How to apply Mobius inversion formula

**mathematicaphoenix** · December 31st 2011, 06:27 PM

How do you apply Mobius inversion formula
' If $f(n)$ and $g(n)$ are arithmetic functions satisfying
$g(n)=\sum_{d\mid n}f(d)$
then
$f(n)=\sum_{d\mid n}\mu (d)g(\frac{n}{d})$ '

How do you get identities like
$\frac{1}{\zeta (s)}=\sum_{n=1}^{\infty }\frac{\mu (n)}{n^{s}}$
or

If
$\ln \zeta (s)=\sum_{n=1}^{\infty }\frac{\zeta_p(sn) }{n}$ (2)
where
$\zeta_p(sn)$
is the prime zeta function. When Mobius inversion formula is applied to (2) gives
$\zeta_p(s)=\sum_{n=1}^{\infty }\frac{\mu (n) }{n}\ln \zeta (sn)$ (3)

Can anyone explain how to get (3) by applying Mobius inversion to (2)

Thanks
Nabigh

**PaulRS** · January 1st 2012, 02:17 PM

$\frac{1}{\zeta(s)} = \sum_{n\geq 1} \frac{\mu(n)}{n^s}$ .

To derive this from Möbius inversion formula as you ask, remember that $\left(\sum_{n\geq 1} \frac{f(n)}{n^s}\right)\cdot \left(\sum_{n\geq 1} \frac{g(n)}{n^s}\right) = \sum_{n\geq 1} \frac{h(n)}{n^s}$ where $h(n) = \sum_{d|n} f(d)\cdot g\left(\tfrac{n}{d}\right)$

So in fact $\left(\sum_{n\geq 1} \frac{\mu(n)}{n^s}\right) \cdot \zeta(s) = \sum_{\n\geq 1} {\tfrac{h(n)}{n^s}}$ where $h(n) = \sum_{d|n}\mu(n)$ (what is this sum equal to? apply Möbius inversion formula

)

Remark: Although this works, the problem is that Möbius inversion formula is first derived from the value of $\sum_{d|n}\mu(n)$ to start with ( another way would be to derive the above result from the fact that $\mu$ is multiplicative -factorizing the sum-, and then prove the inversion).

$\zeta_p(s) = \sum_{n=1}^{\infty} { \frac{\mu(n)}{n} \cdot \log \zeta(s\cdot n) }$ .

Let $L = \sum_{n=1}^{\infty} { \frac{\mu(n)}{n} \cdot \log \zeta(s\cdot n) }$ (to avoid writing it out everywhere)

By (2) we have : $L = \sum_{n=1}^{\infty} { \frac{\mu(n)}{n} \cdot \log \zeta(s\cdot n) } = \sum_{n=1}^{\infty} { \frac{\mu(n)}{n} \cdot \sum_{k=1}^{\infty}{\frac{\zeta_p(s\cdot n\cdot k)}{k}} = \sum_{n=1}^{\infty} { \sum_{k=1}^{\infty}{\frac{\mu(n)}{n} \cdot \frac{\zeta_p(s\cdot n\cdot k)}{k}}$

Here comes the trick:

Let's look at each $\zeta_p(s\cdot n\cdot k)$ . We can distinguish them by the value of $n\cdot k$ .
Note that in $\sum_{n=1}^{\infty} { \sum_{k=1}^{\infty}{\frac{\mu(n)}{n} \cdot \frac{\zeta_p(s\cdot n\cdot k)}{k}}$ we have that $\frac{\mu(n)}{n} \cdot \frac{1}{k}$ adds to the coefficient of $\zeta_p(s\cdot m)$ if and only if $m = n\cdot k$ .
This means that actually $L = \sum_{n=1}^{\infty} { \sum_{k=1}^{\infty}{\frac{\mu(n)}{n} \cdot \frac{\zeta_p(s\cdot n\cdot k)}{k}} = \sum_{m=1}^ {\infty}{ \left( \sum_{n\cdot k = m} \frac{\mu(n)}{n} \cdot \frac{1}{k} \right) \cdot \zeta_p(s\cdot m)}$

But $\sum_{n\cdot k = m} \frac{\mu(n)}{n} \cdot \frac{1}{k} = \sum_{d|m} \frac{\mu (d)}{d }\cdot \frac{1}{(m/d)} = \tfrac{1}{m}\cdot \sum_{d|m} \mu (d)$ and we know how to sum this!

