A344678 Coefficients for normal ordering of (x + D)^n and for the unsigned, probabilist's (or Chebyshev) Hermite polynomials H_n(x+y).
1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 3, 3, 1, 1, 4, 6, 6, 12, 4, 3, 6, 1, 1, 5, 10, 10, 30, 10, 15, 30, 5, 15, 10, 1, 1, 6, 15, 15, 60, 20, 45, 90, 15, 90, 60, 6, 15, 45, 15, 1, 1, 7, 21, 21, 105, 35, 105, 210, 35, 315, 210, 21, 105, 315, 105, 7, 105, 105, 21, 1
Offset: 0
Examples
(x + D)^0 = 1, (x + D)^1 = x + D, (x + D)^2 = x^2 + 2 x D + 1 + D^2, (x + D)^3 = x^3 + 3 x^2 D + 3 x + 3 x D^2 + 3 D + D^3, (x + D)^4 = x^4 + 4 x^3 D + 6 x^2 + 6 x^2 D^2 + 12 x D + 4 x D^3 + 3 + 6 D^2 + D^4. (x + D)^5 = x^5 + 5 x^4 D + 10 x^3 + 10 x^3 D^2 + 30 x^2 D + 10 x^2 D^3 + 15 x + 30 x D^2 + 5 x D^4 + 15 D + 10 D^3 + D^5 H_6(x + y) = x^6 + 6 x^5 y + 15 x^4 + 15 x^4 y^2 + 60 x^3 y + 20 x^3 y^3 + 45 x^2 + 90 x^2 y^2 + 15 x^2 y^4 + 90 x y + 60 x y^3 + 6 x y^5 + 15 + 45 y^2 + 15 y^4 + y^6 H_7(x + y) = x^7 + 7 x^6 y + 21 x^5 + 21 x^5 y^2 + 105 x^4 y + 35 x^4 y^3 + 105 x^3 + 210 x^3 y^2 + 35 x^3 y^4 + 315 x^2 y + 210 x^2 y^3 + 21 x^2 y^5 + 105 x + 315 x y^2 + 105 x y^4 + 7 x y^6 + 105 y + 105 y^3 + 21 y^5 + y^7
Links
- P. Blasiak and P. Flajolet, Combinatorial Models of Creation-Annihilation, arXiv:1010.0354v3 [math.CO], 2011.
- A. Varvak, Rook numbers and the normal ordering problem, arXiv:math/0402376v2 [math.CO], 2004
Crossrefs
Programs
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Mathematica
Last /@ CoefficientRules[#, {x, y}] & /@ Table[Simplify[(-y)^n (-2)^(-n/2) HermiteH[n, (x + 1/y)/Sqrt[-2]]], {n, 0, 7}] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)
Formula
The bivariate e.g.f. is e^{t^2/2} e^{t(x + y)} = Sum_{n >= 0} H_n(x+y) t^n/n! = e^{t H.(x+y)} = e^{t (x + H.(y))}, as described below.
The coefficient of x^k D^m is n! h_{n-k-m} / [(n-k-m)! k! m!] with 0 <= k,m <= n and (k+m) <= n with h_n, as defined in the comments, aerated A001147.
Row lengths, r(n): 1, 2, 4, 6, 9, 12, 16, 20, 25, ... A002620(n).
Row sums: A005425 = 1, 2, 5, 14, 43, 142, ... .
The recursion H_{n+1}(x+y) = (x+y) H_n(x+y) + n H_{n-1}(x+y) follows from the differential raising and lowering operations of the Hermite polynomials.
The Baker-Campbell-Hausdorff-Dynkin expansion leads to the disentangling relation e^{t (x + D)} = e^{t^2/2} e^{tx} e^{tD} from which the formula above for the coefficients may be derived via differentiation with respect to t.
The row bivariate polynomials P_n(x,y) with y a commutative analog of D, or L, have the e.g.f. e^{-xy} e^{t(x + D)} e^{xy} = e^{-xy} e^{t^2/2} e^{tx} e{tD} e^{xy} = e^{t^2/2} e^{t(x + y)} = e^{t(h. + x + y)} = e^{t (x + H.(y))} = e^{t H.(x +y)}, so P_n(x,y) = H_n(x + y) = (x + H.(y))^n, the Hermite polynomials mentioned in the comments along with the umbral composition. The row sums are H_n(2), listed in A005425. For example, P_3(x,y) = (x + H.(y))^3 = x^3 H_0(y) + 3 x^2 H_1(y) + 3 x H_2(y) + H_3(y) = H_3(x+y) = (x+y)^3 + 3(x+y) = x^3 + 3 x^2 y + 3 x + 3 x y^2 + 3 y + y^2.
Alternatively, P_n(x,y) = H_n(x+y) = (z + d/dz)^n 1 with z replaced by (x+y) after the repeated differentiations since (x + D)^n 1 = H_n(x).
With initial index 1, the lengths r(n) of the rows of nonzero coefficients are the same as those for the polynomials given by 1 + (x+y)^2 + (x+y)^4 + ... + (x+y)^n for n even and for those for (x+y)^1 + (x+y)^3 + ... + (x+y)^n for n odd since the Hermite polynomials are even or odd polynomials. Consequently, r(n)= O(n) = 1 + 2 + 4 + ... n for n odd and r(n) = E(n) = 2 + 4 + ... + n for n even, so O(n) = ((n+1)/2)^2 and E(n) = (n/2)((n/2)+1) = n(n+2)/4 = 2 T(n/2) where T(k) are the triangular numbers defined by T(k) = 0 + 1 + 2 + 3 + ... + k = A000217(k). E(n) corresponds to A002378. Additionally, r(n) + r(n+1) = 1 + 2 + 3 + ... + n+1 = T(n+1).
From Tom Copeland, May 31 2021: (Start)
e^{D^2/2} = e^{h.D}, so e^{D^2/2} x^n = e^{h. D} x^n = (h. + x)^n = H_n(x) and e^{D^2/2} (x+y)^n = e^{h. D} (x+y)^n = (h. + x + y)^n = H_n(x+y).
From the Appell Sheffer polynomial calculus, the umbral compositional inverse of the sequence H_n(x+y), i.e., the sequence HI_n(x+y) such that H_n(HI.(x+y)) = (h. + HI.(x+y))^n = (h. + hi. + x + y)^n = (x+y)^n, is determined by e^{-t^2/2} = e^{hi. t}, so hi_n = -h_n and HI_n(x+y) = (-h. + x + y)^n = (-1)^n (h. - x - y)^n = (-1)^n H_n(-(x+y)). Then H_n(-H.(-(x+y))) = (x+y)^n.
In addition, HI_n(x) = (x - D) HI_{n-1}(x) = (x - D)^n 1 = e^{-D^2/2} x^n = e^{hi. D} x^n = e^{-h. D} x^n with the e.g.f. e^{-t^2/2} e^{xt}.
The umbral compositional inversion property follows from x^n = e^{-D^2/2} e^{D^2/2} x^n = e^{-D^2/2} H_n(x) = H_n(HI.(x)) = e^{D^2/2} e^{-D^2/2} x^n = HI_n(H.(x)). (End)
The umbral relations above reveal that H_n(x+y) = (h. + x + y)^n = Sum_{k = 0..n} binomial(n,k) h_k (x+y)^{n-k}, which gives, e.g., for n = 3, H_3(x+y) = h_0 * (x+y)^3 + 3 h_1 * (x+y)^2 + 3 h_2 * (x+y) + h_3 = (x+y)^3 + 3 (x+y), the n-th through 0th rows of the Pascal matrix embedded within the n-th row of the Pascal matrix modulated by h_k. - Tom Copeland, Jun 01 2021
Varvak gives the coefficients of x^(n-m-k) D^{m-k} as n! / ( 2^k k! (n-k-m)! (m-k)! ), referring to them as the Weyl binomial coefficients, and derives them from rook numbers on Ferrers boards. (No mention of Hermite polynomials nor matchings on simplices are made.) Another combinatorial model and equivalent formula are presented in Blasiak and Flajolet (p. 16). References to much earlier work are given in both papers. - Tom Copeland, Jun 03 2021
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