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Math Help - How to apply Mobius inversion formula

  1. #1
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    How to apply Mobius inversion formula

    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
    Last edited by mathematicaphoenix; January 1st 2012 at 12:43 AM.
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  2. #2
    Super Member PaulRS's Avatar
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    Re: How to apply Mobius inversion formula

    • \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!
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  3. #3
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    Re: How to apply Mobius inversion formula

    Quote Originally Posted by PaulRS View Post
    • \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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