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May 25

Propositional Calculus

Sorry if I abuse any terminology in what follows. I took a course in the above last year. Once our teacher had set up the basics, but before we discussed deductive proofs, we discussed a method which he called "Beth Trees". Each node (?) of these trees would consist of two sets of sentences (each possibly empty), one of which was said to be true, the other false. From a node, one could, with reference to truth tables established for functions OR, AND, NOT, IMPLICATION, proceed to another node or nodes. For example, given the node "true: [(NOT p)], false: [empty]" you could proceed to the node "true: [empty], false:[p]".

The idea was that you could start with a sentence and reduce it to a set of valuations on the language which satisfied it. You could also establish that a sentence was a tautology or a logical contradiction. And anything you found out was rock solid because it was directly from truth tables, but also more efficient than using truth table brute force.

Anyway, I can't find reference to Beth Trees anywhere - all I want to know is whether they are more usually known by another name?

Thanks --The Gold Miner 02:44, 25 May 2006 (UTC)[reply]

Perhaps the term of interest is "Truth Tree". Wikipedia does not have an article, but a web search works. (Howard Pospesel says: The method is also known as the "semantic tableau" test. It was conceived independently by E. W. Beth and Jaakko Hintikka in the 1950's.) --KSmrq^T 03:28, 25 May 2006 (UTC)[reply]

Ah, thank you so much, that's the one! Funny, it never occurred to me that Beth would be a person. Hey ho. For anyone interested, I did in fact find a description under Method of analytic tableaux. Thanks again. --The Gold Miner 06:17, 25 May 2006 (UTC)[reply]

why earth orbit is elliptical?

sir my question is why the orbit of the earth is not spherical...but i heard that its elliptical...does this is the reason that the no. of days in the year does not remain same???? — Preceding unsigned comment added by 80.247.152.121 (talk • contribs)

I don't know why it's not spherical (you might want to read up about orbits), but the number of days per year adjustment is due to the fact that there are not exactly 365 days in a year, rather, just less than 365.25. Hence, every once in a while, we make a correction to make up for this. x42bn6 Talk 07:18, 25 May 2006 (UTC)[reply]

Kepler's laws of planetary motion describe how two bodies orbit each other, the planetary orbit will either be an ellipse, hyperbola or parabola. Spherical orbits are a special case of elliptical orbits and require very special conditions to hold, conditions which are not satisfied by the planets. When you look into it the concept of Year gets quite involved. --Salix alba (talk) 07:56, 25 May 2006 (UTC)[reply]

The orbit of a planet around a single star without other major perturbations can never be spherical, but it can be circular. We do not know precisely how the planets formed and assumed their present orbits, but we do know it would be extraordinary to have a perfectly circular orbit. Instead, the stable orbits of the planets are ellipses, differing from circularity by various amounts, measured as eccentricity. The orbit of Earth is only slightly non-circular; the orbit of Mars, more noticeably so. The orbit of Mercury is so close to the Sun that its mostly elliptical orbit shows relativistic precession.

Earth's orbit is such that the orbital period, the time to circle the Sun once, is constant for most practical purposes. The rate at which Earth spins on its axis is nearly constant as well. However, the ratio of the length of a year to the length of a day is not a whole number, but is slightly larger than 365. (The ratio of year to month is also fractional.) Thus a calendar with a fixed number of days in each year will drift noticeably over time, so that instead of (Northern Hemisphere) winter in January it could be in April. Because this was considered undesirable, attempts were made to adjust the calendar by the carefully timed insertion of leap years. (We also have the lesser-known leap seconds to adjust the length of a day.)

So the mathematical reason the number of days in the year changes in our Gregorian calendar is to approximate a fraction. --KSmrq^T 17:41, 25 May 2006 (UTC)[reply]

One consequence of the slight ellipticity of the earth's orbit is that the earliest sunset in winter doesn't occur on the same day as the latest sunrise. If I recall, they're about two weeks apart. (In the northern hemisphere, one is mid-December, the other around New Year, but I forget which is which.) Arbitrary username 21:05, 26 May 2006 (UTC)[reply]

Diagonalizing a matrix.

How can i find the corresponding Eigen vectors for 3X3 matrix with eigen values labda one =labda two, different from labda three. Nkomali

When a matrix can be diagonalized, it has a complete set of eigenvectors which may be chosen mutually perpendicular, and that span to form a basis. However, when two eigenvalues are equal we may be free to choose the eigenvectors as any two perpendicular vectors within a subspace, but we also have to guard against a matrix like

A={\begin{bmatrix}2&1&0\\0&2&0\\0&0&3\end{bmatrix}},

which cannot be diagonalized. From its characteristic polynomial, det(λI−A) = λ³−7λ²+16λ−12, we know that the eigenvalues are 2, 2, and 3; but it does not have a complete set of eigenvectors. Recall that the idea of an eigenvector is a direction in which the action is pure scaling. In this example, that is true for (1,0,0) and (0,0,1), but not for their mutal perpendicular (0,1,0). One further complication is worth mentioning, though it cannot occur with the specific constraints given. A matrix like

R={\begin{bmatrix}0&-1&0\\1&0&0\\0&0&1\end{bmatrix}}

has only one real eigenvalue and associated real eigenvector. Its characteristic polynomial, λ³−λ²+λ−1 = (λ−1)(λ²+1), does not factor into three roots over the reals.

It is common for the matrices in applications to be symmetric, and these can always be diagonalized. Given a matrix, numerical software ordinarily finds eigenvalues and eigenvectors simultaneously, though there are less satisfactory ways of producing an eigenvector from an eigenvalue when possible. --KSmrq^T 19:13, 25 May 2006 (UTC)[reply]

Segmentation Fault

During a 1 user Solaris test the CMS(Central Management Server) crashed. The core file was generated at 2AM, so you check the machine on which the database was running. You find an event logs that says "MS SQL Server: unknown error" and it was generated at around 2AM. You then extract the thread stacks from the core file, and you see that a thread had a segmentation fault (SIGSEGV). At the top of the stack you see that the thread was executing some Database Subsystem code of ours.

Question: What do you think is the most likely cause of the segmentation fault and why?

Probably buggy software. Who dumped core? Dysprosia 11:24, 25 May 2006 (UTC)[reply]

I doubt it's being caused by the MS SQL Server code. If you use the command "file core" on the core dump file, it will tell you the name of the program that dumped core. You can use a debugger (like dbx or gdb to examine the call sequence and the arguments being sent to the last call that failed. Most of the time, a segmentation fault is called by a null pointer. (Apologies if you knew this already.) --Elkman 16:39, 25 May 2006 (UTC)[reply]

Nitpick: In my experience segfaults are more often caused by bad pointers (e.g. to freed memory) than by null pointers. Null pointers are easy to identify, but bad pointers can be sneaky. —Keenan Pepper 19:35, 25 May 2006 (UTC)[reply]

precentages

if a final is 20% of your grade and you get a 95% on it and your grade for the class is 81% what is your new grade?

amanda

Just use weighted average: ${\frac {80\times 81\%+20\times 95\%}{100}}=83.8\%$ --Borbrav 21:56, 25 May 2006 (UTC)[reply]

What about this

$(0.95\times 0.2+0.8\times 0.81)\times 100=83.8$

Same thing. Black Carrot 17:51, 26 May 2006 (UTC)[reply]

May 26

Why isn't mathematics cool?

I study undergraduate mathematics. I like it and I'm proud of it. The other day I was chatting to a new acquaintance, an arts student. I told her that I studied maths and her response was "Aww, gutted!" which round here means "Awww, that must be terrible, you poor thing!" My initial reaction was one of pity for someone cursed with such bad manners, but it did get me thinking.

Now, the briefest exposure to pop culture will reveal that this attitude (i.e. maths is hard, boring and ostracising) prevails throughout the UK, the US and beyond. But I want to know: is there anywhere in the world where mathematics is cool? I mean, is it a cultural thing, or what? --The Gold Miner 13:43, 26 May 2006 (UTC)[reply]

Dunno. I guess that the reason might be that high school-level maths is indeed boring; I mean, all you do are ugly calculations, and no nice reasoning. Cthulhu.mythos 14:28, 26 May 2006 (UTC)[reply]

My impression is that a lot of people think that advanced mathematics is cool. Like "oh, you study infinite-dimensional hyperchaotic transfinite differential matrix supermanifolds? Cool! That sounds so much more interesting than my job as a copyeditor or administrative assistant." (NB: there is no such field of mathematics). On the other hand, most innumerate people view people who excel at math as kind of nerdy. So saying you study math is an admission of nerdiness, and you have to deal with the derision until you get to the hyperchaotic supermanifolds. But the question remains: why are people who excel at math viewed as nerdy? As far as I can tell, it's because they are nerdy. I mean, I myself am not (well, I probably am, but am also unable to recognize it), but I know of lots of math/science/engineering types who are either very socially awkward, bad at sports, physically weak, sexually unattractive, extremely eccentric, obsessive about minutiae, or otherwise nerdy. It seems to me that nerdy people are just a lot more common among the mathematically inclined. -lethe ^{talk +} 14:52, 26 May 2006 (UTC)[reply]

While I don't know for sure why exactly people think like that, I think I have noticed that a disproportionate number of maths and CS students are people-shy and have difficulty maintaining a conversation on topics, such as art and literature, that involve subjective judgments. By "disproportionate" I mean more than you would expect among a random group of age mates of the same sex. This is self-perpetuating, since it discourages people-oriented students to pursue the study of mathematics. I must further say that a lot of maths is taught in an awful and uninspiring way, giving no hint that this is something you might actually love. More puzzling to me is actually why people can proudly proclaim or even acclaim their advanced and terminal innumeracy, as cogently observed by Hofstadter. Whatever the case, I am convinced both phenomena are related to the breakdown of communication written about in The Two Cultures. --Lambiam ^Talk 17:01, 26 May 2006 (UTC)[reply]

Perhaps they've never met the mathematicians described in (ISBN 0809249766) and (ISBN 0151581754) and (ISBN 0253211190). I find a widespread misconception that mathematicians spend their time doing numerical calculations. Mathematicians have spoken and written to the general public about topics such as Symmetry (ISBN 0691023743), Geometry and the Imagination (ISBN 0821819984), and even Journey Through Genius: The Great Theorems of Mathematics (ISBN 014014739X). Few people today have been exposed to these. Television shows and movies paint caricatures that do little to dispel the stereotypes. And, finally, mathematical beauty is unlike the beauty of a natural scene, perhaps a sunset, because what is observed is mostly in the mind. --KSmrq^T 23:33, 26 May 2006 (UTC)[reply]

Many thanks to everyone who responded! --The Gold Miner 22:40, 27 May 2006 (UTC)[reply]

just to answer another part of your question (as a fellow maths undergraduate)- maths is cool in france. i've only had limited discussions with french people but, from that sample, it seems to be that the nation that brought us such un-rigorous and airy-fairy concepts such as human rights and existentialism also values maths. strange but true. (although i probably vould have ran upstairs and told you rather than posted that here, but this was is so much more fun!)

Thanks - you wouldn't be a doll and put the kettle on, would you? --The Gold Miner 18:07, 28 May 2006 (UTC)[reply]

Another reason is that true mathematics requires spending a lot of time in your head, which isn't very conducive to social skills. Da niel (‽) 19:05, 29 May 2006 (UTC)[reply]

That depends entirely on what you spend your time on. Brilliance and careful thought can be, in theory, applied to anything. Of course, since we're talking about career nerds, it's obvious what trains of thought they've chosen to go with. Black Carrot 20:12, 29 May 2006 (UTC)[reply]

Oh, and to answer your question, math reasonably cool where I live, because I go to a magnet high school. A lot of the people I know both are good at math and are perfectly normal highschoolers. I'd be hard to block us out of the social circle in the first place, because there's just too many of us there. Black Carrot 20:16, 29 May 2006 (UTC)[reply]

Math is cool as in it's not hated or anything where I live. But that's mostly because my school is about 70% Asian, and you might know that a lot of Asian parents foster their kids' interest in math. --M1ss1ontomars2k4 ^{(T | C | @)} 20:20, 29 May 2006 (UTC)[reply]

System of seminorms that induces topology weaker than other system, is a 'weaker system'

Hello,

this question has been bugging me for months. My professor hasn't given me a clear proof, and other sources even say it is wrong. I could really use some advice on this problem :

Let V be a vector space (complex) Let P and Q be two systems of seminorms. A system of seminorms is a set of seminorms that has the filtering and separating property.

