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For the Legendre polynomials $P_n$ of degree $n$, normed to $P_n(1)=1$, i found the following formula for its generating function: \begin{equation} \sum_{n=0}^{\infty} P_n(x) z^n = \frac{1}{\sqrt{1-2xz+z^2}}, \end{equation} for $x \in \mathbb{R}, z \in \mathbb{C}$, with $|z|<1$.

Since $|z|<1$, we already know that \begin{equation} 1-2xz+z^2 \neq 0. \end{equation} But i still have a problem with the square root expression, since for complex valued arguments, it is not clear which of the two possible square roots we should take here.

Does anyone know how to interpret this expression?

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This polynomials are defined for $|x| \le 1$, and in this case $1 - 2xz + z^2 \ge 0$. To give you an idea, call $x = \cos\theta$, this is always possible because $|x| \le 1$. Therefore

$$ 1 -2xz + z^2 = 1 - 2z \cos\theta + z^2 = |\mathbf{e} - \mathbf{z}|^2 \ge 0 $$

where $\theta$ is the angle between $\mathbf{z}$ and the unit vector $\mathbf{e}$. The root $(1 - 2xz + z^2)^{1/2}$ then takes only real values

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  • $\begingroup$ I don't really understand your argument. If i take $x=\frac{1}{2}$ and $z=\frac{i}{2}$, then $1-2xz+z^2 = \frac{3}{4} - \frac{1}{2}i$, i.e. a complex valued number. How do you take a real valued root from it? $\endgroup$ – Simul Dec 7 '16 at 17:20
  • $\begingroup$ @Simul My apologies, I think I misunderstood your question the first time. There's no problem with the generating function, if $w = 1 - 2xz + z^2 $ just take the principal argument of $w$ and half it $\endgroup$ – caverac Dec 7 '16 at 17:54

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