A Somewhat Unusual Rational Function

Hello, this is 108Hassium.

Recently, I discovered a somewhat unusual rational function (a function that can be expressed as a polynomial divided by a polynomial).

One of them is $${f(z,c)=\frac{c}{(z-1)(z-c)}}$$, and the Mandelbrot set for this function looks like the following.

※ The definitions of the Mandelbrot set and Julia set in this article are based on the following article.

Looking at the formula and the image,

• $${c}$$ appears in two places

• there is a mesh-like region

• the convergent region extends in the vertical direction

...and various other seemingly unusual points can be found, but what I focused on were not these elements.

First, if you plot the graph of $${f(x)=\frac{c}{(x-1)(x-c)}}$$, it looks like this.

It has a shape that is not particularly unusual for a rational function, gradually approaching a horizontal line as it moves away from the origin.

Next, the graph of $${f(f(x))}$$ looks like this.

It may look different from the previous one, but it is the same in that it flattens out as it moves away from the origin.

So, what happens with $${f(f(f(x)))}$$...?

The slope does not flatten out, and the left and right ends have begun to extend sharply upward.

This is the "somewhat unusual property" that I focused on.

Due to this property, the sequence $${z_{n+1}=\frac{c}{(z_n-1)(z_n-c)}}$$ will diverge to infinity while approaching a 3-period cycle of "0→1→∞→0...".

By the way, the article I cited earlier as the source for the definition of the Mandelbrot set mentions that $${{\frac{c}{z^2-1}+1}}$$ has a similar property.

$${z_{n+1}=\frac{c}{z_n^2-1}+1}$$ diverges by approaching a 2-cycle of "1→∞→1…".

Therefore, I will explain how to find functions that diverge in such a strange way and their properties.

Terminology Definitions

Before the explanation, I will define a few terms necessary for the discussion.

First, we define the degree of a rational function as "degree of numerator - degree of denominator".

For example, in $${{\frac{z}{z^2-1}}}$$, the numerator is degree 1 and the denominator is degree 2, so the degree of this function is -1.

Also, in cases like $${{\frac{1}{z^2}+1}}$$, it can be transformed into $${{\frac{1+z^2}{z^2}}}$$, so it becomes degree 0.

By the way, there seem to be several types of definitions for the degree of a rational function, and the Wikipedia page for "Rational function" uses a different definition. (The definition of the degree itself is not listed, but the English version lists three types of definitions.)

By examining the degree, we can understand the behavior of a rational function as $${z→\infty}$$.

• Degree is 1 or greater → $${f(\infty)=\infty}$$

• Degree is -1 or less → $${f(\infty)=0}$$

*Hereinafter, $${{\displaystyle{\lim_{x→\infty}}f(x)}}$$ is abbreviated as $${f(\infty)}$$. Also, while there were two types of infinity, +∞ and -∞, for real numbers, it seems there is no need to distinguish between them when dealing with complex rational functions. I don't know for sure, though.

In the case of degree 0, $${f(\infty)}$$ can be calculated by transforming it into the form $${g(z)+c}$$ (where $${g(z)}$$ is degree -1 or less).

Next, we will call the following property "n-cycle divergence".

• The degrees of $${f(z),f(f(z)),f(f(f(z))),…f^{n-1}(z)}$$ are all 0 or less

• The degree of $${f^n(z)}$$ is 2 or greater

*$${f^m(z)=\underbrace{f(f(f(…f(}_mz)…)))}$$

For example, $${{\frac{c}{z^2-1}+1}}$$ is,

• $${f(z)=\frac{c+z^2-1}{z^2-1}}$$ is degree 0

• $${f(f(z))=\frac{z^4+c-1}{2z^2+c-2}}$$ is quadratic

Since it has this property, it is a 2-cycle divergence, and the previous $${{\frac{c}{(z-1)(z-c)}}}$$ is a 3-cycle divergence.

Now, I will skip the explanation, but finding a 2-cycle divergence function is easy.

