Exponential Bell polynomials edit

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 edit

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

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

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

Thanks to the first condition on indices, we can rewrite the formula as

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

where we have used the multinomial coefficient.

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.

The exponential Bell polynomial encodes the information related to the ways a set can be partitioned. For example, if we consider a set {A, B, C}, it can be partitioned into two non-empty, non-overlapping subsets, which are also referred to as parts or blocks, in 3 different ways:

{{A}, {B, C}}

{{B}, {A, C}}

{{C}, {B, A}}

Thus, we can encode the information regarding these partitions as

${\displaystyle B_{3,2}(x_{1},x_{2})=3x_{1}x_{2}.}$ 

Here, the subscripts of B_3,2 tell us that we are considering the partitioning of a set with 3 elements into 2 blocks. The subscript of each x_i indicates the presence of a block with i elements (or block of size i) in a given partition. So here, x₂ indicates the presence of a block with two elements. Similarly, x₁ indicates the presence of a block with a single element. The exponent of x_i^j indicates that there are j such blocks of size i in a single partition. Here, the fact that both x₁ and x₂ have exponent 1 indicates that there is only one such block in a given partition. The coefficient of the monomial indicates how many such partitions there are. Here, there are 3 partitions of a set with 3 elements into 2 blocks, where in each partition the elements are divided into two blocks of sizes 1 and 2.

Since any set can be divided into a single block in only one way, the above interpretation would mean that B_n,1 = x_n. Similarly, since there is only one way that a set with n elements be divided into n singletons, B_n,n = x₁ⁿ.

As a more complicated example, consider

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

This tells us that if a set with 6 elements is divided into 2 blocks, then we can have 6 partitions with blocks of size 1 and 5, 15 partitions with blocks of size 4 and 2, and 10 partitions with 2 blocks of size 3.

The sum of the subscripts in a monomial is equal to the total number of elements. Thus, the number of monomials that appear in the partial Bell polynomial is equal to the number of ways the integer n can be expressed as a summation of k positive integers. This is the same as the integer partition of n into k parts. For instance, in the above examples, the integer 3 can be partitioned into two parts as 2+1 only. Thus, there is only one monomial in B_3,2. However, the integer 6 can be partitioned into two parts as 5+1, 4+2, and 3+3. Thus, there are three monomials in B_6,2. Indeed, the subscripts of the variables in a monomial are the same as those given by the integer partition, indicating the sizes of the different blocks. The total number of monomials appearing in a complete Bell polynomial B_n is thus equal to the total number of integer partitions of n.

Also the degree of each monomial, which is the sum of the exponents of each variable in the monomial, is equal to the number of blocks the set is divided into. That is, j₁ + j₂ + ... = k . Thus, given a complete Bell polynomial B_n, we can separate the partial Bell polynomial B_n,k by collecting all those monomials with degree k.

Finally, if we disregard the sizes of the blocks and put all x_i = x, then the summation of the coefficients of the partial Bell polynomial B_n,k will give the total number of ways that a set with n elements can be partitioned into k blocks, which is the same as the Stirling numbers of the second kind. Also, the summation of all the coefficients of the complete Bell polynomial B_n will give us the total number of ways a set with n elements can be partitioned into non-overlapping subsets, which is the same as the Bell number.

In general, 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 edit

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.

Below is a triangular array of the incomplete Bell polynomials ${\displaystyle B_{n,k}(x_{1},x_{2},\dots ,x_{n-k+1})}$ :

Generating function edit

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 edit

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 edit

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 edit

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 edit

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 edit

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 edit

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 edit

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,

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

• ${\displaystyle B_{n,k}(1,2,3,\ldots ,n-k+1)={\binom {n}{k}}k^{n-k}}$  which gives the idempotent number.
• ${\displaystyle B_{n,k}(\alpha \beta x_{1},\alpha \beta ^{2}x_{2},\ldots ,\alpha \beta ^{n-k+1}x_{n-k+1})=\alpha ^{k}\beta ^{n}B_{n,k}(x_{1},x_{2},\ldots ,x_{n-k+1})}$ .
• The complete Bell polynomials satisfy the binomial type relation:

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

  ${\displaystyle B_{n,k}{\Bigl (}{\frac {x_{q+1}}{\binom {q+1}{q}}},{\frac {x_{q+2}}{\binom {q+2}{q}}},\ldots {\Bigr )}={\frac {n!(q!)^{k}}{(n+qk)!}}B_{n+qk,k}(\ldots ,0,0,x_{q+1},x_{q+2},\ldots ).}$ 

This corrects the omission of the factor ${\displaystyle (q!)^{k}}$  in Comtet's book.^[7]

