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Thread: proving the converse theorem

  1. #1
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    Question proving the converse theorem

    Here is another question that I got stuck with. Please see attachment. I typed the question in " Microsoft word" . If you can't load the attachment , please do let me know.

    Thank you very much.
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    Here is another question that I got stuck with. Please see attachment. I typed the question in " Microsoft word" . If you can't load the attachment , please do let me know.

    Thank you very much.
    Did you consider using the Mobius inversion formula?

    If $\displaystyle g(n)=\sum_{d|n} f(d)$is multiplicative then,

    $\displaystyle f(n)=\sum_{d|n} \mu (n/d) \cdot g(d) $

    Now, $\displaystyle g(n)$ is multiplicative function
    And $\displaystyle \mu(n)$ (the Mobius mu function) is also multiplicative.
    Thus,
    $\displaystyle g(n)\cdot \mu(n)$ is also multiplicative.
    Thus,
    $\displaystyle f(n)$ is multiplicative (because of the summation theorem).
    Last edited by ThePerfectHacker; Dec 27th 2006 at 12:22 PM.
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    Hi perfecthacker,

    Thank you very much for your reply. I have never learnt Mobius inversion forumla. I will go through your prove. If I don't understand it, I will come back to you.
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  4. #4
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    Hi perfecthacker,

    Can you show me the other way ( not by mobius inversion forumla) to prove this question? I don't really get your prove because I dont' quite understand the mobius inversion formula. Thank you very much for your help.
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  5. #5
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    Quote Originally Posted by Jenny20 View Post
    Hi perfecthacker,

    Can you show me the other way ( not by mobius inversion forumla) to prove this question? I don't really get your prove because I dont' quite understand the mobius inversion formula. Thank you very much for your help.
    I just looked up the proof in my number theory book and they prove it the same way I did. Thus, perhaps this is the standard way.
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