Prime Dream（3）——Mertens的幾個(gè)漸進(jìn)公式

2022-03-05 08:20 作者:子瞻Louis 0人讀過(guò) | 我要投稿

上一節(jié)里扯到了下面這樣的漸進(jìn)公式：

#） $%5Csum_%7Bp%5Cle%20x%7D%5Cfrac%7B%5Clog%20p%7D%7Bp%7D%3D%5Clog%20x%2B%5Cmathcal%20O(1)$

它被稱為Mertens第一定理，通過(guò)右式發(fā)散便能得到Euclid定理，此外，可以由他出發(fā)得到另外的與素?cái)?shù)有關(guān)的漸進(jìn)公式。本文出現(xiàn)的兩個(gè)公式是由Mertens本人提出的，它們與Mertens第一定理可以被共同稱為Mertens公式，而最后一個(gè)公式又被稱為Mertens第二定理

素?cái)?shù)的倒數(shù)和

引入一個(gè)常數(shù)——Mertens常數(shù)

$M%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Csum_%7Bp%5Cle%20x%7D%5Cfrac1p-%5Clog%5Clog%20x$

它在定義上與Euler常數(shù)十分類似，不過(guò)我們需要說(shuō)明這個(gè)式子是有意義的，即要說(shuō)明右側(cè)確實(shí)是收斂到某個(gè)常數(shù)的。為了簡(jiǎn)便，這里會(huì)采用以下記號(hào)：

$%5Cint_%7Bc-%7D%5Ex%3A%3D%5Clim_%7B%5Cdelta%5Cto%2B0%7D%5Cint_%7Bc-%5Cdelta%7D%5Ex$

首先令

$R(x)%3A%3D%5Csum_%7Bp%5Cle%20x%7D%5Cfrac%7B%5Clog%20p%7Dp-%5Clog%20x$

由Mertens第一定理可知 $R(x)%3D%5Cmathcal%20O(1)$ .根據(jù)Riemann-Stieltjes積分的分部積分，有

$%5Cbegin%7Baligned%7D%5Csum_%7Bp%5Cle%20x%7D%5Cfrac1p%26%3D%5Csum_%7Bp%5Cle%20x%7D%5Cfrac1%7B%5Clog%20p%7D%5Ccdot%5Cfrac%7B%5Clog%20p%7Dp%5C%5C%26%3D%5Cint_%7B2-%7D%5Ex%5Cfrac1%7B%5Clog%20t%7D%5Cmathrm%20d%5Csum_%7Bp%5Cle%20t%7D%5Cfrac%7B%5Clog%20p%7Dp%5C%5C%26%3D%5Cint_%7B2-%7D%5Ex%5Cfrac1%7B%5Clog%20t%7D%5Cmathrm%20d%5Clog%20t%2B%5Cint_%7B2-%7D%5Ex%5Cfrac1%7B%5Clog%20t%7D%5Cmathrm%20dR(t)%5C%5C%26%3D%5Clog%5Clog%20x-%5Clog%5Clog2%2B%5Cfrac%7BR(x)%7D%7B%5Clog%20x%7D-%5Cfrac%7BR(2)%7D%7B%5Clog2%7D%2B%5Cint_%7B2-%7D%5Ex%5Cfrac%7BR(t)%7D%7Bt%5Clog%5E2t%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

于是，可得

$%5Clim_%7Bx%5Cto%5Cinfty%7D%5Csum_%7Bp%5Cle%20x%7D%5Cfrac1p-%5Clog%5Clog%20x%3D1-%5Clog%5Clog2%2B%5Cint_%7B2-%7D%5E%5Cinfty%5Cfrac%7BR(t)%7D%7Bt%5Clog%5E2t%7D%5Cmathrm%20dt$

顯然右式收斂到一個(gè)常數(shù)，即Mertens常數(shù)，將它代回式中，可得：

$%5Csum_%7Bp%5Cle%20x%7D%5Cfrac1p%3D%5Clog%5Clog%20x%2BM%2B%5Cfrac%7BR(x)%7D%7B%5Clog%20x%7D-%5Cint_%7Bx%7D%5E%5Cinfty%5Cfrac%7BR(t)%7D%7Bt%5Clog%5E2t%7D%5Cmathrm%20dt$

