My 2019 Mathematics A To Z: Wallis Products

Today’s A To Z term was suggested by Dina Yagodich, whose YouTube channel features many topics, including calculus and differential equations, statistics, discrete math, and Matlab. Matlab is especially valuable to know as a good quick calculation can answer many questions.

Wallis Products.

The Wallis named here is John Wallis, an English clergyman and mathematician and cryptographer. His most tweetable work is how we follow his lead in using the symbol ∞ to represent infinity. But he did much in calculus. And it’s a piece of that which brings us to today. He particularly noticed this:

$\frac{1}{2}\pi = \frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdot \frac{8}{7}\cdot \frac{8}{9}\cdot \frac{10}{9}\cdot \frac{10}{11}\cdots$

This is an infinite product. It’s multiplication’s answer to the infinite series. It always amazes me when an infinite product works. There are dangers when you do anything with an infinite number of terms. Even the basics of arithmetic, like that you can change the order in which you calculate but still get the same result, break down. Series, in which you add together infinitely many things, are risky, but I’m comfortable with the rules to know when the sum can be trusted. Infinite products seem more mysterious. Then you learn an infinite product converges if and only if the series made from the logarithms of the terms in it also converges. Then infinite products seem less exciting.

There are many infinite products that give us π. Some work quite efficiently, giving us lots of digits for a few terms’ work. Wallis’s formula does not. We need about a thousand terms for it to get us a π of about 3.141. This is a bit much to calculate even today. In 1656, when he published it in Arithmetica Infinitorum, a book I have never read? Wallis was able to do mental arithmetic well. His biography at St Andrews says once when having trouble sleeping he calculated the square root of a 53-digit number in his head, and in the morning, remembered it, and was right. Still, this would be a lot of work. How could Wallis possibly do it? And what work could possibly convince anyone else that he was right?

As it common to striking discoveries it was a mixture of insight and luck and persistence and pattern recognition. He seems to have started with pondering the value of

$\int_0^1 \left(1 - x^2\right)^{\frac{1}{2}} dx$

Happily, he knew exactly what this was: $\frac{1}{4}\pi$. He knew this because of a bit of insight. We can interpret the integral here as asking for the area that’s enclosed, on a Cartesian coordinate system, by the positive x-axis, the positive y-axis, and the set of points which makes true the equation $y = \left(1 - x^2\right)^\frac{1}{2}$. This curve is the upper half of a circle with radius 1 and centered on the origin. The area enclosed by all this is one-fourth the area of a circle of radius 1. So that’s how he could know the value of the integral, without doing any symbol manipulation.

The question, in modern notation, would be whether he could do that integral. And, for this? He couldn’t. But, unable to do the problem he wanted, he tried doing the most similar problem he could and see what that proved. $\left(1 - x^2\right)^{\frac{1}{2}}$ was beyond his power to integrate; but what if he swapped those exponents? Worked on $\left(1 - x^{\frac{1}{2}}\right)^2$instead? This would not — could not — give him what he was interested in. But it would give him something he could calculate. So can we:

$\int_0^1 \left(1 - x^{\frac{1}{2}}\right)^2 dx = \int_0^1 1 - 2x^{\frac{1}{2}} + x dx = 1 - 2\cdot\frac{2}{3} + \frac{1}{2} = \frac{1}{6}$

And now here comes persistence. What if it’s not $x^{\frac{1}{2}}$ inside the parentheses there? If it’s x raised to some other unit fraction instead? What if the parentheses aren’t raised to the second power, but to some other whole number? Might that reveal something useful? Each of these integrals is calculable, and he calculated them. He worked out a table for many values of

$\int_0^1 \left(1 - x^{\frac{1}{p}}\right)^q dx$

for different sets of whole numbers p and q. He trusted that if he kept this up, he’d find some interesting pattern. And he does. The integral, for example, always turns out to be a unit fraction. And there’s a deeper pattern. Let me share results for different values of p and q; the integral is the reciprocal of the number inside the table. The topmost row is values of q; the leftmost column is values of p.

