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(高中生數(shù)學(xué))斯特林公式證明

2023-06-18 02:29 作者:封包  | 我要投稿

高中生自行整理,如有錯(cuò)誤,歡迎指出,不勝感激。

斯特林公式(Stirling's approximation)是一條用來(lái)取n的階乘的近似值的數(shù)學(xué)公式。一般來(lái)說(shuō),階乘的計(jì)算復(fù)雜度為線(xiàn)性。當(dāng)要為某些極大大的n求階乘時(shí),常見(jiàn)的方法復(fù)雜度不可接受。斯特林公式能夠?qū)⑶蠼怆A乘的復(fù)雜度降低到對(duì)數(shù)級(jí)。而且,即使在n很小的時(shí)候,斯特林公式的取值已經(jīng)十分準(zhǔn)確。

斯特林公式如下:

n%20!%3D%5Csqrt%7B2%20%5Cpi%20n%7D%20%5Ccdot%5Cleft(%5Cfrac%7Bn%7D%7B%5Cmathrm%7Be%7D%7D%5Cright)%5En%20%5Ccdot%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B%5Ctheta%7D%7B12%20n%7D%7D%20%5Cquad(0%3C%5Ctheta%3C1)

我們已經(jīng)知道ln(1+x)與ln(1-x)的泰勒展開(kāi):

%5Cln%20(1%2Bx)%3Dx-%5Cfrac%7Bx%5E2%7D%7B2%7D%2B%5Cfrac%7Bx%5E3%7D%7B3%7D-%5Ccdots%2B(-1)%5E%7Bn-1%7D%20%5Cfrac%7Bx%5En%7D%7Bn%7D%2B%5Ccdots%20%5Cquad(-1%3Cx%20%5Cleq%201)

%E5%B0%86x%E6%9B%BF%E6%8D%A2%E4%B8%BA-x%3A%0A%5Cln%20(1-x)%3D-x-%5Cfrac%7Bx%5E2%7D%7B2%7D-%5Cfrac%7Bx%5E3%7D%7B3%7D-%5Ccdots-%5Cfrac%7Bx%5En%7D%7Bn%7D-%5Ccdots%20%5Cquad(-1%3Cx%3C1)

%E4%B8%A4%E8%80%85%E7%9B%B8%E5%87%8F%E5%8E%BB%E5%BE%97%E5%88%B0%EF%BC%9A%0A%5Cln%20%5Cfrac%7B1%2Bx%7D%7B1-x%7D%3D2%20x%5Cleft(1%2B%5Cfrac%7Bx%5E2%7D%7B3%7D%2B%5Cfrac%7Bx%5E4%7D%7B5%7D%2B%5Ccdots%2B%5Cfrac%7Bx%5E%7B2%20m%7D%7D%7B2%20m%2B1%7D%2B%5Ccdots%5Cright)%20%5Cquad(-1%3Cx%3C1)

設(shè)x=1/(2n+1)?,n=1,2,3,···,得:

%5Cfrac%7B1%2Bx%7D%7B1-x%7D%3D%5Cfrac%7B1%2B%5Cfrac%7B1%7D%7B2%20n%2B1%7D%7D%7B1-%5Cfrac%7B1%7D%7B2%20n%2B1%7D%7D%3D%5Cfrac%7B2%20n%2B2%7D%7B2%20n%7D%3D%5Cfrac%7Bn%2B1%7D%7Bn%7D

所以上式子就變?yōu)榱耍?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%5Cln%20%5Cfrac%7Bn%2B1%7D%7Bn%7D%3D%5Cfrac%7B2%7D%7B2%20n%2B1%7D%5Cleft%5B1%2B%5Cfrac%7B1%7D%7B3%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B(2%20n%2B1)%5E%7B2%7D%7D%2B%5Cfrac%7B1%7D%7B5%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B(2%20n%2B1)%5E%7B4%7D%7D%2B%5Ccdots%5Cright%5D" alt="%5Cln%20%5Cfrac%7Bn%2B1%7D%7Bn%7D%3D%5Cfrac%7B2%7D%7B2%20n%2B1%7D%5Cleft%5B1%2B%5Cfrac%7B1%7D%7B3%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B(2%20n%2B1)%5E%7B2%7D%7D%2B%5Cfrac%7B1%7D%7B5%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B(2%20n%2B1)%5E%7B4%7D%7D%2B%5Ccdots%5Cright%5D">

