I'm trying to prove that the Mandelbrot set intersects the X-axis on the interval [-2,.25]. I understand and have proven that the Mandelbrot set lies in a radius of 2. Mostly, I'm wondering how to prove that .25 is the greatest possible point on the X-axis. I've attempted to develop a proof using this argument:

$ z_{n+1}=z_n^2+c $

$\\ 0=z_n^2+c-z_{n+1}$

Upon finding the discriminant, $ \sqrt{1-4c} $, it's clear that the only values for which the equation yields real answers are those with $ c \geq 1/4$. But, given the recursive definition, I'm not sure manipulating the quadratic equation is valid.



It is straightforward for real values of $z_n$: $$ z_{n+1} - z_n = z_n^2 - z_n + c = \left(z_n - \frac12\right)^2 +c -\frac14 \ge c-\frac14, $$ so if $c-\frac14 > 0$ you get by induction that $z_n \ge z_0 + n\left(c-\frac14\right) \to \infty$ for $n \to \infty$.

  • $\begingroup$ Ah, got it. Thanks so much! $\endgroup$ – cloudchamber Feb 5 '15 at 4:26

Let me write $f=f_c(x)=x^2+c$.

The proof that $c\notin M$ for $c>1/4$ was already given by Lukas, although there is an even softer argument: observe (using the quadratic formula) that $f_c(x)=x$ has no solutions for for $c>1/4$, and hence $f_c(x)>x$ for all $x$ (since this is true for large $x$).

So, for every $x_0\in \mathbb{R}$, the corresponding sequence $x_n$ is increasing, and hence either has a finite limit or converges to infinity. But if $x_n\to x\in\mathbb{R}$, then we would have $$ f(x) = f(\lim_{n\to \infty} x_n) = \lim_{n\to\infty} f(x_n) = \lim_{n\to\infty} x_{n+1} = x $$ by continuity, and this is a contradiction to $f$ not having any fixed points.

For $c\leq 1/4$, there is a fixed point $$ \beta(c) = \frac{1 + \sqrt{4-c}}{2},$$ again found using the quadratic formula. (Actually if $c<1/4$, there are exactly two fixed points, but we only care about the larger one.)

Let us change coordinates, for convenience, to the logistic family, given by $$ g_{\lambda}(t) := g(t) := \lambda t (1-t), \qquad \lambda = 2\cdot \beta(c), \qquad t = \frac{1}{2} - \frac{x}{\lambda}.$$ (It is easy to check that this change of coordinate does indeed transform $f$ to the function $g$.)

Now $c\in [-2,1/4]$ corresponds to $\lambda\in [1,4]$, and in this case we have $g([0,1])\subset [0,1]$. (Note that $g(0)=0$, $g(1)=0$, and $g$ takes its maximum at $t=1/2$: $g(1/2)=\lambda/4$.) It follows that the critical point $t=1/2$ (which corresponds to $x=0$) has bounded orbit. This implies that $c\in M$.

Finally, if $c<-2$, then we see, again considering the function $g_{\lambda}$ and setting $t_0=1/2$, that $t_1 = g(t_0) = \lambda/4 > 1$, $t_2 = g(t_1) < 0$, and $g(t_{n+1}) < t_n$ for all $n\geq 2$. As above, it follows that $t_n\to -\infty$, and $c\notin M$.

[If you do not wish to change coordinates, of course you can do the calculations also in the original parameterisation, but it becomes a little awkward.]


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