Orthogonal Polynomials for Bernoulli and Euler Polynomials Lin JIU - PowerPoint PPT Presentation

Orthogonal Polynomials for Bernoulli and Euler Polynomials Lin JIU Dalhousie University Number Theory Seminar Jan. 14th, 2019

Acknowledgment ◮ Diane Shi ◮ Tianjin University

Objects ◮ Bernoulli numbers B n : ◮ Euler numbers E n : ∞ t n 2 e t ∞ t n t � � e t − 1 = B n n !; e 2 t + 1 = E n n !; n = 0 n = 0 ◮ Bernoulli polynomial B n ( x ) : ◮ Euler polynomial E n ( x ) : ∞ B n ( x ) t n ∞ E n ( x ) t n 2 t e t − 1 e xt = e t + 1 e xt = � � n !; n !; n = 0 n = 0 ◮ Bernoulli polynomial of order p ◮ Euler polynomial of order p B ( p ) E ( p ) n ( x ) : n ( x ) : � p � p ∞ n ( x ) t n ∞ n ( x ) z n � � 2 t e xt = e xz = � B ( p ) � E ( p ) n !; n ! ; e z + 1 e t − 1 n = 0 n = 0 B ( 1 ) n ( x ) = B n ( x ) ; B n ( 0 ) = B n ; E ( 1 ) n ( x ) = E n ( x ) ; E n ( 1 / 2 ) = E n / 2 n .

Orthogonal Polynomials Let X be a random variable with density function p ( t ) on R and with moments m n , i.e., � m n = E [ X n ] = t n p ( t ) d t . R Let P n ( y ) be the monic orthogonal polynomials with respect to X (or w. r. t. m n ), i.e., deg P n = n , LC [ P n ] = 1, and � � c n , if m = n ; P m ( t ) P n ( t ) p ( t ) d t = c n δ m , n = 0 , otherwise. R Equivalently, for all 0 ≤ r < n � y r P n ( y ) � = 0 . � � y k = m k P n satisfies a three-term recurrence: P 0 = 1, P 1 = y − s 0 , and P n + 1 ( y ) = ( y − s n ) P n ( y ) − t n P n − 1 ( y ) . Example . 1. Carlitz [3, eq. 4.7] and also with Al-Salam [1, p. 93] gave the monic orthogonal polynomials, denoted by Q n ( y ) , with respect to E n : Q n + 1 ( y ) = yQ n ( y ) + n 2 Q n − 1 ( y ) . 2. Touchard [16, eq. 44] computed the monic orthogonal polynomials with respect to the B n , denoted by R n ( y ) : n 4 � y + 1 � R n + 1 ( y ) = R n ( y ) + 4 ( 2 n + 1 )( 2 n − 1 ) R n − 1 ( y ) . 2

1st Task ◮ Find the orthogonal polynomials with respect to B n ( x ) , denoted by ̺ n ( y ) ; ◮ and the orthogonal polynomials with respect to E n ( x ) , denoted by Ω n ( y ) . Namely, for 0 ≤ r < n , � � y r ̺ n ( y ) � = 0 = y r Ω n ( y ) � . � � � y k = B k ( x ) � y k = E k ( x ) [Question] Why? ◮ Generalization ◮ Probabilistic interpretations: Letting p B ( t ) := π 2 sech 2 ( π t ) and p E ( t ) := sech ( π t ) , ( t ∈ R ) we define two random variables L B and L E with density functions p B and p E , respectively. Then, with i 2 = − 1, � n � � n �� iL B + x − 1 � it + x − 1 � B n ( x ) = E = p B ( t ) d t , [ 5 , eq . 2 . 14 ] 2 2 R � n � � n �� iL E + x − 1 � � it + x − 1 E n ( x ) = E = p E ( t ) d t . [ 9 , eq . 2 . 3 ] 2 2 R

3rd Reason ◮ The generalized Motzkin numbers: 1. Motzkin numbers: M 0 , 0 = 1, M n , k = 0 if k > n or n < 0, and M n + 1 , k = M n , k − 1 + M n , k + M n , k + 1 . M n , k − 1 M n , k M n , k + 1 ց ↓ ւ M n + 1 , k M n , k = # of paths from ( 0 , 0 ) to ( n , k ) . Path: Lattice path on N 2 (0 ∈ N ) and only three types are considered:  α k : ( j , k ) → ( j + 1 , k + 1 ) diagonally up ր   β k : ( j , k ) → ( j + 1 , k ); horizontal →  γ k : ( j , k ) → ( j + 1 , k − 1 ) diagonally down ց  Example .   1 M 0 , k M 1 , k 1 1    ⇒ M 3 , 1 = 5   2 2 1 M 2 , k  M 3 , k 4 5 3 1

3rd Reason ◮ The generalized Motzkin numbers: 2. generalized Motzkin numbers: M n + 1 , k = M n , k − 1 + σ k M n , k + τ k + 1 M n , k + 1

3rd Reason ◮ The generalized Motzkin numbers: 2. generalized Motzkin numbers: M n + 1 , k = M n , k − 1 + σ k M n , k + τ k + 1 M n , k + 1 M n , k = sum of weighted lattice paths from ( 0 , 0 ) to ( n , k ) . ∞ 1 M n , 0 z n = � τ 1 z 2 1 − σ 0 z − n = 0 1 − σ 1 z − τ 2 z 2 ··· Let X be an arbitrary random variable, with moments m n and monic orthogonal polynomials P n ( y ) satisfying the recurrence P n + 1 ( y ) = ( y − s n ) P n ( y ) − t n P n − 1 ( y ) . Then, we have ∞ m 0 m n z n = � . t 1 z 2 1 − s 0 z − n = 0 t 2 z 2 1 − s 1 z − 1 − s 2 z −···

Continued Fractions By letting ( σ k , τ k ) = ( s k , t k ) . If further assuming m 0 = 1, we have M n , 0 = m n = E [ X n ] . Example . The monic orthogonal polynomials with respect to E n , Q n ( y ) , satisfy Q n + 1 ( y ) = yQ n ( y ) + n 2 Q n − 1 ( y ) . Euler numbers E n are given by the weighted lattice paths ( 1 , 0 , − k 2 ) ⇒ horizontal paths are eliminated. Therefore, E n counts the weighted Dyck paths , related to Catalan numbers C n . n = 6: C 3 := 1 � 6 � = 5 4 3 Then, by noting that each diagonally down path from ( j , k ) to ( j + 1 , k − 1 ) has weight − k 2 , we have 3 2 2 2 1 2 + 2 2 2 2 1 2 + 1 2 2 2 1 2 + 2 2 1 2 1 2 + 1 2 1 2 1 2 � − 61 = E 6 = ( − 1 ) 3 � .

1st Task ◮ Find the orthogonal polynomials with respect to B n ( x ) , denoted by ̺ n ( y ) ; ◮ and the orthogonal polynomials with respect to E n ( x ) , denoted by Ω n ( y ) . Namely, for 0 ≤ r < n , � � y r ̺ n ( y ) = 0 = y r Ω n ( y ) � � . � � � y k = B k ( x ) � y k = E k ( x ) NOTE: � n � � n � �� iL B + x − 1 �� iL B − 1 B n ( x ) = E and B n = B n ( 0 ) = E , 2 2 and the monic orthogonal polynomials with respect to the B n , denoted by R n ( y ) , are given by n 4 � y + 1 � R n + 1 ( y ) = R n ( y ) + 4 ( 2 n + 1 )( 2 n − 1 ) R n − 1 ( y ) . 2 Let c be a constant. ∼ P n ( y ) X X + c ∼ ? ∼ ? cX E n = 2 n E n ( 1 / 2 )

