Calculus of the Incalculable - On Circles and Giving Up - Burn Math Class: And Reinvent Mathematics for Yourself (2016)

Burn Math Class: And Reinvent Mathematics for Yourself (2016)

Act II

4. On Circles and Giving Up

4.9. Calculus of the Incalculable

At this point, any attempt to differentiate the machines V and H would seem to be a hopeless delusion. After all, we certainly don’t know how to calculate V (α) or H(α) for any α. We can’t even write down a description of them! We “defined” them by drawing pictures, so how could we possibly differentiate them? We probably can’t. However, the fact that this seems impossible is all the more reason to give it a shot. Since everything we know about these machines so far can be phrased in terms of pictures, let’s try to differentiate them by drawing pictures.

Figure 4.12: Delusionally trying to differentiate V and H, even though we can only describe them with pictures. We’re using to stand for right angles, because seeing over and over can be a bit confusing.

Even though we can only draw pictures, let’s see if we can make some progress. At this point, we draw Figure 4.12. Now, how might we be able to find the derivatives of V and H? Well, let’s do what we always do when we’re trying to differentiate something. We’ve got a machine that we feed some food. Then we change the food by a tiny amount, increasing it from food to food + d(food), and then we look at how the machine’s response changed between the two cases. In this case, our machines are V and H, and the food is the angle α.

So let’s make a tiny change in the angle, increasing it from α to α + , and then look at dHH(α + ) − H(α), and let’s also do the same for V. Figure 4.12 shows what happens when we make a tiny change in the angle. Since H(α) was defined to be an abbreviation for H(ℓ, α) when = 1, we’re making sure that the shortcut distance is equal to 1 both before and after the change in angle. So we’re effectively looking at an infinitely thin slice of the circle. Let’s zoom in infinitely far where all the action is happening and see if we can conclude anything.

If we’ve really only increased the angle by an infinitely small amount , then the two lines on the far left of Figure 4.13 have to be exactly parallel to each other (or if you prefer, infinitely close to being parallel). If this seems counterintuitive, think of it this way: if they weren’t exactly parallel, then there would be some small but measurable angle between them that would be strictly bigger than zero. In principle, we could then measure and say how big it was. This would mean that the tiny increase was just very small, not infinitely small. This kind of reasoning is odd, but let’s run with it and see where it gets us. The line between the words “before” and “after” in Figure 4.13 is the line going from where we were before the angle change to where we are after the angle change. Since we changed the angle without changing the radius, it seems like this line should be perpendicular to the other two lines that shoot off of it down and to the left. It may help to think of a thin slice of pie: the crust should be perpendicular to either of the two sides, in the limit when the slice becomes infinitely thin.

Figure 4.13: Zooming in after having made an infinitely small change in the angle α.

Since we’re writing for right angles, the angle in Figure 4.13 should get a at this point. Now, the line from “before” to “after” in Figure 4.13 is tilted, so if we imagine breaking it up into horizontal and vertical pieces (i.e., drawing the horizontal and vertical lines in Figure 4.13), then we notice something surprising.

The tiny triangle in Figure 4.13 looks astonishingly similar to the original triangle in Figure 4.12. It almost looks like they’re the “same” triangle, in the sense that one is a shrunken-down and rotated version of the other. We could convince ourselves of this if we could convince ourselves that the two triangles had all the same angles. We know that both of them have a right angle, because both triangles arose from us trying to break up a tilted thing into horizontal and vertical pieces. That’s one; what about the other two? Notice that the three angles , , and? on the right side of Figure 4.13 add up to a straight line. But because of the way we’re measuring angles, a straight line is an angle of , or what the textbooks call π. So we can summarize this by saying:

Figure 4.14: For the reasons discussed in the main text, we discover that the angles of this infinitely small triangle have to be the same as the angles of the original triangle. The angle has to be a right angle (i.e., ), and the angles α and are whatever they were in the original triangle.

It seems like the angle? in Figure 4.13 has to be the same as the angle α in Figure 4.12, because the triangles look so similar. How might we be able to convince ourselves of this? Here’s one idea.

Make two copies of the triangle and stack them to build a rectangle. Each of the angles in the rectangle is , and there are four of them, so adding up all the angles inside a rectangle gives you 4. Because of this, the sum of the angles in the original triangle has to be half of this, or 2. But since is a right angle, 2 must be a straight line, which is another way of writing the angle we’re calling . So the sum of all the angles in any triangle has to be . Using this fact on the original triangle from Figure 4.12, we have:

Figure 4.15: Since our two triangles have all the same angles, the tiny one has to be a shrunken version of the original. Because of the way we chose to measure angles, the longest side of the tiny triangle has length . Having found one side of the tiny triangle, we can find the others. This lets us find the derivatives of V and H.

