※ 引述《wwli (想延畢的q我)》之銘言：
: ※ 引述《plover (+oo)》之銘言：
: : n! ~ ( 2 Pi / (n+1) )^(1/2) e^(-(n+1)) (n+1)^(n+1)
: 感謝plover大師, 這就是Stirling公式的完整版,
: 不過原發問者的意思, 好像正是要問說這個公式何
: 以得到.
: 最簡單的一種看法大概是這樣的:
: log(n!) = log(1) + log(2) + ... + log(n)
: 以積分逼近這個和(想法類似於積分審斂法), 可得
: 當n大時, log(n!) "差不多" 相近於
: n
: ∫ log(x)dx = nlog(n) - n + 1
: 1
: 嚴格的估計可以用上下和來夾積分值獲得. 當然,
: 若要得到像plover大師那般精確的估計, 就需要
: 更深的技巧.
再貼一篇:
作者 [email protected] (.................), 看板 math
標題 Re: 請問這公式... 為什麼丫?
時間 清華資訊(楓橋驛站) (Fri May 3 00:40:09 2002)
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※ 引述《[email protected] (Embracing the Wind)》之銘言：
> ln(N!) = N*ln(N) - N
這等號不可能是對的, 應該是這樣的:
ln(x!) ～ x(lnx) - x as x->+∞.
證明如下:
By Stirling Formula: Γ(x+1):= x!～ (x^x)( e^(-x) )( √(2πx) ) as x->+∞.
So,
x!
ln --------------------------- ->0 as x->+∞.
(x^x)( e^(-x) )( √(2πx) )
That is, ln x! - (xln x - x) - ln √(2πx) -> 0 as x ->+∞.
So, given ε>0, there exists a positive M such that as x≧M, we have:
|lnx! - (xln x - x)| < |ln √(2πx)| + ε. (*)
Note that we want to show that ln(x!)/(xlnx - x) ->1 as x ->+∞.
Consider (*), and thus we have: ( for x ≧M )
|lnx! - (xln x - x)| |ln √(2πx)| + ε.
-------------------- < -------------------
|(xlnx - x)| |(xlnx - x)|
Hence, as x->+∞, the above inequality (the right term -> 0). So by
Sandwich theorem, we know that
|lnx! - (xln x - x)|
-------------------- -> 0 as x ->+∞.
|(xlnx - x)|
Therefore, |lnx!|/|(xlnx - x)| ->1 as x->+∞.
So, lnx!/(xlnx - x) -> 1 as x->+∞. That is, ln x! ～ (xlnx - x).
Note that the symbol "～" is called Landau Notation. It means that
if f(x)/g(x) ->1 as x->+∞, we denote it by f～g as x->+∞.
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