That’s Unpossible! The Elusive Dimensions.

Over the past few weeks we’ve discussed the Koch Snowflake, Hexaflakes, infinite boundaries with finite areas, and non-integer dimensions.  You know, cuz we are nerdy like that.  Okay, maybe that’s just me.

Anyway, it got me thinking.  All of the shapes we’ve investigated had non-integer dimension based on the number of self-similar copies created after every iteration, and the scale factor required to return the copies to the original size.  The Koch Snowflake had dimension of \frac{\log{(4)}}{\log{(3)}}, while the Hexaflake had dimension of \frac{\log{(7)}}{\log{(3)}}.

I wondered,

was it possible for a shape to have a particular dimension, or were we limited to a set of specific non-integer values?

That is, was it possible to imagine a shape that would have a specific non-integer dimension such as \pi?  Or perhaps e?  Possibly \phi?

Essentially, this comes down to finding a shape that is constructed by created C copies of the original shape, that need to be scaled up by a factor of S to make each of the copies the same size as the original.  If you recall, that means our dimension D would be


The Golden Dragon Fractal

A little investigation (read: I asked the almighty Google) gave me an answer to at least one of these questions.  Specifically, a dimension of \phi\approx 1.61803.  To come up with this dimension, one simply needs to investigate the golden dragon fractal.

But e and \pi are very different numbers than \phi.  That is, e\approx 2.71828 and \pi\approx 3.14159 are known as transcendental numbers while \phi is known as an algebraic number.  Say what?  Don’t worry, the definitions aren’t really all that complicated.  Essentially, a number is either algebraic, or it is transcendental.  But what does that mean?

An algebraic number is a number that satisfies a non-zero polynomial equation that has integer valued (or equivalently rational valued) coefficients.  In math-speak, we are seeking to find any x that satisfies the following equation


where c_{i} are all integer valued, and x is called the root of the equation.  Transcendental numbers are those numbers that do not satisfy this property.

Huh?  This might be easier to see by example.  Consider the number \sqrt{2}.  This number is known as an algebraic number because it is the solution to the polynomial equation


To see this, we can simply rearrange the equation as follows: x^{2}=2.  Finally, take the square root of both sides to solve for x, and the root becomes clear: x=\sqrt{2}.  Note that the coefficients to the polynomial equation x^{2}-2=0 are 1, 0, and -2 for the x^{2}, x^{1}, and x^{0} terms respectively.  This is easier to see if we write the equation as

\underbrace{1}_{c_{2}}\cdot x^{2}+\underbrace{0}_{c_{1}}\cdot x^{1}+\underbrace{(-2)}_{c_{0}}\cdot x^{0}=0.

That is, the coefficients are all integers, and they aren’t all 0.  Unfortunately, we can’t say the same for \pi or e.  That is, no matter how hard we try, we can’t find polynomial equations with integer coefficients that have either \pi or e as a solution.  In other words, we can’t write

\displaystyle{\sum_{i=0}^{n}}c_{i}\pi^{i}=0, or \displaystyle{\sum_{i=0}^{n}}c_{i}e^{i}=0,

unless we assume that all the c_{i} are identically 0.  Neat-o.

Why do we care about transcendental and algebraic numbers?  Well, without getting into too much gory detail, they are used for example, to show that the circle cannot be squared.  Check out this link for more information on that particular ancient slice of mathematical awesomeness.

So, does the transcendental nature of \pi or e matter?  At first I thought it might.  Transcendental numbers can be strange and wonderful.  But the number line is also full of them (in fact, there are more transcendental numbers than algebraic ones – but I’ll leave that for another post).  Their ubiquity might indicate that it wouldn’t or shouldn’t be a problem.  That is, finding a shape with fractal dimension of \pi or e should be possible.  Maybe I just had to dig a little deeper, look a litter further, Google a little Googler.

Sierpinski triangle, a fractal having a Hausdo...
The Sierpinski Triangle. Image via Wikipedia

And then I found an example of something with non-integer dimension that was in fact a transcendental dimension.  Specifically, the fractal known as the Sierpinski Triangle.  It has a fractal dimension of \frac{\log{(3)}}{\log{(2)}}, which is transcendental.  In fact, based on the Gelfond-Schneider theorem (that’s right, I just dropped theorem on your lap), so long as \frac{\log{(C)}}{\log{(S)}} is not a rational number1, then D must be transcendental.  This gave me hope that I should be able to find something that had the required dimension of \pi or e.

Sadly, at this point I have not found anything on the intertubes that has given me an answer.  I do know that algebraic and transcendental fractal dimensions are possible.  The question now is whether or not I can determine an appropriate C and S such that D=\pi or D=e.  This means that I need to find C and S that satisfy

C=S^{\pi}, or C=S^{e}.

Anyway, since I have yet to figure this out, I’ll end this post here.  Hopefully I’ll be able to present an answer in a future post.  Stay tuned!

1 A rational number is any number than can be expressed as the ratio of two integers, p and q. That is, k is said to be rational if k can be written as \frac{p}{q}, where q\neq 0, and p, q\in\mathbb{Z}.

Related Articles


2 Comments Add yours

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s