User:Virginia-American/Sandbox/Dirichlet convolutions

Generating functions

The following formulas are known:

$A)\;\;\;\;\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\prod _{p}\left(1-{\frac {1}{p^{s}}}\right)^{-1}$

$B)\;\;\;\;\zeta (s-k)=\sum _{n=1}^{\infty }{\frac {n^{k}}{n^{s}}}$

$B1)\;\;\;\zeta (s-1)=\sum _{n=1}^{\infty }{\frac {n}{n^{s}}}$      

$C)\;\;\;\;{\frac {1}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}$      

$D)\;\;\;\;\zeta (s)\zeta (s-k)=\sum _{n=1}^{\infty }{\frac {\sigma _{k}(n)}{n^{s}}}.$      

$D0)\;\;\;\zeta ^{2}(s)=\sum _{n=1}^{\infty }{\frac {d(n)}{n^{s}}}$      

$D1)\;\;\;\zeta (s)\zeta (s-1)=\sum _{n=1}^{\infty }{\frac {\sigma (n)}{n^{s}}}$      

$E)\;\;\;\;{\frac {\zeta (s-1)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\phi (n)}{n^{s}}}$      

$F)\;\;\;\;{\frac {\zeta (2s)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}.$      

$G)\;\;\;\;{\frac {\zeta (s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {|\mu (n)|}{n^{s}}}.$      

$H)\;\;\;\;{\frac {\zeta ^{2}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {2^{\omega (n)}}{n^{s}}}$      

$I)\;\;\;\;{\frac {\zeta ^{3}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {d(n^{2})}{n^{s}}}$      

$J)\;\;\;\;{\frac {\zeta ^{4}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {d^{2}(n)}{n^{s}}}$      

Let χ(n) be the characteristic function of the squares:

$\chi (n)={\begin{cases}&1{\mbox{ if }}n{\mbox{ is a square }}\\&0{\mbox{ if }}n{\mbox{ is not square.}}\end{cases}}$

$K)\;\;\;\;\zeta (2s)=\sum _{n=1}^{\infty }{\frac {\chi (n)}{n^{s}}}$

$L)\;\;\;\;{\frac {\zeta '(s)}{\zeta (s)}}=-\sum _{n=1}^{\infty }{\frac {\Lambda (n)}{n^{s}}}$      

$M)\;\;\;\;\zeta '(s)=-\sum _{n=1}^{\infty }{\frac {\log n}{n^{s}}}$      

Convolutions

$\sum _{\delta \mid n}\mu (\delta )=\sum _{\delta \mid n}\lambda \left({\frac {n}{\delta }}\right)|\mu (\delta )|={\begin{cases}&1{\mbox{ if }}n=1\\&0{\mbox{ if }}n\neq 1.\end{cases}}$       A) × C) = F) × G) = 1.

$\sum _{\delta \mid n}\phi (\delta )=n.$       A) × E) = B1.

$\sum _{\delta \mid n}\phi (\delta )d\left({\frac {n}{\delta }}\right)=\sigma (n).$       E) × D0) = D1.

$\sum _{\delta \mid n}|\mu (\delta )|=2^{\omega (n)}.$       A) × G) = H.

$\sum _{\delta \mid n}2^{\omega (\delta )}=d(n^{2}).$       A) × H) = I.

$\sum _{\delta \mid n}d(\delta ^{2})=d^{2}(n).$       A) × I) = J.

$\sum _{\delta \mid n}d\left({\frac {n}{\delta }}\right)2^{\omega (\delta )}=d^{2}(n).$       D0) × H) = J.

$\sum _{\delta \mid n}\lambda (\delta )={\begin{cases}&1{\mbox{ if }}n{\mbox{ is a square }}\\&0{\mbox{ if }}n{\mbox{ is not square.}}\end{cases}}$       A) × F) = K.

$\sum _{\delta \mid n}\Lambda (\delta )=\log n.$       A) × L) = M.

