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Math Help - Sum of Möbius Function

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Sum of Möbius Function

    Let  \mu be the Möbius function and  k,n\in\mathbb{N} , evaluate

     \displaystyle \sum_{d^k\mid n} \mu(d)
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    We can factor this as \prod_{p^{km}||n}(\mu(1)+\mu(p)+\dots+\mu(p^m)) where p^{km} || n means that p^{km}|n but p^{k(m+1)}\nmid n. Clearly the product is 0 unless p^{0}||n for each p. Hence the sum is 1 if n is not divisible by a k-power (other than 1), and 0 otherwise.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    So, is my solution accepted?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Bruno J. View Post
    So, is my solution accepted?
    Haha! I forgot about this thread!

    It appears you're assuming the sum i posted is multiplicative. Why is this so?
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Well it's more that I'm assuming the reader will have no trouble convincing him/herself of it!

    Each term in the sum is \mu(d), where d=p_1^{a_1}\cdots p_j^{a_j} and d^k = p_1^{ka_1}\cdots p_j^{ka_j} is such that d^k | n. In the product I wrote, the term \mu(d) will be obtained as \mu(p_1^{a_1})\cdots \mu(p_j^{a_j}) = \mu(d) once the product is expanded. Conversely, each term in the expansion of the product can be found in the sum.
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