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I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers

by Grue3 12y ago

I disagree that 0^0 is undefined. I would argue should be 1 and the function 0^x is not continuous at zero. The basic definition of exponentiation for integers is n^m is a product of m instances of n. Because the multiplicative identity is 1, a product of zero numbers is always 1. Therefore 0^0=1.

Xophmeister 12y ago
The limit of x^x does equal one when approached from the right (i.e., when x is positive, reducing to zero). From the left, it also approaches one, but is only continuous over the complex numbers. Anyway, while the limit approaches one from both sides, the function cannot be evaluated at x=0 and so is undefined.
- philh 12y ago
  Why are you considering 0^0 to be the limit of the function x^x, rather than (say) 0^x or (e^(-1/x))^x?
- Grue3 12y ago
  Just because the function has different limits at some point does not mean it is undefined at that point. Consider the signum function. It has limit -1 when approached from negative side, limit 1 when approached from positive side, but is actually defined as 0 at zero.
lutusp 12y ago
It's more of an open question than your reply would seem to indicate: http://mathforum.org/dr.math/faq/faq.0.to.0.power.html http://mathforum.org/dr.math/faq/faq.0.to.0.power.html It's less controversial when the terms are integers, but even there an argument can be made for indeterminacy. As the linked article says, "There is no one definition that always works well for 0^0" I can use 0^0 = 1 to prove that 1 = 2, but I'm sure you can anticipate the argument's form.
- akater 12y ago
  > "There is no one definition that always works well for 0^0" Yes there is. :-) 0^0 = 1. Actually, the only case for claiming it to be an “indeterminate”, comes from so-called “continuous exponents”. Which, arguably, something that never occurs in reality — only in exam sheets by lazy calculus teachers. Whenever you meet an algebraic equation with sum over 0 <= k <= n… and there's some m^n or m^k in it, it's always only true when 0^0 = 1. I don't claim I've seen them all but really, try to find a counterexample. What exactly was that argument for indeterminacy you were talking about? Those equations come from reasoning about meaningful entities, not chimeras of “x^x”, or “x^y”, or worse. Again, try to find, say, a physics paper with x^y in it. I haven't read much physics papers but I'm pretty sure you'll find precisely 0. As one mathematician I knew put it, “hard analysis is the primary source of all obscurantism in mathematics out there”. ;-)
  - lutusp 12y ago
    > What exactly was that argument for indeterminacy you were talking about? http://www.math.vanderbilt.edu/~schectex/commerrs/#Infinity http://www.math.vanderbilt.edu/~schectex/commerrs/#Infinity From the bookmark given above, search for the phrase "That reminds me of a related question that seems to bother many students", followed by a brief and instructive exposition on the indeterminacy of 0^0. Also http://mathforum.org/dr.math/faq/faq.0.to.0.power.html http://mathforum.org/dr.math/faq/faq.0.to.0.power.html, already given. It's nice to be so sure of oneself, but in this case it's misleading.
    - lutusp 12y ago
      Ingenious solution -- faced with contradicting evidence, and rather than debate the topic on its merits, just downvote the post and walk away.
    - akater 12y ago
