Thread: Two more questions

1. 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!

2. It is: $
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) $

\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: $
\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 $
e = \dot k
$
we have $
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 $
e\equiv{1}(\bmod.k)
$
we have $

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 $e$ nor $e-1$ is a multiple of k, thus $
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 $
\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: $
\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 $
\sum\limits_{n = 1}^\infty {\tfrac{{u^{\left[ k \right]} \left( n \right)}}
{{n^s }}} = \zeta \left( {s \cdot k} \right)
$

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

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