I'm looking for the zeroes of $f(k) = e^{\sqrt{k}}[\frac{s}{k} - \frac{d}{\sqrt{k}}] - 1$ on the set $k > 0$. Is there a nice way to describe the set of solutions for given $s$ and $d$? Thanks!
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The trouble with this function is the ${}-1$. If it were simply $f(k) = e^{\sqrt{k}}\left[\frac{s}{k} - \frac{d}{\sqrt{k}}\right]$, then the only root for which $k > 0$ would be $k = \frac{s^2}{d^2}$. With the ${}-1$, I don't believe there is an algebraic method to solve for $k$. – Shaun Ault Mar 10 '12 at 23:09
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4This is equivalent to solving $e^{\sqrt{k}}(s-d\sqrt{k}) = k$; equivalently, $e^x(s-dx) = x^2$. It's unlikely to have a "nice" solution, even with Lambert's W function thrown into the mix. – Arturo Magidin Mar 10 '12 at 23:11
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When the ratio $s/d$ is small, we can use the Lagrange inversion formula (see, e.g., my answer here) to solve $e^x(s-dx) = x^2$ (as per Arturo's comment) for $x$ as a power series in either $s$ or $d$:
$$\begin{align}x &= \frac{s}{d}-\frac{s^2}{d^3}+ \left(\frac{1}{d^4}+\frac{2}{d^5}\right) s^3-\left(\frac{1}{2 d^5}+\frac{5}{d^6}+\frac{5}{d^7}\right) s^4 + \cdots \\ &= \frac{s}{d}-\frac{s^2}{d^3}+\frac{s^3}{d^4}+\frac{4 s^3-s^4}{2 d^5} -\frac{30 s^4-s^5}{6 d^6} + \cdots\end{align}$$
Replacing $x$ by $\sqrt{k}$ will then yield solutions to the original problem.
If $s$ and $d$ are related in some way then perhaps more appropriate estimates could be obtained.
Antonio Vargas
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