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 March 17th, 2019, 09:56 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 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. March 23rd, 2019, 02:34 PM #2 Newbie   Joined: Sep 2017 From: Belgium Posts: 19 Thanks: 7 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. Thanks from Jaket1 Tags convolution, dirichlet, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post MrBibbles Number Theory 4 July 25th, 2013 01:07 PM giga.chalauri Calculus 0 April 9th, 2013 04:28 AM Dougy Number Theory 7 April 6th, 2012 07:27 AM sneaky Algebra 0 October 28th, 2010 10:54 AM vams42 Calculus 1 October 22nd, 2010 05:16 AM

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