A question about Euler's phi function

Bruno J. · #1 $Old$ November 5th, 2009, 10:51 PM

This one is for fun. I just made it up.

Show that $\inf_{n \in \mathbb{N}}\frac{\phi(n)}{n}=0$ .

chisigma · #2 $Old$ November 6th, 2009, 01:03 AM

We can start from the explicit expression of $\varphi (n)$ ...

$\varphi(n)= n\cdot \prod_{p|n} (1-\frac{1}{p})$ (1)

... from which we derive immediately...

$\frac{\varphi(n)}{n} = \prod_{p|n} (1-\frac{1}{p})$ (2)

Now we remember the well known 'Euler's product' written in the 'not very usual form' ...

$\frac{1}{\zeta(s)} = \prod_{p} (1-\frac{1}{p^{s}})$ (3)

... from which, setting $s=1$ we derive immediately...

$\prod_{p} (1-\frac{1}{p})=0$ (4)

The (4) tells us that for any $\varepsilon >0$ we have an integer $k$ so that is...

$\prod_{n=1}^{k} (1-\frac{1}{p_{n}}) < \varepsilon$ (5)

Combining (2) and (5) we conclude that for any $\varepsilon >0$ it exist a k so that setting...

$n=\prod_{i=1}^{k} p_{i}$ (6)

...is...

$\frac{\varphi(n)}{n} < \varepsilon$ (7)

Kind regards

$\chi$ $\sigma$

tonio · #3 $Old$ November 6th, 2009, 06:06 AM

Quote:

Originally Posted by chisigma $View Post$

We can start from the explicit expression of $\varphi (n)$ ...

$\varphi(n)= n\cdot \prod_{p|n} (1-\frac{1}{p})$ (1)

... from which we derive immediately...

$\frac{\varphi(n)}{n} = \prod_{p|n} (1-\frac{1}{p})$ (2)

Now we remember the well known 'Euler's product' written in the 'not very usual form' ...

$\frac{1}{\zeta(s)} = \prod_{p} (1-\frac{1}{p^{s}})$ (3)

... from which, setting $s=1$ we derive immediately...

$\prod_{p} (1-\frac{1}{p})=0$ (4)

Doesn't Riemann's zeta function has a pole in s = 1?

Tonio

The (4) tells us that for any $\varepsilon >0$ we have an integer $k$ so that is...

$\prod_{n=1}^{k} (1-\frac{1}{p_{n}}) < \varepsilon$ (5)

Combining (2) and (5) we conclude that for any $\varepsilon >0$ it exist a k so that setting...

$n=\prod_{i=1}^{k} p_{i}$ (6)

...is...

$\frac{\varphi(n)}{n} < \varepsilon$ (7)

Kind regards

$\chi$ $\sigma$

.

chisigma · #4 $Old$ November 6th, 2009, 06:28 AM

Doesn't Riemann's zeta function has a pole in s = 1?

...

Tonio

... of course it has!... that's why the inverse of the Riemann's zeta function has a zero in s=1, so that is...

$\frac{1}{\zeta(s)}_{s=1}=0$ (1)

... of course!

...

Kind regards

$\chi$ $\sigma$