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 &chi;(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
$$ \sum_{\delta\mid n}\mu(\delta)= \sum_{\delta\mid n}\lambda\left(\frac{n}{\delta}\right)|\mu(\delta)|= \begin{cases} &1\mbox{ if } n=1\\ &0\mbox{ if } n\ne1. \end{cases} $$       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 &zeta;(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{align} \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{align} $$

i.e. a(n) = &mu;(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

$$ \begin{align} a(n) &= \sum_{\delta\mid n}\mu(\delta)\frac{n}{\delta}\\ &=n\left\{1-\sum_{p\mid n}\frac{1}{p}+\sum_{pp'\mid n}\frac{1}{pp'}-\sum_{pp'p\mid n}\frac{1}{pp'p}+\dots\right\}\\ &=n\prod_{p\mid n}\left(1-\frac{1}{p}\right)\\ &=\phi(n) \end{align} $$

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

Consider
 * $$\begin{align}

\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{align} $$

i.e.

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

Now consider
 * $$\begin{align}

\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{align} $$

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\ne1. \end{cases} $$