Now a system like that gives a topology, we say a subset $E\subseteq V$ is open if and only if for each point p in E there is an open semiball around p completely in E: $\forall p\in E:\exists f\in P:\exists r>0:b_{f}(p,r)\subseteq E$

Now we say a topology is weaker than another, if every open set in the first is also an open set in the second. An alternative way of saying this : topology A is weaker than topology B , if the identity map from the second to the first is continuous.

Now, we say a system of seminorms P is weaker than a system Q on the same space, when for every p in P, there is a seminorm q in Q and a constant C>0 , such that $\forall v\in V:p(v)\leq Cq(v)$

Now it is obvious that if a system is weaker than another, the induced topology is weaker.. but what about the converse? According to my professor, the converse is true. But how to prove that? I've done it for two norms on the same space, but those are two very special cases of systems of seminorms.

Really, all replies welcome!

Evilbu 15:26, 26 May 2006 (UTC)[reply]

Given a topology on a locally convex topological vector space, you can arrive at a family of seminorms using the Minkowski gauge. (If the topology is Hausdorff, the family will be separating, but I don't recall at the moment what it means for a family of seminorms to "filter". Can you remind me?). If I were going to attempt to prove your converse, that's how I would do it. -lethe ^{talk +} 15:30, 26 May 2006 (UTC)[reply]

Actually, I think the proof is even easier than that. The topology induced by a family of seminorms is the initial topology with respect to those maps. The comparison condition you cite is nothing but the continuity condition for a locally convex space, so the two conditions are equivalent. In other words, given two families of maps A and B, B is stronger than A if and only if all of A are continuous in the initial topology of B (this is the definition of stronger families of maps). But this is nothing but the statement that the initial topology of A is coarser than the initial topology of B. -lethe ^{talk +} 15:38, 26 May 2006 (UTC)[reply]

But how does that transfer to systems of seminorms? Why is a finer/coarser topology induced by a stronger/weaker system? Not knowing anything about it, here I am thinking aloud. Let topology Ti be induced by system Si, 1 = 1, 2. Assume T1 is strictly finer than T2. If S1 and S2 are comparable as to strength, clearly if S2 is stronger than S1, then T2 is finer than T1, contradicting the assumption on T1 and T2. Since S1 and S2 are comparable, and S2 is not stronger than S1, we conclude that S1 is stronger than S2. So far so good. But what if S1 and S2 are not comparable? Can you show that there are comparable S1' and S2' inducing, respectively, T1 and T2? (I assume that that is good enough to establish the somewhat imprecisely formulated claim.) --Lambiam ^Talk 17:20, 26 May 2006 (UTC)[reply]

As far as I can see, the definitions are the same. One family of maps is stronger than another if the second is continuous in the first's initial topology (def). This is equivalent to that the initial topology of the first family be finer than that of the second. As for your question, if the two topologies are not comparable, then the families of seminorms which induce it are also not comparable, a corollary of the fact that the definitions are the same. -lethe ^{talk +} 21:20, 26 May 2006 (UTC)[reply]

Assume that the topology T(P) is weaker than T(Q) for some systems of seminorms P and Q. For any $f\in P$ and any r>0 $\{v:f(v)<r\}$ is open in T(P), thus it is open in T(Q), so there is $g\in Q$ and r'>0 such that $\{g\leq r'\}\subset \{f<r\}$ . Assume that v is such that g(v)=0, thus 0=ag(v)=g(av)<r', so f(av)<r for any a, therefore f(v)=0. Assume $g(v)\not =0$ , then

g\left({\frac {r'v}{g(v)}}\right)=r'\Rightarrow f\left({\frac {r'v}{g(v)}}\right)<r\Rightarrow f(v)<{\frac {r}{r'}}g(v)

for any v.

(Igny 05:19, 27 May 2006 (UTC))[reply]

Rushing Roullette

What are the that someone would lose Rushing Roullette if you played 2 times in a row? What about 3,4,5 times in a row? More importantly, what equations would one use to solve this?199.201.168.100 15:53, 26 May 2006 (UTC)[reply]

Do you mean Russian roulette ? First, you would need to know how many chambers the revolver had and how many bullets were loaded. If there was 1 bullet and 6 chambers, then the odds of dying on the first shot would be 1/6 (assumming a 100% death rate if the bullet is in the chamber you try to fire). If the player survives and the barrel is spun after each trigger pull, then the odds would be the same (1/6) every time after, as well. Otherwise, if the player survives and barrel is NOT spun, the revolver advances to the next chamber, the odds would be 1/5, then 1/4, 1/3, 1/2, 1/1 (this means, if the first 5 players survived, the 6th would be guaranteed to die). Of course, the chances someone will die before it gets to be your turn go up the later your turn is. If you had the second turn, your chances of having to play (the chance the first player survived) would be 1 - 1/6 or 5/6. This, multiplied by the 1/5 chance of dying if you have to pull the trigger, gives you (5/6)(1/5) or 1/6 chance of dying, which is the same as the first player. In fact, the odds of dying are the same 1/6 for all six trigger pulls if the barrel is not spun between them. Thus, if you plan to fire a maximum of twice (it doesn't matter if it's in a row or not), your chance of dying is 2(1/6) or 2/6, three times is 3/6, four times is 4/6, five times is 5/6 and if you fire six times in a row your chance of dying is 6/6, or 100% guaranteed. I don't suggest that you try to verify this at home. StuRat 16:55, 26 May 2006 (UTC)[reply]

If this is a single player, spinning the cylinder (or whatever it is that revolves on a revolver) each time, the odds of survival in the N+1^th round given survival up to and including the N^th round is 5/6. Because each round is independent, you find then survival odds of (5/6)^N for N rounds, which means fatality odds of 1 – (5/6)^N. For the first few values of N this amounts to:

N = 0: p_fatal = 0.00000

N = 1: p_fatal = 0.16667

N = 2: p_fatal = 0.30556

N = 3: p_fatal = 0.42130

N = 4: p_fatal = 0.51775

N = 5: p_fatal = 0.59812

N = 6: p_fatal = 0.66510

N = 7: p_fatal = 0.72092

N = 8: p_fatal = 0.76743

N = 9: p_fatal = 0.80619

Personally I'd call any player of Russian roulette a loser even before they begin to play. --Lambiam ^Talk 17:38, 26 May 2006 (UTC)[reply]

Postscriptum. I just read the article Russian roulette, and my impression is that the section Odds is at odds with probability theory, or else so unclear as to be misleading. It should be rewritten, which ought to be a cinch for any probabilist out there (or should I say "in here"?). --Lambiam ^Talk 17:45, 26 May 2006 (UTC)[reply]

What is the problem? Just as StuRat shows, it is correct under the assumption that the barrel is not spun after every shot, and it displays the probabilty that you will die on round n, given that you have made it there. -- Meni Rosenfeld (talk) 18:06, 26 May 2006 (UTC)[reply]

I'm not a gun expert, but isn't it the cylinder that is spun (or not)? The article states: "If the cylinder is spun after every shot, the odds of losing remain the same". Still correct? --Lambiam ^Talk 23:23, 26 May 2006 (UTC)[reply]

I have no idea what is spun. And yes, if the whatever is spun after every shot, the probability of losing on round n, given that you have made it there, is 1/6 - The same as in the first round ("remain the same" does not mean "the same as in the previous case", but rather "the same as in the first round" - As opposed to the first case, where the probability increases at each round). -- Meni Rosenfeld (talk) 12:57, 27 May 2006 (UTC)[reply]

Right. You know what it is that it is the same as, I know what it is that it is the same as, but does the reader know? It could very easily be taken to mean: "the same as in the previous case". I think it is also confusing that the probabilities are conditional: "given that you survived this far". The statistically uneducated reader could easily understand the given table to imply that the odds of surviving 2 rounds are 1-1/5 = 4/5 instead of 4/6. --Lambiam ^Talk 14:06, 27 May 2006 (UTC)[reply]

I agree that it could have been clearer and that a rewrite would be useful, but still, I think it's clear enough and that a rewrite is not necessary. -- Meni Rosenfeld (talk) 15:51, 27 May 2006 (UTC)[reply]

That reminds me, I must rent a copy of The Deerhunter and reminisce about these scenes. JackofOz 14:26, 27 May 2006 (UTC)[reply]

I agree that our article wasn't clear on whether they were giving the odds assuming you get to the second (and subsequent) trigger pull or the odds including the probability that the game may not progress that far. StuRat 02:26, 29 May 2006 (UTC)[reply]

Unless someone objects I'll copy this over to the talk page of Russian roulette. --Lambiam ^Talk 09:25, 29 May 2006 (UTC)[reply]

Maths in perspective

I've been thinking about this for a while, and never quite found any good answer, though I'm sure the question is actually extremely simple. The idea is to be able to describe the length of a window on a traincar which is positioned at a certain degree relative to you. I regularly draw, so this is of some practical importance to me.

Imagine a coordinate frame x-y (preferably, keep it so that both coordinates are always positive). Now, the train is represented by a line going from nowhere in particular, but for this example we can choose origo. The window is represented by an AB length somewhere on this line. The line has the function x=y, so it stands at 45 degrees relative to both x and y. As this function changes to 2x=y, 3x=y and so forth, one needs to figure out the amount of degrees that the line is at, relative to either axis of the frame. By doing so, one should in theory easily be able to tell what arclength AB has, and thus, what length the window should be drawn to have on the two-dimensional paper of the artist. Even has the line is moved randomly around on the frame (representing the artist changing his or her point of view), this shouldn't be difficult to calculate. My question is only, how? Thanks. :) Henning 19:37, 26 May 2006 (UTC)[reply]

The width of the window as projected onto your direction of view is going to be the true width of the window multiplied by the cosine of the angle between the direction of view and the normal to the window. If you are trying to relate it to these equations of lines, then remember that the numbers in the equation basically relate to the tan of the angle. To relate these to cos of the angle you want to use the relationship that cos(theta) = 1/sqrt(1+(tan(theta))^2) if the angle is called theta. e.g. for 2x=y, tan(theta) = 2 (if you're looking parallel to the y axis, say from somewhere on the x axis), so cos(theta) = 1/sqrt(5), or about 45% of the window's true width. Arbitrary username 21:16, 26 May 2006 (UTC)[reply]

Wikipedia has several articles related to perspective; perspective (graphical) is a good place to start. --KSmrq^T 05:13, 28 May 2006 (UTC)[reply]

May 27

Real life Block Transfer Computations

block transfer computations [UK television series "Dr. Who"] Computations so fiendishly subtle and complex that they could not be performed by machines. Used to refer to any task that should be expressible as an algorithm in theory, but isn't.

Doctor Who, tiring a bit over his TARDIS' fixed police box exterior, decides maybe his chameleon circuit could use some fixing. However, to fix the circuit, he needs to perform some Block Transfer Computations. These computations are so complex, no computer can handle it. The only way they can be done is using the services of a mathematically adept monk-like people called the Logopolians. The math monks of Logopolis perform these computations by meditation and very loud chanting. An acropolis-like building called the Central Registry handles the inputs and outputs of the math monks.

My question is this: Are there any real life block transfer computations in mathematics?

Ohanian 12:11, 27 May 2006 (UTC)[reply]

What about theorem proving? If I remember correctly, this is possible by computers in only rather basic cases, but not in general. Dysprosia 12:17, 27 May 2006 (UTC)[reply]

[Edit conflict] That depends. Any proof (in some traditional sense) is just a finite string of characters taken from an enumerable set, and is therefore enumerable. A computer with infinite memory seeking to prove a theorem can simply scan through all such strings until it finds one that proves it. If a proof exists, it is bound to find one. If memory is finite, even a PC should, given enough time, outperform humans in finding proofs - But "enough time" would usually be orders of magnitude greater than what a human would need. However, telling whether a given statement is decidable (that is, either provable or refutable) is something that AFAIK a computer cannot do in the general case - But I don't know of any monks who can do that, either. In other words, I doubt there is any computation which humans can do and computers cannot. -- Meni Rosenfeld (talk) 12:50, 27 May 2006 (UTC)[reply]

Is not the problem verifying that a proof found in the way that you describe is in fact the proof? Dysprosia 12:59, 27 May 2006 (UTC)[reply]

No, that is completely technical, with given axioms and inference rules. The proof itself contains information about which of these to use and how. It would AFAIK only require an amount of memory proportional to the length of the proof - Which I believe will be measured in kilobytes in common cases. -- Meni Rosenfeld (talk) 13:04, 27 May 2006 (UTC)[reply]

I've misunderstood something you said earlier. The proof instance must be encoded in terms of the axioms and inference rules. Dysprosia 13:26, 27 May 2006 (UTC)[reply]

In practice, certainly, current automated theorem provers are very limited beasts.

