by computer plotting: The red points are P (Xn ) in the cases (`; r) = (1; 2) ; and the blue curve is Gaussian distribution with n = 10000, centered at its peak.. Figure 5.1 (`; r) = (1; 2) ; n = 10000. 9.

(16) 5.2. Case 2 (`; r) = (1; 3):. We set A(z; u) = exp((1 + u)z + the expected value. n. can be computed as 2. n. z2 z3 + ); 2 3 3. 2. 3. @ [z n ] @u A(z; u)ju=1 [z n ]z exp(2z + z2 + z3 ) [z n 1 ] exp(2z + z2 + z3 ) = = = : 3 3 2 2 [z n ]A(z; 1) [z n ] exp(2z + z2 + z3 ) [z n ] exp(2z + z2 + z3 ). Now we use the Hayman formula to compute the asymptotic formula for the coe¢ cients 2 3 of the formula exp(2z + z2 + z3 ) : let z2 z3 + 2 3. f (z) = exp 2z + then. f 0 (z) = 2 + z + z 2 exp 2z + then a (r) = r. ;. z2 z3 + 2 3. ;. f 0 (r) = 2r + r2 + r3 ; f (r). thus a0 (r) = 2 + 2r + 3r2 : We …rst solve the equation 2rn + rn2 + rn3 = n: The Hayman formula is useful, but we face a problem at …rst: solve the equation a (rn ) = n: It always in the form C1 z 1 + C2 z 2 + + Ck z k = n; but we have no formula to solve it. So we transform the equation to the speci…c form u = t (u), by proper substitution, then by the Lagrange inversion formula : Theorem 9 (Lagrange Inversion Formula, [1], p.117) Suppose F (z) = zG (F (z)) ; G (0) 6= 0: Then [z n ] F (z) =. 1 n z n. G (z)n. 1. Now we can get the asymptotic we want: Let u = (rn ). 1. ;. then 2u. 1. +u. 2. +u. 3. = n;. implies 2u2 + u + 1 = u3 n: 10. (n. 1) :.

(17) Thus. 1 3. 2u2 + u + 1 Let t=n. 1 3. 1. = un 3 :. ( ). (u) = 2u2 + u + 1. ;. 1 3. and note that (0) = 1; then ( ) becomes u = t (u) : Now we applied the Lagrange inversion formula [tn ] u =. 1 n t n. ( (t))n. 1. (n. 1) :. Compute (t) =. 2ut2 + t + 1. 2. (t) =. 2t2 + t + 1. 3. (t) =. 2t2 + t + 1. 4. (t) =. 2t2 + t + 1. Hence u (t) = t +. t 5t2 + + 3 9 2t 11t2 + =1+ + 3 9. 1 3. 2 3 3 3 4 3. =1+. ; ;. = 1 + t + 2t2 ; and 4t 26t2 68t3 + + =1+ + 3 9 81. t2 2t3 17t4 + + + 3 3 81. :. ;. and then 1. rn = u = t. 1. = t. 1. = t. 1 1. = n3. t 2t2 17t3 1+ + + + 3 3 81 t 5t2 16t3 1 + + 3 9 81 1 5t 16t2 + + 3 9 81 1 5 1 16 2 n 3 + n 3 +O n 3 9 81. 1. 1. :. The asymptotic formula for the coe¢ cients of the formula exp(2z + [z n ] exp(2z +. z2 z3 + ) 2 3. f (r ) p n rnn 2 b (rn ) 1. 2. 11. 1. e n e2n3 + 6 n3 p ( )3 n 6n. 11. 11 18. (1. 95 324n. 1 3. z2 2. +. z3 ) 3. + O(n 1 )):. is :.

