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        riddles >> putnam exam (pure math) >> f(f(x)) = 2x
        
(Message started by: Icarus on Nov 30^th, 2005, 7:27pm)

Title: f(f(x)) = 2x
Post by Icarus on Nov 30^th, 2005, 7:27pm

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?

Title: Re: f(f(x)) = 2x
Post by towr on Dec 1^st, 2005, 3:26am

yes, for example

f(0)=0
f(x) = -(2/a)x iff x>0
f(x) = -ax iff x<0
where a > 0

Title: Re: f(f(x)) = 2x
Post by Icarus on Dec 1^st, 2005, 7:14pm

Good trick!

How about if we require f to be analytic and defined on all R, or all C?

Title: Re: f(f(x)) = 2x
Post by towr on Dec 2^nd, 2005, 1:06am

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.

Title: Re: f(f(x)) = 2x
Post by Eigenray on Dec 2^nd, 2005, 3:30am

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.

Title: Re: f(f(x)) = 2x
Post by Eigenray on Dec 2^nd, 2005, 3:45am

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.

Title: Re: f(f(x)) = 2x
Post by paul_h on Dec 21^st, 2005, 7:32pm

on 12/02/05 at 03:45:43, Eigenray wrote:

so the above argument shows g can't be entire and non-polynomial either.

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.

Title: Re: f(f(x)) = 2x
Post by Icarus on Dec 21^st, 2005, 8:02pm

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.