explain xkcd - User contributions [en]

Talk:1305: Undocumented Feature

2014-05-20T19:03:12Z

Varal7: http://www.xkcd.com/test/ now links to installing (1367)

:;Please never edit existing posts at the talk page! Just add your content! And NEVER edit foreign posts! Use the "Sign Button" on top of editor or type this at the END of your post <nowiki>~~~~</nowiki>. This will add the IP or User and a timestamp to the END of your post.--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:53, 18 December 2013 (UTC)

This sound pretty cool... Anyone know if it's real or which tool it's in? [[Special:Contributions/173.245.55.222|173.245.55.222]] 05:53, 18 December 2013 (UTC)

* its real, there are 8 other users, but must stay a secret. {{unsigned ip|108.162.231.233}}

* There is no secret chat room, stop looking for it. It doesn't exist. Look for your own island on the interweb, don't come spoil ours. [[User:scr_admin|scr_admin]]

Okay, let's be honest: how many of us, upon seeing today's comic, immediately went here to see if it was real or not? --[[Special:Contributions/108.162.245.4|108.162.245.4]] 07:47, 18 December 2013 (UTC)

* I honestly did just that. --[[Special:Contributions/173.245.53.137|173.245.53.137]] 08:06, 18 December 2013 (UTC)

* I also just did that... [[Special:Contributions/108.162.231.206|108.162.231.206]] 08:07, 18 December 2013 (UTC)

* I didn't start up my VM to test it, but I came here to see if was real >.< [[Special:Contributions/108.162.216.56|108.162.216.56]] 09:47, 18 December 2013 (UTC)

* I also did that. But I take that, if it is real and someone uncovers it, it may destroy that community... [[Special:Contributions/173.245.53.123|173.245.53.123]] 10:28, 18 December 2013 (UTC)

* Same here. If it is real, I sincerely hope Randall has a) wiresharked it to find out where this chat room resides so he can prod the admin if it ever goes down b) has a backup plan to migrate himself and his friends to some other private chat room. It won't have the same mystery surrounding it, but at least it's something. [[Special:Contributions/108.162.231.222|108.162.231.222]] 10:51, 18 December 2013 (UTC)

:;Please never edit existing posts at the talk page! Just add your content! And NEVER edit foreign posts! Use the "Sign Button" on top of editor or type this at the END of your post <nowiki>~~~~</nowiki>. This will add the IP or User and a timestamp to the END of your post.--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:53, 18 December 2013 (UTC)

It's not about Youtube, but Facebook, which just launched AUTOPLAYING video ads. Look at the title text, it's about Facebook's real name policy. [[Special:Contributions/108.162.231.232|108.162.231.232]] 08:11, 18 December 2013 (UTC)

* I wouldn't limit the scope of this commentary just to Facebook; YouTube's been doing autoplaying video ads for years. YouTube's also been asking for real names recently. [[Special:Contributions/108.162.212.200|108.162.212.200]] 14:26, 18 December 2013 (UTC)

* The video ads thing is definitely related to Facebook, but the title text is probably a reference to Youtube recently asking continuously to switch to the real name of google plus account and not the nickname many used on YouTube. Edited the explanation accordingly, since there was no reference to the title text. Spesknight [[Special:Contributions/108.162.231.216|108.162.231.216]] 09:08, 19 December 2013 (UTC)

I just searched after reading - and found this site! -- {{unsigned ip|141.101.99.247}}

* The real secret place is here! {{unsigned ip|108.162.229.75}}

* So THIS is the secret chat [[Special:Contributions/108.162.229.7|108.162.229.7]] 09:50, 18 December 2013 (UTC)

* One day this place will be forgotten and so will we. --[[Special:Contributions/108.162.231.197|108.162.231.197]] 09:52, 18 December 2013 (UTC)

anyone else recognizes the wonderful tcp-ip explanation movie of Ericsson [http://www.youtube.com/watch?v=hymzoUpM0K0 Dawn of the net] in frames 6 till 10? [User:Tesshavon|Tesshavon]] ([[User talk:Tesshavon|talk]])

* Tesshavon you're in my mind ! Also, the 6th frame is comes from one of the most common Friends posters (see e.g. here : [http://www.infinitydish.com/tvblog/wp-content/uploads/2012/01/Friends-friends-69087_1024_768.jpg Friends] ) [[User:dandraka|dandraka]]

It's true. Small online communities offer a more folksy experience than the online giants. Some of the best places to hang out are BBS's that made it onto the Internet and have been there for 25+ years. {{unsigned ip|216.150.130.111}}

Well there's always IRC... {{unsigned ip|108.162.221.30}}

:;Please never edit existing posts at the talk page! Just add your content! And NEVER edit foreign posts! Use the "Sign Button" on top of editor or type this at the END of your post <nowiki>~~~~</nowiki>. This will add the IP or User and a timestamp to the END of your post.--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:53, 18 December 2013 (UTC)

I've rewritten all the explanation.
As far as I'm concerned, I'd remove the incomplete box.
I just keep it because it's likely that someone else will feel something is missing.
[[Special:Contributions/173.245.53.180|173.245.53.180]] 15:27, 18 December 2013 (UTC)

If you're interested in a tightknit community out on the fringes of the Net, go join a MUD. Some are combat oriented, some are roleplay and chat oriented, all are text-based, and many have largely the same exact userbase as they had twenty years ago. - [[Special:Contributions/108.162.212.228|108.162.212.228]] 15:48, 18 December 2013 (UTC)

* If you want a really small and odd community check out the Plato network, you have to emulate a terminal from the late 70's early 80's to use it. --[[User:DECtape|DECtape]] ([[User talk:DECtape|talk]]) 00:27, 15 April 2014 (UTC)

Hmm i think randall also wants to share his believs in the subcontext of the comic, the reason why we live on erth as a random error, the sysadmin who probably sees it all(=god), the question what will happen after all that is gone (his opinion, that our lives are compelty senseless)..etc. {{unsigned ip|108.162.254.161}}

* Anyone else think of comic 37 when reading the last panel (due to the ambiguity of whether he is talking about fucking "video ads" or "fucking video" ads)? [[Special:Contributions/173.245.50.227|173.245.50.227]] 18:31, 18 December 2013 (UTC)

Yes, of COURSE I came here to see if it really exists! I don't know if there's actually a chatroom as described, but Usenet has become much smaller, has no ads, and doesn't require you to know the secret application to get in. IIf a text experience with no ads appeals, dump FB, come back to Usenet! Tell 'em Sea Wasp sent you! :) [[Special:Contributions/108.162.219.186|108.162.219.186]] 19:15, 18 December 2013 (UTC)

*Shhh! You're forgetting the first rule of Usenet! [[Special:Contributions/173.245.54.6|173.245.54.6]] 17:57, 19 December 2013 (UTC)

:;Please never edit existing posts at the talk page! Just add your content! And NEVER edit foreign posts! Use the "Sign Button" on top of editor or type this at the END of your post <nowiki>~~~~</nowiki>. This will add the IP or User and a timestamp to the END of your post.--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:53, 18 December 2013 (UTC)

It's obviously about life and religion. The sysadmin who never writes anything must be there to keep everything running, because else the chat would stop to exist. Like most religions contribute to a god who is never seen or heard. --[[Special:Contributions/108.162.231.232|108.162.231.232]] 08:03, 19 December 2013 (UTC)

I'm wondering if he got this idea from Starship Titanic. They had a very similar thing happen. [http://www.metafilter.com/98848/The-Post-That-Cannot-Possibly-Go-Wrong#3435156 See this epic MeFi comment from the self-described "main web hacker" behind Starship Titanic's web site.] [[Special:Contributions/199.27.128.119|199.27.128.119]] 17:29, 19 December 2013 (UTC)

I've made several edits to clean up the explanation. Not sure whether I should remove the incomplete tag or not. --[[Special:Contributions/173.245.52.227|173.245.52.227]] 17:57, 19 December 2013 (UTC)
:Please don't do that. A comic at this size isn't complete within one or two days. Removing the incomplete tag is a minor issue, explaining is the major one. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:46, 19 December 2013 (UTC)

I don't really see why the trivia should be there. [[Special:Contributions/108.162.216.45|108.162.216.45]] 20:29, 19 December 2013 (UTC)
:This content was moved from the explain section to a trivia section by me. It still needs some rework but it belongs to "old Windows utilities" like Randall is talking about here at the first panel.--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:46, 19 December 2013 (UTC)

Reminds me of MUDs. I still check in on New Moon [http://eclipse.cs.pdx.edu/] a few times a year. [[Special:Contributions/108.162.236.25|108.162.236.25]] 16:15, 20 December 2013 (UTC)
:I see what you mean. For me it's the Discworld MUD. But it could similarly (i.e. not exactly like the comic suggests) apply to some long-term Usenet groups that I (in)frequent. [[Special:Contributions/141.101.99.229|141.101.99.229]] 16:22, 20 December 2013 (UTC)

It may not be the tool from the comic, but people here might be interested in: http://kurlander.net/DJ/Projects/ComicChat/resources.html {{unsigned|Jvfrmtn}}

If this chatroom was real I'd love to see it. I know ts not though. Of course what if there's a little fridge horror in this comic? Like a chatroom 98 sort of thing? Maybe the sysadmin or the people Cueball and the others are talking to are really ghosts or souls that were sucked into an old server forever doomed to spend their days talking to themselves until another unsuspecting user is sucked in.[[Special:Contributions/108.162.215.36|108.162.215.36]] 02:54, 23 December 2013 (UTC)

Ido: Can someone explain why the URL www.xkcd.com/test reference to this strip? looks like an undocumented feature to me :) {{unsigned ip|141.101.98.220}}

:It doesn't anymore… [[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 19:03, 20 May 2014 (UTC)

I was doing some searching on the internet, and found, in addition to the one/few on this page, some people who said/implied that they have used this chat before, although, like anything on the internet, the claims may not be true. (Links: http://community.spiceworks.com/topic/436369-does-this-actually-exist [see comments 3, 12, and 14], http://pastebin.com/95nGh8Hk [Says it exists, but doesn't elaborate]) [[User:Z|Z]] ([[User talk:Z|talk]]) 22:02, 12 March 2014 (UTC)

Talk:1047: Approximations

2014-05-17T22:06:20Z

Varal7:

They're actually quite accurate. I've used these in calculations, and they seem to give close enough answers. '''[[User:Davidy22|Davidy22]]'''[[User talk:Davidy22|<tt>[talk]</tt>]] 14:03, 8 January 2013 (UTC)

