Bell polynomials

The partial or incomplete exponential Bell polynomials are a triangular array of polynomials given by

${\displaystyle B_{n,k}(x_{1},x_{2},\dots ,x_{n-k+1})=\sum {n! \over j_{1}!j_{2}!\cdots j_{n-k+1}!}\left({x_{1} \over 1!}\right)^{j_{1}}\left({x_{2} \over 2!}\right)^{j_{2}}\cdots \left({x_{n-k+1} \over (n-k+1)!}\right)^{j_{n-k+1}},}$

where the sum is taken over all sequences j₁, j₂, j₃, ..., j_n−k+1 of non-negative integers such that these two conditions are satisfied:

${\displaystyle j_{1}+j_{2}+\cdots +j_{n-k+1}=k,}$

${\displaystyle j_{1}+2j_{2}+3j_{3}+\cdots +(n-k+1)j_{n-k+1}=n.}$

The sum

${\displaystyle B_{n}(x_{1},\dots ,x_{n})=\sum _{k=1}^{n}B_{n,k}(x_{1},x_{2},\dots ,x_{n-k+1})}$

is called the nth complete exponential Bell polynomial.

Likewise, the partial ordinary Bell polynomial is defined by

${\displaystyle {\hat {B}}_{n,k}(x_{1},x_{2},\ldots ,x_{n-k+1})=\sum {\frac {k!}{j_{1}!j_{2}!\cdots j_{n-k+1}!}}x_{1}^{j_{1}}x_{2}^{j_{2}}\cdots x_{n-k+1}^{j_{n-k+1}},}$

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

${\displaystyle j_{1}+j_{2}+\cdots +j_{n-k+1}=k,}$

${\displaystyle j_{1}+2j_{2}+\cdots +(n-k+1)j_{n-k+1}=n.}$

The ordinary Bell polynomials can be expressed in the terms of exponential Bell polynomials:

${\displaystyle {\hat {B}}_{n,k}(x_{1},x_{2},\ldots ,x_{n-k+1})={\frac {k!}{n!}}B_{n,k}(1!\cdot x_{1},2!\cdot x_{2},\ldots ,(n-k+1)!\cdot x_{n-k+1}).}$

In general, Bell polynomial refers to the exponential Bell polynomial, unless otherwise explicitly stated.

For example, we have

${\displaystyle B_{6,2}(x_{1},x_{2},x_{3},x_{4},x_{5})=6x_{5}x_{1}+15x_{4}x_{2}+10x_{3}^{2}}$

because the ways to partition a set of 6 elements as 2 blocks 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,

${\displaystyle B_{6,3}(x_{1},x_{2},x_{3},x_{4})=15x_{4}x_{1}^{2}+60x_{3}x_{2}x_{1}+15x_{2}^{3}}$

because the ways to partition a set of 6 elements as 3 blocks 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.

The exponential partial Bell polynomials can be defined by the double series expansion of its generating function:

${\displaystyle {\begin{aligned}\Phi (t,u)&=\exp \left(u\sum _{j=1}^{\infty }x_{j}{\frac {t^{j}}{j!}}\right)=\sum _{n\geq k\geq 0}B_{n,k}(x_{1},\ldots ,x_{n-k+1}){\frac {t^{n}}{n!}}u^{k}\\&=1+\sum _{n=1}^{\infty }{\frac {t^{n}}{n!}}\sum _{k=1}^{n}u^{k}B_{n,k}(x_{1},\ldots ,x_{n-k+1}).\end{aligned}}}$

In other words, by what amounts to the same, by the series expansion of the k-th power:

${\displaystyle {\frac {1}{k!}}\left(\sum _{j=1}^{\infty }x_{j}{\frac {t^{j}}{j!}}\right)^{k}=\sum _{n=k}^{\infty }B_{n,k}(x_{1},\ldots ,x_{n-k+1}){\frac {t^{n}}{n!}},\qquad k=0,1,2,\ldots }$

The complete exponential Bell polynomial is defined by ${\displaystyle \Phi (t,1)}$, or in other words:

${\displaystyle \Phi (t,1)=\exp \left(\sum _{j=1}^{\infty }x_{j}{\frac {t^{j}}{j!}}\right)=\sum _{n=0}^{\infty }B_{n}(x_{1},\ldots ,x_{n}){\frac {t^{n}}{n!}}.}$

