Shanghai Jiaotong University How to generalize Eulerian polynomials - PowerPoint PPT Presentation

Shanghai Jiaotong University How to generalize Eulerian polynomials via combinatorics and continued fractions Jiang Zeng Universit´ e Claude Bernard Lyon 1, France May 14, 2018, Shanghai

Plan of the talk 1 Eulerian polynomials 2 q -Carlitz-Scoville’s multivariate Eulerian polynomials 3 Eulerian polynomials as moments of orthogonal polynomials 4 Multivariate generalizations of Eulerian polynomials

Eulerian polynomials (Euler 1775) 1 t k = � 1 − t k ≥ 0 For n = 1 , 2 , . . . applying the operators ( tD ) n , with D = d dt , to the above identity we obtain �� � tA n ( t ) k n t k = � ( tD ) n t k = (1 − t ) n +1 . k ≥ 0 k ≥ 0 Thus A 1 ( t ) = 1 , A 2 ( t ) = 1 + t , A 3 ( t ) = 1 + 4 t + t 2 , A 4 ( t ) = 1 + 11 t + 11 t 2 + t 3 .

Exponential generating functions tA n ( t ) x n x n � � n ! (1 − t ) n +1 � k n t k 1 + n ! = 1 + n ≥ 1 n ≥ 1 k ≥ 0 (1 − t ) n ( kx ) n � t k � = 1 + (1 − t ) n ! k ≥ 0 n ≥ 1 t k ( e k (1 − t ) x − 1) � = 1 + (1 − t ) k ≥ 0 1 − t = 1 − te (1 − t ) x . (1) or A n ( t ) x n 1 − t � 1 + n ! = e ( t − 1) x − t . (2) n ≥ 1

Stieltjes continued fraction (1890) The formal Laplace transformation is a linear operator L on K [[ x ]] defined by � ∞ L ( t n , x ) = n ! x n +1 = t n e − t / x dt n ≥ 0 . for 0 n ≥ 0 a n t n n ≥ 0 a n x n +1 . If f ( t ) = � n ! , then L ( f ( t ) , x ) = � � ∞ x 1 − t e y (1 − t ) − t e − y / x dy = x 0 1 − tx 1 − 2 x 1 − 2 tx 1 − 1 − · · ·

Stieltjes continued fraction (1890) 1 − t Noticing that e y (1 − t ) − t is the generating function of Eulerian polynomials the above formula is equivalent to ∞ 1 A n ( t ) x n = � 1 + x n =1 1 − tx 1 − 2 x 1 − 2 tx 1 − 1 − · · · 1 = , 1 · t x 2 1 − 1 · x − 2 2 · t x 2 1 − (2 + t ) · x − 1 − (3 + 2 t ) · x − 3 2 · t x 2 · · · where b k = k (1 + t ) + 1 and λ k = k 2 t . The t = 1 case is due to Euler.

Combinatorial interpretations Let S n be the set of permutations on [ n ] := { 1 , . . . , n } . des σ = # { i ∈ [ n − 1] | σ ( i ) > σ ( i + 1) } , exc σ = # { i ∈ [ n ] | σ ( i ) > i } , wex σ = # { i ∈ [ n ] | σ ( i ) ≥ i } . Proposition 1 (Riordan (1958), MacMahon (1900)) t des σ = t exc σ = � � � t wex σ − 1 . A n ( t ) = σ ∈ S n σ ∈ S n σ ∈ S n

An example for S 3 σ des exc wex 1 2 3 0 0 3 1 3 2 1 1 2 2 1 3 1 1 2 2 3 1 1 2 2 3 1 2 1 1 1 3 2 1 2 1 2 Hence A 3 ( t ) = 1 + 4 t + t 2 . The equidistribution exc ∼ des can be explained by Foata’s transformation. The mapping σ �→ τ defined by τ = σ (2) . . . σ ( n ) σ (1) has the property that wex( σ ) = 1 + exc( τ ).