I only see a use for the liters in a gallon one. The rest are for trolling or simple amusement. The cosine identity bit our math team in the butt at a competition. It was painful. --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 05:27, 17 August 2013 (UTC)

Annoyingly this explanation does not cover 42 properly, it does not say that Douglas Adams got the number 42 from Lewis Carroll, who is more relevant to the page because he was a mathematician named Charles Lutwidge Dodgson. He was obsessed with the number forty-two. The original plate illustrations of Alice in Wonderland drawn by him numbered forty-two. Rule Forty-Two in Alice in Wonderland is "All persons more than a mile high to leave the court", There is also a Code of Honour in the preface of The Hunting of the Snark, an extremely long poem written by him when he was 42 years old, in which rule forty-two is "No one shall speak to the Man at the Helm". The queens in Alice Through the Looking Glass the White Queen announces her age as "one hundred and one, five months and a day", which - if the best possible date is assumed for the action of Through the Looking-Glass - gives a total of 37,044 days. With the further (textually unconfirmed) assumption that both Queens were born on the same day their combined age becomes 74,088 days, which is 42 x 42 x 42. --[[Special:Contributions/139.216.242.254|139.216.242.254]] 02:43, 29 August 2013 (UTC)

:: This explanation covers 42 adequately, and would probably be made slightly worse if such information were added. The very widely known cultural reference is to Adams's interpretation, not Dodgson's original obsession. Adding it would be akin to introducing the MPLM into the explanation for the hijacking of Renaissance artists' names by the TMNT. I definitely concede that it does not cover 42 exhaustively, but I think it can be considered complete and in working order without such an addition. If it really irks you, be bold and add it! --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 00:37, 30 August 2013 (UTC)

"sqrt(2) is not even algebraic in the quotient field of Z[pi]" is not correct. Q is part of the quotient field of Z[pi] and sqrt(2) is algebraic of it. The needed facts are that pi is not algebraic, but the formula implies it is in Q(sqrt(2)). --DrMath 06:47, 7 September 2013 (UTC)

13/15 is a better approximation to sqrt(3)/2 than is e/pi. Continued fraction approximations are great! --DrMath 07:23, 7 September 2013 (UTC)

How could he forget 1 gallon ≈ 0.1337 ft³?! [[Special:Contributions/67.188.195.182|67.188.195.182]] 00:51, 8 September 2013 (UTC)

Worth mentioning that Wolfram Alpha now officially recognizes the [http://www.wolframalpha.com/input/?i=e%5E-%28%281%2B8%5E%281%2F%28e-1%29%29%29%5E%281%2Fpi%29%29 White House switchboard constant] and the [http://www.wolframalpha.com/input/?i=%287%5E%28e-1%2Fe%29-9%29*pi%5E2 Jenny constant]. [[Special:Contributions/86.164.243.91|86.164.243.91]] 18:28, 8 October 2013 (UTC)

Maybe we should add the [Extension:LaTeXSVG LaTeX extension] to make it easier to transcribe these equations. -- [[Special:Contributions/108.162.219.220|108.162.219.220]] 23:02, 16 December 2013 (UTC)

;Protip - Does anyone see the correct equation?
Maybe this is just an other Wolfram Alpha error, like we recently have had here: [[1292: Pi vs. Tau]]. All equations still look invalid to me.
*''√2 = 3/5 + π/(7-π)'': is impossible because √2 is an irrational number and no equation can match.
*''cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2'': could only match if ''cos(x) + cos(3x) + cos(5x) = 1/2'' would be valid, because ''π/7'' is also an irrational number.
*''γ = e/34 + e/5 or γ = e/54 + e/5'': would mean that a sum of two irrational numbers do fit to the Gamma Constant. Impossible.
*''√5 = 13 + 4π / 24 - 4π'': √5 and π are irrational numbers, there is no way to match them in any equation like this.
*''Σ 1/nn = ln(3)e'': doesn't make any sense either.
Maybe [[:Category:Comics featuring Miss Lenhart|Miss Lenhart]] can help.
--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 21:41, 17 December 2013 (UTC)

cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is exactly correct.

Let a=π/7, b=3π/7, and c=5π/7, then
(cosa+cosb+cosc)⋅2sina=2cosasina+2cosbsina+2coscsina=sin2a+sin(b+a)−sin(b−a)+sin(c+a)−sin(c−a)=sin(2π/7)+sin(4π/7)−sin(2π/7)+sin(6π/7)−sin(4π/7)=sin(6π/7)=sin(π/7)=sina

Hence, cos(π/7) + cos(3π/7) + cos(5π/7) = sin(π/7) / 2sin(π/7) = 1/2
[[Special:Contributions/108.162.216.74|108.162.216.74]] 01:57, 16 January 2014 (UTC)

:What is this: sin(6π/7)=sin(π/7) ? A new math is born... --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:49, 16 January 2014 (UTC)

::Actually it does. My proof is geometric: the sines of two supplementary angles (angle a + angle b = π (in radians)) are equivalent because they necessarily have the same x height in a Cartesian plane. Look on a unit circle, or even a sine function. Also, Calculus and most other mathematics use radians over degrees because they make the functions simpler and eliminate irrationality when a trig function shows up, but physics uses degrees because it's easier to understand and taught first. Anonymous 01:27, 13 February 2014 (UTC)
::As an aside, just how far along in math are you? Radian measure is taught in high school (at least the good ones). Anonymous 13:24, 13 February 2014 (UTC)
:::Sure, I was wrong at my last statement. sin(6π/7)=sin(π/7) is correct by using the radian measure. But just change π/7 to π/77 would give a very different result on that formular here. I still can't figure out why PI divided by the number 7 should be that unique, PI divided by 77 should be the same. My fault is: I still can't find the Nerd Sniping here. And we all do know that Randall did use wrong WolframAlpha results here. According to the last question: I'm very well on Math, that's because I want to understand this. This is like 0.999=1. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 22:01, 13 February 2014 (UTC)
::::Ah, I see. I think it has to do with the way e^i*π breaks down, as one of the answers shown in the corresponding link explains, but other answers rely on various angle identities (including the supplementary sines one in the proof above). Anonymous 03:10, 14 February 2014 (UTC) (PS, have you checked [[545]] lately? I answered your question there, too)
:::::As per the derivation from january 16 , you can use any a,b,c that satisfies this set of equations: 2 a = b - a, a + b = c- a, c + a = π - a. This is due to the fact that sin(x) = sin(π-x), and what was derived the 16th. [[Special:Contributions/173.245.53.199|173.245.53.199]] 12:38, 21 February 2014 (UTC)
:::: Dgbrt: If not convinced by the proofs linked to in the "explanation" part, you might want to try this: [http://www.wolframalpha.com/input/?i=sum_%28k%3D0%29%5E38+cos%281%2F77+%282+k%2B1%29+pi%29]. I'm sure you'll find inspiration for similar formulas using PI over [any odd integer]. Your assumption that Randall used WolframAlpha for this very identity is probably wrong. This is a very well-known formula that appears in many high school books, and I am pretty sure it is part of Randall's culture. And this has nothing to do with 1=1. As for your original post,
::::*√2 = (√2-1)/((4-2)π/2-π)+1 : Is this what you call "matching an equation" to √2?
::::*So what you mean is that if an equation is true for an irrational number, then it must be for any real number? Like, (√2)^2 = 2, but because √2 is irrational, then x^2=2 (for all x?)
::::*This one's a bit tough. You will probably agree that γ-√2 is irrational. And so is √2. What about their sum?
::::*Well, maybe it doesn't to you. But is Σ n-2 = π^2/6 any better? Well, this one is true (using Fourier's expansion of the rectangular function).
::::Finally,
::::*√2 = 3/5 + π/(7-π) is false because it would imply that π is an algebraic number
::::*cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is true, and proven by many
::::*γ = e/34 + e/5 seems false. But there doesn't seem to be a quick way to disprove.
::::*Σ 1/nn = ln(3)e seems false, but I can't see why. [[Special:Contributions/108.162.210.234|108.162.210.234]] 09:15, 11 May 2014 (UTC)

;So, still incomplete?

Where's our (in)complete judge? [[Special:Contributions/199.27.128.186|199.27.128.186]] 19:21, 18 December 2013 (UTC)
:The protip is still a mystery. I'm calling for help a few lines above. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 21:16, 18 December 2013 (UTC)
::The cosine one, in radians, is correct [[Special:Contributions/141.101.88.225|141.101.88.225]] 12:54, 28 April 2014 (UTC)

The 'Seconds in a year' ones remind me of one of my favorite quotes: "How many seconds are there in a year? If I tell you there are 3.155 x 10^7, you won't even try to remember it. On the other hand, who could forget that, to within half a percent, pi seconds is a nanocentury" -- Tom Duff, Bell Labs. [[User:Beolach|Beolach]] ([[User talk:Beolach|talk]]) 19:14, 17 April 2014 (UTC)
:Please do not change former discussions. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:57, 17 April 2014 (UTC)

;cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 1/2 ???
Why the hell the divider seven makes the difference?
*cos(pi) + cos(3*pi) + cos(5*pi) = -3
*cos(pi/8) + cos(3*pi/8) + cos(5*pi/8) = 0.92387953251128675612818318939678828682241662586364...
So why the "magic" prime number seven produces this exact result? I know radians and π/7 is just a small part of a circle which is 2π. One prove claims that sin(6π/7) equals to sin(π/7); my best calculator can't show a difference. Of course sin(6π) equals to sin(π), in radians, BUT sin(6π/8) is NOT equal to sin(π/8). So if the number 7 plays a magic rule here this would be "one of the", no... the BIGGEST mystery in mathematics forever. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:03, 16 May 2014 (UTC)
:Dgbrt, please see my answer from 11 May 2014 up there. Any odd integer will do, as long as you sum enough of cos(pi/[thing]).
:*Let's try with 5 : cos(pi/5) + cos (3pi/5) = 1/2.
:*With 9 : cos(pi/9)+ cos(3pi/9) + cos (5pi/9) + cos(7pi/9) = 1/2
: No big mystery around here. Just a beautiful formula :) I think there are similar formulas with cosines and even integers. I'll post them here if I have time. [[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 09:56, 17 May 2014 (UTC)

::You mixing up to different equations and even not prove them. If there is any prove to a mathematician I would accept and include a proper explain for non math people here. We still have to find a prove. And I do not trust my calculators, we just have to explain why even cos(pi/5) + cos (3pi/5) is also nearly the same. This issue is still not explained. So please give us a explain. And a PROTIP: This does not work with Integers, PI is infinite--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 17:55, 17 May 2014 (UTC)

