Wikipedia:Reference desk/Archives/Mathematics/2007 July 18

= July 18 =

Question
Are there any other solutions for x^y = y^x other than y=x? --⁪frotht 19:44, 18 July 2007 (UTC)
 * Yes. For example 2^4 = 4^2 = 16. Try drawing the graph. It looks like one lobe of an hyperbola (but it isn't). --Trovatore 19:47, 18 July 2007 (UTC)


 * Here's how you can find more solutions :
 * $$x^y=y^x \Leftrightarrow y \ln(x) = x \ln(y) \Leftrightarrow \frac{\ln(y)}{y} = \frac{\ln(x)}{x}$$
 * (With x and y both real)
 * This leads to the study of the function f(x) = ln(x)/x
 * For each value of a such that the line of equation f(x) = a cuts the curve f(x) = ln(x)/x in two points, there is a solution to the equation x^y = y^x, as this gives two different values of x for which ln(x)/x is the same.
 * --Xedi 20:35, 18 July 2007 (UTC)


 * It's worth noting that graphing $$x^y=y^x$$ can confuse some numerical methods a bit (the OS X Grapher application makes the curve a bit jaggy as it moves out, but panning/zooming eliminates the weird jagginess). I think, but I haven't proved, that the curved part of the graph intersects the straight line at (e,e). I think one way to show this is to calculate the derivatives of the graph and show that it is undefined at (e,e) but nowhere else on the line y = x. Donald Hosek 22:34, 18 July 2007 (UTC)


 * [[Image:Ln(X)divX.png]] -- SGBailey 09:56, 19 July 2007 (UTC)


 * If you define the function u on the real numbers except 0 by:
 * $$u(z) = \exp\left(\frac{z}{e^z-1}\right)\,,$$
 * all real-valued solutions in nonnegative reals of xy = yx with x ≠ y are given parametrically by:
 * $$(x, y) = (u(z), u(-z)),~z\ne 0\,.$$
 * The solution (2, 4) corresponds to z = 0.69314718... . The point 0 is a removable singularity; just define u(0) = e.
 * In the rationals, (x, y) = (–2, –4) is also a solution. --Lambiam 23:49, 18 July 2007 (UTC)
 * Where does this solution come from ? --Xedi 17:44, 19 July 2007 (UTC)
 * From fiddling around with the equation xy = yx to get a parametric solution, and then some more fiddling to make it symmetric. I'm sorry, I have to confess it is OR, but readily verified. --Lambiam 20:03, 19 July 2007 (UTC)


 * There is a Hungarian language article about this equation: Lóczy Lajos, Mikor kommutatív illetve asszociatív a hatványozás, Középiskolai Matematikai és Fizikai Lapok 2000/1 p. 7&#x2013;16. However, that issue is not in the online archive, and it has no English abstract nor bibliography, so you can probably only use it if you speak Hungarian.  &#x2013; b_jonas 21:27, 23 July 2007 (UTC)
 * I think I've found an English translation: http://www.komal.hu/cikkek/loczy/powers/commpower.e.shtml. The article gives a parametric solution that is equivalent to mine, but not as symmetric. On the other hand, it considers the rational solutions and finds a whole family (for example, x = 117649/46656, y = 823543/279936), and also considers solutions of the "associative case", being the equation (xy)z = x(y z) . --Lambiam 03:52, 24 July 2007 (UTC)
 * Oh, lucky you found it. I didn't look at the webpage because new articles almost never appear there nowdays and didn't think that this old article might be on.  &#x2013; b_jonas 10:44, 24 July 2007 (UTC)