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Using the recurrence relation $$(n+1)P_{n+1}=x(2n+1)P_n(x)-nP_{n-1}(x) \ \ n\geq 1,$$ I've calculated the first four Legendre Polynomials as \begin{align} P_0(x)&=1 \\ P_1(x)&=x \\ P_2(x)&=\frac{3x^2}{2}-\frac{1}{2} \\ P_3(x)&=\frac{5x^3}{2}-\frac{3x}{2}. \end{align}

My question is, if I can show by direct integration (integral over the domain $-1<x<1$ is zero) that $P_0$ is orthogonal to $P_1$ and $P_0$ is orthogonal to $P_2$, does this then imply that $P_1$ is also orthogonal to $P_2$?

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    $\begingroup$ Not in general: On $[-1,1]$, $x$ is orthogonal to $1$ and $x^2$ but $1$ is not orthogonal to $x^2$. This suggests that you probably need to use some other properties of $P_n$. $\endgroup$ – Chee Han Mar 12 '19 at 7:20
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    $\begingroup$ Let me put it this way: if we show that $P_0$ is orthogonal to $P_1$, and also that $P_0$ is orthogonal to $2P_1$, does this imply that $P_1$ is orthogonal to $2P_1$? $\endgroup$ – Ivan Neretin Mar 12 '19 at 7:25
  • $\begingroup$ I see your point. $\endgroup$ – user557493 Mar 12 '19 at 7:38
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No way! You would have to check for each individual combination.

It works just as with vectors on a 2D plane. (1,0) is ortogonal to (0,1), so is (0,1) to (-1,0). However, (1,0) and (-1, 0) are as far of ortogonal as you can possibly get.

If your claim were true, then the Legendre polynomials would be $\{1, x, x^2, x^3\}$ and so on, there would be no need for overcomplicating things

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