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How does the definition of Mandelbrot set rely on the starting point of the iteration?
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Let $f_c(z)=z^2+c$. The Mandelbrot set is the subset of the complex plane given by
$$M=f_c^(n)(0),$$
where $f_c^(n)$ denotes the $n$-th iterate of $f_c$. Let
$$M_a=f_c^(n)(a).$$
How are these sets related?
dynamical-systems fractals complex-dynamics
edited Aug 6 at 12:20
Mark McClure
22.5k34069
asked Aug 6 at 3:27
Nienz
1018
add a comment |Â
up vote
8
down vote
favorite
Let $f_c(z)=z^2+c$. The Mandelbrot set is the subset of the complex plane given by
$$M=f_c^(n)(0),$$
where $f_c^(n)$ denotes the $n$-th iterate of $f_c$. Let
$$M_a=f_c^(n)(a).$$
How are these sets related?
dynamical-systems fractals complex-dynamics
edited Aug 6 at 12:20
Mark McClure
22.5k34069
asked Aug 6 at 3:27
Nienz
1018
2
$c in M_a iff a in J_c$, the filled Julia set of $f_c$.
– lhf
Aug 6 at 10:21
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Let $f_c(z)=z^2+c$. The Mandelbrot set is the subset of the complex plane given by
$$M=f_c^(n)(0),$$
where $f_c^(n)$ denotes the $n$-th iterate of $f_c$. Let
$$M_a=f_c^(n)(a).$$
How are these sets related?
dynamical-systems fractals complex-dynamics
edited Aug 6 at 12:20
Mark McClure
22.5k34069
asked Aug 6 at 3:27
Nienz
1018
Let $f_c(z)=z^2+c$. The Mandelbrot set is the subset of the complex plane given by
$$M=f_c^(n)(0),$$
where $f_c^(n)$ denotes the $n$-th iterate of $f_c$. Let
$$M_a=f_c^(n)(a).$$
How are these sets related?
dynamical-systems fractals complex-dynamics
edited Aug 6 at 12:20
Mark McClure
22.5k34069
asked Aug 6 at 3:27
Nienz
1018
edited Aug 6 at 12:20
Mark McClure
22.5k34069
edited Aug 6 at 12:20
Mark McClure
22.5k34069
edited Aug 6 at 12:20
Mark McClure
22.5k34069
22.5k34069
asked Aug 6 at 3:27
Nienz
1018
asked Aug 6 at 3:27
Nienz
1018
asked Aug 6 at 3:27
Nienz
1018
1018
2
$c in M_a iff a in J_c$, the filled Julia set of $f_c$.
– lhf
Aug 6 at 10:21
add a comment |Â
2
$c in M_a iff a in J_c$, the filled Julia set of $f_c$.
– lhf
Aug 6 at 10:21
2
2
$c in M_a iff a in J_c$, the filled Julia set of $f_c$.
– lhf
Aug 6 at 10:21
$c in M_a iff a in J_c$, the filled Julia set of $f_c$.
– lhf
Aug 6 at 10:21
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
The Mandelbrot set is interesting because the location of a complex number $c$ in the Mandelbrot set yields information about the dynamics of the function $f_c$ and about the geometry of the Julia set $J_c$. This is due to the strong influence of the critical orbit (i.e. the orbit of the critical point zero). Generally, the orbits of of arbitrary points have no such overwhelming influence, so there's really no reason to think the sets $M_a$ to be particularly interesting.
Nonetheless, we might expect some superficial relationships. Here are some hasty claims - hasty in the sense that I've only given these a few minutes of thought.
Similarities
- Like $M$, the set $M_a$ is bounded by $max(2,|a|)$ for every $ainmathbb C$.
- Should follow from the quadratic escape criterium.
- Like $M$, the set $M_a$ is closed for every $ainmathbb C$.
- By continuity of $f_c^n$, the complement of $M_a$ should be open.
- Like $M$, the set $M_a$ is non-empty for every $ainmathbb C$.
- I'm thinking that $a-a^2in M_a$ since, for $c=a-a^2$, we get $$f_c(a) = a^2 + a - a^2 = a.$$ That is, $a$ is a fixed point of $f_c$ so its orbit is certainly bounded.
A difference
- Unlike $M$, the set $M_a$ need not be connected.
- This is just a matter of experimentation. The set $M_1$, for example, is shown below.
