Which is larger $e^\pi$ or $\pi^e$?
November 3, 2008
Bailey Hall 207
Refreshments will be served at 4:15 in Bailey 204
Two years ago, while studying $\pi$, one of my thesis students found an interesting paper addressing this question.
Given positive real numbers $x < y$, which of the following holds: $x^y < y^x$, $x^y > y^x$, or $x^y=y^x$? It is easy to see that all three cases are possible. For example, $2 < 3$ and $2^3 < 3^2$, since $2^3 = 8$ and $3^2 = 9$, but $3 < 4$ and $3^4 > 4^3$, since $3^4 = 81$ and $4^3 = 64$. One can even get equality, e.g., $2^4 = 4^2$, since both are $16$. Of course, $e < \pi$ and both are close to $3$. Does $\pi$ act like $3$ in the first inequality so that $e^\pi < \pi^e$? Does $e$ act like $3$ in the second inequality so that $e^\pi > \pi^e$? Or, do they both act like $3$, so that $e^\pi=\pi^e$?
In this talk, we will use a little calculus to determine the relationship between $x^y$ and $y^x$, for every pair of positive real numbers. As a consequence, we will see that $x^y=y^x$ holds for infinitely many pairs, but $2^4 = 4^2$ is the only case where both are integers.
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Created automatically on: Fri Apr 20 22:05:41 EDT 2018