4th RART, July 17-20 2017, Universidad Complutense de Madrid Horizontal and Vertical Recurrence Relations for Exponential Riordan Matrices and Their Applications Ji-Hwan Jung Sungkyunkwan University, Korea jh56k@skku.edu (Joint work with Gi-Sang Cheon and Paul Barry) July 17, 2017 1/28
Outline I. Introduction 1.1 Motivation 1.2 Exponential Riordan matrix 2. Horizontal and vertical formulas for e-riordan matrices 3. Applications 3.1 Determinant of the Hessenberg matrix 3.2 d-orthogonality of Sheffer polynomials 2/28
1. Introduction 3/28
Motivation Eric Temple Bell (1883.02.07-1960.12.21) is the eponym of the Bell polynomials and the Bell numbers of combinatorics. 4/28
An n n matrix A n =[a i,j ] n 1 i,j=0 is called lower Hessenberg matrix if a i,j =0 for j > i + 1, i.e., A n = E.T. Bell in 1923 a 0,0 a 0,1 0 0 a 1,0 a 1,1 a 1,2 0 a 2,0 a 2,1 a 2,2 0.... a n 2,0 a n 2,1 a n 2,2 a n 2,n 1 a n 1,0 a n 1,1 a n 1,2 a n 1,n 1. Let a(z) = k 0 a kz k F 0 with a 0 =1. Ifa i,j = a i j+1 for j i +1 i.e., A n is the Toeplitz matrix then 1 a(z) =1+ ( 1) k det(a k )z k. k 1 5/28
A. Inselberg, On determinants of Toeplitz-Hessenberg matrices arising in power series, J. Math. Anal. Appl., 63(1978), 347-353. W. B. Gragg, The QR algorithm for unitary Hessenberg matrices, J. Comput. Appl. Math, 16(1986), 1-8. J. L. Stuart, Hessenberg L-matrices, Linear Algebra Appl., 167(1992), 261-265. L. Ching, The maximum determinant of an n n lower Hessenberg (0, 1)-matrix, Linear Algebra Appl., 183(1993), 147-153. M. Benhamadou, On the calculation of the multiplicity of a real eigenvalue of Hessenberg matrix, Adv. Eng. Softw., 30(1999), 201-210. T. Sogabe, A note on a fast numerical algorithm for the determinant of a pentadiagonal matrix, Appl. Math. Comput., 201(2008), 561-564. 6/28
M. Elouafi, and A. D. A. Hadj, A recursion formula for the characteristic polynomial of Hessenberg matrices, Appl. Math. Comput. 208(2009), 177-179. Y. H. Chen, C. Y. Yu, A new algorithm for computing the inverse and the determinant of a Hessenberg matrix, Appl. Math. Comput., 218(2011), 4433-4436. M. Merca, A note on the determinant of a Toeplitz-Hessenberg matrix, Special Matrices, 1(2013), 10-16. Z. Cinkir, A fast elementary algorithm for computing the determinant of Toeplitz matrices, J. Comput. Appl. Math., 255(2014), 353-361. A. Ipek, A. Kamil, On Hessenberg and pentadiagonal determinants related with Fibonacci and Fibonacci-like numbers, Appl. Math. Comput., 229(2014), 433-439. D. A. Bini, L. Robol, Quasiseparable Hessenberg reduction of real diagonal plus low rank matrices and applications, Linear Algebra Appl., 502(2016), 186-213. 7/28
