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Prove that Λ(n) = ∑ μ(n/d)log(d) = -∑ μ(d)log(d) , where the summations are taken over all d|n.?

• Λ(n) is the Mangoldt function: Λ(n) = log(p) if n = p^k, where p is a prime and k>=1 else Λ(n) = 0 μ(n) is Möbius function where μ(n) = 1 if n = 1 else μ(n) = 0 if (p^2)|n for some prime p else μ(n) = (-1)^r if n = (p_1)(p_2)...(p_r) where p_i are distinct primes.

• Answer:

  Heh, it's rather obvious I haven't dealt with number theoretic functions much before...I'm pretty lost on your work there. As for the last step, decompose the logarithm: ∑ μ(d)log(n/d) = ∑ μ(d)*[log(n)-log(d)] = ∑ [μ(d)*log(n)] - ∑ [μ(d)*log(d)] = log(n)*∑ [μ(d)] - ∑ [μ(d)*log(d)]. And we know it is true that ∑ [μ(d)]=0, so we have our result. (For a while I was trying to use n/d as a divisor of n itself; unfortunately μ didn't seem particularly symmetric about this change of summation variable. I just thought I would mention it as food for thought.)