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Find $f$ if $f(f(x))=\sqrt{1-x^2} \land [-1; 1] \subseteq Dom(f)$ $$$$Please give both real and complex functions. Can it be continuous or not (if f is real)

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  • $\begingroup$ $f(f(f(f(x))))=|x|$, $f$ has only one fixpoint $\frac{1}{\sqrt2}$, I also tried substituting $x=sin(y)$ but no succes $\endgroup$ Commented Dec 15, 2013 at 0:40
  • $\begingroup$ OK. Help incoming. $\endgroup$
    – Alec
    Commented Dec 15, 2013 at 0:41
  • $\begingroup$ Do they want ONE function or many? $\endgroup$
    – Igor Rivin
    Commented Dec 15, 2013 at 0:43
  • $\begingroup$ I want (if exist) a continuous real function or just a real one. -> Thanks to Alec we already know a complex solution. $\endgroup$ Commented Dec 15, 2013 at 0:50

3 Answers 3

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I guessed that $f$ is of the form $\sqrt{ax^2+b}$. Then, $f^2$ is $\sqrt{a^2x^2 + \frac{a^2-1}{a-1}b}$. From here on in, it is algebra: $$ a^2 =-1 \implies a = i ~~~~\text{and}~~~~\frac{a^2-1}{a-1}b = 1 \implies b = \frac{1-i}{2} $$ So we get $f(x) = \sqrt{ix^2 + \frac{1-i}{2}}$. I checked using Wolfram, and $f^2$ appears to be what we want.

DISCLAIMER: This is not the only solution.

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  • $\begingroup$ What do you mean by $a^2=1 \implies a=i $? $\endgroup$ Commented Dec 15, 2013 at 0:48
  • $\begingroup$ I simply followed one solution; if you take $a=-i$ you get a $b =\frac{i+1}{2}$ and you get another one. The point of the answer was to sketch out what to do. But yes, as usual, you are right. $\endgroup$
    – Alec
    Commented Dec 15, 2013 at 0:52
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Consider real functions on $[0, 1].$ Notice that $\sqrt{1-x^2}$ is an involution, so think of it as a permutation of of the (infinite) set $[0, 1]$ Its cycle type is a product of disjoint transpositions. Now this permutation is supposed to be a square of some other permutation, and while a transposition is not a square of a (finitely supported) permutation, for sign reasons, a product of two disjoint transpositions is (it's a square of a four-cycle). So, break your transpositions into pairs in your favorite way, and there is your function (which is not very continuous, in general). This argument would seem to indicate that you cannot extend such a function to $[-1, 1],$ though I could be wrong.

EDIT The above, can, in fact, be completed to a function defined on $[-1, 1].$

Take $x<0.$ if $-x$ fits into a four-cycle $(-x, a, \sqrt{1-x^2}, b)$ map $x$ to $a.$

ANOTHER EDIT of course, this also answers the question of uniqueness: there are uncountably many such functions.

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  • $\begingroup$ I was thinking about using permutations but I didn't know whether it works for continuously infinite sets as well. From this point I still don't know how to extend it to $[-1,1]$ because the fact that $f$ won't be an injective function anymore makes the approach with permutations useless but doesn't tell anything about the other possible solutions. $\endgroup$ Commented Dec 15, 2013 at 1:04
  • $\begingroup$ @user115826 I disagree. The point is that the structure of real line is only useful if you are thinking of continuous functions, otherwise the real line is just a set. Since you understand the action on positive part of the set, the negative part is not so far away, see the edit. $\endgroup$
    – Igor Rivin
    Commented Dec 15, 2013 at 1:14
  • $\begingroup$ (user115826 was me) Ok, this gave us a real function but doesn't prove that there isn't any continuous real one. You wants to map x and -x to the same number but this restriction which may cause the absence of a continuous function. Can you explain why a 4 cycle makes it not continuous? $\endgroup$ Commented Dec 15, 2013 at 1:48
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I don't know how to approach the real case, but here's a way to solve it if you allow $f$ to be complex.

Start from the guess $f(x) = \sqrt{1-x^2}$. It almost works, and suggests trying the larger family $$f(x) = \sqrt{a-bx^2}.$$ For what values of $a,b$ does $f(f(x)) = \sqrt{1-x^2}$?

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  • $\begingroup$ See my answer for the real case... $\endgroup$
    – Igor Rivin
    Commented Dec 15, 2013 at 1:18

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