In a homework question, we are asked to show that the Legendre polynomials do indeed solve the Legendre Differential Equation: $$ \frac{d}{dx} \left[ (1 - x^2) \frac{d}{dx} P_n(x) \right] + n (n + 1) P_n(x) = 0. $$ According to Wikipedia, it is sufficient to show that after deriving it n+1 times the following equation must hold true: $$ (x^2 - 1) \frac{d}{dx} (x^2 - 1)^n = 2 n x (x^2 - 1)^n. $$ This was also given as a hint in the problem. Since I don't want to blindly accept the hint, I was trying to figure out how it was deduced, but didn't manage to. My question therefore is, how do we arrive at the equation above (the "hint equation")?

I can see, that the left part of the equation is nearly equal to the first part of the Legendre Differential Equation, except for a missing outer derivative. However, I am somehow missing the steps taken to arrive at the right side: Where does the $2nx$ come from?

  • $\begingroup$ What is your definition of the Legendre polynomials? I'm assuming using Rodrigue's formula? To prove the Hint equation, try using induction. $\endgroup$ – Raghav Nov 26 '13 at 21:44

It is the other way around: if you wish to prove Rodrigues' formula $$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$ where $f^{(n)}$ is symbolic for the $n$th derivative of $f$, then the first step is to prove that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the differential equation. When $$[(1-x^2)y']' + n(n+1)y = 0$$ then obviously also $$(1-x^2)y'' - 2xy' + n(n+1)y = 0$$ So we wish to somehow evaluate $$(1-x^2)[(x^2-1)^n]^{(n+2)} - 2x[(x^2-1)^n]^{(n+1)}$$ Now starting with $$(x^2-1) \cdot [(x^2-1)^n]^{(1)} = (x^2-1) \cdot n(x^2-1)^{(n-1)} \cdot 2x = 2nx(x^2-1)^{n}$$ and taking left and right the $(n+1)$st derivative, we find that $$(x^2-1)[(x^2-1)^n]^{(n+2)} + \binom{n+1}{1}\cdot 2x \cdot [(x^2-1)^n]^{(n+1)} + \binom{n+1}{2}\cdot 2 \cdot [(x^2-1)^n]^{(n)} \\ = 2nx \cdot [(x^2-1)^n]^{(n+1)} + \binom{n+1}{1}\cdot 2n \cdot [(x^2-1)^n]^{(n)}$$ or $$(x^2-1)[(x^2-1)^n]^{(n+2)} + (2x(n+1)-2nx)[(x^2-1)^n]^{(n+1)} + (n(n+1) - 2n(n+1)) [(x^2-1)^n]^{(n)} = 0$$ which leads to $$(x^2-1)F''(x) + 2xF'(x) - n(n+1)F(x) = 0$$ or, equivalently $$[(1-x^2)F'(x)]' + n(n+1)F(x) = 0$$ which shows that $F(x) = [(x^2-1)^n]^{(n)}$ is indeed a solution of the Legendre differential equation. Normalizing $P_n = \alpha \cdot F$ by requiring $P_n(1) = 1$, we then find the factor $\alpha = 1/2^nn!$


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