Hi,
Recently I was thinking a bit why 0 factorial equals 1. I came to a few possible explanations, from very basic (less general) to pretty abstract . Please let me know your thoughts on that.
Factorial as the number of permutation of n distinct objects
Taking into account most basic factorial definition
[; n!=n\times (n-1)\times (n-2)\times \ldots \times 2\times 1 ;]
it is very easy to see that n! counts the possible distinct sequences of n distinct objects.
Let's see the k-permutations of n (sometimes called partial permutations or variations)
The k-permutations of n are the different ordered arrangements of a k-element subset of an n-set. The number of such k-permutations of n is
[; P_k^n = n\times (n-1)\times (n-2)\times\ldots\times \bigg(n-(k-1)\bigg) = \frac{n!}{(n-k)!} ;]
Additionally
[; n! = P_n^n ;]
[; n! = \frac{n!}{(n-n)!} = \frac{n!}{0!} ;]
First insight why 0! = 1
[; 0! \times n! = n! ;]
but this is just a rationale, this is not a proof :-)
Function as a binary relation, meaning function as a subset of a Cartesian product
Function
[; f:A\to B ;]
can be defined as the following binary relation
[; (a,b)\in f \subseteq A\times B \iff f(a)=b ;]
Injective function
Shortly
[; x\neq y \implies f(x) \neq f(y) ;]
Injective function iis a function that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain.
Surjective function
Shortly
[; {\large \displaystyle\forall_{b \in B} \quad\displaystyle\exists_{a\in A}\quad}f(a)=b ;]
Bijective function
Bijective function, or one-to-one correspondence, is a function where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set. There are no unpaired elements. In mathematical terms, a bijective function both injective and surjective mapping of a set A to a set B.
Bijective function vs Permutation
Permutation is a function that returns the order of the set, i.e. if we consider the n-element set {1, 2, ..., n} then permutation will be a function of
[; p:\{1, 2, ..., n\}\to\{1, 2, ..., n\} ;]
satisfying the bijective function condition. By asking about the number of permutations we can equally ask about the number of different bijections from a given set into ourselves.
Empty function
An empty function is every function whose domain is an empty set.
[; f:\emptyset\to B ;]
The empty function "chart" is an empty set, as the Cartesian product
[; \emptyset\times B = \emptyset ;]
The empty function preserves distinctness (is injective), because in the domain (an empty set) there are no two different elements for which the value of the function is equal.
A special case of empty function
Let's analyse a function that maps empty set onto empty set
[; f:\emptyset\to\emptyset ;]
Such a function is a bijection because it is injective function (as shown above) and there is no element in codomain (the codomain is an empty set) that is not in relation to the element of the domain.
Please note that there is exactly one such bijection, which is a results of the fact that the function is a subset of the Cartesian product of domain and codomain. In the case of the
[; f:\emptyset\to\emptyset ;]
The Cartesian products is empty
[; \emptyset\times\emptyset = \emptyset ;]
The empty set has exactly one subset, which is empty set - thus such a bijection is uniquely defined.
0! = 1 vs empty function
I wrote above that the number of permutations of an n-element set equals the number of distinct bijective functions from this set into itself. Following - the permutation of 0-element set correspond to the bijection from an empty set into the empty set - so the special case f empty function! And I presented the proof that there exists only one such a function :-)
The gamma function
In mathematics, the Gamma function is one of the extensions of the factorial function with its argument shifted down by 1, to real and complex numbers.
[; \Gamma(z)=\displaystyle\int_0^{+\infty}t^{z-1}e^{-t}dt ;]
After integration by parts we got
[; \Gamma(z+1)=z\cdot\Gamma(z) ;]
Let's see the value of
[; \Gamma(1) ;]
[; \Gamma(1)=\displaystyle\int_0^{+\infty}e^{-t}dt=\displaystyle\int_{-\infty}^{0}e^{t}dt ;]
[; \Gamma(n+1)=n! ;]
[; 0! = \Gamma(1) = 1 ;]
Number e and factorial are deeply related
This is fascinating
[; e=\displaystyle\sum_{n=0}^\infty\frac{1}{n!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots ;]
Please share your thoughts :-)
Best regards
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from math http://bit.ly/2ZFGo8p
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