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Let $x\in (0,1)$, $$ M(v,x) = \bigl(1+v^x\bigr) \bigl(v^{1/x}-v\bigr) + x \bigl(1+v^{1/x}\bigr) \bigl(v^x-v\bigr) $$ and let $v_0(x)$ be the root of $M(v,x)$ in $(0,1)$. As $x \rightarrow 1$ this root (numerically) approaches a value $0.090776278\ldots$ which seems to be $(\alpha-1)/(\alpha+1)$ where $\alpha = 1.199678\ldots$ is the positive solution to $\coth(\alpha) = \alpha$, a quantity related the the Laplace limit constant (see A033259 and the references therein). I'm having problems proving this, any pointers would be appreciated.

edit: changed $[0,1]$ to $(0,1)$ as the location of the root of $M$: as mentioned in the original version of joriki's answer, both $0$ and $1$ are roots of $M$ (independent of $x$), so the former implies that $v_0$ is not well-defined.

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up vote 4 down vote accepted

I'll assume you meant $(0,1)$ instead of $[0,1]$, as both $0$ and $1$ are also roots, independent of $x$.

Wolfram|Alpha provides the series expansion at $x=1$:

$$M(v,1+\epsilon)=\epsilon^2v\log v\left(2+2v+\log v-v\log v\right)+O\left(\epsilon^3\right)\;.$$

For $\epsilon\to0$, the root must converge to a root of the expression in parentheses. Your transformation $v=(\alpha-1)/(\alpha+1)$ transforms this into


Again the expression in parentheses must vanish, so

$$ \log\frac{\alpha-1}{\alpha+1}=-2\alpha\;, \\ \frac{\alpha-1}{\alpha+1}=\mathrm e^{-2\alpha}\;, \\ \mathrm e^{\alpha}(\alpha-1)=\mathrm e^{-\alpha}(\alpha+1)\;, \\ \alpha\left(\mathrm e^{\alpha}-\mathrm e^{-\alpha}\right)=\mathrm e^{\alpha}+\mathrm e^{-\alpha}\;, \\ \alpha=\coth\alpha\;. $$

The negative root leads to $v\notin(0,1)$, so the root we want is the positive one.

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Oh, that's pretty joriki, very pretty. Many thanks. – n00b May 15 '12 at 22:14
@n00b: You're welcome. – joriki May 16 '12 at 6:16

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