(20 pts)
Locate your personal cubic Julia set in the parameter space of cubic Julia sets. That is:
Find parameter values $a$ and $b$ so that
$$f_{a,b}(z) = z^3 - 3\,a^2 z + b$$
is affinely conjugate to your personal cubic polynomial.
Open this notebook on the cubic parameter space from our class webpage and use the parameterPic command defined there to plot a slice of the cubic parameter space containing your parameters and mark the precise parameter with a point.
Are we posting a slice with $a$ constant or with $b$ constant?
@RedCrayon I think that either an $a$-slice or a $b$-slice would be sufficient.
Let $$g(z)=(2760 + 4/10) - (8164 + 8/10) z + (8051 + 4/10) z^2 - 2646 z^3$$ be my personal cubic. For $f_{a,b}(z)$ to be affinely conjugate to $g(z)$ there must exist $c,d\in\mathbb{C}$ such that $$(f_{a,b}\circ\phi)(z)=(\phi\circ g)(z)$$ where $\phi(z)=cz+d$. Define the functions in Mathematica using
g[z_] := (2760 + 4/10) - (8164 + 8/10) z + (8051 + 4/10) z^2 - 2646 z^3
f[a_, b_][z_] := z^3 - 3 a^2 z + b
\[Phi][z_] := c z + dNext, define a list of equations
eqs = Thread[CoefficientList[\[Phi][g[z]] - f[a, b][\[Phi][z]], z] == {0, 0, 0, 0}]and solve using
Solve[eqs, {a, b, c, d}]I chose the solution where $a=\frac{3i}{5}\sqrt{\frac{3}{2}}$ and $b=\frac{3i}{125}\sqrt{6}$. If we fix $b$ and take a slice of the parameter space using the parameterPic command, we get the image
where the point $a$ is marked in yellow. The code for the image above is
a = 3/5 I Sqrt[3/2]
b = (3 I Sqrt[6])/125
Show[
parameterPic[
"b", b, -1.4 - 1.4 I, 1.4 + 1.4 I, 0.005,
Frame -> True, PlotRangePadding -> None
],
ListPlot[{{Re[a], Im[a]}}, PlotStyle -> Yellow]
]The Julia set for $f_{a,b}(z)$ with my choices for $a$ and $b$ is
The Julia set for $g(z)$ is
My personal cubic polynomial is:
$$g(z) = -531 z^3 + (48321/10) z^2 - (146556/10) z + 148179/10.$$
I wrote it as $g(z)$, rather than $f(z)$ so as not to confuse it with
$$f_{a,b}(z) = z^3 - 3a^2 z + b.$$
I guess we need $\varphi(z) = m z + c$ such that $\varphi\circ g = f_{a,b}\circ \varphi$. In Mathematica code, we need to solve the following equations:
g[z_] = -531 z^3 + (48321/10) z^2 - (146556/10) z + 148179/10;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
phi[z_] = m*z + c;
eqs = CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0};
eqs // InputForm
(* Out:
{-b + c + 3*a^2*c - c^3 + (148179*m)/10,
(-73278*m)/5 + 3*a^2*m - 3*c^2*m,
(48321*m)/10 - 3*c*m^2,
-531*m - m^3} == {0, 0, 0, 0}
*)We can solve these equations as follows:
sols = Solve[eqs, {a, b, m, c}]
(* Out:
{a -> -((I Sqrt[59])/10), b -> (9 I Sqrt[59])/500, m -> 3 I Sqrt[59], c -> -((91 I Sqrt[59])/10)},
{a -> (I Sqrt[59])/10, b -> (9 I Sqrt[59])/500, m -> 3 I Sqrt[59], c -> -((91 I Sqrt[59])/10)},
{a -> -((I Sqrt[59])/10), b -> -((9 I Sqrt[59])/500), m -> -3 I Sqrt[59], c -> (91 I Sqrt[59])/10},
{a -> (I Sqrt[59])/10, b -> -((9 I Sqrt[59])/500), m -> -3 I Sqrt[59], c -> (91 I Sqrt[59])/10}
*)Thus, I guess my personal cubic polynomial is affinely conjugate to $f_{a,b}$, where $a$ and $b$ are given by any of those four solutions. If I choose the second one, with no leading minus signs, I get
ff[z_] = f[a, b][z] /. sols[[2]]
(* Out: (9 I Sqrt[59])/500 + (177 z)/100 + z^3 *)Let's check the two Julia sets:
Grid[{{
JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[g[z], z, ImageSize -> 400,
PlotRange -> {{2.95, 3.12}, {-0.03, 0.03}}]
}}, Spacings -> 5]
Now, I need to locate this in the cubic parameter space using the parameterPic command.
parameterPic["a", (I Sqrt[59])/10, -1 - I, 1 + I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, (9 Sqrt[59])/500}]}]
Our personal cubic polynomial is $P_{Tim}(z)=-513z^3+4668.3z^2-14158.8z+14315.7$. We wish to find parameter values $a$ and $b$ so that the polynomial $f_{a,b}(z)=z^3-3a^2z+b$ is affinely conjugate to $P_{Tim}(z)$.