1.2 Exponential Riordan matrix { } F n = j n f jz j C[[z]] f n 0. An exponential Riordan matrix or e-riordan matrix [r n,k ] n,k 0 is defined by g(z) F 0 and f (z) F 1 s.t. its kth column egf is r n,k z n /n! =g(z)f k (z)/k!. n k As usual, the matrix is denoted by E(g(z), f (z)) or E(g, f ). Let E be the set of all e-riordan matrices. The set E forms a group called the exponential Riordan group under the Riordan multiplication defined by E(g, f ) E(h,l)=E(g h(f ),l(f )). The identity is E(1, z), the usual identity matrix. E(g, f ) 1 = E(1/g(f ), f )wheref (f (z)) = f (f (z)) = z. 8/28
2. Horizontal and vertical formulas for e-riordan matrices 9/28
Theorem 1 (Horizontal formula) An i.l.t. matrix R =[r n,k ], r 0,0 =1,isE(g, f ) iff there exists a horizontal pair {h n ; h n } n 0 of sequences s.t. r n,n = h n 0 0, r n, 1 =0; n k r n+1,k = h 0 r n,k 1 + j=0 (k + j)! k! (h j + k h j+1 ) r n,k+j, (n k 0) where H R = n 0 h nz n = g ( f ) g( f ) and H R = n 0 h n z n = f ( f ). 10 / 28
r 1,0 r 1,1 0 0 0 r 2,0 r 2,1 r 2,2 0 0 r 3,0 r 3,1 r 3,2 r 3,3 0 r 4,0 r 4,1 r 4,2 r 4,3 r 4,4 r 5,0 r 5,1 r 5,2 r 5,3 r 5,4 = r 0,0 0 0 0 0 r 1,0 r 1,1 0 0 0 r 2,0 r 2,1 r 2,2 0 0 r 3,0 r 3,1 r 3,2 r 3,3 0 r 4,0 r 4,1 r 4,2 r 4,3 r 4,4 h 0 h0 0 0 0 1! h 1 1! (h 0 + h 1 ) h0 0 0 2! h 2 1! (h 2! 1 + h 2 ) 2! (h 0 +2 h 1 ) h 0 0 3! h 2 1! (h 2 + h 3! 3 ) 2! (h 3! 1 +2 h 2 ) 3! (h 0 +3 h 1 ) h0 4! h 2 1! (h 3 + h 4! 4 ) 2! (h 4! 2 +2 h 3 ) 3! (h 4! 1 +3 h 2 ) 4! (h 0 +4 h 1 ) 11 / 28
r 1,0 r 1,1 0 0 0 r 2,0 r 2,1 r 2,2 0 0 r 3,0 r 3,1 r 3,2 r 3,3 0 r 4,0 r 4,1 r 4,2 r 4,3 r 4,4 r 5,0 r 5,1 r 5,2 r 5,3 r 5,4 = r 0,0 0 0 0 0 r 1,0 r 1,1 0 0 0 r 2,0 r 2,1 r 2,2 0 0 r 3,0 r 3,1 r 3,2 r 3,3 0 r 4,0 r 4,1 r 4,2 r 4,3 r 4,4 h 0 h0 0 0 0 1! h 1 1! (h 0 + h 1 ) h0 0 0 2! h 2 1! (h 2! 1 + h 2 ) 2! (h 0 +2 h 1 ) h 0 0 3! h 2 1! (h 2 + h 3! 3 ) 2! (h 3! 1 +2 h 2 ) 3! (h 0 +3 h 1 ) h0 4! h 2 1! (h 3 + h 4! 4 ) 2! (h 4! 2 +2 h 3 ) 3! (h 4! 1 +3 h 2 ) 4! (h 0 +4 h 1 ) 11 / 28
Theorem 2 (Vertical formula) An i.l.t. matrix R =[r n,k ], r 0,0 =1,isE(g, f ) iff there exists a vertical pair {v n ;ṽ n } n 0 of sequences s.t. r n,n =ṽ0 n 0; n k n! r n,k 1 =ṽ 0 r n+1,k + (n j)! (v j +(n j)ṽ j+1 ) r n j,k, (n k 1) j=0 where V R = n 0 v nz n = g gf and Ṽ R = n 0 ṽnz n = 1 f. 12 / 28
