# Why leave 0^0 undefined?

1. Jul 11, 2011

### Josh Swanson

$0^0$ (that is, if ^ is the exponentiation function, ^(0, 0)) is sometimes undefined. The only argument I've seen for this is that $0^0$ is an indeterminate form, though I don't accept that as an argument against defining ^(0, 0). It seems that $0^0$ could be interpreted as an indeterminate form or as ^(0, 0) = 1 depending on context without difficulty. On the other hand, I've found four good arguments for letting ^(0, 0) = 1: it makes the binomial theorem work out, it makes our notation for power series work out, it makes sense if m^n is the number of functions from an n-element set to an m-element set, and it follows from a natural definition of exponentiation via repeated products, using the empty product as 1.

So, is there any good argument for leaving it undefined?

2. Jul 11, 2011

### micromass

Staff Emeritus
Hi Josh Swanson!

There are a few reasons why we leave it undefined. For example, consider the function $f(x)=0^x$, then the limit of this function in 0 will be 0. While the limit of the function $f(x)=x^x$ is 1.

In general, the function $f(x,y)=x^y$ has an essential singularity in (0,0). This would be a good reason to leave it undefined.

3. Jul 11, 2011

### Josh Swanson

I'm sorry if I was unclear. I'm aware of that argument (which boils down to the fact that $0^0$ is an indeterminate form) but I don't accept it. Coming at it from another direction, the binomial theorem holds in general in unital commutative rings and (if 0^0 is defined) it implies $0^0 = 1$. Why should the essential singularity of $x^y$ at (0, 0) prevent us from defining that function at (0, 0) when doing so can be seen as natural? Nothing will remove the singularity, but so what? There's reasoning missing with this argument, and I haven't been able to supply the missing steps.

Stripping away the analytic structure of the complex numbers and dealing just with unital commutative rings, the binomial theorem argument remains while the essential singularity argument is gone.

4. Jul 11, 2011

### micromass

Staff Emeritus
When writing $0^0$ in the binomial theorem, then the two zeroes have another roll. This is best seen when looking at the commutative ring case. The lower 0 is the zero element of the ring, but the exponent-zero is not an element of the ring, but an element of the natural numbers. So what we're dealing with is some kind of action (bad word) of the naturals on the commutative ring. In this respect, defining 00=1 seems to be ok.

However, we could also define 00 when both numbers are real (or complex) and it is clear that we cannot use the same arguments here. We must deal with continuity arguments here. Indeed, how do we define $2^\pi$, well as the limit of the sequence

$$2^3,2^{3.1},2^{3.14},2^{3.141},...$$

And it turns out that this choice is very good: it allows us to make a lot of functions continuous, there is no ambiguity.
However, when defining 00 and when looking for continuity arguments, it seems that there is no good way of defining it.

It's not good to tell students all the time that 00=1. But if we later deal with limits, to tell them that this is actually not true. I don't find that correct. It would be better not to define the thing.

Let's just say that we don't define 00 because we want to warn people that we can't just work with it like we want to.

5. Jul 11, 2011

### Studiot

Keep the discussion going chaps - it's fascinating to a non pure mathematician.

6. Jul 11, 2011

### pmsrw3

I have some sympathy with Swanson -- in information theory it is very convenient to define 0 log 0 = 0 (which is equivalent to 0^0 = 1). But it's context dependent. How about just saying defining 0^0 = 1 within certain regions of math, without trying to impose a global ukase?

7. Jul 11, 2011

### micromass

Staff Emeritus
Yes, that's how it's done now, I think. Sometimes defining 00=1 is extremely handy, so it is done in those cases (which are mostly the cases where you deal with integer exponents). In other contexts it's not good to define it as 1, so we don't do that.

8. Jul 11, 2011

### Josh Swanson

@Micromass:
Actually, I don't agree. If we already have 0^0 = 1 in the ring of integers, it's very natural to say 0^0 = 1 in the ring of rationals. This can be extended to the ring of reals naturally as well. Take the reals as the usual completion of the rationals, and take the real number $0 = \{r_1, r_2, ...\}$ where the sequence is a Cauchy sequence of rationals that's equivalent to the sequence $\{0, 0, ...\}$. The real number 0^0 can then be defined as the Cauchy sequence $\{r_1^0, r_2^0, ...\}$. Since we've worked up 0^0 = 1 in the ring of rationals, this sequence is $\{1, 1, ...\} = 1$. This seems eminently reasonable to me. It even avoids certain bits of "ugliness", where defining exponentiation by reals analogously otherwise has to avoid 0^0 terms in the sequences generated.

