# Thread: How to apply Mobius inversion formula

1. ## How to apply Mobius inversion formula

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

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

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

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

Thanks
Nabigh

2. ## Re: How to apply Mobius inversion formula

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

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

So in fact $\displaystyle \left(\sum_{n\geq 1} \frac{\mu(n)}{n^s}\right) \cdot \zeta(s) = \sum_{\n\geq 1} {\tfrac{h(n)}{n^s}}$ where $\displaystyle h(n) = \sum_{d|n}\mu(n)$ (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 $\displaystyle \sum_{d|n}\mu(n)$ to start with ( another way would be to derive the above result from the fact that $\displaystyle \mu$ is multiplicative -factorizing the sum-, and then prove the inversion).

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

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

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

Here comes the trick:

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

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

3. ## Re: How to apply Mobius inversion formula

Originally Posted by PaulRS
• $\displaystyle \frac{1}{\zeta(s)} = \sum_{n\geq 1} \frac{\mu(n)}{n^s}$.

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

So in fact $\displaystyle \left(\sum_{n\geq 1} \frac{\mu(n)}{n^s}\right) \cdot \zeta(s) = \sum_{\n\geq 1} {\tfrac{h(n)}{n^s}}$ where $\displaystyle h(n) = \sum_{d|n}\mu(n)$ (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 $\displaystyle \sum_{d|n}\mu(n)$ to start with ( another way would be to derive the above result from the fact that $\displaystyle \mu$ is multiplicative -factorizing the sum-, and then prove the inversion).

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

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

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

Here comes the trick:

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

But $\displaystyle \sum_{n\cdot k = m} \frac{\mu(n)}{n} \cdot \frac{1}{k} = \sum_{d|m} \frac{\mu (d)}{d }\cdot \frac{1}{(m/d)} = \tfrac{1}{m}\cdot \sum_{d|m} \mu (d)$ and we know how to sum this!
If you could a little more explanation on individual would be nice...

Thanks