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Thread: Stumped again

  1. #1
    Mar 2008

    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}
    (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 }<br />
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)
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  2. #2
    Super Member PaulRS's Avatar
    Oct 2007
    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)


    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)

    (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)
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