Prove that $x^2$ and $x^3$ are topological conjugated on $[0, infty)$.
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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?
real-analysis dynamical-systems
asked Aug 26 at 13:15
Marcinek665
25019
add a comment |Â
up vote
7
down vote
favorite
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?
real-analysis dynamical-systems
asked Aug 26 at 13:15
Marcinek665
25019
If $f$ is defined on $[2,4)$, then you can extend the domain of $f$ to $[2,infty)$ following the rule $f(x^2) := f(x)^3$
– Kenny Lau
Aug 26 at 13:45
So, we are looking for a homeomorphism $f : [0,infty)to [0,infty)$ such that $f(x^2) = f(x)^3$?
– amsmath
Aug 26 at 13:46
Other preliminary stuff: the function must be striclty increasing, $f(0) = 0$, $f(1) = 1$.
– Kenny Lau
Aug 26 at 13:46
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 $f(1) = sqrt2/3$ (which is absurd since we know that $f(1)=1$).
– Sobi
Aug 26 at 13:49
I think, the function is completely determined by its values on $(1-varepsilon,1+varepsilon)$ for any $varepsilon > 0$.
– amsmath
Aug 26 at 13:57
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
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?
real-analysis dynamical-systems
asked Aug 26 at 13:15
Marcinek665
25019
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?
real-analysis dynamical-systems
asked Aug 26 at 13:15
Marcinek665
25019
asked Aug 26 at 13:15
Marcinek665
25019
asked Aug 26 at 13:15
Marcinek665
25019
asked Aug 26 at 13:15
Marcinek665
25019
25019
If $f$ is defined on $[2,4)$, then you can extend the domain of $f$ to $[2,infty)$ following the rule $f(x^2) := f(x)^3$
– Kenny Lau
Aug 26 at 13:45
So, we are looking for a homeomorphism $f : [0,infty)to [0,infty)$ such that $f(x^2) = f(x)^3$?
– amsmath
Aug 26 at 13:46
Other preliminary stuff: the function must be striclty increasing, $f(0) = 0$, $f(1) = 1$.
– Kenny Lau
Aug 26 at 13:46
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 $f(1) = sqrt2/3$ (which is absurd since we know that $f(1)=1$).
– Sobi
Aug 26 at 13:49
I think, the function is completely determined by its values on $(1-varepsilon,1+varepsilon)$ for any $varepsilon > 0$.
– amsmath
Aug 26 at 13:57
add a comment |Â
If $f$ is defined on $[2,4)$, then you can extend the domain of $f$ to $[2,infty)$ following the rule $f(x^2) := f(x)^3$
– Kenny Lau
Aug 26 at 13:45
So, we are looking for a homeomorphism $f : [0,infty)to [0,infty)$ such that $f(x^2) = f(x)^3$?
– amsmath
Aug 26 at 13:46
Other preliminary stuff: the function must be striclty increasing, $f(0) = 0$, $f(1) = 1$.
– Kenny Lau
Aug 26 at 13:46
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 $f(1) = sqrt2/3$ (which is absurd since we know that $f(1)=1$).
– Sobi
Aug 26 at 13:49
I think, the function is completely determined by its values on $(1-varepsilon,1+varepsilon)$ for any $varepsilon > 0$.
– amsmath
Aug 26 at 13:57
If $f$ is defined on $[2,4)$, then you can extend the domain of $f$ to $[2,infty)$ following the rule $f(x^2) := f(x)^3$
– Kenny Lau
Aug 26 at 13:45
If $f$ is defined on $[2,4)$, then you can extend the domain of $f$ to $[2,infty)$ following the rule $f(x^2) := f(x)^3$
– Kenny Lau
Aug 26 at 13:45
So, we are looking for a homeomorphism $f : [0,infty)to [0,infty)$ such that $f(x^2) = f(x)^3$?
– amsmath
Aug 26 at 13:46
So, we are looking for a homeomorphism $f : [0,infty)to [0,infty)$ such that $f(x^2) = f(x)^3$?
– amsmath
Aug 26 at 13:46
Other preliminary stuff: the function must be striclty increasing, $f(0) = 0$, $f(1) = 1$.
