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Math Help - exponents

  1. #1
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    exponents

    For n>1, define E(n) oto be the highest exponent to which a prime divides it. For instance, E(12)=E(36)=2. Prove that \lim_{N \to \infty} \frac{1}{N} \sum\limits_{n=2}^{N}E(n)exists.
    Last edited by CaptainBlack; June 13th 2009 at 12:34 AM. Reason: minor correction
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  2. #2
    Super Member PaulRS's Avatar
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    Previous Result: The cardinal of: <br />
\left\{ {x \in \mathbb{Z}^ +  /x \leqslant n{\text{ and }}p^j \not | x{\text{ }}\forall p{\text{ prime}}} \right\}<br />
is <br />
\sum\limits_{k = 1}^\infty  {\left\lfloor {\tfrac{n}<br />
{{k^j }}} \right\rfloor  \cdot \mu \left( k \right)} <br />
where <br />
\left\lfloor x \right\rfloor <br />
is the floor function and \mu(n) is the Möbius function.

    Proof

    Consider: A=<br />
\bigcup\limits_{p{\text{ prime}}} {\left\{ {x \in \mathbb{Z}^ +  /x \leqslant n{\text{ and }}\left. {p^j } \right|x} \right\}} <br />
, the cardinal of this set is, by inclusion-exclusion: <br />
\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 ...<br />

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

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

    That is: <br />
\sum\limits_{k = 1}^\infty  {\left\lfloor {\tfrac{n}<br />
{{k^{j + 1} }}} \right\rfloor  \cdot \mu \left( k \right)}  - \sum\limits_{k = 1}^\infty  {\left\lfloor {\tfrac{n}<br />
{{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|<br />

    Thus we get: <br />
\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}<br />
{{k^{j + 1} }}} \right\rfloor  - \left\lfloor {\tfrac{n}<br />
{{k^j }}} \right\rfloor } \right) \cdot \mu \left( k \right)} } \right) \cdot j} <br />
(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: <br />
\mathop {\lim }\limits_{n \to  + \infty } \tfrac{1}<br />
{n} \cdot \sum\limits_{k = 1}^n {E\left( k \right)}  =<br />
{\tfrac{1}<br />
{{\zeta \left( {2} \right)}}}<br />
+ \sum\limits_{j = 2}^\infty  {\left( {\tfrac{1}<br />
{{\zeta \left( {j + 1} \right)}} - \tfrac{1}<br />
{{\zeta \left( j \right)}}} \right) \cdot j} <br />
-remember <br />
\sum\limits_{k = 1}^\infty  {\tfrac{{\mu \left( k \right)}}<br />
{{k^s }}}  = \tfrac{1}<br />
{{\zeta \left( s \right)}}<br />
and <br />
\sum\limits_{k = 1}^\infty  {\left\lfloor {\tfrac{n}<br />
{k}} \right\rfloor  \cdot \mu \left( k \right)}  = \sum\limits_{k = 1}^n {\sum\limits_{\left. d \right|k} {\mu \left( d \right)} }  = 1<br />
-
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