Thursday, March 14, 2013

Pi Day! (Pi Randomness, and the Case for Tau) [Pre-Release Text]

[This is really just my script for a video that I wanted to do, but I couldn't in time. (Update: I know I said I'd have it the day after, but that has not come to pass. It should be up soon, but I don't have a specific date yet.) Some things may be hard to understand without visual aids and mathematical formatting.]

Happy Pi Day! In case you didn't figure, Pi Day comes from the first three digits of pi: 3.14. This yields the date 3/14 in month-day form, or March 14.

Those who don't go with the month-day form, and go with day-month instead can still have their day with Pi Approximation Day, 22 July, which looks just like the approximation 22/7 for pi in day-month form.

But let's get back to Pi Day. The only way this works is in decimal form, which most of us use. But what if we tried a different base, say 2? Binary is just as important in our modern (computerized) world, so let's go with that.

Pi begins with the binary expansion 11.00100100. Of course, we could pick out different numbers of digits to get the date we want, say, the first two, 1 and 1, and celebrate Binary Pi Day on New Year's. But that doesn't capture the essence, just as 3/1 doesn't do it for Pi Day.

So we'll maximize the number of digits to take. A month in the Gregorian Calendar requires 4 binary digits, with range 0 to 15, to be represented. A day requires 5 binary digits, with range 0 to 31.

So let's do that. If we use month-day, we have 1100 and 10010, resulting in 12 and 18 in decimal. Mark your calendars for December 18! Or, if we use day-month, we have 11001 and 0010, or 25 and 2. We've barely passed 25 February. Set your sights for next year!

There's a lot more to be said about pi, but I think what I might say has been said before. I might do something soon about it. But now, let's set our sights on a related constant: tau.

Tau is equal to 2 pi. Equivalently, it is the ratio of a circle's circumference to its radius. That formula for circumference, C = 2 pi r, becomes C = tau r.

Before we continue, I'd like to say that, though Tau Day (6/28) is not the same as Pi Day, they both agree for binary. Since tau is just pi multiplied by 2, the binary representation is simply shifted over one place value: 110.0100100... So December 18 and 25 February should really be Binary Pi/Tau Days!

Moving on. There are lots of reasons why tau works better than pi. You can go ahead and read "The Tau Manifesto" (which should be linked) [http://tauday.com/tau-manifesto], or continue reading. (I do have to admit that much of this is taken from there.)

Well, if tau works better then pi, why do we use pi in the first place? The answer is practicality. Put simply, it's much easier to measure a circle's diameter than its radius. Engineering specifications for circular forms are usually given as a diameter, with its own special symbol (okay, not really special as it resembles a few other symbols with distinct uses).

But that's why we use pi. But enter modern mathematics, where circles are considered by their radii, not their diameters. (Seriously, if you see "d" in math class, you're either dealing with distance or differential calculus [and possibly both]).

So we should use tau. It clears up formulae, as we've already seen with circumference: C = tau r. The period of the sine, cosine, secant, and cosecant functions is... 2 pi? How about just tau! As a consequence, we can write angular frequency omega = tau frequency f. As a consequence of that, here's one for the physicists: no more dealing with that pesky formula for the reduced Planck's constant. Reduced Planck's constant h-bar = Planck's constant h / tau! Much simpler!

Next we return to what brought us here in the first place: circles. Radians are excellent for measuring angles since they simply represent the ratio of the arc length to the radius. That is, radian angle theta = arc length s / radius r. So, how many radians does a full circle subtend? 2 pi radians. How about tau radians! So a quadrant of a circle subtends 1/4 tau radians, which reveals the quarter aspect. 1/2 pi radians? What?

Oh, the grand Euler equation: e^(pi i) + 1 = 0. The famous equation relating five of the greatest numbers in mathematics, derived from Euler's formula: e^(xi) = cos x + i sin x. Pi still has its stronghold... Not exactly. This is just a more elegant way of saying e^(pi i) = -1, which doesn't look too good with that negative sign. Even better, we can say e^(tau i) = 1. Okay, the zero is gone, but it was only there in the original one to get rid of the negative sign. We can re-introduce it by considering Euler's formula: e^(tau i) = 1 + 0i.

And finally, we come to the most contentious formula in the battle between tau and pi: area of a circle. A = pi r^2. A = 1/2 tau r^2? Doesn't look too good, does it?

Well, that's true, but... Tell me, in physics, how far does an object travel, starting at rest with a constant acceleration, in a given amount of time? Answer: distance x = 1/2 acceleration a (time t)^2. Similarly, what is its kinetic energy? energy E = 1/2 mass m (velocity v)^2. These two formulae are clues to what's going on.

To fully answer this, we're going to need calculus. Let's start with the distance problem. Acceleration is the second derivative of position, so d^2x/dt^2 = a, which is a constant as established earlier. Integrating, we have dx/dt = at + C. Since velocity is just the first derivative of position, we have v = at + C. To find the value of C, note that the object is initially at rest, so v(0) = 0. 0 = a(0) + C, so C = 0. We're left with dx/dt = at. Integrating again, we have x = 1/2 at^2 + C. We'll assume that the object's initial position is 0, so C = 0. Finally, we have x = 1/2 a t^2.

We can use this to derive the formula for kinetic energy, and we will see that it retains the 1/2 factor from the formula we just derived for position. Kinetic energy is essentially the work needed to bring an object to a certain velocity. Work W = force F distance x. We'll assume a constant force to simplify things, so that implies a constant acceleration. Force F = mass m acceleration a, and we already have x = 1/2 a t^2, so W = (ma)(1/2 a t^2) = 1/2 m a^2 t^2. But since v = at for an object initially at rest and undergoing constant acceleration, the formula can be simplified to W = 1/2 m v^2. Since kinetic energy is equal to this work, E = 1/2 m v^2.

Now, how does this apply to the area of a circle? It works with one proof for the formula for the area of a circle. We'll break up the circle into concentric rings (or annuli, singular annulus, if you will).

Let's consider one of these rings (annuli). Let's call the radius of this ring little r (to distinguish it from big R, the radius of the whole circle) and the width dr. The area is roughly equal to that of a rectangle of length that is the circumference of the ring (it doesn't matter inside or outside; since this is calculus, we'll make these rings infinitesimally thin), and height dr. Remember that tau is the ratio of circumference to radius, so the length is tau r. So the area of this ring is tau r dr.

Summing up these circles with integral (0 to big R) tau little r dr, we have 1/2 tau R^2. As "The Tau Manifesto" states at the end of the section exploring this formula: "If you were still a pi partisan at the beginning of this section, your head has now exploded."