Editing Talk:217: e to the pi Minus pi

Asserting that the programmers' algorithms truncated to three decimal digits is an unsupported and unnecessary extrapolation.  Most floating-point implementations use binary, not decimal, and 19.999099979 ''looks'' very much like a rounding error in binary floating-point that has accumulated over several operations. [[User:Daddy|Daddy]] ([[User talk:Daddy|talk]]) 12:39, 29 April 2013 (UTC)
:Fixed. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 22:57, 16 August 2013 (UTC)

The "not good at math" might be too harsh, if they've (tried to) read the floating point spec.  Depending on precision and rounding regime and order of operations, I could easily imagine the "equation" to be true ... and therefore a test that you were rounding "properly", even when it wasn't intuitive.

The third bullet-point above needs changing... (9^2+(19^2/22))=97.4090909091 which is close to pi to the fourth power, so it should be (as noted in the text) (9^2+(19^2/22))^1/4  [[User:Squirreltape|Squirreltape]] ([[User talk:Squirreltape|talk]]) 19:27, 25 February 2014 (UTC)
:Actually, in-case you didn't notice, it says "∜(9² + 19²/22)", not just the sum on its own. I checked the sum on my calculator, and it is equal to what the page is saying. "∜(9² + 19²/22)" means "4th root of (9^2+19^2/22)" (What the title text is saying), or on Windows Calculator, "(9^2+19^2/22) yroot(4)" (Basically what the sum is saying). So, the 3rd bullet point is correct. --[[User:Katavschi|Katavschi]] ([[User talk:Katavschi|talk]]) 22:48, 23 April 2014 (UTC)

It says above that (π + 20)^i ≈ -i, but this should be (π + 20)^i ≈ -1. Proof: π + 20 ≈ e^π => (π + 20)^i ≈ (e^π)^i = e^(πi) = -1.

The ACM competitions are famous for being under tight time pressure. Making your own team waste time would absolutely get you kicked out (and make enemies) [[User:Danshoham|Mountain Hikes]] ([[User talk:Danshoham|talk]]) 04:40, 23 September 2015 (UTC)

;"If they thought about the mathematics"

hm, are you saying it is obvious that e^ pi - pi is not 20? How would you know without approximating it? The sum of two irrationals is not necessarily irrational. [[Special:Contributions/162.158.34.194|162.158.34.194]] 01:58, 26 October 2015 (UTC)
:approximate e^pi using slightly bigger numbers than e and pi (say e: 2.7183 and pi: 3.1416) and subtract a value that is slightly smaller than pi (say 3.1415). The result is less than 20 and a upper limit for e^pi - pi [[Special:Contributions/141.101.93.49|141.101.93.49]] 19:59, 22 August 2016 (UTC)

the [title](https://myighack.com/) text was close; the real identity is e^(π - 2) = π [[Special:Contributions/173.245.52.165|173.245.52.165]] 05:39, 7 April 2021 (UTC)

The approximation in the title text(the first quantity) is an approximation provided by Ramanujan. [[User:Sarah the Pie(yes, the food)|Sarah the Pie(yes, the food)]] ([[User talk:Sarah the Pie(yes, the food)|talk]]) 21:09, 22 February 2022 (UTC)

== Not a coincidence ==

In the "Explanation" section, it says that "the title text pokes fun at ''another'' coincidence". This implies that what came before was a coincidence (without ever actually explicitly saying so). As already mentioned in the trivia section, this was indeed what was believed when the comic was published, but it is no longer true (an explanation for e<sup>π</sup> − π ≈ 20 was found in 2023; it is a simple consequence of π ≈ 22/7). The text should be updated accordingly. Relevant discussion about this can be found on {{w|Talk:Mathematical_coincidence#Explaining_e^%CF%80_%E2%88%92_%CF%80_%E2%89%88_20|this Wikipedia talk page}}, [https://pastebin.com/VzqPG5Gk here], and in the update eventually made to the [https://mathworld.wolfram.com/AlmostInteger.html Mathworld website] (the original source for much of this article). [[Special:Contributions/172.71.250.8|172.71.250.8]] 13:57, 22 April 2024 (UTC)