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Be $\mu(z_1, \ldots, z_L)$ the only positive real solution to the equation

\begin{equation} \sum_{l=1}^L z_l \mu^l = 1 \end{equation}

With $z_1 = 1$, $z_l \geq 0 \forall l$. Clearly, varying the parameters $z_l$ varies the solution $\mu$. Now, is $\mu(z_1, \ldots, z_L)$ an analytic function? Can this be proved easily? Are there any references in literature I can read up?

I thought using the analytic implicit function theorem but I can't seem to find any reference suiting my case.

Matteo Monti
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  • I started to give a proof, allowing the $z_l$ to be complex, and assuming for simplicity that $\mu(z_1, \ldots, z_L)$ is a simple zero of $\sum_{l=1}^L z_l \mu^l - 1$, so Rouche's theorem ensures that the function $\mu$ is well defined in a neighbourhood of $(z_1, \ldots, z_L)$, and it is enough to prove that it is continuous and holomorphic in $z_1, \ldots, z_L$ separately. But the answers to this question may already answer yours. If not, I suggest you add tags 'complex-analysis', 'several-complex-variables', and wait for an expert to appear. – Calum Gilhooley Mar 11 '16 at 23:27
  • A similar question has just appeared here, where someone has also pointed to this relevant question. – Calum Gilhooley Mar 12 '16 at 00:55

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