Taken together, the above two equations tell us that the angle? has to be α. Okay, so in trying to convince ourselves that the tiny triangle in Figure 4.13 has all the same angles as the original triangle in Figure 4.12, we’ve convinced ourselves that (i) each triangle has a right angle , and (ii) each triangle has an angle α inside it. But we just showed that the sum of the angles in both triangles has to be . This tells us that the third angle in the tiny triangle from figure Figure 4.13 must be , another one of the angles from the original triangle we started with. At this point, we can summarize everything we know by drawing another picture, shown in Figure 4.14.

So the tiny triangle in Figure 4.13 is just a shrunken-down and rotated version of the triangle in Figure 4.12. Do we know anything else? Well, we know that the tiny angle increase we made was , but because of the way we’re measuring angles, a full turn is . However, is also the distance around a circle of radius 1. We’re effectively dealing with such a circle, although we’re only looking at an infinitely thin slice of this circular pie. So on this circle with radius 1, the word “angle” just means “distance.” This helps us tremendously, in that it tells us that the line between the two copies of in Figure 4.14 must have length , because that’s the angle it spans.

Now, the original triangle in Figure 4.12 had sides of length H(α), V (α), and 1. But because the tiny triangle is just a shrunken version of the original, it must have sides of length H(α), V (α), and , respectively. Now we can draw Figure 4.15 to summarize everything we’ve found.

The goal of all this was to see if we could figure out the derivatives of V and H, even though at this point we can only describe them graphically. Using what we found above, we want to determine

where dV ≡ VafterVbefore and dHHafterHbefore. The V and H here refer to the horizontal and vertical lengths of the triangle from Figure 4.12 before and after we make the tiny change in the angle. So dV and dH are just the side lengths of the tiny triangle — the lengths we’ve just figured out how to write. When we increased α a tiny amount, the vertical length of the large triangle increased by the small amount dV, but from Figure 4.15, we have dV = H(α), or to say the same thing another way:

Whoa! We actually found the derivative of V. Let’s see if we can do the same thing for H. When we increased α a tiny amount, the horizontal length of the large triangle decreased a tiny amount dH, but from Figure 4.15, we have dH = −V (α), where the negative sign shows up because lengths are always positive, but we’re talking about a decrease in length, so the change is negative. To say the same thing another way:

This is great! Even though we have no idea how to write down a description of the machines V and H without using pictures, we managed to figure out their derivatives. Luckily, their derivatives are almost just each other, sometimes with a minus sign. If the truth of the matter hadn’t been so elegant, there’s no way that this attempt to differentiate V and H would have worked.

This is yet another example of what we’ve been saying from the start. These “prerequisites” to calculus are surprisingly difficult, and typically require calculus itself before we can fully understand them. The difficult problem, which we still haven’t solved, is how to calculate V (α) and H(α) when someone hands us an arbitrary angle α. Strangely, we managed to figure out the derivatives of these machines with an argument whose basic logic is fairly simple, though we had to be a bit verbose, since the format of a book makes it difficult to show different snapshots of a picture throughout the course of an argument. But the verbosity is my own shortcoming. The argument itself is not complicated at all, as you’ll see if you can manage to re-create it for yourself. As always, it’s just zooming in, feeding a machine some food, making tiny changes in the food, and seeing what changed.

Recall that the standard textbooks use the names “sine” and “cosine” for what we’re calling V and H, so in their language, our discovery would be written:

We’ll see very soon how to solve the basic dilemma with which this section began: the problem of completely describing the machines V and H in a way that actually lets us use them. Only once we know — at least in principle — how to determine V (α) and H(α) for an arbitrary angle α will we finally understand the (supposedly simple, non-)subject of trigonometry.

4.9.1Burying Tangent

Having figured out the derivatives of V and H above, we can very briefly show how to avoid memorizing strange facts you might have heard in mathematics courses, such as the fact that the derivative of tangent is “1 plus tangent squared,” or the fact that the derivative of secant is “who knows I can’t even remember that nonsense.” Recall that tan(x) was the textbooks’ name for , or what we’re calling . And sec(x) was a completely unnecessary name for , or , or H−1. Let’s prevent ourselves from having to remember either of those facts above by building them using things we already built. Abbreviating “tangent” as , we’ll try to figure out . Using our hammer for multiplication from Chapter 3:

What’s that [H−1]′ piece? It’s easier to see by lying and correcting (essentially just using our reabbreviation hammer from Chapter 3). This gives

In a sense, we’ve just computed the derivative of “secant” (i.e., H−1) by accident. Now forget it forever. Tossing this back into where we got stuck in computing T(x), we get

So we invented the fact that the derivative of “tangent” is “1 plus tangent squared.” Now bury that fact somewhere deep underground, and tangent along with it. Hopefully we’ll never have to use it again. . .