Irrationality of pi

1. Irrationality of pi
2. Irrationality of pi: the unpossible function
3. Irrationality of pi: derivatives of f
4. Irrationality of pi: curiouser and curiouser
5. Irrationality of pi: the impossible integral

We’re getting close! Last time, we defined a new function and showed that and are both integers, and that . So, consider the following:

The first step uses the product rule for differentiation (recalling that and ); the last step is what we showed last time. Now we see the point of defining : it’s just so that we have a convenient way to talk about the antiderivative of . We could just do everything directly in terms of alternating sums of derivatives of … but it’s much clearer this way, don’t you agree?

Now that we know the antiderivative of , we can use the Fundamental Theorem of Calculus to compute the following integral:

Note that the value of this integral is an integer, since both and are integers. But next time we’ll show that it is also strictly between 0 and 1 (for a suitable choice of ), which is clearly nonsense!

Hat tip to Phil Wadler.

1. The players pick a positive integer.
2. The two players work together to write down all the divisors of the chosen integer (being sure to include 1 and the integer itself).
3. The players now alternate moves as follows: on a player’s turn, she must choose one of the divisors , and then cross out that divisor as well as all of the other listed numbers which are divisible by .
4. On subsequent turns, players may only choose numbers which are not yet crossed out.
5. Whoever is forced to choose 1 (because it is the only number left) is the loser!

For example, suppose the chosen number is 12. We write down the divisors of 12:

1, 2, 3, 4, 6, 12.

Now suppose the first player chooses 4 (actually, this is a bad move; I’ll let you figure out why); they then cross out 4 and 12, since 12 is divisible by 4. The game now looks like

1, 2, 3,  4 , 6,  12 .

Now it’s the other player’s turn; suppose they pick 3 (this is actually a bad move too…!), so they cross out 3 and 6. Now the game looks like

1, 2,  3 ,  4 ,  6 ,  12 .

The first player now crosses out 2, and the second player is forced to choose 1, so the first player wins.

The kids thought this was a lot of fun and it leads to all kinds of interesting discussions. First, of course, you have to figure out how to write down all the divisors of the starting number (how do you know when you’ve listed them all? what are some systematic strategies for listing the divisors?). Then you can talk about strategies for playing the game. I might talk about some of these things in some future posts. For now I will just note that this actually has some deep connections to the theory of posets (we are basically just using each integer as an abbreviation for its poset of divisors). Although I’ve played around with it a bit I don’t yet know of a general strategy — although any particular starting integer necessarily gives a winning strategy for ONE of the two players, and it’s not too hard to figure it out by working backwards. More on this later, I suppose.

In the meantime, have fun playing!

Irrationality of pi

1. Irrationality of pi
2. Irrationality of pi: the unpossible function
3. Irrationality of pi: derivatives of f
4. Irrationality of pi: curiouser and curiouser
5. Irrationality of pi: the impossible integral

I’ve been remiss in posting here lately, which I will attribute to Christmas and New Year travelling and general craziness, and then starting a new semester craziness… but things have settled down a bit, so here we go again!

Since it’s been a while since my last post in this series, here’s a quick recap: I’m presenting a proof by Ivan Niven that is irrational, that is, that it cannot be represented as the ratio of two integers (and hence its decimal expansion goes on forever without repeating). My first post just gave some background and an outline of the general argument. In my second post, we began by assuming that is rational, and defined the function

(really, a family of functions, one for each value of ) where and are the “numerator” and “denominator” of . We then showed that , and in fact that is symmetric, with . In my third post, we showed that all the derivatives of take on integer values when evaluated at both 0 and . We’re about halfway there! Today we’ll continue by defining a new function in terms of , and show some of its properties. Recall too our overall plan: we’re going to wind up with an integral which is strictly greater than 0, strictly less than 1, and also an integer! Since this is clearly nonsense (there are no integers between 0 and 1) we will conclude that our initial assumption—that is rational—was bogus, and that must be irrational after all.

So without further ado, here’s our new function . Actually, this too is technically a family of functions , one for each ; but again, everything we prove about it will be true no matter what is.

In words, is the alternating sum of all the even derivatives of . (I say “all” because, as noted in my last post, any derivative of higher than is zero.) Using Sigma notation, we can also write this more concisely as

There are a few things to note. First, think what happens when we evaluate : since all the derivatives of take on integer values at 0, and is just a sum of a bunch of derivatives of , must be an integer too. Of course, the same thing goes for .

Next, consider

Since the derivative of a sum is the sum of the derivatives, we can compute as

That is, turns into , turns into , and so on. “But wait a minute,” you say. “Shouldn’t the at the end of turn into in ?” In fact, it does—but as noted before, is zero, so that term just goes away. Now we note that every term of has a corresponding term in of the opposite sign, except , which has no corresponding term. So when we add and , everything cancels except :

Astute readers will note a funny resemblance between the definition of and the Taylor series for … and indeed, next time we’ll start making some connections with our old trigonometric friends, and .

You’ve heard the story of Rumpled Stiltsken, right? You know, the one where the topologist’s daughter is locked in the grad student lounge and forced to turn coffee into theorems by morning? …what’s that, you say you haven’t heard that one? Funny, I thought everyone knew that story. Well, it’s a fascinating and sobering tale full of insight into life and the nature of… oops, wait, those were my notes for The Kite Runner. Rumped Stiltsken is… well, just read it, OK?

