Golden Ratio | Teaching Calculus

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 $-2h<-h<0<h<2h$ that we will consider.

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

$f\left( -2h \right)=f\left( h \right)$ 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.
${f}'\left( -2h \right)={f}'\left( 2h \right)$ 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.

Mathematics more often tends to delight when it exhibits an unanticipated result rather than conforming to … expectations. In addition, the pleasure derived from mathematics is related in many cases to the surprise felt upon the perception of totally unexpected relationships and unities.

– Mario Livio, The Golden Ratio

I have a theorem named after me. I did not name it, but I did prove it – well more like I tripped over it. It is a calculus related idea. Here is how it came about. I say “came about” because as you will see I did not set out to prove it. I just sort of fell in my lap as I was working on something else.

I was trying to do an animation of an idea that I had heard about: If you have a fourth degree, or quartic, polynomial with a “W” shape it has two points of inflection. If you draw a line through the points of inflection three regions enclosed by the line and the polynomial’s graph are formed. The areas of these regions are in the ratio of 1:2:1. In order to make the animation work I needed the general coordinates of the four points where the line intersects the quartic.

The straightforward way to proceed would be to write a general fourth degree polynomial,

$f\left( x \right)={{c}_{4}}{{x}^{4}}+{{c}_{3}}{{x}^{3}}+{{c}_{2}}{{x}^{2}}+{{c}_{1}}x+{{c}_{0}}$

differentiate it twice to find the second derivative. Then find the zeros of the second derivative (by the quadratic formula), write the equation of the line through them, and then find where else the line intersects the quartic. Without even starting I realized that even with a CAS the algebra and equation solving was going to be really fun (Not!). So, I decided on an alternative approach.

I decided to let the zeros of the second derivative be x = a and x = b, then at least they would be easy to work with. Then the second derivative is $\displaystyle {{f}'}'\left( x \right)=12{{c}_{4}}\left( x-a \right)\left( x-b \right)$ where the ${{c}_{4}}$ is the leading coefficient of the quartic and the 12 comes from differentiating twice.

I integrated to get the first derivative and added ${{c}_{1}}$ , the coefficient of the linear term, as the constant of integration. I integrated again and added ${{c}_{0}}$ , the constant term. as the constant of integration. This resulted in the original quartic function:

$\displaystyle f\left( x \right)={{c}_{4}}{{x}^{4}}-2\left( a+b \right){{c}_{4}}{{x}^{3}}+6ab{{c}_{4}}{{x}^{2}}+{{c}_{1}}x+{{c}_{0}}$

Then I wrote the equation of y(x) the line through the points of inflection. It is too long to copy, but you may see it in the screen capture at the end of the post. (That is what is nice about a CAS: you really do not have to worry about how complicated things are.)

Then I solved the equation $f\left( x \right)=y\left( x \right)$ . Two of the solutions are x = a and x = b as I expected. (This means that you could do synthetic division by hand since you know two of the roots.) The other two I did not expect. They are:

$\displaystyle {{x}_{1}}=\frac{1+\sqrt{5}}{2}a+\frac{1-\sqrt{5}}{2}b$ and $\displaystyle {{x}_{2}}=\frac{1+\sqrt{5}}{2}b+\frac{1-\sqrt{5}}{2}a$

And that’s when I stopped astonished! Those numbers are the Golden Ratio $\Phi =\frac{1+\sqrt{5}}{2}$ , and its reciprocal $\phi ={{\Phi }^{-1}}=\frac{1-\sqrt{5}}{2}$ . So the roots are $\Phi a+\phi b$ and $\Phi b+\phi a$ . How did they get there?

To this day I have no idea why the Golden Ratio should be so involved with quartic polynomials, but there they are in every quartic!

There were no assumptions made about a and b – they could be Complex numbers. In that case there are no points of inflection, but the “line” and the quartic still will have the same value at the four points.

October 16, 2022 Update: If the solutions of $\displaystyle {y}''=0$ are the Complex conjugates $\displaystyle a=\alpha +\beta i$ and $\displaystyle b=\alpha -\beta i$ then $\displaystyle {{x}_{1}}=\alpha +\sqrt{5}\beta i$ and $\displaystyle {{x}_{2}}=\alpha -\sqrt{5}\beta i$ . When graphed on an Argand diagram the four points are collinear on the vertical line at $\displaystyle x=\alpha$

This was all in 2013 and until just this year I never checked the ratio of the areas. (They check.)

Here is a CAS printout of the entire computation.

An interactive Desmos demo of this can be found here

October 16, 2022, Update: If the solutions of $\displaystyle {y}''=0$ are the Complex conjugates $\displaystyle a=\alpha +\beta i$ and $\displaystyle b=\alpha -\beta i$ then $\displaystyle {{x}_{1}}=\alpha +\sqrt{5}\beta i$ and $\displaystyle {{x}_{2}}=\alpha -\sqrt{5}\beta i$ . When graphed on an Argand diagram the four points are collinear on the vertical line at $\displaystyle x=\alpha$

November 10, 2020, Update: I received an email this week form Dominique Laurain, a computer science and applied math engineer from France, who describes himself as a mathematics hobbyist. He discovered another interesting relationship between the coordinates of the x-coordinates of the four points on the line described above. The four points on the line through the points of inflection of a fourth degree polynomial with Real roots in the drawing above from left to right are p, q, r, and s. The coordinates of the points of inflection are a and b as in the post above.

One, of several, cross-ratios of four points with x-coordinates of p, q r, and s is defined as

$\displaystyle (p,s;r,q)=\frac{{\left( {r-p} \right)\left( {q-s} \right)}}{{\left( {r-s} \right)\left( {q-p} \right)}}$ .

Dominique Laurin discovered that $\displaystyle (p,s;r,q)=\frac{{\left( {r-p} \right)\left( {q-s} \right)}}{{\left( {r-s} \right)\left( {q-p} \right)}}={{\Phi }^{4}}$

The computation is shown in the figure below.

There are 24 other cross-ratios depending on the order of the points. In groups of 4, the 24 possible orders are equal to 6 related values. See the cross-ratio link above above. For example, the cross-ratio in the order (s, p; r, q) is ${{\varphi }^{4}}$