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Math Help - Dirichlet Inverse Problem

  1. #1
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    Dirichlet Inverse Problem

    The Dirichlet Convolution is this:

    (f * g)(n) = \sum\limits_{d \mid n} f(d)g(\frac{n}{d})

    The \iota(n) function:

    \iota(n) = \begin{cases} 1 & \text{if } n = 1 \\ 0 & \text{if } n > 1\\ \end{cases}

    The arithmetic function g is said to be the inverse of the arithmetic function f if f * g = g * f = \iota. Show
    that the arithmetic function f has an inverse if and only if f(1) \neq 0. Show that if f has an inverse it is
    unique. (Hint: When f(1) \neq 0, find the inverse f^{-1} of f by calculating f^{-1}(n) recursively, using the
    fact that \iota(n) = \sum\limits_{d \mid n} f(d) f^{-1}(\frac{n}{d}).)

    I'm really stumped on how to approach this one. Obviously the inverse doesn't exist when f(1) = 0, but beyond that, how do I tackle this one?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Dirichlet Inverse Problem

    By definition, given an arithmetical function f(n) with f(1) \ne 0, its inverse f^{-1} (n) obeys to the relation...

    f^{-1} * f = \sum_{d|n} f^{-1} (d)\ f(\frac{n}{d}) = \iota(n) (1)

    For n=1 the (1) gives us...

    f^{-1}(1)\ f(1) = 1 \implies f^{-1}(1) = \frac{1}{f(1)}} (2)

    For n>1 we have...

     \sum_{d|n} f^{-1} (d)\ f(\frac{n}{d}) = f(1)\ f^{-1} (n) + \sum_{d|n, d<n} f^{-1} (d)\ f(\frac{n}{d}) = 0 (3)

    ... so that from (2) and (3) we derive...

    f^{-1} (n) = - \frac{1}{f(1)}\ \sum_{d|n, d<n} f^{-1} (d)\ f(\frac{n}{d}) (4)

    ... and this formula permits us to compute the f^{-1}(n) recursively...

    Kind regards

    \chi \sigma
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