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Math Help - About Euler's constant (4)...

  1. #1
    MHF Contributor chisigma's Avatar
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    About Euler's constant (4)...

    The proposed 'chellenge' is retired because it has been badly impostated!...

    I beg your pardon for the time You have wasted!...


    Kind regards

    \chi \sigma
    Last edited by chisigma; November 12th 2009 at 07:00 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chisigma View Post
    The proposed 'chellenge' retired becuase it has been badly impostated!...

    I beg your pardon for the time You have wasted!...


    Kind regards

    \chi \sigma
    Prove that \liminf_{n\to\infty}\frac{\varphi(n)\ln\ln(n)}{n}=  e^{-\gamma}, whre \varphi is Euler's totient function. Try that on for size.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Prove that \liminf_{n\to\infty}\frac{\varphi(n)\ln\ln(n)}{n}=  e^{-\gamma}, whre \varphi is Euler's totient function....
    In order to avoid misunderstanding a preliminary question must be answered...

    In the following previous thread...

    http://www.mathhelpforum.com/math-help/number-theory/112730-question-about-eulers-phi-function.html

    ... starting from the basic explicit formula for \varphi(n)...

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

    ... first we derived ...

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

    ... and from which we demonstrated that...

    \frac{\varphi(n)}{n} \ge \prod _{p \le n} (1-\frac{1}{p}) (3)

    Now from (3) I realize that is...

    \inf \{\frac{\varphi(n)}{n}\} = \prod _{p \le n} (1-\frac{1}{p}) (4)

    The question: is all that correct?...

    Kind regards

    \chi  \sigma
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Prove that \liminf_{n\to\infty}\frac{\varphi(n)\ln\ln(n)}{n}=  e^{-\gamma}, whre \varphi is Euler's totient function. Try that on for size.

    In order to avoid any possible confusion I recast the previous question in more general terms...

    The possibilities that the expression  \liminf_{n \rightarrow \infty} x (n) does have any sense are two...

    a) given a sequence x_{n} at it is asociated one and only one f(n) = inf\{x(n)\} and we are interested to the \lim_{n \rightarrow \infty} f(n). In this case we need to have an acceptable definition of  f(n) ...

    b) given a sequence x_{n} it is said to have a 'lower bound' a if it doesn't exist any n_{0} \in \mathbb{N} for which is x(n_{0}) < a and , given a \varepsilon >0, it exist at least one n_{0} \in \mathbb{N}, for which is a \le x(n_{0}) \le a +\varepsilon. In this case however the expression \liminf_{n \rightarrow \infty} is a nonsense...

    The question : which is true?... a) or b)?... or something else?...

    Kind regards

    \chi \sigma
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chisigma View Post
    In order to avoid any possible confusion I recast the previous question in more general terms...

    The possibilities that the expression  \liminf_{n \rightarrow \infty} x (n) does have any sense are two...

    a) given a sequence x_{n} at it is asociated one and only one f(n) = inf\{x(n)\} and we are interested to the \lim_{n \rightarrow \infty} f(n). In this case we need to have an acceptable definition of  f(n) ...

    b) given a sequence x_{n} it is said to have a 'lower bound' a if it doesn't exist any n_{0} \in \mathbb{N} for which is x(n_{0}) < a and , given a \varepsilon >0, it exist at least one n_{0} \in \mathbb{N}, for which is a \le x(n_{0}) \le a +\varepsilon. In this case however the expression \liminf_{n \rightarrow \infty} is a nonsense...

    The question : which is true?... a) or b)?... or something else?...

    Kind regards

    \chi \sigma
    I'm not exactly sure what the confusion here is. If we let S be the set of all subsequential limits of \left\{\frac{\varphi(n)\ln\ln n}{n}\right\}_{n=1}^{\infty} (i.e. \lim_{n\to\infty}\frac{\varphi(a_n)\ln \ln a_n}{a_n} for any, increasing a_n) then \liminf_{n\to\infty}\frac{\varphi(n)\ln \ln n}{n}=\inf\text{ }S
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