Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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

$$\frac2{\alpha+1}\left(2\alpha+\log\frac{\alpha-1}{\alpha+1}\right)\;.$$

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.