Stirling's Formula
筆者在另一篇證明用常態分佈近似二項分佈的文章中,用到了這個Stirling's Formula,這裡證明一下。主要內容來自於page 552, section 11, chapter 11 in Boas's Mathematical Methods in The Physical Sciences 3rd edition. 黑色字體是課文,藍色字體是我的補充說明。
n!∼nne−n√2πn
Proof:
Start with p!=∫∞0xpe−xdx=∫∞0e(plnx)−xdx.
Remark: I show that p!=∫∞0xpe−xdx by integral by parts.xp | e−x | |
\+ | ||
pxp−1 | −e−x | |
\− | ||
p(p−1)xp−2 | e−x | |
\+ | ||
p(p−1)(p−2)xp−3 | −e−x | |
\− | ||
⋮ | ⋮ | ⋮ |
p(p−1)(p−2)⋯2x | ±e−x | |
\± | ||
p! | ±e−x | |
\± | ||
0 | ---∫± | ±e−x |
Substitute a new variable y such that x=p+y\sqrt{p}.
Then dx=\sqrt{p}dy
x=0 \text{ corresponds to }y=-\sqrt{p},
and (11.2) becomes p!=\int_{-\sqrt{p}}^{\infty}e^{p\ln{(p+y\sqrt{p})}-p-y\sqrt{p}}\sqrt{p}dy.\tag{11.3}
For large p, the logarithm can be expanded in the following power series: \ln{(p+y\sqrt{p})}=\ln{p}+\ln{\left(1+\frac{y}{\sqrt{p}}\right)}=\ln{p}+\frac{y}{\sqrt{p}}-\frac{y^2}{2p}+\cdots.\tag{11.4}
since \frac{1}{1-x}=1+x+x^2+\cdots and \ln{(1-x)}=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots by integrating both sides and \ln{(1+x)}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots.
I don't know why we can omit the later terms in the following deduction.
Substituting (11.4) into (11.3), we get \begin{array}{rcl} p! &\sim& \int_{-\sqrt{p}}^{\infty} e^{(p\ln{p})+y\sqrt{p}-(y^2/2)-p-y\sqrt{p}}\sqrt{p}dy\\ &\stackrel{\text{cancel out }y\sqrt{p}}{=}& e^{(p\ln{p})-p}\sqrt{p}\int_{-\sqrt{p}}^{\infty}e^{-y^2/2}dy \\ &=& p^p e^{-p}\sqrt{p}\left[\int_{-\infty}^{\infty}e^{-y^2/2}dy-\int_{-\infty}^{-\sqrt{p}}e^{-y^2/2}dy\right]. \end{array}
The first integral is easily shown to be \sqrt{2\pi} (Problem 9.4). The second integral tends to zero as p\to \infty, and we have p!\sim p^{p} e^{-p} \sqrt{2\pi p}
which is (11.1).
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