– Kenny Lau
Aug 26 at 13:46
Other preliminary stuff: the function must be striclty increasing, $f(0) = 0$, $f(1) = 1$.
– Kenny Lau
Aug 26 at 13:46
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 $f(1) = sqrt2/3$ (which is absurd since we know that $f(1)=1$).
– Sobi
Aug 26 at 13:49
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 $f(1) = sqrt2/3$ (which is absurd since we know that $f(1)=1$).
– Sobi
Aug 26 at 13:49
I think, the function is completely determined by its values on $(1-varepsilon,1+varepsilon)$ for any $varepsilon > 0$.
– amsmath
Aug 26 at 13:57
I think, the function is completely determined by its values on $(1-varepsilon,1+varepsilon)$ for any $varepsilon > 0$.
– amsmath
Aug 26 at 13:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
We can consider the function $f:[0,infty)to[0,infty)$ defined by
$$f(x)=
begincases
e^log(x)^log(3)/log(2)&x>1\
e^-(-log(x))^log(3)/log(2)&0<xleq 1\
0&x=0
endcases$$
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 $xto x^log(3)/log(2)$ is a conjugation between $2x$ and $3x$.
More precisely, we have the commutative diagram below where $h(x)=operatornamesign(x)|x|^log(3)/log(2)$
answered Aug 26 at 13:59
Fabio Lucchini
6,17911126
1
Very nice, Fabio! It's because of $3^fraclog 3log 2 = 3^log_3(2) = 2$.
– amsmath
Aug 26 at 14:06
1
And for $xin (0,1)$, what about $f(x)=e^(-log(x))^log(3)/log(2)$ ? Seems to work.
– amsmath
Aug 26 at 14:12
I put another minus sign to make $f$ a bijection.
– Fabio Lucchini
Aug 26 at 14:31
Fabio: Sure. Fair enough. ;o)
– amsmath
Aug 26 at 14:33
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We can consider the function $f:[0,infty)to[0,infty)$ defined by
$$f(x)=
begincases
e^log(x)^log(3)/log(2)&x>1\
e^-(-log(x))^log(3)/log(2)&0<xleq 1\
0&x=0
endcases$$
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 $xto x^log(3)/log(2)$ is a conjugation between $2x$ and $3x$.
More precisely, we have the commutative diagram below where $h(x)=operatornamesign(x)|x|^log(3)/log(2)$
answered Aug 26 at 13:59
Fabio Lucchini
6,17911126
1
Very nice, Fabio! It's because of $3^fraclog 3log 2 = 3^log_3(2) = 2$.
– amsmath
Aug 26 at 14:06
1
And for $xin (0,1)$, what about $f(x)=e^(-log(x))^log(3)/log(2)$ ? Seems to work.
– amsmath
Aug 26 at 14:12
I put another minus sign to make $f$ a bijection.
– Fabio Lucchini
Aug 26 at 14:31
Fabio: Sure. Fair enough. ;o)
– amsmath
Aug 26 at 14:33
add a comment |Â
up vote
3
down vote
accepted
We can consider the function $f:[0,infty)to[0,infty)$ defined by
$$f(x)=
begincases
e^log(x)^log(3)/log(2)&x>1\
e^-(-log(x))^log(3)/log(2)&0<xleq 1\
0&x=0
endcases$$
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 $xto x^log(3)/log(2)$ is a conjugation between $2x$ and $3x$.
More precisely, we have the commutative diagram below where $h(x)=operatornamesign(x)|x|^log(3)/log(2)$
answered Aug 26 at 13:59
Fabio Lucchini
6,17911126
1
Very nice, Fabio! It's because of $3^fraclog 3log 2 = 3^log_3(2) = 2$.
– amsmath
Aug 26 at 14:06
1
And for $xin (0,1)$, what about $f(x)=e^(-log(x))^log(3)/log(2)$ ? Seems to work.
– amsmath
Aug 26 at 14:12
I put another minus sign to make $f$ a bijection.
– Fabio Lucchini
Aug 26 at 14:31
Fabio: Sure. Fair enough. ;o)
– amsmath
Aug 26 at 14:33
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We can consider the function $f:[0,infty)to[0,infty)$ defined by
$$f(x)=
begincases
e^log(x)^log(3)/log(2)&x>1\
e^-(-log(x))^log(3)/log(2)&0<xleq 1\
0&x=0
endcases$$
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 $xto x^log(3)/log(2)$ is a conjugation between $2x$ and $3x$.
