Gamma函數(shù)的Stirling公式

2022-04-30 16:20 作者:子瞻Louis 0人讀過(guò) | 我要投稿

Stirling公式最早是對(duì)正整數(shù)的階乘的一個(gè)漸進(jìn)公式，由亞伯拉罕·棣莫弗發(fā)現(xiàn)，形式為：

$n!%5Csim%20C%5Csqrt%20n%5Cleft(%5Cfrac%20ne%5Cright)%5En%2C%5Cquad%20n%5Cto%5Cinfty$

而Stirling計(jì)算出了其中的常數(shù)? $C%3D%5Csqrt%7B2%5Cpi%7D$ ，因此我們現(xiàn)在才稱(chēng)它為Stirling公式，它也可以利用Gamma函數(shù)推廣到復(fù)數(shù)，這也正是本文的內(nèi)容

經(jīng)典的Stirling公式

從階乘對(duì)數(shù)來(lái)考慮

$%5Clog%20N!%3D%5Csum_%7Bn%5Cle%20N%7D%5Clog%20n$

用Euler-Maclaurin求和公式可將右式寫(xiě)為

$%5Cbegin%7Baligned%7D%5Csum_%7Bn%5Cle%20N%7D%5Clog%20n%26%3D%5Cint_%7B1%7D%5EN%5Clog%20t%5Cmathrm%20d%5Bt%5D%5C%5C%26%3D%5Cint_%7B1%7D%5EN%5Clog%20t%5Cmathrm%20dt%2B%5Cfrac12%5Clog%20N%2B%5Cint_%7B1%7D%5EN%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cleft(N%2B%5Cfrac12%5Cright)%5Clog%20N-N%2B1%2B%5Cint_1%5EN%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

其中? $%5Ctilde%20B_k(x)%3A%3DB_k(%5C%7Bx%5C%7D)$ ， $B_k(t)$ ?為第k個(gè)伯努利多項(xiàng)式，令? $M%3EN$ ，用一次分部積分，可得

$%5Cint_N%5EM%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%7D%5Cmathrm%20dt%3D%5Cfrac%7B%5Ctilde%20B_2(M)%7D%7B2M%7D-%5Cfrac%7B%5Ctilde%20B_2(N)%7D%7B2N%7D%2B%5Cint_N%5E%5Cinfty%5Cfrac%7B%5Ctilde%20B_2(t)%7D%7B2t%5E2%7D%5Cmathrm%20dt%5Cll%5Cfrac1N$

故當(dāng)? $N%5Cto%5Cinfty$ ?時(shí)式中的積分收斂，由此不妨設(shè)

$%5Cint_1%5E%5Cinfty%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%7D%5Cmathrm%20dt%3DC$

則

$%5Cint_1%5EN%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%7D%5Cmathrm%20dt%3DC-%5Cint_N%5E%5Cinfty%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%7D%5Cmathrm%20dt%3DC%2B%5Cmathcal%20O%5Cleft(%5Cfrac1N%5Cright)$

代回到式中，得

$%5Clog%20N!%3D%5Cleft(N%2B%5Cfrac12%5Cright)%5Clog%20N-N%2B1%2BC%2B%5Cmathcal%20O%5Cleft(%5Cfrac1N%5Cright)$

再根據(jù)? $%5Cexp%5Cmathcal%20O%5Cleft(%5Cfrac1N%5Cright)%3D1%2B%5Cmathcal%20O%5Cleft(%5Cfrac1N%5Cright)$ ，得

$N!%3De%5E%7BC%2B1%7D%5Csqrt%20N%5Cleft(%5Cfrac%20Ne%5Cright)%5EN%5Cleft(1%2B%5Cmathcal%20O%5Cleft(%5Cfrac1N%5Cright)%5Cright)$

