# Two more questions (cont.)

• April 8th 2008, 02:35 PM
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.

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

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

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

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

$p_k{(p^e)}$= $\left\{\begin{array}{cc}1,&\mbox{ if }e

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

(c) $p_k * \lambda_k = I$

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

Any help/hints/suggestions would be greatly appreciated. Thanks!
• April 8th 2008, 05:41 PM
PaulRS
Again $
\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 $
\sum\limits_{\left. d \right|n} {\mu ^{\left[ k \right]} \left( n \right)} = p_k \left( n \right)
$

Thus: $
\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 $
\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: $
\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 $
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 $
{\mu \left( n \right)}
$
(see whether our function is multiplicative) or use the series in (b) and remember/prove that $
\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)}}$