:::Okay. If I understood what you said.
:::* I mix up different topics. -> True. From now on, we'll just focus on the cosine one.
:::* You ask for a proof/explanation. -> My opinion is those are two different requests. Maybe that's why you use the distinction between math people/not math people. For a proof, please read further. What I exposed above are just other "fun experiments" we could do. e.g : [http://www.wolframalpha.com/input/?i=cos%28pi%2F11%29%2Bcos+%283pi%2F11%29+%2B+cos+%285pi%2F11%29+%2B+cos+%287pi%2F11%29%2Bcos+%289pi%2F11%29].
:::* You do not trust your calculators -> Great. I don't either. (Well more accurately, I trust mine to 10^-8, so I would definitely not use it to prove any of the discussed equations in PROTIP). That's why we'll prove the formulas we assert.
:::* "This does not work with integers" -> Well, I got myself misunderstood. It would probably have been better if I had said: the following formula is true for all integer n. sum_{k=0}^{n-1}{cos((2k+1)*pi/(2n+1)). But It's harder to read, so just say. Choose any odd integer, say N=2n+1. Then start the following sum. cos(pi/N) + cos(3pi/N) + … and stop when the numerator is cos((N-2)pi/N). Then the result is 1/2. And that's what we'll prove, a few lines down from here.
:::*"Pi is infinite" -> That's a common misconception. What you mean is, Pi is irrational. (Fun fact: Pi is a transcendental number. Quite difficult theorem. Lindeman proved it in 1882. Hence, if we identify the real number x with the Q-vector space Q[x], it would make sense to say that "x is infinite" because, the Q-vector space Q[x] is indeed of infinite dimension. But then, that's not what mathematicians do). I think Vi Hart made a video where she addresses this issue (or was it someone else?). Anyway, I might come to that point some other time in the future.
:::Okay, so now let's first prove the protip formula. Well first, here is the link that the explainxkcd wiki points to: [http://math.stackexchange.com/questions/140388/how-can-one-prove-cos-pi-7-cos3-pi-7-cos5-pi-7-1-2]. Most of them are correct. Some are more ugly than others. I'll adapt the last one.
::: We need a [http://en.wikipedia.org/wiki/Complex_number complex numbers]. (I choosed this because I think explainxkcd readers are fine with this. See comic [http://explainxkcd.com/179/ 179]). I will be using dots to show the steps of my proof. Please allow me an extra level of indent for clarity's sake.
:::'''Proof'''
:::: *Let z be a primitive 14-th [http://en.wikipedia.org/wiki/Root_of_unity root of unity] (the reader doesn't need to understand the 3 last words). Just say z = exp(i*pi/7) = cos(pi/7) + i sin(pi/7). Using [http://en.wikipedia.org/wiki/Euler%27s_formula Euler's formula].
:::: *We have z^14-1 = (exp(i*pi/7))^14-1 = exp(i*2pi) - 1 = 0. Using exponential law for integer powers, as seen in this article: [http://en.wikipedia.org/wiki/De_Moivre%27s_formula De Moivre's formula].
:::: *Now let's factor: z^14-1 = (z^7-1)(z^7+1) = (z^7-1)(z+1)(z^6-z^5+z^4-z^3+z^2-z+1) = (z^7-1)(z+1)*Phi_14(z). where Phi_14(X)= X^6-X^5+X^4-X^3+X^2-X+1, (see [http://en.wikipedia.org/wiki/Cyclotomic_polynomial cycltomic polynomial]). Now, because z^7-1 = (exp(i*pi/7))^7-1 = exp(i*pi)-1 = -2. And because z is not -1, the two first factors are not 0 so, Phi_14(z) = 0, which is already a pretty awesome equality.
:::: *Note that exp(i*pi/7)*exp(i*6pi/7)= exp(i*pi)=-1. So the inverse of z is -exp(i*6pi/7). But we also know that it is exp(-i*pi/7). Well. That was just a fancy way to prove that exp(-i*pi/7) = - exp(i*6pi/7). Good enough. The same holds for exp(-i*3pi/7) = exp(i*14pi/7)*exp(-i*3pi/7)=exp(i*11pi/7)=exp(i*7pi/7)*exp(i*4pi/7)=-exp(i*4pi/7). And the exact same calculation shows that exp(-i*5pi/7)=-exp(i*2pi/7). Alright.
:::: *Now, use that for any x, we have cos(x) = (exp(ix)+exp(-ix))/2. See [http://en.wikipedia.org/wiki/Euler%27s_formula#Relationship_to_trigonometry here]. Let's calculate twice the sum of the left hand side. 2(cos(pi/7)+cos(3pi/7)+cos(5pi/7))= exp(i*pi/7) + expi(-i*pi/7) + exp(3pi/7) + exp(-3pi/7) + exp(5pi/7) +exp(-5pi/7) = exp(i*pi/7)-exp(i*2pi/7)+exp(i*3pi/7)-exp(i*4pi/7)+exp(i*5pi/7)-exp(i*6pi/7) = -Phi_14(z) +1 = 1.
:::: * So dividing both sides by 2, we get what we want. Pfew.
::: '''Why is 7 so special? Well it isn't.''' Let's prove it for 9.
::::* Let z = exp(i*pi/9) = cos(pi/9) + i sin(pi/9). We have z^18-1 = 0, and z^9-1 and z+1 are not 0, so using the same factorisation, Phi_18(z) = z^8-z^7+z^6-z^5+z^4-z^3+z^2-z+1 = 0.
::::* Hence, the conclusion follow from: 2(cos(pi/9) + cos(3pi/9) + cos(5pi/9) + cos(7pi/9)) = exp(i*pi/9) + exp(-i*pi/9) + exp(i*3pi/9) + exp(-i*3pi/9) + exp(i*5pi/9) + exp(-i*5pi/9) + exp(i*7pi/9) + exp(-i*7pi/9) = -Phi_18(z)+1 = 1.
::: Well, well. I hope you kinda see the pattern. Dgbrt, I know you hate typos, and I'm pretty sure that in this long text lay many of them. So I apologize, and I will correct them later. The following paragraph was posted after I started my text but before I finished mine. It wasn't signed so I will just leave it down there. It's another valid straightforward proof. Oh. And Friendly TIP: Don't say protip when you're not pro. [[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 21:50, 17 May 2014 (UTC)

The valid identity cos(pi/7)+cos(3pi/7)+cos(5pi/7)=1/2 was correctly proved by the writer at 108.162.216.74 above. For a different proof, consider the complex number z = cos(pi/7)+i sin(pi/7) corresponding to rotation of the complex plane by pi/7 radians, i.e., 1/14th of a full rotation. It satisfies z^{14} -1 = 0 (z to the fourteenth is one). Dividing by z-1 gives z^{13} + z^{12} + ... + z + 1 = 0. The same argument, starting with z^2 corresponding to 1/7th of a full rotation, gives z^{12} + z^{10} + ... z^2 + 1 = 0. Taking the difference, we get z^{13} + z^{11} + ... + z^3 + z = 0. Looking only at the real parts, we get cos(13pi/7) + cos(11pi/7) + cos(9pi/7) + cos(7pi/7) + cos(5pi/7) + cos(3pi/7) + cos(pi/7) = 0. Here cos(13pi/7) = cos(pi/7), cos(11pi/7) = cos(3pi/7) and cos(9pi/7) = cos(5pi/7), since cos is even and 2pi-periodic. Finally cos(7pi/7) = -1, so 2(cos(pi/7) + cos(3pi/7) + cos(5pi/7)) - 1 = 0, which you can rewrite as the desired identity. All of this can be clearly visualized using a regular 14-gon, so a proof with pictures is possible.

Talk:1047: Approximations

2014-05-17T22:05:47Z

Varal7:

They're actually quite accurate. I've used these in calculations, and they seem to give close enough answers. '''[[User:Davidy22|Davidy22]]'''[[User talk:Davidy22|<tt>[talk]</tt>]] 14:03, 8 January 2013 (UTC)

I only see a use for the liters in a gallon one. The rest are for trolling or simple amusement. The cosine identity bit our math team in the butt at a competition. It was painful. --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 05:27, 17 August 2013 (UTC)

Annoyingly this explanation does not cover 42 properly, it does not say that Douglas Adams got the number 42 from Lewis Carroll, who is more relevant to the page because he was a mathematician named Charles Lutwidge Dodgson. He was obsessed with the number forty-two. The original plate illustrations of Alice in Wonderland drawn by him numbered forty-two. Rule Forty-Two in Alice in Wonderland is "All persons more than a mile high to leave the court", There is also a Code of Honour in the preface of The Hunting of the Snark, an extremely long poem written by him when he was 42 years old, in which rule forty-two is "No one shall speak to the Man at the Helm". The queens in Alice Through the Looking Glass the White Queen announces her age as "one hundred and one, five months and a day", which - if the best possible date is assumed for the action of Through the Looking-Glass - gives a total of 37,044 days. With the further (textually unconfirmed) assumption that both Queens were born on the same day their combined age becomes 74,088 days, which is 42 x 42 x 42. --[[Special:Contributions/139.216.242.254|139.216.242.254]] 02:43, 29 August 2013 (UTC)

:: This explanation covers 42 adequately, and would probably be made slightly worse if such information were added. The very widely known cultural reference is to Adams's interpretation, not Dodgson's original obsession. Adding it would be akin to introducing the MPLM into the explanation for the hijacking of Renaissance artists' names by the TMNT. I definitely concede that it does not cover 42 exhaustively, but I think it can be considered complete and in working order without such an addition. If it really irks you, be bold and add it! --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 00:37, 30 August 2013 (UTC)

"sqrt(2) is not even algebraic in the quotient field of Z[pi]" is not correct. Q is part of the quotient field of Z[pi] and sqrt(2) is algebraic of it. The needed facts are that pi is not algebraic, but the formula implies it is in Q(sqrt(2)). --DrMath 06:47, 7 September 2013 (UTC)

13/15 is a better approximation to sqrt(3)/2 than is e/pi. Continued fraction approximations are great! --DrMath 07:23, 7 September 2013 (UTC)

How could he forget 1 gallon ≈ 0.1337 ft³?! [[Special:Contributions/67.188.195.182|67.188.195.182]] 00:51, 8 September 2013 (UTC)

Worth mentioning that Wolfram Alpha now officially recognizes the [http://www.wolframalpha.com/input/?i=e%5E-%28%281%2B8%5E%281%2F%28e-1%29%29%29%5E%281%2Fpi%29%29 White House switchboard constant] and the [http://www.wolframalpha.com/input/?i=%287%5E%28e-1%2Fe%29-9%29*pi%5E2 Jenny constant]. [[Special:Contributions/86.164.243.91|86.164.243.91]] 18:28, 8 October 2013 (UTC)

Maybe we should add the [Extension:LaTeXSVG LaTeX extension] to make it easier to transcribe these equations. -- [[Special:Contributions/108.162.219.220|108.162.219.220]] 23:02, 16 December 2013 (UTC)

;Protip - Does anyone see the correct equation?
Maybe this is just an other Wolfram Alpha error, like we recently have had here: [[1292: Pi vs. Tau]]. All equations still look invalid to me.
*''√2 = 3/5 + π/(7-π)'': is impossible because √2 is an irrational number and no equation can match.
*''cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2'': could only match if ''cos(x) + cos(3x) + cos(5x) = 1/2'' would be valid, because ''π/7'' is also an irrational number.
*''γ = e/34 + e/5 or γ = e/54 + e/5'': would mean that a sum of two irrational numbers do fit to the Gamma Constant. Impossible.
*''√5 = 13 + 4π / 24 - 4π'': √5 and π are irrational numbers, there is no way to match them in any equation like this.
*''Σ 1/nn = ln(3)e'': doesn't make any sense either.
Maybe [[:Category:Comics featuring Miss Lenhart|Miss Lenhart]] can help.
--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 21:41, 17 December 2013 (UTC)

cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is exactly correct.

