Consider real roots to the equation
$$ \frac{2}{1- \varepsilon x^{2}} = e^{x} $$
as $\varepsilon \to 0$.
From the monotonicity of $e^{x}$ together with qualitative properties of $\frac{2}{1-\varepsilon x^{2}}$, we conclude that there exist two roots $x_1, x_2$ with $x_1 \sim \ln(2), x_2 \sim \varepsilon^{-1/2}$.
I have no problem setting up an iteration scheme for $x_1$ to extract higher order terms, but have been stuck on the asymptotic expansion of $x_2$, as I have been unable to tame the singularity of the rational function.
We can rescale through $x \equiv \varepsilon^{-1/2} y$, hoping to develop an equation of the form
$$ y = 1 + \phi(y) \hspace{10 mm} \phi(y)=o(1)$$
Upon rescaling, our equation becomes
$$ \frac{2}{1- y^{2}} = e^{\varepsilon^{-1/2} y} $$
for which I still get stuck and am unable to extract an equation of the form above.
I would appreciate any help with tackling this problem, and the use of asymptotics for transcendental equations in general!
Best Answer
Why not consider that we look for the zeros of function $$f(x)=(1-\epsilon x^2)e^x-2$$ and use series expansion followed by series reversion.
This would give with $L=\log(2)$, $$x_1=L+\frac{\epsilon L^2 }{1-(2L+L^2) \epsilon}+\cdots \tag 1$$ and with $K=\epsilon^{-\frac 12}$, $$x_2=K- K e^{-K}+\cdots \tag 2$$
which are equivalent to the first iterate of Newton method. For sure, we could do better adding more terms in the initial expansion.
Playing with various values of $\epsilon$, these seem to be more than decent.
For $x_2$, a better approximation could be $$x_2=K+\frac{2 K}{2 K-2 e^K+1}$$