In theory, however, the question of whether there are questions that can be answered by the human mind, but not by a conventional computer, is one that has attracted some philosophical debate, and is closely related to the Strong AI question. Roger Penrose has written two almost impossibly dense books on the topic; the general conclusion in AI community (and mine, as much as I can figure out what the hell he's trying to say) is that Penrose has gotten it wrong. Have a look at halting problem for some related to discussion. --Robert Merkel 12:38, 27 May 2006 (UTC)[reply]

At Wikipedia:Reference desk archive/Mathematics/May 2006#The Revenge of the UTM you find the construction of a sentence that a machine can prove, but you can not. Stronger, any competent mathematician should agree that it is a valid proof, except you (or else you have shown yourself incompetent). --Lambiam ^Talk 13:40, 27 May 2006 (UTC)[reply]

There are certainly problems that are known to be hard for computers and most humans alike, like the decision problem in Presburger arithmetic, which is known to need more than exponential time, and a similar decision problem for real numbers that requires exponential space. To the best of my knowledge these problems have not been tried out on Logopolians. You might try Wikipedians instead, with the additional advantage that they don't chant that loud. --Lambiam ^Talk 13:49, 27 May 2006 (UTC)[reply]

No, we have no reason to believe that a computing device based on carbon has any qualitative theoretical advantage over a device based on silicon. If such a reason emerged, we would eventually find a way to build a machine to exploit the advantage. This plot idea is yet another instance of the conceit that homo sapiens is neither an animal nor a machine, but something "special" and separate from the rest of the natural world, despite all evidence to the contrary. (However, the Logopolis episode itself is special, though a natural part of the Doctor Who world.) --KSmrq^T 19:27, 27 May 2006 (UTC)[reply]

What about integrations? Human can perform integrations on mathematical expressions but computers cannot (they can only perform numeric integrations not symbolic integrations). Ohanian 22:15, 27 May 2006 (UTC)[reply]

On the contrary. The best symbolic computer algebra systems use the Risch algorithm for symbolic integration, which is (in theory) complete but a bit too complicated for humans to use without mechanical support. --Lambiam ^Talk 22:25, 27 May 2006 (UTC)[reply]

On this planet, computers do a fine job of symbolic integration. Here are indefinite integrals a computer algebra system should be able to do almost immediately; see how quickly you can find an answer to each.

\int {\frac {x}{\sqrt {1-x^{3}}}}\,dx

\int {\frac {3{\sqrt[{3}]{x+e^{x}}}+(2x^{2}+3x)e^{x}+5x^{2}}{x{\sqrt[{3}]{x+e^{x}}}}}\,dx

Try these and more at integrals.wolfram.com, which is a free online demonstration of what a computer symbolic integrator is capable of handling. --KSmrq^T 00:06, 28 May 2006 (UTC)[reply]

The same idea I described for finding a proof to a given statement, can be used to find an antiderivative to a given function. Under certain assumptions, if the antiderivative has a closed form, it will eventually be found. Of course, this is extremely inefficient in practice. -- Meni Rosenfeld (talk) 17:13, 28 May 2006 (UTC)[reply]

May 28

Video editing program

A month or two ago I asked about a program that I can use to cut out and edit audio files. I was refered to Audacity and it has worked great for me. Now what I was wondering if there was a program like Audacity but for video. I would really like if it were free like audacity. Thanks. schyler 03:18, 28 May 2006 (UTC)[reply]

List of video editing software--Frenchman113 on wheels! 14:08, 28 May 2006 (UTC)[reply]

Computer / Network Worms

i would like to know more on a new computer worm called NEMATODES — Preceding unsigned comment added by 195.229.242.54 (talk • contribs) 10:50, 2006 May 28 (UTC)

(Please sign your post with four tildes, ~~~~, even if not logged in.) Perhaps you're thinking of the "good worms" idea proposed by David Aitel at Immunity. If so, background papers are available for download. --KSmrq^T 02:09, 31 May 2006 (UTC)[reply]

variational methods

Sorry for asking here, Wikipedia, but during the last exam period I asked a question here which was answered in a really good way, and I totally understood the reasoning, much better explained than by my personal tutor (and in the end I got 76% for that exam, with 15% of that thanks to Wikipedia). So I thought I may use you good volunteers again. This question is about a part of variational methods (I was away during the relevant lectures, and haven't any notes on this, you see), and how to go about answering a problem of this sort:

Take a question like the following one appearing on an old exam paper:

Find the extremals x(t) of the integral I(x), where
I(x)= integral between 1 and 0 of (4x² + t²)dt; subject to the constraint
Constraint = [integral between 1 and 0 of x²dt]=2, with x(0)=x(1)=0.

I don't think my lecturer likes it when I went straight to solving it via Euler's Equation, so what would be the best way to go about solving equations of this nature? I assume it is relatively straightforward (for final year university maths at least!). Thanks in advance, Wikipedians. --Knotted 13:41, 28 May 2006 (UTC)[reply]

That is a special case: any function satisfying the integral constraint has the same value of I(x). In general, I'd solve variational problems with an integral constraint by adding the constraint with a Lagrange multiplier to I(x). If you can't get notes, it's probably best to look it up in a text book on Calculus of Variations. -- Jitse Niesen (talk) 03:22, 29 May 2006 (UTC)[reply]

Removing bloatware for WinXP

I was wondering if there was a way to remove unneccessary software that is built into Windows (like IE and WMP) without paying $35 for a copy of XPLite?--Frenchman113 on wheels! 14:10, 28 May 2006 (UTC)[reply]

Internet explorer is just about impossible to remove completely. IE is really just an interface to a lot of DLL's which lots of other applications use. On microsofts web site there is an article explaining how to remove internet explorer itself, but this doesn't get rid of the bloat. I once came across a web page that explained how to remove it completely. The procedure was quite involved, and looked to me like a maintenance nightmare. I would also expect to see other applications malfunctioning after doing this. I was unable to retrieve that web page now in a quick google search. --vibo56 14:50, 28 May 2006 (UTC)[reply]

Great idea, but Microsoft has business and legal reasons for wanting to get in your way, so don't look to their web site for help. You may find what you need in nLite; though I haven't tried it. Any computer that can run XP can also run alternatives; GNU/Linux, FreeBSD, and Mac OS are all decendents of Unix that may better serve your needs. --KSmrq^T 17:47, 28 May 2006 (UTC)[reply]

This was already asked on the reference desk, although it's quite difficult to search for it: Wikipedia:Reference_desk_archive/Science/February_2006#Internet_Explorer_Removal. – b_jonas 08:12, 30 May 2006 (UTC)[reply]

Area of a "square" on the surface of a sphere

This question was originally posted in the science section, but belongs here. The original questioner has stated clearly that it is not a homework question. It was formulated as follows:

"How do one find the area of a square drawn on a sphere? A square with the side of 10 cm, and draw loci (10cm) on each corners (quarter of a circle in a square to give the "square")"

Based on the discussion that followed, I think what the questioner had in mind is the area illustrated in yellow in the drawing below:

The red circles are supposed to be two pairs of great circles. The angle between the first pair of great circles, expressed in radians, is 10cm/R, where R is the radius of the sphere. The angle between the second pair of great circles is equal to the angle between the first pair. The plane defined by the axes corresponding to the first pair of great circles is perpendicular to the plane defined by the axes corresponding to the second pair of great circles.

The question is how to express the yellow area in terms of R, the radius of the sphere. Obviously, as R → ∞, the area → 100 cm².

-

I am not a mathematician, but felt that it "ought to" be possible to express this area in terms of R, and decided to try to find the necessary information.

I found Girard's theorem, which states that the area of a triangle on a sphere is (A + B + C - π) × R², where A, B and C are the angles between the sides of the triangle, as illustrated in the second drawing. I also found the law of sines for triangles on a sphere, which relates the angles A, B and C to the angles a, b and c which define the sides of the triangle: ${\frac {\sin a}{\sin A}}={\frac {\sin b}{\sin B}}={\frac {\sin c}{\sin C}}.$

I then attempted to divide the square into two triangles, and compute the area of one of these, but am stuck because I don't know the diagonal. Since this is spherical geometry, I doubt that it is as simple as ${\sqrt {2}}\times 10cm$ . I would appreciate if somebody told me if I am on the right track, and, if so, how to complete the calculations. If my presentation of the problem reveals that I have misunderstood some of the theory, please explain. --vibo56 14:11, 28 May 2006 (UTC)[reply]

The natural way I suspect the question should presumably be answered is to take the square on the flat plane and use Jacobians to transform it onto the sphere. Those with a firmer grip of analysis would probably want to fill in the details at this point... Dysprosia 15:28, 28 May 2006 (UTC)[reply]

An easier way to tackle this might be to exploit the symetry of the situation. Slice the sphere into 4 along z=0 and x=0. This will give four identical squares with four right angles and two sides of length 5. The cut the squares along x+z=0, x-z=0 giving eight triangles, each with one 45 degree angles, one right angle and one side of length 5. --Salix alba (talk) 15:44, 28 May 2006 (UTC)[reply]

I think vibo is on the right track. You can use the law of sines to calculate the length of the diagonal. -lethe ^{talk +} 15:44, 28 May 2006 (UTC)[reply]

The law of cosines for spherical trig gives cos c = cos² a. -lethe ^{talk +} 16:06, 28 May 2006 (UTC)[reply]

From which I get using the spherical law of sines that sin A = sin a/ √(1 – cos⁴ a). A = B and C = π/2, so I have the triangle, and hence the square. -lethe ^{talk +} 16:11, 28 May 2006 (UTC)[reply]

To lethe: How can you say that C = π/2? This is spherical geometry, and the four "right" angles in the "square" in the first drawing add up to more than 2π, don't they, or am I missing something? --vibo56 16:49, 28 May 2006 (UTC)[reply]

You may be right, I cannot assume that the angles are right angles. Let me mull it over some more. -lethe ^{talk +} 16:58, 28 May 2006 (UTC)[reply]

OK, I think the right assumption to make is that C = 2A. I can solve this triangle as well, but it's quite a bit messier. Lemme see if I can clean it up, and then I'll post it. -lethe ^{talk +} 17:20, 28 May 2006 (UTC)[reply]

This makes my final answer

2R^{2}\left(2\sin ^{-1}\left({\frac {\sin a}{\sqrt {1-\cos ^{4}a}}}\right)-{\frac {\pi }{2}}\right)

For the square. Now I just have to see whether this answer works. -lethe ^{talk +} 16:14, 28 May 2006 (UTC)[reply]

And now I'm here to tell you that Mathematica assures me that this function approaches s² as the curvature goes to zero. From the series, I can say that to leading two orders of correction, area = s² + s⁴/6R² + s⁶/360R⁴. -lethe ^{talk +} 16:25, 28 May 2006 (UTC)[reply]

I got the following for the diagonal angle c of the big square from "first principles" (just analytic geometry in 3D): cos(c/2) = 1 / sqrt(1+2τ²), where a = 10cm/R and τ = tan(a/2). --Lambiam ^Talk 16:02, 28 May 2006 (UTC)[reply]

I'm afraid I didn't understand (I'm not a mathematician :-) ). If we let (uppercase) C be the "right" (i.e. 90°+something) angle in the triangle in the second figure, and (lowercase) c be the diagonal that we are trying to calculate, could you please show the steps leading to this result (or rephrase it, if I misinterpreted your choice of which of the angles A,B,C that was the "right" one)? --vibo56 18:07, 28 May 2006 (UTC)[reply]

For simplicity, let's put R = 1, since you can divide all lengths first by R, and multiply the area afterwards by R². Then the equation of the sphere is x² + y² + z² = 1. Take the point nearest to the spectator in the first image to be (x,y,z) = (0,0,1), so z decreases when receding. Take the x-axis horizontal and the y-axis vertical. A grand circle is the intersection of a plane through the sphere's centre (0,0,0) with the sphere. The equation of the plane that gives rise to the grand circle whose arc segment gives the top side of the "square" is y = tan(a/2) × z = τz (think of it as looking sideways along the x-axis). At the top right corner of the "square" we have x = y. Solving these three equations (sphere, plane, x = y) for z, using z > 0, gives us z = 1 / sqrt(1+2τ²). Now if c is the angle between the rays from the centre of the sphere to this corner and its opposite (which, if R = 1, is also the length of the diagonal), so c/2 is the angle between one of these rays and the one through (0,0,1), then z = cos(c/2). Combining this with the other equation for z gives the result cos(c/2) = 1 / sqrt(1+2τ²). Although I did not work out the details, I think you can combine this with Salix alba's "cut in eight" approach and the sines' law to figure out the missing angle and sides. --Lambiam ^Talk 19:59, 28 May 2006 (UTC)[reply]

new calculation

As vibo correctly points out above, the square will not have right angles, so my calculation is not correct. Here is my new calculation. Assuming all angles of the square are equal, label this angle C. Then draw the diagonal, and the resulting triangle will be equilateral with sides a and angles A, and 2A = C. The law of sines tells me

{\frac {\sin a}{\sin A}}={\frac {\sin c}{\sin 2A}}\,\!

from which I have

\sin c=2\cos A\sin a.\,\!