(18) Thus n. [z n 1 ] exp(2z +. =. z2 2. +. 2. [z n ] exp(2z + z2 + 1 1 1 = n3 + +O n 3 3. z3 ) 3. z3 ) 3. and 2. 2 n. @ @ [z n ] @u A(z; u)ju=1 2 A(z; u)ju=1 @u = + [z n ]A(z; 1) [z n ]A(z; 1). =. [z n 2 ] exp(2z + [z n ] exp(2z 1. = n3. +. z2 2. z2 2. +. z3 ) 3. +. 1 + +O n 3. z3 ) 3. 1 3. +. (. n. @ [z n ] @u A(z; u)ju=1 [z n ]A(z; 1). !2. 2 n). :. In this case, h0n (1). n. and h0n (1) + h00n (1) Here pn (u) =. n X k=0. n! k n;k u = 2 i. 2 n. ! 1 as n ! 1:. Z exp (1 + u)z +. z2 2. +. z3 3. dz;. z n+1. C. where C is the counterclockwise oriented contour within center 0 and radius R. Note that jpn (u)j =. n! 2 i n! 2 n! 2 n! 2. Z. Z exp (1 + u)z +. z2 2. +. exp (1 + u)z +. z2 2. +. dz. z n+1. C. z3 3. dz. z n+1. C. Z exp. (1 + u) z + jzjn+1. C. Z exp (1 + u) jzj + jzjn+1. C. 2. =. z3 3. z2 2. +. jzj2 2. +. = n!g (u; R) ; 12. dz. jzj3 3. dz. 3. R R n! exp (1 + u) R + 2 + 3 2 Rn+1 2 3 n! exp (1 + u) R + R2 + R3. Rn. z3 3. 2 R.

(19) bounded with respect to n and R, so does p0n (u); p00n (u); and p000 n (u): Since 3 p000 p0n (u) p0n (u)p00n (u) n (u) h000 (u) = + 2 3 = G (u; R) ; n pn (u) pn (u) (pn (u))2 h000 n (u) (h0n (1) + h00n (1)). 3 2. =O n. 3 2. ! 0 as n ! 1:. Hence by generalized quasi-powers theorem, we have 1;3 ;the leading statistics Theorem 10 On Sn+1. has the mean law.. n. p 3. n. 1 3. Xn =. 1. and the variance. 2 n. 1 p 3. n. 1 ; 3. and it admits a limit Gaussian. We verify above consequences by computer plotting: The red points are P (Xn ) in the cases (`; r) = (1; 3) ; and the blue curve is Gaussian distribution with n = 10000, centered at its peak.. Figure 5.2 (`; r) = (1; 3) ; n = 10000. 13.

(20) 5.3. Case 3 (`; r) = (2; 2) :. We set A(z; u) = exp (1 + u) z + 1 + u2 the expected value. n. z2 2. ;. can be computed as n. @ A(z; u)ju=1 [z n ] @u [z n ]A(z; 1) n [z ] (z + z 2 ) exp(2z + z 2 ) = : [z n ] exp(2z + z 2 ). =. Here we use the Heyman formula again to get the asymptotic formula for the coe¢ cients of the formula exp(2z + z 2 ) : let f (z) = exp 2z + z 2 ; then f 0 (z) = (2 + 2z) exp 2z + z 2 ; then a (r) = r. f 0 (r) = 2r + 2r2 ; f (r). thus a0 (r) = 2 + 4r: We solve the equation at …rst, 2rn + 2rn2 = n; we get. rn =. 2+ p. 2n = 2 p 2n = 2. p 4. 4 + 8n. =. 1+. p. 1 + 2n. 2. 1 1 1+ + 4n 32n2 p p 1 2 2 + p + 2 8 n 64n 23. 14. 1+ = 1 2. :. p. 2n 2. q. 1+. 1 2n.

(21) Hence rnn = = = =. !n p p 2n 1 2 2 + p + 2 2 8 n 64n 32 p !n n 2n 1 1 1 p + 2 2n 4n ! p n 2n 1 1 exp n log 1 p + 2 2n 4n ! ( p n 2n 1 1 1 p + exp n 2 2 2n 4n ! p n p 2n n p : exp 2 2 p. 1 1 p + 2n 4n. 2. +O n. 3 2. !). Also b (r) = ra0 (r) = 2r + 4r2 ; then b (rn ) = 2rn + 4rn2. 4rn2. 2n (n ! 1). and f (rn ) = exp 2rn + rn2 = exp. 1 (n + 2rn ) 2. ! p 1 n 2n 1 = exp exp +O n 2 2 2 2 ! p 1 2n 1 n exp + +O n 2 : 2 2 2 Finally, we have. an. rnn =. f (r ) p n 2 b (rn ). 2e n. n 2. exp. p. p. 2n. 4n. 1 2. 1 + O(n. 1 2. ) :. The asymptotic formula for the coe¢ cients of the formula exp(2z + z 2 ) can also be computed by computer as following: ! p p n 1 2 exp 2n 2e 2 2 p [z n ] exp(2z + z 2 ) 1 + p + O(n 1 ) : n 3 n 4n. 15.

