In the course of the craziness about the whole *you know what* (see the links in the sidebar), it somehow came up that 0^0 (which is to say, 0 to the 0-eth power) is a tricky calculation. Several commenters gave their opinions, and so, for fun, here's mine. Consider the following chart:

Note that negative and zero exponents are pretty much defined to make each column's pattern (successive division by the base for that column) continue. And when you do define it that way, lo and behold, the standard laws of exponents hold, and we're therefore happy with that definition. That's how math works, right? We find patterns and see if our codifications of them continue to make sense after they get more and more abstract.

The problem with 0^0 is that it seems to have to satisfy two patterns at once. The column pattern is that the right hand side is always 0. The row pattern is that the right hand side is always 1. So what goes in the green box? 1 or 0? Both continue a pattern.

Take a look, though, at the entries in the blue column *below* the green box. Since 0 can't appear in a denominator, those entries are missing. Thus, crucially, the pattern in the blue column *will have to end anyway*. So if our goal for this definition is to preserve as many patterns as possible, we don't lose much pattern by saying that 0^0 is 1. That preserves the horizontal yellow pattern, and breaks the vertical blue pattern, but...hey, it was broken soon anyway.

So that's my vote and my reason. 0^0 should be defined as 1.

You might be a geek if....you devote a whole blog post to this.

Hi,

if I remember correctly, the expression X^0 is (or, better said, has always been) *defined* as 1, for any X (real, complex). 0 therefore fits as X, which should yield satisfying 0^0 == 1, patterns or not.

Posted by: ondra z | July 28, 2006 at 01:00 AM

Playing around, Fermat's Little Theorem says that, for a prime P and a natural number A, that

A^P === A (mod P)

Modifying that

A^P === A^1 (mod P)

Divide by A

A^(P-1) === A^0 (mod P)

Extend to whole numbers, and let A = 0

0^(P - 1) === 0^0 (mod P)

P, being prime, must be greater than 1. Thus we can assume that P - 1 is greater than 0. That being so, 0^(P-1) = 0

0 === 0^0 (mod P)

For that to be true, 0^0 would have to equal 0.

But, looking back, this all does require division by 0, which would seem to indicate that it is flawed, which in turn probably supports the idea that 0^0 = 1.

Posted by: Foxy | July 28, 2006 at 01:22 AM

Sheesh, it's been too long since y'all have taken calculus. You're essentially putting forth an argument that we should define 0^0=1 by continuity, which doesn't work in general.

Let f(x) be a function such that f(0)=0^0. I can rig this function so that the limit of f(x) as x approaches 0 is *anything I like.* This reflects the fact that 0^0 is one of the indeterminate forms (like 0/0, 1^(infinity), etc.) which require the application of L'Hopital's rule.

Posted by: Davis | July 28, 2006 at 03:36 AM

By the way, here's a stupid example of a function that lets you get a lot of different "values" for 0^0.

Pick a positive number a. Now let f(x)=[e^(-a/x^2)]^(x^2). Let's say we fail to notice that we can simplify this. Then, as x->0, f(x)->0^0. However, it should be straightforward to see that f(x)=e^(-a) everywhere it's defined, i.e., everywhere but x=0.

There are probably many more interesting examples to contrive.

Posted by: Davis | July 28, 2006 at 03:46 AM

I would just like to point out that, combinatorially, there is a good reason for 0^0 = 1. Specifically, the combinatorial interpretation of a^b is "the number of functions from a set with b objects to a set with a objects." Thus 0^0 is the number of functions from the empty set to itself, namely one--the empty function! Unfortunately, as Davis explains, analysts don't like assigning any value to 0^0 since no such assignment can make x^y a continous function.

Posted by: Cat lover | July 28, 2006 at 09:25 AM

I'm with Cat lover on this one, I think -- if you're thinking about exponentiation as an operation on integers, especially in the context of combinatorics, you really really want 0^0 = 1 or you get lots of terrible special cases everywhere. If you're thinking about exponentation as function on the reals and doing analysis, you really want to think of 0^0 as an indeterminate form.

Posted by: glasser | July 28, 2006 at 09:38 AM

"...analysts don't like assigning any value to 0^0..."

I hope you're not implying that I'm an analyst; as an algebraist I would consider that a grievous insult. :)

Posted by: Davis | July 28, 2006 at 10:57 AM

0^0 is 1 cause it's a product of zero numbers, which is always 1, no matter the numbers (there are zero of them, remember, so who cares what they are?).

Continuity has no effect on this. It's just so regardless what continuity says. To say that 0^0 is 0 "because 0^x is usually 0" you first prove that function 0^x is continuous. To say that 0^0 is 1 "because x^0 is usually 1" you first prove that x^0 is continuous. There's no way you can prove this, I guess. So, forget continuity, just use definition of power. And there's no way you can get anything but 1. And (!) you get no contradiction with 1.

Now what's really funny, is that I just started this topic on my blog several days ago :-) I'm going to have a big article (I'm geek, yes), but it's gonna be in Russian.

Posted by: Ilya Birman | July 28, 2006 at 04:24 PM

Ilya, though I agree that it's probably most sane and reasonable to define 0^0=1, it's significant to note that pretty much everything about exponents is *defined*, rather than being "just so." And many of the facts we take for granted about exponents actually require a bit of work to define properly.

