Consider the following generating function for the Lagrange polinomials:
$$g(x,t) = (1-2xt+t^2)^{-1/2} = \sum_{n=0}^{\infty}P_n(x)t^n$$
If we differentiate:
$$\frac{\partial g(t,x)}{\partial x} = \frac{t}{(1-2xt-t^2)^{3/2}} = \sum_{n=0}^{\infty}P_n'(x)t^n$$
which leads to:
$$(1-2xt+t^2)\sum_{n=0}^{\infty}P_n'(x)t^n-t\sum_{n=0}^{\infty}P_n(x)t^n = 0\tag{1}$$
The coefficient of each power of $t$ is set to equal to $0$ and we obtain:
$$P_{n+1}'(x) + P_{n-1}'(x) = 2xP_n'(x) + P_n(x)$$
However, I'm having trouble equating these coefficients to $0$. We get from (1):
$$\sum_{n=0}^{\infty}P_n'(x)t^n+\sum_{n=0}^{\infty}-2xP_n'(x)t^{n+1}+\sum_{n=0}^{\infty}P_n'(x)t^{n+2} - \sum_{n=0}^{\infty}P_n(x)t^{n+1} = 0$$
Now what exactly it means to make the coefficients to be equal to $0$? I should take the coefficients from the left side that have no $t$ in them and equate to $0$? But there are series which always have $t$. What am I supposed to do?
@Guerlando . If you open the summation, you can see that for coefficient of $t^{2}$ we have $$ \left( P_{2}'(x) -2xP_{1}'(x) +P_{0}'(x) -P_{1}(x) \right) t^{2} $$ and setting it to 0 will get the desired $$ P_{2}'(x) +P_{0}'(x) = 2xP_{1}'(x) +P_{1}(x)$$ this will happen for $t^{m}, \: m \ge 2$. For $m<2$, you can think of the polynomials $P_{n}(x)$ with negative $n$ are equal to zero, since at the beginning you dont have those polynomials.
The coefficients is set to 0 so that we can have a relation of the polynomials that satisfy the original equation : $$(1-2xt+t^{2})^{-1/2} = \sum P_{n}(x)t^{n} $$ Regards, Arief.