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COMMENTS
| Matrix T begins
0;
1,0;
0,2,0;
0,0,3,0;
0,0,0,4,0;
Let M(t) = exp(t*T) = limit [1 + t*T/n]^n as n tends to infinity.
Pascal matrix = [ binomial(n,k) ] = M(1) = exp(T), truncating the series gives the n X n submatrices.
Inverse Pascal matrix = M(-1) = exp(-T) = matrix for inverse binomial transform.
A(j) = T^j / j! equals the matrix [bin(n,k) * delta(n-k-j)] where delta(n) = 1 if n=0 and vanishes otherwise (Kronecker delta); i.e. A(j) is a matrix with all the terms 0 except for the j-th lower (or main for j=0) diagonal which equals that of the Pascal triangle. Hence the A(j)'s form a linearly independent basis for all matrices of the form [binomial(n,k) d(n-k)] which include as a subset the invertible associated matrices of the list partition transform (LPT) of A133314.
For sequences with b(0) = 1, umbrally,
M[b(.)] = exp(b(.)*T) = [ binomial(n,k) * b(n-k) ] = matrices associated to b by LPT.
[M[b(.)]]^(-1) = exp(c(.)*T) = [ binomial(n,k) * c(n-k) ] = matrices associated to c, where c = LPT(b) . Or,
[M[b(.)]]^(-1) = exp[LPT(b(.))*T] = LPT[M(b(.))] = M[LPT(b(.))] = M[c(.)] .
This is related to xDx, the iterated Laguerre transform and the general Euler transformation of a sequence through the comments in A132013 and A132014 and the relation [sum(k=0,...,n) binomial(n,k) * b(n-k) * d(k)] = M(b)*d, (n-th term). See also A132382.
If b(n,x) is a binomial type Sheffer sequence, then M[b(.,x)]*s(y) = s(x+y) when s(y) = (s(0,y),s(1,y),s(2,y),...) is an array for a Sheffer sequence with the same delta operator as b(n,x) and [M[b(.,x)]]^(-1) is given by the formulae above with b(n) replaced by b(n,x) as b(0,x)=1 for a binomial type Sheffer sequence.
T = I - A132013 and conversely A132013 = I - T, which is the matrix representation for the iterated mixed order Laguerre transform characterized in A132013 (and A132014).
(I-T)^m generates the group [A132013]^m for m= 0,1,2,.. discussed in A132014.
The inverse is 1/(I-T) = I+T+T^2+T^3+... = [A132013]^(-1) = A094587 with the associated sequence (0!,1!,2!,3!,...) under the LPT.
And 1/(I-T)^2 = I+2*T+3*T^2+4*T^3+... = [A132013]^(-2) = A132159 with the associated sequence (1!,2!,3!,4!,...) under the LPT.
The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or e.g.f.'s EA(x) and EB(x).
1) b(0) = 0, b(n) = n * a(n-1),
2) B(x) = xDx A(x)
3) B(x) = x * Lag(1,-:xD:) A(x)
4) EB(x) = x * EA(x) where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j and Lag(n,x) is the Laguerre polynomial.
So the exponentiated operator can be characterized as
5) exp(t*T) A(x) = exp(t*xDx) A(x) = [sum(n=0,1,...) (t*x)^n * Lag(n,-:xD:)] A(x) = [exp{[t*u/(1-t*u)]*:xD:} / (1-t*u) ] A(x) (eval. at u=x) = A[x/(1-t*x)]/(1-t*x), a generalized Euler transformation for an o.g.f.,
6) exp(t*T) EA(x) = exp(t*x)*EA(x) = exp[(t+a(.))*x], gen. Euler trf. for an e.g.f.
7) exp(t*T) * a = M(t) * a = [sum(k=0,...,n) binomial(n,k) * t^(n-k) * a(k)] .
The umbral extension of formulae 5, 6 and 7 gives formally
8) exp[c(.)*T] A(x) = exp(c(.)*xDx) A(x) = [sum(n=0,1,...) (c(.)*x)^n * Lag(n,-:xD:)] A(x) = [exp{[c(.)*u/(1-c(.)*u)]*:xD:} / (1-c(.)*u) ] A(x) (eval. at u=x) = A[x/(1-c(.)*x)]/(1-c(.)*x), where the umbral evaluation should be applied only after a power series in c is obtained,
9) exp[c(.)*T] EA(x) = exp(c(.)*x)*EA(x) = exp[(c(.)+a(.))*x]
10) exp[c(.)*T] * a = M[c(.)] * a = [sum(k=0,...,n) binomial(n,k) * c(n-k) * a(k)] .
The n X n principal submatrix of T is nilpotent, in particular, [Tsub_n]^(n+1) = 0, n=0,1,2,3,....
Note (xDx)^n = x^n D^n x^n = x^n n! (:Dx:)^n/n! = x^n n! Lag(n,-:xD:) .
The operator xDx is an important, classical operator explored by among others Dattoli, Al-Salam, Carlitz and Stokes and even earlier investigators.
Contribution from Karol A. Penson (penson(AT)lptl.jussieu.fr), Sep 15 2009: For a recent treatment of xDx, DxD and more general operators see the paper "Laguerre-type derivatives: Dobinski relations and combinatorial identities"
See Copeland's link for generalized Laguerre functions and connection to fractional differ-integrals in exercises through (:Dx:)^a/a!=(D^a x^a)/a! - Tom Copeland, Nov 17 2011
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