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Show that the Mandelbrot set is contained within the closed disc of radius 2 around the origin.

How do I show this?

Nick B
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2 Answers2

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Suppose the point in question $c$ is outside the disc, at $|c| = 2+\epsilon$, where $\epsilon$ is real and $\epsilon >0$.

Then we will show by induction that when $z_0 = 0$ and $z_{n+1} = z_n^2 + c$, with $|c|=2+\epsilon$, for $\epsilon > 0$, that for all integer $n > 0$ $$ |z_n| \geq 2+ (2^n-1) \epsilon $$ The basis is trivial: $|z_1| = |c| = 2+(2^1-1)\epsilon$.

Now assume $|z_n| \geq 2+(2^n-1)\epsilon$. Then $$ |z_n|^2 = \left( 2+(2^n-1)\epsilon \right)^2 = 4 + 2^{n+2}\epsilon -4\epsilon+ \epsilon^2 > 4 + 2^{n+1}\epsilon + \epsilon $$ because for $n\geq 1$, $2^{2n+2} - 4 >2^{2n+1} + 1$.

And now using the triangle equality in the form $|a+b| \geq |a|-|b|$ we have

$$ |z_{n+1}| = \left| z_n^2 + c \right| \geq |z_n|^2 - |c| \geq 4 + 2^{n+1}\epsilon + \epsilon - 2 - \epsilon = 2 + (2^{n+1}-1) \epsilon $$ So $|z_{n+1}| \geq 2 + (2^{n+1}-1) \epsilon$ and induction is established.

Finally, take $|c|= 2+ \epsilon$, and consider any $L$ however large:
Choosing $n > \log_2( L/\epsilon + 1)$ we have $|z_{n+1}| > L$.

So if $c$ is outside the disc fof radius 2, the sequence is unbounded and $c$ is outside the Mandelbrot set.

Mark Fischler
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A complex number $c$ is in the Mandelbrot set if repeated squaring and adding $c$ doesn't diverge to infinity. If you're further away than $2$ from the origin, then squaring increases the absolute value more than adding $c$ could hope to make up for, even if the two operations are directly opposed.

Arthur
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    Doesn't this answer gloss over the need to prove that squaring increases the absolute value more than adding $c$ can make up for? That is not trivial to show. – Mark Fischler Oct 06 '14 at 22:51
  • @MarcFischler First point: a brief question deserves a brief answer. Some times I give more, but not always. Second point: since $|x^2|=|x|^2$, the complex argument of the number in question is irrelevant, and we can assume wlog that a given $z_n$ is real. In that case I do find it a rather trivial matter to prove that $|z_n|>2\implies |z_n|<|z_{n+1}|$, given $|c|\leq|z_n|$. – Arthur Oct 06 '14 at 23:18
  • Your first point is well taken. But though true, it is non-obvious that we can assume $z_n$ is real. Let $z_n = 2.01 i$, and $c = 2.1$, then |z_n^2 - c| is in fact a lot less than 2. The point is that you could never get to that sort of value of $z_n$. – Mark Fischler Oct 06 '14 at 23:57
  • @MarkFiscler And if you let $n$ be the first natural such that $|z_n|\gt 2$, you're obviously in the clear from your problem, and from there the absolute value is easily seen to continue to rise. – Arthur Oct 07 '14 at 08:40