Practice Final 5a
rsehrlich
(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.
${1, x, x^2}$ 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.
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.
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.
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.
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.
Yeah I realized my error. You guys have the right answer!