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The Mandelbrot Set is an extremly complex object that shows new structure at all magnifications. It is the set of complex numbers $c$ for which the iteration indicated nearby remains bounded. $$z_0=c$$ $$z_{n+1}=z_n^2+c$$

  • what is $c$ in Mandelbrot set?
  • isn't $c$ complex number
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It's true, most mathematics can be subsumed into set theory. This question, however, is disjoint of any of the [set-theory] related tags on this site. –  Asaf Karagila Jul 31 '12 at 17:41
What problems do you see with $c$ being a complex number? –  Arthur Fischer Jul 31 '12 at 17:43
@Asaf Karagila ok but there is no Mandelbrot set tag –  Salvador Jul 31 '12 at 17:44
The symbol $c$ is quantified in the definition of the Mandelbrot set: $\mathcal{M} = \{c\in \mathbb{C}\ |\ z_n(c)\mbox{ is bounded }\}$. –  Neal Jul 31 '12 at 17:45
@Arthur Fischer i know that $c$ in short for constant but i dont know what constant, it could be just 1? –  Salvador Jul 31 '12 at 17:46

4 Answers 4

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When you look at an image of the Mandlbrot set it shows a region of the complex plane. The starting ones usually run from $-2$ to $\frac 12$ on the real axis or so and from $-2i$ to $2i$ or so on the imaginary axis. For each point in the region (at the desired grid spacing) we take the value $c$ and see if the iteration stays bounded. We then plot at $c$ black if it does stay bounded and a color that represents how quickly it goes to infinity if it does not. So for $c=1$ the iteration gives $1, 2, 5$ and we know if it ever gets bigger than $2$ it diverges, so we quit here and plot the color for 2 or 3 iterations. For $c=i$ we get $i,-1+i,-i, -1+i, $ etc and it stays bounded, so we plot black.

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can you proof For example, letting c = 1 gives the sequence 0, 1, 2, 5, 26,…, which tends to infinity. As this sequence is unbounded, 1 is not an element of the Mandelbrot set –  Salvador Jul 31 '12 at 18:05
@Salvador: If an iteration is greater than 2 in absolute value, the square is greater than 4, so even adding in $c$ leaves the next value larger in absolute value. It therefore goes to infinity. –  Ross Millikan Jul 31 '12 at 18:22

$c$ in the above formulas is the complex number for which you want to determine if it is or isn't in the Mandelbrot set. In images of the Mandelbrot set, the real part of $c$ is typically mapped to the $x$ axis and the imaginary part to the $y$ axis. In other words, each point of the image corresponds to a different value of $c$. Most pictures of the Mandelbrot set additionally use colours to indicate how quickly the iteration diverges for points not in the Mandelbrot set. That is, the colourful images you usually see for the Mandelbrot set are in some sense images for the complement of the Mandelbrot set: That's where all the structure lies. The Mandelbrot set itself, on the other hand, usually is shown in an uniform colour (most often black, sometimes white, rarely other colours).

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isn't this relation better $z_{n+1}=z_n^2+z_o$? –  Salvador Jul 31 '12 at 17:57
@Salvador: As defined in the question, $z_0=c$, therefore both would be equivalent. However, another (and I think the more usual) starting point would be $z_0=0$ (which then implies $z_1=c$), and in that case $z_{n+1}=z_n^2+z_0$ would be clearly wrong. However, I don't see the advantage of using $z_0$ instead of $c$ even if starting with $z_0=c$. –  celtschk Jul 31 '12 at 18:03
i see mandelbrot set should remains bounded –  Salvador Jul 31 '12 at 18:11

The symbol c is quantified in the definition of the Mandelbrot set: if we put $f_c(z)=z^2 + c$, $$ \mathcal{M} = \{ c\in\mathbb{C}\ |\ f_c^n(0) \mbox{ is bounded }\}.$$ The Mandelbrot set is a subset of the complex plane, $\mathbb{C}$. The "$c$" in the formula is just a tool to help formally express the condition for a complex number to belong to the Mandelbrot set.

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Yes $c$ is a complex number. For the Mandelbrot set, we take a quadratic polynomial (which may be taken of the form $z^2+c$ by a linear change of variables), then investigate its properties under iteration. There is a theorem (which Mandelbrot knew) saying that to analyze the iterative properties, we do not need to try all starting points, but can use only the critical points of the polynimial. For the polynomial $z^2+c$, the critical point is $0$. So we begin the iteration at $0$.

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