Prime Dream（1）——素?cái)?shù)計(jì)數(shù)函數(shù)與它的階

2022-01-30 00:12 作者:子瞻Louis 0人讀過 | 我要投稿

許多數(shù)論的研究大多都是圍繞素?cái)?shù)展開的，而與之相關(guān)的自然是素?cái)?shù)計(jì)數(shù)函數(shù)：

$%5Cpi(x)%3D%5Csum_%7Bp%5Cle%20x%7D1$

上和號(hào)是對(duì)不大于x的素?cái)?shù)求和，也就是說該函數(shù)用來表示不大于x的素?cái)?shù)個(gè)數(shù)，開始之前還要再引入兩個(gè)函數(shù)——切比雪夫theta、psi函數(shù)：（ $x%3E1$ ）

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

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

Dirichlet卷積的部分和

令*表示Dirichlet卷積，設(shè)

$F(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)%2CG(x)%3D%5Csum_%7Bn%5Cle%20x%7Dg(n)$

考慮以下和式，

$%5Cbegin%7Baligned%7DS(x)%26%3D%5Csum_%7Bn%5Cle%20x%7Df*g(n)%5C%5C%26%3D%5Csum_%7Bn%5Cle%20x%7D%5Csum_%7Bn%3Dab%7Df(a)g(b)%5C%5C%26%3D%5Csum_%7Bab%5Cle%20x%7Df(a)g(b)%5Cend%7Baligned%7D$

可以在上式的求和中先固定a，對(duì)所有b求和，再對(duì)所有a求和，得到

$%5Cbegin%7Baligned%7DS(x)%26%3D%5Csum_%7Ba%5Cle%20x%7Df(a)%5Csum_%7Bb%5Cle%5Cfrac%20xa%7Dg(b)%5C%5C%26%3D%5Csum_%7Bn%5Cle%20x%7Df(n)G%5Cleft(%5Cfrac%20xn%5Cright)%5Cend%7Baligned%7D$

同理，也可得

$S(x)%3D%5Csum_%7Bn%5Cle%20x%7Dg(n)F%5Cleft(%5Cfrac%20xn%5Cright)$

特別地取 $f(n)%3D1(n)%5Cequiv1$ ，得到

$%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3D%5Csum_%7Bn%5Cle%20x%7D%5Ctilde%7Bf%7D%20(n)$

其中 $%5Ctilde%20f(n)$ 為 $f(n)$ 的Mobius變換，將Von?Mongoldt函數(shù)代入，可得

1） $%5Clog%20%5Bx%5D!%3D%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D$

取 $x%3Dm$ 為整數(shù)，將右邊改寫一下

$%5Cbegin%7Baligned%7D%5Clog%20m!%26%3D%5Csum_%7Bp%5Ek%5Cle%20m%7D%5Clog%20p%5Cleft%5B%5Cfrac%20m%7Bp%5Ek%7D%5Cright%5D%5C%5C%26%3D%5Csum_%7Bp%5Cle%20m%7D%5Csum_%7Bk%3D1%7D%5E%5Cinfty%5Cleft%5B%5Cfrac%20m%7Bp%5Ek%7D%5Cright%5D%5Clog%20p%5Cend%7Baligned%7D$

最后一個(gè)等式內(nèi)層的求和總是只有限項(xiàng)不為零的，因此可以得到階乘的素因子分解

2） $m!%3D%5Cprod_%7Bp%5Cle%20x%7Dp%5E%7BA(m%2Cp)%7D%2CA(m%2Cp)%3D%5Csum_%7Bk%3D1%7D%5E%5Cinfty%5Cleft%5B%5Cfrac%20x%7Bp%5Ek%7D%5Cright%5D$

具有特殊階的部分和

對(duì)階乘的對(duì)數(shù)用Able求和公式可得

$%5Cbegin%7Baligned%7D%5Clog%5Bx%5D!%26%3D%5Csum_%7Bn%5Cle%20x%7D%5Clog%20n%3D%5Bx%5D%5Clog%20x-%5Cint_1%5Ex%5Cfrac%7B%5Bt%5D%7Dt%5Cmathrm%20dt%5C%5C%26%3Dx%5Clog%20x-x%2B1-%5C%7Bx%5C%7D%5Clog%20x%2B%5Cint_1%5Ex%5Cfrac%7B%5C%7Bt%5C%7D%7Dt%5Cmathrm%20dt%5C%5C%26%3Dx%5Clog%20x-x%2B%5Cmathcal%20O(%5Clog%20x)%5Cend%7Baligned%7D$

