Your favorite Cubic Polynomial

(10 pts)

I’m sure you couldn’t have made it this far without having a favorite cubic polynomial! What is it?

Your response should include the following:

• Your favorite cubic polynomial nicely typeset using the variable z.
• A graph of your cubic as a real function.
• A list of numerical approximations to the roots of your cubic obtained via some piece of mathematical software.

Example:

My favorite cubic polynomial is f(z)=z^3-3z-1. I can graph f with Mathematica as follows:

Plot[z^3 - 3 z - 1, {z, -2, 2}]

Looks like there are three real roots. I can find them like so:

NSolve[z^3 - 3 z - 1 == 0, z]

(* Out: 
  {{z -> -1.53209}, {z -> -0.347296}, {z -> 1.87939}}
*)

Awesome question!!!

My favorite cubic polynomial is f(z)=z^3-1. I think I’ll graph it with Desmos:

desmos-graph.png800x800 7.8 KB

It looks like z=1 is the only real root!

My favorite cubic polynomial (at the moment) is f(z) = z^3 + \sqrt{2}z^2 + \sqrt{3}z + \sqrt{5}. Here is f graphed using Mathematica:

Plot[z^3 + Sqrt[2] z^2 + Sqrt[3] z + Sqrt[5], {z, -2, 1}]

There appears to be one real root. We can find its value in Mathematica:

NSolve[z^3 + Sqrt[2] z^2 + Sqrt[3] z + Sqrt[5] == 0, z, Reals]

Out: {{z -> -1.35437}}

As far as cubic polynomials go, I’ve always been a fan of f(z)=\frac{1}{6}z^3 + \frac{1}{2}z^2 + z + 1. I can graph this in MATLAB using the following:

z = linspace(-3,2,50);
f = 1 + z + (z.^2)/2 + (z.^3)/6;
plot(z,f)

Fave_Cub_Poly.jpeg1182x743 29.9 KB

a = [1/6 1/2 1 1];
r = roots(a)

r =

  -0.7020 + 1.8073i
  -0.7020 - 1.8073i
  -1.5961 + 0.0000i

As a fan of simple polynomials and the number 3, my favorite cubic expression would have to be f(z) = z^3 +9. Finding the roots of this polynomial is essentially the same as finding the cube root of -9. First I used the following python code to explore this graphically:

def f(z): return z**3 + 9
t = np.linspace(-4,4,100)
plt.plot(t, f(t))
plt.grid(True) 

which produced the graph:

It looks like there is one real root, which should be negative, That means there are two complex roots whose real part is positive, but lets use python to double check this:

coeffs = [1,0,0,9]
np.roots(coeffs)

which produced the following answer:

array([-2.08008382+0.        j,  1.04004191+1.80140543j,
    1.04004191-1.80140543j])

Looks like the real root is about -2.08 which agrees with my graph, but this is complex variables and our awesome professor has a really cool online tool to explore complex roots using Newton’s Method lets see if we can see the other roots graphically:

root two.png700x700 76.8 KB

Wow, neat picture, and if you check the smallest dot in the center of the large red patch, that will be our real root located on the real axis, and in the center of the large blue and tan patches are our two complex roots which are complex conjugates just as we expected to find. Who knew that complex numbers could be so artistic.

My favorite cubic polynomial is f(z) = -\sqrt{2}z^3 + \frac{1+\sqrt{5}}{2}z^2 + e^\sqrt{3}z-\pi. We can use Mathematica to depict the function, f.

Plot[-Sqrt[2] * z^3 + 0.5 * (1 + Sqrt[5]) * z^2 + Exp[Sqrt[3]] * z - Pi, {z, -2, 3}]

From the graph we can see that there are three real roots. Using Mathematica’s “Solve” function, we can find (approximately) these values as shown below:

Solve[-Sqrt[2] * z^3 + 0.5 * (1 + Sqrt[5]) * z^2 + Exp[Sqrt[3]] * z - Pi ==   0, z]

{{z -> -1.78736}, {z -> 0.514147}, {z -> 2.41733}}

I like simple things, so I figured I’d go with the cubic polynomial f(z) = z^3 - z. I plotted it using Mathematica: Plot[z^3 - z, {z, -2, 2}]

From the graph it looks like there are three real roots. I found them using Mathematica:
Solve[z^3 - z == 0, z]
{{z -> -1}, {z -> 0}, {z -> 1}}

I couldn’t seem to get the code block to work, so I apologize for the last bit of code, the four spaces didn’t seem to do it for me. Let me know if anyone has insight on that.

