# The Silly π Question

This was inspired by the Quora question “What is Pi backwards?” It seemed like a good early post to test out my ability to port Mathematica code and LaTeX to WordPress.

The concise answer to the question is “That question is meaningless,” but why is this the case? One explanation is that π is an irrational number with unlimited digits, and something without an end cannot be reversed. Or one could say that “reversing” any number really doesn’t have much meaning in the first place. Yet, this question gives us a good opportunity to explore a few concepts like summations, limits, and significant digits.

One way to look at π is as a summation. This one is equivalent to the arctangent function. $$4*ArcTan(1) = \pi$$

$$4 \sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{2 n-1}=4 \left(1 \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\text{…}\right)=\pi$$

If you look at that sum, it has infinite terms, and every term is smaller than the previous one. Thus, every term we evaluate brings us closer to the actual value of π. We can approximate π by going out to 100 terms.

$$4 \sum _{n=1}^{100} \frac{(-1)^{n-1}}{2 n-1} = 3.13159$$

This gets us in the neighborhood. But we can get closer by going to 1000 terms.

$$4 \sum _{n=1}^{1000} \frac{(-1)^{n-1}}{2 n-1} = 3.14059$$

Now π is starting to look like the number we recognize! If we could go forever, the summation converges to π itself. Now you probably have some idea of why approximating π to a certain number of digits is such a popular problem to throw at big computers.

This sets us up to see why “π backwards” ends up being meaningless. The number has no end, so there is nowhere to “start” the reverse-order number. We might as well ask “What are the first sights we’d see on a return trip from an infinitely far away place?” We could never get to the end of our trip, so we can’t really conceive of what the return would look like.

Just for fun, though, we can interpret “π backwards” as an excuse to switch the starting and ending place of the summation.

$$4 \sum _{n=\infty }^1 \frac{(-1)^{n-1}}{2 n-1}=\pm 4\left(\frac{1}{2 (\infty -2)-1}-\frac{1}{2 (\infty -1)-1}+\frac{1}{2 \infty -1}-\frac{1}{2 (\infty -3)-1}\text{…}\right)$$

Notice that we are starting at infinity, so we will also never reach a definite answer to this summation either. But it’s worse than that. We are effectively dividing by infinity in every term, so every term is practically zero. We can’t even figure out if the “first” term is positive or negative! (Not that it really matters when every term is zero.)

$$4 \sum _{n=\infty }^1 \frac{(-1)^{n-1}}{2 n-1}= \pm 4\left(\frac{1}{\infty }+\frac{1}{\infty }-\frac{1}{\infty }-\frac{1}{\infty }\text{…}\right) =\pm (0+0+0+0\text{…})$$

So we have a mess. Every term is effectively zero, yet infinitesimally larger than the previous term, we can’t even figure out the sign of the terms, and there is an infinite number of these infinitely insignificant terms to add up. We will never get to a term that isn’t zero because infinity is infinitely far away from our ending place of 1.

So even if this summation approach gave us a way to look at “π backward,” all we saw were an infinite pile of ambiguous-signed fractions that were so small that they’re indistinguishable from zero. At least looking at π forwards gave us a reasonable approximation of the actual number because it prioritized the most significant terms.

So if any readers were wondering what it means to have a never-ending number, this might help clear things up. Such numbers can be pictured as summations, where each term is smaller and less significant than the last.

Film is one of the three universal languages, the other two: mathematics and music.

Frank Capra