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Topic: f(f(x)) = 2x (Read 1735 times) |
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Icarus
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f(f(x)) = 2x
« on: Nov 30th, 2005, 7:27pm » |
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In the more recent of the two "f(f(x)) = x^2 - 2" threads, the following question has arisen:
How many continuous functions f are there which satisfy f(f(x)) = 2x for all x in their domain?
What if the f is required to be analytic instead?
Two solutions have already been noted: f(x) = sqrt(2)x and f(x) = -sqrt(2)x. However, notice that the similar equation g(g(x)) = x has g(x) = x, -1, or a/x (a <> 0), all as solutions.
Do other solutions exist for f(f(x)) = 2x as well?
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towr
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Re: f(f(x)) = 2x
« Reply #1 on: Dec 1st, 2005, 3:26am » |
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yes, for example
f(0)=0
f(x) = -(2/a)x iff x>0
f(x) = -ax iff x<0
where a > 0
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Icarus
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Re: f(f(x)) = 2x
« Reply #2 on: Dec 1st, 2005, 7:14pm » |
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Good trick!
How about if we require f to be analytic and defined on all R, or all C?
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| « Last Edit: Dec 1st, 2005, 7:21pm by Icarus » |
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towr
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Re: f(f(x)) = 2x
« Reply #3 on: Dec 2nd, 2005, 1:06am » |
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That's a lot more difficult.
I can't get further than f'(f(x)) * f'(x) = 2, which easily gives f'(x) = +/- sqrt(2), but doesn't give any other obvious solution.
hmm.. maybe my other function could be adapted if instead of a fixed a, you take a function a(x)..
It'd be sqrt(2) around 0, and a at -inf, 2/a at +inf
I'll have to be a bit more awake to characterize it further though.
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| « Last Edit: Dec 2nd, 2005, 1:09am by towr » |
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Eigenray
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Re: f(f(x)) = 2x
« Reply #4 on: Dec 2nd, 2005, 3:30am » |
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If f is entire, and not a polynomial, then f(1/z) has an essential singularity at z=0; by Casorati-Weierstrass, { f(z) : |z|>1 } is dense, and therefore intersects the open set { f(z) : |z|<1 }. Thus f can't be injective.
But if f(f(z))=2z, then f is clearly injective. So the only entire solutions are polynomial, which gives the two known solutions.
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| « Last Edit: Dec 2nd, 2005, 3:32am by Eigenray » |
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Eigenray
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Re: f(f(x)) = 2x
« Reply #5 on: Dec 2nd, 2005, 3:45am » |
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Actually, suppose g(g(z))=z2-2. If g(z)=g(w), then w2=z2, so the above argument shows g can't be entire and non-polynomial either.
In fact, suppose h(h(z))=p(z), for some polynomial p, or more generally, any entire function with |p(z)| -> oo as |z| -> oo. Pick R such that
|z|>R implies |p(z)| > sup { |p(w)| : |w| < 1 }.
If h is entire and non-polynomial, { h(z) : |z|>R } is dense, so intersects the open set { h(w) : |w|<1 }. But h(z)=h(w) with |z|>R, |w|<1, means p(z)=h(h(z))=h(h(w))=p(w), contradicting choice of R.
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paul_h
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on Dec 2nd, 2005, 3:45am, Eigenray wrote: so the above argument shows g can't be entire and non-polynomial either.
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I am not quite as fast as you guys but I want to understand this. What above argument are you talking about? The argument above appears to be about f(f(z)) = 2z and not g(g(z)) = z^2 - 2.
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Icarus
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Re: f(f(x)) = 2x
« Reply #7 on: Dec 21st, 2005, 8:02pm » |
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He means that the same argument can be used to show that g is either polynomial, or not entire.
If g(z) = g(w), then g(g(z)) = g(g(w)), so z2 + 2 = w2 + 2, so z2 = w2, and z = +- w. Thus g is "almost" injective, in that g-1(u) can have at most two elements.
But if g(z) is entire, and is not a polynomial, then infinity is an essential singularity. By the stronger version of Picard's theorem, the image of every neighborhood of infinity is the entire complex plane less possibly 2 points. This would mean that for almost all values of a, the equation g(z) = a has infinitely many solutions, not just two. So either g is not entire, or else it is a polynomial.
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