How to find $$\min_{a,b,c\in\mathbb{C}}{\int_0^{\infty}} |a+bx+cx^2+x^3|^2 e^{-x} dx = ?$$ Thanks in advance.
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What is $C$? and do you have some idea of how to solve it? – Shuhao Cao Jun 09 '13 at 21:12
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$\mathbb{C}$ is set of complex numbers. – alans Jun 09 '13 at 21:15
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I think that this integral can be recognised as a norm of function $a+b x+c x^2+x^3$ in space $L^2((0,\infty),e^{-x}dx)$, but I'm kind of stuck now. – alans Jun 09 '13 at 21:19
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$\langle f,g \rangle = \int\limits_0^{\infty} f \bar{g}e^{-x}dx$ is scalar product (for $f,g \in L^2(e^{-x}dx)$), so we have to find squared distances of vector $x^3$ from subspace generated by $1,x,x^2$ (it is true that $1,x,x^2,x^3 \in L^2(e^{-x}dx)$).
Let $x^3=k+lx+mx^2+f, f \perp \mathcal{L}\{1,x,x^2\}$. Scalar multiplication by $1,x,x^2$ it follows that $k+l+2m=6,k+2l+6m=24,2k+6l+24m=120$, so $k=6, l=-18, m=9$ and the required minimum is square norm vector $x^3-6+18x-9x^2$ in the upper scalar product, that is $$\min_{a,b,c \in \textbf{C}} \int\limits_{0}^{\infty} |a+bx+cx^2+x^3|^2e^{-x}dx=\int\limits_{0}^{\infty} |x^3-6+18x-9x^2|^2e^{-x}dx=36.$$
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Hint:
Define the inner product on the vector space $\mathbb C[x]$ by $$(f|g)=\int_0^\infty f(x)\overline{g(x)}e^{-x}dx$$
then $$\min_{a,b,c\in\mathbb{C}}{\int_0^{\infty}} |a+bx+cx^2+x^3|^2 e^{-x} dx = \min_{P\in \mathbb C_2[x]}||x^3-P||^2$$
hence we calculate the orthogonal projection $P$ of $x^3$ on the subspace $\mathbb C_2[x]$ and the desired value is $||x^3-P||^2$