其中

$%5Cleft%7C%5Cint_%7Bx%7D%5E%5Cinfty%5Cfrac%7BR(t)%7D%7Bt%5Clog%5E2t%7D%5Cmathrm%20dt%5Cright%7C%5Cle%20%5Cleft%7C%5Cint_%7Bx%7D%5E%5Cinfty%5Cfrac1%7Bt%5Clog%5E2t%7D%5Cmathrm%20dt%5Cright%7C%3D%5Cfrac1%7B%5Clog%20x%7D$

因此，我們得到素?cái)?shù)倒數(shù)和的漸進(jìn)公式：

1） $%5Csum_%7Bp%5Cle%20x%7D%5Cfrac1p%3D%5Clog%5Clog%20x%2BM%2B%5Cmathcal%20O%5Cleft(%5Cfrac1%7B%5Clog%20x%7D%5Cright)$

在上式中令 $x%3De%5Et$ ，則有

$%5Csum_%7Bp%5Cle%20e%5Et%7D%5Cfrac1p%3D%5Clog%20t%2BM%2B%5Cmathcal%20O%5Cleft(%5Cfrac1t%5Cright)$

對(duì)比自然數(shù)的倒數(shù)和

$%5Csum_%7Bn%5Cle%20t%7D%5Cfrac1n%3D%5Clog%20t%2B%5Cgamma%2B%5Cmathcal%20O%5Cleft(%5Cfrac1t%5Cright)$

發(fā)現(xiàn)他們異常相似。沒(méi)錯(cuò)，數(shù)學(xué)就是這么神奇！

Mertens公式

考慮zeta函數(shù)的歐拉乘積的對(duì)數(shù)，分離出素zeta函數(shù)

$%5Cbegin%7Baligned%7D%5Clog%5Czeta(s)%26%3D%5Csum_%7Bp%7D%5Clog%5Cfrac1%7B1-p%5E%7B-s%7D%7D%5C%5C%26%3D%5Csum_%7Bp%7D%5Cfrac1%7Bp%5E%7Bs%7D%7D%2B%5Csum_%7Bp%7D%5Cleft(%5Clog%5Cfrac1%7B1-p%5E%7B-s%7D%7D-%5Cfrac1%7Bp%5Es%7D%5Cright)%5Cend%7Baligned%7D$

那么分離出來(lái)它有什么用呢？我們都知道這個(gè)等式在s=1處是發(fā)散的，但是，第二個(gè)和式在s=1時(shí)是收斂的：

$%5Cbegin%7Baligned%7D%5Csum_%7Bp%3Ex%7D%5Cleft(%5Clog%5Cfrac1%7B1-p%5E%7B-1%7D%7D-%5Cfrac1%7Bp%7D%5Cright)%26%3D%5Csum_%7Bp%3Ex%7D%5Csum_%7Bn%3D2%7D%5E%5Cinfty%5Cfrac1%7Bnp%5E%7Bn%7D%7D%5C%5C%26%5Cle%5Cfrac12%5Csum_%7Bp%3Ex%7D%5Csum_%7Bn%3D2%7D%5E%5Cinfty%5Cfrac1%7Bp%5E%7Bn%7D%7D%5C%5C%26%3D%5Cfrac12%5Csum_%7Bp%3Ex%7D%5Cleft(%5Cfrac1%7Bp-1%7D-%5Cfrac1p%5Cright)%5C%5C%26%3C%5Csum_%7Bn%3Ex%20%7D%5Cleft(%5Cfrac1%7Bn-1%7D-%5Cfrac1n%5Cright)%3D%5Cfrac1%7B%5Bx%5D%7D%5Cend%7Baligned%7D$