這也就是說

3） $%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3Dx%5Clog%20x-x%2B%5Cmathcal%20O(%5Clog%20x)$

將左式分為素?cái)?shù)和素?cái)?shù)的乘方，

$%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3D%5Csum_%7Bp%5Cle%20x%7D%5Clog%20p%5Cleft%5B%5Cfrac%20xp%5Cright%5D%2B%5Csum_%7Bp%5Cle%20x%7D%5Csum_%7Bk%3D2%7D%5E%5Cinfty%5Cleft%5B%5Cfrac%20x%7Bp%5Ek%7D%5Cright%5D%5Clog%20p$

來康康第二項(xiàng)，對(duì)它放縮一下

$%5Cbegin%7Baligned%7D%5Csum_%7Bp%5Cle%20x%7D%5Csum_%7Bk%3D2%7D%5E%5Cinfty%5Cleft%5B%5Cfrac%20x%7Bp%5Ek%7D%5Cright%5D%5Clog%20p%26%3C%20%5Csum_%7Bp%5Cle%20x%7D%5Csum_%7Bk%3D2%7D%5E%5Cinfty%20%5Cfrac%20x%7Bp%5Ek%7D%5Clog%20p%5C%5C%26%3Dx%5Csum_%7Bp%5Cle%20x%7D%5Cfrac%7B%5Clog%20p%7D%7Bp(p-1)%7D%5C%5C%26%3Cx%5Csum_%7Bn%3D2%7D%5E%5Cinfty%5Cfrac%7B%5Clog%20n%7D%7Bn(n-1)%7D%5Cend%7Baligned%7D$

可以驗(yàn)證最后一個(gè)級(jí)數(shù)是收斂到某個(gè)不大于 $%5Clog4$ 的常數(shù)：

$%5Cbegin%7Baligned%7D%5Csum_%7Bn%3D2%7D%5E%5Cinfty%5Cfrac%7B%5Clog%20n%7D%7Bn(n-1)%7D%26%3C%5Csum_%7Br%3D1%7D%5E%5Cinfty%5Csum_%7B2%5Er%3Cm%5Cle%202%5E%7Br%2B1%7D%7D%5Cfrac%7B%5Clog%202%5Er%7D%7Bm(m-1)%7D%5C%5C%26%3D%5Csum_%7Br%3D1%7D%5E%5Cinfty%7Br%5Clog%202%7D%5Csum_%7B2%5Er%3Cm%5Cle%202%5E%7Br%2B1%7D%7D%5Cfrac1%7Bm-1%7D-%5Cfrac1m%5C%5C%26%3D%5Csum_%7Br%3D1%7D%5E%5Cinfty%5Cfrac%7Br%5Clog2%7D%7B2%5Er%7D%3D%5Clog4%5Cend%7Baligned%7D$

因此可以得到

$%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3D%5Csum_%7Bp%5Cle%20x%7D%5Clog%20p%5Cleft%5B%5Cfrac%20xp%5Cright%5D%2B%5Cmathcal%20O(x)$

4） $%5Csum_%7Bp%5Cle%20x%7D%5Clog%20p%5Cleft%5B%5Cfrac%20xp%5Cright%5D%3Dx%5Clog%20x%2B%5Cmathcal%20O(x)$

Sharpiro Tauber型定理

既然這兩個(gè)部分和的階是差不多的，那我們直接來研究以下形式的和吧

$G(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3Dx%5Clog%20x%2B%5Cmathcal%20O(x)$
$F(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)$
$S(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn$

我們想利用1.式來得到2.式的階，首要任務(wù)就是解決掉取整的部分，注意到當(dāng) $x%2F2%3Cn%3Cx$ 時(shí)，都有 $%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3D1$ ，因此通過以下作差可以就可以讓2.式出現(xiàn)

