OFFSET
1,6
COMMENTS
This entry and the references differ slightly among themselves in the order of coefficients for higher order terms. Table on p. 167 of Comtet has at least one index error.
Let F[FI(x)] = FI[F(x)] = x (i.e., F and FI are a compositional inverse pair) about x=0 with F(0)=FI(0)=0. Define f(x) = 1/[dFI(x)/dx], then for w(x) analytic about the origin, exp[t f(x)d/dx] w(x) = w{F[t+FI(x)]} = q(t,x) with q{t,F[s+FI(x)]} = q(t+s,x). See A145271 for w(x)=x and note that A145271 is embedded in A139605. E.g.f. of the binomial Sheffer sequence associated to F(x) is exp[x f(z)d/dz] exp(t*z)= exp{t*F[x+FI(z)]} evaluated at z=0. - Tom Copeland, Oct 19 2011
dq(t,x)/dt - f(x)dq(t,x)/dx = 0, so (1,-f(x)) gives the components of a vector orthogonal to the gradient of q and therefore tangent to the contour of q at (t,x). - Tom Copeland, Oct 26 2011
The formula exp[t f(x)d/dx] w(x)= w{F[t+FI(x)]} above is implicit in the chain rule formulas on pages 10 and 12 of Mathemagical Forests. Another derivation is alluded to in the Dattoli reference in A080635 (repeated below). - Tom Copeland, Nov 28 2011
Let f(x) and g(x) be two infinitely differential functions. Denote f_0 = f(x), f_1 = df_0/dx, f_2 = df_1/dx, and so on. Same with g_0 = g(x). Define the linear operator L(u(x)) = g(x) * du(x)/dx. Denote F_1 = L(f(x)), F_2 = L(F_1), and so on. When n>0, F_n is a linear combination of f_1, ..., f_n where each f_k is multiplied by a homogeneous polynomial (P(n,k)) of degree n in g_0, ..., g_{n-1}. The triangle of the sum of the coefficients of P(n,k) is A130534. - Michael Somos, Mar 23 2014
Triangle with row n length A000070(n+1) and row n consists of the coefficients: P(n,1), ..., P(n,n). The order of coefficients in P(n,k) is Abramowitz and Stegun order for partitions of n-k with parts g_1, ..., g_{n-k}. - Michael Somos, Mar 23 2014
A130534(n,k) gives the number of rooted trees with (k+1) trunks that are associated with D^(k+1) in the forest of "naturally grown" rooted trees with (n+2) nodes, or vertices, that are associated with R^(n+1) in the example below. Cf. MF link. - Tom Copeland, Mar 23 2014
These partition polynomials appeared in 1823 in a dissertation by Heinrich Scherk. See p. 76 of Blasiak and Flajolet. - Tom Copeland, Jul 14 2021
Schröder made use of iterated derivatives, or iterated infinitesimal generators (IGs), ((1/f') D)^n in his investigations of functional iteration, or iterated functional composition, related to extensions of Newton's method of finding zeros of an equation. He constructs the series, in terms of the IGs, for finv[t+f(z)] evaluated at t = -f(z), giving z_1 = finv(0) although he doesn't present his analysis this way. - Tom Copeland, Jul 19 2021
REFERENCES
F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-like Structures, (1997), Cambridge University Press, p. 386.
H. Davis, The Theory of Linear Operators, (1936), The Principia Press, p. 13.
T. Mansour and M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, Chapman and Hall/CRC, 2015.
LINKS
P. Blasiak and P. Flajolet, Combinatorial Models of Creation-Annihilation, arXiv:1010.0354 [math.CO], 2011.
Louis Comtet, Une formule explicite pour les puissances successives de l'opérateur de dérivation de Lie, Comptes Rendus Acad. Sc. Paris, Serie A tome 276 (1973), pp. 165 - 168.
Tom Copeland, Mathemagical Forests v2, 2008.
Tom Copeland, Infinigens, the Pascal Triangle, and the Witt and Virasoro Algebras, 2012.
