关于0!=1 ?

我知道数学中存在一些规则或规定，但1!=1的同时0!=1，让我感到十分困惑，这是不是一种内部的不一致性？

热心回应 · 2019-8-13 00:41:51

阶乘只对正整数定义。问0的阶乘等于几的时候，顺便把1.1的阶乘，-π的阶乘，2+3i的阶乘都算出来吧。

首先来算积分：以下积分的区间都是0到+∞
∫x*e^(-x)dx=1=1!
∫x^2*e^(-x)dx=2=2!
∫x^3*e^(-x)dx=6=3!
∫x^4*e^(-x)dx=24=4!
……
可以归纳证明：对于正整数n，
∫x^n*e^(-x)dx=n!

这样就很好了，我们干脆重新定义
n!=∫x^n*e^(-x)dx
或者对于任何复数z，定义
z!=∫x^z*e^(-x)dx

这样，不但与原有的定义相容，且可以算
0!=∫e^(-x)dx=1
0的阶乘就有了~
(1.1)!=∫x^(1.1)*e^(-x)dx
任何正实数的阶乘就有了~
甚至对于一个复数z，只要它的实部大于-1，这个积分∫x^z*e^(-x)dx就收敛，所以阶乘的定义可以扩展到{z∈C:Re(z+1)>0}

如果不收敛也没关系，我们有解析延拓呀~

实际上上面的定义是Gamma函数的定义。

热心回应 · 2019-8-13 00:41:52

设
为
个元素的全排列数, 那么容易知道

根据组合意义, 显然
个元素的全排列数
, 在上面的递推关系中令
即有
.

在组合数学中, 很多问题在
时的解是依赖递推关系给出的, 只有这样才能保证内部的一致性. 我们再举一个例子, 比如Fibonacci数
满足

所以我们有
, 甚至还可以定义
,
等等. 递推关系可以让我们把数列脚标从正整数取值推广到整数取值, 从而在某些时候带来解决问题的便利.

热心回应 · 2019-8-13 00:41:53

0确实不等于1啊，，，这句话没问题，，，c++党的良好修养，，，

热心回应 · 2019-8-13 00:41:54

There appears to be a lot of muddle-headness in online lectures in the sense that the starting points of the proofs are not clear. Given that
Definition #1:

  =
,  where
and
no valid reason can be given as to why
.

解释 1. 阶乘是全排列数, 零个元素的全排列数是  1。  There is only one way to arrange zero objects. This only gives you a reason to speculate that
should be
, which cannot be deduced from the above definition.

解释 2.
Proof:

Source: 0! = 1 prove in detail.
Comment: Muddleheaded. Starting points are not clear. According to Definition #1, this proof is not valid unless
  and
are defined beforehand.

解释 3 .
Proof:

Source: Zero Factorial
Comment: Muddleheaded. Again, relying on Definition #1 alone is inadequate. This proof is not valid unless
, and
is  defined beforehand.

解释 4.
Proof:

当
时 ,

   (1)
In virtue of the product rule,
                                                      (2)
(1).(2)

Source: Proof of 0! = 1.
Comment: Muddleheaded. This proof appears to be valid, but the assumption
is hidden in the proof of the permutation theorem, which is not valid until
is predefined.

结论:
A clever definition goes  a long way. It is possible to deduce
from a precise  definition of
such that
Definition #2:

                                                Df;

         Df, where
In this case,
can be deduced from definition #2, and a separate definition of
  is redundant, and
becomes a consequence of definition #2, not a definition itself. The semantics of this definition is that there are two steps to permutate
objects: the first step is to select one out of
and put it in the first slot; the second step is to fill up the remaining
slots. If
is defined by means of permutation, even defining
becomes unnecessary.
Note:
The symbol Df means the expression to the left is a definition.

热心回应 · 2019-8-13 00:41:55

这个是从大一点的数反推回去的，但是不能再推到负数了
（不谈解析延拓）

关于0!=1 ?

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