# More Gold

Last week I discussed Lin McMullin’s Theorem (that would be me). The theorem finds where the Golden Ratio shows up in any fourth-degree polynomial.  Shortly after that began making the rounds, I got an e-mail from David Tschappat who was then a senior in college. He is now a doctoral student working for an online learning solutions provider.

He related the discovery of a similar appearance of the Golden Ratio in cubic polynomials! Here is his theorem and two other related facts about cubic polynomials (lemmas really) as developed by me.

All cubic polynomials are symmetric to their point of inflection. Therefore, we can simplify matters by translating any cubic so that its point of inflection is at the origin. The general equation will then be $f\left( x \right)=k{{x}^{3}}-3{{h}^{2}}kx$.

I choose this rather unusual form so that the x-coordinates of the extreme points will be “nice” variables, namely h and –h the zeros of ${f}'\left( x \right)=3k{{x}^{2}}-3{{h}^{2}}k=3k\left( x-h)(x+h \right)$. The leading coefficient k will eventually divide out and not be a concern.

Now consider two other values x = –2h and x = 2h. So, assuming h > 0, there are five evenly spaced values $-2h<-h<0 that we will consider.

The first two results I will leave as an exercise, as they say:

1. $f\left( -2h \right)=f\left( h \right)$ and $f\left( -h \right)=f\left( 2h \right)$. This means that the extreme values are the same as a point on the graph 3h units away on the other side of the y-axis. This is mildly interesting in itself. The reason it is mentioned id that we will use these points soon.
2. ${f}'\left( -2h \right)={f}'\left( 2h \right)$ which is really true for any points at equal distances on opposite sides of the origin. This is obvious from the symmetry of the cubic.

Now comes the interesting and strange part. An equation of a line through one extreme point with a slope equal to that at the point with the same y-coordinate on the other side of the origin (or for that matter the point h units farther away on the same side of the y-axis) is

$y=f\left( -h \right)+{f}'\left( 2h \right)\left( x-\left( -h \right) \right)$

$y=\left( k{{\left( -h \right)}^{3}}-3{{h}^{2}}k\left( -h \right) \right)+\left( 3k{{\left( 2h \right)}^{2}}-3{{h}^{2}}k \right)\left( x+h \right)$

$y=k\left( 11{{h}^{3}}+9{{h}^{2}}x \right)$

Now find where this line intersect the original cubic by solving $y=f\left( x \right)$. The equation can be solved by hand. We know that x = h is one solution. Synthetic division will give a quadratic and the quadratic formula will do the rest.

$k\left( 11{{h}^{3}}+9{{h}^{2}}x \right)=k{{x}^{3}}-3{{h}^{2}}kx$

$0={{x}^{3}}-12{{h}^{2}}x-11{{h}^{3}}$

$0=\left( x-h \right)\left( {{x}^{2}}+hx-11{{h}^{2}} \right)$

$\displaystyle x=h,\frac{1+3\sqrt{5}}{2}h,\frac{1-3\sqrt{5}}{2}h$

What? You were expecting the Golden Ratio? Not to worry; it’s there!

$\displaystyle \Phi \left( -h \right)+\phi \left( 2h \right)=\frac{1+\sqrt{5}}{2}\left( -h \right)+\frac{1-\sqrt{5}}{2}\left( 2h \right)=\frac{1-3\sqrt{5}}{2}h$

$\displaystyle \Phi \left( 2h \right)+\phi \left( -h \right)=\frac{1+\sqrt{5}}{2}\left( 2h \right)+\frac{1-\sqrt{5}}{2}\left( -h \right)=\frac{1+3\sqrt{5}}{2}h$

Where $\displaystyle \Phi =\frac{1+\sqrt{5}}{2}$ is the Golden Ratio and $\displaystyle \phi ={{\Phi }^{-1}}=\frac{1-\sqrt{5}}{2}$ is the reciprocal of the Golden Ratio.

A similar computation using the other extreme point gives similar results.

Yes, the Golden Ratio is there; I don’t know why it should be there, but it is.

Thank you, David.