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  1. #1
    Member
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    Problem

    Hi--

    Please help me out in solving this following problem:


    Let $\displaystyle \phi_{1}(n)= n \sum\limits_{d \mid n} |\mu(d)|/d$, then prove that $\displaystyle \phi_{1}(n)= n \prod\limits_{d \mid n} (1+ \frac{1}{p})$
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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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    Quote Originally Posted by Chandru1 View Post
    Hi--

    Please help me out in solving this following problem:


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

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

    Therefore if $\displaystyle n=p_1^{\alpha_1}p_2^{\alpha_2}\cdot\cdot\cdot p_k^{\alpha_k} $, then $\displaystyle \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@; Jun 3rd 2010 at 08:45 PM.
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