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Math Help - Dirichlet

  1. #1
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    Dirichlet

    I am sorry for being rude in my previous thread, I would like to apologize and resubmit my question with all the correct notation and hopefully present it more clearer. My text is Elementary Number Theory by Jones and Jones. I do not understand Dirichlet product and series well at all. I am sure the definitions and lemmas in my text would be enough to tackle these problems, but I cannot seem to grasp them. Any help would be greatly appreciated, and I thank you for your time.

    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}

    Suppose h is completely multiplicative. Define f^{*k} inductively by f^{*1}= fand f^{*(k+1)}= f*f^{*k}.

    Prove that:
    (a) (fh)*(gh) = (f*g)h
    (b) (fh)*(gh)^{*k} = (f*g^{*k})h
    (c) (fh)*h = (f*u)h
    (d) (fh)*h^{*k} = (f*u^{*k})h
    (e) h^{*k} = u^{*k}h
    (f) If h is non-zero, then (\mu^{*k}h) * h^{*k} =I
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  2. #2
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    Quote Originally Posted by shadow_2145 View Post
    (a) (fh)*(gh) = (f*g)h
    How are you defining fh? Do you mean f\circ h? The reason why I am afraid to assume you mean f\circ h is because f,h are arithmetic functions, which can mean, f,h:\mathbb{N}\mapsto \mathbb{C} and therefore composition would not make sense. Unless, here your arithmetic functions have their range as a subset of the naturals.
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  3. #3
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    Are you saying that everything would make sense if it was in the form of f\circ h? Or not?
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    Quote Originally Posted by shadow_2145 View Post
    Are you saying that everything would make sense if it was in the form of f\circ h? Or not?
    I am asking how do you define fh? I said it it reasonable to define fh = f\circ h as longs as the ranges of f,h are subsets of \mathbb{N} otherwise the composition will not make sense. . So is it okay to assume that f,g:\mathbb{N}\mapsto \mathbb{N}?
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  5. #5
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    Yes, they are subsets of \mathbb{N}. It is ok to assume what you stated. f,g:\mathbb{N}\mapsto \mathbb{N} is correct. Sorry!
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    Quote Originally Posted by shadow_2145 View Post
    (a) (fh)*(gh) = (f*g)h
    To show these functions are the same you need to show for any n we have [(f\circ h)*(g\circ h)](n) = [(f*g)\circ h](n).

    [(f\circ h)*(g\circ h)](n) = f(h(n))*g(h(n)) = \sum_{d|n} f(h(d))g(h(n/d)).
    [(f*g)\circ h](n) = \sum_{d|n}f(h(n))g(h(n/d)).

    Thus, they are the same. Try doing the other ones in the same way.
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  7. #7
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    Thank you! I will try, but (b) and (d) are also confusing me with the ^{*k}
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