Legendre Polynomial (anti)symmetry proof

[Tetrinity]

Homework Statement

Let P_{n}(x) denote the Legendre polynomial of degree n, n = 0, 1, 2, ... . Using the formula for the generating function for the sequence of Legendre polynomials, show that:

P_{n}(-x) = (-1)^{n}P_{n}(x)

for any x \in [-1, 1], n = 0, 1, 2, ... .

Homework Equations

Generating function for the sequence of Legendre polynomials:

\sum P_{n}(x)r^{n} = (1 - 2rx + r^{2})^{-\frac{1}{2}}

The Attempt at a Solution

I guess I don't really know where to begin with this. I tried differentiating both sides of the generating function with respect to r to obtain:

\sum nP_{n}(x)r^{n-1} = (x - r)\sum P_{n}(x)r^{n}

but I don't see how this might move closer to the desired result. I imagine differentiating with respect to x would lead to similar difficulties.


I then tried just directly substituting -x into the generating function, giving

(1 + 2rx + r^{2})^{-\frac{1}{2}}

as a generating function for P_{n}(-x), but again I don't really see where this might lead me.

Perhaps there's a general method I'm missing; I'm still very new to the concept of Legendre polynomials and generating functions in general. My attempts at scouring the Internet for a proof of this specific identity have turned up a sole reference to a textbook I have no access to. I'm not really looking for a complete proof to be given to me on a plate, but a nudge in the right direction would certainly be appreciated.

 

[Dick]

Try substituting both -x for x and -r for r into the generating function expression. What does that tell you?

 

[Tetrinity]

Ahh, figures it'd be something as simple as that.

\sum P_{n}(-x)(-r)^{n} = (1 + 2rx + r^{2})^{\frac{1}{2}} = \sum P_{n}(x)r^{n}

\Rightarrow \sum P_{n}(x)r^{n} = \sum P_{n}(-x)(-1)^{n}r^{n}

and then by applying the identity theorem for power series we get P_{n}(x) = (-1)^{n}P_{n}(x) as required.

Thanks for your help. :)