Let a=π/7, b=3π/7, and c=5π/7, then
(cosa+cosb+cosc)⋅2sina=2cosasina+2cosbsina+2coscsina=sin2a+sin(b+a)−sin(b−a)+sin(c+a)−sin(c−a)=sin(2π/7)+sin(4π/7)−sin(2π/7)+sin(6π/7)−sin(4π/7)=sin(6π/7)=sin(π/7)=sina

Hence, cos(π/7) + cos(3π/7) + cos(5π/7) = sin(π/7) / 2sin(π/7) = 1/2
[[Special:Contributions/108.162.216.74|108.162.216.74]] 01:57, 16 January 2014 (UTC)

:What is this: sin(6π/7)=sin(π/7) ? A new math is born... --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:49, 16 January 2014 (UTC)

::Actually it does. My proof is geometric: the sines of two supplementary angles (angle a + angle b = π (in radians)) are equivalent because they necessarily have the same x height in a Cartesian plane. Look on a unit circle, or even a sine function. Also, Calculus and most other mathematics use radians over degrees because they make the functions simpler and eliminate irrationality when a trig function shows up, but physics uses degrees because it's easier to understand and taught first. Anonymous 01:27, 13 February 2014 (UTC)
::As an aside, just how far along in math are you? Radian measure is taught in high school (at least the good ones). Anonymous 13:24, 13 February 2014 (UTC)
:::Sure, I was wrong at my last statement. sin(6π/7)=sin(π/7) is correct by using the radian measure. But just change π/7 to π/77 would give a very different result on that formular here. I still can't figure out why PI divided by the number 7 should be that unique, PI divided by 77 should be the same. My fault is: I still can't find the Nerd Sniping here. And we all do know that Randall did use wrong WolframAlpha results here. According to the last question: I'm very well on Math, that's because I want to understand this. This is like 0.999=1. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 22:01, 13 February 2014 (UTC)
::::Ah, I see. I think it has to do with the way e^i*π breaks down, as one of the answers shown in the corresponding link explains, but other answers rely on various angle identities (including the supplementary sines one in the proof above). Anonymous 03:10, 14 February 2014 (UTC) (PS, have you checked [[545]] lately? I answered your question there, too)
:::::As per the derivation from january 16 , you can use any a,b,c that satisfies this set of equations: 2 a = b - a, a + b = c- a, c + a = π - a. This is due to the fact that sin(x) = sin(π-x), and what was derived the 16th. [[Special:Contributions/173.245.53.199|173.245.53.199]] 12:38, 21 February 2014 (UTC)
:::: Dgbrt: If not convinced by the proofs linked to in the "explanation" part, you might want to try this: [http://www.wolframalpha.com/input/?i=sum_%28k%3D0%29%5E38+cos%281%2F77+%282+k%2B1%29+pi%29]. I'm sure you'll find inspiration for similar formulas using PI over [any odd integer]. Your assumption that Randall used WolframAlpha for this very identity is probably wrong. This is a very well-known formula that appears in many high school books, and I am pretty sure it is part of Randall's culture. And this has nothing to do with 1=1. As for your original post,
::::*√2 = (√2-1)/((4-2)π/2-π)+1 : Is this what you call "matching an equation" to √2?
::::*So what you mean is that if an equation is true for an irrational number, then it must be for any real number? Like, (√2)^2 = 2, but because √2 is irrational, then x^2=2 (for all x?)
::::*This one's a bit tough. You will probably agree that γ-√2 is irrational. And so is √2. What about their sum?
::::*Well, maybe it doesn't to you. But is Σ n-2 = π^2/6 any better? Well, this one is true (using Fourier's expansion of the rectangular function).
::::Finally,
::::*√2 = 3/5 + π/(7-π) is false because it would imply that π is an algebraic number
::::*cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is true, and proven by many
::::*γ = e/34 + e/5 seems false. But there doesn't seem to be a quick way to disprove.
::::*Σ 1/nn = ln(3)e seems false, but I can't see why. [[Special:Contributions/108.162.210.234|108.162.210.234]] 09:15, 11 May 2014 (UTC)

;So, still incomplete?

Where's our (in)complete judge? [[Special:Contributions/199.27.128.186|199.27.128.186]] 19:21, 18 December 2013 (UTC)
:The protip is still a mystery. I'm calling for help a few lines above. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 21:16, 18 December 2013 (UTC)
::The cosine one, in radians, is correct [[Special:Contributions/141.101.88.225|141.101.88.225]] 12:54, 28 April 2014 (UTC)

The 'Seconds in a year' ones remind me of one of my favorite quotes: "How many seconds are there in a year? If I tell you there are 3.155 x 10^7, you won't even try to remember it. On the other hand, who could forget that, to within half a percent, pi seconds is a nanocentury" -- Tom Duff, Bell Labs. [[User:Beolach|Beolach]] ([[User talk:Beolach|talk]]) 19:14, 17 April 2014 (UTC)
:Please do not change former discussions. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:57, 17 April 2014 (UTC)

;cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 1/2 ???
Why the hell the divider seven makes the difference?
*cos(pi) + cos(3*pi) + cos(5*pi) = -3
*cos(pi/8) + cos(3*pi/8) + cos(5*pi/8) = 0.92387953251128675612818318939678828682241662586364...
So why the "magic" prime number seven produces this exact result? I know radians and π/7 is just a small part of a circle which is 2π. One prove claims that sin(6π/7) equals to sin(π/7); my best calculator can't show a difference. Of course sin(6π) equals to sin(π), in radians, BUT sin(6π/8) is NOT equal to sin(π/8). So if the number 7 plays a magic rule here this would be "one of the", no... the BIGGEST mystery in mathematics forever. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:03, 16 May 2014 (UTC)
:Dgbrt, please see my answer from 11 May 2014 up there. Any odd integer will do, as long as you sum enough of cos(pi/[thing]).
:*Let's try with 5 : cos(pi/5) + cos (3pi/5) = 1/2.
:*With 9 : cos(pi/9)+ cos(3pi/9) + cos (5pi/9) + cos(7pi/9) = 1/2
: No big mystery around here. Just a beautiful formula :) I think there are similar formulas with cosines and even integers. I'll post them here if I have time. [[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 09:56, 17 May 2014 (UTC)

::You mixing up to different equations and even not prove them. If there is any prove to a mathematician I would accept and include a proper explain for non math people here. We still have to find a prove. And I do not trust my calculators, we just have to explain why even cos(pi/5) + cos (3pi/5) is also nearly the same. This issue is still not explained. So please give us a explain. And a PROTIP: This does not work with Integers, PI is infinite--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 17:55, 17 May 2014 (UTC)