Thus, the n-th complete Bell polynomial is given by

${\displaystyle B_{n}(x_{1},\ldots ,x_{n})=\left.\left({\frac {\partial }{\partial t}}\right)^{n}\exp \left(\sum _{j=1}^{n}x_{j}{\frac {t^{j}}{j!}}\right)\right|_{t=0}.}$

Likewise, the ordinary partial Bell polynomial can be defined by the generating function

${\displaystyle {\hat {\Phi }}(t,u)=\exp \left(u\sum _{j=1}^{\infty }x_{j}t^{j}\right)=\sum _{n\geq k\geq 0}{\hat {B}}_{n,k}(x_{1},\ldots ,x_{n-k+1})t^{n}{\frac {u^{k}}{k!}}.}$

Or, equivalently, by series expansion of the k-th power:

${\displaystyle \left(\sum _{j=1}^{\infty }x_{j}t^{j}\right)^{k}=\sum _{n=k}^{\infty }{\hat {B}}_{n,k}(x_{1},\ldots ,x_{n-k+1})t^{n}.}$

See also generating function transformations for Bell polynomial generating function expansions of compositions of sequence generating functions and powers, logarithms, and exponentials of a sequence generating function. Each of these formulas is cited in the respective sections of Comtet.[1]

The complete Bell polynomials can be recurrently defined as

${\displaystyle B_{n+1}(x_{1},\ldots ,x_{n+1})=\sum _{i=0}^{n}{n \choose i}B_{n-i}(x_{1},\ldots ,x_{n-i})x_{i+1}}$

with the initial value ${\displaystyle B_{0}=1}$.

The partial Bell polynomials can also be computed efficiently by a recurrence relation:

${\displaystyle B_{n+1,k+1}(x_{1},\ldots ,x_{n-k+1})=\sum _{i=0}^{n-k}{\binom {n}{i}}x_{i+1}B_{n-i,k}(x_{1},\ldots ,x_{n-k-i+1})}$

where

${\displaystyle B_{0,0}=1;}$

${\displaystyle B_{n,0}=0{\text{ for }}n\geq 1;}$

${\displaystyle B_{0,k}=0{\text{ for }}k\geq 1.}$

In addition:[2]

${\displaystyle B_{n,k_{1}+k_{2}}(x_{1},\ldots ,x_{n-k_{1}-k_{2}+1})={\frac {k_{1}!\,k_{2}!}{(k_{1}+k_{2})!}}\sum _{i=0}^{n}{\binom {n}{i}}B_{i,k_{1}}(x_{1},\ldots ,x_{i-k_{1}+1})B_{n-i,k_{2}}(x_{1},\ldots ,x_{n-i-k_{2}+1}).}$

The complete Bell polynomials also satisfy the following recurrence differential formula:[3]

${\displaystyle {\begin{aligned}B_{n}(x_{1},\ldots ,x_{n})={\frac {1}{n-1}}\left[\sum _{i=2}^{n}\right.&\sum _{j=1}^{i-1}(i-1){\binom {i-2}{j-1}}x_{j}x_{i-j}{\frac {\partial B_{n-1}(x_{1},\dots ,x_{n-1})}{\partial x_{i-1}}}\\[5pt]&\left.{}+\sum _{i=2}^{n}\sum _{j=1}^{i-1}{\frac {x_{i+1}}{\binom {i}{j}}}{\frac {\partial ^{2}B_{n-1}(x_{1},\dots ,x_{n-1})}{\partial x_{j}\partial x_{i-j}}}\right.\\[5pt]&\left.{}+\sum _{i=2}^{n}x_{i}{\frac {\partial B_{n-1}(x_{1},\dots ,x_{n-1})}{\partial x_{i-1}}}\right].\end{aligned}}}$

The partial derivatives of the complete Bell polynomials are given by[4]

${\displaystyle {\frac {\partial B_{n}}{\partial x_{i}}}(x_{1},\ldots ,x_{n})={\binom {n}{i}}B_{n-i}(x_{1},\ldots ,x_{n-i}).}$

