6 ms·
> "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”, com
by 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 agoIngenious solution -- faced with contradicting evidence, and rather than debate the topic on its merits, just downvote the post and walk away.
- akater 12y agoI'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.
- lutusp 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). I guess that would explain why it's located in a litany of common student errors compiled by math educators, as well as the other reference I provided. But you know what? I'm not interested in posting to a thread that downvotes posts with a probability proportional to their accuracy and relevance. From: http://www.wolframalpha.com/input/?i=0%5E0 http://www.wolframalpha.com/input/?i=0%5E0 Result: 0^0 = indeterminate