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A Proof of Stirling's Formula

A Proof of Stirling's Formula

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!nnen2πn

Proof:

Start with p!=0xpexdx=0e(plnx)xdx.

Remark: I show that p!=0xpexdx by integral by parts.
xpex
\+
pxp1ex
\
p(p1)xp2ex
\+
p(p1)(p2)xp3ex
\
p(p1)(p2)2x±ex
\±
p!±ex
\±
0---±±ex
0xpexdx=limb[xpexpxp1exp(p1)xp2exp(p1)(p2)2xexp!ex]b0=p!,
where limxxcex=0
by L'Hospital Rule.

Substitute a new variable y such that x=p+yp.

Then dx=pdy

x=0 corresponds to y=p,

and (11.2) becomes p!=pepln(p+yp)pyppdy.

For large p, the logarithm can be expanded in the following power series: ln(p+yp)=lnp+ln(1+yp)=lnp+ypy22p+.

since 11x=1+x+x2+ and ln(1x)=xx22x33 by integrating both sides and ln(1+x)=xx22+x33.

I don't know why we can omit the later terms in the following deduction.

Substituting (11.4) into (11.3), we get p!pe(plnp)+yp(y2/2)pyppdycancel out yp=e(plnp)pppey2/2dy=ppepp[ey2/2dypey2/2dy].

The first integral is easily shown to be 2π (Problem 9.4). The second integral tends to zero as p, and we have p!ppep2πp

which is (11.1).

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