Similarly, the partial derivatives of the partial Bell polynomials are given by

${\displaystyle {\frac {\partial B_{n,k}}{\partial x_{i}}}(x_{1},\ldots ,x_{n-k+1})={\binom {n}{i}}B_{n-i,k-1}(x_{1},\ldots ,x_{n-i-k+2}).}$

If the arguments of the Bell polynomials are one-dimensional functions, the chain rule can be used to obtain

${\displaystyle {\frac {d}{dx}}\left(B_{n,k}(a_{1}(x),\cdots ,a_{n-k+1}(x))\right)=\sum _{i=1}^{n-k+1}{\binom {n}{i}}a_{i}'(x)B_{n-i,k-1}(a_{1}(x),\cdots ,a_{n-i-k+2}(x)).}$

The value of the Bell polynomial B_n,k(x₁,x₂,...) on the sequence of factorials equals an unsigned Stirling number of the first kind:

${\displaystyle B_{n,k}(0!,1!,\dots ,(n-k)!)=c(n,k)=|s(n,k)|=\left[{n \atop k}\right].}$

The sum of these values gives the value of the complete Bell polynomial on the sequence of factorials:

${\displaystyle B_{n}(0!,1!,\dots ,(n-1)!)=\sum _{k=1}^{n}B_{n,k}(0!,1!,\dots ,(n-k)!)=\sum _{k=1}^{n}\left[{n \atop k}\right]=n!.}$

The value of the Bell polynomial B_n,k(x₁,x₂,...) on the sequence of ones equals a Stirling number of the second kind:

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

The sum of these values gives the value of the complete Bell polynomial on the sequence of ones:

${\displaystyle B_{n}(1,1,\dots ,1)=\sum _{k=1}^{n}B_{n,k}(1,1,\dots ,1)=\sum _{k=1}^{n}\left\{{n \atop k}\right\},}$

which is the nth Bell number.

${\displaystyle B_{n,k}(1!,2!,\ldots ,(n-k+1)!)={\binom {n-1}{k-1}}{\frac {n!}{k!}}=L(n,k)}$

which gives the Lah number.

Touchard polynomial ${\displaystyle T_{n}(x)=\sum _{k=0}^{n}\left\{{n \atop k}\right\}\cdot x^{k}}$ can be expressed as the value of the complete Bell polynomial on all arguments being x:

${\displaystyle T_{n}(x)=B_{n}(x,x,\dots ,x).}$

If we define

${\displaystyle y_{n}=\sum _{k=1}^{n}B_{n,k}(x_{1},\ldots ,x_{n-k+1}),}$

then we have the inverse relationship

${\displaystyle x_{n}=\sum _{k=1}^{n}(-1)^{k-1}(k-1)!B_{n,k}(y_{1},\ldots ,y_{n-k+1}).}$

More generally,[5][6] given some function ${\displaystyle f}$ admitting an inverse ${\displaystyle g=f^{-1}}$,

${\displaystyle y_{n}=\sum _{k=0}^{n}f^{(k)}(a)\,B_{n,k}(x_{1},\ldots ,x_{n-k+1})\quad \Leftrightarrow \quad x_{n}=\sum _{k=0}^{n}g^{(k)}{\big (}f(a){\big )}\,B_{n,k}(y_{1},\ldots ,y_{n-k+1}).}$

The complete Bell polynomial can be expressed as determinants:

${\displaystyle B_{n}(x_{1},\dots ,x_{n})=\det {\begin{bmatrix}x_{1}&{n-1 \choose 1}x_{2}&{n-1 \choose 2}x_{3}&{n-1 \choose 3}x_{4}&\cdots &\cdots &x_{n}\\\\-1&x_{1}&{n-2 \choose 1}x_{2}&{n-2 \choose 2}x_{3}&\cdots &\cdots &x_{n-1}\\\\0&-1&x_{1}&{n-3 \choose 1}x_{2}&\cdots &\cdots &x_{n-2}\\\\0&0&-1&x_{1}&\cdots &\cdots &x_{n-3}\\\\0&0&0&-1&\cdots &\cdots &x_{n-4}\\\\\vdots &\vdots &\vdots &\vdots &\ddots &\ddots &\vdots \\\\0&0&0&0&\cdots &-1&x_{1}\end{bmatrix}}}$