%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%20%5Cln%20%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%3D%5Cleft%5B1%2B%5Cfrac%7B1%7D%7B3%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B(2%20n%2B1)%5E%7B2%7D%7D%2B%5Cfrac%7B1%7D%7B5%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B(2%20n%2B1)%5E%7B4%7D%7D%2B%5Ccdots%5Cright%5D

而右邊的式子<1%2B%5Cfrac%7B1%7D%7B3%7D%5Cleft(%5Cfrac%7B1%7D%7B(2%20n%2B1)%5E%7B2%7D%7D%2B%5Cfrac%7B1%7D%7B(2%20n%2B1)%5E%7B4%7D%7D%2B%5Cldots%5Cright)%3D1%2B%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B(2%20n%2B1)%5E%7B2%7D%7D%7D%7B1%20%5Cfrac%7B1%7D%7B(2%20n%2B1)%5E%7B2%7D%7D%7D%3D1%2B%5Cfrac%7B1%7D%7B3%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B4%20n%5E%7B2%7D%2B4%20n%7D%3D1%2B%5Cfrac%7B1%7D%7B12%20n(n%2B1)%7D

因?yàn)樽笫斤@然>1,所以有1%3C%5Cleft(n%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%20%5Cln%20%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%3C1%2B%5Cfrac%7B1%7D%7B12%20%5Cln%20(n%2B1)%7D

取一下指數(shù):%5Cleft.e%3C%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5E%7Bn%2B%5Cfrac%7B1%7D%7B2%7D%7D%3Ce%5E%7B1%2B%5Cfrac%7B1%7D%7B2%20n(n%2B1)%7D%7D%20%5CLeftrightarrow%201%3C%5Cfrac%7B%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5E%7Bn%2B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7Be%7D%3Ce%5E%7B%5Cfrac%7B1%7D%7B2%20n(n%2B1)%201%7D%7D%5Cright%5C%7D

設(shè)一下:a_%7Bn%7D%3D%20%5Cfrac%7B%20n%20!%20e%5E%7Bn%7D%7D%7Bn%5E%7Bn%2B%5Cfrac%7B1%7D%7B2%7D%7D%7D那么%5Cfrac%7Ba_%7Bn%7D%7D%7Ba_%7Bn%2B1%7D%7D%3D%5Cfrac%7Bn%20!%20e%5E%7Bn%7D(n%2B1)%5E%7Bn%2B1%2B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7Bn%5E%7Bn%2B%5Cfrac%7B1%7D%7B2%7D%7D(n%2B1)%20!%20e%5E%7Bn%2B1%7D%7D%3D%5Cfrac%7B%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5E%7Bn%2B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7Be%7D

由上面的不等式知道:%5Cfrac%7Ba_%7Bn%7D%7D%7Ba_%7Bn%2B1%7D%7D%3E1 所以該數(shù)列其實(shí)是單調(diào)遞減的,且顯然一定有下界0

那么該數(shù)列的極限一定存在,不妨設(shè)為a

%5Cfrac%7Ba_%7Bn%7D%7D%7Ba_%7Bn%2B1%7D%7D%3D%5Cfrac%7B%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5E%7Bn%2B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B%5Cmathrm%7Be%7D%7D%3C%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7B12%20n(n%2B1)%7D%7D%3D%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7B12%7D%5Cleft(%5Cfrac%7B1%7D%7Bn%7D-%5Cfrac%7B1%7D%7Bn%2B1%7D%5Cright)%7D%3D%5Cfrac%7B%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7B12%20n%7D%7D%7D%7B%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7B12(n%2B1)%7D%7D%7D

最終得到:a_%7Bn%7D%20%5Cmathrm%7Be%7D%5E%7B-%5Cfrac%7B1%7D%7B12%20n%7D%7D%3Ca_%7Bn%2B1%7D%20%5Cmathrm%7Be%7D%5E%7B-%5Cfrac%7B1%7D%7B12(n%2B1)%7D%7D我們能看出來(lái)左右兩個(gè)數(shù)列a_%7Bn%7D%20e%5E%7B-%5Cfrac%7B1%7D%7B12%20n%7D%7D是單調(diào)遞增的。

%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7D%20a_%7Bn%7D%20e%5E%7B-%5Cfrac%7B1%7D%7B12%20n%7D%7D%3Da,所以a_%7Bn%7D%20e%5E%7B-%5Cfrac%7B1%7D%7B12%20n%7D%7D%3Ca%3Ca_%7Bn%7D

-1%3C-%5Ctheta%3C0%20%5CRightarrow-%5Cfrac%7B1%7D%7B12%20n%7D%3C-%5Cfrac%7B%5Ctheta%7D%7B12%20n%7D%3C0

兩邊取下指數(shù):e%5E%7B-%5Cfrac%7B1%7D%7B12%20n%7D%7D%3Ce%5E%7B-%5Cfrac%7B%5Ctheta%7D%7B12%20n%7D%7D%3C1%20%5CRightarrow%20a_%7Bn%7D%20e%5E%7B%5Cfrac%7B1%7D%7B12%20n%7D%7D%3Ca_%7Bn%7D%20e%5E%7B-%5Cfrac%7B%5Ctheta%7D%7B12%20n%7D%7D%3Ca_%7Bn%7D

那么游戲就進(jìn)行到了一半,現(xiàn)在可以得出結(jié)論:a%3Da_%7Bn%7D%20e%5E%7B-%5Cfrac%7B%5Ctheta%7D%7B12%20n%7D%7D%2C%20%5Ctheta%20%5Cin(0%2C1)

把前面的an帶進(jìn)去:n%20!%3D%5Cfrac%7Ba%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B%5Ctheta%7D%7B12%20n%7D%7D%20n%5E%7Bn%2B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B%5Cmathrm%7Be%7D%5E%7Bn%7D%7D%3Da%20%5Csqrt%7Bn%7D%20%5Ccdot%5Cleft(%5Cfrac%7Bn%7D%7B%5Cmathrm%7Be%7D%7D%5Cright)%5E%7Bn%7D%20%5Ccdot%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B%5Ctheta%7D%7B12%20n%7D%7D%2C%20%5Cquad%200%3C%5Ctheta%3C1

Wallis圓周率無(wú)窮乘積公式:%5Cfrac%7B%5Cpi%7D%7B2%7D%3D%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7B2%20n%2B1%7D%5Cleft%5B%5Cfrac%7B2%20n%20!%20!%7D%7B(2%20n-1)%20!%20!%7D%5Cright%5D%5E%7B2%7D

%5Cfrac%7B2%20n%20!%20!%7D%7B(2%20n-1)%20!%20!%7D%3D%5Cfrac%7B(2%20n%20!%20!)%5E%7B2%7D%7D%7B2%20n%20!%20!(2%20n-1)%20!%20!%7D%3D%5Cfrac%7B(2%20n%20!%20!)%5E%7B2%7D%7D%7B2%20n%20!%7D%3D%5Cfrac%7B%5Cleft(2%5E%7Bn%7D%20%5Ctimes%20n%20!%5Cright)%5E%7B2%7D%7D%7B2%20n%20!%7D%3D%5Cfrac%7B2%5E%7B2%20n%7D(n%20!)%5E%7B2%7D%7D%7B2%20n%20!%7D

將前面的結(jié)論n變成2n:(2%20n)%20!%3Da%20%5Csqrt%7B2%20n%7D%20%5Ccdot%5Cleft(%5Cfrac%7B2%20n%7D%7B%5Cmathrm%7Be%7D%7D%5Cright)%5E%7B2%20n%7D%20%5Ccdot%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B%5Ctheta%5E%7B%5Cprime%7D%7D%7B24%20n%7D%7D%2C%20%5Cquad%200%3C%5Ctheta%5E%7B%5Cprime%7D%3C1

