1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Least squares and integration problem

  1. Nov 26, 2005 #1
    Question states
    Consider the vector space C[-1,1] with an inner product defined by
    <f,g> = the integral from 1 to -1 of f(x)g(x) dx
    a)
    Show that
    u1(x)= 1/(2^.5) u2(x)= ((6^.5)/2)x
    form an orthonormal set of vectors
    b)
    Use the result from a) to find the best least squates approximation to
    h(x)= x^(1/3) + x^(2/3)
    by a linear function.
    For part a) I have shown that u1 and u2 each have an inner product of zero and a length of one.
    I've been trying to find a solution to b) in the form
    p(x) = (c1)(u1(x)) + (c2)(u2(x))
    where ci = the integral from -1 to 1 of (ui(x))h(x)
    evaluating this integral for c1 produces
    (1/(2^.5))[(3/4)x^(4/3) + (3/5)x^(5/3)] evaluated from -1 to 1
    Now I've finally got to the problem
    Putting -1 in would produce complex numbers and I don't know how to proceed.
    I'm not sure weather the problem here is that I'm not aproaching the least squares problem correctly or weather I'm not approaching the integral correctly. Any help would be appreciated, thanks.
     
  2. jcsd
  3. Nov 26, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Putting -1 in for x? That doesn't produce complex numbers. The cube root of -1 is -1. Its value at x= 1 is [itex]\frac{1}{\sqrt{2}}\left(\frac{3}{4}+\frac{3}{5}\right)[/itex] and its value at x= -1 is [itex]\frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{3}{5}\right)[/itex] Subtracting gives just [itex]\frac{1}{\sqrt{2}}\frac{6}{5}[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Least squares and integration problem
  1. Least squares problem (Replies: 4)

Loading...