Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

August 22

arithmetic progressions and primes

Resolved

I want to show that any arithmetic progression if continued long enough wouldn't contain primes. How should I proceed. Thanks-Shahab (talk) 07:46, 22 August 2010 (UTC)

[personal attack removed] If it contained primes over and over you'd clearly get a contradiction. Think about it and maybe you'll see it —Preceding unsigned comment added by 114.72.236.126 (talk) 08:10, 22 August 2010 (UTC)

Do you mean that any arithmetic progression eventually contains a number which is not prime? This follows from the prime number theorem. -- Meni Rosenfeld (talk) 08:53, 22 August 2010 (UTC)

Ya don't need any [profanity removed] prime number theory to see it. Besides, it doesn't follow from th eprime number theorem since ya could get primes appearing in the sequence over and over but less frequently as n goes to INFINITY WITHOUT contradicting any bullcrap prime number theory. So no, think harder Meni Rosenfeld and Shahb. —Preceding unsigned comment added by 114.72.192.74 (talk) 09:02, 22 August 2010 (UTC)

Please be polite. Using words such as "bullcrap" is not acceptable, especially in the context of a public forum. I think that you need to think harder; see the post below by Meni Rosenfeld. PST 13:58, 22 August 2010 (UTC)

Yes I did mean so Meni, like the AP 5,11,17,23,29 has primes only but then the next term is non-prime. Uptil now the biggest such prime AP that has been found has around 20 terms. Not sure how the prime number theorem applies. -Shahab (talk) 09:05, 22 August 2010 (UTC)

Prove by contradiction. If there was a sequence with only primes, what would a lower bound for $\lim _{n\to \infty }{\frac {\pi (n)}{n}}$ be? What is the limit in light of the PNT? -- Meni Rosenfeld (talk) 09:19, 22 August 2010 (UTC)

[irrelevant vulgarities and personal attack removed] —Preceding unsigned comment added by 110.20.58.220 (talk) 10:25, 22 August 2010 (UTC)

[personal attacks removed] —Preceding unsigned comment added by 114.72.192.74 (talk) 09:09, 22 August 2010 (UTC)

Because if an arithmetic progressions with difference d contained only primes then the asymptotic density of primes would be at least 1/d, which contradicts the prime number theorem. Gandalf61 (talk) 09:19, 22 August 2010 (UTC)

For a constructive proof, recall that an arithmetic progression looks like $a+bn$ for some choice of $a$ and $b$. You want a choice of $n$ that will make this be obviously non-prime. Since all you have to work with are $a$ and $b$, try choosing one of those for $n$ and seeing what you can get. —Preceding unsigned comment added by 203.97.79.114 (talk) 09:20, 22 August 2010 (UTC)

... but make sure your proof covers the case a = 1. Gandalf61 (talk) 09:30, 22 August 2010 (UTC)

Thanks. I selected n=b^2 for a=1 and n=a otherwise. This seems so much easier then PNT.-Shahab (talk) 12:31, 22 August 2010 (UTC)

See also Primes in arithmetic progression (created by me) whics says: "an AP with common difference a cannot contain more consecutive prime terms than the value of the smallest prime that does not divide a." PrimeHunter (talk) 01:46, 24 August 2010 (UTC)

Divergent Series

OK, so one of the most important lemmas on convergent series sequences is that if a series sequence converges to a limit then any subsequence also converges to that limit. This is not true for divergent series sequences, since you can take the sum of the reciprocals of the natural numbers, which diverges, and, as a subsequence, the sum of the reciprocals of the squares, which converges. My question is, if you have a sequence and you're investigating convergence, does finding a divergent subsequence imply that the sequence itself is divergent? Thanks asyndeton talk 11:38, 22 August 2010 (UTC)

You seem to be confusing series with sequences. For sequences, if the sequence converges, then every subsequence converges to the same limit. Hence if any subsequence diverges, then the sequence itself diverges. No such results hold for series. Algebraist 12:05, 22 August 2010 (UTC)

Well, if a series absolutely converges, then any subseries does as well, though not necessarily to the same value. -- 1.46.157.19 (talk) 12:17, 22 August 2010 (UTC)

Sorry for being so casual on naming; I've amended my question for posterity. And thanks for your answer, it's just what I needed. asyndeton talk 12:22, 22 August 2010 (UTC)

Let me point out that you actually answered your own question. "If a sequence converges, every subsequence converges" and "If a subsequence diverges, the sequence diverges" are contrapositives. —Preceding unsigned comment added by 203.97.79.114 (talk) 12:30, 22 August 2010 (UTC)

