Results 1 to 2 of 2

Thread: exponents

  1. #1
    Member
    Joined
    Feb 2009
    From
    Chennai
    Posts
    148

    exponents

    For $\displaystyle n>1$, define $\displaystyle E(n)$ oto be the highest exponent to which a prime divides it. For instance, $\displaystyle E(12)=E(36)=2$. Prove that $\displaystyle \lim_{N \to \infty} \frac{1}{N} \sum\limits_{n=2}^{N}E(n)$exists.
    Last edited by CaptainBlack; Jun 12th 2009 at 11:34 PM. Reason: minor correction
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Previous Result: The cardinal of: $\displaystyle
    \left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}p^j \not | x{\text{ }}\forall p{\text{ prime}}} \right\}
    $ is $\displaystyle
    \sum\limits_{k = 1}^\infty {\left\lfloor {\tfrac{n}
    {{k^j }}} \right\rfloor \cdot \mu \left( k \right)}
    $ where $\displaystyle
    \left\lfloor x \right\rfloor
    $ is the floor function and $\displaystyle \mu(n)$ is the Möbius function.

    Proof

    Consider: $\displaystyle A=
    \bigcup\limits_{p{\text{ prime}}} {\left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}\left. {p^j } \right|x} \right\}}
    $ , the cardinal of this set is, by inclusion-exclusion: $\displaystyle
    \sum\limits_{p{\text{ prime}}} {\left| {\left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}\left. {p^j } \right|x} \right\}} \right|} - \sum\limits_{p > q{\text{ primes}}} {\left| {\left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}\left. {\left( {pq} \right)^j } \right|x} \right\}} \right|} \pm ...
    $

    But: $\displaystyle
    \left| {\left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}\left. m \right|x} \right\}} \right| = \left\lfloor {\tfrac{n}
    {m}} \right\rfloor
    $ thus note that we can write: $\displaystyle
    A = - \sum\limits_{k = 2}^\infty {\left\lfloor {\tfrac{n}
    {{k^j }}} \right\rfloor \cdot \mu \left( k \right)}
    $ but we want $\displaystyle
    {\bar A}
    $ and the rest follows. $\displaystyle \square$

    At this point, note that every number that is free of squares, is free of cubes ... and so on. Further, if a number $\displaystyle x$ is free of cubes, but not of squares, then $\displaystyle E(x)=2$

    That is: $\displaystyle
    \sum\limits_{k = 1}^\infty {\left\lfloor {\tfrac{n}
    {{k^{j + 1} }}} \right\rfloor \cdot \mu \left( k \right)} - \sum\limits_{k = 1}^\infty {\left\lfloor {\tfrac{n}
    {{k^j }}} \right\rfloor \cdot \mu \left( k \right)} = \left| {\left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}E\left( x \right) = j} \right\}} \right|
    $

    Thus we get: $\displaystyle
    \sum\limits_{k = 1}^n {E\left( k \right)} = \sum\limits_{j = 1}^\infty {\left( {\sum\limits_{k = 1}^\infty {\left( {\left\lfloor {\tfrac{n}
    {{k^{j + 1} }}} \right\rfloor - \left\lfloor {\tfrac{n}
    {{k^j }}} \right\rfloor } \right) \cdot \mu \left( k \right)} } \right) \cdot j}
    $ (This is already very suggestive)

    Mmmm, here I've not been able to finish it formally -it gets quite nasty-, but I do think that: $\displaystyle
    \mathop {\lim }\limits_{n \to + \infty } \tfrac{1}
    {n} \cdot \sum\limits_{k = 1}^n {E\left( k \right)} =
    {\tfrac{1}
    {{\zeta \left( {2} \right)}}}
    + \sum\limits_{j = 2}^\infty {\left( {\tfrac{1}
    {{\zeta \left( {j + 1} \right)}} - \tfrac{1}
    {{\zeta \left( j \right)}}} \right) \cdot j}
    $ -remember $\displaystyle
    \sum\limits_{k = 1}^\infty {\tfrac{{\mu \left( k \right)}}
    {{k^s }}} = \tfrac{1}
    {{\zeta \left( s \right)}}
    $ and $\displaystyle
    \sum\limits_{k = 1}^\infty {\left\lfloor {\tfrac{n}
    {k}} \right\rfloor \cdot \mu \left( k \right)} = \sum\limits_{k = 1}^n {\sum\limits_{\left. d \right|k} {\mu \left( d \right)} } = 1
    $ -
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Exponents and Log Help!!
    Posted in the Algebra Forum
    Replies: 17
    Last Post: Aug 14th 2011, 10:21 AM
  2. Exponents
    Posted in the Algebra Forum
    Replies: 8
    Last Post: Apr 24th 2010, 04:30 PM
  3. exponents
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 17th 2010, 08:37 AM
  4. Exponents
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Aug 18th 2008, 09:51 PM
  5. exponents
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 28th 2008, 01:53 PM

Search Tags


/mathhelpforum @mathhelpforum