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

mark

5 pts

In class, you should have received your very own cubic polynomial. The purpose of this problem is just to practice using the site by sharing your cubic polynomial with everyone. So please respond to this question with the following information:

• The fixed points (if any) of your polynomial, and
• A graph of your polynomial, along with the line $y=x$.
audrey

My personal cubic polynomial is
$$f(x) = -531 x^3+4832.1 x^2-14655.6 x+14817.9.$$

I generated a graph of my cubic (together with the line $y=x$) with the following code:

code goes here:
and here
and here

It looks like it's got three fixed points between $3$ and $3.1$.

RedCrayon

My cubic polynomial is
$$f(x)=2760.4 - 8164.8 x + 8051.4 x^2 - 2646 x^3$$

I generated my cubic using Mathematica with the following code:

f[x_] := 2760.4 - 8164.8 x + 8051.4 x^2 - 2646 x^3
cubic = Plot[f[x], {x, 0.96, 1.06}, PlotRange -> {0.96, 1.06},
AxesOrigin -> {0.96, 0.96}, AspectRatio -> Automatic,
PlotStyle -> Black];
line = Plot[x, {x, 0.8, 1.2}, PlotStyle -> Black];
Show[cubic, line]

It has three fixed points between $1$ and $1.05$.

SomeCallMeTim

My personal cubic polynomial is $f(x)=14315.7 - 14158.8 x + 4668.3 x^2 - 513 x^3 .$ I generated the following graph of my cubic, with the graph of $y=x$ overlaid:

My personal polynomial has three fixed points, located at $x=3, x=3.02653,$ and $x=3.07347$.

Captain_Flapjack

My personal cubic polynomial was:
$$f(x) = -397.9 - 1234.8 x - 1278.9 x^2 - 441 x^3$$

After plotting, there seem to be 3 fixed points, one at $x=-1$, and the other 2 between there and $x=-.92$. I solved $f(x)-x=0$ to find all three points at $x=-1$, $x=-0.9652$, and $x=-0.9348$:

In other news, someone left their umbrella in the classroom, I brought it to the math lab where it will be safe and sound...Hopefully

mark

@RedCrayon - That looks awesome! I did edit it so that your code is typeset as code.

Levente

My personal cubic is $$f(x) = 115.5 x^2 - 2695 x^3$$

I plotted this graph alongside $g(x) = x$ and found that they intersected at $x = 0$, $x = 0.012041$ and $x= 0.0308161$

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

I generated a graph of my cubic function together with the line y=x using this code:

f[x_] := f (x) = 63.4 - 172.8 x + 158.4 x^2 - 48 x^3
cubic = Plot[f[x], {x, .9, 1.25}, PlotRange -> {0.8, 1.3},
AxesOrigin -> {.8, .8}, AspectRatio -> Automatic, PlotStyle -> Black];
line = Plot[x, {x, .9, 1.5}, PlotStyle -> Black];
Show[cubic, line]

The graph has 3 fixed points between 0.9 and 1.2.

mark

@Yousername Excellent start! Notice the edits that I made, though. I slapped some dollar signs around your function so that it would have some nice mathematical typesetting. I indented your code four spaces so that it would look like code. Now, try to upload an image of your plot so I can dish out a like. We can figure out how to save the image out of Mathematica, if we need.

Person

My personal cubic polynomial is
$$f(x)= - 2842 x^3 + 17173.8 x^2 - 34591.2x +23225.2$$

I defined my functions as

    f[x_] := 23225.2 - 34591.2 x + 17173.8 x^2 - 2842 x^3
l[x_] := x

I generated a large scale graph using the code

    pf := Plot[f[x], {x, -36, 36}];
pl := Plot[l[x], {x, -36, 36}];
Show[pf, pl]

I generated a zoomed in graph using the code

    pf := Plot[f[x], {x, 1.99, 2.04}];
pl := Plot[l[x], {x, 0, 3}];
Show[pf, pl]

I found three fixed points namely $x= 2$, $x=2.01107$, $x=2.03179$.

Rick

My personal cubic polynomial is f(x) = 19621.6 - 29223.6 x + 14508.9 x^2 - 2401 x^3.

.

So bad-ass in fact, there is 3 fixed points between 2.00 and 2.03

WillHeDoTheAssignmen

My personal cubic polynomial is $f(x) = -406 - 1260 x - 1305 x^2 - 450 x^3$.
with the graph being created by:

f[x_] := -406 - 1260 x - 1305 x^2 - 450 x^3
cubic = Plot[f[x], {x, -1.1, -.9}, PlotRange -> {-1.1, -.9},
AxesOrigin -> {-1, -1}, AspectRatio -> Automatic,
PlotStyle -> Black];
line = Plot[x, {x, -1.1, .9}, PlotStyle -> Black];
Show[cubic, line]

I found three fixed points by using Mathematica's Solve[f[x] ==x ,x] to find the values of $x$ where $f(x)$ is just equal to $x$, and the three points are $x= -1, x = \frac{-14}{15},$ and $x= \frac{-29}{30}$.

mark

@WillHeDoTheAssignmen Very cool! I edited it a bit to make the code look like code.

notneds

My personal cubic is
$f(x) = 1220.7 - 1180.8 x + 381.3 x^2 - 41 x^3$

I plotted it using

g[z_] = -41*z^3 + 381.3*z^2 - 1180.8*z + 1220.7;
graph = Plot[g[z], {z, 2.8, 3.5}, PlotRange -> {2.5, 4},
AxesOrigin -> {2.7, 2.8}, AspectRatio -> Automatic,
PlotStyle -> Black];
line = Plot[x, {x, -100, 100}, PlotStyle -> Black];
Show[graph, line]

It has a single fixed point at 3.