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

mark

10 pts

Continuing with your personal cubic polynomial $f$, define
$$F_c(x) = f(x) + c.$$
1. Find a value of $c$ such that $F_c$ has a super-attractive fixed point.
2. Find a value of $c$ such that $F_c$ has a super-attractive orbit of period 2.

Note, you'll probably need to do this on the computer.

RedCrayon

1.

Define a cubic function $$f(x)=2760.4-8164.8x+8051.4x^2-2646x^3$$ Then $$F_c(x)=2760.4-8164.8x+8051.4x^2-2646x^3+c$$ We know a super-attractive fixed point must be both a critical point and a fixed point of the function $F_c(x)$ for some value of $c$.

Note that $F_c^{'}(x)=0$ when $x=1$. So $x_0=1$ is a critical point for all $c$.

Also, $F_0(x)=x$ when $x=1$. So $x_1=1$ is a fixed point when $c=0$.

Therefore, $x_0=x_1=1$ is a super-attractive fixed point when $c=0$.

2.

We know a super-attractive orbit must contain a critical point.

Additionally, we know that for all $c$, we will have a critical point at $x_0=1$.

For an orbit of period $2$ to contain $x_0$, $c$ must satisfy $$F_c^2(x_0)=x_0$$ This occurs for $c_1=-0.007504327196$, $c_2=0$ and $c_3=0.05036147005$.

Definitions used: 1.2. A point $x_0 ∈ R$ is a fixed point of $f$ if $f(x_0) = x_0$.
1.3. Suppose that the orbit $(x_n)$ satisfies
$x_0 → x_1 → x_2 · · · → x_n−1 → x_0$
and $x_n = x_0$. Such an orbit is called a periodic orbit and the points themselves
are called periodic points. If $x_k$ does not equal $x_0$ for $k = 1, 2, . . . , n − 1$, then $n$ is called the period of the orbit.

Definition 3.4. Let $f : R → R$ be continuously differentiable and suppose
that $x_0 ∈ R$ is a fixed point of $f$. Then we classify $x_0$ as
1. attractive, if $0 < |f'(x_0)| < 1$,
2. super-attractive, if $|f'(x_0)| =0$
3. repulsive or repelling, if $|f'(x_0)| > 1$, or
4. neutral, if $|f'(x_0)| = 1$

My personal polynomial is $f(x)=63.4−172.8x+158.4x^2−48x^3$.

Here is my graphed personal cubic polynomial from .9<x<1.25.
You can see in this graph that $|f'(x_0)|=0$ when $x_0=1$. $x_0$ is also a fixed point because $f(x_0) = x_0$, so $x_0$ here is a super attractive fixed point when $c=0$. $F_c$ has a super-attractive fixed point when $c=0$.

Using this code:
f[x_] = f(x)=63.4−172.8x+158.4x^2−48x^3;
Solve[f'[x] == 0, x]
(* Out:
{{x -> 1.}, {x -> 1.2}}
*)

I found the critical points x=1 and x=1.2.

Using this code:
F[c_][x_] = f[x] + c;
FF[c_][x_] = Nest[F[c], x, 2];
Solve[FF[c][1] == 1, c]

I got values {c -> -0.0581666}, {c -> 0.}, and {c -> 0.358167} as values for c, so I will test a non zero c value for the attractive orbit of period 2.

Using this code:
NestList[F[.358167], 1, 6]

I got {1, 1.35817, 0.999997, 1.35817, 0.999997, 1.35817, 0.999997}, and found an attractive orbit of period 2 for the value of c=0.358167.

audrey

My personal cubic polynomial is

$$f(x) = -531 x^3 + 4832.1 x^2 - 14655.6 x + 14817.9.$$

I can find the critical points of $f$ with Mathematica like this:

f[x_] = -531 x^3 + 4832.1 x^2 - 14655.6 x + 14817.9;
Solve[f'[x] == 0, x]
(* Out:
{{x -> 3.}, {x -> 3.06667}}
*)

