形式Dirichlet級(jí)數(shù)

2022-01-18 20:48 作者:子瞻Louis 0人讀過 | 我要投稿

已收錄至：《雜文集》

積性數(shù)論中有一個(gè)十分重要的工具——Dirichlet級(jí)數(shù)（也叫Dirichlet生成函數(shù)），它通常是為以下形式：

$%5Cmathcal%20D(s%2Cf)%3A%3DF(s)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bf(n)%7D%7Bn%5Es%7D$

本期并不會(huì)過深入的研究，而是只指出它的有關(guān)數(shù)論函數(shù)的一些代數(shù)性質(zhì)

取 $f(n)%3D1(n)%5Cequiv1$ ，就可以得到著名的zeta函數(shù)：

$%5Cmathcal%20D(s%2C1)%3D%5Czeta(s)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bn%5Es%7D$

Euler乘積公式

先從一個(gè)著名的乘積開始：（ $%5CRe(s)%3E1$ ）

$%5Czeta(s)%3D%5Cprod_p%5Cleft(1-%5Cfrac1%7Bp%5Es%7D%5Cright)%5E%7B-1%7D$

該公式可以由以下更廣義的結(jié)論推出：

若 $f(n)$ 是積性函數(shù)且 $%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20f(n)$ 絕對(duì)收斂，則

$%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20f(n)%3D%5Cprod_p(1%2Bf(p)%2Bf(p%5E2)%2B%E2%80%A6)$

?因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=f(n)" alt="f(n)">是積性函數(shù)，所以 $f(1)%3D1$ ，根據(jù)算術(shù)基本定理可將每個(gè)n唯一分解為若干素?cái)?shù)的乘積，當(dāng)n遍歷所有整數(shù)時(shí)，分解出的乘積素?cái)?shù)將遍歷所有素?cái)?shù)，這里就簡單驗(yàn)證下：

$%5Cprod_%7Bp%5Cle%20N%7D(1%2Bf(p)%2Bf(p%5E2)%2B%E2%80%A6)%3D%5Csum_%7Bn%3D1%7D%5EN%20f(n)%2BR(N)$

其中 $%7CR(N)%7C%5Cle%5Cleft%7C%5Csum_%7Bn%3EN%7Df(n)%5Cright%7C%5Cle%5Csum_%7Bn%3EN%7D%7Cf(n)%7C$

因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20f(n)" alt="%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20f(n)">絕對(duì)收斂，所以當(dāng) $N%5Crightarrow%5Cinfty$ 時(shí) $R(N)%5Crightarrow0$

因此該公式成立?

特別地，若 $f(n)$ 是完全積性函數(shù)，則

$%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20f(n)%3D%5Cprod_p(1-f(p))%5E%7B-1%7D$

顯然可以由幾何級(jí)數(shù)直接推出

因此在上訴公式中取 $f(n)%3D%5Cfrac1%7Bn%5Es%7D$ 可得

$%5Czeta(s)%3D%5Cprod_p%5Cleft(1-%5Cfrac1%7Bp%5Es%7D%5Cright)%5E%7B-1%7D$

在上式中令 $s%5Crightarrow1$ ，因?yàn)橛覀?cè)是發(fā)散的，所以右側(cè)也是發(fā)散的，即右側(cè)必須有無窮項(xiàng)，由此便可得著名的Euclid定理——素?cái)?shù)有無窮多個(gè)

幾個(gè)有用的性質(zhì)：

設(shè) $%5Cmathbb%20A$ 是所有數(shù)論函數(shù)的集合，顯然有

$%5Cmathcal%20D(s%2Cf%2Bg)%3D%5Cmathcal%20D(s%2Cf)%2B%5Cmathcal%20D(s%2Cg)%2Cf%2Cg%5Cin%5Cmathbb%20A$
$%5Cmathcal%20D(s%2Ckf)%3Dk%5Cmathcal%20D(s%2Cf)%2Ck%5Cin%20%5Cmathbb%20C%2Cf%5Cin%5Cmathbb%20A$

令 $*$ 表示Dirichlet卷積，有

$%5Cbegin%7Baligned%7D%5Cmathcal%20D(s%2Cf)%5Cmathcal%20D(s%2Cg)%26%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bf(n)%7D%7Bn%5Es%7D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bg(n)%7D%7Bn%5Es%7D%5C%5C%26%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Csum_%7Bm%3D1%7D%5E%5Cinfty%5Cfrac%7Bf(n)g(m)%7D%7B(mn)%5Es%7D%5Cend%7Baligned%7D$

