# Mandelbrot set of $c \cdot \cos(z)$

I'm given a task to write a program, that determines if a given point $c \in \mathbb{C}$ is in the Mandelbrot set of the function $$f_c(z) = c \cdot \cos (z)$$ That is if the set $\{z_n = f_c^n (0) : n \in \mathbb{N}\}$ is bounded.

With the standard formula $f_c (z) = z^2 + c$, I know that, if the at some iteration $n$ the value $z_n$ is such that $|z_n| > 2$, then $|f_c^n(0)| \rightarrow \infty$, as $n \rightarrow \infty$ and $c$ is not in the Mandelbrot set. But with the function, I'm given, I don't have a clue. Is there a similar criteria for it?

• No, there is not. Also, as far as I know, most researchers in the field would believe that hyperbolicity is dense in this cosine family, which would mean that for an open and dense set of parameter values $c$, the orbit of $0$ would be attracted by an attracting periodic cycle, and so would be bounded. If this is true, your "Mandelbrot set" would be dense in the plane and you would probably not see anything on your picture. May 22, 2015 at 20:06

The Mandelbrot set is essentially the bifurcation locus for the quadratic family $f_c(z)=z^2+c$. There are aspects of the iteration of quadratic functions that ease the study of this particular bifurcation locus but this is the correct level of generality in which to work, if you want to extend it to other families.

When iterating a polynomial, it turns out that the orbits of the critical points (roots of the derivative) dominate the global dynamics. If all critical orbits tend to $\infty$, then the Julia set is totally disconnected. If all critical orbits are bounded, then the Julia set is connected. When studying the iteration of the quadratic family, we start from the point $z_0=0$ because this is the only critical point. Knowledge of that one critical orbit tells us a lot about the corresponding Julia set.

From a broader perspective, the point at $\infty$ is an example of a super-attractive fixed point and the dichotomy above is quite coarse - either, the critical orbit converges to infinity or it does not. At a somewhat finer level, we might ask what types of periodic behaviors can the critical orbit display? That is, we could iterate from the critical point zero and stop iterating if periodicity is detected. Note that this includes the possibility of convergence to $\infty$. We then shade the point $c$ according to the detected periodicity and the amount of time that it took to find that periodicity. This leads to an image of the Mandelbrot set that looks like so: Now, you are studying a different family of functions, namely $g_c(z)=c\cos(z)$. Your function, of course, is not a polynomial; rather, it is an entire transcendental function. That is, it is differentiable on the entire complex plane but not at infinity. Iteration of such functions has been studied fairly well by researchers such as Devaney, Bergweiler, and Eremenko. There are similarities to polynomials but also crucial differences. In particular, the point at $\infty$ is an essential singularity; there's no reason to consider convergence to $\infty$ as important. While boundedness of the critical orbits might be relevant, this is mainly a byproduct of the fact that we are interested in attractive periodic orbits which are, of course, bounded. Also, there are infinitely many critical points. Fortunately, there are only two critical values, $\pm c$, which (by the evenness of the cosine) happen to have identical futures. Thus, you could simply iterate $g_c$ from zero and classify $c$ according to any periodicity that you detect. It's likely that there will be a large region where no periodicity is detected and you need a default color to indicate this situation. That's exactly how I generated the following image using red as the non-periodic color: Note that this question has quite a lot in common with this one, which I also answered. Another answer to that question (which I think was quite good) suggested the use of Lyapunov exponents to classify the stability of the critical orbit. Applying that idea to your function, I came up with a picture like so: • This is a great answer! I'm puzzled by "Your function, of course, is not a polynomial; rather, it is an entire function." Every polynomial function is entire!
– Unit
May 25, 2015 at 17:36
• @Unit Thank you. Your observation is correct, of course. I guess that entire is a broader class of functions, though. Somethings could be true of entire functions in general that are not true of polynomials. Importantly, $c\cos(z)$ is transcendental but, really, I was just following the nomenclature that is common in the research literature. May 25, 2015 at 17:43