Define [math]\Lambda(n):=\log(p)[/math] if n is a power of a prime p

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

 

Prove that [math]\Lambda(n)=\sum_{d|n}\mu(\tfrac{n}{d})\log(d)[/math]

 

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

 

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

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.

New question:

[math]\sigma_k(n)=\sum_{d|n} d^k[/math]

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.

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

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

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

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.

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

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