https://www.explainxkcd.com/wiki/api.php?action=feedcontributions&user=162.158.63.28&feedformat=atomexplain xkcd - User contributions [en]2021-01-20T22:29:42ZUser contributionsMediaWiki 1.30.0https://www.explainxkcd.com/wiki/index.php?title=Talk:2000:_xkcd_Phone_2000&diff=157985Talk:2000: xkcd Phone 20002018-05-30T18:04:26Z<p>162.158.63.28: </p>
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Huh, it's not a milestone comic like 1000 was. [[Special:Contributions/172.68.58.191|172.68.58.191]] 16:16, 30 May 2018 (UTC)<br />
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:Yeah, I'm slightly disappointed honestly [[Special:Contributions/172.68.54.46|172.68.54.46]] 16:22, 30 May 2018 (UTC)<br />
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:Gotta wait for an actual round number, like 0b100000000000 [[Special:Contributions/162.158.111.127|162.158.111.127]] 16:38, 30 May 2018 (UTC)<br />
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:Well it is the phone 2000 and he does mention in passing asking if 2000 is a good number to choose [[User:Zachweix|Zachweix]] ([[User talk:Zachweix|talk]]) 16:39, 30 May 2018 (UTC)<br />
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:What am I waiting for more: 2018 or 2048? [[User:SilverMagpie|SilverMagpie]] ([[User talk:SilverMagpie|talk]]) 16:53, 30 May 2018 (UTC)<br />
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:Waiting for 2018 personally. A comic that has looked at the calendar so many times should be able to see the comic number match the year. [[User:Lukeskylicker|Lukeskylicker]] ([[User talk:Lukeskylicker|talk]]) 17:36, 30 May 2018 (UTC)<br />
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: I think it is pretty clearly a milestone comic. Previous xkcd Phones were timed near iPhone releases. This is very specifically for the 2000th comic. I think it no coincidence that the xkcd Phone 2000 was released for the 2000th one, and think it should be mentioned. [[Special:Contributions/162.158.63.28|162.158.63.28]] 18:04, 30 May 2018 (UTC)<br />
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It might be worth noting, that mouse cursors were a thing on BlackBerry smartphones. [[Special:Contributions/162.158.202.100|162.158.202.100]] 17:00, 30 May 2018 (UTC)<br />
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They actually still work on Android if you pair a mouse with the phone (at least, last I tried it). This was actually annoying - my Apple Magic Mouse prefers to pair to my Android Phone than to my Macbook Pro! [[User:Fluppeteer|Fluppeteer]] ([[User talk:Fluppeteer|talk]]) 17:59, 30 May 2018 (UTC)<br />
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I would buy it just to be able to plug a keyboard into the type A port. I hate the USB host-peripheral thing... My phone is more than capable of handling external devices. [[User:Linker|Linker]] ([[User talk:Linker|talk]])<br />
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Don't you like USB OTG or type C adaptors? Bluetooth keyboards should actually work with many Android (or Windows Mobile) devices. [[User:Fluppeteer|Fluppeteer]] ([[User talk:Fluppeteer|talk]]) 17:59, 30 May 2018 (UTC)</div>162.158.63.28https://www.explainxkcd.com/wiki/index.php?title=1946:_Hawaii&diff=1514931946: Hawaii2018-01-24T14:12:04Z<p>162.158.63.28: /* Explanation */</p>
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<div>{{comic<br />
| number = 1946<br />
| date = January 24, 2018<br />
| title = Hawaii<br />
| image = hawaii.png<br />
| titletext = Ok, I've got it, just need to plug in my security key. Hmm, which way does the USB go? Nope, not that way. I'll just flip it and- OH JEEZ IT FELL INTO THE VENT.<br />
}}<br />
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==Explanation==<br />
{{incomplete|Created by a BOT - Please change this comment when editing this page. Do NOT delete this tag too soon.}}<br />
On January 13, 2018 an emergency alert for the state of Hawaii was sent out warning of an incoming ballistic missile attack. The message was specifically noted to NOT be a drill, and this caused widespread panic and fear amongst the island residents. It was eventually determined that the alert was sent in error -- the explanation is that a technician accidentally sent out the "real" version when they were supposed to just be testing the system during an end-of-shift changeover -- but the fact that it took half an hour for the correction to be sent drew widespread criticism. Many people had already begun panic reactions, and there were follow-up confirmations from local entities who thought the original warning was real.<br />
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On January 23, it was revealed that the reason it took so long for a correction to be sent out was that the governor of Hawaii -- who knew the alert was a false alarm only two minutes after it was sent -- had forgotten his login information for his twitter account.<br />
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This comic shows Cueball, representing the governor, frantically trying to log in to twitter and being stymied by the security features. Off-panel, another person is screaming at him that people are beginning to panic and warning sirens are going off, underscoring the need to get the correction out as fast as possible.<br />
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==Transcript==<br />
{{incomplete transcript|Do NOT delete this tag too soon.}}<br />
