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Given the 2 functions $$ g(x)= \sum_{n=1}^{\infty}f\left(\frac{x}n\right)\log(n)\;, $$ how can I use Möbius inversion to recover $f$ from $g$?? I believe that

$$ f(x)= \sum_{n=1}^{\infty}\mu (n)g\left(\frac{x}n\right)\log(n)\;. $$ Here 'mu' is the Möbius function.

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Suppose $f(x)=x^{10}$. Then $$g(x)=\sum_1^{\infty}x^{10}n^{-10}\log n=Cx^{10}$$ where $C=\sum_1^{\infty}n^{-10}\log n$ is a very small positive constant. Then $$\sum_1^{\infty}\mu(n)g(x/n)\log n=Cx^{10}\sum_1^{\infty}\mu(n)n^{-10}\log n=CDx^{10}$$ where $D=\sum_1^{\infty}\mu(n)n^{-10}\log n$ is a very small constant. We can't have $CD=1$, so we can't have $$f(x)=\sum_1^{\infty}\mu(n)g(x/n)\log n$$

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I was just perusing old posts and I'm confused here - why can't we have $CD=1$? It looks to me like $C=-\zeta'(10)$ and $D$ is the reciprocal of this (based on the Dirichlet series for $\zeta'(10)^{-1}$). –  anon Mar 21 '12 at 18:02
    
@anon, my thinking was that $C$ is surely much smaller than 1, and $|D|$ is surely even smaller, so their product must be smaller still. Is that wrong? –  Gerry Myerson Mar 21 '12 at 22:19
    
You're right; looks like I was confusing the order of differentiation and taking the reciprocal in my Dirichlet series thinking. –  anon Mar 21 '12 at 22:24
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