Is $x^{-0}$ defined?

Question

In the form of mathematics that most of humanity is taught, the following operation is undefined:

$\Large{\frac{x}{0}}$

But, how about the following operation?

$\Large{x^{-0}}$

Is the following statement true?

$\Large{x^{-0}=\frac{1}{x^0}=\frac{1}{1}=1}$

@BillDubuque - That's awesome! Math is one tight dolphin. :O — StudentsTea, Jul 31 '16 at 20:28
@DonAntonio $0^0=1$. It is just that $x_n\to 0$ and $y_n\to 0$ fails to imply $x_n^{y_n}\to 1$ — Hagen von Eitzen, Jul 31 '16 at 20:30
@HagenvonEitzen From the right. From the left there are big problems with $;x^x;$ : it isn't defined in any right neighborhood of zero. — DonAntonio, Jul 31 '16 at 20:32
I usually tend to define $;0^0:=1;$, just making sure that real exponential functions require a positive base. — DonAntonio, Jul 31 '16 at 20:35
My comment can be read in : https://fr.scribd.com/doc/14709220/Zero-puissance-zero-Zero-to-the-Zero-th-Power — JJacquelin, Aug 01 '16 at 05:27

parsiad · Answer 1 · 2016-07-31T20:31:41.780

7

Yes. $-0=0$, hence $x^{-0}=x^0=1$.

Addendum: see DonAntonio's comment below.

edited Jul 31 '16 at 20:31

answered Jul 31 '16 at 20:27

parsiad

25,154

4

As long as $;x\neq0;$ ... – DonAntonio Jul 31 '16 at 20:28
@DonAntonio: true. Though $\lim_{x\rightarrow 0^+} x^x=1$, so in most contexts the definition $0^0=1$ is harmless. – parsiad Jul 31 '16 at 20:29
3

Yes, but in fact it must be the right sided limit $;x\to 0^+;$ , as the left one doesn't exist as the function $;x^x;$ isn't defined in any left neighborhood of zero – DonAntonio Jul 31 '16 at 20:30
Ah yes, you are correct. +1 – parsiad Jul 31 '16 at 20:32

H. H. Rugh · Answer 2 · 2016-08-01T12:00:01.887

4

A bit off discussion, but it may be dangerous to assign a value to $0^0$ as two non-commuting limits are involved. Any positive real number is possible. Given $a>0$ and 'define' (which I wouldn't recommend doing, though): $$ 0^0 = \lim_{x\rightarrow 0+} x^{y(x)}=\lim_{x\rightarrow 0+} \exp(\log x \; y(x)),$$ where $y(x)$ goes to zero as $x$ goes to zero. Choosing $y(x)=\log a/\log x$ (which indeed goes to zero as $x$ goes to zero): $$ 0^0 = \lim_{x\rightarrow 0+} x^{y(x)} = \exp\left(\log x \frac{\log a}{ \log x}\right) = a.$$ Later edit: This suggests better leave this case undefined when context indicates that powers are in ${\Bbb R}$ or ${\Bbb C}$. There is a broad general concensus (of algebraic and practical nature, binomial formula, Taylor series,...) of defining $x^0 =1$ for any real or complex $x$ whenever the context indicates that powers are in ${\Bbb N}$.

edited Aug 01 '16 at 12:00

answered Jul 31 '16 at 20:46

H. H. Rugh

35,236

1

There need not be any limits at all involved. Indeed, in a lot of contexts, the only exponents that appear are integers. – celtschk Jul 31 '16 at 20:49
1

True. If we are dealing with integer exponents then $z^0=1$ for $z>0$ admits analytic continuation to the complex plane, so that gives an argument in favor. – H. H. Rugh Jul 31 '16 at 20:56
1

