# 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

$D=\frac{\log{(C)}}{\log{(S)}}.$

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

$\displaystyle{\sum_{i=0}^{n}}c_{i}x^{i}=c_{n}x^{n}+\ldots+c_{2}x^{2}+c_{1}x^{1}+c_{0}x^{0}=0,$

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

$x^{2}-2=0.$

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.

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 $p, q\in\mathbb{Z}$.

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# That’s Unpossible! Final Act: A Squishy Dimension

Several days ago we went on an adventure exploring the Koch Snowflake.  During our exploration, I told you to trust me when I wrote that

$\displaystyle{\sum_{i=0}^{\infty}} \left(\frac{4}{9}\right)^{i}=\frac{9}{5}$.

A few days later, we learned that if $|p|<1$, we could write

$\displaystyle{\sum_{i=0}^{\infty}}p^{i}=\frac{1}{1-p}$.

And since in our example $p=\frac{4}{9}<1$, my original equation was correct.

Of course, I also stated that the snowflake somehow existed in a space that wasn’t quite 1 dimensional, but it also wasn’t quite 2 dimensional.  That is, it existed in a non-integer dimension.  What the what?  That’s crazy talk, you say.  Probably.  But let’s explore this further.

Most of you have probably heard of a Fractal.  The basic idea here is that you have an object that has a property known as self-similarity.  In a nutshell, if I were to zoom in on any part of the object, it would look like the original object.  If you think about it, one could easily get lost in a fractal because every region looks like every other region under some magnification.  It’s almost trippy if you think about it too long.

Regardless, the important point is that fractals are self-similar.  Can we say the same about other shapes?  And if so, what might this tell us about dimension?

Consider something in a 1 dimensional world: a line of length 1.  If we magnify the line 2 times, clearly we would end up with 2 copies of the original line.  That is, a 2x magnification gives us 2 copies of the original object.  If we magnified the line 3 times, we’d have 3 copies.  In fact, if we magnified the line $n$ times, we’d have $n$ copies.

What about something in 2 dimensions such as a square with side length of 1?  If we magnify the square by a factor of 2, you can imagine that we would end up with 4 copies of the original square.  In this case, a 2 times magnification gives us 4 copies of the original object.  A 3 times magnification would give us 9 copies, and in general, an $n$ times magnification would give us $n^{2}$ copies.

In a similar manner, a regular object in 3 dimensions would give us 8 copies if the original were magnified 2 times.  And in general, an $n$ times magnification would give us $n^{3}$ copies.

To recap:

• Scaling up something by a factor of $n$ in 1 dimension gives $n$ copies of the original,
• Scaling up something by a factor of $n$ in 2 dimensions gives $n^{2}$ copies of the original, and
• Scaling up something by  a factor of $n$ in 3 dimensions gives $n^{3}$ copies of the original.

In general, if we scale something up by a factor of $n$ in an $m$-dimensional space, we will end up with $n^{m}$ copies.  Cool!  That is, our scaling value $n$ raised to the dimension $m$ gives us the number of copies $s$.  In mathy terms

$s=n^{m}$.

We’re going to rearrange this in order to determine the dimension of a fractal.  That is, we need to solve for $m$.  Taking the $\log$ of both sides and isolating $m$ we get

$m=\frac{\log{(s)}}{\log{(n)}}$.

So, the dimension $m$ of an object is simply the ratio of the $\log$ of the number of copies $s$ one gets by scaling the object by a factor of $n$, to the $\log$ of the scaling factor $n$.  Sweet freaking awesome.

In the case of the Koch Snowflake, remember that we started with a line of length 1.  We then removed the middle third and inserted 2 lines of equal length (both being one-third the length of the original line).  If we were to magnify the line by a factor of 3 so that each of the line segments of length one-third looked like the original line segment of length 1, you’d see that we’d have 4 copies of the original line.  That is, $s=4$ while $n=3$.  This gives us a dimension of

$m=\frac{\log{(4)}}{\log{(3)}}\approx 1.262$.

Cripes on a cracker - an integer dimension.  That is, the Koch Snowflake isn’t quite 1 dimensional, and it isn’t quite 2 dimensional either.  Wild stuff!

For those that think this is all bunk, consider this: the Koch Snowflake starts with a line.  We then remove the middle third and add 2 line segments measuring one-third the original.  The result is a bent line.  If we continue this forever, we still end up with a bent line.  More specifically, an infinitely bent line.  So no matter how much we zoom in, it’s always going to look fuzzy (due to the infinite bending).   We won’t ever see crisp lines – things that exist in the 1 dimensional world.  But by the same token, we can’t say that the fuzziness that we see represents something that exists in the 2-dimensional world either.  Technically the lines have only length, but no width.  So clearly this object exists in some sort of nexus between these dimensions.  How mind-blowing-ly awesome is that?  Personally, I find the idea of a non-integer dimension completely satisfying.

Happy Mathematics Awareness Month all y’all!