Next, my $F_c$ is
$$F_c(x) = -531 x^3 + 4832.1 x^2 - 14655.6 x + 14817.9 + c.$$
Sticking with the easier critical point, namely $3$, I guess we need to know a value of $c$ for which
$F_c(F_c(3)) = 3$ but $F_c(3)\neq 3$. We can find find this like so:

F[c_][x_] = f[x] + c;
FF[c_][x_] = Nest[F[c], x, 2];
Solve[FF[c][3] == 3, c]
(* Out:
{{c -> -0.016206}, {c -> 0.}, {c -> 0.116206}}
*)

Now, the 0 doesn't count, because 3 is already a fixed point of $f$. So let's try one of the others:

NestList[F[0.116206], 3, 4]
(* Out:
{3, 3.11621, 3., 3.11621, 3.}
*)

Looks good!

Person

My polynomial is

$$f(x)=23225.2 - 34591.2 x + 17173.8 x^2 - 2842 x^3$$$$\Rightarrow F_c(x)=23225.2 - 34591.2 x + 17173.8 x^2 - 2842 x^3+c$$

defined my functions as such in mathematica:

f[x_] := 23225.2 - 34591.2 x + 17173.8 x^2 - 2842 x^3
F[c_][x_] := f[x] + c;

First to find the critical points, I used Solve[f'[x] == 0, x] to find where $f'(x)=0$ which got me $2.$ and $2.02857$.

I then used the code

F[c_][x_] := f[x] + c;
FF[c_][x_] = Nest[F[c], x, 2];
Solve[FF[c][2] == 2, c]

to find $c=-0.00705034$ and $c=0.0499075$.

Then to test if $c=0.0499075$ has a supper atractive orbit of period 2, I used the code

NestList[F[0.049907485687264004], 2, 4]

which gave me an output of { $2, 2.04991, 2., 2.04991, 2.$}.

SomeCallMeTim

My Polynomial is $f(x)=14315.7-14158.8x+4668.3x^2-513x^3$
$\Rightarrow F_c(x)=14315.7-14158.8x+4668.3x^2-513x^3+c$

I begin by defining my functions in Mathematica:

f[x_] := 14315.7 - 14158.8 x + 4668.3 x^2 - 513 x^3
F[c_][x_] := f[x] + c

I next solve for super attractive points and test if the point is fixed:

Solve[f'[x] == 0, x]

Which gives us the output {{x -> 3.}, {x -> 3.06667}}, which we test with

f[3];
f[3.0666666666666753];

The output of $f(3)=3$, but $f(3.0666666666666753)\neq 3.0666666666666753$,
and so we have $3$ as a super-attractive fixed point when $c=0$.

Next, we seek to find a value of $c$ such that $F_c(x)$ has a super attractive orbit of period $2$.

We already know $3$ is a super attractive point for $F$, so we again turn to mathematica, define our twice iterated function, and solve for a c which produces the required behavior:

FF[c_][x_] = Nest[F[c], x, 2];
Solve[FF[c][3] == 3, c]

Which gives out the output {{c -> -0.0167032}, {c -> 0.}, {c -> 0.116703}}.
We already know the solution for $c=0$ corresponds to the fixed point solution, so we need only test the behavior of the two non-zero solutions.

NestList[F[-0.01670320638112171], 3, 4]
NestList[F[0.11670320634838026], 3, 4]

Which produces the outputs {3, 2.9833, 3., 2.9833, 3.} and {3, 3.1167, 3., 3.1167, 3.} respectively. These are both cycles of period $2$, and in both cases $F'_c(3)$=0, so $c=-0.01670320638112171$ and $c=0.11670320634838026$ are super-attractive orbits of period $2$.

