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.

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: