ProductLog Sorry, I thought I explained, but obviously I didn't. ProductLog is the inverse function of x*e^x, so ProductLog(p) gives you the number x such that x*e^x = p. See: http://functions.wolfram.com/ElementaryFunctions/ProductLog2/ Greetings, Han.
I wish I had Mathematica, but cannot afford it. I have MathCad 7 and PSI Plot 5, both of which will do surface plots, as will Mathematica. I never use these capabilities and do not want to spend the time now on this problem. Perhaps a Mathematica owner might want to use the surface plot capabilities to check for all real solutions. Superimpose the following three surface plots. z = x^y z = y^x The plane y = x Each surface intersects the plane at possible solutions. If both surfaces intersect at the same height off the XY-Plane, you have a solution. It should not be difficult to pick out possible solutions, verify them numerically, and be pretty sure you found them all. BTW: Has anybody considered complex solutions? I have not studied all the posts carefully. Perhaps this has already been done. If not, it is an interesting extension of the problem.
Well, x,y,z, are N, but it really is interesting... OKAY EVERYBODY!! UPDATE!!!! X,Y and Z ARE COMPLEX!!! Thanx, Dinosaur Please Register or Log in to view the hidden image!
you can't use complex numbers, as given in the problem statement anyways, the formal solution is anything that satisfies the following system k=1+ ln(k)/ln(x) ; where y=k*x; and y and x are non-zero integers (hence is k) , as that is a trivial solution space anyways given that this system is underspecified, only iterative solution seeking can work, as evidenced earlier
Complex powers Is x^y defined for every x and y? According to my math-book (Thomas M. Apostol) x^y is defined to be exp(y*Log(x)) for x,y>0. The Exp and Log funtion are defined for complex numbers, so maybe x^y can be defined for complex x and y. Does anyone know this? Greetings, Han.
i take it you don't realize your question answered itself? Please Register or Log in to view the hidden image!
Correct! I think you mean that x^y is defined to be Exp(y*Log(x)) for Complex x and y where x,y <>0. I suspected so, but I couldn't get any confirmation. Well, if that is the case the solution to the problem x^y = y^x is as follows: x^y = y^x = { definition of x^y, x, y<>0} Exp(y*Log(x)) = Exp(x*Log(y)) = { Exp(x) = Exp(y) <=> x=y } y*Log(x) = x*Log(y) = { x,y <>0, division by x and by y} 1/x*Log(x) = 1/y*Log(y) = { Log(x) = -Log(1/x)} 1/x*Log(1/x) = 1/y*Log(1/y) = { x = Exp(Log(x)) } Exp(Log(1/x)))*Log(1/x) = 1/y*Log(1/y) = { Definition of ProductLog: x*Exp(x) = y <=> x=ProductLog(y) } Log(1/x) = ProductLog(1/y*Log(1/y)) = { Exp(x) = Exp(y) <=> x=y, x = Exp(Log(x)) } 1/x = Exp(ProductLog(1/y*Log(1/y))) = { 1/x = Exp(y) <=> x=Exp(-y) } x=Exp(-ProductLog(1/y*Log(1/y))) Please Register or Log in to view the hidden image! There has to be a step in the proof where a '<=' should be instead of '='. I cannot find it. This has to be the case since the last formula doesn't describe all the solutions, since clearly x=y is also a solution. Greetings, Han.