Captain_Flapjack

As many have stated, a $\textit{super attractive}$ fixed point of a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is a point $x_0$ such that $f(x_0)=x_0$ $\textbf{AND}$ $F'(x_0)=0$. Using the following Mathematica code, I found two of my function's three fixed points to be supper attractive if I choose $c=0$ :

In[1]:= f[x_] = -397.9 - 1234.8 x - 1278.9 x^2 - 441 x^3;
NSolve[f'[x] == 0, x]
Out[2]= {{x -> -1.}, {x -> -0.933333}}

So now, using the critical point $x_0=-1$, I need to find a $c$ such that $f^2_c(-1)=-1$ but $f_c(-1)\neq-1$:

In[3]:= F[c_][x_] = f[x] + c;
G[c_][x_] = Nest[F[c], x, 2];
NSolve[G[c][-1] == -1, c]
Out[4]= {{c -> -0.0190476}, {c -> 0.}, {c -> 0.119048}}

The middle point actually just corresponds to the super attractive fixed point at $c=0$. Just to make sure, I plugged each point in and iterated:

In[5]:= NestList[F[-0.01904761904760373], -1, 4]
NestList[F[0.11904761904771746], -1, 4]
NestList[F[0], -1, 4]
Out[6]= {-1, -1.01905, -1., -1.01905, -1.}
Out[7]= {-1, -0.880952, -1., -0.880952, -1.}
Out[8]= {-1, -1., -1., -1., -1.}

Looks like 2 period 2 orbits and a fixed point!

WillHeDoTheAssignmen

With my function of $-406 - 1260 x - 1305 x^2 - 450 x^3$, I found 3 super attractive fixed points (points where $f(x_0) = x_0$ and $f'(x_0) = x_0 )$. Using my code

f[x_]  := -406 - 1260 x - 1305 x^2 - 450 x^3;
NSolve [f'[x] == 0, x]

I got

  {{x -> -1.}, {x -> -0.933333}}

Then using

F[c_][x_] = f[x] + c;
G[c_][x_] = Nest[F[c], x, 2];
NSolve[G[c][-1] == -1, c]

I got

{{c -> -0.0187184}, {c -> 0.}, {c -> 0.118718}}

Testing each of these values with NestList of each yields:

NestList[F[-0.018718427093627678], -1, 4]
{-1, -1.01872, -1., -1.01872, -1.}
NestList[F[0], -1, 4]
{-1, -1, -1, -1, -1}
NestList[F[0.11871842709362768], -1, 4]
{-1, -0.881282, -1., -0.881282, -1.}

Lo and behold, these points are super-attractive with periodicity 2.

Levente

My function was $f(x) = 115.5x^2 - 2695x^3$, we can easily take its derivative and find the critical points, namely we get $z = 0$ and $z= 0.0285714$

Now we have $F_c(x) = 115.5x^2 - 2695x^3 + c$
We could use $z = 0$ since it easier to work with,

So now we define our twice iterated function and since super attractive orbits attract critical points all we need to do is solve for a fixed point of the twice iterated function.

FF[c_][x_] = Nest[F[c], x, 2];
Solve[FF[c][0] == 0, c]

//output
{{c -> -0.00738533}, {c -> 0.}, {c -> 0.0502425}}

We know that 0 will not work for us since it is results in a fixed point of the original $f$, so we test the other two
NestList[F[-0.007385334717316609], 0, 4]
which gives a orbit of period 2 ( if we exclude the numeric error happening here )
{0, -0.00738533, -1.73472*10^-18, -0.00738533, -1.73472*10^-18},

and we try

NestList[F[.05024247757445947], 0, 4]

which also gives us an orbit of period 2 (again excluding the numeric error)

{0, 0.0502425, -2.08167*10^-17, 0.0502425, -2.08167*10^-17}

Thus we successfully found a value of $c$ so that $f(x) + c$ has a superattractive orbit of period 2.

notneds

My personal Cubic is
$f(x) = 1220.7 - 1180.8 x + 381.3 x^2 - 41 x^3$
To find a $c$ so that $F_c$ has a super-attractive fixed point, first we solve

g[z_] = -41*z^3 + 381.3*z^2 - 1180.8*z + 1220.7;
NSolve[g[x] == x, x]

Which returns

{{x -> 3.}, {x -> 3.2}}

Checking this with

g'[x]

we see that 3 is a super attractive fixed point!