This is Andy Novocin. I was creating a comment on Bill Wood’s post . After several paragraphs it seemed long enough to turn into a blog post.

Bill, thank you for your article. It has made me think all morning about better \(\pi\) representations. We are all so human-centered and our beloved math aspires not to be.

I was struck by how \(\phi\) can so nicely be represented as the continued fraction \([1; 1, 1, \ldots]\) while \(\pi\) is not nice as a continued fraction \([3; 7, 15, 1, 292, \ldots]\). It is tempting to think (prior to any research) that we’ve spotted a sub-classification of irrationals. Namely, take the rules we were taught for classifying rationals and irrationals by examining their decimal representation and apply them to continued fraction representations. We’ve just created two new classes: “repeating continued fractions” and “non-repeating (infinite) continued fractions”. \(\phi\) is in the first class and \(\pi\) is in the second.

Both \(\phi\) and \(\pi\) appear throughout nature and culture. Is \(\pi\) fundamentally more complex than \(\phi\)? Well, yes, actually. \(\phi\) is a solution to a simple polynomial \(x^2 – x -1\) with rational coefficients. That means that some people call it an algebraic number. \(\pi\) on the other hand is a transcendental number, meaning that no polynomial with only rational coefficients has \(\pi\) as a root. That is a very difficult fact to prove. Could it be that continued fractions with infinite repeating integers are algebraic and those without are transcendental? That could make for some simpler proofs of transcendentality.

Well, I suspect that it won’t be that simple. It seems that the simple sum \(e + \pi\) has never been proved to be transcendental nor proved algebraic. While the numbers \(e\) and \(\pi\) are both known transcendentals.

In math we often run into problems which are easy to state and hard to prove. My technique with that sort of problem is to explore it until I can spot the true difficulty. That usually means finding a one or two sentence sub problem which, if solved, would solve the hard problem. Then I stop putting time into it for a while. Whenever a new technique pops into my life I’ll make a point to test the new technique on each difficult problem I remember to see if I can make any progress. This method is not a sure-fire problem solving technique, but it is a decent time management technique. We have to be willing to tackle the most difficult problems but wise about how we spend our time on them.

Let’s try to explore whether or not patterned continued fractions are algebraic, like \(\phi\), while non patterned continued fractions are transcendental, like \(\pi\). Whether or not we succeed we should certainly learn.

I figure that exploring another transcendental all-star, \(e\) the base of the natural log, might disprove the claim straight off without having to do any actual hard work. To compute the continued fraction representation (CFR from here on out), I turned to SAGE (an open source computation platform which is essentially the python language but with powerful math). The CFR for \(e\) looks like this:

\([2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, 14, 1, 1, 16, \ldots ]\)

Do you see the pattern? \(1,1,2k\) and it continues for as far as I wanted to compute. Well that convinced me that a repeated sequence was not going to take place. Which still supports our claim. On a side note, it looks a bit nicer than \(\pi\) did (recall that \(\pi\) was \([3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, \ldots]\)). Maybe \(e\) is simpler than \(\pi\), but I’ll have to explore that later, we’re still not done with \(\phi\).

Now to pen and paper. Perhaps a good first step is to show that any repeating CFR can solve some polynomial. If we succeed then we’ll know that all repeating continued fractions are algebraic. After that we would just need to show that solutions to polynomials can be written with a repeating CFR.

So I tried to reverse engineer the polynomials which are solved by \([r0; r1, r2, \mbox{repeating}]\) and \([r0; r1, r2, r3, \mbox{repeating}]\). A method similar to Bill’s approach with \([1; 1, 1, \ldots]\) and a few pages of pencil scratch later and I found only quadratic polynomials. This was odd, I was kind of hoping that as the pattern got longer we’d see higher degree polynomials. Since that would give me hope that all polynomials would have repeating continued fraction expansions.

At this point I turned to the internet and found such high quality names as Euler, Lagrange, and Galois exploring this problem. That made me feel good that 1) the problem was in fact worth exploring and 2) the problem is not going to be easy.

Euler had taken my approach and showed that all repeating continued fractions are solutions to quadratic polynomials! So if you have something which has a repeating continued fraction then it is not transcendental, thus \(\pi\) will never begin some infinite repeating pattern in its continued fraction representation, no matter how much you look!

This also means that \(\sqrt[3]{5}\) will never repeat as a CFR, otherwise it would be the solution of a quadratic equation. Of course it is an algebraic number as it solves \(x^3 – 5\). That means that our claim is false! Hooray!

So is that the end of the road? What about a nice representation of \(\pi\)? Well, maybe there is some other representation of real numbers which would let our rule determine whether numbers are roots of cubics, quartics, and so on? Apparently that is an open question posed by Hermite in 1848, so it is worth looking into further.

So I’ll leave you with some other representations of \(\pi\) which seem nice and patterned to me. The only use integers and basic arithmetic (although infinitely). One is the infinite series approach (Calc II):

$$\pi = 4\sum_{i=0}^{\infty} \frac{(-1)^{i}}{1+2i}$$

Another type is like continued fractions but with some flare:

$$\pi = 3+\frac{1}{6+\frac{9}{6+\frac{25}{6+\frac{49}{6+\cdots}}}}$$

Thanks for meandering with me.