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MObius transform sum

  1. Aug 23, 2006 #1
    let be the sum (over all the divisors d of n):

    [tex] f(n)= \sum_{d|n} \mu (n/d)g(d) [/tex] my question is if n=prime then you have only 2 numbers 1 and p that are divisors so you get:

    [tex] f(p)= \mu (p)g(1) + \mu (1) g(p) [/tex] is that correct?...now the question is to know what's the value of mu(x) function for x=1 or p. :uhh:
  2. jcsd
  3. Aug 23, 2006 #2


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    Step #1 when trying to learn about mobius inversion and such:

    Look at the definition of the mobius function.

    Complete this step and [tex]\mu(1)[/tex] and [tex]\mu(p)[/tex] will be apparant.
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