Wikipedia:Reference desk/Mathematics and Reviews on the Run: Difference between pages

Add links
From Wikipedia, the free encyclopedia
< Wikipedia:Reference desk
(Difference between pages)
Content deleted Content added
VisualWikitext
 
 
Line 1: Line 1:
{{Refimprove|date=September 2008}}
[[Category:Non-talk pages that are automatically signed]]{{Wikipedia:Reference desk/header|WP:RD/MA}}
{{Importance|date=September 2007}}
[[Image:Reviewsontherun.png|250px|thumb|right|The original ''Reviews on the Run'' logo, for stations outside the US. Used from 2002-2007]]
[[Image:JD whole.png|250px|thumb|right|The only-in-America ''Judgment Day'' logo. The show ran from 2002-2006 on G4.]]


'''''Reviews on the Run''''' (formerly known as '''''Judgment Day''''' in the [[United States]]) is a [[video game]] review [[television program|TV show]] hosted by [[Victor Lucas]] and [[Tommy Tallarico]], and produced by Lucas' production company, Greedy Productions. The two hosts rate games independently on a scale of .5 point increments from 0 through 10, with 0 being the lowest and 10 being the highest. The show started as a segment on ''[[The Electric Playground]]'' called "Reviews on the Run", but was spun off into its own show in 2002. As of December 2007, ''Reviews on the Run'' airs in [[Canada]] on [[G4techTV Canada]], [[CITY-TV|Citytv Toronto]], and [[CKVU-TV|Citytv Vancouver]], with pre-2006 episodes airing on [[G4 (TV channel)|G4]] in the United States. The show was previously shown in Canada on [[A-Channel]], [[Space (TV channel)|Space]], [[Razer]], and [[OMNI.1]].
{{Wikipedia:Reference_desk/Archives/Mathematics/2008 September 29}}


The show is filmed on location at several different locales including [[Vancouver]], [[British Columbia]]. Episodes are also occasionally recorded in other cities. The hosts stand in frame as video game footage is projected onto an object in the background such as a [[Billboard (advertising)|billboard]] or the side of a building. The show is filmed over the course of several hours and later edited to fit the show's thirty minute time frame.{{Fact|date=September 2008}}
{{Wikipedia:Reference_desk/Archives/Mathematics/2008 September 30}}


==History==
{{Wikipedia:Reference_desk/Archives/Mathematics/2008 October 1}}
On [[January 13]], [[2006]], Lucas announced on the G4 forums that ''Judgment Day'' was no longer going to be produced for [[G4 (TV channel)|G4]]. The program will still run internationally (including on [[G4techTV Canada]]) as the original ''Reviews on the Run''.<ref>http://forums.g4tv.com/messageview.cfm?catid=22&threadid=545731</ref> Specific details about the negotiations have not been disclosed. As of July 2008, the show has not returned to U.S. airwaves with new episodes.


On [[December 31]], [[2006]], Lucas announced on ''The Electric Playground'' forums that Greedy Productions had cancelled its contract with [[CHUM Limited|CHUM television]], which broadcast ''Electric Playground'' and ''Reviews on the Run'' on [[Space (TV channel)|Space]] and [[A-Channel]], and signed a two year exclusive deal with [[Rogers Communications]], to broadcast the shows on G4techTV Canada and then additionally on other Rogers owned TV stations.<ref>{{cite web |url=http://bb.elecplay.com/viewtopic.php?t=9336 |title=Huge 2007 News For ROTR and EP!!! Need Your Help! :) |accessdate=2008-09-08 |author=[[Victor Lucas]] |date=2006-12-31 |work=[[The Electric Playground]] Forum |publisher=The Electric Playground <!-- |archivedate=2008-01-19 |archiveurl=http://web.archive.org/web/20080119192850/http://bb.elecplay.com/viewtopic.php?t=9336 -->}}</ref> ''Reviews on the Run'' aired on [[OMNI.1]] from [[January 28]], [[2007]] to [[September 3]] [[2007]]. On [[December 15]] [[2007]], ''Reviews on the Run'' premiered on [[CITY-TV|Citytv Toronto]] and [[CKVU-TV|Citytv Vancouver]]. On [[June 2]], [[2008]], G4 started airing episodes of ''Judgment Day'' as a part of their G4 Rewind block.
= October 2 =


==Cast==
== Question related to .999... ==
[[Victor Lucas]] and [[Tommy Tallarico]] have been both co-workers and good friends since their first meeting at [[E3]] in 1995. Onscreen the two enjoy plenty of brotherly scraps, with Tallarico typically playing the role of a more aggressive, even gutter-minded gamer and Lucas being more businesslike and sedate. (A typical exchange - Tallarico: "Come on, you're saying that you never shot any of those guys in the crotch, even once?" Lucas: "No, I did NOT!") During the "Hardware" segment of ''Reviews on the Run'', Tallarico has occasionally taken the video gaming hardware (console controllers for example) that they were reviewing, and put it down the front of his pants he was wearing, to the disgust of Lucas.


Due to commitments with [[Video Games Live]], Tallarico was absent for six episodes from the 2006 season, and was the co-host in three episodes of the show in the last two seasons. [[Geoff Keighley]] co-hosted with Lucas in all of the 2006 episodes Tallarico missed. The 2007 and 2008 seasons of the show have featured a number of guest hosts, including Scott Jones, Tom Russo, [[Marc Saltzman]], Jose "Fubar" Sanchez, Ben Silverman, and Steve Tilley.
While I understood long ago that .999... = 1, I was wanting to ask a question that is implicatively the inverse of the .999... theorem: is 0.000...{infinity}...001 = 0? As an aside, I realise this question is quite trivial and simple; I ask only as a matter of interest, and extend thanks to anyone kind enough to offer me an answer. :) &mdash;<strong>[[User:Anonymous Dissident|<span style="font-family:Script MT Bold;color:DarkRed">Anonymous Dissident</span>]]</strong>[[User_talk:Anonymous Dissident|<sup><span style="font-family:Verdana;color:Gray">Talk</span></sup>]] 03:36, 2 October 2008 (UTC)


During the Hardware segment of the show, Hardware Girls show off pieces of video gaming hardware, while Lucas and Tallarico review the product in question. Hardware Girls are usually shown wearing T-shirts that have messages relating to video games and gaming (for example: "I've got collision detection"). ''[[Lost (TV series)|Lost]]'' cast member [[Evangeline Lilly]]'s first job in television was as a "Hardware Girl" for ''Reviews on the Run'', which was referenced when the Lost [[video game]] was reviewed on the show.
:I don't see how 0.000...{infinity}...001 is a meaningful expression. If it is has infinite zeros then it can't terminate in a one. [[User:Dragons flight|Dragons flight]] ([[User talk:Dragons flight|talk]]) 03:47, 2 October 2008 (UTC)
::Language peeve alert -- please, I think you mean infinitely ''many'' zeroes, not that there are zeroes that are individually infinite.
::On the content -- it depends on what you mean by "meaningful". There is nothing wrong, syntactically, with having an infinite string of zeroes (this time ''infinite'' as an adjective is fine, since it modifies ''string'' rather than ''zeroes''), then followed by a one. However this string is not the decimal representation of any [[real number]]. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 03:53, 2 October 2008 (UTC)
:::I understand it is not real; I just was wondering if infinity could be confined in any sort of way so that we could have something on the other side. While ∞+1makes no sense, my real question was whether we could have an (∞) and then something else on the other side, in a decimal context. &mdash;<strong>[[User:Anonymous Dissident|<span style="font-family:Script MT Bold;color:DarkRed">Anonymous Dissident</span>]]</strong>[[User_talk:Anonymous Dissident|<sup><span style="font-family:Verdana;color:Gray">Talk</span></sup>]] 04:05, 2 October 2008 (UTC)
::::Sure, why not? But it's just a string of symbols (albeit an infinitely large string). It's not a decimal representation. It doesn't ''denote'' anything (at least, not in any standard way). --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 04:12, 2 October 2008 (UTC)
:::::No, it's worse than that. It suggests you should construct a string of symbols that could never be written down, is logically impossible, and can't be connected to the normal meaning of such symbols. It's like saying "write the 32nd letter of the English alphabet". It's not even false, because what it attempts to refer to is non-existent. In short it's mathematical gibberish. [[User:Dragons flight|Dragons flight]] ([[User talk:Dragons flight|talk]]) 04:23, 2 October 2008 (UTC)
::::::Can't write down? But you can't write down, say, a string of a quintillion symbols, either. That doesn't prevent us from studying such strings.
::::::Logically impossible? Please provide a proof of a statement and its negation, from the assumption that such a string of symbols exists, using logic alone (that is, you are not allowed any other assumptions whatsoever). Logical impossibility is an extremely strong condition.
::::::There is nothing wrong whatsoever with infinite strings of symbols. They give you a good way to think about (for example) the [[Cantor space]] and the [[Baire space (set theory)|Baire space]]. A very nice elementary example from the theory of [[Borel equivalence relation]]s is doubly-infinite words on some finite alphabet, where you [[forgetful functor|forget]] where the origin is. None of this is "mathematical gibberish"; it's actually excellent mathematics. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 04:37, 2 October 2008 (UTC)
:::::::One can add definitions to mathematics, and create a system where that symbol might refer to something, but that's the same as saying I can define the 32 letter in English to be "[[Image:smile.png|10px]]". Yes, there are many contexts where infinite sequences exist and are useful. But can you give any examples where the infinite sequences are terminated by a finite number of elements? To say an infinite sequence is terminated is a logically impossibility for any normal definition of "infinite sequence" and "terminated". The above notation doesn't make sense within the normal framework of decimal notation, and trying to bootstrap meaning onto a post-infinite notation simply to respond to Anonymous Dissident is bad mathematics. [[User:Dragons flight|Dragons flight]] ([[User talk:Dragons flight|talk]]) 05:49, 2 October 2008 (UTC)
::::::::Believe it or not we do this (put something at the end of an infinite sequence) all the time. You should look at the articles [[ordinal number]] and [[transfinite induction]]. I see you have an undergrad degree in math but unless you took a course specifically in math logic you likely didn't get to these. In first-year graduate real analysis you'd likely have run across them in passing, say in the construction of the [[Borel hierarchy]].
::::::::Now, as for it not making sense in the "normal framework of decimal notation", I said as much. I said that this infinitely long symbol doesn't, in any ordinary way, denote anything. But this is a separate issue from whether it makes sense to have an infinite sequence with something at the end -- it most definitely ''does'' make sense and is ''not'' bad mathematics. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 07:17, 2 October 2008 (UTC)
:::::::::You can write the .00000.....1 as (0,1) instead and then you'll find it all works out quite nicely as in [[nonstandard arithmetic]], plus it's shorter to write. That article could do with a bit of expansion. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 07:46, 2 October 2008 (UTC)
::::::::::Um, this one is not clear to me; I'm not sure what your notation means. What I was telling Anonymous Dissident was that you could make the string of symbols, not that there was any natural arithmetic to do with those strings. If you're talking about nonstandard models of arithmetic (and/or analysis) that's another kettle of fish—the positions in the decimal expansions of "reals" in the sense of those models are themselves nonstandard naturals, and some care has to be taken to make sense of it all. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 09:34, 2 October 2008 (UTC)


==Segments==
:::::::'''Dragons flight''' asked: 'can you give any examples where the infinite sequences are terminated by a finite number of elements?' <br /> Please see the [[Sharkovskii's theorem]], which gives an example of such infinite chain, that has a known infinite sequence as its beginning, a known infinite sequence as a tail, an infinite number of infinite sub-sequences inside itself and exhausts positive integers putting them in a [[total order]]. --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 12:29, 2 October 2008 (UTC)


*The majority of ''Reviews on the Run'' consists of review segments for individual games on a current major console or on PC. Prior to December 2007, both hosts gave a recap of the "hits" (positives) and "misses" (negatives) after the review ended, with the host who gave the higher rating stating the positive side of the game. Three of these review segments occur in most episodes.
::::Have you already read the article on [[0.999...]]? While it doesn't actually use the 0.000...01 notation, the section on "Alternative number systems" covers some of the ideas of what you have to break in the real numbers to let you have a non-zero value equal to 1 - 0.999... The talk page and associated arguments page (linked from talk) go a little more into it, and the fact is that you ''can'' break out [[transfinite numbers]] or [[surreal numbers]] or something to extend decimal expansions, but you can't then just assume that they'll work like normal decimals, and you have to determine a new set of rules for working with them. Since the decimal system is designed to represent real numbers, something's gotta give. [[User:ConMan|Confusing Manifestation]]<small>([[User talk:ConMan|Say hi!]])</small> 04:17, 2 October 2008 (UTC)


*'''Pocket Reviews''' - Lucas or Tallarico reviews a game on a handheld console or device. Two Pocket Review segments happen each episode, with Lucas doing both when Tallarico is unavailable. Save for the occasional "Versus" segment, handheld games are rarely given a normal review by themselves. G4techTV Canada infrequently airs Pocket Reviews during commercial breaks under the title "Reviews on the Run: Quickplay Edition"
:::::Thanks very much for that pointer to [[0.999...]]. The bit about alternative number systems says all I wanted to say about [[infinitesimal calculus]] and I was just thinking of looking for that bit of game theory with [[Hackenbush]] where the value of one position can be greater than that of another but not by any finite amount. The introduction is good, I suppose it's okay to be dogmatic about .999 being 1 and erroneous intuitions about infinitesimals even if that really depends on using standard real numbers as defined using [[Dedekind cuts]] or suchlike. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 22:10, 2 October 2008 (UTC)


*'''Buried Treasures''' - Lucas or Tallarico talks about a game they felt was great, but was overlooked by consumers. This segment was held in an [[EB Games]] or a [[Rogers Video]] store. A Buried Treasure game is not rated, only recommended. Occasionally, games reviewed here would be featured on a later or previous episode in a normal review format. Buried Treasures was removed from the show in 2007, and replaced by "Download Games".
===Infinite sequences with something after them===
OK, this is not strictly in the spirit of the refdesk, but I'd really like to say something about this strange meme that nothing can follow an infinite sequence, or that you can't do infinitely many things and then do something else. You see this in lots of otherwise mathematically well-informed folks, and it has a pernicious effect on their mathematical reasoning, sometimes leading them away from the most natural arguments.


