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Math Help - Problem

  1. #1
    Member
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    Problem

    Hi--

    Please help me out in solving this following problem:


    Let \phi_{1}(n)= n \sum\limits_{d \mid n} |\mu(d)|/d, then prove that \phi_{1}(n)= n \prod\limits_{d \mid n} (1+ \frac{1}{p})
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  2. #2
    Junior Member
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    I’m not sure but I think you can use the Möbius inversion formula.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Champaign, Illinois
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    Quote Originally Posted by Chandru1 View Post
    Hi--

    Please help me out in solving this following problem:


    Let \phi_{1}(n)= n \sum\limits_{d \mid n} |\mu(d)|/d, then prove that \phi_{1}(n)= n \prod\limits_{d \mid n} (1+ \frac{1}{p})
    Theorem: if  f(n) is a multiplicative arithmetic function, then so is  \sum_{d\mid n}f(d) . Let me know if you would like to see the proof of this.

     |\mu(n)| is the square indicator function, so one can easily deduce it's multiplicative. Since  \frac{1}{n} is too, so is  f(n)=\frac{|\mu(n)|}{n} .

    Therefore if  n=p_1^{\alpha_1}p_2^{\alpha_2}\cdot\cdot\cdot p_k^{\alpha_k} , then  \phi_1(n)= n\prod_{i=1}^k \phi_1(p_i^{\alpha_i}) = \prod_{i=1}^k p_i^{\alpha_i}\sum_{d\mid p_i^{\alpha_i}} \frac{|\mu(d)|}{d} = n\prod_{p\mid n}\left(1+\frac{1}{p}\right) .
    Last edited by chiph588@; June 3rd 2010 at 09:45 PM.
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