Note that this image suggests a gap on the real axis around $c=-2.5$. Indeed, if you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-2.5$, you'll find that the it diverges to $infty$. If, however, you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-3$, you'll find that it is just the bounded two cycle $1to-2to1$. Of course, the orbit of $1$ is fixed under $f_c(z)=z^2$. This suggests that the gap is genuine.
answered Aug 7 at 1:59
Mark McClure
22.5k34069
2
Good answer. The figure for $M_1$ does not convince me that it is not connected. I would be tempted to try some larger constant, like $M_4$. It seems like it might be obvious from that plot.
– Ross Millikan
Aug 7 at 4:10
@RossMillikan Thanks. I added some more detail to the answer indicating that there is a genuine gap on the real axis around $c=-2.5$. Of course, the set might be connected but not simply connected but it doesn't look like it. :)
– Mark McClure
Aug 7 at 4:30
Is it known whether the Mandelbrot set is simply connected?
– DanielWainfleet
Aug 7 at 7:22
1
@DanielWainfleet Yes, it certainly is. Even simpler, it's quite easy to show that the intersection of the Mandelbrot set with the real axis is an interval, which is clearly not the case for $M_1$.
– Mark McClure
Aug 7 at 12:03
add a comment |Â
up vote
0
down vote
How are these sets related?
They are related by $M = M_0$ :)
Sorry, I couldn't resist that answer. But you probably want to know how $M_a$ is related to $M$ more generally. Given $f_c$, define $g_c,a(x) := f_c(x+a) - a = (x+a)^2 + (c-a).$ Then we have
$$
f^n_c(a) = g_c,a^n(0).
$$
Therefore, $M_a$ is the equivalent of the Mandelbrot set for quadratic polynomials with a linear coefficient included (note nolinear coefficient is present in the quadratic polynomial $f_c$, but $g_c,a$ has linear coefficient $2a$).
One could define a four-dimensional version of the Mandelbrot set where we consider the behavior of all possible polynomials $boldsymbolz^2 + bx + c$ iterated starting from $z = 0$. This would be four-dimensional because there are now two complex variables, $a$ and $b$. Then $M_a$ would be the slice of this four-dimensional set where $b = 2a$, and $M$ would be the slice where $b = 0$.
answered Aug 7 at 5:15
6005
34.7k750123
I'm not sure that it makes any more sense to iterate $z^2+bx+c$ from 0 than it does to iterate $z^2+c$ from $a$.
– Mark McClure
Aug 7 at 11:40
@MarkMcClure I know, it's not a particularly illuminating observation, and of course does not compare to your answer :). I just thought it was worth pointing out the connection between starting from an arbitrary point $a$ and using a general quadratic rather than just $z^2 + c$.
– 6005
Aug 7 at 15:41
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The Mandelbrot set is interesting because the location of a complex number $c$ in the Mandelbrot set yields information about the dynamics of the function $f_c$ and about the geometry of the Julia set $J_c$. This is due to the strong influence of the critical orbit (i.e. the orbit of the critical point zero). Generally, the orbits of of arbitrary points have no such overwhelming influence, so there's really no reason to think the sets $M_a$ to be particularly interesting.
Nonetheless, we might expect some superficial relationships. Here are some hasty claims - hasty in the sense that I've only given these a few minutes of thought.
Similarities
- Like $M$, the set $M_a$ is bounded by $max(2,|a|)$ for every $ainmathbb C$.
- Should follow from the quadratic escape criterium.
- Like $M$, the set $M_a$ is closed for every $ainmathbb C$.
- By continuity of $f_c^n$, the complement of $M_a$ should be open.
- Like $M$, the set $M_a$ is non-empty for every $ainmathbb C$.
- I'm thinking that $a-a^2in M_a$ since, for $c=a-a^2$, we get $$f_c(a) = a^2 + a - a^2 = a.$$ That is, $a$ is a fixed point of $f_c$ so its orbit is certainly bounded.
A difference
- Unlike $M$, the set $M_a$ need not be connected.
- This is just a matter of experimentation. The set $M_1$, for example, is shown below.
Note that this image suggests a gap on the real axis around $c=-2.5$. Indeed, if you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-2.5$, you'll find that the it diverges to $infty$. If, however, you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-3$, you'll find that it is just the bounded two cycle $1to-2to1$. Of course, the orbit of $1$ is fixed under $f_c(z)=z^2$. This suggests that the gap is genuine.
answered Aug 7 at 1:59
Mark McClure
22.5k34069
2
Good answer. The figure for $M_1$ does not convince me that it is not connected. I would be tempted to try some larger constant, like $M_4$. It seems like it might be obvious from that plot.
