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Thread: Dirichlet Inverse Problem

  1. #1
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    Dirichlet Inverse Problem

    The Dirichlet Convolution is this:

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

    The $\displaystyle \iota(n)$ function:

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

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

    I'm really stumped on how to approach this one. Obviously the inverse doesn't exist when $\displaystyle f(1) = 0$, 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 $\displaystyle f(n)$ with $\displaystyle f(1) \ne 0$, its inverse $\displaystyle f^{-1} (n)$ obeys to the relation...

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

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

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

    For n>1 we have...

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

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

    $\displaystyle 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 $\displaystyle f^{-1}(n)$ recursively...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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