Mixing
Preimage of circle with critical point and its image inside is simple and closed
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I am reading Devaney's "An Introduction to Chaotic Dynamical Systems" and I am trying to convince myself of a claim made there (Section 3.6, proposition 6.2). The proposition concerns showing that the Julia set of $f(z)=z^2+c$ is simple and closed if $|c|<1/4$. The claim is
If the critical point $0$ and its image $c$ is contained in a circle $Gamma$ then the preimage of $Gamma$ is a simple closed curve.
I am stuck on how to prove this. Would I need to construct a continuous limiting curve as in the proof for the Julia set?
complex-dynamics
asked Feb 1 at 13:33
user30523user30523
564510
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add a comment |
$begingroup$
I am reading Devaney's "An Introduction to Chaotic Dynamical Systems" and I am trying to convince myself of a claim made there (Section 3.6, proposition 6.2). The proposition concerns showing that the Julia set of $f(z)=z^2+c$ is simple and closed if $|c|<1/4$. The claim is
If the critical point $0$ and its image $c$ is contained in a circle $Gamma$ then the preimage of $Gamma$ is a simple closed curve.
I am stuck on how to prove this. Would I need to construct a continuous limiting curve as in the proof for the Julia set?
complex-dynamics
asked Feb 1 at 13:33
user30523user30523
564510
$endgroup$
add a comment |
$begingroup$
I am reading Devaney's "An Introduction to Chaotic Dynamical Systems" and I am trying to convince myself of a claim made there (Section 3.6, proposition 6.2). The proposition concerns showing that the Julia set of $f(z)=z^2+c$ is simple and closed if $|c|<1/4$. The claim is
If the critical point $0$ and its image $c$ is contained in a circle $Gamma$ then the preimage of $Gamma$ is a simple closed curve.
I am stuck on how to prove this. Would I need to construct a continuous limiting curve as in the proof for the Julia set?
complex-dynamics
asked Feb 1 at 13:33
user30523user30523
564510
$endgroup$
I am reading Devaney's "An Introduction to Chaotic Dynamical Systems" and I am trying to convince myself of a claim made there (Section 3.6, proposition 6.2). The proposition concerns showing that the Julia set of $f(z)=z^2+c$ is simple and closed if $|c|<1/4$. The claim is
If the critical point $0$ and its image $c$ is contained in a circle $Gamma$ then the preimage of $Gamma$ is a simple closed curve.
I am stuck on how to prove this. Would I need to construct a continuous limiting curve as in the proof for the Julia set?
complex-dynamics
complex-dynamics
asked Feb 1 at 13:33
user30523user30523
564510
asked Feb 1 at 13:33
user30523user30523
564510
asked Feb 1 at 13:33
user30523user30523
564510
asked Feb 1 at 13:33
user30523user30523
564510
asked Feb 1 at 13:33
user30523user30523
564510
564510
add a comment |
add a comment |
1 Answer
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In fact, we can write down a pretty simple parameterization for this simple closed curve. Suppose that our circle $Gamma$ is parametrized by
$$
gamma(t) = z_0 + r e^{it}, text{ for } : 0leq t leq 2pi.
$$
Then, the parameterization of $f^{-1}(Gamma)$ will be
$$
p(t) = pm sqrt{(z_0-c) + r e^{it}}.
$$
The fact that $c$ is inside $Gamma$ implies that the shifted circle $Gamma_c = (z_0-c) + r e^{it}$ contains the origin. Thus it boils down to showing that the inverse image of any circle under the square function contains the origin. This is a simple consequence of the fact that the square root function maps the circle to the right half-plane, the negative square root function maps the circle to the left half-plane, and the endpoints match up.
answered Feb 3 at 10:17
Mark McClureMark McClure
23.9k34472
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$begingroup$
How do the facts $0inGamma_c$ and $0 in (z^2)^{-1}(text{circle})$ show $p(t)$ is closed and simple?
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– user30523
Feb 19 at 19:13
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In fact, we can write down a pretty simple parameterization for this simple closed curve. Suppose that our circle $Gamma$ is parametrized by
$$
gamma(t) = z_0 + r e^{it}, text{ for } : 0leq t leq 2pi.
$$
Then, the parameterization of $f^{-1}(Gamma)$ will be
$$
p(t) = pm sqrt{(z_0-c) + r e^{it}}.
$$
The fact that $c$ is inside $Gamma$ implies that the shifted circle $Gamma_c = (z_0-c) + r e^{it}$ contains the origin. Thus it boils down to showing that the inverse image of any circle under the square function contains the origin. This is a simple consequence of the fact that the square root function maps the circle to the right half-plane, the negative square root function maps the circle to the left half-plane, and the endpoints match up.
answered Feb 3 at 10:17
Mark McClureMark McClure
23.9k34472
$endgroup$
$begingroup$
How do the facts $0inGamma_c$ and $0 in (z^2)^{-1}(text{circle})$ show $p(t)$ is closed and simple?
