Infinite dimensions and matrices

1. Jul 31, 2011

Lolsauce

1. The problem statement, all variables and given/known data

Find 2 more orthonormal polynomials on the interval [-2,1] up to degree 2 given that the first polynomial p(x) = 1/√3. ( Note: Take the highest coefficient to be positive and enter your answer as a decimal.)

2. Relevant equations

This is a web assign equation so the answer format is in something like this, where I enter in the solutions:

degree 1 = (something)x + something
degree 2 = (something)x^2+(something)x+something

3. The attempt at a solution

I'm trying to understand this problem here. What exactly does the 1 over square root three give us? How does it help?

I was told by my professor that to do this use the dot product of two functions f and g then find integral of f(x)g(x)dx over the boundary conditions. I'm not exactly sure what this means but I followed some examples from the homeworks, as we hadn't really learned this in lecture yet.

So I make up two equations:

f(x) = a + bx
g(x) = x(a+bx)

(1) I integrate both equations on the given boundaries: Int[-2,1] (a + bx) dx= 3a - (3/2)b

(2) INT [-2,1]x(a+bx) dx = -(3/2)a + 3b

After this step I have NO IDEA what to do. I have a system of two equations. I can make a matrix
| 3 -(3/2) |
|-(3/2) 3 |

But how does this help me? If anyone could give me guidance, please do. Thank you very much

Last edited: Jul 31, 2011
2. Jul 31, 2011

Ray Vickson

Label the three polynomials as p0(x), p1(x) and p2(x) of respective degrees 0, 1 and 2. You are told that the interval is [-2,1] and that p0(x) = 1/sqrt(3), hence int_{x=-2..1} p0(x)^2 dx = 1. That seems to be saying that the polynomials should also be "orthonormal", that is, have "squared norm" = 1. Anyway, you need int_{x=-2..1} p0(x)*p1(x) dx = 0 (p0 orthogonal to p1) and int_{x=-2..1} p1(x)^2 dx = 1 norm condition), so you can determine both constants a and b in the formula p1(x) = a + b*x. Now, with p2(x) = e + f*x +g*x^2, you can determine e, f and g from the conditions int_{x=-2..1} p0(x)*p2(x) dx = 0, int_{x=-2..1} p1(x)*p2(x) dx = 0 and int_{x=-2..1} p2(x)^2 dx = 1. You will have three simple linear equations in the three unknowns e,f and g.

RGV

3. Jul 31, 2011

Lolsauce

Thank for for your response. I have a question though, I was able to solve the first degree equation of a + bx, but I can't seem to get the second one. So you set the first two equations p0p2 and p1p2 to zero correct?

Last edited: Jul 31, 2011
4. Jul 31, 2011

Lolsauce

Never mind I got it, thanks!