Is Fractal perimeter always infinite?

Looking for information on fractals through google I have read several time that one characteristic of fractals is :

• finite area
• infinite perimeter

Although I can feel the area is finite (at least on the picture of fractal I used to see, but maybe it is not necessarly true ?), I am wondering if the perimeter of a fractal is always infinite ?

If you think about series with positive terms, one can find :

• divergent series : harmonic series for example $\sum_0^\infty{\frac{1}{n}}$
• convergent series : $\sum_0^\infty{\frac{1}{2^n}}$

So why couldn't we imagine a fractal built the same way we build the Koch Snowflake but ensuring that at each iteration the new perimeter has grown less than $\frac{1}{2^n}$ or any term that make the whole series convergent ?

What in the definition of fractals allows or prevent to have an infinite perimeter ?

Solution 1:

You can certainly define a fractal analogous to the Koch snowflake with finite perimeter.

The iterative rule for the standard Koch snowflake is to replace the middle third of each line segment with two line segments of equal length and meeting the original segment and each other at $60^\circ$ angles. This applies the rule $x \mapsto 4x/3$ to the overall length of the curve, so applying it repeatedly produces a divergent sequence of lengths.

A modified rule might be to start with a red line segment and iteratively replace the middle half of each red line segment with the top of a trapezoid made of one red and two blue line segments (all of equal length and again making $60^\circ$ angles). Then the overall red/blue lengths $(x,y)$ are subjected to the rule $(x,y)\mapsto(3x/4, y + x/2)$, which is convergent when iterated. It is a nice exercise to show that this process, starting from a red segment of length $1$, produces an entirely blue curve of length $2$; and starting from an equilateral red triangle of perimeter $3$ produces a closed blue curve of perimeter $6$. (Calculating the area inside the curve is another nice exercise. Moreover, since the trapezoid can be constructed either red-blue-blue or blue-red-blue at each step, there are any number of different fractals with the same perimeter and area.)

Solution 2:

The "finite area" probably refers to the area enclosed by some closed fractal curve (where actually the curve is the fractal, not the enclosed area, and the area of the curve itself is $0$ — which admittedly is also finite).

Basically, an object of fractal dimension $f$ has the property that

• if you calculate its measure in dimension $<f$, you always get $\infty$.
• if you calculate its measure in dimension $>f$, you always get $0$.

For example, the Koch Snowflake curve has dimension $1 < \ln 4/\ln 3 < 2$, therefore its length (measure of dimension 1) is infinite, but its area (measure of dimension 2) is $0$.

Note however that the "filled snowflake" is an ordinary 2-dimensional object (although its border has fractal dimension — it's the Koch Snowflake curve, after all). Therefore the 2d measure (area) of that is not $0$. However it is not true that you always have to have a finite area; imagine that instead of putting 6 Koch curves to a snowflake forn, you'd just added them linearly one after the other. The border would still be a fractal (although it now is no longer astonishing that its length is infinite because it goes to infinity on both sides), but the area would be the area of one half of the plane, which is of course infinite as well.

On the other hand, the fractal dimension of the Cantor set is $\ln 2/\ln 3$, therefore even its 1d measure is $0$. Nevertheless it also is a fractal, although it doesn't fit the characterization you gave.

Solution 3:

The problem is: What is the perimeter of a point set after all? However, in the case of curve-like fractals like thesnowflake, we produce successive approximations by starting with a line segment and then repeatedly replace a line segment by a sequence of line segments which lengthens the path. In the case of the Koch snowflake, the lengtheing is by a constant factor of $\frac43>1$, hence the lengths of the approximating polylines grow without bound.

If on the other hand you have some set $S$ that is maybe quite zig-zag, but can be said to be of length $L$ in a suitable manner (that is: We can map $f:S\to \mathbb [0,L]$ injectively and with dens image, such that $|f(x)-f(y)|\ge |x-y|$), then its dimension is not fractal. Indeed, if $\epsilon>0$ is given, you can select the $\approx\frac L\epsilon$ points along the curve at length offsets $0, \epsilon, 2\epsilon, \ldots$ and observe that the curve is covered by $\frac L\epsilon$ balls of radius $\epsilon$, hence the Hausdroff dimension is $1$ and not fractal.

Solution 4:

Yes and No.

Yes, you can generate an everywhere-bending curve like the Koch snowflake but with finite length, in exactly the way you describe.

It will be a fractal in some ways but not others, so whether it is technically a fractal will depend on who you ask.

Quoting from Wikipedia:

There is some disagreement among mathematicians about how the concept of a fractal should be formally defined. [...] In 1982 Mandelbrot stated that "A fractal is by definition a set for which the Hausdorff–Besicovitch dimension strictly exceeds the topological dimension." Later, seeing this as too restrictive, he simplified and expanded the definition to: "A fractal is a shape made of parts similar to the whole in some way."

The curve you are looking for will not satisfy the first definition, but it can satisfy the second. (Similarly, the answer to the title of this question is "Yes" for the first definition but "No" for the second.)

Note that these are just two of many possible definitions. The informality of the second definition corresponds to the fact that mathematics has no real need to formally classify objects as fractals or not -- the objects themselves are the more interesting thing to study.

Properties of your curve

Your curve will necessarily have the property that as you zoom in on a random part of it, it will look more and more straight. This is unavoidable, as otherwise the curve would indeed have infinite length. However, it can still have "bends" everywhere, just like the Koch snowflake. They just need to be minor bends, getting less and less sharp as you zoom in.

Related to this, your curve will have a fractal dimension of one, since you are asking for its one dimensional length to be finite. This is why it will not satisfy the first definition above.

Another property of fractals mentioned on the Wikipedia page is that they are usually not differentiable anywhere. Your curve, however, will be differentiable almost everywhere. On the other hand, unlike a smooth curve, its derivative will still have discontinuities in every part of it, no matter how much you zoom in, due to its fractal nature.

An example

One easy way to make such a curve is to alter the Koch snowflake recipe so that the angles between the four segments become less sharp at each stage. As you suggest, any convergent series will do.

So that the terms don't decrease too quickly, let's use $\sum_1^n \frac{1}{n^2}$. Each iteration of the construction will increase the length by some multiplicative factor, so we want the logs of these factors to correspond to the terms of the convergent series, so at step $n$ we want the length to increase by a factor of $e^{1/n^2}$.

Since the terms drop quickly at the beginning, I'll skip the first 10 terms, and I'll scale the remaining terms (of the arithmetic series) by 34.81 so that the first term is roughly 4/3, like the Koch snowflake. This gives an increase in length of $e^{\frac{34.81}{(n+10)^2}}$ at the $n$'th step, and results in an overall length of about 3.3 times the straight-line distance between the endpoints of the curve.

Here are some views of the resulting curve:
(The first one is the whole curve.)

Of course, if you would zoom in exactly on one of the "elbows" of the curve, then you would see a bend at that angle no matter how far you zoom in.