More definitions Definition 1 For σ ∈ S n , let σ (0) = σ ( n + 1) = n + 1. Then any entry σ ( i ) ( i ∈ [ n ]) can be classified according to one of the four cases: a peak if σ ( i − 1) < σ ( i ) and σ ( i ) > σ ( i + 1); a valley if σ ( i − 1) > σ ( i ) and σ ( i ) < σ ( i + 1); a double ascent if σ ( i − 1) < σ ( i ) and σ ( i ) < σ ( i + 1); a double descent if σ ( i − 1) > σ ( i ) and σ ( i ) > σ ( i + 1). Let val ( σ ) , pk ( σ ) , da ( σ ), and dd ( σ ) be the numbers of valleys, peaks, double ascents and double descents in σ , respectively.

Carlitz and Scoville’s formula For σ ∈ S n with σ (0) = σ ( n + 1) = n + 1 let w ( σ ) = u val ( σ ) u pk ( σ ) u da ( σ ) u dd ( σ ) . 1 2 3 4 Carlitz and Scoville (1974) proved the following generalization of Euler’s formula: e α 2 x − e α 1 x x n � � w ( σ ) = u 1 α 2 e α 1 x − α 1 e α 2 x n ! n ≥ 1 σ ∈ S n = u 1 x + u 1 ( u 3 + u 4 ) x 2 � x 3 � ( u 3 + u 4 ) 2 + 2 u 1 u 2 2! + u 1 3! + · · · , where u 4 + u 3 = α 1 + α 2 , u 1 u 2 = α 1 α 2 .

q -analogs of exponential function e u n =0 (1 − aq n ) and ( a ; q ) n = � n − 1 Let ( a ; q ) ∞ = � ∞ k =0 (1 − aq n ). Euler (1748) u n (1 − uq n ) − 1 = � � e q ( u ) := ; ( q ; q ) n n ≥ 0 n ≥ 0 u n q ( n 2 ) � � (1 + uq n ) = E q ( u ) := . ( q ; q ) n n ≥ 0 n ≥ 0 Cauchy (1843) u n ( au ; q ) ∞ � = ( a ; q ) n . ( u ; q ) ∞ ( q ; q ) n n ≥ 0

q -analogs (combinatorial version) An inversion of the permutation σ ∈ S n is a pair ( σ i , σ j ) such that i < j and σ i > σ j . Let inv( σ ) be the number of inversions of σ . For n ∈ N let [ n ] q = 1 + q + · · · + q n − 1 = 1 − q n 1 − q . Then it is known that q inv( π ) = n ! q := [1] q [2] q . . . [ n ] q . � (3) π ∈ S n The two combinatorial versions of q -exponential function are ∞ ∞ x n 2 ) x n q ( n � � exp q ( x ) = Exp q ( x ) = , and n ! q n ! q n =0 n =0 where 0! q = 1.

q -differential operator d The q -analog δ of the derivative dx for f ( x ) ∈ C [[ x ]] is defined by δ x ( f ( x )) = f ( x ) − f ( qx ) . (1 − q ) x Thus δ x (1) = 0 and for n > 0, δ x ( x n ) = [ n ] q x n − 1 . Also, for f ( x ) , g ( x ) ∈ C [[ x ]], δ x ( f ( x ) g ( x )) = δ x ( f ( x )) g ( x ) + f ( qx ) δ x ( g ( x )) , and � f ( x ) � = δ x ( f ( x )) g ( x ) − f ( x ) δ x ( g ( x )) δ x . g ( x ) g ( x ) g ( qx )

Stanley’s inv q -analog Let A n ( t , q ) be the combinatorial q -analog of Eulerian polynomials defined by � A inv t 1+des( σ ) q inv( σ ) . n ( t , q ) = σ ∈ S n Then Stanley (1976) proved the following q -analogue of Euler’s generating function formula ∞ n ( t , q ) x n 1 − t � A inv 1 + = 1 − t exp q ( x (1 − t )) . (4) n ! q n =1

q -analog of Carlitz-Scoville’s formula Theorem 2 (Pan-Z., 2018) Let � u val ( σ ) u pk ( σ ) u da ( σ ) u dd ( σ ) q inv( σ ) . P n ( u 1 , u 2 , u 3 , u 4 , q ) = (5) 1 2 3 4 σ ∈ S n Then P n ( u 1 , u 2 , u 3 , u 4 , q ) x n � n ! q n ≥ 1 � � � � exp q ( α 2 − u 4 ) x − exp q ( α 1 − u 4 ) x = u 1 � , (6) � � � α 2 exp q ( α 1 − u 4 ) x − α 1 exp q ( α 2 − u 4 ) x where u 4 + u 3 = α 1 + α 2 , u 1 u 2 = α 1 α 2 .

