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Consider $P_{2}(\mathbb{R})$ together with inner product:

$\langle p(x), q(x)\rangle = \int_{0}^{1} p(x)q(x) \, dx$

I am trying to come up with an orthogonal basis with respect to this inner product for

$p(x) = a + bx + cx^2$


Let $\alpha = \{ 1, x, x^2 \}$ standard basis for $P_{2}(\mathbb{R})$

Using the Gram-Schmidt process:

Let $v_1 = \alpha_1 = 1$

Let

$v_{2} = \alpha_{2} - P_{v_{1}}(\alpha_{2}) = \alpha_{2} - \frac{\langle\alpha_{2},v_{1}\rangle}{\langle v_{1}, v_{1}\rangle} \cdot v_{1} = x - \frac{\langle x, 1\rangle}{\langle 1,1\rangle} \cdot 1$

Since $\langle p(x), q(x)\rangle = \int_0^1 p(x)q(x) \, dx$,

$\langle x, 1\rangle = \int_0^1 x \cdot 1 \, dx = \int_0^1 x \, dx = \frac{x^2}{2}$

$\langle 1, 1\rangle = \int_0^1 1 \cdot 1 \, dx = \int_0^1 1 \, dx = x$

$\Rightarrow v_{2} = x - \frac{x^2}{2} \cdot \frac{1}{x} \cdot 1 = x - \frac{x}{2}$


I'm positive this is not correct, as the answer should be $v_2 = x - \frac{1}{2}$

What am I doing wrong?

  • 0
    Sweet, Thanks!!2012-04-25

1 Answers 1

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You forgot to plug in the values after integrating. Note that:

$\int_0^1 x\,dx=\frac{1}{2}$

and

$\int_0^1 1\,dx=1$

After these modifications you're done.

  • 0
    There are some good free online integration tutor materials if you google, and most definitely some excellent textbooks available too. Too bad I learned these things from textbooks of my native language (which isn't English), so I don't know any books of this topic to recommend. :/2012-04-25