and

${\displaystyle B_{n}(x_{1},\dots ,x_{n})=\det {\begin{bmatrix}{\frac {x_{1}}{0!}}&{\frac {x_{2}}{1!}}&{\frac {x_{3}}{2!}}&{\frac {x_{4}}{3!}}&\cdots &\cdots &{\frac {x_{n}}{(n-1)!}}\\\\-1&{\frac {x_{1}}{0!}}&{\frac {x_{2}}{1!}}&{\frac {x_{3}}{2!}}&\cdots &\cdots &{\frac {x_{n-1}}{(n-2)!}}\\\\0&-2&{\frac {x_{1}}{0!}}&{\frac {x_{2}}{1!}}&\cdots &\cdots &{\frac {x_{n-2}}{(n-3)!}}\\\\0&0&-3&{\frac {x_{1}}{0!}}&\cdots &\cdots &{\frac {x_{n-3}}{(n-4)!}}\\\\0&0&0&-4&\cdots &\cdots &{\frac {x_{n-4}}{(n-5)!}}\\\\\vdots &\vdots &\vdots &\vdots &\ddots &\ddots &\vdots \\\\0&0&0&0&\cdots &-(n-1)&{\frac {x_{1}}{0!}}\end{bmatrix}}.}$

For sequences x_n, y_n, n = 1, 2, ..., define a convolution by:

${\displaystyle (x{\mathbin {\diamondsuit }}y)_{n}=\sum _{j=1}^{n-1}{n \choose j}x_{j}y_{n-j}.}$

The bounds of summation are 1 and n − 1, not 0 and n .

Let ${\displaystyle x_{n}^{k\diamondsuit }\,}$ be the nth term of the sequence

${\displaystyle \displaystyle \underbrace {x{\mathbin {\diamondsuit }}\cdots {\mathbin {\diamondsuit }}x} _{k{\text{ factors}}}.\,}$

Then[2]

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

For example, let us compute ${\displaystyle B_{4,3}(x_{1},x_{2})}$. We have

${\displaystyle x=(x_{1}\ ,\ x_{2}\ ,\ x_{3}\ ,\ x_{4}\ ,\dots )}$

${\displaystyle x{\mathbin {\diamondsuit }}x=(0,\ 2x_{1}^{2}\ ,\ 6x_{1}x_{2}\ ,\ 8x_{1}x_{3}+6x_{2}^{2}\ ,\dots )}$

${\displaystyle x{\mathbin {\diamondsuit }}x{\mathbin {\diamondsuit }}x=(0\ ,\ 0\ ,\ 6x_{1}^{3}\ ,\ 36x_{1}^{2}x_{2}\ ,\dots )}$

and thus,

${\displaystyle B_{4,3}(x_{1},x_{2})={\frac {(x{\mathbin {\diamondsuit }}x{\mathbin {\diamondsuit }}x)_{4}}{3!}}=6x_{1}^{2}x_{2}.}$

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

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

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

${\displaystyle f(x)=\sum _{n=1}^{\infty }{a_{n} \over n!}x^{n}\qquad {\text{and}}\qquad g(x)=\sum _{n=1}^{\infty }{b_{n} \over n!}x^{n}.}$

Then

${\displaystyle g(f(x))=\sum _{n=1}^{\infty }{\frac {\sum _{k=1}^{n}b_{k}B_{n,k}(a_{1},\dots ,a_{n-k+1})}{n!}}x^{n}.}$

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

${\displaystyle \exp \left(\sum _{i=1}^{\infty }{a_{i} \over i!}x^{i}\right)=\sum _{n=0}^{\infty }{B_{n}(a_{1},\dots ,a_{n}) \over n!}x^{n},}$

which also represents the exponential generating function of the complete Bell polynomials on a fixed sequence of arguments ${\displaystyle a_{1},a_{2},\dots }$.

