Laws of physics are often shockingly compact. A force law, an equation of motion, a conservation rule: sometimes the exact mathematical principle governing millions of real-world phenomena fits on one line. And yet the world built from those rules is all but compact or simple. Coastlines wrinkle. Ferns repeat themselves leaf inside leaf. Clouds and turbulent gas form structure inside structure. Even DNA, branching trees, lightning, river networks, and snowflakes seem to share the same feature: repeat a simple instruction many times, and simplicity starts to look complicated.

Fractals live in that same tension between simple rules and complicated outcomes.

So let us begin with the least glamorous machine imaginable. Pick a number $C$, start at $x_0=0$, and repeat

$$ x_{n+1}=x_n^2+C. $$

That is the whole rule - just one number being fed back into the same equation over and over again.

And already something interesting happens.

For some values of $C$, the sequence settles down.
For others, it stays bounded but never settles to a single value.
And for others, it runs away to infinity.

A first warning that simple does not mean boring

A few sample choices already show the split:

$C$	behavior
$-2.1$	diverges quickly
$-2$	settles to 2
$-1$	bounded, oscillatory
$-0.5$	converges
$0.25$	converges slowly
$0.26$	diverges slowly

Instead of checking a few hand-picked values, we can scan a whole interval.

Iterating $x_{n+1}=x_n^2+C$ from $x_0=0$. Outside $[-2,1/4]$, the orbit escapes. Between $-3/4$ and $1/4$, it converges to a fixed point. Between $-2$ and $-3/4$, it remains bounded but becomes more interesting.

This picture already contains the central fractal lesson:

bounded does not necessarily mean simple.

The interval $[-3/4,1/4]$ is the tame part: the orbit converges to a fixed value.

The interval $[-2,-3/4)$ is still bounded, but no longer tame. Here we begin to see oscillation, period-doubling, and the first hints of chaos.

Another way to see the same transition is to plot selected iterates $x_n(C)$:

Selected functions $C \mapsto x_n(C)$. In the convergent region, late iterates lie nearly on top of one another. In the bounded oscillatory region, they split into branches.

AsideWhy these real intervals appear Show proof Hide proof

Let

$$ f_C(x) = x^2 + C, \qquad x_0 = 0, \qquad x_{n+1} = f_C(x_n). $$

A convergent orbit must approach a fixed point $L$, which must satisfy $L = f_C (L)$, so

$$ L = L^2 + C. $$

In other words, if the orbit lands exactly on $L$, it stays there forever.

We can rewrite this relation for $C$,

$$ C = L - L^2. $$

The convergence towards the fixed point can be examined by considering a value slightly off $L$:

$$ x_n = L + \varepsilon_n $$

where $\varepsilon_n$ is a tiny error from the fixed point $L$.

The next iterate is

$$ x_{n+1} = f_C (L + \varepsilon_n) $$

Which, for small $\varepsilon_n$ can be expanded into Taylor series in $\varepsilon_n$:

$$ x_{n+1} \approx f_C (L) + f^\prime_C (L) \varepsilon_n $$

However, since $L$ is a fixed point of $f_C$, then $f_C (L) = L$ and thus

$$ x_{n+1} \approx L + f^\prime_C (L) \varepsilon_n $$

Now, we assume that $x_{n+1}$ can also be written as $L$ with a small correction, $x_{n+1} = L + \varepsilon_{n+1}$. This leaves us with the following equation for the error propagation:

$$ \varepsilon_{n+1} \approx f^\prime_C (L) \varepsilon_n $$

For the convergence to happen, the fixed point needs to be attractive, i.e. $\lvert \varepsilon_{n+1} \rvert < \lvert \varepsilon_n \rvert$, which happens when

$$ \lvert f_C'(L) \rvert = \lvert 2L \rvert < 1, $$

with the endpoint cases $\lvert 2L \rvert=1$ giving the slower borderline behavior. Thus the attracting range is governed by

$$ -\frac{1}{2} \leq L \leq \frac{1}{2}. $$

On that interval, the function $C=L-L^2$ ranges from $-3/4$ to $1/4$. That gives the fixed-point convergence interval

$$ -\frac{3}{4} \leq C \leq \frac{1}{4}. $$

This explains the fixed-point convergence interval, but boundedness is slightly wider. The fixed-point argument already gives boundedness on $[-3/4,1/4]$, so it remains to check the left interval $[-2,-3/4]$.