– Ross Millikan
Aug 7 at 4:10
@RossMillikan Thanks. I added some more detail to the answer indicating that there is a genuine gap on the real axis around $c=-2.5$. Of course, the set might be connected but not simply connected but it doesn't look like it. :)
– Mark McClure
Aug 7 at 4:30
Is it known whether the Mandelbrot set is simply connected?
– DanielWainfleet
Aug 7 at 7:22
1
@DanielWainfleet Yes, it certainly is. Even simpler, it's quite easy to show that the intersection of the Mandelbrot set with the real axis is an interval, which is clearly not the case for $M_1$.
– Mark McClure
Aug 7 at 12:03
add a comment |Â
up vote
5
down vote
accepted
The Mandelbrot set is interesting because the location of a complex number $c$ in the Mandelbrot set yields information about the dynamics of the function $f_c$ and about the geometry of the Julia set $J_c$. This is due to the strong influence of the critical orbit (i.e. the orbit of the critical point zero). Generally, the orbits of of arbitrary points have no such overwhelming influence, so there's really no reason to think the sets $M_a$ to be particularly interesting.
Nonetheless, we might expect some superficial relationships. Here are some hasty claims - hasty in the sense that I've only given these a few minutes of thought.
Similarities
- Like $M$, the set $M_a$ is bounded by $max(2,|a|)$ for every $ainmathbb C$.
- Should follow from the quadratic escape criterium.
- Like $M$, the set $M_a$ is closed for every $ainmathbb C$.
- By continuity of $f_c^n$, the complement of $M_a$ should be open.
- Like $M$, the set $M_a$ is non-empty for every $ainmathbb C$.
- I'm thinking that $a-a^2in M_a$ since, for $c=a-a^2$, we get $$f_c(a) = a^2 + a - a^2 = a.$$ That is, $a$ is a fixed point of $f_c$ so its orbit is certainly bounded.
A difference
- Unlike $M$, the set $M_a$ need not be connected.
- This is just a matter of experimentation. The set $M_1$, for example, is shown below.
Note that this image suggests a gap on the real axis around $c=-2.5$. Indeed, if you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-2.5$, you'll find that the it diverges to $infty$. If, however, you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-3$, you'll find that it is just the bounded two cycle $1to-2to1$. Of course, the orbit of $1$ is fixed under $f_c(z)=z^2$. This suggests that the gap is genuine.
answered Aug 7 at 1:59
Mark McClure
22.5k34069
2
Good answer. The figure for $M_1$ does not convince me that it is not connected. I would be tempted to try some larger constant, like $M_4$. It seems like it might be obvious from that plot.
– Ross Millikan
Aug 7 at 4:10
@RossMillikan Thanks. I added some more detail to the answer indicating that there is a genuine gap on the real axis around $c=-2.5$. Of course, the set might be connected but not simply connected but it doesn't look like it. :)
– Mark McClure
Aug 7 at 4:30
Is it known whether the Mandelbrot set is simply connected?
– DanielWainfleet
Aug 7 at 7:22
1
@DanielWainfleet Yes, it certainly is. Even simpler, it's quite easy to show that the intersection of the Mandelbrot set with the real axis is an interval, which is clearly not the case for $M_1$.
– Mark McClure
Aug 7 at 12:03
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The Mandelbrot set is interesting because the location of a complex number $c$ in the Mandelbrot set yields information about the dynamics of the function $f_c$ and about the geometry of the Julia set $J_c$. This is due to the strong influence of the critical orbit (i.e. the orbit of the critical point zero). Generally, the orbits of of arbitrary points have no such overwhelming influence, so there's really no reason to think the sets $M_a$ to be particularly interesting.
Nonetheless, we might expect some superficial relationships. Here are some hasty claims - hasty in the sense that I've only given these a few minutes of thought.
Similarities
- Like $M$, the set $M_a$ is bounded by $max(2,|a|)$ for every $ainmathbb C$.
- Should follow from the quadratic escape criterium.
- Like $M$, the set $M_a$ is closed for every $ainmathbb C$.
- By continuity of $f_c^n$, the complement of $M_a$ should be open.
- Like $M$, the set $M_a$ is non-empty for every $ainmathbb C$.