*'''Download Games''' - Lucas reviews a game downloaded from the a major console's online service. Download Games were introduced in 2007, replacing the Buried Treasures segment.
Take as an example the [[Cantor-Bendixson theorem]], which says that if you have a [[closed set]] of real numbers (or closed set in some other [[Polish space]]), then you can remove countably many points and have remaining a (possibly empty) [[perfect set]]; that is, a closed set with no [[isolated point]]s. What's the natural way to show this? Well, delete all the isolated points, of course (easy to see there can be only countably many such).


*'''Classic Reviews''' - A flashback from an older episode is shown, with a condensed review of an older title. In most cases, the game being reviewed ties in with a game reviewed on the same episode.
Unfortunately, some points that were not isolated before may have ''become'' isolated, because you've removed a sequence of points that accumulated to them. So you remove these new isolated points as well. And you keep going, if necessary.


*'''Hardware''' - The two hosts review a video game peripheral device, system, add-on, or controller. Occasionally, a game can also be reviewed here if it requires the hardware in question to play it. "Hardware Girls" are shown demonstrating and modeling the hardware in question during the review. Instead of rating the device on a scale, the hosts will either "recommend" or "not recommend" the device. Occasionally, the hardware device will be "not recommended with an asterisk" (e.g. the device is not worth its current price, or not recommended if you currently own an older model). Hardware reviews initially featured a "hits and misses" recap, which was later dropped from the show.
Now, after infinitely many iterations (that is, taking the intersection of everything you got at a finite iteration), do you still have isolated points? Possibly. You could have removed one point at stage 0, another at stage 1, another at stage 2, and so on, and these accumulated to some point that becomes isolated only at this first infinite stage. No problem; we keep on going beyond that.


*'''The Hit List''' - Lucas gives a quick rundown of the top five games in a particular category, such as a particular genre, or a given system. From 2006 onward, a variation called the "Sh_t List" has occasionally appeared in place of The Hit List, which has the bottom five games in a given category.
There are a few details to check, but it's a fact that this process must close off at some countable ordinal number. At each stage you've removed only countably many points, and the countable union of countable sets is countable, so you're done, proof complete.


*'''Versus''' - The two hosts review and compare two or three games of a similar genre or subject. They typically break down the review by comparing the graphics, audio, and gameplay of the games in question, with both hosts naming which of the games was better in each category. The games are scored by both hosts, with the game with the highest cumulative score being named the winner of the Versus segment. On occasion, the Versus segment is replaced by both hosts giving quick reviews on two or three unrelated new titles, with no head to head competition in mind.
This is by far the most natural and perspicacious proof of the theorem. But lots of mathematicians, perversely, prefer an all-at-once proof involving looking at the set of [[condensation point]]s. of the original set. Their proof, I will concede, is not really ''harder''. But it's much less ''enlightening''. The only reason I can see to prefer it is if you're not comfortable with transfinite induction. But there's no excuse not to be comfortable with transfinite induction! It's an absolutely basic technique and every mathematician needs to know it. I think that this unfortunate notion that nothing can follow infinitely many things may be at the root of their discomfort. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 19:42, 3 October 2008 (UTC)


*After the versus segment, Lucas recaps all of the games featured (excluding Pocket, Buried Treasure, Download Games, Classic, and Hardware reviews). He goes over quickly, the positive and negative features of each.
:Interesting. But I'm not convinced the prejudice against your proof has anything to do with transfinite induction. Here's a similar example that doesn't even reach ω:
:*Problem: Given ''n'' "red" points and ''n'' "blue" points, a "pairing" of the points is a set of ''n'' line segments with one red and one blue endpoint, different for each segment. Show that any set of ''n'' red and ''n'' blue points on the plane in general position has a pairing with no intersections.
:**Proof 1: Start with any pairing and uncross intersecting segments (by swapping which blue point corresponds to which red) until you can't any more. This process must terminate because each uncrossing operation decreases the total length of the segments, by the triangle inequality, and there are only finitely many possible pairings.
:**Proof 2: Pick a pairing of minimum total length. It can't contain any intersections because you could then get a shorter pairing by uncrossing them.
:I find the first proof more intuitive, but the mathematician in me prefers the second one. I think the prejudice in your problem and mine is against proofs that invoke time and change. The answer to the problem is there in the Platonic realm, so better (the thinking goes) to grab it directly than to muck around with other objects that don't answer the question. -- [[User:BenRG|BenRG]] ([[User talk:BenRG|talk]]) 08:29, 4 October 2008 (UTC)


*'''Rogers Video Game of the Week''' The highest rated game featured on that show (including titles in the Pocket and Download Games reviews) is awarded the title of Rogers Video Game Of The Week.
::But an argument in favour of Proof 1 is that it is more general because it depends only on the topological properties of the underlying space, whereas Proof 2 assumes that the underlying space is a metric space. So Proof 1 can be generalised to cases where the connections between the points are not necessarily line segments (and maybe there is not even a definition of "straight line" in the space), whereas Proof 2 cannot be so generalised. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 09:52, 4 October 2008 (UTC)
:::I may be missing something, but I don't think Proof 1 is so easily generalized. You need to show that the uncrossing process eventually terminates, and the way that's done above uses the metric properties of the space. -- [[User:BenRG|BenRG]] ([[User talk:BenRG|talk]]) 17:03, 4 October 2008 (UTC)


*Before the credits, a few bloopers or deleted scenes from the episode are shown.
So Ben, I don't really think your example is quite analogous. Your two proofs are close enough that I would call them the "same" proof; the inductive procedure in proof 1 is limited to showing that every finite partial order has a minimal element, which the second proof simply takes for granted. Other than that the two proofs are identical, unless I've missed something.


*Every December, an episode is aired where the hosts rate the current major video game consoles, and the PC, on the titles released that year for that platform. The episode was called "Holiday Reviews on the Run" in 2006 and "Year in Review" in 2007.
In my example, on the other hand, it's trivial to see that there's a subset of the original set that's left unchanged by one iteration; it's the empty set. But that doesn't help, because what we want to know is that we have to delete only countably many points. The all-at-once proof ''does'' have the merit of characterizing the points that are kept, but the inductive proof shows the structure of what has to be removed (and in particular the structure of countable closed sets in general).


==Video game reviews of note==
Another class of examples of my point is the way that algebra classes, following Bourbaki, bundle all uses of the axiom of choice into the clunky and intuitively obscure [[Zorn's lemma]], applications of which can almost always be replaced by a simple and clear transfinite induction in which you make a choice at each step. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 19:12, 4 October 2008 (UTC)
As of June 2008, only eight games on the show have been given 10/10 by both hosts. Five of them are the [[multiplatform]] titles: ''[[Prince of Persia: The Sands of Time]]'' (which ended with Tommy running around the set [which was a hockey rink] shouting "Ten! Ten! Ten!" with the scoreboard flashing "Tommy gives it a 10!"), ''[[Burnout 3: Takedown]]'', ''[[Burnout Revenge]]'' (which Lucas actually wanted to give an 11/10 before finally agreeing to a 10), ''[[The Orange Box]]'', and ''[[Grand Theft Auto IV]]; with the three other games being ''[[God of War II]]'' for the [[PlayStation 2]], ''[[Halo 3]]'' for the [[Xbox 360]], and ''[[Metal Gear Solid 4: Guns of the Patriots]]'' for the [[PlayStation 3]].


The lowest rated game ever on the show was ''[[Chicken Shoot]]'' ([[Wii]]) which Lucas gave 0/10 and guest host Jose "Fubar" Sanchez gave -2.0/10, making it the first game to score less than 0 on the show. Fubar thought Vic was mad at him so he made him review this game. The first 0/10 was given by Tallarico for the game ''[[High Heat Major League Baseball 2003]]''; he hated the game so much he stormed off partway through the review, followed by sounds of a car screeching away. The only other game to receive a 0 was [[Wacky Races]] for the Nintendo DS, which Tommy reviewed alone.
==Logic problem: Working out Wendy's birthdate==
I wrote an article on [[Wendy Whiteley]] the other day, and later I started looking at what little evidence I've mustered, in order to narrow down when she was born.

I thought I'd got it to between 7 April and mid-October 1941, but when I added in the known age (17) of her future husband [[Brett Whiteley]] when they first met, and knowing exactly when he was born (7 April 1939), I was led to conclude she must have been born ''before'' 7 April 1941. It can't have been both before ''and'' after 7 April 1941. The possibility that they had the same birthday (7 April), exists, but I'm sure that coincidence would have been mentioned many times in the literature, and I've never seen any such references.

Can some kind soul take a glance over my thinking at [[Talk:Wendy Whiteley]] and tell me what logical errors I've made or what wrong assumptions I've assumed. Or, maybe I've made no logical errors, but the evidence is inherently contradictory and my analysis has simply revealed this to be the case. I'm pretty sure everything you need to know is there, but if not, just ask.

Thanks for any assistance you can provide. -- [[User:JackofOz|JackofOz]] ([[User talk:JackofOz|talk]]) 03:55, 2 October 2008 (UTC)

: Bear in mind that being a "round number", 'now 65' might mean 'now over 65' and thus anything from 65.0 to 69.9 -- [[User:SGBailey|SGBailey]] ([[User talk:SGBailey|talk]]) 07:57, 2 October 2008 (UTC)
::That's possible, but I think unlikely. The full sentence is ''"Standing on the balcony of the Whiteley house, gazing out at the water through the sinuous Moreton Bay fig and three cabbage tree palms that Brett so often depicted in his harbour paintings, Wendy, now 65, is, as ever, an eye-catching walking work of art herself"''. If she was really 67 at the time, this would be a very misleading statement. -- [[User:JackofOz|JackofOz]] ([[User talk:JackofOz|talk]]) 19:58, 2 October 2008 (UTC)

== Continuity on the complex plane ==

Hi. I'm in a complex variables class, and I'm struggling with a homework problem. I'd like a hint, or a push in the right direction.

We're given that D is the domain obtained by removing the interval <math>[0, \infty)</math> of the real line from the complex plane. We want to show that a branch of the cube root function on this domain is given by:

:<math>g(z) = \begin{cases} \sqrt[3]{z} & \mbox{if } Re(z) \ge 0 \\ \frac{-1 + i \sqrt{3}}{2} \sqrt[3]{z} & \mbox{if } Re(z) < 0 \end{cases} </math>

To be a branch of an inverse function, g(z) must be a right inverse of f(z) (where f(z)=z<sup>3</sup>), and it must be continuous. Showing that it's a right inverse is no problem, but the continuity is giving me trouble. My instinct tells me that, when choosing <math>\delta</math>, I'll want to make it a minimum of two options, one for the line near the origin, and another for the rest of the line. I'm also pretty confident that I want to start out with:

<math>|\sqrt[3]{z} - \sqrt[3]{z_0}| = \frac{|z - z_0|}{|z^{2/3} + z^{1/3}z_0^{1/3} + z_0^{2/3}|} </math>

...and then I want to bound that denominator from below, but I'm stuck right there. Can someone please throw me a bone? -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 15:56, 2 October 2008 (UTC)

:The equation x<sup>3</sup>=z has 3 complex solutions, each of which can be called a 'cube root of z'. So your function g(z) is not really well defined. If x is one of the solutions, the other two are xa and xa<sup>2</sup> where a=e<sup>2&pi;i/3</sup> is a primitive cube [[root of unity]], (a&ne;1, a<sup>2</sup>&ne;1, a<sup>3</sup>=1). Assume that z moves continuously around the unit circle from 1 to 1. (z=e<sup>i&phi;</sup>, 0<&phi;<2&pi;), then the root x=e<sup>i&phi;/3</sup> moves continuously from 1 to a, and the root xa moves continuously from a to a<sup>2</sup>, and the root xa<sup>2</sup> moves continuously from a<sup>2</sup> to 1. I hope this is helpful to your understanding. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 20:12, 2 October 2008 (UTC).
::I probably should have said that, in the notation we've been using, <math>\sqrt[3]{z}</math> is defined as <math>\sqrt[3]{|z|} \cdot e^\frac{Arg(z)}{3}</math>, where the first factor is a real number and where <math>Arg(z)</math> denotes the [[principal argument]] of z, which is always in the interval <math>(-\pi , \pi ]</math>. In other words, <math>\sqrt[3]{z}</math> denotes the ''principal'' cube root of z, which satisfies <math>-\frac{\pi}{3} < Arg \left( \sqrt[3]{z} \right) \leq \frac{\pi}{3}</math>. With that clarification, the function <math>g(z)</math> is well-defined, and I still don't know how to show that it's continuous along the negative real axis. <p> Thanks for pointing out the ambiguity in my notation, though. -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 22:11, 2 October 2008 (UTC)