From the law of cosines I have that

\cos c=\cos ^{2}a+\sin ^{2}a(2\cos ^{2}A-1).\,\!

My goal here is to eliminate c. First I substitute cos A:

\cos c=\cos ^{2}+\sin ^{2}a\left({\frac {\sin ^{2}c}{2\sin ^{2}a}}-1\right)\,\!

which reduces to the quadratic equation

\cos ^{2}c+2\cos c-1=2\cos 2a.\,\!

So I have

\cos c=-1\pm {\sqrt {2+2\cos 2a}}\,\!

and using cos A = sin c/2sin a, I am in a position to solve the triangle

A=\cos ^{-1}\left({\frac {1}{2}}{\sqrt {1-\left(-1+{\sqrt {2+2\cos 2a}}\right)^{2}}}\csc a\right).\,\!

I'm pretty sure this can be simplified quite a bit, but the simplification I got doesn't agree with the one Mathematica told me. Anyway, the expansion also has the right limit of s². -lethe ^{talk +} 20:16, 28 May 2006 (UTC)[reply]

Despite the figure, which is only suggestive (and not quite correct), are we agreed on the definition of a "square on a sphere"? The question stipulates equal side lengths of 10 cm. To avoid a rhombus we should also stipulate equal interior angles at the vertices, though we do not have the luxury of stipulating 90° angles. Food for thought: Is such a figure always possible, even on a small sphere? (Suppose the equatorial circumference of the sphere is itself less than 10 cm; what then?) Even if it happens that we can draw such a figure, is it clear what we mean by its area? Or would we prefer to stipulate a sufficiently large sphere? (If so, how large is large enough?) Figures can be a wonderful source of inspiration and insight, but we must use them with a little care. --KSmrq^T 20:40, 28 May 2006 (UTC)[reply]

The figure was drawn by hand, and is obviously not quite correct, but doesn't the accompanying description:

The red circles are supposed to be two pairs of great circles. The angle between the first pair of great circles, expressed in radians, is 10cm/R, where R is the radius of the sphere. The angle between the second pair of great circles is equal to the angle between the first pair. The plane defined by the axes corresponding to the first pair of great circles is perpendicular to the plane defined by the axes corresponding to the second pair of great circles.

resolve the ambiguity with respect to the rhombus, provided that the area of the square is less than half of the area of the sphere? --vibo56 21:51, 28 May 2006 (UTC)[reply]

What is meant by "the angle between the … circles"? That's not really the same as the arclength of a side as depicted. Also note that the orginal post suggests that the side might be a quarter of a circle. If that is true, then the "square" is actually a great circle! Each angle will be 180°, and the area "enclosed" will be a hemisphere of a sphere with radius 20 cm/π, namely 2π(20 cm/π)² = 800 cm²/π, approximately 254.65 cm².

By a series of manipulations I came up with

\cos ^{2}A={\frac {\cos a}{1+\cos a}},

where a is 10 cm/R, the side length as an angle. The angle of interest is really C = 2A, for which

\cos C=-\tan ^{2}{\frac {a}{2}}.

For the hemisphere case, a = π/2 produces C = π; while for the limit case, a = 0 produces C = π/2.

The original question was about the area, so we should conclude with that: (4C−2π)R². --KSmrq^T 04:43, 29 May 2006 (UTC)[reply]

By "the angles between a pair of great circles", I meant the angle between the plane P₁ in which the first great circle lies, and the plane P₂ in which the second great circle lies. The arc length depicted was intended to represent the intersection between the surface of the sphere, and a plane P₃, which is orthogonal to P₁ and P₂, and which passes through the centre of the sphere. As previously stated, I have little mathematical training. I therefore made a physical model by drawing on the surface of a ball, before making the first image. I convinced myself that such a plane is well-defined, and that this length of arc on a unit sphere would be identical to the angle between P₁ and P₂. Please correct me if I am mistaken, or confirm if I am right. --vibo56 20:19, 29 May 2006 (UTC)[reply]

Every great circle does, indeed, lie in a well-defined plane through the center of the sphere. Between two such planes we do have a well-defined dihedral angle. The problem arises when we cut with a third plane. If we cut near where the two planes intersect we get a short arc; if we cut far from their intersection we get a longer arc. In other words, the dihedral angle between the two planes does not determine the arclength of the "square" side.

Instead, use the fact that any two distinct points which are not opposite each other on the sphere determine a unique shortest great circle arc between them, lying in the plane containing the two points and the center. Our value a is the angle between the two points, as measured at the center of the sphere. Were we to pick two opposite points, we'd have a = π, which is half the equatorial circumference of a unit sphere. For a sphere of radius R, the circumference is 2πR. We are told that the actual distance on the sphere is exactly 10 cm, but we are not told the sphere radius. The appearance of the "square" depends a great deal on the radius, and so does its area. When the radius is smaller, the sides "bulge out" to enclose more area, the corner angles are greater, and the sphere bulges as well. As the sphere radius grows extremely large, the square takes up a negligible portion of the surface, the sides become straighter, the angles approach perfect right angles, and the sphere bulges little inside the square.

We do not have a handy rule for the area of a square on a sphere. Luckily, the area of a triangle on a sphere follows a powerful and surprisingly simple rule, based on the idea of angular excess. Consider a triangle drawn on a unit sphere, where the first point is at the North Pole (latitude 90°, longitude irrelevant), the second point drops straight down onto the equator (latitude 0°, longitude 0°), and the third point is a quarter of the way around the equator (latitude 0°, longitude 90°). This triangle has three perfect right angles for a total of 270° (or 3π/2), and encloses exactly one octant — one eighth of the surface area — of the sphere. The total surface area is 4π, so the triangle area is π/2. This area value is exactly the same as the excess of the angle sum, 3π/2, compared to a flat right triangle, π. The simple rule is, this is true for any triangle on a unit sphere. If instead the sphere radius is R, the area is multipled by R².

Thus we simplify our area calculation by two strategies. First, we divide out the effect of the radius so that we can work on a unit sphere. Second, we split the "square" into two equal halves, two equilateral triangles, by drawing its diagonal. Of course, once we find the triangle's angular excess we must remember to double the value (undoing the split) and scale up by the squared radius (undoing the shrink).

Notice that this mental model assumes the sphere radius is "large enough", so that at worst the square becomes a circumference. We still have not considered what we should do if the sphere is smaller than that. It seems wise to ignore such challenges for now. --KSmrq^T 21:23, 29 May 2006 (UTC)[reply]

Thank you. I really appreciate your taking the time to explain this to me which such detail and clarity. --vibo56 23:34, 29 May 2006 (UTC)[reply]

Coordinate Transform

What if we perform a simple coordinate transform to spherical coordinates and perform a 2-dimensional integral in phi and theta (constant r = R). Then, dA = r^2*sin(theta)*dphi*dtheta, and simply set the bounds of phi and theta sufficient to make the lengths of each side 10 cm. Nimur 18:11, 31 May 2006 (UTC)[reply]

Calculation completed

Thanks a million to the users who have put a lot of work in explaining this to me, and in showing me the calculations necessary. I started out based on the work of lethe. Armed with a table of trigonometric identities, I went carefully through the calculations, and am happy to report that I feel that I understood every single step. I was not able to simplify the last expression much further, the best I can come up with is

\cos A={\frac {1}{2}}\csc a{\sqrt {2{\sqrt {2+2\cos 2a}}-2\cos 2a-2}}\,.\!

You should probably make use of the identity

\cos 2a=2\cos ^{2}a-1

here, it simplifies this expression quite a bit. -lethe ^{talk +} 02:01, 30 May 2006 (UTC)[reply]

Since the r.h.s. is based on a only, which is a known constant when the radius and length of arc are given (a=10cm/R for the given example), let us substitute $G=g(a)\,\!$ for the r.h.s. Note that the function is undefined at a=0°±180° because of the sine function in the denominator. There is a graph of g(a) on my user page.

We can now calculate the area of the triangle, and that of the square.

\cos A=G.\,\!

\cos 2A=2\cos ^{2}A-1=2G^{2}-1\,.\!

According to Girard's formula, we then have

area_{triangle}=(A+B+C-\pi )\times R^{2}\!

area_{triangle}=\left(2\cos ^{-1}G+\cos ^{-1}(2G^{2}-1)-\pi \right)\times R^{2}\!

area_{square}=2\times \left(2\cos ^{-1}G+\cos ^{-1}(2G^{2}-1)-\pi \right)\times R^{2}\!

I calculated the behaviour of this area function on a unit sphere when a is in the range (0°...180°):

Seems reasonable up to 90°. The value at 90° corresponds to the "square" with four corners on a great circle that KSmrq mentions above, i.e. to a hemisphere, and the area, 2π is correct. in the interval [90°..180°), the function returns the smaller of the two areas. I also notice that the function looks suspiciously elliptical. Are we computing a much simpler function in a roundabout way?

I next studied how the formula given by KSmrq works out:

area_{triangle}=\left(2\cos ^{-1}\left({\sqrt {\frac {\cos a}{1+\cos a}}}\,\right)+\cos ^{-1}(-\tan ^{2}{\frac {a}{2}})-\pi \right)\times R^{2}\!

area_{square}=2\times \left(2\cos ^{-1}\left({\sqrt {\frac {\cos a}{1+\cos a}}}\,\right)+\cos ^{-1}(-\tan ^{2}{\frac {a}{2}})-\pi \right)\times R^{2}\!

I computed the area, and found that in the range (0°..90°], the formulae of lethe and KSmrq yield identical results, within machine precision. Above 90°, the formula of KSmrq leads to numerical problems (nans).

Finally, I would like to address the question of the orignial anonymous user who first posted this question on the science desk. Let us see how the area of the square behaves as R increases, using 10 cm for the length of arc in each side of the "square". The smallest "reasonable" value of R is 20cm/π ≈ 6.366 cm, which should lead to a surface area of approximately 254.65 cm², as KSmrq points out. Driven by curiosity, I will start plotting the function at lower values than the smallest reasonable one (in spite of KSmrq's advice to "ignore such challenges for now").

Here is the graph:

Unsurprisingly, the function behaves weirdly below the smallest reasonable value of R, but from R ≈ 6.366 cm and onwards, the function behaves as predicted, falling rapidly from 254.65 cm², and approaching 100 cm² asymptotically. In case anybody is interested in the calculations, I have put the program on my user page.

Again, thank you all. --vibo56 23:54, 29 May 2006 (UTC)[reply]

Well done. It does appear that you overlooked my simple formula for the area, which depends on C alone. Recall that when the square is split, the angle A is half of C, so the sum of the angles is A+A+C, or simply 2C. This observation applies to User:lethe's results as well, where we may use simply 4A. So, recalling that a = 10 cm/R, a better formula is

$\mathrm {area} _{\mathrm {square} }\,\!$	${}=\left(4\cos ^{-1}(-\tan ^{2}{\frac {a}{2}})-2\pi \right)\times R^{2}\,\!$
	${}=\left(4\cos ^{-1}(-\tan ^{2}{\frac {10\ \mathrm {cm} }{2R}})-2\pi \right)\times R^{2}.\,\!$

For the arccosine to be defined, its argument must be between −1 and +1, and this fails when the radius goes below the stated limit. (A similar problem occurs with the formula for A, where a quantity inside a square root goes negative.) Both algebra and geometry are telling us we cannot step carelessly into the domain of small radii. Try to imagine what shape the "square" may take when the circumference of the sphere is exactly 10 cm; both ends of each edge are the same point! Not only do we not know the shape, we do not know what to name and measure as the "inside" of the square.

