Delete comment from: Computational Complexity
You can just look at the partial derivatives of f(x,y) = x^y. You can use these to see that f grows faster by changing y when x and y are both at least e.
More formally, f_x = x^y ln x and f_y = y x^(y-1). If a>e, then ln a < a/e, so f_x(a,e) < f_y(e,a). So by integrating over a on both sides from e to pi, you get f(e,pi)>f(pi,e). A similar calculation should give you something like Boris's claim.
Feb 17, 2010, 4:10:13 AM
Posted to e to the pi vs pi to the e