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As in the topic. I have to prove that the functions $x ^ 2$ and $x ^ 3$ are topologically conjugated. I tried to write it out by definition: $f (x ^ 2) = f (x) ^ 3$ and choose $f (x) = x ^ a$, but unfortunately it doesn't work. It's my beginnings in this field, so I do not have much experience yet.

Do you have any hints?

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  • $\begingroup$ If $f$ is defined on $[2,4)$, then you can extend the domain of $f$ to $[2,\infty)$ following the rule $f(x^2) := f(x)^3$ $\endgroup$
    – Kenny Lau
    Aug 26, 2018 at 13:45
  • $\begingroup$ So, we are looking for a homeomorphism $f : [0,\infty)\to [0,\infty)$ such that $f(x^2) = f(x)^3$? $\endgroup$
    – amsmath
    Aug 26, 2018 at 13:46
  • $\begingroup$ Other preliminary stuff: the function must be striclty increasing, $f(0) = 0$, $f(1) = 1$. $\endgroup$
    – Kenny Lau
    Aug 26, 2018 at 13:46
  • $\begingroup$ If $f$ is also differentiable (not that it needs to be, just presenting an observation), then $2xf'(x^2) = 3f(x)^2f'(x),$ and inserting $x=1,$ we have $2f'(1)=3f(1)^2f'(1)$, so we either have $f'(1) = 0$, or otherwise $f(1) = \sqrt{2/3}$ (which is absurd since we know that $f(1)=1$). $\endgroup$
    – MSDG
    Aug 26, 2018 at 13:49
  • $\begingroup$ I think, the function is completely determined by its values on $(1-\varepsilon,1+\varepsilon)$ for any $\varepsilon > 0$. $\endgroup$
    – amsmath
    Aug 26, 2018 at 13:57

1 Answer 1

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We can consider the function $f:[0,\infty)\to[0,\infty)$ defined by $$f(x)= \begin{cases} e^{\log(x)^{\log(3)/\log(2)}}&x>1\\ e^{-(-\log(x))^{\log(3)/\log(2)}}&0<x\leq 1\\ 0&x=0 \end{cases}$$

To find this function, can be useful to note that $x^2$ is topologically conjugated to $2x$ through $\log$ function and similarly $x^3$ is conjugated to $3x$. Moreover $x\to x^{\log(3)/\log(2)}$ is a conjugation between $2x$ and $3x$. More precisely, we have the commutative diagram below where $h(x)=\operatorname{sign}(x)|x|^{\log(3)/\log(2)}$

enter image description here

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    $\begingroup$ Very nice, Fabio! It's because of $3^{\frac{\log 3}{\log 2}} = 3^{\log_3(2)} = 2$. $\endgroup$
    – amsmath
    Aug 26, 2018 at 14:06
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    $\begingroup$ And for $x\in (0,1)$, what about $f(x)=e^{(-\log(x))^{\log(3)/\log(2)}}$ ? Seems to work. $\endgroup$
    – amsmath
    Aug 26, 2018 at 14:12
  • $\begingroup$ I put another minus sign to make $f$ a bijection. $\endgroup$ Aug 26, 2018 at 14:31
  • $\begingroup$ Fabio: Sure. Fair enough. ;o) $\endgroup$
    – amsmath
    Aug 26, 2018 at 14:33

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