Prime Dream（2）——素數(shù)定理的推論

2022-02-04 17:27 作者:子瞻Louis 0人讀過 | 我要投稿

簡概

自然數(shù)的乘法賦予了自然數(shù)集一種特殊的結(jié)構(gòu)，從而誕生了整除這一關(guān)系，而有一類數(shù)，無法找到除了1與它本身外能夠整除它的整數(shù)，這類數(shù)就叫做素數(shù)（prime?number）或質(zhì)數(shù)。它們的分布是十分不規(guī)則的，因此對它的研究可以說是十分艱難的，從幾千年前的歐幾里得，到如今，有關(guān)素數(shù)的難題出現(xiàn)了許許多多，諸如哥德巴赫猜想，孿生素數(shù)猜想，世界七大難題之一的黎曼假設(shè)等，盡管如此，數(shù)學(xué)家們也始終執(zhí)著地追求著看清這其中的奧秘

之前有一期專欄中有提到過Euclid定理——素數(shù)有無窮多個，即

$%5Cpi(x)%5Crightarrow%5Cinfty%20%5Cqquad%20(x%5Crightarrow%5Cinfty)$

這也令對素數(shù)的研究更加復(fù)雜了，畢竟如果是有限個素數(shù)那研究起來多方便。實際上最大的困難之處還是在于素數(shù)的分布太雜亂了，要得到一個精確的公式是十分不容易的。本期只展示素數(shù)定理的等價形式及推論

Tchebyshev的兩個函數(shù)與素數(shù)定理

首先從切比雪夫的兩個函數(shù)出發(fā)：

$%5Cvartheta(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5Clog%20p$

$%5Cpsi(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)$

利用素數(shù)示性函數(shù)：

$%5Cmathrm%20p(n)%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%0A1%20%2C%26%20n%5Cin%5Cmathbb%20P%20%5C%5C%200%2C%20%20%26%20n%5Cnotin%5Cmathbb%20P%0A%5Cend%7Barray%7D%5Cright.$

發(fā)現(xiàn)若將n取不大于x的所有整數(shù)并求和剛好就是不大于x的素數(shù)個數(shù)，即

$%5Cpi(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5Cmathrm%20p(n)$

$%5CRightarrow%20%5Cmathrm%20p(n)%3D%5Cpi(n)-%5Cpi(n-1)$

將其代入到Tchebyshev?theta函數(shù)中，因為當 $%5Cdelta%3E0$ 時，在 $%5B2-%5Cdelta%2C2%5D$ 的范圍內(nèi)沒有素數(shù)，所以可將求和域改為2-δ到x，利用Able求和公式，可得

$%5Cbegin%7Baligned%7D%5Cvartheta(x)%26%3D%5Csum_%7B2-%5Cdelta%3Cn%5Cle%20x%7D%5Cmathrm%20p(n)%5Clog%20n%5C%5C%26%3D%5Cpi(x)%5Clog%20x-%5Cpi(2-%5Cdelta)%5Clog(2-%5Cdelta)-%5Cint_%7B1-%5Cdelta%7D%5Ex%5Cpi(t)%5Cmathrm%20d%5Clog%20t%5C%5C%26%3D%5Cpi(x)%5Clog%20x-%5Cint_%7B2-%5Cdelta%7D%5Ex%5Cfrac%7B%5Cpi(t)%7Dt%5Cmathrm%20dt%5Cend%7Baligned%7D$

接下來令 $%5Cdelta%5Crightarrow0%5E%2B$ ，得到

$%5Cpi(x)%5Clog%20x%3D%5Cvartheta(x)%2B%5Cint_%7B2%5E-%7D%5Ex%5Cfrac%7B%5Cpi(t)%7D%7Bt%7D%5Cmathrm%20dt$

為了方便采用記號? $%5Cint_%7B2%5E-%7D%5Ex%3A%3D%5Clim_%7B%5Cdelta%5Cto0%5E%2B%7D%5Cint_%7B2-%5Cdelta%7D%5Ex$

