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The binomial Sheffer sequence of Bell / Touchard / exponential polynomials $\phi_n (x) $, whose coefficients are the Stirling numbers of the second kind, have the representation

$(RL)^n=\phi_n (:RL:) $

where $R $ and $L $ are the raising and lowering operators of any sequence of Sheffer polynomials and $:RL:^n=R^n \; L^n$ by definition (a notational convenience).

Are there other polynomial functions of the ladder operators $K(L,R)$ such that

$ K^n(L,R) = u_n(B.(L,R))$

for $u_n(x)$ a Sheffer sequence and $(B.(L,R))^n = B_n(L,R)$ a sequence of operator polynomials?

(Other than for the trivial case $u_n(x) = x^n$.)

Note that $R^nL^n= (RL)_n = (RL)!/(RL-n)!$, the falling factorial polynomials, whose coefficients are the signed Stirling numbers of the first kind.

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  • $\begingroup$ For those interested in this question, I'll probably delete it soon after I incorparate it into a blog post on general umbral derivatives. $\endgroup$ – Tom Copeland Oct 28 '16 at 23:13
  • $\begingroup$ Blog post Compositional Inverse Operators and Sheffer Sequences tcjpn.wordpress.com/2016/11/06/… $\endgroup$ – Tom Copeland Nov 7 '16 at 4:57
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Well, upon further reflection, the answer is that any pair of Sheffer sequences that are an umbral inverse pair suffice to construct such an arrangement. Let $u_n (x)$ and $v_n (x)$ be such a pair, i.e.,

$u_n (v.(x)) = x^n=v_n(u.(x))$.

Then

$K^n = u_n(v.(K))$ and $B_n(K) = v_n (K) $.

So, the Stirling pair is not unique.

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  • $\begingroup$ The uniqueness of the Stirling pair is in the reduction of powers of the Euler / state number op into linear sums of the homogeneous normal-ordered ops. $\endgroup$ – Tom Copeland Oct 28 '16 at 21:42

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