Computing the inverse of $n^{n^2}e^{-n^2}$

Question

What is the method for computing the inverse of this function?

I'm interested in asymptotic approximation. We can consider that $n \in \mathbb{R}$.

Thank you very much.

Answer 1

The problem is to solve $\quad y^{y^2}e^{-y^2}=x\quad$ for $x$ in order to express $y(x)$.

Let $y=e^z \quad\to\quad (e^{z})^{e^{2z}}e^{-e^{2z}}=e^{(z-1)e^{2z}}=x$ $$(z-1)e^{2z}=\ln(x)$$ $2(z-1)e^{2z}=2\ln(x)$

$2(z-1)e^{2(z-1)}=2e^{-2}\ln(x)$

With $\begin{cases}W=2(z-1)\\X=2e^{-2}\ln(x)\end{cases} \quad\to\quad We^W=X$ $$2(z-1)=W\left(2e^{-2}\ln(x) \right)$$ $W(X)$ is the Lambert W function.

$z=1+\frac{1}{2}W\left(2e^{-2}\ln(x) \right)$

The inverse function of $y^{y^2}e^{-y^2}=x$ is : $$y=e^{1+\frac{1}{2}W\left(2e^{-2}\ln(x) \right)}$$ HINT:

The asymptotic behaviour of $W(X)$ is : $$W(X)\sim \ln(X)-\ln(\ln(X)) $$

http://mathworld.wolfram.com/LambertW-Function.html

Answer 2

$$y=n^{n^2}e^{-n^2}=\left( \frac{n}{e} \right)^{n^2} \\ \ln(y)= n^2 (\ln(n)-1) \\ \sqrt{ \ln (y) } =n \sqrt{\ln(n)-1}$$

This implies that for $n \geq 4$. $$n \leq \sqrt{ \ln(y)}$$ and hence $$f^{-1}(y) \leq \sqrt{\ln(y)}$$

Also, $$\sqrt{ \ln (y) } =n \sqrt{\ln(n)-1} \leq n \sqrt{ \sqrt{ \ln(y)}-1} \Rightarrow \\ n \geq \frac{\sqrt{ \ln (y) }}{\sqrt{ \sqrt{ \ln(y)}-1} } $$

This gives $$\frac{\sqrt{ \ln (y) }}{\sqrt{ \sqrt{ \ln(y)}-1} } \leq f^{-1}(y) \leq \sqrt{\ln(y)} \,.$$