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T he Saddle point method in combinatorics asymptotic analysis: - PowerPoint PPT Presentation

The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum T he Saddle point method in combinatorics asymptotic analysis: successes and failures (A personal view) Guy Louchard


  1. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum To check the effect of the correction, we first give in Figure 2, for n = 60, the comparison between J n ( j ) and the asymptotics (7), without the 1 / n term. Figure 3 gives the same comparison, with the constant term − 51 / (50 n ) in the correction. Figure 4 shows the quotient of J n ( j ) and the asymptotics (7), with the constant term − 51 / (50 n ). The “hat” behaviour, already noticed by Margolius, is apparent. Finally, Figure 5 shows the quotient of J n ( j ) and the asymptotics (7), with the full correction. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  2. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 0.005 0.004 0.003 0.002 0.001 0 700 800 900 1000 1100 j Figure 2: J n ( j ) (circle) and the asymptotics (7) (line), without the 1 / n term, n = 60 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  3. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 0.005 0.004 0.003 0.002 0.001 0 700 800 900 1000 1100 j Figure 3: J n ( j ) (circle) and the asymptotics (7) (line), with the constant in the 1 / n term, n = 60 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  4. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 1.01 1 0.99 0.98 0.97 0.96 700 800 900 1000 1100 Figure 4: Quotient of J n ( j ) and the asymptotics (7), with the constant in the 1 / n term, n = 60 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  5. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 1 0.98 0.96 0.94 0.92 0.9 700 800 900 1000 1100 Figure 5: Quotient of J n ( j ) and the asymptotics (7), with the full 1 / n term, n = 60 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  6. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Case j = n − k It is easy to see that here, we have z ∗ = 1 / 2. We obtain, to first order, [ C 1 , n − 2 j − 2 + 2 n ] + [ C 2 , n − 4 j − 4 − 4 n ] ε = 0 with C 1 + O (2 − n ) , = C 1 , n ∞ − 2 i � = 2 i − 1 = − 5 . 48806777751 . . . , C 1 i =1 C 2 + O (2 − n ) , = C 2 , n ∞ 4 i ( i 2 i − 2 i + 1) � = = 24 . 3761367267 . . . . C 2 (2 i − 1) 2 i =1 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  7. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Set j = n − k . This shows that, asymptotically, ε is given by a Laurent series of powers of n − 1 , starting with ( k − 1 + C 1 / 2) / (4 n ). We next obtain [ C 1 − 2 j − 2 + 2 n ] + [ C 2 − 4 j − 4 − 4 n ] ε + [ C 3 + 8 n − 8 j − 8] ε 2 = 0 for some constant C 3 . More generally, powers ε 2 k lead to a O (1) · ε 2 k term, powers ε 2 k +1 lead to a O ( n ) · ε 2 k +1 term. This gives ε = ( k − 1+ C 1 / 2) / (4 n )+(2 k − 2+ C 1 )(4 k − 4+ C 2 ) / (64 n 2 )+ O (1 / n 3 ) . Now we derive z ) = ln( Q ) − C 1 ( k − 1 + C 1 / 2) / (4 n ) + O (1 / n 2 ) S 1 (˜ with Q := � ∞ i =1 (1 − 1 / 2 i ) = . 288788095086 . . . . Similarly z ) = 2 ln(2) n +(1 − k ) ln(2)+( − k 2 / 2+ k − 1 / 2+ C 2 1 / 8) / (2 n )+ O (1 / n 2 ) S 2 (˜ Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  8. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum and so z ) = ln( Q )+2 ln(2) n +(1 − k ) ln(2)+( A 0 + A 1 k − k 2 / 4) / n + O (1 / n 2 ) S (˜ with − ( C 1 − 2) 2 / 16 , := A 0 A 1 := ( − C 1 / 2 + 1) / 2 . Finally, we obtain Q I n ( n − k ) ∼ e 2 ln(2) n +(1 − k ) ln(2) (2 π S (2 , 1) ) 1 / 2 × ( A 0 + 1 / 8 + C 2 / 16) + ( A 1 + 1 / 4) k − k 2 / 4 / n + O (1 / n 2 ) �� � � × exp . (8) Figure 6 shows, for n = 300, the quotient of I n ( n − k ) and the asymptotics (8). Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  9. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 1.001 1 0.999 0.998 0.997 0.996 0.995 0.994 0.993 285 290 295 300 305 310 Figure 6: Quotient of I n ( n − k ) and the asymptotics (8), n = 300 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  10. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Case j = α n − x , α > 0 Of course, we must have that α n − x is an integer. For instance, we can choose α , x integers. But this also covers more general cases, for instance I α n + β ( γ n + δ ), with α , β , γ , δ integers. We have here z ∗ = α/ (1 + α ). We derive, to first order, C 1 , n ( α ) − ( j + 1)(1 + α ) /α + (1 + α ) n ] + [ C 2 , n ( α ) − ( j + 1)(1 + α ) 2 /α 2 − (1 + α ) 2 n ] ε = 0 with, setting ϕ ( i , α ) := [ α/ (1 + α )] i , C 1 ( α ) + O ([ α/ (1 + α )] − n ) , C 1 , n ( α ) = ∞ i (1 + α ) ϕ ( i , α ) � C 1 ( α ) = α [ ϕ ( i , α ) − 1] , i =1 C 2 ( α ) + O ([ α/ (1 + α )] − n ) , C 2 , n ( α ) = ∞ � ϕ ( i , α ) i (1 + α ) 2 ( i − 1 + ϕ ( i , α )) / [( ϕ ( i , α ) − 1) 2 α 2 ] . C 2 ( α ) = i =1 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  11. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum ∞ ∞ � α � i � α � � � ˆ � � Q ( α ) := (1 − ϕ ( i , α )) = 1 − = Q . 1 + α 1 + α i =1 i =1 We finally derive I n ( α n − x ) ∼ e [ − ln(1 / (1+ α )) − α ln( α/ (1+ α ))] n +( x − 1) ln( α/ (1+ α )) × ˆ Q ( α ) − (1 + 3 α + 4 α 2 − 12 α 2 C 1 + 6 C 2 1 α 2 �� (2 π S (2 , 1) ) 1 / 2 exp + α 4 + 3 α 3 − 6 C 2 α 2 − 12 C 3 1 α ) / [12 α (1 + α ) 3 ] + x (2 α 2 − 2 C 1 α + 3 α + 1) / [2 α (1 + α ) 2 ] − x 2 / [2 α (1 + α )] / n + O (1 / n 2 ) � � . (9) Figure 7 shows, for α = 1 / 2, n = 300, the quotient of I n ( α n − x ) and the asymptotics (9). Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  12. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 1.04 1.02 1 0.98 0.96 130 140 150 160 170 Figure 7: Quotient of I n ( α n − x ) and the asymptotics (9), α = 1 / 2, n = 300 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  13. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The moderate Large deviation, j = m + xn 7 / 4 Now we consider the case j = m + xn 7 / 4 . We have here z ∗ = 1. We observe the same behaviour as in in the Gaussian limit for the coefficients of ε in the generalization of (5). Finally we obtain e − 18 x 2 √ n − 2916 / 25 x 4 × J n ∼ x 2 ( − 1889568 / 625 x 4 + 1161 / 25) / √ n � × exp + ( − 51 / 50 − 1836660096 / 15625 x 8 + 17637426 / 30625 x 4 ) / n � + O ( n − 5 / 4 ) / (2 π n 3 / 36) 1 / 2 . (10) Note that S (3) (˜ z ) does not contribute to the correction and that this correction is equivalent to the Gaussian case when x = 0. Of course, the dominant term is null for x = 0. The exponent 7 / 4 that we have chosen is of course not sacred; any fixed number below 2 could also have been considered. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  14. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 1.07 1.06 1.05 1.04 1.03 1.02 1.01 1 600 700 800 900 1000 1100 1200 Figure 8: Quotient of J n ( j ) and the asymptotics (10), with the 1 / √ n and 1 / n term, n = 60 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  15. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Large deviations, j = α n ( n − 1), 0 < α < 1 / 2 Here, again, z ∗ = 1. Asymptotically, ε is given by a Laurent series of powers of n − 1 , but here the behaviour is quite different: all terms of the series generalizing (5) contribute to the computation of the coefficients. It is convenient to analyze separately S (1) and 1 S (1) 2 . This gives, by substituting z := 1 − ε, j = α n ( n − 1) , ε = a 1 / n + a 2 / n 2 + a 3 / n 3 + O (1 / n 4 ) , ˜ and expanding w.r.t. n , (1 / a 1 − α ) n 2 + ( α − α a 1 − a 2 / a 2 S (1) 2 (˜ z ) ∼ 1 ) n + O (1) , n − 1 S (1) � 1 (˜ z ) ∼ f ( k ) , k =0 − ( k + 1)(1 − ε ) k / [1 − (1 − ε ) k +1 ] f ( k ) := − ( k + 1)(1 − [ a 1 / n + a 2 / n 2 + a 3 / n 3 + O (1 / n 4 )]) k = / { 1 − (1 − [ a 1 / n + a 2 / n 2 + a 3 / n 3 + O (1 / n 4 )]) k +1 } . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  16. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum This immediately suggests to apply the Euler-Mac Laurin summation formula, which gives, to first order, � n f ( k ) dk − 1 S (1) 1 (˜ z ) ∼ 2( f ( n ) − f (0)) , 0 so we set k = − un / a 1 and expand − f ( k ) n / a 1 . This leads to � n f ( k ) dk 0 � − a 1 ue u � 1 (1 − e u ) n 2 ∼ − a 2 0 e u [2 a 2 1 − 2 e u a 2 1 − 2 u 2 a 2 − u 2 a 2 1 + 2 e u ua 2 1 ] � + n du 2 a 3 1 (1 − e u ) 2 O (1) − 1 + 2( f ( n ) − f (0)) . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  17. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum This readily gives � n f ( k ) dk ∼ − dilog ( e − a 1 ) / a 2 1 n 2 0 1 e − a 1 + a 4 1 e − a 1 − 4 a 2 dilog ( e − a 1 ) + 4 a 2 dilog ( e − a 1 ) e − a 1 [2 a 3 + 1 e − a 1 − 2 a 2 1 ( e − a 1 − 1)] n + O (1) . 2 a 2 a 2 1 + 2 a 2 1 e − a 1 ] / [2 a 3 + Combining S (1) z ) + S (1) 1 (˜ 2 (˜ z ) = 0, we see that a 1 = a 1 ( α ) is the solution of − dilog ( e − a 1 ) / a 2 1 + 1 / a 1 − α = 0 . We check that lim α → 0 a 1 ( α ) = ∞ , lim α → 1 / 2 a 1 ( α ) = −∞ . Similarly, a 2 ( α ) is the solution of the linear equation α − α a 1 − a 2 / a 2 1 + e − a 1 / [2(1 − e − a 1 )] − 1 / (2 a 1 ) 1 e − a 1 + a 4 1 e − a 1 + 4 a 2 dilog ( e − a 1 )( e − a 1 − 1) [2 a 3 + 1 e − a 1 − 2 a 2 1 ( e − a 1 − 1)] 2 a 2 a 2 1 + 2 a 2 1 e − a 1 ] / [2 a 3 + = 0 and lim α → 0 a 2 ( α ) = −∞ , lim α → 1 / 2 a 2 ( α ) = ∞ . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  18. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum We could proceed in the same manner to derive a 3 ( α ) but the computation becomes quite heavy. So we have computed an approximate solution ˜ a 3 ( α ). Finally, J n ( α n ( n − 1)) 1 ∼ e [1 / 72 a 1 ( a 1 − 18+72 α )] n + C 0 (2 π n 3 / 36) 1 / 2 × 1 / 72 a 2 �� × exp 2 + 1 / 36 a 1 a 3 − 1 / 4 a 3 + 1 / 4 a 2 − 1 / 2 a 1 + a 3 α + a 1 a 2 α + 1 / 3 a 3 1 α − a 2 α − 1 / 2 a 2 1 α − 5 / 24 a 1 a 2 + 1 / 24 a 2 1 a 2 + 1139 / 18000 a 2 1 − 1 / 16 a 3 / n + O (1 / n 2 ) � � 1 + 87 / 25 − 18 α . (11) C 0 := 1 / 72 a 3 1 − 1 / 4 a 2 + 1 / 4 a 1 − a 1 α − 5 / 48 a 2 1 + 1 / 36 a 1 a 2 + a 2 α + 1 / 2 a 2 1 α Note that, for α = 1 / 4, the 1 / n term gives − 51 / 50, again as expected. Figure 9 shows, for n = 80 and α ∈ [0 . 15 .. 0 . 35], the quotient of J n ( α n ( n − 1)) and the asymptotics (11). Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  19. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 1 0.9 0.8 0.7 0.6 0.5 0.4 1000 1200 1400 1600 1800 2000 2200 Figure 9: Quotient of J n ( α n ( n − 1)) and the asymptotics (11), n = 80 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  20. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A The distribution studied by Zheng, p ( n ), depending on two parameters, A , k , is defined as follows. Set θ := 1 / A . We have � n − 1 � 1 1 1 1 � ξ ( n ) = − n + 1 + j ξ ( j ) , n ≥ 1 , 1 + θ n ( n − j )( n − j + 1) 1 ξ (0) = ln(1 + 1 /θ ) . (12) Note that n − 1 1 ( n − j )( n − j + 1) ∼ 1 − ln n 1 � n . j n 1 Also, with some extra integer parameter k , we set n p ( n ) = − k 1 � j ξ ( j ) p ( n − j ) , n ≥ 1 , p (0) = (1 + 1 /θ ) k . n 1 A picture of p ( n ) ( A = 20 , k = 2) is given in Figure 10. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  21. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 0.003 0.0025 0.002 0.0015 0 50 100 150 200 250 300 Figure 10: p ( n ) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  22. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The analysis We know that the GF of ξ ( n ) is given by ∞ � 1 − 1 � ξ ( n ) z n = ln � G ( z ) = θ ( − 1 + ϕ ( z )) , G (1) = 0 , (13) 0 with ∞ z n � 1 � � ϕ ( z ) = n ( n + 1) = 1 + z − 1 ln(1 − z ) . 1 Set ∞ � z n p ( n ) . F ( z ) := 1 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  23. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Now ∞ ∞ n � � � zF ′ z ) = z n np ( n ) = − k z j j ξ ( j ) z n − j p ( n − j ) 1 n =1 j =1 ∞ ∞ ∞ � � � z n n ξ ( n ) p (0) − k z j j ξ ( j ) z n − j p ( n − j ) = − k 1 j =1 n = j +1 = − kzG ′ ( z ) p (0) − kzG ′ ( z ) F ( z ) . Solving, this gives �� z − u − 1 − k (ln(1 − u ) + u ) k [ θ u − ln(1 − u ) + u ln(1 − u )] k − 1 F ( z ) = 0 � � − k � θ + 1 z k � ( θ z − ln(1 − z ) + z ln(1 − z )) k . du + C θ Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  24. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum For instance, for k = 2, we have F ( z ) = − [2 z 2 ln(1 − z ) θ − 2 θ z 2 − 2 z ln(1 − z ) θ + ln(1 − z ) 2 − z 2 − 2 z ln(1 − z ) + z 2 ln(1 − z ) 2 ] θ 2 � [(1 + θ ) 2 [ θ z − ln(1 − z ) + z ln(1 − z )] 2 ] . Of course p (0) + F (1) = 1 . The GF of � ∞ n p ( j ) is given by ∞ ∞ z � � z n 1 − z [1 − p (0) − F ( z )] . FD ( z ) = p ( j ) = 1 n Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  25. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum How to find n ∗ such that [ z n ∗ ] FD ( z ) = 1 / 2 . Of course, we can’t use the singularity of FD ( z ) at z = 1, as it gives [ z n ] FD ( z ) for large n , but FIXED θ . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  26. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The Saddle point We have e ln( FD ( z )) 1 � FD ( z ) 1 � [ z j ] FD ( z ) = z j +1 dz = dz . (14) z j +1 2 π i 2 π i Ω Ω It appears, after some experiments, that the values of j such that [ z j ] FD ( z ) = 1 / 2 are such that j ∼ α ∗ A ln( A ) = − α ∗ ln( θ ) , θ for some constant α ∗ . Note that this does not confirm Zheng’s conjecture. For instance, k = 1 leads to α ∗ = 0 . 92 . . . , k = 2 to α ∗ = 0 . 2 . 52 . . . , k = 3 to α ∗ = 4 . 25 . . . . For now on, we set for instance k = 2. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  27. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Set − α ∗ ln( θ ) � � G ( z ) := ln( FD ( z )) − + 1 ln( z ) . θ Set z to C and expand G ′ ( z ) into θ . This gives, to first order, α ln( θ ) 1 + 1 − C = 0 , C θ or θ α ln( θ ) + O ( θ 2 ) . C = 1 + So we try z = 1 + βθ ˜ ln( θ ) . (15) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  28. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum This gives, after some complicated manipulations (done by Maple) z ) ∼ [ − β + αβ 2 + 2 α + 3 βα − 3] ln( θ ) G ′ (˜ . (1 + β )(2 + β ) θ Solving G ′ (˜ z ) = 0 gives √ 1 + 6 α + α 2 β = 1 − 3 α + . 2 α But this is impossible: β is decreasing as expected, but PROBLEM 1: β (3 / 2) = 0!. The same problem appears for all k . To be convinced that (15) is the correct asymptotics, we have computed numerically the solution zn of G ′ ( zn ) = 0 for different values of θ and α and we have extracted the corresponding values of β from (15). For instance, for θ = 1 / 1000, we obtain in Figure 11 the graph of β vs α . And the values of θ : 1 / 100 , 1 / 500 , 1 / 1500 give the same function, with some remarkable fit. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  29. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0.5 1 1.5 2 2.5 3 Figure 11: β vs α , θ = 1 / 1000, k = 2 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  30. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Independently of PROBLEM 1, it appears, after some experiments, that the next term would be z = 1 + βθ ln( θ ) + γθ ln( − ln( θ )) ˜ . ln( θ ) 2 This leads to some equation ϕ ( γ, α ) = 0 . Also, we obtain, to first order, z ) ∼ f ( β )ln( θ ) 2 7 + 6 β + β 2 G ′′ (˜ , f ( β ) = (2 + β ) 2 (1 + β ) 2 , θ 2 and � 2 + β � z ) ∼ − ln( θ )+ g ( β )+ln( − ln( θ ))+ αβ, G (˜ g ( β ) = ln . (1 + β ) 2 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  31. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Hence, by standard manipulations, e αβ + g ( β ) [ z j ] FD ( z ) ∼ √ = φ ( α ) say , � 2 π f ( β ) independent of θ . If we knew β ( α ), this would give α ∗ from φ ( α ∗ ) = 1 / 2. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  32. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Sum of positions of records in random permutations The statistic srec is defined as the sum of positions of records in random permutations. The generating function (GF) of srec is given by n ( z i + i − 1) , � G ( z ) = (16) i =1 and the probability generating function (PGF) is given by i =1 ( z i + i − 1) � n Z ( z ) = . (17) n ! This statistic has been the object of recent interest in the litterature. Kortchemski obtains the the GF (16) and also proves that J n ( j ) := z j [ G ( z )] ∼ e n ln( n ) y , where j = n ( n + 1) x = 1 − y 2 , x , 2 (18) with an error O (1 / ln( n )). Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  33. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The large deviation j = n ( n +1) (1 − y 2 ) 2 By Cauchy’s theorem, 1 � G ( z ) J n ( j ) = z j +1 dz 2 π i Ω 1 � e S ( z ) dz , = (19) 2 π i Ω where Ω is inside the analyticity domain of the integrand and encircles the origin and n � n ( n + 1) � ln( z i + i − 1) − � (1 − y 2 ) + 1 S ( z ) = ln( z ) . 2 i =1 Set S ( i ) := d i S dz i . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  34. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum First we must find the solution of S (1) (˜ z ) = 0 with smallest module. This leads to n z i � n ( n + 1) � i ˜ � (1 − y 2 ) + 1 z i + i − 1 − = 0 . (20) ˜ 2 i =1 z = z ∗ + ε , plug into In some previous sections , we simply tried ˜ (20), and expanded into ε . Here it appears that we cannot get this expansion. So we expand first (20) itself. But we must be z ˜ i = ˜ careful. There exists some ˜ i such that ˜ i . Some numerical experiments suggest that ˜ i = O ( n ). So we set ˜ i = α n , 0 < α < 1 and we must now compute α . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  35. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum z = e ξ , with We obtain ˜ ξ = L + ln( α ) , α n where here and in the sequel, L := ln( n ). Note that this leads to � L + ln( α ) � z n = exp = n 1 /α α 1 /α . ˜ α Using the classical splitting of the sum technique (20) leads to (we provide in the sequel only a few terms in the expansions, but Maple knows and uses more) ˜ i z i i ˜ � Σ 1 := z i + i − 1 , ˜ i =1 n z i i ˜ � n ( n + 1) � � (1 − y 2 ) + 1 Σ 2 := z i + i − 1 − . ˜ 2 i =˜ i +1 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  36. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum As z i − 1 z ˜ i − 1 z ˜ i ˜ < ˜ < ˜ = 1 , i < ˜ i , ˜ ˜ i i i we have ˜ ˜ z i − 1 z i − 1 i i � � 2 � z i � ˜ ˜ 1 − ˜ � � z i Σ 1 = = ˜ + + . . . 1 + ˜ z i − 1 i i i =1 i i =1 (21) The first summation is immediate ˜ i z ˜ i +1 − 1 z i = ˜ ˜ z � − ˜ z − 1 . ˜ z − 1 ˜ i =1 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  37. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum For the next summations, we again use Euler-Mac Laurin summation formula. First of all, the correction (to first order) is given by ˜ 1 2 + 1 = 1 2 + 1 2 − 1 / ( α n ) ∼ 1 α n i 4 α n . (22) 1 + ˜ 2 2 i − 1 ˜ i Next, we must compute integrals such as � ˜ z i − 1 � k i � ˜ z i ˜ di . (23) i 1 But we know that � ˜ � ˜ z i − 1 � e ξ i − 1 i i � ˜ � � z i e ξ i ˜ di = di = Ei (1 , − 2 ξ ) − Ei (1 , − ξ ) i i 1 1 + Ei (1 , − ˜ i ξ ) − Ei (1 , − 2˜ i ξ ) = Ei (1 , − 2 ξ ) − Ei (1 , − ξ ) + Ei (1 , − ( L + ln( α ))) − Ei (1 , − 2( L + ln( α ))) , where Ei ( x ) is the exponential integral. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  38. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Setting L 1 := L + ln( α ), we have ℜ ( Ei (1 , − ξ )) = − γ − ln( ξ ) − ξ − ξ 2 4 + . . . , � � � � − 1 − 1 − 1 − 1 ℜ ( Ei (1 , − L 1 )) = e L 1 + . . . = α n + . . . . L 2 L 2 L 1 L 1 1 1 We use similar expansions for terms like (23). This finally leads, with (22), to � 7 α 2 12 L + α 2 ( − 84 ln( α ) − 31) � Σ 1 = n 2 + . . . + O ( n ) . 144 L 2 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  39. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Now we turn to Σ 2 . As ˜ ˜ i − 1 i + 1 − 1 i z i = 1 , i > ˜ < < i , z i z ˜ ˜ ˜ i +1 ˜ we have n � n ( n + 1) � i � (1 − y 2 ) + 1 − Σ 2 = 1 + i − 1 2 z i i =˜ ˜ i +1 n � � 2 � 1 − i − 1 � i − 1 � n ( n + 1) � � (1 − y 2 ) + 1 = i + + . . . − z i ˜ ˜ z i 2 i =˜ i +1 2 − α 2 2 − 47 α 2 � 1 � = n 2 60 L + . . . + O ( n ) + n 3 α � n ( n + 1) � � � (1 − y 2 ) + 1 α 1 /α L + . . . − + . . . n 1 /α 2 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  40. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum So 2 − α 2 2 − α 2 � 1 � S ′ (˜ z ) = Σ 1 + Σ 2 = n 2 5 L + . . . − (1 − y 2 ) / 2 + O ( n ) + n 3 α � � α 1 /α L + . . . + . . . = 0 (24) n 1 /α Putting the coefficient of n 2 to 0, and solving wrt α gives 5 L + − 3199 y / 1800 + ln( y ) y / 5 α ∗ = y − y + . . . (25) L 2 Now we must consider the other terms of (24). First we must n 3 compare n with n 1 /α . n 3 − 1 /α > n and vice-versa. The most interesting case If α > 1 2 , is the case α > 1 2 (the other one can be treated by similar method). Note hat there are also other terms in (24) of order n k − ( k − 2) /α , k ≥ 4. This is greater than n if α > ( k − 2) / ( k − 1). Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  41. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Returning to (24), we first compute n 1 /α = n 1 / y ϕ ( y , L ), with ϕ ( y , L ) = e L (1 /α − 1 / y ) = e 1 / (5 y ) − e 1 / (5 y ) ( − 3271 + 360 ln( y ) + . . . 1800 yL (26) So we obtain from (24) the term n 3 α � � α 1 /α L + . . . , n 1 / y ϕ ( y , L ) and with (25), n 3 � � y y 1 / y e 1 / (5 y ) L + . . . . n 1 / y Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  42. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Now we set α = α ∗ + Cn n 1 / y , plug into (24) (ignoring the O ( n ) term), and expand. The n 2 term must theoretically be 0. Actually, it is given by a series of large powers of 1 / L as we only use a finite n 3 number of terms in (25). Solving the coefficient of n 1 / y wrt C , we obtain C = e − 1 / (5 y ) y (3 y − 1) / y + . . . Ly 3 and α = α ∗ + Cn n 1 / y + . . . (27) e S (˜ z ) ˜ J n ( j ) = (28) � 2 π S ′′ (˜ z ) In Figure 12, we give, for n = 150, a comparison between ln( J n ( j )) (circle) and ln(˜ J n ( j )) (line). The fit is quite good, but when y is near 1. But j is then small and our symptotics are no more very efficient. We also show the first approximation (18): nLy (line blue) which is only efficient for very large n . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  43. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 700 600 500 400 300 200 100 0.2 0.4 0.6 0.8 Figure 12: Comparison between between ln( J n ( j )) (circle) and ln(˜ J n ( j )) (line), n = 150. Also it shows the first approximation (1): nLy (line blue) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  44. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The central region j = yn From (17), we see that, as expected, Z (1) = 1. Moreover n iz i − 1 � Z ′ ( z ) = z i + i − 1 Z ( z ) , i =1 and i [ z i − 2 i 2 − 2 z i − 2 − z 2 i − 2 + z i − 2 ] n � Z ′′ ( z ) = Z ( z ) ( z i + i − 1) 2 i =1 � n � 2 iz i − 1 � + Z ( z ) . z i + i − 1 i =1 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  45. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum So E ( srec ) = Z ′ (1) = n , and the variance is given by n ( i − 2) + n 2 + n − n 2 = n ( n − 1) V ( srec ) = Z ′′ (1) + n − n 2 = � . 2 i =1 Of course Z ( z ) corresponds to a sum of independent non identically distributed random variables, but it is clear that the Lindeberg-L´ evy conditions are not satified here. The distribution is not asymptotically Gaussian. As will be clear later on, it is convenient to separate the cases j ≥ 1 and j < 1. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  46. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The case y ≥ 1 Again, by Cauchy’s theorem, 1 � Z ( z ) P ( srec = j ) = z j +1 dz 2 π i Ω 1 � e S ( z ) dz , = (29) 2 π i Ω where Ω is inside the analyticity domain of the integrand and encircles the origin and n ln( z i + i − 1) − ( yn + 1) ln( z ) − ln( n !) . � S ( z ) = i =1 First we must find the solution of S (1) (˜ z ) = 0 with smallest module. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  47. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum This leads to n z i i ˜ � z i + i − 1 − ( yn + 1) = 0 . (30) ˜ i =1 z := z ∗ + ε , where, here, it is easy to check that z ∗ = 1. Set Set ˜ j = yn . This leads, to first order (keeping only the ε term in (30)), to + − 4 + 10 y − 4 y 2 ε := 2( y − 1) + O (1 / n 3 ) . n 2 n This shows that, asymptotically, ε is given by a Laurent series of powers of n − 1 . To obtain more precision, we set again j = yn , expand (30) into powers of ε (we use 9 terms), set 4 ε = a 1 a i � n + n i i =2 expand in powers of n − 1 , and equate each coefficient to 0. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  48. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum This gives, for the coefficient of n 1 − y + 1 / 120 a 4 1 + 1 / 2 a 1 + 1 / 720 a 5 1 + 1 / 362880 a 8 1 + 1 / 6 a 2 1 + 1 / 5040 a 6 1 + 1 / 3628800 a 9 1 + 1 / 24 a 3 1 + 1 / 40320 a 7 1 = 0 . We observe that all terms of the expansion of (30) into ε contribute to the computation of the coefficients. We have already encountered this sitution in analyzing the number in inversions in permutations. So we must turn to another approach. Setting i = k + 1, (30) becomes n − 1 � f ( k ) − ( yn + 1) = 0 , (31) k =0 where z k +1 f ( k ) = ( k + 1)˜ . z k +1 + k ˜ Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  49. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum This gives, always by Euler-Mac Laurin summation formula, to first order, � n f ( k ) dk − 1 2( f ( n ) − f (0)) − ( yn + 1) = 0 , 0 so we set k = un / a 1 , ˜ z = 1 + a 1 / n and expand f ( k ) n / a 1 . This leads to � a 1 e u ndu − yn + O (1) = 0 , a 1 0 or e a 1 = ya 1 + 1 . (32) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  50. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Note that a 1 ( y ) ∼ ln( y ) , y → ∞ , a 1 (1) = 0 . The explicit solution of (5) is easily found, we have e a 1 = ya 1 + 1 = y [ a 1 − 1 / y ] = e a 1 +1 / y e − 1 / y , − e − [ a 1 +1 / y ] [ a 1 + 1 / y ] = − e − 1 / y / y , − [ a 1 + 1 / y ] = W ( − 1 , − e − 1 / y / y ) , a 1 = − 1 / y − W ( − 1 , − e − 1 / y / y ) , where W ( − 1 , x ) is the suitable branch of the Lambert equation: W ( x ) e W ( x ) = x . PROBLEM 2: THE THIRD DERIVATIVE IS OF ORDER n 3 ALTHOUGH THE SECOND DERIVATIVE IS OF ORDER n 2 . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  51. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Finally, by standard technique, (29) should lead to e F ( y ) P ( srec = j ) ∼ , (33) � 2 π n 2 F 2 ( y ) and 1 P ( srec = n ) ∼ . (34) � 2 π n 2 / 2 Also P ( srec = j ) ∼ e − y ln( y ) , y → ∞ . � 2 π n 2 y Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  52. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The case y < 1 We have j n − 1 P ( srec = j ) = 1 [ z i + i − 1] n ![ z j ] � � u , i =1 u = j and, if j is large, by (34), j [ z i + i − 1] ∼ j ! 1 � [ z j ] √ π j . i =1 So n − 1 P ( srec = j ) ∼ 1 ( j − 1)! 1 � √ π n √ π. u = n ! u = y For j large enough, P ( srec = j ) is constant and given by (34). We have made a numerical comparison of P ( srec = j ) , n = 200 , j = 1 .. 3 n with (34) and (33). This is given in Figure 13. This is quite satisfactory. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  53. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 0.003 0.0025 0.002 0.0015 0.001 0.0005 0 0.5 1 1.5 2 2.5 3 Figure 13: Comparison between P ( srec = j ) , n = 200 , j = 1 .. 