I am trying to understand the following part of the paper https://doi.org/10.1137/20M1389522 where the author argues about the inefficiency of Gauss-Hermite quadrature. I think I get the gist of the argument the author is trying to make, but there are typos that I cannot fix by myself.
The author claims that the right choice of the domain truncation for integration when the integrand decays like $\exp(-x^2)$ is $[-O(n^{1/3}), O(n^{1/3})]$:
In exact arithmetic, we can get convergence of $I_n(f)$ to $I(f)$ with $I_n$ defined by quadrature over a finite interval, provided the interval grows as $n \to \infty$. The right choice for analytic functions $f$ bounded on $(-\infty, \infty)$ is to truncate to an interval of size $[-O(n^{1/3}), O(n^{1/3})]$, which balances a domain-truncation error of order $\exp(-n^{2/3})$ and a discretization error of the same order since the sample step size will be $h = O(n^{-2/3})$.
I follow this argument. Then the author continues as follows, which I do not follow. Bold texts below are by me.
Note that $[-O(n^{1/3}), O(n^{1/3})]$ is much narrower than the interval $[-O(n^{1/2}), O(n^{1/2})]$ sampled by the Gauss-Hermite formula. Comparing the two enables us to quantify the inefficiency of Gauss-Hermite quadrature. In effect, only a fraction of order $n^{1/3}/n^{1/2} = n^{2/3}$ of the Gauss-Hermite nodes is utilized, meaning that Gauss-Hermite quadrature employs more nodes than necessary by a factor of order $n^{1/3}$.
The part "$n^{1/3}/n^{1/2} = n^{2/3}$" is a typo because $n^{1/3}/n^{1/2}=n^{-1/6}$. This affects the part that follows: meaning that Gauss-Hermite quadrature employs more nodes than necessary by a factor of order what? What would be the implication for the inefficiency in terms of the order of convergence?
This also affects the next sentence, which goes
The ratio increases to nearly order $n^{1/2}$ for nonanalytic functions $f$, where intervals growing just logarithmically rather than algebraically with $n$ are appropriate for balancing domain-truncation and discretization errors.
Which ratio should this be? Is it $n^{1/3}/n^{1/2}=n^{-1/6}$? Where does $n^{1/2}$ come from?
