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August 13

Mathematicians and Math (What's the deal?)

OK,So I'm going to ask a few questions about what mathematicians know and don't know.I'd appreciate answers. Thanks guys.Can someone learn math without knowing linear algebra or complex analysis?How about the following. Is it possible to ...

(i)Learn differential geometry without knowing what a top. space is?

(ii)Learn algebraic geometry without knowing what a group is?

(iii)Learn harmonic analysis without knowing what a Hilbert space is?

(iv)Learn fucntional analyss without knowing what a Fourier series is?

(v)Learn ring theory without knowing what a field is?

(vi)Learn set tehory and logic without knowing what anything in math is?

(vii)Learn algebraic number theory without knowing elementary number theory?

(viii)Learn statistics without knowing what a measure space is?

(ix)Learn f.g.t without knowing what a character theory is?

(x(THE MONSTER!)Learn any area out of algebra, analysis, topology, number theory etcetera without knowing any of th eother areas(absolutely NOTHING in fact out of the other areas.)

And lastly,what's the deal with the fake proof of P = NP on the internet?And lastly again,why are mathematicians so uneducated to the point that algebra dudes don't know what a top. space is,top. dudes don't know what a Hilbert space is,analysis dudes don't know what a HOMOTOPY GROUP IS. (I mean how dumb can you get!?)Why do math dudes specialize so much into one thing or the other? I mean,WHY? Why don't they try to do EVERY ARAE? (I couldn't help using "dude" btw ...)I have a typsetting bug btw, hence the punctuation errors. —Preceding unsigned comment added by 114.72.245.4 (talk) 06:42, 13 August 2010 (UTC)

You don't need to know all of maths to get useful work done any more than a person studying insects needs to know all that much about fish, and would they do so much good work if they spent more time studying? And why do you refer to a 'fake proof'? It looks like a quite valiant attempt to me and fake implies wrongdoing. And anyway what gives you the idea a professional mathematician would be completely unfamiliar with something in your list? I'm not a professional mathematician but I've read up on all the things you've mentioned. This sounds like one of those company exhortations that says things like 'we must concentrate on improving on all fronts' ;-) Dmcq (talk) 08:22, 13 August 2010 (UTC)

thanks dmqc but can you give me answers to my questions as well? Thanks. What about other people's opinions? Can others give their opinions as well? I'd like a broad range of opinions from the math community? Thanks.

In the guidance information at the top of this Project Page it states The reference desk does not answer requests for opinions ... Dolphin (t) 07:59, 16 August 2010 (UTC)

Which question did I not answer? Dmcq (talk) 08:44, 13 August 2010 (UTC)

Why do you want to know? I am confused by your questions. Sometimes it is fertile to combine information from seemingly very different areas of science. But your central question "Why don't they try to do EVERY ARAE? " seems strangely naïve to me. Time is limited. There is a time for reading and a time for problem solving and a time for teaching and a time for original research. There is a time for seeking general orientation and a time for seeking knowledge for a specific purpose. As the amount of litterature grows faster than your speed of reading, reading everything is not possible. Bo Jacoby (talk) 09:01, 13 August 2010 (UTC).

At my school every student in the math PhD program had to pass a "qualifying examination" showing they had basic knowledge of all the topics you mention. They would then concentrate more narrowly for their thesis research. There was one loophole: if you wanted to become a logician without studying a lot of math, you could get a logic degree from the philosophy department instead of the math department. In that program you would still have to take a bunch of mathematical logic courses in the math department from math professors, but not much topology or anything like that. 67.122.209.167 (talk) 09:51, 13 August 2010 (UTC)

I think a bit of topology should be a prerequisite nowadays for logic, there's various thinks like the axiom of determinacy or large cardinals which need enough to at least understand things like Baire categories. Dmcq (talk) 12:05, 13 August 2010 (UTC)

Perhaps the above answers need to be "decoded" so that you can understand them. The answer to most of your questions (i)->(x) is "yes". By this I mean, one can learn the subjects mentioned without knowing what you have described. E.g., one can learn what a ring is without knowing what a field is (in fact, I do not know of anyone who learnt this the other way around), one can learn a good deal of statistics without knowing what "measure theory" is, one can learn functional analysis without knowing what a Fourier series is etc.

The answers to some of your other questions is, to some extent, "no". E.g., learning a little algebraic number theory without knowing elementary number theory is possible if you stick to the abstract concepts such as Dedekind domains, DVR's etc. and know in your own mind what a prime ideal is without needing to motivate it from elementary number theory, but there will come a time when intuitions from elementary number theory play a role. Similarly, one can learn basic Fourier analysis without knowing, at least abstractly, what a Hilbert space is, but concrete Hilbert spaces occur throughout Fourier analysis and so one needs to, at least at some point, learn how to identify an abstract Hilbert space. Also, it is a must to know what a ring is before one does algebraic geometry, and I assume, at least I hope that I do not need to "assume" this, that all students who can identify rings can also identify groups. Even in a more concrete sense, group schemes and algebraic groups are central to arithmetic geometry.

On the other hand, "learning" and "doing research" are different things to a certain extent. One can learn finite group theory (I presume that is what you mean by "f.g.t") very deeply without knowing much about characters. But seeing as character theory has provided efficient solutions to problems across finite group theory for which no purely group-theoretic proofs are known, "avoiding" character theory would significantly narrow your research interests. Furthermore, even though it is possible to do elementary differential geometry without having familiarity with the abstract definition of a topological space, many important more advanced techniques in differential geometry rely on other fields such as algebraic geometry where abstract concepts such as topological spaces play a role.

I agree with Bo Jacoby and Dmcq here. Although it is extremely important for a research mathematician to broaden his interests as much as possible, the amount of mathematics in existence is currently too great for a single individual to learn in a lifetime. I suppose you can at least "see" most of the central areas of mathematics to some extent, and indeed, many first-rate mathematicians have worked on several areas in their lifetime (e.g., Michael Atiyah, Jean-Pierre Serre and Alexander Grothendieck). But just because someone specializes in an area does not mean that that person cannot do good work; nearly every mathematician specializes in some area or the other!

Finally, I would like to draw an analogy with the "world map". Supose we take an (immortal) individual with no knowledge of the "global structure" of the Earth: where other lands are located, what the shape of the continents are, succinctly, absolutely no knowledge whatsoever of the world map. Then we put this individual somewhere randomly in the world with no tools (perhaps except for ample amount of food and water, and a ship to travel between lands) and asked him/her to draw the entire world map as accurately as possible after travelling across the world. (And by this I mean, nearly as accurately as people have depicted the world map today using advanced technology.) The amount of time this would take probably does not even compare to how much time it would take to learn all of the mathematics currently in existence. But collectively, all individuals on Earth do know the structure of the world map upon communication with each other. The situation is identical with mathematics; no-one knows every area of mathematics at a research level. That is why people collaborate with each other. I think that is the answer to your question: not every mathematician has the time or energy to learn many branches of mathematics to some extent; perhaps these are examples of the mathematicians you have seen. But together, mathematicians have been able to do remarkable things. PST 10:24, 13 August 2010 (UTC)

The user 119.72.245.4 should be directed to the summary recently written up by Charles Matthews at the conclusion of a similar discussion at WPM. Tkuvho (talk) 11:04, 13 August 2010 (UTC)

114.72.245.4 (aka 110.20.24.219), is there any particular reason that you have twice modified the IP address in SineBot's sig of your initial post to read 119.72.245.4? Do you wish you were in Tokyo? -- 1.46.68.59 (talk) 13:08, 13 August 2010 (UTC)

permutations and combination

There are four sections in a test each carrying 45 marks.In how many ways a student can pass if the pass or cut-off mark is 90 marks? —Preceding unsigned comment added by Imteyazmohsin (talk • contribs) 11:12, 13 August 2010 (UTC)

Basically, how many ways can you select at least n items from 2n? The number of ways to select any number of items would be 2²ⁿ. Half of that would be the number of ways to select more than n plus half of the number of ways to select exactly n.--RDBury (talk) 12:46, 13 August 2010 (UTC)