The amended question is still confused about sequences versus series. The question about the sum of the reciprocals is about series, not sequences, whereas the lemma you mention is about sequences, not series. Michael Hardy (talk) 19:25, 22 August 2010 (UTC)

Every sequence is a series and every series is a sequence. If an is a sequence and bn = an - an-1 and b1 = a1 then bn is a series and an can be a series. If bn is a series then its partial sums form a sequence. Every theory about sequence has one theory about series. And Meni Rosenfeld, don't dob on me mate. If I can't correct someone for the good of humanity, what can I do? And please don't say I have attacked anyone. I just asked Meni to not "dob on me mate". That's not a personal attack. I just want to better understand what I've done wrong and unless you tell me, I can't understand that. Blocking me won't help me. —Preceding unsigned comment added by 110.20.6.240 (talk) 22:57, 22 August 2010 (UTC)

Yes, sequence and series are related via partial sums, but care must be taken not to conflate the two. The lemma mentioned by the OP was of the convergence of subsequences, where an arbitrary subseries is guaranteed convergence only if the original series is absolutely convergent. With a conditionally convergent series of real values, you may choose any arbitrary real number and there will be a subseries which converges to that value. (Actually, there are infinitely many such subseries for every real number.) The Riemann series theorem addresses permutations of conditionally convergent series; is there a name for the corresponding statement regarding subseries? -- 1.46.93.58 (talk) 00:22, 23 August 2010 (UTC)

Re the other issue, please see Wikipedia's policy on civility. Editors intelligent enough to contribute answers at the mathematics reference desk should also be able to understand the civility policy and follow it without a lot of additional explanation from other users. Therefore, any failure to do this should be treated as presumptive trolling and result in blocks. 67.122.209.167 (talk) 01:59, 23 August 2010 (UTC)

Sorry it won't happen again. I've removed "bulltwang" from my posts. Sorry again.

August 23

Logic puzzles

Complete grid, without copyright problems

Hi. I recently came across some logic puzzles in my textbook. I don't have the book with me now, but I'll try to describe them. They are the kind where there are 4 cats, each with his own toy, sleeping space, age, and name. Then you get some clues like "Rocky doesn't like the 10 year old cat" or "The cat that sleeps on the floor doesn't play with the rubber mouse", and others. Then you have to match the cat to the sleeping spot to the age to the toy (the things like name/age/sleeping spot/toy change between problems of this kind, of course). I'm told there's an algebraic way to solve this kind of problem, but how? 76.228.196.92 (talk) 01:02, 23 August 2010 (UTC)

You can often make a list of the relations and then use a resolution-like procedure to look for a satisfying assignment. 67.122.209.167 (talk) 02:02, 23 August 2010 (UTC)

See Logic puzzle#Logic grid puzzles and Logic-Grid Brain Teasers. -- 114.128.212.28 (talk) 06:17, 23 August 2010 (UTC)

I'm sorry that no one else has responded, as I was hoping someone would provide a link to instructions on using the grid. I've run across these a few times on standardized tests, and each time I've improvised a solution, taking more time than I feel is appropriate. I've not known how the typical small grids are supposed to help; with the one illustrating Logic puzzle#Logic grid puzzles, I know what to do when told that Nikolas doesn't like champagne, but not if told that the person who likes feta doesn't like champagne. -- ToE^T 12:06, 24 August 2010 (UTC)

You could try [1] which has tutorials on how to use the grid and how to solve it. --Salix (talk): 12:37, 24 August 2010 (UTC)

Thanks Salix. Interestingly, it uses the very same grid that appears in our article. Despite the instructions saying, "The grid allows you to cross-reference every possible option in every category.", I still don't see what to do with that truncated grid if told that the person who likes feta doesn't like champagne. The first full puzzle it offered me included not a truncated grid, but a full block triangular grid with n(n-1)/2 blocks (for n attributes), giving nC2 blocks. There I understand what to do! -- ToE^T 17:01, 24 August 2010 (UTC)

Looks like the grid has been truncated for display purposes. The real grid would have three large box on the top row, two on the second and one in a third row. This would give space to add the feta/champagne info. You need the full grid to be able to solve the puzzle. It might be an idea to change the picture, and there is a chance it might be a copyright violation.--Salix (talk): 18:59, 24 August 2010 (UTC)

Hmm. I've seen truncated grids before (containing n-1 blocks so that all attributes are shown, but not all correlations) that I thought were intended for use, but perhaps they too were only for illustration. Regarding the copyright issue of the image, the uploader has properly revealed the source for all three puzzle images they have uploaded, but also claims to be the copyright holder. The three websites are related, so it is possible that the uploader runs the sites. As the uploader has not been active in a year and a half, is the proper action to contact the site? -- 1.47.203.216 (talk) 00:40, 25 August 2010 (UTC)