作代換 $k%3Dmn$ ，可得

$%5Cbegin%7Baligned%7D%5Cmathcal%20D(s%2Cf)%5Cmathcal%20D(s%2Cg)%26%3D%5Csum_%7Bk%3D1%7D%5E%5Cinfty%5Cfrac%7B1%7D%7Bk%5Es%7D%5Csum_%7Bk%3Dmn%7Df(n)g(m)%5C%5C%26%3D%5Csum_%7Bk%3D1%7D%5E%5Cinfty%5Cfrac%7Bf*g(k)%7D%7Bk%5Es%7D%5Cend%7Baligned%7D$

$%5CRightarrow%20%5Cmathcal%20D(s%2Cf)%5Cmathcal%20D(s%2Cg)%3D%5Cmathcal%20D(s%2Cf*g)$

以 $%5Ctilde%7Bf%7D%20(n)$ 表示 $f(n)$ 的Mobius變換，則有

$%5Cmathcal%20D(s%2C%5Ctilde%7Bf%7D)%3D%5Czeta(s)%5Cmathcal%20D(s%2Cf)$

一個(gè)應(yīng)用

取mobius函數(shù) $%5Cmu(n)$ ，根據(jù)廣義Euler乘積公式

$%5Cmathcal%20D(s%2C%5Cmu)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5Cmu(n)%7D%7Bn%5Es%7D%3D%5Cprod_p%5Cleft(1%2B%5Cfrac%7B%5Cmu(p)%7D%7Bp%5Es%7D%2B%5Cfrac%7B%5Cmu(p%5E2)%7D%7Bp%5E%7B2s%7D%7D%2B%E2%80%A6%5Cright)%3D%5Cprod_p%5Cleft(1-%5Cfrac1%7Bp%5Es%7D%5Cright)$

因此，可以得到

$%5Cmathcal%20D(s%2C%5Cmu)%3D%5Cfrac1%7B%5Czeta(s)%7D$

再根據(jù)Dirichlet級(jí)數(shù)乘積性質(zhì)，就再次推出了Mobius反演公式，又有

$1%3D%5Cmathcal%20D(s%2C%5Cvarepsilon)$

其中 $%5Cvarepsilon(n)$ 是單位示性函數(shù)，因此也可得1與Mobius函數(shù)的Dirichlet關(guān)系

再從zeta函數(shù)出發(fā)，對(duì)它取一階導(dǎo)數(shù)，可得

$%5Czeta'(s)%3D-%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5Clog%20n%7D%7Bn%5Es%7D%3D-%5Cmathcal%20D(s%2C%5Clog)$

進(jìn)一步，若取k階導(dǎo)數(shù)，有

$%5Cfrac%7B%5Cmathrm%20d%5Ek%7D%7B%5Cmathrm%20ds%5Ek%7D%5Czeta(s)%3D(-1)%5Ek%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5Clog%5Ek%20n%7D%7Bn%5Es%7D%3D(-1)%5Ek%5Cmathcal%20D(s%2C%5Clog%5Ek)$

對(duì)zeta函數(shù)的Euler乘積取對(duì)數(shù)

$%5Cbegin%7Baligned%7D%5Cln%5Czeta(s)%26%3D-%5Csum_p%5Cln%5Cleft(1-%5Cfrac1%7Bp%5Es%7D%5Cright)%5C%5C%26%3D%5Csum_p%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bnp%5E%7Bsn%7D%7D%5Cend%7Baligned%7D$

再取導(dǎo)數(shù)，得到

$%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)%3D-%5Csum_p%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5Clog%20n%7D%7Bp%5E%7Bsn%7D%7D$

這里以 $%5Cfrac%20%7Bf'%7Df(s)$ 表示 $%5Cfrac%7Bf'(s)%7D%7Bf(s)%7D$

注意到此和式實(shí)際上就是遍歷所有素?cái)?shù)的乘方，因此可以利用Von mangoldt函數(shù)將它改寫為

$%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)%3D-%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5CLambda(n)%7D%7Bn%5Es%7D%3D-%5Cmathcal%20D(s%2C%5CLambda)$

由此及乘積性質(zhì)可以建立Von?mangoldt函數(shù)與自然對(duì)數(shù)間的Mobius變換關(guān)系

$%5Csum_%7Bd%7Cn%7D%5CLambda(d)%3D%5Clog%20n$

Selberg等式

關(guān)于該等式，利用Dirichlet級(jí)數(shù)可以給出一個(gè)十分簡潔的證明，有

$%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)%3D-%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5CLambda(n)%7D%7Bn%5Es%7D%3D-%5Cmathcal%20D(s%2C%5CLambda)$

對(duì)它取個(gè)導(dǎo)數(shù)吧

$%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20ds%7D%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5CLambda(n)%5Clog%20n%7D%7Bn%5Es%7D%3D%5Cmathcal%20D(s%2C%5CLambda%5Ccdot%5Clog)$

而又有

$%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20ds%7D%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)%3D%5Cfrac%7B%5Czeta''%7D%7B%5Czeta%7D(s)-%5Cleft(%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)%5Cright)%5E2$