If I am reading this correctly, it is asking how many ways can you get >=90 when adding up four integers each with a value 0 to 45 inclusive. 2,271,181. -- kainaw™ 13:37, 13 August 2010 (UTC)

That's certainly quicker than counting up to 811,753,894,769,360,571,756,961,179,474,567,246,637,861,540,684,562,488. -- 117.47.211.213 (talk) 14:29, 13 August 2010 (UTC)

Two ways that I can see. Either do the work learning and revising or cheat. I'd advise the former. :) Dmcq (talk) 18:57, 13 August 2010 (UTC)

Perhaps the intended problem was: how many different combinations of scores are there (from the four sections) that constitute a passing grade? In other words, how many combinations of integers a, b, c, and d are there such that each is in the range [0,45] and a + b + c + d >= 90? -- Tom N (tcncv) talk/contrib 00:04, 19 August 2010 (UTC)

Unsolved problems

How much math would you have to learn before you could reasonably begin trying to solve the famous unsolved problems in math? I mean this both in terms of years of education and level of math (i.e., calculus, analysis, number theory, etc.) In theory you could try without knowing much beyond high school maths but you probably wouldn't succeed that way. 99.137.221.46 (talk) 14:49, 13 August 2010 (UTC)

A somewhat related question was asked a short while ago, and can be seen here. -- Meni Rosenfeld (talk) 15:29, 13 August 2010 (UTC)

Going by the history of famous problems that have been solved up to now, a lot, think post-doctorate. It would also seem to take a good deal of inborn talent and a certain amount of luck. Usually famous unsolved problems are famous because the greatest mathematical minds of the day have tried to solved them and failed.--RDBury (talk) 17:47, 13 August 2010 (UTC)

You may be interested in the (oft-repeated) story of George Dantzig, who, as a graduate student, solved two famously unsolved problems when he came into class late and mistook them for homework. -- 140.142.20.229 (talk) 01:52, 14 August 2010 (UTC)

Topological 'closeness'

I've learned an embarrassing amount of topology without really ever getting clear in my head some of the foundational concepts, so I thought it was time to (at least partially) remedy that. I'm hoping if I get these points cleared up, something will finally click, and I'll have a complete idea of how the subject works.

I understand the progression from metric spaces to topological spaces, since a lot of the proofs for metric spaces just generalise when you talk about 'open sets' which satisfy some of the same properties, but as soon as we drop the notion of distance and start dealing with abstract topological spaces, I sort of lose focus. I've heard the notion of 'closeness' talked about with topological spaces - obviously, with metric spaces, the distance function allows for a clear visualisation of this, but how do you talk about points x and y of a topological space as being 'close' (obviously relative to some standard of 'closeness'). Is it to do with the number of open sets in which both points appear together? Thanks, Icthyos (talk) 15:10, 13 August 2010 (UTC)

The basic relation of a topological space can indeed be seen as a closeness relation, but between points and sets, not between points and points: a point x is close to a set A if x is in the closure of A. Algebraist 15:14, 13 August 2010 (UTC)

(edit conflict)When you generalise to topology, you inter alia drop scale factors. A "very small" metric space (say, with a diametre of 0.001), and a "large" space (say, with the diametre 1000) certainly are different as metric spaces, but may be essentially equivalent (i. e., homeomorphic) as topological spaces. Say, that the first and the second space actually have the same underlying point set X, but that all distances are a million times larger with the metric (distance function) in the second space. In other words, say that the relation between the metrics only is a matter of a rescaling:

$d_{2}(x,y)=1000000d_{1}(x,y)$ for all x,y ∈ X.

Note, that the same subsets of X are open, if you choose the one or the other metric. Thus, indeed, the topology is the same in both cases.

However, assume that there are two points x and y in X, such that $d_{1}(x,y)=0.0001$. Are they "close" or not in the topology? 0.0001 is not much, but if we rescale, then we find that $d_{2}(x,y)=100$, which is a good bit more. And this was just using one scale factor; we could rescale by means of any positive real number, and get a different metric on X, with the same topology. Thus, the question whether just two points x and y are "close" or not is meaningless in an (ordinary) topological space.

As Algebraist just explained, there are a lot of other questions about closeness which are meaningful. If x is a point in X, and S is an (infinite) subset of X, then x is close to S, if there are elements in S arbitrarily close to x, with respect to any one of the aforementioned metrics. Meditate a bit over this; you ought to see that the scale factor doesn't matter for x being close to S in this sense. Arbitrarily close means that for any ε > 0 there should be a y ∈ S with $d_{*}(x,y)<\epsilon$; and you should see that you may counter a rescaling by a judicious choice of an "auxiliary" ε.

I hope this is to some help with your trouble with the intuition. Since this seems to have been your problem, I limited my arguments to such topologies as may be derived from metrics; in particular, these are Hausdorff. The nice thing with reformulation in terms of open sets (only), is that the results work not only independently of the metric, but also for topological spaces where no compatible metric at all exists. However, that may be a later step in abstraction... JoergenB (talk) 20:21, 13 August 2010 (UTC)

To elaborate slightly on this notion: a point x is close to a set A when there are points in the set arbitrarily close to x. What does arbitrarily close mean? In a metric space, it means we can get within any ε of x. In a topology, it means we can get within any open set containing x. Open sets determine what "arbitrarily close" means in a topology.

Note that the notion of relative closeness does not exist in Topology; one point cannot usually be said to be closer or further from x than another point (I may be wrong about this, but I think the only time you can really argue such a statement is if the space is disconnected, then points in a different connected component might be considered to be further than points in the same connected component, but even that isn't really true). Rather, we have a more global notion of points that are arbitrarily close to x. --COVIZAPIBETEFOKY (talk) 19:49, 13 August 2010 (UTC)

I think you could justifiable describe y as closer to x than z is close to x if any open set containing x and z also contains y and there is an open set containing y and x but not z. I can't think of any space where that holds, other than ones constructed specially for the purpose, though. --Tango (talk) 20:20, 13 August 2010 (UTC)

This is a nice idea from Tango. I can't think of a counterexample. Moreover, I can think of an example that goes against standard metric topology. Think of the real line, and the points x = 0, z = 1, and y = 2. With the metric topology, x is closer to z than it is to y, i.e. | x − z | < | x − y |. With the particular point topology, where the open sets (along with R and ∅) are the sets containing y. Every open set containing x and z must also contain y. But there is an open set containing x and y, namely {x,y}, that doesn't contain z. So in this topology, in Tango's sense, x is closer to y. — Fly by Night (talk) 00:20, 14 August 2010 (UTC)

Well, yes, that just proves that you can have completely different topologies on the same underlying set. The only thing the real line with the standard topology and your space have in common, really, is the cardinality of the underlying set. --Tango (talk) 00:44, 14 August 2010 (UTC)

It's even deeper than that. The underlying set, probably, has little to do with anything. Any set can be given many, many different topologies. All of which give that set very, very different properties. The question that we ought to ask ourselves is this: what is the cardinality of the topologies on a given set? — Fly by Night (talk) 01:07, 14 August 2010 (UTC)

Well, as far as topologies are concerned, a set is completely defined by its cardinality. That cardinality is very important, but other than that sets only differ by the names we give the elements. --Tango (talk) 02:11, 14 August 2010 (UTC)

Not quite. The topology of a set depends exactly upon its topolgy (I would worry if the topology of a set did not depend upon the set's topology). Take as a counterexample the real line with the trivial topology and the real line with the discrete topology. The real line and the real line have the same cardinality, but they are not homeomorphic when they carry these two topolgies. (Although, two sets with the trivial topology are homeomorphic if and only if they have the same cardinality.) — Fly by Night (talk) 03:20, 14 August 2010 (UTC)

I wasn't talking about topological spaces; I was talking about sets. You can define exactly the same topology on the real line as on the Euclidean place as on the complex plane as on the real projective line, etc., and get exactly the same topological space. All those sets are essentially the same because they all have the same cardinality. --Tango (talk) 09:27, 14 August 2010 (UTC)

OP, from my limited experience the whole point of using open sets instead of distance function is for the ease of it. It's much more time consuming to define metrics on sets, when all we do is talk about the open sets. So I think of R^n as a product space and not a metric space. Furthermore, some topologies are not metrizable (there's no distance function that give precisely the same open sets), for example the long line and the Zariski topology. Also, when you said "I've learned an embarrassing amount of topology without really ever getting clear in my head some of the foundational concepts", this is why you don't rush math. Money is tight (talk) 04:33, 14 August 2010 (UTC)

The original question by Icthyos seemingly concerned such topologies as may be derived from metric spaces. There are numerous other, and rather interesting ones, too. Consider the spectrum of a (commutative and unitary) ring R,

X = $\mathrm {Spec} \;R:=\{{\text{prime ideals in }}R\}$,

together with its Zariski topology. Note, that if p and q are elements in X, i. e., prime ideals in R, such that p is a proper subset of q, then every open set that contains q also will contain p, but not vice versa. This indicates that one should be a little careful in defining "closeness between two points" in a general topological space; beware of coming up with a definition, such that "p is close to q, but q is not close to p".

