by John on (#5GGG1)
Is 0 a stable fixed point of f(x) = 4x (1-x)? If you set x = 0 and compute f(x) you will get exactly 0. Apply f a thousand times and you’ll never get anything but zero. But this does not mean 0 is a stable attractor, and in fact it is not stable. It’s […]The post Mathematical stability vs numerical stability first appeared on John D. Cook.