P_n(x) = (2 - 1/n) x P_{n-1}(x) - (1 - 1/n) P_{n-2}(x)
This is numerically stable if x is in [0,1], but I haven't validated that you can get to very high n using floating point.
P_n(x) = (2 - 1/n) x P_{n-1}(x) - (1 - 1/n) P_{n-2}(x)
This is numerically stable if x is in [0,1], but I haven't validated that you can get to very high n using floating point.