proving the converse theorem

**Jenny20** · December 26th 2006, 08:43 AM

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.

**ThePerfectHacker** · December 26th 2006, 08:55 AM

Originally Posted by Jenny20

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.

Did you consider using the Mobius inversion formula?

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

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

Now, $g(n)$ is multiplicative function
And $\mu(n)$ (the Mobius mu function) is also multiplicative.
Thus,
$g(n)\cdot \mu(n)$ is also multiplicative.
Thus,
$f(n)$ is multiplicative (because of the summation theorem).

**Jenny20** · December 26th 2006, 08:59 AM

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.

**Jenny20** · December 26th 2006, 09:49 AM

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.

**ThePerfectHacker** · December 26th 2006, 12:00 PM

Originally Posted by Jenny20

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.

I just looked up the proof in my number theory book and they prove it the same way I did. Thus, perhaps this is the standard way.

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