1. ## Stumped again

Man.. I would love it if this stuff made sense to me, lol.

$\displaystyle k$ 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

$\displaystyle \hat{f}(s)$ = $\displaystyle \sum_{n=1}^{\infty}{f(n)/n^s}$
1.
(a) Compute $\displaystyle \widehat{fN^k}$ in terms of $\displaystyle \hat{f}.$ (Here $\displaystyle N(n)=n)$
(b) Compute $\displaystyle \widehat{N^k}$

2. Define the arithmetic function $\displaystyle f^{[k]}$ by
$\displaystyle f^{[k]}(n)=$ $\displaystyle \left\{\begin{array}{cc}f(m),&\mbox{ if } n=m^k \mbox { for some m} \in \mathbb{Z}^+\\0, & \mbox{ otherwise }\end{array}\right.$

Show that:
(a) $\displaystyle f^{[k]}*g^{[k]} = (f*g)^{[k]}$

(b) If $\displaystyle f*g = I, \mbox{ then } f^{[k]}*g^{[k]} = I$

(c) $\displaystyle \widehat{f^{[k]}}$$\displaystyle (s) = \displaystyle \hat{f}$$\displaystyle (ks)$

2. 1
(a)
Let $\displaystyle F(s)=\sum_{n=1}^{\infty}{\frac{f(n)}{n^s}}$

Note that: $\displaystyle F(s-k)=\sum_{n=1}^{\infty}{\frac{f(n)}{n^{s-k}}}=\sum_{n=1}^{\infty}{\frac{f(n)\cdot{n^k}}{n^{ s}}}=\sum_{n=1}^{\infty}{\frac{f(n)\cdot{[N(n)]^k}}{n^{s}}}$

THus the DSGF of $\displaystyle f(n)\cdot{[N(n)]^k}$ is $\displaystyle F(s-k)$

(b)

Let $\displaystyle f(n)=1$ for n=1,2,... thus: $\displaystyle F(s)=\sum_{n=1}^{\infty}{\frac{1}{n^s}}=\zeta(s)$

It follows from (a) that the DSGF of $\displaystyle 1\cdot{[N(n)]^k}=[N(n)]^k$ is $\displaystyle \zeta(s-k)$

2.
(c) $\displaystyle \sum_{n=1}^{\infty}{\frac{f^{[k]}(n)}{n^s}}=\sum_{n=1}^{\infty}{\frac{f^{[k]}(n^{k})}{n^{s\cdot{k}}}}=\sum_{n=1}^{\infty}{\fra c{f(n)}{n^{s\cdot{k}}}}=F(s\cdot{k})$

The first equality is true since all the other terms are zero.

(a) Let $\displaystyle G(s)=\sum_{n=1}^{\infty}{\frac{f(n)}{n^s}}$

Note that: $\displaystyle F(s\cdot{k})\cdot{G(s\cdot{k})}=\sum_{n=1}^{\infty }{\frac{\sum_{d|n}f(d)\cdot{g\left(\frac{n}{d}\rig ht)}}{n^{s\cdot{k}}}}=\sum_{n=1}^{\infty}{\frac{(f *g)^{[k]}(n)}{n^s}}$ by (c)

Applying (c) again:$\displaystyle \sum_{n=1}^{\infty}{\frac{f^{[k]}(n)}{n^s}}=F(s\cdot{k})$ and $\displaystyle \sum_{n=1}^{\infty}{\frac{g^{[k]}(n)}{n^s}}=G(s\cdot{k})$

Multplying: $\displaystyle F(s\cdot{k})\cdot{G(s\cdot{k})}=\sum_{n=1}^{\infty }{\frac{\sum_{d|n}f^{[k]}(d)\cdot{g^{[k]}\left(\frac{n}{d}\right)}}{n^{s}}}$

Therefore we must have: $\displaystyle [f^{[k]}*g^{[k]}](n)=\sum_{d|n}f^{[k]}(d)\cdot{g^{[k]}\left(\frac{n}{d}\right)}=(f*g)^{[k]}(n)$