接著研究左側(cè)式子，令 $s%3D1%2B%5Cepsilon%2C(0%3C%5Cepsilon%3C1)$ ，有

$%5Cbegin%7Baligned%7D%5Czeta(1%2B%5Cepsilon)-%5Cfrac1%5Cepsilon%26%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bn%5E%7B1%2B%5Cepsilon%7D%7D-%5Cint_1%5E%5Cinfty%5Cfrac1%7Bu%5E%7B1%2B%5Cepsilon%7D%7D%5Cmathrm%20du%5C%5C%26%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cint_n%5E%7Bn%2B1%7D%5Cfrac1%7Bn%5E%7B1%2B%5Cepsilon%7D%7D-%5Cfrac1%7Bu%5E%7B1%2B%5Cepsilon%7D%7D%5Cmathrm%20du%5C%5C%26%5Cle%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bn%5E%7B1%2B%5Cepsilon%7D%7D-%5Cfrac1%7B(n%2B1)%5E%7B1%2B%5Cepsilon%7D%7D%5Cle1%5Cend%7Baligned%7D$

由此得到

$%5Clog%5Czeta(1%2B%5Cepsilon)%3D%5Clog%5Cleft(%5Cfrac1%5Cepsilon%2B%5Cmathcal%20O(1)%5Cright)%3D%5Clog%5Cfrac1%5Cepsilon%2B%5Clog(1%2B%5Cmathcal%20O(%5Cepsilon))%3D%5Clog%5Cfrac1%5Cepsilon%2BO(%5Cepsilon)$

又有：

$%5Clog%5Cfrac%7B1-e%5E%7B-%5Cepsilon%7D%7D%7B%5Cepsilon%7D%3D%5Clog(1%2B%5Cmathcal%20O(%5Cepsilon))%3D%5Cmathcal%20O(%5Cepsilon)$

于是

$%5Cbegin%7Baligned%7D%5Clog%5Czeta(1%2B%5Cepsilon)%26%3D%5Clog%5Cfrac1%7B1-e%5E%7B-%5Cepsilon%7D%7D%2B%5Clog%5Cfrac%7B1-e%5E%7B-%5Cepsilon%7D%7D%7B%5Cepsilon%7D%2B%5Cmathcal%20O(%5Cepsilon)%5C%5C%26%3D%5Clog%5Cfrac1%7B1-e%5E%7B-%5Cepsilon%7D%7D%2B%5Cmathcal%20O(%5Cepsilon)%5Cend%7Baligned%7D$

代入到分離素zeta函數(shù)后的歐拉乘積中，

$%5Clog%5Cfrac1%7B1-e%5E%7B-%5Cepsilon%7D%7D%2B%5Cmathcal%20O(%5Cepsilon)%3D%5Csum_%7Bp%7D%5Cfrac1%7Bp%5E%7B1%2B%5Cepsilon%7D%7D%2B%5Csum_%7Bp%7D%5Cleft(%5Clog%5Cfrac1%7B1-p%5E%7B-1-%5Cepsilon%7D%7D-%5Cfrac1%7Bp%5E%7B1%2B%5Cepsilon%7D%7D%5Cright)$

圍繞上式，接下來(lái)考慮除收斂部分外的兩項(xiàng)的差，記

$H(t)%3A%3D%5Csum_%7Bn%5Cle%20t%7D%5Cfrac1n%2CP(t)%3A%3D%5Csum_%7Bp%5Cle%20t%7D%5Cfrac1p$

由Riemann-Stieltjes積分的分部積分，

$%5Cbegin%7Baligned%7D%5Clog%5Cfrac1%7B1-e%5E%7B-%5Cepsilon%7D%7D%26%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Be%5E%7B-%5Cepsilon%20n%7D%7Dn%5C%5C%26%3D%5Cint_%7B0%7D%5E%5Cinfty%20e%5E%7B-%5Cepsilon%20t%7D%5Cmathrm%20dH(t)%5C%5C%26%3D%5Cepsilon%5Cint_%7B0%7D%5E%5Cinfty%20H(t)e%5E%7B-%5Cepsilon%20t%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

