Legendre Equation Properties Is there a nice way to derive, starting from the Legendre differential equation, the generating function, the recurrence relation, the Rodrigues differential form & the Schlafli integral form without memorizing crazy generating functions or stuff that magically works? Every time I try to study this stuff I'm given formula's that are supposed to work, how do you derive all this from scratch?
 A: For deriving Rodrigues' form, see my post at (Proof: Legendre Polynomials Solving the Corresponding Differential Equation).
For deriving the recurrence relation
$$(n+1)P_{n+1} = (2n+1)xP_n - nP_{n-1}$$
an elegant proof makes use of some special characteristics of the polynomials $P_n$.
First of all, it is not difficult to see that each polynomial of degree m can be decomposed into a linear combination of $P_1, P_2, ..., P_m$.
So, $xP_n$ is a linear combination of $P_1, P_2, ..., P_{n+1}$.
The inner product
$$(f,g) = \int_{-1}^1fgdx$$
defined for functions $f, g$ on $[-1, 1]$ is special in the sense that $(xf, g) = (f, xg)$.
And thus $$(xP_n, P_m) = (P_n, xP_m)$$
Now $xP_m$ has degree $(m+1)$ so it is a linear combination of $P_1, P_2, ..., P_{m+1}$.
Because of the fact that $(P_i, P_j) = 0$ if $i \not= j$, it follows that if $m+1 < n$, then $(P_n, xP_m) = 0$. Hence also $(xP_n, P_m) = 0$.
It follows that $xP_n$ is a linear combination of $P_{n+1}, P_n, P_{n-1}$.
But we can do even better than that: because $xP_n$ contains only terms with powers of $x$ equal to $n+1, n-1, ...$, and $P_n$ contains only terms with "wrong" powers of $x$, namely $n, n-2, ...$, in fact $xP_n$ is a linear combination of $P_{n+1}$ and $P_{n-1}$.
So we conclude that $xP_n = \alpha P_{n+1} +\beta P_{n-1}$.
In order to determine $\alpha$ and $\beta$, we first look at $x=1$: all $P_n$ have been normalized to $P_n(1) = 1$. So $\alpha +\beta = 1$.
Next, we look at the term with the highest power of $x$, namely $x^{n+1}$.
In another post (Choice of the First Term in Legendre Polynomials) I proved (only using Rodrigues' formula) that
$$P_n(x) = \frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}^2(x+1)^{n-k}(x-1)^k$$
hence it follows that
$$\frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}^2 = \alpha \cdot \frac{1}{2^{n+1}}\sum_{k=0}^{n+1}\binom{n+1}{k}^2$$
or
$$\frac{1}{2^n}\binom{2n}{n} = \alpha\frac{1}{2^{n+1}}\binom{2n+2}{n+1}$$
so
$$\alpha = 2 \cdot \frac{(2n)!}{n!^2} \cdot \frac{(n+1)!^2}{(2n+2)!} = \frac{2(n+1)^2}{(2n+2)(2n+1)} = \frac{n+1}{2n+1}$$
and consequently $$\beta = 1 - \alpha = \frac{n}{2n+1}$$
So we arrive at the recurrence relation
$$xP_n = \frac{n+1}{2n+1}P_{n+1} + \frac{n}{2n+1}P_{n-1}$$
or
$$(2n+1)xP_n = (n+1)P_{n+1} + nP_{n-1}$$
Q.E.D.
For the Schlafli integral you will need some knowledge of complex function theory obviously, but not much. A simple consequence of Cauchy's integral formula
$$f(a) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z-a}dz$$
is
$$f^{(n)}(a) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z-a)^{n+1}}dz$$
where $f^{(n)}$ is the $n^{th}$ derivative of $f$.
Plugging in Rodrigues' formula $P_n(z) = \frac{1}{2^nn!}[(z^2-1)^n]^{(n)}$, for which $f(z) = (z^2-1)^n$we get Schlafli's integral:
\begin{equation}
\begin{split}
P_n(a)&=\frac{1}{2^nn!}[(z^2-1)^n]^{(n)}\left.\right|_{z=a}
\\ &=\frac{1}{2^nn!} \cdot \frac{n!}{2\pi i}\oint_C \frac{(z^2-1)^n}{(z-a)^{n+1}}dz
\\ &=\frac{1}{2\pi i}\oint_C \frac{(z^2-1)^n}{2^n(z-a)^{n+1}}dz
\end{split}
\end{equation}
From this, it is again a small step to the generating function, but I will add that step later on.