0 1 2 3 4 5 6 7
0 1 1 1 1 1 1 1 1
1 1 2 3 4 5 6 7 7
2 1 3 6 10 15 21 28 36
3 1 4 10 20 35 56 84 120
4 1 5 15 35 70 126 210 330
5 1 6 21 56 126 252 462 792
6 1 7 28 84 210 462 924 1716
7 1 8 36 120 330 792 1716 3432

There is a deep pattern here, although I’m not sure Wallis noticed that one. Look along the diagonals, running from lower-left to upper-right. These are the coefficients of the binomial expansion. Yang Hui’s triangle, if you prefer. Pascal’s triangle, if you prefer that. Let me call the term in row p, column q of this table $a_{p, q}$. Then

$a_{p, q} = \frac{(p + 1)!}{p! q!}$

Great material, anyway. The trouble is that it doesn’t help Wallis with the original problem, which — in this notation — would have $p = \frac12$ and $q = \frac12$. What he really wanted was the Binomial Theorem, but western mathematicians didn’t know it yet. Here a bit of luck comes in. He had noticed there’s a relationship between terms in one column and terms in another, particularly, that

$a_{p, q} = \frac{p + q}{q} a_{p, q - 1}$

So why shouldn’t that hold if p and q aren’t whole numbers? … We would today say why should they hold? But Wallis was working with a different idea of mathematical rigor. He made assumptions that it turned out in this case were correct. Of course, had he been wrong, we wouldn’t have heard of any of this and I would have an essay on some other topic.

With luck in Wallis’s favor we can go back to making a table. What would the row for $p = \frac12$ look like? We’ll need both whole and half-integers. $p = \frac12, q = 1$ is easy; its reciprocal is 1. $p = \frac12, q = \frac12$ is also easy; that’s the insight Wallis had to start with. Its reciprocal is $\frac{4}{\pi}$. What about the rest? Use the equation just up above, relating $a_{p, q}$ to $a_{p, q - 1}$; then we can start to fill in:

0 1/2 1 3/2 2 5/2 3 7/2
1/2 1 $\frac{4}{\pi}$ $\frac{3}{2}$ $\frac{4}{3}\frac{4}{\pi}$ $\frac{3\cdot 5}{2\cdot 4}$ $\frac{2\cdot 4}{5}\frac{4}{\pi}$ $\frac{3\cdot 5\cdot 7}{2\cdot 4\cdot 6}$ $\frac{2\cdot 2\cdot 4\cdot 4}{5\cdot 7}\frac{4}{\pi}$

Anything we can learn from this? … Well, sure. For one, as we go left to right, all these entries are increasing. So, like, the second column is less than the third which is less than the fourth. Here’s a triple inequality for you:

$\frac{4}{\pi} < \frac{3}{2} < \frac{4}{3}\frac{4}{\pi}$

Multiply all that through by, on, $\frac{2}{\pi}$. And then divide it all through by $\frac{3}{2}$. What have we got?

$\frac{2\cdot 2}{3} < \frac{\pi}{2} < \frac{2\cdot 2}{3}\cdot \frac{2\cdot 2}{3}$

I did some rearranging of terms, but, that’s the pattern. One-half π has to be between $\frac{2\cdot 2}{3}$ and four-thirds that.

Move over a little. Start from the row where $q = \frac32$. This starts us out with

$\frac{4}{3}\frac{4}{\pi} < \frac{3}{2} < \frac{2\cdot 4}{5}\frac{4}{\pi}$

Multiply everything by $\frac{\pi}{4}$, and divide everything by $\frac{3}{2}$ and follow with some symbol manipulation. And here’s a tip which would have saved me some frustration working out my notes: $\frac{\pi}{4} = \frac{\pi}{2}\cdot\frac{3}{6}$. Also, 6 equals 2 times 3. Later on, you may want to remember that 8 equals 2 times 4. All this gets us eventually to

$\frac{2\cdot 2\cdot 4\cdot 4}{3\cdot 3\cdot 5} < \frac{\pi}{2} < \frac{2\cdot 2\cdot 4\cdot 4}{3\cdot 3\cdot 5}\cdot \frac{6}{5}$

Move over to the next terms, starting from $q = \frac52$. This will get us eventually to

$\frac{2\cdot 2\cdot 4\cdot 4 \cdot 6 \cdot 6}{3\cdot 3\cdot 5\cdot 5\cdot 7} < \frac{\pi}{2} < \frac{2\cdot 2\cdot 4\cdot 4 \cdot 6 \cdot 6}{3\cdot 3\cdot 5\cdot 5\cdot 7}\cdot \frac{8}{7}$

You see the pattern here. Whatever the value of $\frac{\pi}{2}$, it’s squeezed between some number, on the left side of this triple inequality, and that same number times … uh … something like $\frac{10}{9}$ or $\frac{12}{11}$ or $\frac{14}{13}$ or $\frac{1,000,000,000,002}{1,000,000,000,001}$. That last one is a number very close to 1. So the conclusion is that $\frac{\pi}{2}$ has to equal whatever that pattern is making for the number on the left there.

