Let f(x) = 9, g(x) = -9x + 3 and h(x) = 9x^2. Consider the...

Let f(x) = 9, g(x) = -9x + 3 and h(x) = 9x^2. Consider the inner product ?p, q? = ??¹ p(x)q(x) dx in the vector space C?[0, 1] of continuous functions on the domain [0, 1]. Use the Gram-Schmidt process to determine an orthonormal basis for the subspace of C?[0, 1] spanned by the functions f(x), g(x), and h(x).

Transcript

00:01 Okay, we're given three functions.

00:04 You notice that one of them is linear.

00:07 Well, one of them is constant.

00:08 The second one is linear, and the third one is quadratic.

00:12 They're defined on the interval 0 .1.

00:15 We want to find a basis for the function space on 01 that will basically accommodate these three functions.

00:26 We're not actually supposed to write these functions in terms of the basis.

00:30 We're just supposed to find the basis, which means we need three quadratic functions that are orthanormal as a basis.

00:53 And we're defining our so -called dot product or scalar product like this.

01:05 Okay.

01:07 And we're told, what's actually fairly obvious, f1, f0 is 1.

01:14 So the thing of order 0 is 1.

01:18 And if i integrate f0 squared from 0 to 1, i get 1.

01:31 And then i'm going to define my other two functions that i'm interested in like this, where a, b, c, d, and e are constants that i'm going to determine.

01:43 So the gramschmidt process is basically to keep evaluating those scalar products of the different functions and then making the requirement that they be orthogonal if they're different and normalized if they're the same.

02:02 So, for instance, the scalar product of f0 and f1, we put those two in there.

02:10 We do those integrals, evaluated from zero to one, set it equal to zero.

02:17 We have one -half a plus b equals zero, so that a is minus 2b.

02:26 And then we can normalize it.

02:32 So calculate f1 with f1.

02:36 That has to equal one.

02:45 The problem with the gram -schmidt procedure, and there's nothing better, is that it could be very tedious.

03:13 Okay.

03:14 And so we get from f -1 with f -1, we get third a -squared plus a -b -b -b -squared.

03:21 And then we can substitute in a from before, a equals minus 2b.

03:29 And then what we'll do is we'll get an equation for b.

03:37 And that tells us that b squared, b is square of 3, and a is minus 2 times the square of 3.

04:03 So we have f0 is 1, and f1 is a square root of 3 times minus 2x plus 1.

04:17 Okay...

Question

Please give Ace some feedback