Here is another question that I got stuck with. Please see attachment. I typed the question in " Microsoft word" . If you can't load the attachment , please do let me know.

Thank you very much.

Printable View

- Dec 26th 2006, 08:43 AMJenny20proving the converse theorem
Here is another question that I got stuck with. Please see attachment. I typed the question in " Microsoft word" . If you can't load the attachment , please do let me know.

Thank you very much. - Dec 26th 2006, 08:55 AMThePerfectHacker
Did you consider using the Mobius inversion formula?

If $\displaystyle g(n)=\sum_{d|n} f(d)$is multiplicative then,

$\displaystyle f(n)=\sum_{d|n} \mu (n/d) \cdot g(d) $

Now, $\displaystyle g(n)$ is multiplicative function

And $\displaystyle \mu(n)$ (the Mobius mu function) is also multiplicative.

Thus,

$\displaystyle g(n)\cdot \mu(n)$ is also multiplicative.

Thus,

$\displaystyle f(n)$ is multiplicative (because of the summation theorem). - Dec 26th 2006, 08:59 AMJenny20
Hi perfecthacker,

Thank you very much for your reply. I have never learnt Mobius inversion forumla. I will go through your prove. If I don't understand it, I will come back to you. - Dec 26th 2006, 09:49 AMJenny20
Hi perfecthacker,

Can you show me the other way ( not by mobius inversion forumla) to prove this question? I don't really get your prove because I dont' quite understand the mobius inversion formula. Thank you very much for your help. - Dec 26th 2006, 12:00 PMThePerfectHacker