1st Task Lemma . [L. Jiu and D. Shi] random variable moments monic orthogoal polynomial X m n P n ( y ) : P n + 1 ( y ) = ( y − s n ) P n ( y ) − t n P n − 1 ( y ) n � n � � m k c n − k ¯ P n + 1 ( y ) = ( y − s n − c ) ¯ ¯ P n ( y ) − t n ¯ X + c P n ( y ) : P n − 1 ( y ) k k = 0 ˜ P n + 1 ( y ) = ( y − Cs n ) ˜ ˜ P n ( y ) − C 2 t n ˜ C n m n CX P n ( y ) : P n − 1 ( y ) Proof. ¯ P n ( y ) := C n P n ( y / C ) . ˜ P n ( y ) := P n ( y − c ) and Theorem . [L. Jiu and D. Shi] n 4 � y + 1 � R n + 1 ( y ) = R n ( y ) + 4 ( 2 n + 1 )( 2 n − 1 ) R n − 1 ( y ) B n 2 n 4 � y − x + 1 � B n ( x ) ̺ n + 1 ( y ) = ̺ n ( y ) + 4 ( 2 n + 1 )( 2 n − 1 ) ̺ n − 1 ( y ) 2 Q n + 1 ( y ) = yQ n ( y ) + n 2 Q n − 1 ( y ) E n Ω n ( y ) + n 2 y − x + 1 E n ( x ) Ω n + 1 ( y ) = � � 4 Ω n − 1 ( y ) 2

2nd Task ◮ Find the orthogonal polynomials with respect to B ( p ) n ( x ) , denoted by ̺ ( p ) n ( y ) ; ◮ and the orthogonal polynomials with respect to E ( p ) ( x ) , denoted by Ω ( p ) n ( y ) . n Recall that ◮ Bernoulli polynomial B n ( x ) : ◮ Euler polynomial E n ( x ) : ∞ B n ( x ) t n ∞ E n ( x ) t n 2 t e t − 1 e xt = e t + 1 e xt = � � n ! ; n ! ; n = 0 n = 0 ◮ Bernoulli polynomial of order ◮ Euler polynomial of order p p B ( p ) E ( p ) n ( x ) : ( x ) : n � p ∞ n ( x ) t n � p ∞ ( x ) z n � � 2 t e xt = B ( p ) e xz = E ( p ) � � n ! ; n ! ; n e t − 1 e z + 1 n = 0 n = 0 ◮ ◮ � n � � n � �� iL B + x − 1 �� iL E + x − 1 B n ( x ) = E E n ( x ) = E 2 2 B ( p ) E ( p ) n ( x ) = E [?] ( x ) = E [?] n

Sum of random variables Let X and Y be two independent variables, with moments m n and m ′ n , respectively. In addition, define the moment generating functions: ∞ ∞ t n t n � e tX � � e tY � � � F ( t ) := E = and G ( t ) := E = m ′ m n n n ! n ! n = 0 n = 0 Then, n � n � � E [( X + Y ) n ] = m k m ′ n − k k k = 0 and � e t ( X + Y ) � E = F ( t ) G ( t ) . Consider a sequence of independent and identically distributed (i. i. d. ) random � p � j = 1 with each L E j ∼ L E (sech( t )) . Then E ( p ) variables L E j ( x ) is the n th moment of a n certain random variable: n     p iL E j − p E ( p ) �  . ( x ) = E  x + n   2 j = 1 � p � j = 1 with L B j ∼ L B ( π sech 2 ( π t ) / 2 ) , Similarly, given an i. i. d. L B j n     p iL B j − p B ( p ) �  . n ( x ) = E  x +   2 j = 1

Hankel Determinant Given a sequence a = ( a n ) ∞ n = 0 , the n th Hankel determinant of a is defined by · · ·  a 0 a 1 a 2 a n  a 1 a 2 a 3 · · · a n + 1   ∆ n ( a ) := det . . . .  .  ...  . . . .   . . . .  a n a n + 1 a n + 2 · · · a 2 n Recall that given a sequence of numbers/moments m := ( m n ) ∞ n = 0 , let P n ( y ) be the monic orthogonal polynomials with respect to m n . Namely, for all 0 ≤ r < n � y r P n ( y ) � = 0 . � y k = m k � Suppose P n satisfies the three-term recurrence: P n + 1 ( y ) = ( y − s n ) P n ( y ) − t n P n − 1 ( y ) . Theorem . (1) If m 2 k + 1 = 0 for all k ∈ N , then, s n = 0; Recall The monic orthogonal polynomials with respect to E n , Q n ( y ) , satisfy Q n + 1 ( y ) = yQ n ( y ) + n 2 Q n − 1 ( y ) . And � ∞ t n 2 e t � � E 2 k + 1 = 0 . n ! = e 2 t + 1 = sech( t ) . E n n = 0