## COVID-19

COVID-19 is officially a pandemic. It is up to all of us to do what we can to limit the spread of this disease. An earlier post shows how herd immunity can stop an outbreak of disease by making the effective reproductive rate, $$R$$, below 1. In that post, we looked at measles. In measles, the basic reproductive rate $$R_0$$ is around 15, but when over about 93% of the population is immune (either by vaccination or by having survived the disease) we get an $$R$$ below 1, meaning the likelihood of an outbreak is low. Please read that post to see how all that works.

Most estimates I’ve seen for COVID-19 estimate its $$R_0$$ at about 2 or 3 (I have been cautioned that there are wildly divergent estimates of this number) but since there is as yet no vaccine and few people have had the disease, without other interventions, $$R = R_0$$, and the disease continues to spread. Until there’s a vaccine, is there some other way to reduce $$R$$ below 1?

Recall from that earlier post that $$R_0$$ represents the average number of people that one infected person will infect. This number depends on many factors, including biological ones (the probability that an infected person infects someone in close contact, that I’ll call $$p_0$$) and also sociological ones (how many people the average person comes into close contact with, that I’ll call $$n_0$$). Assuming there are no other factors, $$R_0 = p_0 n_0$$.

Until there’s a vaccine or until there’s a significant fraction of the population that’s immune, there’s not much we can do about $$p_0$$. That means $$R = p_0 n$$, where $$n$$ is the average number of people we come into close contact with after we change our behavior. How much do we have to change our behavior? We want $$R < 1$$: $\begin{eqnarray*} R & < & 1 \\ R & < & \frac{R_0}{R_0} \\ p_0 n & < & \frac{p_0 n_0}{R_0} \\ n & < & \frac{n_0}{R_0} \end{eqnarray*}$

So here’s the good news: In theory, if each of us reduces the probability of coming into close contact with another person by over a factor of $$\frac{1}{R_0}$$, we can stop this epidemic. If we estimate $$R_0 = 3$$, that means we just need to come into close contact $$\frac{1}{3}$$ as much as we usually do. We don’t have to go into total isolation or do perfect hygiene. We just need to limit our interactions with other people and to use better hygiene.

In practice, of course, it’s much more complicated. A friend pointed me to this excellent and extremely detailed article, which recommends a “hammer” phase, in which $$R$$ must go much lower than 1, to get the exponential decay to happen much more quickly, and a later “dance” phase in which we can bring $$R$$ back closer to, but still below, one. The “hammer” phase, according to the author, will require much more stringent restrictions.

## That new quadratic equation solving technique

A New York Times article describes a new technique for solving quadratic equations which emerged from an article by Poh-Shen Loh. At first I was skeptical, but I decided to work through it enough to get a real feel for it. Though this technique does appear to be old information in a new guise, it does have some practical advantages over some of the traditional techniques (e.g. factoring, completing the square, the Quadratic Formula). I plan to use some of my students as guinea pigs…

By the way, kudos to reporters Kenneth Chang and Jonathan Corum for getting real math into the New York Times.

## Background

Loh’s technique emerges from a few well-known facts. First, any quadratic equation can be written as $a x^2 + bx + c = 0$ In this article, we will assume that the equation is monic (i.e. $a=1$) so the equation simplifies to $x^2 + bx + c = 0$ (If the equation is not monic, just divide both sides by $a$ and adjust the coefficients.)

Second, in that equation, $-b$ is the sum of the roots and $c$ is the product of the roots, since if the roots are $r$ and $s$, the factored form will be $(x-r)(x-s) = 0$ Multiply the left side out and you’ll see that $x^2 – (r+s)x + r s = 0$ so $-b=r+s$ and $c=r s$. Notice also that $- \frac{b}{2}$ is the average of the roots, which is also the $x$-coordinate of the line of symmetry.

## Loh’s method

The new method focuses on $u$, which is the distance between each root and the average of the roots. Thus the roots are $-\frac{b}{2} \pm u$.

It’s easy to find $u$ by realizing that $c$ is the product of the roots. What’s cool here is that the product of the roots is a factored form of a difference of squares so it multiplies cleanly, leaving no middle term:

$$\begin{eqnarray} c & = & \left( – \frac{b}{2} – u \right) \left( – \frac{b}{2} + u \right)\newline c & = & \frac{b^2}{4} – u^2\newline u^2 & = & \frac{b^2}{4} – c\newline u & = & \sqrt{\frac{b^2}{4} – c} \end{eqnarray}$$

Thus the roots are

$$\begin{eqnarray} x & = & – \frac{b}{2} \pm u\newline & = & – \frac{b}{2} \pm \sqrt{\frac{b^2}{4} – c} \end{eqnarray}$$

This looks suspiciously like the Quadratic Formula…

## Example

Let’s solve $x^2-4x-12=0$ using this method.

