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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 Miner13:43, 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.mythos14:28, 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. -lethetalk+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. --LambiamTalk17:01, 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. --KSmrqT23:33, 26 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!)
Another reason is that true mathematics requires spending a lot of time in your head, which isn't very conducive to social skills. Daniel (‽) 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 Carrot20:12, 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 Carrot20:16, 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
is open if and only if for each point p in E there is an open semiball around p completely in 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
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.
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. -lethetalk+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. -lethetalk+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.) --LambiamTalk17:20, 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. -lethetalk+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 and any r>0 is open in T(P), thus it is open in T(Q), so there is and r'>0 such that . 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 , then
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?199.201.168.10015:53, 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. StuRat16:55, 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+1th round given survival up to and including the Nth 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:
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"?). --LambiamTalk17:45, 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? --LambiamTalk23:23, 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. --LambiamTalk14:06, 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]
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. StuRat02:26, 29 May 2006 (UTC)[reply]
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. :) Henning19:37, 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 username21:16, 26 May 2006 (UTC)[reply]
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?
What about theorem proving? If I remember correctly, this is possible by computers in only rather basic cases, but not in general. Dysprosia12:17, 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]
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. Dysprosia13:26, 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 Merkel12:38, 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. --LambiamTalk13:49, 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.) --KSmrqT19:27, 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). Ohanian22:15, 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.
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. schyler03:18, 28 May 2006 (UTC)[reply]
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. --KSmrqT02:09, 31 May 2006 (UTC)[reply]
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. --Knotted13:41, 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]
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. --vibo5614:50, 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. --KSmrqT17:47, 28 May 2006 (UTC)[reply]
There isn't much you can do to remove IE, at least. As previously mentioned, it has a bunch of core components that many other programs use. Removing those components would probably cripple/break too many other things. The real "bloatware" to watch out for is malware such as viruses. DoomBringer17:44, 2 June 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
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 . 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. --vibo5614:11, 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... Dysprosia15:28, 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]
From which I get using the spherical law of sines that sin A = sin a/ √(1 – cos4a). A = B and C = π/2, so I have the triangle, and hence the square. -lethetalk+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? --vibo5616:49, 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. -lethetalk+17:20, 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. -lethetalk+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). --LambiamTalk16:02, 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)? --vibo5618:07, 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. --LambiamTalk19:59, 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
from which I have
From the law of cosines I have that
My goal here is to eliminate c. First I substitute cos A:
which reduces to the quadratic equation
So I have
and using cos A = sin c/2sin a, I am in a position to solve the triangle
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. -lethetalk+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. --KSmrqT20:40, 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? --vibo5621:51, 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
where a is 10 cm/R, the side length as an angle. The angle of interest is really C = 2A, for which
For the hemisphere case, a = π/2 produces C = π; while for the limit case, a = 0 produces C = π/2.
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. --vibo5620:19, 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. --KSmrqT21:23, 29 May 2006 (UTC)[reply]
Thank you. I really appreciate your taking the time to explain this to me which such detail and clarity. --vibo5623:34, 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. Nimur18:11, 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
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
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.
According to Girard's formula, we then have
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:
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.
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
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. --KSmrqT02:36, 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. --vibo5619:40, 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. --HappyCamper03:38, 29 May 2006 (UTC)[reply]
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. --HappyCamper04:32, 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
, where k = 0, 1, 2.
So, your solutions are of the form
Now, let's just substitute in the k...these are your values for z
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 . 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 , where i is the imaginary unit? Hm.... --HappyCamper05:22, 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 Pepper05:46, 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]
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
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
has a zero when x is 1⁄2, but that is not a whole number and so may not be acceptable. Or consider
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
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. --KSmrqT05:56, 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 was presented in a somewhat non-conventional way. More conventional is , which makes clear that the occurrence of a is not under the square root's vinculum, which would have been . So b2 = 3a2 (which is where we came from), and not 3a. --LambiamTalk10:19, 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?
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
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
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
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
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
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! 12.183.203.18408:17, 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 .
