Question

Determine uma base ortonormal para R[x]2, munido do produto interno = integral de 0 a 1 p(x)q(x) dx, aplicando o processo de Gram-Schmidt à base canônica { 1,x,x² }.

Answer 1

Sea
                $p_1(x)=1\\p_2(x)=x\\p_3(x)=x^2$

Primero hallemos una base ortogonal

(*) $u_1(x)=1$

(**) $u_2 =p_2-\dfrac{\langle p_2,u_1\rangle}{\langle u_1,u_1\rangle}\cdot u_1$

Cálculo:
            
            $u_2 =x-\dfrac{\langle x,1\rangle}{\langle 1,1\rangle}\cdot 1\\ \\ u_2 = x -\dfrac{\int _{0}^1xdx}{\int_{0}^11dx}\\ \\ \\ \boxed{u_2=x-\dfrac{1}{2}}$

(***) $u_3 =p_3-\dfrac{\langle p_3,u_1\rangle}{\langle u_1,u_1\rangle}\cdot u_1-\dfrac{\langle p_3,u_2\rangle}{\langle u_2,u_2\rangle}\cdot u_2$

Cálculo

    $u_3 =x^2-\dfrac{\langle x^2,1\rangle}{\langle 1,1\rangle}\cdot 1-\dfrac{\langle x^2,x-1/2\rangle}{\langle x-1/2,x-1/2\rangle}\cdot (x-1/2)\\ \\ \\ u_3=x^2-\dfrac{\int_{0}^1x^2dx}{\int_{0}^1dx}-\dfrac{\int_{0}^1x^2(x-1/2)dx}{\int_{0}^1(x-1/2)^2dx}(x-1/2)\\ \\ \\ u_3=x^2-\dfrac{1}{3}-\left(x-\dfrac{1}{2}\right)\\ \\ \\ \boxed{u_3=x^2-x+\dfrac{1}{6}}$

Entonces la base ORTOGONAL es

              $B=\left\{1,x-\dfrac{1}{2},x^2-x+\dfrac{1}{6}\right\}$

Ahora hallemos la NORMA de cada polinomio

$\|1\|=\langle 1,1\rangle^{1/2} =\sqrt{\int_{0}^1 1^2dx}=1\\ \\ \|x^2-1/2\|=\langle x^2-1/2,x^2-1/2\rangle^{1/2}=\sqrt{\int_{0}^1(x^2-1/2)^2dx}=\dfrac{1}{2\sqrt3}\\ \\ \|x^2-x+\frac{1}{6}\|=\langle x^2-x+\frac{1}{6},x^2-x+\frac{1}{6}\rangle^{1/2}=\sqrt{\int_{0}^1(x^2-x+\frac{1}{6})^2dx}=\dfrac{1}{6\sqrt{5}} \\ \\.$

Por lo tanto la base ORTONORMAL es:

         $\boxed{H=\left\{1\;,\;2\sqrt{3}x-\sqrt{3}\;,\;6\sqrt5x^2-6\sqrt5x+\sqrt5\right\}}$