:::Okay. If I understood what you said.
:::* I mix up different topics. -> True. From now on, we'll just focus on the cosine one.
:::* You ask for a proof/explanation. -> My opinion is those are two different requests. Maybe that's why you use the distinction between math people/not math people. For a proof, please read further. What I exposed above are just other "fun experiments" we could do. e.g : [http://www.wolframalpha.com/input/?i=cos%28pi%2F11%29%2Bcos+%283pi%2F11%29+%2B+cos+%285pi%2F11%29+%2B+cos+%287pi%2F11%29%2Bcos+%289pi%2F11%29].
:::* You do not trust your calculators -> Great. I don't either. (Well more accurately, I trust mine to 10^-8, so I would definitely not use it to prove any of the discussed equations in PROTIP). That's why we'll prove the formulas we assert.
:::* "This does not work with integers" -> Well, I got myself misunderstood. It would probably have been better if I had said: the following formula is true for all integer n. sum_{k=0}^{n-1}{cos((2k+1)*pi/(2n+1)). But It's harder to read, so just say. Choose any odd integer, say N=2n+1. Then start the following sum. cos(pi/N) + cos(3pi/N) + … and stop when the numerator is cos((N-2)pi/N). Then the result is 1/2. And that's what we'll prove, a few lines down from here.
:::*"Pi is infinite" -> That's a common misconception. What you mean is, Pi is irrational. (Fun fact: Pi is a transcendental number. Quite difficult theorem. Lindeman proved it in 1882. Hence, if we identify the real number x with the Q-vector space Q[x], it would make sense to say that "x is infinite" because, the Q-vector space Q[x] is indeed of infinite dimension. But then, that's not what mathematicians do). I think Vi Hart made a video where she addresses this issue (or was it someone else?). Anyway, I might come to that point some other time in the future.
:::Okay, so now let's first prove the protip formula. Well first, here is the link that the explainxkcd wiki points to: [http://math.stackexchange.com/questions/140388/how-can-one-prove-cos-pi-7-cos3-pi-7-cos5-pi-7-1-2]. Most of them are correct. Some are more ugly than others. I'll adapt the last one.
::: We need a [http://en.wikipedia.org/wiki/Complex_number complex numbers]. (I choosed this because I think explainxkcd readers are fine with this. See comic [http://explainxkcd.com/179/ 179]). I will be using dots to show the steps of my proof. Please allow me an extra level of indent for clarity's sake.
:::'''Proof'''
:::: *Let z be a primitive 14-th [http://en.wikipedia.org/wiki/Root_of_unity root of unity] (the reader doesn't need to understand the 3 last words). Just say z = exp(i*pi/7) = cos(pi/7) + i sin(pi/7). Using [http://en.wikipedia.org/wiki/Euler%27s_formula Euler's formula].
:::: *We have z^14-1 = (exp(i*pi/7))^14-1 = exp(i*2pi) - 1 = 0. Using exponential law for integer powers, as seen in this article: [http://en.wikipedia.org/wiki/De_Moivre%27s_formula De Moivre's formula].
:::: *Now let's factor: z^14-1 = (z^7-1)(z^7+1) = (z^7-1)(z+1)(z^6-z^5+z^4-z^3+z^2-z+1) = (z^7-1)(z+1)*Phi_14(z). where Phi_14(X)= X^6-X^5+X^4-X^3+X^2-X+1, (see [http://en.wikipedia.org/wiki/Cyclotomic_polynomial cycltomic polynomial]). Now, because z^7-1 = (exp(i*pi/7))^7-1 = exp(i*pi)-1 = -2. And because z is not -1, the two first factors are not 0 so, Phi_14(z) = 0, which is already a pretty awesome equality.
:::: *Note that exp(i*pi/7)*exp(i*6pi/7)= exp(i*pi)=-1. So the inverse of z is -exp(i*6pi/7). But we also know that it is exp(-i*pi/7). Well. That was just a fancy way to prove that exp(-i*pi/7) = - exp(i*6pi/7). Good enough. The same holds for exp(-i*3pi/7) = exp(i*14pi/7)*exp(-i*3pi/7)=exp(i*11pi/7)=exp(i*7pi/7)*exp(i*4pi/7)=-exp(i*4pi/7). And the exact same calculation shows that exp(-i*5pi/7)=-exp(i*2pi/7). Alright.
:::: *Now, use that for any x, we have cos(x) = (exp(ix)+exp(-ix))/2. See [http://en.wikipedia.org/wiki/Euler%27s_formula#Relationship_to_trigonometry here]. Let's calculate twice the sum of the left hand side. 2(cos(pi/7)+cos(3pi/7)+cos(5pi/7))= exp(i*pi/7) + expi(-i*pi/7) + exp(3pi/7) + exp(-3pi/7) + exp(5pi/7) +exp(-5pi/7) = exp(i*pi/7)-exp(i*2pi/7)+exp(i*3pi/7)-exp(i*4pi/7)+exp(i*5pi/7)-exp(i*6pi/7) = -Phi_14(z) +1 = 1.
:::: * So dividing both sides by 2, we get what we want. Pfiou.
::: '''Why is 7 so special? Well it isn't.''' Let's prove it for 9.
::::* Let z = exp(i*pi/9) = cos(pi/9) + i sin(pi/9). We have z^18-1 = 0, and z^9-1 and z+1 are not 0, so using the same factorisation, Phi_18(z) = z^8-z^7+z^6-z^5+z^4-z^3+z^2-z+1 = 0.
::::* Hence, the conclusion follow from: 2(cos(pi/9) + cos(3pi/9) + cos(5pi/9) + cos(7pi/9)) = exp(i*pi/9) + exp(-i*pi/9) + exp(i*3pi/9) + exp(-i*3pi/9) + exp(i*5pi/9) + exp(-i*5pi/9) + exp(i*7pi/9) + exp(-i*7pi/9) = -Phi_18(z)+1 = 1.
::: Well, well. I hope you kinda see the pattern. Dgbrt, I know you hate typos, and I'm pretty sure that in this long text lay many of them. So I apologize, and I will correct them later. The following paragraph was posted after I started my text but before I finished mine. It wasn't signed so I will just leave it down there. It's another valid straightforward proof. Oh. And Friendly TIP: Don't say protip when you're not pro. [[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 21:50, 17 May 2014 (UTC)

The valid identity cos(pi/7)+cos(3pi/7)+cos(5pi/7)=1/2 was correctly proved by the writer at 108.162.216.74 above. For a different proof, consider the complex number z = cos(pi/7)+i sin(pi/7) corresponding to rotation of the complex plane by pi/7 radians, i.e., 1/14th of a full rotation. It satisfies z^{14} -1 = 0 (z to the fourteenth is one). Dividing by z-1 gives z^{13} + z^{12} + ... + z + 1 = 0. The same argument, starting with z^2 corresponding to 1/7th of a full rotation, gives z^{12} + z^{10} + ... z^2 + 1 = 0. Taking the difference, we get z^{13} + z^{11} + ... + z^3 + z = 0. Looking only at the real parts, we get cos(13pi/7) + cos(11pi/7) + cos(9pi/7) + cos(7pi/7) + cos(5pi/7) + cos(3pi/7) + cos(pi/7) = 0. Here cos(13pi/7) = cos(pi/7), cos(11pi/7) = cos(3pi/7) and cos(9pi/7) = cos(5pi/7), since cos is even and 2pi-periodic. Finally cos(7pi/7) = -1, so 2(cos(pi/7) + cos(3pi/7) + cos(5pi/7)) - 1 = 0, which you can rewrite as the desired identity. All of this can be clearly visualized using a regular 14-gon, so a proof with pictures is possible.

Talk:1047: Approximations

2014-05-17T21:52:41Z

Varal7:

They're actually quite accurate. I've used these in calculations, and they seem to give close enough answers. '''[[User:Davidy22|Davidy22]]'''[[User talk:Davidy22|<tt>[talk]</tt>]] 14:03, 8 January 2013 (UTC)

I only see a use for the liters in a gallon one. The rest are for trolling or simple amusement. The cosine identity bit our math team in the butt at a competition. It was painful. --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 05:27, 17 August 2013 (UTC)

Annoyingly this explanation does not cover 42 properly, it does not say that Douglas Adams got the number 42 from Lewis Carroll, who is more relevant to the page because he was a mathematician named Charles Lutwidge Dodgson. He was obsessed with the number forty-two. The original plate illustrations of Alice in Wonderland drawn by him numbered forty-two. Rule Forty-Two in Alice in Wonderland is "All persons more than a mile high to leave the court", There is also a Code of Honour in the preface of The Hunting of the Snark, an extremely long poem written by him when he was 42 years old, in which rule forty-two is "No one shall speak to the Man at the Helm". The queens in Alice Through the Looking Glass the White Queen announces her age as "one hundred and one, five months and a day", which - if the best possible date is assumed for the action of Through the Looking-Glass - gives a total of 37,044 days. With the further (textually unconfirmed) assumption that both Queens were born on the same day their combined age becomes 74,088 days, which is 42 x 42 x 42. --[[Special:Contributions/139.216.242.254|139.216.242.254]] 02:43, 29 August 2013 (UTC)

:: This explanation covers 42 adequately, and would probably be made slightly worse if such information were added. The very widely known cultural reference is to Adams's interpretation, not Dodgson's original obsession. Adding it would be akin to introducing the MPLM into the explanation for the hijacking of Renaissance artists' names by the TMNT. I definitely concede that it does not cover 42 exhaustively, but I think it can be considered complete and in working order without such an addition. If it really irks you, be bold and add it! --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 00:37, 30 August 2013 (UTC)

"sqrt(2) is not even algebraic in the quotient field of Z[pi]" is not correct. Q is part of the quotient field of Z[pi] and sqrt(2) is algebraic of it. The needed facts are that pi is not algebraic, but the formula implies it is in Q(sqrt(2)). --DrMath 06:47, 7 September 2013 (UTC)

13/15 is a better approximation to sqrt(3)/2 than is e/pi. Continued fraction approximations are great! --DrMath 07:23, 7 September 2013 (UTC)

How could he forget 1 gallon ≈ 0.1337 ft³?! [[Special:Contributions/67.188.195.182|67.188.195.182]] 00:51, 8 September 2013 (UTC)

Worth mentioning that Wolfram Alpha now officially recognizes the [http://www.wolframalpha.com/input/?i=e%5E-%28%281%2B8%5E%281%2F%28e-1%29%29%29%5E%281%2Fpi%29%29 White House switchboard constant] and the [http://www.wolframalpha.com/input/?i=%287%5E%28e-1%2Fe%29-9%29*pi%5E2 Jenny constant]. [[Special:Contributions/86.164.243.91|86.164.243.91]] 18:28, 8 October 2013 (UTC)

Maybe we should add the [Extension:LaTeXSVG LaTeX extension] to make it easier to transcribe these equations. -- [[Special:Contributions/108.162.219.220|108.162.219.220]] 23:02, 16 December 2013 (UTC)

;Protip - Does anyone see the correct equation?
Maybe this is just an other Wolfram Alpha error, like we recently have had here: [[1292: Pi vs. Tau]]. All equations still look invalid to me.
*''√2 = 3/5 + π/(7-π)'': is impossible because √2 is an irrational number and no equation can match.
*''cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2'': could only match if ''cos(x) + cos(3x) + cos(5x) = 1/2'' would be valid, because ''π/7'' is also an irrational number.
*''γ = e/34 + e/5 or γ = e/54 + e/5'': would mean that a sum of two irrational numbers do fit to the Gamma Constant. Impossible.
*''√5 = 13 + 4π / 24 - 4π'': √5 and π are irrational numbers, there is no way to match them in any equation like this.
*''Σ 1/nn = ln(3)e'': doesn't make any sense either.
Maybe [[:Category:Comics featuring Miss Lenhart|Miss Lenhart]] can help.
--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 21:41, 17 December 2013 (UTC)

cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is exactly correct.

