Derive asymptotic behavior of inverse of the normal cdf with respect to 2^n

Question

I have a normal distribution $\mu = 0$ and $\sigma = 0.58n$ where $n > 0 $ and I am trying to derive the asymptotic behavior of the following equation: $$\Phi\left(\frac{x}{0.58n}\right)\;=\;2^{1-n}$$ Follows: $$ \DeclareMathOperator\inverfc{inverfc} x = -0.58 \sqrt{2}n \inverfc{({2^{2 - n}})} $$ So I want to find $O( n\inverfc{({2^{2 - n}})}) $. More specifically I want to confirm my suspicion that it is in $O(n \log{n})$. However since Inverf is a special function I can't wrap my mind around to analyse it. I gave complete context since another derivation might be more helpful here.

Answer 1

When $z\to+\infty$, $\Phi(-z)\sim1/(z\mathrm e^{z^2/2}\sqrt{2\pi})$. Since $2^{1-n}\to0$, this is the regime of interest. The solution of $\Phi(-z_n)=b\mathrm e^{-cn}$ with $b=2$ and $c=\log2$ solves $b\sqrt{2\pi}z_n\mathrm e^{z_n^2/2}\sim\mathrm e^{cn}$, that is, $$ z_n^2+2\log z_n=2cn-\log(2\pi)-2\log b+o(1), $$ in particular $z_n\sim\sqrt{2cn}$. The question asks about $x_n=-anz_n$ with $a=0.58$ hence $x_n\sim -an\sqrt{2cn}$, in particular $x_n=\Theta(n\sqrt{n})$ hence $x_n=\Omega(n\sqrt{n})$ and $x_n=O(n\sqrt{n})$.

If need be, the equivalent in the first paragraph yields more precise estimates, for example, one has $z_n=\sqrt{2cn}-\log n/\sqrt{8cn}+o(\log n/\sqrt{n})$ hence, introducing $\alpha=a\sqrt{2c}$ and $\beta=a/\sqrt{8c}$, $$ x_n=-\alpha n\sqrt{n}+\beta\sqrt{n}\log n+o(\sqrt{n}\log n). $$