Can anybody explain how this comes about?
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22It's simple: don't trust machines blindly!...and neither people, btw. – DonAntonio Apr 05 '13 at 20:21
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7$0^0$ is $1$. It is an indeterminate form, but it has a value. Not sure why WA lists its digits that way, however. – Thomas Andrews Apr 05 '13 at 20:21
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6Probably a leftover joke of some kind... – Zhen Lin Apr 05 '13 at 20:26
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4@ThomasAndrews, I'm not sure all would accept $,0^0=1,$ and even less under all circumstances. For example there's a continuity argument: $$x^x=e^{x\log x}\xrightarrow[x\to 0^+]{}1$$ but it doesn't work nicely for negative values... – DonAntonio Apr 05 '13 at 20:33
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1It is indeterminate in the usual sense, and that can be proved, and moreover it is equal to $1$, and that can also be proved. See my answer below. – Michael Hardy Apr 05 '13 at 21:23
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1Wolfram|$\alpha$ is not to be trusted. It frequently gives answers that are completely insane. Here is a fourth example. – MJD Apr 05 '13 at 21:32
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@DonAntonio I am a clear advocate of the fact that $a^b$ is defined for $a=b=0$, whereas $\exp(b\ln(a))$ is not and that writing $a^b$ with non-integer $b$ is a matter (or sometimes abuse) of notation. – Hagen von Eitzen Apr 05 '13 at 21:44
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1Well, it's a matter of agreement, I believe. I don't think "it can be proved" that $,0^0=1,$ anymore that "can be proved" that $,0!=1,$: it's just a matter of convenience to make some things work nicely. Both the analytic explanation in the power series for the exponential (Michael's answer) and the combinatoric one (Dominic's answer) are pretty compelling for me, but just as well someone else can come up with an argument against. Agreement, ladies and gentlemen: plain, good'ol agreement about something that works nice...or not. – DonAntonio Apr 05 '13 at 23:39
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@MJD Of course, no software should be trusted blindly - we should always check results. Looking through the links you mention, I don't see any serious problems with the results currently given by WolframAlpha. Perhaps those were just bugs (like the "0^0" result clearly is) which have been fixed (which the "0^0" result certainly will be. – Mark McClure Apr 06 '13 at 16:17
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1@MJD In fact, it's fixed already. :) – Mark McClure Apr 12 '13 at 14:07
4 Answers
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$0^0$ is "indeterminate" in the sense that if $f(x),g(x)\to0$ as $x\to\text{something}$, then $f(x)^{g(x)}$ could be any positive number or $0$ or $\infty$, depending on which functions $f$ and $g$ are. But the limit is $1$ if $f$, $g$ are both analytic [Apparently I'm missing a hypothesis here . . . ], and it's $1$ if $(f(x),g(x))$ approaches $(0,0)$ from within a sector bounded by two lines of positive slope.
But also, $0^0=1$. This is seen in things like $$ e^z = \sum_{n=0}^\infty \frac{z^n}{n!}, $$ where the first term is $\dfrac{z^0}{0!}$, and that term will fail to be $1$ when $z=0$ unless $0^0$ is $1$. That $0^0=1$ is a fact arising in combinatorics, set theory, and probability from the fact that $0^0$ is an empty product, i.e. a product of no numbers at all; hence is equal to $1$ since not multiplying by anything is the same as multiplying by $1$.
(But I don't know why Wolfram Alpha expressed it in that odd form.)
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+1 Thanks! I put the top arrow in there inadvertently, and, as a fortunate consequence, learned a few things. However, the question is really about the $...0000000001$ bit. :) – Řídící Apr 05 '13 at 21:35
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"But the limit is $1$ if $f,g$ are both analytic..." What about $f(x)=0$, $g(x)=x$? I'm not sure what the missing condition is. – Owen Biesel Apr 12 '13 at 14:33
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Well in set theory $0^0$ is defined and not indeterminate. $0^0$ is the number of functions from the empty set to the empty set. There is exactly 1 function from the empty set to the empty set.
In Analysis $0^0$ is often not not defined as there are limits of the form $0^0$ which are indeterminate. Sometimes $0^0$ is defined to be $1$ in analysis as for example $$\lim_{x\to 0} x^x =1$$
In my personal opinion I guess it is a joke and that last few digits and stands for $$00000000\dots 00001$$
The definition $0^0$ is very convenient as mentioned in comments, it avoids a lot of special cases.
