I ran into a transcendental equation of the following form:
$$116.2e^{-2t}-16t+12570=0$$
and was naively thinking that I could turn this into a perturbation problem by changing the problem into
$$(116+\epsilon)e^{-2t}-16t+12570=0$$
and then assuming $t=a_{0}+\epsilon a_{1}+\epsilon^2 a_{2}+...$. At the very end I would simply plug in $\epsilon=.2$ and see if the series gave a decent approximation. The point is to solve for $a_{0}$, $a_{1}$, and so on at each power of epsilon.
I realize that my choice of integer powers of epsilon might not work, but I wanted to see if they did anyway just in case I got lucky.
The problem I had when doing it this way was that I KNEW, by checking wolfram that there was only one solution to this equation. However I was not recovering it, so I wanted to know where I was going wrong, and maybe someone can shove me in the right direction as for getting a perturbation expansion out of this thing? Maybe I didn't expand $e^{-2t}$ far out enough in the Taylor expansion?
Also, yes, I know that you can simply solve this by using a Lambert-W function, that's how I was planning on checking my answer numerically.
Thanks.
The "small" part of this, if $t>10$ or so, is the $116.2 e^{-2t}$.
So write it as $\epsilon e^{-2t} - 16 t + 12570$ (don't worry that $116.2$ doesn't look small, it's small enough in this context). For $\epsilon = 0$ the solution is $t = 12570/16 = 785.625$. Call this $t_0$. If $t = t_0 + \epsilon t_1 + \epsilon^2 t_2 + \ldots$, $\epsilon e^{-2t} = e^{-2 t_0} (\epsilon - 2 t_1 \epsilon^2 + \ldots)$, and we get $$ \eqalign{ \exp(-1571.25)-16 t_1 &= 0\cr -2 \exp(-1571.25) t_1-16 t_2 &= 0\cr \text{etc}\cr}$$