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March 17th, 2019, 09:56 AM   #1
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Dirichlet Convolution question

if p is prime and k>1 then let $f(p^{k})=log(p)$, for all other n let f(n)=0.

Prove that$(f*u)(n)=log(n)$ for all n

where u(n)=1 for all n.

Ok so for this question i have started by saying

\[(f*u)(n)=\sum_{j|n}f(j)*u(\frac{n}{j})=f(1)*u(p^k) +f(p)*u(p^{k-1})+...f(p^k)*u(1)\]

but i don't know how i can go from here since i can not see how i get to the end answer. I was thinking maybe i could various values of d which divide n expressed in the prime factorisation?? Thanks to anyone who can solve this.

Last edited by Jaket1; March 17th, 2019 at 10:03 AM.
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March 23rd, 2019, 02:34 PM   #2
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Did you meant "and $k\geqslant1$"? with Von Mangoldt $\Lambda * 1 = log$

Well, if you look at $n=p_{i_1}^{j_1}\cdot p_{i_2}^{j_2}....$ its prime factorization, and knowing $log(ab)=log(a)+log(b)$ or $log(a^b)=b\cdot log(a)$, you should be fine with sums of logs.
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