The recurrence $a_n = 2a_{n-1} - 2a_{n-2}$ has characteristic roots $r = 1 \pm i$. In polar form, $r = \sqrt{2} e^{\pm i\pi/4}$. The general solution is $a_n = \sqrt{2}^n [A\cos(n\pi/4) + B\sin(n\pi/4)]$.

Given $a_0 = 1$ and $a_1 = 1$, what is $a_2$?