Prove that if f is multiplicative and the sum of f(d) (where the sum extends over all positive divisors of n) = n for all n, then f = Phi ( where the phi function is defined as the the number of positive integers not exceeding n that are relatively prime to n). As related to the Mobius Inversion Formula.
Can anyone get me going on this proof?
Dec 9th 2009, 08:25 PM
Well suppose you have a function such that . Then by the inversion formula you have , which is precisely .
You actually don't even need the hypothesis that is multiplicative.
Dec 9th 2009, 09:01 PM
I don't see how it is the same as phi(n). I understand the mechanics of the mobius inversion formula - but I'm not seeing the relationship to the phi function. Can you help me connect them?
Dec 9th 2009, 09:02 PM
Sure! Apply the inversion formula to . What do you get? (Giggle)
Dec 9th 2009, 09:14 PM
I get Phi=Summation of M-function(n/d) (d) which is also equal to n? Also, how do I access the math symbols?
If you apply the inversion formula you get (or , which is the same). Now compare this with above...
Dec 9th 2009, 09:23 PM
You're the best(Rofl) I can't express how thankful I am that you are willing to help me.
Dec 9th 2009, 09:34 PM
Hi. I'm working on a few other questions as well. If you be so kind, I'd appreciate your input on this problem.
Suppose n>0, p/n, p is prime, p^2/n, Show that there exists integers a and b such that n=a^3b^2.
Now, it seems to me that all exponents of the prime factors of n have to be at least power of 2 which means that the odd exponents have power at least 3 which would satisfy the above statement. But I'm not sure how to show this symbolically - perhaps the explanation is enough?
Dec 10th 2009, 05:04 AM
Actually, you do not need to know Möbius formula for this.
If you think of it, having: for a given function , determines for all positive integers . (In fact is linear combination of for )
Now since on the other post we proved that satisfies it follows that for all positive integers .
PS: But I have to say that proving the assertion above is as long (if not longer) than proving Möbius inversion formula. (Rofl) - although it seems a pretty obvious fact-