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{{comic discussion}}</div>162.158.63.28https://www.explainxkcd.com/wiki/index.php?title=Talk:1925:_Self-Driving_Car_Milestones&diff=148849Talk:1925: Self-Driving Car Milestones2017-12-07T01:24:17Z<p>162.158.63.28: </p>
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This page is, without offense to the creator, a mess. We're gonna need a table for this. [[Special:Contributions/172.68.47.78|172.68.47.78]] 19:14, 6 December 2017 (UTC)<br />
* Or at least a list. I have created one, but it could use fleshing out.[[User:WingedCat|WingedCat]] ([[User talk:WingedCat|talk]]) 19:55, 6 December 2017 (UTC)<br />
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I'm going to go with a [citation needed] on that "sex in a self-driving car has probably already happened." Are there stats suggesting the amount of coitus per vehicle in the relevant counties?<br />
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"This a joke about Boolean satisfiability, as evaluating an arbitrarily complex bumper sticker and determining whether to honk is NP-complete." What? Determining whether to honk has nothing to do with the satisfiability problem; this is more of a joke about getting a computer to evaluate the truth of Boolean expressions that it may have no information about. [[User:Checkmate|Checkmate]] ([[User talk:Checkmate|talk]]) 22:07, 6 December 2017 (UTC)Checkmate<br />
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I believe the "Autonomous canyon jumping" is related to the self-loathing; a self-loathing car is likely to autonomously jump off a cliff. [[Special:Contributions/108.162.212.179|108.162.212.179]] 22:30, 6 December 2017 (UTC)<br />
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"As of 2017, self-driving cars require a human to be able to take over just in case, but any such trip where the human never actually took control would qualify for this milestone."<br />
I seems like not all places require a human backup driver: https://www.theverge.com/2017/11/7/16615290/waymo-self-driving-safety-driver-chandler-autonomous<br />
[[Special:Contributions/172.69.22.146|172.69.22.146]] 23:19, 6 December 2017 (UTC)<br />
Time to start printing "Honk if this statement evaluates as 'do not honk!'" bumper stickers! [[Special:Contributions/162.158.63.28|162.158.63.28]] 01:24, 7 December 2017 (UTC)</div>162.158.63.28https://www.explainxkcd.com/wiki/index.php?title=1908:_Credit_Card_Rewards&diff=1471551908: Credit Card Rewards2017-10-28T06:28:48Z<p>162.158.63.28: /* Explanation */ Fixed grammar and spelling mistakes.</p>
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<div>{{comic<br />
| number = 1908<br />
| date = October 27, 2017<br />
| title = Credit Card Rewards<br />
| image = credit_card_rewards.png<br />
| titletext = I should make a list of all the things I could be trying to optimize, prioritized by ... well, I guess there are a few different variables I could use. I'll create a spreadsheet ... eh<br />
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==Explanation==<br />
{{incomplete|VERY basic explanation.}}<br />
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A credit card, at it's most basic form, is a loan contract to an individual from a bank. Like all contracts, the bank will offer several different types in an attempt to appeal to a large number of individuals. Unlike traditional loans which focus on a single item (car, house, boat, etc), a credit card is an unsecured loan geared towards daily and weekly transactions. Because these transactions cover a wide variety of items, credit cards can be further tweaked towards offering benefits in certain areas. Gas purchases, or even gas purchases through a single retail chain can offer higher rewards on one type of plan vs other plans.<br />
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These benefits, typically called rewards, have several different options. "Cash back" is a reward where the individual is given money back when they make a purchase that follows certain rules spelled out in the contract. "No interest" is a reward where the individual is not charged interest on their purchases if they pay the loaned money back within a specified amount of time. "Points" are similar to the cash back program, but are typically reserved towards purchasing a single large item or plan. Points towards a vacation is a popular option. Besides these three types of rewards, the number of actual rewards to pick from are limited only by the creativity and fiscal limitations, and of the issuing bank's CEO.<br />
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Cueball is trying to choose the optimal credit card program (the one that will result in biggest savings with the yearly fiscal median (YFM) he has). He realizes that he has to subtract the cost of him spending time on optimizing, so he wants to optimize the time needed to do the optimizing. But in order to to that efficiently, he first has to optimize the time spent on optimizing the time.<br />
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Hairy notices a hidden assumption that Cueball will spend his time on something more productive than this, e.g. that his time has value. Cueball responds that he can "fail to optimize so many better things." This means that Cueball is aware of the big flaw in his reasoning.<br />
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The title text further expands the idea. Cueball wants to present a list of things to optimize to Hairy. However, he still needs to optimize the priorities of that list, before optimizing the list itself.<br />