This raises an important general point about the teaching, understanding, and application of mathematics. Statements in mathematics are always delimited by their range of applicability. Every function has a stated domain; every theorem has preconditions; every proof depends on specific axioms and rules of inference. Once upon a time, we manipulated every series with freedom, with no regard to convergence; to our chagrin, that sometimes produced nonsense results. Once it was supposed that every geometry was Euclidean, and that every number of interest was at worst a ratio of whole numbers; we now make regular use of spherical geometry and complex numbers. When we state the Pythagorean theorem, we must include the restriction of the kind of geometry in which it applies. When we integrate a partial differential equation, the boundary conditions are as important as the equation itself. It is all too easy to fall into the careless habit of forgetting the relevance of limitations, but we do so at our peril. --KSmrq^T 02:36, 30 May 2006 (UTC)[reply]

Yes, I did overlook the (now painfully obvious) fact that the sum of the angles was 2C. Your final point is well taken. I understood that the reason for the NaN's was a domain error, but thanks for pointing out the exact spots. --vibo56 19:40, 30 May 2006 (UTC)[reply]

Test Statistic in KPSS test of stationarity

What are the values of the test statistic in KPSS test for stationarity of a variable? Example : While running a KPSS test using Gretl 1.5.1 to check the stationarity of my variable (TotalProduction) i recieve the following output.

KPSS test for TotalProd (including trend)

Lag truncation parameter = 4
Test statistic = 0.281322
                   10%      5%    2.5%      1%
Critical values: 0.119   0.146   0.176   0.216

I understand that the value to examine is the test statistic but what are the values of it that indicate that my examined time-series is non-stationary? If the test statistic is 0.05 is my series stationary?

As your test statistics, 0.281322, is greater that the critical value for 1% then it passes the test for being stationary at the 1% level, which is pretty good. On the otherhand a test statistic of 0.05 is less than the 10% level so it would not be safe to assume its stationary. Its worth having a think about whether you would expect low or high values of the statistic to indicate a stationary variable. --Salix alba (talk) 18:33, 28 May 2006 (UTC)[reply]

May 29

Roots

recently my maths teacher gave me a problem to solve, but i have had trouble with it. i need to find out what z is, in the equation z^3=1, the obvious answer is 1, but he said that it was more difficult than the previous question he gave me, solving the square root of i. He also asked me to find n in the equation, z^n = 1, and he then said there were n answers. any help would be much appreciated, sorry if this is too simple.

This has to do with the beautiful roots of unity. Essentially, the full solution involves complex numbers. If you plot these solutions out in the complex plane, they always end up on a unit circle centered around the origin. --HappyCamper 03:38, 29 May 2006 (UTC)[reply]

They are also equally spaced around this circle. —Bkell (talk) 03:39, 29 May 2006 (UTC)[reply]

THANKS GUYS

No problem! Actually, I think we can add more stuff to the article, like a diagram of a circle along with the position of the roots. Feel free to come back if you have more questions. --HappyCamper 04:32, 29 May 2006 (UTC)[reply]

ok i've had some problems. To solve z^3=1, do you just replace z with (a+bi)^3=1? Because then i get 1 = a^3 - b^3i - 3ab^2 + 3a^2bi and i dont know where to go from here. thanks.

The trick is not to use the representation z = a + bi, but rather z = r * exp(i θ). Now, I'll just give a colloquial sketch of what is going on. Because you know it's a root of unity, r = 1. So, you know that z = exp(i θ).

Now, to go around a circle, you need to traverse 2π radians. Since you are trying to find z^3=1, you look at the exponent and take a note that there is a 3. What angles (in radians) give multiples of 2π when multiplied by 3?

One of them is obviously 2π/3. There's also 4π/3. There's also plain old zero. Note how these can all be written like

\theta =k{\frac {2\pi }{3}}

, where k = 0, 1, 2.

So, your solutions are of the form

z=\exp(i{\frac {k2\pi }{3}})

Now, let's just substitute in the k...these are your values for z

z=\exp(0),\exp(i{\frac {2\pi }{3}}),\exp(i{\frac {4\pi }{3}})

Finally, use Euler's formula,

e^{i\theta }=\cos \theta +i\sin \theta \,\;

and you get your answers

z=1,-{1 \over 2}+i{\frac {\sqrt {3}}{2}},-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}},

Does this help? --HappyCamper 05:02, 29 May 2006 (UTC)[reply]

yeah, thanks a lot. Do u have a maths degree?

On second thought, I think your question can be answered much much simpler, although I haven't worked it out entirely yet.

Take the approach you had before, set z = a + bi, and just expand. You get a³ - ib³ - 3ab² + 3ia²b = 1. Now, you invoke this insight: The real and imaginary parts on boths sides must be equal! So, you get two simultaneous equations for a and b:

a³ - 3ab² = a(a² - 3 b²) = 1

3a²b - b³ = b(3a² - b²) = 0

From the second equation, you get that either b = 0, or

b=\pm {\sqrt {3}}a

. You see where this is going? Substitute these into the first and solve for a, and you get the same solutions as above. Much much easier than having to wave magic wands with complex exponentials and such. However, there is a lot of insight you can get from understanding the complex exponentials. I'd encourage you to revisit this problem again when you get the chance to. You might also wonder about...what is the quantity

i^{i}

, where i is the imaginary unit? Hm.... --HappyCamper 05:22, 29 May 2006 (UTC)[reply]

Right, that's a very important princple: any equality of complex numbers corresponds to two independent equalities of real numbers. —Keenan Pepper 05:46, 29 May 2006 (UTC)[reply]

Is there an article with a formal proof of this on Wikipedia? --HappyCamper 05:47, 29 May 2006 (UTC)[reply]

That's the definition of equality of complex numbers: Two complex numbers are equal if and only if their imaginary parts are equal and their real parts are equal. —Bkell (talk) 05:51, 29 May 2006 (UTC)[reply]

This was missing from complex number! [1] --HappyCamper 05:54, 29 May 2006 (UTC)[reply]

The problem as posed makes a hidden assumption. Any time we look for zeros (roots) of a polynomial, we must first know what kinds of values we are allowed to substitute for the unknown. For example, the polynomial

x+1\,\!

has a zero when x is −1, but that may not be admissible if we require a positive whole number. If we allow a negative whole number then we have exactly one root. Similarly, the polynomial

2x-1\,\!

has a zero when x is ¹⁄₂, but that is not a whole number and so may not be acceptable. Or consider

x^{2}-2\,\!

with a zero when x is √2, which is not a fraction! (This is the length of the diagonal of a square by the Pythagorean theorem, and the discovery that it was not a "proper" number deeply disturbed ancient Greek mathematicians.) Notice that all these polynomials are extremely simple, and all have integer coefficients.

Now look at the polynomial

x^{2}+1\,\!

which might seem to have no roots at all. For, the numbers we are accustomed to using in arithmetic have the property that any non-zero number squares to a positive number, yet we require a number whose square is −1. Is there no end to the need to enlarge our concept of number, from natural numbers to integers to rational numbers to real numbers to these new beasts? To our great relief, when we expand from real numbers to complex numbers by including a value that squares to −1, and commonly denoted by ⅈ, along with all multiples and sums with it, we find that no polynomial can ever again stop us. Hurray! Complex numbers make our life simple.

The usual picture is that real numbers extend smoothly left and right along a linear continuum, and complex numbers extend above and below that line as well. To turn 1 into −1 requires a 180° reversal; to do that in two steps (squaring) requires a 90° turn, which is what ⅈ accomplishes. Notice that in this case the magnitude of the number — its distance from zero — remains one before, during, and after the turn. Observe that 1 itself has two numbers that square to it: +1, which involves either a 0° or 360° turn, and −1, which involves two 180° turns. Also observe that −1 has two square roots: ⅈ, which involves two 90° turns, and −ⅈ, which involves turns of 270° or −90°.

With a little imagination we can picture three options for a turn that, when repeated three times, gives (an integer multiple of) 360°. These would be "cube roots of unity". We note that paired with every "positive" turn there will be a matching "negative" turn; this corresponds to replacing ⅈ by −ⅈ, an operation called complex conjugation. --KSmrq^T 05:56, 29 May 2006 (UTC)[reply]

It should be pointed out that the reason that squaring numbers like 1, −1, i, and −i does not change the number's magnitude is that the magnitude of all these numbers is 1, and 1² = 1. In general, when you square a complex number z, the magnitude of the resulting number will be the square of the magnitude of z. —Bkell (talk) 06:11, 29 May 2006 (UTC)[reply]

well. that was a pretty hard question considering the last thing we did in maths was inverse functions and we havent even been taught complex numbers yet.

ok i get up to where you said, but when trying to solve a, how does a^3 + 9a^2 - 1 give you -1/2 for a? and also with a^3 - 9a^2 - 1 = -1/2?

HappyCamper's formula

b=\pm {\sqrt {3}}a

was presented in a somewhat non-conventional way. More conventional is

b=\pm a{\sqrt {3}}

, which makes clear that the occurrence of a is not under the square root's vinculum, which would have been

b=\pm {\sqrt {3a}}

. So b² = 3a² (which is where we came from), and not 3a. --Lambiam ^Talk 10:19, 29 May 2006 (UTC)[reply]

Thank you to everyone who contributed in answering this question.

Math Question 1

If there is a ticket raffle, and the prize is worth $1,000, with 100 tickets outstanding I should clearly purchase tickets at $1 each, as they are worth more than that, but as I do, I increase the # of tickets oustanding, and at some point (1000 tickets I think) it is no longer advantagous to purchase tickets. (is this correct?)

But lets say I own 5 of the 100 oustanding tickets. These tickets would be valued at $10 each ( I think). What if instead of ONE $1000 price, there are 5 $200 prizes. How would this effect the "intrinsic value" of each ticket if:

-after a ticket is drawn it is put back into the pool and can be drawn again for one of the other prizes

-After a ticket is drawn it isnt put back but the ticketholder can win on the other tickets.

-After a ticket is drawn the ticketholder is ineligable to win other prizes regardless of his remaning tickets?

Anyone? 12.183.203.184

If there are 100 tickets outstanding, and you buy one, then there is a 1/101 chance you will win $1000, and a 100/101 chance you will win nothing. So the expected value of your winnings is

{1 \over 101}\times \$1000+{100 \over 101}\times \$0\approx \$9.90.

So for that first ticket, you should be willing to pay up to $9.90. Suppose you buy it. Now you consider whether to buy a second ticket. Since there are now 101 tickets outstanding, the chance that your second ticket will win $1000 is 1/102. So now the expected value of your winnings from the second ticket is

{1 \over 102}\times \$1000\approx \$9.80.

So it seems that you should pay up to $9.80 for that second ticket.

The problem, though, is that buying the second ticket changes the odds, and so the calculation for the first ticket is no longer valid. If you buy two tickets, then you have two chances in 102 of winning the thousand dollars, so the expected value of your winnings is

{2 \over 102}\times \$1000\approx \$19.61,

so if you are planning to buy two tickets the total amount you should pay should be no greater than $19.61. (This of course assumes that no one else will be buying any other tickets; if you buy your two tickets for $19.61, and then someone else buys a hundred tickets, then you will have overpaid.)

You can continue this line of thought to figure out how many tickets you should buy for a dollar each. If you buy n tickets, then the expected value of your winnings is

{n \over 100+n}\times \$1000.

You're going to have to spend n dollars to buy n tickets, so if this expected value is less than n, you shouldn't buy so many. —Bkell (talk) 02:34, 30 May 2006 (UTC)[reply]

Now suppose that you have five of the 100 outstanding tickets and there are five $200 prizes. If tickets are put back into the pool after being drawn, then in each drawing you will have five chances in 100 of winning $200. This means that the expected value of your winnings is

{5 \over 100}\times \$200+{5 \over 100}\times \$200+{5 \over 100}\times \$200+{5 \over 100}\times \$200+{5 \over 100}\times \$200=\$50,

which means that your five tickets together are worth $50 (that is, $10 each). In fact, as long as the tickets are returned to the pool after each drawing, it doesn't matter how the $1000 is divided up. There could be a hundred thousand drawings for a penny each, and your expected winnings would still be $50. (You will find, however, that as the $1000 is divided ever more finely, you have a greater chance of winning closer to $50; that is, it becomes less and less likely that you will win $0 or $1000, and more and more likely that you will win $50, or $45, or $55.) —Bkell (talk) 02:53, 30 May 2006 (UTC)[reply]

Math probibilty question 2

Say a casino sells the "oppertunity" to roll a 6-sided dice. You win the number rolled x 100. e.g. 2 = $200. The value of this appears to be $350. If the casino allowed you to reroll at your choice, what would the value be? If you could reroll twice (3 rolls total) what would the value be, and what number should you take the reroll option if the dice number turned up is less than? I think this should be higher on the 1st reroll than on the 2nd option. Is that right? How would one calcualate that? Thanks! 12.183.203.184 08:17, 29 May 2006 (UTC)[reply]

With one reroll, you would reroll if the expected value of the reroll is more than the amount you have already won: that is, if the die currently shows a 1, 2 or 3 ($100, $200 or $300). Thus the expected value of that situation is

{\begin{matrix}{\frac {1}{6}}\end{matrix}}\left(\$350+\$350+\$350+\$400+\$500+\$600\right)=4{\begin{matrix}{\frac {1}{4}}\end{matrix}}\times \$100

.