Using the mentioned notebook as a reference, we modify the code to calculate solutions for our parameters.
phi[z_] = m*z + d;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
g[z_] = -513*z^3 + 4668.3*z^2 + -14158.8*z + 14315.7;
eqs = Thread[
CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0}]
Solve[eqs, {a, b, d, m}]This gives us several solutions, we select $$a=0.7549834435270749i,\; b=-0.10569768209379049i$$ $$d=68.70349336096382i,\; m=-22.64950330581225i.$$
Thus, we have $$f(z)=z^3-3(0.754983i)^2z+(-.105698209379049i)=z^3+1.71z-.105698209379049i$$
which is conjugate to $P_{Tim}(z)$ by the conjugacy $$\phi(z)=(-22.64950330581225i) z +68.70349336096382i.$$
We look back at our previous consideration of our personal julia set, and compare that image to the following image, generated in mathematica with our new $f(z)$.
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}]}
}
]
cubicJuliaPic[0.7549834435270749*I, -0.10569768209379049*I, 100, 600]We note that their shape appears the same, aside from a shift, stretch, and rotation, as expected for conjugate functions.
We are now asked to utilize the parameterPic command to plot a slice of the cubic parameter space corresponding to our $a$ and $b$, with a plot marking the precise value of the varied parameter.
We first produce a fixed $a$ slice, and mark our particular $b$:
parameterPic["a", 0.7549834435270749*I, -2 - I, 2 + I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {{PointSize[Medium], Yellow,
Point[{{0, -0.10569768209379049}}]}}]
Next, we fix our $b$ value, and mark our particular value of $a$.
parameterPic["b", -0.10569768209379049*I, -2 - I, 2 + I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {{PointSize[Medium], Yellow,
Point[{{0, 0.7549834435270749}}]}}]
My function is
$$g(z)=2842x3+17173.8x2−34591.2x+23225.2$$
We want to find $\phi(z)=mz+c$ such that $\phi\circ g=f_{a,b}\circ\phi$ where $f{a,b}=z^3-3a^2z+b$. So we will use the code
g[z_] = \[Minus]2842 z^3 + 171738/10 z^2 \[Minus] 345912/10 z + 232252/10;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
phi[z_] = m*z + c;
eqs = CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0};
eqs // InputForm
sols = Solve[eqs, {a, b, m, c}]to get a list of coefficents which is
\begin{align*}
b \rightarrow -c (-1 - 3 a^2 + c^2),\ m \rightarrow 0, a \rightarrow -(1/5)i \sqrt{29/2}\\
My function is
$$g(z)=2842x3+17173.8x2−34591.2x+23225.2$$
We want to find $\phi(z)=mz+c$ such that $\phi\circ g=f_{a,b}\circ\phi$ where $f{a,b}=z^3-3a^2z+b$. So we will use the code
g[z_] = \[Minus]2842 z^3 + 171738/10 z^2 \[Minus] 345912/10 z + 232252/10;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
phi[z_] = m*z + c;
eqs = CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0};
eqs // InputForm
sols = Solve[eqs, {a, b, m, c}]to get
{{b -> -c (-1 - 3 a^2 + c^2), m -> 0}, {a -> -(1/5) I Sqrt[29/2],
b -> (2 I Sqrt[58])/125, m -> 7 I Sqrt[58],
c -> -(141/5) I Sqrt[29/2]}, {a -> 1/5 I Sqrt[29/2],
b -> (2 I Sqrt[58])/125, m -> 7 I Sqrt[58],
c -> -(141/5) I Sqrt[29/2]}, {a -> -(1/5) I Sqrt[29/2],
b -> -((2 I Sqrt[58])/125), m -> -7 I Sqrt[58],
c -> 141/5 I Sqrt[29/2]}, {a -> 1/5 I Sqrt[29/2],
b -> -((2 I Sqrt[58])/125), m -> -7 I Sqrt[58],
c -> 141/5 I Sqrt[29/2]}We then use the following command to pull out the most interesting coefficients and name them.
