# Proving that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator

Prove that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator.

I did in the following way. Are there other ways?

Proof : Let $f(x)=e\pi\frac{\ln x}{x}$. Then, $$e^{\pi}-{\pi}^e=e^{f(e)}-{e}^{f(\pi)}\tag1$$ Now, $$f'(x)=\frac{e\pi(1-\ln x)}{x^2},\quad f''(x)=\frac{e\pi (2\ln x-3)}{x^3},\quad f'''(x)=\frac{e\pi (11-6\ln x)}{x^4}.$$ Since $f'(x)\lt 0$ for $e\lt x\lt\pi$, one has $f(e)\gt f(\pi)$. By Taylor's theorem, there exists a point $c$ in $(e,\pi)$ such that \begin{align}f(\pi)&=f(e)+(\pi-e)f'(e)+\frac{(\pi-e)^2}{2}f''(c)\\&\gt f(e)+(\pi-e)\cdot 0+\frac{(\pi-e)^2}{2}\cdot \frac{e\pi(2\ln e-3)}{e^3}\\&=f(e)-\frac{\pi(\pi-e)^2}{2e^2}\tag2\end{align} because $f'''(x)\gt 0\ (e\lt x\lt \pi)$ implies $f''(c)\gt f''(e)$.

By the mean value theorem and $(2)$, $$e^{f(e)}-e^{f(\pi)}\lt (f(e)-f(\pi))e^{f(e)}\lt \frac{\pi (\pi-e)^2}{2e^2}\cdot e^{\pi}=\frac{e\pi(\pi-e)^2}{2}\cdot e^{\pi-3}\tag3$$

Since $e^x\lt \frac{1}{1-x}\ (0\lt x\lt 1)$ and $0\lt \pi-3\lt 1$, $$e^{\pi-3}\lt\frac{1}{4-\pi}\tag4$$

From $(1)(3)(4)$, $$e^{\pi}-{\pi}^e\lt \frac{e\pi(\pi-e)^2}{2}\cdot\frac{1}{4-\pi}\lt\frac{3\times\frac{22}{7}\left(\frac{22}{7}-2.718\right)^2}{2}\cdot\frac{1}{4-\frac{22}{7}}\lt 0.993\lt 1$$

• I deleted an answer involving numerical approximations, but thought I should mention something of the difficulty involved there. Since the actual difference between the two numbers is already $\ \approx \ 0.68 \$ , each of the two values (about 23) must be estimated to a precision of order 1% so that the estimate of the difference remains less than 1 . This turns out to be extremely difficult to arrange for transcendental powers of transcendental numbers if one is using rational numbers to represent sums of series when one must obtain convincing results without a calculator. – colormegone Aug 30 '15 at 2:24
• @mathlove If I may make a suggestion, I have been bothered by the last line, since it uses a decimal approximation for $\ e \$ , leading to a difference squared which would seem to require a calculator to demonstrate that the value is less than 0.993 . You also use $\ 3 \$ as a bound on $\ e \$ which makes this coarser than necessary. Since $\ \frac{19}{7} \ < \ e \ \frac{20}{7} \$ , you can use $\ \frac{20}{7} \$ as the first factor and $\ \frac{19}{7} \$ in the difference-squared. This leads to (continued) – colormegone Aug 30 '15 at 2:41
• $$\frac{1}{2} \ \cdot \ \frac{20}{7} \ \cdot \ \frac{22}{7} \ \cdot \ ( \frac{22}{7} - \frac{19}{7})^2 \ \cdot \left(\frac{1}{4 - \frac{22}{7}} \right) \ = \ \frac{20 \cdot 22 \cdot 9 \cdot 7}{2 \cdot 7 \cdot 7 \cdot 49 \cdot 6} \ = \ \frac{330}{343} \ < \ 1 \ \ .$$ The factors cancel in a way that makes simplification possible without a calculator. [This is an omission-typo in the previous comment: the inequality was meant to read $\ \frac{19}{7} \ < \ e \ < \ \frac{20}{7} \$ . ] – colormegone Aug 30 '15 at 2:44
• Fantastic method of proof! – Chandler Watson Sep 4 '15 at 15:03