# User:Virginia-American/Sandbox/Dirichlet convolutions

## Generating functions

The following formulas are known:

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

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

${\displaystyle B1)\;\;\;\zeta (s-1)=\sum _{n=1}^{\infty }{\frac {n}{n^{s}}}}$       [1][2]

${\displaystyle C)\;\;\;\;{\frac {1}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}}$       [3][4]

${\displaystyle D)\;\;\;\;\zeta (s)\zeta (s-k)=\sum _{n=1}^{\infty }{\frac {\sigma _{k}(n)}{n^{s}}}.}$       [5][6]

${\displaystyle D0)\;\;\;\zeta ^{2}(s)=\sum _{n=1}^{\infty }{\frac {d(n)}{n^{s}}}}$       [7][8]

${\displaystyle D1)\;\;\;\zeta (s)\zeta (s-1)=\sum _{n=1}^{\infty }{\frac {\sigma (n)}{n^{s}}}}$       [9]

${\displaystyle E)\;\;\;\;{\frac {\zeta (s-1)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\phi (n)}{n^{s}}}}$       [10][11]

${\displaystyle F)\;\;\;\;{\frac {\zeta (2s)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}.}$       [12][13]

${\displaystyle G)\;\;\;\;{\frac {\zeta (s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {|\mu (n)|}{n^{s}}}.}$       [14][15]

${\displaystyle H)\;\;\;\;{\frac {\zeta ^{2}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {2^{\omega (n)}}{n^{s}}}}$       [16][17]

${\displaystyle I)\;\;\;\;{\frac {\zeta ^{3}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {d(n^{2})}{n^{s}}}}$       [18]

${\displaystyle J)\;\;\;\;{\frac {\zeta ^{4}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {d^{2}(n)}{n^{s}}}}$       [19][20]

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

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

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

${\displaystyle L)\;\;\;\;{\frac {\zeta '(s)}{\zeta (s)}}=-\sum _{n=1}^{\infty }{\frac {\Lambda (n)}{n^{s}}}}$       [21][22]

${\displaystyle M)\;\;\;\;\zeta '(s)=-\sum _{n=1}^{\infty }{\frac {\log n}{n^{s}}}}$       [23]

## Convolutions

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

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

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

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

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

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

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

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

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

## Derivations

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

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

and thus

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

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

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

From the Euler product

{\displaystyle {\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

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

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

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

Setting k to 0 and 1 gives the special cases

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

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

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

{\displaystyle {\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}}}

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

Consider

{\displaystyle {\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.

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

Now consider

{\displaystyle {\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.

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

${\displaystyle {\frac {\zeta ^{2}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {2^{\omega (n)}}{n^{s}}}}$
${\displaystyle {\frac {\zeta ^{3}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {d(n^{2})}{n^{s}}}}$
${\displaystyle {\frac {\zeta ^{4}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {d^{2}(n)}{n^{s}}}}$
${\displaystyle {\frac {\zeta (2s)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}}$

Since A) times C) is one,

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

1. ^ Hardy & Wright, § 17.2
2. ^ Titchmarsh, § 1.1
3. ^ Hardy & Wright, Thm. 287
4. ^ Titchmarsh, Eq. 1.1.4
5. ^ Hardy & Wright, Th. 291
6. ^ Titchmarsh, Eq. 1.3.1
7. ^ Hardy & Wright, Thm. 289
8. ^ Titchmarsh, Eq. 1.2.1
9. ^ Hardy & Wright, Thm. 290
10. ^ Hardy & Wright, Thn. 288
11. ^ Titchmarsh, Eq. 1.2.12
12. ^ Hardy & Wright, Thm. 300
13. ^ Titchmarsh, Eq. 1.2.11
14. ^ Hardy & Wright, Th. 302
15. ^ Titchmarsh, Eq. 1.2.7
16. ^ Hardy & Wright, Thm. 301
17. ^ Titchmarsh, Eq. 1.2.8
18. ^ Titchmarsh, § 1.2.9
19. ^ Hardy & Wright, Thm. 304
20. ^ Titchmarsh, Eq. 1.2.10
21. ^ Hardy & Wright, Thm. 294
22. ^ Titchmarsh, Eq. 1.1.8
23. ^ Hardy & Wright, Thm. 294