### 0^0 still wishes it were equal to 1 (but it still isn’t)

Here’s a little supplement to the last post. I mentioned before that there are several reasons somebody might try to define 0^0 := 1. The big reason before (flawed though it turned out to be) involved calculus and limits. But there’s an entirely different way to interpret the expression *y*^*x* which puts 0^0 in an intriguing different light.

If *X* and *Y* are sets, with |*X*|= x and |*Y*| = y (*X* has *x* many members and *Y* has *y* many), then the number of distinct functions f:*X* -> *Y* is given by *y*^*x*. Think about it in the case *X* just has 2 elements: the first element has *y* different places to which it could be mapped, and likewise for the second element, leading to *y^*2 many functions.

So then 0^0 should count the number of functions from the empty set to itself.

There are two possible replies to this neat observation:

(1) Of course there is exactly one such function - it takes in nothing and sends nothing nowhere. A useless, trivial, but perfectly understandable function. Hence 0^0 = 1.

(2) The very idea of a function rests on sending elements to elements. This idea is meaningless when it comes to the empty set - there are no elements! So 0^0 is nonsensical.

Now here’s a reply back to (2): Mathematicians actually do allow functions to be defined on the empty set - they’re called empty functions and they don’t do anything interesting, but they’re required for making certain definitions in category theory work nicely. Any two empty functions f: { } -> *Y* and g: { } -> *Y* (with *Y* nonempty) are identical, and this reflects the fact that *y*^0 = 1.

(See http://en.wikipedia.org/wiki/Empty_function)

I can’t find any math literature that says there’s no way of making sense of a function from the empty set to itself. It seems to me such a function wouldn’t break anything. In other words, I dig response (1) above. But unfortunately, this still isn’t any kind of hard proof that 0^0 = 1.

### 0^0 wishes it were equal to 1 (but it’s not)

Hey folks. It’s been a long time. I’ve had lots of things on the brain, along with lots of new responsibilities, and I just kind of fell out of the posting spirit around here (though I’ve still been reading others’ blogs regularly).

Anyway, maybe a short and easy math post will get me back into the swing of things (no promises on that). Here’s something irksome. It’s common knowledge that for a real number *x*, we have *x*^0 = 1. Well, not just any real number - *x* can’t itself be zero. People since the time of Cauchy have called 0^0 an “indeterminate form”. Ugly, ugly.

0^0 wants to equal 1, and here’s one of several compelling reasons why:

Nevermind the orange graph - for almost all negative *x*, we need complex numbers to make sense of *x^x*, but for simplicity we can just restrict our attention to positive *x*.

Well, I won’t go through the computation of this limit - you can just go to Wolfram Alpha and click “show steps”. But doesn’t the fact that this limit equals 1 suggest to us, at the top of its lungs, so to speak, that we should go ahead and define 0^0 := 1?

Unfortunately doing so would be hasty and lead to problems. The key point is we have only demonstrated *one instance* in which 1 is a sensible definition for 0^0. Although *x^x* admittedly provides us a very nice model for thinking about 0^0, it’s not the only available one, and by no means is it special, natural, canonical, universal or other such adjectives. It’s merely pulled out of thin air.

There’s still a gleam of hope in favor of defining 0^0 := 1, though! Perhaps if we take any two functions f(*x*) and g(*x*) such that both have limits of 0 as *x* approaches 0, then we’ll find the limit as *x* goes to 0 of f(*x*)^g(*x*) equals 1. This would be tremendous evidence in favor of 0^0 := 1.

Sadly, it doesn’t work. I invite you to come up with two such functions f and g which break our goal. I’ll reveal a solution below. But first, a couple examples of menacing looking pairs f and g which *do* work as nicely as we’d like. First we take f(*x*) := tan(*x*) and g(*x*) := *x*^3 - *x*^2 - *x*, both of which go to 0 as *x* does.

Splendid. Now let’s try taking f(*x*) := ln(1 + x^2) and g(*x*) := sin(x^(1/3)). Again, both functions go to 0 as *x* goes to 0.

Wow! If our hopeful identity holds for such gnarly functions, it must take a really hideous pair to break it, right?

Yes and no. We will use one somewhat hideous function, taking our f(*x*) to be *e*^(-1/x). As *x* approaches 0 through positive numbers, the limit is 0 (since -1/x is going to negative infinity). The left-sided limit comes out different (namely to infinity), which means f has no hope of being continuous at 0, but for us that doesn’t matter. Now take g(*x*) := *x*, as nice and simple as can be. We look at (*e*^(-1/*x*))^*x* - but by simplifying, this is just equal to *e*^(-1) for nonzero *x*. And so the limit as *x* approaches 0 for this pair is 1/*e*.

…ugly, ugly. Nobody told poor 0^0 about such freakish occurrences back when it first dreamed of equaling 1.

It seems we have no choice but to leave 0^0 undefined, since on some occasions it wants to be 1 and on others it wants to be 1/*e* (and you can be sure there will be other times it wants to be something else). And after all, there are times we want to use l’Hôpital’s rule to evaluate a limit of the form 0^0 and get some nontrivial function of *x* as the answer; it would sure be stupid and break a lot of calculus if such limits were automatically just equal to 1.

