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Math Help - A question about Euler's phi function

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    MHF Contributor Bruno J.'s Avatar
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    A question about Euler's phi function

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

    Show that \inf_{n \in \mathbb{N}}\frac{\phi(n)}{n}=0.
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    MHF Contributor chisigma's Avatar
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    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
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    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
    .
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  4. #4
    MHF Contributor chisigma's Avatar
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    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
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