r 1,0 r 1,1 0 0 0 r 2,0 r 2,1 r 2,2 0 0 r 3,0 r 3,1 r 3,2 r 3,3 0 r 4,0 r 4,1 r 4,2 r 4,3 r 4,4 r 5,0 r 5,1 r 5,2 r 5,3 r 5,4 1! 1! (v 0 +ṽ 1 ) ṽ 0 0 0 0 2! 1! (v 2! 1 +ṽ 2 ) 2! (v 0 +2ṽ 1 ) ṽ 0 0 0 3! = 1! (v 3! 2 +ṽ 3 ) 2! (v 3! 1 +2ṽ 2 ) 3! (v 0 +3ṽ 1 ) ṽ 0 0 4! 1! (v 4! 3 +ṽ 4 ) 2! (v 4! 2 +2ṽ 3 ) 3! (v 4! 1 +3ṽ 2 ) 4! (v 0 +4ṽ 1 ) ṽ 0 5! 1! (v 5! 4 +ṽ 5 ) 2! (v 5! 3 +2ṽ 4 ) 3! (v 5! 2 +3ṽ 3 ) 4! (v 5! 1 +4ṽ 1 ) 5! (v 0 +5ṽ 1 ) r 1,1 0 0 0 0 r 2,1 r 2,2 0 0 0 r 3,1 r 3,2 r 3,3 0 0 r 4,1 r 4,2 r 4,3 r 4,4 0 r 5,1 r 5,2 r 5,3 r 5,4 r 5,5 13 / 28
r 1,0 r 1,1 0 0 0 r 2,0 r 2,1 r 2,2 0 0 r 3,0 r 3,1 r 3,2 r 3,3 0 r 4,0 r 4,1 r 4,2 r 4,3 r 4,4 r 5,0 r 5,1 r 5,2 r 5,3 r 5,4 1! 1! (v 0 +ṽ 1 ) ṽ 0 0 0 0 2! 1! (v 2! 1 +ṽ 2 ) 2! (v 0 +2ṽ 1 ) ṽ 0 0 0 3! = 1! (v 3! 2 +ṽ 3 ) 2! (v 3! 1 +2ṽ 2 ) 3! (v 0 +3ṽ 1 ) ṽ 0 0 4! 1! (v 4! 3 +ṽ 4 ) 2! (v 4! 2 +2ṽ 3 ) 3! (v 4! 1 +3ṽ 2 ) 4! (v 0 +4ṽ 1 ) ṽ 0 5! 1! (v 5! 4 +ṽ 5 ) 2! (v 5! 3 +2ṽ 4 ) 3! (v 5! 2 +3ṽ 3 ) 4! (v 5! 1 +4ṽ 1 ) 5! (v 0 +5ṽ 1 ) r 1,1 0 0 0 0 r 2,1 r 2,2 0 0 0 r 3,1 r 3,2 r 3,3 0 0 r 4,1 r 4,2 r 4,3 r 4,4 0 r 5,1 r 5,2 r 5,3 r 5,4 r 5,5 14 / 28
r 1,0 r 1,1 0 0 0 r 2,0 r 2,1 r 2,2 0 0 r 3,0 r 3,1 r 3,2 r 3,3 0 r 4,0 r 4,1 r 4,2 r 4,3 r 4,4 r 5,0 r 5,1 r 5,2 r 5,3 r 5,4 1! 1! (v 0 +ṽ 1 ) ṽ 0 0 0 0 2! 1! (v 2! 1 +ṽ 2 ) 2! (v 0 +2ṽ 1 ) ṽ 0 0 0 3! = 1! (v 3! 2 +ṽ 3 ) 2! (v 3! 1 +2ṽ 2 ) 3! (v 0 +3ṽ 1 ) ṽ 0 0 4! 1! (v 4! 3 +ṽ 4 ) 2! (v 4! 2 +2ṽ 3 ) 3! (v 4! 1 +3ṽ 2 ) 4! (v 0 +4ṽ 1 ) ṽ 0 5! 1! (v 5! 4 +ṽ 5 ) 2! (v 5! 3 +2ṽ 4 ) 3! (v 5! 2 +3ṽ 3 ) 4! (v 5! 1 +4ṽ 1 ) 5! (v 0 +5ṽ 1 ) r 1,1 0 0 0 0 r 2,1 r 2,2 0 0 0 r 3,1 r 3,2 r 3,3 0 0 r 4,1 r 4,2 r 4,3 r 4,4 0 r 5,1 r 5,2 r 5,3 r 5,4 r 5,5 14 / 28
Corollary 3 Let R = E(g, f )bee-riordan matrix with generating functions H R and H R for a horizontal pair {h n ; h n }. Then the generating functions V R and Ṽ R for a vertical pair {v n ;ṽ n } are given by V R = H R(f ) H R (f ) and Ṽ R = 1 H R (f ). Corollary 4 Let R be an e-riordan matrix with the horizontal pair {h n ; h n } n 0 and the vertical pair {v n ;ṽ n } n 0.ThenT = R 1 iff the horizontal pair of T is {v n ;ṽ n } n 0 and the vertical pair of T is {h n ; h n } n 0. Corollary 5 An e-riordan matrix R is an involution iff its horizontal pair and vertical pair are same. 15 / 28
Example 1. Consider S 2 = E(1, e z 1) = 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 3 1 0 0 0 1 7 6 1 0 0 1 15 25 10 1 Let g =1andf = e z 1. Since by f =ln(1+z) we obtain H S2 = g ( f ) g( f ) =0 and H S2 = f ( f )=1+z,. from the Theorem 1 the horizontal pair of S 2 is {0,...;1, 1, 0,...}, i.e. S(n +1, k) =S(n, k 1) + ks(n, k). 16 / 28
