Integral with conditions

Question

Compute

$$\displaystyle\min_{a,b,c} \displaystyle\int_{0}^{\infty} \left | x^3-a-bx-cx^2 \right |^2e^{-x}\, dx$$

Please, any suggestions are welcome.

Thanks you all.

Answer 1

Hint: with the scalar product (for instance, of the space of polynomial functions defined on $(0,\infty )$) $$ f,g \to \int_0^\infty f(x)g(x) e^{-x}dx $$ $x\to a+bx+cx^2$ is the orthogonal projection of $x\to x^3$ on $\text{span }\{ x\to 1, x\to x, x\to x^2 \}$

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to solve this, see that if $\pi f$ is the orthogonal projection of $f$, \begin{align} \langle \pi f - f, x\to x^2\rangle &=0\\ \langle \pi f - f, x\to x\rangle &=0\\ \langle \pi f - f, x\to 1\rangle &=0\\ \end{align} which are three equations for 3 unknowns:

\begin{align} \int_0^\infty e^{-x}[x^3 - a - bx - cx^2] x^2dx &=0\\ \int_0^\infty e^{-x}[x^3 - a - bx - cx^2] x dx &=0\\ \int_0^\infty e^{-x}[x^3 - a - bx - cx^2] dx &=0\\ \end{align}

which you can simplify using $$ \int_0^\infty x^n e^{-x}dx = n! $$to give

\begin{align} 120 - 2a - 6b- 24c &=0\\ 24 - a - 2b- 6c &=0\\ 3 - a - b- 2c &=0\\ \implies a=-3, b=-9, c&= 7.5 \end{align}

Answer 2

Expand the square (the absolute value disappears), and you will obtain a polynomial times a negative exponential. Use the result on the Gamma function for integer arguments.

You'll get

$$6!-2c5!+c^24!-2b4!+2bc3!-2a3!+b^22!+2ac2!+2ab1!+a^20!.$$

Cancel the gradient and solve the $3\times 3$ linear system

$$2a+2b+4c-12=0$$ $$2a+4b+12c-48=0$$ $$4a+12b+48c-240=0$$