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Consider the function $f:(0,\infty)\to\mathbf{R}$ given by

$$ f(\alpha) = 1-\prod_{k=0}^{n} (1-Ae^{-\alpha k})$$

where $0<A<1$ is a fixed constant, and $n\in\mathbf{N}$ is fixed. For $A<1$, this function is invertible (for $A=1$, $f(\alpha)=1\ \forall\ \alpha$; for $A>1$ it is not injective).

My question is, is there an identity or something that might help me get an analytic expression for the inverse of this function? Or, does someone have an explanation of why it wouldn't exist?

I have tried expanding the product by hand but it got very messy and intractable-looking.

David M.
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  • For $n>5$, Galois theory is probably going to tell you there's no solution in radicals, since the right-hand side is then a polynomial in $e^{-\alpha}$ of degree greater than $5$. – Chappers Jul 20 '17 at 14:18
  • Setting $e^{-\alpha}$, you get a polynomial of degree $n(n+1)/2$. Chances are high that it doesn't lead to a closed form solution. –  Jul 20 '17 at 14:19
  • I was worried about something like that. Thank you for the helpful comments! – David M. Jul 20 '17 at 14:19
  • @Chappers: anyway, the polynomial has a special form (it just depends on two independent parameters), so we can't be affirmative. ($x^{543}-1=0$ does have closed-form solutions.) –  Jul 20 '17 at 14:21

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