A Class of Differential Operators and the Stirling Numbers

The differential operator $\displaystyle (x^{1+y} \; D)^n$ with $\displaystyle D=d/dx$ can easily be expanded in terms of the operators $\displaystyle (:xD:)^n = x^n \; D^n$ by considering its action on $\displaystyle x^s \; .$

First expand it in terms of the number operator $\displaystyle xD \; :$

$\displaystyle (x^{1+y} \; D)^n \; x^s = s(s+y)(s+2y) \cdots (s+(n-1)y) \; x^{s+ny}$

$\displaystyle = x^{ny} \; y^n \; n! \; \binom{\frac{s}{y}-1+n}{n} \; x^s = x^{ny} \; (-y)^n \; n! \; \binom{\frac{-s}{y}}{n} \; x^s$

$\displaystyle = x^{ny} \; y^n \; n! \; \binom{\frac{xD}{y}-1+n}{n} \; x^s = x^{ny} \; (-y)^n \; n! \; \binom{\frac{-xD}{y}}{n} \; x^s \; \; .$

(For more on this expansion, cf. OEIS-A094638.)

Now consider the generalized Dobinski relation for the Bell / Touchard polynomials defined operationally by $\displaystyle \phi_n(:xD:)= (xD)^n$ with $\displaystyle (:xD:)^n = x^n \; D^n \;$ in conjunction with an Euler transformation. Umbral composition of these polynomial operators with an analytic function formally gives

$\displaystyle f(\phi.(x))=e^{-x} \; f(\phi.(:xD:)) \; e^x = e^{-x} \; f(xD) \; e^x$

$\displaystyle = e^{-x} \; \sum_{n \ge 0} f(n) \frac{x^n}{n!}= e^{-x} \; e^{a.x} = e^{-(1-a.)x}=\sum_{n \ge 0} (-1)^n [\sum_{k=0}^n (-1)^k \binom{n}{k} a_k] \; \frac{x^n}{n!}$

$\displaystyle = \sum_{n \ge 0} (-1)^n \triangle_k^n a_k \; \frac{x^n}{n!} = \sum_{n \ge 0} (-1)^n \triangle_k^n f(k) \; \frac{x^n}{n!} \; .$

Summarizing,

$\displaystyle f(\phi.(x)) = \sum_{n \ge 0} (-1)^n \triangle_k^n f(k) \; \frac{x^n}{n!} \; ,$

so

$\displaystyle f(xD) = f(\phi.(:xD:)) = \sum_{n \ge 0} (-1)^n \triangle_k^n f(k) \; \frac{:xD:^n}{n!} \; .$

Note that action on $\displaystyle x^s$ of the differential op generates the Newton series interpolation

$\displaystyle f(s) = \sum_{n \ge 0} (-1)^n \binom{s}{n} \; \triangle_k^n f(k) = \triangle_n^s \; \triangle_k^n f(k) \; .$

The same result is achieved by umbral substitution of the falling factorial $\displaystyle (s)_n = \frac{s!}{(s-n)!}$ into the functional series since the falling factorials and the Bell polynomials are an umbral compositional inverse pair, i.e.,

$\displaystyle \phi_n((s).)=s^n=(\phi.(s))_n \; .$

We can derive the well-known formula for the coefficients of the Bell polynomials, i.e., the Stirling numbers of the second kind  $\displaystyle S2_{n,k} \; ,$ by choosing $\displaystyle f(x) = x^n \; ,$ then

$\displaystyle \phi_n(x) = (\phi .(x))^n = \sum_{j \ge 0} (-1)^j \triangle_k^j k^n \; \frac{x^j}{j!} = \sum_{j =0}^n S2_{n,j} \; x^j \; ,$

and we can identify

$\displaystyle S2_{n,j}=\frac{(-1)^j}{j!} \triangle_k^j \; k^n \; .$

Applying the general operadic Dobinski relation to the binomial expressions for our op at the top of the page, we obtain

$\displaystyle (x^{1+y} \; D)^n = x^{ny} \; (-y)^n \; n! \; \binom{\frac{-xD}{y}}{n} = x^{ny} \sum_{j \ge 0} (-1)^j [\triangle_k^j (-y)^n \; n! \; \binom{\frac{-k}{y}}{n}\;] \; \frac{:xD:^j}{j!} \; .$

Now using the relation between the Stirling numbers of the first kind $S1_{n,k}$ and the falling factorials

