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It's known that the Legendre Polynomials follow the recursion:

$$P_n(x)=\frac{2n-1}{n}xP_{n-1}(x)-\frac{n-1}{n}P_{n-2}(x)$$

with $$P_0(x) = 1, P_1(x)=x$$

Now I am finding an elementary method to prove the orthogonality of the series only based on the recursion.

By the induction I can get that the polynomial is even when n is even, and odd when n is odd.

Now the only problem is that I can't verify that the inner product of $P_n$ and $P_{n-2}$ is zero. Is there any simple method which can prove it only by the recursion without using the theory of differential equation?

Any hint will be helpful.

Thank you in advance!

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  • $\begingroup$ @coffeemath $[ -1,1 ]$ is the interval. See en.wikipedia.org/wiki/… $\endgroup$
    – Donald Splutterwit
    Commented Apr 11 at 9:59
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    $\begingroup$ It's my fault. The inner product is $\int_{-1}^{1}f(x)g(x)dx$. $\endgroup$
    – Xinhan Yuan
    Commented Apr 11 at 10:06
  • $\begingroup$ I'm sorry to give a wrong recursion because of the typo of my book and I correct it now. $\endgroup$
    – Xinhan Yuan
    Commented Apr 11 at 11:45
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    $\begingroup$ The following is my idea: To verify that $<Pn, P_{n-2}>=0$, according to the Schmit Orthogonalization, I need to prove $||P_n||=\frac{2}{2n+1}$ or $\frac{<P_n,P_n>}{<P_{n-1},P_{n-1}}=\frac{2n-1}{2n+1}$ But in the induction, when I deal with the norm, I need to use the result that the inner product of $P_{n+1}$ and $P_{n-1}$ is zero, which hasn't been proved in this step. That's what blocks me. $\endgroup$
    – Xinhan Yuan
    Commented Apr 11 at 11:56

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