arXiv:2107.08436v1 [math.CO] 18 Jul 2021 quence eeaigfunction generating oShebr n de,wihsae htasqec ( sequence a that states which Edrei, and Schoenberg to sT.W a htafiiesequence finite a that say We TP. is 00MSC: 2010 sttlypstv,te oi h ira array Riordan the Keywords: is so then positive, totally is ´ lafeunysequence P´olya frequency eso hti h matrix the if that show We ioso l resaenneaie nifiienonnegativ infinite An nonnegative. are orders all of minors Let Abstract rpitsbitdt LAA to submitted Preprint [email protected] ∗ mi addresses: Email author. Corresponding olwn aln[] nifiiemti scalled is matrix infinite an [8], Karlin Following R b a e nte rtro o h oa oiiiyo ira ar Riordan of positivity total the for criterion another Yet colo ahmtc n ttsis ins omlUnive Normal Jiangsu Statistics, and Mathematics of School colo ahmtclSine,Dla nvriyo Tech of University Dalian Sciences, Mathematical of School = a 0 R a , ( d 1 ira ra,Ttlypstv arx ´lafeunys P´olya frequency matrix, positive Totally array, Riordan 54,1B6 15B05 15B36, 15B48, ( a , . . . , t ) h , ( Y Wang) (Yi [email protected] t n )b ira ra,where array, Riordan a be )) , 0 . . . , o hrl,P) fisTelt matrix Toeplitz its if PF), shortly, (or [ a sP.Afnaetlcaatrzto o Fsqecsi du is sequences PF for characterization fundamental A PF. is iniMao Jianxi i n X − ≥ j 0 ] i,j a n ≥ t 0 d d d n . . . 1 0 2 = = a a, 0 h h h Ct iiMu Lili , Jax Mao), (Jianxi . . . a , 1 0 2 a a a a . . . k 1 3 2 1 0 e a , . . . , h h 0 0 γt 0 1 a a a Q Q 2 1 0 R h j 0 j b, d 0 ≥ ≥ . n a a ( ∗ 0 0 t iWang Yi , 1 0 1+ (1 sP ftecrepnigifiiese- infinite corresponding the if PF is = ) (1 [email protected] · · · oal positive totally . . a . − 0 ooy ain162,PR China P.R. 116024, Dalian nology, P α st,Xzo 216 ..China P.R. 221116, Xuzhou rsity, β j j n a . t t ≥ . ) ) n eune( sequence e . 0 , ) a, n d ≥ n 0 t n sP fadol fits if only and if PF is and o hrl,T) fits if TP), shortly, (or equence Ll Mu), (Lili h a ( n t = ) ) n ≥ 0 P uy2,2021 20, July rays scle a called is n ≥ 0 h n t n e . N where C > 0, k ∈ , αj, βj , γ ≥ 0, and j≥0(αj + βj) < +∞ (see [8, p. 412] for instance). In this case, the generating function isP called a P´olya frequency formal power series. We refer the reader to [1, 5, 9, 11, 13] for the total positivity of matrices. Our concern in this note is the total positivity of Riordan arrays. Riordan arrays play an important unifying role in enumerative combinatorics [10, 12]. n n Let d(t)= n≥0 dnt and h(t)= n≥0 hnt be two formal power series. A Riordan array, denoted byPR(d(t), h(t)), is an infiniteP matrix whose generating function of the kth column is d(t)hk(t) for k ≥ 0. Chen and Wang [4, Theorem 2.1] gave the following criterion for the total positivity of Riordan arrays. Theorem 1 ([4, Theorem 2.1]). Let R = (d(t), h(t)) be a Riordan array. If both d(t) and h(t) are P´olya frequency formal power series, then R is totally positive.
We say that R(d(t), h(t)) is proper if d0 =6 0, h0 = 0 and h1 =6 0. In this case, R(d(t), h(t)) is an infinite lower triangular matrix. It is well known that a proper Ri- ordan array R = [rn,k]n,k≥0 can be characterized by two sequences (an)n≥0 and (zn)n≥0 such that r0,0 = 1, rn+1,0 = zjrn,j, rn+1,k+1 = ajrn,k+j Xj≥0 Xj≥0 for n, k ≥ 0 (see [6, 7] for instance). Call (an)n≥0 and (zn)n≥0 the A- and Z-sequences of R respectively. Chen et al. [2, Theorem 2.1 (i)] gave the following criterion for the total positivity of Riordan arrays. Theorem 2 ([2, Theorem 2.1 (i)]). Let R be the proper Riordan array with the A- and Z-sequences (an)n≥0 and (zn)n≥0. If the product matrix
z0 a0 0 0 · · · z1 a1 a0 0 P = z a a a 2 2 1 0 . . . . . .. is totally positive, then so is R. In this note we establish a new criterion for the total positivity of Riordan arrays, which can be viewed as a dual version of Theorem 2 in a certain sense. n Theorem 3. Let R = (d(t), h(t)) be a Riordan array, where d(t)= n≥0 dnt and h(t)= n n≥0 hnt . If the Hessenberg matrix P P d0 h0 0 0 · · · d1 h1 h0 0 H = d h h h (1) 2 2 1 0 . . . . . .. is totally positive, then so is R.