:Consider a small positive angle &epsilon;.
::<math>\lim_{\epsilon\rightarrow0^+}\sqrt[3]{-e^{i\epsilon}}=\lim_{\epsilon\rightarrow0^+}e^{\epsilon-\pi i\over 3}=e^{-\pi i\over 3}
</math>
:while
::<math>\sqrt[3]{ \lim_{\epsilon\rightarrow 0^+}-e^{i\epsilon}}=\sqrt[3]{-1}=e^{+\pi i\over 3}</math>
:so the ''principal'' cube root is not continuous at &minus;1. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 10:39, 3 October 2008 (UTC).
:::I know the principal cube root is discontinuous at -1; the point of this problem is to define a different branch, g(z), that ''is'' continuous at -1, and which is discontinuous at 1. The function g(z) is not the principal cube root; it's the notation <math>\sqrt[3]{z}</math> that is taken to mean the principal cube root. That's how the function is well-defined, because we're not using "<math>\sqrt[3]{z}</math>" to mean just any old cube root. As you can see, g(z) is ''different from that'' in the bottom half-plane. It's different by an argument of 2π/3. -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 20:38, 4 October 2008 (UTC)
::Indeed. As given, ''g'' is discontinuous along the negative axis. Moreover, it cannot be fixed by changing only the second clause of the definition: principal <math>\sqrt[3]z</math> is continuous along the ''positive'' real axis, hence if such a variant of ''g'' were continuous, it would have a holomorphic extension to all of <math>\mathbb C\setminus\{0\}</math>, which is impossible. —&nbsp;[[User:EmilJ|Emil]]&nbsp;[[User talk:EmilJ|J.]] 11:15, 3 October 2008 (UTC)
:::Ok, apparently I'm not communicating this clearly at all. This problem comes straight out of the textbook, and I don't think it's a misprint. <p> I am quite well aware that the principal cube root is continuous on the positive real axis, and discontinuous on the negative real axis. This problem is about defining a ''different'' cube root function - not the principal cube root - that is continuous on the negative real axis, and discontinuous on the positive real axis. Nobody is trying to define a cube root that's holomorphic on <math>\mathbb C\setminus\{0\}</math>; I agree that's impossible. What we're doing is moving the discontinuity to the positive side, which is ''not'' impossible. <p> The function g(z), given above, is very, very discontinuous on the positive real axis, and it's very, very continuous on the negative real axis. Just think about plugging in a number whose principal argument is just short of π - the function g will return a number just short of π/3. Now plug in a number just below the negative real axis, so its principal argument is very close to -π. The function g, as given, will return a value slightly greater than -π/3, multiplied by <math>\frac{-1 + i \sqrt{3}}{2}</math>, which puts us just past π/3. <p> The function g, as given, is equal to the principal cube root ''only in the top half-plane'', and it's equal to a ''different'' cube root in the bottom half-plane, one chosen so that it will be continuous on the negative real axis, and the discontinuity is moved to the positive real axis. The range of g(z) is the segment of the complex plane with principal argument ranging from 0 to <math>2\pi/3</math>. It's not impossible; in fact, this is a quite standard problem in defining different branches of inverse functions. What I really want is some help with the epsilon-delta argument. Can anyone help me with that, or do I have to settle for people telling me that my textbook, my professor, and my clear understanding are wrong, and that g(z) is either not well-defined, or is the principal cube root, when I know damn well that it isn't? This is frustrating. Can someone who understands branches of inverse functions please help me with the epsilons and deltas? -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 20:38, 4 October 2008 (UTC)
::::I haven't read all this discussion, but I think that in your original definition of ''g'', you wanted to split the definition according to the sign of the ''imaginary'' part of ''z'', not the ''real'' part. [[User talk:Algebraist|Algebraist]] 22:19, 4 October 2008 (UTC)
:::::Holy cow; you're right. That's precisely what I meant to do. Now I feel stupid, and I'd really like to know how to prove continuity on the negative real line for the function defined:
::::::<math>g(z) = \begin{cases} \sqrt[3]{z} & \mbox{if } Im(z) \ge 0 \\ \frac{-1 + i \sqrt{3}}{2} \sqrt[3]{z} & \mbox{if } Im(z) < 0 \end{cases} </math>
:::::I apologize for screwing up so royally. Now I know why this discussion has been so incredibly frustrating. If anyone can look past my stupidity and help me, I would be very, very grateful. -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 00:34, 5 October 2008 (UTC)
:Just show that <math>\lim_ {\epsilon\rightarrow 0^+} (g(-ae^{i\epsilon}))=g(\lim_{\epsilon\rightarrow 0^+}(-ae^{i\epsilon}))</math> for positive ''a''. And don't call yourself stupid. [[Errare humanum est]]. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 08:15, 5 October 2008 (UTC).
::That would only show that the limit is the same coming from two directions; to show continuity, we must show that it is the same from any direction. In particular, I have to show that, for every <math>z_0</math> on the negative real line, and given any small <math>\epsilon</math>, there is some <math>\delta</math> so that, whenever <math>|z - z_0| < \delta</math>, we have <math>|g(z) - g(z_0)| < \epsilon</math>. There are examples of complex functions with limits existing when a point is approached in two, four, or any number of directions, which still fail to be continuous. I know that there is an argument beginning in the manner I suggested in my original post, and I'd very much like to find it. <p> I'm not stupid, but I did something stupid, and then compounded it by failing to re-read my original post and discover the error. Don't worry about my self-esteem; I know that I'm quite bright, but I've also got stupidity in me. Similarly, I'm generous, but I've got greed in me; I'm modest, but I've got pride in me; I'm patient and thoughtful, but I've got impetuousness in me. I'm okay with all of that. Thanks though, for reminding me that I'm allowed my errors. -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 22:31, 5 October 2008 (UTC)
:You are perfectly right. But take one step at a time. The g-function is composed of functions that are continuous except at the real axis. That proves continuity except at the real axis. The second step is the one above. The third step is to release the constraint that ''&epsilon;'' is a positive real. (For variable ''z'' and fixed ''z''<sub>0</sub>&ne;0 consider ''z''/''z''<sub>0</sub>&minus;1). The fourth step is the case ''z''<sub>0</sub>=0. Making mistakes is an unavoidable and actually desirable component of the process of learning. If you are less stupid today than yesterday, then you have learned something. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 07:25, 6 October 2008 (UTC).
::Wow, this is very frustrating. I feel as if I'm being talked down to. Please drop the condescending tone. I don't feel bad about making mistakes, nor do I need reassurance about the learning process. I've been teaching mathematics for about 15 years, and learning it for about twice that long. I've made thousands and thousands of mistakes, and I'm grateful for every one of them. The mistake in this case was a fucking ''stupid'' typo (though I'm not stupid), and then a failure to scroll up and read what I'd typed. I'm grateful for that stupid mistake; thank you. (And no, I don't use words like "stupid" except when I'm among colleagues who know that we're all qualified and not really stupid. I thought this was such a context - it has been before.)(<sub>Insertion: I am sorry. I had no intention of talking down to anybody. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 12:27, 7 October 2008 (UTC).</sub>)<p> I know that we've got continuity except on the negative real line - "step one" was done before I ever posted here. The "second step" you indicated is easy. I can do it in my sleep: When <math>\epsilon \rightarrow 0^-</math>, there's nothing to do; just let <math>\epsilon</math> go to 0. When <math>\epsilon > 0</math>, then we have: <math>g(z_0 e^{i\epsilon}) = \sqrt[3]{z_0 e^{i\epsilon}} \frac{-1 + i\sqrt{3}}{2} = \sqrt[3]{|z_0|e^{i(-\pi + \epsilon)}} \frac{-1 + i\sqrt{3}}{2} = \sqrt[3]{|z_0|} e^{i(-\frac{\pi}{3} +\frac{\epsilon}{3})} e^{\frac{2\pi i}{3}} = \cdots</math> <math>\cdots = \sqrt[3]{|z_0|} e^{i\left(\frac{\pi}{3} + \frac{\epsilon}{3}\right)} = \sqrt[3]{|z_0|} e^{i\frac{\pi}{3}} e^{i\frac{\epsilon}{3}} = \sqrt[3]{z_0} e^{i\frac{\epsilon}{3}} = g(z_0) e^{i\frac{\epsilon}{3}}</math>, and again we can just let <math>\epsilon</math> go to 0. Ok? <p> That's easy because we're only comparing z<sub>0</sub> with other numbers that have the same modulus. If I write up what I just did, my professor will ask why I'm screwing around with the easy case instead of just going straight to the general one. What you're calling "step 3" makes your "step 2" redundant and unnecessary. It's a good exercise for beginners, but this is a graduate level course. You're right that <math>z_0 = 0</math> is a separate case, and I've not asked for help with that one; all I need there is to take <math>\delta < \epsilon^3</math>. because the cube root is a strictly increasing function on the positive reals. That's a piece of cake. <p> Sigh. <p> Your hint about looking at <math>\frac{z}{z_0} - 1</math> may work. I don't yet see how it will give me a string of inequalities beginning with <math>|g(z) - g(z_0)|</math> and ending with <math>\epsilon</math>, but I'll work on that. Thank you for that hint. It's already too late to turn this in, but I'm going to work out the details anyway, for my own learning. Thanks again - I mean that. I am irritated with this interaction on other levels, but I do appreciate the math help. <p> I've used this resource in the past without feeling that I've been treated like a child; I'm not sure what went wrong this time. If I ask for help here in the future, I'll try to more clearly indicate my level of understanding, so I don't get handed Continuity 101 and the "it's ok to make mistakes" lecture. I hope you can understand what I find exasperating about this exchange. -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 15:50, 6 October 2008 (UTC)

:::You don't need to compute any more limits this far in the game. Let <math>z_0</math> be a point on the negative real line, and <math>U</math> its circular open neighborhood avoiding the origin. Let <math>g_1,g_2,g_3</math> be the unique branches of the cube root defined on <math>U</math> such that <math>g_1(z_0)=g(z_0)</math>, <math>g_2(z)=g(z)</math> for <math>{\rm Im}z>0</math>, and <math>g_3(z)=g(z)</math> for <math>{\rm Im}z<0</math>. All three functions are continuous by definition. Since you already proved <math>g(z_0)=\lim_{\epsilon\to0}g(z_0e^{i\epsilon})</math>, and <math>g_2(z_0)=\lim_{\epsilon\to0+}g_2(z_0e^{i\epsilon})=\lim_{\epsilon\to0+}g(z_0e^{i\epsilon})</math> by continuity, we obtain <math>g_2(z_0)=g(z_0)</math>, hence <math>g_1=g_2</math>. By a symmetrical argument, <math>g_1=g_3</math>, hence <math>g=g_1=g_2=g_3</math> is continuous in <math>U</math>. —&nbsp;[[User:EmilJ|Emil]]&nbsp;[[User talk:EmilJ|J.]] 16:24, 6 October 2008 (UTC)
::::I don't see how I can assume that there exist continuous branches of the cube root function on U. This problem is supposed to ''establish'' that one such branch exists. From the definition of a "branch of an inverse function on U" (a right inverse function that is continuous on U), it's not clear that one exists when U is an open set that straddles the negative real line.... or is it? How do we know that there exist continuous functions <math>g_1,g_2,g_3</math> satisfying the conditions you describe? What am I missing? <p> Isn't there an epsilon-delta argument that someone can help me put together? Do we ''have to'' avoid that? I'm not married to the epsilon-delta format, but it's what I used for the square root function when I proved that it was uniformly continuous on certain domains, and it seems to be the flavor of proof that we usually use. <p> On a side note, the argument at z=0 doesn't have to be made, because g is defined on the domain <math>\mathbb C\setminus\left[ 0,\infty\right)</math> -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 19:47, 6 October 2008 (UTC)

:::::I may be mistaken, but isn't there some general theorem which guarantees the existence of branches of inverse functions on simply connected domains? Anyway, in this particular case, it is easy to define the branches explicitly. In fact, now that I think about it, it is possible to define directly a branch of cube root on the whole of <math>\mathbb C\setminus[0,\infty)</math>, giving a completely limit-free argument avoiding the "step 2" above. (Sorry, I'm not in an epsilon-delta mood.) The argument is as follows.