- I'm thinking that $a-a^2in M_a$ since, for $c=a-a^2$, we get $$f_c(a) = a^2 + a - a^2 = a.$$ That is, $a$ is a fixed point of $f_c$ so its orbit is certainly bounded.
A difference
- Unlike $M$, the set $M_a$ need not be connected.
- This is just a matter of experimentation. The set $M_1$, for example, is shown below.
Note that this image suggests a gap on the real axis around $c=-2.5$. Indeed, if you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-2.5$, you'll find that the it diverges to $infty$. If, however, you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-3$, you'll find that it is just the bounded two cycle $1to-2to1$. Of course, the orbit of $1$ is fixed under $f_c(z)=z^2$. This suggests that the gap is genuine.
answered Aug 7 at 1:59
Mark McClure
22.5k34069
The Mandelbrot set is interesting because the location of a complex number $c$ in the Mandelbrot set yields information about the dynamics of the function $f_c$ and about the geometry of the Julia set $J_c$. This is due to the strong influence of the critical orbit (i.e. the orbit of the critical point zero). Generally, the orbits of of arbitrary points have no such overwhelming influence, so there's really no reason to think the sets $M_a$ to be particularly interesting.
Nonetheless, we might expect some superficial relationships. Here are some hasty claims - hasty in the sense that I've only given these a few minutes of thought.
Similarities
- Like $M$, the set $M_a$ is bounded by $max(2,|a|)$ for every $ainmathbb C$.
- Should follow from the quadratic escape criterium.
- Like $M$, the set $M_a$ is closed for every $ainmathbb C$.
- By continuity of $f_c^n$, the complement of $M_a$ should be open.
- Like $M$, the set $M_a$ is non-empty for every $ainmathbb C$.
- I'm thinking that $a-a^2in M_a$ since, for $c=a-a^2$, we get $$f_c(a) = a^2 + a - a^2 = a.$$ That is, $a$ is a fixed point of $f_c$ so its orbit is certainly bounded.
A difference
- Unlike $M$, the set $M_a$ need not be connected.
- This is just a matter of experimentation. The set $M_1$, for example, is shown below.
Note that this image suggests a gap on the real axis around $c=-2.5$. Indeed, if you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-2.5$, you'll find that the it diverges to $infty$. If, however, you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-3$, you'll find that it is just the bounded two cycle $1to-2to1$. Of course, the orbit of $1$ is fixed under $f_c(z)=z^2$. This suggests that the gap is genuine.
answered Aug 7 at 1:59
Mark McClure
22.5k34069
edited Aug 7 at 4:29
answered Aug 7 at 1:59
Mark McClure
22.5k34069
answered Aug 7 at 1:59
Mark McClure
22.5k34069
answered Aug 7 at 1:59
Mark McClure
22.5k34069
22.5k34069
2
Good answer. The figure for $M_1$ does not convince me that it is not connected. I would be tempted to try some larger constant, like $M_4$. It seems like it might be obvious from that plot.
– Ross Millikan
Aug 7 at 4:10
@RossMillikan Thanks. I added some more detail to the answer indicating that there is a genuine gap on the real axis around $c=-2.5$. Of course, the set might be connected but not simply connected but it doesn't look like it. :)
– Mark McClure
Aug 7 at 4:30
Is it known whether the Mandelbrot set is simply connected?
– DanielWainfleet
Aug 7 at 7:22
1
@DanielWainfleet Yes, it certainly is. Even simpler, it's quite easy to show that the intersection of the Mandelbrot set with the real axis is an interval, which is clearly not the case for $M_1$.
– Mark McClure
Aug 7 at 12:03
add a comment |Â
2
Good answer. The figure for $M_1$ does not convince me that it is not connected. I would be tempted to try some larger constant, like $M_4$. It seems like it might be obvious from that plot.
– Ross Millikan
Aug 7 at 4:10
@RossMillikan Thanks. I added some more detail to the answer indicating that there is a genuine gap on the real axis around $c=-2.5$. Of course, the set might be connected but not simply connected but it doesn't look like it. :)
– Mark McClure
Aug 7 at 4:30
Is it known whether the Mandelbrot set is simply connected?
– DanielWainfleet
Aug 7 at 7:22
1
@DanielWainfleet Yes, it certainly is. Even simpler, it's quite easy to show that the intersection of the Mandelbrot set with the real axis is an interval, which is clearly not the case for $M_1$.
– Mark McClure
Aug 7 at 12:03
2
2
Good answer. The figure for $M_1$ does not convince me that it is not connected. I would be tempted to try some larger constant, like $M_4$. It seems like it might be obvious from that plot.
