Results 1 to 2 of 2

Math Help - Two more questions

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    99

    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.

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

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

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

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

    \lambda_k{(p^e)}= \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) \widehat{\lambda_k}(s)= \frac{\zeta{(ks)}}{\zeta{(s)}}

    Any help would be appreciated, thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    It is: <br />
u^{\left[ k \right]} \left( n \right) = \left\{ \begin{gathered}<br />
  1{\text{ if }}n = m^k {\text{ for some }}m \in \mathbb{Z}^ +   \hfill \\<br />
  0{\text{ otherwise}} \hfill \\ <br />
\end{gathered}  \right.<br />
, right?

    (a) <br /> <br />
\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}<br />
{d}} \right)}  = \sum\limits_{j = 0}^e {\mu \left( {p^j } \right) \cdot u^{\left[ k \right]} \left( {p^{e - j} } \right)} <br /> <br />

    Thus: <br />
\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)<br />

    Now consider the different cases
    • If <br />
e = \dot k<br />
we have <br />
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<br />
    • If <br />
e\equiv{1}(\bmod.k)<br />
we have <br /> <br />
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<br /> <br />
    • Otherwise neither e nor e-1 is a multiple of k, thus <br />
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<br />
    (b)
    Multiplying <br />
\left( {\sum\limits_{n = 1}^\infty  {\tfrac{{u^{\left[ k \right]} \left( n \right)}}<br />
{{n^s }}} } \right) \cdot \left( {\sum\limits_{n = 1}^\infty  {\tfrac{{\mu \left( n \right)}}<br />
{{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}<br />
{d}} \right)} }}<br />
{{n^s }}} <br /> <br />

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

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

    Recall the identity: <br />
\sum\limits_{n = 1}^\infty  {\tfrac{{\mu \left( n \right)}}<br />
{{n^s }}}  = \frac{1}<br />
{{\zeta \left( s \right)}}<br />

    Therefore: <br />
\sum\limits_{n = 1}^\infty  \frac{\lambda_k \left( n \right)}<br />
{n^s }  = \frac{{\zeta \left( {s \cdot k} \right)}}<br />
{{\zeta \left( s \right)}}<br />
    Last edited by PaulRS; April 9th 2008 at 05:01 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 05:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 05:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 04:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 08:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 02:53 AM

Search Tags


/mathhelpforum @mathhelpforum