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I want to find a two-term asymptotic expansion of the equation above. For the small root, I know the first term may be $\epsilon$, but I don't know how to find the second term. Moreover, I don't know how to find the root tending to infinite either.

ZYZmoonlight
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1 Answers1

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(I was taught to do asymptotics the heuristic way, so please forgive the handwaviness below. There are more sophisticated ways to do things if you want more than two terms.)

Small root

We have seen that the first term is $x \sim \epsilon$. For the next term, suppose that $$ x \sim \epsilon + a \epsilon^n $$ where $n > 1$.

Noting that $x \mathrm{e}^{-x} \sim x (1 - x)$, we have

\begin{align} (\epsilon + a \epsilon^n) (1 - \epsilon - a \epsilon^n) &\sim \epsilon \\ \epsilon - \epsilon^2 + a \epsilon^n &\sim \epsilon, \end{align}

where we have discarded terms smaller than $\epsilon^n$. We see that $n = 2$ (which is indeed greater than $1$) and $a = 1$. Hence, to two terms,

$$ x \sim \epsilon + \epsilon^2. $$

Large root

The large root is not as straightforward. Taking logarithms we have $$ \log x - x = \log\epsilon. $$

Now, if $x$ is exceedingly large, then we may discard $\log x \ll x$, giving the very crude estimate $$ x \sim \log\frac{1}{\epsilon}. $$

For the next term, suppose that $$ x \sim \log\frac{1}{\epsilon} + x_1 $$ where $x_1 \ll \log(1/\epsilon)$. Substituting into the log expression, we get

\begin{align} \log \left( \log\frac{1}{\epsilon} + x_1 \right) - \log\frac{1}{\epsilon} - x_1 &\sim \log\epsilon \\ x_1 &\sim \log \left( \log\frac{1}{\epsilon} + x_1 \right) \\ &\sim \log \left( \log\frac{1}{\epsilon} \right), \end{align} which, for exceedingly exceedingly small $\epsilon$, is indeed much less than $\log(1/\epsilon)$.

Hence to two terms, $$ x \sim \log\frac{1}{\epsilon} + \log \log\frac{1}{\epsilon}. $$

yawnoc
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