Why is 0^0 Undefined in Mathematics?

[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?

 

[micromass]

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.

 

[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.

 

[micromass]

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.

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.

 

[Studiot]

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

 

[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?

 

[micromass]

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?

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.

 

[Josh Swanson]

@Micromass:

However, we could also define $0^0$ 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.

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:

It's not good to tell students all the time that $0^0=1$. 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.

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?

 

[micromass]

@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.

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.

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 $0^0 = 1$.

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

 

[pmsrw3]

@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?

Not that I personally have run into, no.

 

[micromass]

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.

 

[Josh Swanson]

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

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.

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."

I don't find it very natural to define this in the ring of the rationals.

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.

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

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).

Furthermore, it keeps people from making arguments like

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

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

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. $0 = 0^1 = 0^{2-1} = 0^2/0^1 = 0/0$.

 

[micromass]

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

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

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?

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.

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."

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.

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.

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!

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. $0 = 0^1 = 0^{2-1} = 0^2/0^1 = 0/0$.

[/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!)

 

[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

 

[pmsrw3]

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.

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.

 

[micromass]

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.

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!

 

[JJacquelin]

Hello chaps !

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

Seriously, let us not delude ourselves: we have not seen the last of the Power Less monster

 

[Mute]

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.

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. =)

 

[jambaugh]

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

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.

 

[Josh Swanson]

@micromass:

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.

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.

There are many things we only define on integers and not on the rationals

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.

 

[Dickfore]

$0^0$ (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?

Suppose:
$$0^{0} = x$$

Then, taking the logarithm:
$$0 \cdot \log{0} = \log{x}$$
But, the l.h.s. is an indefinite form $0 \times \infty$, and so is the r.h.s. and the original expression.

 

[jambaugh]

...
@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.

It is more than an indeterminate form. sin(x)/x is indeterminate at 0 but has limit of 1 at x=0. (likewise sin(y)/x has limit 1 at (x,y)=(0,0) independent of direction.)

Limits are the way we deal with indeterminate forms, we can generalize to directional limits (e.g. for 1 variable left limits and right limits).

The problem with trying to define a value especially when the limits don't exist/agree, is that the value should be usable in all contexts, not just the one(s) at hand.

Note. If you constructively define what x^y means in a narrower context than real exponentiation, i.e. the examples given in the OP, then it is perfectly OK to define 0^0 = 1 (or whatever is appropriate) within that context. Often when generalizing a topic you cannot extend all formulas and relations though they are perfectly valid in the narrower context. There isn't just one definition for powers applying to all subjects.

 

[Josh Swanson]

It is more than an indeterminate form. sin(x)/x is indeterminate at 0 but has limit of 1 at x=0. (likewise sin(y)/x has limit 1 at (x,y)=(0,0) independent of direction.)

I've been using the term "indeterminate form" in its technical sense (eg. see the http://en.wikipedia.org/wiki/Indeterminate_form" on the topic for a more formal description of what I mean). In that context, sin(x)/x's behavior is irrelevant, though of course it provides a nice example that 0/0 can be interpreted as 1. Your parenthetical statement is incorrect. The limit of sin(y)/x as (x,y) approaches (0,0) is not always 1. Approach (0,0) along the x-axis, with y=0, to get that the limit in this case is 0.

The problem with trying to define a value especially when the limits don't exist/agree, is that the value should be usable in all contexts, not just the one(s) at hand.

I don't agree. "All or nothing" doesn't seem to be necessary. The function $x^y$ can be used with the domain $\mathbb{R} \times \mathbb{R} \setminus \{0\} \times (-\infty, 0)$ or $\mathbb{R} \times \mathbb{R} \setminus \{0\} \times (-\infty, 0]$. In the first case, it's continuous everywhere. In the second case, it's not continuous. Changing the domain according to context prevents the value at (0,0) from being used in contexts where it wouldn't make sense.

Often when generalizing a topic you cannot extend all formulas and relations though they are perfectly valid in the narrower context. There isn't just one definition for powers applying to all subjects.

Certainly some generalizations conflict with each other, but I have difficulty thinking of another example where things in a narrower context need to be retroactively undefined to work in a more general context.

 

[micromass]

Certainly some generalizations conflict with each other, but I have difficulty thinking of another example where things in a narrower context need to be retroactively undefined to work in a more general context.

Well, the function

$$f(x)=e^{\frac{-1}{x}}$$

on $]0,+\infty[$ is often defined as 0 in 0. This function is of great importance as it provides an example of a bump function.
However, if we extend this domain (for example if we would allow complex numbers), then defining it 0 in 0 will not work. Indeed, the function f has an essential singularity in 0.

There are a lot of things that work in a limited case, but which become awkward in more general context. For example, when working in sequences in $\mathbb{R}$, then we might say that the sequence

$$x_n=n^2$$

diverges to infinity. We might even use the notation $n^2\rightarrow +\infty$. But when working with sequences in $\mathbb{R}^2$, we don't use "divergence to infinity" anymore (I've never seen such a thing anyway). We can probably make sense of it, but it doesn't seem to be commonly used.

I admit, these are pretty lame examples. But all I want to make clear is that it's not unnatural for me to define things in narrow context and not define it in general context.

