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Thread: Euler function phi(n) using the Moubilus function

  1. November 18th 2008, 08:28 AM #1
    bamby
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    Cool Euler function phi(n) using the Moubilus function

    Hi,
    How can I do that?
    Express Euler function phi(n) using the Moubilus function miu(n)
    Thank you!
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  2. November 18th 2008, 11:42 AM #2
    ThePerfectHacker
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    Quote Originally Posted by bamby View Post
    Hi,
    How can I do that?
    Express Euler function phi(n) using the Moubilus function miu(n)
    Thank you!
    Consider the group \mathbb{Z}_n. Every element has order a divisor of n. Also if d|n then the number of elements that has order d is \phi (d). Therefore, we have that \sum_{d|n}\phi (d) = n.

    Since this is true for all n\geq 1 by Mobius inversion formula we see that, \phi (n) = n\left( \sum_{d|n} \frac{\mu (d)}{d} \right).
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