Is there a closed form for the nontrivial solutions of $x^y = y^x$?

Question

It can be shown $\forall x \in \Bbb{R}^+ \exists \{y_1, y_2\} : x^y = y^x.$, and the first of these two numbers is trivial, $y_1 = x$. The second is nontrivial, and I cannot find a closed form for all the nontrivial solutions. I have found a good approximation, $y_2$ ~ $\frac{(e-1)^2}{x-1} +1$. Is there a function for all the values of $y_2$, or can it be shown $y_2$ is or is not a rational function?

An example solution is $2^4 = 4^2$, the only integer solution.

Answer 1

Note: A Google search for "rational solutions to x^y = y^x" came up with, among others, this: http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/on-the-rational-solutions-of-xy-yx

Summary: "The author provides a new way of finding all rational solutions of $x^y=y^x$ as well as a historical survey of this problem."

Here is the history, from the article (equation (1) is $x^y = y^x$):

In a letter by Daniel Bernoulli to Goldbach (1728) [l], equation (1) is mentioned with the statement that (x, y) = (2,4) (or (4,2)) is the only integer solution. In his answer Goldbach gives the general solution of (1) by writing $y =ax$, hence, $x^ {ax} = ( ax)^x$ and after simplification, and ignoring the trivial case when $a = 1$, $x=a^{1/(a-1)}$ and $y=a^{a/(a-1)}$

Equation (1) is also discussed in some detail by Euler in [2].

"REFERENCES"

"1. Correspondence Math. Phys. (edited by Fuss), Vol. 2, pp. 262 and 280."

"2. Leonhard Euler, lntroductio in analysin infinitorum, Tom II, Cap. 21 §519."

"3. L. E. Dickson, History of the Theory of Numbers, New York, Vol. II, Geneva and Lausanne, 1748, p.687."

"4. E. J. Moulton, The real function defined by $X^y=Y^x$, Amer. Math. Monthly 23 (1916), 233-237."

"5. A. Hausner, Algebraic number fields and the Diophantine equation $m^n = n^m$, Amer. Math. Monthly 68 (1961), 856-861."

My favorite parameterization:

Let $y = rx$. Then

$\begin{array}\\ x^y = y^x \iff &x^{rx} = (rx)^x\\ \iff &x^{r} = rx \qquad\text{(take the x-th root)}\\ \iff &x^{r-1} = r \qquad\text{(divide by x)}\\ \iff &x = r^{1/(r-1)} \qquad\text{(take the r-1th root)}\\ \text{and} &y = rx = r^{r/(r-1)}\\ \end{array} $

Answer 2

Just as Shailesh commented, the second root expresses in terms of Lambert function which corresponds to $z=W(z)\, e^{W(z)}$.

Applied to the case of the equation $x^y = y^x$, the solution for the second root is given by $$y_2=-\frac{x }{\log (x)}W\left(-\frac{\log (x)}{x}\right)$$ which will be real valued if $x \geq e$.

In the Wikipedia page, you will find nice approximations of $W(z)$ for small and large values of the argument.

Applied to the specific case of the equation $x^y = y^x$, for large values of $x$, a series expansion would give $$y=1+\frac{\log(x)}{x}+\frac{3 \log ^2(x)}{2 x^2}+\frac{8 \log ^3(x)}{3 x^3}+\frac{125 \log ^4(x)}{24 x^4}+O\left(\left(\frac{1}{x}\right)^5\right)$$

Concerning your approximation, I would like to congratulate you. For integer values of $x$ between $3$ and $10$, I curve fitted (nonlinear regression) the model $$y_2=a+\frac b {x-c}$$ and obtained $$y_2=1.06074 +\frac{2.79877}{x-1.02478}$$ which is almost perfect.

Answer 3

Suppose we want to find a pair $(x,y)$ so that $x^y=y^x$. Let us write $x=t^a$ and $y=t^b$. Then $$ t^{\large at^b}=x^y=y^x=t^{\large bt^a}\tag{1} $$ Whether $t\gt1$ or $t\lt1$, $t^x$ is monotonic; therefore, we must have $$ at^b=bt^a\tag{2} $$ which is the same as $$ \frac ab=t^{a-b}\tag{3} $$ Now we can solve $a-b=1$ and $\frac ab=t$, which gives $a=\frac t{t-1}$ and $b=\frac1{t-1}$.

Therefore, we can use $$ \bbox[5px,border:2px solid #C0A000]{(x,y)=\left(\lower{3pt}{t^{\large\frac{t}{t-1}},t^{\large\frac1{t-1}}}\right)}\tag{4} $$ Notice that $t=\frac12$ gives $(2,4)$ and $t=2$ gives $(4,2)$. In fact, $t$ and $\frac1t$ give the same pair in reverse order, so we can restrict either to $t\in(0,1)$ or to $t\in(1,\infty)$.

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$t=\frac13$ gives $\raise{1pt}{\left(\lower{2pt}{3^{\frac12},3^{\frac32}}\right)}$, and indeed $$ \left(\lower{2pt}{3^{\frac12}}\right)^{3^{\frac32}}=3^{\frac12\cdot3^{\frac32}}=3^{\frac32\cdot3^{\frac12}}=\left(\lower{2pt}{3^{\frac32}}\right)^{3^{\frac12}} $$

$t=\frac14$ gives $\raise{1pt}{\left(\lower{2pt}{4^{\frac13},4^{\frac43}}\right)}$, and indeed $$ \left(\lower{2pt}{4^{\frac13}}\right)^{4^{\frac43}}=4^{\frac13\cdot4^{\frac43}}=4^{\frac43\cdot4^{\frac13}}=\left(\lower{2pt}{4^{\frac43}}\right)^{4^{\frac13}} $$