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I am trying to prove the following property of Legendre polynomials. $$nP_n(x)=x{P_n^\prime(x)} - P^\prime_{n-1}(x)$$ My guess is that I somehow have to use the Bonnets recursion formula $$(n+1)P_{n+1}(x)=(2n+1)xP_n(x)-nP_{n-1}(x)$$ which is proved using the generating function of the legendre polynomials. However, I am not being able to eliminate the derivative of $P_{n+1}$ from this formula. I am not being able to do problems of a similiar nature, of recursion relations between the derivative of legendre polynomials.

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2 Answers 2

up vote 2 down vote accepted

If $f(x,t)=\sum P_n(x)t^n=(1-2tx+t^2)^{-1/2}$ is the generating function then you want to show that $$t\frac{\partial}{\partial t}f=x\frac{\partial}{\partial x}f-t\frac{\partial}{\partial x}f$$ and it is true: $$(t\frac{\partial}{\partial t}+(t-x)\frac{\partial}{\partial x})(1-2tx+t^2)=0.$$

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Nice answer. Waiting for alternatives. –  ramanujan_dirac Oct 2 '12 at 12:05

I found another proof, using Legendre's equation:

$$\frac{d}{dt}\left[(1-t^2)P'_n \right] +n(n+1)P_n=0$$

Assuming: $P'_n=\frac{nP_{n-1}-ntP_n}{1-t^2}$

$$\frac{d}{dt}\left(nP_{n-1}-ntP_n\right) +n(n+1)P_n=0$$

You can check that it will reduce to your desired equation.


Using your proposed method:

$$tP_{n}^{\prime}-P_{n-1}^{\prime}\overset{(1)}{=}t\frac{n}{t^{2}-1}(tP_{n}-P_{n-1})-\frac{(n-1)}{t^{2}-1}(tP_{n-1}-P_{n-2}) =\frac{nt^{2}P_{n}-ntP_{n-1}-(n-1)tP_{n-1}+(n-1)P_{n-2}}{t^{2}-1}=\frac{nt^{2}P_{n}+(1-2n)tP_{n-1}+(n-1)P_{n-2}}{t^{2}-1}\overset{(2)}{=} \frac{nt^{2}P_{n}-nP_{n}}{t^{2}-1}=\frac{n(t^{2}-1)P_{n}}{t^{2}-1}=nP_{n}$$

(1): Using $P_{n}^{\prime}=\frac{n}{t^{2}-1}(tP_{n}-P_{n-1})$

(2): Using $(n+1)P_{n+1}=(2n+1)tP_{n}-nP_{n-1}$ we can get:

$nP_{n}=(2n-1)tP_{n-1}-(n-1)P_{n-2}\implies-nP_{n}=(1-2n)tP_{n-1}+(n-1)P_{n-2}$

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