Tom Copeland, Pre-Lie algebras, Cayley's analytic trees, and mathemagical forests, 2018.
G. Datolli, P. L. Ottaviani, A. Torre and L. Vazquez, Evolution operator equations: integration with algebraic and finite differences methods.[...], La Rivista del Nuovo Cimento 20,2 (1997) 1-133. eq. (I.2.18).
D. Grinberg, Commutators, matrices, and an identity of Copeland, arXiv:1908.09179 [math.RA], 2019.
Guo-Niu Han and Shi-mei Ma, Eulerian polynomials and the g-indices of Young tableaux, Proc. Amer. Math. Soc., (2023).
Kentaro Ihara, Derivations and automorphisms on non-commutative power series, Journal of Pure and Applied Algebra, Volume 216, Issue 1, January 2012, Pages 192-201.
T. Mansour and M. Shork, The generalized Touchard polynomials revisited, Journal of Applied Mathematics and Computation, Volume 219, Issue 19, June 2013, Pages 9978-9991.
MathOverflow, Important formulas in combinatorics: The combinatorics underlying iterated derivatives (infinitesimal Lie generators) for compositional inversion and flow maps for vector fields, answer by Tom Copeland to an MO question posed by Gil Kalai, 2015.
MathOverflow, Expansions of iterated, or nested, derivatives, or vectors--conjectured matrix computation, a MO question posed by Tom Copeland, answered by Darij Grinberg, 2019.
Mathoverflow, Formula for n-th iteration of dx/dt=B(x), a question on MathOverflow posed by the user resolvent and answered by Tom Copeland, 2021.
H. Scherk, De evolvenda functione (yd.yd.yd...ydX/dxn) disquisitiones nonnullae analyticae, PhD thesis, Berlin, 1823. Scanned copy at Göttinger Digitalisierungszentrum (GDZ).
E. Schröder, Ueber unendlich viele Algorithmen zur Auflösung der Gleichungen, Mathematische Annalen vol. 2, 317-365, 1870.
G. Stewart, On infinitely many algorithms for solving equations, 1993, (translation into English of Schröder's paper above).
FORMULA
Equivalent matrix computation: Multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by f_n = (d/dx)^n f(x) to obtain the matrix VP with VP(n,k) = binomial(n,k) f_(n-k). Then R^n = (1, 0, 0, 0,..) [VP * S]^n (1, D, D^2, ..)^T, where S is the shift matrix A129185, representing differentiation in the basis x^n/n!. Cf. A145271. - Tom Copeland, Jul 17 2016
A formula for the coefficients of this matrix is presented in Ihara, obtained from Comtet. - Tom Copeland, Mar 25 2020
Elaborating on my 2011 comments: Let exp[x F(t)] = exp[t p.(x)] be the e.g.f. for the binomial Sheffer sequence of polynomials (p.(x))^n = p_n(x). Then, evaluated at x = t = 0, the coefficient p_(n,k) = (D_x^k/k!) p_n(x) = D_t^n [F(t)]^k/k! = (f(x)D_x)^n x^k/k! = R^n x^k/k!, and so p_(n,k) is the coefficient of D^k of the operator R^n below evaluated at x=0. - Tom Copeland, May 14 2020
Per earlier comments, the action of the differentials of this entry on an exponential is exp(x g(u)D_u) e^(ut) = e^(t H^{(-1)}(H(u)+x)) with g(x) = 1/D(H(x)) and H^{(-1)}(x) the compositional inverse of H(x). With H^{(-1)}(x) = -log(1-x), the inverse about x=0 is H(x) = 1-e^(-x), giving g(x) = e^x and the resulting action e^(-t log(1-x)) = (1-x)^(-t) for u=0, an e.g.f. for the unsigned Stirling numbers of the first kind (see A008275, A048994, and A130534). Consequently, summing the coefficients of this entry over each associated derivative gives these Stirling numbers. E.g., the fifth row in the examples reduces to (1+4+1) D + (7+4) D^2 + 6 D^3 + D^4 = 6 D + 11 D^2 + 6 D^3 + D^4. The Connes-Moscovici weights A139002 are a refinement of this entry. - Tom Copeland, Jul 14 2021
EXAMPLE
Let R = f(x)d/dx = f(x)D and (j,k) = [(d/dx)^j f(x)]^k ; then
R^0 = 1.
