A calculus problem by ابراهيم فقرا

The functions $\varphi_n(x) \; = \; \frac{1}{\sqrt{2^n n!} \pi^{\frac14}}H_n(x) e^{-\frac12x^2}$ form an orthonormal sequence in $L^2(\mathbb{R})$ , where $H_n(x)$ is the $n$ Hermite polynomial. But then, for any real valued $f \in L^2(\mathbb{R})$ and $\alpha_0,\alpha_1,\alpha_2 \in \mathbb{R}$ , we have by the standard theory of orthonormal sequences, that $\int_{-\infty}^\infty\left| f(x) - \sum_{j=0}^2\alpha_j\varphi_j(x)\right|^2\,dx \; = \; \int_{-\infty}^\infty |f(x)|^2\,dx - \sum_{j=0}^2 \left|\int_{-\infty}^\infty f(x)\varphi_j(x)\,dx\right|^2 + \sum_{j=0}^2 \left[\alpha_j - \int_{-\infty}^\infty f(x)\varphi_j(x)\,dx\right]^2$ and so, putting $f(x) = x^4e^{-\frac12x^2}$ , the minimum value of $\int_{-\infty}^\infty \big|x^4 - a - bx - cx^2\big|^2e^{-x^2}\,dx$ as $a,b,c$ range over all reals is $\int_{-\infty}^\infty x^8e^{-x^2}\,dx - \sum_{j=0}^2 \left(\int_{-\infty}^\infty x^4e^{-\frac12x^2}\varphi_j(x)\,dx\right)^2 \; = \; \tfrac32\sqrt{\pi} = \boxed{2.658680776}$

I took an elementary approach, which produces the answer quickly. Noting that $x$ is orthogonal to the three other monomials, we want $b=0$ . Using the given values, we find $\int_{-\infty}^{\infty}(x^4-a-cx^2)^2e^{-x^2}dx=a^2I_0+2acI_2+(c^2-2a)I_4-2cI_6+I_8=\frac{\sqrt \pi}{2}\left(\frac{1}{8}(4a+2c-3)^2+(c-3)^2+3\right)$ where $I_n=\int_{-\infty}^{\infty}x^n e^{-x^2}dx$ . The minimum (and infimum) is $\frac{3\sqrt{\pi}}{2}\approx \boxed{2.659}$ , attained when $a=-\frac{3}{4}, b=0$ and $c=3$ .