Asymptotic behavior of Lambert function

Question

In my calculation, I got $$x^{3/2}e^{-x}=C$$ $$ \frac{3}{2} \log{[x]}-x=\log{C}$$ Where $C<<1$, I know that the general solutions to this equation are Lambert Functions, but considering $C$ value I was hoping maybe I can use a simpler, asymptotic behavior of $x$ to make the rest of my calculation easier.
Can anyone help me with that?

Answer 1

Since you know about Lambert function, you know that the solution of $$ x^{3/2}\,e^{-x}=C\implies x=-\frac{3}{2} W\left(-\frac{2 C^{2/3}}{3}\right)$$ For simplicity, let use as notation $a=-\frac{2 C^{2/3}}{3}$.

Now, since $C <<1$, you could use the series expansion built around $a=0$ $$W(a)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}a^n =a-a^2+\frac{3}{2}a^3-\frac{8}{3}a^4+\frac{125}{24}a^5+O\left(a^6\right)$$ which is given here.

There is also Padé approximants which are quite good (better than series) such as $$W(a)\approx \frac{\frac{4 a^2}{3}+a}{\frac{5 a^2}{6}+\frac{7 a}{3}+1}$$ $$W(a)\approx\frac{\frac{451 a^3}{340}+\frac{228 a^2}{85}+a}{\frac{133 a^3}{204}+\frac{1193 a^2}{340}+\frac{313 a}{85}+1}$$