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I have been struggling to show that for the following equation, where $\psi \in (0,1)$ and $S\in \{2,3,\dots\}$

$$\ln\left(\frac{1}{1-\sqrt{\psi}}\right) + \ln\left[\left(1 - \frac{\sqrt{\psi}}{S}\right)^{S}\right] - \psi = 0$$

is such that the solution $\psi^{\ast}$ that solves this equation is decreasing in $S$.

What I do is:

  1. define $F(\psi,S) = 0$ just as above
  2. Apply implicit function theorem to get $\frac{\partial \psi}{\partial S}$.

But I do not get a clear-cut solution. I think that's because of the root $\psi=0$ which I would like to ignore.

Are there better ways of doing this?

Thank you!

danny dan
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1 Answers1

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Hint

Simplify the problem using $\psi=x^2$, expand the second logarithm and consider that you are looking for the zero of function $$f(x)=-\log(1-x)+S \log \left(1-\frac{x}{S}\right)-x^2$$

What you need to notice is that, if $S\to \infty$, the equation is $$f(x)\to -x (x+1)-\log (1-x)$$ which has a solution close to $x_\infty \sim 0.684$.

So, you need to show if $x_\infty$ is or not an upper bound of $x_S$.

Claude Leibovici
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