      I'm sorry but your links lead to the same continuity example that has nothing to do with real world (and real mathematics, as well). Can you prove (0^0 = 1) -> (1 = 2), as you claimed above? And no, we don't need “temporary definitions”. For any combinatorial investigation imaginable leads unambiguously to 0^0 = 1: https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zero https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer... The only concept a general expression of the form x^y could possibly represent is the space of mappings y -> x. There's one and only one such mapping when x and y represent empty set. Arguing this to be wrong as in “but it can't be 1 because there are no mappings to empty space from non-empty sets!” ( = “0^x = 0”) is plain ridiculous. This definition simply does not fall apart. 0^0 is not a special case for it in any way. Enumerating numbers one can represent with 0 digits put in a string of length 0 leads to the same conclusion. It's clear that there is one and only one string of length 0, unless you demand it to contain more than 0 digits, in which case there are none. [However, this combinatorial problem is not an independent one: it's equivalent to enumerating mappings from space of strings to space of chars.] In other words, not only there are definitions that work well for all cases, including 0^0, there's actually only one such definition.
leni536 12y ago
The main argument that it shouldn't be defined because there will be some situations where any definition will fall short. Your definition only works for situations where you are dealing with integers.
titter 12y ago
The debate goes beyond that point, as illustrated by the lengthy Wikipedia section: https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zero https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer... From an undergraduate math point of view, it's wise to understand that 0^0 is not well defined.
- Grue3 12y ago
  This argument assumes that the function f(x,y)=x^y is continuous near 0 and therefore lim (f(x)^g(x)) = (lim f(x))^(lim g(x)) at zero. But since it's not continous, that formula does not necessarily hold true.
hansy 12y ago
I posted an article about 0^0 to HN sometime back: http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/ http://www.askamathematician.com/2010/12/q-what-does-00-zero... HN discussion: https://news.ycombinator.com/item?id=7519827 https://news.ycombinator.com/item?id=7519827
RiderOfGiraffes 12y ago
It goes deeper than that. Many things in mathematics are defined in a manner that is consistent and convenient. Defining 0! to be 1 is a similar case. In doing so nothing goes wrong, and the binomial theorem becomes simple and convenient to state. Without defining 0! as 1, it's a dreadful mish-mash of special cases. Similarly with 0^0. Considering x^y where x and y are complex numbers, there is no consistent single value as x and y each approach 0. So in the reals and the complex numbers we leave 0^0 as undefined. However, in the case of natural numbers there is a case for declaring 0^0 to be 1. That's to make it convenient to talk about the set A^B as the collection of functions from B to A. When we do that we get |A^B| = |A|^|B|. There is more than expected in mathematics that's defined for convenience.
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- darkxanthos 12y ago
  Just completing Real Analysis and an interesting addition to your point is the decision to even call zero a natural number. It's really opened my eyes to just how much of the definitions around the math I group up with is due to convenience more than anything.