令? $C_0%3De%5E%7BC%2B1%7D$ ?就是Stirling公式的雛形了，下面就來(lái)計(jì)算其中常數(shù)的值了

根據(jù)正弦函數(shù)的無(wú)窮乘積展開(kāi)，

$%5Csin%7B%5Cpi%20z%7D%3D%5Cpi%20z%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1-%5Cfrac%7Bz%5E2%7D%7Bn%5E2%7D%5Cright)$

取? $z%3D%5Cfrac12$ ?，可得大名鼎鼎的Wallis乘積：

$%5Cbegin%7Baligned%7D%5Cfrac2%5Cpi%26%3D%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1-%5Cfrac%7B1%7D%7B(2n)%5E2%7D%5Cright)%5C%5C%26%3D%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1-%5Cfrac1%7B2n%7D%5Cright)%5Cleft(1%2B%5Cfrac1%7B2n%7D%5Cright)%5C%5C%26%3D%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B2n-1%7D%7B2n%7D%5Ccdot%5Cfrac%7B2n%2B1%7D%7B2n%7D%5Cend%7Baligned%7D$

$%5CRightarrow%5Cfrac2%5Cpi%3D%5Cfrac12%5Ccdot%5Cfrac32%5Ccdot%5Cfrac34%5Ccdot%5Cfrac54%5Ccdot%5Cfrac56%5Ccdot%5Cfrac76%5Ccdot%5Cdots$

用雙階乘符號(hào)：

$n!!%3A%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%201%5Ccdot3%5Ccdot5%5Cdots(n-2)%5Ccdot%20n%20%26%20n%E6%98%AF%E5%A5%87%E6%95%B0%5C%5C2%5Ccdot4%5Ccdot6%5Cdots(n-2)%5Ccdot%20n%20%26%20n%E6%98%AF%E5%81%B6%E6%95%B0%20%5Cend%7Barray%7D%20%5Cright.$

可將乘積公式寫(xiě)為

$%5Cfrac2%5Cpi%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7B(2N-1)!!%5E2(2N%2B1)%7D%7B(2N)!!%5E2%7D$

又不難注意到

$(2N-1)!!%3D%5Cfrac%7B1%5Ccdot2%5Cdots%20(2N-1)%5Ccdot2N%7D%7B2%5Ccdot4%5Cdots(2N-2)%5Ccdot2N%7D%3D%5Cfrac%7B(2N)!%7D%7B(2N)!!%7D$

$(2N)!!%3D2%5ENN!$

由此可得

$%5Cfrac2%5Cpi%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7B(2N)!%5E2(2N%2B1)%7D%7B2%5E%7B4N%7DN!%5E4%7D$

將之前得到的漸進(jìn)公式代入，可得

$%5Cbegin%7Baligned%7D%5Cfrac2%5Cpi%26%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7BC_0%5E2%5Ccdot2N%5Ccdot%5Cleft(%5Cfrac%20%7B2N%7De%5Cright)%5E%7B4N%7D(2N%2B1)%7D%7BC_0%5E4%5Ccdot%20N%5E2%5Ccdot%5Cleft(%5Cfrac%20%7B2N%7De%5Cright)%5E%7B4N%7D%7D%5C%5C%26%3D%5Cfrac2%7BC_0%5E2%7D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7B2N%2B1%7D%7BN%7D%5C%5C%26%3D%5Cfrac4%7BC_0%5E2%7D%5Cend%7Baligned%7D$

即? $C_0%3D%5Csqrt%7B2%5Cpi%7D$ ，于是便得到：

（經(jīng)典Stirling公式）對(duì)足夠大的正整數(shù) N

$N!%3D%5Csqrt%20%7B2%5Cpi%20N%7D%5Cleft(%5Cfrac%20Ne%5Cright)%5EN%5Cleft(1%2B%5Cmathcal%20O%5Cleft(%5Cfrac1N%5Cright)%5Cright)$

或者

$%5Clog%20N!%3D%5Cleft(N%2B%5Cfrac12%5Cright)%5Clog%20N-N%2B%5Cfrac12%5Clog%20%7B2%5Cpi%7D%2B%5Cmathcal%20O%5Cleft(%5Cfrac1N%5Cright)$