$%5CRightarrow%20%5Cmathcal%20D(s%2C%5CLambda%5Ccdot%5Clog)%3D%5Cmathcal%20D(s%2C%5Clog%5E2)%5Cmathcal%20D(s%2C%5Cmu)-%5Cmathcal%20D%5E2(s%2C%5CLambda)$

$%5CRightarrow%5Cmathcal%20D(s%2C%5CLambda%5Ccdot%5Clog)%3D%5Cmathcal%20D(s%2C%5Clog%5E2%5Ccirc%5Cmu-%5CLambda%5Ccirc%5CLambda)$

$%5CLambda(n)%5Clog%20n%3D%5Csum_%7Bd%7Cn%7D%5Cmu%5Cleft(%5Cfrac%20nd%5Cright)%5Clog%5E2d-%5Csum_%7Bd%7Cn%7D%5CLambda(d)%5CLambda%5Cleft(%5Cfrac%20nd%5Cright)$

雜燴

不妨試試把一些數(shù)論函數(shù)揉進(jìn)Dirichlet級(jí)數(shù)里：

（以下 $%5Comega(n)$ 表示n的不同素因子個(gè)數(shù)， $%5COmega(n)$ 表示所有素因子個(gè)數(shù)， $%5Comega(1)%3D%5COmega(1)%3D0$ ）

由于 $d(n)%3D%5Csum_%7Bd%7Cn%7D1%3D1*1(n)$ ，因此

$%5Cmathcal%20D(s%2Cd)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bd(n)%7D%7Bn%5Es%7D%3D%5Czeta%5E2(s)$
更進(jìn)一步，根據(jù)Dirichlet卷積的另一種定義，有

$%5Cunderbrace%7B1*1*%E2%80%A6*1%7D_%7Bk%E4%B8%AA%7D%3D%5Csum_%7Bn%3Da_1a_2%E2%80%A6a_k%7D1$

右式可以看做將n分解為k個(gè)數(shù)相乘的方法種數(shù)，

設(shè) $%5Ctau_k(n)%3D%5Csum_%7Bn%3Da_1a_2%E2%80%A6a_k%7D1$

$%5Cbegin%7Baligned%7D%5CRightarrow%20%5Cmathcal%20D(s%2C%5Ctau_k)%26%3D%5Cmathcal%20D(s%2C%5Cunderbrace%7B1*1*%E2%80%A6*1%7D_%7Bk%E4%B8%AA%7D)%5C%5C%26%3D%5Cmathcal%20D%5Ek(s%2C1)%3D%5Czeta%5Ek(s)%5Cend%7Baligned%7D$
還用另一種方式推廣：設(shè)

$%5Csigma_k(n)%3D%5Csum_%7Bd%7Cn%7Dd%5Ek%3D1*%5Cmathrm%20%7Bid%7D%5Ek(n)$

$%5CRightarrow%20%5Cmathcal%20D(s%2C%5Csigma_k)%3D%5Czeta(s)%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bn%5Ek%7D%7Bn%5Es%7D%3D%5Czeta(s)%5Czeta(s-k)$
根據(jù)Mobius函數(shù)的性質(zhì)，Euler函數(shù)

$%5Cvarphi(n)%3Dn%5Cprod_%7Bp%7Cn%7D%5Cleft(1-%5Cfrac1p%5Cright)%3D%5Csum_%7Bd%7Cn%7D%5Cmu(d)%5Ccdot%5Cfrac%20nd%3D%5Cmu*%5Cmathrm%20%7Bid%7D(n)$

因此，其Dirichlet生成函數(shù)為：

$%5Cmathcal%20D(s%2C%5Cvarphi)%3D%5Cfrac1%7B%5Czeta(s)%7D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bn%7D%7Bn%5Es%7D%3D%5Cfrac%7B%5Czeta(s-1)%7D%7B%5Czeta(s)%7D$
再由zeta函數(shù)在偶數(shù)處的值，我們得到一個(gè)形式上的zeta函數(shù)在奇數(shù)處的公式：

$%5Czeta(2n-1)%3D(-1)%5E%7Bn%2B1%7D%5Cfrac%7BB_%7B2n%7D(2%5Cpi)%5E%7B2n%7D%7D%7B2(2n)!%7D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5Cvarphi(n)%7D%7Bn%5Es%7D$
這個(gè)公式并不能用來計(jì)算在奇數(shù)處的值，也就是說它其實(shí)沒啥鳥用(
因Mobius函數(shù)僅在n無平方因子時(shí)不為零且絕對(duì)值都是1，因此取它的絕對(duì)值或平方即可表示對(duì)無平方因子整數(shù)的示性函數(shù)