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==Transcript==<br />
{{incomplete transcript|Do NOT delete this tag too soon.}}<br />
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[Cueball sits at a desk and is on his laptop. Hairy stands behind him.]<br />
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:Cueball: I'm trying to figure out which of these credit card rewards programs is best given my spending.<br />
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[Cueball leans backwards.]<br />
:Cueball: But at some point, the cost of the time it takes me to understand the options outweighs their difference in value.<br />
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[Close-up of Cueball's head and torso.]<br />
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:Cueball: So I need to figure out where that point is, and stop before I reach it.<br />
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[Cueball has his arms outstretched.]<br />
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:Cueball: But... when I factor in the time to calculate <i>THAT</i>, it changes the overall answer.<br />
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:Hairy: I question the assumption that you'd otherwise be spending your time on something more valuable.<br />
:Cueball: Come on, I could be failing to optimize so many better things!<br />
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{{comic discussion}}<br />
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[[Category:Comics featuring Hairy]]<br />
[[Category:Comics featuring Cueball]]</div>162.158.63.28https://www.explainxkcd.com/wiki/index.php?title=Talk:1888:_Still_in_Use&diff=145284Talk:1888: Still in Use2017-09-12T18:53:23Z<p>162.158.63.28: </p>
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<s>Just for all writers: The {{w|Garbage collection (computer science)|Garbage collection}} prominently belongs to {{w|Java (programming language)|Java}}. Microsoft had adopted this only in C# and it's NOT used in file systems.--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 14:47, 11 September 2017 (UTC)</s> EDIT by --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:47, 11 September 2017 (UTC)<br />
:Note: Java adopted garbage collection over 30 years after it had been used in Lisp. I would question the use of 'prominently belongs'. Any user of Gnu Emacs will be aware of what happens when garbage collection hits unexpectedly... [[Special:Contributions/141.101.98.112|141.101.98.112]] 16:13, 11 September 2017 (UTC)<br />
::I have to admit that my first comment doesn't belong to the content of this comic. But, even when you are right (Garbage collection was invented by John McCarthy around 1959 to simplify manual memory management in Lisp.), who really uses Lisp compared to Java? So when trying to explain the GC I would use the most common language using this feature and compare it to the vast majority of other common languages like C, Delphi/Pascal, or scripting languages like Perl or PHP. And we can lisp, sorry list, many more languages in this latter context. In the middle there are object-oriented programming languages without GC like C++, a destructor must be explicitly called which than removes everything belonging to a particular instance on an object. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 20:47, 11 September 2017 (UTC)<br />
:::Both Perl and PHP are using garbage collection. So do Python, Ruby and Javascript. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 23:49, 11 September 2017 (UTC)<br />
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Seams like a pretty clear reference to not being able to empty the computers Trash because files are in use. Normally files in the Trash can't be opened, and files can't be moved to the Trash if opened but weird things can happen. The real rub here is that the computer does know '''exactly''' what process has each file open and is intentionally designed and told not to tell you the user the remedy that it already knows nor to show an option to remedy the problem itself leaving you in the lurch. [[Special:Contributions/172.68.206.58|172.68.206.58]] 15:37, 11 September 2017 (UTC)<br />
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It could be that another user is the one that used the paper towel last. I was actually running into an issue where I couldn't delete a file from a network share that I had used last. The dialog would tell me what program was still using it (Source Tree, which had unceremoniously crashed and didn't close out properly) but my coworker simply got the "Is in use by another program" message. [[User:Bpendragon|Bpendragon]] ([[User talk:Bpendragon|talk]]) 15:43, 11 September 2017 (UTC)<br />
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I've also come across the issue where the process using the file is the file manager utility itself; I opened the trash to see what was in there, it started trying to make preview images of all the items, and of course when you close the window it doesn't release whatever filehandle it was currently trying to make a preview for. -- [[User:Angel|Angel]] ([[User talk:Angel|talk]]) 16:12, 11 September 2017 (UTC)<br />
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You could always power the house down and restart it! [[Special:Contributions/162.158.78.28|162.158.78.28]] 16:37, 11 September 2017 (UTC)<br />
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Part of the joke is how unhelpful windows tend(ed/s) to be in helping you identify the application that is still hanging onto the file.--[[User:Henke37|Henke37]] ([[User talk:Henke37|talk]]) 17:08, 11 September 2017 (UTC)<br />
