1 node Gauss–Hermite quadrature?

Question

In this page, we have Gauss–Hermite quadrature for more than 2 nodes. But can we get one node and weight for Gauss–Hermite quadrature?

Answer 1

This can be done, but it's not very useful. The weights for $N^\textrm{th}$ come from the tower equations $$ \sum_{i=0}^{N-1} w_iH_j(x_i) = \delta_{j,0} $$ and the abscissa are the zeros of the $N^\textrm{th}$ Hermite polynomial.

Nothing stops you from setting $N=1$, with result $w_0=1$, $x_0=0$ since $H_1(x)=x$.

What I mean by not useful is that only first order polynomials are integrated exactly, which is not a very big class of functions. Further, the error is proportional to the second derivative.

What this does is evaluates the function at the average of the Gaussian distribution you are integrating against. For example $$ I=\int_{-\infty}^\infty f(z) \exp\left(-\frac{(z-\mu)^2}{2\sigma^2}\right)\frac{dz}{\sqrt{2\pi\sigma^2}} $$ can be put into a form amenable to Gauss-Hermite quadrature by the transformation $x=\frac{z-\mu}{\sigma}$ $$ I=\int_{-\infty}^\infty f(\sigma x + \mu) \exp\left(-\frac{x^2}{2}\right)\frac{dx}{\sqrt{2\pi}}. $$ Applying first order Gauss-Hermite quadrature leads to $$ I\approx f(\mu). $$