復(fù)Stirling公式

從Gamma函數(shù)的Weierstrass公式來(lái)考慮

$%5Cfrac1%7B%5CGamma(s)%7D%3Dse%5E%7B%5Cgamma%20s%7D%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1%2B%5Cfrac%20sn%5Cright)e%5E%7B-s%2Fn%7D$

對(duì)它取對(duì)數(shù)

$-%5Clog%5CGamma(s)%3D%5Clog%20s%2B%5Cgamma%20s%2B%5Clim_%7BN%5Cto%5Cinfty%7D%5Csum_%7Bn%3D1%7D%5EN%5Clog(n%2Bs)-%5Cfrac%20sn-%5Clog%20N!$

再次用Euler-Maclaurin公式，有

$%5Cbegin%7Baligned%7D%5Csum_%7B0%3Cn%5Cle%20N%7D%5Clog(n%2Bs)%26%3D%5Cint_0%5EN%5Clog(t%2Bs)%5Cmathrm%20dt%2B%5Cfrac12(%5Clog(N%2Bs)-%5Clog%20s)%2B%5Cint_0%5EN%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%2Bs%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cleft(s%2BN%2B%5Cfrac12%5Cright)%5Clog%20(N%2Bs)-N-%5Cleft(s%2B%5Cfrac12%5Cright)%5Clog%20s%2B%5Cint_0%5EN%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%2Bs%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

其中積分項(xiàng)當(dāng)? $s%5Cin%5Cmathbb%20C%5Cbackslash%20%5Cmathbb%20R%5E-$ ?時(shí)收斂。由經(jīng)典的估計(jì)

$%5Csum_%7Bn%5Cle%20N%7D%5Cfrac%20sn%3Ds%5Clog%20N%2B%5Cgamma%20s%2B%5Cmathcal%20O%5Cleft(%5Cfrac1N%5Cright)$

將他們相減并再減去經(jīng)典的Stirling公式，有

$%5Cbegin%7Baligned%7D%26%5Csum_%7Bn%3D1%7D%5EN%5Clog(n%2Bs)-%5Cfrac%20sn-%5Clog%20N!%5C%5C%3D%26%5Cleft(s%2BN%2B%5Cfrac12%5Cright)%5Clog%20(N%2Bs)-N-%5Cleft(s%2B%5Cfrac12%5Cright)%5Clog%20s%2B%5Cint_0%5EN%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%2Bs%7D%5Cmathrm%20dt%5C%5C%26-s%5Clog%20N-%5Cgamma%20s-%5Cleft(N%2B%5Cfrac12%5Cright)%5Clog%20N%2BN-%5Cfrac12%5Clog%20%7B2%5Cpi%7D%2B%5Cmathcal%20O%5Cleft(%5Cfrac1N%5Cright)%5C%5C%3D%26-%5Cgamma%20s-%5Cleft(s%2B%5Cfrac12%5Cright)%5Clog%20s-%5Cfrac12%5Clog%202%5Cpi%2B%5Cint_0%5EN%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%2Bs%7D%5Cmathrm%20dt%5C%5C%26%2B%5Cleft(s%2BN%2B%5Cfrac12%5Cright)%5Clog%5Cleft(1%2B%5Cfrac%20sN%5Cright)%2B%5Cmathcal%20%0A%20O%5Cleft(%5Cfrac1N%5Cright)%5Cend%7Baligned%7D$

令? $N%5Cto%5Cinfty$ ，可得

$%5Cbegin%7Baligned%7D%26%5Clim_%7BN%5Cto%5Cinfty%7D%5Csum_%7Bn%3D1%7D%5EN%5Clog(n%2Bs)-%5Cfrac%20sn-%5Clog%20N!%5C%5C%3D%26-%5Cgamma%20s-%5Cleft(s%2B%5Cfrac12%5Cright)%5Clog%20s%2Bs-%5Cfrac12%5Clog%202%5Cpi%2B%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%2Bs%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

