The Mandelbrot set is the set of all complex numbers $c$ that cause the function $z_{n+1} = z_n^2 + c$ to remain bounded within a circle of radius 2 when iterated from $z_0 = 0$. Looking at only the real numbers, all values between positive $\frac{1}{4}$ and negative 2 are part of the set. If we plot the attractive fixed point value on the vertical axis against the input $c$ value on the horizontal axis we get a classic bifurcation diagram. Going from right to left, the diagram starts with the function $\frac{1 – \sqrt{-4c + 1}}{2}$ where $\frac{1}{4} ≥ c ≥ -\frac{3}{4}$. It then splits into two branches as $\frac{-1 ± \sqrt{-4c + 3}}{2}$ where $-\frac{3}{4} ≥ c ≥ -\frac{5}{4}$. Both branches divide again and again, over increasingly narrow intervals, until the plot dissolves into chaos around $-\frac{7}{5}$.
Are there closed-form equations for any of the divisions beyond the first two I've listed? For example, the period-4 region or the period-8 region? They don't seem to line up with quadratic expressions or any other expressions I've tried. I also can't find closed-form expressions for the end-points of the intervals for each division, that is, the $c$ value for each bifurcation point.
The closest I've been able to get so far for the topmost period-four section is $\frac{(-4c – 5)^{\frac{1}{2.05}}} {2.5} – \frac{1 – \sqrt{2}} {2}$, which is pretty darn close, but it's not exact.
I do know that the sweeping curves visible in the chaotic portion of the bifurcation diagram, the critical curves, are of the form $(c² + c)² + c$, $((c² + c)² + c)² + c$, $(((c² + c)² + c)² + c)² + c$, and so on, but these curves don't line up with the bifurcating parts either.
Best Answer
If we let $$f_c(z)=z^2+c$$ then the lines on that diagram (in the column corresponding to $c$) are exactly at attractive cycles of $f_c$ - which is to say that they are certain solutions of $$f_c^n(z)=z$$ where $f_c^n$ is the $n$-fold composition of $f_c$ with itself - and where $n$ will be the number of lines seen on the plot. More specifically, they are the solutions for which the derivative $|(f_c^n)'(z)| \leq 1$.
So, the curves are described as some of the roots of the following degree $2^n$ polynomial in $z$ (but note that cycles of every length appear in that plot - so certain pieces will require arbitrarily large $n$ to describe): $$f_c^n(z)-z.$$ You can slightly reduce this by noting that if $m$ divides $n$ then $f_c^m(z)-z$ divides $f_c^n(z)-z$, which at least lets us factor the polynomial a little bit. As far as I know, the only $n$ for which the solutions of this have a form with radicals is $n=1$ (the fixed point) and $n=2$ (the two-cycle).
You can certainly, however, solve these equations using numerical methods for solving polynomials - and even something simple such as Newton's method is going to converge a lot faster than finding the cycles just by iterating a single point (as is how bifurcations diagrams are usually made) under $f_c$ itself.