Two more questions (cont.)

**shadow_2145** · April 8th 2008, 02:35 PM

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 $e \in N:$

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

(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!

**PaulRS** · April 8th 2008, 05:41 PM

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)}}$

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