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Thread: Two more questions

  1. #1
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    Two more questions

    I will seperate my next 2 questions into 2 separate threads so that it does not get cluttered. I don't know if the following definition is needed, but I will include it.

    $\displaystyle k$ is a positive integer and $\displaystyle f,g$ and $\displaystyle h$ are arithmetic functions. $\displaystyle \hat{f}$ denotes the Dirichlet series for $\displaystyle f$, so

    $\displaystyle \hat{f}(s)$ = $\displaystyle \sum_{n=1}^{\infty}{f(n)/n^s}$

    (1) Define $\displaystyle \lambda_k$=$\displaystyle \mu * u^{[k]}$. Show that

    (a) If $\displaystyle k>1$ then for all primes $\displaystyle p$ and all $\displaystyle e$ $\displaystyle \in N:$

    $\displaystyle \lambda_k{(p^e)}$=$\displaystyle \left\{\begin{array}{cc}1,&\mbox{ if }e \equiv 0 \mbox{ mod } k\\-1, & \mbox{ if }e \equiv 1 \mbox{ mod } k\\0, & \mbox{ otherwise } \end{array}\right.$

    (b)$\displaystyle \widehat{\lambda_k}$(s)=$\displaystyle \frac{\zeta{(ks)}}{\zeta{(s)}}$

    Any help would be appreciated, thanks!
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  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
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    It is: $\displaystyle
    u^{\left[ k \right]} \left( n \right) = \left\{ \begin{gathered}
    1{\text{ if }}n = m^k {\text{ for some }}m \in \mathbb{Z}^ + \hfill \\
    0{\text{ otherwise}} \hfill \\
    \end{gathered} \right.
    $ , right?

    (a) $\displaystyle

    \lambda _k \left( {p^e } \right) = \sum\limits_{\left. d \right|p^e } {\mu \left( d \right) \cdot u^{\left[ k \right]} \left( {\tfrac{n}
    {d}} \right)} = \sum\limits_{j = 0}^e {\mu \left( {p^j } \right) \cdot u^{\left[ k \right]} \left( {p^{e - j} } \right)}

    $

    Thus: $\displaystyle
    \lambda _k \left( {p^e } \right) = \mu \left( 1 \right) \cdot u^{\left[ k \right]} \left( {p^e } \right) + \mu \left( p \right) \cdot u^{\left[ k \right]} \left( {p^{e - 1} } \right) = u^{\left[ k \right]} \left( {p^e } \right) - u^{\left[ k \right]} \left( {p^{e - 1} } \right)
    $

    Now consider the different cases
    • If $\displaystyle
      e = \dot k
      $ we have $\displaystyle
      u^{\left[ k \right]} \left( {p^e } \right) = 1{\text{ and }}u^{\left[ k \right]} \left( {p^{e - 1} } \right) = 0 \Rightarrow \lambda _k \left( {p^e } \right) = 1
      $
    • If $\displaystyle
      e\equiv{1}(\bmod.k)
      $ we have $\displaystyle

      u^{\left[ k \right]} \left( {p^e } \right) = 0{\text{ and }}u^{\left[ k \right]} \left( {p^{e - 1} } \right) = 1 \Rightarrow \lambda _k \left( {p^e } \right) = - 1

      $
    • Otherwise neither $\displaystyle e$ nor $\displaystyle e-1$ is a multiple of k, thus $\displaystyle
      u^{\left[ k \right]} \left( {p^e } \right) = 0{\text{ and }}u^{\left[ k \right]} \left( {p^{e - 1} } \right) = 0 \Rightarrow \lambda _k \left( {p^e } \right) = 0
      $
    (b)
    Multiplying $\displaystyle
    \left( {\sum\limits_{n = 1}^\infty {\tfrac{{u^{\left[ k \right]} \left( n \right)}}
    {{n^s }}} } \right) \cdot \left( {\sum\limits_{n = 1}^\infty {\tfrac{{\mu \left( n \right)}}
    {{n^s }}} } \right) = \sum\limits_{n = 1}^\infty {\tfrac{{\sum\limits_{\left. d \right|n} {\mu \left( d \right) \cdot u^{\left[ k \right]} \left( {\tfrac{n}
    {d}} \right)} }}
    {{n^s }}}

    $

    Thus: $\displaystyle
    \left( {\sum\limits_{n = 1}^\infty {\tfrac{{u^{\left[ k \right]} \left( n \right)}}
    {{n^s }}} } \right) \cdot \left( {\sum\limits_{n = 1}^\infty {\tfrac{{\mu \left( n \right)}}
    {{n^s }}} } \right) = \sum\limits_{n = 1}^\infty {\tfrac{{\lambda _k \left( n \right)}}
    {{n^s }}}
    $

    Now by this post http://www.mathhelpforum.com/math-help/number-theory/33568-stumped-again.html
    we can say that $\displaystyle
    \sum\limits_{n = 1}^\infty {\tfrac{{u^{\left[ k \right]} \left( n \right)}}
    {{n^s }}} = \zeta \left( {s \cdot k} \right)
    $

    Recall the identity: $\displaystyle
    \sum\limits_{n = 1}^\infty {\tfrac{{\mu \left( n \right)}}
    {{n^s }}} = \frac{1}
    {{\zeta \left( s \right)}}
    $

    Therefore: $\displaystyle
    \sum\limits_{n = 1}^\infty \frac{\lambda_k \left( n \right)}
    {n^s } = \frac{{\zeta \left( {s \cdot k} \right)}}
    {{\zeta \left( s \right)}}
    $
    Last edited by PaulRS; Apr 9th 2008 at 04:01 AM.
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