:::::Define the function
::::::<math>h(z)=\frac{1+i\sqrt3}2\sqrt[3]{-z}</math>
:::::for <math>z\in\Omega=\mathbb C\setminus[0,\infty)</math>. Obviously <math>(h(z))^3=z</math>, and being a composition of continuous functions, ''h'' is continuous, thus it is a branch of the cube root on <math>\Omega</math>. Now it suffices to show that ''g'' = ''h''. Consider <math>z\in\Omega</math> such that <math>{\rm Im}(z)\ge0</math>. As <math>{\rm Arg}(z)\in(0,\pi]</math>, we compute easily <math>{\rm Arg}(g(z)),{\rm Arg}(h(z))\in(0,\pi/3]</math>. As <math>(g(z))^3=(h(z))^3</math>, the difference <math>{\rm Arg}(g(z))-{\rm Arg}(h(z))</math> is an integral multiple of <math>2\pi/3</math>, hence in fact <math>{\rm Arg}(g(z))={\rm Arg}(h(z))</math>. Also <math>{|g(z)|}={|h(z)|}</math>, thus <math>g(z)=h(z)</math>. The case <math>{\rm Im}(z)<0</math> is similar. —&nbsp;[[User:EmilJ|Emil]]&nbsp;[[User talk:EmilJ|J.]] 10:16, 7 October 2008 (UTC)
:::Not in an epsilon-delta mood, eh? No one seems to be. I guess if I want to use this problem to practice that, then I'm on my own. Whatever. <p> I talked to a few people yesterday, and came up with a couple of alternative strategies. The one you suggest is elegant, and works. Another option, which has a nice geometric appeal, is to define a different branch of the argument function, say arg(z) (lower-case 'a') where the range is the interval [0,2π), and establish that ''it'' is continuous. Then we could define <math>h(z) = \sqrt[3]{|z|}\cdot e^\frac{arg(z)}{3}</math>, and show that g=h. <p> Anyway, I suspect this horse is dead, and I thank everyone for their help. -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 14:50, 7 October 2008 (UTC)
:::Oh, and that theorem about branches on simply connected domains may exist, but we haven't got it yet in our class, so I probably shouldn't use it. -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 15:12, 7 October 2008 (UTC)

== Expressions ==

Hi. I've wondered for a long time, but never bothered asking anyone, is whether or not the expression <math>\frac{x}{x}</math> is always equal to 1 or <math>\frac{x^2}{x}</math> is always equal to x. My basic algebraic skills are quite strong and I can easily simplify the afore mentioned expressions but my problem lies in what happens to these expressions, as well as others of a similar nature, when you let x equal zero. Can anyone shed some light on the matter please? Thanks [[Special:Contributions/92.3.238.32|92.3.238.32]] ([[User talk:92.3.238.32|talk]]) 16:28, 2 October 2008 (UTC)
:When ''x'' equals 0 they are formally undefined. However, for a particular application, one can define them to be whatever is convenient or useful. Typically this would be their respective limits as ''x'' approached 0. Thus in your cases, it could be reasonable to define them as 1 and 0 respectively, although note that that is merely a convention you would choose to maintain. In other words, you cannot ''prove'' it, unless you did so circularly without realizing it. [[User:Baccyak4H|Baccyak4H]]&nbsp;([[User talk:Baccyak4H|Yak!]]) 16:47, 2 October 2008 (UTC)

:These expressions have a [[removable singularity]] at <math>x = 0</math>. (Unfortunately, that article as currently written makes little sense to anyone not already familiar with [[complex analysis]].) —[[User:Ilmari Karonen|Ilmari Karonen]] <small>([[User talk:Ilmari Karonen|talk]])</small> 20:31, 2 October 2008 (UTC)
::So to avoid having to learn complex analysis, the graph <math>y=\frac{x}{x}</math> would look just like <math>y=1</math> except there would be an open dot at (0,1) which would not be solution of the equation. <math>y=\frac{x^2}{x}</math> would act like <math>y=x</math> with a discontinuity at(0,0). [[User:Paragon12321|<span style="color:blue">Paragon</span>]][[User talk:Paragon12321|<span style="color:red"><sup>12321</sup></span>]] 02:59, 3 October 2008 (UTC)

:To answer this question with precision we need to distinguish between [[Expression (mathematics)|formal expression]]s and [[Function (mathematics)|function]]s. As formal expressions, <math>\frac{x^2}{x}</math> is algebraically equal to <math>x</math> because we can convert the first expression into the second following the formal rules of algebra - in this context, <math>x</math> is a formal variable and takes no values. Similarly the ''expression'' <math>\frac{x^2-x}{x-1}</math> is also algebraically equal to the ''expression'' <math>x</math>. However, once we start to talk about specific values of <math>x</math>, we enter the realm of functions. And as ''functions'', <math>x</math>, <math>\frac{x^2}{x}</math> and <math>\frac{x^2-x}{x-1}</math> are three different functions of <math>x</math> - the first is defined everywhere, the second is not defined at 0 and the third is not defined at 1. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 09:36, 3 October 2008 (UTC)

== Find an analytic function f(z) such that f(f(z)) = exp(z) ==

[[User:Count Iblis|Count Iblis]] ([[User talk:Count Iblis|talk]]) 23:11, 2 October 2008 (UTC)

:I asked this a few years back. [[Wikipedia:Reference desk archive/Mathematics/February 2006#"square root" of exponential function]]. I asked on the newsgroup sci.math and sci.math.research, also. — [[User:Arthur Rubin|Arthur Rubin]] [[User talk:Arthur Rubin|(talk)]] 23:43, 2 October 2008 (UTC)
: [http://groups.google.com/group/sci.math/browse_thread/thread/322d675b66f31f91/11a142d9699ab9cf?lnk=st&q=rubin+arthur+group%3Asci.math+exp#11a142d9699ab9cf] (Google groups, 1992) states that
:* C Horowitz "Iterated Logarithms of Entire Functions" ISRAELI JOURNAL OF MATHEMATICS, v29, n1, pp. 31-42 (1978).
:provides an entire solution of h(z+1)=exp(h(z)), meaning that f(z)=h(h<sup>-1</sup>(z)+.5) will work. I've never actually checked it out, though. — [[User:Arthur Rubin|Arthur Rubin]] [[User talk:Arthur Rubin|(talk)]] 00:28, 3 October 2008 (UTC)
::That analysis seems to assume that "h" is 1-1, which is clearly not possible for an entire function. Sorry about that. It appears I wasn't paying attention to the answer in 1992, either. — [[User:Arthur Rubin|Arthur Rubin]] [[User talk:Arthur Rubin|(talk)]] 00:34, 3 October 2008 (UTC)

:There must be hundreds of people who've spent odd moments on this, it really deserves its own article. I came to the conclusion years ago no analytic solution was possible though I'm willing to be proved wrong as I've not read anything about it - it seemed more the sort of thing to just mess around with. An interesting path it led me to was considering things like (x+f(x)/n) iterated n times where n tends to infinity and mixing two such functions up in various ways. But I bet there's probably been named and got papers on it too by now. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 08:57, 3 October 2008 (UTC)

::Arthur, Thanks anyway, that article looks interesting! Dmcq, so, perhaps we need to think in terms of conformal algebra. The question is then: what is the generator of the exponential function? [[User:Count Iblis|Count Iblis]] ([[User talk:Count Iblis|talk]]) 15:50, 3 October 2008 (UTC)

:We have an article called [[half iterate]] that seems to be about this, although it has almost no content. --[[Special:Contributions/128.97.245.105|128.97.245.105]] ([[User talk:128.97.245.105|talk]]) 21:18, 3 October 2008 (UTC)
::We also have a tiny article [[functional square root]]. I can't currently see a reason to keep both. [[User talk:Algebraist|Algebraist]] 01:22, 4 October 2008 (UTC)
:::They look like exactly the same thing to me. The latter is the more intuitive name, so I'd merge the former into it (I really should be packing, though, so I won't do it now!). --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 12:19, 4 October 2008 (UTC)
::::Merged! I don't think I missed anything of substance (I dropped one red link); if someone wants to double-check me, cool. -[[User:GTBacchus|GTBacchus]]<sup>([[User talk:GTBacchus|talk]])</sup> 15:02, 7 October 2008 (UTC)

:::I just had a look on the web and I see Babbage is credited with being the first to make a study of all the solutions of f<sup>n</sup>(x) = x. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 14:18, 4 October 2008 (UTC)

= October 3 =

== Imaginary Numbers ==

Hello. Does i<sup>4a</sup> = 1 because (i<sup>4</sup>)<sup>a</sup> = 1<sup>a</sup> = 1? There are no parameters onto which set of numbers ''a'' must belong. Is there any amibugity? Thanks in advance. --[[User:Mayfare|Mayfare]] ([[User talk:Mayfare|talk]]) 00:10, 3 October 2008 (UTC)

:1<sup>(3/2)</sup> = (e<sup>(2iπ)</sup>)<sup>(3/2)</sup> = -1 (just kidding ;), 1<sup>a</sup> = 1 for '''any''' a. [[User:Hydnjo|hydnjo]] [[User talk:Hydnjo|talk]] 02:16, 3 October 2008 (UTC)

:(e/c) i<sup>4a</sup>=1 only if a is an integer. Going from i<sup>4a</sup> to (i<sup>4</sup>)<sup>a</sup> does not follow order of operations. If a= 3/4, we get i<sup>4•3/4</sup>=i<sup>3</sup>=-i. [[User:Paragon12321|<span style="color:blue">Paragon</span>]][[User talk:Paragon12321|<span style="color:red"><sup>12321</sup></span>]] 02:53, 3 October 2008 (UTC)

::Parentheses matter. [[User:Hydnjo|hydnjo]] [[User talk:Hydnjo|talk]] 03:05, 3 October 2008 (UTC)

::This is another question where wikipedia ought really have a special article. I just had a look for a paradox based on it by [[Thomas Clausen (mathematician)]] because I was feeling evil, but I can even find it using google. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 09:19, 3 October 2008 (UTC)
::Actually, there's nothing wrong with i<sup>4•3/4</sup>=(i<sup>4</sup>)<sup>3/4</sup>=1<sup>3/4</sup>. You just have to remember that the complex exponential is multivalued, and you have to take the right value of the right-hand-side. The values of 1<sup>3/4</sup> are 1, -1, i and -i. [[User talk:Algebraist|Algebraist]] 09:23, 3 October 2008 (UTC)

::Hmm perhaps I'll just put in a version of that paradox by Clausen:
:::e<sup>2πin</sup> = 1 for all integer n
:::e<sup>1+2πin</sup> = e
:::(e<sup>1+2πin</sup>)<sup>1+2πin</sup> = e
:::e<sup>1+4πin-4π<sup>2</sup>n<sup>2</sup></sup> = e
:::e.e<sup>-4π<sup>2</sup>n<sup>2</sup></sup> = e
:::e<sup>-4π<sup>2</sup>n<sup>2</sup></sup> = 1
::This is plainly false for any integer n except 0 but the original line was true for all integer n [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 09:47, 3 October 2008 (UTC)

== Margin of error question...? ==

If I ask a question to a simple random sample of 8 individuals out of a population of 40, and 1 out of these 8 say yes, then the percentage of yesses is 12.5% But how do I calculate the margin of error for, let's say, 90% confidence.? What formula could I plug all those numbers in?--[[User:Sonjaaa|Sonjaaa]] ([[User talk:Sonjaaa|talk]]) 14:59, 3 October 2008 (UTC)

:Statistics isn't my area, so I don't feel confident giving you a definite answer, but [[Sample size#Estimating proportions]] looks like the place to start. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 15:28, 3 October 2008 (UTC)

The population size is <math> N= 40 </math> and the sample size is <math> n = 8 </math>.
The sample number of yeas is <math> i=1 </math> and the sample number of nays is <math> n-i=7 </math>.
The population number of yeas, <math> I </math>, assumes values from <math> (i =) </math> 1 to <math> (N-n+i =) </math> 33, and
the population number of nays, <math> N-I </math>, assumes values from <math> (n-i =) </math> 7 to <math> (N-i =)</math> 39. The number of ways to obtain <math> (N, n, I, i) </math> is the product of [[binomial coefficient]]s <math>\tbinom I i \tbinom{N-I}{n-i}=\tbinom I 1 \tbinom{40-I}{7}.</math> These 33 numbers are 15380937 25240512 30886416 33390720 33622600 32277696 29904336 26926848 23666175 20358000 17168580 14208480 11544390 9209200 7210500 5537664 4167669 3069792 2209320 1550400 1058148 700128 447304 274560 160875 89232 46332 22176 9570 3600 1116 256 33. Their sum is 350343565, which happens to be equal to <math>\tbinom {41} 9.</math> The general distribution function for <math> I </math> is actually <math>\tbinom I i \tbinom{N-I}{n-i}/\tbinom{N+1}{n+1}</math> but I am not going to prove it here. The accumulated values <math>\sum_{k=1}^I\tbinom k i \tbinom{N-k}{n-i}/\tbinom{N+1}{n+1}</math> are 0.04 0.12 0.2 0.3 0.4 0.49 0.57 0.65 0.72 0.78 0.82 0.86 0.9 0.92 0.94 0.96 0.97 0.98 0.99 0.99 0.99 1 1 1 1 1 1 1 1 1 1 1 1 for <math> I= </math> 1..33. So a 90% confidence interval is <math> 2\le I \le 15 </math> or <math> 5% \le I/N \le 37.5% </math>. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 19:21, 4 October 2008 (UTC).

