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Showing functions form a basis

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that the functions po(t)=1, p1(t)=t, p2(t)=1/2(3t2-1), and p3(t)=(3/2)*[(5/3)t3-t) also form a basis for the vector space P3(R) ... "R" meaning all real numbers

    2. Relevant equations

    I know these polynomials are the first four Legendre polynomials


    3. The attempt at a solution

    I know that proving functions form a basis involves proving that each funciton has a unique representation as a linear combination... I'm not certain on what this means exactly but I'm trying my best to figure it out... heres my work so far on the problem

    a(1) + b(t) + c/2(3t2-1) + d/2(5t3-3t) = a + bt + ct2 + dt3

    at t=0 ; -c/2 = 0 ; so c=0

    a + b(t) + d/2(5t3-3t)=0
    at t=0 ; a=0
    at t=1 ; b=-d
    at t=2 ; -2d + d/2(34) = 15d = 0 ; d=0 ; b=0

    so the legendre polynomials are a linearly independent set at; 1/2 + t - 3/2t + 3/2t2 +5/3t3 = 1/2-1/2t+3/2t2+5/3t3
    ; a polynomial in the span of P(R) that forms a basis

    thanks in advance for any advice you have to offer me :smile:
     
  2. jcsd
  3. Jan 28, 2010 #2

    vela

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    To show linear independence, you want to set the righthand side to zero and show that requires a=b=c=d=0.

    You also need to show the four polynomials span P3(R). This time, you want the righthand side to be an arbitrary 3rd-degree polynomial -- so don't use a, b, c, and d that appear on the lefthand side on the righthand side -- and show you can solve for a, b, c, and d.
     
  4. Jan 28, 2010 #3

    Dick

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    Ok, so you want to show a+b*t+c*t^2+d*t^3 has a unique representation of the form A*p0(t)+B*p1(t)+C*p2(t)+D*p3(t). That means you just want to find A, B, C, and D in terms of a, b, c and d, right? You really don't have to put t values in. You know a+b*t+c*t^2+d*t^3=a'+b'*t+c'*t^2+d'*t^3 only if a=a', b=b', c=c' and d=d', yes? So just start at the top. What is the relation between d and D?
     
  5. Jan 29, 2010 #4
    is it D = dt2/[(5/2)t2-(3/2)]
     
  6. Jan 29, 2010 #5

    Dick

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    Nooo. D is supposed to be constant. Expand the Legendre side and equate the cubic terms.
     
  7. Jan 29, 2010 #6

    HallsofIvy

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    Since [itex]\{ 1, x, x^2\}[/itex] form a basis for P3, you know that it has dimension 3. That, in turn, tells you that any independent set of three functions will be a basis. After you have shown that the given functions are independent, you don't have to show that they span the space.
     
    Last edited: Jan 30, 2010
  8. Jan 29, 2010 #7
    ooo alright, thanks for all the helpful explanations =D
     
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