Gram Schmidt process to generalize the orthogonal sequence $(-1)^ne^{x^2/2} \frac{d^n}{dx^n}(e^{-x^2})$

Question

Let us consider a sequence of functions $\{f_n(x)\}_{0}^{\infty}$ given by $$f_n(x)=x^{n}e^{-x^2/2},~~~~\forall~ x \in \mathbb R,~~n \geq 0.$$ Then the sequence is basically, $~\{e^{-x^2/2},~xe^{-x^2/2},~x^2e^{-x^2/2},~\cdots\}.$ Now notice that $f_n \in L_2(\mathbb R)$ for all $n \geq 0$.
Now using Gram Schmidt (GS) orthogonalization process we will find an orthogonal sequence $v_n$ where $$v_n(x)=(-1)^ne^{x^2/2} \frac{d^n}{dx^n}(e^{-x^2}),~~~n \ge 0.$$ I know $~\displaystyle H_n(x)=(-1)^ne^{x^2} \frac{d^n}{dx^n}(e^{-x^2}),~~~n \ge 0~ $ is called the Hermite polynomials of degree $n$ and by our context $~~v_n(x)=e^{-x^2/2}H_n(x).$
My attempt: Let $\{u_n\}_{1}^{\infty}$ be the orthogonal set be obtained by applying GS-process. Then we have, $$ \begin{align*} &u_0=f_0\\ &u_1=f_1-\frac{\langle u_0,f_1\rangle}{\|u_0\|^2}u_0=xe^{-x^2/2}-\frac{\int_{-\infty}^{\infty}xe^{-x^2/2}\cdot e^{-x^2/2}dx }{\int_{-\infty}^{\infty}e^{-x^2/2}\cdot e^{-x^2/2}dx}e^{-x^2/2}=xe^{-x^2/2}-\frac{0}{2\pi}e^{-x^2/2}=xe^{-x^2/2} \end{align*} $$ I have also computed $u_2,~u_3$ by $v_n \neq u_n$ by my calculation. Here also $v_1=2xe^{-x^2/2}$ but $u_1=xe^{-x^2/2}$. And also I am not able to find the closed form $v_n(x)=(-1)^ne^{x^2/2} \frac{d^n}{dx^n}(e^{-x^2}),~~~n \ge 0$.
Please help me to solve this.

Answer 1

The issue is that the original formula $$ u_n(x) = (-1)^ne^{-\frac{x^2}{2}}\frac{d^n}{dx^n}e^{-x^2} $$ does not constitute an orthogonal series. The Rodriguez formula shows which polynomials of this form that are orthogonal $$ p_n(x) = \frac{c_n}{w(x)}\frac{d^n}{dx^n}\left[B^n(x)w(x)\right] $$ where $B(x)$ is some polynomial of order at most 2 (https://en.wikipedia.org/wiki/Rodrigues%27_formula). Note that your formula is not of this form, and therefore will not result in orthogonal polynomials.

To do GS, you should choose a domain and a weighting function and you will find the orthogonal series explicitly.