More precisely, we have the commutative diagram below where $h(x)=operatornamesign(x)|x|^log(3)/log(2)$
answered Aug 26 at 13:59
Fabio Lucchini
6,17911126
We can consider the function $f:[0,infty)to[0,infty)$ defined by
$$f(x)=
begincases
e^log(x)^log(3)/log(2)&x>1\
e^-(-log(x))^log(3)/log(2)&0<xleq 1\
0&x=0
endcases$$
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 $xto x^log(3)/log(2)$ is a conjugation between $2x$ and $3x$.
More precisely, we have the commutative diagram below where $h(x)=operatornamesign(x)|x|^log(3)/log(2)$
answered Aug 26 at 13:59
Fabio Lucchini
6,17911126
edited Aug 26 at 14:29
answered Aug 26 at 13:59
Fabio Lucchini
6,17911126
answered Aug 26 at 13:59
Fabio Lucchini
6,17911126
answered Aug 26 at 13:59
Fabio Lucchini
6,17911126
6,17911126
1
Very nice, Fabio! It's because of $3^fraclog 3log 2 = 3^log_3(2) = 2$.
– amsmath
Aug 26 at 14:06
1
And for $xin (0,1)$, what about $f(x)=e^(-log(x))^log(3)/log(2)$ ? Seems to work.
– amsmath
Aug 26 at 14:12
I put another minus sign to make $f$ a bijection.
– Fabio Lucchini
Aug 26 at 14:31
Fabio: Sure. Fair enough. ;o)
– amsmath
Aug 26 at 14:33
add a comment |Â
1
Very nice, Fabio! It's because of $3^fraclog 3log 2 = 3^log_3(2) = 2$.
– amsmath
Aug 26 at 14:06
1
And for $xin (0,1)$, what about $f(x)=e^(-log(x))^log(3)/log(2)$ ? Seems to work.
– amsmath
Aug 26 at 14:12
I put another minus sign to make $f$ a bijection.
– Fabio Lucchini
Aug 26 at 14:31
Fabio: Sure. Fair enough. ;o)
– amsmath
Aug 26 at 14:33
1
1
Very nice, Fabio! It's because of $3^fraclog 3log 2 = 3^log_3(2) = 2$.
– amsmath
Aug 26 at 14:06
Very nice, Fabio! It's because of $3^fraclog 3log 2 = 3^log_3(2) = 2$.
– amsmath
Aug 26 at 14:06
1
1
And for $xin (0,1)$, what about $f(x)=e^(-log(x))^log(3)/log(2)$ ? Seems to work.
– amsmath
Aug 26 at 14:12
And for $xin (0,1)$, what about $f(x)=e^(-log(x))^log(3)/log(2)$ ? Seems to work.
– amsmath
Aug 26 at 14:12
I put another minus sign to make $f$ a bijection.
– Fabio Lucchini
Aug 26 at 14:31
I put another minus sign to make $f$ a bijection.
– Fabio Lucchini
Aug 26 at 14:31
Fabio: Sure. Fair enough. ;o)
– amsmath
Aug 26 at 14:33
Fabio: Sure. Fair enough. ;o)
– amsmath
Aug 26 at 14:33
add a comment |Â
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If $f$ is defined on $[2,4)$, then you can extend the domain of $f$ to $[2,infty)$ following the rule $f(x^2) := f(x)^3$
– Kenny Lau
Aug 26 at 13:45
So, we are looking for a homeomorphism $f : [0,infty)to [0,infty)$ such that $f(x^2) = f(x)^3$?
– amsmath
Aug 26 at 13:46
Other preliminary stuff: the function must be striclty increasing, $f(0) = 0$, $f(1) = 1$.
– Kenny Lau
Aug 26 at 13:46
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 $f(1) = sqrt2/3$ (which is absurd since we know that $f(1)=1$).
– Sobi
Aug 26 at 13:49
I think, the function is completely determined by its values on $(1-varepsilon,1+varepsilon)$ for any $varepsilon > 0$.
– amsmath
Aug 26 at 13:57