# 数学代写|组合学代写Combinatorics代考|Second approach

## 数学代写|组合学代写Combinatorics代考|Second approach

We now show another approach that is simpler but, at the same time, it gives a weaker result. Begin with the (2.43) horizontal generating function:
$$\sum_{m=0}^n\left{\begin{array}{l} n \ m \end{array}\right} x(x-1)(x-2) \cdots(x-m+1)=x^n .$$
Substitute $x=k$ and rewrite the falling factorial so that
$$\sum_{m=0}^k\left{\begin{array}{l} n \ m \end{array}\right} \frac{k !}{(k-m) !}=k^n$$
From here, by separating the last term, and dividing by $k$ ! we get
$$\sum_{m=0}^{k-1}\left{\begin{array}{l} n \ m \end{array}\right} \frac{1}{(k-m) !}+\left{\begin{array}{l} n \ k \end{array}\right}=\frac{k^n}{k !}$$
The right-hand side of (7.3) follows since the sum is positive on the left-hand side is positive. On the other hand, by using this last relation again:
$$\left{\begin{array}{l} n \ k \end{array}\right}=\frac{k^n}{k !}-\sum_{m=0}^{k-1}\left{\begin{array}{l} n \ m \end{array}\right} \frac{1}{(k-m) !}>\frac{k^n}{k !}-\frac{1}{(k-1) !} \sum_{m=0}^{k-1}\left{\begin{array}{l} n \ m \end{array}\right} \frac{(k-1) !}{(k-1-m) !}=$$ $$\frac{k^n}{k !}-\frac{1}{(k-1) !} \sum_{m=0}^{k-1}\left{\begin{array}{c} n \ m \end{array}\right}(k-1)^{m}=\frac{k^n}{k !}-\frac{(k-1)^n}{(k-1) !}$$
This is nothing else but the left-hand side of (7.3).
That this is indeed weaker than the Bonferroni inequality comes from the fact that it is equivalent to (7.3), and this is the $l=0$ and $l=1$ particular cases of Bonferroni. Higher values of $l$ give better approximations.

More about the approximation of $\left{\begin{array}{l}n \ k\end{array}\right}$ can be found in the Outlook of this chapter.

## 数学代写|组合学代写Combinatorics代考|The Euler Gamma functionA

The Euler Gamma function generalizes the $n$ ! function to real numbers (and even to complex numbers if one wants).

Definition 7.3.1. The Euler Gamma function is defined by the improper integral
$$\Gamma(x)=\int_0^{\infty} t^{x-1} e^{-t} d t$$
where $x>0$.
It can be seen that $\Gamma(1)=1=0$ !, and by partial integration it follows that $\Gamma(2)=1=1 !, \Gamma(3)=2=2$ !, and, in general,
$$\Gamma(n)=(n-1) !$$
Partial integration says actually much more, it comes easily that $\Gamma$ satisfies the functional equation
$$\Gamma(x+1)=x \Gamma(x),$$
Although the above integral is divergent when $x \leq 0$, but the functional equation (7.5) still enables us to extend $\Gamma$ to negative (non-integer) values. Indeed, for example,
$$-\frac{1}{2} \Gamma\left(-\frac{1}{2}\right)=\Gamma\left(-\frac{1}{2}+1\right)$$
whence
$$\Gamma\left(-\frac{1}{2}\right)=-2 \Gamma\left(\frac{1}{2}\right)$$
On the other hand, the integral definition of the $\Gamma$ function is valid and makes sense not only for integers but for real or even complex numbers when the real part is positive. Therefore, the $\Gamma$ function (and thus the factorial) is extensible to the whole complex plane (except the non-positive integers) ${ }^7$.

# 组合学代写

## 数学代写|组合学代写Combinatorics代考|Second approach

$\backslash$ sum_{ ${m=0}^{\wedge} n \backslash \backslash$ eft ${$ begin ${$ array $\left.} \mid}\right} n \backslash m \backslash$ |end{array} $}$ right $} x(x-1)(x-2) \backslash c d o t s(x-m+1)=x^{\wedge} n$ 。

$\left\lfloor\right.$ sum ${m=0}^{\wedge} k \backslash$ left $\backslash$ begin $\left.{a r r a y}{1}\right} n \backslash m \backslash$ end ${a r r a y} \backslash \mid$ ight $} \backslash$ frac $\left.{k !}(k m) !\right}=k^{\wedge} n$

(7.3) 的右侧如下，因为左侧的总和为正。另一方面，通过再次使用最后一个关系:
$\$ \$$\left.\backslash f r a c\left{k^{\wedge} n\right}{k !}-\backslash f r a c{1}(k-1) !\right} \backslash sum {m=0}^{\wedge}{k-1} \backslash left { n m \backslash right \left.}(k-1)^{\wedge}{m}=\mid f r a c{k \wedge n} k k !\right}-\mid f r a c\left{(k-1)^{\wedge} n\right}((k-1) !} \ \$$

## 数学代写|组合学代写Combinatorics代考|The Euler Gamma functionA

Euler Gamma 函数概括了 $n$ ！对实数起作用（如果需要，甚至可以对复数起作用）。

$$\Gamma(x)=\int_0^{\infty} t^{x-1} e^{-t} d t$$

$$\Gamma(n)=(n-1) !$$

$$\Gamma(x+1)=x \Gamma(x)$$

$$-\frac{1}{2} \Gamma\left(-\frac{1}{2}\right)=\Gamma\left(-\frac{1}{2}+1\right)$$

$$\Gamma\left(-\frac{1}{2}\right)=-2 \Gamma\left(\frac{1}{2}\right)$$另一方面，积分的定义 $\Gamma$ 函数是有效的，并且不仅对整数有意义，而且当实部为正时对 实数甚至复数也有意义。因此，「函数 (以及阶乘) 可扩展到整个复平面 (非正整数除 外 ${ }^7$.

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