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Legendre polynomial problem

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    The Legendre polynomials [tex]P_l(x)[/tex] are a set of real polynomials orthogonal in the interval [tex]-1< x <1[/tex] , [tex] l\neq l' [/tex]

    [tex]\int dx P_l(x)P_l'(x)=0, -1<x<1[/tex] The polynomial [tex]P_l(x)[/tex] is of order l
    , that is, the highest power of x is [tex]x^l[/tex]. It is normalized to [tex]P_l(x)=1 [/tex]

    Starting with the set of functions ,

    [tex] \varphi_l(x)=x^l, l=0,1,2,...,[/tex] used the orthogonalization procedure to derive the polynomials [tex] P_0,P_1,P_2, and P_3[/tex]
    2. Relevant equations




    3. The attempt at a solution


    I have no idea what my book (Peebles) means by orthogonalization procedure. But I looked at Griffifth book on QM , and perhaps they are talking about Rodrigues formula on p. 136 eqn. 4.28?
     
  2. jcsd
  3. Sep 22, 2009 #2
  4. Sep 22, 2009 #3
    Still a little confused .so for l= 0 , would look like this: [tex]|e_0>=e_0/||e_0||[/tex]
     
  5. Sep 22, 2009 #4
    P_n(x) are the original functions, the orthonormal functions are denoted by R_n(x). Normalize the first:

    R_0(x) = P_0(x)/sqrt[<P_0|P_0>]


    Take the second function, and subtract the component in the direction of R_0(x):

    Q_1(x) = P_1(x) - <P_1|R_0> R_0(x)

    The inner product of Q_1 with R_0 is clearly zero. Normalize Q_1 by dividing it by the square root of its inner product with itself to get R_1. Then you compute Q_2 by subtracting from P_2 the components in the direction of

    Q_2(x) = P_2(x) - <P_2|R_1> R_1(x) - <P_2|R_0> R_0(x)

    Normalize Q_2 to obtain R_2.

    Then when you're done, you renormalize the R_n to make the coefficients of the highest powers equal to 1.
     
  6. Sep 22, 2009 #5
    sorry , this might take me a while to absorb in:


    [tex]R_0=P_0/(P_0,P_0),(P_0,P_0)=1?[/tex]; [tex]Q_2=P_2(x)[/tex] since [tex]<P_2|R_1>=0, <P_2|R_0>=0[/tex]

    [tex]\varphi_0=1,\varphi_1=x,\varphi_2=x^2[/tex]; When do all 3 phi's and the integral expression come into play? How do I obtain [tex]P_0, P_1,P_2[/tex]
     
  7. Sep 23, 2009 #6
    I think i got it! [tex]P_0=c_0,P_1=d_1*x+d_0,P_2=e_2*x^2+e_1*x+e_0,P_3=f_3*x^3+f_2*x^2+f_1*x+f_0,P_0(1)=1=>c_0=1,P_1(1)=1=d_1*x+d_0,P_2(1)=1=e_2*x^2+e_1*x+e_0,P_3(1)=1=f_3*x^3+f_2*x^2+f_1*x+f_0[/tex]Using integration I need to show that:

    [tex]\int P_0*P_1 dx=0[/tex] ,[tex]\intP_0*P_2 dx= 0[/tex], [tex]\int P_0*P_3=0[/tex],[tex]\int P_1*P_2=[/tex],[tex]\int P_1*P_3=0[/tex],[tex]\int P_2*P_3=0[/tex] all on interval -1<x<1[/tex](look at my latex code, output might not my input accurately)

    calculated [tex]c_o[/tex] ,not sure how to calculate [tex]d_1,d_0,e_0,e_1,e_2,f_0,f_1,f_2,f_3[/tex] I think all of the terms will be equal to zero or one though.
     
    Last edited: Sep 23, 2009
  8. Sep 23, 2009 #7
    hard time reading my solution again
     
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