Is 0 a stable fixed point of *f*(*x*) = 4*x* (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 easy to get mathematical stability and numerical stability confused, because **the latter often illustrates the former**. In this post I point out that the points on a particular unstable orbit cannot be represented exactly in floating point numbers, so iterations that start at the floating point representations of these points will drift away.

In the example here, 0 *can* be represented exactly, and so we do have a *computationally* stable fixed point. But 0 is not a stable fixed point in the mathematical sense. Any starting point close to 0 but not exactly 0 will eventually go all over the place.

If we start at some very small ε > 0, then *f*(ε) ≈ 4ε. Every iteration multiplies the result by approximately 4, and eventually the result is large enough that the approximation no longer holds.

For example, suppose we start with *x* = 10^{-100}. After 100 iterations we have

*x* = 4^{100} 10^{-100} = 1.6 × 10^{-40}.

If we were to plot this, we wouldn’t see any movement. But by the time we iterate our function 200 times, the result is chaotic. I get the following: