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Homework Help: Linear Algebra - Finding coordinates of a set

  1. Feb 20, 2017 #1
    1. The problem statement, all variables and given/known data

    Find the coordinates of each member of set S relative to B.
    B = {1, cos(x), cos2(x), cos3(x), cos4(x), cos5(x)}
    S = {1, cos(x), cos(2x), cos(3x), cos(4x), cos(5x)}

    I am to do this using Mathematica software. Each spanning equation will need to be sampled at six separate points to create a 6x6 system that solves for the coordinates of each element of S.

    2. Relevant equations

    Ax = b
    If I was given a simpler problem, such as "Find the coordinate vector of w = (1, 0) relative to the basis S = {u, v}, u = (2, 2) and v = (1, 1)" my equation would start:

    (1, 0) = k1(2, 2) + k2(1, 1)

    and I'd solve for k1 and k2

    3. The attempt at a solution

    I started with the equation S = kB, where k is a constant, but I was told to build a 6x6 matrix by sampling each member of B at six points, so I knew that's wasn't right. I built a 6x6 matrix from B with columns 1-6 filled with evaluations of B with x = {0, pi/6, pi/4, pi/3, pi/2, pi}, but I don't think that's right either, because I didn't create any spanning equations. I tried using a matrix {{1,0,0,0,,00},{0,cos(x),0,0,0,0}...{0,0,0,0,0,cos5(x)}}, to test for span, but that didn't feel right either. How do I create spanning equations to sample to build the matrix?
  2. jcsd
  3. Feb 20, 2017 #2


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    Homework Helper

    Given that [tex]
    \cos (n \pm 1)x = \cos nx \cos x \mp \sin nx \sin x[/tex] we obtain [tex]
    \cos (n+1)x = 2\cos x \cos nx - \cos(n-1)x.[/tex] From this recurrence relation you can express [itex]\cos nx[/itex] as a polynomial in [itex]\cos x[/itex], which gives you a matrix equation [tex]
    (1, \cos x, \dots, \cos 5x)^{T} = A (1, \cos x, \dots, \cos^5 x)[/tex] and your problem is to invert [itex]A[/itex].
  4. Feb 20, 2017 #3


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    Homework Helper

    So, what you actually need to do is write all the elements of S as linear combinations of elements of B. For example:

    ##1 = 1*1 + 0*cos(x) + 0*cos^2(x) + 0*cos^3(x) + 0*cos^4(x) + 0*cos^5(x)##

    Thus, the coordinates of ##1## relative to ##B## is ##(1,0,0,0,0,0)##.

    Another example:

    ##cos(2x) = cos^2(x) - sin^2(x) = cos^2(x) - (1 - cos^2(x)) = -1 + 2cos^2(x)##

    Thus, the coordinates of ##cos(2x)## relative to ##B## is ##(-1,0,2,0,0,0)##

    Now, try the rest of the exercise.

    EDIT: I didn't read carefully about the mathematica part. You can ignore this post.
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