– Ross Millikan
Aug 7 at 4:10
Good answer. The figure for $M_1$ does not convince me that it is not connected. I would be tempted to try some larger constant, like $M_4$. It seems like it might be obvious from that plot.
– Ross Millikan
Aug 7 at 4:10
@RossMillikan Thanks. I added some more detail to the answer indicating that there is a genuine gap on the real axis around $c=-2.5$. Of course, the set might be connected but not simply connected but it doesn't look like it. :)
– Mark McClure
Aug 7 at 4:30
@RossMillikan Thanks. I added some more detail to the answer indicating that there is a genuine gap on the real axis around $c=-2.5$. Of course, the set might be connected but not simply connected but it doesn't look like it. :)
– Mark McClure
Aug 7 at 4:30
Is it known whether the Mandelbrot set is simply connected?
– DanielWainfleet
Aug 7 at 7:22
Is it known whether the Mandelbrot set is simply connected?
– DanielWainfleet
Aug 7 at 7:22
1
1
@DanielWainfleet Yes, it certainly is. Even simpler, it's quite easy to show that the intersection of the Mandelbrot set with the real axis is an interval, which is clearly not the case for $M_1$.
– Mark McClure
Aug 7 at 12:03
@DanielWainfleet Yes, it certainly is. Even simpler, it's quite easy to show that the intersection of the Mandelbrot set with the real axis is an interval, which is clearly not the case for $M_1$.
– Mark McClure
Aug 7 at 12:03
add a comment |Â
up vote
0
down vote
How are these sets related?
They are related by $M = M_0$ :)
Sorry, I couldn't resist that answer. But you probably want to know how $M_a$ is related to $M$ more generally. Given $f_c$, define $g_c,a(x) := f_c(x+a) - a = (x+a)^2 + (c-a).$ Then we have
$$
f^n_c(a) = g_c,a^n(0).
$$
Therefore, $M_a$ is the equivalent of the Mandelbrot set for quadratic polynomials with a linear coefficient included (note nolinear coefficient is present in the quadratic polynomial $f_c$, but $g_c,a$ has linear coefficient $2a$).
One could define a four-dimensional version of the Mandelbrot set where we consider the behavior of all possible polynomials $boldsymbolz^2 + bx + c$ iterated starting from $z = 0$. This would be four-dimensional because there are now two complex variables, $a$ and $b$. Then $M_a$ would be the slice of this four-dimensional set where $b = 2a$, and $M$ would be the slice where $b = 0$.
answered Aug 7 at 5:15
6005
34.7k750123
I'm not sure that it makes any more sense to iterate $z^2+bx+c$ from 0 than it does to iterate $z^2+c$ from $a$.
– Mark McClure
Aug 7 at 11:40
@MarkMcClure I know, it's not a particularly illuminating observation, and of course does not compare to your answer :). I just thought it was worth pointing out the connection between starting from an arbitrary point $a$ and using a general quadratic rather than just $z^2 + c$.
– 6005
Aug 7 at 15:41
add a comment |Â
up vote
0
down vote
How are these sets related?
They are related by $M = M_0$ :)
Sorry, I couldn't resist that answer. But you probably want to know how $M_a$ is related to $M$ more generally. Given $f_c$, define $g_c,a(x) := f_c(x+a) - a = (x+a)^2 + (c-a).$ Then we have
$$
f^n_c(a) = g_c,a^n(0).
$$
Therefore, $M_a$ is the equivalent of the Mandelbrot set for quadratic polynomials with a linear coefficient included (note nolinear coefficient is present in the quadratic polynomial $f_c$, but $g_c,a$ has linear coefficient $2a$).
One could define a four-dimensional version of the Mandelbrot set where we consider the behavior of all possible polynomials $boldsymbolz^2 + bx + c$ iterated starting from $z = 0$. This would be four-dimensional because there are now two complex variables, $a$ and $b$. Then $M_a$ would be the slice of this four-dimensional set where $b = 2a$, and $M$ would be the slice where $b = 0$.
answered Aug 7 at 5:15
6005
34.7k750123
I'm not sure that it makes any more sense to iterate $z^2+bx+c$ from 0 than it does to iterate $z^2+c$ from $a$.
– Mark McClure
Aug 7 at 11:40
@MarkMcClure I know, it's not a particularly illuminating observation, and of course does not compare to your answer :). I just thought it was worth pointing out the connection between starting from an arbitrary point $a$ and using a general quadratic rather than just $z^2 + c$.
