# Iterative solution of an implicit equation

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?

• hint: if the $\ln$ is large, then $\xi$ is large and $1\over \xi$ is small . Mar 27, 2021 at 20:15
• 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 $\xi_\alpha$? Mar 27, 2021 at 20:22
• 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 $\frac{\beta^2}{\alpha^2} = (1-\frac{1}{m \xi_\alpha}) \frac{1}{(m \xi_\alpha)^\frac{b}{m^2}}$ Mar 27, 2021 at 20:43

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 $$t=0$$
$$W(t)=t-t^2+\frac{3 }{2}t^3-\frac{8 }{3}t^4+O\left(t^5\right)$$