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In combinatorial mathematics, the Bell polynomials, named in honor of Eric Temple Bell, are a triangular array of polynomials given by

\( B_{n,k}(x_1,x_2,\dots,x_{n-k+1}) \)

\( =\sum{n! \over j_1!j_2!\cdots j_{n-k+1}!} \left({x_1\over 1!}\right)^{j_1}\left({x_2\over 2!}\right)^{j_2}\cdots\left({x_{n-k+1} \over (n-k+1)!}\right)^{j_{n-k+1}}, \)

where the sum is taken over all sequences j1, j2, j3, ..., jn−k+1 of non-negative integers such that

\( j_1+j_2+\cdots = k\quad\mbox{and}\quad j_1+2j_2+3j_3+\cdots=n. \)

Complete Bell polynomials

The sum

\( B_n(x_1,\dots,x_n)=\sum_{k=1}^n B_{n,k}(x_1,x_2,\dots,x_{n-k+1}) \)

is sometimes called the nth complete Bell polynomial. In order to contrast them with complete Bell polynomials, the polynomials Bn, k defined above are sometimes called "partial" Bell polynomials.

The complete Bell polynomials satisfy the following identity

\( B_n(x_1,\dots,x_n) = \det\begin{bmatrix}x_1 & {n-1 \choose 1} x_2 & {n-1 \choose 2}x_3 & {n-1 \choose 3} x_4 & {n-1 \choose 4} x_5 & \cdots & \cdots & x_n \\ \\ -1 & x_1 & {n-2 \choose 1} x_2 & {n-2 \choose 2} x_3 & {n-2 \choose 3} x_4 & \cdots & \cdots & x_{n-1} \\ \\ 0 & -1 & x_1 & {n-3 \choose 1} x_2 & {n-3 \choose 2} x_3 & \cdots & \cdots & x_{n-2} \\ \\ 0 & 0 & -1 & x_1 & {n-4 \choose 1} x_2 & \cdots & \cdots & x_{n-3} \\ \\ 0 & 0 & 0 & -1 & x_1 & \cdots & \cdots & x_{n-4} \\ \\ 0 & 0 & 0 & 0 & -1 & \cdots & \cdots & x_{n-5} \\ \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ \\ 0 & 0 & 0 & 0 & 0 & \cdots & -1 & x_1 \end{bmatrix}. \)

Combinatorial meaning

If the integer n is partitioned into a sum in which "1" appears j1 times, "2" appears j2 times, and so on, then the number of partitions of a set of size n that collapse to that partition of the integer n when the members of the set become indistinguishable is the corresponding coefficient in the polynomial.
Examples

For example, we have

\( B_{6,2}(x_1,x_2,x_3,x_4,x_5)=6x_5x_1+15x_4x_2+10x_3^2 \)

because there are

6 ways to partition a set of 6 as 5 + 1,
15 ways to partition a set of 6 as 4 + 2, and
10 ways to partition a set of 6 as 3 + 3.

Similarly,

\( B_{6,3}(x_1,x_2,x_3,x_4)=15x_4x_1^2+60x_3x_2x_1+15x_2^3 \)

because there are

15 ways to partition a set of 6 as 4 + 1 + 1,
60 ways to partition a set of 6 as 3 + 2 + 1, and
15 ways to partition a set of 6 as 2 + 2 + 2.

Properties

\( B_{n,k}(1!,2!,\dots,(n-k+1)!) = \binom{n}{k}\binom{n-1}{k-1} (n-k)! \)

Stirling numbers and Bell numbers

The value of the Bell polynomial \(B_{n,k}(x_1 ,x_2 ,\dots) \) when all xs are equal to 1 is a Stirling number of the second kind:

\( B_{n,k}(1,1,\dots)=S(n,k)=\left\{{n\atop k}\right\}. \)

The sum

\( \sum_{k=1}^n B_{n,k}(1,1,1,\dots) = \sum_{k=1}^n \left\{{n\atop k}\right\} \)

is the nth Bell number, which is the number of partitions of a set of size n.
Convolution identity

For sequences xn, yn, n = 1, 2, ..., define a sort of convolution by:

\( (x \diamondsuit y)_n = \sum_{j=1}^{n-1} {n \choose j} x_j y_{n-j}. \)

Note that the bounds of summation are 1 and n − 1, not 0 and n .

Let \( x_n^{k\diamondsuit}\, \) be the nth term of the sequence

\( \displaystyle\underbrace{x\diamondsuit\cdots\diamondsuit x}_{k\ \mathrm{factors}}.\, \)

Then

\( B_{n,k}(x_1,\dots,x_{n-k+1}) = {x_{n}^{k\diamondsuit} \over k!}.\, \)

For example, let us compute \( B_{4,3}(x_1,x_2) \) . We have

\( x = ( x_1 \ , \ x_2 \ , \ x_3 \ , \ x_4 \ , \dots ) \)

\( x \diamondsuit x = ( 0,\ 2 x_1^2 \ ,\ 6 x_1 x_2 \ , \ 8 x_1 x_3 + 6 x_2^2 \ , \dots ) \)

\( x \diamondsuit x \diamondsuit x = ( 0 \ ,\ 0 \ , \ 6 x_1^3 \ , \ 36 x_1^2 x_2 \ , \dots ) \)

and thus,

\( B_{4,3}(x_1,x_2) = \frac{ ( x \diamondsuit x \diamondsuit x)_4 }{3!} = 6 x_1^2 x_2. \)

Applications of Bell polynomials
Faà di Bruno's formula
Main article: Faà di Bruno's formula

Faà di Bruno's formula may be stated in terms of Bell polynomials as follows:

\( {d^n \over dx^n} f(g(x)) = \sum_{k=0}^n f^{(k)}(g(x)) B_{n,k}\left(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\right). \)

Similarly, a power-series version of Faà di Bruno's formula may be stated using Bell polynomials as follows. Suppose

\( f(x)=\sum_{n=1}^\infty {a_n \over n!} x^n \qquad \mathrm{and} \qquad g(x)=\sum_{n=1}^\infty {b_n \over n!} x^n. \)

Then

\( g(f(x)) = \sum_{n=1}^\infty {\sum_{k=1}^{n} b_k B_{n,k}(a_1,\dots,a_{n-k+1}) \over n!} x^n. \)

In particular, the complete Bell polynomials appear in the exponential of a formal power series:

\( \exp\left(\sum_{n=1}^\infty {a_n \over n!} x^n \right) = \sum_{n=0}^\infty {B_n(a_1,\dots,a_n) \over n!} x^n. \)

Moments and cumulants

The sum

\( B_n(\kappa_1,\dots,\kappa_n)=\sum_{k=1}^n B_{n,k}(\kappa_1,\dots,\kappa_{n-k+1}) \)

is the nth moment of a probability distribution whose first n cumulants are (\ kappa_1,\dots,\kappa_n \). In other words, the nth moment is the nth complete Bell polynomial evaluated at the first n cumulants.
Representation of polynomial sequences of binomial type

For any sequence \( a_, a_, a_, \dots \) of scalars, let

\( p_n(x)=\sum_{k=1}^n B_{n,k}(a_1,\dots,a_{n-k+1}) x^k. \)

Then this polynomial sequence is of binomial type, i.e. it satisfies the binomial identity

\( p_n(x+y)=\sum_{k=0}^n {n \choose k} p_k(x) p_{n-k}(y) \)

for n ≥ 0. In fact we have this result:

Theorem: All polynomial sequences of binomial type are of this form.

If we let

\( h(x)=\sum_{n=1}^\infty {a_n \over n!} x^n \)

taking this power series to be purely formal, then for all n,

\( h^{-1}\left( {d \over dx}\right) p_n(x) = n p_{n-1}(x). \)

Software

Bell polynomials, complete Bell polynomials and generalized Bell polynomials are implemented in Mathematica as BellY.