Stirling‘s Formula

Stirling‘s Formula"/>

Stirling‘s Formula

文章目录

• • Stirling's Formula

Keith Conrad. Stirling’s Formula.

Stirling’s Formula

lim ⁡ n → ∞ n ! ( n n / e n ) 2 π n = 1. \lim_{n \rightarrow \infty} \frac{n!}{(n^n/e^n)\sqrt{2\pi n}} =1. n→∞lim​(nn/en)2πn ​n!​=1.

Proof:

n ! = ∫ 0 ∞ x n e − x d x = ∫ − n ∞ ( n + n t ) n e − ( n + n t ) n d t = n n n e n ∫ − n ∞ ( 1 + t n ) n e − n t d t . = n n n e n ∫ − ∞ ∞ f n ( t ) d t , \begin{array}{ll} n! &= \int_{0}^\infty x^n e^{-x} \mathrm{d}x \\ &= \int_{-\sqrt{n}}^\infty (n+\sqrt{n}t)^n e^{-(n+\sqrt{n}t)} \sqrt{n} \mathrm{d}t \\ &= \frac{n^n \sqrt{n}}{e^n} \int_{-\sqrt{n}}^{\infty} (1+\frac{t}{\sqrt{n}})^n e^{-\sqrt{n}t} \mathrm{d}t. \\ &= \frac{n^n \sqrt{n}}{e^n} \int_{-\infty}^{\infty} f_n(t) \mathrm{d}t, \end{array} n!​=∫0∞​xne−xdx=∫−n ​∞​(n+n ​t)ne−(n+n ​t)n ​dt=ennnn ​​∫−n ​∞​(1+n ​t​)ne−n ​tdt.=ennnn ​​∫−∞∞​fn​(t)dt,​
其中
f n ( t ) = { 0 t < n ( 1 + t n ) n e − n t t ≥ n f_{n}(t) = \left \{ \begin{array}{ll} 0 & t< \sqrt{n} \\ (1+\frac{t}{\sqrt{n}})^n e^{-\sqrt{n}t} & t\ge \sqrt{n} \end{array} \right. fn​(t)={0(1+n ​t​)ne−n ​t​t<n ​t≥n ​​
接下来证明 f n ( t ) f_n(t) fn​(t)趋于 e − t 2 2 e^{-\frac{t^2}{2}} e−2t2​,
ln ⁡ f n ( t ) = n ln ⁡ ( 1 + t n ) − n t , t ≥ n , \ln f_n(t) = n \ln (1+ \frac{t}{\sqrt{n}}) - \sqrt{n}t , t \ge \sqrt{n}, lnfn​(t)=nln(1+n ​t​)−n ​t,t≥n ​,
ln ⁡ ( 1 + x ) = 0 + x − x 2 2 + o ( x 2 ) , \ln (1+x) = 0 + x - \frac{x^2}{2} + o(x^2), ln(1+x)=0+x−2x2​+o(x2),
当 n n n足够大的时候
ln ⁡ f n ( t ) = n t − t 2 / 2 + n t + o ( t 2 / n ) = − t 2 2 + o ( t 2 / n ) , \ln f_n(t) = \sqrt{n}t -t^2/2+\sqrt{n}t+o(t^2/n)=-\frac{t^2}{2}+o(t^2/n), lnfn​(t)=n ​t−t2/2+n ​t+o(t2/n)=−2t2​+o(t2/n),
故 f n ( t ) → e − t 2 / 2 f_n(t) \rightarrow e^{-t^2/2} fn​(t)→e−t2/2.
观察( t ≥ − n t \ge -\sqrt{n} t≥−n ​)
d d t ( ln ⁡ f n + 1 ( t ) − ln ⁡ f n ( t ) ) = n t n + t − n + 1 t n + 1 + t = ( n − n + 1 ) t 2 ( n + t ) ( n + 1 + t ) ≤ 0 , \begin{array}{ll} \frac{\mathrm{d}}{\mathrm{d}t}(\ln f_{n+1}(t) - \ln f_n(t) ) &= \frac{\sqrt{n}t}{\sqrt{n}+t} - \frac{\sqrt{n+1}t}{\sqrt{n+1}+t} \\ &= \frac{(\sqrt{n}-\sqrt{n+1})t^2}{(\sqrt{n}+t)(\sqrt{n+1}+t)} \le 0, \end{array} dtd​(lnfn+1​(t)−lnfn​(t))​=n ​+tn ​t​−n+1 ​+tn+1 ​t​=(n ​+t)(n+1 ​+t)(n ​−n+1 ​)t2​≤0,​
又 f n ( 0 ) = 0 f_n(0)=0 fn​(0)=0, 故
f n + 1 / f n ≥ 1 , t ∈ [ n , 0 ) , f_{n+1} /f_n \ge 1, \quad t \in [\sqrt{n},0), fn+1​/fn​≥1,t∈[n ​,0),
f n + 1 / f n ≤ 1 , t ∈ [ 0 , + ∞ ) . f_{n+1} /f_n \le 1, \quad t \in [0, +\infty). fn+1​/fn​≤1,t∈[0,+∞).
又 f n ( t ) f_n(t) fn​(t)非负, 故根据单调收敛定理和优解控制定理可知
lim ⁡ n → ∞ ∫ − ∞ + ∞ f n ( t ) d t = ∫ − ∞ + ∞ lim ⁡ n → ∞ f n ( t ) d t = ∫ − ∞ + ∞ e − t 2 2 d t = 2 π . \lim_{n\rightarrow \infty} \int_{-\infty}^{+\infty} f_n(t) \mathrm{d}t = \int_{-\infty}^{+\infty} \lim_{n\rightarrow \infty} f_n(t) \mathrm{d}t = \int_{-\infty}^{+\infty}e^{-\frac{t^2}{2}}\mathrm{d} t=\sqrt{2 \pi}. n→∞lim​∫−∞+∞​fn​(t)dt=∫−∞+∞​n→∞lim​fn​(t)dt=∫−∞+∞​e−2t2​dt=2π ​.

证毕.