$%5Cbegin%7Baligned%7DG(x)-2G%5Cleft(%5Cfrac%20x2%5Cright)%26%3D%5Csum_%7Bn%5Cle%20x%2F2%7Df(n)%5Cleft(%5Cleft%5B%5Cfrac%20xn%5Cright%5D-2%5Cleft%5B%5Cfrac%20x%7B2n%7D%5Cright%5D%5Cright)%2B%5Csum_%7Bx%2F2%3Cn%5Cle%20x%7Df(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%5C%5C%26%3D%5Csum_%7Bn%5Cle%20x%2F2%7Df(n)%5Cleft(%5Cleft%5B%5Cfrac%20xn%5Cright%5D-2%5Cleft%5B%5Cfrac%20x%7B2n%7D%5Cright%5D%5Cright)%2BF(x)-F%5Cleft(%5Cfrac%20x2%5Cright)%5Cend%7Baligned%7D$

對(duì)任意 $t%3E0$ ， $%5Bt%5D-2%5Bt%2F2%5D$ 或?yàn)?或?yàn)?，所以得到

$G(x)-2G%5Cleft(%5Cfrac%20x2%5Cright)%5Cge%20F(x)-F%5Cleft(%5Cfrac%20x2%5Cright)$

又根據(jù)1.式G(x)的階可知

$%5Cbegin%7Baligned%7DG(x)-2G%5Cleft(%5Cfrac%20x2%5Cright)%26%3Dx%5Clog%20x-2%5Ccdot%5Cfrac%20x2%5Clog%5Cfrac%20x2%2B%5Cmathcal%20O(x)%5C%5C%26%3Dx%5Clog%20x-x%5Clog%20x%2B%5Cmathcal%20O(x)%5C%5C%26%3D%5Cmathcal%20O(x)%5Cend%7Baligned%7D$

這也就是說存在某個(gè)常數(shù)C，使得

$F(x)-F%5Cleft(%5Cfrac%20x2%5Cright)%5Cle%20G(x)-2G%5Cleft(%5Cfrac%20x2%5Cright)%5Cle%20Cx$

不斷用 $%5Cfrac%20x2%2C%5Cfrac%20x4%2C%E2%80%A6$ 替換掉x，得到

$%5Cbegin%7Baligned%7DF%5Cleft(%5Cfrac%20x2%5Cright)-F%5Cleft(%5Cfrac%20x4%5Cright)%26%5Cle%20%5Cfrac12Cx%5C%5CF%5Cleft(%5Cfrac%20x4%5Cright)-F%5Cleft(%5Cfrac%20x8%5Cright)%26%5Cle%20%5Cfrac14Cx%5C%5C%26%E2%80%A6%5Cend%7Baligned%7D$

把這些不等式依次加起來，便有

$F(x)%5Cle%5Cleft(1%2B%5Cfrac12%2B%5Cfrac14%2B%E2%80%A6%5Cright)Cx%3D2Cx$

于是可以得到

5） $%5Csum_%7Bn%5Cle%20x%7Df(n)%3D%5Cmathcal%20O(x)$

下面來看另一個(gè)和，因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%5Bt%5D%3Dt%2B%5Cmathcal%20O(1)" alt="%5Bt%5D%3Dt%2B%5Cmathcal%20O(1)">，所以

$%5Cbegin%7Baligned%7D%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%26%3Dx%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%2B%5Cmathcal%20O%5Cleft(%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cright)%5C%5C%5CRightarrow%20x%5Clog%20x%2B%5Cmathcal%20O(x)%26%3Dx%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%2B%5Cmathcal%20O(x)%5Cend%7Baligned%7D$

6） $%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%3D%5Clog%20x%2B%5Cmathcal%20O(1)$

現(xiàn)在我們?cè)O(shè) $S(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%3D%5Clog%20x%2BR(x)$ ，由6）式可知存在常數(shù) $A$ 使得 $%7CR(x)%7C%5Cle%20A$ ，某一常數(shù) $0%3Ca%3C1$ ，使得

