-log(1 + b(1) x + b(2) x^2 + ...) = Sum_{n>=1} F(n,b(1),...,b(n)) * x^n/n.
-d(1 + b(1) x + b(2) x^2 + ...)/dx / (1 + b(1) x + b(2) x^2 + ...) = Sum_{n>=1} F(n,b(1),...,b(n)) x^(n-1).
F(n,b(1),...,b(n)) = -n*b(n) - Sum_{k=1..n-1} b(n-k)*F(k,b(1),...,b(k)).
Umbrally, with B(x) = 1 + b(1) x + b(2) x^2 + ..., B(x) = exp[log(1-F.x)] and 1/B(x) = exp[-log(1-F.x)], establishing a connection to the e.g.f. of
A036039 and the symmetric polynomials.
The Stirling partition polynomials of the first kind St1(n,b1,b2,...,bn;-1) = IF(n,b1,b2,...,bn) (cf. the Copeland link Lagrange a la Lah, signed
A036039, and p. 184 of Airault and Bouali), i.e., the cyclic partition polynomials for the symmetric groups, and the Faber polynomials form an inverse pair for isolating the indeterminates in their definition, for example, F(3,IF(1,b1),IF(2,b1,b2)/2!,IF(3,b1,b2,b3)/3!)= b3, with bk = b(k), and IF(3,F(1,b1),F(2,b1,b2),F(3,b1,b2,b3))/3!= b3.
The polynomials specialize to F(n,t,t,...) = (1-t)^n - 1.
See Newton Identities on Wikipedia on relation between the power sum symmetric polynomials and the complete homogeneous and elementary symmetric polynomials for an expression in multinomials for the coefficients of the Faber polynomials.
(n-1)! F(n,x[1],x[2]/2!,...,x[n]/n!) = - p_n(x[1],...,x[n]), where p_n are the cumulants of
A127671 expressed in terms of the moments x[n]. -
Tom Copeland, Nov 17 2015
-(n-1)! F(n,B(1,x[1]),B(2,x[1],x[2])/2!,...,B(n,x[1],...,x[n])/n!) = x[n] provides an extraction of the indeterminates of the complete Bell partition polynomials B(n,x[1],...,x[n]) of
A036040. Conversely, IF(n,-x[1],-x[2],-x[3]/2!,...,-x[n]/(n-1)!) = B(n,x[1],...,x[n]). -
Tom Copeland, Nov 29 2015
For a square matrix M, determinant(I - x M) = exp[-Sum_{k>0} (trace(M^k) x^k / k)] = Sum_{n>0} [ P_n(-trace(M),-trace(M^2),...,-trace(M^n)) x^n/n! ] = 1 + Sum_{n>0} (d[n] x^n), where P_n(x[1],...,x[n]) are the cycle index partition polynomials of
A036039 and d[n] = P_n(-trace(M),-trace(M^2),...,-trace(M^n)) / n!. Umbrally, det(I - x M)= exp[log(1 - b. x)] = exp[P.(-b_1,..,-b_n)x] = 1 / (1-d.x), where b_k = tr(M^k). Then F(n,d[1],...,d[n]) = tr[M^n]. -
Tom Copeland, Dec 04 2015
Given f(x) = -log(g(x)) = -log(1 + b(1) x + b(2) x^2 + ...) = Sum_{n>=1} F(n,b(1),...,b(n)) * x^n/n, action on u_n = F(n,b(1),...,b(n)) with
A133932 gives the compositional inverse finv(x) of f(x), with F(1,b(1)) not equal to zero, and f(g(finv(x))) = f(e^(-x)). Note also that exp(f(x)) = 1 / g(x) = exp[Sum_{n>=1} F(n,b(1),...,b(n)) * x^n/n] implies relations among
A036040,
A133314,
A036039, and the Faber polynomials. -
Tom Copeland, Dec 16 2015
The Dress and Siebeneicher paper gives combinatorial interpretations and various relations that the Faber polynomials must satisfy for integral values of its arguments. E.g., Eqn. (1.2) p. 2 implies [2 * F(1,-1) + F(2,-1,b2) + F(4,-1,b2,b3,b4)] mod(4) = 0. This equation implies that [F(n,b1,b2,...,bn)-(-b1)^n] mod(n) = 0 for n prime. -
Tom Copeland, Feb 01 2016
With the elementary Schur polynomials S(n,a_1,a_2,...,a_n) = Lah(n,a_1,a_2,...,a_n) / n!, where Lah(n,...) are the refined Lah polynomials of
A130561, F(n,S(1,a_1),S(2,a_1,a_2),...,S(n,a_1,...,a_n)) = -n * a_n since sum_{n > 0} a_n x^n = log[sum{n >= 0} S(n,a_1,...,a_n) x^n]. Conversely, S(n,-F(1,a_1),-F(2,a_1,a_2)/2,...,-F(n,a_1,...,a_n)/n) = a_n. -
Tom Copeland, Sep 07 2016
See Corollary 3.1.3 on p. 38 of Ardila and Copeland's two MathOverflow links to relate the Faber polynomials, with arguments being the signed elementary symmetric polynomials, to the logarithm of determinants, traces of powers of an adjacency matrix, and number of walks on graphs. -
Tom Copeland, Jan 02 2017
The umbral inverse polynomials IF appear on p. 19 of Konopelchenko as partial differential operators. -
Tom Copeland, Nov 19 2018
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