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Let $p, q \in P_3(\Bbb R)$. Define the inner product by

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

What is the reflection of $q(x) = x ^ 2$ on the subspace $P_1(\Bbb R)$?

My result is $R(q)(x) = \frac{2}{3} + \frac{3x}{2} - x^2 $

But the answer in the book is $ R(q)(x) = -\frac{1}{3} + 2x - x^2$

I don't what is the right procedure to this problem.

JoOkuma
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  • Can you show us the steps which carry you to your result $R(q)(x) = \frac{2}{3} + \frac{3x}{2} - x^2$? – GNUSupporter 8964民主女神 地下教會 Dec 12 '16 at 12:35
  • I used Reflection = 2 * projection of q(x) - q(x), like this $$R(q)(x) = 2(\frac{\langle q(x), 1\rangle}{\langle 1,1 \rangle}1 + \frac{\langle q(x), x\rangle}{\langle x,x \rangle}*x) - q(x)$$ – JoOkuma Dec 12 '16 at 12:43
  • If you include steps like this, we will be able to spot out where you get stuck, and can show others that you've really worked on the problem. – GNUSupporter 8964民主女神 地下教會 Dec 12 '16 at 12:48
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    $$R(q)(x) = 2(\frac{\langle x^2, 1\rangle }{\langle 1, 1\rangle} 1 + \frac{\langle x^2, x \rangle}{\langle x, x \rangle} * x )- x^2$$

    $$ R(q)(x) = 2(\frac{\int_0^1x^ 2dx}{\int_0^11dx} + \frac{\int_0^1x^3dx}{\int_0^1x^2dx}x) - x^2$$

    $$R(q)(x) = 2 * (\frac{\frac{1}{3}}{1} + \frac{\frac{1}{4}}{\frac{1}{3}}*x) - x^2$$

    $$R(q)(x) = 2* (\frac{1}{3} + \frac{3}{4}x) - x^2$$

    $$R(q)(x) = \frac{2}{3} + \frac{3}{2}x-x^2$$

     – JoOkuma Dec 12 '16 at 13:05
  • I'm sorry for a late response. I've figured out why. I'm going to type the answer. – GNUSupporter 8964民主女神 地下教會 Dec 15 '16 at 11:36

1 Answers1

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You have to start with an orthogonal base for $P_1(\Bbb R)$ in order to calculate the reflection.

  1. We take $1$ as the first element of our orthogonal base for $P_1(\Bbb R)$.
    • $\lVert1\rVert^2 = \langle1,1\rangle = 1$ is obvious.
  2. We apply the Gram-Schmidt process on $x$ and $x^2$ to obtain the second and third element in this basis. At this stage, we'll have the projection of $q(x)=x^2$ on $P_1(\Bbb R)$, denoted by $\pi(q(x))=\pi(x^2)$.
    • $x-\dfrac{\langle x,1 \rangle}{\lVert1\rVert^2}\cdot 1 = x - \dfrac{\int_0^1 x\,\mathrm{d}x}{1} = x - \dfrac12$, which is the second element in our orthogonal basis.
    • Find $\pi(q(x))$. \begin{align} & \pi(q(x)=\pi(x^2) \\ =& \frac{\langle x^2, 1\rangle}{\langle 1, 1\rangle} \cdot 1 + \frac{\langle x^2, x - \frac12 \rangle}{\langle x - \frac12, x - \frac12 \rangle} \cdot \left(x - \frac12\right) \\ =& \frac{\int_0^1 x^2\,\mathrm{d}x}{1} \cdot 1 + \frac{\int_0^1\left(x^3-\frac12 x^2\right)\,\mathrm{d}x}{\int_0^1 \left(x - \frac12\right)^2\,\mathrm{d}x} \cdot \left(x - \frac12\right) \\ =& \frac{\frac13}{1}\cdot 1 + \frac{\left.\frac14 x^4 - \frac16 x^3 \right\rvert_0^1}{\left.\frac13\left(x-\frac12\right)^3\right\rvert_0^1}\cdot \left(x - \frac12\right) \\ =& \frac13 + \frac{\frac14-\frac16}{\frac23 \left(\frac12\right)^3} \cdot \left(x - \frac12\right) \\ =& x - \frac16 \end{align}
  3. We calculate \begin{align} R(q)(x) =& 2\pi(q(x))-x^2 \\ =& 2\pi(x^2)-x^2 \\ =& 2 \left(x - \frac16\right)-x^2 \\ =& -x^2 + 2x -\frac13. \end{align}
GNUSupporter 8964民主女神 地下教會
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