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I have the following equation:

$$\ln \left(\frac{\alpha^2}{\beta^2}\right) = \frac{1}{m \xi_\alpha} + \frac{b\ln(m \xi_\alpha)}{m^2}$$

where $\xi_\alpha := \xi(\alpha^2)$. The objective is to solve for $\xi_\alpha$ given that $\ln (\frac{\alpha^2}{\beta^2}) \gg 1$.

Here is my attempt:

$$ \ln\left(\frac{\alpha^2}{\beta^2}\right) = \frac{1}{m \xi_\alpha} + \frac{b\ln(m \xi_\alpha)}{m^2}$$

Exponentiating both sides, we get,

$$\frac{\alpha^2}{\beta^2} = e^\frac{1}{m\xi_\alpha} (m\xi_\alpha)^\frac{b}{m^2}$$

At this point, I am completely stuck. I am not sure how to use the fact that $\ln (\frac{\alpha^2}{\beta^2}) \gg 1$. I tried reading some materials online and there were some suggestions using iteratively solving the equation, but I am not sure how to proceed. Could someone help me with this?

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  • $\begingroup$ hint: if the $\ln$ is large, then $\xi$ is large and $1\over \xi$ is small . $\endgroup$
    – user619894
    Mar 27, 2021 at 20:15
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    $\begingroup$ Ah okay. So I can see that I can take the last equation from my work and inverse it and then the left hand quantity will be $ <<1 $. So would it mean that I can taylor expand the exponential and try to solve for $\xi_\alpha$? $\endgroup$
    – time12
    Mar 27, 2021 at 20:22
  • $\begingroup$ I am still not sure how taylor expanding the exponential term would help. For instance, if I try to cut the taylor expansion at first order, I am getting something like $\frac{\beta^2}{\alpha^2} = (1-\frac{1}{m \xi_\alpha}) \frac{1}{(m \xi_\alpha)^\frac{b}{m^2}}$ $\endgroup$
    – time12
    Mar 27, 2021 at 20:43

1 Answer 1

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Why do you want an iterative method since there is an explicit solution ?

Let $$a =\log\left(\frac{\alpha^2}{\beta^2}\right) \qquad \text{and} \qquad y=m \xi_\alpha $$ to make the equation $$a=\frac 1 y+\frac b{m^2} \log(y)$$ the solution of which being $$y=-\frac {m ^2}{b\,W(t)}\qquad \text{where} \qquad t=-\frac{m^2}{b}e^{-\frac{a m^2}{b}}$$ where $W(t)$ is Lambert function.

To make even simpler, let $k=-\frac {m^2}b$ which makes $$y=\frac{k}{W\left(k e^{a k}\right)}$$

Now, if $a$ is large and then $t$ is small, you can use the expansion of $W(t)$ around $t=0$

$$W(t)=t-t^2+\frac{3 }{2}t^3-\frac{8 }{3}t^4+O\left(t^5\right)$$

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  • $\begingroup$ Vraiment impressionne! +1 $\endgroup$
    – Svyatoslav
    Mar 28, 2021 at 20:16

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