However, my advice to Icthyos is: Ignore non-metrisable topologies for the time being (but remember that the general topology you try to learn will cover more than the examples you choose to consider right now)! JoergenB (talk) 13:23, 14 August 2010 (UTC)

Thank you all for the detailed responses. I believe something has clicked! Icthyos (talk) 00:04, 15 August 2010 (UTC)

August 14

Cardinality of Topology

Given a set X, what is the cardinality of the set of subsets of X which form a topology for X? In other words: what is the cardinality of the topologies of X? — Fly by Night (talk) 01:07, 14 August 2010 (UTC)

P.S. It's not just the cardinality of the power set P(X). The topology whose open sets are all subsets is the discrete topology. So this topology counts as a single member of the set of topologies. — Fly by Night (talk) 01:22, 14 August 2010 (UTC)

The cardinality of P(P(X)) (which is $2^{2^{|X|}}$, with an appropriate interpretation of infinite exponents where necessary) is clearly an upper bound. A topology is an element of that power set of the power set of X. Some of those elements will not be valid topologies, of course. The question is what proportion actually are valid topologies. The cardinality of P(X) (ie. $2^{|X|}$) is an obvious lower bound (for any subset A of X, there is a distinct topology where a set is open iff it is empty or contains A). Where inbetween those bounds it actually is, I don't know. Some small finite X, we can just work it out by enumeration, of course. --Tango (talk) 02:25, 14 August 2010 (UTC)

Every filter is a topology. There are $2^{2^{|X|}}$ many filters (see above for a proof by Trovatore). —Preceding unsigned comment added by 203.97.79.114 (talk) 11:31, 14 August 2010 (UTC)

Thanks for the credit, but the proof is actually due to Pospíšil. --Trovatore (talk) 07:18, 16 August 2010 (UTC)

How can this hold for a finite set without implying that every collection of subsets is a topology? -- 1.47.99.181 (talk) 23:44, 14 August 2010 (UTC)

It doesn't hold for finite sets. The reason is in infinite sets "very sparse subsets" (like cantor set in the real numbers) can still have the same cardinality as the whole set Money is tight (talk) 04:29, 15 August 2010 (UTC)

Oh, to be picky, every filter is almost a topology. Every topology has to have the empty set as an element (that is, the empty set is open), whereas no nontrivial filter has the empty set as an element. But that's OK; just adjoin the empty set as an element, and you have a topology. That's fine for cardinality purposes (it gives you an injective map from nontrivial filters to topologies). --Trovatore (talk) 07:41, 16 August 2010 (UTC)

On finite X, topologies are in 1-1 correspondence with preorders, and counting them appears to be a difficult enumeration problem. See [1].—Emil J. 13:34, 16 August 2010 (UTC)

Rings, subfields and integral domains

Hi all - I'm working on the following problem and was looking for some guidance:

"Let R be a ring, and K a subring of R which is a field. Show that if R is an integral domain and $dim_{K}(R)<\infty$ then R is a field."

The problems I'm working on are from a course lectured at my university about 10 years ago, so I don't have an explicit definition of the notation, but I presume $dim_{K}(R)$ is the dimension of R as a vector space over K - however, my first query is, is it necessarily the case that if K and R are related as described above, then R can -definitely- be treated as a vector space over K? I don't need a proof or anything, I just want to check that's definitely the case.

Secondly and more pertinently, where should I start? I don't want a complete answer and indeed I'd rather not have one, I want to work through it myself - but where to begin? I considered writing an n-dimensional basis for R; $e_{1},\,e_{2},...$ and then expanding a general $r\in R$ as $r=\sum _{1}^{n}r_{i}e_{i},\,r_{i}\in K$, then finding a formula for the inverse, but I'm not sure if that's feasible or even a sensible way to go about the problem. Could anyone suggest a place to start?

Many thanks, 86.30.204.236 (talk) 19:45, 14 August 2010 (UTC)

You're over thinking the proof, Let r be in R and consider the set {1, r, r², ...}. R is finite dimensional so this set can't be linearly independent and there is a relation of the form a + br + cr² + ... + krⁿ. Assume n is minimal in this relation. Then divide by r once to get an expression for 1/r in R.--RDBury (talk) 21:46, 14 August 2010 (UTC)

Forgive me, but I'm not convinced that proof is entirely rigorous; I don't think you can just "divide by r", as you put it! In "dividing by r" you are multiplying by the inverse of r, which is the very thing you are trying to prove exists. This is pedantic, but instead you should state that as K is a field, you can take WLOG the constant term to be the multiplicative identity, then factorise to get something of the form 1 = r(b + cr + ... + kr²) to show that r has such an inverse before using it. Also, worth adding that the proof fails when R is not an ID as {1, r, r², ...} need not be an infinite set.--86.165.252.185 (talk) 21:30, 18 September 2010 (UTC)

Ah, I have a habit of going in over the top, whoops! Thankyou very much, that's great 86.30.204.236 (talk) 23:53, 14 August 2010 (UTC)

That R is a vector space over K is very simple too. A vector space is just an additive group with multiplication by elements of the underlying field. Since R is a ring, it is an additive group, and obviously you can multiple elements of R by elements of a subring of R. We don't immediately have a basis for R over K, but RDBury's proof doesn't require a basis. I'm not sure if there is any general form for a basis of a vector space of that type - does anyone know of one? --Tango (talk) 20:13, 15 August 2010 (UTC)

Well for the finite dimensional case, pick any element e₁ in R for your first basis element, and then pick any element e₂ not in Ke₁ for your second one, then pick any element not in the span of e₁ and e₂ for the third one, etc. Rckrone (talk) 06:07, 16 August 2010 (UTC)

Well, yes, that's the standard algorithm for finding a basis of any finite dimensional vector space. I was wondering if there was any more definite result for the special case of a ring over a subring that is a field. --Tango (talk) 14:13, 16 August 2010 (UTC)

There is no general way of describing the basis explicitly if the dimension is infinite. Think cases like R = R, K = Q. Any such basis is nonmeasurable, and cannot be proven to exist without the axiom of choice.—Emil J. 13:30, 16 August 2010 (UTC)

August 15

Math Child Prodigies of 21st Century

Can we systematically list and rank the math child prodigies of 2010 or of the 21st century?For some odd reason, we don't anymore hear of math prodigies,the lsat being Terence Tao about 30-40 years ago.Why has the waterfall of child prodigies stopped?Has math become less popular.If it hasn't,and if there's concrete evidence,I'd like a list please of all math child prodigies.Thanks guys.