Assume that $-2 \leq C \leq -3/4$. We will show, by induction, that for all $n\geq 1$,

$$ C \leq x_n \leq C^2+C. $$

The base case holds because

$$ x_1=C, \qquad x_2=C^2+C. $$

Now assume that

$$ C \leq x_n \leq C^2+C. $$

Then

$$ 0 \leq x_n^2 \leq \max\left(C^2,\,(C^2+C)^2\right). $$

But

$$ (C^2+C)^2 = C^4+2C^3+C^2 = C^2 + C^3(C+2). $$

For $-2\leq C\leq -3/4$, we have $C^3\leq 0$ and $C+2\geq 0$, so

$$ C^3(C+2)\leq 0. $$

Therefore

$$ (C^2+C)^2 \leq C^2, $$

and hence

$$ 0 \leq x_n^2 \leq C^2. $$

Adding $C$ everywhere gives

$$ C \leq x_n^2+C \leq C^2+C. $$

Since $x_{n+1}=x_n^2+C$, this is exactly

$$ C \leq x_{n+1} \leq C^2+C. $$

So the interval $[C,C^2+C]$ traps the orbit, and the orbit is bounded for every $C\in[-2,-3/4]$.

Outside $[-2,1/4]$, the orbit escapes. If $C>1/4$, then the equation $L=L^2+C$ has no real solution, so the orbit cannot settle to a real fixed point; starting from $0$, it eventually increases without bound. If $C<-2$, then $x_1=C<-2$ and $x_2=C^2+C>2$, after which the squaring term drives the orbit to larger and larger positive values.

So, for the real one-dimensional map:

$$ \text{bounded orbit from }0: \quad -2 \leq C \leq \frac{1}{4}, $$ $$ \text{fixed-point convergence:} \quad -\frac{3}{4} \leq C \leq \frac{1}{4}. $$

The interval $[-2,-3/4)$ is the fun warning sign: bounded does not necessarily mean convergent.

From the real line to the complex plane

A real number can only move left and right. To get the famous picture we’re all waiting for, we now let the starting value and the parameter live in the complex plane.

The rule becomes

$$ z_{n+1}=z_n^2+c, \qquad z_0=0. $$

For each complex value $c$, we ask the same question as before:

does the orbit starting at $z = 0$ remain bounded, or does it escape?

This one yes/no question sorts the complex plane into two sets. The bounded values form the Mandelbrot set.

To build a first picture, we do not need anything clever. We simply sample a grid of points in the complex plane, iterate the rule from $z_0=0$, and color each point black if it does not appear to escape within the allowed iteration limit.

This is crude, but it works.

Black/white escape test for the Mandelbrot set. Black points did not escape within the iteration limit; white points did. Current grid: 64 × 64.

Even at low resolution, the shape is already recognizable. There is a large central body, a big round bulb on the left, smaller bulbs attached to it, and an increasingly rough boundary.

Refining the grid does not change the object, it only reveals it more faithfully.

The interior is not uniform

The black region we just saw forming is not just one blob of equally boring stability.

Inside the main cardioid-shaped region, the orbit converges to a single attracting value.

Inside the large circular bulb on the left, the orbit settles into a 2-cycle: it alternates between two values forever.

Inside other bulbs, it settles into 3-cycles, 4-cycles, 5-cycles, and so on.

So the Mandelbrot set is not merely the set of bounded points. It is also a map of different kinds of stable periodic behavior.

The main cardioid corresponds to period 1.
The large disk on the left corresponds to period 2.
Smaller bulbs correspond to higher periods.

Details of this periodic behavior will be explored in a future post.

We started with a recurrence simple enough to fit on one line, and from it we got a complicated parameter-space object whose interior is organized into infinitely many regions of stability.

But black and white still hide a lot. Not all escaping points escape equally fast. Some run away immediately. Others hover near the boundary for a long time before leaving. The outside has structure too.

In the next part, we will stop asking only whether a point escapes, and start asking how it escapes.