It's a bit funny that Wolfram Alpha doesn't reproduce this one here
Dominic Michaelis
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it is the distinction between being an "indeterminate form" and being "undefined." $0^0=1$ is defined, but if $x,y$ are only known to be close to zero, we cannot say anything about $x^y$. – Thomas Andrews Apr 05 '13 at 20:24
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2Also, in analysis, we actually use $0^0=1$ implicitly a lot, when we evaluate at power series $\sum_{i=0}^\infty a_ix^i$ at $x=0$ to get $a_0$. In this we are dealing with the function $x^0$ where the $0$ is known to be an integer, not an approximation of $0$. :) – Thomas Andrews Apr 05 '13 at 20:26
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1@ThomasAndrews i said it is often not defined in analysis, but some define $0^0$ to be 1 but I know some who don't define it. – Dominic Michaelis Apr 05 '13 at 20:28
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2Yeah, my point was, even when we don't define it, we often use it implicitly in this particular case (evaluation of power series.) – Thomas Andrews Apr 05 '13 at 20:32
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Thanks to both of you for clearing up the indeterminate bit. But where does the $...0000000001$ bit come from? – Řídící Apr 05 '13 at 20:32
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1That's just a weird WA bug (or a joke) @Gugg. The question about $0^0$ comes up a lot, so it could easily be a joke, although not particularly funny. – Thomas Andrews Apr 05 '13 at 20:32
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1In combinatorics it needs to be 1 to avoid adding special cases to things like the binomial theorem. Concrete Mathematics even has an index entry for $0^0$ pointing to the page which discusses this. – Peter Taylor Apr 05 '13 at 20:44
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yeah I think the convention or definition or however you want to call it $0^0=1$ is very convenient and i am using it personally – Dominic Michaelis Apr 05 '13 at 20:47
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+1 I learned something. But I'm not convinced (yet) about the joke part. Note that the $...0000000001$ is the main question (being in the title). – Řídící Apr 05 '13 at 20:58
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@Gugg or this one well but i think it is an uncommon format of the jokes in wolfram alpha – Dominic Michaelis Apr 05 '13 at 21:11
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@DominicMichaelis I tried $42$ and I didn't get a joke back. I'm sorry, I can't accept the joke-theory for $0^0$. :) – Řídící Apr 05 '13 at 21:18
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Could you give some reference on "$0^0$ is the number of functions from the empty set to the empty set"? I'm not familiar with this. Thanks! – qazwsx Apr 06 '13 at 19:13
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The thing about "analysis" is ... they may evaluate $0^0$ as $1$ when the exponent is the integer $0$. But consider $0.^{0.}$ as undefined (and indeterminate) when the exponent is the real number zero. – GEdgar Apr 12 '13 at 14:24
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Wolfram Alpha calculates "last few digits" if you input large numbers such as 12^12^12, but not if you input a small number such as 2^2^2. I think it simply preemptively mistakes 0^0 for a large number, and calculates its last few digits for you, and since 0^0 is defined as 1, you simply get ...000001.
Wolfram Alpha gets really confused if you enter 0^0^0, where it seems to evaluate at least one instance of 0^0 as the limit Indeterminate rather than the number 1, and it replies with the raw code ChineseRemainder[{Indeterminate, Indeterminate}, {0000001024, 0009765625}].
Samuel
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+1 Not too convinced, as of yet, of the anthropomorphic explanation though. But perhaps I'm underestimating WA. :) – Řídící Apr 05 '13 at 22:54
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1@Gugg: I could imagine what's happening behind the scenes is: before calculating anything, WA first analyzes the input to determine what it should calculate, and it recognizes that it is $a^b$ for some $a,b$, which triggers the calculation of the "last few digits" of the number, in case it happens to be large. Then it calculates the number $a^b$, and if the result is an integer with less than 100 digits or so, WA would just print all of them and throw away the "last few digits" information as redundant, but since Indeterminate was returned, this is not triggered in this case. – Samuel Apr 05 '13 at 23:58
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This was clearly a bug. It's been fixed, as have the higher order towers.
Mark McClure
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@GEdgar Ha! You'll just have to try. Fortunately, there's a character limit on the input, so it can't go forever. :) – Mark McClure Apr 12 '13 at 14:29