*Mid-twentieth century computing device - The slide rule. The more expensive slide rules were crafted from bamboo. Mid-priced units were aluminum. Most students had plastic versions, such as the one pictured. (Photograph by Roger McLassus, via Wikimedia Commons. Click for larger image.)*

When I reached college, simple electronic calculators became available. These were called

*I've retained my HP-33C Scientific Programmable Calculator, no longer working, as a keepsake.This calculator had a display with an eight digit mantissa and two digit exponent formed from red light-emitting diodes.Nobel Physics Laureate, Kenneth G.Wilson (1936-2013), about whom I wrote in an earlier article (Kenneth G. Wilson, June 21, 2013), remarked about his purchase of an HP pocket calculator."I buy this thing and I can't take my eyes off it, and I have to figure out something that I can actually do that would somehow enable me to have fun with this calculator."[1](Photo by author.)*

Now to the topic of this article, whether e

eA mathematician would scoff at such a crude resolution of the question, since no proof is involved. An article by Presh Talwalkar gives a few such proofs, most of which throw a lot of mathematics at the problem.[2-3] However, one such proof involving a Taylor series expansion is simple; so, I'll give it here. The Taylor series expansion of e^{π}= 23.14069...

π^{e}= 22.45915...

eIn a series such as this, when^{x}= 1 + x + x^{2}/2! + x^{3}/3! + ...

eIf we select as our value for x, (π/e - 1), we find that^{x}> 1 + x

esince^{π/(e - 1)}> (π/e - 1) + 1

(π/e - 1) + 1 = π/eand

ewe have^{-1}= (1/e)

(1/e) eand^{π/e}> π/e

eThen, raising both side of the inequality to the power of e,^{π/e}> π

eAnd, as they say, quod erat demonstrandum (QED). A recent paper on arXiv by Andrés Vallejo and Italo Bove of the Universidad de la República (Montevideo, Uruguay) gives a different proof of this proposition.[4] This solution is based on the second law of thermodynamics, the entropy law.^{π}> π^{e}

*The thermal environment for the thought experiment demonstrating that e ^{π} > π^{e}. (Created using Inkscape)*

The thought experiment demonstrating that e

ΔSThermal equilibrium of the reservoir and solid is achieved by exchanging heat Q and energy U.^{A}= C ln(T_{2}/T_{1}) = C(1 − ln(π))

QThe entropy change of the reservoir and the total change of entropy of the system are calculated.^{B}= −Q^{A}= −∆U^{A}= C(T_{1}- T_{2}) = C(π − e)

∆S = Q^{B}/T^{B}= C(π/e - 1)

ΔSAccording to the second law of thermodynamics, the total entropy change must be positive.^{total}= ΔS^{A}+ ΔS^{B}

ΔS^{A}+ ΔS^{B}= C[(π/e) - ln(π)]

π/e − log(π) ≥ 0 =>Then, by making both sides of the inequality a power of e, we get.

π ≥ log(π^{e})

e^{π}> π^{e}

- Martin Weil, "Kenneth Wilson, Nobel winner who explained nature's sudden shifts, dies in Maine at 77," Bangor Daily News, June 19, 2013.
- Presh Talwalkar, "Monday puzzle: what is greater: e^pi or pi^e?," MindYourDecisions Blog, August 5, 2013.
- Comparing pi^e and e^pi without calculating them, Mathematics Stack Exchange, Oct 26, 2010.
- Andrés Vallejo and Italo Bove, "Which is greater: e
^{π}or π^{e}? An unorthodox solution to a classic puzzle," arXiv, September 9, 2023, https://doi.org/10.48550/arXiv.2309.10826.