Find boundary conditions that allow the recursion $f(k)=\frac{1}{2} f(k-1) + \frac{1}{2} f(k+1)$ for $-B
The boundary conditions $f(A) = 1$ and $f(-B)=0$ uniquely determine $f$, but I am not certain as to why these values give the unique solution. I've encountered this in several books and I always took it for granted, but never quite figured out the intuition behind the solution.