– 6005
Aug 7 at 15:41
add a comment |Â
up vote
0
down vote
up vote
0
down vote
How are these sets related?
They are related by $M = M_0$ :)
Sorry, I couldn't resist that answer. But you probably want to know how $M_a$ is related to $M$ more generally. Given $f_c$, define $g_c,a(x) := f_c(x+a) - a = (x+a)^2 + (c-a).$ Then we have
$$
f^n_c(a) = g_c,a^n(0).
$$
Therefore, $M_a$ is the equivalent of the Mandelbrot set for quadratic polynomials with a linear coefficient included (note nolinear coefficient is present in the quadratic polynomial $f_c$, but $g_c,a$ has linear coefficient $2a$).
One could define a four-dimensional version of the Mandelbrot set where we consider the behavior of all possible polynomials $boldsymbolz^2 + bx + c$ iterated starting from $z = 0$. This would be four-dimensional because there are now two complex variables, $a$ and $b$. Then $M_a$ would be the slice of this four-dimensional set where $b = 2a$, and $M$ would be the slice where $b = 0$.
answered Aug 7 at 5:15
6005
34.7k750123
How are these sets related?
They are related by $M = M_0$ :)
Sorry, I couldn't resist that answer. But you probably want to know how $M_a$ is related to $M$ more generally. Given $f_c$, define $g_c,a(x) := f_c(x+a) - a = (x+a)^2 + (c-a).$ Then we have
$$
f^n_c(a) = g_c,a^n(0).
$$
Therefore, $M_a$ is the equivalent of the Mandelbrot set for quadratic polynomials with a linear coefficient included (note nolinear coefficient is present in the quadratic polynomial $f_c$, but $g_c,a$ has linear coefficient $2a$).
One could define a four-dimensional version of the Mandelbrot set where we consider the behavior of all possible polynomials $boldsymbolz^2 + bx + c$ iterated starting from $z = 0$. This would be four-dimensional because there are now two complex variables, $a$ and $b$. Then $M_a$ would be the slice of this four-dimensional set where $b = 2a$, and $M$ would be the slice where $b = 0$.
answered Aug 7 at 5:15
6005
34.7k750123
answered Aug 7 at 5:15
6005
34.7k750123
answered Aug 7 at 5:15
6005
34.7k750123
answered Aug 7 at 5:15
6005
34.7k750123
34.7k750123
I'm not sure that it makes any more sense to iterate $z^2+bx+c$ from 0 than it does to iterate $z^2+c$ from $a$.
– Mark McClure
Aug 7 at 11:40
@MarkMcClure I know, it's not a particularly illuminating observation, and of course does not compare to your answer :). I just thought it was worth pointing out the connection between starting from an arbitrary point $a$ and using a general quadratic rather than just $z^2 + c$.
– 6005
Aug 7 at 15:41
add a comment |Â
I'm not sure that it makes any more sense to iterate $z^2+bx+c$ from 0 than it does to iterate $z^2+c$ from $a$.
– Mark McClure
Aug 7 at 11:40
@MarkMcClure I know, it's not a particularly illuminating observation, and of course does not compare to your answer :). I just thought it was worth pointing out the connection between starting from an arbitrary point $a$ and using a general quadratic rather than just $z^2 + c$.
– 6005
Aug 7 at 15:41
I'm not sure that it makes any more sense to iterate $z^2+bx+c$ from 0 than it does to iterate $z^2+c$ from $a$.
– Mark McClure
Aug 7 at 11:40
I'm not sure that it makes any more sense to iterate $z^2+bx+c$ from 0 than it does to iterate $z^2+c$ from $a$.
– Mark McClure
Aug 7 at 11:40
@MarkMcClure I know, it's not a particularly illuminating observation, and of course does not compare to your answer :). I just thought it was worth pointing out the connection between starting from an arbitrary point $a$ and using a general quadratic rather than just $z^2 + c$.
– 6005
Aug 7 at 15:41
@MarkMcClure I know, it's not a particularly illuminating observation, and of course does not compare to your answer :). I just thought it was worth pointing out the connection between starting from an arbitrary point $a$ and using a general quadratic rather than just $z^2 + c$.
– 6005
Aug 7 at 15:41
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2
$c in M_a iff a in J_c$, the filled Julia set of $f_c$.
– lhf
Aug 6 at 10:21