$%5Cbegin%7Baligned%7D%5Csum_%7Bp%7D%5Cfrac1%7Bp%5E%7B1%2B%5Cepsilon%7D%7D%26%3D%5Cint_%7B1%7D%5E%5Cinfty%5Cfrac1%7Bu%5E%5Cepsilon%7D%5Cmathrm%20dP(u)%5C%5C%26%3D%5Cint_%7B0%7D%5E%5Cinfty%20e%5E%7B-%5Cepsilon%20t%7D%5Cmathrm%20dP(e%5Et)%5C%5C%26%3D%5Cepsilon%5Cint_%7B0%7D%5E%5Cinfty%20P(e%5Et)e%5E%7B-%5Cepsilon%20t%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

由此可得

$%5Clog%5Cfrac1%7B1-e%5E%7B-%5Cepsilon%7D%7D-%5Csum_%7Bp%7D%5Cfrac1%7Bp%5E%7B1%2B%5Cepsilon%7D%7D%3D%5Cepsilon%5Cint_%7B0%7D%5E%5Cinfty%20(H(t)-P(e%5Et))e%5E%7B-%5Cepsilon%20t%7D%5Cmathrm%20dt$

根據(jù)H(t)與P(t)的漸進(jìn)公式，得

$H(t)-P(e%5Et)%3D%5Cgamma-M%2B%5Cmathcal%20O%5Cleft(%5Cfrac1t%5Cright)$

于是

$%5Cbegin%7Baligned%7D%5Clog%5Cfrac1%7B1-e%5E%7B-%5Cepsilon%7D%7D-%5Csum_%7Bp%7D%5Cfrac1%7Bp%5E%7B1%2B%5Cepsilon%7D%7D%26%3D%5Cgamma-M%2B%5Cmathcal%20O%5Cleft(%5Cepsilon%5Cint_%7B0%7D%5E%5Cinfty%20%5Cfrac1%7Bt%2B1%7D%20e%5E%7B-%5Cepsilon%20t%7D%5Cmathrm%20dt%5Cright)%5C%5C%26%3D%5Cgamma-M%2B%5Cmathcal%20O%5Cleft(%5Cepsilon%5Clog%20%5Cfrac1%5Cepsilon%5Cright)%5Cend%7Baligned%7D$

再次代回到歐拉乘積中，并令 $%5Cepsilon%5Cto%200$ ，得到

$%5Cgamma-M%3D%5Csum_%7Bp%7D%5Cleft(%5Clog%5Cfrac1%7B1-p%5E%7B-1%7D%7D-%5Cfrac1%7Bp%7D%5Cright)$

將右式分為不大于x與大于x的兩部分和，由前文的

$%5Csum_%7Bp%3Ex%7D%5Cleft(%5Clog%5Cfrac1%7B1-p%5E%7B-1%7D%7D-%5Cfrac1%7Bp%7D%5Cright)%5Cll%5Cfrac1x$

以及素?cái)?shù)倒數(shù)和的漸進(jìn)公式，可得

$%5Cbegin%7Baligned%7D%5Csum_%7Bp%5Cle%20x%7D%5Clog%5Cleft(1-%5Cfrac1p%5Cright)%26%3DM_0-%5Cgamma-%5Csum_%7Bp%5Cle%20x%7D%5Cfrac1p%2BO%5Cleft(%5Cfrac1%7Bx%7D%5Cright)%5C%5C%26%3D-%5Cgamma-%5Clog%5Clog%20x%2B%5Cmathcal%20O%5Cleft(%5Cfrac1%7B%5Clog%20x%7D%5Cright)%5Cend%7Baligned%7D$

取e的冪，根據(jù)

$%5Cexp%7B%5Cmathcal%20O%5Cleft(%5Cfrac1%7B%5Clog%20x%7D%5Cright)%7D%3D1%2B%5Cmathcal%20O%5Cleft(%5Cfrac1%7B%5Clog%20x%7D%5Cright)$

就能得到大名鼎鼎的Mertens公式了：

2） $%5Cprod_%7Bp%5Cle%20x%7D%5Cleft(1-%5Cfrac1p%5Cright)%3D%5Cfrac%7Be%5E%7B-%5Cgamma%7D%7D%7B%5Clog%20x%7D%5Cleft(1%2B%5Cmathcal%20O%5Cleft(%5Cfrac1%7B%5Clog%20x%7D%5Cright)%5Cright)$

