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In a paper I am trying to understand the following

For any arithmetic function $f$, its Dirichlet transform is defined by $$\hat{f} :=\sum_{d|n}f(d)$$ and its Möbius transform is defined as $$\check{f} := \sum_{d|n} \mu(n/d)f(d)$$

It then says, for any $f$, we have $f = \check{\hat{f}} = \hat{\check{f}}$. I am not sure how this follows from the Möbius Inversion formula. I am sure its a trivial matter, but I don't see how definitions (:=) show the inverse property.

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1 Answer 1

The "Dirichlet transform" takes a Dirichlet series $\sum \frac{f(n)}{n^s}$ and multiplies it (formally) by the Riemann zeta series $\sum \frac{1}{n^s}$. The inverse of this operation is to multiply by the reciprocal of the zeta series. The formula with the Moebius function is multiplication of a series by $\sum \mu(n)n^{-s}$, and in order for this to reverse the Dirichlet transform this series must be the multiplicative inverse to zeta.

The zeta series can be inverted using the Euler product over primes, $(\sum n^{-s})^{-1} = \prod (1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \dots)^{-1} = \prod (1 - \frac{1}{p^s})$.

That product, written as a Dirichlet series, is $\sum \frac{g(n)}{n^s}$ for the function $g(n)$ that is multiplicative, has $g(1)=1$, $g(p)=-1$, and $g(p^n)=0$ for $n > 1$. Which is the definition of the Moebius function.

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Your last paragraph says that the product can be written as a Dirichlet series. Is there a reason why? –  Tyler Hilton Apr 5 '13 at 21:57
    
I am confused by your answer. It would be appreciated if you can expand on it if you have time. –  Tyler Hilton Apr 5 '13 at 22:09
    
@TylerHilton see en.wikipedia.org/wiki/… –  anon Apr 5 '13 at 23:21
    
The product (for the inverse of zeta) can be written as Dirichlet series because for every $n$, only a finite number of terms in the product contain $p^s$ with $p$ dividing $n$, so that these (ie, the product of their coefficients) can be assigned unambiguously to the $1/n^s$ term of a Dirichlet series. –  zyx Apr 5 '13 at 23:47
    
For things that are unclear there are many people here such as @anon (thanks!) who can answer, either a new and better formulation or an expansion of the one I gave, and you might get a faster reply by posting a new question. But if there is something specific about the answer that is confusing, I can try to elaborate it here, as well. –  zyx Apr 5 '13 at 23:49

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