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What does 0^0 equal? Why do mathematicians and high school teachers disagree?

askamathematician.com

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Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#101
post #84

Earlier quoted context omitted.

This is not the case with 0^0=1, which is inconsistent with many limits. So what? It's only a problem if you want the exponentation function to be continuous, so you escape the problem by leaving it undefined. You could place similar unbased requirements on complex exponentiation to make it seem incorrect. For instance, real exponentiation always gives a positive value for positive base, while complex does not, so e^…

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 evalu…

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 school teachers disagree?

#102
post #36

Earlier quoted context omitted.

What is "indeterminate form"? What does it mean for expression to "have a specific solution"? You see, 0^0 = 1, and it's obvious to a mathematician. The only problem is that the function f: [0, \infty) x R -> R, f(x, y) = x^y is discontinuous in (0, 0) and that's what causes problems -- for instance, this is the source of the whole "indeterminate form" notion. If a function f is continuous in (a, b), then for every t…

Mathematicians don't argue about what an expression "really is" (or at least, real mathematics doesn't involve this). They define functions and use axioms to prove theories about them. "No really" . Mathematics just isn't concerned with this stuff. Sometimes infinity it defined as single point making the real number compact, sometimes a "positive infinity" and a "negative infinity" are defined. Sometimes you add poin…

Anthropomorphizing math as having a "concern" is also fuzzy thinking.

>If you want to know what something "really is", consult philosophy or something.

You mean like the "foundations of mathematics"? http://en.wikipedia.org/wiki/Foundations_of_mathematics

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#103
post #99
post #85

Earlier quoted context omitted.

0^-1 is definitely infinity x^(-1) doesn't have an upper bound as x approaches zero.

Try approaching 0 from both the positive and negative sides.

So you get a +∞ or -∞.

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#104

> How would you explain to a 10-year old why 3^0 = 1 You draw the line 3^x. It "passes through" 1 when x = 0. So don't think about the point, think about the line. It's not rigorous but it's intuitive. http://fooplot.com/index.php?q0=3^x edit: added link and fixed typos

This is precisely the problem, though.

The same 10 year old draws two lines: 0^x, and x^0.

They clearly do not meet.

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#105
post #36
post #11

Technically, 0^0 is an indeterminate form and has no specific solution. Accurate but unhelpful. Practically, 0^0 highlights the issue that most of us don't have a good conceptual model for what exponents really do. How would you explain to a 10-year old why 3^0 = 1 beyond "it's necessary to make the algebra of powers work out". I use an "expand-o-tron" analogy http://betterexplained.com/articles/understanding-exponen…

What is "indeterminate form"? What does it mean for expression to "have a specific solution"? You see, 0^0 = 1, and it's obvious to a mathematician. The only problem is that the function f: [0, \infty) x R -> R, f(x, y) = x^y is discontinuous in (0, 0) and that's what causes problems -- for instance, this is the source of the whole "indeterminate form" notion. If a function f is continuous in (a, b), then for every t…

Yes. Tying to define f(0) as 0 or 1 won't make it continuous, as approaching from the lines x = 0 and y = 0 will make the limits differ (the definition of lack of a limit).

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#106
post #86

Earlier quoted context omitted.

But my point is, if 0^0 = 1 is _the_ answer in discrete setting, and it's _an_ answer in continuous setting, why don't we just agree that 0^0 = 1 and stop creating confusing situation where sometimes it's defined and sometimes it's not. 0^0 = 1 does not make calculus theorems more complicated to state or prove with modern language. It was a problem in XIX century, when mathematicians did not have a solid foundation w…

> "why don't we just agree that 0^0 = 1" Because sometimes it's better not to. Sometimes it's inconsistent with our definitions. Just like sometimes we agree that you can't divide by zero, and sometimes we agree that you can. Sometimes infinity is an actual value (say, in the extended reals), and sometimes it's just a symbol for "unbounded". Sometimes we agree that you can't take the square root of a negative number,…

> Because sometimes it's better not to. Sometimes it's inconsistent with our definitions.

I'd love to see even one example of 0^0=1 being inconsistent with a definition. The closest I've ever seen is that it bothers people that for reasons of their own had their hearts set on (x,y) -> x^y having no discontinuities...

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#107
post #101

Earlier quoted context omitted.

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 evalu…

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, b…

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 school teachers disagree?

#108
post #101

Earlier quoted context omitted.

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, b…

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…

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 school teachers disagree?

#109
post #60
post #11

Technically, 0^0 is an indeterminate form and has no specific solution. Accurate but unhelpful. Practically, 0^0 highlights the issue that most of us don't have a good conceptual model for what exponents really do. How would you explain to a 10-year old why 3^0 = 1 beyond "it's necessary to make the algebra of powers work out". I use an "expand-o-tron" analogy http://betterexplained.com/articles/understanding-exponen…

>most of us don't have a good conceptual model for what exponents really do Instead of matching math to real world objects (1= one banana, 2 = two bananas, 1+2 = 3 bananas etc. ) and building up to exponentiation, multiplication etc. thereby introducing all sorts of paradoxes, group theory dodges all that and treats the whole thing as a very consistent rule-based system. Things fall into place quickly once the rules…

You're right, apart from the fact that real exponentiation does not really belong to group theory -- or even abstract algebra, for that matter.

Re: What does 0^0 equal? Why do mathematicians and high school teachers disagree?

#110
post #96

Earlier quoted context omitted.

Please, treat exponentiation just like every other function out there. I don't get the whole limit argument at all. Given any function f: R x R -> R, if it happens that a_n -> a, b_n -> b, but lim f(a_n, b_n) != f(a, b), people just say that f is not continuous in (a, b), and the case is over. However, if f happens to be exponentiation function, people instead argue that f should not be defined in (a, b), forgetting…

Consider this related case: if you evaluate a limit and you get 0/0, you recognize that you need to do more work to find the actual limit. It could be 1, -1, 0, infinite, etc. depending on how you reached it (say, sin(x)/x versus sin(x)/x^2). The issue is not the continuity of x/x; the issue is whether setting a convention for 0/0 would give you the right value for a limit. Since it doesn't always, we call it "indete…

Consider this related case: if you evaluate a limit and you get 0/0

What do you mean by "getting 0/0" in the process of evaluating limits?

The issue is not the continuity of x/x; the issue is whether setting a convention for 0/0 would give you the right value for a limit.

Please, tell me - what is the relation between lim f(a_n) and f(lim a_n) ?

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