The future of math is in the hands of child prodigies and we need to start appreciating that IMO.So who's the greatest child prodigy right now?BTW,Typsetting Bug!And can we document any unsuccessful child prodigies?I'd like to hear examples. —Preceding unsigned comment added by 110.20.19.24 (talk) 01:43, 15 August 2010 (UTC)

Mathematics is not in the hands of child prodigies. Sure Terry Tao was a prodigy and one of the best mathematicians alive. But you only hear the successful child prodigies, because many of the so called "prodigies" just work very hard and appear to be very knowledge while still in their teens, and are unable to produce results later on. Money is tight (talk) 04:34, 15 August 2010 (UTC)

Even Tao wasn't doing original research from an early age, from what I can tell. He got his PhD aged 20, which is significantly earlier than most, but only by 4 or 5 years, and his Wikipedia article doesn't mention any papers he published before his PhD. Professional mathematics (in academia) is all about original research (and teaching). You get child prodigies that are brilliant and learning and doing existing mathematics, but hold no particular talent for original research. Maths is in the hands of people that do lots of great original research and starting 4 or 5 years earlier isn't going to make much difference to how much you can do in a career. So, it's the quality of the mathematician, not their speed of development, that matters. --Tango (talk) 20:34, 15 August 2010 (UTC)

110.20.19.24 (aka 110.20.18.61), why are you again changing the IP address in SineBot's sig of your initial post, replacing the first octet with 119? Do you now wish you were in China? (I'd be happy to be in NSW.) -- 115.67.79.109 (talk) 06:26, 15 August 2010 (UTC)

No.There's a annoying typsetting bug on my computer that's making all these changes.As you can see,my punctuation is typed like I'm a 5 year old.It's annoying but I can't do anything about it.But thanks for correcting it for me.

My I suggest registering a username if you wish to hide your ip? Doing things like you did draw more attention than doing nothing at all. There is a List of child prodigies but I wouldn't think it is very complete and personally I can't see much point in the list. Dmcq (talk) 09:22, 15 August 2010 (UTC)

I don't wish to hide my fucking IP!Sheesh,at least fucking ask me before jumping to damn conclusions as the IP above politely did.

The changes you made were explicitly to change the ip. You gave misleading edit comments each time you did it. It had nothing to do with a typesetting bug as the only other thing you did was stick in a new line between two paragraphs. Please do not start saying 'fucking' or 'damn' about reasonable advice or I will remove the question on WP:CIVIL grounds. Dmcq (talk) 09:55, 15 August 2010 (UTC)

Can you explain exactly what the typesetting bug does? The only problem I see with your punctuation is that there are no spaces after punctuation marks. -- Meni Rosenfeld (talk) 12:21, 15 August 2010 (UTC)

Yes I was wondering about that. I guess they must be using a mobile phone with some textbox input method which suppresses the space for some reason or they go onto a new line and it removes the space at the end of a line. In which case perhaps not going onto a new line or sticking a space at the beginning of the new line would help. I'd be interested in who make the browser. Dmcq (talk) 13:06, 15 August 2010 (UTC)

I have heard of Lenhard Ng (I saw him called Lenny Ng). I heard he was the youngest to ever score an 800 on the math portion of the SAT and also that he went on to be a 3 time Putnam fellow (and he only had 3 years of college (4 time Putnam fellows are weird)). And, the article mentions many other feats including 4 time perfect on AHSME and 2 time gold medalist/1 time silver medalist in IMO. Also, I found a web page with a list of his publications and he has quite a few, though not an insane number or anything. And, number of publications is not necessarily that important. Quality is also important. I don't know quality/difficulty of his work so I can't say anything. I don't know if it's groundbreaking or anything. StatisticsMan (talk) 02:29, 17 August 2010 (UTC)

According to our article on him, he got his PhD aged 25, which is a normal age. So, he doesn't seem to have started research early either. --Tango (talk) 07:00, 18 August 2010 (UTC)

25 is really pretty early. Generally you get your Bachelor's at age 22, and five years is pretty fast for a math or pure-science PhD in the States (engineers sometimes get it done a little faster). Certainly 25 is not exceptionally early, but it's definitely fast. --Trovatore (talk) 09:10, 19 August 2010 (UTC)

Partial fractions

Hi,

I'm having difficulty in solving (to me) a fiendish partial fraction:-

(8x^3 - 22x^2 + 6x - 4) / (x - 2)^2 (x^2 + x +2)

whatever I try, I end up with multiple unknowns - I've found 'B', but after that I end up with 'A', 'C' or 'D' in the same equation. Is it a case of solving the later parts simultaneously?

Thanks for any helpVs60t (talk) 10:08, 15 August 2010 (UTC)

Yes, solving a system of linear equations is the normal way to find partial fractions. It's certainly easier than going out of your way to try avoid it. -- Meni Rosenfeld (talk) 11:28, 15 August 2010 (UTC)

I try to avoid using that method. I call that method the "stupid high school method" :) . My preferred method works by finding the poles of the function in the complex plane. Then you compute the truncated Laurent expansion fp(x) around each pole p that contains precisely all the singular terms and none of the regular terms. The sum of all the fp(x) is then the partial fraction expansion, if the degree of the numerator is smaller than the dgree of the denominator. If that's not the case, you can simply perform a long division in the usual way. Note that doing that amounts to adding the singular terms from the expansion around infinity to the sum of the fp, so this fits in nicely with the picture that a partial fraction expansion is simply the sum of all the singular parts of all expansions around all the singular points.

Proof. If f(x) is a rational function we want to expand in patial fractions, then g(x) = f(x) - sum of fp(x) is also a rational function, but without any singularities. Hence it is a polynomial. If the degree of the numerator is less than the degree of the denominator g(x) will tend to zero for x to infinity, so g(x) must be identical to zero. If the degree of the numerator is larger than or equal to that of the nhumerator, we can compute the polynomial g(x) by expanding f(x) around infinity (i.e. in powers of w = 1/x). Only the coefficients of the negative powers of the expansion parameter w will be nonzero (otherwise g(x) would not be a polynomial). Count Iblis (talk) 14:56, 15 August 2010 (UTC)

Would you like to show us how to compute the principal part of the Laurent series at x = ½(–1 ± i√7), without using Matlab, Mathematica, Maple, or any other computer algebra package? I just tried and I ran out of paper very quickly. — Fly by Night (talk) 18:54, 15 August 2010 (UTC)

We can write the function as:

$f(x)={\frac {8x^{3}-22x^{2}+6x-4}{(x-2)^{2}(x-a)(x-b)}}\,$

with

${\begin{aligned}a&={\frac {1}{2}}\left(-1+i{\sqrt {7}}\right)\\b&={\frac {1}{2}}\left(-1-i{\sqrt {7}}\right)\end{aligned}}\,$

The principal part of the Laurent expansion at x = a is:

$f(x)=\left({\frac {8a^{3}-22a^{2}+6a-4}{(a-2)^{2}(a-b)}}\right){\frac {1}{x-a}}\,$

The question is then how to simplify the coefficient of 1/(x-a) efficiently. A simple method is to exploit the fact that since:

$a^{2}+a+2=0\,$

you can reduce all algebraic expressions Modulo(a^2 + a + 2).

So, we can write:

$8a^{3}-22a^{2}+6a-4=-30a^{2}-10a-4=20a+56$

And

$(a-2)^{2}=a^{2}-4a+4=-5a+2$

With these simplifications it is now trivial to insert the value for a in the expression and to simply it to get the square roots out of the numerator.

Then the expansion around x = b is simply the complex conjugate of the above expansion (taking x real), so we don't have to do that expansion. Count Iblis (talk) 21:21, 15 August 2010 (UTC)

OP - thanks for the help and adviceVs60t (talk) 11:44, 16 August 2010 (UTC)

Rearranging a formula - getting confused!

