<s>[INST] Asymptotic expansion computation I found a paper in which the following  expression
$$\log\left(1 - \frac{1}{x}\right) + \frac{1}{x} + \frac{A}{x^2}+\epsilon x = 0$$
Where $\epsilon$ is a constant of order $10^{-2}$ to $10^{-5}$, $A$ is a constant of order unity, is approximated for $\frac{1}{x}<<1$ as
$$\epsilon x^4 + \left(A-\frac{1}{2}\right) x - \frac{1}{3} + O\left(\frac{1}{x}\right) = 0$$
I am unable to derive this result.
Going back to the first equation, I denote $\frac{1}{x} = y$.
Then by using the Taylor expansion
$$\log(1-y) = y -\frac{1}{2}y^2 +\frac{2}{3}y^3 +...$$
And substitute I obtain
$$2y +\left(a-\frac{1}{2}\right)y^2 + \frac{2}{3}Y^3 + \frac{\epsilon}{y} +O(x^4)= 0$$
And it is not even close. Any hint on my error would be very appreciated, thanks
 [/INST] $$\begin{align}0 &=\ln\left(1 - \frac{1}{x}\right) + \frac{1}{x} + \frac{A}{x^2}+\epsilon x \\
&= -\frac1x - \frac1{2x^2}-\frac1{3x^3} +\cdots+\frac1x+\frac a{x^2} + \epsilon x  \\
&=\epsilon x^4+(a-1/2)x -1/3+\cdots  \end{align} $$
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