I've now made a complete grid which is all my own work so no copyright problems. A slight simpler 4x4 puzzle.--Salix (talk): 09:10, 25 August 2010 (UTC)

That looks really good. Thanks. -- ToE^T 12:05, 25 August 2010 (UTC)

The Cantor Cube

Hi, I try to understand the Cantor cube and something is missing. If I understand correctly, it is a topological group. Does anybody know what is the topology defined on it? I mean, It's said that it is the product topology. But what is the topology defined on each coordinate. I mean, Suppose that our space is $\{0,1\}^{\mathbb {N} }$, then do we take the discrete topology on $\{0,1\}$? and if we do, isn't the product topology simply the discrete topology? Which seems not logical to me because then it would be more or less like saying almost nothing on this group would't it? Can someone maybe describe the open sets? Thanks! Topologia clalit (talk) 06:57, 23 August 2010 (UTC)

No, the product topology is not the discrete topology. The box topology would be the discrete topology. The product topology is different because the rectangles that serve as basic open sets must have all but finitely many factors equal to the whole space. --Trovatore (talk) 06:59, 23 August 2010 (UTC)

I hope I get what you mean.. Do you mean that an open set is determind by a finite number of coordinates? for example, is the set $\{(a_{1},a_{2},a_{3},...):a_{1}=1\}$ open in $\{0,1\}^{\mathbb {N} }$? —Preceding unsigned comment added by Topologia clalit (talk • contribs) 09:45, 23 August 2010 (UTC) Topologia clalit (talk) 09:47, 23 August 2010 (UTC)

Right. --Trovatore (talk) 09:50, 23 August 2010 (UTC)

Thanks! Topologia clalit (talk) 10:16, 23 August 2010 (UTC)

An open set is a union of sets of that kind. Michael Hardy (talk) 02:05, 25 August 2010 (UTC)

PhD in Math

Hi guys. Apologies for my previous bad behaviour. I've understood what I've done wrong and put my word that it won't happen again. I'll be a nice bloke from now on ...

My question is about a PhD in math. I'm looking to get a PhD in math. I've just completed my undergrad. and I'm looking to do a PhD somewhere in the States. But I've heard about the so-called qualifying exams the no.1 nightmare for math PhD students. I've heard that they're notoriously hard, very time constrained, and you like need to know everything imaginable in a number of very unrelated fields in math.

So how are these qualifying exams really? What sorts of topics do they cover? I've seen a couple from some pretty top tier places, and they look pretty hard to me so how are they in general? Feel free to comment even if you haven't particpited in one cause they're probably similar things in Aus, UK and other countries. My question is like: Which University has the easiest quals.? How do I strike a balance between good university and easy wuals? And lastly, how do I choose a university where the quals. don't require so much math background. Thanks guys ... —Preceding unsigned comment added by 110.20.55.3 (talk) 08:17, 23 August 2010 (UTC) , 23 August 2010 (UTC)

Qualifying exams at a good research university are not easy, but they're probably the easiest phase of your PhD studies. The hard part is the original research. The quals are there to make sure you have enough background to be successful at research. Having the background is usually a necessary condition, but not at all sufficient.

So basically I think you should forget looking for a school with easy quals. Look for one with potential advisors who do stuff you think is interesting. It's not going to advantage you much if you sail through the qualifying exams, but then can't find anyone who can guide you into a productive line of research. --Trovatore (talk) 08:32, 23 August 2010 (UTC)

Thanks for the inspiring answer Trevor! Much appreciated mate. But I'm still kind of scratching my head with the quals. Yeah I definitely agree I should go where there're lots of good mathematicians who like the same things I like. But what are the quals. exactly? What kind of animals are they? :) I mean an example: [2]. And UChicago requires you know this much math! [3]. It's seems to much for an average guy like me. But I've heard if I don't go to these places (or similar top tier places) for my phD I'll never get jobs anywhere unless my thesis is spectacular, because the guys who offer jobs look at where you've got PhD from: two guys with the same quality thesis, one from Harvard, one from some country village university, even if they've done similarly spectacular work, the Harvard guy will get the thumbs up. Is this true in the math community? I see you've got a PhD so you'll probably be an expert on these matters. How have you found it when it comes to getting jobs? —Preceding unsigned comment added by 110.20.55.3 (talk) 08:45, 23 August 2010 (UTC)