First, $-\frac{b}{2} = -\frac{-4}{2} = 2$ is the average of the roots.

Now $c=-12$ is the product of the roots so we solve for $u$:

$$\begin{eqnarray} -12 & = & ( 2 – u ) ( 2 + u)\newline -12 & = & 2^2 – u^2\newline u^2 & = & 2^2 + 12\newline u & = & \sqrt{2^2 + 12}= 4\newline \end{eqnarray}$$

So the roots are $2-4 = -2$ and $2+4=6$

## Relationship to the Quadratic Formula

The Quadratic Formula $x = \frac{- b \pm \sqrt{b^2 – 4 ac}}{2 a}$ gives the roots of a standard form quadratic equation of the form $a x^2+b x+c = 0$.

We are working with a monic quadratic (in which $a=1$) so that simplifies to

$$\begin{eqnarray} x & = & \frac{- b \pm \sqrt{b^2 – 4 c}}{2}\newline & = & \frac{- b}{2} \pm \frac{\sqrt{b^2 – 4 c}}{2}\newline & = & – \frac{b}{2} \pm \sqrt{\frac{b^2-4c}{4} }\newline & = & – \frac{b}{2} \pm \sqrt{\frac{b^2}{4} – c}\newline & = & – \frac{b}{2} \pm u \end{eqnarray}$$

You’ll notice that this is exactly what we get in the end of the “Loh’s Method” section.

## Is this new or useful?

It’s clear that Loh’s method is equivalent to the monic version of the Quadratic Formula. What I dislike about the Quadratic Formula, though, is that most students just memorize it without understanding, and even those of us who do understand its derivation still use it by rote. Loh’s method is simple enough to derive each time, summarized below, and the calculations seem a bit simpler:

1. If necessary, make the equation monic by dividing by $a$
2. Realize that $-\frac{b}{2}$ is the average of the roots
3. Let $u$ represent the distance between $-\frac{b}{2}$ and each root, so the roots are $-\frac{b}{2} \pm u$
4. Realize that $c$ is the product of the roots, i.e. $c = ( – \frac{b}{2} – u ) ( – \frac{b}{2} + u )$
5. Solve that equation for $u$, giving the roots as $-\frac{b}{2} \pm u$

Furthermore, like the Quadratic Formula and completing the square, Loh’s method can solve any monic quadratic equation. Unlike traditional factoring, it requires no guess-and-check (what multiplies to $c$ and adds to $-b$?) It works either in the real or complex fields. In the real case, of course, some equations cannot be solved, e.g. $x^2+1=0$.

## Vaccines and herd immunity

Many of us have heard of “herd immunity,” but few of us understand it. I didn’t either until I got curious, did a little reading, and figured some stuff out. The basics aren’t hard to understand if you understand exponential functions. With vaccination rates going down and disease rates going up, it’s important that we understand this stuff so we can make rational choices. Remember asking, “When are we going to use this stuff?” Now’s the time. Let’s go.

For our example, let’s look at measles.

1. A typical person with measles will infect about 15 other people while infected, assuming none of those people are immune. That number, about 15 for measles, is called the basic reproductive rate, or R0, of the disease. (R0 depends on how easily the disease is transmitted and how many people a typical infected person comes into contact with.) Most of those infected will be infected within about two weeks. To keep things simple, let’s assume that each infected person infects exactly 15 people in exactly two weeks. Let’s also assume that after infecting 15 new people, an infected person ceases to be infected. Of course, each of those 15 newly infected people will infect 15 other people in the next two weeks, and so on. Assuming nobody is immune, after one person is infected, how many people will be infected after 4 weeks (2 two-week periods)? 8 weeks (four two-week periods)?
2. Define a function f(t) which gives the number of people infected after t weeks, assuming nobody is immune.
3. Still assuming nobody is immune, how many weeks would it take to infect the entire population of the United States, around 350,000,000? (Hint: if you know about logarithms, use ’em! Otherwise, try graphing software or a graphing calculator. Or use the brute force approach: keep multiplying by 15, representing two more weeks, until you top 350,000,000.) (Of course, the more people who are infected, the more unrealistic our assumption that nobody is immune. Still, those are sobering numbers!)
4. The average number of people actually infected by each person with the disease, called the effective reproductive rate, or R, is less than R0 if a fraction of the population is immune (either vaccinated or already had the disease). Let H be the fraction of the population that is immune (so if, say, 2/3 of the population were immune, H would be 2/3). Based on R0 and H, what is a formula for R? (Hint: what fraction is not immune?)
5. No longer assuming everybody is immune, now based on R, define a function g(t) which gives the number of people infected after t weeks.
6. What values of R make g(t) an increasing function? A decreasing function? What does that question have to do with the spread of the disease?
7. At least what fraction, which we’ll call H, of the population must be immune for g(t) to be a decreasing function?
8. That fraction is called the herd immunity threshold for the disease, or HIT. Do you see how the herd immunity threshold is important for public health?