From this point on, we can remove the $100 factor, because it just serves to confuse things. This means that after the first roll of three, the remaining two rolls are worth

4{\begin{matrix}{\frac {1}{4}}\end{matrix}}

if exercised; thus they will be exercised if the current value of the die is less than

4{\begin{matrix}{\frac {1}{4}}\end{matrix}}

i.e. on a 1, 2, 3 or 4. So the expected value of 3 rolls is

{\begin{matrix}{\frac {1}{6}}\end{matrix}}\left(4\times 4{\begin{matrix}{\frac {1}{4}}\end{matrix}}+5+6\right)=4{\begin{matrix}{\frac {2}{3}}\end{matrix}}

. EdC 14:13, 29 May 2006 (UTC)[reply]

In other words, denote the value of n rolls by

v_{n}

. What Edc meant was that this value satisfies a recurrent equation (see Conditional expectation)

v_{n+1}=E(x|x\geq v_{n})P(x\geq v_{n})+v_{n}P(x<v_{n})

and forms a sequence 3.5,4.25,4.66,4.94,5.13,5.27,...(Igny 16:52, 29 May 2006 (UTC))[reply]

Or, equivalently but conceptually simpler:

v_{n+1}={\frac {1}{6}}\sum _{i=1}^{6}\max(i,v_{n}).

For n ≥ 5 a closed form expression is given by v_n = 6 − (6768/3125)(5/6)ⁿ. --Lambiam ^Talk 17:19, 29 May 2006 (UTC)[reply]

While we just provided a correct answer, I wonder if the proof is needed that

v_{n}

is indeed the maximal value of n rolls, and our strategy is optimal. (Igny 18:05, 29 May 2006 (UTC))[reply]

I think it's obvious. But on demand a proof can be supplied. --Lambiam ^Talk 23:04, 29 May 2006 (UTC)[reply]

Salary Statement

How to prepare a SALARY STATEMENT of any type in Ms Excel?--86.62.239.145 10:55, 29 May 2006 (UTC)[reply]

You will probably like Microsoft InfoPath (even although it is used to make XML forms, it can be used for receipts; anyway, if you dont have that you should just lay it out as it appears in most wage slips. Kilo-Lima|^(talk) 16:50, 29 May 2006 (UTC)[reply]

Windows Media Audio

Hi, I went and ripped some files from a bought music CD and it was extracted as .wma format. I then put it into WMP library. I then wanted to burn them to a CD, so I put in a blank disk and I keep getting the same error that it is not blank—I can assure you, it is blank! Does anybody know what's happening here? Thanks, Kilo-Lima|^(talk) 16:46, 29 May 2006 (UTC)[reply]

Ugggh...now you know why I like iTunes (no WMA crap). But anyways...maybe your CD is messed up (defective, etc.)--could somebody have "borrowed" when you weren't looking? --M1ss1ontomars2k4 ^{(T | C | @)} 21:41, 29 May 2006 (UTC)[reply]

Yeah, iTunes is totally free of DRM and everything, too. Dysprosia 22:20, 29 May 2006 (UTC)[reply]

May 30

Cell CPU

Hey ppl, i hope this is the right place to ask about the Cell CPU thats gonna be used in the Playstation 3. I'v asked SOOO many ppl and they aren't quite sure how it works either. Please, can someone explain to me how all the Synergistic Processing Elements (SPE's) work together to process the info and why game developers developing games for the PS3 say its so hard to program for 7 cores? Even the main wikipedia article on the Cell says its too complicated. Thx! KittenKiller

It's mostly because it's a very uncommon architecture, so almost nobody has any experience with it.

Historically, consumer-level computers have had a single general-purpose processor. Programming in such a situation is very well-understood: it's simply a matter of dividing up the available time between tasks. (For example: five milliseconds for AI, one for checking for user input, three for updating the sound effects, one for updating the game logic, twenty-five for updating the screen, repeat)

More recently, computers have started being produced with multiple identical general-purpose processors. Writing programs for this sort of system is fairly well-understood in theory: you divide up the tasks in such a way that the workload can be balanced among the processors. In practice, it's a bit harder, because some tasks can't be divided up, and you need to make sure the tasks don't step on each other's toes, but it's still familiar ground. (For example, the game AI would be one or more tasks, physics would be a task, generating sound effects would be a task, playing the background music would be a task, checking for user input would be a task, updating the graphics would be a task, and so on)

Then, there's the Cell CPU. It's got one general-purpose processor and seven special-purpose processors (the SPEs) that are extremely efficient for certain types of tasks, and very slow for others. Nobody's quite sure how to handle such a situation, or what the seven SPEs should be used for. It's known that they're good for graphics, but the PS3 has a dedicated graphics processor that's even better. They're probably good for sound effects, but that won't take the effort of more than one SPE. They're not good for AI, or for user input, or for background music. They might be good for physics, but nobody knows -- simulating physics is cutting-edge research right now. --Serie 22:48, 30 May 2006 (UTC)[reply]

Series

What is this series of numbers called:

1, 11, 21, 1211, 111221....?

Looks like the see-and-say sequence. Dysprosia 08:11, 30 May 2006 (UTC)[reply]

In mathematical jargon this is not a series. Goes to show that mathematicians are strange people. --Lambiam ^Talk 09:58, 30 May 2006 (UTC)[reply]

There's a difference between series and sequences. OEIS A005150 : Look and Say sequence: describe the previous term! (method A - initial term is 1 - Formerly M4780) --DLL 20:32, 30 May 2006 (UTC)[reply]

Of course, but I think it's quite obvious what the poster above meant, so I don't think there's need for excessive concern about this. Dysprosia 22:19, 30 May 2006 (UTC)[reply]

Noo, I'm more concerned about the mathematicians ;-) Lambiam ^Talk 02:50, 31 May 2006 (UTC)[reply]

Well telling someone once is fine, but twice is getting to be a bit onerous ;) Dysprosia 03:09, 31 May 2006 (UTC)[reply]

Watermellon. Black Carrot 00:23, 1 June 2006 (UTC)[reply]

I don't get it. Dysprosia 02:01, 1 June 2006 (UTC)[reply]

Font

for some unknown reason, the font in which i view wikipedia has changed to tahoma, and i would like to revert it. When editing articles my font is times new roman, but after i have edited them, it appears in tahoma. Can anyone shed any light on my problem?

Character encoding ? In Firefox the menu offers this entry with plenty of encoding possibilities. In other brothels it may be worse, brother. --DLL 20:28, 30 May 2006 (UTC)[reply]

Browser settings, font availability, and custom Wikipedia monobook.css can all affect what's displayed. Usually the text typed in the edit window appears in a fixed-width font like Courier, not Times New Roman; and usually the text for the preview or regular display of an article appears in a sans-serif font like Arial. The Tahoma font would be a reasonable substitute for Arial, but are you sure the editing font is proportionally spaced? --KSmrq^T 01:50, 31 May 2006 (UTC)[reply]

quantization

2 questions that keep cropping up in old exam papers, which will hopefully get me a few extra marks:

Define a scalar quantizer with M interval
Define a scalar quantization problem

I'm a bit rusty on these, plus Wikipedia has no articles on them. Any help would be appreciated. --Talented Wikipedian File:Kiss.png 10:42, 30 May 2006 (UTC)[reply]

Could you help us by giving a hint what field the exam is about: signal processing, perhaps more specifically image processing, sound processing, and music, or is it perhaps about physics? --Lambiam ^Talk 13:55, 30 May 2006 (UTC)[reply]

It is part of a data compression module, and is some kind of lossy compression, I can't be sure whether it is to do with sound files or image files. --Talented Wikipedian File:Kiss.png 16:36, 30 May 2006 (UTC)[reply]

I suspect that "M interval" is not standard terminology. Could the meaning be "M intervals" (or "levels"), as in 16 intervals, 256 intervals, 65,536 intervals, ..., M intervals? Here are a few links that may be helpful to you:

--Lambiam ^Talk 18:19, 30 May 2006 (UTC)[reply]

Perhaps this is meant in contrast to vector quantization? --KSmrq^T 22:37, 30 May 2006 (UTC)[reply]

astronomy-orbit of planets

What branch of calculus do I need to understand the position of a planet orbiting the sun at any given time?

Thank you, Denis

Assuming that you are happy enough with Newton's law of universal gravitation and do not require General relativity, what you need is the area of Ordinary differential equations, also known as ODEs, together with a general understanding of Newton's laws of motion and their role in Mechanics. --Lambiam ^Talk 18:29, 30 May 2006 (UTC)[reply]

And if that's a bit much for you, try geometry, specifically conic section geometry. StuRat 22:38, 30 May 2006 (UTC)[reply]

That will tell you where, but you'll need Kepler's laws to know when where. I see now that the Kepler article shows how to derive these laws from the ODE's given by Newton's law. --Lambiam ^Talk 02:58, 31 May 2006 (UTC)[reply]

Rope strength

If a single strand manilla rope has a working load of 100 lbs., does that mean that a triple strand rope of the same material has a working load of 300 lbs.? Is there a formula for figuring this out? Thanks, Pdpdpd1

If the box in Exercise 5.14 weighs 20 kg and has a coeficent of friction of .2, what is the minimum strength rope that can be used to pull it?

A rule of thumb for question 1 would say that there is always one of the rope that bears more weight. So do not triple it. But as the rope itself is made of strings, the standard load might be over . So try to triple it and do not stay under the load. --DLL 20:24, 30 May 2006 (UTC)[reply]

Question 1 is a yes, as long as it still obeys Hooke's law at that point. At least I think so. Question 2 deals with Hooke's Law and the Young modulus, but I don't understand what the "strength" refers to. But you could find out what the frictional force is and find the force required to pull the box. Remember, we don't do your homework. x42bn6 Talk 07:56, 31 May 2006 (UTC)[reply]

why is in locally pathconnected topological space, path connected component same as connected component

http://en.wikipedia.org/wiki/Connected_space#Local_connectedness

defines the concept of a locally path connected topological space. Now I wanna prove that in a locally pathconnected topological space, path connected components and connected components are the same.

Now I do know that this is a relevant fact : in a locally pathconnected space, every open set U has (in its induced topology) open pathconnected components.

But how to use this?

I have gotten this far : let C be the component of x, and P the strictly contained in C pathconnected component. Now I now C will be closed. I cannot say that of P. I need to find contradiction though...

Thanks,

Evilbu 23:08, 30 May 2006 (UTC)[reply]

Well, I hope this is not for homework. What you need to prove is that every connected component is also path connected. To do this, fix a connected component

C

and choose

p\in C

. If you show that the set of all points in

C

that can be connected to

p

by a path is the whole

C

itself, you are done.

To do this, you show that it is both open and closed: as

C

is connected by definition, and nonempty because it trivially contains

p

, you are done.

Openness: suppose

q

can be connected to

p

by a path. Let

U

be a neighbourhood of

q

which is path connected (it exists because the space is locally pathconnected). Then any point in

U

can be connected to

p

by a path simply by going to

q

first and then to

p

. Hence, your set is open.

Closedness: consider the subset of

C

that cannot be connected to

p

by a path, and choose

q

as such. Then choose a neighbourhood

U

as above: if any point of

U

could be connected to

p

by a path, then by following the two paths as before you could also connect

q

to

p

. This means that this set is also open, i.e. the set of points that can be connected to

p

by a path is closed.

