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Re:What does 0^0 equal? Why do mathematicians and high schoo... by lotharbot on Wednesday August 24, 2011 @09:40PM (#2923425)
It has nothing to do with wanting exponentiation to be continuous. It's simply a recognition that limits of the form 0^0 are indeterminate, which means a convention that 0^0=1 is not appropriate in the context of evaluating limits. This isn't an argument that it "seems" incorrect, like your bizarre argument about real vs complex exponentiation; it's an argument that it IS incorrect in that context. If you're evaluating limits, you have to treat 0^0 as indeterminate, not as 1.

Even the original article noted that we don't choose the 0^0 convention because it's "correct", but because it's "nice" -- which is why we define it that way in the contexts where it makes sense to define it that way. If you're working with combinatorics, 0^0=1. If you're working with cardinal exponentiation, 0^0=1. If you're taking limits, or working in the hyperreals, or in certain other contexts, the convention doesn't apply. In some circumstances, 0^0 isn't even a valid statement -- like if you're working directly with the field axioms of R.

Recognize what context you're working in, and what assumptions or conventions apply in that context. That's just good mathematics.

Re:What does 0^0 equal? Why do mathematicians and high schoo... by xyzzyz on Wednesday August 24, 2011 @10:41PM (#2923535)
It's simply a recognition that limits of the form 0^0 are indeterminate, which means a convention that 0^0=1 is not appropriate in the context of evaluating limits.

But the whole point of distinguishing some "forms" as "indeterminate" is to work around the discontinuity of elementary functions! My absolutely first sentence in this thread is asking, what exactly the "indeterminate form" is. I'm asking this question, because this not a formal notion and you will not find any formal definition of it. Its existence is rooted in the fact that for no reason other than the tradition (and convenience) we use special notation for some functions. Instead of +: R x R -> R, +(2, 3) = 5, we write 2 + 3 = 5. The same goes for ^: [0, \infty] x R -> R. The only reason we have all those fancy limit evaluating laws is because these function are continuous most of the time. For instance, + is continuous everywhere, so lim +(a_n, b_n) = +(lim a_n, lim b_n), if both lim a_n and lim b_n make sense. Similarly, /: R x R - {0} -> R is also continuous everywhere, so lim /(a_n, b_n) = /(lim a_n, lim b_n), if the right hand expression makes sense. If it does not make sense, for instance when both lim a_n and lim b_n are equal to zero, we need cannot approach this problem in such a simple way. Now, some people would call /(0, 0) an "indeterminate form", which makes for me no more sense than calling f(0, 0) an indeterminate form, where f(x, y) = log_(1/x) (y) -- while f is continuous everywhere where defined, you cannot extend its domain to contain (0, 0) for it to stay continuous, just like you cannot do it with / function.

As I repeated many times, the whole affair is because ^ seem to be more familiar than beta function (we have a special notation for it, for instance), people want it to behave nicely, so that for instance it conforms to some arbitrary limit evaluating laws, missing the whole underlying concept of continuity.

Re:What does 0^0 equal? Why do mathematicians and high schoo... by lotharbot on Wednesday August 24, 2011 @11:39PM (#2923681)
The whole point of distinguishing some forms as "indeterminate" is to work around the fact that you're trying to conduct operations on the real numbers that are not defined under the field axioms of the real numbers (or the axioms of the extended reals [-inf,inf]). That's where its essence is rooted; that's why this whole affair exists -- the fact that 0/0, 0^0, 0xinf, inf-inf, etc. are not well defined by our axioms.

This actually relates to all three examples I've presented where the 0^0=1 convention fails. It should be treated as an indeterminate form in limits because it's not well-defined by the axioms of the real numbers; it's also not well-defined by the axioms of the hyperreals, but division of infinitesimals is well-defined in the hyperreals, which gives us an alternate method of computing limits that avoids the "indeterminate form" entirely.

Let me reiterate: 0^0 is not defined under the field axioms of the real numbers. The choice to define it as 1 is a convention which makes certain math easier, in certain areas of mathematics. It is by no means a universal convention; it is by no means the one and only correct definition of 0^0. You continue to argue for the convention, but miss the larger point that it is a convention which is chosen for convenience, and which is not always appropriate.

Re:What does 0^0 equal? Why do mathematicians and high schoo... by xyzzyz on Thursday August 25, 2011 @12:34AM (#2923775)
Exponential function is not defined by the axioms of real numbers, as opposed to addition and multiplication, so this point is irrelevant -- you can define it in any way you want. There's no inconvenience in defining 0^0 = 1, apart from misunderstanding the concept of limits by some people. Defining 0^0 = 1 is universal convention -- people either do it like this, or do not define 0^0 at all, which I'm fighting against.

The whole point of distinguishing some forms as "indeterminate" is to work around the fact that you're trying to conduct operations on the real numbers that are not defined

I am not. Are you? Let me reiterate: the whole concept of "indeterminate forms" (which, I repeat, is not formal at all) stems from misunderstanding the process of taking limits.

Re:What does 0^0 equal? Why do mathematicians and high schoo... by lotharbot on Thursday August 25, 2011 @12:49AM (#2923799)
> "people either do it like this, or do not define 0^0 at all, which I'm fighting against."

I think it's silly to fight against it. There are circumstances in which leaving it undefined is good, and in which trying to define it as 1 would lead to either misunderstandings (in the case of beginners doing limits, a case you are too quick to dismiss) or actually incorrect (an equivalent problem in the hyperreals could violate the transfer principle).

It's a broad convention, but it is not universal, and it shouldn't be.