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This is problem only on MS Windows, where most file-access is '''blocking'''; Linux allows deleting file even if some process is accessing it - said process would see old version of the file, while all other would see it deleted. This has its own problems (you delete files, but you don't recover free space), but I think it is less annoying. --[[User:JakubNarebski|JakubNarebski]] ([[User talk:JakubNarebski|talk]]) 18:03, 11 September 2017 (UTC)<br />
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*More specific, Windows tracks open files through their pathname, linux tracks open files through their numeric ID. This means that you could create file A, open it, delete the file, create another file B with the same filename, open it, delete the file, create another file C with the same filename, open it, delete the file. Those three files would still be present on the disk, each of the apps that has the file open would see different contents (which they could write to and change), but you would never see any of the files through a directory listing (but it would take up disk space until the files were closed). I believe linux viruses delete themselves to make them more difficult to discover, this also explains why linux system updates don't require rebooting the computer afterwards (although if you just changed the system kernel it's likely recommended)[[User:Odysseus654|Odysseus654]] ([[User talk:Odysseus654|talk]]) 18:26, 11 September 2017 (UTC)<br />
:Note that the numeric ID is called inode - and yes, it's better behaviour for updating. Regarding viruses, well they may do it but it would be bad strategy, as it would remove them on reboot and make them still visible in commands as lsof. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 23:49, 11 September 2017 (UTC)<br />
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No one though about Docker? That was lit(t)erally my first thought: He must be talking about Docker. Ever tried to find out which container is still using a volume? --[[User:AndreKR|AndreKR]] ([[User talk:AndreKR|talk]]) 18:30, 11 September 2017 (UTC)<br />
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I figured this was about program removal (i.e. "uninstall"). Sometimes one file is still in use (sometimes the program's folder instead of a file), but the rest are deleted as expected. -- '''BigMal''' // [[Special:Contributions/108.162.216.166|108.162.216.166]] 18:58, 11 September 2017 (UTC)<br />
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"...However, the operating system gives no indication of which application(s) have open file(s)..." Which operating systems do this? (Xubuntu tells you which application(s) are the culprit(s). I gather Windows doesn't. OS/X? Other Linuxes? Maybe just say "some operating systems give"? [[User:Bill Gray1|Bill Gray1]] ([[User talk:Bill Gray1|talk]]) 19:41, 11 September 2017 (UTC)<br />
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This is one of the most annoying things windows does, and a huge contribution to why I use macOS and hate Windows! It seems that just about every time I try to do anything significant on windows, I run into this or a similarly frustrating inane problem I never have with a Mac! The one I hate most is when I can't delete an empty folder because either it or a hidden thumbs.db in it is "in use" by the exact same app (windows explorer) as I'm trying to use to delete it! That this is still a problem with a commercially successful OS made in the 21st century is unbelievable to me! [[User:PotatoGod|PotatoGod]] ([[User talk:PotatoGod|talk]]) 22:09, 11 September 2017 (UTC)<br />
:Agreed, one of the worst features on windows[[User:Dontknow|Dontknow]] ([[User talk:Dontknow|talk]]) 23:39, 11 September 2017 (UTC)<br />
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I removed the irrelevant commentary that Windows doesn't use the Unix inode data structures, as the object manager provides the functionality allowing file locks to be made, queried, and released. http://m.windowsitpro.com/systems-management/inside-nts-object-manager<br />
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Windows also support asynchronous deletion or moving of in-use files.<br />
https://docs.microsoft.com/en-us/sysinternals/downloads/movefile<br />
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[[http://www.linkedin.com/in/Comet Comet]] 23:01, 11 September 2017 (UTC)<br />
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This is NOT a Windows-only-problem. The very same happens if you try to u(n)mount a partition in Linux. It will fail if a program still has a file on this filesystem open, but Linux does NOT tell you which file is open – you have to use other programs like ''lsof'' to find out. --[[User:DaB.|DaB.]] ([[User talk:DaB.|talk]]) 23:28, 11 September 2017 (UTC)<br />
:Asynchronous deletion is nowhere near as useful as the ability to just remove the file and keep just the inode open. But you are right that while Linux doesn't have this problem with files, it does have them with filesystems (partitions) and it can be very annoying. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 23:49, 11 September 2017 (UTC)<br />
