Method of Least Squares question

[clope023]

Homework Statement

for vector space C[-1,1] with L^2 inner product

<f,g> = \intf(x)g(x)dx

find the best least squares approximation for function x^(1/3) on [-1,1] by a quadratic function q(x) = c0 + c1x + c2x^2

Homework Equations

s+r = n

<t^s, t^r> = \intt^ndt = { 2/(n+1) if n is even
0 if n is odd }

The Attempt at a Solution

q(x) = c0*1 + c1*x + c2*x^2

take inner product of functions of q(x)

||1|| = sqrt(2)
||x|| = sqrt(2/3)
||x^2|| = sqrt(2/5)

normalize vectors in the basis

\hat{u1} = 1/sqrt(2)
\hat{u2} = x/sqrt(2/3)
\hat{u3} = x^2/sqrt(2/5)

find coefficients by taking integrals of unit vectors with function x^1/3

c1 = (1/sqrt(2))\intx^1/3dx = \stackrel{3}{4sqrt(2)}

c2 = (1/sqrt(2/3))\intx^4/3dx = \stackrel{3}{7sqrt(2/3)}

c3 = (1/sqrt(2/5))\intx^7/3dx = \stackrel{3}{10sqrt(2/5)}

therefore p(x) = c1\hat{u1} + c2\hat{u2} + c3\hat{u3}

just wanting to confirm my answer, thanks for any and all help anyone can give and I'll write back this time, lol

 

[lanedance]

are your basis functions orthogonal? i think that might help...

 

[clope023]

are your basis functions orthogonal? i think that might help...

I did test for that, 1 and x were orthogonal, x and x^2 were orthogonal, however 1 and x^2 were not orthogonal and neither were any of the functions with each other, to try and remidy this I changed the functions such that I added a variable to make them orthogonal or make their integral equal to zero, so

\int1dx = \intadx = 0

\intx^2dx = \int(x-a)^2dx = 0

\intx^2x^2dx = \intx^4dx = \int(x-a)^4dx = 0

however many of my solutions except the first one turned into some horrible monster with complex numbers and wasn't so sure that was correct, was my reasoning correct to do this?

 

[lanedance]

or the least squares method to miinimise the error, I'm pretty sure your functions need to be orthonormal (actualy orthogonal, but as the normlaisation helps), the way to do it is thorugh gram schimdt type process

so for the zeroth order function, pick the most general constant
f_0(x) = a
test nomalisation
\int_{-1}^{1} dx (f_0(x))^2 = 2a^2
f_0(x) = \frac{1}{\sqrt{2}}

and again for the next, f1
f_1(x) = b + cx
test orthognality
\int_{-1}^{1} dx (f_0(x).f_1(x)) <br /> = \int_{-1}^{1} dx \frac{1}{\sqrt{2}} (b+cx) <br /> = \frac{1}{\sqrt{2}}(bx+cx^2) _{-1}^{1}<br /> = \frac{1}{\sqrt{2}}(b(1-(-1)) +c(1^2-(-1)^2)) <br /> = \frac{1}{\sqrt{2}}(2b) = 0 <br />

hence b = 0 (as you found), then do the normalisation for c, and for the last, start from
d + ex +fx^2