Results 1 to 2 of 2

Thread: Stumped again

  1. #1
    Mar 2008

    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

    $\displaystyle \hat{f}(s)$ = $\displaystyle \sum_{n=1}^{\infty}{f(n)/n^s}$
    (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)$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Oct 2007
    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 $\displaystyle F(s-k)$


    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 $\displaystyle \zeta(s-k)$

    (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: $\displaystyle [f^{[k]}*g^{[k]}](n)=\sum_{d|n}f^{[k]}(d)\cdot{g^{[k]}\left(\frac{n}{d}\right)}=(f*g)^{[k]}(n)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Stumped... cosθ=sin(90-θ)
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Jul 24th 2010, 12:17 PM
  2. Stumped
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 14th 2010, 08:21 AM
  3. Help, I'm stumped!
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Sep 11th 2009, 06:44 AM
  4. Stumped
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Sep 9th 2008, 04:38 PM
  5. Stumped by this one
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Feb 23rd 2006, 01:48 AM

Search Tags

/mathhelpforum @mathhelpforum