(10 pts)

Plot the Julia set of your personal poynomial and indicate the location of any attractive behavior that you see.

An archived instance of discourse for discussion in undergraduate Complex Dynamics.

mark

(10 pts)

Plot the Julia set of your personal poynomial and indicate the location of any attractive behavior that you see.

audrey

Here's the code I used to generate my Julia set:

```
f[z_] = -531z^3 + (48321/10)z^2-(146556/10)z+148179/10;
pic = JuliaSetPlot[f[z],z,ImageResolution->1000,
PlotRange -> {{2.95,3.15},{-0.03,0.03}},
Epilog -> {
{PointSize[Medium],Green,Point[{{3,0},{3.075,0}}]},
{PointSize[Medium],Red,Point[{3.025,0}]}}]
```

I found the fixed points using this code: `NSolve[f[z] == z, z]`

RedCrayon

Here's the code I used to generate my Julia set:

```
f[z_] := (2760 + 4/10) - (8164 + 8/10) z + (8051 + 4/10) z^2 - 2646 z^3;
JuliaSetPlot[
f[z],
z,
PlotRange -> {
{0.97, 1.06},
{-0.02, 0.02}
},
ImageResolution -> 1100,
Epilog -> {
{PointSize[Medium], Green, Point[{z0[[1]], 0}]},
{PointSize[Medium], Red, Point[{z0[[2]], 0}]},
{PointSize[Medium], White, Point[{z0[[3]], 0}]}
}
]
```

The fixed points are $z_1=0$, $z_2=1.01241$ and $z_3=1.03044$.

$z_1$ is a super-attractive fixed point since $f'(z_1)=0$.

$z_2$ is a repulsive fixed point since $|f'(z_2)|=1.5922>1$.

$z_3$ is an attractive fixed point since $0<|f'(z_3)|=0.452204<1$.

EDIT: Here's the picture.

Person

Here is the code I used to generate my picture

```
f[x_] := 23225.2 - 34591.2 x + 17173.8 x^2 - 2842 x^3
JuliaSetPlot[f[x], x, PlotRange -> {{1.98, 2.050}, {-0.015, 0.015}},
ImageResolution -> 1000,
Epilog -> {
{PointSize[Medium], Blue, Point[{{2, 0}, {2.031788049679393`, 0}}]},
{PointSize[Medium], Red, Point[{2.011069093162045`, 0}]}}]
```

and here is my picture

Yousername

I used this code `Solve[f[x] == x, x]`

to get the fixed points x=1, 1.10918, and 1.19082.

I used the following code to generate my Julia Set:

`f[x_] := 63.4 \[Minus] 172.8 x + 158.4 x^2 \[Minus] 48 x^3;`

, 0}]}}]`

JuliaSetPlot[f[x], x, PlotRange -> {{0.8, 1.4}, {-0.16, 0.16}},

ImageResolution -> 1100,

Epilog -> {{PointSize[Medium], Blue,

Point[{{1, 0}, {1.19082, 0}}]}, {PointSize[Medium], Red,

Point[{1.10918

Here is the picture generated by Mathematica:

SomeCallMeTim

My Personal (complex) Polynomial is $f(z)=14315.7-14158.8z+4668.3z^2-513z^3$.

I used the code

`NSolve[f[z] == z, z]`

To find the fixed points $3$,$3.02653$, and $3.07347$. To classify these points, I calculated

$|f'(3)|=0 \Rightarrow 3$ is a super attractive fixed point,

$|f'(3.02653)|=1.63877>1$, so $3.02653$ is a repulsive fixed point,

$|f'(3.07347)|=.769255$, so $3.07347$ is an attractive fixed point.

