Im looking at the Legendre Polynomials and trying to recreate the first few terms
$${d \over dx} \left[ (1-x^2) {d \over dx} P_n(x) \right] + n(n+1)P_n(x) = 0.$$
What exactly is P_n(x)? and could you give an example of how you would produce the Legendre polynomial if we set n to 3 or 4?
Also, maybe a separate question but does $nP_n$ make any sense as an idividual statement in this context?
This is the differential equation the Legendre polynomials satistky; you may use the recursive relation, known as Bonnet's recursion formula, in order to recreate them:
$(n+1)P_{n+1}(x) = (2n+1)xP_n(x) - nP_{n-1}(x)$
You can use this recursion in order to get them, as long as you get the first 2 terms: you can find them using explicitly the differential equation:
For $n=0$, the equation becomes $\frac{d}{dx}\Big[ (1-x^2) \frac{d}{dx}P_0(x) \Big] = 0$ This clearly has $P_0(x)=$ constant as a solution, and if you use the normalization
$P_n(1)=1$ for every $n$
you get $P_0(x)=1$.
For $n=1$, the equation becomes $\frac{d}{dx}\Big[ (1-x^2) \frac{d}{dx}P_1(x) \Big] +2P_1(x)= 0$. This can seem a little messy: let's just try polynomials of the simplest form (non-constant), that is, of the form $Ax+B$ (linear); we get
$-2Ax + 2Ax + 2B=0$
which clearly tells us that the polynomial $P_1(x) = Ax$ is a solution; using again the normalization condition, we get $P_1(x) = x$.
From now on, is trivial to get the other ones, using the above recursion formula: for example for the 2nd polynomial we have
$2P_2(x) = 3xP_1(x)-P_0(x) = 3x^2 - 1$