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 if exercised; thus they will be exercised if the current value of the die is less than i.e. on a 1, 2, 3 or 4. So the expected value of 3 rolls is . EdC14:13, 29 May 2006 (UTC)[reply]
While we just provided a correct answer, I wonder if the proof is needed that is indeed the maximal value of n rolls, and our strategy is optimal. (Igny18:05, 29 May 2006 (UTC))[reply]
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]
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. --Serie22:48, 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) --DLL20:32, 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. Dysprosia22:19, 30 May 2006 (UTC)[reply]
Nothing. It was just that several posts had gone by without anyone saying anything, at all. I was taking that a bit farther. Black Carrot02:13, 2 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. --DLL20:28, 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? --KSmrqT01:50, 31 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:
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. --LambiamTalk02:58, 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 45.359237 kilograms. So try to triple it and do not stay under the load. --DLL20:24, 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. x42bn6Talk07:56, 31 May 2006 (UTC)[reply]
why is in locally pathconnected topological space, path connected component same as connected component
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...
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 and choose . If you show that the set of all points in that can be connected to by a path is the whole itself, you are done.
To do this, you show that it is both open and closed: as is connected by definition, and nonempty because it trivially contains , you are done.
Openness: suppose can be connected to by a path. Let be a neighbourhood of which is path connected (it exists because the space is locally pathconnected). Then any point in can be connected to by a path simply by going to first and then to . Hence, your set is open.
Closedness: consider the subset of that cannot be connected to by a path, and choose as such. Then choose a neighbourhood as above: if any point of could be connected to by a path, then by following the two paths as before you could also connect to . This means that this set is also open, i.e. the set of points that can be connected to by a path is closed.
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.
Evilbu10:45, 31 May 2006 (UTC)[reply]
May 31
Integration of sin and cos
What happens when you integrate sin(), -sin(), cos() and -cos()
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). Nimur18:06, 31 May 2006 (UTC)[reply]
A sophisticated answer uses Euler's formula, cos ϑ+ⅈsin ϑ = ⅇⅈϑ, and the fact that ⅇz is an eigenfunction for the linear operation of integration. For example,
so
A more geometric approach observes a unit circle parameterized by arclength,
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. --KSmrqT18:42, 31 May 2006 (UTC)[reply]
very exact definition of "orientable topological 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.
thanks, but could you explain what you mean by 'it preserves orientation"?
<Mount Soapbox>
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.
Now fix Bn, 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. --KSmrqT23:36, 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? -lethetalk+04:21, 1 June 2006 (UTC)[reply]
I just mentioned some of the additional conditions that often appear, without regard for independence. For example, it is not uncommon for a paracompact space to be required to be Hausdorff. Anyway, the paracompact space article mentions that "Every locally compact second-countable space is paracompact." And the locally compact space article points out that "Topological manifolds share the local properties of Euclidean spaces and are therefore also all locally compact." (There may be a caveat about Hausdorffness in there.) Fortunately, we can discuss orientability without concern for these extra conditions. --KSmrqT06:35, 2 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. -lethetalk+04:24, 1 June 2006 (UTC)[reply]
And also this one: a differentiable manifold is not orientable if and only if it admits a loop (with base point ) such that if you choose a base of having as its first vector and then drag it along using charts, when you came back to at time you have a base of whose orientation is reversed w.r.t. the base you started from. Cthulhu.mythos08:56, 2 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 , because of the boundary) is two-sided if it has a tubular neighbourhood which is just a product of a copy of itself with , and it is one-sided if its tubular neighbourhood is a twisted -bundle over the same surface. Example (call the Moebius strip once and for all): in the manifold the levels are two-sided, whereas the torus (where is the core of the Moebius strip) is one-sided. Cthulhu.mythos10:15, 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. -lethetalk+17:37, 1 June 2006 (UTC)[reply]
So the article on embedding didn't mention proper embeddings. It does now, although I'm not sure how to make it clearer. A proper embedding preserves boundaries; for example, embedding in as a chord is proper, but embedding it as a radius is not. So the Môbius strip can't be properly embedded in , because the Môbius strip has a boundary () and doesn't. (This raises the question of what subspaces of the Môbius strip can be properly embedded in. I think it's obvious that the boundary has to have positive genus, though I'm not sure how to prove it.) -- EdC04:16, 2 June 2006 (UTC)[reply]
Oh, right. Assume that M is embeddable in R3 less a contractible subspace A. Then ∂M (= S1) is contained in the boundary of A, so its contraction S* is D2. So M ∪ S* is an embedding of the projective plane in R3, which is a contradiction (there is no such embedding). -- EdC04:29, 2 June 2006 (UTC)[reply]
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 "ankaNamn. space between two beams or pillars in a house." Whichever we pick, the question makes no sense. --KSmrqT23:58, 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 Miner23:35, 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 Pepper05:39, 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. 18.228.1.11313:34, 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."