:Linux will not tell you when using the command line (and you'll have to run lsof), but any modern distro will tell you when using the GUI. Specifically, on Ubuntu 16.04LTS you'll get a window telling you which applications are using the filesystem and will also give you the option to force unmount anyway (at your own risk). -- [[Special:Contributions/162.158.63.28|162.158.63.28]] 18:53, 12 September 2017 (UTC)<br />
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The thing is, the use of a garbage can here makes for an obvious parallel to emptying the Recycling Bin of Windows ("Trash Bin" before the last decade or so, before Microsoft decided to suck up to environmentalists), except this problem can't appear like that. Programs can't use files or folders in the Bin, they'd have to be restored first. So you can't be blocked from emptying the virtual Bin like this. I can think of two ways to get such an error: Trying to delete a file / SEND it to the Recycling Bin, or trying to disconnect an external drive. Either you can't dispose of a file because it's in use (a file being in the Bin counts as already disposed of), or you can't disconnect the drive the file is on because it's in use. I suspect the same can be said of other operating systems that have a Trash or Recycling Bin (I believe this is indeed true of Macs, that you can't use files in the Trash). [[User:NiceGuy1|NiceGuy1]] ([[User talk:NiceGuy1|talk]]) 04:43, 12 September 2017 (UTC)<br />
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This reminds me more of when I try to eject an external drive (memory stick, portable hd, whatever). If Windows has a file open on it, you can't eject it, and you have to guess what is keeping it locked. --[[User:Lou Crazy|Lou Crazy]] ([[User talk:Lou Crazy|talk]]) 16:39, 12 September 2017 (UTC)<br />
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Holy crap, this read as an uneducated user Windows hate piece (until I changed it). Resource Monitor has been available since Vista (which is why the SysInternals utility is no longer being developed), and Windows 7 and up has a link straight from the Task Manager. It's also not a Windows exclusive problem. Windows trash bin behaviour is to lock it off from the rest of the system, and is always located on the same drive so "delete" (really a move/rename command) operations are virtually instant; you'd have to be trying REALLY hard to get a locked error from a "send to recycle" bin command, mitigated if the OS/app simply retried a second later.<br />
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Here's a FreeBSD link for the same problem: https://lists.freebsd.org/pipermail/freebsd-questions/2012-June/241627.html , for example<br />
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For those needing / wanting to educate themselves with file locking and the problems all OSes have, go here: https://en.wikipedia.org/wiki/File_locking --[[User:Someone asdf|Someone asdf]] ([[User talk:Someone asdf|talk]]) 16:40, 12 September 2017 (UTC)</div>162.158.63.28https://www.explainxkcd.com/wiki/index.php?title=Talk:1047:_Approximations&diff=139595Talk:1047: Approximations2017-05-05T13:41:13Z<p>162.158.63.28: </p>
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<div>The world population estimate is still accurate to within .1 billion. [[Special:Contributions/162.158.63.28|162.158.63.28]] 13:41, 5 May 2017 (UTC)<br />
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They're actually quite accurate. I've used these in calculations, and they seem to give close enough answers. '''[[User:Davidy22|<span title="I want you."><u><font color="purple" size="2px">David</font><font color="green" size="3px">y</font></u><sup><font color="indigo" size="1px">22</font></sup></span>]]'''[[User talk:Davidy22|<tt>[talk]</tt>]] 14:03, 8 January 2013 (UTC)<br />
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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)<br />
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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)<br />
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:: 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)<br />
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"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)<br />
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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)<br />
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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)<br />
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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)<br />
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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)<br />
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;Protip - Does anyone see the correct equation?<br />
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.<br />
*''√2 = 3/5 + π/(7-π)'': is impossible because √2 is an irrational number and no equation can match.<br />
*''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.<br />
*''γ = e/3<sup>4</sup> + e/5 or γ = e/54 + e/5'': would mean that a sum of two irrational numbers do fit to the Gamma Constant. Impossible.<br />
*''√5 = 13 + 4π / 24 - 4π'': √5 and π are irrational numbers, there is no way to match them in any equation like this.<br />
*''Σ 1/n<sup>n</sup> = ln(3)<sup>e</sup>'': doesn't make any sense either.<br />
Maybe [[:Category:Comics featuring Miss Lenhart|Miss Lenhart]] can help.<br />
--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 21:41, 17 December 2013 (UTC)<br />
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cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is exactly correct. <br />
