# Arithmetical functions

• December 7th 2009, 12:38 AM
Chandru1
Arithmetical functions
Let $P(n)$ be the product of positive integers which are $\leq n$ and relatively prime to $n$. Prove that $\displaystyle P(n) =n^{\phi(n)} \prod\limits_{d \mid n} \biggl(\frac{d!}{d^d} \biggr)^{\mu(n/d)}$
• December 7th 2009, 03:01 AM
PaulRS
Let $F( n) = \sum_{(k,n)=1;1\leq k\leq n} {\log(\tfrac{k}{n})}$

Then $\sum_{d|n}F(d) =\log(\tfrac{1}{n})+...+\log(\tfrac{n}{n})=\log\le ft(\frac{n!}{n^n}\right)$ (see here )

Thus, by Möbius inversion formula: $F(n) = \sum_{d|n}\log\left(\frac{d!}{d^d}\right)\cdot \mu\left(\frac{n}{d}\right) = \log\left(\prod_{d|n}\left(\frac{d!}{d^d}\right)^{ \mu\left(\frac{n}{d}\right) }\right)$ (1)

Now: $F(n) = \sum_{(k,n)=1;1\leq k\leq n} {\log(k)} -\phi(n)\cdot \log( n) = \log(P(n))-\log(n^{\phi(n)})$

Then taking exponentials in (1) your identity follows.
• December 8th 2009, 02:47 AM
Chandru1
superb
Hi--

Paul, that was a superb answer. I really liked the proof. By the way i am just a beginner in analytic number theory and i am finding this a bit difficult as i am learning it on my own. Difficulty not in the proofs or theorems but in solving problems.

Like, i couldn never have considered the way you taken F(n). I mean how did u get the idea...

Pls help..i really want to master number theory as its such a wonderful subject....