Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 11 down vote accepted

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 $c \gt \frac 14$

share|cite|improve this answer
Is there a proof? – Carl Brannen Jul 22 '12 at 21:14
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. – gnasher729 Jun 10 '14 at 23:27
@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 – Ross Millikan Jun 10 '14 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 $[-2,0.7]\times[-1.2,1.2]$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.