Q.E.D. Cthulhu.mythos 09:12, 31 May 2006 (UTC)[reply]

Hmm, yes I see now. I guess one could not from the start that in a locally pathconnected space, all pathconnected components are open, right? Nope, this is not homework, yet it is not unrelated to university work, we did it quickly in class and I wanted to clear things up for myself. Evilbu 10:45, 31 May 2006 (UTC)[reply]

May 31

Integration of sin and cos

What happens when you integrate sin(), -sin(), cos() and -cos()

Well, what happens when you differentiate any of them? Notice something interesting? Then just apply the Fundamental Theorem of Calculus and you're done. If that's not enough, check List of trigonometric identities#Calculus. Confusing Manifestation 12:01, 31 May 2006 (UTC)[reply]

Also check out Table of integrals#Trigonometric functions. -- Meni Rosenfeld (talk) 15:23, 31 May 2006 (UTC)[reply]

Implications

The fact that the differentiation of trigonometric functions (sine and cosine) results in linear combinations of the same two functions is of fundamental importance to many fields of mathematics, including differential equations and fourier transformations. (I added this tidbit at the above article). Nimur 18:06, 31 May 2006 (UTC)[reply]

A sophisticated answer uses Euler's formula, cos ϑ+ⅈsin ϑ = ⅇ^ⅈϑ, and the fact that ⅇ^z is an eigenfunction for the linear operation of integration. For example,

\cos \vartheta ={\frac {1}{2}}e^{i\vartheta }+{\frac {1}{2}}e^{-i\vartheta },\,\!

so

\int \cos \vartheta \,d\vartheta ={\frac {i}{2}}e^{i\vartheta }-{\frac {i}{2}}e^{-i\vartheta }+C.\,\!

A more geometric approach observes a unit circle parameterized by arclength,

(x,y)=(x(\vartheta ),y(\vartheta ))=(\cos \vartheta ,\sin \vartheta ),\,\!

and considers that a tangent to the circle is always a unit vector perpendicular to the radius. Thus starting from the initial condition (x₀,y₀) = (0,−1) and integrating (cos ϑ,sin ϑ) traces out a counterclockwise unit circle parameterized by arclength, (cos(ϑ−^π⁄₂),sin(ϑ−^π⁄₂)). This serves to remind us of the important role of constants of integration, here a consequence of the initial condition.

The best approach is to pay attention in class, read the textbook, and ask the instructor or teaching assistant for homework help. --KSmrq^T 18:42, 31 May 2006 (UTC)[reply]

very exact definition of "orientable topological manifold"?

From http://en.wikipedia.org/wiki/Manifold

" Overlapping charts are not required to agree in their sense of ordering, which gives manifolds an important freedom. For some manifolds, like the sphere, charts can be chosen so that overlapping regions agree on their "handedness"; these are orientable manifolds. For others, this is impossible. "

Okay, but strictly speaking, this is still not really exact. I would like to have a very explicit definition of "agreeing on handedness". I have some very fundamental experience with charts of (topological varieties), and I know that in a vector space of an ordered field, two bases are considered having the same orientation if their matrix has positive determinant.

So how can it make this precise?

Thanks,

Evilbu 13:28, 31 May 2006 (UTC)[reply]

The problem here is whether we want the article to report the formal definition or not. If this is the case, suppose

\phi _{U}:U\rightarrow \mathbb {R} ^{n}

and

\phi _{V}:V\rightarrow \mathbb {R} ^{n}

are two charts. Then

\phi _{V}|_{U\cap V}\cdot \phi _{U}|_{U\cap V}^{-1}:\mathbb {R} ^{n}\rightarrow \mathbb {R} ^{n}

is a homeomorphism: if for every choice of

U

and

V

it preserves orientation, then the manifold is said to be orientable. Cthulhu.mythos 16:44, 31 May 2006 (UTC)[reply]

thanks, but could you explain what you mean by 'it preserves orientation"?

Keep in mind that the manifold article is an overview of all different kinds of manifold, so deliberately does not dive into all the details. The topological manifold article should include a precise definition of orientability; however, these specialized articles have been starved for attention as numerous editors have nitpicked the survey article.

</Mount Soapbox>

The definition of a topological n-manifold states that it is a topological space in which every point has an open neighborhood homeomorphic to an open n-ball, Bⁿ. (Typically, the space must also be Hausdorff, paracompact, and second-countable; but those are technical details that needn't concern us here.)

Now fix B^{n^{, and choose a list of n+1 points (vertices) in general position within it to form a simplex (or choose an ordered basis of n vectors for the Rⁿ in which it lives). The choice imposes one of two possible orientations on Bⁿ, which we may call "right-handed" or "left-handed". If we exchange two vertices in a right-handed simplex we obtain a left-handed simplex, and these are topologically distinct in the sense that we cannot deform one into the other. For example, the triangle ABC in a planar disc cannot be deformed to ACB without violating its topology.}}

By the axioms of topology, if two open neighborhoods have a non-null intersection, then the intersection is an open set. And since a homeomorphism is a continuous map with a continuous inverse, we can create a composite map from Bⁿ to itself by way of the two neighborhood homeomorphisms. By a suitable deformation we can pass the simplex through this map and compare the image to the original. Thus we can decide if the overlapping homeomorphisms imply the same or opposite orientations.

The key observation is that we can cover some manifolds with neighborhoods so that overlapping neighborhoods always agree in orientation; these are orientable manifolds. Significantly, this is not always possible.

A basic example of a non-orientable manifold is the projective plane, RP². We can think of this as a sphere with opposite points on the surface identified, or as a disc with opposite points on the circular boundary identified. A 2-simplex is merely a triangle, and a triangle in B² maps to a three-sided region in RP², which might as well be a triangle itself. But because the manifold is non-orientable, there is no essential difference between ABC and ACB. For example, put C on the boundary of the disc, with A and B inside; then the identification of opposite boundary points means we cannot distinguish the two orientations of the triangle. We might see this more directly with the sphere model: Draw the triangle as ABC in the northern hemisphere, then use the identification of opposite points to transfer it to an oppositely oriented triangle in the southern hemisphere.

With an orientable manifold, such as an ordinary sphere, S², this kind of reversal is impossible. We can consistently define a right-handed triangle as distinct from a left-hand triangle, and never the twain shall meet. One point of caution is that the dimension of the simplex must be the same as the dimension of the space. For example, we can easily flip a triangle (a 2-simplex) in 3-space; we must work with a tetrahedron (a 3-simplex) to explore orientability in 3-space. --KSmrq^T 23:36, 31 May 2006 (UTC)[reply]

You mention that a manifold must be paracompact. Isn't this redundant for a second countable space locally homeomorphic to Rⁿ? -lethe ^{talk +} 04:21, 1 June 2006 (UTC)[reply]

Evilbu: it is worth mentioning that manifolds with additional structure admit alternate definitions for orientability, which of course agree with the topological definition. A differentiable manifold if it admits an atlas whose transition functions have positive determinant on the tangent bundle. Equivalently, it has a nowhere zero differential n-form. -lethe ^{talk +} 04:24, 1 June 2006 (UTC)[reply]

I would also remark that orientability is an intrinsec property, and does not depend on how the manifold can be embedded in some euclidean space. For example, it occurs frequently to hear that "the Moebius strip is not orientable because it has only one side and one boundary, whereas an annulus is orientable because it has two sides and two boundaries". But that is not correct. A compact surface properly embedded in a 3-manifold (and the Moebius strip is NOT properly embedded in

\mathbb {R} ^{3}

, because of the boundary) is two-sided if it has a tubular neighbourhood which is just a product of a copy of itself with

[-1,1]

, and it is one-sided if its tubular neighbourhood is a twisted

I

-bundle over the same surface. Example (call

M

the Moebius strip once and for all): in the manifold

M\times S^{1}

the levels

M\times \{p\}

are two-sided, whereas the torus

\gamma \times S^{1}

(where

\gamma

is the core of the Moebius strip) is one-sided. Cthulhu.mythos 10:15, 1 June 2006 (UTC)[reply]

Seems like I'll create one-sided (submanifold) and two-sided (submanifold) soon... Cthulhu.mythos 10:17, 1 June 2006 (UTC)[reply]

I don't understand why the Möbius strip isn't properly embedded in R³. On the contrary, I'm pretty sure the Möbius strip can be embedded in R³. I made one in third grade with scissors and paper, and it wasn't on the day we took the field trip to R⁶. -lethe ^{talk +} 17:37, 1 June 2006 (UTC)[reply]

htaccess for Zeus Web Server

I need this htaccess file written for Apache to be re-writen for Zeus Web Server

      Options FollowSymLinks
      RewriteEngine On
      RewriteCond %{REQUEST_FILENAME} !-f
      RewriteCond %{REQUEST_FILENAME} !-d
      RewriteRule ^(.+)$ /index.php?title=$1 [L,QSA]

Thanks, Gerard Foley 21:49, 31 May 2006 (UTC)[reply]

Nellore Telugu Measurement

I would like to know exactly how many feet are there in an "Ankanam" based on Nellore region, AndhraPradesh, India. Thank You

Maybe the question is flawed. One site listing Nellore housing says "One Ankanam =4 sq.yds.", while an online Telugu-English dictionary translates "ankaNam n. space between two beams or pillars in a house." Whichever we pick, the question makes no sense. --KSmrq^T 23:58, 31 May 2006 (UTC)[reply]

Real Analysis - teaching and learning

What is (or should be) gained by undertaking an introductory course in Real Analysis? What are the implications of this for the way it's taught? Thanks. --The Gold Miner 23:35, 31 May 2006 (UTC)[reply]

I always thought real analysis was just "calculus again, but for serious this time". You go over the definition of limit again, use fancy topology words like compact, and focus on pathological examples like differentiable functions whose derivatives are not continuous. —Keenan Pepper 05:39, 1 June 2006 (UTC)[reply]

The main benefit of a real analysis course is not so much the content, as the mathematical maturity gained by taking the course. Quite often RA courses are the student's first serious introduction to writing detailed proofs and working with high-level (i.e. Rudin) mathematics textbooks. 18.228.1.113 13:34, 1 June 2006 (UTC)js[reply]

Vast numbers of first-year university students are required to learn calculus, for use in a variety of different fields. Therefore the typical calculus textbooks and courses are designed to serve this genuine need for the broad community. However, this is rather different from training mathematicians. Weierstraß and others went to a great deal of trouble to build solid foundations under calculus, which is how real and complex analysis began, but the journeyman calculus applications rarely trouble themselves with such details. Complex analysis, which builds on real analysis, provides an indispensable set of tools for a great deal of modern mathematics. One last observation, admittedly a little biased: I don't recall ever hearing anyone, not even a non-mathematician, complain about knowing too much mathematics. In fact, I'll close with two quotations from Albert Einstein:

In 1943 he answered a letter from a little girl who had difficulties in school with mathematics: "…Do not worry about your difficulties in Mathematics. I can assure you mine are still greater."
"Any man who can drive safely while kissing a pretty girl is simply not giving the kiss the attention it deserves."

The first is well-known; the second shows his true genius. ;-) --KSmrq^T 06:48, 1 June 2006 (UTC)[reply]

Oh, we're listing choice quotes, are we? Here's an interesting anecdote. When Weisskopf was asked "How much mathematics does a theoretical physicist need to know?", he answered simply "more." -lethe ^{talk +} 17:47, 1 June 2006 (UTC)[reply]

Thanks to everyone who responded. My reason for asking is that I recently took such a course and found it a demoralising experience. Students at my University come from a wide variety of backgrounds and their levels of ability vary considerably. Success depends largely on diligent preparation for the end of semester examination. In a perfect world success in exams should, I think, depend on having developed a strong working intuition for the subject, and on being sharp and imaginative on the day. However, because of the way this particular course was taught, the exam positively encouraged cramming and parrot-fashion regurgitation of results and proofs, rather than said intuition and imagination. This rendered the course unfortunately somewhat pointless. I just wonder, given the subject matter and students of mixed ability, if it could ever have been otherwise... --The Gold Miner 15:26, 1 June 2006 (UTC)[reply]

June 1

Problem with complex numbers

Hello,

You are given that the complex number alpha = 1 + j satisfies the equation z^3 + 3z^2 + pz + q = 0, where pand qare real constants. (i) Find and in the form a + bj. Hence show that and p = -8 and q = 10 [6] (ii) Find the other two roots of the equation. [3] (iii) Represent the three roots on an Argand diagram. [2]

Thanks guys as always. DR Jp.

This sounds like a homework problem. Is there a particular thing you don't understand? Maybe you should take a look at complex number and complex plane. —Bkell (talk) 00:47, 1 June 2006 (UTC)[reply]

Thanks, but I'm currently writing a book and I would like someone to check these problems are doable. If they could put their working so I could see how you are thinking that'd be even better.

How much time is needed to travel around the world for the following modes of transportaion -- the space shuttle, a jet airliner, a cruise ship

Please make sure that your proxy server is either configured correctly, or you refrain from using them, since each time you edit you introduce backslashes before "'"s, breaking the markup. Dysprosia 01:02, 1 June 2006 (UTC)[reply]

Do you take us all for idiots? Do your own homework. (People who genuinely write books are typically careful with language, and have students, colleagues, and paid reviewers — not to mention an editorial staff — to check their work. Also, it is not helpful to see how a trained mathematician approaches a problem if you want to know how a student would think.) --KSmrq^T 03:51, 1 June 2006 (UTC)[reply]

Heh, writing a book, that's a good one. —Keenan Pepper 05:36, 1 June 2006 (UTC)[reply]

Yes, the problem is doable. I would start with part (ii) - find the other two roots of the equation. You know the value of α. Once you know that p and q are real, a second root is immediately obvious. And you can find the third root because you know the co-efficient of z². Once you have all three roots, you can find the values of p and q. Gandalf61 08:50, 1 June 2006 (UTC)[reply]

Starting with (i) is also easy enough, though. --Lambiam ^Talk 14:15, 1 June 2006 (UTC)[reply]

How do I tell if I've chosen an appropriate statistical distribution

I have a group of 12 observations. I'd like to predict what my observations will be in the future. I also need the distribution to apply Bayes Theorem.

Right now, I'm using the normal distribution but I don't know if that's the right choice. I've calculated the skewness and kurtosis of the data, but I don't have any idea what they're supposed to be! I mean, I know if my observations were truly normally distributed, the skewness would be zero, but I don't know if my skewness of 1.65 is "close enough" or what. Are there rules of thumb for this? moink 05:50, 1 June 2006 (UTC)[reply]

Under the assumption of normal distribution, the probability that a sample of 12 observations has a skewness whose absolute value is at least 1.65 is about 0.002. That is fairly low, and normally ground to reject the null hypothesis of normalcy. What is the source of the observations and how critical is the accuracy of the estimated distribution? Often the physical or other origin of the data suggests a plausible crude model for the distribution that is good enough in practice. --Lambiam ^Talk 06:36, 1 June 2006 (UTC)[reply]

It is not particularly critical. I was actually kinda hoping not to have to share the type of data, but since it's apparently a very poor fit to the normal distribution, I guess I will. It's the length of my menstrual cycle. Now all the boys on the math RD can get all grossed out. :) I like to know if I should carry tampons on me, and the Bayes' theorem thing... well, if you're very bright you may be able to figure it out but I will not provide an explanation. Here's the data: 32, 29, 28, 28, 26, 27, 27, 29, 36, 25, 26, 28. moink 07:41, 1 June 2006 (UTC)[reply]

You know, you could use a neural network for precisely this task. Neural networks can be used to predict the length of menstrual cycles as well as stock market values or other things. Choose an encoding for the lengths, train some sort of recurrent network on the data you have, and then get it to generate predictions. If I get time, and I am sufficiently bored, I might even try this for you. Dysprosia 07:49, 1 June 2006 (UTC)[reply]

Sounds cool but beyond my abilities. Right now, though, I'm less interested in predicting exactly the length of the next cycle, and more in knowing the approximate probability that it is at least some length so I can apply Bayes' theorem. moink 07:56, 1 June 2006 (UTC)[reply]

Using your data and the formula at skewness, I find a skewness of 1.27, which is still significantly different from the null hypothesis but less so. Looking at the data, the problem appears to be the outliers at the high side. If you censor the data by discarding values > 30, you get a good agreement with a normal distribution. Given the application censoring at the high side is acceptable, since you want confidence at the low side. The sample is still a bit small, though, to really confidently assume the low end behaves normally, without outliers. --Lambiam ^Talk 14:38, 1 June 2006 (UTC)[reply]

So much for trusting my spreadsheet software. I thought about dropping the large ones, but it's in the higher range that I'm most interested in the probability, and since it seems that it does occasionally get that large, and not that rarely, I wanted to take that into account. moink 15:28, 1 June 2006 (UTC)[reply]

Well, I'm by no means suggesting that this is what you are trying to calculate, but just for the sake of argument: if A=pregnant, and B=menstruation has not yet occurred, and one were interested in P(A|B), then P(B|A) would of course be very close to unity, but what value should be used for P(A)? Would the age specific fertility rate be correct? --vibo56 15:01, 1 June 2006 (UTC)[reply]

Addendum: P(A) would obviously have to be either zero, or a lot higher... --vibo56 15:11, 1 June 2006 (UTC)[reply]

Why would you say that? I mean, it could be zero, but it could be the small numbers you'd get using the failure rates of certain contraceptives. Even with several instances of unprotected sex in a month, it will generally not go above 25-30%. moink 15:24, 1 June 2006 (UTC)[reply]

Agreed. You are right. --vibo56 16:35, 1 June 2006 (UTC)[reply]

Chi squared might be your answer to whether the data is normal or not. Basically this works by dividing up the domain in to a number of boxes, you then count how many of your data items fall into each box and compare with the number predicted from the normal distribution. Add up the square of differences and compare with the approptite Chi-squared statistic. This should give a confidence interval as to whether the difference is significant or not. I suspect with only twelve points you don't really have enough data to meaningfully talk about skew. --Salix alba (talk) 15:10, 1 June 2006 (UTC)[reply]

Sigh. Ok, so I'm transparent. My prior distribution in this context is from a record of instances of penetrative sexual intercourse along with the underlying numbers used by this site combined with a pdf of the date of ovulation using the pdf above and the possibly quite poor assumption of a constant luteal phase of 14 days. moink 15:14, 1 June 2006 (UTC)[reply]

If the goal is to get pregnant, I'd start off by measuring my body temperature, to get a more precise estimate of the time of ovulation. After one year with no success, I would definitely go see a gynecologist. If, on the other hand, the goal is not to get pregnant, and you want a statistical tool to tell you when to start worrying, I'm afraid your approach won't work. Biological distributions tend to have very heavy tails, and you simply do not have enough data to make a sensible estimate of the distribution. With a limited dataset, however, you could make control charts. Here's a link to a how-to (powerpoint), courtesy of the British NHS Modernisation Agency. --vibo56 17:18, 1 June 2006 (UTC)[reply]

Newton was right after all

Ok, here is this little idea I had today. It has the potential of solving all problems of theoretical physics in one stroke of genius. Basically, the idea is that the interactions in the world are subject to very simple rules, namely Newtonian mechanics. But hey, I hear you say, wasn't there a guy called Einstein who has proven Newton wrong? Well, he did, but it would be Newton who will be having the last laugh.

The world of Newtonian mechanics is very simple: a Euclidean space and a few differential equations, solutions of which are nicely differentiable curves. There is one problem with this world: it is continuous, which makes its "implementation" extremely hard. There are no continuous things in our world: the space is discrete, the time is discrete, and this brings us to the next point: the world is a finite state machine. The world is, basically, a computer. The bit-twiddling aspect of our world is studied by quantum mechanics.

N-body problem is a classical problem of mechanics. There are a few particles floating around in space under the Newtonian law of gravity. It translates into a system of differential equations, which in general is not solvable by mathematical means. We'll try to simulate this problem on a computer. When n=2 theory states that these two bodies would have elliptic orbits (well, not exactly, the orbits may also be hyperbolic or parabolic, but we assume that they are ellptic). What happens when we simulate this on computer? At first everything seems okay, the smaller planet rotates around the bigger one. Let's modify the program so the path of the smaller planet is visible. Then, after a few rotations we'll see a strange effect: the elliptical orbit is slowly turning! More interesting is that the same thing happens in real life: a result predicted by General Relativity theory. But isn't it strange, that Newton's law of gravity when emulated on a computer produces the same effect?

There are several numerical methods for solving differential equations. They all have the same flaw: when you run the method for a long time, round-off errors and discretization errors build up and the result strays off from the right solution. The same thing happened in our simulation - during the rotation the error builds up as we do little discrete time steps, and it rotates the orbit. The same thing, I suspect, happens in our real world, which faithfully tries to solve Newton's differential equation by discrete means.

Comments are welcome. Grue 12:36, 1 June 2006 (UTC)[reply]

It appears unlikely that this theory can be tweaked to give a quantitative agreement with the observations. Using methods such as Runge-Kutta integration, it is easy enough to get a precision that is able to differentiate between Newtonian and Einsteinian gravitation. And since Wolfram we all know that the universe is a cellular automaton. --Lambiam ^Talk 14:08, 1 June 2006 (UTC)[reply]

I think there's a simple explanation for your observation. Newtonian mechanics predicts precession of planetary orbits due to small pertubations caused by other planets. In your computer models, I suspect that the rounding errors introduce similar pertubations into a 2-body situation, and so give rise to qualitatively similar precession. Both Newtonian mechanics and general relativity predicted a precession of Mercury's orbit - the difference was that the Newtonian prediction of the rate of precession did not agree with the observed value, whereas the GR prediction did agree (within the limits of observational error) - see Tests of general relativity. Gandalf61 15:33, 1 June 2006 (UTC)[reply]

One way hash/encryption function that is guaranteed unique

Any crypto fans here? I need to generate object IDs from other object IDs to mask the true object with one that is a plausible replacement.

Say I have a unique object ID, is there an encryption algorithm I can use against it to produce a different (cyphertext) object ID that is one way (can't easily get the original plaintext Object ID from the cyphertext Object ID, either not at all, or at least short of a computationally expensive attack), and unique, meaning that no 2 original plaintext object IDs produce the same cyphertext object ID and that I never get different cyphertext object IDs from the same plaintext object ID. The text lengths need not be the same I guess, although obviously the cyphertext one can't be shorter. I have looked at Message_digest and it speaks of cypher/hash functions that are unlikely to have different plaintexts go to the same cyphertext. I need guaranteed... the cyphertext length need not be a hash, it can be as long or longer as the original (exactly the same size would be convenient, though). thanks! ++Lar: t/c 18:34, 1 June 2006 (UTC)[reply]

PS... another way of saying this is that I need Collision_resistance that is not just hard but impossible, or at least practically impossible. ++Lar: t/c 18:36, 1 June 2006 (UTC)[reply]

What do you call this type of notation?

${\begin{matrix}{}_{n}\\{}*{}\\{}^{k=1}\end{matrix}}k$ or ${{}*{} \atop {}^{k\in S}}{k \atop {}}$

As far as I can remember, ∗ is always been one of Σ, Π, ∩, ∪, or ∐ (coproduct), but theoretically couldn't it be extended to apply to any binary operation? Or maybe even to any function?

I don't understand why this ubiquitous notation does not seem to have a name.

@@ Line 874: / Line 874: @@
 Say I have a unique object ID, is there an encryption algorithm I can use against it to produce a different (cyphertext) object ID that is one way (can't easily get the original plaintext Object ID from the cyphertext Object ID, either not at all, or at least short of a computationally expensive attack), and unique, meaning that no 2 original plaintext object IDs produce the same cyphertext object ID and that I never get different cyphertext object IDs from the same plaintext object ID. The text lengths need not be the same I guess, although obviously the cyphertext one can't be shorter. I have looked at [[Message_digest]] and it speaks of cypher/hash functions that are unlikely to have different plaintexts go to the same cyphertext. I need guaranteed... the cyphertext length need not be a hash, it can be as long or longer as the original (exactly the same size would be convenient, though). thanks! '''<font color="green">[[User:Lar/Esperanza|+]]</font><font color="blue">[[Special:Emailuser/Lar|+]]</font>'''[[User:Lar|Lar]]: [[User_talk:Lar|t]]/[[Special:Contributions/Lar|c]] 18:34, 1 June 2006 (UTC)
 :PS... another way of saying this is that I need [[Collision_resistance]] that is not just hard but impossible, or at least practically impossible. '''<font color="green">[[User:Lar/Esperanza|+]]</font><font color="blue">[[Special:Emailuser/Lar|+]]</font>'''[[User:Lar|Lar]]: [[User_talk:Lar|t]]/[[Special:Contributions/Lar|c]] 18:36, 1 June 2006 (UTC)
+What do you call this type of notation?
+<math>\begin{matrix} {}_n \\ {} * {} \\ {}^{k=1} \end{matrix} k</math> or
+<math>{{} * {} \atop {}^{k \in S}} {k \atop {}}</math>
+As far as I can remember, &#x2217; is always been one of Σ, Π, &#x2229;, &#x222a;, or &#x2210; (coproduct), but theoretically couldn't it be extended to apply to any binary operation? Or maybe even to any function?
+I don't understand why this ubiquitous notation does not seem to have a name.