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Math Help - Legendre proofs

  1. #1
    Super Member Deadstar's Avatar
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    Legendre proofs

    I have two Legendre questions, can anyone help me with either of them?
    Suppose 0 \le k < n. Let p be a polynomial. Show that

    (\frac{d}{dx})^k ((x^2 -1)^n p(x)) = (x^2 -1)^{n-k} q(x)

    for some polynomial q.

    Taking p(x) = 1 this implies

    (\frac{d}{dx})^k ((x^2 -1)^n) = (x^2 -1) q(x)

    for some polynomial q.

    and the other one is...

    Use integration by parts to show that

    (L_n , x^k) = 0

    for all 0 \le k < n. Deduce that (L_0, L_1, ... ,L_n) is an orthogonal basis of P_n.

    L_n may be L_n(x) = \frac{(2n)!}{{2^n}(n!)^2}x^n +... or it may be L_n(x) = \frac{1}{2^n n!}(\frac{d}{dx})^n ((x^2 - 1)^n)... Im not sure really which is why im stuck! Both are used in my handout
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    I have two Legendre questions, can anyone help me with either of them?
    Suppose 0 \le k < n. Let p be a polynomial. Show that

    (\frac{d}{dx})^k ((x^2 -1)^n p(x)) = (x^2 -1)^{n-k} q(x)

    for some polynomial q.

    [snip]
    Are you allowed to use proof by induction? Or do you have to derive (ha ha) the result?
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  3. #3
    Super Member Deadstar's Avatar
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    i thought about that as they had the p(x) = 1 example so thought that would be the base case. I suppose you could yeah, it doesnt specify that you have to derive it.
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  4. #4
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    Quote Originally Posted by Deadstar View Post
    I have two Legendre questions, can anyone help me with either of them?
    Suppose 0 \le k < n. Let p be a polynomial. Show that

    (\frac{d}{dx})^k ((x^2 -1)^n p(x)) = (x^2 -1)^{n-k} q(x)

    for some polynomial q.

    Taking p(x) = 1 this implies

    (\frac{d}{dx})^k ((x^2 -1)^n) = (x^2 -1) q(x)

    for some polynomial q.

    and the other one is...

    Use integration by parts to show that

    (L_n , x^k) = 0 Mr F says: The integration is over the interval (-1, 1).

    for all 0 \le k < n. Deduce that (L_0, L_1, ... ,L_n) is an orthogonal basis of P_n. Mr F says: On the interval (-1, 1).

    L_n may be L_n(x) = \frac{(2n)!}{{2^n}(n!)^2}x^n +... or it may be L_n(x) = \frac{1}{2^n n!}(\frac{d}{dx})^n ((x^2 - 1)^n)... Im not sure really which is why im stuck! Both are used in my handout

    Mr F says: They are equivalent expressions. The former is the series solution to Legendre's Equation. The latter is Rodrigues' Formula.
    ..
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  5. #5
    Super Member Deadstar's Avatar
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    right so what im integrating is;

    \int_{-1}^{1} \frac{(2n)!}{2^n(n!)^2} x^n x^k dx

    can the \frac{(2n)!}{2^n(n!)^2} be taken outside the integral as a constant? My guess is no which means im nowhere new. I already knew the integral was between 1 and -1, I just dont know how to integrate that. Do I use integration by parts? Substitution?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Deadstar View Post
    right so what im integrating is;

    \int_{-1}^{1} \frac{(2n)!}{2^n(n!)^2} x^n x^k dx

    can the \frac{(2n)!}{2^n(n!)^2} be taken outside the integral as a constant? My guess is no which means im nowhere new. I already knew the integral was between 1 and -1, I just dont know how to integrate that. Do I use integration by parts? Substitution?
    I'm not sure what you are doing or if it is right but:

    \int_{-1}^{1} \frac{(2n)!}{2^n(n!)^2} x^n x^k dx=<br />
\frac{(2n)!}{2^n(n!)^2}\int_{-1}^{1}  x^{n+k} dx

    RonL
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  7. #7
    Super Member Deadstar's Avatar
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    yeah that was where i'd got too. should be able to solve from there, thanks.
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