Talk:1856: Existence Proof

Explain xkcd: It's 'cause you're dumb.
Revision as of 01:37, 29 December 2022 by Nitpicking (talk | contribs) (that "=" symbol means equality)
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Whoever added the citation needed got more of a laugh out of me then Randall did this morning. Well done. -- 17:32, 28 June 2017 (UTC)

I hope you enjoy the joke just as much the second time. And the third. And the fourth. And the fifth. And the sixth. And the... 00:56, 29 June 2017 (UTC)
[citation needed][citation needed][citation needed][citation needed][citation needed][citation needed][citation needed] 03:54, 2 July 2017 (UTC)
I don't re-read old pages of explain xkcd so often it would stop being funny. -- Hkmaly (talk) 04:11, 29 June 2017 (UTC)
I disagree, I must have seen 5 or 6 crazy "citation needed"s in recent memory, and for me it never stops being funny. :) A couple of faves have been that a baby could not plan and execute a jewel heist [citation needed] and 5 million years is longer than the average lifespan [citation needed]. :) I might have to start collecting these. NiceGuy1 (talk) 04:22, 30 June 2017 (UTC)
<AOL>Me too!</AOL> RoyT (talk) 07:10, 29 June 2017 (UTC)
[ Citation needed ]Mathmannix (talk) 16:23, 18 July 2017 (UTC)

Does the function have any special hidden meaning, or is it just some random function? Thawn (talk) 20:23, 28 June 2017 (UTC)

Yeah - I wondered that too. But I'm not sure if there is enough information to know. SteveBaker (talk) 21:28, 28 June 2017 (UTC)
Without knowing what the functions are, there's no way to tell. Gmcgath (talk) 23:52, 28 June 2017 (UTC)
Unless I'm way off-base, There are an infinite number of solutions. For example, let's assume f(x)=2x and G(x)=x+1. X can, in this example, be literally any number because G(f(0)) = G(2*0) = G(0) = 0+1 = 1. As long as G(x) takes the result of f(0) and makes it equal to 1, it doesn't matter what f(x) is. 13:25, 29 June 2017 (UTC)
Well, f(x) would also have to be equal to 1, not just G(f(0)). 14:14, 5 July 2017 (UTC)
They're defined to be equal. It's right there on the board! Nitpicking (talk) 01:37, 29 December 2022 (UTC)
Mitchell Feigenbaum's study of the universality of period-doubling ratios involved a function that solved f(0)=1 and af(x)=f(x/a) (IIRC). The equation in the comic reminded me of this, though it's not quite right. 16:55, 29 June 2017 (UTC)

"I'm finally in the right one," made me laugh more than usual. It added character to Offscreen Student #2, something that the comic usually lacks HisHighestMinion (talk) 03:56, 29 June 2017 (UTC)

Could be Black Hat? Observer of the Absurd (talk) 12:23, 6 May 2019 (UTC)

According to "existence proof" means a non-constructive proof. Such proofs are annoying to some mathematicians as they claim existence of something but do not show how to find it. So I fully understand the teacher that she wants to grab a sword and finally find it. 08:54, 29 June 2017 (UTC)

Jokes on her. The number is a | nonstandard integer. 10:12, 29 June 2017 (UTC)

To me, the comic reads (especially with the title text) with the implication that the teacher is encouraging the students to help her actually fight real numbers in real life, as if the platonic idea of numbers was "realer" than we think. 10:42, 29 June 2017 (UTC)

The sentence "There exists some number x such that f(x)=G(f(0))=1." boils down to "There is an x such that f(x)=1". The part with G(f(0)) is only a way to arrive at 1. For some reason there is an x that satisfies f(x)=G(f(0)), and since G(f(0))=1, it is equivalent to f(x)=1. (talk) (please sign your comments with ~~~~)