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." -lethetalk+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 Miner15:26, 1 June 2006 (UTC)[reply]
Yes, it could have been different. I was a physics major who had only taken calculus type courses until my senior year, when I took advanced calculus. For the first time, my understanding was limited only by the time I spent thinking about things, and not by the vagueness of definitions, etc. Now, I'm a mathematician, and it is very satisfying (both research and teaching). Here's a quote from a recent student: "I maintain that taking Advanced Calculus was the turning point of my undergraduate career. It had reminded me how beautiful math is, how much it means to me, and how much I was willing to work in order to succeed in it." (Cj6720:12, 1 June 2006 (UTC))[reply]
It's bad enough that a first-year calculus course might be taught so poorly; it's tragic that this should have been afflicted on aspiring mathematicians. But I must warn you that mathematics is not a field in which participants self-select because of their superior social skills (like teaching ability). Some mathematicians are awesome teachers; others are horrid. The ability to see a subject clearly for oneself is not at all the same as being able to communicate that insight clearly to others. So it has always been. This is one reason for Stigler's law of eponymy. --KSmrqT01:57, 2 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, 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. Dysprosia01:02, 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.) --KSmrqT03:51, 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. Gandalf6108:50, 1 June 2006 (UTC)[reply]
Heh, it looks like an IB Math problem. He put the possible marks next to each part of the question "(ii) Find the other two roots of the equation. [3]". --Codeblue8719:58, 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? moink05:50, 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. --LambiamTalk06:36, 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. moink07:41, 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. Dysprosia07:49, 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. moink07:56, 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. --LambiamTalk14:38, 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. moink15:28, 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? --vibo5615:01, 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%. moink15:24, 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. moink15:14, 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. --vibo5617:18, 1 June 2006 (UTC)[reply]
When reading my previous comment: forgetting to mention this was maybe a male freudian slip, but anyway: if the goal is getting pregnant, it would be a good idea to have your partner checked as well. --vibo5621:33, 1 June 2006 (UTC)[reply]
Well, the goal is complicated. It is one, the other, or both of the above, in addition to saving on costs of Human chorionic gonadotropin tests (which have high false negative rates, especially when used too early) by using them at the right time. For example, applying an additional Bayesian update rule, a negative test with a sensitivity of 25 mIU of hCG would reduce my probability by a factor of nearly three if I used it today, while it would reduce the probability by a factor of eight if used tomorrow. And buying a basal thermometer would negate those cost savings. :) The other main goal is the fun of overanalyzing these things. :) moink04:12, 2 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.
