Results 1 to 2 of 2

Thread: Two more questions (cont.)

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    99

    Two more questions (cont.)

    This is the 2nd part to my questions:

    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

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


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

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

    $\displaystyle p_k{(p^e)}$=$\displaystyle \left\{\begin{array}{cc}1,&\mbox{ if }e<k\\0, & \mbox{ if } e \geq k \end{array}\right.$

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

    (c) $\displaystyle p_k * \lambda_k = I$

    (d) $\displaystyle p_2 = |\mu| = \mu^2$.

    Any help/hints/suggestions would be greatly appreciated. Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Again $\displaystyle
    \mu ^{\left[ k \right]} \left( n \right) = \left\{ \begin{gathered}
    \mu \left( n \right){\text{ if }}n = m^k {\text{ for some }}m \in \mathbb{Z}^ + \hfill \\
    0{\text{ otherwise}} \hfill \\
    \end{gathered} \right.
    $

    (b) We have$\displaystyle
    \sum\limits_{\left. d \right|n} {\mu ^{\left[ k \right]} \left( n \right)} = p_k \left( n \right)
    $

    Thus: $\displaystyle
    \left( {\sum\limits_{n = 1}^\infty {\frac{{\mu ^{\left[ k \right]} \left( n \right)}}
    {{n^s }}} } \right) \cdot \left( {\sum\limits_{n = 1}^\infty {\frac{1}
    {{n^s }}} } \right) = \sum\limits_{n = 1}^\infty {\frac{{\sum\limits_{\left. d \right|n} {\mu ^{\left[ k \right]} \left( n \right)} }}
    {{n^s }}} = \sum\limits_{n = 1}^\infty {\frac{{p_k \left( n \right)}}
    {{n^s }}}

    $

    But as you know by http://www.mathhelpforum.com/math-he...ped-again.htmlwe have $\displaystyle
    \sum\limits_{n = 1}^\infty {\frac{{\mu ^{\left[ k \right]} \left( n \right)}}
    {{n^s }}} = \sum\limits_{n = 1}^\infty {\frac{{\mu \left( n \right)}}
    {{n^{s \cdot k} }}} = \frac{1}
    {{\zeta \left( {s \cdot k} \right)}}
    $

    Thus: $\displaystyle
    \sum\limits_{n = 1}^\infty {\frac{{p_k \left( n \right)}}
    {{n^s }}} = \frac{{\zeta \left( s \right)}}
    {{\zeta \left( {s \cdot k} \right)}}

    $

    (c) Hint: Multiply the DSGF of $\displaystyle
    p_k \left( n \right){\text{ and }}\lambda _k \left( n \right)
    $
    (a) Consider the different cases
    (d) Use part (a) and the definition of $\displaystyle
    {\mu \left( n \right)}
    $ (see whether our function is multiplicative) or use the series in (b) and remember/prove that $\displaystyle
    \sum\limits_{n = 1}^\infty {\frac{{\left| {\mu \left( n \right)} \right|}}
    {{n^s }}} = \frac{{\zeta \left( s \right)}}
    {{\zeta \left( {2 \cdot s} \right)}}$
    Last edited by PaulRS; Apr 9th 2008 at 04:27 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Abstract questions(cont)
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 27th 2011, 11:56 PM
  2. PEMDAS Help Cont.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Feb 21st 2011, 03:53 PM
  3. conversions cont.
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: Aug 12th 2010, 07:19 PM
  4. right triangle trig cont. w/ questions
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Aug 2nd 2010, 02:50 PM
  5. trig cont.
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Jul 25th 2009, 06:48 AM

Search Tags


/mathhelpforum @mathhelpforum