ff[z_] = f[a, b][z] /. sols[[2]]
b = 2 I Sqrt[58]/125Now we will use
Grid[{{JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}], JuliaSetPlot[g[z], z]}}, Spacings -> 5]to get
Now we will find $g(z)$ in the cubic parameter space by using the command
parameterPic["a", b, -1.5 - 1.5 I, 1.5 + 1.5 I, 0.005, Frame -> True,
PlotRangePadding -> None, Epilog -> {Yellow, PointSize[Large], Point[{0, Re[b]}]}]to get
My personal polynomial is $g(z)=63.4−172.8z+158.4z^2−48z^3.$
We need $φ(z)=mz+c$ such that $φ∘g=f_{a,b}∘φ$.
In Mathematica, I solved the following equations:
g[z_] =63.4−172.8z+158.4z^2−48z^3;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
phi[z_] = m*z + c;
eqs = CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0};
eqs // InputForm
(* Out:
{-b + c + 3*a^2*c - c^3 + 63.4*m,
-172.8*m + 3*a^2*m - 3*c^2*m,
158.4*m - 3*c*m^2, -48*m - m^3} == {0, 0, 0, 0}
*)To solve this I used:
sols = Solve[eqs, {a, b, m, c}]
(* Out:
{{b -> -1. c (-1. - 3. a^2 + c^2), m -> 0.},
{a -> 0. - 0.69282 I, b -> 0. - 0.0277128 I, m -> 0. + 6.9282 I, c -> 0. - 7.62102 I},
{a -> 0. + 0.69282 I, b -> 0. - 0.0277128 I, m -> 0. + 6.9282 I, c -> 0. - 7.62102 I},
{a -> 0. - 0.69282 I, b -> 0. + 0.0277128 I, m -> 0. - 6.9282 I, c -> 0. + 7.62102 I},
{a -> 0. + 0.69282 I, b -> 0. + 0.0277128 I, m -> 0. - 6.9282 I, c -> 0. + 7.62102 I}}
*)So my personal cubic polynomial is affinely conjugate to $f_{a,b}$, where $a$ and $b$ are given by any of those four solutions. If I use the second solution with no leading minus sign to check the conjugacy, I get:
ff[z_] = f[a, b][z] /. sols[[2]]
(* Out: (0. - 0.0277128 i) + (1.44 + 0. i) z + z^3*)To check the Julia sets I used:
Grid[{{JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[g[z], z, ImageSize -> 400,
PlotRange -> {{2.95, 3.12}, {-0.03, 0.03}}]}}, Spacings -> 5]But I could only get the vertical Julia set to appear. Does anyone know what I did wrong here to get only one solid picture instead of two? It is probably something simple that I am overlooking.
Using the parameterPic command I will locate this in the cubic parameter space and create a cool picture using the following code in Mathematica:
parameterPic["a", 0.6928203230275509 I, -1 - I, 1 + I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, (9 Sqrt[59])/500}]}]And I got this image:
The resolution isn't the best because I had to copy and paste the image onto a new Mathematica document to get the image to save for some reason, but it still looks groovy.
My guess for the grid is that you need to change your plot range for the 2nd JuliaSePlot in order to see what you want to see.