Let a=π/7, b=3π/7, and c=5π/7, then
(cosa+cosb+cosc)⋅2sina=2cosasina+2cosbsina+2coscsina=sin2a+sin(b+a)−sin(b−a)+sin(c+a)−sin(c−a)=sin(2π/7)+sin(4π/7)−sin(2π/7)+sin(6π/7)−sin(4π/7)=sin(6π/7)=sin(π/7)=sina

Hence, cos(π/7) + cos(3π/7) + cos(5π/7) = sin(π/7) / 2sin(π/7) = 1/2
[[Special:Contributions/108.162.216.74|108.162.216.74]] 01:57, 16 January 2014 (UTC)

:What is this: sin(6π/7)=sin(π/7) ? A new math is born... --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:49, 16 January 2014 (UTC)

::Actually it does. My proof is geometric: the sines of two supplementary angles (angle a + angle b = π (in radians)) are equivalent because they necessarily have the same x height in a Cartesian plane. Look on a unit circle, or even a sine function. Also, Calculus and most other mathematics use radians over degrees because they make the functions simpler and eliminate irrationality when a trig function shows up, but physics uses degrees because it's easier to understand and taught first. Anonymous 01:27, 13 February 2014 (UTC)
::As an aside, just how far along in math are you? Radian measure is taught in high school (at least the good ones). Anonymous 13:24, 13 February 2014 (UTC)
:::Sure, I was wrong at my last statement. sin(6π/7)=sin(π/7) is correct by using the radian measure. But just change π/7 to π/77 would give a very different result on that formular here. I still can't figure out why PI divided by the number 7 should be that unique, PI divided by 77 should be the same. My fault is: I still can't find the Nerd Sniping here. And we all do know that Randall did use wrong WolframAlpha results here. According to the last question: I'm very well on Math, that's because I want to understand this. This is like 0.999=1. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 22:01, 13 February 2014 (UTC)
::::Ah, I see. I think it has to do with the way e^i*π breaks down, as one of the answers shown in the corresponding link explains, but other answers rely on various angle identities (including the supplementary sines one in the proof above). Anonymous 03:10, 14 February 2014 (UTC) (PS, have you checked [[545]] lately? I answered your question there, too)
:::::As per the derivation from january 16 , you can use any a,b,c that satisfies this set of equations: 2 a = b - a, a + b = c- a, c + a = π - a. This is due to the fact that sin(x) = sin(π-x), and what was derived the 16th. [[Special:Contributions/173.245.53.199|173.245.53.199]] 12:38, 21 February 2014 (UTC)
:::: Dgbrt: If not convinced by the proofs linked to in the "explanation" part, you might want to try this: [http://www.wolframalpha.com/input/?i=sum_%28k%3D0%29%5E38+cos%281%2F77+%282+k%2B1%29+pi%29]. I'm sure you'll find inspiration for similar formulas using PI over [any odd integer]. Your assumption that Randall used WolframAlpha for this very identity is probably wrong. This is a very well-known formula that appears in many high school books, and I am pretty sure it is part of Randall's culture. And this has nothing to do with 1=1. As for your original post,
::::*√2 = (√2-1)/((4-2)π/2-π)+1 : Is this what you call "matching an equation" to √2?
::::*So what you mean is that if an equation is true for an irrational number, then it must be for any real number? Like, (√2)^2 = 2, but because √2 is irrational, then x^2=2 (for all x?)
::::*This one's a bit tough. You will probably agree that γ-√2 is irrational. And so is √2. What about their sum?
::::*Well, maybe it doesn't to you. But is Σ n-2 = π^2/6 any better? Well, this one is true (using Fourier's expansion of the rectangular function).
::::Finally,
::::*√2 = 3/5 + π/(7-π) is false because it would imply that π is an algebraic number
::::*cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is true, and proven by many
::::*γ = e/34 + e/5 seems false. But there doesn't seem to be a quick way to disprove.
::::*Σ 1/nn = ln(3)e seems false, but I can't see why. [[Special:Contributions/108.162.210.234|108.162.210.234]] 09:15, 11 May 2014 (UTC)

;So, still incomplete?

Where's our (in)complete judge? [[Special:Contributions/199.27.128.186|199.27.128.186]] 19:21, 18 December 2013 (UTC)
:The protip is still a mystery. I'm calling for help a few lines above. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 21:16, 18 December 2013 (UTC)
::The cosine one, in radians, is correct [[Special:Contributions/141.101.88.225|141.101.88.225]] 12:54, 28 April 2014 (UTC)

The 'Seconds in a year' ones remind me of one of my favorite quotes: "How many seconds are there in a year? If I tell you there are 3.155 x 10^7, you won't even try to remember it. On the other hand, who could forget that, to within half a percent, pi seconds is a nanocentury" -- Tom Duff, Bell Labs. [[User:Beolach|Beolach]] ([[User talk:Beolach|talk]]) 19:14, 17 April 2014 (UTC)
:Please do not change former discussions. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:57, 17 April 2014 (UTC)

;cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 1/2 ???
Why the hell the divider seven makes the difference?
*cos(pi) + cos(3*pi) + cos(5*pi) = -3
*cos(pi/8) + cos(3*pi/8) + cos(5*pi/8) = 0.92387953251128675612818318939678828682241662586364...
So why the "magic" prime number seven produces this exact result? I know radians and π/7 is just a small part of a circle which is 2π. One prove claims that sin(6π/7) equals to sin(π/7); my best calculator can't show a difference. Of course sin(6π) equals to sin(π), in radians, BUT sin(6π/8) is NOT equal to sin(π/8). So if the number 7 plays a magic rule here this would be "one of the", no... the BIGGEST mystery in mathematics forever. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:03, 16 May 2014 (UTC)
:Dgbrt, please see my answer from 11 May 2014 up there. Any odd integer will do, as long as you sum enough of cos(pi/[thing]).
:*Let's try with 5 : cos(pi/5) + cos (3pi/5) = 1/2.
:*With 9 : cos(pi/9)+ cos(3pi/9) + cos (5pi/9) + cos(7pi/9) = 1/2
: No big mystery around here. Just a beautiful formula :) I think there are similar formulas with cosines and even integers. I'll post them here if I have time. [[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 09:56, 17 May 2014 (UTC)

::You mixing up to different equations and even not prove them. If there is any prove to a mathematician I would accept and include a proper explain for non math people here. We still have to find a prove. And I do not trust my calculators, we just have to explain why even cos(pi/5) + cos (3pi/5) is also nearly the same. This issue is still not explained. So please give us a explain. And a PROTIP: This does not work with Integers, PI is infinite--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 17:55, 17 May 2014 (UTC)

:::Okay. If I understood what you said.
:::* I mix up different topics. -> True. From now on, we'll just focus on the cosine one.
:::* You ask for a proof/explanation. -> My opinion is those are two different requests. Maybe that's why you use the distinction between math people/not math people. For a proof, please read further. What I exposed above are just other "fun experiments" we could do. e.g : [http://www.wolframalpha.com/input/?i=cos%28pi%2F11%29%2Bcos+%283pi%2F11%29+%2B+cos+%285pi%2F11%29+%2B+cos+%287pi%2F11%29%2Bcos+%289pi%2F11%29].
:::* You do not trust your calculators -> Great. I don't either. (Well more accurately, I trust mine to 10^-8, so I would definitely not use it to prove any of the discussed equations in PROTIP). That's why we'll prove the formulas we assert.
:::* "This does not work with integers" -> Well, I got myself misunderstood. It would probably have been better if I had said: the following formula is true for all integer n. sum_{k=0}^{n-1}{cos((2k+1)*pi/(2n+1)). But It's harder to read, so just say. Choose any odd integer, say N=2n+1. Then start the following sum. cos(pi/N) + cos(3pi/N) + … and stop when the numerator is cos((N-2)pi/N). Then the result is 1/2. And that's what we'll prove, a few lines down from here.
:::*"Pi is infinite" -> That's a common misconception. What you mean is, Pi is irrational. (Fun fact: Pi is a transcendental number. Quite difficult theorem. Lindeman proved it in 1882. Hence, if we identify the real number x with the Q-vector space Q[x], it would make sense to say that "x is infinite" because, the Q-vector space Q[x] is indeed of infinite dimension. But then, that's not what mathematicians do). I think Vi Hart made a video where she addresses this issue (or was it someone else?). Anyway, I might come to that point some other time in the future.
:::Okay, so now let's first prove the protip formula. Well first, here is the link that the explainxkcd wiki points to: [http://math.stackexchange.com/questions/140388/how-can-one-prove-cos-pi-7-cos3-pi-7-cos5-pi-7-1-2]. Most of them are correct. Some are more ugly than others. I'll adapt the last one.
::: We need a [http://en.wikipedia.org/wiki/Complex_number complex numbers]. (I choosed this because I think explainxkcd readers are fine with this. See comic [http://explainxkcd.com/179/ 179]). I will be using dots to show the steps of my proof. Please allow me an extra level of indent for clarity's sake.
:::'''Proof'''
:::: *Let z be a primitive 14-th [http://en.wikipedia.org/wiki/Root_of_unity root of unity] (the reader doesn't need to understand the 3 last words). Just say z = exp(i*pi/7) = cos(pi/7) + i sin(pi/7). Using [http://en.wikipedia.org/wiki/Euler%27s_formula Euler's formula].
:::: *We have z^14-1 = (exp(i*pi/7))^14-1 = exp(i*2pi) - 1 = 0. Using exponential law for integer powers, as seen in this article: [http://en.wikipedia.org/wiki/De_Moivre%27s_formula De Moivre's formula].
:::: *Now let's factor: z^14-1 = (z^7-1)(z^7+1) = (z^7-1)(z+1)(z^6-z^5+z^4-z^3+z^2-z+1) = (z^7-1)(z+1)*Phi_14(z). where Phi_14(X)= X^6-X^5+X^4-X^3+X^2-X+1, (see [http://en.wikipedia.org/wiki/Cyclotomic_polynomial cycltomic polynomial]). Now, because z^7-1 = (exp(i*pi/7))^7-1 = exp(i*pi)-1 = -2. And because z is not -1, the two first factors are not 0 so, Phi_14(z) = 0, which is already a pretty awesome equality.
:::: *Note that exp(i*pi/7)*exp(i*6pi/7)= exp(i*pi)=-1. So the inverse of z is -exp(i*6pi/7). But we also know that it is exp(-i*pi/7). Well. That was just a fancy way to prove that exp(-i*pi/7) = - exp(i*6pi/7). Good enough. The same holds for exp(-i*3pi/7) = exp(i*14pi/7)*exp(-i*3pi/7)=exp(i*11pi/7)=exp(i*7pi/7)*exp(i*4pi/7)=-exp(i*4pi/7). And the exact same calculation shows that exp(-i*5pi/7)=-exp(i*2pi/7). Alright.
:::: *Now, use that for any x, we have cos(x) = (exp(ix)+exp(-ix))/2. See [http://en.wikipedia.org/wiki/Euler%27s_formula#Relationship_to_trigonometry here]. Let's calculate twice the sum of the left hand side. 2(cos(pi/7)+cos(3pi/7)+cos(5pi/7))= exp(i*pi/7) + expi(-i*pi/7) + exp(3pi/7) + exp(-3pi/7) + exp(5pi/7) +exp(-5pi/7) = exp(i*pi/7)-exp(i*2pi/7)+exp(i*3pi/7)-exp(i*4pi/7)+exp(i*5pi/7)-exp(i*6pi/7) = -Phi_14(z) +1 = 1.
:::: * So dividing both sides by 2, we get what we want. Pfiou.
::: '''Why is 7 so special? Well it isn't.''' Let's prove it for 9.
::::* Let z = exp(i*pi/9) = cos(pi/9) + i sin(pi/9). We have z^18-1 = 0, and z^9-1 and z+1 are not 0, so using the same factorisation, Phi_18(z) = z^8-z^7+z^6-z^5+z^4-z^3+z^2-z+1 = 0.
::::* Hence, the conclusion follow from: 2(cos(pi/9) + cos(3pi/9) + cos(5pi/9) + cos(7pi/9)) = exp(i*pi/9) + exp(-i*pi/9) + exp(i*3pi/9) + exp(-i*3pi/9) + exp(i*5pi/9) + exp(-i*5pi/9) + exp(i*7pi/9) + exp(-i*7pi/9) = -Phi_18(z)+1 = 1.
::: Well, well. I hope you kinda see the pattern. Dgbrt, I know you hate typos, and I'm pretty sure that in this long text lay many of them. So I apologize, and I will correct them later. The following paragraph was posted after I started my text but before I finished mine. It wasn't signed so I will just leave it down there. It's another valid straightforward proof.[[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 21:50, 17 May 2014 (UTC)

