# Orthogonality of the First Four Legendre Polynomials

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 $$-1 is zero) that $$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$$?

• 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 $P_n$. – Chee Han Mar 12 '19 at 7:20
• 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 $2P_1$? – Ivan Neretin Mar 12 '19 at 7:25
• I see your point. – user557493 Mar 12 '19 at 7:38

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