Having found these characteristics of our set, we now use the following code to generate the following image:

```
pic = JuliaSetPlot[f[z], z, ImageResolution -> 1000,
PlotRange -> {{2.95, 3.11}, {-0.03, 0.03}},
Epilog -> {
{PointSize[Medium], Green, Point[{3.0000000000036087, 0}]},
{PointSize[Medium], Red, Point[{3.02653, 0}]},
{PointSize[Medium], White, Point[{3.07347, 0}]}
}
]
```

Levente

My polynomial is $f(z) = 115.5 z^2 - 2695 z^3$, I plotted its Julia set with the following Mathematica Code

```
f[z_] = 115.5 z^2 - 2695 z^3;
pic = JuliaSetPlot[f[z], z, ImageResolution -> 1000,
PlotRange -> {{-.02, .05}, {-0.02, 0.02}},
Epilog -> {{PointSize[Medium], Green,
Point[{{0, 0}, {0.012041, 0}}]}, {PointSize[Medium], Red,
Point[{0.0308161, 0}]}}]
```

The plot can be seen below with a super attractive fixed point at $0$ an attractive fixed point at $x =.0308161$

and a repulsive fixed point at $x =0.012041$

Rick

I used this code to generate my Julia set:

```
f[z_] = 19621.6 - 29223.6 z + 14508.9 z^2 - 2401 z^3;
JuliaSetPlot[f[z], z, PlotRange -> {{1.98, 2.05}, {-.015, .015}},<img
ImageResolution -> 1100,
Epilog -> {{PointSize[Medium], Green,
Point[{{2.01489, 0}, {2.02796, 0}}]}, {PointSize[Medium], Red,
Point[{2.00, 0}]}}]
```

This pretty thang has a super attractive fixed point at

z=2, a repulsive fixed point at z = 2.01489, and an attractive fixed point at

z = 2.02796.

WillHeDoTheAssignmen

I used this code to generate my Julia Set:

```
f[z_] = -406 - 1260 z - 1305 z^2 - 450 z^3
pic = JuliaSetPlot[f[z], z, ImageResolution -> 1000,
PlotRange -> {{-1.05, -.89}, {-0.03, 0.03}},
Epilog -> {{PointSize[Medium], Green,
Point[{-1, 0}]}, {PointSize[Medium], Red,
Point[{-0.9666666666667537, 0}]}, {PointSize[Medium], White,
Point[{-0.9333333333332109, 0}]}}]
```

The point (-1,0) is a superattractive fixed point, as f'[-1] = 0.

The point (-0.96666,0) is a repulsive fixed point, as f'[-0.96666] = 1.5, which is greater than 1.

the point (-0.93333,0) is a superattractive fixed point, as f'[-0.9333333333332109] = 0.

Captain_Flapjack

I checked my 2 fixed points for behavior, one was super attractive, one attractive, and one repulsive and one.The super attractive point is at $z=-1$. The attractive point $z=-0.934754$ has multiplier $\lambda=0.122672$, and is quite close to the critical point at $z=-.933\bar{3}$. The repulsive point at $z=-0.965246$ has multiplier $\lambda=1.46733$. I use the following code:

```
f[x_] = -3979/10 - 12348/10 x - 12789/10 x^2 - 441 x^3;
NSolve[f'[x] == 0, x]
NSolve[f[x] == x, x]
f'[-0.9652455338982358`]
f'[-0.9347544661016687`]
f[-0.9333333333333309`]
```

Then I used the following code to generate a julia set plot with a green dot at the super attractive point, a blue dot at the attractive point, and a red dot at the repulsive one:

```
JuliaSetPlot[f[x], x, ImageSize -> Large,
PlotRange -> {{-1.045, -.89}, {-.05, .05}},
Epilog -> {{PointSize[Medium], Green,
Point[{{-1, 0}}]}, {PointSize[Medium], Blue,
Point[{{-0.9347544661016687`, 0}}]}, {PointSize[Medium], Red,
Point[{-0.9652455338982358`, 0}]}}]
```

For some fun stuff:

notneds

Using

```
g[z_] = -41*z^3 + 381.3*z^2 - 1180.8*z + 1220.7;
JuliaSetPlot[g[z], z,
ImageSize -> Large,
Epilog -> {{PointSize[Medium], Green,
Point[{{3, 0}}]}, {PointSize[Medium], Blue, Point[{{3.2, 0}}]}}]
```

I generated the picture

There are two attractive fixed points, one is located at 3.2 and the other is located at 3 and is super-attractive!

mark

@notneds That's crazy looking!