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. --LambiamTalk14:08, 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. Gandalf6115:33, 1 June 2006 (UTC)[reply]
Methods exist for integrating differential equations while respecting conservation laws, and special perturbation methods have been devised for orbital simulations. When JPL computes long-term ephemerides and certain critical spacecraft trajectories, they find it necessary not only to be extraordinarily careful with their numerical methods, but also to include the effects of general relativity. Some tests of general relativity effects require meticulous care to distinguish the sought effect from numerous other sources of perturbation; a notable example is Gravity Probe B. Other effects, like gravitational lenses and black holes, are not subtle at all, and are not only thoroughly observed but also dramatically different from Newtonian predictions. To speak bluntly, it is absurdly arrogant to imagine that your uninformed "little idea" is more insightful and clever than the work of large numbers of trained professional physicists. We are convinced Einstein's theory will eventually be replaced by a more integrated theory, something Einstein himself attempted (unsucessfully) in his later years; but we can never go back to Newton. --KSmrqT23:50, 1 June 2006 (UTC)[reply]
You all seem to miss the point. It is true that there are high-precision numerical methods for solving differential equations. It is not true that the world is using these methods to calculate trajectories. It is using the simplest one, that is, Euler integration. With Planck time being a very short interval, the error would be noticeably high. "Continuous" Newton mechanics doesn't take this error into account. Relativity does. Grue 07:10, 2 June 2006 (UTC)[reply]
I said and still maintain that it appears unlikely that this theory can be tweaked to give a quantitative agreement with the observations. How so is that beside the point? The onus of showing that you can get a good agreement is on you. Einstein predicted a difference of 43 arc-seconds per century with Newtonian theory; what does your theory predict? See further Scientific method. For the rest, Euler integration? – now if you could explain away action at a distance, that might make it interesting even if zany. --LambiamTalk14:35, 2 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/c18:34, 1 June 2006 (UTC)[reply]
Well, the whole notion of hash functions is that they're practically good enough, but if you want to reduce the possibility of collisions further, just encrypt the original object. There aren't (AFAIK) formal methods for one-way-only full-length encryption, but you can get the effect by generating a public/private key pair (say, via PGP / GPG). Encrypt normally via the public key, and then hide or destroy the decrypting private key as required. Public-key encryption is preferred here since knowledge of the only in-use key can't expose the data, but there's no reason any other encryption won't work so long as you protect the key. — LomnTalk19:13, 1 June 2006 (UTC)[reply]
a further note on "practically good enough": the SHA-1 hash function will have its first random collision, on average, after about 1024 entries. Do you really anticipate needing that sort of collision resistance? — LomnTalk19:20, 1 June 2006 (UTC)[reply]
It's a customer requirement, and their answer is yes. Mine would be no though! Thanks for your help on this. I'm thinking public key/private key (with a destroyed private key) is the way to go, the object ID is short enough that encrypting it is fine, should not be too compute intensive ++Lar: t/c20:02, 1 June 2006 (UTC)[reply]
One other option to consider would be sticking to a standard hash but appending some additional nonce such as date-of-entry alongside the hash output. Of course, if hash function collision rates aren't "practically impossible" enough for the customer, I don't know that this is really any better (unless it simply looks stronger to a corporate perspective). Another consideration, if you have short object IDs, is the possibility that object contents will collide more frequently than the cryptographic process -- for instance, I'm quite certain you'd find the MD5 for "password" far more frequently in a password hash file than mere chance would suggest because the source entries aren't random. This may necessitate a nonce in its own right (and, of course, it could be appended prior to hash/encryption as well). — LomnTalk20:27, 1 June 2006 (UTC)[reply]
Right. But these aren't passwords and I have to be able to reproducably and repeatably produce the same encrypted OID from the cleartext one without having a salt value stored. Whether THEIR algorithm for making an OID is unique or not is a different matter. ++Lar: t/c01:37, 2 June 2006 (UTC)[reply]
A method that gives good scramble of an N-bit string is to XOR it with a randomly chosen bit string, apply a randomly chosen permutation to the bits of the string, multiply modulo 2N with another random but odd bit string, and XOR+permute again (using different choices). Since each step is invertible, you are guaranteed to have no collisions. For additional security the procedure can be repeated. I have no idea how safe this is against various cryptanalytic attacks; for all I know there is a way of breaking the scheme that is obvious to more devious minds than mine. Use at your own risk. --LambiamTalk14:53, 2 June 2006 (UTC)[reply]
In general, any block cipher should satisfy the conditions you specify, as long as you never reveal the key. If there's a limit on the block size, you can always construct your own ad hoc cipher, which is pretty much what the previous suggestion does. To be safe, however, you probably want to construct your custom cipher around an established cryptographic hash function, using a well-studied construction such as a Feistel cipher. For my own earlier take on a similar problem, see [2]. (Note that the code actually rather overdoes it, using 2*5 = 10 rounds where, per Luby and Rackoff, 4 should suffice.) —Ilmari Karonen (talk) 17:27, 2 June 2006 (UTC)[reply]
notation
What do you call this type of notation?
or
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.