Start with a^x for positive integers x. It seems clear enough what that should mean. But it's not clear what a^0 should mean until you notice that a^x a^y= a^(x+y). With this in mind, it becomes natural to define a^0=1 so this still holds. Once you have a^0=1, you can then define a^x for negative integers

Posted by: Davis | July 28, 2006 at 05:37 PM

Bah, hit post by accident. Continuing:

you can define a^x for negative integers by requiring the addition rule still hold: a^x a^(-x)=a^0=1.

You can then proceed to figure out what a^x should be when x is rational, by applying the same reasoning to the multiplication rule, (a^x)^y=a^(xy).

The place where it gets weird is when you want to define a^x for irrational x. Think about this for a minute -- what the heck does 2^(pi) *mean*? I have no idea what it should mean, but we'd really like a^x to be continuous, so we define a^(irrational) so that this is the case. (This is actually non-trivial, and relies on the fact that the rational numbers are dense in the reals.)

Sorry to go off like that, but I sort of feel like taking an oversimplified view of this actually hides some of the coolness and complexity of it all.

Posted by: Davis | July 28, 2006 at 05:42 PM

I like the pattern idea, but if you have to fill a value in for it, I'd say take the result of the limit. But Davis (above) points out, that really doesn't make sense.

Using 1 is really clean, but math isn't always that way. :)

So I'd vote for for no answer.

Posted by: Trey Jackson | July 28, 2006 at 05:59 PM

Davis, much of the stuff you have written I have also written for my article on the subject which is not published yet, which is quite funny :-)

My point is that "coolness and complexity" comes into play when there's no way you get do it otherwise. Like, there's no way you can define 2^pi without it. But 0^0 is not the case, cause it's just a product of no numbers. It's one. Logic and "contradictionless". Why not?

You use continuity, when there's no other way, but when there is one, and it's more intuitive and straightforward, why bother?

Posted by: Ilya Birman | July 28, 2006 at 06:26 PM

"...it's just a product of no numbers."

The problem is that this has no rigorous meaning -- again, saying x^0=1 is a definition, and not something that is a priori obvious. Thus when you say 0^0=1, you're making a definition, rather than drawing a conclusion. I definitely disagree that this is intuitive or straightforward (otherwise, students would not have such a difficult time remembering that x^0=1).

Ultimately, there are two reasons why I think this should be left undefined. The first is that it forces exceptions: it's common to take a limit of f(x) as x->a, try to compute this as f(a), and get 0^0. However, in this context you *must* consider this indeterminate; if 0^0 has a definition, this means adding an exception to the typical method used for computing limits.

The second is that it's not especially useful. The only places I have seen 0^0 arise are in limits as above, and in some combinatorial formulas. So if anything, having a definition for 0^0 makes things more complicated, and not simpler.

Posted by: Davis | July 29, 2006 at 01:39 PM

I'm disappointed by the lack of vitriol in this conversation. Isn't anybody ready to call anybody else stupid?

Posted by: Archie | July 31, 2006 at 10:56 AM

Archie! You're just plain STOOPID!

0^0 = exp(log(0^0)) = exp(0*log(0)) = x

If x=1, then doesn't that imply 0*log(0) = 0?

Maybe saying y = exp(log(y)) is not valid when y = 1? (kinda like division by 0?)

Posted by: Twinkle | July 31, 2006 at 11:37 AM

(BTW, I think, actually, that 0^0 = 0.999...)

Posted by: Twinkle | July 31, 2006 at 11:54 AM

Twinkle -- cood catch on that inconsistency. Essentially you've connected the 0^0 indeterminacy to the 0 times infinity indeterminacy (the infinity comes from log(0)). Which means it can also be connected to 0/0, and probably 1^(infinity) and all the other indeterminate forms.

Posted by: Davis | July 31, 2006 at 03:56 PM

see http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

Posted by: bd | August 03, 2006 at 03:44 PM

The Wikipedia article on this is actually quite good. :)

http://en.wikipedia.org/wiki/Empty_product#Technical_justification

Posted by: Porges | August 16, 2006 at 02:55 AM

Agreed; after reading the Wikipedia entry I'm close to convinced that we should define 0^0 to be 1, except when it's not.

Posted by: Davis | August 20, 2006 at 03:29 AM

2Davis.

As x approaches zero, y=0^x indeed approaches zero also, which "contradicts" with 0^0=1 in some way. However, no one said 0^x must be continuous. Now, even if you have 0^0="undefined" you still have a bug in y=0^x at x=0! It's not continuous at x=0 anyway, so you solve no single problem by this.

However, 0^x is the only problematic thing regarding 0^0 that I know of (maybe you show me something else?). And as you can see in Wikipedia article for example, there are many cases where 0^0 = 1 is useful.

So I see no reason "defining" 0^0 as "undefined". It does not give any advantages.

Oh, and really, I don't understand how people don't get that x^0 = 1, but DO get that x*0 = 0 :-)

Posted by: Ilya Birman | August 25, 2006 at 12:33 PM

Wikipedia's proof using the power series won my vote, but I don't think that means I won't be teaching 0^0 as an indeterminate form in calculus anymore. :p

Posted by: Angelsnite | October 07, 2006 at 06:50 PM