Derivations

We have the series and Euler product formulas for ζ(s):

$A)\;\;\;\;\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\prod _{p}\left(1-{\frac {1}{p^{s}}}\right)^{-1}$

and thus

$B)\;\;\;\;\zeta (s-k)=\sum _{n=1}^{\infty }{\frac {n^{k}}{n^{s}}}$

$B1)\;\;\;\zeta (s-1)=\sum _{n=1}^{\infty }{\frac {n}{n^{s}}}$

Let ${\frac {1}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {a(n)}{n^{s}}}.$

From the Euler product

${\begin{aligned}{\frac {1}{\zeta (s)}}&=\prod _{p}\left(1-{\frac {1}{p^{s}}}\right)\\&=\left(1-{\frac {1}{2^{s}}}\right)\left(1-{\frac {1}{3^{s}}}\right)\left(1-{\frac {1}{5^{s}}}\right)\dots \\&=1-{\frac {1}{2^{s}}}-{\frac {1}{3^{s}}}-{\frac {1}{5^{s}}}+{\frac {1}{6^{s}}}-{\frac {1}{7^{s}}}+{\frac {1}{10^{s}}}\dots \end{aligned}}$

i.e. a(n) = μ(n), or

$C)\;\;\;\;{\frac {1}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}$

Let A) times B) be    $\zeta (s)\zeta (s-k)=\sum _{n=1}^{\infty }{\frac {a(n)}{n^{s}}}.$   Then   $a(n)=\sum _{\delta \mid n}\delta ^{k}=\sigma _{k}(n),$   giving

$D)\;\;\;\;\zeta (s)\zeta (s-k)=\sum _{n=1}^{\infty }{\frac {\sigma _{k}(n)}{n^{s}}}.$

Setting k to 0 and 1 gives the special cases

$D0)\;\;\;\zeta ^{2}(s)=\sum _{n=1}^{\infty }{\frac {d(n)}{n^{s}}}$

$D1)\;\;\;\zeta (s)\zeta (s-1)=\sum _{n=1}^{\infty }{\frac {\sigma (n)}{n^{s}}}$

Let B1) times C) be   ${\frac {\zeta (s-1)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {a(n)}{n^{s}}}.$ Then

${\begin{aligned}a(n)&=\sum _{\delta \mid n}\mu (\delta ){\frac {n}{\delta }}\\&=n\left\{1-\sum _{p\mid n}{\frac {1}{p}}+\sum _{pp'\mid n}{\frac {1}{pp'}}-\sum _{pp'p''\mid n}{\frac {1}{pp'p''}}+\dots \right\}\\&=n\prod _{p\mid n}\left(1-{\frac {1}{p}}\right)\\&=\phi (n)\end{aligned}}$

$E)\;\;\;\;{\frac {\zeta (s-1)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\phi (n)}{n^{s}}}$

Consider

${\begin{aligned}{\frac {\zeta (s)}{\zeta (2s)}}&=\prod _{p}{\frac {1-p^{2s}}{1-p^{s}}}\\&=\prod _{p}\left(1+p^{s}\right)\\&=\left(1+{\frac {1}{2^{s}}}\right)\left(1+{\frac {1}{3^{s}}}\right)\left(1+{\frac {1}{5^{s}}}\right)\dots \\&=\sum _{n=1}^{\infty }{\frac {|\mu (n)|}{n^{s}}}\end{aligned}}$

i.e.

$F)\;\;\;\;{\frac {\zeta (s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {|\mu (n)|}{n^{s}}}.$

Now consider

${\begin{aligned}{\frac {\zeta (2s)}{\zeta (s)}}&=\prod _{p}{\frac {1-p^{s}}{1-p^{2s}}}\\&=\prod _{p}\left(1+p^{s}\right)^{-1}\\&=\left(1-{\frac {1}{2^{s}}}+{\frac {1}{4^{s}}}-{\frac {1}{8^{s}}}+\dots \right)\left(1-{\frac {1}{3^{s}}}+{\frac {1}{9^{s}}}-{\frac {1}{27^{s}}}+\dots \right)\left(1-{\frac {1}{5^{s}}}+\dots \right)\dots \\&=\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}\end{aligned}}$

i.e.

$G)\;\;\;\;{\frac {\zeta (2s)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}.$

${\frac {\zeta ^{2}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {2^{\omega (n)}}{n^{s}}}$

${\frac {\zeta ^{3}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {d(n^{2})}{n^{s}}}$

${\frac {\zeta ^{4}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {d^{2}(n)}{n^{s}}}$

${\frac {\zeta (2s)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}$

Since A) times C) is one,

$\sum _{\delta \mid n}\mu (\delta )={\begin{cases}&1{\mbox{ if }}n=1\\&0{\mbox{ if }}n\neq 1.\end{cases}}$