3 n (circle) and the asymptotics (34) and (33) (line) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  54. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Merten’s theorem for toral automorphisms Let n � (1 − z k )(1 − z − k ) ϕ n ( z ) := 1 What is the symptotic behaviour of [ z j ] ϕ n ( z ) , in particular the asymptotic value of [ z 0 ] ϕ n ( z ) . A plot of [ z j ] ϕ n ( z ), n = 15 is given in Figure 14. This seems to have a Gaussian envelope. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  55. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 200 100 –100 –50 50 100 –100 –200 Figure 14: [ z j ] ϕ n ( z ), n = 15 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  56. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum But n � (ln(1 − z k ) + ln(1 − z − k ) − ln( z ) S ( z ) := ln( ϕ n ( z )) − ln( z ) = 1 doesn’t appear to posses zeroes, two plots of | S ( z ) | 2 given in Figures 15, 16, reveal a quite irregular behaviour. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  57. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 100000 80000 60000 40000 20000 0 –1.5 –1.5 –1 –1 –0.5 –0.5 0 0 x y 0.5 0.5 1 1 1.5 1.5 Figure 15: | S ( z ) | 2 , n = 15 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  58. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 5000 4000 3000 2000 1000 0 –1.5 –1.5 –1 –1 –0.5 –0.5 0 0 x y 0.5 0.5 1 1 1.5 1.5 Figure 16: | S ( z ) | 2 , n = 15 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  59. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Representations of numbers as � n k = − n ε k k We consider the number of representations of m as � n k = − n ε k k , where ε k ∈ { 0 , 1 } . For m = 0, this is sequence A000980 in Sloane’s encyclopedia. This problem has a long history. Here, we extend previous ranges a bit, to O ( n 3 / 2 ). But we improve at the same time the quality of the approximation The generating function of the number of representations for fixed n is given by n � (1 + z k )(1 + z − k ) , C n (1) = 2 · 4 n . C n ( z ) = 2 k =1 By normalisation, we get the probability generating function of a random variable X n : n � F n ( z ) = 4 − n (1 + z k )(1 + z − k ) , k =1 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  60. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The Gaussian limit We obtain mean M and variance V : σ 2 := V ( n ) = n ( n + 1)(2 n + 1) M ( n ) = 0 , . 12 We consider values j = x σ , for x = O (1) in a neighbourhood of the mean 0. The Gaussian limit of X n can be obtained by using the Lindeberg-L´ evy conditions, but we want more precision. We know that 1 � e S ( z ) dz P n ( j ) = 2 π i Ω where S ( z ) := ln( F n ( z )) − ( j + 1) ln z with n � [ln(1 + z i ) + ln(1 + z − i ) − ln 4] . ln( F n ( z )) = i =1 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  61. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum z := z ∗ − ε , where, here, z ∗ = 1. Set ˜ Finally this leads to P n ( j ) ∼ e − x 2 / 2 · exp � � [ − 39 / 40+9 / 20 x 2 − 3 x 4 / 40] / n + O ( n − 3 / 2 ) / (2 π n 3 / 6) (35) Note again that S (3) (˜ z ) does not contribute to the 1 / n correction. Note also that, unlike in the instance of the number of inversions in permutations, we have an x 4 term in the first order correction. Figure 17 shows, for n = 60, Q 3 : the quotient of P n ( j ) and the asymptotics (35), with the constant term − 39 / (40 n ) and the x 2 term 9 x 2 / (20 n ), and Q 4 : the quotient of P n ( j ) and the full asymptotics (35). Q 4 gives indeed a good precision on a larger range. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  62. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 1 0.9998 0.9996 0.9994 0.9992 0.999 0.9988 0.9986 –400 –200 0 200 400 Figure 17: Q 3 (blue) and Q 4 (red) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  63. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The case j = n 7 / 4 − x Now we show that our methods are strong enough to deal with probabilities that are far away from the average; viz. n 7 / 4 − x , for fixed x . Of course, they are very small, but nevertheless we find asymptotic formulæ for them. Later on, n 7 / 4 − x will be used as an integer This finally gives P n ( j ) ∼ e − 3 n 1 / 2 − 27 / 10 � 1 + 369 / 175 / n 1 / 2 + 931359 / 245000 / n + 1256042967 / 471625000 / n 3 / 2 + 4104 x / 175 / n 7 / 4 − 9521495156603 / 2145893750000 / n 2 + 7561917 x / 122500 / n 9 / 4 + ( − 235974412468129059 / 341392187500000 + 18 x 2 ) / n 5 / 2 + · · · � � (2 π n 3 / 6) 1 / 2 . (36) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  64. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 5.4 5.2 5 4.8 4.6 4.4 4.2 6420 6422 6424 6426 6428 6430 6432 6434 6436 6438 i Figure 18: P n ( j ) (circle) and the full asymptotics (36) up to the n − 5 / 2 term,(line), n = 150, scaling= 10 21 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  65. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The q -Catalan numbers The q -Catalan numbers C n ( q ) are defined as 1 − q ( q ; q ) 2 n C n ( q ) = , 1 − q n +1 ( q ; q ) n ( q ; q ) 2 n with ( q ; q ) n = (1 − q )(1 − q 2 ) . . . (1 − q n ). Note that C n = C n (1), a Catalan number, and the polynomial F n ( q ) = C n ( q ) / C n is the probability generating function of a distribution X n that we call the Catalan distribution. Is has been shown recently that this distribution is asymptotically normal. Since we attack F n ( q ) and C n ( q ) mostly with analytic methods, we find it more appropriate to replace the letter q by z ; note that n − 1 1 − z n + i +1 � C n ( z ) = 1 − z i +1 . i =1 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  66. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum It is easy to get the mean m and variance σ 2 of the random variable X n characterized by P n ( i ) := P ( X n = i ) := [ z i ] C n ( z ) , C n namely m = n ( n − 1) , 2 σ 2 = n ( n 2 − 1) . 6 (In the full paper we sketch how these moments and higher ones can be computed quite easily.) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  67. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The Gaussian limit n ( n 2 − 1) / 6. � Set j = m + x σ , with m = n ( n − 1) / 2 and σ = Then, for fixed x , the following approximation holds: P n ( j ) ∼ e − x 2 / 2 · exp ( − 9 / 40+9 / 20 x 2 − 3 x 4 / 40) / n � �� (2 π n 3 / 6) 1 / 2 . (37) Note that, like in the Representations of numbers as � n k = − n ε k k , we have a x 4 term in the first order correction. Figure 19 shows, for n = 60, Q 3 : the quotient of P n ( j ) and the asymptotics (37), with the constant term − 9 / (40 n ) and the x 2 term 9 x 2 / (20 n ) and Q 4 : the quotient of P n ( j ) and the full asymptotics (37). Q 4 gives indeed a good precision on a larger range. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  68. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 1 0.9995 0.999 0.9985 0.998 0.9975 1200 1400 1600 1800 2000 2200 Figure 19: Q 3 (blue) and Q 4 (red) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  69. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The case j = n − k We deal with probabilities that are far away from the average; viz. j = n − k , for fixed k . Again, they are very small, but nevertheless we find asymptotic formulæ for them. We have z ∗ = 1 and, as we will see, ε is again given by a Puiseux series of powers of n − 1 / 2 . But the approach of previous Sections is doomed to failure: all terms of the generalization of (5) contribute to the computation of the coefficients. So we have to turn to another technique. We have n − 1 1 − z n + i +1 � P n ( n − k ) = [ z n − k ] � C n . 1 − z i +1 i =1 So we set S := S 1 + S 2 + S 3 + S 4 , Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  70. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum with n − 1 � ln(1 − z n + i +1 ) , S 1 := i =1 n − 1 n � � ln(1 − z i +1 ) = − ln(1 − z i ) , S 2 := − i =1 i =2 S 3 := − ( n + 1 − k ) ln z , S 4 := − ln( C n ) . z = z ∗ − ε = 1 − ε . We must have S ′ (˜ Set ˜ z ) = 0, with S ′ ( z ) = S ′ 1 ( z ) + S ′ 2 ( z ) + S ′ 3 ( z ) , n − 1 − ( n + i + 1) z n + i � S ′ 1 ( z ) = , 1 − z n + i +1 i =1 n − 1 ( i + 1) z i S ′ � 2 ( z ) = 1 − z i +1 , i =1 3 ( z ) = − n + 1 − k S ′ . z Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  71. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum It is clear that we must have either S ′ z ) = O ( n ) or S ′ 1 (˜ 2 (˜ z ) = O ( n ). Set ε = f ( n ) n , for a function still to be determined. Let us first (roughly) solve S ′ 1 (˜ z ) = n . We have z n z ∼ n 2 e − f ( n ) z ) ∼ n ˜ S ′ 1 (˜ . 1 − ˜ f ( n ) This leads to the equation n 2 e − f ( n ) = n , i.e. e f ( n ) f ( n ) = n , f ( n ) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  72. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum which is solved by f ( n ) = W ( n ) , where W is the Lambert function. But we know that � ln ln( n ) � W ( n ) ∼ ln( n ) − ln ln( n ) + O . ln( n ) So ε ∼ ln( n ) . n But then n 2 1 S ′ z ) ∼ z ) 2 ∼ 2 (˜ ln( n ) 2 = Ω( n ) . (1 − ˜ Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  73. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum So we must turn to the other choice, i.e. � n 2 � 1 � � S ′ 2 (˜ z ) ∼ O ∼ O . z ) 2 f ( n ) 2 (1 − ˜ The equation n 2 f ( n ) 2 = n leads now to f ( n ) = √ n , and z ) ∼ n 2 e −√ n = n 3 / 2 e −√ n , S ′ √ n 1 (˜ which is exponentially negligible. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  74. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum This rough analysis leads to the more precise asymptotics n 1 / 2 + a 2 a 1 n + a 3 ε = n 3 / 2 + · · · (38) and we must solve S ′ z ) + S ′ 2 (˜ 3 (˜ z ) = 0 , i.e. n − 1 z j ( j + 1)˜ z j +1 − n + 1 − k � = 0 . 1 − ˜ z ˜ j =1 We rewrite this equation as n z j j ˜ � z j = M , 1 − ˜ j =2 and will later replace M by n + 1 − k . It is not hard to see that we can solve ∞ z j j ˜ � z j = M , 1 − ˜ j =2 since the extra terms introduce an exponentially small error. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  75. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum z by e − t : Now we replace ˜ ∞ je − jt � g ( t ) = 1 − e − jt , j =2 and compute its Mellin transform: g ∗ ( s ) = ( ζ ( s − 1) − 1) ζ ( s )Γ( s ) . (39) The original function can be recovered by a contour integral: � 3+ i ∞ 1 ( ζ ( s − 1) − 1) ζ ( s )Γ( s ) t − s ds . g ( t ) = 2 π i 3 − i ∞ Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  76. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum This integral can be approximately (in a neighbourhood of t = 0) solved by taking residues into account: g ( t ) = π 2 12 + t 3 t 5 6 t 2 − 3 2 t + 13 24 − t 30240 + O ( t 7 ) . 720 − Now we solve π 2 12 + t 3 t 5 6 t 2 − 3 2 t + 13 24 − t 720 − 30240 ∼ M , and find √ 4 M − 1 + (81 + 13 π 2 ) 32 + π 2 t ( M ) ∼ π M − 1 / 2 − 3 6 � 13 M − 3 / 2 − � M − 2 √ 288 π 144 6 To simplify the next expressions, we will set n = w 2 . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  77. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Now, since 1 − ε = e − t , we set M = n + 1 − k , expand and reorganize: √ 6(4 π 4 + 3 π 2 + 72 π 2 k + 243) � π 2 ε ∼ π 12 + 3 w − 1 − � w − 2 + w − 3 . √ 4 864 π 6 (40) Note that k appears (linearly) only in the coefficient of w − 3 . The quadratic term in k appears in the coefficient of w − 5 . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  78. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum This finally leads to � 1 − 2 1 / 2 3 1 / 2 216 + 13 π 2 + 24 π 2 k P n ( n − k ) ∼ e T 4 2 − 1 / 2 π 3 / 2 12 144 π w + 19008 π 2 + 20736 π 2 k + 31104 + 217 π 4 + 624 π 4 k + 576 π 4 k 2 6912 π 2 w 2 − 3 1 / 2 2 1 / 2 [ − 229635 + 11104128 π 2 + 1907712 π 4 k + 11197440 π 2 k + 4069 π 6 + 771984 π 4 + 1244160 π 4 k 2 + 15624 π 6 k � +13824 π 6 k 3 + 22464 π 6 k 2 ] / [2985984 π 3 w 3 ] √ 6 π w T 4 := − 2 ln(2) w 2 + . 3 The quality of the approximation is given in Figure 20, with the w − 2 term (and k 2 contribution) and in Figure 21, with the w − 2 and w − 3 terms (and k 3 contribution). The fit is rather good: the curves cover the exact graph above and below. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  79. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 5e–28 4e–28 3e–28 2e–28 1e–28 0 –10 –8 –6 –4 –2 2 4 6 8 10 k Figure 20: P n ( n − k ) , n = 60 exact (circle), asymptotics (line), with w − 2 term Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  80. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 5e–28 4e–28 3e–28 2e–28 1e–28 0 –10 –8 –6 –4 –2 2 4 6 8 10 k –1e–28 Figure 21: P n ( n − k ) , n = 60 exact (circle), asymptotics (line), with w − 2 and w − 3 terms Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  81. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum A simple case of the Mahonian statistic Canfield, Janson and Zeilberger have analyzed the Mahonian distribution on multiset permutations: classic permutations on m objects can be viewed as words in the alphabet { 1 , . . . , m } . If we allow repetitions, we can consider all words with a 1 occurences of 1, a 2 occurences of 2 , . . . , a m occurences of m . Let J m denote the number of inversions. Assuming that all words are equally likely, the probability generating function of J m is given, setting N = a 1 + · · · + a m , by � m i =1 a i ! � N i =1 (1 − z i ) φ a 1 ,..., a m ( z ) = i =1 (1 − z i ) . � a j N ! � m j =1 The mean µ and variance σ 2 are given by = ( e 1 + 1) e 2 − e 3 σ 2 ( J m ) µ ( J m ) = e 2 ( a 1 , . . . , a m ) / 2 , , 12 where e k ( a 1 , . . . , a m ) is the degree k elementary symmetric function. Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  82. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum Let a ∗ = max j a j and N ∗ = N − a ∗ . Recently it was proved that, if N ∗ → ∞ then the sequence of normalized random variables J m − µ ( J m ) σ ( J m ) tends to the standard normal distribution. The authors also conjecture a local limit theorem and prove it under additional hypotheses. In this talk, we analyze simple examples of the Mahonian statistic, for instance, we consider the case m = 2 , a 1 = an , a 2 = bn , n → ∞ . Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  83. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum We analyze the central region j = µ + x σ and one large deviation j = µ + xn 7 / 4 . The exponent 7 / 4 that we have chosen is of course again not sacred, any fixed number below 2 and above 3 / 2 could also have been considered. We have here ( an )!( bn )! � ( a + b ) n (1 − z i ) i =1 φ ( z ) = , (( a + b ) n )! � an i =1 (1 − z i ) � bn i =1 (1 − z i ) µ = abn 2 , 2 σ 2 = ab ( a + b + 1 / n ) n 3 . 12 Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  84. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The Gaussian limit To compute S 1 (˜ z ), we first compute the asymptotics of the i term, this leads to a ln( i ) contribution, which will be cancelled by the factorials. We obtain Z 2 ( j ) ∼ e − x 2 / 2 · − 3 a 2 + 13 ab + 3 b 2 + 3( a 2 + b 2 + ab ) x 2 �� �� � n + O ( n − 3 / 2 ) exp 20 ab ( a + b ) 10 ab ( a + b ) � ( π ab ( a + b ) n 3 / 6) 1 / 2 . (41) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  85. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum The Large deviation, j = µ + xn 7 / 4 We derive ab ( a + b ) n 1 / 2 − 36( a 2 + ab + b 2 ) x 4 6 x 2 � Z 2 ( j ) ∼ exp − 5[ ab ( a + b )] 3 + C 9 ( x , a , b ) / n 1 / 2 + C 10 ( x , a , b ) / n + . . . � ( π ab ( a + b ) n 3 / 6) 1 / 2 . � (42) To check the effect of the correction, we first give in Figure 22, for n = 50 , a = b = 1 / 2 and x ∈ [0 .. 0 . 2], the comparison between Z 2 ( j ) and the asymptotics (42). Figure 23 shows the quotient of Z 2 ( j ) and the asymptotics (42) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  86. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 0.006 0.004 0.002 0 320 340 360 380 400 420 440 460 480 500 i Figure 22: The comparison between Z 2 ( j ) and the asymptotics (42). Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

  87. The number of inversions in permutations Median versus A ( A large) for a Luria-Delbruck-like distribution, with parameter A Sum 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 320 340 360 380 400 420 440 460 480 500 Figure 23: The quotient of Z 2 ( j ) and the asymptotics (42) Guy Louchard The Saddle point method in combinatorics asymptotic analysis:

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