$endgroup$
– user30523
Feb 19 at 19:13
add a comment |
$begingroup$
In fact, we can write down a pretty simple parameterization for this simple closed curve. Suppose that our circle $Gamma$ is parametrized by
$$
gamma(t) = z_0 + r e^{it}, text{ for } : 0leq t leq 2pi.
$$
Then, the parameterization of $f^{-1}(Gamma)$ will be
$$
p(t) = pm sqrt{(z_0-c) + r e^{it}}.
$$
The fact that $c$ is inside $Gamma$ implies that the shifted circle $Gamma_c = (z_0-c) + r e^{it}$ contains the origin. Thus it boils down to showing that the inverse image of any circle under the square function contains the origin. This is a simple consequence of the fact that the square root function maps the circle to the right half-plane, the negative square root function maps the circle to the left half-plane, and the endpoints match up.
answered Feb 3 at 10:17
Mark McClureMark McClure
23.9k34472
$endgroup$
$begingroup$
How do the facts $0inGamma_c$ and $0 in (z^2)^{-1}(text{circle})$ show $p(t)$ is closed and simple?
$endgroup$
– user30523
Feb 19 at 19:13
add a comment |
$begingroup$
In fact, we can write down a pretty simple parameterization for this simple closed curve. Suppose that our circle $Gamma$ is parametrized by
$$
gamma(t) = z_0 + r e^{it}, text{ for } : 0leq t leq 2pi.
$$
Then, the parameterization of $f^{-1}(Gamma)$ will be
$$
p(t) = pm sqrt{(z_0-c) + r e^{it}}.
$$
The fact that $c$ is inside $Gamma$ implies that the shifted circle $Gamma_c = (z_0-c) + r e^{it}$ contains the origin. Thus it boils down to showing that the inverse image of any circle under the square function contains the origin. This is a simple consequence of the fact that the square root function maps the circle to the right half-plane, the negative square root function maps the circle to the left half-plane, and the endpoints match up.
answered Feb 3 at 10:17
Mark McClureMark McClure
23.9k34472
$endgroup$
In fact, we can write down a pretty simple parameterization for this simple closed curve. Suppose that our circle $Gamma$ is parametrized by
$$
gamma(t) = z_0 + r e^{it}, text{ for } : 0leq t leq 2pi.
$$
Then, the parameterization of $f^{-1}(Gamma)$ will be
$$
p(t) = pm sqrt{(z_0-c) + r e^{it}}.
$$
The fact that $c$ is inside $Gamma$ implies that the shifted circle $Gamma_c = (z_0-c) + r e^{it}$ contains the origin. Thus it boils down to showing that the inverse image of any circle under the square function contains the origin. This is a simple consequence of the fact that the square root function maps the circle to the right half-plane, the negative square root function maps the circle to the left half-plane, and the endpoints match up.
answered Feb 3 at 10:17
Mark McClureMark McClure
23.9k34472
answered Feb 3 at 10:17
Mark McClureMark McClure
23.9k34472
answered Feb 3 at 10:17
Mark McClureMark McClure
23.9k34472
answered Feb 3 at 10:17
Mark McClureMark McClure
23.9k34472
23.9k34472
$begingroup$
How do the facts $0inGamma_c$ and $0 in (z^2)^{-1}(text{circle})$ show $p(t)$ is closed and simple?
$endgroup$
– user30523
Feb 19 at 19:13
add a comment |
$begingroup$
How do the facts $0inGamma_c$ and $0 in (z^2)^{-1}(text{circle})$ show $p(t)$ is closed and simple?
$endgroup$
– user30523
Feb 19 at 19:13
$begingroup$
How do the facts $0inGamma_c$ and $0 in (z^2)^{-1}(text{circle})$ show $p(t)$ is closed and simple?
$endgroup$
– user30523
Feb 19 at 19:13
$begingroup$
How do the facts $0inGamma_c$ and $0 in (z^2)^{-1}(text{circle})$ show $p(t)$ is closed and simple?
$endgroup$
– user30523
Feb 19 at 19:13
add a comment |
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