== Please help in German–English maths translation ==

[[Borsuk's conjecture#Conjecture status]] waits for someone to verify a translation from German. --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 15:01, 3 October 2008 (UTC)

I can provide you the final English translation of that quotation on Saturday (tomorrow) if you answer my stats question above today... :) --[[User:Sonjaaa|Sonjaaa]] ([[User talk:Sonjaaa|talk]]) 15:21, 3 October 2008 (UTC)

:: I do not trade my skills here. And I really don't care what you can provide me, but would be glad if you can – and want to! – help to improve the article. <br/>Anyway I <big>'''[[hate]]'''</big> statistics. [<span title="This place intentionally left blank.">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>] --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 06:26, 7 October 2008 (UTC)

== Grading on a Curve: Percents, Percentiles, and Z-Scores ==

Let us say that there is a college class with 100 students, and this group is being co-taught by two professors. Philosophically, Professor X and Professor Y agree that ideal grade distributions are: 10% A, 20 % B, 40 % C, 20% D, and 10 % F. At the end of the year, each professor independently grades the class and this is how they grade. Professor X rank-orders the students' scores, high to low, from 1 to 100. Then, Professor X assigns an "A" grade to students ranked 1 through 10; a "B" grade to students ranked 11 through 30; a "C" grade to students ranked 31 through 70; a "D" grade to students ranked 71 through 90; and an "F" grade to students ranked 91 through 100. Professor Y approaches the task differently. Professor Y calculates a z-score for each student. A student whose z-score is at or above the 90th percentile (z = 1.28) earns an "A" grade. A student whose z-score is at or below the 10th percentile (z = -1.28) earns an "F" grade. And so forth with the appropriate z-score cut-offs for the "B" and "C" and "D" students. (To make the conversation easier, let us assume that there are no tie-scores and no tie-ranks at all.) Question 1: Will Professor X and Professor Y ultimately have the same final grades for the same students ... or will each method give different results? Question 2: If the results are different, why is that exactly? Question 3: If the class scores are (or are not) normally distributed, does that make any difference or not? Thanks. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 16:58, 3 October 2008 (UTC))

:It all depends on the distribution of marks. Consider the case where 1 student gets 100%, 1 gets 99.99%, 1 gets 99.98% and so on for the first 50 down to 99.51% and the other 50 have a similar distribution in the range 0% to 0.49%. Prof X will get the ideal grade distributions, but Prof Y will get half B's and half D's (if I've calculated it correctly). The difference is because Prof X forces the grades to be exactly the ideal distribution whereas Prof Y calculates the grades based on individual marks and weights it so that the ''expected'' distribution for a normally distributed set of marks will be the ideal distribution. The actual grades will depend on the actual marks, just because a variable is normally distributed doesn't mean a sample will follow that distribution exactly (or even at all for a small sample). I would say Prof Y has the better method, since how good a student you are doesn't actually depend on how good your peers are - some years are just cleverer than other years due to all kinds of influences (including random fluctuation). As long as you try and keep the exam difficulty the same from year to year (not always easy, admittedly), then you'll get fair grades. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 00:11, 4 October 2008 (UTC)

:: Thank you. Three follow-ups to your above comment. (1) If indeed the particular scores of this class were normally distributed, then Professor X's method and Professor Y's method will and should yield the exact same results ... is that correct? (2) To paraphrase, are you saying something along the lines of the following? Professor Y is basing his system on an idealized (normal) distribution, and then superimposing his actual student performance over that theoretical student performance (normal distribution). Professor X is artificially assuming that his group is indeed ideal (normal), whether they are or not. Is that what you are in essence saying? (3) You make the statement that: "Prof Y has the better method, since how good a student you are doesn't actually depend on how good your peers are." Under Professor Y's system, your entire grade (via your z-score) is indeed based on the performance of your peers, no? That is the very definition of your z-score ... comparing you with those being evaluated along with you. A student's z-score derives from the overall class mean and standard deviation. And thus, how well I do (via my z-score) is indeed contingent upon my comparison to my peers and how well they do. Is that not correct? Thanks. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 01:19, 4 October 2008 (UTC))
:::1) Yes, I think so (to the extent that it makes sense to say that a sample is normally distributed - I think it's better to say that it's representative of a normal distribution, although that may just be me). 2) Not really, Prof X's method doesn't involve a normal distribution in any way. 3) Yes, good point. It does depend on your peers, but only via the class performance as a whole rather than individual peers, which is why it is better. It would be better still to calculate the z-scores compared to the last few years data which would account for differences between cohorts (although it requires consistent exams, which can be hard to achieve, especially if the syllabus changes). --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 12:17, 4 October 2008 (UTC)

:Re (2). No, professor X is making no assumption whether the distribution is a [[normal distribution]] or not. He simply forces the 'ideal grade distribution' to be true each year, without regard to whether the students this year are better or worse than the students last year. If the answers and grades from two years are compared, the results may be unfair. The ambitious student selects a class of stupid peers in order to obtain a high grade. Basicly, both professors are cheeting. The grade should reflect the competence of the student, and not that of his peers. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 04:38, 4 October 2008 (UTC).
::An interesting case of normalizing scores happens with setting up [[IQ]] tests where the normalization involves ranking the scores for a test sample of people and projecting them onto a normal distribution. So the z-score will be exactly the same for that sample. The distribution of types of questions can also be altered so different groups of people get the same average - this is to avoid bias! [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 12:23, 4 October 2008 (UTC)
:::That's too what I was trying to avoid thinking about. IQ tests are done like that also. They grade people like prof Y. For example, the IQ test which I'm studying and have trouble accepting, is vulnerable to the scenario mentioned. It would be possible for the clusters of scores to skew the Z-scores in a way that makes no intuitive sense. [[User:Sentriclecub|Sentriclecub]] ([[User talk:Sentriclecub|talk]]) 13:18, 4 October 2008 (UTC)
::::For an IQ test to work they need to use a large enough sample for the normalisation so that the risk of clusters like that is minimal. The IQ reported is basically a z-score, just expressed in a different way (z-scores are arranged so the mean is 0 and the standard deviation is 1, for an IQ the mean is 100 and the standard deviation is fixed, although what's it's fixed to varies from test to test). --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 15:19, 4 October 2008 (UTC)

=== Follow up ===

Thanks to all for the above comments. Here is my follow-up question. In the world of education (specifically, higher education) ... or even in the world of math / statistics / testing / evaluation / etc. ... is there any accepted "standard" of grade distribution? In the above example, I simply (and conveniently) "made up" the distribution of 10%-20%-40%-20%-10% for the A-B-C-D-F grade ranges. Are there any statistically sound or generally accepted distributions? Thanks. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 14:46, 4 October 2008 (UTC))
:Not really. See [[grade inflation]] for one thing that stops such a distribution from existing. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 15:19, 4 October 2008 (UTC)

:If you want a mathematically justified choice of grade distribution, you could choose a uniform distribution (i.e., 20%-20%-20%-20%-20%). This choice of distribution maximizes the [[entropy (information theory)|information]] conveyed by a single grade. However, this choice ignores the non-mathematical issues which dominate the choice of a grade distribution. Eric. [[Special:Contributions/131.215.159.210|131.215.159.210]] ([[User talk:131.215.159.210|talk]]) 10:20, 5 October 2008 (UTC)

:: I guess that I would like to know ... what percent of a distribution would (statistically) be considered "exceptional", what percent would be considered "above average", "average", "below average", etc. Alas ... it is probably circular. The grade of A ("exceptional") means whatever X% the professor considers to be exceptional. Just wondered if there were any standards across fields of statistics / testing / psychology / measurements & evaluation / etc. Thanks. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 21:47, 5 October 2008 (UTC))

:::I don't think there is a standard definition. You may fine [[outlier]] interesting, it gives some of the definitions people use. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 12:58, 6 October 2008 (UTC)

:::: Thanks. I will look at that article. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 14:57, 6 October 2008 (UTC))

: My impression, based on a few years of teaching nights at a local college, is that there is a three part distribution; the A students identify themselves and so do the F students; the B-C-D folks might follow a Gaussian distribution. [[User:Gzuckier|Gzuckier]] ([[User talk:Gzuckier|talk]]) 15:40, 6 October 2008 (UTC)

Thanks to all for your input on my question. Much appreciated. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 01:10, 7 October 2008 (UTC))

= October 4 =

== request for solution for maths question ==

Upon solving the system of equation x15y17=r and x2y7=s for x and y, the answers always turns out to be of the form x=r^a/s^b and y=s^c/r^d where a, b, c and d are positive integers. Calculate the sum of a + b + e + d .

Ans: 31
[[User:Invisiblebug590|Invisiblebug590]] ([[User talk:Invisiblebug590|talk]]) 10:24, 4 October 2008 (UTC)

:Assuming you mean:
::<math>x^{15}y^{17}=r</math>
::<math>x^2y^7=s</math>
::<math>x=r^as^{-b}</math>
::<math>y=s^cr^{-d}</math>
:then I get the following simultaneous equations:
::<math>15a-17d=1</math>
::<math>17c-15b=0</math>
::<math>2a-7d=0</math>
::<math>7c-2b=1</math>
:which has a unique solution, but ''not'' one where ''a'', ''b'', ''c'' and ''d'' are integers, neither is their sum 31. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 11:00, 4 October 2008 (UTC)

== maths question ==

the 2 squares in the diagram have side length 1cm, find the area of the slanted rectangle ABCD.

<gallery>
Image:Diagram-maths-question.jpg|pls help
</gallery>

:Sounds like homework which is a no no here. But as a hint your teacher was probably recently moving triangles around or else was doing similar triangles or else was doing the [[Pythagorean theorem]]. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 12:01, 4 October 2008 (UTC)

::You'll find that the little right-angled triangle at the right is similar to the one whose hypotenuse is the diagonal of the original (two-square) rectangle, which should enable you to find the length of the sides of the slanted rectangle.…[[Special:Contributions/81.132.235.170|81.132.235.170]] ([[User talk:81.132.235.170|talk]]) 14:24, 4 October 2008 (UTC)

There's a quicker way: (1) snip off the triangle with vertices B and C and the third point on the vertical line downward from B, and paste it on the left edge of the figure, getting a triangle below the original rectangle having one vertical side on the left and one horizontal side along the bottom of the original rectangle. (2) Now take the latter-mentioned triangle and turn it over and place it on top of the triangle that is the part of the original rectangle that is above the depicted diagonal. You see that the new rectangle has the SAME area as the original rectangle. [[User:Michael Hardy|Michael Hardy]] ([[User talk:Michael Hardy|talk]]) 21:57, 4 October 2008 (UTC)

: ...or yet another way: Consider the triangle where the two rectangles overlap. The area of that triangle is exactly half of the area of '''either''' of the two rectangles; therefore the two rectangles have equal areas. [[User:Michael Hardy|Michael Hardy]] ([[User talk:Michael Hardy|talk]]) 23:12, 4 October 2008 (UTC)

:You can add more and more rectangles going round in a spiral like in the [[Spiral of Theodorus]] which approximates the [[Archimedean spiral]]. Any idea what this one would approximate or does it stop going round? Just daydreaming, thought that was better than doing homework. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 09:38, 5 October 2008 (UTC)

::A nice problem. Translate the geometry into algebra. Take the rectangle as unit of area. Represent the sides by complex numbers, ''a'' and ''b''. The direction of ''a'' is ''â'' and the length of ''a'' is |''a''|, so ''a'' = ''â''&middot;|''a''|. The direction of ''b'' is i''â'' and the length is |''a''|<sup>&minus;1</sup>, so ''b'' = i''â''&middot;|''a''|<sup>&minus;1</sup> = i''a''&middot;|''a''|<sup>&minus;2</sup> = i''a''&middot;(''aa*'')<sup>&minus;1</sup> = i&middot;(''a*'')<sup>&minus;1</sup> where ''a''* is the complex conjugate of ''a''. The proces of replacing a rectangle by the new one is ''a'' &rarr; ''a''+''b'' = ''a''+ i(''a*'')<sup>&minus;1</sup>. A numeric experiment shows that the first turn is after replacement number 28, the second turn is after replacement number 96, and the third turn is after replacement number 203. It doesn't look like if it stops turning, but it is too early to make a conclusion. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 22:16, 5 October 2008 (UTC).
::Continuing. If ''a''<sub>''n''</sub> =10, then ''a''<sub>''n''+1</sub> = ''a''<sub>''n''</sub>+ i(''a<sub>''n''</sub>*'')<sup>&minus;1</sup> = 10+0.1i = 10&middot;e<sup>0.00005+i&middot;0.01</sup>. One turn is made after about 2&pi;/0.01 = 628 steps and ''a''<sub>''n+628''</sub> = 10&middot;(e<sup>0.00005+i&middot;0.01</sup>)<sup>628</sup> = 10.32. So the spiral is much more tightly wound than the Archimedian spiral. I doubt if |''a''<sub>''n''</sub>| &rarr; ∞. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 12:06, 7 October 2008 (UTC).