Hi folks,

How can you get from:

${n[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times n[{\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}]}$

to here:

${\sqrt {[n(\sum _{i=1}^{n}{X_{i}}^{2})-(\sum _{i=1}^{n}{X_{i}})^{2}]\times [n(\sum _{i=1}^{n}{Y_{i}}^{2})-(\sum _{i=1}^{n}{Y_{i}})^{2}]}}$

I'm stuck! One thing I know is that:

${\bar {X}}={{\sum _{i=1}^{n}{X_{i}}} \over n}$

and similarly:

${\bar {Y}}={{\sum _{i=1}^{n}{Y_{i}}} \over n}$

Anyone have any suggestions? - 114.76.235.170 (talk) 12:59, 15 August 2010 (UTC)

Should have noted that I've tried the following:

${n[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times n[{\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}]}$

$={n[{\sqrt {\sum _{i=1}^{n}(X_{i}-{{\sum _{i=1}^{n}{X_{i}}} \over n})^{2}}}]\times n[{\sqrt {\sum _{i=1}^{n}(Y_{i}-{{\sum _{i=1}^{n}{Y_{i}}} \over n})^{2}}}]}$

But I totally got stuck - I tried expanding the formula to see if I can see a pattern, and it's just not happening for me :( 114.76.235.170 (talk) 13:03, 15 August 2010 (UTC)

Did you get as far as

$\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}=\sum _{i=1}^{n}{X_{i}^{2}}-\sum _{i=1}^{n}{{\bar {X}}^{2}}$

Dmcq (talk) 13:28, 15 August 2010 (UTC)

By the way I think the n should be squared going under the square root, are you sure you haven't copied something wrong about n? Dmcq (talk) 13:39, 15 August 2010 (UTC)

That will only be half a fix, note there is another summand. -- Meni Rosenfeld (talk) 13:47, 15 August 2010 (UTC)

There's a small error in your problem statement; it should have been

${{\sqrt {n\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}\times {\sqrt {n\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}}.$

This equality actually comes from the equality

$n\sum _{i}(X_{i}-{\bar {X}})^{2}=n\sum _{i}X_{i}^{2}-\left(\sum _{i}X_{i}\right)^{2}$

applied once for X and once for Y. It is equivalent to a useful way to compute variance, $\mathbb {V} [X]=\mathbb {E} [X^{2}]-\mathbb {E} [X]^{2}$. To prove it, start with

$n\sum _{i}(X_{i}-{\bar {X}})^{2}=n\sum _{i}(X_{i}^{2}-2X_{i}{\bar {X}}+{\bar {X}}^{2}),$

And remember what we taught you about terms that don't depend on i. -- Meni Rosenfeld (talk) 13:45, 15 August 2010 (UTC)

By the way, I see that in Talk:Pearson product-moment correlation coefficient you managed to transform the numerator. You have a few mistakes there and you wrote it in a way more complicated than necessary, but you got the main ideas right, and what you have here is very similar.

Another hint: Writing ${\bar {X}}$ is easier than ${{\sum _{i=1}^{n}{X_{i}}} \over n}$, so you'll want to turn the latter into the former, rather than the other way around, as long as possible. -- Meni Rosenfeld (talk) 13:57, 15 August 2010 (UTC)

Whoa, you are both correct! I made an error in the way I copied in the TeX... sorry, I'm really new to this! You are both so excellent, Meni, let me read and digest your answer(s) :-) And yeah, I'm trying to understand how to get from one forumla to another from the Pearson product-moment correlation coefficient :-)

One day I hope to be able to contribute to Wikipedia articles, or even to this reference desk :-) For now, it's a journey of discovery...

Incidentally, I'm not doing a Uni/College course - I'm trying to (slowly) teach myself statistics. It's heavy going... - 114.76.235.170 (talk) 14:01, 15 August 2010 (UTC)

Another attempt... still a bit stuck

OK, I'm trying again... this time I've gotten quite a bit further, but still can't quite get to the next stage...

The goal

Prove that:

$r={n(\sum {{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}}) \over {\sqrt {[n(\sum _{i=1}^{n}{X_{i}}^{2})-(\sum _{i=1}^{n}{X_{i}})^{2}]\times [n(\sum _{i=1}^{n}{Y_{i}}^{2})-(\sum _{i=1}^{n}{Y_{i}})^{2}]}}}$

is the same as:

$r={\frac {\sum _{i=1}^{n}(X_{i}-{\bar {X}})(Y_{i}-{\bar {Y}})}{{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}}}$

Proof

OK, so the formula here is:

$r={\frac {\sum _{i=1}^{n}(X_{i}-{\bar {X}})(Y_{i}-{\bar {Y}})}{{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}}}$

Now if you multiple the numerator and the denominator by n then this gives you:

$r={\frac {n({\sum _{i=1}^{n}(X_{i}-{\bar {X}})(Y_{i}-{\bar {Y}})})}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

Then you expand the numerator:

$r={\frac {n({\sum _{i=1}^{n}({X_{i}}{Y_{i}}-{\bar {X}}{Y_{i}}-{\bar {Y}}{X_{i}}-{\bar {X}}{\bar {Y}}}))}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

Which is the same as:

$r={\frac {n({\sum _{i=1}^{n}({X_{i}}{Y_{i}}-({\sum _{i=1}^{n}X_{i} \over n})\times {Y_{i}}-({\sum _{i=1}^{n}Y_{i} \over n})\times {X_{i}}-({\sum _{i=1}^{n}X_{i} \over n})({\sum _{i=1}^{n}Y_{i} \over n})}))}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

Which is:

$r={\frac {n({\sum _{i=1}^{n}({X_{i}}{Y_{i}}-({\sum _{i=1}^{n}{X_{i}}{Y_{i}} \over n})-({\sum _{i=1}^{n}{Y_{i}}{X_{i}} \over n})-({\sum _{i=1}^{n}X_{i} \over n})({\sum _{i=1}^{n}Y_{i} \over n})}))}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

This cancels a term...

$r={\frac {n({\sum _{i=1}^{n}({X_{i}}{Y_{i}}-({\sum _{i=1}^{n}X_{i} \over n})({\sum _{i=1}^{n}Y_{i} \over n})}))}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

Leading to the simplified numerator...

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

Now Meni suggests that I don't expand the X-bar to sum notation... so I'll try it without. It leads me to:

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}{X_{i}}^{2}-2.{X_{i}}.{\bar {X}}+{\bar {X}}^{2}}}\times {\sqrt {\sum _{i=1}^{n}{Y_{i}}^{2}-{Y_{i}}.{\bar {Y}}+{\bar {Y}}^{2}}}\right]}}$

Well... I think I do need to expand the X-bar and Y-bar now.

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}\left({X_{i}}^{2}-2.{X_{i}}.{{\sum _{i=1}^{n}{X_{i}}} \over n}+{{\sum _{i=1}^{n}{X_{i}}} \over n}^{2}\right)}}\times {\sqrt {\sum _{i=1}^{n}\left({Y_{i}}^{2}-2.{Y_{i}}.{{\sum _{i=1}^{n}{Y_{i}}} \over n}+{{\sum _{i=1}^{n}{Y_{i}}} \over n}^{2}\right)}}\right]}}$

Which is:

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}\left({X_{i}}^{2}-{{\sum _{i=1}^{n}2.{X_{i}}^{2}} \over n}+{{\sum _{i=1}^{n}{X_{i}}} \over n}^{2}\right)}}\times {\sqrt {\sum _{i=1}^{n}\left({Y_{i}}^{2}-{{\sum _{i=1}^{n}2.{Y_{i}}^{2}} \over n}+{{\sum _{i=1}^{n}{Y_{i}}} \over n}^{2}\right)}}\right]}}$

This leads to:

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}\left(n.{X_{i}}^{2}-{\sum _{i=1}^{n}2.{X_{i}}^{2}}+{\sum _{i=1}^{n}{X_{i}}}^{2}\right)}}\times {\sqrt {\sum _{i=1}^{n}\left(n.{Y_{i}}^{2}-{\sum _{i=1}^{n}2.{Y_{i}}^{2}}+{\sum _{i=1}^{n}{Y_{i}}}^{2}\right)}}\right]}}$

or also:

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}\left(n.{X_{i}}^{2}-{\sum _{i=1}^{n}{X_{i}}}^{2}\right)}}\times {\sqrt {\sum _{i=1}^{n}\left(n.{Y_{i}}^{2}-{\sum _{i=1}^{n}{Y_{i}}}^{2}\right)}}\right]}}$

Am I on the right track? Seems to be that way... any help getting to the next step would be appreciated. :-) 114.76.235.170 (talk) 15:32, 15 August 2010 (UTC)

No, you are on wrong track. After expanding the numerator (correcting the sign error) you should notice that

$\sum _{i=1}^{n}({X_{i}}{Y_{i}}-{\bar {X}}{Y_{i}}-{\bar {Y}}{X_{i}}+{\bar {X}}{\bar {Y}})$

$=(\sum _{i=1}^{n}{X_{i}}{Y_{i}})-(\sum _{i=1}^{n}{\bar {X}}{Y_{i}})-(\sum _{i=1}^{n}{\bar {Y}}{X_{i}})+(\sum _{i=1}^{n}{\bar {X}}{\bar {Y}})$

$=(\sum _{i=1}^{n}{X_{i}}{Y_{i}})-{\bar {X}}(\sum _{i=1}^{n}{Y_{i}})-{\bar {Y}}(\sum _{i=1}^{n}{X_{i}})+{\bar {X}}{\bar {Y}}(\sum _{i=1}^{n}1)$

$=(\sum _{i=1}^{n}{X_{i}}{Y_{i}})-{\bar {X}}(n{\bar {Y}})-{\bar {Y}}(n{\bar {X}})+{\bar {X}}{\bar {Y}}(n)$

$=(\sum _{i=1}^{n}{X_{i}}{Y_{i}})-n{\bar {X}}{\bar {Y}}$

Bo Jacoby (talk) 06:30, 16 August 2010 (UTC).

