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There are short and sweet proofs of various forms of Stirling's approximation. But even the sweetest among them don't instill the same conviction in the reader as a direct bijective proof.

Computer scientists can often get away with a very weak form of Stirling's approximation:

$$(n/2)^{n/2} \leq n! \leq n^n$$

From this follows

$$(n/2)\ log(n) - (n/2)\ log(2) \leq log(n!) \leq n\ log(n)$$

and therefore $log(n!) = \Theta(n\ log(n))$. This is sufficient to establish the lower bound on comparison-based sorting algorithms and many other asymptotic bounds.

Does anyone know a natural bijective proof of $(n/2)^{n/2} \leq n! \leq n^n$?

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Why it's important to have a bijective proof?! I would be very happy to hear some motivation. Any way, if you wish to have some combinatorics involved, you should clarify how to interpret $(n/2)^{n/2}$ for $n$ odd. –  Wadim Zudilin Aug 2 '10 at 5:31
    
For $n$ even, $n/2$ of the factors of $n!$ are at least $n/2$. So, the natural interpretation for $n$ odd is to round down $n/2$. But I would be happy with a bijective proof that only directly encompasses the even case. As for why I want a bijective proof? The usual reason of wanting to see the underlying combinatorial structure. –  Per Vognsen Aug 2 '10 at 5:39
    
For a stronger lower bound which implies yours, see mathoverflow.net/questions/27912/… . –  Wadim Zudilin Aug 2 '10 at 5:40
1  
By the way, maybe I should mention my original intuition that the lower bound should have something to do with the existence of fixed-point-free involutions on sets of even cardinality. –  Per Vognsen Aug 2 '10 at 5:56
    
Btw, as to elementary forms of the Stirling inequality, note also $e^n\geq \frac{n^n}{n!}$ from the exponential series. –  Pietro Majer Aug 2 '10 at 8:00

2 Answers 2

up vote 6 down vote accepted

Think of all maps from the first $n/2$ elements of {$1,...,n$} to the last $n/2$. Say, let $a_1< \ldots < a_k \to z$. Make a cycle $a_1 \to a_2 \to \ldots \to a_k \to z \to a_1$. Do this for all $z$. The details are straightforward. This proves the lower bound.

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Beautiful. Thanks! This also works in the odd $n$ case by fixing the median element. –  Per Vognsen Aug 2 '10 at 8:57

n! counts one-to-one functions from {1,2,...,n} to itself while $n^n$ counts all such functions. For a bijective proof of the lower bound one would likely want n=2m, Then the lower number is the $m^m$ functions g from {1,...,m} to itself . I thought I had an easy bijection to a class of partial one-to-one functions but the on I had isn't right...

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Yeah, the upper bound poses no problem. I too have had a couple of false starts on a bijective proof of the lower bound, so I'm sympathetic to your mistake. :) –  Per Vognsen Aug 2 '10 at 5:19
    
OK, its not so pretty but one-to-one functions from {1,...,m} to {1,...,n} number n*(n-1)*...*(m+1) and all m terms are greater than m. So take any of the m^m functions g from {1..m} to itself and make a function f by assigning in turn f(1),f(2),... where f(i) is the g(i)th in order from the things not already assigned. –  Aaron Meyerowitz Aug 2 '10 at 5:32

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