== The logics of significance and insignificance ==

So my question is not as interesting as the headline might suggest, but here goes:

Is it correct to say that the statements "X is not insignificant" and "x is significant" are not logically equivalent? And would it be correct to say that this is so because the opposite of "significant" is "not significant", and not "insignificant"? Also: is there a way to show this using predicate logic? (Don't worry: this is not a homework, I just had a discussion about it today with a lawyer-friend). <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/83.250.105.19|83.250.105.19]] ([[User talk:83.250.105.19|talk]]) 17:58, 4 October 2008 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

: Yes, I agree with your conclusion ... and I also agree with your reasoning behind it. Perhaps a more intuitive example helps us to understand the question you have raised. Statement A ("The odor is pleasant.") means something different than Statement B ("The odor is not unpleasant."). I think that the main distinction is that there can exist some middle neutral ground ... a continuum somewhat along the lines of: unpleasant odor -----> neutral odor (or non-odorous at all, even) -----> pleasant odor. Thus, an odor that is "pleasant" is pleasant. However, an odor that is "not unpleasant" may ''either'' be (a) pleasant ... '''or''' ... (b) neutral. If an odor is "not unpleasant", that does not necessarily mean that it must belong to category "a" (pleasant odors), since it can also belong to category "b" (neutral odors). And items in category "b" are items different than those called "pleasant odors" (namely, they are neutral odors). Thanks. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 18:15, 4 October 2008 (UTC))

::It's just a matter of definitions. If you strictly define insignificant as "not significant", then it's trivial. If it's not binary, however -- if there's a neutral choice, neither significant nor insignificant -- then it's a different issue. --[[User:Jpgordon|jpgordon]]<sup><small>[[User talk:Jpgordon|&#8711;&#8710;&#8711;&#8710;]]</small></sup> 18:06, 4 October 2008 (UTC)
:::''Not insignificant'' is an example of a rhetorical figure known as [[litotes]]. Ask about it on the language desk. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 18:16, 4 October 2008 (UTC)
:::::I don't see what the neutral choice between significance and insignificance could be... --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 19:12, 4 October 2008 (UTC)
:What if "significant" means "known to be significant" and "insignificant" means "known to be not significant" etc? <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/65.110.174.74|65.110.174.74]] ([[User talk:65.110.174.74|talk]]) 21:09, 4 October 2008 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->
::That's a bit philosophical for me... "If the probability of falsely rejecting the null hypothesis is less than 5% but nobody knows it, is it significant?" (cf. [[If a tree falls in a forest]]). --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 21:16, 4 October 2008 (UTC)

::: I took the original question to be generic for any adjective ... not specific to the adjective "significant". I assumed that the original poster was asking whether or not the adjectives "X" and "not un-X" (or "not in-X") are logically equivalent ... regardless of which adjective "X" is considered. Granted, some adjectives are more prone to having a neutral intermediate than others. ([[User:Joseph A. Spadaro|Joseph A. Spadaro]] ([[User talk:Joseph A. Spadaro|talk]]) 23:55, 4 October 2008 (UTC))
:::: Yes, that was exactly my question. Thanks a lot for everybody's input. <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/83.250.105.19|83.250.105.19]] ([[User talk:83.250.105.19|talk]]) 07:24, 5 October 2008 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

:::::Don't know if you're interested in this side of it, but a fair bit of research has been done in logical systems with more than two truth values. One of the results of this research is [[fuzzy logic]]. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 07:27, 5 October 2008 (UTC)

:::::Incidentally, I'm not sure I'd say that the "unpleasant", "not unpleasant", "pleasant", and "not pleasant" lie on any kind of a continuum. Unpleasantness describes the existence of something disagreeable. Pleasantness describes the existence of something agreeable. Each lies on a continuum of its own, from "very X" to "not very X". I would say that "not significant" and "insignificant", however, are describing different parts of the same range from "very significant" to "not very significant". So that, for instance, in designing a fuzzy logic description these three would each have a rating from 0 to 1 attached to them. [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 07:51, 5 October 2008 (UTC)

== Definite matrix ==

I am reading a paper on the work of [[Wilhelm Cauer]] which uses the phrase "positive definite matrices" in regard to the ''n''&nbsp;x&nbsp;''n'' matrices '''''L''''', '''''R''''' and '''''D''''' in this equation,

:<math>\boldsymbol{A}=\lambda^2 \boldsymbol{L} + \lambda \boldsymbol{R} + \boldsymbol{D} \,\!</math>

I get the "positive" bit (I think) but am not so sure about "definite". The paper is written in English by a German so there may possibly be a language issue. I thought at first it just means "positive real" but later on we have -

"A key problem is the simultaneous principle axis transformation of two quadratic ''semi''-definite forms (as opposed to the standard case where at least one form is non-degenerate)."

Which implies that I actually haven't understood a thing and "definite" means something else. Anyone know what? [[User:Spinningspark|<font style="background:#FFF090;color:#00C000">'''Sp<font style="background:#FFF0A0;color:#80C000">in<font style="color:#C08000">ni</font></font><font style="color:#C00000">ng</font></font><font style="color:#2820F0">Spark'''</font>]] 18:29, 4 October 2008 (UTC)
:You can't separate "positive" from "definite"; it's a single term, ''positive definite''. See [[positive-definite matrix]]. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 18:43, 4 October 2008 (UTC)

::Ta very much, quick answer, slow to look at it. [[User:Spinningspark|<font style="background:#FFF090;color:#00C000">'''Sp<font style="background:#FFF0A0;color:#80C000">in<font style="color:#C08000">ni</font></font><font style="color:#C00000">ng</font></font><font style="color:#2820F0">Spark'''</font>]] 21:36, 4 October 2008 (UTC)

== "Undiscovered integer"? ==

In one of his books(I think The Demon-Haunted World) Carl Sagan suggests the thought experiment: "What if there is an undiscovered integer between six and seven?"(followed up by a question about the possibility of an undiscovered chemical element whose atomic number is that of the new integer). Does the idea of an undiscovered integer have any sort of meaning at all, or is this more likely intended to suggest, like many paradoxical thought experiments, that our grip on reality may not be as good as we think? [[Special:Contributions/69.107.248.106|69.107.248.106]] ([[User talk:69.107.248.106|talk]]) 20:52, 4 October 2008 (UTC)

:Seven is usually defined to be the next integer after six (see [[Peano axioms]]), so in that sense there can't be an integer inbetween. However, there is a difference between the mathematical concept of integers and the real life concept of collections of discrete objects. We model those discrete objects as integers, but that model could conceivably be flawed (well, I can't conceive how that would work myself, but the point stands that mathematical models are not absolute in the same way that mathematics is [if you ignore Gödel, anyway...]). --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 21:13, 4 October 2008 (UTC)

:You mean like the extra hours in a day that only managers know about? :) [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 21:17, 4 October 2008 (UTC)
:More seriously for people with [[Dyscalculia]] often really wouldn't notice if a few numbers were missing. In fact I think most probably all people have blindnesses of one sort or another. Some people just can't give a greater weight to something that's bigger or see that a lot of small bits can add to a large amount. Probabilities are a complete haze. Ask what shape you get if you stick a corner of a cube into some plasticine and many people will say four. Personally I can see colours perfectly well but the colour of my front door is something I have explicitly learned because it seemed so silly not to know it when asked. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 21:40, 4 October 2008 (UTC)

:[http://people.cs.uchicago.edu/~dinoj/msglef.html Microsoft Supercomputer Discovers New Integer Between 5 and 6 ] - [[User:Fredrik|Fredrik Johansson]] 22:18, 4 October 2008 (UTC)

:In the Internal Set Theory approach to non-standard analysis the natural numbers turn out to contain "non-standard" elements, even though the set of naturals is exactly the same as we usually think of it. In some sense these non-standard naturals are undiscovered and undiscoverable - it's not possible for 379274, or any other integer you might think of, to be non-standard, because it is "uniquely defined by a classical formula". To quote Robert's book Nonstandard Analysis (in Dover, everyone should get one), "These nonstandard integers have a certain ''charm'' that prevents us from really grasping them! Their existence is axiomatic". [[Special:Contributions/163.1.148.158|163.1.148.158]] ([[User talk:163.1.148.158|talk]]) 11:03, 5 October 2008 (UTC)

:In first-order Peano arithmetic, [[Non-standard arithmetic|there exist models]] in which nonstandard elements exist. They have the properties of natural numbers required by the axioms, but they aren't natural numbers (generally, they live 'above' the natural numbers, and succeed each other in various patterns). They're sometimes called ''alien intruders,'' because "they live among us". [[User:Maelin|Maelin]] <small>([[User talk:Maelin|Talk]] | [[Special:Contributions/Maelin|Contribs]])</small> 11:36, 5 October 2008 (UTC)
::According to our article (and my limited understanding), there can't be any nonstandard elements between six and seven—they always come after all the standard natural numbers. Of course, no one ever said you have to believe the Peano axioms in the first place. People have from time to time [http://www.math.dartmouth.edu/~doyle/docs/three/three.pdf expressed doubts] that a function with the properties ascribed to S actually exists, or that Peano induction is valid. -- [[User:BenRG|BenRG]] ([[User talk:BenRG|talk]]) 12:50, 5 October 2008 (UTC)

::There certainly isn't a non-standard integer between 6 and 7 --- such a number wouldn't be an integer. I brought up nsa in response to the second part of the question. [[Special:Contributions/163.1.148.158|163.1.148.158]] ([[User talk:163.1.148.158|talk]]) 13:49, 5 October 2008 (UTC)

:::Oh, agreed. I didn't mean to suggest that those nonstandard elements could lie between the standard natural numbers. I was just mentioning them out of interest. It would be a pretty crummy arithmetic system indeed if it allowed elements to lie between six and seven. [[User:Maelin|Maelin]] <small>([[User talk:Maelin|Talk]] | [[Special:Contributions/Maelin|Contribs]])</small> 01:16, 6 October 2008 (UTC)

:What if there's an undiscovered (or [[Middle Bronze Age alphabets|forgotten]]) letter between [[C]] and [[D]] and nutrition deficient in the [[vitamin]] for that letter is the cause of lots of illnesses in humans? &#x2013; [[User:b_jonas|b_jonas]] 13:26, 5 October 2008 (UTC)
::Then we're all ''doomed'', '''doomed!''' --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 16:31, 5 October 2008 (UTC)
:::[[Contraposition|And since we're not doomed . . .]] [[Special:Contributions/Zain Ebrahim111|Zain Ebrahim]] ([[User talk:Zain Ebrahim111|talk]]) 13:20, 6 October 2008 (UTC)

:::[http://www.youtube.com/watch?v=fqcn_TPu4qQ] [http://icanhascheezburger.files.wordpress.com/2007/07/doomed.jpg] [http://babelfish.yahoo.com/translate_url?doit=done&tt=url&intl=1&fr=bf-home&trurl=http%3A%2F%2Fdoooooooooooooooomed.ytmnd.com%2F&lp=fr_de&btnTrUrl=Translate] [[User:Black Carrot|Black Carrot]] ([[User talk:Black Carrot|talk]]) 03:22, 6 October 2008 (UTC)

:If you like this sort of thing you might like reading some of [[Gregory Chaitin]]'s stuff, like his number that is defined but can't be computed or some theorems being random. I think there's probably some really simple things which are inaccessible to our way of thinking and imagination, perhaps computers will tell us some of them. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]])

= October 5 =

== Complexity of solving a sparse linear system ==

I am interested in estimating the asymptotic average run-time complexity of solving a sparse [[system of linear equations]] using typical [[iterative method]]s. Neither of these articles seem to provide a definitive answer. I assume the matrix ''A'' is ''n'' × ''n'', with up to ''k'' arbitrarily positioned nonzero entries in every row and column, and we are looking for an approximate solution with tolerance &epsilon;. Thanks for any insight or references regarding the matter. [[Special:Contributions/77.126.196.252|77.126.196.252]] ([[User talk:77.126.196.252|talk]]) 13:54, 5 October 2008 (UTC)

:Hi. Are you specifically ''not'' reducing the [[Sparse_matrix#Bandwidth|bandwidth]]? Once that is done the problem is much more deterministic. [[User:Saintrain|Saintrain]] ([[User talk:Saintrain|talk]]) 15:25, 5 October 2008 (UTC)

::I haven't considered that. The article mentions bandwidth reduction for ''symmetric'' matrices, which mine isn't, but I do not see why symmetry is necessary. If applicable, I do not exclude bandwidth reduction, or any other robust preprocessing. What would the complexity look like in this case? Thanks. [[Special:Contributions/77.126.196.252|77.126.196.252]] ([[User talk:77.126.196.252|talk]]) 19:59, 5 October 2008 (UTC)

:(Took me a while to remember back to the old days when things like this were important; modern PCs can handle "''n''"s of hundreds of thousands.) Not quite what you asked, and if I remember correctly (HA!), for a direct solution, the solution time will be about proportional to the sum of the ''A'' matrix column "heights" (up to the top-most non-zero element). [[User:Saintrain|Saintrain]] ([[User talk:Saintrain|talk]]) 23:31, 6 October 2008 (UTC)
:(P.s. none of our business, but what is it?)

= October 6 =

== Solution manual ==

is there an online solution manual for for A first course in Probability by Sheldon Ross (in downloadable electronic format)? <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/203.128.4.231|203.128.4.231]] ([[User talk:203.128.4.231|talk]]) 04:20, 6 October 2008 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

== [[Julia set]] in nature ==

Is there anything in nature that has the structure of a [[Julia set]]? [[User:Mr.K.|Mr.K.]] [[User_talk:Mr.K.|(talk)]] 11:58, 6 October 2008 (UTC)

:How about [http://www.aanda.org/index.php?option=article&access=standard&Itemid=129&url=/articles/aa/full/2003/13/aa2175/aa2175.right.html Fractal Structure of the Horsehead Nebula]. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 12:49, 6 October 2008 (UTC)

::Nice example, and something more earthly?[[User:Mr.K.|Mr.K.]] [[User_talk:Mr.K.|(talk)]] 13:05, 6 October 2008 (UTC)

:Are you looking for the Julia set in particular, or fractals in general? If it's the latter, the Fractal article has a section [[Fractal#In nature]]. -- [[Special:Contributions/128.104.112.147|128.104.112.147]] ([[User talk:128.104.112.147|talk]]) 00:05, 8 October 2008 (UTC)

= October 7 =

==Name==
What's the name given to numeric 6's and 9's that have their thingos swinging out, like the ones in this font? [[User:February 15, 2009|February 15, 2009]] ([[User talk:February 15, 2009|talk]]) 00:23, 7 October 2008 (UTC)

:What font? Wikipedia text is shown in your browser's default [[sans-serif]] font, which is probably not the same as ''my'' browser's default sans-serif font. Anyway, perhaps you're referring to [[text figures]]? —[[User:Ilmari Karonen|Ilmari Karonen]] <small>([[User talk:Ilmari Karonen|talk]])</small> 04:23, 7 October 2008 (UTC)
Sometimes, the two hosts will have completely differing opinions on a game, with one giving way different scores than the other. (An example of this was the 2005 review of ''[[We Love Katamari]]''; while Lucas enjoyed the gameplay and pointed out the good of the game, Tallarico hated the game, particularly its music. The result: Lucas gave it an 8/10 while Tallarico gave it a 3/10.)
::I believe that "seriff" is the "feet" on letters such as ''i, l, t '' etc... But I believe the rule might be similar to whether or not a lowercase Q or a lowercase J gets curved at bottom, or just a straight line. Guessing from memory (and yes I had to learn this for a computer class) it might be Times, but there's also a rule of whether or not a W gets the same amount of width, as a lowercase L. Its interesting stuff, but your question about 6's for example, is probably the same answer as to the stipulation about lowercase Q and lowercase J. That might help you find it easier, since you can expand your question, i.e. a google search. [[User:Sentriclecub|Sentriclecub]] ([[User talk:Sentriclecub|talk]]) 10:45, 7 October 2008 (UTC)

::Try the article on [[font]]. [[User:Sentriclecub|Sentriclecub]] ([[User talk:Sentriclecub|talk]]) 10:46, 7 October 2008 (UTC)

:::[[Swash (typography)]] that should be it. Each typeface has its own rules of whether or not to give swashes onto a 6 or 9. [[User:Sentriclecub|Sentriclecub]] ([[User talk:Sentriclecub|talk]]) 10:56, 7 October 2008 (UTC)

== Compass and Midpoint ==

I recently encountered a problem related to geometry drawing, but did not find an article on it.

