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I have this Legendre polynomial:

$p_k(x) = \frac{1}{2^k k!} \frac{d^k}{dx^k} ((x^2 - 1)^k) $

I have calculated that: $p_0 = 1, p_1 = x, p_2 = \frac{1}{3}(3x^2-1), p_3 = \frac{1}{2}(5x^3-3x) $

Now I have to show that $<p_2,p_1> \; = 0$.

I have then integrated $p_2 $ and $ p_1$ from [-1,1] and get 0 and 0. And <0,0> = 0 is true.

My question is, am I on the right track? Is there 'more' I have to show?

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1 Answer 1

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You got the definition of $\langle p_2, p_1 \rangle$ wrong. $\langle p_2, p_1 \rangle$ is not $(\int p_1(x)dx )(\int p_2(x)dx)$.

$\langle p_2, p_1 \rangle=\int_{-1}^{1}p_1(x)p_2(x)dx=\int_{-1}^{1} \frac x 3 (3x^{2}-1)dx= 0 -0=0$.

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  • $\begingroup$ Thank you for your reply. But this gives me <p2,p1> = 1/2. Shouldn't it give me 0? $\endgroup$
    – bestmate21
    Commented Apr 5, 2021 at 12:17
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    $\begingroup$ $\langle p_2, p_1 \rangle=0$ - the integrand is antisymmetric over $x\to-x$ $\endgroup$
    – Svyatoslav
    Commented Apr 5, 2021 at 12:39

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