Since by Corollary 3 we obtain V S2 = H S 2 (f ) H S2 (f ) =0 and Ṽ S2 = 1 H S2 (f ) = e z = ( 1) n 1 n! z n, n 0 the vertical pair of S 2 is { 0,...; 1 0!, 1 1!, 1 2!,...},i.e. S(n, k 1) = n k+1 j=0 ( 1) j ( n j ) S(n j +1, k). 17 / 28
Let S 1 = S 1 2 = E(1, ln(1 + z)) = 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 2 3 1 0 0 0 6 11 6 1 0 0 24 50 35 10 1. From Corollary 4, the horizontal pair of S 1 is { 0,...; 1 0!, 1 1!, 1 2!,...} and the vertical pairs of S 1 is {0,...;1, 1, 0,...}, i.e. and s(n +1, k) = ( ) k + j 1 ( 1) j s(n, k + j 1) j n k+1 j=0 s(n, k 1) = s(n +1, k)+ns(n, k). 18 / 28
E(g, f ) Horizontal and vertical pairs Comments {h n} = {r( 1) n+1 (n +1)} ( ( ) r E 1 2, 1+ ) { h n} = { {( 1) n } } 1+2z 1+2z {v n} = r, r, r 3!! 5!!, r { 2! 3!,... } {ṽ n} = 1, 1, 1, 3!! 3!, 5!! { 4!,... } {h n} = r, r, r 3!! 5!!, r ( ) 2! 3!,... E (1 + z) r, z + z2 { h n} = {1, 1, 1, 3!! 2 3!, 5!! 4!,...} {v n} = {r( 1) n+1 (n +1)} {ṽ n} = {( 1) n } } {h n} = { r ( m)n ( ( ) r ) { n! E 1 m, ln(1 + mz) m 1 { h n} = ( m) n } 1+mz n! {v n} = {r, 0, 0,...} {ṽ n} = {1, m, 0...} {h n} = {r, 0, 0,...} ( ) { h n} = {1, m, 0...} E e rz, emz 1 } m {v n} = { r ( m)n { n! ( m) n } {ṽ n} = n! ( ( ) 2r ) {h n} = {2r, 2rm, 0,...} E 1 m, z { h n} = { 1, 2m, m 2, 0,...} 1 mz 1 mz {v n} = {2r, 2rm, 0,...} {ṽ n} = { 1, 2m, m 2, 0,...} r-bessel numbers of the 1st kind r-bessel numbers of the 2nd kind r-whiteny numbers of the 1st kind r-whiteny numbers of the 2nd kind r-whiteny-lah numbers 19 / 28
3. Applications 20 / 28
3.1 Determinant of the Hessenberg matrix For a vertical pair {v n ;ṽ n } n 0 of an e-riordan matrix E(g, f ), let v 0 ṽ 0 0 0 v 1 v 0 +ṽ 1 2ṽ 0 0 v 2 v 1 +ṽ 2 v 0 +2ṽ 1 0 V n = and.... v n 2 v n 3 +ṽ n 3 v n 4 +2ṽ n 3 (n 1)ṽ 0 v n 1 v n 2 +ṽ n 1 v n 3 +2ṽ n 2 v 0 +(n 1)ṽ 1 ṽ 1 ṽ 0 0 0 ṽ 2 ṽ 1 ṽ 0 0 ṽ 3 ṽ 2 ṽ 1 0 W n = n!.... 0 ṽ n 1 ṽ n 2 ṽ n 2 ṽ 0 ṽ n ṽ n 1 ṽ n 2 ṽ 1 Remark. The matrix D n V n Dn 1 is the n n leading principle matrix of Stieltjes transform of E(g, f ) 1 where D n = diag(0!, 1!,...,(n 1)!), i.e., (E(g, f )E(g, f ) 1 ) n = D nv n D 1 n. 21 / 28
Theorem 6 Let R = E(g, f )beane-riordan matrix with a vertical pair {v n ;ṽ n } n 0. Then g =1+ n 1 g n z n n! and f = n 1 f n z n n! can be determined by g n = ( 1 ) n ( det(v n ) and f n+1 = 1 ) n+1 det(w n ) ṽ 0 ṽ 0 where det(v 0 )=det(w 0 ) = 1. In particular, if f = zg then det(v n )= 1 n +1 det(w n). 22 / 28