• When ${\displaystyle 1\leq a<n}$ ,

  ${\displaystyle B_{n,n-a}(x_{1},\ldots ,x_{a+1})=\sum _{j=a+1}^{2a}{\frac {j!}{a!}}{\binom {n}{j}}x_{1}^{n-j}B_{a,j-a}{\Bigl (}{\frac {x_{2}}{2}},{\frac {x_{3}}{3}},\ldots ,{\frac {x_{2(a+1)-j}}{2(a+1)-j}}{\Bigr )}.}$ 

• Special cases of partial Bell polynomials:

${\displaystyle {\begin{aligned}B_{n,1}(x_{1},\ldots ,x_{n})={}&x_{n}\\B_{n,2}(x_{1},\ldots ,x_{n-1})={}&{\frac {1}{2}}\sum _{k=1}^{n-1}{\binom {n}{k}}x_{k}x_{n-k}\\B_{n,n}(x_{1})={}&x_{1}^{n}\\B_{n,n-1}(x_{1},x_{2})={}&{\binom {n}{2}}x_{1}^{n-2}x_{2}\\B_{n,n-2}(x_{1},x_{2},x_{3})={}&{\binom {n}{3}}x_{1}^{n-3}x_{3}+3{\binom {n}{4}}x_{1}^{n-4}x_{2}^{2}\\B_{n,n-3}(x_{1},x_{2},x_{3},x_{4})={}&{\binom {n}{4}}x_{1}^{n-4}x_{4}+10{\binom {n}{5}}x_{1}^{n-5}x_{2}x_{3}+15{\binom {n}{6}}x_{1}^{n-6}x_{2}^{3}\\B_{n,n-4}(x_{1},x_{2},x_{3},x_{4},x_{5})={}&{\binom {n}{5}}x_{1}^{n-5}x_{5}+5{\binom {n}{6}}x_{1}^{n-6}(3x_{2}x_{4}+2x_{3}^{2})+105{\binom {n}{7}}x_{1}^{n-7}x_{2}^{2}x_{3}\\&+105{\binom {n}{8}}x_{1}^{n-8}x_{2}^{4}.\end{aligned}}}$ 

The first few complete Bell polynomials are:

${\displaystyle {\begin{aligned}B_{0}={}&1,\\[8pt]B_{1}(x_{1})={}&x_{1},\\[8pt]B_{2}(x_{1},x_{2})={}&x_{1}^{2}+x_{2},\\[8pt]B_{3}(x_{1},x_{2},x_{3})={}&x_{1}^{3}+3x_{1}x_{2}+x_{3},\\[8pt]B_{4}(x_{1},x_{2},x_{3},x_{4})={}&x_{1}^{4}+6x_{1}^{2}x_{2}+4x_{1}x_{3}+3x_{2}^{2}+x_{4},\\[8pt]B_{5}(x_{1},x_{2},x_{3},x_{4},x_{5})={}&x_{1}^{5}+10x_{2}x_{1}^{3}+15x_{2}^{2}x_{1}+10x_{3}x_{1}^{2}+10x_{3}x_{2}+5x_{4}x_{1}+x_{5}\\[8pt]B_{6}(x_{1},x_{2},x_{3},x_{4},x_{5},x_{6})={}&x_{1}^{6}+15x_{2}x_{1}^{4}+20x_{3}x_{1}^{3}+45x_{2}^{2}x_{1}^{2}+15x_{2}^{3}+60x_{3}x_{2}x_{1}\\&{}+15x_{4}x_{1}^{2}+10x_{3}^{2}+15x_{4}x_{2}+6x_{5}x_{1}+x_{6},\\[8pt]B_{7}(x_{1},x_{2},x_{3},x_{4},x_{5},x_{6},x_{7})={}&x_{1}^{7}+21x_{1}^{5}x_{2}+35x_{1}^{4}x_{3}+105x_{1}^{3}x_{2}^{2}+35x_{1}^{3}x_{4}\\&{}+210x_{1}^{2}x_{2}x_{3}+105x_{1}x_{2}^{3}+21x_{1}^{2}x_{5}+105x_{1}x_{2}x_{4}\\&{}+70x_{1}x_{3}^{2}+105x_{2}^{2}x_{3}+7x_{1}x_{6}+21x_{2}x_{5}+35x_{3}x_{4}+x_{7}.\end{aligned}}}$ 

Faà di Bruno's formula edit

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

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

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

${\displaystyle f(x)=\sum _{n=1}^{\infty }{a_{n} \over n!}x^{n}\qquad {\text{and}}\qquad g(x)=\sum _{n=0}^{\infty }{b_{n} \over n!}x^{n}.}$ 

Then

${\displaystyle g(f(x))=\sum _{n=1}^{\infty }{\frac {\sum _{k=0}^{n}b_{k}B_{n,k}(a_{1},\dots ,a_{n-k+1})}{n!}}x^{n}.}$ 

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 edit

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 edit

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 edit

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:

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

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