0
$\begingroup$

The problem: Find a polynomial $q \in P_2(\mathbb{R})$ such that

$$ p(1/2)= \int_0^1 p(x)q(x) \, dx $$

for every $p \in P_2(\mathbb{R})$.

So far, I established that the inner product on $P_2(\mathbb{R})$ is

$$\left< p, q \right> = \int_0^1 p(x)q(x) \, dx $$

and that

$$p(1/2) = \varphi(p) = \left< p,q \right> = \int_0^1 p(x)q(x) \, dx $$

and that

$$ e_1 = 1, \\ e_2 = \sqrt(3)(2x-1),\\ e_3 = \sqrt(5)(6x^2-6x+1) $$

and finally that

$$ q= \varphi(e_1)e_1 + ... + \varphi(e_n)e_n.$$

The final answer is given as $q= -3/2 + 15x - 15x^2$.

My situation is that I do not understand how the inner product $\left< p, q \right>$, the orthonormal basis $(e_1,e_2,e_3)$ and the formula for $q$ relate to give me the final answer.

Thank you.

1 Answers 1

0

You just need to plug everything in the last formula:

$$ q = \varphi(e_1)e_1 + \varphi(e_2)e_2 + \varphi(e_3)e_3 = \\ 1 \cdot 1 + \sqrt{3} (1 - 1) \cdot \sqrt{3}(2x - 1) + \sqrt{5} \left( \frac{6}{4} - \frac{6}{2} + 1 \right) \sqrt{5} \left( 6x^2 - 6x + 1 \right) =\\ 1 - \frac{5}{2}(6x^2 - 6x + 1).$$

  • 0
    Oh wow, thank you so much! I was confused how to apply it, but that makes complete sense, thank you again! – 2017-02-06