2 Proof. Let R[n] be the submatrix consisting of the first n + 1 columns of R. Clearly, R is TP if and only if all submatrices R[n] are TP for n ≥ 0. So it suffices to show that R[n] is TP for all n ≥ 0. We proceed by induction on n. k Let R = [rn,k]n,k≥0. Since the generating function of the kth column of R is d(x)h (x), we have
r0,0 d0 r0,k h0 0 0 · · · r0,k−1 r1,0 d1 r1,k h1 h0 0 r1,k−1 r = d , r = h h h r 2,0 2 2,k 2 1 0 2,k−1 . . . . . . . . . . .. . for k ≥ 1. It follows that
r0,0 r0,1 · · · r0,n+1 d0 h0 0 0 · · · 1 0 0 · · · 0 r1,0 r1,1 · · · r1,n+1 d1 h1 h0 0 0 r0,0 r0,1 · · · r0,n r r · · · r = d h h h 0 r r · · · r , 2,0 2,1 2,n+1 2 2 1 0 1,0 1,1 1,n . . . . . . . . . . . . .. . . . or equivalently, 1 0 R[n +1] = H . (2) 0 R[n] The first matrix H on the right-hand side of (2) is TP by assumption, which implies that all dn are nonnegative, and the matrix R[0] is therefore TP. Assume now that the matrix 1 0 R[n] is TP for n ≥ 0. Then the second matrix on the right-hand side of (2) 0 R[n] is also TP. It is well known that the product of TP matrices is still TP by the classic Cauchy-Binet formula. Thus the matrix R[n + 1] on the left-hand side of (2) is TP. The matrix R is therefore TP by induction, and the proof is complete.
k Example 4. Consider Lucas polynomials Ln(x)= k Ln,kx defined by P Ln+1(x)= Ln(x)+ xLn−1(x) (3) with L0(x)=2 and L1(x) = 1. Lucas matrix is the lower triangular infinite matrix 2 1 1 2 1 3 1 4 2 L = [Ln,k]n,k≥0 = . 1 5 5 16 9 2 1 7 14 7 . . . ..
3 L n Let k(t)= n≥0 Ln,kt denote the generating function of the kth column of L for k ≥ 0. Clearly, L0(tP) = (2 − t)/(1 − t). On the other hand, we have Ln,k = Ln−1,k + Ln−2,k−1 for L t2 L n>k> 0 by (3). It follows that n(t)= 1−t n−1(t) for n ≥ 1. Thus L is a Riordan array:
2 − t t2 L = R , . 1 − t 1 − t
The corresponding Hessenberg matrix is
20 00 · · · 10 00 11 00 H = 11 10 , 11 11 . . . . . .. which is clearly TP, and so is L by Theorem 3. However, the total positivity of L can be followed neither from Theorem 1 since d(t)= (2 − t)/(1 − t) is not PF, nor from Theorem 2 since L is improper.
Remark 5. We can show that Theorem 3 implies Theorem 1. Consider first several important classes of proper Riordan arrays R = R(d(t), h(t)).
(i) Let h(t)= t. Then R is a Toeplitz-type Riordan array, which is precisely the Toeplitz matrix of the sequence (dn)n≥0. If d(t) is PF, then R(d(t),t) is TP. (ii) Let h(t)= td(t). Then R is a Bell-type Riordan array. In this case, the corresponding Hessenberg matrix (1) is the Toeplitz matrix of (dn)n≥0. If d(t) is PF, i.e., h(t) is PF, then R(h(t)/t, h(t)) is TP by Theorem 3.
(iii) Let d(t) = 1. Then R is a Lagrange-type Riordan array. Note that
1 0 R(1, h(t)) = . 0 R (h(t)/t, h(t))
If h(t) is PF, then R(h(t)/t, h(t)) is TP, and so is R(1, h(t)).
It is well known [10] that every proper Riordan array can be decomposed into the product of a Toeplitz-type Riordan array and a Lagrange-type Riordan array:
R(d(t), h(t)) = R(d(t),t) ·R(1, h(t)).
We conclude that if both d(t) and h(t) are PF, then R is TP. In other words, Theorem 1 follows from Theorem 3.
4 Acknowledgement
This work was supported in part by the National Natural Science Foundation of China (Grant Nos. 11771065, 11701249).
References
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