Do you always write $e^x=1+\sum_{n=1}^{\infty}\frac{x^n}{n!}$ instead of $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$? – egreg Jul 31 '16 at 21:08
No, I must admit ;-) But sometimes when dealing with (square) matrices or operators, I write: $\exp(A) = {\bf 1} + \sum_{n\geq 1} A^n/n!$. Anyway I agree that if the context is integer exponents it seems reasonable to say $x^0=1$ for any $x$. – H. H. Rugh Jul 31 '16 at 21:33
This "answer" (though it's not answering anything the OP has asked) is simply wrong. The limiting behaviour of $f(x,y) = x^y$ as $x,y \to 0$ is irrelevant (or you can have $y$ be a function of $x$, it doesn't matter). We want the value at the point. That's it. If the function is not continuous at $(0,0)$, so what? $0^0 = 1$, the empty product; this has been the consensus among mathematicians for quite some time. – MathematicsStudent1122 Aug 01 '16 at 04:25
Edit to my previous comment: according to the other answers and Wiki, there are some in the mathematical community who say $0^0$ is undefined. So there is no consensus. Nevertheless I find it odd and the reasoning is rather unconvincing. – MathematicsStudent1122 Aug 01 '16 at 04:37
See this answer for my take on the issue. – celtschk Aug 01 '16 at 06:45

score 3 · Answer 3 · edited Apr 13 '17 at 12:20

3

$x^0=1$ for all $x$ except $x=0$, which some conventions have undefined while others set equal to $1$ (see Zero to the zero power - is $0^0=1$?). This means that $x^{-0}=1$ for all $x$ except $x=0$, for which it is (depending on convention) either undefined or $1$. This has nothing to do with the fact that $x\over 0$ is always undefined.

I should add that I am strongly in favor of "$0^0=1$" - to my mind there's no question that this is the "right" definition - but there are definitely people out there who disagree, and I have seen some professional mathematicians treat $0^0$ as undefined, so I'm erring on the side of caution above.

edited Apr 13 '17 at 12:20

Community

1

answered Jul 31 '16 at 20:28

Noah Schweber

245,398

Mathematics shouldn't be about opinion. There is only truth and untruth. If some mathematicians leave $0^0$ as undefined, why are they not refuted? You would think after all these years we'd have a consensus. – MathematicsStudent1122 Aug 01 '16 at 04:33
The statement "There is only truth and untruth" is false. There is also undecidability. For example, Gödel's incompleteness theorems. Also, one cannot say $x=y$ is true or not, if $x$ and $y$ are not defined and/or if the context of the equation is not fully known. – JJacquelin Aug 01 '16 at 06:07

score 1 · Answer 4 · answered Jul 31 '16 at 20:31

1

$x^y=e^{ylog(x)}$ so $x^0=x^{-0}$ are defined for $x>0$.

answered Jul 31 '16 at 20:31

Tsemo Aristide

87,475

1

Thanks! What if $x=0$? What if $x < 0$? – StudentsTea Jul 31 '16 at 20:33
2

I think bringing logs into this overcomplicates things, since it makes the $x<0$ case seem mysterious. – Noah Schweber Jul 31 '16 at 20:33
1

The case $x<0$, $y\neq 0$ is quite non-trivial unless $y$ is an integer. – H. H. Rugh Jul 31 '16 at 20:57

score 1 · Answer 5 · answered Jul 31 '16 at 21:20

There are two principles: "anything to the power $0$ is $1$" and "$0$ raised to any positive power is $0$". It makes sense to allow the first principle to define $0^0:=1$, because there is no reason for the second principle to extend to $0^0$, as it clearly does not extend to any negative power, and there are abundant examples, such as $\lim_{x\to0+}x^x=1$, which would violate it. That said, one can easily find examples where such a limit may not be $1$: Consider $\lim_{x\to0+}y^x$, which is $0$ in the case that $y=\exp(-1/x^2)$ although both $x$ and $y$ go to $0$. It is probably because of cases like this that some mathematicians are uncomfortable with the convention that $0^0=1$, and prefer to consider it undefined.

Is $x^{-0}$ defined?

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