代回Weierstrass公式的對(duì)數(shù)中，得

$-%5Clog%5CGamma(s)%3D-%5Cleft(s-%5Cfrac12%5Cright)%5Clog%20s%2Bs-%5Cfrac12%5Clog%202%5Cpi%2B%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%2Bs%7D%5Cmathrm%20dt$

即

（復(fù)Stirling公式）對(duì)? $s%5Cin%5Cmathbb%20C%5Cbackslash%5Cmathbb%20R%5E-$ ?

$%5Clog%5CGamma(s)%3D%5Cleft(s-%5Cfrac12%5Cright)%5Clog%20s-s%2B%5Cfrac12%5Clog%202%5Cpi-%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%2Bs%7D%5Cmathrm%20dt$

也可以將它取導(dǎo)數(shù)，即

$%5Cfrac%7B%5CGamma'%7D%7B%5CGamma%7D(s)%3D%5Clog%20s-%5Cfrac1%7B2s%7D%2B%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7B(t%2Bs)%5E2%7D%5Cmathrm%20dt$

且由? $%7C%5Ctilde%20B(t)%7C%5Cle1$ 可得

$%5Cfrac%7B%5CGamma'%7D%7B%5CGamma%7D(s)%3D%5Clog%20s%2B%5Cmathcal%20O%5Cleft(%5Cfrac1%7B%7Cs%7C%7D%5Cright)$

帶型區(qū)域上的估階

接下來(lái)考慮Gamma函數(shù)在復(fù)平面上的帶型區(qū)域中的狀態(tài)，之所以要這樣考慮，是因?yàn)樵谟懻揜iemann?zeta函數(shù)的函數(shù)方程時(shí)， $s$ ?的實(shí)部并不會(huì)太大，反而虛部有可能會(huì)很大.?那么就假定有? $a%5Cle%5CRe(s)%5Cle%20b$ ?，對(duì)? $s%3D%5Csigma%2Bi%5Ctau(%5Csigma%2C%5Ctau%5Cin%5Cmathbb%20R)$ ?的情況，我們來(lái)對(duì)? $%5CGamma$ ?的復(fù)Stirling公式一項(xiàng)一項(xiàng)看，

$%5Clog%7Cs%7C%3D%5Cfrac12%5Clog(%5Csigma%5E2%2B%5Ctau%5E2)%3D%5Clog%7C%5Ctau%7C%2B%5Cfrac12%5Clog%5Cleft(1%2B%5Cfrac%7B%5Csigma%5E2%7D%7B%5Ctau%5E2%7D%5Cright)$

因此當(dāng)? $%5Ctau%5Cge%5Cmax(a%2Cb)$ ?時(shí)，有

$%5Clog%7Cs%7C%3D%5Clog%7C%5Ctau%7C%2B%5Cmathcal%20O%5Cleft(%5Cfrac1%7B%5Ctau%5E2%7D%5Cright)$

并且

$%5Cbegin%7Balign%7D%5Carg%20s%26%3D%5CIm%5Clog%20(%5Csigma%2Bi%5Ctau)%5C%5C%26%3D%5CIm%5Clog(i%5Ctau)%2B%5CIm%5Clog%5Cleft(1%2B%5Cfrac%7B%5Csigma%7D%7Bi%5Ctau%7D%5Cright)%5C%5C%26%3D%5Cfrac%5Cpi2%5Ctext%7Bsgn%7D(%5Ctau)-%5Cfrac%7B%5Csigma%7D%7B%5Ctau%7D%2B%5Cleft(%5Cfrac%7B1%7D%7B%5Ctau%5E2%7D%5Cright)%5Cend%7Balign%7D$

