2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
7
$\begingroup$

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 $\alpha = 1/2^nn!$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.