This was inspired by the Quora question “What is Pi backwards?” It seemed like a good early post to try converting Mathematica code and LaTeX to WordPress.
The concise answer to the question is “That question is meaningless,” but why is this the case? After all we can write 123 “backwards” as 321, why not π?
One explanation is that π is an irrational number with unlimited digits, and something without an end cannot be reversed. Or something with no end, written backwards, becomes a thing with no beginning. 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 what we mean when we describe π as an irrational number.
One way to look at π is as a summation, like this summation 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 adds up an 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. But, we never run out of terms, so we never find the exact value of π. We just get closer and closer forever.
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. (Note that I rounded to 7 digits.) 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 I kept adding more terms, then I would need to round to a larger number of digits to see the change. 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. Just for fun, though, we can interpret “π backwards” as an excuse to switch the starting and ending place of the summation.
$$4 \sum _{n=\infty }^1 \frac{(-1)^{n-1}}{2 n-1}=\pm 4\left(\frac{1}{2 \infty -1}-\frac{1}{2 \infty -3}+\frac{1}{2 \infty -5}\text{…}\right)$$
Notice that we are starting at infinity, so we will 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.)
$$4 \sum _{n=\infty }^1 \frac{(-1)^{n-1}}{2 n-1}= \pm 4\left(\frac{1}{\infty }-\frac{1}{\infty }+\frac{1}{\infty }\right) =\pm (0-0+0\text{…})$$
So we do have an answer for what “Pi backwards” might look like. 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 by starting at the “infinity” term, we put ourselves infinitely far away the “end” of our summation.
So “Pi backwards” looks like an infinite summation of zeroes. At least looking at π forwards gave us a reasonable approximation of the actual number because it prioritized the largest, most significant terms.
Film is one of the three universal languages, the other two: mathematics and music.
Frank Capra