#） $%5Cfrac%7B%5Cpi(x)%5Clog%20x%7Dx%3D%5Cfrac%7B%5Cvartheta(x)%7Dx%2B%5Cfrac1x%5Cint_%7B2%5E-%7D%5Ex%5Cfrac%7B%5Cpi(t)%7D%7Bt%7D%5Cmathrm%20dt$

因此可以通過Tchebyshev theta函數(shù)間接對素數(shù)分布進行研究

素數(shù)定理（prime?number?theorem）描述的正是

1） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpi(x)%5Clog%20x%7D%7Bx%7D%3D1%5Cqquad%5Cleft(%5Cpi(x)%5Csim%5Cfrac%7Bx%7D%7B%5Clog%20x%7D%5Cright)$

當然這只是它最簡單的形式。我們可以由此推出與它等價的兩種形式

從#式出發(fā)，上一期專欄中我們證明了存在兩個常數(shù) $C_1%2CC_2$ ，使得

$C_1%5Cle%5Cfrac%7B%5Cpi(x)%5Clog%20x%7D%7Bx%7D%5Cle%20C_2$

從而有

$%5Cfrac%7BC_1%7D%7B%5Clog%20x%7D%5Cle%5Cfrac%7B%5Cpi(x)%7D%7Bx%7D%5Cle%20%5Cfrac%7BC_2%7D%7B%5Clog%20x%7D$

于是#）式中的積分項

$%5Cfrac%7BC_1%7Dx%5Cint_2%5Ex%5Cfrac1%7B%5Clog%20t%7D%5Cmathrm%20dt%5Cle%5Cfrac1x%5Cint_2%5Ex%5Cfrac%7B%5Cpi(t)%7D%7Bt%7D%5Cmathrm%20dt%5Cle%5Cfrac%7BC_2%7Dx%5Cint_2%5Ex%5Cfrac1%7B%5Clog%20t%7D%5Cmathrm%20dt$

而又有

$0%3C%5Cfrac%7BC%7Dx%5Cint_2%5Ex%5Cfrac1%7B%5Clog%20t%7D%5Cmathrm%20dt%3D%5Cfrac%20C%7B%5Clog%20x%7D-%5Cfrac%20%7B2C%7D%7Bx%5Clog2%7D%2B%5Cfrac%20Cx%5Cint_2%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%5E2t%7D%5Cxrightarrow%7Bx%5Cto%5Cinfty%7D0$

所以可得以下等式：

2） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpi(x)%5Clog%20x%7D%7Bx%7D%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cvartheta(x)%7D%7Bx%7D$

若能證明右側(cè)的極限為1，則就能間接得到素數(shù)定理，至此就得到了素數(shù)定理的一個等價表述

現(xiàn)在將注意力轉(zhuǎn)到Tchebyshev psi函數(shù)上，它是對不超過x的素數(shù)的乘方求和，

$%5Cpsi(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%3D%5Csum_%7Bm%3D1%7D%5E%5Cinfty%5Csum_%7Bp%5Em%5Cle%20x%7D%5Clog%20p$

因為不超過x的p乘方總是有限的，所以最右邊是一有限和，不難將它與theta函數(shù)聯(lián)系起來

$%5Cpsi(x)%3D%5Csum_%7Bm%3D1%7D%5E%5Cinfty%5Csum_%7Bp%5Cle%5Csqrt%5Bm%5D%20x%7D%5Clog%20p%3D%5Csum_%7Bm%3D1%7D%5E%5Cinfty%5Cvartheta(%5Csqrt%5Bm%5Dx)$

