Difference between revisions of "Talk:217: e to the pi Minus pi"

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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.
 
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.
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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)

Revision as of 04:40, 23 September 2015

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. Daddy (talk) 12:39, 29 April 2013 (UTC)

Fixed. Xhfz (talk) 22:57, 16 August 2013 (UTC)

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 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. --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) Mountain Hikes (talk) 04:40, 23 September 2015 (UTC)