Pedagogical arguments are fine with me:
I agree that this is unfortunate. Only calculus (or higher) students should ever encounter the ambiguity, so I don't think this is important enough to ignore the 4 good reasons I found for defining $0^0 = 1$.

@Studiot: I also find it interesting :). I found the question elsewhere online and had never seriously considered it myself. I had assumed that defining 0^0 would break one of the usual laws of exponents or rules of addition/multiplication, but I was only able to show that 0^0 must be 0 or 1 if it is defined. I actually started thinking that 0^0 "should" be undefined, but I wasn't able to find a reason that convinced me.

@pmsrw3: I don't see where defining 0^0 = 1 would actually cause problems. If 0^0 is an indeterminate form and not "1", that should be clear from context. Do you have an example of a case where defining 0^0 = 1 would cause problems?

9. Jul 11, 2011

### micromass

Staff Emeritus
I don't find it very natural to define this in the ring of the rationals. Already with the rational numbers, we need to be aware of continuity problems. Completion of the rationals is unimportant here.

Only calculus students will ever encounter your 4 good reasons anyway, so I see no real merit in defining it.

Defining 00 would always pose problems when dealing with continuity, which is important enough not to define it

10. Jul 11, 2011

### pmsrw3

Not that I personally have run into, no.

11. Jul 11, 2011

### micromass

Staff Emeritus
Don't get me wrong Josh, I understand your arguments and I agree with them to some extent. It's just that you seem to find that the whole continuity argument is not to the points, but I can't really find a reason why it would not be to the point. All other operations are continuous, so it's natural to expect exponentiotian to be continuous too!!

Furthermore, it keeps people from making arguments like

$0^0=1~\text{thus}~0=\log(0^0)=0\log(0)=0.(-\infty)$

Or $0^0=0^{1-1}=0/0$.

Which would still be wrong if 00 were defined, but it would be harder to explain why.

12. Jul 11, 2011

### Josh Swanson

Thank you for discussing this with me. I find it very interesting.

That a function can't be made continuous at a point isn't enough to prevent me from defining it at that point anyway. The argument "$x^y$ cannot be made continuous at (0, 0) so it shouldn't be defined at (0, 0)" is incomplete. Why is it helpful to not define it? In either case it's not continuous on the whole plane, but is continuous on the punctured plane. Perhaps the rest of the argument says "$0^0 = 1$ seems to imply that $0^0$ is not an indeterminate form, i.e. that $x^y$ is continuous at (0, 0)"? That's not enough for me, since it sacrifices a convenient notation in every area I've thought of for a day or two of possible confusion. If it happened that, for $x^y$ to be continuous at (0, 0) we need $0^0 = 7$, then I could understand continuity being an important consideration. Continuity is broken either way, though--I say "oh well, let's not break the binomial theorem too."

It seems artificial to me to define 0^0 = 1 in the ring of integers, but not to define 0^0 in the ring of rationals, of which the integers are a subring.

The fourth--the empty product being 1--might be low enough level to teach earlier, but the others are definitely too high level for most people (even though I love the function counting definition) There are other bits of math that are too advanced to be explained adequately at a low level, though they're not left undefined (for instance, the power rule from calculus is valid for real exponents, but the proof generally uses the exponential function which is too advanced when the power rule is first encountered).

I like those, especially the 0^0 = 0/0 one. The log one doesn't bother me much, since a similar argument is also wrong, eg. $\log(1) = \log((-1)^2) = 2 \log(-1) = 0$. That the base of the exponent is > 0 for the rule to be applied covers both. After thinking about it for a little longer, the second one is also close to a similar argument that is also wrong, eg. $0 = 0^1 = 0^{2-1} = 0^2/0^1 = 0/0$.

13. Jul 11, 2011

### micromass

Staff Emeritus
I certainly agree with this. You have some fine points, and you're in good company here. Knuth also said what you said.

It's helpful because I feel it may be more honest. Defining it is basically saying that there are no problems, but there are. I have never seen a function defined in an essential singularity: we never define $\sin(1/x)$ in 0 or $e^{1/x}$ in 0. Basically every value can be defined here. So, from continuitic point-of-view, defining 00 is as bad as a definition as 00=7.