注意到 $%5Csqrt%5Bm%5Dx%3C2$ 即 $m%3E%5Clog_2x$ 時外層的求和是空的，將這些空的去掉，得到

$%5Cpsi(x)%3D%5Csum_%7Bm%5Cle%5Clog_2x%7D%5Cvartheta(%5Csqrt%5Bm%5Dx)$

下面來康康這兩個函數(shù)的差吧

$0%5Cle%5Cpsi(x)-%5Cvartheta(x)%3D%5Csum_%7B2%5Cle%20m%5Cle%5Clog_2x%7D%5Cvartheta(%5Csqrt%5Bm%5Dx)$

根據(jù)Tchebyshev theta函數(shù)的定義，又可得對 $m%5Cge2$ ，

$%5Cvartheta(%5Csqrt%5Bm%5Dx)%3D%5Csum_%7Bp%5Cle%5Csqrt%5Bm%5Dx%7D%5Clog%20p%5Cle%5Csqrt%5Bm%5Dx%5Clog%5Csqrt%5Bm%5Dx%5Cle%5Csqrt%20x%5Clog%20x$

$%5Cbegin%7Baligned%7D%5CRightarrow%20%5Cpsi(x)-%5Cvartheta(x)%26%5Cle%5Csum_%7B2%5Cle%20m%5Cle%5Clog_2x%7D%5Csqrt%20x%5Clog%20x%5C%5C%26%5Cle%5Csqrt%20x%5Clog%20x%5Clog_2x%3D%5Cfrac%7B%5Csqrt%20x%5Clog%5E2x%7D%7B%5Clog2%7D%5Cend%7Baligned%7D$

經(jīng)過上面這些粗略的放縮，得到

$0%5Cle%5Cfrac%7B%5Cpsi(x)%7Dx-%5Cfrac%7B%5Cvartheta(x)%7Dx%5Cle%5Cfrac%7B%5Clog%5E2x%7D%7B%5Csqrt%20x%5Clog2%7D$

當x足夠大時他們的差越來越小且最終會趨于0，因此上訴不等式就是告訴了我們

3） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpsi(x)%7Dx%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cvartheta(x)%7Dx$

再根據(jù)2）式就能得到素數(shù)定理的另一等價表述，

素數(shù)定理的推論

假設(shè)素數(shù)定理已經(jīng)成立，對1）式取自然對數(shù)，得到

$%5Clim_%7Bx%5Cto%5Cinfty%7D(%5Clog%5Cpi(x)%2B%5Clog%5Clog%20x-%5Clog%20x)%3D0$

再變一下，就是

$%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cleft%5B%5Clog%20x%5Cleft(%5Cfrac%7B%5Clog%5Cpi(x)%7D%7B%5Clog%20x%7D%2B%5Cfrac%7B%5Clog%20%5Clog%20x%7D%7B%5Clog%20x%7D-1%5Cright)%5Cright%5D%3D0$

用epsilon語言來說就是 $%5Cexists%20X%5Cin%5Cmathbb%20R_%2B%2Cx%3EX$ 時， $%5Cforall%20%5Cepsilon%3E0$

$%5Cleft%7C%5Clog%20x%5Cleft(%5Cfrac%7B%5Clog%5Cpi(x)%7D%7B%5Clog%20x%7D%2B%5Cfrac%7B%5Clog%20%5Clog%20x%7D%7B%5Clog%20x%7D-1%5Cright)%5Cright%7C%3C%5Cepsilon$

$%5CRightarrow%5Cleft%7C%5Cfrac%7B%5Clog%5Cpi(x)%7D%7B%5Clog%20x%7D%2B%5Cfrac%7B%5Clog%20%5Clog%20x%7D%7B%5Clog%20x%7D-1%5Cright%7C%3C%5Cfrac%5Cepsilon%7B%5Clog%20x%7D$

所以由此可得

$%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cleft(%5Cfrac%7B%5Clog%5Cpi(x)%7D%7B%5Clog%20x%7D%2B%5Cfrac%7B%5Clog%20%5Clog%20x%7D%7B%5Clog%20x%7D-1%5Cright)%3D0$

又因為

$%5Cfrac%7B%5Clog%5Clog%20x%7D%7B%5Clog%20x%7D%5Cxrightarrow%7Bx%5Crightarrow%5Cinfty%7D0$

