How can we prove $e^{\pi}-\pi\simeq 20$ geometrically? [closed]

Using a calculator we can easily check that $$\color{Green}{e^{\pi}-\pi}=19.999\cdots\color{Green}{\approx 20}$$ This article and this one provides some details about this almost near identity, but no explanation.

Also we know that $\pi$ plays an important role in geometry.
(not only in Euclidean geometry but also in trigonometry, complex numbers and almost all other geometries.) My question is :

Can we obtain the above almost equality geometrically?

closed as unclear what you're asking by Did, UserX, martini, Mankind, Hagen von EitzenOct 6 '15 at 15:34

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• The first source you quoted states but no satisfying explanation as to "why" $e^{\pi}-\pi \approx 20$ is true has yet been discovered". Are you asking there are any progress in this "open" problem or what? – achille hui Oct 6 '15 at 10:13
• Why on earth should you expect a geometric explanation of the fact that $\sum_{j=2}^{\infty} \frac{\pi^{j}}{j!}$ is close to $19$ (it's not that close in any case)? – Geoff Robinson Oct 6 '15 at 11:27
• This answer shows a geometric interpretation for the terms of the power series for sine and cosine; consequently, $e^\theta$ is simply the total length of the polygonal spiral $P_0P_1P_2P_3\dots$. I don't see an obvious connection between the $e^\pi$ spiral and $20$, but that could just be a failure of imagination on my part. It might be worth noting that the not-very-good approximation $\sqrt{2}+\sqrt{3}\approx \pi$ has a geometric "explanation", so this question isn't completely unreasonable. – Blue Oct 6 '15 at 15:56
• The point for that $\sqrt2+\sqrt3$ is that both of them are "constructible" (is that a word?). Personally I don't know a geometrical approximation of $e$, let alone $e^\pi$. – Quang Hoang Oct 7 '15 at 6:19
• Seems unlikely that a pure geometric solution can be found, as there is no ruler and compass construction for the exponential. The number $e$ itself doesn't seem to mate well with geometry. (And by the way, we have no explanation at all regarding the quasi-equality.) – Yves Daoust Oct 7 '15 at 6:28