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That doesn’t help in understanding why this thing is numerically quite stable. It only gives some confidence that it is.
by Someone 4y ago
That doesn’t help in understanding why this thing is numerically quite stable. It only gives some confidence that it is.
- ataylor284_ 4y agoAll the visualizations of the mandelbrot set I've seen use a threshold of iterations that define the edge. I think that threshold is more significant than any instability due to floating point numbers. A more formal analysis of numerical stability comes down to whether the iterated function is considered well-conditioned. I don't have the math to state it formally, but my intuition is that the mandelbrot function is, in that a small change to the function input will have a small effect on the number of iterations except at the very edge where the threshold will define the boundary anyway.
- kaba0 4y agoYeah, I didn’t intend it as a formal proof, but it is an easy argument to make. I’m sure an error limit can be given, as it uses a quite basic arithmetics only.