Also in case my first polynomial wasn’t interesting enough I also decided to tack on f(z) = z^3 + z. Also plotted in Mathematica: Plot[z^3 + z, {z, -2, 2}]

One real root, two complex roots:
Solve[z^3 + z == 0, z]
{{z -> 0}, {z -> -I}, {z -> I}}

My favorite polynomial is f(z) = 3 z^3 - 10 z ^2 + z + 6. I graphed f with Mathematica using the code

Plot[3 z^3 - 10 z^2 + z + 6, {z, -3, 4}]

I found the three real roots using the code

NSolve[3  z^3 - 10 z ^2 + z + 6 == 0, z] 

which gave me the three roots of

{{z -> -0.666667}, {z -> 1.}, {z -> 3.}}

I also did f(z) = 3 z^3 - 2 z ^2 + z to see a polynomial with imaginary roots
which gave me

using the code

Plot[3  z^3 - 2 z ^2 + z, {z, -3, 4}].  

f has one real root and two imaginary and with Mathematica’s help I was able to identify the roots using

NSolve[3  z^3 - 2 z ^2 + z == 0, z]
out: {{z -> 0.}, {z -> 0.333333 - 0.471405 I}, {z -> 0.333333 + 0.471405 I}}

@Sidneykidney That looks really good! I’d really like to see the code formatted as code in mono-spaced font. You can make it so by indenting your code block four spaces. Have a look at this post for more details.

My favorite 3rd degree polynomial is f(z) = z^3 + 2z^2 + 3z +4 . I just learned how to graph it using Mathematica:

Plot[z^3 + 2 z^2 + 3 z + 4 , {z, -5, 5}]

Looks like there is only one real root, but I can’t pick it out, so I learned how to use Mathematica to estimate it.

NSolve[z^3 + 2 z^2 + 3 z + 4 == 0, z]
Out: {{z -> -1.65063}, {z -> -0.174685 - 1.54687 I}, {z -> -0.174685 + 
1.54687 I}}

We see that the real root is close to -1.65 and that there are two other complex roots not in the reals, like we expect.

My favorite cubic polynomial is 8x^3-4x^2+2x-1.
Using Desmos I can graph this polynomial to find that it has one real root at .5 :

desmos-graph.png800x800 17.8 KB

My favorite cubic polynomial is f(x)=-1/2 x^3 + (1/3) x - 1/4 . I can graph this function in Mathematica like this:

f[x_] := -1/2 x^3 + (1/3) x - 1/4
Plot[f[x], {x, -2, 2}]

We can see from the plot that f(x) has one real root, and we can calculate them as so:

NSolve[f[x] == 0, x, Reals]
Out[]= {{z -> -1.35437}}

Wow.

My favorite cubic polynomial is

I graphed it using desmos.

desmos-graph.png800x800 26.1 KB

It has three real roots

which have exact representations in the complex plane.

What I like about this polynomial are it’s points of inflection and it’s minima and maxima.

My Favorite cubic polynomial is f(z)=4z^3+3z^2+2z+1. I graphed it in Mathematica with the following
Plot[4 z^3 + 3 z^2 + 2 z + 1, {z, -2, 2}]

It appears there is one real root and two complex roots. They can be found with…
NSolve[4 z^3 + 3 z^2 + 2 z + 1 == 0, z]
{{z -> -0.60583}, {z -> -0.0720852 - 0.638327 I}, {z -> -0.0720852 +
0.638327 I}}

My favourite cubic polynomial uses my birthday as the coefficients. (Hey, I usually celebrate my birthday by taking multiple final exams – I have to show it some love at some point!) I messed around with addition and subtraction signs until I got something kind of attractive. The end result was the equation

f(z) = 5z^3 + 3z^2 -9z -6

And I just realized that this could be taken as some clever pun on the “generation z” term the news media like to throw around. Ugh.

Anyway, I graphed it in Desmos (because I’ve been working with high school students and my default method of solving ANYTHING is to graph it in Desmos) which revealed that this thing has three real roots:

image.png1038x1040 30.3 KB

Desmos indicates that the roots are around -1.274, -.691 (my graphing calculator’s nSolve function gave the more precise -.690708, but I couldn’t work out how to make it give me any other roots), and 1.364

My favorite cubic polynomial would have to be f(z) = -\frac{1}{9}z^3 + \frac{1}{2}z^2 - z . To plot this in Mathematica I used the following code.

f[z_] := -3/27*z^3  + 1/2 z^2 - z ;
Plot[f[x], {x, -10, 10}, PlotRange -> {-10, 10}]

This cubic has one real root at z = 0 , to find the other two complex roots I used Mathematica’s NSolve function,

roots = z /. NSolve[f[z] == 0, z, 4]
out: {0, 2.250 - 1.984 I, 2.250 + 1.984 I}

.