So, following basically the same steps as everyone else, but with perhaps some slightly different naming. My personal cubic is:
$$f(z)=-441z^3-1278.9z^2-1234.8z-397.9$$
So I created two new functions $\phi(z)$ and $g_{a,b}(z)$ such that my polynomial is affinely conjugate to $g$ for some $a,b$ via $\phi$ for some $c,d$:
$$
\begin{array}{rcl}
g_{a,b}(z)&=&z^3-3a^2z+b\\
&\text{and}&\\
\phi(z)&=&cz+d
\end{array}
$$
I used Mackdaddica to solve:
$$(g_{a,b}\circ\phi )(z)=(\phi\circ g)(z)$$
for $a,b,c,$ and $d$:
f[z_] := -3979/10 - 12348/10 z - 12789/10 z^2 - 441 z^3
g[a_, b_][z_] := z^3 - 3 a^2*z + b
ϕ [z_] := c*z + d
eqs := CoefficientList[ϕ [f[z]] - g[a, b][ϕ [z]], z] == {0, 0,
0, 0}
eqs // InputForm;
sols = Solve[eqs, {a, b, c, d}]
(* Out:
{{b -> -d (-1 - 3 a^2 + d^2), c -> 0},
{a -> -((7 I)/10), b -> (7 I)/500, c -> -21 I, d -> -((203 I)/10)},
{a -> (7 I)/10, b -> (7 I)/500, c -> -21 I, d -> -((203 I)/10)},
{a -> -((7 I)/10), b -> -((7 I)/500), c -> 21 I, d -> (203 I)/10},
{a -> (7 I)/10, b -> -((7 I)/500), c -> 21 I, d -> (203 I)/10}}
*)Like many others, I chose the solution where the values of $a,b$ were both positive. For me, this was where $a=\frac{7i}{10}$ and $b=\frac{7i}{500}$. Then I compared the Julia sets and, like any others, found my conjugated Julia set to be rotated, translated, and streched, but the same shape:
ff[z_] = g[a, b][z] /. sols[[3]]
Grid[{{JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[f[z], z, ImageSize -> 400,
PlotRange -> {{-1.05, -.88}, {-0.03, 0.03}}]}}, Spacings -> 5]
So now I had to test my two parameterPic diagrams, one with $a$ fixed, and one with $b$ fixed
parameterPic["a", (7 I)/10, -1 - I, 1 + I, 0.005, Frame -> True,
PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, 7 /500}]}]
parameterPic["b", (7 I)/500, -1 - I, 1 + I, 0.005, Frame -> True,
PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, 7 /10}]}]
My personal cubic polynomial was $f(z) = 115.5 z^2 - 2695 z^3 $, we search for a conjugation $\phi(z) = c*z + d$ so that $f$ is affinely conjugate to $g_{a,b}(z)$.
To do this I used the following code in Mathematica
f[z_] := 115.5 z^2 - 2695 z^3
p[z_] := c*z + d
g[a_, b_][z_] := z^3 - 3 a^2*z + b
eqs = Thread[
CoefficientList[p[f[z]] - g[a, b][p[z]], z] == {0, 0, 0, 0}]
Solve[eqs, {a, b, c, d}]One of the solutions that I found was $a = -0.74162 i, b = 0.074162 i, c = 51.9134 i, d =
-0.74162 i$
This means $g_{a,b}(z) = z^3 -1.6500006732 z + .074162i$ and $\phi(z) = 51.934i*z - .74162i$
Now we backtrack to the Julia set of our original $f(z)$, utilizing Mathematica we can generate the Julia set for $f$ and $g$ and compare them
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}]}}]
and now for $g$ we use the same code and just change the PlotRange parameters and the function to get something that does correspond to a stretch of magnitude $51.934$ a rotation of $\pi / 2 $ and a shift by $-.74162 i$ 
Now we use the parameter pic plot command to show the diagrams with 'b' fixed and 'a' changing, in yellow we have highlighted our value of 'a' namely $a = .74162i$
parameterPic["b", -.074162 I, -1.5 - 1.5 I, 1.5 + 1.5 I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, 0.74162}]}]
@mark Thanks for a good class 
My personal cubic is:
$f(x) = 1220.7 - 1180.8 x + 381.3 x^2 - 41 x^3$
To find parameter values so and b so that
$f_{a,b}[z] = z^3 - 3 a^2*z + b;$
is affinity conjugate to my personal cubic we need to solve
hi[z_] = m*z + d;
f[a_, b_][z_] = z^3 - 3 a^2*z + b;
g[z_] = -41*z^3 + 381.3*z^2 - 1180.8*z + 1220.7;
eqs = Thread[CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0}]Which was taken from the mathematica notebook.
When doing so we get 4 solutions
solutions = {{b -> -1.` d (-1.` - 3.` a^2 + d^2),
m -> 0.`}, {a -> 0.` - 0.6403124237432849` I,
b -> 0.` - 0.11525623627379128` I, d -> 0.` - 19.84968513604183` I,
m -> 0.` + 6.4031242374328485` I}, {a ->
0.` + 0.6403124237432849` I, b -> 0.` - 0.11525623627379128` I,
d -> 0.` - 19.84968513604183` I,
m -> 0.` + 6.4031242374328485` I}, {a ->
0.` - 0.6403124237432849` I, b -> 0.` + 0.11525623627379128` I,
d -> 0.` + 19.84968513604183` I,
m -> 0.` - 6.4031242374328485` I}, {a ->
0.` + 0.6403124237432849` I, b -> 0.` + 0.11525623627379128` I,
d -> 0.` + 19.84968513604183` I, m -> 0.` - 6.4031242374328485` I}}I used the fourth to get
ff[z_] = f[a, b][z] /. solutions[[4]]
out = (0. + 0.115256 I) + (1.23 + 0. I) z + z^3which gives the Julia set
compared with the julia set of my personal cubic it looks correct!
