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Showing orthogonal polynomials are unique

  1. Sep 17, 2012 #1
    1. The problem statement, all variables and given/known data

    We are given that a set of polynomials on [-1,1] have the following properties and have to show they are unique by induction. I have a way to show they are unique, but is not what he is looking for. I honestly have never seen it presented this way.

    P_n(x) = Ʃa_in*x^i
    All P_n are mutually orthogonal
    P_n(1) = 1

    I know they are Legendre polynomials and I've derived them using Gram-Schmidt, but not sure what to try here.

    We have to find P_0 and P_1 which are 0 and x respectively and then assume there are some P_n and P_n* that both satisfy the definition above and we show they are identical.

    2. Relevant equations

    See above.

    3. The attempt at a solution

    His only hint was that through showing P_n is unique, you would need P_n-1 and should use induction. All I've ever done is take a difference, you get a polynomial of lower degree and after a bit of algebra, you get that the difference in the two must be zero so they are identical. That does only apply to monic polynomials, but it is easy to redefine P_n so that is monic.
     
  2. jcsd
  3. Sep 17, 2012 #2

    Dick

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    You know P_0(x) is unique, because it's just a constant so P_0(x)=1 since P_0(1)=1. Now think about P_1(x). It's a linear polynomial. The space of linear polynomials is 2 dimensional and P_1(x) lies in the linear subspace of polynomials orthogonal to P_0(x). So that subspace must have dimension 1. Since you also know P_1(0)=1, that gives you a unique choice for P_1(x). Can you see how to extend this to an induction argument for all n? It's pretty much Gram-Schmidt without the gory details.
     
    Last edited: Sep 17, 2012
  4. Sep 18, 2012 #3
    Hmmm, that basically sounds like finding the formula which he said wasn't necessary, but it was the only idea I had to begin with. He just made it sound like Gram-Schmidt wasn't necessary. We have to some how show that a P_n and P_n* are identical as well and just finding the formula doesn't do that.

    Some how, through showing uniqueness, this also can be used to show that legendre polynomials are even when n is even and odd when n is odd. Really the only thing I have to go on so guess that'll be it. His hints are rather vague. Thanks.
     
    Last edited: Sep 18, 2012
  5. Sep 18, 2012 #4
    So I was able to get a little more info and I think it is enough for me to show uniqueness, but not by induction like I'm supposed to, unless it is induction and I'm not seeing it. Also, this still doesn't get me to the point that P_n is even if n is even, P_n is odd if n is odd.

    We create a system of n+1 unknowns and n+1 equations. We know the sum of the coefficients is all 1 by the fact that P(1)=1. Since P_n is an orthogonal set, we also know that

    Ʃa_{in}*<x^n,P_m> = 0 for m = 0,1,...,n-1

    That is all fine and dandy, create your system, you know the rows are linearly independent of eachother based on the fact that <x^n,P_m>=0 if m > n and <x^n,P_m>≠0. So I know there is a unique solution (or it appears that way). Now, if I plug 1 and -1 into the matrix for x, all that happens is the signs on every other row change. Maybe there is something in matrix theory that I don't remember, but not sure how that shows P_n is even/odd for n even/odd.
     
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