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I am trying to determine if $(0,0)$ a solution to $x^y-y^x=0$. My hunch is that it is undefined since $0^0$ is an indeterminate form. To attempt to prove this, I have tried the usual "different paths give different limits" trick with

$(x,ax^n)\rightarrow(0,0)$

$(x,\sin(x))\rightarrow(0,0)$

$(x,e^x-1)\rightarrow(0,0)$

None of the above accomplished my goal.

I did find this older post ($x^y = y^x$ for integers $x$ and $y$) which included an answer by "Yuval Filmus" which states $y=x=0$ is $\it{trivially}$ a solution. He accomplished this by defining $0^0$ to some value and moves on. I would like to see something more rigorous, if it exists.

Any hints on how to proceed?

Edited for wording.

Edit: Stefan Smith confirms my suspicion that the comment of Yuval Filmus cannot be made rigorous.

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NoClue
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  • For $y=x$ we have that $f(x,x)=0$. Now,is $f$ continuous on $0$? – Haha Dec 14 '13 at 00:37
  • @NoClue : you are really asking if $0^0$ has a well-defined value. Paths and limits seem irrelevant to me here. You are asking if one particular point satisfies one equation. – Stefan Smith Dec 14 '13 at 03:47
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    @StefanSmith The question stemmed from a discussion of whether the graph of $x^y=y^x$ contains the origin. Knowing the process of showing $0^0$ is an indeterminate form, I hoped to follow a similar procedure using paths/limits. – NoClue Dec 14 '13 at 04:05
  • @NoClue : yes, the answer to the question seems to depend on who's asking, and whether they consider $0^0$ to have a well-defined value. There are several recent questions on this here. I asked one myself. In combinatorics, it seems people always want $0^0=1$, and it seems to lead to no contradictions using algebra. In analysis, often people want $0^0$ to be undefined. – Stefan Smith Dec 14 '13 at 22:03

3 Answers3

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This might sound unsatisfying, but if you define $0^0$ to be any finite number, then $(0,0)$ is a solution, otherwise it isn't. I personally like $0^0=1$, but some people disagree. This site has a lot to say about the issue of $0^0$, and you can search for that. It is ultimately a matter of definition.

Dylan Yott
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  • I searched for "0^0" on this site and found nothing. I cannot find how defining $0^0=1$ results in any contradiction if you use this definition in algebra or arithmetic. – Stefan Smith Dec 14 '13 at 03:42
  • Spot on. The only question is whether $0^0$ is defined. If it is, $x=y=0$ is a solution. If not, it is not. Whether it is defined is a matter of convience-if you give a good definition of $x^y$ I can answer the question. I would like to award a bounty on this, but haven't figured out how. +100 – Ross Millikan Dec 14 '13 at 05:33
  • This answer and comment by Ross most closely represent my expected answer. Namely, it is a matter of how we are interpreting $x^y$. – NoClue Dec 14 '13 at 15:00
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First of all, this function may not be well-defined if $x<0$ or $y<0$ (for example, if $x=-1$ and $y=1/2$, then $x^y=i$, which is not real). Therefore, I suggest you do the following:

  • Define the function only on $\mathbb R_{++}^2\equiv\{(x,y)\in\mathbb R^2\,|\,x>0,\, y>0\}$
  • Show that for any sequence $(x_n,y_n)_{n\in\mathbb Z_+}\subset\mathbb R_{++}^2$ that converges to $(0,0)$, you have that $f(x_n,y_n)$ converges to $0$.

In this way, you can continuously extend the function to $(0,0)$ by defining $f(0,0)\equiv0$.

Too bad this cannot be done! To see this, let

\begin{align*} (x_n,y_n)\equiv\left(\frac 1 n,\frac{1}{\ln(n+1)}\right)\quad\forall n\in\mathbb Z_+ \end{align*} Clearly, $(x_n,y_n)\to 0$ as $n\to\infty$. However, \begin{align*} f(x_n,y_n)=\left(\frac{1}{n}\right)^{1/\ln(n+1)}-\left(\frac{1}{\ln (n+1)}\right)^{1/n} \end{align*} converges to $1/e-1$ as $n\to\infty$. Because of this, $0$ is not a good definition of $f(0,0)$, since $0$ may not be approached by the function $f$ when evaluated close to $(0,0)$.

In terms of how you put it originally, try

$$\left(x,\ln\left(\dfrac{x+1}{x}\right)^{-1}\right)$$

This converges to $(0,0)$ as $x\searrow0$, but the function value converges to $1/e-1$.

triple_sec
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The short answer is "no". The reason is that there is no universal, always-agreed-upon definition of what $0^0$ is supposed to equal. And if $0^0$ is undefined, so is $0^0-0^0$, and $(0,0)$ is not a solution of $x^y-y^x=0$.

Having said that, there are many situations in which it is convenient to allow $0^0$ to take a value, as long as you are extremely careful, you have specified what number system you are using (reals? integers? nonnegative integers? complex numbers?) and what rules you are operating under. In most such situations I have seen, that value is $1$.

Your question does not provide any such context. There is a (diophantine-equations) tag on your question, but your question does not mention Diophantine equations, and whether $x$ and $y$ are integers. So one can't really make a good case that $(0,0)$ solves your equation.

Stefan Smith
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    So $e^x$ isn't any longer $\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$. Somewhat sad, isn't it? – Michael Hoppe Dec 14 '13 at 01:55
  • @MichaelHoppe: yes, when we use power series, we use "$0^0 = 1$" all the time and and no one seems to notice. I admit it doesn't seem to cause a problem in practice. – Stefan Smith Dec 14 '13 at 03:24
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    @WillJagy: See e.g. the section on Wikipedia and references linked therefrom: there are many good reasons (combinatorial and otherwise) to define $0^0=1$, and only one silly reason not to: namely, that if you have a limit of form $f(t)^{g(t)}$ where both $f(t)$ and $g(t)$ approach $0$, then you can't just "plug in" the individual limits to conclude that the limit is $1$. I consider this a rule about calculating limits (that you can't make that substitution), and not to do with defining $0^0$. – ShreevatsaR Dec 14 '13 at 07:09
  • @ShreevatsaR : I just posted a question whether there is a good, algebraic reason not to define $0^0=1$. If this never leads to contradictions, then the only reason not to I can think of is if defining $0^0=0$ never leads to contradictions, either. – Stefan Smith Dec 14 '13 at 17:29