(22) Thus n. [z n ] (z + z 2 ) exp(2z + z 2 ) [z n ] exp(2z + z 2 ) [z n 1 ] exp(2z + z 2 ) + [z n 2 ] exp(2z + z 2 ) = [z n ] exp(2z + z 2 ) 1 n+O n 1 2. =. and 2. @ @ [z n ] @u A(z; u)ju=1 2 A(z; u)ju=1 @u = + [z n ]A(z; 1) [z n ]A(z; 1). 2 n. =. @ [z n ] @u A(z; u)ju=1 [z n ]A(z; 1). !2. [z n 4 ] exp(2z + z 2 ) [z n 3 ] exp(2z + z 2 ) [z n 2 ] exp(2z + z 2 ) + 2 + 2 + [z n ] exp(2z + z 2 ) [z n ] exp(2z + z 2 ) [z n ] exp(2z + z 2 ) p 1 2n n + O (1) : 2 4. n. In this case, h0n (1). n. and h0n (1) + h00n (1) Here pn (u) =. n X k=0. n! k n;k u = 2 i. 2 n. ! 1 as n ! 1;. Z exp (1 + u) z + (1 + u2 ) z2 2 z n+1. C. dz;. where C is the counterclockwise oriented contour within center 0 and radius R. Note that jpn (u)j =. n! 2 i n! 2 n! 2 n! 2. Z. Z exp (1 + u) z + (1 + u2 ) z2 2. dz. z n+1. C. 2. exp (1 + u) z + (1 + u2 ) z2. dz. z n+1. C. Z exp. 2. (1 + u) z + (1 + u2 ) z2 jzjn+1. C. dz. Z exp (1 + u) jzj + (1 + u2 ) jzj2 2 jzjn+1. C. 2. 2 R n! exp (1 + u) R + (1 + u ) 2 2 Rn+1 n! exp (1 + u) + (1 + u2 ) R2 = Rn 1 = n!g (u; R) ;. 16. dz. 2 R. (. 2 n).

(23) bounded with respect to n and R, so does p0n (u); p00n (u); and p000 n (u): Since 3 p000 p0n (u) p0n (u)p00n (u) n (u) h000 (u) = + 2 3 = G (u; R) ; n pn (u) pn (u) (pn (u))2 h000 n (u) (h0n (1) + h00n (1)). 3 2. =O n. 1 2. ! 0 as n ! 1:. Hence by generalized quasi-powers theorem, we have 2;2 ;the leading statistics Theorem 11 On Sn+1. Xn = has the mean law.. n. n 2. and the variance. 1. 1 1 n 2. 2 n. p. 2n ; 4. and it admits a limit Gaussian. We verify above consequences by computer plotting. The red points are P (Xn ) in the cases (`; r) = (2; 2) ; and the blue curve is Gaussian distribution with n = 10000, centered at its peak.. Figure 5.3 (`; r) = (2; 2) ; n = 10000. 17.

(24) 6. Conclusion and Further Discussion. In this thesis, we have already use di¤erent ways to prove that the leading statistic of 1;2 1;3 2;2 three special cases Sn+1 ; Sn+1 ; and Sn+1 , converge to normal distributions. For general 2;3 cases, such like Sn+1 , we believe that they also converge to normal distributions in the same way. We will keep verifying these cases. In the future, a good direction to similar question is to prove other statistics, such like the second statistic, the third statistic, are all converge to normal distributions, or other types of distributions.. References [1] M. Aigner, A Course in Enumeration, Springer-Verlag Berlin Heidelberg 2007 [2] Sen-Peng Eu, Yen-Chi R. Lin, Yuan-Hsun Lo, Bounded deviated permutations, preprint [3] Philippe Flajolet, Robert Sedgewick, Analytic Combinatorics, Cambridge University Press 2009 [4] Hayman, Walter, A generalisation of Stirling’s formula, Journal für die reine und angewandte Mathematik 196 (1956), 67-95 [5] Herbert S. Wilf, Generatingfunctionology, 2nd edition, 1994. 18.

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