# Challenge 16: About the equation x^y = y^x

1. May 27, 2014

### micromass

Staff Emeritus
The idea of this challenge is to investigate the equation

$$x^y = y^x$$

Prove the following parts:

• If $0<x<1$ or if $x=e$, then there is a unique real number $y$ such that $y^x = x^y$.
• However, if $x>1$ and $x\neq e$, then there is precisely one number $g(x)\neq x$ such that $g(x)^x = x^{g(x)}$.
• If $1<x<e$, then $g(x)>e$. And if $x>e$, then $g(x)<e$.
• The function $g$ defined on $1<x<e$ is differentiable.

Please list all the sources you consulted when solving this question! Using google is considered cheating :tongue:

Points will be given as follows:
1) First person to post a correct solution to one of the above points will receive 3 points (So if you solved 3 and 4, you will receive 6 points).

2) Anybody to post a (original) correct solution to one of the above points will receive 1 point.

3) Anybody posting a (nontrivial) generalization of some of the above questions will receive 2 points

4) The person who posts a solution with the least advanced mathematical machinery will receive 1 extra point for his solution (for example, somebody solving this with basic calculus will have an "easier" solution than somebody using singular homology), if two people use the same mathematical machinery, then we will look at how complicated the proof is.

5) The person with the most elegant solution will receive 1 extra point for his solution (I decide whose solution is most elegant)

Please thank this post if you think this is an interesting challenge (no, I'm not doing this to get more thanks, I just want to see if I should post more things like this and the "thanks" system is the easiest way for this).

Private messages with questions, problem suggestions, etc. are always welcome!

Last edited: May 29, 2014
2. May 27, 2014

### Pranav-Arora

Rewrite the equation as $y^{1/y}=x^{1/x}$.

Consider the plot of $f(x)=x^{1/x}$ for $x>0$.

http://www4a.wolframalpha.com/Calculate/MSP/MSP32551i2aca58c8f32d1c00004h9b230ih5d68ee8?MSPStoreType=image/gif&s=64&w=300.&h=186.&cdf=RangeControl [Broken]

The maximum occurs at $x=e$ and as $x\rightarrow \infty$, $x^{1/x}\rightarrow 1$.

If $0<k<1$ then $0<k^{1/k}<1$. This means that if a line $y=k^{1/k}$ is drawn, it would intersect the graph at only one point showing that there is only value of $x$ which achieves this value. Also, the line $y=e^{1/e}$ touches the graph at only one point. This completes the first part.

If $k>1$ and $k\neq e$, it is clear that the line $y=k^{1/k}$ will intersect the curve at two points which shows that there are exactly two values of $x$ resulting in the same $x^{1/x}$ value.

The third part is trivial. :tongue:

$$g'(x)=(x^{1/x})'=\frac{1-\ln x}{x^2}$$
and g'(x) is defined everywhere in $1<x<e$ completing the proof.

I understand that the above statements are very loose and I am not sure about the right words. Please let me know if anything is unclear.

Last edited by a moderator: May 6, 2017
3. May 27, 2014

### micromass

Staff Emeritus
You seem to use $g(x) = x^{1/x}$, but that's not true, is it?

4. May 27, 2014

### Pranav-Arora

No. Sorry, that was a very silly mistake.

$$g(x)^x =x^{g(x)} \Rightarrow x\ln g(x)=g(x) \ln x \Rightarrow \frac{\ln g(x)}{g(x)}=\frac{\ln x}{x}$$
$$\Rightarrow \frac{g'(x)-g'(x)\ln g(x)}{g^2(x)}=\frac{1-\ln x}{x^2}$$
$$\Rightarrow g'(x)=\left(\frac{g(x)}{x}\right)^2\left(\frac{1-\ln x}{1-\ln g(x)}\right)$$
The above is defined everywhere in $1<x<e$.

BTW, are the proofs for other parts correct?

5. May 27, 2014

### micromass

Staff Emeritus
You do obtain an ODE that is correct, but only if $g$ has been proven to be differentiable

Yes.

6. May 28, 2014

### jostpuur

I'll try to deal with the differentiability issue. First we have the mapping

$$f:[0,\infty [\to [0,e^{\frac{1}{e}}],\quad f(x)=x^{\frac{1}{x}}$$

Then we define new mappings

$$f^A: ]1,e[\to ]1,e^{\frac{1}{e}}[$$
$$f^B :]e,\infty [\to ]1,e^{\frac{1}{e}}[$$

by taking restrictions of the original $f$. The new mappings $f^A$ and $f^B$ can be proven to be bijections (by examining their derivatives and limits), so their inverses exist, unlike for the $f$.

Then the $g$, which is suppose to satisfy $f(x)=f(g(x))$ and $x\neq g(x)$ for $1<x$ and $x\neq e$, can be written as

$$g(x) = (f^B)^{-1}(f^A(x))\quad\quad 1<x<e$$
$$g(x) = (f^A)^{-1}(f^B(x))\quad\quad e<x$$

It is clear that $f$ is differentiable. The final task is to prove that also $(f^A)^{-1}$ and $(f^B)^{-1}$ are differentiable. Once that has been accomplished, the differentiability of $g$ is clear by chain rule. Since it is difficult to remember the basic results, at this point we turn to google, and get some references:

http://math.stackexchange.com/quest...oof-for-differentiability-of-inverse-function

So Spivak's Calculus book contains a result like this (I hope this hasn't been distorted in the internet. I don't have the book myself...): Assume that $f$ is continuous bijection defined on some interval, and also assume that $f$ is differentiable at some point $x$ with a non-zero derivative. Then $f^{-1}$ is differentiable at $f(x)$.

