I could "prove" that $P_n(1)=1=-1$. But I'm not sure where is my mistake.
If I use the generating function:
$$\frac{1}{\sqrt{1-2z+z^2}}=\sum_{n=0}^{\infty}P_n(1)z^n$$
Since: $\sqrt{1-2z+z^2 } = \sqrt{(1-z)^2}=\sqrt{(z-1)^2}=z-1$
$$\frac{1}{z-1}=-\frac{1}{1-z}=-\sum_{n=0}^{\infty}z^n$$
So: $P_n (1)=-1$, instead of $P_n (1)=1$.
Does this mean that the generating function doesn't define uniquely Legendre polynomials?
Best Answer
Use the recurrence for Legendre polynomials and it should be trivial to obtain $P_n(1)$ and $P_n(-1)$. $$(n+1)P_{n+1}(x) = (2n+1)xP_n(x) - nP_{n-1}(x)$$
In your case, note that $$\sqrt{(1-z)^2} = \left\vert 1- z \right\vert$$ Since the valid domain for the Legendre polynomials is $[-1,1]$, you need to have $$\sqrt{(1-z)^2} = 1- z$$