##### Practice Final #5
tedallen
Given: find an orthonormal basis for $P_{2}(R)$ where $\langle f,g\rangle = \int\limits_0^2 fg dx$. Solution: first of all, for $P_{n}$, $\langle p,q \rangle$ = p(-1)q(-1) + p(0)q(0) + p(1)q(1). let f(x)=1. then $g(x) = a + bx + cx^{2}$. then $\langle 1,a + bx + cx^{2} \rangle$ = 1(a-b+c) + 1(a) + 1(a+b+c) = 3a + 2c. evaluate $\int\limits_0^2 1(a + bx + cx^{2}) dx$ and you get $2a + 2b + \frac{8}{3}c$. Set them equal and simplify you get 3a - 2c = 6b. (remember that 3a + 2c must be 0 for the basis to be orthogonal). final answer: $\beta = \big[ \frac{1}{3}, \frac{3x^{2}-2}{6}\big]$
CarlHarmon: May 31, 2015, 10:42 a.m.
\newline Um... this is definitely not correct. At the very least, the answer at the end needs to have three terms. I'm not sure where that definition of inner product at the beginning is coming from. \hfill \break \newline The way I approached this problem by using {$1,x,x^2$} as a basis for $P_2(R)$. Then I tried caculating $u_1, u_2$ and $u_3$ using these vectors. But by the time you get to $u_3$ the math becomes hideous, so hope there is a better way.\hfill \break
CarlHarmon: May 31, 2015, 10:45 a.m.
And it looks like 5b is exactly the same as 7...
CarlHarmon: June 1, 2015, 5:55 p.m.
Okay, here's my answer: {$v_1, v_2, v_3$}={$1, x-1, x^2-2x+2/3$}
ckenny9739: June 1, 2015, 7:54 p.m.
Actually, that answer is not exactly right because it asks for an orthonormal basis. That is the correct orthogonal basis. My answer is {v1,v2,v3}={$\frac{1}{\sqrt{2}}$,(x−1)*($\frac{2}{3}^{-1/2}$, $\frac{(x^{2}−2x+2/3)}{\sqrt{8/45}}$}
ckenny9739: June 1, 2015, 7:55 p.m.
Add in an extra parenthesis after the 2/3 to the -1/2...
Brianlui: June 1, 2015, 7:56 p.m.
I checked and I also got this as my answer {v1, v2, v3} = { $\frac{1}{\sqrt{2}}$, $\frac{x-1}{\sqrt{2/3}}$, $\frac{x^{2}-2x+2/3}{\sqrt{8/45}}$} The orthonormal basis is found by dividing by the square root of the inner product of itself.
yulduzkhonbruin: June 1, 2015, 10:43 p.m.
I also have the same answer!
tedallen: June 1, 2015, 11:11 p.m.
Sorry everyone. Brainlui's answer is correct. I re-solved it. please no one look at my original answer
lilb: June 2, 2015, 1 a.m.
can someone post their work for this? i don't understand how to do it
carlintou: June 2, 2015, 11:30 a.m.
Why are the denominators in the final answer a square root?