Editing Bell polynomials (section)

==Properties==

===Generating function===
The exponential partial Bell polynomials can be defined by the double series expansion of its generating function:
:<math> 
\begin{align}
\Phi(t,u) &= \exp\left( u \sum_{j=1}^\infty x_j \frac{t^j}{j!} \right) = \sum_{n\geq k \geq 0} B_{n,k}(x_1,\ldots,x_{n-k+1}) \frac{t^n}{n!} u^k\\
 &= 1 + \sum_{n=1}^\infty \frac{t^n}{n!} \sum_{k=1}^n u^k B_{n,k}(x_1,\ldots,x_{n-k+1}).
\end{align}
</math>

In other words, by what amounts to the same, by the series expansion of the ''k''-th power:
:<math> \frac{1}{k!}\left( \sum_{j=1}^\infty x_j \frac{t^j}{j!} \right)^k = \sum_{n=k}^\infty B_{n,k}(x_1,\ldots,x_{n-k+1}) \frac{t^n}{n!}, \qquad k = 0, 1, 2, \ldots  </math>
 
The complete exponential Bell polynomial is defined by <math>\Phi(t,1)</math>, or in other words:
:<math> \Phi(t,1) = \exp\left( \sum_{j=1}^\infty x_j \frac{t^j}{j!} \right) = \sum_{n=0}^\infty B_n(x_1,\ldots, x_n) \frac{t^n}{n!}.</math>

Thus, the ''n''-th complete Bell polynomial is given by
:<math> B_n(x_1,\ldots, x_n) = \left. \left(\frac{\partial}{\partial t}\right)^n \exp\left( \sum_{j=1}^n x_j \frac{t^j}{j!} \right) \right|_{t=0}. </math>

Likewise, the ''ordinary'' partial Bell polynomial can be defined by the generating function
:<math> \hat{\Phi}(t,u) = \exp \left( u \sum_{j=1}^\infty x_j t^j \right) = \sum_{n\geq k\geq 0} \hat{B}_{n,k}(x_1,\ldots,x_{n-k+1}) t^n \frac{u^k}{k!}.</math>

Or, equivalently, by series expansion of the ''k''-th power:
:<math>\left(\sum_{j=1}^\infty x_j t^j\right)^k = \sum_{n=k}^\infty \hat{B}_{n,k}(x_1, \ldots, x_{n-k+1}) t^n. </math>

See also [[Generating function transformation#Powers of an OGF and composition with functions|generating function transformations]] for Bell polynomial generating function expansions of compositions of sequence [[generating functions]] and [[Exponentiation|powers]], [[logarithm]]s, and [[exponential function|exponentials]] of a sequence generating function. Each of these formulas is cited in the respective sections of Comtet.{{Sfn|Comtet|1974}}

===Recurrence relations===

The complete Bell polynomials can be [[Recurrence relation|recurrently]] defined as
:<math> B_{n+1}(x_1, \ldots, x_{n+1}) = \sum_{i=0}^n {n \choose i} B_{n-i}(x_1, \ldots, x_{n-i}) x_{i+1}</math>
with the initial value <math>B_0 = 1</math>.

The partial Bell polynomials can also be computed efficiently by a recurrence relation:
:<math> B_{n+1,k+1}(x_1, \ldots, x_{n-k+1}) = \sum_{i=0}^{n-k} \binom{n}{i} x_{i+1} B_{n-i,k}(x_1, \ldots, x_{n-k-i+1})</math>
where
:<math> B_{0,0} = 1; </math>
:<math> B_{n,0} = 0 \text{ for } n \geq 1; </math>
:<math> B_{0,k} = 0 \text{ for } k \geq 1. </math>

In addition:{{sfn|Cvijović|2011}}

:<math>B_{n, k_1 + k_2}(x_1, \ldots, x_{n-k_1-k_2+1}) = \frac{k_1! \, k_2!}{(k_1 + k_2)!} \sum_{i=0}^n \binom{n}{i}  B_{i, k_1}(x_1, \ldots, x_{i-k_1+1}) B_{n-i, k_2}(x_1, \ldots, x_{n-i-k_2+1}).</math>

When <math>1 \le a < n</math>,
:<math>B_{n, n-a}(x_1, \ldots, x_{a+1}) = \sum_{j = a+1}^{2a}\frac{j!}{a!}\binom{n}{j}x_1^{n-j} B_{a, j-a}\Bigl(\frac{x_2}{2}, \frac{x_3}{3}, \ldots, \frac{x_{2(a+1)-j}}{2(a+1)-j}\Bigr).</math>

The complete Bell polynomials also satisfy the following recurrence differential formula:{{Sfn|Alexeev|Pologova|Alekseyev|2017|loc=sect. 4.2}}
: <math>
\begin{align}
B_n(x_1, \ldots, x_n) = \frac{1}{n-1} \left[ \sum_{i=2}^n \right. & \sum_{j=1}^{i-1} (i-1) \binom{i-2}{j-1} x_j x_{i-j}\frac{\partial B_{n-1}(x_1,\dots,x_{n-1})}{\partial x_{i-1}} \\[5pt]
& \left. {} + \sum_{i=2}^n \sum_{j=1}^{i-1} \frac{x_{i+1}}{\binom i j} \frac{\partial^2 B_{n-1}(x_1,\dots,x_{n-1})}{\partial x_j \partial x_{i-j}} \right. \\[5pt]
& \left. {} + \sum_{i=2}^n x_i \frac{\partial B_{n-1}(x_1,\dots,x_{n-1})}{\partial x_{i-1}} \right].
\end{align}
</math>

===Derivatives===

The partial derivatives of the complete Bell polynomials are given by{{Sfn|Bell|1934|loc=identity (5.1) on p. 266}}
: <math> \frac{\partial B_{n}}{\partial x_i} (x_1, \ldots, x_{n}) = \binom{n}{i} B_{n-i}(x_1, \ldots, x_{n-i}).</math>

Similarly, the partial derivatives of the partial Bell polynomials are given by
: <math> \frac{\partial B_{n,k}}{\partial x_i} (x_1, \ldots, x_{n-k+1}) = \binom{n}{i} B_{n-i,k-1}(x_1, \ldots, x_{n-i-k+2}).</math>

If the arguments of the Bell polynomials are one-dimensional functions, the chain rule can be used to obtain
: <math> \frac{d}{dx} \left(B_{n,k}(a_1(x), \cdots, a_{n-k+1}(x))\right) = \sum_{i=1}^{n-k+1} \binom{n}{i} a_i'(x) B_{n-i,k-1}(a_1(x), \cdots, a_{n-i-k+2}(x)).</math>


===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>

===Convolution identity===

For sequences ''x''<sub>''n''</sub>, ''y''<sub>''n''</sub>, ''n'' = 1, 2, ..., define a [[convolution]] by:

:<math>(x \mathbin{\diamondsuit} y)_n = \sum_{j=1}^{n-1} {n \choose j} x_j y_{n-j}.</math>

The bounds of summation are 1 and ''n''&nbsp;−&nbsp;1, not 0 and ''n'' .

Let <math>x_n^{k\diamondsuit}\,</math> be the ''n''th term of the sequence

:<math>\displaystyle\underbrace{x\mathbin{\diamondsuit}\cdots\mathbin{\diamondsuit} x}_{k \text{ factors}}.\,</math>

Then{{Sfn|Cvijović|2011}}

:<math>B_{n,k}(x_1,\dots,x_{n-k+1}) = {x_n^{k\diamondsuit} \over k!}.\,</math>

For example, let us compute <math> B_{4,3}(x_1,x_2) </math>.  We have

:<math> x = ( x_1 \ , \ x_2 \ , \ x_3 \ , \ x_4 \ , \dots ) </math>

:<math> x \mathbin{\diamondsuit} x = ( 0,\  2 x_1^2 \ ,\  6 x_1 x_2 \ , \  8 x_1 x_3 + 6 x_2^2 \ , \dots ) </math>

:<math> x \mathbin{\diamondsuit} x \mathbin{\diamondsuit} x = (  0 \ ,\ 0 \  , \ 6 x_1^3 \ , \ 36 x_1^2 x_2 \ , \dots ) </math>

and thus,

:<math> B_{4,3}(x_1,x_2) = \frac{ ( x \mathbin{\diamondsuit} x \mathbin{\diamondsuit} x)_4 }{3!} = 6 x_1^2 x_2. </math>