2 ms·
I'm not sure that even works in 1 dimension...? I don't think you necessarily have g(a) >= 0 and g(b) <= 0. Finding the starting points for binary search might
by ChrisKnott 5y ago
I'm not sure that even works in 1 dimension...?
I don't think you necessarily have g(a) >= 0 and g(b) <= 0. Finding the starting points for binary search might be essentially as hard as finding the fixed point.
- im3w1l 5y agoa <= f(x) <= b Hence a <= f(a), so 0 <= f(a) - a = g(a)