We can make this more rigorous. Like, we don’t have to just talk about squeezing the number we want between two nearly-equal values. We can rely on the use of the … Squeeze Theorem … to prove this is okay. And there’s much we have to straighten out. Particularly, we really don’t want to write out expressions like

$\frac{2\cdot 2 \cdot 4\cdot 4\cdot 6\cdot 6\cdot 8\cdot 8 \cdot 10\cdot 10 \cdots}{3\cdot 3\cdot 5\cdot 5 \cdot 7\cdot 7 \cdot 9\cdot 9 \cdot 11\cdot 11 \cdots}$

Put that way, it looks like, well, we can divide each 3 in the denominator into a 6 in the numerator to get a 2, each 5 in the denominator to a 10 in the numerator to get a 2, and so on. We get a product that’s infinitely large, instead of anything to do with π. This is that problem where arithmetic on infinitely long strings of things becomes dangerous. To be rigorous, we need to write this product as the limit of a sequence, with finite numerator and denominator, and be careful about how to compose the numerators and denominators.

But this is all right. Wallis found a lovely result and in a way that’s common to much work in mathematics. It used a combination of insight and persistence, with pattern recognition and luck making a great difference. Often when we first find something the proof of it is rough, and we need considerable work to make it rigorous. The path that got Wallis to these products is one we still walk.

There’s just three more essays to go this year! I hope to have the letter X published here, Thursday. All the other A-to-Z essays for this year are also at that link. And past A-to-Z essays are at this link. Thanks for reading.

Something Cute I Never Noticed Before About Infinite Sums

This is a trifle, for which I apologize. I’ve been sick. But I ran across this while reading Carl B Boyer’s The History of the Calculus and its Conceptual Development. This is from the chapter “A Century Of Anticipation”, developments leading up to Newton and Leibniz and The Calculus As We Know It. In particular, while working out the indefinite integrals for simple powers — x raised to a whole number — John Wallis, whom you’ll remember from such things as the first use of the ∞ symbol and beating up Thomas Hobbes for his lunch money, noted this:

$\frac{0 + 1}{1 + 1} = \frac{1}{2}$

Which is fine enough. But then Wallis also noted that

$\frac{0 + 1 + 2}{2 + 2 + 2} = \frac{1}{2}$

And furthermore that

$\frac{0 + 1 + 2 + 3}{3 + 3 + 3 + 3} = \frac{1}{2}$

$\frac{0 + 1 + 2 + 3 + 4}{4 + 4 + 4 + 4 + 4} = \frac{1}{2}$

$\frac{0 + 1 + 2 + 3 + 4 + 5}{5 + 5 + 5 + 5 + 5 + 5} = \frac{1}{2}$

And isn’t that neat? Wallis goes on to conclude that this is true not just for finitely many terms in the numerator and denominator, but also if you carry on infinitely far. This seems like a dangerous leap to make, but they treated infinities and infinitesimals dangerously in those days.

What makes this work is — well, it’s just true; explaining how that can be is kind of like explaining how it is circles have a center point. All right. But we can prove that this has to be true at least for finite terms. A sum like 0 + 1 + 2 + 3 is an arithmetic progression. It’s the sum of a finite number of terms, each of them an equal difference from the one before or the one after (or both).

Its sum will be equal to the number of terms times the arithmetic mean of the first and last. That is, it’ll be the number of terms times the sum of the first and the last terms and divided that by two. So that takes care of the numerator. If we have the sum 0 + 1 + 2 + 3 + up to whatever number you like which we’ll call ‘N’, we know its value has to be (N + 1) times N divided by 2. That takes care of the numerator.

The denominator, well, that’s (N + 1) cases of the number N being added together. Its value has to be (N + 1) times N. So the fraction is (N + 1) times N divided by 2, itself divided by (N + 1) times N. That’s got to be one-half except when N is zero. And if N were zero, well, that fraction would be 0 over 0 and we know what kind of trouble that is.

It’s a tiny bit, although you can use it to make an argument about what to expect from $\int{x^n dx}$, as Wallis did. And it delighted me to see and to understand why it should be so.

A bit more about Thomas Hobbes

You might remember a post from last April, Thomas Hobbes and the Doing of Important Mathematics, timed to the renowned philosopher’s birthday. I talked about him because a good bit of his intellectual life was spent trying to achieve mathematical greatness, which he never did.

Recently I’ve had the chance to read Douglas M Jesseph’s Squaring The Circle: The War Between Hobbes And Wallis, about Hobbes’s attempts to re-build mathematics on an intellectual foundation he found more satisfying, and the conflict this put him in with mainstream mathematicians, particularly John Wallis (algebra and calculus pioneer, and popularizer of the ∞ symbol). The situation of Hobbes’s mathematical ambitions is more complicated than I realized, although the one thing history teaches us is that the situation is always more complicated than we realized, and I wanted to at least make my writings about Hobbes a bit less incomplete. Jesseph’s book can’t be fairly reduced to a blog post, of course, and I’d recommend it to people who want to really understand what the fuss was all about. It’s a very good idea to have some background in philosophy and in 17th century English history going in, though, because it turns out a lot of the struggle — and particularly the bitterness with which Hobbes and Wallis fought, for decades — ties into the religious and political struggles of England of the 1600s.

