0

Informally, I have heard for a function to be considered as a fractal - the function can never be "smooth". This means that if you keep "zooming in", you will never have a "smooth" section. Smoothness is an important property for derivatives - if a function is never smooth, this means that you can not take it's derivative (I have always accepted this fact as a "intuitive" fact, but I have never understood the pure mathematical logic behind why a function needs to be smooth in order for it to be differentiable).

But is there a mathematical definition that can be used to decide whether a given function can be considered as a "Fractal" or "Not a Fractal"?

I tried looking for such a mathematical function, but the closest thing I could find was the following link (https://en.wikipedia.org/wiki/Fractal):

According to Falconer, fractals should be only generally characterized by a gestalt of the following features:

  • Self-similarity, which may include:
  • Exact self-similarity: identical at all scales, such as the Koch snowflake
  • Quasi self-similarity: approximates the same pattern at different scales; may contain small copies of the entire fractal in distorted and degenerate forms; e.g., the Mandelbrot set's satellites are approximations of the entire set, but not exact copies.
  • Statistical self-similarity: repeats a pattern stochastically so numerical or statistical measures are preserved across scales; e.g., randomly generated fractals like the well-known example of the coastline of Britain for which one would not expect to find a segment scaled and repeated as neatly as the repeated unit that defines fractals like the Koch snowflake.

-Qualitative self-similarity

  • Multifractal scaling: characterized by more than one fractal dimension or scaling rule
  • Fine or detailed structure at arbitrarily small scales. A consequence of this structure is fractals may have emergent properties. -Irregularity locally and globally that cannot easily be described in the language of traditional Euclidean geometry other than as the limit of a recursively defined sequence of stages.

For example, an example of a famous Fractal as well as one of the earliest defined Fractals is the Weierstrass Function (https://en.wikipedia.org/wiki/Weierstrass_function) :

enter image description here

https://download.ericduminil.com/weierstrass_zoom.gif

I have heard that the Weierstrass Function is non-differentiable - this is because even though the Weierstrass Function itself converges, it can be shown that the derivative of the Weierstrass Function at any point is "infinity" (i.e. does not converge) : this effectively means that the Weierstrass Function does not have a derivative.

We say that a function that is a fractal is non-differentiable - but how exactly do we know if a function can be considered as a "fractal" or "not a fractal"? As a facetious question - why are y = x^2 and y = sin(x) not considered as "fractals", but the Weierstrass Function is considered as a "fractal"?

Thanks!

Note: It seems to me that the Weierstrass Function is non-differentiable primarily because of the mathematical proof involving its derivatives not-converging - and less to do with it being considered as a fractal? Or does these properties go hand-in-hand?

References:

stats_noob
  • 3,112
  • 4
  • 10
  • 36
  • 2
    A fractal is typically considered to be a figure whose Hausdorff dimension strictly exceeds its topological dimension. In paticular the Weierstrass function has Hausdorff dimension $3/2$ which is strictly greater than $1$, which is the topological dimension. – CyclotomicField Mar 05 '22 at 16:31
  • 1
    Regarding that Weierstrauss function. Note that $\frac{1}{a^m}f((x-x_0)/b^m) = f(x-x_0)+C(a,b,m)$. So ignoring a vertical shift, if you zoom in at $x_0$ horizontally by a factor of $b^m$ and vertically by a factor of $a^m$, you have the same graph. So you have lots of self-similarity. And it being non-differentiable makes it nontrivial, like say a constant function which trivially has self-similarity. You do not have a relation like this with $x^2$ or $\sin(x)$. – 2'5 9'2 Mar 05 '22 at 16:47
  • In my comment above I incorrectly imply that the vertical shift $C$ does not depend on $x$ or $x_0$. In fact $C=\sum_{n=-m}^{-1}a^nf(b^n\pi(x-x_0))$. But this is a smooth function. So it only "corrupts" the relation $\frac{1}{a^m}f((x-x_0)/b^m) = f(x-x_0)$ in a smooth way, and the self-similarity of the non-differentiable jaggedness of the curve remains. – 2'5 9'2 Mar 05 '22 at 18:25
  • https://math.stackexchange.com/a/2677204/468350 – Xander Henderson Mar 07 '22 at 19:59
  • Paraphrasing Justice Stewart, "I'll know it when I see it." There is not a universally agreed upon definition of the term "fractal" in mathematics, but most of us who work in the parts of mathematics which deals with fractals have a sense of what we mean by the word. – Xander Henderson Mar 07 '22 at 20:04

0 Answers0