%3D%5Cfrac%7B2%5E%7B2%20n%7D(n%20!)%5E%7B2%7D%7D%7B2%20n%20!%7D%3D%5Cfrac%7B2%5E%7B2%20n%7D%5Cleft%5Ba%20%5Csqrt%7Bn%7D%20%5Ccdot%5Cleft(%5Cfrac%7Bn%7D%7Be%7D%5Cright)%5E%7Bn%7D%20%5Ccdot%20e%5E%7B%5Cfrac%7B%5Ctheta%7D%7B12%20n%7D%7D%5Cright%5D%5E%7B2%7D%7D%7Ba%20%5Csqrt%7B2%20n%7D%20%5Ccdot%5Cleft(%5Cfrac%7B2%20n%7D%7B%5Cmathrm%7Be%7D%7D%5Cright)%5E%7B2%20n%7D%20%5Ccdot%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B%5Ctheta%5E%7B%5Cprime%7D%7D%7B24%20n%7D%7D%7D%3Da%20%5Cfrac%7B%5Csqrt%7Bn%7D%7D%7B%5Csqrt%7B2%7D%7D%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B4%20%5Ctheta-%5Ctheta'%7D%7B24%20n%7D%7D

帶回到極限中:%5Cfrac%7B%5Cpi%7D%7B2%7D%3D%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7B2%20n%2B1%7D%5Cleft%5Ba%20%5Cfrac%7B%5Csqrt%7Bn%7D%7D%7B%5Csqrt%7B2%7D%7D%20e%5E%7B%5Cfrac%7B4%20%5Ctheta-%5Ctheta%5E%7B%5Cprime%7D%7D%7B24%20n%7D%7D%5Cright%5D%5E%7B2%7D%3D%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7B2%20n%2B1%7D%20%5Ccdot%20a%5E%7B2%7D%20%5Ccdot%20%5Cfrac%7Bn%7D%7B2%7D%20%5Ccdot%20e%5E%7B%5Cfrac%7B4%20%5Ctheta-%5Ctheta%5E%7B%5Cprime%7D%7D%7B12%20n%7D%7D%3D%5Cfrac%7Ba%5E%7B2%7D%7D%7B4%7D

所以a%3D%5Csqrt%7B2%20%5Cpi%7D

大功告成,代回我們之前的結(jié)論:n%20!%3D%5Csqrt%7B2%20%5Cpi%20n%7D%20%5Ccdot%5Cleft(%5Cfrac%7Bn%7D%7B%5Cmathrm%7Be%7D%7D%5Cright)%5E%7Bn%7D%20%5Ccdot%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B%5Ctheta%7D%7B12%20n%7D%7D%20%5Cquad(0%3C%5Ctheta%3C1)即斯特林公式。

n很大的時(shí)候:%5Clim%20_%7Bn%20%5Crightarrow%2B%5Cinfty%7D%20%5Cfrac%7Bn%20!%7D%7B%5Csqrt%7B2%20%5Cpi%20n%7D%5Cleft(%5Cfrac%7Bn%7D%7Be%7D%5Cright)%5E%7Bn%7D%7D%3D1

所以n%20!%20%5Capprox%20%5Csqrt%7B2%20%5Cpi%20n%7D%5Cleft(%5Cfrac%7Bn%7D%7Be%7D%5Cright)%5E%7Bn%7D

斯特林公式在理論和應(yīng)用上都具有重要的價(jià)值,對(duì)于概率論的發(fā)展也有著重大的意義。在數(shù)學(xué)分析中,大多都是利用Г函數(shù)、級(jí)數(shù)和含參變量的積分等知識(shí)進(jìn)行證明或推導(dǎo),很為繁瑣冗長(zhǎng)。近年來(lái),一些國(guó)內(nèi)外學(xué)者利用概率論中的指數(shù)分布、泊松分布、χ2分布證之。

以下是一個(gè)計(jì)算斯特林公式的近似值的代碼。(使用Python語(yǔ)言):

請(qǐng)注意,斯特林公式是一個(gè)近似公式,對(duì)于較大的n,它的近似值更加準(zhǔn)確。


(高中生數(shù)學(xué))斯特林公式證明的評(píng)論 (共 條)

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