$%5Cbegin%7Baligned%7DS(x)-S(ax)%26%3D%5Clog%20x%2BR(x)-%5Clog%20ax-R(ax)%5C%5C%26%3D-%5Clog%20a%2BR(x)-R(ax)%5C%5C%26%5Cge%20-%5Clog%20a-%7CR(x)%7C-%7CR(ax)%7C%5C%5C%26%5Cge-%5Clog%20a-2A%3D1%5Cend%7Baligned%7D$

也就是說 $a%3De%5E%7B-2A-1%7D$ 。因?yàn)镾(x)是定義在 $x%3E1$ 上的，所以其中要求了 $ax%5Cge1$ 。另一方面，有

$%5Cbegin%7Baligned%7DS(x)-S(ax)%26%3D%5Csum_%7Bax%3Cn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%5C%5C%26%5Cle%5Cfrac1%7Bax%7D%5Csum_%7Bax%3Cn%5Cle%20x%7Df(n)%5C%5C%26%5Cle%5Cfrac1%7Bax%7DF(x)%5Cend%7Baligned%7D$

結(jié)合上面兩個(gè)式子與5），就得到：對(duì)于 $%7CR(x)%7C%5Cle%20A$ ，當(dāng) $x%5Cge%20e%5E%7B2A%2B1%7D$ 時(shí)，都有存在兩個(gè)常數(shù) $C_1%2CC_2$ ，（其中 $C_1%3De%5E%7B-2A-1%7D$ 是已經(jīng)確定的）

7） $C_1x%5Cle%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cle%20C_2x$

可以用大theta符號(hào)改寫一下：對(duì)于 $x%5Cge%20e%5E%7B2A%2B1%7D$ ，都有

7') $%5Csum_%7Bn%5Cle%20x%7Df(n)%3D%5CTheta(x)$

將上面得到的所有結(jié)論放在一起，便組成了Shaprio?Tauber型定理:若

$G(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3Dx%5Clog%20x%2B%5Cmathcal%20O(x)$

則當(dāng)x足夠大時(shí)，以下兩個(gè)漸進(jìn)公式成立：

$S(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%3D%5Clog%20x%2B%5Cmathcal%20O(1)$
$C_1x%5Cle%20F(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cle%20C_2x$

其中 $C_1%3De%5E%7B-2A-1%7D$ ，A為1.式中大O符號(hào)的絕對(duì)值上界

素?cái)?shù)計(jì)數(shù)函數(shù)的階

將4）式代入到Sharprio定理中，可得

8） $%5Csum_%7Bp%5Cle%20x%7D%5Cfrac%7B%5Clog%20p%7Dp%3D%5Clog%20x%2B%5Cmathcal%20O(1)$
9） $%5Cvartheta(x)%3D%5Csum_%7Bp%5Cle%20x%7D%5Clog%20p%3D%5CTheta(x)$

其中8）式就被稱為Mertens第一定理，根據(jù)右側(cè)當(dāng) $x%5Cto%5Cinfty$ 時(shí)是發(fā)散的，又能再次說明素?cái)?shù)有無窮多個(gè)。同樣將4）式代入也可得

10） $%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7B%5CLambda(n)%7Dn%3D%5Clog%20x%2B%5Cmathcal%20O(1)$
11） $%5Cpsi(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%3D%5CTheta(x)$

接下來就是要得到素?cái)?shù)計(jì)數(shù)函數(shù)的階了，可以利用Able求和公式作以下變換：

$%5Cbegin%7Baligned%7D%5Cpi(x)%26%3D%5Csum_%7B2-%5Cdelta%3Cp%5Cle%20x%7D%5Cfrac%7B%5Clog%20p%7D%7B%5Clog%20p%7D%5Cqquad(%5Cdelta%3E0)%5C%5C%26%3D%5Cfrac%7B%5Cvartheta(x)%7D%7B%5Clog%20x%7D%2B%5Cint_%7B2-%5Cdelta%7D%5Ex%5Cfrac%7B%5Cvartheta(t)%7D%7Bt%5Clog%5E2t%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

將9）式代入，因?yàn)榉e分

$0%5Cle%5Cint_2%5E%7Bx%7D%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%5E2%20t%7D%3C%5Cfrac%7Bx%7D%7B%5Clog%5E2x%7D%3D%5Cmathcal%20O%5Cleft(%5Cfrac%20x%7B%5Clog%20x%7D%5Cright)$