1. Four weeks: R04/2 = 152 = 225. Eight weeks: R08/2 = 154 = 50625.
2. f(t) = R0t/2 = 15t/2
3. Solve for $$t$$: $\begin{eqnarray*} f (t) & = & 350, 000, 000\\ R_0^{\frac{t}{2}} & = & 350, 000, 000\\ \frac{t}{2} & = & \log_{R_0} (350, 000, 000)\\ t & = & 2 \log_{R_0} (350, 000, 000)\\ & \approx & 14.5 \end{eqnarray*}$
That’s 14.5 weeks to infect all of the US! Sobering.
4. R = R0(1-H) (Note that 1-H is the fraction of the population that is not immune.)
5. g(t) = Rt/2 = (R0(1-H))t/2. Notice that this is the same function as f, but using the effective reproductive rate R instead of the basic reproductive rate R0.
6. Increasing: R > 1. Decreasing: R < 1.
7. We need to solve R0(1-H) < 1. So
$\begin{eqnarray*} R_0 (1 – H) & < & 1\\ 1 - H & < & \frac{1}{R_0}\\ - H & < & \frac{1}{R_0} - 1\\ H & > & 1 – \frac{1}{R_0}\\ H & > & \frac{14}{15} \\ H & > & 93 \% \end{eqnarray*}$
The lower bound for the value H, in this case 14/15 or about 93%, is the herd immunity threshold (HIT) for measles.
8. When more than the HIT, about 93%, are immune, R < 1 so each infected person infects, on average, fewer than one person, so disease outbreaks are unlikely. When fewer than 93% are immune, R > 1 so each infected person infects, on average, more than one person, so the number of infected people grows over time. And increasing exponential functions are relentless.

## Building a stone wall video: How does it work?

Here’s an amazing video in which a long row of stone blocks is arranged on-edge on top of a stone wall. Spoiler alert! I recommend watching the video before reading further!

Here’s what happens: Somebody pushes the leftmost block over, which knocks down the one to its right, which knocks down the one to its right, and so on in a chain reaction. This is more or less what you’d expect, just like a row of dominoes.

We notice that once the blocks fall, they don’t fall all the way down; the top of a block ends up resting on the bottom of the next block, overlapping a bit. That’s also not too surprising.

What is surprising is what happens once all the blocks fall. Almost instantly comes the amazing second chain reaction. Starting from the right, all the blocks that were resting on each other fall down flat, with virtually no space between them, making a perfect row of blocks laid end-to-end.

When students normally are exposed to pseudo-real-world problems, often called “word problems,” they are normally given exactly the information needed to solve it. But in the actual real world, we encounter actual problems and are given no hints about what facts are relevant. When I first saw this video I just wondered what was going on. I was flummoxed at first but gradually realized what’s happening. This is much closer to what really happens when we try to use mathematics to understand the world. But I’m not going to tell, at least not yet. You could do as I did, and just draw some pictures, make some assumptions, figure out what’s relevant, and figure it out. But to make it a little more, er, concrete, here are some starting points. Imagine the blocks are 12″ tall and 3″ wide when standing on edge (or 3″ high and 12″ long when lying flat). Here are some questions to think about:

1. How far apart are the blocks initially? (Say, measured from the left side of one to the left side of the next.)
2. Why don’t they fall to horizontal during the first chain reaction?
3. Why do they all fall to horizontal after the rightmost one falls and starts the second chain reaction?

And, for extra credit, a few more that require trigonometry:

1. What is the angle of a falling block as it hits the one to its right?
2. What is the eventual resting angle of the blocks before the final chain reaction flattens them all?

## That ancient Babylonian “trigonometry” tablet isn’t really trigonometry

The new math craze on the Internet seems to be that ancient “Plimpton 322” (P322) Babylonian tablet in which trigonometry is invented millennia before it was first thought to have been. The authors of the paper themselves, University of New South Wales mathematicians Daniel Mansfield and Norman Wildberger, are responsible for much of the hyperbole, when they say things like, “It’s a trigonometric table that’s so unfamiliar and advanced that in some respects it’s superior even to modern trigonometry.” Or this gem: “It’s actually trigonometry, but a different kind of trigonometry…this is a ratio-based trigonometry.” They even call it “Babylonian Exact Sexagesimal Trigonometry.”

