## Practice Final 5a |
rsehrlich |
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(For reference, I looked at the first part of example 7 in section 5.5 of the book.) I chose 1, x, $x^{2}$ as my basis for $P_{2}$(R), since we already know that this is an orthogonal basis. All we have to do is turn it into an orthonormal basis. Using the procedures from the book, I got my basis as $\frac{\sqrt{2}}{2}$, $\frac{\sqrt3}{\sqrt8}x$, $\frac{\sqrt5}{\sqrt32}x^2$. |

CarlHarmon: June 1, 2015, 5:49 p.m. |
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${1, x, x^2}$ is not an orthongal basis. For a basis to be orthogonal, each pair of vectors has to have inner product 0. For 1 and x, the inner product would be 2. |

hsienchih: June 1, 2015, 8:15 p.m. |
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Does anyone know how to solve this problem? I think in order for the inner product to be 0, f or g has to be 0, but then it is impossible to find an orthonormal basis |

hsienchih: June 1, 2015, 9:27 p.m. |
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I think I got it. The basis I got is <$\frac{1}{\sqrt{2}}$>, $\sqrt{\frac{3}{2}}$(x-1), $\sqrt{\frac{45}{8}}$($x^{2}$-2x+$\frac{2}{3}$)>. Does anyone get the same thing |

tedallen: June 1, 2015, 11:18 p.m. |
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my answer agrees with hsienchih. I got $\Big\langle \sqrt{\frac{1}{2}}, \sqrt{\frac{3}{2}}\big(x-1\big), \sqrt{\frac{45}{8}}\big(x^{2}-2x+\frac{2}{3}\big)\Big\rangle$ |

alexwang0518: June 2, 2015, 4:09 p.m. |
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I also got that basis. I think the good thing to note about this problem is that the magnitude is also related to the inner product, so both the numerator and denominator change in the projection. |

rsehrlich: June 2, 2015, 8:28 p.m. |
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Yeah I realized my error. You guys have the right answer! |