My point is that the only time we need the binomial theorem work, we need the exponents to be integers. This is because the zero in the exponent and the lower zero have another status (in my view). So in these cases, I would say that 00=1 is good.

I don't see why it is artificial. There are many things we only define on integers and not on the rationals. And again, only looking at the rationals as a ring is a limitation. If you only look at the rationals as a ring, then 00=1 is ok. But the rationals are so much more than just a ring, they also have a metric space structure!!

[/QUOTE]

The log one is very bothersome since it is actually the definition of complex exponentiation:

$$a^b=e^{b\log(a)}$$

(note that the complex logarithm is multivalued however). And this definition makes sense everywhere, but for 0, we get

$$e^{0(-\infty)}$$

which is undefined. So defining 90=1 in the complex numbers is quite artificial, since it does not agree with the definition in all the other numbers!!)

14. Jul 11, 2011

### Studiot

I wish I had something useful to contribute but am still finding the discussion fascinating.

I have never been in the camp where if I had five equations/definitions/axioms I would not be happy until I had reduced their number to four or even three.

I would be happy to keep my five because in my experience you usually have to expand compact notation or theorems to do anything useful anyway.

go well

15. Jul 11, 2011

### pmsrw3

It's not ONLY the binomial theorem, or even only integers, though. In the information theory case I mentioned, P log P is entropy, and P is a real number in [0, 1]. (As you might expect from the name, there's a similar application in thermodynamics.) This wants to be zero because 0 is a big sort of zero but log 0 is a small sort of infinity, so 0 is able to beat log 0.

16. Jul 11, 2011

### micromass

Staff Emeritus
This also happens in measure theory and with integrals where $0(+\infty)=0$ is defined. But that it works in these particular instances, doesn't mean that we should define it generally!!

17. Jul 12, 2011

### JJacquelin

Hello chaps !

aren't you tired to brood over always the same old song ?

18. Jul 12, 2011

### Mute

I'm sure there's one instance in which you have seen a function defined to have a value at an essential singularity: $\exp(-1/x^2)$ is defined to be zero at $x=0$ in one famous example involving Taylor series. ;) Of course, the point of the example is to show that bad things like the Taylor series being zero can happen when you try to patch up a non-analytic function like that. So, this is merely to remind you of this one case where a function is given a definition at an essential singularity, and not to argue against any points you're making. =)

19. Jul 12, 2011

### jambaugh

As mentioned already when you consider the function $f(x,y)=x^y$ has an essential singularity at (0,0).
But it is in fact "messier" than that.

Consider the limit for the values of $f(x,y)$ as you follow a curve ending at (0,0). You can consider the limit of the value as you follow this path.

As it turns out the limit depends on how you approach the point. Along the line y=0 you get 1, along the line x=0 you get 0. You can get other values by approaching along different paths so the answer to the question of "which value?" has an ambiguous answer.

20. Jul 12, 2011

### Josh Swanson

@micromass:
If I understand correctly, you would define $$\text{^}(x, y): \mathbb{R} \times \mathbb{Z}^+_0 \to \mathbb{R}~\text{with}~\text{^}(0, 0) = 1$$ and $$\text{^}(x, y): \mathbb{R} \times \mathbb{R} \setminus \{0\} \times \mathbb{Z}^-_0 \to \mathbb{R}$$? [It occurred to me that $x^y$ is actually not continuous or defined when x is 0 and y<0 anyway, so the continuity statements of my previous post were slightly incorrect.] To get power series to work out, we need the base to be real (or complex; I'm using reals for simplicity) numbers and not just integers. This might be sufficient, though. It makes power series and the binomial theorem work out, while also not defining $x^y$ at (0, 0) when y is real (analogously, complex). It doesn't make $0 \log(0)$ work out, unfortunately.

It's probably my lack of imagination, but I can't think of any examples (besides perhaps this).

@jambaugh: Yes. That behavior is part of why I've preferred to say $0^0$ is an indeterminate form, rather than saying $x^y$ has an essential singularity at (0, 0). But the fact that the function can't be made continuous at (0, 0) isn't enough to prevent me from defining it at (0, 0) anyway. There must be more reasoning--as micromass put it, "It's helpful because I feel it may be more honest. Defining it is basically saying that there are no problems, but there are." While I don't quite agree with this reasoning, it seems to be such a matter of opinion that debating it isn't worthwhile. I can understand the argument and try to respect it.