# 数学代写|组合学代写Combinatorics代考|MA1510

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## 数学代写|组合学代写Combinatorics代考|The sum and product of the zeros of B

We know from the last section that we can write
$$B_n(x)=\left(x-t_1^{(n)}\right)\left(x-t_2^{(n)}\right) \cdots\left(x-t_n^{(n)}\right)$$
where $t_i^{(n)}$ are the zeros of the Bell polynomials $(i=1, \ldots, n)$, and that these numbers are negative, except one zero which is $x=0$. We choose this to be $t_1^{(n)}$, that is, $t_1^{(n)}=0$

By Viète’s ${ }^3$ formulas we can have a bit more knowledge about these zeros. One of these formulas says that if we have a polynomial
$$p(x)=a_n x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0,$$
such that the zeros are $x_1, \ldots, x_n$ (these can be real or complex), that is,
$$p(x)=a_n\left(x-x_1\right)\left(x-x_2\right) \cdots\left(x-x_n\right),$$
then the sum of these zeros is
$$x_1+x_2+\cdots+x_n=-\frac{a_{n-1}}{a_n} .$$
Another formula evaluates the product of the zeros:
$$x_1 x_2 \cdots x_n=(-1)^n \frac{a_0}{a_n}$$
These can easily be seen by comparing the coefficients of $x^{n-1}$ and $x^0$ on the right-hand sides of (3.5) and (3.6), respectively ${ }^4$.

These formulas result in the following sum and product formulas with respect to the zeros of the Bell polynomials:
$$t_1^{(n)}+t_2^{(n)}+\cdots+t_n^{(n)}=-\frac{\left{\begin{array}{c} n \ n-1 \end{array}\right}}{\left{\begin{array}{l} n \ n \end{array}\right}}=-\left(\begin{array}{l} n \ 2 \end{array}\right)=-\frac{n(n-1)}{2} \quad(n \geq 1),$$
and (taking $B_n(x) / x$ instead)
$$t_2^{(n)} \cdots t_n^{(n)}=-\frac{\left{\begin{array}{l} n \ 1 \end{array}\right}}{\left{\begin{array}{l} n \ n \end{array}\right}}=-1 \quad(n \geq 2) .$$

## 数学代写|组合学代写Combinatorics代考|The irreducibility of Bn(x)

Whether the Bell polynomials ever have rational zeros $(n \geq 3)$ other than the trivial $t_1^{(n)}=0$ is unknown. If a non-constant polynomial has no rational zeros, we say that it is irreducible over the rationals.

In 1983 J. W. Layman and C. L. Prather [360] gave five different statements equivalent to the irreducibility of the Bell polynomials $B_n(x) / x$. One of these is the non-vanishing property of the Bell polynomials at $x=-1$. This is actually easy to see once we recall the rational root theorem [574, p. 109].
The rational root theorem says that if a polynomial
$$p(x)=a_n x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0$$
has a rational zero $x^$, i.e., it is of the form $x^=\frac{a}{b}$ (where $a$ and $b$ have no divisors in common), then $a$ must divide $a_0$, and $b$ must divide $a_n$.

Since the constant coefficient and the highest order coefficient in $B_n(x) / x$ are both one, the only possible rational zeros are \pm 1 . Of course, $B_n(x)>0$ if $x>0$, so the only possible rational root is minus one. Thus, we arrive at the following statement: $B_n(x) / x$ is irreducible over the rationals if and only if $B_n(-1) \neq 0$. Since $B_2(x)=1+x$, we have that $B_2(-1)=0$, but this seems to be the only one case. In [360] this was checked up to $n \leq 110$ (up to 900 in [359]), and we checked by computer up to $n=10000$ but none of $B_n(-1)$ numbers vanishes $(n \neq 2)$. Layman further studied the arithmetical properties of the $B_n(-1)$ numbers $[359]$.

After these initial studies, many achievements have been made. It is known now, due to T. Amdeberhan, V. De Angelis, and V. H. Moll [18], that there is at most one $n>2$ such that $B_n(-1)=0$. See [29] for an “economical account” of the proof, and for historical remarks. The paper [589] contains many additional arithmetical properties of this sequence. The conjecture that $B_n(-1) \neq 0$ for all $n>2$ is attributed to $\mathrm{H}$. Wilf, and it is often referred to as Wilf’s conjecture. The source of this attribution is seemingly the paper of Y. Yang [608]. In this paper and also in [572], a detailed asymptotic analysis is carried out for this sequence. Among others, it is known that
$$\limsup _{n \rightarrow \infty} \frac{\log \left|B_n(-1)\right|}{n \log n}=1$$

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|The sum and product of the zeros of B

$$B_n(x)=\left(x-t_1^{(n)}\right)\left(x-t_2^{(n)}\right) \cdots\left(x-t_n^{(n)}\right)$$

$$p(x)=a_n x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0,$$

$$p(x)=a_n\left(x-x_1\right)\left(x-x_2\right) \cdots\left(x-x_n\right),$$

$$x_1+x_2+\cdots+x_n=-\frac{a_{n-1}}{a_n} .$$

$$x_1 x_2 \cdots x_n=(-1)^n \frac{a_0}{a_n}$$

$t_{-} 1 \wedge{(n)}+t_{-} 2 \wedge{(n)}+\mid c d o t s+t_{-} n \wedge{(n)}=-$ frac ${\backslash$ eft ${\backslash$ begin ${$ array

## 数学代写|组合学代写Combinatorics代考|The irreducibility of Bn(x)

1983 年，JW Layman 和 CL Prather [360] 给出了五个不 同的等价于贝尔多项式不可约性的陈述 $B_n(x) / x$. 其中 之一是贝尔多项式的非零性质 $x=-1$. 一旦我们回忆起 有理根定理 [574， p. 574]，这实际上就很容易看出。 109]。

$$p(x)=a_n x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0$$

$B_2(x)=1+x$ ，我们有 $B_2(-1)=0$ ，但这似乎是唯 一的一个案例。在 [360] 中，这被检查到 $n \leq 110$
([359] 中最多 900 个)，我们通过计算机检查了最多 $n=10000$ 但没有一个 $B_n(-1)$ 数字消失 $(n \neq 2)$.
Layman 进一步研究了 $B_n(-1)$ 数字 $[359]$.

$$\limsup _{n \rightarrow \infty} \frac{\log \left|B_n(-1)\right|}{n \log n}=1$$

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