Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anybody explain how this comes about?

enter image description here

share|cite|improve this question
It's simple: don't trust machines blindly!...and neither people, btw. – DonAntonio Apr 5 '13 at 20:21
$0^0$ 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 5 '13 at 20:21
Probably a leftover joke of some kind... – Zhen Lin Apr 5 '13 at 20:26
@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 5 '13 at 20:33
It 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 5 '13 at 21:23
up vote 4 down vote accepted

This was clearly a bug. It's been fixed, as have the higher order towers.

share|cite|improve this answer
And when they fix '0^0^0', will the but migrate to '0^0^0^0' ?? – GEdgar Apr 12 '13 at 14:20
@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

$0^0$ 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.)

share|cite|improve this answer
+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. :) – Keep these mind Apr 5 '13 at 21:35
"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

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

share|cite|improve this answer
Also, 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 5 '13 at 20:26
@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 5 '13 at 20:28
Yeah, 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 5 '13 at 20:32
That'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 5 '13 at 20:32
@Gugg I would have thought $0^0=1=0000....00001$. – Jp McCarthy Apr 5 '13 at 20:36

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}].

share|cite|improve this answer
+1 Not too convinced, as of yet, of the anthropomorphic explanation though. But perhaps I'm underestimating WA. :) – Keep these mind Apr 5 '13 at 22:54
@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 5 '13 at 23:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.