### A solution

When I first heard the submarine puzzle in the post below, I wasn’t able to solve it until I got some hints. The trick, as with many puzzles and even serious math problems, is to replace the problem at hand with one easy enough you can solve it, but which still captures the essential difficulty of the original problem. This tends to be easier said than done, unless you have a great deal of experience and so-called “mathematical maturity”. Sometimes you have to work your way through a series of gradually harder problems until you settle the whole matter.

**Step 1: Assume the submarine starts at position 0 and is traveling with positive speed, i.e. from left to right.**

Now it would be pretty useless to say “Just bomb 0 immediately!” in this situation, as this provides you no insight to the original problem. Instead you have to decide what you would do if you *missed* on the first bombing attempt. Well, suppose the submarine’s speed is as simple as possible: 1. Then the strategy would be to bomb position 1 on the next possible occasion. If this bombing fails, we can conclude that the submarine does not travel with unit speed. Perhaps, then, its speed is 2? If so it would have been at position 2 when you bombed position 1, so the next spot you should target is position 4. If this bombing fails, we conclude the speed is different from 2.

Now you should have the idea of what to do in this simplified scenario: the bombing sequence goes (1,4,9,16,25,…,*v*^2,…). This is guaranteed to eventually hit the submarine, specifically on the *v*th bombing, at location *v*^2, where *v* is the sub’s speed.

**Step 2: Assume the submarine starts at position 0 without putting any constraints on the direction of travel.**

This is handled with a small modification to the above strategy: we alternatively bomb to the right and the left of the origin. The bombing sequence goes (1,-2,6,-8,15,-18,28,-32,…). This is messier than the last sequence, but not harder to understand; the negative terms are where the sub would be assuming right-to-left travel.

**Step 3: Assume the submarine starts at position m, where m is any integer.**

Then we simply translate the last bombing sequence by *m*. There was nothing special about 0 in the above argument.

Now for the truly tricky and interesting part. For convenience, we can represent the situation of the sub starting at position *m* with velocity *v* as an ordered pair (*m*,*v*). Such an ordered pair is often called a lattice point (or Gaussian integer, when the plane is thought of as a geometric manifestation of the complex numbers).

**Step 4: Remember that the lattice points of the plane are countable.**

Informally this just means the set of points (*m*,*v*) can be listed out in a systematic way that doesn’t miss or double-count a single member. Here’s one way to do it helpful in our situation, using a spiral winding its way out from (1,1):

In this picture, I didn’t bother plotting any points of the form (*m,*0), because we know the submarine has non-zero velocity.*

So, what’s the bombing plan for the most general situation? Employ the listing of lattice points illustrated above, and when you get to point (*m*,*v*) in the list, bomb the location where the submarine would currently be if it had started with velocity *v* at point *m* (a situation we fully understand thanks to Step 3 above). This is guaranteed to hit the sub in finite time - though it may take a rather long time if (*m*,*v*) is very far from (1,1)!

By the way, the winding spiral above also proves that the set of all rational numbers is countable, if we merely observe that (*m*,*v*) may be identified with the fraction *m*/*v*, since *v* is not zero!

*Actually, if the original problem were modified to include the possibility of a motionless submarine, we would still be fine because of this argument.

### A puzzle

It’s the Cold War, and you command a missile station tasked with bombing an enemy submarine. Intelligence has provided you with these facts:

**1) The submarine travels with some constant integer velocity, due either east or west (i.e., travels along the standard number line, either to the right or left). **

**2) At any integral time, the submarine’s position on the number line is an integer.**

For example, it could be that at time 1, the sub is at position 5, at time 2 it’s at position 3, at time 3 it’s at position 1, at time 4 it’s at position -1, and so forth. In this case the constant integer velocity is -2.

Your missile station is able to fire one missile at each integral time value. So if in the above example you had tried the sequence of bombings (0,1,2,3) - the ordered numbers here representing positions on the number line - you would have missed every time. But if you just so happened to bomb the location -3 on the fifth try, you would have a lucky hit.

Unfortunately the above is just one of many possible situations, and Intelligence has supplied you no information about the submarine’s particular velocity and current location. Nevertheless, can you develop a bombing plan that will eventually find your target?

—

Toru Takemitsu, *Yume to Kazu*, p.21; translation from Taniyama, ‘The Development of Toru Takemitsu’s Musical Philosophy’, p.90.

(via Peter Burt’s ‘The Music of Toru Takemitsu’, p.251.)

### Geometrically Finding the Area of a Trapezoid

I hope the picture says it all. The last step uses the fact that the area of a parallelogram of base *b* and height *h* is simply *bh*, which can easily be proved geometrically by slicing off a triangle and sliding it to the other side to form a rectangle with the same base and height.

Sorry for the lack of posts recently, but they should perk up as the school year begins and I have lots of interesting things to think about all the time.