Example 2. Let g = C and f = zc where C = 1 1 4z 2 is the ordinary generating function of the Catalan numbers C n = n+1( 1 2n ) n.since V R = g gf = C = n 0 C n z n and ṼR = 1 f = 1 4z = n 0 1 1 2n ( ) 2n z n, n ByTheorem6,( 1) n n!c n = det(v n )= 1 n+1 det(w n)wherev n = C n and ṽ n = 1 ( 2n ) 1 2n n. For instance, when n =4 4!C 4 = det 1 1 0 0 1 3 2 0 2 3 5 3 5 6 5 7 = 4! 5 det 2 1 0 0 2 2 1 0 4 2 2 1 10 4 2 2. 23 / 28
3.2 d-orthogonality of Sheffer polynomials Definition (P. Maroni in 1989) Let d be an arbitrary positive integer. The PS {P n (x)} Pis a d- orthogonal if there exists a d-dimensional vector of linear functionals, U =(u 0,...,u d 1 ) T, such that { uk, P r (x)p n (x) =0 if r > nd + k, (1) u k, P n (x)p nd+k (x) 0 if n N for each integer k {0, 1,...,d 1}. The orthogonality conditions (1) are equivalent to satisfying the following (d + 2)-term recurrence relation in the monic form: P n+1 (x) =(x β n )P n (x) d λ n,n j P n j (x), λ n,n d 0 j=1 where P 0 (x) =1andP n (x) =0ifn < 0. 24 / 28
Let A n be the n n Hessenberg matrix of the form A n = β 0 1 0 0 λ 1,0 β 1...... λ 2,0 λ 2,1........... λ 3,1............. λ d,0................. 0 λ d+1,1............... 0.................... 1 0 0 λ n 1,n 1 d λ n 1,n 3 λ n 1,n 2 β n 1. Then it is known that P n (x) =det(xi n A n ) for n 1andP 0 (x) =1iff the sequence {P n (x)} n 0 is monic d-orthogonal. 25 / 28
Using the Horizontal formula, we obtain that {P n (x)} is Sheffer for (g, f ) iff there exists a horizontal pair {h n ; h n } s.t. P 0 (x) =1 and P n+1 (x) = n (h j + h j x) d j dx j P n(x) (n 0) j=0 where H R = g ( f ) and H g( f ) R = f ( f ). Similarly, using the vertical formula we obtain that {P n (x)} is Sheffer for (g, f ) iff there exists a vertical pair {v n ;ṽ n } s.t. P 0 (x) =1, ṽ 0 P 1 (x) =x v 0 ṽ 0 P n+1 (x) =(x v 0 nṽ 1 )P n (x) n j=1 n! (n j)! (v j +(n j)ṽ j+1 )P n j (x) where V R = g gf and Ṽ R = 1 f. In this section, we only consider monic polynomials. From now on, let h 0 =ṽ 0 =1. 26 / 28
Theorem 7 Let the sequence {P n (x)} n 0 be Sheffer with a vertical pair {v n ;ṽ n }. Then {P n (x)} n 0 is d-orthogonal iff v d mṽ d+1 and v d+i =ṽ d+i+1 = 0 for all i 1andm 0. Corollary 8 Let the sequence {P n (x)} n 0 be Sheffer for (g, f ) with a vertical pair {v n ;ṽ n }.Then{P n (x)} is d-orthogonal iff there exist g and f with g(0) = 1 and f (0) = 0 such that VR ln(g) = dz and f = dz Ṽ R 1ṼR where V R = d n 0 v nz n and Ṽ R = d+1 n 0 ṽnz n. 27 / 28
1. 감사합니다. 2. Grazie. 3. Thank you. 4. Gracias.