So, I will explain how to find a 3-cycle divergence function.

How to find a 3-cycle divergence function

Since it is difficult to search through all rational functions, I will narrow the search range to functions where the denominator is quadratic.

First, let $${f(z)=\frac{dz^2+ez+g}{az^2+bz+c}}$$ and calculate the degree of $${f(f(f(z)))}$$ straightforwardly.

First, calculate $${f(f(z))}$$.

$${f(f(z))=f(\frac{dz^2+ez+g}{az^2+bz+c})\\=(d(\frac{dz^2+ez+g}{az^2+bz+c})^2+e(\frac{dz^2+ez+g}{az^2+bz+c})+g)÷(a(\frac{dz^2+ez+g}{az^2+bz+c})+b(\frac{dz^2+ez+g}{az^2+bz+c})+c)\\=\frac{d(dz^2+ez+g)^2+e(dz^2+ez+g)(az^2+bz+c)+g(az^2+bz+c)^2}{a(dz^2+ez+g)^2+b(dz^2+ez+g)(az^2+bz+c)+c(az^2+bz+c)^2}\\=\frac{(d^2+ade+a^2g)z^4+(2d^2e+bde+ae^2+2abg)z^3+(de^2+2d^2g+cde+be^2+aeg+b^2g+2acg)z^2+(2deg+ce^2+beg+2bcg)z+dg^2+ceg+c^2g}{(ad^2+abd+a^2c)z^4+(2ade+b^2d+abe+2abc)z^3+(ae^2+2adg+bcd+b^2e+abg+b^2c+2ac^2)z^2+(2aeg+bce+b^2g+2bc^2)z+ag^2+bcg+c^3}}$$

For $${f(f(z))}$$ to be quadratic, the coefficients of the quartic and cubic terms in the denominator of this expression must be zero, so you can find a 2-cycle divergence function by solving such a system of equations. (Actually, there is an easier way)

Next, calculate $${f(f(f(z)))}$$.

$${f(f(f(z)))\\=f(\frac{(d^2+ade+a^2g)z^4+(2d^2e+bde+ae^2+2abg)z^3+(de^2+2d^2g+cde+be^2+aeg+b^2g+2acg)z^2+(2deg+ce^2+beg+2bcg)z+dg^2+ceg+c^2g}{(ad^2+abd+a^2c)z^4+(2ade+b^2d+abe+2abc)z^3+(ae^2+2adg+bcd+b^2e+abg+b^2c+2ac^2)z^2+(2aeg+bce+b^2g+2bc^2)z+ag^2+bcg+c^3})}$$

...This is a pain!

Just expanding $${f(f(f(z)))}$$ is incredibly tedious, and considering solving the system of equations after expanding, this approach is hopeless.

So, after some trial and error, I discovered the following method.

First, let $${f(z)=\frac{cz+d}{(z-a)(z-b)}+e}$$.

This setup has the following characteristics:

• The numerator is linear, so it is easy to calculate

• The value of $${f(\infty)}$$ is easy to understand ($${=e}$$)

• The values of $${z}$$ where $${f(z)=\infty}$$ are easy to understand ($${=a,b}$$)

Next, let's consider the value $${f(f(f(a)))}$$.

• $${f(a)=\infty}$$

• $${f(f(a))=f(\infty)=e}$$

• $${f(f(f(a)))=f(e)=\frac{ce+d}{(e-a)(e-b)}+e}$$

Ultimately, it did not result in a clean value, but if $${f(f(f(a)))=a}$$ holds, the value of $${f(f(f(\infty)))}$$ becomes as follows.

$${f(f(f(\infty)))\\=f(f(e))\\=f(a)\\=\infty}$$

The significance of this is that $${f(f(f(\infty)))=\infty}$$ implies that the degree of $${f(f(f(z)))}$$ is at least 1.

To find a 3-periodic divergent function, it must be of degree 2 or higher, not degree 1, but in conclusion, it can be managed.

