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Topic: trying to get an upper bound on 1/log(n/m) when (n,m) = 1 and
n >m > 0 , n & m integers

Replies: 19   Last Post: Apr 16, 2012 3:28 PM

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 Pubkeybreaker Posts: 1,290 Registered: 2/12/07
Re: trying to get an upper bound on 1/log(n/m) when (n,m) = 1 and n
>m > 0 , n & m integers

Posted: Apr 13, 2012 5:42 PM

On Apr 13, 8:53 am, David Bernier <david...@videotron.ca> wrote:
> (a) n > m > 0
> (b) n and m are positive integers
> (c) (n, m) = 1.

The problem statement says "upper bound".
I interpret this to mean "upper bound over all possible m for
a given n such that (n,m) = 1". This will happen when n/m is as
close to 1 as possible;

i.e. when m = n-1.

estimating 1/log( n/(n-1)) should not be hard.
It is just 1/log(1 + 1/n - 1/n^2 + ....) , just use the Taylor
series for log(1+epsilon)....

Date Subject Author
4/13/12 David Bernier
4/13/12 Pubkeybreaker
4/14/12 barker
4/14/12 quasi
4/14/12 Pubkeybreaker
4/15/12 barker
4/15/12 quasi
4/15/12 J. Antonio Perez M.
4/15/12 quasi
4/15/12 quasi
4/15/12 barker
4/15/12 quasi
4/16/12 Pubkeybreaker
4/16/12 Pubkeybreaker
4/16/12 quasi
4/16/12 Pubkeybreaker
4/16/12 quasi
4/16/12 Pubkeybreaker
4/16/12 Pubkeybreaker
4/14/12 Frederick Williams