 

[Josh Swanson]

However, if we extend this domain (for example if we would allow complex numbers), then defining it 0 in 0 will not work. Indeed, the function f has an essential singularity in 0.

Well, I again disagree that defining it as 0 at 0 won't work just because it has an essential singularity there. But I understand that the specific property of continuity is maintained in the narrower case while it's lost in the more general case, so to maintain the more general version of continuity one needs to "undefine" the function at 0.

We might even use the notation $n^2\rightarrow +\infty$

I've always secretly wished for a notation like $n^2 \rightarrow \infty_{(1)}$ or $-n^2 \rightarrow \infty_{(-1)}$; in more dimensions, $(n^2, n^2) \rightarrow\infty_{(1, 1)}$, where the thing after the infinity is a vector giving the direction that the sequence is headed in. $+\infty$ and $-\infty$ are then just shorthand for the more general notaion, which does generalize well.

 

[micromass]

I've always secretly wished for a notation like $n^2 \rightarrow \infty_{(1)}$ or $-n^2 \rightarrow \infty_{(-1)}$; in more dimensions, $(n^2, n^2) \rightarrow\infty_{(1, 1)}$, where the thing after the infinity is a vector giving the direction that the sequence is headed in. $+\infty$ and $-\infty$ are then just shorthand for the more general notaion, which does generalize well.

Well, it's not impossible to do such a thing. But for some reasons, people never do it (or I haven't seen it anyways). So this was a bad example

 

[Hurkyl]

I've always secretly wished for a notation like $n^2 \rightarrow \infty_{(1)}$ or $-n^2 \rightarrow \infty_{(-1)}$; in more dimensions, $(n^2, n^2) \rightarrow\infty_{(1, 1)}$, where the thing after the infinity is a vector giving the direction that the sequence is headed in. $+\infty$ and $-\infty$ are then just shorthand for the more general notaion, which does generalize well.

The idea you want is "compactification". For the real line, there are two standard ways to compactify it:

• The projective real numbers, which adds a single element $\infty$. This compactification is particularly useful for algebraic operations and analytic functions with poles. We can extend various operations continuously to projective real values, such as $1/0 = \infty$ or $\tan (-\pi / 2) = \infty$
• The extended real numbers, which adds two elements $+\infty$ and $-\infty$. This is useful for other purposes, such as expression $\arctan(-\infty) = -\pi / 2$.

When you're considering a plane, there are more commonly useful things to choose from. We can compactify it with a single point (think Riemann sphere), we could consider the projective plane (very good for algebra and poles of analytic functions), we could add a circle at infinity, we could think of a plane whose coordinates are projective real numbers, or a plane whose coordinates are extended real numbers, ...

Incidentally, it is fairly standard to use the Riemann sphere when doing complex analysis, and to use the projective plane when doing algebra in two independent variables.

 

[jambaugh]

I've been using the term "indeterminate form" in its technical sense [...]Your parenthetical statement is incorrect.

Quite right and my apologies for my sloppiness. Indeterminate forms are indeterminate for exactly the ambiguity. I guess I was trying to misread "indeterminate" as "undefined" and what I said was just plain wrong.

Some additional comments: I find it important in teaching to emphasize to the students that when we change or restrict domains of functions we are in fact defining a distinct function. The idea of "undefining" when expanding context is not a good way of stating things. It makes mathematics appear much more ad hoc than it is.

 

[Josh Swanson]

The idea of "undefining" when expanding context is not a good way of stating things.

I agree. "Restricting" is more proper.

The idea you want is "compactification".

Adding a circle at infinity is a much more elegant approach than my ad-hoc notation, even if they amount to the same thing. I had considered using projective points instead of vectors but I wanted $+\infty$ and $-\infty$ to be distinct.


Right now I'm leaning toward the following default convention: If the exponents in a formula are integers, 0^0 = 1. If they aren't, 0^0 is undefined. Perhaps a similar convention works to say that 0log(0) = 0 in the information theory example. This isn't a silver bullet, unfortunately: the generalized binomial theorem uses real exponents, but still uses the 0^0=1 convention.

So far the discussion has focused on more abstract reasons for leaving 0^0 undefined ("it's more honest"; it's an indeterminate form). From a purely practical standpoint, I have difficulty finding real applications where leaving 0^0 undefined is helpful, while I've found several where 0^0 = 1 is helpful. 0^0 is currently a point of confusion when learning about exponents, so it's not clear to me that either convention is better there. I don't recall being confused in calculus class by the notation 0^0 (as an indeterminate form), though I do dimly remember thinking that it fits that 0^0 is undefined, just like 0/0 is undefined. If 0^0 was 1, I doubt it would have caused anything more than mild interest into why 0^0 = 1 yet 0^0 is an indeterminate form. I wonder if 0^0 = 1 is a common convention somewhere.

 

[Hurkyl]

From a purely practical standpoint, I have difficulty finding real applications where leaving 0^0 undefined is helpful, while I've found several where 0^0 = 1 is helpful.

If you are doing complicated manipulations with limits and related tasks (e.g. deriving approximations), it would be rather dangerous to adopt $0^0=1$.