Editing Bell polynomials (section)

===Stirling numbers and Bell numbers===

The value of the Bell polynomial ''B''<sub>''n'',''k''</sub>(''x''<sub>1</sub>,''x''<sub>2</sub>,...) on the sequence of [[factorial]]s equals an unsigned [[Stirling number of the first kind]]:
:<math>B_{n,k}(0!,1!,\dots,(n-k)!)=c(n,k)=|s(n,k)| = \left[{n\atop k}\right].</math>
The sum of these values gives the value of the complete Bell polynomial on the sequence of factorials:
:<math>B_n(0!,1!,\dots,(n-1)!)=\sum_{k=1}^n B_{n,k}(0!,1!,\dots,(n-k)!) = \sum_{k=1}^n \left[{n\atop k}\right] = n!.</math>

The value of the Bell polynomial ''B''<sub>''n'',''k''</sub>(''x''<sub>1</sub>,''x''<sub>2</sub>,...) on the sequence of ones equals a [[Stirling number of the second kind]]:
:<math>B_{n,k}(1,1,\dots,1)=S(n,k)=\left\{{n\atop k}\right\}.</math>
The sum of these values gives the value of the complete Bell polynomial on the sequence of ones:
:<math>B_n(1,1,\dots,1)=\sum_{k=1}^n B_{n,k}(1,1,\dots,1) = \sum_{k=1}^n \left\{{n\atop k}\right\},</math>
which is the ''n''th [[Bell number]].
:<math>B_{n,k}(1!,2!,\ldots,(n-k+1)!) = \binom{n-1}{k-1} \frac{n!}{k!} = L(n,k)</math>
which gives the [[Lah number]].

===Touchard polynomials===

{{main|Touchard polynomials}}

Touchard polynomial <math>T_n(x) = \sum_{k=0}^n \left\{{n\atop k}\right\}\cdot x^k</math> can be expressed as the value of the complete Bell polynomial on all arguments being ''x'':
: <math>T_n(x) = B_n(x,x,\dots,x).</math>

===Inverse relations===
If we define

:<math>y_n = \sum_{k=1}^n B_{n,k}(x_1,\ldots,x_{n-k+1}),</math>

then we have the inverse relationship

:<math>x_n = \sum_{k=1}^n (-1)^{k-1} (k-1)! B_{n,k}(y_1,\ldots,y_{n-k+1}).</math>
More generally,<ref>{{Cite journal |last1=Chou |first1=W.-S. |last2=Hsu |first2=Leetsch C. |last3=Shiue |first3=Peter J.-S. |date=2006-06-01 |title=Application of Faà di Bruno's formula in characterization of inverse relations |journal=Journal of Computational and Applied Mathematics |language=en |volume=190 |issue=1–2 |pages=151–169 |doi=10.1016/j.cam.2004.12.041|doi-access=free }}</ref><ref>{{Cite journal |last=Chu |first=Wenchang |date=2021-11-19 |title=Bell Polynomials and Nonlinear Inverse Relations |url=https://www.combinatorics.org/ojs/index.php/eljc/article/view/v28i4p24 |journal=The Electronic Journal of Combinatorics |volume=28 |issue=4 |doi=10.37236/10390 |issn=1077-8926|doi-access=free }}</ref> given some function <math>f</math> admitting an inverse <math>g = f^{-1}</math>,<blockquote><math>y_n = \sum_{k=0}^n f^{(k)}(a) \, B_{n,k}(x_1, \ldots, x_{n-k+1}) \quad \Leftrightarrow \quad x_n = \sum_{k=0}^n g^{(k)}\big(f(a)\big) \, B_{n,k}(y_1, \ldots, y_{n-k+1}). </math></blockquote>

===Determinant forms===

The complete Bell polynomial can be expressed as [[determinant]]s:

:<math>B_n(x_1,\dots,x_n) = \det\begin{bmatrix}
x_1 & {n-1 \choose 1} x_2 & {n-1 \choose 2}x_3 & {n-1 \choose 3} x_4 & \cdots & \cdots & x_n \\  \\
-1 & x_1 & {n-2 \choose 1} x_2 & {n-2 \choose 2} x_3 &  \cdots & \cdots & x_{n-1} \\  \\
0 & -1 & x_1 & {n-3 \choose 1} x_2 & \cdots & \cdots & x_{n-2} \\  \\
0 & 0 & -1 & x_1 & \cdots  & \cdots & x_{n-3} \\  \\
0 & 0 & 0 & -1 & \cdots & \cdots & x_{n-4} \\  \\
\vdots & \vdots & \vdots &  \vdots & \ddots & \ddots & \vdots  \\  \\
0 & 0 & 0 & 0 & \cdots & -1 & x_1  \end{bmatrix}</math>

and

:<math>B_n(x_1,\dots,x_n) = \det\begin{bmatrix}
\frac{x_1}{0!} & \frac{x_2}{1!} & \frac{x_3}{2!} & \frac{x_4}{3!} & \cdots & \cdots & \frac{x_n}{(n-1)!} \\  \\
-1 & \frac{x_1}{0!} & \frac{x_2}{1!} & \frac{x_3}{2!} &  \cdots & \cdots & \frac{x_{n-1}}{(n-2)!} \\  \\
0 & -2 & \frac{x_1}{0!} & \frac{x_2}{1!} & \cdots & \cdots & \frac{x_{n-2}}{(n-3)!} \\  \\
0 & 0 & -3 & \frac{x_1}{0!} & \cdots  & \cdots & \frac{x_{n-3}}{(n-4)!} \\  \\
0 & 0 & 0 & -4 & \cdots & \cdots & \frac{x_{n-4}}{(n-5)!} \\  \\
\vdots & \vdots & \vdots &  \vdots & \ddots & \ddots & \vdots  \\  \\
0 & 0 & 0 & 0 & \cdots & -(n-1) & \frac{x_1}{0!} \end{bmatrix}.</math>