by Grue3 12y ago

Xophmeister 12y ago

The limit of x^x does equal one when approached from the right (i.e., when x is positive, reducing to zero). From the left, it also approaches one, but is only continuous over the complex numbers. Anyway, while the limit approaches one from both sides, the function cannot be evaluated at x=0 and so is undefined.

philh 12y ago
Why are you considering 0^0 to be the limit of the function x^x, rather than (say) 0^x or (e^(-1/x))^x?
Grue3 12y ago
Just because the function has different limits at some point does not mean it is undefined at that point. Consider the signum function. It has limit -1 when approached from negative side, limit 1 when approached from positive side, but is actually defined as 0 at zero.

lutusp 12y ago

It's more of an open question than your reply would seem to indicate: http://mathforum.org/dr.math/faq/faq.0.to.0.power.html http://mathforum.org/dr.math/faq/faq.0.to.0.power.html It's less controversial when the terms are integers, but even there an argument can be made for indeterminacy. As the linked article says, "There is no one definition that always works well for 0^0" I can use 0^0 = 1 to prove that 1 = 2, but I'm sure you can anticipate the argument's form.

akater 12y ago
> "There is no one definition that always works well for 0^0" Yes there is. :-) 0^0 = 1. Actually, the only case for claiming it to be an “indeterminate”, comes from so-called “continuous exponents”. Which, arguably, something that never occurs in reality — only in exam sheets by lazy calculus teachers. Whenever you meet an algebraic equation with sum over 0 <= k <= n… and there's some m^n or m^k in it, it's always only true when 0^0 = 1. I don't claim I've seen them all but really, try to find a counterexample. What exactly was that argument for indeterminacy you were talking about? Those equations come from reasoning about meaningful entities, not chimeras of “x^x”, or “x^y”, or worse. Again, try to find, say, a physics paper with x^y in it. I haven't read much physics papers but I'm pretty sure you'll find precisely 0. As one mathematician I knew put it, “hard analysis is the primary source of all obscurantism in mathematics out there”. ;-)
- lutusp 12y ago
  > What exactly was that argument for indeterminacy you were talking about? http://www.math.vanderbilt.edu/~schectex/commerrs/#Infinity http://www.math.vanderbilt.edu/~schectex/commerrs/#Infinity From the bookmark given above, search for the phrase "That reminds me of a related question that seems to bother many students", followed by a brief and instructive exposition on the indeterminacy of 0^0. Also http://mathforum.org/dr.math/faq/faq.0.to.0.power.html http://mathforum.org/dr.math/faq/faq.0.to.0.power.html, already given. It's nice to be so sure of oneself, but in this case it's misleading.
  - lutusp 12y ago
    Ingenious solution -- faced with contradicting evidence, and rather than debate the topic on its merits, just downvote the post and walk away.
  - akater 12y ago
    I'm sorry but your links lead to the same continuity example that has nothing to do with real world (and real mathematics, as well). Can you prove (0^0 = 1) -> (1 = 2), as you claimed above? And no, we don't need “temporary definitions”. For any combinatorial investigation imaginable leads unambiguously to 0^0 = 1: https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zero https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer... The only concept a general expression of the form x^y could possibly represent is the space of mappings y -> x. There's one and only one such mapping when x and y represent empty set. Arguing this to be wrong as in “but it can't be 1 because there are no mappings to empty space from non-empty sets!” ( = “0^x = 0”) is plain ridiculous. This definition simply does not fall apart. 0^0 is not a special case for it in any way. Enumerating numbers one can represent with 0 digits put in a string of length 0 leads to the same conclusion. It's clear that there is one and only one string of length 0, unless you demand it to contain more than 0 digits, in which case there are none. [However, this combinatorial problem is not an independent one: it's equivalent to enumerating mappings from space of strings to space of chars.] In other words, not only there are definitions that work well for all cases, including 0^0, there's actually only one such definition.

leni536 12y ago

The main argument that it shouldn't be defined because there will be some situations where any definition will fall short. Your definition only works for situations where you are dealing with integers.

titter 12y ago

The debate goes beyond that point, as illustrated by the lengthy Wikipedia section: https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zero https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer... From an undergraduate math point of view, it's wise to understand that 0^0 is not well defined.

Grue3 12y ago
This argument assumes that the function f(x,y)=x^y is continuous near 0 and therefore lim (f(x)^g(x)) = (lim f(x))^(lim g(x)) at zero. But since it's not continous, that formula does not necessarily hold true.

hansy 12y ago

I posted an article about 0^0 to HN sometime back: http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/ http://www.askamathematician.com/2010/12/q-what-does-00-zero... HN discussion: https://news.ycombinator.com/item?id=7519827 https://news.ycombinator.com/item?id=7519827

RiderOfGiraffes 12y ago

It goes deeper than that. Many things in mathematics are defined in a manner that is consistent and convenient. Defining 0! to be 1 is a similar case. In doing so nothing goes wrong, and the binomial theorem becomes simple and convenient to state. Without defining 0! as 1, it's a dreadful mish-mash of special cases. Similarly with 0^0. Considering x^y where x and y are complex numbers, there is no consistent single value as x and y each approach 0. So in the reals and the complex numbers we leave 0^0 as undefined. However, in the case of natural numbers there is a case for declaring 0^0 to be 1. That's to make it convenient to talk about the set A^B as the collection of functions from B to A. When we do that we get |A^B| = |A|^|B|. There is more than expected in mathematics that's defined for convenience.

deleted 12y ago
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- deleted 12y ago
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darkxanthos 12y ago
Just completing Real Analysis and an interesting addition to your point is the decision to even call zero a natural number. It's really opened my eyes to just how much of the definitions around the math I group up with is due to convenience more than anything.