# Number Theory : Mobius Inversion

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Define $\Lambda(n):=\log(p)$ if n is a power of a prime p

and 0 if n = 1 or n is a composite number

Prove that $\Lambda(n)=\sum_{d|n}\mu(\tfrac{n}{d})\log(d)$

The hint says to look at $\sum_{d|n}\Lambda(d)$ and apply the Mobius inversion formula.

So far I have got $\sum_{d|n}\Lambda(d)= \sum_{i=1}^r \log(p_i)= \log(\prod_{i=1}^r p_i)$

assuming that n has r distinct primes in its expansion.

So help

Don't mind the above, I have figured it out. I will post more questions if any in this thread instead.

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New question:

$\sigma_k(n)=\sum_{d|n} d^k$

It was asked to show that sigma is multiplicative, which I have done.

I have to find a formula for it which is where I am stuck.

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Why not break down n into: [imath]n=p_1^{e_1} \cdots p_n^{e_n}[/imath]?

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Trouble is that there are so many possible divisors of n, I have trouble keeping track of all of them.

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You've shown $\sigma_k$ is mutiplicative so it's enough to evaluate it on prime powers, which is just a geometric series.

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Yep, I figured it out. I didn't realise sigma k was just sum of divisor function for k=1 for which we have already done the formula.

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Ahh, good. k=0 should be familiar as well.

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Sorry, I should probably have said that I meant each pi to be prime

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