Now to locate this in the cubic parameter space we will use
parameterPic["a", 0.6403124237432849 I, -1 - I, 1 + I, 0.005,
Frame -> True, PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large],
Point[{0, 0.11525623627379128}]}]which yields 
My function is
g[z_] = -406 - 1260 z - 1305 z^2 - 450 z^3;We want to find $\phi(z) = mz+c$ such that $\phi \circ g = f_{a,b} \circ \phi$. In mathematica, solving
phi[z_] = m*z + d;
f[a_, b_][z_] = z^3 + 3 a^2*z + b;
g[z_] = -406 - 1260 z - 1305 z^2 - 450 z^3;
eqs = Thread[
CoefficientList[phi[g[z]] - f[a, b][phi[z]], z] == {0, 0, 0, 0}]
Solve[eqs, {a, b, d, m}]This gives us a list of 4 possible coefficients each for a, b, d, and m. Choosing the combination with all positive values of $a= \frac{1}{\sqrt{2}} = 0.707106, b = 0, d = \frac{29i}{\sqrt{2}} = 20.5061i,$ and $m = 15i\sqrt{2} = 21.2132i$.
Therefore our new $f(z)$ function is $f(z) = z^3 - 3(0.707106)^2z + 0$ = $z^3 +1.5z$. This is conjugate to $\phi(z) = 21.2132i*z + 20.5061i$.
Checking the 2 Julia sets gives us
Grid[{{JuliaSetPlot[ff[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[g[z], z, ImageSize -> 400,
PlotRange -> {{-1.05, -.85}, {-.1, .1}}]}}, Spacings -> 5]
We can loate this in the cubic parameter space with the parameterPic command
parameterPic["a", 1/Sqrt[2], -2 - I, 2 + I, 0.005, Frame -> True,
PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{1/Sqrt[2], 0}]}]
Lets take another look at this bad-ass cubic:
$$g(z) = 19621.6 - 29223.6 z + 14508.9 z^2 - 2401 z^3$$.
We want to find $a$ and $b$ such that $f_{a,b}(z) = z^3 - 3\,a^2 z + b$ such that $f_{a,b}$ is affinely conjugate to $f$. Thus, $f_{a,b}(\phi(z))=\phi(g(z))$ where $\phi(z)=cz+d$.
We then implement the following mathematica code:
g[z_] := 19621.6 - 29223.6 z + 14508.9 z^2 - 2401 z^3
f[a_, b_][z_] := z^3 - 3 a^2 z + b
\[Phi][z_] := c z + d
eqs = CoefficientList[\[Phi][g[z]] - f[a, b][\[Phi][z]], z] == {0, 0,
0, 0};
eqs // InputFormYielding,
{-b + 19621.6*c + d + 3*a^2*d - d^3, -29223.6*c + 3*a^2*c - 3*c*d^2, 14508.9*c - 3*c^2*d,
-2401*c - c^3} == {0, 0, 0, 0}We then solve this equation using
sols = Solve[eqs, {a, b, c, d}],yielding
{a -> 0. - 0.7 I, b -> 0. - 0.014 I, c -> 0. + 49. I, d -> 0. - 98.7 I},
{a -> 0. + 0.7 I, b -> 0. - 0.014 I, c -> 0. + 49. I, d -> 0. - 98.7 I},
{a -> 0. - 0.7 I, b -> 0. + 0.014 I, c -> 0. - 49. I, d -> 0. + 98.7 I},
{a -> 0. + 0.7 I, b -> 0. + 0.014 I, c -> 0. - 49. I, d -> 0. + 98.7 I}}Thus, my personal cubic polynomial is affinely conjugate to $f_{a,b}$ whenever $a$ and $b$ are one of those solutions. We will now compare the julia set of the first solution with our initial cubic using the code:
h[z_] = f[a, b][z] /. sols[[2]]
Grid[{{JuliaSetPlot[h[z], z, ImageSize -> {Automatic, 400}],
JuliaSetPlot[g[z], z, ImageSize -> 400,
PlotRange -> {{1.98, 2.05}, {-0.015, 0.015}}]}}, Spacings -> 5]Looking pretty bad-ass i'd say.
Awesome. Now we can locate this is the cubic parameter space with the code:
parameterPic["a", -0.7I, -2 - I, 2 + I, 0.005, Frame -> True,
PlotRangePadding -> None,
Epilog -> {Yellow, PointSize[Large], Point[{0, -0.07}]}]