The valid identity cos(pi/7)+cos(3pi/7)+cos(5pi/7)=1/2 was correctly proved by the writer at 108.162.216.74 above. For a different proof, consider the complex number z = cos(pi/7)+i sin(pi/7) corresponding to rotation of the complex plane by pi/7 radians, i.e., 1/14th of a full rotation. It satisfies z^{14} -1 = 0 (z to the fourteenth is one). Dividing by z-1 gives z^{13} + z^{12} + ... + z + 1 = 0. The same argument, starting with z^2 corresponding to 1/7th of a full rotation, gives z^{12} + z^{10} + ... z^2 + 1 = 0. Taking the difference, we get z^{13} + z^{11} + ... + z^3 + z = 0. Looking only at the real parts, we get cos(13pi/7) + cos(11pi/7) + cos(9pi/7) + cos(7pi/7) + cos(5pi/7) + cos(3pi/7) + cos(pi/7) = 0. Here cos(13pi/7) = cos(pi/7), cos(11pi/7) = cos(3pi/7) and cos(9pi/7) = cos(5pi/7), since cos is even and 2pi-periodic. Finally cos(7pi/7) = -1, so 2(cos(pi/7) + cos(3pi/7) + cos(5pi/7)) - 1 = 0, which you can rewrite as the desired identity. All of this can be clearly visualized using a regular 14-gon, so a proof with pictures is possible.

Talk:1047: Approximations

2014-05-17T21:50:14Z

Varal7:

They're actually quite accurate. I've used these in calculations, and they seem to give close enough answers. '''[[User:Davidy22|Davidy22]]'''[[User talk:Davidy22|<tt>[talk]</tt>]] 14:03, 8 January 2013 (UTC)

I only see a use for the liters in a gallon one. The rest are for trolling or simple amusement. The cosine identity bit our math team in the butt at a competition. It was painful. --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 05:27, 17 August 2013 (UTC)

Annoyingly this explanation does not cover 42 properly, it does not say that Douglas Adams got the number 42 from Lewis Carroll, who is more relevant to the page because he was a mathematician named Charles Lutwidge Dodgson. He was obsessed with the number forty-two. The original plate illustrations of Alice in Wonderland drawn by him numbered forty-two. Rule Forty-Two in Alice in Wonderland is "All persons more than a mile high to leave the court", There is also a Code of Honour in the preface of The Hunting of the Snark, an extremely long poem written by him when he was 42 years old, in which rule forty-two is "No one shall speak to the Man at the Helm". The queens in Alice Through the Looking Glass the White Queen announces her age as "one hundred and one, five months and a day", which - if the best possible date is assumed for the action of Through the Looking-Glass - gives a total of 37,044 days. With the further (textually unconfirmed) assumption that both Queens were born on the same day their combined age becomes 74,088 days, which is 42 x 42 x 42. --[[Special:Contributions/139.216.242.254|139.216.242.254]] 02:43, 29 August 2013 (UTC)

:: This explanation covers 42 adequately, and would probably be made slightly worse if such information were added. The very widely known cultural reference is to Adams's interpretation, not Dodgson's original obsession. Adding it would be akin to introducing the MPLM into the explanation for the hijacking of Renaissance artists' names by the TMNT. I definitely concede that it does not cover 42 exhaustively, but I think it can be considered complete and in working order without such an addition. If it really irks you, be bold and add it! --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 00:37, 30 August 2013 (UTC)

"sqrt(2) is not even algebraic in the quotient field of Z[pi]" is not correct. Q is part of the quotient field of Z[pi] and sqrt(2) is algebraic of it. The needed facts are that pi is not algebraic, but the formula implies it is in Q(sqrt(2)). --DrMath 06:47, 7 September 2013 (UTC)

13/15 is a better approximation to sqrt(3)/2 than is e/pi. Continued fraction approximations are great! --DrMath 07:23, 7 September 2013 (UTC)

How could he forget 1 gallon ≈ 0.1337 ft³?! [[Special:Contributions/67.188.195.182|67.188.195.182]] 00:51, 8 September 2013 (UTC)

Worth mentioning that Wolfram Alpha now officially recognizes the [http://www.wolframalpha.com/input/?i=e%5E-%28%281%2B8%5E%281%2F%28e-1%29%29%29%5E%281%2Fpi%29%29 White House switchboard constant] and the [http://www.wolframalpha.com/input/?i=%287%5E%28e-1%2Fe%29-9%29*pi%5E2 Jenny constant]. [[Special:Contributions/86.164.243.91|86.164.243.91]] 18:28, 8 October 2013 (UTC)

Maybe we should add the [Extension:LaTeXSVG LaTeX extension] to make it easier to transcribe these equations. -- [[Special:Contributions/108.162.219.220|108.162.219.220]] 23:02, 16 December 2013 (UTC)

;Protip - Does anyone see the correct equation?
Maybe this is just an other Wolfram Alpha error, like we recently have had here: [[1292: Pi vs. Tau]]. All equations still look invalid to me.
*''√2 = 3/5 + π/(7-π)'': is impossible because √2 is an irrational number and no equation can match.
*''cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2'': could only match if ''cos(x) + cos(3x) + cos(5x) = 1/2'' would be valid, because ''π/7'' is also an irrational number.
*''γ = e/34 + e/5 or γ = e/54 + e/5'': would mean that a sum of two irrational numbers do fit to the Gamma Constant. Impossible.
*''√5 = 13 + 4π / 24 - 4π'': √5 and π are irrational numbers, there is no way to match them in any equation like this.
*''Σ 1/nn = ln(3)e'': doesn't make any sense either.
Maybe [[:Category:Comics featuring Miss Lenhart|Miss Lenhart]] can help.
--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 21:41, 17 December 2013 (UTC)

cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is exactly correct.

Let a=π/7, b=3π/7, and c=5π/7, then
(cosa+cosb+cosc)⋅2sina=2cosasina+2cosbsina+2coscsina=sin2a+sin(b+a)−sin(b−a)+sin(c+a)−sin(c−a)=sin(2π/7)+sin(4π/7)−sin(2π/7)+sin(6π/7)−sin(4π/7)=sin(6π/7)=sin(π/7)=sina

Hence, cos(π/7) + cos(3π/7) + cos(5π/7) = sin(π/7) / 2sin(π/7) = 1/2
[[Special:Contributions/108.162.216.74|108.162.216.74]] 01:57, 16 January 2014 (UTC)

:What is this: sin(6π/7)=sin(π/7) ? A new math is born... --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:49, 16 January 2014 (UTC)

::Actually it does. My proof is geometric: the sines of two supplementary angles (angle a + angle b = π (in radians)) are equivalent because they necessarily have the same x height in a Cartesian plane. Look on a unit circle, or even a sine function. Also, Calculus and most other mathematics use radians over degrees because they make the functions simpler and eliminate irrationality when a trig function shows up, but physics uses degrees because it's easier to understand and taught first. Anonymous 01:27, 13 February 2014 (UTC)
::As an aside, just how far along in math are you? Radian measure is taught in high school (at least the good ones). Anonymous 13:24, 13 February 2014 (UTC)
:::Sure, I was wrong at my last statement. sin(6π/7)=sin(π/7) is correct by using the radian measure. But just change π/7 to π/77 would give a very different result on that formular here. I still can't figure out why PI divided by the number 7 should be that unique, PI divided by 77 should be the same. My fault is: I still can't find the Nerd Sniping here. And we all do know that Randall did use wrong WolframAlpha results here. According to the last question: I'm very well on Math, that's because I want to understand this. This is like 0.999=1. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 22:01, 13 February 2014 (UTC)
::::Ah, I see. I think it has to do with the way e^i*π breaks down, as one of the answers shown in the corresponding link explains, but other answers rely on various angle identities (including the supplementary sines one in the proof above). Anonymous 03:10, 14 February 2014 (UTC) (PS, have you checked [[545]] lately? I answered your question there, too)
:::::As per the derivation from january 16 , you can use any a,b,c that satisfies this set of equations: 2 a = b - a, a + b = c- a, c + a = π - a. This is due to the fact that sin(x) = sin(π-x), and what was derived the 16th. [[Special:Contributions/173.245.53.199|173.245.53.199]] 12:38, 21 February 2014 (UTC)
:::: Dgbrt: If not convinced by the proofs linked to in the "explanation" part, you might want to try this: [http://www.wolframalpha.com/input/?i=sum_%28k%3D0%29%5E38+cos%281%2F77+%282+k%2B1%29+pi%29]. I'm sure you'll find inspiration for similar formulas using PI over [any odd integer]. Your assumption that Randall used WolframAlpha for this very identity is probably wrong. This is a very well-known formula that appears in many high school books, and I am pretty sure it is part of Randall's culture. And this has nothing to do with 1=1. As for your original post,
::::*√2 = (√2-1)/((4-2)π/2-π)+1 : Is this what you call "matching an equation" to √2?
::::*So what you mean is that if an equation is true for an irrational number, then it must be for any real number? Like, (√2)^2 = 2, but because √2 is irrationnal, then x^2=2 (for all x?)
::::*This one's a bit tough. You will probably agree that γ-√2 is irrationnal. And so is √2. What about their sum?
::::*Well, maybe it doesn't to you. But is Σ n-2 = π^2/6 any better? Well, this one is true (using Fourier's expansion of the rectangular function).
::::Finally,
::::*√2 = 3/5 + π/(7-π) is false because it would imply that π is an algebraic number
::::*cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is true, and proven by many
::::*γ = e/34 + e/5 seems false. But there doesn't seem to be a quick way to disprove.
::::*Σ 1/nn = ln(3)e seems false, but I can't see why. [[Special:Contributions/108.162.210.234|108.162.210.234]] 09:15, 11 May 2014 (UTC)

;So, still incomplete?

