Gram-Schmidt Orthogonalization Process

[Mr Davis 97]

Homework Statement

Find an orthogonal basis for $\operatorname{span} (S)$ if $S= \{1,x,x^2 \}$, and $\langle f,g \rangle = \int_0^1 f(x) g(x) \, dx$

Homework Equations

The Attempt at a Solution

So we start by the normal procedure.

Let $v_1 = 1$. Then $\displaystyle v_2 = x - \frac{\langle x,1 \rangle}{\| 1 \|^2}(1) = x - \frac{1}{2}$.
Then $\displaystyle v_3 = x^2 - \frac{\langle x^2,1 \rangle}{\| 1 \|^2}(1) - \frac{\langle x^2,x \rangle}{\| x \|^2}(x) = x^2 - \frac{1}{3} - \frac{3}{4}x$.

But this is not correct, because if I calculate $\displaystyle \langle 1, x^2 - \frac{1}{3} - \frac{3}{4}x\rangle = \int_0^1 x^2 - \frac{1}{3} - \frac{3}{4}x \, dx = -\frac{3}{2} \ne 0$.

What am I doing wrong?

 

[FactChecker]

Shouldn't the v₃ definition use v₂ = x-1/2 in the third term instead of x?

 

[Ray Vickson]

Homework Statement

Find an orthogonal basis for $\operatorname{span} (S)$ if $S= \{1,x,x^2 \}$, and $\langle f,g \rangle = \int_0^1 f(x) g(x) \, dx$

Homework Equations

The Attempt at a Solution

So we start by the normal procedure.

Let $v_1 = 1$. Then $\displaystyle v_2 = x - \frac{\langle x,1 \rangle}{\| 1 \|^2}(1) = x - \frac{1}{2}$.
Then $\displaystyle v_3 = x^2 - \frac{\langle x^2,1 \rangle}{\| 1 \|^2}(1) - \frac{\langle x^2,x \rangle}{\| x \|^2}(x) = x^2 - \frac{1}{3} - \frac{3}{4}x$.

But this is not correct, because if I calculate $\displaystyle \langle 1, x^2 - \frac{1}{3} - \frac{3}{4}x\rangle = \int_0^1 x^2 - \frac{1}{3} - \frac{3}{4}x \, dx = -\frac{3}{2} \ne 0$.
_
What am I doing wrong?

Wrong normalization.
$$v_2=\frac{u_2}{|| u_2 ||}, \\
u_2 = x - <x, v_1> v_1 $$