Friday, December 05, 2008

Some updates: Rhyno, a quite successful math major, is correct. I did make an assumption of continuity, which, though warranted, is easily demonstrated through proof of differentiability. Thank you for adding to the proofiness of my proof (with a vocabulary like mine, I'm gonna be the next Colbert).

And thank you, Katharine, for pointing out the VERY interesting fact that the minimum value for f(x) = x^x (between x=0 and x=1) occurs at x = 1/e. I wonder why that is. How did you figure that out?

I'm starting to wonder, even if there are infinite pairs of x and y such that x^x = y^y and x=/=y, how many of those pairs are rational? Is there a finite number of rational values for which it holds true? What is that number? How does one find them?

2 comments:

Anonymous said...

The reason why the minimum is 1/e is that the minimum value occurs when the derivative is 0. When f'(x) = 0, x^x(ln[x]+1) = 0. The only way this is true is if x = 1/e (recall ln[1/e] = -1), because x^x =/= 0 at any time.

-A

PiFry said...

Yeah...I know local minimums occur at first derivative zeroes...I guess I was just hoping for something a little more, I dunno, cosmic and grand. Maybe I'm expecting too much from a puzzle, but I'm hoping the meaning of life is spelled out clearly through this endeavor.