Gram-Schmidt?

Question

Use Gram-Schmidt to find the first three orthogonal polynomials Yo, Y1, Y2 for the weight function:

w(x)=(1-x^2)^(1/2) for -1

Answer 1

If f, g are polynomials then their inner product is

= INT sqrt(1-x^2) f(x) g(x) dx

over the interval (-1,1). For the first function we pick y0(x) = 1, and find for the square norm

= INT sqrt(1-x^2) dx = pi/2
(after all, the integral describes the are of half the unit circle)

Next we take y1(x) = x, and find
= INT sqrt(1-x^2) x dx = 0
because the integrand is an odd function. So y0 and y1 are already orthogonal. By the way,

= INT sqrt(1-x^2) x^2 dx
... = INT (cos^2 a) (sin^2 a) da
(with x = sin a, a from -pi/2 to +pi/2)
... = 1/4 INT sin^2 (2a) da
... = 1/8 INT [1 - 1/8 cos (4a)] da
... = pi/8

Now y2(x) = x^2. We find

= INT sqrt(1 - x^2) x^3 = 0
because the integrand is an odd function.

= INT sqrt(1 - x^2) x^2 = pi/8
because the integral is the same as above.

Therefore, we must replace y2 by the linear combination y2 - a y0:

= 0
- a = 0
pi/8 - a * pi/2 = 0
a = 1/4

so we choose
y2'(x) = x^2 - 1/4

The orthogonal system is therefore
y0(x) = 1
y1(x) = x
y2(x) = x^2 - 1/4

Answer 2

dammit, now i got a headache ! thanks a lot !