Now, first we solve the equation $${\frac{ce+d}{(e-a)(e-b)}+e=a}$$, but since there are 5 variables for a single equation, it seems fine to substitute appropriate constants for some of the variables.

However, to avoid breaking the previous consideration regarding the degree, the following restrictions apply.

• $${e\neq a}$$ (because it would result in $${f(e)=f(a)}$$)

• $${ac+d\neq 0}$$ (because it would not result in $${f(a)=\infty}$$)

For now, let's substitute $${(a,c,e)=(1,0,0)}$$.

$${\frac{ce+d}{(e-a)(e-b)}+e=a}$$
$${\frac{0×0+d}{(0-1)(0-b)}+0=1}$$
$${\frac{d}{b}=1}$$
$${d=b}$$

Substituting the values of $${a,c,d,e}$$ into $${f(z)}$$ gives $${{\frac{b}{(x-1)(x-b)}}}$$ and if we rewrite $${b}$$ as $${c}$$, it becomes the function $${{\frac{c}{(x-1)(x-c)}}}$$ introduced at the beginning.

I have already explained that this function has 3-periodic divergence, but what about other patterns?

For example, if we set $${(a,c,e)=(2,0,1)}$$, then $${d=b-1}$$, and another function $${f(z)=\frac{b-1}{(z-2)(z-b)}+1}$$ is generated.

And when we plot the graph of $${f(f(f(z)))}$$, we can see that it has a quadratic form.

Actually, it seems that for functions obtained by this calculation method, $${f(f(f(z)))$$ is always quadratic if $${c=0}$$.

Also, setting $${c=0}$$ provides another convenience.

To draw the Mandelbrot set, one needs to find the critical points of $${f(z)}$$, but when $${c=0}$$, the critical point can be expressed by the relatively simple formula $${(a+b)/2}$$.

Actually, there are combinations where the critical point can be expressed with a clean formula even when $${c\neq 0}$$ and the degree is 2 or higher, but I would like to omit the explanation regarding that this time.

Relationship with Mandelbrot and Julia Sets

The reason I thought about this in the first place is, of course, to draw interesting fractal figures.

So, I will introduce some Mandelbrot and Julia sets for 3-periodic divergent functions.

This is the one I introduced at the beginning.

This uses the formula I gave as an example in the calculation method section, but if you look closely, it appears to be just a translation of the previous $${c/(z-1)(z-c)}$$.

This is what you get by substituting $${(a,c,e)=(-b,0,1)}$$.

It looks just like the previous one, but since the shape of the large hole in the center is different, it doesn't seem to be similar.

Next, I will introduce the Julia sets.

Now, these fractal figures have one strange thing in common.

Every figure has a shape with holes, but if you look closely, there are many places where three holes meet at a single point.

On the other hand, if you observe the Mandelbrot set and Julia set of the 2-cycle divergent function c/(z^2-1)+1, you can see that the number of holes meeting at a single point is two.

I am not sure if this property is related to the divergence period, but as far as the Julia set is concerned, it can be mostly explained.

In the first place, the "holes" in a Julia set are a collection of points where z_{n+1}=f(z_n) diverges to infinity, but in the case of a 3-cycle divergent function, diverging to infinity is the same as converging to a 3-cycle.

This means that just as the 3-cycle region of a Julia set becomes a trefoil shape, it is not at all strange for the region that diverges to infinity to also become a trefoil shape.

By the way, trefoil-shaped holes do not necessarily form in the Julia set of a 3-cycle divergent function either.

I suspect that these cases correspond to the "enclaves" in the Julia set of z^2+c.

Furthermore, I have no idea why similar characteristics occur in the case of the Mandelbrot set.

Finally

I have already investigated topics such as "what happens if f(f(f(z))) is of the first degree" and "how to find a 4-cycle divergent function," but since I haven't obtained any very interesting results, I would like to end here for now.

If I get some good results, I will add them to this article or summarize them in a new article.

So, goodbye.

Postscript: I have written the continuation.