**mathematicaphoenix** · January 2nd 2012, 07:28 AM

Originally Posted by PaulRS

$\frac{1}{\zeta(s)} = \sum_{n\geq 1} \frac{\mu(n)}{n^s}$ .

To derive this from Möbius inversion formula as you ask, remember that $\left(\sum_{n\geq 1} \frac{f(n)}{n^s}\right)\cdot \left(\sum_{n\geq 1} \frac{g(n)}{n^s}\right) = \sum_{n\geq 1} \frac{h(n)}{n^s}$ where $h(n) = \sum_{d|n} f(d)\cdot g\left(\tfrac{n}{d}\right)$

So in fact $\left(\sum_{n\geq 1} \frac{\mu(n)}{n^s}\right) \cdot \zeta(s) = \sum_{\n\geq 1} {\tfrac{h(n)}{n^s}}$ where $h(n) = \sum_{d|n}\mu(n)$ (what is this sum equal to? apply Möbius inversion formula

)

Remark: Although this works, the problem is that Möbius inversion formula is first derived from the value of $\sum_{d|n}\mu(n)$ to start with ( another way would be to derive the above result from the fact that $\mu$ is multiplicative -factorizing the sum-, and then prove the inversion).

$\zeta_p(s) = \sum_{n=1}^{\infty} { \frac{\mu(n)}{n} \cdot \log \zeta(s\cdot n) }$ .

Let $L = \sum_{n=1}^{\infty} { \frac{\mu(n)}{n} \cdot \log \zeta(s\cdot n) }$ (to avoid writing it out everywhere)

By (2) we have : $L = \sum_{n=1}^{\infty} { \frac{\mu(n)}{n} \cdot \log \zeta(s\cdot n) } = \sum_{n=1}^{\infty} { \frac{\mu(n)}{n} \cdot \sum_{k=1}^{\infty}{\frac{\zeta_p(s\cdot n\cdot k)}{k}} = \sum_{n=1}^{\infty} { \sum_{k=1}^{\infty}{\frac{\mu(n)}{n} \cdot \frac{\zeta_p(s\cdot n\cdot k)}{k}}$

Here comes the trick:

Let's look at each $\zeta_p(s\cdot n\cdot k)$ . We can distinguish them by the value of $n\cdot k$ .
Note that in $\sum_{n=1}^{\infty} { \sum_{k=1}^{\infty}{\frac{\mu(n)}{n} \cdot \frac{\zeta_p(s\cdot n\cdot k)}{k}}$ we have that $\frac{\mu(n)}{n} \cdot \frac{1}{k}$ adds to the coefficient of $\zeta_p(s\cdot m)$ if and only if $m = n\cdot k$ .
This means that actually $L = \sum_{n=1}^{\infty} { \sum_{k=1}^{\infty}{\frac{\mu(n)}{n} \cdot \frac{\zeta_p(s\cdot n\cdot k)}{k}} = \sum_{m=1}^ {\infty}{ \left( \sum_{n\cdot k = m} \frac{\mu(n)}{n} \cdot \frac{1}{k} \right) \cdot \zeta_p(s\cdot m)}$

But $\sum_{n\cdot k = m} \frac{\mu(n)}{n} \cdot \frac{1}{k} = \sum_{d|m} \frac{\mu (d)}{d }\cdot \frac{1}{(m/d)} = \tfrac{1}{m}\cdot \sum_{d|m} \mu (d)$ and we know how to sum this!

If you could a little more explanation on individual would be nice...

Thanks

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