因此

$%5Clog%20s%3D%5Clog%7Cs%7C%2Bi%5Carg%20s%3D%5Clog%7C%5Ctau%7C%2Bi%5Cleft%5C%7B%5Cfrac%5Cpi2%5Ctext%7Bsgn%7D(%5Ctau)-%5Cfrac%7B%5Csigma%7D%7B%5Ctau%7D%5Cright%5C%7D%2B%5Cleft(%5Cfrac%7B1%7D%7B%5Ctau%5E2%7D%5Cright)$

從而

$%5Cbegin%7Balign%7D%5Cleft(s-%5Cfrac12%5Cright)%5Clog%20s%3D%26%5Cleft(%5Csigma%2Bi%5Ctau-%5Cfrac12%5Cright)%5Cleft(%5Clog%7C%5Ctau%7C%2Bi%5Cleft%5C%7B%5Cfrac%5Cpi2%5Ctext%7Bsgn%7D(%5Ctau)-%5Cfrac%7B%5Csigma%7D%7B%5Ctau%7D%5Cright%5C%7D%5Cright)%2B%5Cleft(%5Cfrac%7B1%7D%7B%7C%5Ctau%7C%7D%5Cright)%5C%5C%3D%26%5Cleft(%5Csigma-%5Cfrac12%5Cright)%5Cln%7C%5Ctau%7C-%5Cfrac%7B%5Cpi%7C%5Ctau%7C%7D2%2B%5Csigma%2BiK_%5Csigma(%5Ctau)%2B%5Cmathcal%20O%5Cleft(%5Cfrac1%7B%7C%5Ctau%7C%7D%5Cright)%5Cend%7Balign%7D$

其中

$K_%5Csigma(%5Ctau)%3D%5Ctau%5Cln%7C%5Ctau%7C%2B%5Cfrac%5Cpi2%5Ctext%7Bsgn%7D(%5Ctau)%5Cleft(%5Csigma-%5Cfrac12%5Cright)%5Cin%5Cmathbb%20R$

然后因? $%7C%5Ctilde%20B_2(t)%7C%5Cle1$ ?，所以積分

$%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Ctilde%20B_1(t)%7D%7Bt%2Bs%7D%5Cmathrm%20dt%3D%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Ctilde%20B_2(t)%7D%7B2(t%2Bs)%5E2%7D%5Cmathrm%20dt%5Cll%5Cfrac1%7B%7Cs%7C%7D%5Cle%5Cfrac1%7B%7C%5Ctau%7C%7D$

由此將它們代入Stirling公式中，得

$%5Cbegin%7Balign%7D%5Clog%5CGamma(%5Csigma%2Bi%5Ctau)%3D%26%5Cleft(%5Csigma-%5Cfrac12%5Cright)%5Cln%7C%5Ctau%7C-%5Cfrac%7B%5Cpi%7C%5Ctau%7C%7D2%2B%5Clog%20%5Csqrt%7B2%5Cpi%7D%5C%5C%26%2Bi(K_%5Csigma(%5Ctau)-%5Ctau)%2B%5Cmathcal%20O%5Cleft(%5Cfrac1%7B%7C%5Ctau%7C%7D%5Cright)%5Cend%7Balign%7D$

最后取個(gè)冪，得

$%5CGamma(%5Csigma%2Bi%5Ctau)%3D%5Cleft%5C%7B1%2B%5Cmathcal%20O%5Cleft(%5Cfrac1%5Ctau%5Cright)%5Cright%5C%7D%5Csqrt%7B2%5Cpi%7D%7C%5Ctau%7C%5E%7B%5Csigma-%5Cfrac12%7De%5E%7B-%5Cpi%7C%5Ctau%7C%2F2%7De%5E%7Bi(K_%5Csigma(%5Ctau)-%5Ctau)%7D$

好了，差不多了，溜了溜了

標(biāo)簽：斯特林公式 Gamma函數(shù)數(shù)學(xué)階乘

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Gamma函數(shù)的Stirling公式

經(jīng)典的Stirling公式

復(fù)Stirling公式

帶型區(qū)域上的估階