Sketch of Proof Let P n := P n ( u 1 , u 2 , u 3 , u 4 , q ). Lemma 3 We have P 1 = u 1 and for n ≥ 1 , n − 1 � n � � P n +1 = ( q n u 4 + u 3 )P n + q k u 2 P n − k P k , (7) k q k =1 � n � where the q-binomial coefficients q are defined by k � n � n ! q = (0 ≤ k ≤ n ) . k ! q ( n − k )! q k q Proof The identity is straightforward by considering the position of n + 1 in the permutations of S n +1 as in the case of Eulerian polynomials.

n ≥ 1 P n x n Let P( x ) := � n ! q . Then the above equation is equivalent to the q -differential equation δ x (P( x )) = u 3 P( x ) + u 4 P( qx ) + u 2 P( qx )P( x ) + u 1 . (8) Lemma 4 n ≥ 0 a n x n be a series with complex Let 0 < q < 1 and f ( x ) = � coefficients. Then δ x ( f ( x )) = 0 if and only if f ( x ) = f (0) , δ x ( f ( x )) = f ( x ) if and only if f ( x ) = f (0) exp q ( x ) . The result follows then by solving the q -differential equation (8).

A recent formula of Zhuang (2017) Let P (inv , pk , des) � q inv( π ) y pk ( π )+ 1 t des( π )+1 . ( q , y , t ) := n π ∈ S n Then ∞ ( q , y , t ) x n � P (inv , pk , des) n n ! q n =1 � � � � u (1 − v ) 1 − v Exp q 1+ uv x exp q 1+ uv x − 1 = v (1 + u ) � , (9) � � � 1 + uv u (1 − v ) 1 − v 1 − v Exp q exp q 1+ uv x 1+ uv x where � � � 1 + t 2 − 2 yt − (1 − t ) (1 + t ) 2 − 4 yt u = / (2(1 − y ) t ) , � � � (1 + t ) 2 − 2 yt − (1 + t ) (1 + t ) 2 − 4 yt v = / (2 yt ) .

Proof. Note that P (inv , pk , des) ( q , y , t ) = ytP n (1 , yt , 1 , t , q ). Hence the n generating function in (9) is equal to yt ( P ( x ) − 1) with u 1 = u 3 = 1, u 2 = yt and u 4 = t . The generating function is � � � � Exp q ( α 2 − 1) x exp q ( α 2 − t ) x − 1 α 1 1 − α 1 � � � � ( α 2 − 1) x exp q ( α 2 − t ) x α 2 Exp q with α 1 + α 2 = 1 + t and α 1 α 2 = yt . Then, Zhuang’s result follows by choosing α 1 = v (1 + u ) / (1 + uv ); α 2 = (1 + u ) / (1 + uv ) .

S-type and J-type continued fractions If ( a n ) n ≥ 0 is a sequence of combinatorial numbers or polynomials with a 0 = 1, it is often fruitful to seek to express its ordinary generating function (OGF) as a continued fraction of either Stieltjes type (S-type), ∞ 1 a n t n = � , α 1 t n =0 1 − α 2 t 1 − 1 − · · · or Jacobi type (J-type), ∞ 1 a n t n = � , β 1 t 2 n =0 1 − γ 0 t − β 2 t 2 1 − γ 1 t − 1 − · · ·

Contraction formulae of an S-fraction to a J-fraction Both sides of these expressions are to be interpreted as formal power series inthe inderterminate x . 1 1 = . α 1 α 2 x 2 α 1 x 1 − 1 − α 1 x − 1 − ( α 2 + α 3 ) x − α 3 α 4 x 2 1 − α 2 x · · · · · · i.e., γ 0 = α 1 γ n = α 2 n + α 2 n +1 for n ≥ 1 β n = α 2 n − 1 α 2 n .

Euler’s continued fraction formulae 1 n ! x n = � 1 x n ≥ 0 1 − 1 x 1 − 2 x 1 − 1 − 2 x · · · 1 = 1 2 x 2 1 − x − 1 − 3 x − 2 2 x 2 · · · with coefficients α 2 k − 1 = k , α 2 k = k .