Legendre polynomial with argument zero $P_n ( 0 )$ I want to find the expression for $P_n(x)$ with $x = 0$, ie $P_n(0)$ for any $n$. The first few non-zero legendre polynomials with $x=0$ are 
$P_0(0) = 1$, $P_2(0) = -\frac{1}{2}$, $P_4(0) = \frac{3}{8}$, $P_6(0) = -\frac{5}{16}$, $P_8(0) = \frac{35}{128}$ but I can't find a relationship between them to write as an equation for arbitrary $n$. Any help is appreciated.


If you look here, the recurrence relation is  $$(n+1) P_{n+1}(x) = (2n+1) x P_n(x) - n P_{n-1}(x)$$ So $$(n+1) P_{n+1}(0) =  - n P_{n-1}(0)$$ In the same Wikipedia page, you will find that, from Bonnet recursion formula,
$$P_n(x) = \sum_{k=0}^n (-1)^k \begin{pmatrix} n \\ k \end{pmatrix}^2 \left( \frac{1+x}{2} \right)^{n-k} \left( \frac{1-x}{2} \right)^k$$ So the result as given in achille hui's answer. 
