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    Post Closed as "too localized" by José Figueroa-O'Farrill, Gjergji Zaimi, Felipe Voloch, Emil Jeřábek, Andy Putman

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I have a nice problem which is related to algebra and polynomials or even operator theory. I would be very grateful if any of you could solve it. Here is the problem: Consider the function $f_0(x) =x(1-x)$ and for $n \geq 0$ define $$f_{n+1}(x) =\frac 12 (f_n(x^2) + f_n(1-(1-x)^2)).$$ f_n((1-x)^2)).$$ Now, looking more closely at $f_0(x)$, we see that it is increasing on $[0,\frac 12]$ and decreasing on $[\frac 12, 1]$ . The problem is to prove that such a property holds for all the $f_n$'s. More precisely, prove that each $f_n (x)$ is increasing on $x \in [0,\frac 12]$ and decreasing on $x \in [\frac 12, 1]$ . I would be very thankful if any of you could come up with a solution.

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A nice problem that I am unable to solve

I have a nice problem which is related to algebra and polynomials or even operator theory. I would be very grateful if any of you could solve it. Here is the problem: Consider the function $f_0(x) =x(1-x)$ and for $n \geq 0$ define $$f_{n+1}(x) =\frac 12 (f_n(x^2) + f_n(1-(1-x)^2)).$$ Now, looking more closely at $f_0(x)$, we see that it is increasing on $[0,\frac 12]$ and decreasing on $[\frac 12, 1]$ . The problem is to prove that such a property holds for all the $f_n$'s. More precisely, prove that each $f_n (x)$ is increasing on $x \in [0,\frac 12]$ and decreasing on $x \in [\frac 12, 1]$ . I would be very thankful if any of you could come up with a solution.