1

Show that: $$e^{\pi}-\pi^{e} >\frac12$$

I am not sure, if anything can be done using elementary methods. Because $e$ and $\pi$ are not algebraic numbers. Therefore, I find it impossible to prove this with elementary techniques. Maybe using the elementary formula $\left(1+\frac 1n\right)^n\rightarrow e$ for $e$ would work. But there is no such formula for $\pi$. I guess there is no other way than using series, right?

User
  • 1,671

1 Answers1

4

This solution isn’t elegant, but the fact that $\pi$ and $e$ aren’t algebraic doesn’t mean we can’t bound this expression easily. You can use finite decimal approximations. Since $\pi$ and $e$ are computable, we can compute their decimal expansions $\pi = 3.14159 \cdots$ and $e = 2.71828 \cdots$

Using these expansions, we can bound: $3.141 < \pi < 3.142$ and $2.718 < e < 2.719$

If we have that $a < b < c$ and $x < y < z$, we may bound $a^x < b^y < c^z$. Thus we may bound:

$$ \begin{align*} & e^{\pi} > 2.718^{3.141} > 23.11 \\ & \pi^e < 3.142^{2.719} < 22.49 \end{align*}$$

Since these are rational numbers, there are no issues in bounding these expressions. Simply compute the first few digits by hand.

Then it follows that $e^\pi - \pi^e > 23.11 - 22.49 = .62 > .5$

Joe
  • 2,794
  • "Simply compute the first few digits by hand" How...? Is there a way to generally compute $a^b$ when $b$ is a non-integral rational? – TheBestMagician Aug 15 '22 at 21:28
  • 2
    Yeah of course. I mean it’s not fun, but it’s possible. Separate $a^{n/m} = a^n a^{1/m}$. $a^n$ is trivial, and $a^{1/m}$ can be approximated digit-by-digit with a simple algorithm by hand. Like I said, this solution isn’t elegant at all. – Joe Aug 15 '22 at 21:42