所以可以得到

4） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Clog%5Cpi(x)%7D%7B%5Clog%20x%7D%3D1$

這雖然有些許不可思議——不大于給定數(shù)的素數(shù)個數(shù)總是遠小于這個數(shù)的，但是仔細想想因為對數(shù)函數(shù)發(fā)散的速度很慢，以至于彌補了它們之間的差距，所以造成它們的比值極限為1

結(jié)合4）式，可知素數(shù)定義亦等價于

5） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpi(x)%5Clog%5Cpi(x)%7D%7Bx%7D%3D1$

在上式中令x沿素數(shù)集趨向無窮，即

$%5Clim_%7B%5Cmathbb%20P%5Cni%20x%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpi(x)%5Clog%5Cpi(x)%7D%7Bx%7D%3D1$

設(shè) $x%3Dp_n$ 為第n個素數(shù)，而 $p_n%5Cto%5Cinfty%5CRightarrow%20n%5Cto%5Cinfty$ ，且又有 $%5Cpi(p_n)%3Dn$ ，代入上式即得

6） $%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Bn%5Clog%20n%7D%7Bp_n%7D%3D1$

素數(shù)定理的更精確形式

利用小o符號，由

$%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cvartheta(x)%7Dx%3D1$

可以寫出

$%5Cvartheta(x)%3Dx%2Bo(x)%2C%20%5Cquad%20x%5Cto%5Cinfty$

又由Abel求和公式，有

$%5Cbegin%7Baligned%7D%5Cpi(x)%26%3D%5Csum_%7Bp%5Cle%20x%7D%5Cfrac%7B%5Clog%20p%7D%7B%5Clog%20p%7D%5C%5C%26%3D%5Cfrac%7B%5Cvartheta(x)%7D%7B%5Clog%20x%7D%2B%5Cint_%7B2%5E-%7D%5Ex%5Cfrac%7B%5Cvartheta(t)%7D%7Bt%5Clog%5E2%20t%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

由此可得

$%5Cpi(x)%3D%5Cfrac%7Bx%7D%7B%5Clog%20x%7D%2B%5Cint_%7B2%7D%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%5E2t%7D%2Bo%5Cleft(%5Cfrac%20x%7B%5Clog%20x%7D%5Cright)$

又由分部積分得

$%5Cbegin%7Baligned%7D%5Cint_%7B2%7D%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%5E2t%7D%26%3D-%5Cint_%7B2%7D%5Ext%5Cmathrm%20d%5Cfrac1%7B%5Clog%20t%7D%5C%5C%26%3D%5Cfrac2%7B%5Clog2%7D-%5Cfrac%20x%7B%5Clog%20x%7D%2B%5Cint_%7B2%7D%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%20t%7D%5Cend%7Baligned%7D$

記? $%5Cmathrm%7Bli%7D(x)%3D%5Cint_2%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%20t%7D$ ?，則

7） $%5Cpi(x)%3D%5Cmathrm%7Bli%7D(x)%2Bo%5Cleft(%5Cfrac%20x%7B%5Clog%20x%7D%5Cright)%2C%5Cquad%20x%5Cto%5Cinfty$

一些情況下，使用下限為0的積分是非常方便的，因此引入

$%5Cmathrm%7BLi%7D(x)%3D%5Cint_0%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%20t%7D%3D%5Cmathrm%7Bli%7D(x)%2B%5Cmathrm%7BLi%7D(2)$

其中? $%5Cmathrm%7BLi%7D(2)$ ?取Cauchy主值，

$%5Cmathrm%7BLi%7D(2)%3D%5Clim_%7B%5Cdelta%5Cto0%5E%2B%7D%5Cint_0%5E%7B1-%5Cdelta%7D%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%20t%7D%2B%5Cint_%7B1%2B%5Cdelta%7D%5E2%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%20t%7D%5Capprox%201.04%5Cdots$