Colin Adams entertains us with this and many other humorously mathematical (mathematically humorous?) stories in his new book, Riot at the Calc Exam and Other Mathematically Bent Stories. Tips on how to avoid RERI (Repetitive Eye Roll Injury), advice from a mathematical ethicist, stories about everyone’s favorite Principal Investigator, Dirk Mangum, P. I., a transcript from the hit radio show Math Talk with Plug and Chug… the list goes on. Some are funnier than others, of course (by the end, the conceit of anthromorphizing/metaphorizing mathematical theorems and the process or proving them had gotten particularly old), but on balance my Funny-o-Meter was definitely pointing somwhere between “amusing” and “hilarious”. This book would make a great gift for that special person in your life who likes to read funny stories about math while they are in the bathroom, or for anyone who likes reading funny stories about math in general, or anyone who likes funny stories, or who likes math. This book would not make a good gift for grumpy people who hate math. Don’t say I didn’t warn you.

Full disclosure: the AMS kindly sent me a free review copy of this book. Also, Colin Adams was actually one of my professors in college, which you might think would make me somewhat biased, which is probably true, but it also means that I happen to know that he really is quite funny, and also that he is the Fastest Draw(er of 3D surfaces with colored chalk) in the West(ern Massachusetts). Also also, this morning for breakfast I ate a bowl of shredded wheat cereal.

Nevertheless, I was intrigued, especially since my correspondent claimed that this website was endorsed by the eminent mathematician and educator Marcus du Sautoy. So I visited the site and tried playing a few games… and was pleasantly surprised! The games are fairly high-quality and humorous (I actually spent twenty minutes or so playing the first game I tried, even though it was rather easy for me), and the site promises to track points and accomplishments for students who register (a definite requirement if you want to get students hooked on the games).

On the flip side, the commercial status of the site isn’t completely clear—you can play all the games for free but it claims this is “for a limited time”, so I’m not sure what happens after the limited time is up. The site also appears to have very little to do with Manga, so the title is a bit odd. But these are minor considerations at the moment.

I’m still not sold on the idea of interactive games for teaching math—but if you’re looking for such things, MangaHigh.com seems like one of the best sites currently out there.

Irrationality of pi

1. Irrationality of pi
2. Irrationality of pi: the unpossible function
3. Irrationality of pi: derivatives of f
4. Irrationality of pi: curiouser and curiouser
5. Irrationality of pi: the impossible integral

In my previous post in this series, we defined the function

and showed that . Today we’ll show the surprising fact that, for every positive integer , although and are not necessarily zero, they are always integers. (The notation means the th derivative of ; that is, take the derivative of , then the derivative of that, then the derivative of that, … times.) Put more succinctly: every derivative of takes on integer values at and .

Why might this be surprising? It’s surprising because of the in the denominator of . For example, consider the function (which I just made up):

It’s easy to see that . But let’s take the derivative: so , which is clearly not an integer. For the derivatives of to always give an integer at (let alone at ) there must be some fancy canceling going on!

For now we will consider only (we’ll come back to later). Of course, substituting for causes every term containing to disappear, so is just the constant term of . Hence, we must show that the constant term of is always an integer.

Consider the numerator of , that is,

Note that , when expanded out, is a polynomial of the form , where the ellipsis contains a bunch of terms with integer coefficients and powers of between 1 and . (In fact, we could use the Binomial Theorem to compute the precise coefficients—but it really doesn’t matter; all we will care about is that they are integers.) Multiplying by , we see that

so is a polynomial with terms of degree through , and hence so is , since dividing by changes the coefficients but not the exponents. (Note that has no constant term, so —but we already knew that.)

Recall that the derivative of is , so taking the derivative of a polynomial reduces each of the exponents by one. So the first derivative of is a polynomial with terms of degree through (and hence a constant term of zero); the second derivative has terms of degree through (still no constant term); and so on. We can see that none of the first derivatives of will have a constant term, so (which is certainly an integer) for . What about the th derivative and higher? This is where the fancy canceling comes in!

As we noted above, when expanded out is a sum of a bunch of terms of the form

where and is some integer. When we take the derivative, this term will turn into ; if we take the derivative again, it will become ; another derivative gives us , and so on. Do you see what is happening? After taking the derivative exactly times, we will end up with the constant term

and here’s our fancy canceling: is clearly divisible by , so this is some integer times , which is also an integer. Voila! Said a different way, and more succinctly: since each term of has degree at least , by the time we have taken the derivative enough times for it to yield a constant term, the will be canceled from the denominator, since we will have taken the derivative at least at each power of from down to .

Finally, if we take the derivative of more than times, we get , so no problems there.

Great, so is always an integer. But what about ? Well, remember, last time we showed that . If we take the derivative of both sides with respect to (being careful to use the chain rule on the left side, noting that the derivative of with respect to is ), we get

We can repeat this process to find that (the two negatives cancel on the left side), , and so on. But the extra negative sign for odd derivatives doesn’t really matter: in either case, , which is an integer.

Getting closer! Next time, we will define another special function in terms of and its derivatives; this function will help us compute —which (if you recall the punchline) will turn out to be an integer strictly between and (which is impossible).

The Numberland News runs personal ads. 21 was looking for a new friend and put an ad in.

Two-digit, semi-prime, triangular, Fibonacci number seeks same. I’m a binary palindrome, what about you?

Will 21 find a friend?

A semi-prime is a number with exactly two prime factors, like 6. See this post for a definition of triangular number, this post for some hints on how to figure out a general formula for computing triangular numbers, and this one for the solution. Fibonacci numbers are discussed here. Finally, a palindrome is a number (or word, or phrase) which is the same forwards and backwards; a binary palindrome is a number which is a palindrome when expressed in base two.