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What is "indeterminate form"? http://en.wikipedia.org/wiki/Indeterminate_form http://en.wikipedia.org/wiki/Indeterminate_form You see, 0^0 = 1, and it's obvio

by Dove 15y ago

What is "indeterminate form"? http://en.wikipedia.org/wiki/Indeterminate_form http://en.wikipedia.org/wiki/Indeterminate_form You see, 0^0 = 1, and it's obvious to a mathematician . . . we define 0^0 = 1, to be consistent with exponentiation rules Well, you're going to be inconsistent with them no matter how you define it, since, as you point out, x^y should be zero if you approach (0,0) along the x=0 axis, and it should be one if you approach along the y=0 axis. 0^0 is simply an expression that doesn't make sense. There isn't an answer, and there certainly isn't something we could agree to define it as. It is gibberish, nothing more, nothing less. One cannot assume just because there are mathematical symbols on paper that they make sense.

xyzzyz 15y ago
By "exponentiation rules" I mean algebraic equalities, like a^x * a^y = a^(x+y). Most of them work no matter if you define 0^0 = 1 or 0, but some of them are cleaner with 0^0 = 1. It's also consistent with cardinal and ordinal exponentiation (look it up). "Approaching along x axis" is not algebraic notion, it's analytic one. 0^0 makes no less sense than, say, -e^(i pi). They're both 1 because we define them like this. If you think that -e^(i pi) makes more sense than 0^0, please, explain me why. Also, mathematicians agree in this, seriously. Go and ask one.
- lotharbot 15y ago
  I haven't seen any reason why -e^(i pi)=1 could be considered incorrect. It's consistent with the Taylor Series expansion of e^x. It's consistent with the view of complex exponentiation as rotation. I don't know of any particular problems that arise from taking -e^(i pi)=1. This is not the case with 0^0=1, which is inconsistent with many limits. That's why 0^0=1 is an agreed-upon convention sometimes. http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero... has a fairly nice summary of the issues involved in defining it.
  - xyzzyz 15y ago
    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^(i pi) = -1 is wrong. I agree that this is ridiculous requirement, but leaving 0^0 undefined because the math is not as we want it to be (e.g. exponentation is not continuous) looks just as ridiculous and silly to me. On the other hand, putting 0^0 = 1 makes it consistent many combinatoric formulas, and is also consistent with cardinal exponentation, where nobody objects to 0^0 = 1, when you look at 0 as the cardinal number.
    - lotharbot 15y ago
      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.
      - xyzzyz 15y ago
        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.
- JadeNB 15y ago
  > Also, mathematicians agree in this, seriously. Go and ask one. Mathematician here; we do not. See http://math.stackexchange.com/questions/11150/zero-to-zero-power http://math.stackexchange.com/questions/11150/zero-to-zero-p.... More precisely, as Arturo Magidin points out at http://math.stackexchange.com/questions/11150/zero-to-zero-power/11155 http://math.stackexchange.com/questions/11150/zero-to-zero-p..., if we view exponentiation in the 'discrete setting', then $0^0$ must be $1$; whereas, if we view it in the continuous setting, there is simply no good answer—unlike $e^{i\pi}$, which also lives in the continuous setting, but has a perfectly good, unambiguous answer. (lotharbot gives a nice explanation below of the ways that this is consistent with existing mathematics; but it can also be derived from the definition of the exponential function, with no further arbitrary conventions needed.)
  - xyzzyz 15y ago
    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 with concepts like limit or continuity, but it's over now. Mathematician here; we do not. It seems I was a little too bold with my claim. All the mathematicians I know (and I'm a mathematician as well) agree with 0^0 = 1. It's a folklore specific thing, I guess.
    - Dove 15y ago
      I'm a mathematician as well As am I, by training if not by profession. As is lotharbot. You're in a thread full of mathematicians. :) Which is what I would expect on this site, actually. I'm always timid making technical claims here unless I'm sure I'm correct; it seems to be a place frequented by arbitrarily large fish. To answer your question, 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. I'm not persuaded it is always the answer. I think the fact that it is an indeterminate form in limits is a forceful enough demonstration of that. It all depends on context. If I came across a 0^0 in, say, an engineering context, my first instinct would be to check whether the formula was defined in that case, not to just assume that 1 would work. I mean, it's like 1/0. If you're working in R, that's simply illegal. If you're working in R*, it's the infinite point. If you're taking a limit, it means "unbounded". If you're working in my favorite field, the hyperreals, it could be any number of flavors of infinity depending on the flavor of zero it was. It would be foolhardy to try to define the symbol; without a context to supply some sort of sense, it is nonsense. And that is how I feel about 0^0 as well.
- maththrowaway 15y ago
  > Also, mathematicians agree in this, seriously. Go and ask one. As an erstwhile mathematician, I don't know of anyone who would suggest that -e^(i pi) is defined as 1. It falls out, rather elegantly, from the definitions of the constants involved.
rfurmani 15y ago
> x^y should be zero if you approach (0,0) along the x=0 axis, and it should be one if you approach along the y=0 axis. Technically, 0^y is 0 only if you approach it from the right: y>0. To the left of 0 it is indeterminate or infinity depending on how you look at it. x^0, however, makes sense for all x and is always 1