Hobbes’s project, I better understand now, was not merely the squaring of the circle or the solving of other ancient geometric problems like the doubling of the cube or the trisecting of an arbitrary angle, although he did claim to have various proofs or approximate proofs of them. He seems to have been interested in building a geometry on more materialist grounds, more directly as models of the real world, instead of the pure abstractions that held sway then (and, for that matter, now). This is not by itself a ridiculous thing to do: we are almost always better off for having multiple independent ways to construct something, because the differences in those ways teaches us not just about the thing, but about the methods we use to discover things. And purely abstract constructions have problems also: for example, if a line can be decomposed into nothing but an enormous number of points, and absolutely none of those points has any length, then how can the line have length? You can answer that, but it’s going to require a pretty long running start.

Trying to re-build the logical foundations of mathematics is an enormously difficult thing to do, and it’s not surprising that someone might fail to do so perfectly. Whole schools of mathematicians might be needed just to achieve mixed success. And Hobbes wasn’t able to attract whole schools of mathematicians, in good part because of who he was.

Hobbes achieved immortality as an important philosopher with the publication of Leviathan. What I had not appreciated and Jesseph made clear was that in the context of England of the 1650s, Hobbes’s views on the natures of God, King, Society, Law, and Authority managed to offend — in the “I do not know how I can continue to speak with a person who holds views like that” — pretty much everybody in England who had any strong opinion about anything in politics, philosophy, or religion. I do not know for a fact that Hobbes then went around kicking the pet dogs of any English folk who didn’t have strong opinions about politics, philosophy, or religion, but I can’t rule it out. At least part of the relentlessness and bitterness with which Wallis (and his supporters) attacked Hobbes, and with which Hobbes (and his supporters) attacked back, can be viewed as a spinoff of the great struggle between the Crown and Parliament that produced the Civil War, the Commonwealth, and the Restoration, and in that context it’s easier to understand why all parties carried on, often quibbling about extremely minor points, well past the point that their friends were advising them that the quibbling was making themselves look bad. Hobbes was a difficult person to side with, even when he was right, and a lot of his mathematics just wasn’t right. Some of it I’m not sure ever could be made right, however many ingenious people you had working to avoid flaws.

An amusing little point that Jesseph quotes is a bit in which Hobbes, making an argument about the rights that authority has, asserts that if the King decreed that Euclid’s Fifth Postulate should be taught as false, then false it would be in the kingdom. The Fifth Postulate, also known as the Parallel Postulate, is one of the axioms on which classical Greek geometry was built and it was always the piece that people didn’t like. The other postulates are all nice, simple, uncontroversial, common-sense things like “all right angles are equal”, the kinds of things so obvious they just have to be axioms. The Fifth Postulate is this complicated-sounding thing about how, if a line is crossed by two non-parallel lines, you can determine on which side of the first line the non-parallel lines will meet.

It wouldn’t be really understood or accepted for another two centuries, but, you can suppose the Fifth Postulate to be false. This gives you things named “non-Euclidean geometries”, and the modern understanding of the universe’s geometry is non-Euclidean. In picking out an example of something a King might decree and the people would have to follow regardless of what was really true, Hobbes picked out an example of something that could be decreed false, and that people could follow profitably.

That’s not mere ironical luck, probably. A streak of mathematicians spent a long time trying to prove the Fifth Postulate was unnecessary, at least, by showing it followed from the remaining and non-controversial postulates, or at least that it could be replaced with something that felt more axiomatic. Of course, in principle you can use any set of axioms you like to work, but some sets produce more interesting results than others. I don’t know of any interesting geometry which results from supposing “not all right angles are equal”; supposing that the Fifth Postulate is untrue gives us general relativity, which is quite nice to have.

Again I have to warn that Jesseph’s book is not always easy reading. I had to struggle particularly over some of the philosophical points being made, because I’ve got only a lay understanding of the history of philosophy, and I was able to call on my love (a professional philosopher) for help at points. I imagine someone well-versed in philosophy but inexperienced with mathematics would have a similar problem (although — don’t let the secret out — you’re allowed to just skim over the diagrams and proofs and go on to the explanatory text afterwards). But for people who want to understand the scope and meaning of the fighting better, or who just want to read long excerpts of the wonderful academic insulting that was current in the era, I do recommend it. Check your local college or university library.