When you add a new sum, try giving it a new index (i.e. j, k, l) so that you are sure to to confuse the i in the outer sum with the i in the inner sum. You can not mix them and this mistake appears several times in your algebra. 0¹⁸ (talk) 17:35, 16 August 2010 (UTC)

Pivoting

Is the pivoting operation in the simplex algorithm the same as the pivoting operation in Gauss–Jordan elimination? Yaris678 (talk) 21:33, 15 August 2010 (UTC)

Nearly. In terms of the actual row operations they are the same. In Gauss-Jordan you're trying to write all the variables in terms of a subset of them but in simplex you already have that and are making a new choice of which subset is to be used. This might make a difference in how you store the matrices when you implement the two algorithms.--RDBury (talk) 12:10, 16 August 2010 (UTC)

Brilliant explanation. Thanks. Yaris678 (talk) 05:28, 17 August 2010 (UTC)

August 16

trying to compare homicides

I need some raw data to try some different things and comparisons...

How many murders have been committed in Minneapolis, Minnesota in year 2010?

If available the same data for Hennepin County and the state as a whole

How many murders have been committed in the Greater London Area (boroughs plus The Square Mile) in the year 2010?

How many in England & Wales?

I am really curious as to WHERE to find this data so I can directly do my own research in the future. My intent is to play around with the population and educational level variables and just experiment with this rather depressing data. Thanks I have a reference question (talk) 05:25, 16 August 2010 (UTC)

The miscellaneous reference desk would be better for a question like that. Also have you tried searching using Google or a suchlike search engine? Dmcq (talk) 08:37, 16 August 2010 (UTC)

Crime statistics for Greater London are available from the Metropolitan Police here - for example, there were 125 homicides (murder, manslaughter, corporate manslaughter and infanticide) in Greater London in the 12 months to July 2010. Crime statistics for England and Wales are available from the Home Office - their report on Crime in England and Wales 2009/2010 is available here. Gandalf61 (talk) 12:39, 16 August 2010 (UTC)

Laurent Series for ln(z) in annulus around z=0?

Hello, is there a laurent series expanded around z=0 for ln(z)? Thanks.Rich (talk) 11:54, 16 August 2010 (UTC)

ln(z) around 0 is multi valued or discontinuous, while laurent series are single valued and continuous. So no. Bo Jacoby (talk) 12:58, 16 August 2010 (UTC).

Hmm thanks that's interesting. Does that mean ln(z) has an essential singularity at 0?-Rich Peterson24.7.28.186 (talk) 13:36, 17 August 2010 (UTC)

No, because that would require the function to be holomorphic in a neighbourhood of 0 minus 0 itself, and as Bo explained, this condition fails for whatever branch of ln(z) you choose. If the function had an essential singularity in 0, it would also have a Laurent series around 0.—Emil J. 13:43, 17 August 2010 (UTC)

This type of singularity is called a "branch point singularity". Note that functions with such a singularity at some point can have well defined values at that point. The word "singularity" refers to non-analytic behavior. So, e.g. the function z^p for positive non-integer p has a branch point singularity at z = 0, even though the function is well defined at z = 0. Count Iblis (talk) 14:43, 17 August 2010 (UTC)

We have an article titled branch point (I haven't looked at it recently, so I can't promise anything about what it says. Michael Hardy (talk) 15:02, 17 August 2010 (UTC)

Thanks everyone.Rich (talk) 02:46, 19 August 2010 (UTC)

SO(n) vs Spin group

I think I understand this discussion of why SO(3) is not simply connected. I gather that Spin(3) is simply connected, but I'm curious what this exactly means. As I understand it, it should be possible, with any simply connected manifold, to continuously deform any path between two points on the manifold to any other path with the same end points. So, suppose I represent a rotation changing over time using Slerp. Suppose further that I use Slerp, or an extension like Squad to mediate a rotation of $2\pi$ about some axis. Is it possible to continuously "deform" this rotation until it represents no rotation over time whatsoever? Thanks.--Leon (talk) 13:51, 16 August 2010 (UTC)

Slerp does not explain why the circle S¹ is not simply connected but the spheres Sⁿ for n>1 are simply connected. Note that Spin(3) is the 3-sphere. The main point you should try to understand is the difference between the circle and the 2-sphere with regard to simple connectivity. Tkuvho (talk) 14:23, 16 August 2010 (UTC)

I'm not too clear on what you mean. And I'm not sure why the circle S¹ is relevant.--Leon (talk) 14:31, 16 August 2010 (UTC)

Thinking in terms of Slerp, what would be your guess concerning the circle: does Slerp suggest it is simply connected, or not? Tkuvho (talk) 14:39, 16 August 2010 (UTC)

Well I believe the answer is "no", but I'm still not entirely sure what you mean.

What I'm after is an intuitive geometrical understanding of the difference between the two groups when used to represent rotations in $\mathbb {R} ^{3}$, and for a theoretical engineering application. Also, does the statement

"Surprisingly, if you run through the path twice, i.e., from north pole down to south pole and back to the north pole so that φ runs from 0 to 4π, you get a closed loop which can be shrunk to a single point"

mean that if I were to construct a rotation of $4\pi$ about some axis, I could continuously deform such a rotation to no rotation at all?--Leon (talk) 14:48, 16 August 2010 (UTC)

Exactly. More precisely, you can deform such a family of rotations, to the trivial (i.e. constant) family. You are unlikely to be able to do that with something as canonical as the Slerp deformations, but perhaps you could. Tkuvho (talk) 15:01, 16 August 2010 (UTC)

To summarize what's involved: rotations of 3-space can be represented by unit quaternions q. Two "opposite" quaternions, q and -q, represent the same rotation of 3-space. Hence the space of rotations is the quotient of the unit quaternions by the antipodal map. Thus, while unit quaternions form a 3-sphere (simply-connected), the rotations form a manifold which is its antipodal quotient, and therefore non-simply connected. Tkuvho (talk) 15:05, 16 August 2010 (UTC)

Cool, thanks. But, with regards to the earlier bit, is there any "intuitive, geometrical" difference between the two groups when used to represent rotations? To me it seems that the significance is one can continuously deform a path described by elements of Spin(3) to any other such path, keeping the same end points the same, and this isn't true of SO(3). As a $2\pi$ rotation in Spin(3) is not an example of an element going to itself (quaternions: 1->i->-1 is a $2\pi$ rotation but it isn't an element travelling some path to end where it started) it couldn't be contracted to a point in any case. Or am I missing something...--Leon (talk) 15:13, 16 August 2010 (UTC)

The path you have described illustrates the point very nicely. As you mentioned, the path is not closed. However, under the antipodal quotient, it does descend to a closed path in the quotient space, namely SO(3)=RP³. This is because 1 and -1 define the same rotation, namely the trivial one. By general covering space theory, whenever a non-closed path descends to a loop, the resulting loop cannot be contracted to a point. Tkuvho (talk) 15:18, 16 August 2010 (UTC)

Really, I struck it out because I felt like an idiot for writing it! In any case, thanks for your help.--Leon (talk) 15:21, 16 August 2010 (UTC)

Some notions in topology

Hi, Does anybody know hwat is the meaning of the $\pi -weight$ and $\pi -base$ ? —Preceding unsigned comment added by 212.199.96.133 (talk) 13:56, 16 August 2010 (UTC) Topologia clalit (talk) 13:58, 16 August 2010 (UTC)

A π-base is a collection B of nonempty open sets such that every nonempty open set contains some member of B. I guess that the π-weight of a space is the cardinality of its smallest π-base.—Emil J. 14:01, 16 August 2010 (UTC)

Since it may not be immediately obvious what is the difference between a base and a π-base, here's some basic info: any base is also a π-base, but not vice versa. Consider the Sorgenfrey line S. Since every nonempty basic set in S contains a nonempty open set in the standard topology of the reals, the set {(p,q): p,q in Q, p < q} is a π-base of S, and in particular, S has countable π-weight. However, this set is not a base. In fact, if B is any base of S, then for each real a there is a set U_a in B such that $a\in U_{a}\subseteq [a,a+1)$. Since this makes a = min(U_a), all these sets have to be distinct, hence S has weight $2^{\aleph _{0}}$.—Emil J. 15:10, 16 August 2010 (UTC)

Thanks! Topologia clalit (talk) 07:19, 18 August 2010 (UTC)

correlation conundrum

Hi there - I hope someone can help me out - it's been a long time since college stats classes...

