This is a very hard functional equation.
the problem is this :
find all functions $ f : \mathbb{R} \rightarrow \mathbb{R} $ such that : $ f(f(x))=x^2-2 $
to solve it i have no idea! can we solve it with highschool olympiad education?
please help : )
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2http://math.stackexchange.com/questions/832792/if-f-is-a-strictly-increasing-function-with-ffx-x22-then-f3#comment1719142_832792 – Dario Jul 02 '14 at 06:34
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3Well, $f$ cannot be injective, surjective, or weakly monotone increasing, for these sets of functions are closed under composition. Also, $f$ cannot be a polynomial by degree arguments. – Jul 02 '14 at 07:06
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1@Dr.AKA You should look at the answers there. – M. Vinay Jul 02 '14 at 08:58
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Yes, there is the sesquicentennial solution of Schroeder. – Cosmas Zachos Jun 02 '17 at 15:14
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As shown by Gottfried Helms in a linked question, a solution over $[-1,1]$ is given by a function defined over $(-2,+\infty)$: $$ 2\cdot T_{\sqrt{2}}(x/2) $$ where $T_n$ is a solution to the Chebyshev differential equation $$ (1-x^2)\frac{d^2 y}{dx^2}-x\frac{dy}{dx}+ n^2 y = 0.$$ The first terms of the Taylor series in zero are: $$2 \cos\left(\frac{\pi }{\sqrt{2}}\right)+\sqrt{2}\,\sin\left(\frac{\pi }{\sqrt{2}}\right) x-\frac{1}{2} \cos\left(\frac{\pi }{\sqrt{2}}\right) x^2-\frac{1}{12\sqrt{2}}\sin\left(\frac{\pi }{\sqrt{2}}\right) x^3-\frac{1}{48} \cos\left(\frac{\pi }{\sqrt{2}}\right) x^4+\ldots$$
Jack D'Aurizio
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I don't understand, on the duplicate question, the answer said that there is no such $f$, does I miss something ? – Oct 13 '15 at 14:16
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Nah, at the high-school level you can explicitly verify 362193 works. It was discovered in 1870 by Ernst Schroeder. – Cosmas Zachos Jun 02 '17 at 15:11