Tchebyshev定理

在之前一期專欄中得到了

$A%5Cle%5Cfrac%7B%5Cpi(x)%5Clog%20x%7D%7Bx%7D%5Cle%20B$

這里就來(lái)研究一下中間函數(shù)的上確界與下確界吧，設(shè)

$l%3A%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cinf%5Cfrac%7B%5Cpi(x)%5Clog%20x%7Dx%2CL%3A%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Csup%5Cfrac%7B%5Cpi(x)%5Clog%20x%7Dx$

對(duì) $%5Cepsilon%3E0$ ，存在 $x_0%5Cge2$ ，使得 $t%3Ex_0$ 時(shí)，

$l-%5Cepsilon%5Cle%5Cfrac%7B%5Cpi(t)%5Clog%20t%7D%7Bt%7D%5CRightarrow%5Cpi(t)%5Cge(l-%5Cepsilon)%5Cfrac%20t%7B%5Clog%20t%7D$

利用素?cái)?shù)計(jì)數(shù)函數(shù)，有

$%5Cbegin%7Baligned%7D%5Csum_%7Bp%5Cle%20x%7D%5Cfrac1p%5Cge%5Cint_%7Bx_0%7D%5Ex%5Cfrac%7B%5Cmathrm%20d%20%5Cpi(t)%7Dt%26%3D%5Cfrac%7B%5Cpi(x)%7Dx-%5Cfrac%7B%5Cpi(x_0)%7D%7Bx_0%7D%2B%5Cint_%7Bx_0%7D%5Ex%5Cfrac%7B%5Cpi(t)%7D%7Bt%5E2%7D%5Cmathrm%20dt%5C%5C%26%5Cge%20o(1)-1%2B(l-%5Cepsilon)%5Cint_%7Bx_0%7D%5Ex%5Cfrac1%7Bt%5Clog%20t%7D%5Cmathrm%20dt%5C%5C%26%3D(l-%5Cepsilon)(%5Clog%5Clog%20x-%5Clog%5Clog%20x_0)-1%2Bo(1)%5Cend%7Baligned%7D$

根據(jù)素?cái)?shù)倒數(shù)和的漸進(jìn)公式，可知 $l-%5Cepsilon%5Cle1$ 又由 $%5Cepsilon$ 任意小，得到 $l%5Cle1$ ，類似的又有

$%5Csum_%7Bp%5Cle%20x%7D%5Cfrac1p%5Cle(L%2B%5Cepsilon)(%5Clog%5Clog%20x-%0A%5Clog%5Clog%20x_0)%2B%5Cmathcal%20O(1)$

從而 $l%5Cle1%5Cle%20L$ ，即

3） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cinf%5Cfrac%7B%5Cpi(x)%5Clog%20x%7Dx%5Cle1%5Cle%5Clim_%7Bx%5Cto%5Cinfty%7D%5Csup%5Cfrac%7B%5Cpi(x)%5Clog%20x%7Dx$

此時(shí)的結(jié)論已經(jīng)夠強(qiáng)了，如果能證明 $%5Cfrac%7B%5Cpi(x)%5Clog%20x%7D%7Bx%7D$ 的極限存在，那么其必收斂到1，然而素?cái)?shù)定理的證明難就難在其極限存在性僅用初等方法是不容易確定的，因此人們會(huì)尋找更強(qiáng)更有力的方法來(lái)解決它——解析方法。不出意外的話下一期就要進(jìn)入漫漫解析路了

參考

《解析與概率數(shù)論導(dǎo)引》by G.特倫鮑姆
Meten定理與素?cái)?shù)定理 by?TravorLZH:https://zhuanlan.zhihu.com/p/338578631

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Prime Dream（3）——Mertens的幾個(gè)漸進(jìn)公式

素?cái)?shù)的倒數(shù)和

Mertens公式

Tchebyshev定理

參考