# Legendre Polynomial (anti)symmetry proof

1. Mar 5, 2012

### Tetrinity

1. The problem statement, all variables and given/known data

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, ... .

2. Relevant equations

Generating function for the sequence of Legendre polynomials:

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

3. 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.

2. Mar 5, 2012

### Dick

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

3. Mar 5, 2012

### 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.