Using this result we see that $(f^A)^{-1}$ and $(f^B)^{-1}$ are differentiable in their domains.

7. May 28, 2014

### economicsnerd

Define $h:\mathbb R \to \mathbb R$ via $h(t)=te^{-t}$, an analytic function. Notice that $x^y=y^x\iff y\log x = x\log y \iff \dfrac{\log x}{x}=\dfrac{\log y}{y} \iff h(\log x) = h(\log y)$.

As $\log: \mathbb R_{++}\to\mathbb R$ is bijective and smooth, the given statements are equivalent to:
1) If $t<0$ or $t=1$, then there is a unique real number $u$ such that $h(u)=h(t)$.
2) If $t\in \mathbb R_{++}\setminus\{1\}$, there is a unique real number $k(t)\neq t$ such that $h(k(t))=h(t)$.
3) $k((0,1))\subseteq (1,\infty)$ and $k((1,\infty))\subseteq (0,1)$ [Actually, the given question asks only for $k((1,\infty))\subseteq (-\infty,1)$, but we prove the intended latter statement.]
4) $k|_{(0,1)}$ is differentiable.

For reference:
- $h'(t)= e^{-t}(1-t)$
- $h'(t)=0$ only for $t=1$, at which $h$ is maximized.
- $h'(u) \text{ is } \begin{cases} <0 & \text{ for } u<0 \\ >0 & \text{ for } u\in(0,1) \\ <0 & \text{ for } u>1 \end{cases}$, so that its restriction to each of the three regions is (strictly monotone, and therefore) injective.
- $h(0)=0$, $h(1)=\frac1e$, and $\lim_{t\to\infty}h(t)=0$. Intermediate value theorem, along with monotonicity of $h$ on the regions, gives $h((0,1))=h((1,\infty))=\left(0,\frac1e\right)$.

1) First consider the case of $t<0$. Notice that $h$ is sign-preserving, so that any such $u$ would have to be negative as well. However, $h|_{(-\infty,0)}$ is injective. Thus the only such $u$ is $t$ itself. The result for $t=1$ holds because $1$ is the unique maximizer of $h$.

2) Existence follows immediately from $h((0,1))=h((1,\infty))$. Uniqueness follows from injectivity of $h$ on the two intervals, and from the fact that $h((-\infty,0))\subseteq (-\infty,0)$ is disjoint from each of $h((0,1)),h((1,\infty))$.

3) This then follows immediately from injectivity of $h$ on the two intervals.

4) $k|_{(0,1)}= \left(h|_{(1,\infty)}\right)^{-1}\circ h|_{(0,1)}$, a composition of differentiable functions by inverse function theorem, thus differentiable.

8. May 29, 2014

### D H

Staff Emeritus
Sources: wolframalpha.com, mathworld.wolfram.com and wikipedia.com.

WolframAlpha suggests using the Lambert W function. WA links to mathworld pages, but those mathworld pages can be rather dense. Wikipedia has a nice search capability. For example, go to wikipedia's main page and type "Lambert W function" in the search field and voila, there's the wiki page on the Lambert W function.

Not cheating! :tongue:

Taking the log of both sides of $x^y=y^x$ and rearranging yields $\frac {\ln x} x = \frac {\ln y} y$. This suggests looking at the function $f(x)=\frac {\ln x} x$. This function is negative for x<1 and positive for x>1, and is monotonically increasing for x in (0,e) and monotonically decreasing for x in (e,∞). This immediately gives the answer to part 1 of the challenge:
• Since ln(x)/x is negative only for x in (0,1) and since ln(x)/x is monotonic in this interval, there is only one solution to xy=yx for x in (0,1).
[*]Since ln(x)/x reaches a global minimum at x=e, there is only one solution to xy=yx at x=e.

The answer to part 3 is also obvious from the behavior of $f(x)=\frac {\ln x} x$. This function is monotonically increases from 0+ to 1/e for x in (1,e), and monotonically decreases from 1/e to 0+ for x in (e,∞). This means that for every x in (1,e), there is a exactly one y>e such that ln(y)/y = ln(x)/x, and similarly for every x in (e,∞), there is a exactly one y in (1,e) such that ln(y)/y = ln(x)/x.

To solve part 2 of the challenge, it helps to transform the equation once again, this time with substitutions. Defining $u = \ln(1/x) = -\ln x$ and $v = \ln(1/y) = -\ln y$ transforms $- \frac {ln x} x = - \frac {\ln y} y$ to $ue^u = ve^v \equiv w$. The distinct solutions to $x^y=y^x$ lie on different branches of the Lambert W function. In particular, the solutions are given by $y_0(x) = \frac {W_0(-\ln(x)/x)}{-\ln(x)/x}$ and $y_1(x) = \frac {W_1(-\ln(x)/x)}{-\ln(x)/x}$. The desired solution is y1(x) for x in (1,e), y0(x) for x in (e,∞).

Finally, since all branches of the Lambert W function are differentiable except at -1/e, the function y1(x) is differentiable for all x in (1,e).

9. May 29, 2014

### jostpuur

I can defend my posting being the most optimal, since I checked what had been proven above, and only filled in the missing piece

10. May 29, 2014

### micromass

Staff Emeritus
Thank you everybody for your very nice and elegant solutions! It's nice to see how everybody decided to look at this question from a slightly different perspective and then ended up with a slightly different (but correct) solution.

So congratulations to everybody for solving this question!