None of this is really true.

Trigonometry at its most basic level is about the relationships between angles in a triangle and the ratios of the side lengths. They tout the newness of “ratio-based trigonometry” though trig has always been ratio-based. For example, in a 30-60-90 right triangle, the ratio of the leg opposite the 30° angle to the hypotenuse is 1/2. (To see why, start with an equilateral triangle and cut it in half. Notice that each half is a 30-60-90 triangle. Now think some more!) In modern trigonometry, that 1/2 ratio is called the sine of the 30° angle or the cosine of the 60° angle (the complement of the 30° angle). How do we determine this using P322 “trigonometry”? We don’t, because P322 contains no idea of angle. Triangle ratios is trigonometry like Dick Smothers is the Smothers Brothers.

What P322 actually contains is a series of “Pythagorean triples,” along with a squared ratio of either the short leg to the long leg or of the hypotenuse to the long leg (we’re not sure since the left side has broken off and the only difference would be whether they start with a one or a zero). These fractions are written as sexagesimals, which are like decimals but in base 60 rather than base 10. A Pythagorean triple is a sequence of integers like 3,4,5, in which the squares of the first two add up to the square of the third, making them legs and hypotenuse of a right triangle. In this case, 3²+4²=5². Assuming the left column contains squared ratios of the short to long legs, we would get (3/4)² or 9/16. We could leave it at that, but let’s convert it to decimal: .5625. That’s the exact squared ratio as a decimal. So far so good. The next Pythagorean triple is 5,12,13. If we compute the same squared ratio, we get (5/12)², or 25/144. That’s still exact, but if we try to convert it to decimal we get a repeating decimal, .1736111…, which we can’t write exactly if we limit ourselves to a finite number of digits. That’s where the sexagesimal (base 60) numeral system of the Babylonians wins out, because many more fractions are terminating sexagesimals than are terminating decimals. (I call it a numeral system rather than the more natural number system to emphasize that these aren’t new numbers, they are just new, to most of us, ways of writing numbers, i.e. numerals.) So many more fractions are terminating sexagesimals than are terminating decimals. That’s cool, but far from Earth-shattering. Maybe that was very useful to the Babylonians, but it’s not trigonometry. And it certainly won’t help us at all today.

Depending on which ratios were in the original table, these would either be the squared tangents or squared cosecants of the angle opposite the short side, which sounds a lot like trigonometry except that P322 is completely silent about angles! So all we really have is a table containing Pythagorean triples and one of their (squared) ratios.

What of the claim that the Babylonians knew the Pythagorean Theorem  a millennium before Pythagoras? That’s not news either. For example the Encyclopedia Brittanica already knew this. The origin (or origins) of the theorem and its proof are lost to history, and it may have been the Babylonians, but being aware of it, as the authors of P322 surely were, is not the same as having proved it.

What was P322 good for? If you know two sides of a right triangle, including the long leg, you can find the squared ratio of the side you know to the long leg, and find the closest match in the table. I’m not even sure what that’s good for (since you already know the other ratio since the squared ratio of the hypotenuse to the long leg is just the squared ratio of the short leg to the long leg plus one). There are some theories, but it’s not trigonometry.

Is the “exact” nature of it new? No, just look up any list of Pythagorean triples and compute any ratios you want. Or you can just use the Pythagorean Theorem directly, e.g. if you know the two legs of a right triangle are 6 and 8, just compute the square root of 6²+8² and get 10 for the hypotenuse. Nothing to see here, move along!

As final “proof” that this is not trigonometry, I give you an example of a typical problem given to beginning trig students: If a 10′ ladder (teachers of trig are obsessed with ladders) is placed at a 70° angle to the ground against a wall, how high is the top of the ladder from the ground? P322 is completely useless here, but using modern trigonometry, the sine of 70°, about .94, is the ratio of the side opposite that angle to the hypotenuse. Multiply that by the hypotenuse of 10′ and you get the length of the opposite side, which in this case is the height, about 9.4′. Of course, if you also knew the distance from the bottom of the ladder to the wall along the ground, you could also figure this out with P322 or the Pythagorean Theorem, but calling it trigonometry is still a big stretch. Incidentally, I’m not alone. Evelyn Lamb in Scientific American is also critical of the hype, with much more detail, though she doesn’t seem as concerned as I am about the complete absence of angles in this “trigonometry.”