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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?

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    $\begingroup$ In the generating function, $|z| < 1 \implies \sqrt{1-2z+z^2} = 1 - z$. $\endgroup$ Apr 12, 2014 at 17:27

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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 $$\sqrt{(1-z)^2} = 1- z$$

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  • $\begingroup$ Sorry, still not got it. Does "should be trivial" means induction or other more straight ways? $\endgroup$
    – MathArt
    Mar 9 at 15:52

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