So basically, you have a compass. No ruler or anything else. This means you cannot draw a straight line. So given a line segment, use the compass to find the midpoint of it.

Anyways if there is any relevance in this, I thought it might be important to write an article related to geometry drawing? Any tips on the problem let me know ;).

--[[User:ElectricRush|electricRush]]&nbsp;(<small>[[User_talk:ElectricRush|T]]&nbsp;[[Special:Contributions/ElectricRush|C]]</small>) [[Image:USA-China.JPG|20px]] [[Image:Wikipedia Rollback.png|20px]] 03:23, 7 October 2008 (UTC)

:Are you absolutely sure that you do not have access to a straightedge? Because if you did, you could draw two congruent circles centered on the endpoints of the line segment. Then you would draw a line between the points of intersection between these circles, and then that line would intersect with the original segment at its midpoint. [[Special:Contributions/206.117.40.10|206.117.40.10]] ([[User talk:206.117.40.10|talk]]) 03:55, 7 October 2008 (UTC)

::Yes, for this problem you may not use a straightedge. All you can do is use the compass to draw circles. This problem is supposed to be very hard though... --[[User:ElectricRush|electricRush]]&nbsp;(<small>[[User_talk:ElectricRush|T]]&nbsp;[[Special:Contributions/ElectricRush|C]]</small>) [[Image:USA-China.JPG|20px]] [[Image:Wikipedia Rollback.png|20px]] 04:08, 7 October 2008 (UTC)
:::Can you access [http://homeweb2.unifr.ch/hungerbu/pub/Mohr/mohr.ps this file] (PS)? It describes how to perform the operation. It also speaks of the (for me unknown until tonight) [[Mohr-Mascheroni theorem]]. [[User:Pallida_Mors_76|<span style="background:#000;border:#c3c0bf;color: #fff;border:1px solid #999">Pallida&nbsp;</span>]][[User talk:Pallida_Mors_76|<span style="background:#fff;border:#c3c0bf;color:#000;border:1px solid #999">&nbsp;Mors</span>]] 06:52, 7 October 2008 (UTC)
:::In case you can't handle the postcript, here there's [http://www.cut-the-knot.org/do_you_know/compass14.shtml a web site]. [[User:Pallida_Mors_76|<span style="background:#000;border:#c3c0bf;color: #fff;border:1px solid #999">Pallida&nbsp;</span>]][[User talk:Pallida_Mors_76|<span style="background:#fff;border:#c3c0bf;color:#000;border:1px solid #999">&nbsp;Mors</span>]] 07:04, 7 October 2008 (UTC)

::Even more amazingly see {{mathworld|MatchstickConstruction|Matchstick Construction}}, you only need matchsticks to do anything a normal compass and straightedge can do. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 07:55, 7 October 2008 (UTC)

== Infinite increase ==

If I have an integer, let's take 1, and I increase that integer at an infinite speed (or rate), does my number become infinite in a zero time period? &mdash;<strong>[[User:Anonymous Dissident|<span style="font-family:Script MT Bold;color:DarkRed">Anonymous Dissident</span>]]</strong>[[User_talk:Anonymous Dissident|<sup><span style="font-family:Verdana;color:Gray">Talk</span></sup>]] 07:40, 7 October 2008 (UTC)
:It's hard to increase an integer; they're pretty stubborn about staying just what they are. (An integer ''variable'' is another matter.)
:Extrapolating from your question, it's quite possible for a function to have infinite first derivative at a point and yet be pretty well behaved generally. (OK, technically, the derivative won't be defined at that point, but it's pretty obvious what "infinite first derivative" means.) An example is the cube-root function, whose derivative becomes infinite at zero, but (the original function) takes only finite values on the real line. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 07:54, 7 October 2008 (UTC)
::From the concept of a [[massless string]], the answer is yes. If you apply a net force to a massless string, it will accelerate infinitely and reach an infinite velocity, instantly. But you can't apply a force, before the force is applied, which is why your Q is a bit tricky. As soon as the instant you apply your force, say at t=2.7 seconds, then yes it instantly reaches infinite velocity. Oops, upon second thought, you would need an infinite acceleration, not an infinite speed (or rate). But since you're asking about the number reaching an infinite "position" (which i'm butchering your question now), then yes if you apply an infinite velocity to an object, it will instantly reach an infinite position. Just the same that a tiny net-force applied to a massless string will cause it to reach an instant infinite velocity, then the reasoning holds true about your original question. So the answer is ''yes'' but you could better clearify the last 5 words, since the word "instantly" is what you are asking, right? [[User:Sentriclecub|Sentriclecub]] ([[User talk:Sentriclecub|talk]]) 10:39, 7 October 2008 (UTC)
:::Actually the answer is "no" (or more exactly, "not necessarily"). That's what my first answer said, but apparently I didn't say it clearly enough. --[[User:Trovatore|Trovatore]] ([[User talk:Trovatore|talk]]) 12:50, 7 October 2008 (UTC)

== Correlation between shapes ==

I am looking for a formula which determines a correlation coefficient which measures the correlation between two shapes in a two-dimensional space. Each shape is given by connecting the heads of six vectors all pointing in independent directions with angles of 2*pi/6 radians between them. There are two shapes. If the two shapes are identical then the correlation should be one. If the shapes are pointing in opposite directions we would expect the correlation to be negative. The correlation coefficient should between minus one and plus one. Does anybody know how to construct such a formula? [[User:HowiAuckland|HowiAuckland]] ([[User talk:HowiAuckland|talk]]) 10:34, 7 October 2008 (UTC)

: I'm afraid you can't. Think again what 'the shapes are pointing in opposite directions' means.
: Imagine a regular [[hexagon]], centered at the coordinate system's origin. Let P and Q be two opposite vertices of the figure. Make two new shapes by moving P and Q, respectively, a bit outside from their original positions. I suspect you would call it 'shapes pointing in opposite directions', thus assigning a correlation of &minus;1 to that pair. However if the vertices displacement is small enough, the shapes stay 'almost' identical to each other (and to the original hexagon) so their 'correlation' ought to be close to +1.
: So we have a pair of shapes which deserve a correlation of &minus;1 and +1 at the same time. How are you going to solve it?
: [[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 13:28, 7 October 2008 (UTC)

:: Hi CiaPan. Thanks for your reply and you may very well be correct in your assertion. However, in your example just noted the second polygon isn't actually pointing in an opposite direction as the first polygon. In fact, the second polygon is equally pointing, the same amount, in both opposite directions, which is very similar to the first polygon, so I would expect the correlation to be close to +1. Your example has no assymetry, and therefore I expect the correlation to be close to +1. The negative correlation should come from an asymmetry between the two polygons. An example of two polygons which I think should give a negative asymmetry is if you took your example, and instead of stretching out both vertices P and Q equally, you simply stretch one side, say P. Now I would think this would have a negative correlation between these two shapes.
:: A quandary that I have come across which has made me think that the problem might not be well-defined is if you take as your first polygon, a polygon with P stretch outward significantly in one direction, and the second polygon stretched outward at Q in the opposite direction, then you should get a negative correlation. However if you then look at the limit as the P and Q vertices are shrunk back to the same length as the other vertices from the center, then the correlation should get progressively negative but closer to zero. The quandary is that in that limit one might also expect the correlation to approach +1 since in that limit the two shapes are identical. Does anybody think that this example also indicates the possible non-well-posedness of the problem? [[User:HowiAuckland|HowiAuckland]] ([[User talk:HowiAuckland|talk]]) 20:53, 7 October 2008 (UTC)

:You could calculate the mean of the six points and measure the angle between the means for each shape (considered in polar co-ordinates). That would give you a number between 0 and pi, call it ''x''. Then consider 1-2''x''/pi, does that satisfy your requirements? It would consider two shapes that are just simple scalings of eachother to be identical, is that a problem? --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 23:09, 7 October 2008 (UTC)
:: This quantity would be undefined when the means for either shape add up to zero. [[User:HowiAuckland|HowiAuckland]] ([[User talk:HowiAuckland|talk]]) 23:36, 7 October 2008 (UTC)
:::That's a good point, but I think that's due to your question. You talk about whether or not they are pointing in the same direction, but a regular hexagon doesn't point in any direction. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 23:49, 7 October 2008 (UTC)
::::Very true. Regular hexagon wouldn't point in any direction. However, the correlation between two identical shapes should always be +1, and if they are very similar it should be close to +1. (I think, subject to the quandary I listed above, which I can't resolve.) [[User:HowiAuckland|HowiAuckland]] ([[User talk:HowiAuckland|talk]]) 23:58, 7 October 2008 (UTC)

:If you're willing to skip regular hexagons, you could just literally use the [[correlation]] of the two sextuples of vector lengths. Then any other shape would have correlation 1 with itself and would typically have negative correlation with itself rotated or reflected. But you're not going to get perfect -1 in general. --[[User:Tardis|Tardis]] ([[User talk:Tardis|talk]]) 00:37, 8 October 2008 (UTC)
::Yes, I think this might be a good idea, except it seems to give strange results. As an example, in the case where you have all the lengths zero except one vector equal to 1 in the first shape, and then in the second shape you have all lengths to be zero except for the opposite direction with length of 1. Then this gives a correlation of &minus;0.2. Similarly, if you take all lengths to be zero except for a single entry, regardless of position, to be one, you will always get a correlation of &minus;0.2, even when the vectors in the two shapes are pointing nearly in the same direction. I would expect this case to give a positive correlation. Even though the sign is curious, certainly they shouldn't all give the same value of the correlation. They are entirely different shapes. [[User:HowiAuckland|HowiAuckland]] ([[User talk:HowiAuckland|talk]]) 01:35, 8 October 2008 (UTC)

== side, edge of a circle ==

what is the difference between or the meaning of side of a circle and edge of a circle...
How many sides does a circle have?
How many edges does a circle have?

[[Special:Contributions/116.75.181.164|116.75.181.164]] ([[User talk:116.75.181.164|talk]]) 10:43, 7 October 2008 (UTC)
:The [[boundary]] of a circle is neither called ''side'' nor ''edge'', but ''[[circumference]]''. So the short answer to your question is: zero sides and zero edges. However it does make some sense to say that the circle has an infinite number of sides, as it is approximated sufficiently well by a polygon having sufficiently many sides. [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 11:21, 7 October 2008 (UTC).

== The 1-D Heat Equation ==

Help! I don't understand this Heat Equation stuff.

I have a problem with a semi-infinite cylinder of metal on the positive x-axis with sides and face at x=0 insulated. The heat will therefore diffuse according to the 1-D heat equation. The initial temperature distribution is <math>u(x,0)=u_0 \exp(-x/L)</math>, the end is insulated which indicates that <math>u_x=(0,t)=0</math>.

I'm trying to solve this using Green's function, extended to the infinite axis as an even function, and then changing variables to get the error function. But I have been given the solution and no matter what I try I can't get to it. I don't think I understand exactly how to incorporate the initial value function as part of the integrand.

The solution given is <math>u(x,t)=(1/2)u_0exp(c^2t/L^2)[exp(-x/L{1+erf((x-2c^2t/L)/\sqrt(4c^2t)}+exp(x/L){1-erf((x+2c^2t/L)/\sqrt{4c^2t}}]
</math>

If anyone can explain where this solution comes from I would really appreciate it. <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/122.108.235.248|122.108.235.248]] ([[User talk:122.108.235.248|talk]]) 13:49, 7 October 2008 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->
:I suppose you mean <math>u_x(0,t)=0</math>. The ''c'' is unexplained. There are some missing parentheses in the formula which might have been
::<math>u(x,t)=(1/2)u_0\exp(c^2t/L^2)[\exp(-x/L)(1+\operatorname{erf}((x-2c^2t/L)/\sqrt{4c^2t}))+\exp(x/L)(1-\operatorname{erf}((x+2c^2t/L)/\sqrt{4c^2t}))]</math>
:where
::<math>\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt</math>
:is the [[error function]].
:Simplify the computations by choosing the unit of temperature to be <math>u_0</math>, and the unit of length to be <math>L</math>, and the unit of time such that <math>c=1</math>. Then the formula looks like this:
::<math>u(x,t)=e^t[e^{-x}(1+\operatorname{erf}((x-2t)/\sqrt{4t}))+e^x(1-\operatorname{erf}((x+2t)/\sqrt{4t}))]/2</math>.
:The initial condition at ''t''&rarr;0<sup>+</sup> is
::<math>u(x,0^+)=[e^{-x}(1+\operatorname{erf}(x/0^+))+e^{x}(1-\operatorname{erf}(x/0^+))]/2\,</math>
::<math>=[e^{-x}(1+\operatorname{erf}(+\infty))+e^{x}(1-\operatorname{erf}(+\infty))]/2</math>
::<math>=[e^{-x}(1+1)+e^{x}(1-1)]/2=e^{-x}.\,</math>.
:OK! Then check the other initial condition
::<math>u_x(0,t)=0\,</math>
:and the [[heat equation]].
::<math>u_t-u_{xx}=0\,</math>
:[[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 20:02, 7 October 2008 (UTC).

==Simple statistics question==

(related to [[Talk:Mongolia]]): Assuming that 800 children are born in a hospital in one year, and assuming that the probability that a newborn is a boy is 0.5 (feel free to calculate with a more correct value, but then tell me so), what is the probability of more than 430 of the newborn being girls?

My impression is that this should be roughly 0.025 (normal distribution, E=400, sigma square = 200). Anything wrong about this? [[User:Yaan|Yaan]] ([[User talk:Yaan|talk]]) 18:50, 7 October 2008 (UTC)

:Using a normal approximation should get a pretty good answer for n=800 (I haven't checked to see if you calculated it right, but the method is good). You could also do it directly as a [[binomial distribution]] (that article should have the formulae you need). The normal approximation is easier, though, if you don't need an absolutely precise answer. --[[User:Tango|Tango]] ([[User talk:Tango|talk]]) 18:59, 7 October 2008 (UTC)

::From [[human sex ratio]], ''"The natural sex ratio at birth is estimated to be close to 1.1 males/female"'', so you need to take that into account. But your stat is the reverse!? [[User:Saintrain|Saintrain]] ([[User talk:Saintrain|talk]]) 19:48, 7 October 2008 (UTC)

The standard deviation is 200<sup>1/2</sup>=14.1421. The deviation of the observation from the mean is 430&minus;400=30. And 30/14.1421=2.12133. The probability that a normally distributed observation is ''greater than the mean + 2.12133 standard deviations'' is 0.016947, or roughly 1/59, (as compared to 0.025=1/40). [[User:Bo Jacoby|Bo Jacoby]] ([[User talk:Bo Jacoby|talk]]) 21:17, 7 October 2008 (UTC).

: If you're going to claim so much accuracy, you should do a [[continuity correction]] using the fact that in this context, ">&nbsp;430" is the same as "≥&nbsp;430.5". So
:: <math> \frac{430.5 - 400}{\sqrt{200}} = 2.1567, </math>
: and the probability that the standard normal exceeds that is about 0.015515, or about 1/64. [[User:Michael Hardy|Michael Hardy]] ([[User talk:Michael Hardy|talk]]) 01:08, 8 October 2008 (UTC)

= October 8 =


==In popular culture==
== Minimizing first-time comparisons ==
* In the series ''[[Mega64]]'', ''Judgement Day'' is parodied as ''Judgmental Night'', hosted by Tommy Tallarico and "Luke Victorious". Tommy Tallarico appears in the skit as Hairy Gary the Feral Child; however, the Tommy Tallarico character in the same skit is portrayed by Derrick Acosta.


== References ==
What's the best sorting algorithm that minimizes the number of first-time comparisons? This generally involves data that is expensive to initially compare (or otherwise can't be algorithmically compared), but once the initial comparison is made, the result can be cached.
<references/>


==External links==
After a new comparison is made, my system looks for things similar to A < B < C and ensures that A < C. It then searches the cache to find which pair eliminates the most combinations. Even though this works, I feel this may be the bottleneck, since a full search (probably O(n^4+)) could make the sorting much longer than it could, and a limited search (currently O(n^3)) may be less accurate.
*{{imdb title|id=0367347|title=Judgment Day / Reviews on the Run}}


[[Category:G4 television series]]
I'm aware of a monte-carlo algorithm, but I'm not sure whether to go for that or the full-search in the comparison matrix. --[[User:Sigma 7|Sigma 7]] ([[User talk:Sigma 7|talk]]) 02:29, 8 October 2008 (UTC)
[[Category:Television programs about video games]]
[[Category:2000s American television series]]
[[Category:Video game culture]]
[[Category:Space: The Imagination Station network shows]]

Revision as of 03:07, 10 October 2008

 ??? 

File:Reviewsontherun.png
The original Reviews on the Run logo, for stations outside the US. Used from 2002-2007
File:JD whole.png
The only-in-America Judgment Day logo. The show ran from 2002-2006 on G4.

Reviews on the Run (formerly known as Judgment Day in the United States) is a video game review TV show hosted by Victor Lucas and Tommy Tallarico, and produced by Lucas' production company, Greedy Productions. The two hosts rate games independently on a scale of .5 point increments from 0 through 10, with 0 being the lowest and 10 being the highest. The show started as a segment on The Electric Playground called "Reviews on the Run", but was spun off into its own show in 2002. As of December 2007, Reviews on the Run airs in Canada on G4techTV Canada, Citytv Toronto, and Citytv Vancouver, with pre-2006 episodes airing on G4 in the United States. The show was previously shown in Canada on A-Channel, Space, Razer, and OMNI.1.

The show is filmed on location at several different locales including Vancouver, British Columbia. Episodes are also occasionally recorded in other cities. The hosts stand in frame as video game footage is projected onto an object in the background such as a billboard or the side of a building. The show is filmed over the course of several hours and later edited to fit the show's thirty minute time frame.[citation needed]

History

On January 13, 2006, Lucas announced on the G4 forums that Judgment Day was no longer going to be produced for G4. The program will still run internationally (including on G4techTV Canada) as the original Reviews on the Run.[1] Specific details about the negotiations have not been disclosed. As of July 2008, the show has not returned to U.S. airwaves with new episodes.

On December 31, 2006, Lucas announced on The Electric Playground forums that Greedy Productions had cancelled its contract with CHUM television, which broadcast Electric Playground and Reviews on the Run on Space and A-Channel, and signed a two year exclusive deal with Rogers Communications, to broadcast the shows on G4techTV Canada and then additionally on other Rogers owned TV stations.[2] Reviews on the Run aired on OMNI.1 from January 28, 2007 to September 3 2007. On December 15 2007, Reviews on the Run premiered on Citytv Toronto and Citytv Vancouver. On June 2, 2008, G4 started airing episodes of Judgment Day as a part of their G4 Rewind block.

Cast

Victor Lucas and Tommy Tallarico have been both co-workers and good friends since their first meeting at E3 in 1995. Onscreen the two enjoy plenty of brotherly scraps, with Tallarico typically playing the role of a more aggressive, even gutter-minded gamer and Lucas being more businesslike and sedate. (A typical exchange - Tallarico: "Come on, you're saying that you never shot any of those guys in the crotch, even once?" Lucas: "No, I did NOT!") During the "Hardware" segment of Reviews on the Run, Tallarico has occasionally taken the video gaming hardware (console controllers for example) that they were reviewing, and put it down the front of his pants he was wearing, to the disgust of Lucas.

Due to commitments with Video Games Live, Tallarico was absent for six episodes from the 2006 season, and was the co-host in three episodes of the show in the last two seasons. Geoff Keighley co-hosted with Lucas in all of the 2006 episodes Tallarico missed. The 2007 and 2008 seasons of the show have featured a number of guest hosts, including Scott Jones, Tom Russo, Marc Saltzman, Jose "Fubar" Sanchez, Ben Silverman, and Steve Tilley.

During the Hardware segment of the show, Hardware Girls show off pieces of video gaming hardware, while Lucas and Tallarico review the product in question. Hardware Girls are usually shown wearing T-shirts that have messages relating to video games and gaming (for example: "I've got collision detection"). Lost cast member Evangeline Lilly's first job in television was as a "Hardware Girl" for Reviews on the Run, which was referenced when the Lost video game was reviewed on the show.

Segments

  • The majority of Reviews on the Run consists of review segments for individual games on a current major console or on PC. Prior to December 2007, both hosts gave a recap of the "hits" (positives) and "misses" (negatives) after the review ended, with the host who gave the higher rating stating the positive side of the game. Three of these review segments occur in most episodes.
  • Pocket Reviews - Lucas or Tallarico reviews a game on a handheld console or device. Two Pocket Review segments happen each episode, with Lucas doing both when Tallarico is unavailable. Save for the occasional "Versus" segment, handheld games are rarely given a normal review by themselves. G4techTV Canada infrequently airs Pocket Reviews during commercial breaks under the title "Reviews on the Run: Quickplay Edition"
  • Buried Treasures - Lucas or Tallarico talks about a game they felt was great, but was overlooked by consumers. This segment was held in an EB Games or a Rogers Video store. A Buried Treasure game is not rated, only recommended. Occasionally, games reviewed here would be featured on a later or previous episode in a normal review format. Buried Treasures was removed from the show in 2007, and replaced by "Download Games".
  • Download Games - Lucas reviews a game downloaded from the a major console's online service. Download Games were introduced in 2007, replacing the Buried Treasures segment.
  • Classic Reviews - A flashback from an older episode is shown, with a condensed review of an older title. In most cases, the game being reviewed ties in with a game reviewed on the same episode.
  • Hardware - The two hosts review a video game peripheral device, system, add-on, or controller. Occasionally, a game can also be reviewed here if it requires the hardware in question to play it. "Hardware Girls" are shown demonstrating and modeling the hardware in question during the review. Instead of rating the device on a scale, the hosts will either "recommend" or "not recommend" the device. Occasionally, the hardware device will be "not recommended with an asterisk" (e.g. the device is not worth its current price, or not recommended if you currently own an older model). Hardware reviews initially featured a "hits and misses" recap, which was later dropped from the show.
  • The Hit List - Lucas gives a quick rundown of the top five games in a particular category, such as a particular genre, or a given system. From 2006 onward, a variation called the "Sh_t List" has occasionally appeared in place of The Hit List, which has the bottom five games in a given category.
  • Versus - The two hosts review and compare two or three games of a similar genre or subject. They typically break down the review by comparing the graphics, audio, and gameplay of the games in question, with both hosts naming which of the games was better in each category. The games are scored by both hosts, with the game with the highest cumulative score being named the winner of the Versus segment. On occasion, the Versus segment is replaced by both hosts giving quick reviews on two or three unrelated new titles, with no head to head competition in mind.
  • After the versus segment, Lucas recaps all of the games featured (excluding Pocket, Buried Treasure, Download Games, Classic, and Hardware reviews). He goes over quickly, the positive and negative features of each.
  • Rogers Video Game of the Week The highest rated game featured on that show (including titles in the Pocket and Download Games reviews) is awarded the title of Rogers Video Game Of The Week.
  • Before the credits, a few bloopers or deleted scenes from the episode are shown.
  • Every December, an episode is aired where the hosts rate the current major video game consoles, and the PC, on the titles released that year for that platform. The episode was called "Holiday Reviews on the Run" in 2006 and "Year in Review" in 2007.

Video game reviews of note

As of June 2008, only eight games on the show have been given 10/10 by both hosts. Five of them are the multiplatform titles: Prince of Persia: The Sands of Time (which ended with Tommy running around the set [which was a hockey rink] shouting "Ten! Ten! Ten!" with the scoreboard flashing "Tommy gives it a 10!"), Burnout 3: Takedown, Burnout Revenge (which Lucas actually wanted to give an 11/10 before finally agreeing to a 10), The Orange Box, and Grand Theft Auto IV; with the three other games being God of War II for the PlayStation 2, Halo 3 for the Xbox 360, and Metal Gear Solid 4: Guns of the Patriots for the PlayStation 3.

The lowest rated game ever on the show was Chicken Shoot (Wii) which Lucas gave 0/10 and guest host Jose "Fubar" Sanchez gave -2.0/10, making it the first game to score less than 0 on the show. Fubar thought Vic was mad at him so he made him review this game. The first 0/10 was given by Tallarico for the game High Heat Major League Baseball 2003; he hated the game so much he stormed off partway through the review, followed by sounds of a car screeching away. The only other game to receive a 0 was Wacky Races for the Nintendo DS, which Tommy reviewed alone.

Sometimes, the two hosts will have completely differing opinions on a game, with one giving way different scores than the other. (An example of this was the 2005 review of We Love Katamari; while Lucas enjoyed the gameplay and pointed out the good of the game, Tallarico hated the game, particularly its music. The result: Lucas gave it an 8/10 while Tallarico gave it a 3/10.)

In popular culture

  • In the series Mega64, Judgement Day is parodied as Judgmental Night, hosted by Tommy Tallarico and "Luke Victorious". Tommy Tallarico appears in the skit as Hairy Gary the Feral Child; however, the Tommy Tallarico character in the same skit is portrayed by Derrick Acosta.

References

  1. ^ http://forums.g4tv.com/messageview.cfm?catid=22&threadid=545731
  2. ^ Victor Lucas (2006-12-31). "Huge 2007 News For ROTR and EP!!! Need Your Help! :)". The Electric Playground Forum. The Electric Playground. Retrieved 2008-09-08.

External links

Retrieved from "https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Mathematics&oldid=244288607"