Don't forget \bigwedge (wedge product or join), \bigvee (meet) \bigoplus (direct sum), \bigotimes (tensor product), and I'm sure there are more lesser used examples. It's not binary operators that you do this with, but n-ary operators (especially infinitary operations). Of course any associative binary operation can be extended by induction to an n-ary operation. I think it'd be more risky to try to do this with a nonassociative operation, and of course the notation really earns its lunch for infinitary operations, which do not arise from binary operations by induction. As for what the notation is called, well I don't view it as a problem that the notation doesn't have a name. Lots of notations don't have names. What's the dx symbol that sits next to the integrand called in an integration? But if you have to have a name, I think the AMS-LaTeX guide calls them cumulative operators, which might serve you. -lethetalk+19:19, 1 June 2006 (UTC)[reply]
So could a non-binary operation be extended to an n-ary operation with this notation? I'm particularly thinking of the hyper operator.
No, because the hyper operators (past multiplication) are nonassociative. The notation could conceivably be extended to nonassociative operators, but order of evaluation would need to be provided. -- EdC00:50, 2 June 2006 (UTC)[reply]
But could this notation be defined for a function that would, without it, take 3 or more arguments? For example, if you have:
can you define the following?
Ugh. No. Absolutely not. No way. Look, you have to consider that these cumulative operators are essentially functions defined on multisets (or, at a pinch, sequences); the point is that an associative binary operator (with range a subset of its domain) extends to a function on finite sequences (if it's commutative, on finite multisets). You can't do that with operators of greater arity; for one, the length of the sequence (size of the multiset) is constrained (for ternary operators) to be an odd number ≥ 3. (And where do you grow the evaluation tree?) You'd do better to package the sequence as pairs ((n-1)-tuples) and define a F on sequences of pairs:
Does differential mean infinitesimally small number? If so, then yes, it was true in Newton's day that the dx symbol in an integration was an infinitesimal number, but it's not true today. Nonstandard analysis puts the infinitesimal on rigorous footing, but even that still does not allow you to interpret the symbol under the integral sign as an infinitesimal. You may interpret the symbol as a differential form in some cases. So I ask you, what does the word "differential" mean to you? The WP page is just disambiguation. -lethetalk+21:44, 1 June 2006 (UTC)[reply]
Statistical process control
In statistical process control using control charts, I have noticed that presenters often recommend calculating the standard deviation in a, so to speak, nonstandard way. The recommended procedure is to calculate a mean moving range, i.e. , using a relatively small dataset, and then divide the mean moving range by the magic number 1.128. If you google for "(1.128 and calculate)" and are feeling lucky today, you will find a such a presentation. The number 1.128 is often represented by the symbol d2. Does anybody know the maths behind this non-standard estimator of the standard deviation? --vibo5619:00, 1 June 2006 (UTC)[reply]
Indices
Can anyone help me with the problem below. I am aware that it is a homework question, but that is why I don't want you to answer it! The question is to simplify it and I assume that you have to multiply out the parenthesises:
It looks like your title may be your problem, as those appear to be exponents rather than indices. Given that, you're right that you need to multiply out; now you just need to check the rules for ax * ay = a?. — LomnTalk20:18, 1 June 2006 (UTC)[reply]
Ok, I won't answer it. You are correct in assuming that you have to multiply out the parentheses. I am fairly confident that your maths textbook will provide all the information you need to solve it. Just do it step-by-step, slowly and carefully. Write down each line, and be sure to know what rules you are applying in getting from one line to the next. Good luck with your homework. --vibo5620:24, 1 June 2006 (UTC)[reply]
Also remember that multiplication is distributive. The first equation of the distributivity article should help you; you can ignore the more complicated stuff after that. moink22:40, 1 June 2006 (UTC)[reply]
This question is a study in arithmetic properties of exponents. We first meet exponents in the limited form of positive integer powers, such as a3 = a·a·a; but here we are challenged to extend our understanding to fractional powers.