R^1 = (0,1)D.
R^2 = (0,1)(1,1)D + (0,2)D^2.
R^3 = [(0,1)(1,2) + (0,2)(2,1)]D + 3 (0,2)(1,1)D^2 + (0,3)D^3.
R^4 = [(0,1)(1,3) + 4 (0,2)(1,1)(2,1) + (0,3)(3,1)]D +
[7 (0,2)(1,2) + 4 (0,3)(2,1)]D^2 + 6 (0,3)(1,1)D^3 + (0,4)D^4. - Tom Copeland, Jun 12 2008
R^5 = [(0,1)(1,4) + 11 (0,2)(1,2)(2,2) + 4 (0,3)(2,2) + (0,4)(4,1) + 7 (0,3)(1,1)(3,1)]D + [15 (0,2)(1,3) + 30 (0,3)(1,1)(2,1) + 5 (0,4)(1,3)]D^2 + [25 (0,3)(1,2) + 10 (0,4)(2,1) + 25 (0,3)(1,2)]D^3 + 10 (0,4)(1,1)D^4 + (0,5)D^5. - Tom Copeland, Jul 17 2016
R^6 = [(0,1)(1,5) + 26 (0,2)(1,3)(2,1) + 34 (0,3)(1,1)(2,2) + 32 (0,3)(1,2)(3,1) + 11 (0,4)(1,1)(4,1) + 15 (0,4)(2,1)(3,1) + (0,5)(1,5)]D + [31 (0,2)(1,4) + 146 (0,3)(1,2)(2,1) + 57 (0,4)(1,1)(3,1) + 34 (0,4)(2,2) + 6 (0,5)(4,1)]D^2 + [90 (0,3)(1,3) + 120 (0,4)(1,1)(2,1) + 15 (0,5)(3,1)]D^3 + [65 (0,4)(1,2) + 20 (0,5)(2,1)]D^4 + 15 (0,5)(1,1)D^5 + (0,6)D^6. - Tom Copeland, Jul 17 2016
------------
F_1 = (1*g_0) * f_1, F_2 = (1*g_0*g_1) * f_1 + (1*g_0^2) * f_2, F_3 = (1*g_0*g_1^2 + 1*g_0^2*g_2) * f_1 + (3*g_0^2*g_1) * f_2 + (1*g_0^3) * f_3. - Michael Somos, Mar 23 2014
P(4,2) = 4*g0^3*g2 + 7*g0^2*g1^2. P(5,2) = 5*g0^4*g3 + 30*g0^3*g1*g2 + 15*g0^2*g1^3. - Michael Somos, Mar 23 2014
1
1 , 1
1 1 , 3 , 1
1 4 1 , 4 7 , 6 , 1
1 7 4 11 1, 5 30 15 , 10 25 , 10 , 1
1 11 15 32 34 26 1 , 6 57 34 146 31 , 15 120 90 , 20 65 , 15 , 1
MATHEMATICA
row[n_] := With[{pn = CoefficientRules[Nest[g[x] D[#, x] &, f[x], n], Derivative[#][f][x] & /@ Range[n]][[;; , 2]] /. {Derivative[k_][g][x] :> h[k], g[x] -> 1}}, Table[Coefficient[pn[[k]], Product[h[x], {x, p}]], {k, n - 1}, {p, Sort[Sort /@ IntegerPartitions[n - k]]}]~Join~{{1}}];
Table[row[n], {n, 7}] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Tom Copeland, Jun 12 2008
EXTENSIONS
Title expanded by Tom Copeland, Mar 17 2014
Sequence terms rearranged in Abramowitz and Stegun order by Michael Somos, Mar 23 2014
Title expanded by Tom Copeland, Jul 14 2021
STATUS
approved