$%5Cmu%5E2(n)%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%0A1%20%2C%26%20%5Cnexists%20p%2Cp%5E2%7Cn%20%5C%5C%200%2C%20%20%26%20%5Ctext%7Botherwise%7D%0A%5Cend%7Barray%7D%5Cright.$

Mobius函數(shù)是積性的，因此其平方也是積性，將它揉進(jìn)Dirichlet級(jí)數(shù)里并利用Euler乘積：

$%5Cbegin%7Baligned%7D%5Cmathcal%20D(s%2C%5Cmu%5E2)%26%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5Cmu%5E2(n)%7D%7Bn%5Es%7D%5C%5C%26%3D%5Cprod_p%5Cleft(1%2B%5Cfrac%7B%5Cmu%5E2(p)%7D%7Bp%5Es%7D%2B%5Cfrac%7B%5Cmu%5E2(p%5E2)%7D%7Bp%5E%7B2s%7D%7D%E2%80%A6%5Cright)%5C%5C%26%3D%5Cprod_p%5Cleft(1%2B%5Cfrac1%7Bp%5Es%7D%5Cright)%5C%5C%26%3D%5Cprod_p%5Cfrac%7B%5Cleft(1%2B%5Cfrac1%7Bp%5Es%7D%5Cright)%5Cleft(1-%5Cfrac1%7Bp%5Es%7D%5Cright)%7D%7B%5Cleft(1-%5Cfrac1%7Bp%5Es%7D%5Cright)%7D%5C%5C%26%3D%5Cprod_p%5Cfrac%7B%5Cleft(1-%5Cfrac1%7Bp%5E%7B2s%7D%7D%5Cright)%5E%7B-1%7D%7D%7B%5Cleft(1-%5Cfrac1%7Bp%5E%7Bs%7D%7D%5Cright)%5E%7B-1%7D%7D%5Cend%7Baligned%7D$

$%5CRightarrow%20%5Cmathcal%20D(s%2C%5Cmu%5E2)%3D%5Cfrac%7B%5Czeta(2s)%7D%7B%5Czeta(s)%7D$
引入Liouville函數(shù) $%5Clambda(n)%3D(-1)%5E%7B%5COmega(n)%7D$

$%5Cforall%20m%2Cn%5Cin%5Cmathbb%20N%2C%5COmega(mn)%3D%5COmega(m)%2B%5COmega(n)%5CRightarrow%20%5Clambda(mn)%3D%5Clambda(m)%5Clambda(n)$
于是Liouville函數(shù)是完全積性的，因此利用Euler乘積，
$%5Cbegin%7Baligned%7D%5Cmathcal%20D(s%2C%5Clambda)%26%3D%5Cprod_p%5Cleft(1-%5Cfrac%7B%5Clambda(p)%7D%7Bp%5Es%7D%5Cright)%5E%7B-1%7D%5C%5C%26%3D%5Cprod_p%5Cleft(1%2B%5Cfrac1%7Bp%5Es%7D%5Cright)%5E%7B-1%7D%5Cend%7Baligned%7D$
于是如法炮制地得到
$%5Cmathcal%20D(s%2C%5Clambda)%3D%5Cfrac%7B%5Czeta(2s)%7D%7B%5Czeta(s)%7D%3D%5Cfrac1%7B%5Cmathcal%20D(s%2C%5Cmu%5E2)%7D$
因此Liouville函數(shù)與Mobius函數(shù)的平方是互為Dirichlet逆的
$%5Czeta(2s)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B1%7D%7Bn%5E%7B2s%7D%7D$
注意到該和式只有正整數(shù)的平方參與，因此可以設(shè)
$%5Ctext%7Bsqrt%7D(n)%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%0A1%20%2C%26%20%5Cexists%20m%5Cin%5Cmathbb%20N%2Cn%3Dm%5E2%20%5C%5C%200%2C%20%20%26%20%5Ctext%7Botherwise%7D%0A%5Cend%7Barray%7D%5Cright.$
則在其絕對(duì)收斂的情況下，
$%5Czeta(2s)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5Ctext%7Bsqrt%7D(n)%7D%7Bn%5Es%7D$
由omega函數(shù)的定義可知

$%5Comega(n)%3D%5Csum_%7Bp%7Cn%7D1%3D1*%5Cmathrm%20p(n)$

其中 $%5Cmathrm%20p(n)%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%0A1%20%2C%26%20n%5Cin%20%5Cmathbb%20P%20%5C%5C%200%2C%20%20%26%20n%5Cnotin%5Cmathbb%20P%0A%5Cend%7Barray%7D%5Cright.$ 為素?cái)?shù)的示性函數(shù)

$%5Cmathcal%20D(s%2C%5Comega)%3D%5Czeta(s)%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5Cmathrm%20p(n)%7D%7Bn%5Es%7D%3D%5Czeta(s)%5Csum_p%5Cfrac1%7Bp%5Es%7D$