Topologically conjugate function

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  • #1
Mentallic
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Homework Statement


Are [itex]f(x)=2x, x\in R[/itex] and [itex]g(x)=x^2, x>0[/itex] topologically conjugate?

i.e. does there exist an h(x) such that

[tex]h(g(x))=f(h(x))[/tex]


The Attempt at a Solution



My professor gave one example in class about finding such a function h which was by guessing it to be equal to xn and subsequently solving and finding the value of n. However, when I tried to apply the same idea to this problem, I come off short.

[tex]h(g(x)) = h(x^2)[/tex]

[tex]f(h(x)) = 2h(x)[/tex]

If we let [itex]h(x)=x^n[/itex] then we want to solve for n in

[tex]x^{2n}=2x^n[/tex]

[tex]x^n=2 = h(x)[/tex]

Hence I find h(x)=2 but this doesn't work. Are there other guesses I could make? Or better yet, is there a more systematic approach to these sorts of problems? Does there even exist an h?
 

Answers and Replies

  • #2
pasmith
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Homework Statement


Are [itex]f(x)=2x, x\in R[/itex] and [itex]g(x)=x^2, x>0[/itex] topologically conjugate?

i.e. does there exist an h(x) such that

[tex]h(g(x))=f(h(x))[/tex]


The Attempt at a Solution



My professor gave one example in class about finding such a function h which was by guessing it to be equal to xn and subsequently solving and finding the value of n. However, when I tried to apply the same idea to this problem, I come off short.

[tex]h(g(x)) = h(x^2)[/tex]

[tex]f(h(x)) = 2h(x)[/tex]

If we let [itex]h(x)=x^n[/itex] then we want to solve for n in

[tex]x^{2n}=2x^n[/tex]

[tex]x^n=2 = h(x)[/tex]

Hence I find h(x)=2 but this doesn't work. Are there other guesses I could make? Or better yet, is there a more systematic approach to these sorts of problems? Does there even exist an h?

You need [itex]h(x^2) = 2h(x)[/itex] for all [itex]x \in \mathbb{R}[/itex]. The left hand side is an even function, so [itex]h[/itex] must itself be even. We can ensure that by taking [itex]h(x) = p(|x|)[/itex] for some [itex]p : \{x \in \mathbb{R} : x > 0\} \to \mathbb{R}[/itex].

Do you know of any functions defined on [itex]x > 0[/itex] with the property that [itex]p(x^\alpha) = \alpha p(x)[/itex]?
 
  • #3
Mentallic
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You need [itex]h(x^2) = 2h(x)[/itex] for all [itex]x \in \mathbb{R}[/itex]. The left hand side is an even function, so [itex]h[/itex] must itself be even. We can ensure that by taking [itex]h(x) = p(|x|)[/itex] for some [itex]p : \{x \in \mathbb{R} : x > 0\} \to \mathbb{R}[/itex].

Thanks! This analysis helped a lot!

Do you know of any functions defined on [itex]x > 0[/itex] with the property that [itex]p(x^\alpha) = \alpha p(x)[/itex]?

Ahh well when you put it that way, evidently [itex]p(x)=log_k(x)[/itex] for k>0 would work.

So does it suffice to choose h to be

[tex]h(x) = |\log_k(x)| : x\in \mathbb{R} \backslash {0} , h(x) = 0 : x=0[/tex]
 
  • #4
Mentallic
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Actually, I only need to find one such h so ln(x) would suffice.
 

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