$\displaystyle n! \binom{s}{n} = (s)_n = \sum_{k=0}^n S1_{n,k} \; s^k$

and the formula for the Stirling numbers of the second kind, we obtain

$\displaystyle (x^{1+y} \; D)^n = x^{ny} \; \sum_{j=0}^n \; [\sum_{m=0}^n S1_{n,m}\; (-y)^{n-m} \; S2_{m,j}] \; :xD:^j$

$\displaystyle = x^{ny} \; \sum_{j=0}^n \;T_{n,j}(y) \; x^j \; D^j\; .$

Written as lower triangular matrices,

$\displaystyle T(y) = M(y)\; S1 \; M^{-1}(y)\; S2 = M(y) \; S1 \; M^{-1}(y) \; S1^{-1}= M(y) \; S1 \; [S1 \; M(y)]^{-1}$
$\displaystyle = M(y) \; S2^{-1} \; M^{-1}(y) \; S2 = [S2 \; M^{-1}(y)]^{-1} \; M^{-1}(y) \;S2 \; ,$

with the diagonal matrix $\displaystyle M(y) = Diagonal[1,-y,y^2,(-y)^3,...] \; .$

The first few rows of the Stirling matrices are

(A048994 and A008275)

$\displaystyle S1 = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & 0 & 0 \\ 0 & 2 & -3 & 1 & 0 & 0 \\ 0 & -6 & 11 & -6 & 1 & 0 \\ 0 & 24 & -50 & 35 & -10 & 1 \\ \end{bmatrix}$

and   (A008277 and A048993)

$\displaystyle S2 = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 3 & 1 & 0 & 0 \\ 0 & 1 & 7 & 6 & 1 & 0 \\ 0 & 1 & 15 & 25 & 10 & 1 \\ \end{bmatrix}$ .

Examples of the matrix $\displaystyle T(n) \;$ and associated OEIS entries

OEIS-A000369:

$\displaystyle T(-4) = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & -3 & 1 & 0 & 0 & 0 \\ 0 & 21 & -9 & 1 & 0 & 0 \\ 0 & -231 & 111 & -18 & 1 & 0 \\ 0 & 3465 & -1785 & 345 & -30 & 1 \\ \end{bmatrix}$

$\displaystyle T(-3) = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & -2 & 1 & 0 & 0 & 0 \\ 0 & 10 & -6 & 1 & 0 & 0 \\ 0 & -80 & 52 & -12 & 1 & 0 \\ 0 & 880 & -600 & 160 & -20 & 1 \\ \end{bmatrix}$

Coefficients of Bessel polynomials, A001497 and A132062:

$\displaystyle T(-2) = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & 0 & 0 \\ 0 & 3 & -3 & 1 & 0 & 0 \\ 0 & -15 & 15 & -6 & 1 & 0 \\ 0 & 105 & -105 & 45 & -10 & 1 \\ \end{bmatrix}$
inverse is A122848.

Identity matrix

$\displaystyle T(-1) = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$

$\displaystyle S2$, A008277 and A048993:

$\displaystyle T(0) = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 3 & 1 & 0 & 0 \\ 0 & 1 & 7 & 6 & 1 & 0 \\ 0 & 1 & 15 & 25 & 10 & 1 \\ \end{bmatrix}$
inverse is $\displaystyle S1$.

Lah numbers, A105278 and unsigned A111596:

$\displaystyle T(1) = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 \\ 0 & 6 & 6 & 1 & 0 & 0 \\ 0 & 24 & 36 & 12 & 1 & 0 \\ 0 & 120 & 240 & 120 & 20 & 1 \\ \end{bmatrix}$
inverse is A111596.

$\displaystyle T(2) = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 3 & 1 & 0 & 0 & 0 \\ 0 & 15 & 9 & 1 & 0 & 0 \\ 0 & 105 & 87 & 18 & 1 & 0 \\ 0 & 945 & 975 & 285 & 30 & 1 \\ \end{bmatrix}$

$\displaystyle T(3) = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 4 & 1 & 0 & 0 & 0 \\ 0 & 28 & 12 & 1 & 0 & 0 \\ 0 & 280 & 160 & 24 & 1 & 0 \\ 0 & 3640 & 2520 & 520 & 40 & 1 \\ \end{bmatrix}$

$\displaystyle T(4) = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 5 & 1 & 0 & 0 & 0 \\ 0 & 45 & 15 & 1 & 0 & 0 \\ 0 & 585 & 255 & 30 & 1 & 0 \\ 0 & 9945 & 5175 & 825 & 50 & 1 \\ \end{bmatrix}$  .

Related umbral compositional identities:

The inverse of $\displaystyle T(y)$ is simple to derive by inspection. An alternate method incorporating umbral compositional inversion is illustrated below.

Define the signed Lah polynomials through

$\displaystyle Lah_n(x)= n! \; \sum_{k=0}^n (-1)^k \binom{n-1}{k-1} \; \frac{x^k}{k!}.$

These polynomials form a self-inverse set under umbral composition; that is,

$\displaystyle Lah_n(Lah.(x))= n! \; \sum_{k=0}^n (-1)^k \binom{n-1}{k-1} \; \frac{Lah_k(x)}{k!}= x^n \;$

since

$\displaystyle n! \; \sum_{k=0}^n (-1)^k \binom{n-1}{k-1} \; \sum_{j=0}^k (-1)^j \binom{k-1}{j-1} \; \frac{x^j}{j!}= n! \sum_{j=0}^\infty (-1)^j \frac{x^j}{j!} \; \sum_{k=0}^n (-1)^k \binom{n-1}{k-1} \; \binom{k-1}{j-1} \;$

$\displaystyle = n!\; \sum_{j=0}^\infty (-1)^j \frac{x^j}{j!} \; \binom{n-1}{j-1} \; \sum_{k=0}^n (-1)^k \binom{n-j}{n-k} \;$

$\displaystyle = n!\; \sum_{j=0}^\infty (-1)^j \frac{x^j}{j!} \; \binom{n-1}{j-1} (-1)^n(1-1)^{n-j} = x^n \; .$

Then with the falling factorial polynomials

$\displaystyle ((x)_{\frac{\bullet}{}})^n = (x)_{\frac{n}{}}= \frac{(x)!}{(x-n)!}$

and the rising factorial polynomials

$\displaystyle ((x)_{\frac{}{\bullet}})^n=(x)_{\frac{}{n}}= \frac{(x-1+n)!}{(x-1)!}=(-1)^n \frac{(-x)!}{(-x-n)!}=(-(-x)_{\frac{\bullet}{}})^n \; ,$

we have, from the Vandermonde-Chu identity, the umbral identity

$\displaystyle (x)_{\frac{}{n}}= Lah_n(-(x)_{\frac{\bullet}{}}) = ((x)_{\frac{}{\bullet}})^n \; \; \; (Eqn. I),$

which, from the sign relations between the factorials, implies

$\displaystyle (-1)^n (x)_{\frac{n}{}}= Lah_n((x)_{\frac{}{\bullet}}) = (-(x)_{\frac{\bullet}{}})^n \; \; \; (Eqn. II).$

The last relation follows also from the self-inverse property of the Lah polynomials:

$\displaystyle Lah_n(Lah.(-(x)_{\frac{\bullet}{}}))=(-(x)_{\frac{\bullet}{}})^n \; .$

Substituting $\phi.(-x)$ in the last expression restores the original self-inverse formula for the Lah polynomials since the falling factorials and the Bell polynomials are an inverse pair under umbral composition, as mentioned above.

A formula for the Lah polynomials that reflects the formula for $T(1)$ comes from substituting $\phi.(x)$ into Eqn. I:

$\displaystyle (\phi.(x))_{\frac{}{n}}= Lah_n(-x) \; ,$

implying

$\displaystyle Lah_n(-x)= (-1)^n (-\phi.(x))_{\frac{n}{}} = \sum_{k=0}^n S1_{n,k} (-1)^{n-k} \sum_{j=0}^k S2_{k,j} \; x^j = \sum_{j=0}^n \;T_{n,j}(1) \; x^j \;.$

The formula for $\displaystyle T(-1)$ can be derived from the inverse relation between the Bell and falling factorials $x^n = (\phi.(x))_{\frac{n}{}}\; ,$

and the general relation is

$\displaystyle [-y(-\phi.(x)/y)_{\frac{\bullet}{}}]^n = \sum_{k=0}^n S1_{n,k} (-y)^{n-k} \sum_{j=0}^k S2_{k,j} \; x^j = \sum_{j=0}^n \;T_{n,j}(y) \; x^j = RT_n(x;y)\;,$

the row polynomials for $T(y)$, which by the generalized Dobinski relation also gives

$\displaystyle \sum_{j=0}^n \;T_{n,j}(y) \; x^j = (-y)^n e^{-x} \sum_{k=0}^\infty (-k/y)_{\frac{n}{}} \; \frac{x^k}{k!} = RT_n(x;y) \; .$

The umbral inverse of this expression, giving the matrix inverse for $T(y)$, is

$\displaystyle [(-y\phi.(-x/y) )_{\frac{\bullet}{}}]^n = (-y\phi.(-x/y) )_{\frac{n}{}} = \sum_{k=0}^n S1_{n,k} (-y)^k \sum_{j=0}^k S2_{k,j} \; (-x/y)^j$

$\displaystyle = \sum_{j=0}^n \; \sum_{k=0}^n S1_{n,k} (-y)^{k-j} \; S2_{k,j} \; x^j = \sum_{j=0}^n \;T^{-1}_{n,j}(y) \; x^j = RT_n^{inv}(x;y)\; .$

This is the umbral compositional inverse of the row polynomials (ordinary generating function), i.e.,

$\displaystyle RT_n(RT.^{inv}(x;y);y)=x^n= RT^{inv}_n(RT.(x;y);y) \; .$

The generalized Dobinski formula gives

$\displaystyle RT_n^{inv}(x;y)= e^{x/y} \sum_{k=0}^\infty (-ky)_{\frac{n}{}} \; \frac{(-x/y)^k}{k!} =e^{x/y} \sum_{k=0}^\infty (-1)^n \frac{(ky-1+n)!}{(ky-1)!} \; \frac{(-x/y)^k}{k!} \;$

$\displaystyle = e^{x/y} \sum_{k=0}^\infty \frac{(-ky)!}{(-ky-n)!} \; \frac{(-x/y)^k}{k!} = \sum_{k=0}^n \triangle_j^k \; (-jy)_{\frac{n}{}} \frac{(x/y)^k}{k!} \;.$

Then accumulating the coefficients from the above formulas,

$\displaystyle T_{n,k}^{-1}(y) = \sum_{j=0}^n S1_{n,j} (-y)^{j-k} \; S2_{j,k} =\frac{y^{-k}}{k!} \; \triangle_j^k \; (-jy)_{\frac{n}{}} = (-1)^n \frac{y^{-k}}{k!} \; \triangle_j^k \; (jy)_{\frac{}{n}} \; ,$

with the matrix rep

$\displaystyle T^{-1}(y)= S1 \; M(y) \; S2 \; M^{-1}(y) \; ,$

and

$\displaystyle T_{n,k}(y) = \sum_{j=0}^n S1_{n,j} (-y)^{n-j} \; S2_{j,k} = \frac{(-1)^k}{k!} \; (-y)^n \; \triangle_j^k \; (-j/y)_{\frac{n}{}} = \frac{(-1)^k}{k!} \; y^n \; \triangle_j^k \; (j/y)_{\frac{}{n}} \;,$

with the matrix rep

$\displaystyle T(y) = M(y)\; S1 \; M^{-1}(y)\; S2 \; .$

An equivalent method using the generalized differential shift op

A way to represent umbral substitution is with the generalized shift operators, which, for me, provide some intuition for the simple Euler transformations and the Newton series interpolation as well.

Consider the action of the diff op on the integer power basis:

$\displaystyle (x^{1++y}D)^n x^m = x^{ny} \; (-y)^n (-m/y)_{\frac{n}{}} \; x^m \; ,$

so

$\displaystyle (x^{1++y}D)^n = x^{ny} \; (-y)^n \; \sum_{k \ge 0} (-k/y)_{\frac{n}{}} \; \frac{x^k D^k_{x=0}}{k!} = x^{ny} \; (-y)^n \; e^{a.\; :xD_{x=0}:}$

acting on functions analytic at the origin.

Then formally

$\displaystyle e^{a.\; :xD_{x=0}:} \; f(x) = f(a.x) = f(x-(1-a.)x) = e^{-(1-a.):xD_x:} \; f(x) = \sum_{k \ge 0} (-1)^k \triangle_j^k \; a_j \; \frac{x^k D_x^k}{k!} \; f(x).$

For the diff op we are investigating, $\displaystyle (a.)^k = a_k$ is a polynomial in $\displaystyle k$ of degree $\displaystyle n$; therefore, the finite differences higher than that degree vanish, and the last infinite series is truncated to a polynomial in the op basis. That polynomial op may act on functions analytic at $\displaystyle x$ other than the origin, such as $x^s$, which generates a Newton series interpolating the coefficients $a_k$. Acting on $\displaystyle e^x$ generates the Euler transformation used above.

Summarizing,

$\displaystyle (x^{1++y}D)^n = x^{ny} \; (-y)^n \; \sum_{k \ge 0} (-k/y)_{\frac{n}{}} \; \frac{x^k D^k_{x=0}}{k!} = x^{ny} \; (-y)^n \; \sum_{k=0}^n (-1)^k \triangle_j^k \; (-j/y)_{\frac{n}{}} \; \frac{x^k D_x^k}{k!} \; .$

Related stuff:

Joni, Rota, and Sagan in From Sets to Functions: Three Elementary Examples have combinatorial derivations for some of these identities.