The Mandelbrot set is defined over the complex numbers and is quite complicated. It's defined by the complex numbers $c$ that remain bounded under the recursion: $$ z_{n+1} = z_n^2 + c,$$ where $z_1 = 0$.

If $c$ is real, then above recursion will remain real. So for what values of $c$ does the recursion remain bounded?

  • $\begingroup$ real slice of Mandelbrot set $\endgroup$
    – Adam
    Apr 3 at 19:51

2 Answers 2


The Wikipedia page gives the intersection of the set with the real axis as $[-2,0.25]$

Added: You can verify that $-2$ is in the set easily, and that any more negative number decreases each iteration without bound. For the positive end, each iteration is greater than the one before. To hit a limit, you must have $z=z^2+c$, which has the solution $z=\frac{1+\sqrt{1-4c}}2$, which becomes imaginary at $c \gt \frac 14$

  • $\begingroup$ Is there a proof? $\endgroup$ Jul 22, 2012 at 21:14
  • $\begingroup$ Behaviour of iterations when x is slightly larger than 0.25 is quite interesting. It can take very long for the iterations to escape (value > 2), and if you use floating-point arithmetic, c = 0.25 + eps will not escape at all for very small eps, due to rounding errors. $\endgroup$
    – gnasher729
    Jun 10, 2014 at 23:27
  • 1
    $\begingroup$ @gnasher729: If $c=\frac 14+\epsilon, c^2+c=\frac 5{16}+\frac 12 \epsilon + \epsilon^2$ You have to take $\epsilon$ very small to get killed by roundoff $\endgroup$ Jun 10, 2014 at 23:58

If z is a complex number whose distance to origin is bigger than $|c|$ and 2 then z is a point than scape for the iteration of the function $z^2+c$. It's easy to demostrate this, then the recursion remain bounded inside the closed ball of radio 2, but we can find the mandelbrot set inside $[-2,0.7]\times[-1.2,1.2]$.


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