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> Generally you can't figure out which point is fixed. So just thinking out loud here but like in the 1d case... The set will be an interval [a,b] Let g(x) = f
by im3w1l 5y ago
> Generally you can't figure out which point is fixed.
So just thinking out loud here but like in the 1d case... The set will be an interval [a,b] Let g(x) = f(x) - x
g(a) >= 0 and g(b) <= 0. Now we can use the intermediate value value theorem and binary search to find (get arbitrary close) a point with g(x) = 0 (i.e. fix point of f).
Not possible to do such searches in higher dimensions?
- ChrisKnott 5y agoI'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)
- SquishyPanda23 5y agoYou can generally approximate the fixed point. You can possibly do this to arbitrary precision, but I'm not sure. This is useful for practical applications, but it's not the same as knowing the fixed point for mathematical applications. E.g. you may want to prove that some function preserves the irrational number pi. But you may not be able to prove it, only show that computations make your conjecture more and more likely.