I have five business groups, for each one I have a performance metric (a percentage score). I am looking for factors which might explain that metric (I understand that correlation <> causation!)

The groups have offices (between 20 and 90, each group has a different number), they have revenue (varies with each group), and the offices are in different locations. Some of the offices are in locations that are on a list of 20 places that are problematic - we'll call this the bad location list. All the metrics are for the group as a whole, not the individual offices.

OK, so, number of offices correlates well with performance - r2 - .87. Percentage of offices that are on the list correlates well (but negatively) about -.8. So far so good - the more locations, the better, and the fewer of them that are on the bad list the better. What I'm worried about is that the larger the number of offices, the lower the percentage on the bad list CAN be - that is to say there are only 20 'bad' locations, so if you have more than 20 office locations you are automatically going to have a lower percentage on the bad list - how can I deal with this? I want to disagregate the effect of size per-se from the effect of bad locations.

Thanks so much! —Preceding unsigned comment added by 97.115.64.81 (talk) 14:54, 16 August 2010 (UTC)

5 data points is very few, so whatever results you'll find will probably be highly insignificant. When there's more data, one thing you can do is to look at all points where the total number of offices is some fixed value, and see how the performance varies with number of bad offices.

Perhaps you can use some form of ANOVA where you do multiple regression for various choices of variables and see which contribute the most to explaining the variance. If you use a baseline model using only the total number of offices, and a second model including both that and the number of bad offices, then a large difference in the RSS between them will tell you that the number of bad offices indeed correlates with performance.

This is of course assuming that a linear model is at all appropriate. -- Meni Rosenfeld (talk) 07:40, 19 August 2010 (UTC)

Identifying Multiples of low primes

It's easy to identify multiples of 2 (last digit even), 3 (digits sum to 3 or a multiple of 3), 5 (last digit 0 or 5) and 11 (alternate digits sum to 11). I seem to remember that there's also a trick to spotting multiples of 7, but I can't remember what it is. Can anyone help? Thanks Rojomoke (talk) 16:26, 16 August 2010 (UTC)

Divisibility rule#Divisibility by 7. Algebraist 16:38, 16 August 2010 (UTC)

The powers of 10 are 1, 10, 100, 1000 and so on. Modulo 7 these powers are 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, and so on. So the rule is: if the ones plus thrice the tenths plus twice the hundreds minus the thousands minus thrice the tenthousands minus twice the hundredthousands and so on, is a multiple of seven, then so is the original number. Bo Jacoby (talk) 16:52, 16 August 2010 (UTC).

If you do the same thing as for 11 except with thousands you can make the checks for 7, 11 and 13 easier. For instance with 134,768,345,558 add 134+345 and 768+558 to give 479 and 1326. Subtract the smaller from the larger to give 847 and then just check this number 847 instead of the original. It is divisible by 7 but not by 11 or 13 so the original number is divisible by 7 but not 11 or 13. This is because 1001 = 7.11.13 Dmcq (talk) 19:12, 16 August 2010 (UTC)

I see there is an article about all this Divisibility rule Dmcq (talk) 19:18, 16 August 2010 (UTC)

August 17

Starting values for Root-finding algorithms (polynomials)

Are there any algorithms for finding appropriate starting values for algorithms such as Brent's method? I think what would be needed would be 1. some approximation of the highest and lowest possible roots, and 2. what resolution to search at to make sure you aren't skipping any roots. —Preceding unsigned comment added by 70.162.15.58 (talk) 00:54, 17 August 2010 (UTC)

In general, there is no easy way to do this, but in case of polynomial roots, you can always use Sturm's theorem. (Igny (talk) 03:14, 17 August 2010 (UTC))

See Durand-Kerner method. Bo Jacoby (talk) 14:59, 17 August 2010 (UTC).

integral of sinx cosx

I am exposing my ignorance of trig here... When I do u substitution with u = sinx I get (1/2)(sinx)^2 but my ti89 gives me -(1/2)(cosx)^2. What am I missing here? are these two results equivalent? -- 99.20.118.197 (talk) 04:02, 17 August 2010 (UTC)

What happens when you do the substitution u = cosx? What is sin²x + cos²x (the most important trig identity)? How does C (the constant of integration) fit into this? -- 111.84.234.215 (talk) 05:05, 17 August 2010 (UTC)

The easiest check is to differentiate both of them and with both of them you end up with $\sin {x}\cos {x}$. We can therefore safely say they are both integrals of $\sin {x}\cos {x}$. Are they equivalent? Let's plug in some numbers. ${\frac {1}{2}}\sin ^{2}{0}=0$.   $-{\frac {1}{2}}\cos ^{2}{0}=-{\frac {1}{2}}$. So clearly they are not equivalent. Look at the coefficient of integration next and see if you spot anything. Readro (talk) 08:21, 17 August 2010 (UTC)

The integral can be looked at in a number of different ways.

• Substitute u = sin x, with (du/dx) dx = cos x dx:

$\int \sin x\cos xdx=\int udu={\frac {1}{2}}u^{2}+C={\frac {1}{2}}\sin ^{2}x+C.$

• Substitute u = cos x, with (du/dx) dx = -sin x dx:

$\int \sin x\cos xdx=-\int udu=-{\frac {1}{2}}u^{2}+C=-{\frac {1}{2}}\cos ^{2}x+C=-{\frac {1}{2}}(1-\sin ^{2}x)+C={\frac {1}{2}}\sin ^{2}x-{\frac {1}{2}}+C.$

• Multiply by 2, take 0.5 outside the integral, and convert to sin 2x:

$\int \sin x\cos xdx={\frac {1}{2}}\int 2\sin x\cos xdx={\frac {1}{2}}\int \sin 2xdx={\frac {1}{2}}(-{\frac {1}{2}}\cos 2x)+C={\frac {1}{4}}(2\sin ^{2}x-1)+C={\frac {1}{2}}\sin ^{2}x-{\frac {1}{4}}+C.$

However you look at it, though, the antiderivative is the same up to a constant – and this difference is accounted for by the constant of integration C. —Anonymous Dissident^Talk 10:39, 17 August 2010 (UTC)

Can't somebody mention the trigonometric identity that says

$\sin x\cos x={\frac {\sin(2x)}{2}}$?

Michael Hardy (talk) 14:57, 17 August 2010 (UTC)

OK, I see someone did mention that. But why wasn't it the first thing mentioned here? To anyone who knows trigonometry, that would be the first thing that comes to mind. Michael Hardy (talk) 14:59, 17 August 2010 (UTC)

Probably because this isn't what the OP asked. He asked how to reconcile his (correct) computation for the integral with the output of the calculator. -- Meni Rosenfeld (talk) 17:50, 17 August 2010 (UTC)

Thanks alot everyone. I worked out your examples on paper and read the article on C. Before I posted my question I did get the feeling that I could do some pythagorean substitution to change sine to cosine ... but the piece of the puzzle I was missing was an understanding of +C. Thanks again -- 99.20.118.197 (talk) 15:31, 17 August 2010 (UTC)

Simple math problem

Hi. The rate of my dorm for 38 weeks is 1900, so it gives 50 every week (it's even written on the page that it is 50). I have to pay each month 237,5. On the average, a month has 30,42 days, which gives a month 4,35 weeks. 4,35 weeks times 50 is 217,5. Where are those 20?! (I didn't give the currency, but I think it's ok). 83.31.113.33 (talk) 14:05, 17 August 2010 (UTC)

It looks like you are right. Either you're paying for some extra service, there's an additional tax, there's a mistake in their billing or published numbers, or they're just ripping you off. -- Meni Rosenfeld (talk) 14:31, 17 August 2010 (UTC)

237.5 is exactly one eighth of 1900. 38 weeks is just under 9 months. So maybe it is simpler for your bank and whoever is letting the dorm to process 8 equal monthly payments than it would be to process 8 equal payments and then an odd amount for the "short" month at the end. Sensible thing is to ask whoever is letting the dorm to clarify how many monthly payments you will be making. Gandalf61 (talk) 14:40, 17 August 2010 (UTC)

Hm, I think you are right. If anyone wants to check their information, it's here: [2] (Lyon St, twin). 83.31.113.33 (talk) 14:42, 17 August 2010 (UTC)

If you read the rates details [3] it says you get the 38 weeks for 8 monthly payemnts of £237.50. It says that the weekly amount is for info only. It all matches up because you aren't paying for 8.7 months. -- SGBailey (talk) 16:29, 18 August 2010 (UTC)

August 18

Optics: 35mm focal length equivalent

Hobbyist filmer here. I'd like to make mind-blowing ultra wide-angle images like Terry Gilliam, but in the more economical format of Super 8 mm film. What I do know is that I'm not even remotely looking like Gilliam with any focal length above 18mm...but that's my desired upper length in 35mm only. The focal length to achieve a particular angle of view (which is the thing that makes for the mind-blowing images with wide-angle images) is different with any format and sensor size you use, hence there's articles such as 35 mm equivalent focal length and crop factor. In other words, if you change the format (i. e. size of your film or sensor) but wanna have the same angle of view, you need a different focal length.

Now, I have a chance of acquiring a lens (for a Super8 camera made by the Austrian Eumig brand) which is labeled as ultra wide-angle, according to trade press this lens is guaranteed to be entirely rectilinear (no barrel aka fish-eye distortion, as I don't want this), and its focal length in Super8 is 4mm.

So what I'd like to know is, what's the 35mm equivalent of these 4mm in Super8, according to crop factor? Or in other words: If my desired upper limit is an 18mm focal length in 35mm, what equivalent focal length would that be in Super8?

I guess what might help are the dimensions of the Super8 frame area: 5.97mm horizontal x 4.01mm vertical, compared to 22mm horizontal x 16mm vertical in 35mm.

My second choice would be a 3CCD miniDV with a 1" chip size. What's the equivalent to 18mm there? --79.193.41.61 (talk) 06:50, 18 August 2010 (UTC)

I don't think this is the right forum for the question, unless there is a mathie who happens to be photography hobbyist as well. You might try the science desk.--RDBury (talk) 21:11, 18 August 2010 (UTC)

I've responded at science. Thegreenj 22:42, 18 August 2010 (UTC)

Asymptotic order

I have always been absolutely terrible at this type of problem. Any hints for making it easier to solve would be appreciated. The basic problem is, given two functions, which is asymptotically greater than the other? In other words, given function f and g, to be asymptotically greater than g, it must be proven that f is O(g). Here is an example and how I solve it (which is surely the most difficult way to solve it):

Given $f=2^{\sqrt {\lg {n}}}$ and $g=n(\lg {n})^{3}$.

I solve this by estimating a value for ${2^{\sqrt {\lg {n}}}} \over {n(\lg {n})^{3}}$. I will show my steps because I am sure I do not remember the log rules correctly:

${{2^{\sqrt {\lg {n}}}} \over {n(\lg {n})^{3}}}={{\lg {2^{\sqrt {\lg {n}}}}} \over {\lg {(n(\lg {n})^{3})}}}={{\sqrt {\lg {n}}} \over {\lg {n}+\lg {((\lg {n})^{3})}}}$

Now, if I look at this as ${\sqrt {x}} \over {x+lg(x^{3})}$, it is obvious that result will be less than 1. Therefore, I claim that g = O(f) and g is asymptotically greater than f. Correct? -- kainaw™ 20:48, 18 August 2010 (UTC)

Looks OK to me.--RDBury (talk) 21:05, 18 August 2010 (UTC)

Be careful,

${\frac {x}{y}}$ and ${\frac {\lg {x}}{\lg {y}}}$ are not always equal. What you have proven is that $\,\!\lg {f}=O(\lg {g})$. This in general does not imply that $\,\!f=O(g)$ (consider for example, $\,\!f=x$ and $\,\!g=x^{1/2}$).

In any case, your problem is a bit clearer if you substitute $\,\!x=\lg {n}$. Invrnc (talk) 03:45, 19 August 2010 (UTC)

If by "asymptotically greater" you mean "greater for every value starting at some point", then this is not what big O notation is. $f=O(g)$ means that f is less than (not greater than, as you wrote) g up to a constant. In your example, the (absolute value of the) ratio between f and g is bounded (it doesn't have to be less than 1), so $f=O(g)$. This is equivalent to showing that its logarithm is bounded from above. The correct log rule is $\lg \left({\frac {x}{y}}\right)=\lg x-\lg y$, so you just need to show that $\lg {{2^{\sqrt {\lg {n}}}} \over {n(\lg {n})^{3}}}={{\lg {2^{\sqrt {\lg {n}}}}}-{\lg {(n(\lg {n})^{3})}}}<M$ eventually (this follows from the fact that it goes to $-\infty$, which is easy to see). -- Meni Rosenfeld (talk) 07:27, 19 August 2010 (UTC)

Space Curves

Consider a smooth space curve, parametrised by arc length. The Frenet frame {T,N,B} defines a rigid body at each point of the curve. This rigid body is the cube with sides T, N and B. The infinitesimal axis of symmetry of this rigid body is generated by τT + κB where κ is the curvature, and τ is the torsion, of the space curve. This means that infinitesimally the cube is rotating about the line spanned by τT + κB. Can someone show me how to calculate the infinitesimal angular frequency of the cube about this infinitesimal axis of symmetry? — Fly by Night (talk) 22:55, 18 August 2010 (UTC)

August 19

Best Mathematicians of 2010

Does anyone know who won the fields medal? I can't seem to find this information anywhere! ... If it hasn't been released, when will it be? I heard that it would be released on 19 August,2010 but I don't want to be waiting till midnight for the declaration of the world's best mathematicians! Please guys, put me out of my misery and tell me how many more hourse I need to wait? I've waited for 4 years already ... Thanks! Signed:THE DUDE. (BTW, I fixed my typsetting bug!) —Preceding unsigned comment added by THE DUDE (talk) 06:33, 19 August 2010 (UTC)

I think I know now ... Damn, why do applied mathematicians always get the awards? Where are the algebraists, topologists and geometers (and number theorists for that matter)? And why on Earth hasn't anyone received it for logic and set theory yet? (FOrget the the first question in above paragraph, I think I know who won it.) Thanks guys ... Signed:THE DUDE. —Preceding unsigned comment added by THE DUDE (talk) 06:56, 19 August 2010 (UTC)

There is already a Wikipedia User:THE DUDE, which apparently is not you. Both impersonating other users and pretending to be a user which doesn't really exist is forbidden. Your continued failure to sign your posts, followed by vandalizing your automatically-supplied signatures, is disruptive. Please stop. You can create an account if you wish. -- Meni Rosenfeld (talk) 08:16, 19 August 2010 (UTC)

Meni Rosenfeld, I just don't want to be known to the public. I want to be anonymous. You have a name, Meni Rosenfeld. If THE DUDE is taken, can I have another name. I never vandalized signatures. I just didn't want to be made public. Isn't that a simple rquest? Or are the people here at Wikipedia so serious. What's a change in signature between friends?