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Let a=π/7, b=3π/7, and c=5π/7, then <br />
(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<br />
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Hence, cos(π/7) + cos(3π/7) + cos(5π/7) = sin(π/7) / 2sin(π/7) = 1/2<br />
[[Special:Contributions/108.162.216.74|108.162.216.74]] 01:57, 16 January 2014 (UTC)<br />
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: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)<br />
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::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)<br />
::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)<br />
:::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)<br />
::::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)<br />
:::::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)<br />
:::: 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,<br />
::::*√2 = (√2-1)/((4-2)π/2-π)+1 : Is this what you call "matching an equation" to √2?<br />
::::*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?)<br />
::::*This one's a bit tough. You will probably agree that γ-√2 is irrational. And so is √2. What about their sum?<br />
::::*Well, maybe it doesn't to you. But is Σ n<sup>-2</sup> = π^2/6 any better? Well, this one is true (using Fourier's expansion of the rectangular function). <br />
::::Finally,<br />
::::*√2 = 3/5 + π/(7-π) is false because it would imply that π is an algebraic number<br />
::::*cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is true, and proven by many<br />
::::*γ = e/3<sup>4</sup> + e/5 seems false. But there doesn't seem to be a quick way to disprove.<br />
::::*Σ 1/n<sup>n</sup> = ln(3)<sup>e</sup> seems false, but I can't see why. [[Special:Contributions/108.162.210.234|108.162.210.234]] 09:15, 11 May 2014 (UTC)<br />
::Dgbrt, yes, sin(6π/7)=sin(π/7). Simple proof: sin(6π/7)=sin(π-π/7)=sin(π)cos(-π/7)+cos(π)sin(-π/7)=0*cos(-π/7)+(-1)*(-sin(π/7))=0+sin(π/7)=sin(π/7) [[Special:Contributions/108.162.215.89|108.162.215.89]] 02:34, 20 May 2014 (UTC) <br />
;So, still incomplete?<br />
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Where's our (in)complete judge? [[Special:Contributions/199.27.128.186|199.27.128.186]] 19:21, 18 December 2013 (UTC)<br />
: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)<br />
::The cosine one, in radians, is correct [[Special:Contributions/141.101.88.225|141.101.88.225]] 12:54, 28 April 2014 (UTC)<br />
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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)<br />
:Please do not change former discussions. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:57, 17 April 2014 (UTC)<br />
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;cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 1/2 ???<br />
Why the hell the divider seven makes the difference?<br />
*cos(pi) + cos(3*pi) + cos(5*pi) = -3<br />
*cos(pi/8) + cos(3*pi/8) + cos(5*pi/8) = 0.92387953251128675612818318939678828682241662586364...<br />
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)<br />
: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]). <br />
:*Let's try with 5 : cos(pi/5) + cos (3pi/5) = 1/2.<br />
:*With 9 : cos(pi/9)+ cos(3pi/9) + cos (5pi/9) + cos(7pi/9) = 1/2<br />
: 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)<br />
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::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)<br />
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:::Okay. If I understood what you said.<br />
:::* I mix up different topics. -> True. From now on, we'll just focus on the cosine one.<br />
:::* 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].<br />
:::* 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.<br />
:::* "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.<br />
:::*"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.<br />
:::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.<br />
::: 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.<br />
:::'''Proof'''<br />
:::: *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].<br />
:::: *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].<br />
:::: *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.<br />
:::: *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.<br />
:::: *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.<br />
:::: * So dividing both sides by 2, we get what we want. Pfew. <br />
::: '''Why is 7 so special? Well it isn't.''' Let's prove it for 9. <br />
::::* 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. <br />
::::* 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. <br />
::: 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)<br />
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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. {{unsigned ip|141.101.81.216}}<br />
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;99 is sexual reference?<br />
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In first explanation it says: "99^8 and 69^8 are sexual references". 69 I understand, but what would 99 refer too? <br />
--[[Special:Contributions/173.245.53.167|173.245.53.167]] 17:38, 18 May 2014 (UTC)<br />
: see [[487: Numerical Sex Positions]][[Special:Contributions/141.101.70.181|141.101.70.181]] 15:33, 20 July 2014 (UTC)<br />
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I'd add pi = (9^2 + (19^2)/22)^(1/4) [[Special:Contributions/198.41.230.73|198.41.230.73]] 02:41, 13 May 2015 (UTC)</div>162.158.63.28