Exponential Bell polynomials

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

${\displaystyle {\begin{aligned}B_{n,k}(x_{1},x_{2},\dots ,x_{n-k+1})&=\sum {n! \over j_{1}!j_{2}!\cdots j_{n-k+1}!}\left({x_{1} \over 1!}\right)^{j_{1}}\left({x_{2} \over 2!}\right)^{j_{2}}\cdots \left({x_{n-k+1} \over (n-k+1)!}\right)^{j_{n-k+1}}\\&=n!\sum \prod _{i=1}^{n-k+1}{\frac {x_{i}^{j_{i}}}{(i!)^{j_{i}}j_{i}!}},\end{aligned}}}$

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

${\displaystyle {\begin{aligned}B_{n}(x_{1},\dots ,x_{n})&=\sum _{k=1}^{n}B_{n,k}(x_{1},x_{2},\dots ,x_{n-k+1})\\&=n!\sum _{1j_{1}+\ldots +nj_{n}=n}\prod _{i=1}^{n}{\frac {x_{i}^{j_{i}}}{(i!)^{j_{i}}j_{i}!}}\end{aligned}}}$

is called the nth complete exponential Bell polynomial.

Ordinary Bell polynomials

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.

Examples

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.

Generating function

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]

Recurrence relations

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]

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

Derivatives

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

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

Stirling numbers and Bell numbers

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 polynomials

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:

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

Inverse relations

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 ${\displaystyle g=f^{-1}}$,

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

Determinant forms

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

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

Convolution identity

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

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 ${\displaystyle a_{1},a_{2},\dots }$.

Reversion of series

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 ${\displaystyle g_{1}={\frac {1}{f_{1}}}.}$

Asymptotic expansion of Laplace-type integrals

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:

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

Symmetric polynomials

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:

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

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