Let two functions f and g be expressed in formal power series as

${\displaystyle f(w)=\sum _{k=0}^{\infty }f_{k}{\frac {w^{k}}{k!}},\qquad {\text{and}}\qquad g(z)=\sum _{k=0}^{\infty }g_{k}{\frac {z^{k}}{k!}},}$

such that g is the compositional inverse of f defined by g(f(w)) = w or f(g(z)) = z. If f₀ = 0 and f₁ ≠ 0, then an explicit form of the coefficients of the inverse can be given in term of Bell polynomials as[8]

${\displaystyle g_{n}={\frac {1}{f_{1}^{n}}}\sum _{k=1}^{n-1}(-1)^{k}n^{\bar {k}}B_{n-1,k}({\hat {f}}_{1},{\hat {f}}_{2},\ldots ,{\hat {f}}_{n-k}),\qquad n\geq 2,}$

with ${\displaystyle {\hat {f}}_{k}={\frac {f_{k+1}}{(k+1)f_{1}}},}$ and ${\displaystyle n^{\bar {k}}=n(n+1)\cdots (n+k-1)}$ is the rising factorial, and ${\displaystyle g_{1}={\frac {1}{f_{1}}}.}$

Consider the integral of the form

${\displaystyle I(\lambda )=\int _{a}^{b}e^{-\lambda f(x)}g(x)\,\mathrm {d} x,}$

where (a,b) is a real (finite or infinite) interval, λ is a large positive parameter and the functions f and g are continuous. Let f have a single minimum in [a,b] which occurs at x = a. Assume that as x → a⁺,

${\displaystyle f(x)\sim f(a)+\sum _{k=0}^{\infty }a_{k}(x-a)^{k+\alpha },}$

${\displaystyle g(x)\sim \sum _{k=0}^{\infty }b_{k}(x-a)^{k+\beta -1},}$

with α > 0, Re(β) > 0; and that the expansion of f can be term wise differentiated. Then, Laplace–Erdelyi theorem states that the asymptotic expansion of the integral I(λ) is given by

${\displaystyle I(\lambda )\sim e^{-\lambda f(a)}\sum _{n=0}^{\infty }\Gamma {\Big (}{\frac {n+\beta }{\alpha }}{\Big )}{\frac {c_{n}}{\lambda ^{(n+\beta )/\alpha }}}\qquad {\text{as}}\quad \lambda \rightarrow \infty ,}$

where the coefficients c_n are expressible in terms of a_n and b_n using partial ordinary Bell polynomials, as given by Campbell–Froman–Walles–Wojdylo formula:

${\displaystyle c_{n}={\frac {1}{\alpha a_{0}^{(n+\beta )/\alpha }}}\sum _{k=0}^{n}b_{n-k}\sum _{j=0}^{k}{\binom {-{\frac {n+\beta }{\alpha }}}{j}}{\frac {1}{a_{0}^{j}}}{\hat {B}}_{k,j}(a_{1},a_{2},\ldots ,a_{k-j+1}).}$

The elementary symmetric polynomial ${\displaystyle e_{n}}$ and the power sum symmetric polynomial ${\displaystyle p_{n}}$ can be related to each other using Bell polynomials as:

${\displaystyle {\begin{aligned}e_{n}&={\frac {1}{n!}}\;B_{n}(p_{1},-1!p_{2},2!p_{3},-3!p_{4},\ldots ,(-1)^{n-1}(n-1)!p_{n})\\&={\frac {(-1)^{n}}{n!}}\;B_{n}(-p_{1},-1!p_{2},-2!p_{3},-3!p_{4},\ldots ,-(n-1)!p_{n}),\end{aligned}}}$

${\displaystyle {\begin{aligned}p_{n}&={\frac {(-1)^{n-1}}{(n-1)!}}\sum _{k=1}^{n}(-1)^{k-1}(k-1)!\;B_{n,k}(e_{1},2!e_{2},3!e_{3},\ldots ,(n-k+1)!e_{n-k+1})\\&=(-1)^{n}\;n\;\sum _{k=1}^{n}{\frac {1}{k}}\;{\hat {B}}_{n,k}(-e_{1},\dots ,-e_{n-k+1}).\end{aligned}}}$

These formulae allow one to express the coefficients of monic polynomials in terms of the Bell polynomials of its zeroes. For instance, together with Cayley–Hamilton theorem they lead to expression of the determinant of a n × n square matrix A in terms of the traces of its powers:

${\displaystyle \det(A)={\frac {(-1)^{n}}{n!}}B_{n}(s_{1},s_{2},\ldots ,s_{n}),~\qquad {\text{where }}s_{k}=-(k-1)!\operatorname {tr} (A^{k}).}$

The cycle index of the symmetric group ${\displaystyle S_{n}}$ can be expressed in terms of complete Bell polynomials as follows: