# 数学代写|组合学代写Combinatorics代考|The Bernoulli polynomials

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## 数学代写|组合学代写Combinatorics代考|The Bernoulli polynomials

Expression of power sums via the Bernoulli polynomials
The Bernoulli polynomials are defined as
$$B_n(x)=\sum_{m=0}^n\left(\begin{array}{l} n \ m \end{array}\right) x^m B_{n-m} .$$
Clearly, $B_n=B_n(0)$. The Bernoulli polynomials are of high importance in the analytic theory of numbers ${ }^4$.

From our present viewpoint, the Bernoulli polynomials are useful, because (5.11) can be expressed in a very compact form with them. Note that the right-hand side of (5.11) is almost a Bernoulli polynomial:
$$\sum_{m=1}^{p+1} n^m\left(\begin{array}{c} p+1 \ m \end{array}\right) B_{p+1-m}=B_{p+1}(n)-B_{p+1}$$

Thus, we get yet another expression (and the simplest looking one) for our power sums:
$$1^p+2^p+\cdots+(n-1)^p=\frac{1}{p+1}\left(B_{p+1}(n)-B_{p+1}\right) .$$
This gives another reason why it is better to sum up to $n-1$ : the argument of the Bernoulli polynomial is the simplest one in this case.

Power sums of arithmetic progressions and the Bernoulli polynomials

The initial point in the proof that the power sums can be expressed by the Stirling numbers was (5.1).

## 数学代写|组合学代写Combinatorics代考|The Cauchy numbers of the first and second kind

Definition 5.3.1. The Cauchy numbers of the first and second kind ${ }^6$ are defined, respectively, by the definite integrals
\begin{aligned} c_n & =\int_0^1 x(x-1)(x-2) \cdots(x-n+1) d x, \ C_n & =\int_0^1 x(x+1)(x+2) \cdots(x+n-1) d x . \end{aligned}

Both kinds of Cauchy numbers are rational numbers (they are definite integrals of polynomials), and $C_n$ is always positive.

Note that we can rewrite the definition of the Cauchy numbers in a shorter form if we take the binomial coefficients into account:
\begin{aligned} c_n & =\int_0^1 x^n d x=n ! \int_0^1\left(\begin{array}{l} x \ n \end{array}\right) d x, \ C_n & =\int_0^1 x^{\bar{n}} d x=(-1)^n n ! \int_0^1\left(\begin{array}{c} -x \ n \end{array}\right) d x=n ! \int_0^1\left(\begin{array}{c} x+n-1 \ n \end{array}\right) d x . \end{aligned}
(For the definition of the binomial coefficients for real upper parameters see $(2.51)$.

These numbers are connected to the Stirling numbers. This connection immediately comes if we expand the falling and rising factorials by using the Stirling numbers:
\begin{aligned} & c_n=\int_0^1 x(x-1)(x-2) \cdots(x-n+1) d x=\sum_{k=0}^n \overline{\left[\begin{array}{l} n \ k \end{array}\right]} \frac{1}{k+1}, \ & C_n=\int_0^1 x(x+1)(x+2) \cdots(x+n-1) d x=\sum_{k=0}^n\left[\begin{array}{l} n \ k \end{array}\right] \frac{1}{k+1} . \ & \end{aligned}
The generating function of $c_n$ and $C_n$ comes easily:
\begin{aligned} \sum_{n=0}^{\infty} c_n \frac{x^n}{n !} & =\frac{x}{\log (1+x)} \ \sum_{n=0}^{\infty} C_n \frac{x^n}{n !} & =\frac{x}{(x-1) \log (1-x)} \end{aligned}

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|The Bernoulli polynomials

$$B_n(x)=\sum_{m=0}^n\left(\begin{array}{l} n \ m \end{array}\右）x^m B_{nm} 。$$

$$\sum_{m=1}^{p+1} n^m\left(\begin{array}{c} p+1 \ m \end{array}\right) B_{p+1-m}=B_{p+1}(n)-B_{p+1}$$

$$1^p+2^p+\cdots+(n-1)^p=\frac{1}{p+1}\left (B_{p+1}(n)-B_{p+1}\右)。$$

## 数学代写|组合学代写Combinatorics代考|The Cauchy numbers of the first and second kind

bbegin ${$ aligned
c_n \& = lint_ $0^{\wedge} 1 x(x-1)(x-2) \backslash \operatorname{cdots}(x-n+1) d x, 1$
C_n \& $=\backslash$ int_ $0^{\wedge} 1 x(x+1)(x+2) \backslash \operatorname{cdots}(x+n-1) d x$ 。
、结束 ${$ 对齐 $}$
$\$ \$$两种柯西数都是有理数（它们是多项式的定积分）， \ C_{-} n \$$ 总是正数。

Vbegin{aligned} \& c_n=lint_ $0^{\wedge} 1 x(x-1)(x-2) \backslash$ |cdots $(x-$ $\mathrm{n}+1) \mathrm{d} x=\mid$ sum_{ ${k=0}^{\wedge} n$ loverline ${\backslash$ left[lbegin ${$ array}$} \mid}$
$\mathrm{n} \backslash \mathrm{k}$ lend ${$ array $} \backslash$ right $]} \backslash f r a c{1} \mathrm{k}+1}, \backslash \&$
C_n=lint_ $0^{\wedge} 1 \mathrm{x}(\mathrm{x}+1)(x+2) \backslash \operatorname{cots}(x+n-1)$
$\mathrm{d}=$ Isum_{k=0}^ $n$ left $[$ |begin ${$ array $} \mid} n \backslash k$
$$c_n=\int_0^1 x(x-1)(x-2) \cdots(x-n+1) d x=\sum_{k=0}^n$$

$$\sum_{n=0}^{\infty} c_n \frac{x^n}{n !}=\frac{x}{\log (1+x)} \sum_{n=0}^{\infty} C_n \frac{x^n}{n !}=\frac{}{(x-1)}$$

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