It is a property of all real numbers, however obtained, that multiplication distributes over addition: r(s+t) = rs+rt. (This is also a property of complex numbers.) Therefore, if we follow your hunch and "multiply out", we obtain
Still being thoughtful and cautious, we observe that multiplication is both associative and commutative, so we may slightly rearrange to get
Lacking experience with the fancier forms of exponentiation, it may not be obvious that anything we have done so far is particularly helpful. However, it does bring factors ax, for various x, next to each other. In fact, this is helpful, for it brings us to the crux of the problem. But to simplify any further we must begin to understand the meaning of fractional exponents, and learn the rules that apply to them. In other words, just as we have used rules of multiplication and addition — such as distributivity, associativity, and commutativity — to get this far, we must use analogous rules involving exponents to proceed.
The big picture will take time to see, but eventually we are able to construct a formal correspondence between expressions like
on the one hand, and
on the other, so long as a is positive. At that point (which most of us have reached) the last steps of simplification will be obvious. The purpose of this question is to help you get to that point. So, read the article on exponentiation, and enjoy. --KSmrqT00:59, 2 June 2006 (UTC)[reply]
A common blunder among people first learning this stuff is that exponentiation distributes over addition. It does not. In other words, . Don't make that mistake. To figure out the right way to deal with exponentiation of sums, consult binomial formula. -lethetalk+01:14, 2 June 2006 (UTC)[reply]
how to run a Monte Carlo simulation on a DEM using ArcGIS
PLEASE I WOULD LIKE TO KNOW HOW THE Monte Carlo simulation techniques IS USED to evaluate the impact of DEM error on viewshed analyses.
I have a question involving circular motion. Visualise a roller coaster's path. At the start, a straight part of length x, at an angle of 10° above the horizontal exists. A particle rolls down this, and meets an arc of radius 8m sloping downwards, and declines to a straight path 40° below the horizontal. The path is smooth and is only affected by gravity. I have to find the maximum value of x so that the particle stays on the path. I've only ever encountered particles with a mass, and this one has none specified, so I assume it cancels somewhere.
Anyway,
If the particle leaves the path, would it leave as soon as it encounters the curved component? (i.e. it takes off and misses the first part of the arc.) Or is it possible that it leaves the path during the arc?
Does weightlessness have anything to do with it when a particle is on the verge of leaving a path?
Once the particle leaves the path, there will be no upward force acting on the particle any more, so the only force acting on the particle will be gravity. What shape of path will this particle take? If you need help, take a look at the article about trajectory. I think how you want to solve this problem is to find out when this "natural" trajectory lies above the path, so that the particle will tend to "take off" instead of following the path. —Bkell (talk) 11:07, 2 June 2006 (UTC)[reply]
A particle follows a path only as long as the resultant force can supply the acceleration needed to keep it on that path. In this case, the resultant force comes from gravity; to stay on the path the centripetalcomponent of the gravitational vector must be at least as great as the centripetal acceleration of a particle taking that path. Now, the centripetal acceleration depends solely on velocity and arc radius; velocity depends solely on height lost (assuming a frictionless path i.e. total conversion of gravitational potential energy into kinetic energy).
This implies that the particle becomes more prone to leave the path as it proceeds round the arc: the velocity increases, and the centripetal component of the gravitational vector decreases. So the maximum x is that which has centripetal acceleration equal to centripetal component of gravitational vector at the end of the arc. Quite beautifully, both mass and gravity cancel out. See here to check your solution.
1. First of all, pick the number of times a week that you would like to have chocolate
(more than once but less than 10)
2. Multiply this number by 2 (just to be bold)
3. Add 5
4. Multiply it by 50 -- I'll wait while you get the calculator
5. If you have already had your birthday this year add 1756 .... If you haven't, add 1755.
6. Now subtract the four digit year that you were born.
You should have a three digit number
The first digit of this was your original number (i.e, how many times you want to have chocolate each week).
The next two numbers are:
YOUR AGE! (Oh YES, it is!!!!!)
See elementary algebra. If you want chocolate n times a week and you were born in year y, then you get
50 (2n + 5) + 1755 - y if you haven't had your birthday this year.
50 (2n + 5) + 1756 - y if you have had your birthday this year.
This rearranges to:
2005 - y + 100n (if you haven't had your birthday)
2006 - y + 100n (if you have had your birthday)
Hopefully it's now obvious that the (2005 - y) or (2006 - y) bit is your age, and the 100n bit gives n in the hundreds column (given the range you specified for n).