by Dove 15y ago

xyzzyz 15y ago

By "exponentiation rules" I mean algebraic equalities, like a^x * a^y = a^(x+y). Most of them work no matter if you define 0^0 = 1 or 0, but some of them are cleaner with 0^0 = 1. It's also consistent with cardinal and ordinal exponentiation (look it up). "Approaching along x axis" is not algebraic notion, it's analytic one. 0^0 makes no less sense than, say, -e^(i pi). They're both 1 because we define them like this. If you think that -e^(i pi) makes more sense than 0^0, please, explain me why. Also, mathematicians agree in this, seriously. Go and ask one.

lotharbot 15y ago
I haven't seen any reason why -e^(i pi)=1 could be considered incorrect. It's consistent with the Taylor Series expansion of e^x. It's consistent with the view of complex exponentiation as rotation. I don't know of any particular problems that arise from taking -e^(i pi)=1. This is not the case with 0^0=1, which is inconsistent with many limits. That's why 0^0=1 is an agreed-upon convention sometimes. http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero... has a fairly nice summary of the issues involved in defining it.
- xyzzyz 15y ago
  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^(i pi) = -1 is wrong. I agree that this is ridiculous requirement, but leaving 0^0 undefined because the math is not as we want it to be (e.g. exponentation is not continuous) looks just as ridiculous and silly to me. On the other hand, putting 0^0 = 1 makes it consistent many combinatoric formulas, and is also consistent with cardinal exponentation, where nobody objects to 0^0 = 1, when you look at 0 as the cardinal number.
  - lotharbot 15y ago
    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.
    - xyzzyz 15y ago
      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.
JadeNB 15y ago
> Also, mathematicians agree in this, seriously. Go and ask one. Mathematician here; we do not. See http://math.stackexchange.com/questions/11150/zero-to-zero-power http://math.stackexchange.com/questions/11150/zero-to-zero-p.... More precisely, as Arturo Magidin points out at http://math.stackexchange.com/questions/11150/zero-to-zero-power/11155 http://math.stackexchange.com/questions/11150/zero-to-zero-p..., if we view exponentiation in the 'discrete setting', then $0^0$ must be $1$; whereas, if we view it in the continuous setting, there is simply no good answer—unlike $e^{i\pi}$, which also lives in the continuous setting, but has a perfectly good, unambiguous answer. (lotharbot gives a nice explanation below of the ways that this is consistent with existing mathematics; but it can also be derived from the definition of the exponential function, with no further arbitrary conventions needed.)
- xyzzyz 15y ago
  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 with concepts like limit or continuity, but it's over now. Mathematician here; we do not. It seems I was a little too bold with my claim. All the mathematicians I know (and I'm a mathematician as well) agree with 0^0 = 1. It's a folklore specific thing, I guess.
  - Dove 15y ago
    I'm a mathematician as well As am I, by training if not by profession. As is lotharbot. You're in a thread full of mathematicians. :) Which is what I would expect on this site, actually. I'm always timid making technical claims here unless I'm sure I'm correct; it seems to be a place frequented by arbitrarily large fish. To answer your question, 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. I'm not persuaded it is always the answer. I think the fact that it is an indeterminate form in limits is a forceful enough demonstration of that. It all depends on context. If I came across a 0^0 in, say, an engineering context, my first instinct would be to check whether the formula was defined in that case, not to just assume that 1 would work. I mean, it's like 1/0. If you're working in R, that's simply illegal. If you're working in R*, it's the infinite point. If you're taking a limit, it means "unbounded". If you're working in my favorite field, the hyperreals, it could be any number of flavors of infinity depending on the flavor of zero it was. It would be foolhardy to try to define the symbol; without a context to supply some sort of sense, it is nonsense. And that is how I feel about 0^0 as well.
maththrowaway 15y ago
> Also, mathematicians agree in this, seriously. Go and ask one. As an erstwhile mathematician, I don't know of anyone who would suggest that -e^(i pi) is defined as 1. It falls out, rather elegantly, from the definitions of the constants involved.

rfurmani 15y ago

> x^y should be zero if you approach (0,0) along the x=0 axis, and it should be one if you approach along the y=0 axis. Technically, 0^y is 0 only if you approach it from the right: y>0. To the left of 0 it is indeterminate or infinity depending on how you look at it. x^0, however, makes sense for all x and is always 1