Orthonormal Set spanning the subspace (polynomials)

Cassi
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Homework Statement


In the linear space of all real polynomials with inner product (x, y) = integral (0 to 1)(x(t)y(t))dt, let xn(t) = tn for n = 0, 1, 2,... Prove that the functions y0(t) = 1, y1(t) = sqrt(3)(2t-1), and y2 = sqrt(5)(6t2-6t+1) form an orthonormal set spanning the same subspace as {x0, x1, x2}.

Homework Equations

The Attempt at a Solution


I was attempting to use the Legendre Polynomials Rules to show that these polynomials form the basis but when I devise the Legendre Polynomials, they are different than those given.
 
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Cassi said:

Homework Statement


In the linear space of all real polynomials with inner product (x, y) = integral (0 to 1)(x(t)y(t))dt, let xn(t) = tn for n = 0, 1, 2,... Prove that the functions y0(t) = 1, y1(t) = sqrt(3)(2t-1), and y2 = sqrt(5)(6t2-6t+1) form an orthonormal set spanning the same subspace as {x0, x1, x2}.

Homework Equations

The Attempt at a Solution


I was attempting to use the Legendre Polynomials Rules to show that these polynomials form the basis but when I devise the Legendre Polynomials, they are different than those given.

What's stopping you? Can you prove they are orthogonal? Orthonormal? Span the same space?
 
What are the "Legendre Polynomials Rules"?
 
So x_0= 1, x_1= t, and x_2= t^2. What subspace do those span?

y_0(t) = 1, y_1(t) = \sqrt{3}(2t-1), and y_2 = \sqrt{5}(6t^2-6t+1). Show that these span the same subspace as the above. To show that they are orthonormal (which would also show that they are independent) you need to do 6 integrals.

For example, \int_0^1 (y_0(t))^2 dt must be equal to 1 while \int_0^1 y_0y_1 dt must be equal to 0.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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