Where's our (in)complete judge? [[Special:Contributions/199.27.128.186|199.27.128.186]] 19:21, 18 December 2013 (UTC)
:The protip is still a mystery. I'm calling for help a few lines above. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 21:16, 18 December 2013 (UTC)
::The cosine one, in radians, is correct [[Special:Contributions/141.101.88.225|141.101.88.225]] 12:54, 28 April 2014 (UTC)

The 'Seconds in a year' ones remind me of one of my favorite quotes: "How many seconds are there in a year? If I tell you there are 3.155 x 10^7, you won't even try to remember it. On the other hand, who could forget that, to within half a percent, pi seconds is a nanocentury" -- Tom Duff, Bell Labs. [[User:Beolach|Beolach]] ([[User talk:Beolach|talk]]) 19:14, 17 April 2014 (UTC)
:Please do not change former discussions. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:57, 17 April 2014 (UTC)

;cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 1/2 ???
Why the hell the divider seven makes the difference?
*cos(pi) + cos(3*pi) + cos(5*pi) = -3
*cos(pi/8) + cos(3*pi/8) + cos(5*pi/8) = 0.92387953251128675612818318939678828682241662586364...
So why the "magic" prime number seven produces this exact result? I know radians and π/7 is just a small part of a circle which is 2π. One prove claims that sin(6π/7) equals to sin(π/7); my best calculator can't show a difference. Of course sin(6π) equals to sin(π), in radians, BUT sin(6π/8) is NOT equal to sin(π/8). So if the number 7 plays a magic rule here this would be "one of the", no... the BIGGEST mystery in mathematics forever. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:03, 16 May 2014 (UTC)
:Dgbrt, please see my answer from 11 May 2014 up there. Any odd integer will do, as long as you sum enough of cos(pi/[thing]).
:*Let's try with 5 : cos(pi/5) + cos (3pi/5) = 1/2.
:*With 9 : cos(pi/9)+ cos(3pi/9) + cos (5pi/9) + cos(7pi/9) = 1/2
: No big mystery around here. Just a beautiful formula :) I think there are similar formulas with cosines and even integers. I'll post them here if I have time. [[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 09:56, 17 May 2014 (UTC)

::You mixing up to different equations and even not prove them. If there is any prove to a mathematician I would accept and include a proper explain for non math people here. We still have to find a prove. And I do not trust my calculators, we just have to explain why even cos(pi/5) + cos (3pi/5) is also nearly the same. This issue is still not explained. So please give us a explain. And a PROTIP: This does not work with Integers, PI is infinite--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 17:55, 17 May 2014 (UTC)

:::Okay. If I understood what you said.
:::* I mix up different topics. -> True. From now on, we'll just focus on the cosine one.
:::* You ask for a proof/explanation. -> My opinion is those are two different requests. Maybe that's why you use the distinction between math people/not math people. For a proof, please read further. What I exposed above are just other "fun experiments" we could do. e.g : [http://www.wolframalpha.com/input/?i=cos%28pi%2F11%29%2Bcos+%283pi%2F11%29+%2B+cos+%285pi%2F11%29+%2B+cos+%287pi%2F11%29%2Bcos+%289pi%2F11%29].
:::* You do not trust your calculators -> Great. I don't either. (Well more accurately, I trust mine to 10^-8, so I would definitely not use it to prove any of the discussed equations in PROTIP). That's why we'll prove the formulas we assert.
:::* "This does not work with integers" -> Well, I got myself misunderstood. It would probably have been better if I had said: the following formula is true for all integer n. sum_{k=0}^{n-1}{cos((2k+1)*pi/(2n+1)). But It's harder to read, so just say. Choose any odd integer, say N=2n+1. Then start the following sum. cos(pi/N) + cos(3pi/N) + … and stop when the numerator is cos((N-2)pi/N). Then the result is 1/2. And that's what we'll prove, a few lines down from here.
:::*"Pi is infinite" -> That's a common misconception. What you mean is, Pi is irrational. (Fun fact: Pi is a transcendental number. Quite difficult theorem. Lindeman proved it in 1882. Hence, if we identify the real number x with the Q-vector space Q[x], it would make sense to say that "x is infinite" because, the Q-vector space Q[x] is indeed of infinite dimension. But then, that's not what mathematicians do). I think Vi Hart made a video where she addresses this issue (or was it someone else?). Anyway, I might come to that point some other time in the future.
:::Okay, so now let's first prove the protip formula. Well first, here is the link that the explainxkcd wiki points to: [http://math.stackexchange.com/questions/140388/how-can-one-prove-cos-pi-7-cos3-pi-7-cos5-pi-7-1-2]. Most of them are correct. Some are more ugly than others. I'll adapt the last one.
::: We need a [http://en.wikipedia.org/wiki/Complex_number complex numbers]. (I choosed this because I think explainxkcd readers are fine with this. See comic [http://explainxkcd.com/179/ 179]). I will be using dots to show the steps of my proof. Please allow me an extra level of indent for clarity's sake.
:::'''Proof'''
:::: *Let z be a primitive 14-th [http://en.wikipedia.org/wiki/Root_of_unity root of unity] (the reader doesn't need to understand the 3 last words). Just say z = exp(i*pi/7) = cos(pi/7) + i sin(pi/7). Using [http://en.wikipedia.org/wiki/Euler%27s_formula Euler's formula].
:::: *We have z^14-1 = (exp(i*pi/7))^14-1 = exp(i*2pi) - 1 = 0. Using exponential law for integer powers, as seen in this article: [http://en.wikipedia.org/wiki/De_Moivre%27s_formula De Moivre's formula].
:::: *Now let's factor: z^14-1 = (z^7-1)(z^7+1) = (z^7-1)(z+1)(z^6-z^5+z^4-z^3+z^2-z+1) = (z^7-1)(z+1)*Phi_14(z). where Phi_14(X)= X^6-X^5+X^4-X^3+X^2-X+1, (see [http://en.wikipedia.org/wiki/Cyclotomic_polynomial cycltomic polynomial]). Now, because z^7-1 = (exp(i*pi/7))^7-1 = exp(i*pi)-1 = -2. And because z is not -1, the two first factors are not 0 so, Phi_14(z) = 0, which is already a pretty awesome equality.
:::: *Note that exp(i*pi/7)*exp(i*6pi/7)= exp(i*pi)=-1. So the inverse of z is -exp(i*6pi/7). But we also know that it is exp(-i*pi/7). Well. That was just a fancy way to prove that exp(-i*pi/7) = - exp(i*6pi/7). Good enough. The same holds for exp(-i*3pi/7) = exp(i*14pi/7)*exp(-i*3pi/7)=exp(i*11pi/7)=exp(i*7pi/7)*exp(i*4pi/7)=-exp(i*4pi/7). And the exact same calculation shows that exp(-i*5pi/7)=-exp(i*2pi/7). Alright.
:::: *Now, use that for any x, we have cos(x) = (exp(ix)+exp(-ix))/2. See [http://en.wikipedia.org/wiki/Euler%27s_formula#Relationship_to_trigonometry here]. Let's calculate twice the sum of the left hand side. 2(cos(pi/7)+cos(3pi/7)+cos(5pi/7))= exp(i*pi/7) + expi(-i*pi/7) + exp(3pi/7) + exp(-3pi/7) + exp(5pi/7) +exp(-5pi/7) = exp(i*pi/7)-exp(i*2pi/7)+exp(i*3pi/7)-exp(i*4pi/7)+exp(i*5pi/7)-exp(i*6pi/7) = -Phi_14(z) +1 = 1.
:::: * So dividing both sides by 2, we get what we want. Pfiou.
::: '''Why is 7 so special? Well it isn't.''' Let's prove it for 9.
::::* Let z = exp(i*pi/9) = cos(pi/9) + i sin(pi/9). We have z^18-1 = 0, and z^9-1 and z+1 are not 0, so using the same factorisation, Phi_18(z) = z^8-z^7+z^6-z^5+z^4-z^3+z^2-z+1 = 0.
::::* Hence, the conclusion follow from: 2(cos(pi/9) + cos(3pi/9) + cos(5pi/9) + cos(7pi/9)) = exp(i*pi/9) + exp(-i*pi/9) + exp(i*3pi/9) + exp(-i*3pi/9) + exp(i*5pi/9) + exp(-i*5pi/9) + exp(i*7pi/9) + exp(-i*7pi/9) = -Phi_18(z)+1 = 1.
::: Well, well. I hope you kinda see the pattern. Dgbrt, I know you hate typos, and I'm pretty sure that in this long text lay many of them. So I apologize, and I will correct them later. The following paragraph was posted after I started my text but before I finished mine. It wasn't signed so I will just leave it down there.[[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 21:50, 17 May 2014 (UTC)

The valid identity cos(pi/7)+cos(3pi/7)+cos(5pi/7)=1/2 was correctly proved by the writer at 108.162.216.74 above. For a different proof, consider the complex number z = cos(pi/7)+i sin(pi/7) corresponding to rotation of the complex plane by pi/7 radians, i.e., 1/14th of a full rotation. It satisfies z^{14} -1 = 0 (z to the fourteenth is one). Dividing by z-1 gives z^{13} + z^{12} + ... + z + 1 = 0. The same argument, starting with z^2 corresponding to 1/7th of a full rotation, gives z^{12} + z^{10} + ... z^2 + 1 = 0. Taking the difference, we get z^{13} + z^{11} + ... + z^3 + z = 0. Looking only at the real parts, we get cos(13pi/7) + cos(11pi/7) + cos(9pi/7) + cos(7pi/7) + cos(5pi/7) + cos(3pi/7) + cos(pi/7) = 0. Here cos(13pi/7) = cos(pi/7), cos(11pi/7) = cos(3pi/7) and cos(9pi/7) = cos(5pi/7), since cos is even and 2pi-periodic. Finally cos(7pi/7) = -1, so 2(cos(pi/7) + cos(3pi/7) + cos(5pi/7)) - 1 = 0, which you can rewrite as the desired identity. All of this can be clearly visualized using a regular 14-gon, so a proof with pictures is possible.

Talk:1047: Approximations

2014-05-17T10:01:50Z

Varal7:

Talk:1047: Approximations

2014-05-17T10:00:51Z

Varal7:

Talk:1047: Approximations

2014-05-17T09:56:19Z

Varal7:

Talk:1047: Approximations

2014-05-17T09:55:55Z

Varal7:

Talk:1047: Approximations

2014-05-17T09:53:51Z

Varal7: