# Math Help - Stumped again

1. ## Stumped again

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

$k$ 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.
(a) Compute $\widehat{fN^k}$ in terms of $\hat{f}.$ (Here $N(n)=n)$
(b) Compute $\widehat{N^k}$

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

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

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

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

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

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

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

(b)

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

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

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

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

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

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

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

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

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