The job market in academic math is hard. No sugar-coating that. If an academic job is your aim, you need to understand that you have set yourself a difficult goal. To me that means, if you get into a place like Chicago (with sufficient support to live on), go there and work as hard as you need to. (I'm not saying I always did that; it's what I should have done. :-)

As to how hiring committees work, I've really never been on that side of it, so I can't say how they judge based on the quality of the school, but a school with a good name will certainly not hurt, and the better school might get better work out of you in the first place. Also, and I hesitate to mention this because there's an argument for not thinking too much about your safety net, if you don't make it all the way through, a Master's from Chicago is better than a Master's from a no-name. And if you do get the PhD but don't get a tenure-track job, you'll get more interviews in industry if your PhD says Chicago on it. --Trovatore (talk) 08:58, 23 August 2010 (UTC)

(Edit Conflict) I would say that the qualifying exams offered at Harvard are by far the easiest among many similar ranked universities. Here "easy" is the measure of the "difficulty" of the exams. Clearly you need to have a relatively broad background in mathematics to be able to answer these questions but trust me, once you have that background the exams will look far, far easier. There is no need to panic just yet if you have recently completed your undergraduate education. In fact, typically the first two years in graduate school is spent to prepare for these qualifying exams so you will have plenty of time to learn the necessary background material. (Some students who already have most of the required background generally sit the exam earlier - but this is not mandatory and generally with some universities you can take the qualifying exam multiple times.)

I am not sure what you mean by "country village university" but I can assure you that if your PhD thesis is of a very high quality, it should not matter from where you have obtained your PhD. (But of course, as Trovatore (his name is not "Trevor"!) mentioned, you should look for universities where there are people with similar interests to yours.) In any case, doing your PhD at Harvard (for example) will not have any value unless it is of a high quality.

And please do not be under the impression that Harvard and Chicago are the only places for mathematics! There are so many other places with excellent mathematicians and you should consider your options carefully. In particular, there are certain areas in mathematics in which you might wish to pursue a PhD, for which Harvard is not the best place. However, if you are interested in topology/geometry or representation theory I would strongly recommend Harvard. PST 09:08, 23 August 2010 (UTC)

(edit conflict) In addition to what Trovatore said, you should keep in mind that grad schools at the PhD level generally want their students to do well. I can't speak to math directly, but my experience in physics was the following. Quals consisted of four parts, two 3-hour written exams and two 1.5-hour oral exams. Each part was scored independently, and any part could be repeated up to three times. Only about 25% of students passed all four parts on the first try. Hence nearly everyone repeated at least one exam, and many people repeated more than one. After the first not passing grade, the department generally required students to take supplemental coursework in the area(s) they were weak. (The department wants to help you pass, even though the whole process can feel like torture.) In most cases (perhaps 80%), the second attempt was successful. In part this was because the faculty are more lenient on second attempts. A borderline effort the first time tends to fail, but a similar effort the second or third time will tend to pass. In the few cases where individuals did fail round two, a significant fraction decided not to continue trying to get a PhD. By the time someone has taken the quals at least twice they have already had a significant amount of coursework and an opportunity to consider research specialties and career prospects. I suspect that many of the people who left after a second failed qual had reached the conclusion that they wouldn't be competitive in an academic marketplace and decided they were better off spending their time elsewhere. In general though, don't despair. The entire process is a giant pain in the ass, but most people will make it through, and the department wants to help their students succeed. Dragons flight (talk) 09:12, 23 August 2010 (UTC)

One thing you should do before you pick one is make sure that "most people" do indeed make it through in the program. This was not the case where I did my PhD, and doesn't make for a great environment. Staecker (talk) 14:15, 23 August 2010 (UTC)

Looking at those exam questions from Harvard, I have to agree with PST. They aren't that hard at all. Once you know the definitions of the objects then you are almost at a solution. Although, it makes me wonder about the point of them. If you have just finished a postgraduate course with many lectures on many subjects then you will easily pass those exams; especially if you can repeat multiple times! But how many research fellows have all of that material at their finger tips? How many people, after ten years of specialised research, could pass those exams without studying again? They want to check that you're a Jack of all trades when, indeed, you may be master of none. I understand that they need a way to filter their candidates, but this doesn't seem to be an ideal way of doing. Take as a hypothetical example, which I know is extreme, a candidate may have an insane topological intuition and may have published 10 papers. But because they have concentrated soley on topology, they will never pass the Qualification Exams and they will never be accepted to do a PhD. — Fly by Night (talk) 15:00, 23 August 2010 (UTC)

There was a (true) story I once heard of a mathematician - she specialized in von Neumann algebras but when her advisor gave her a paper on the cohomology of the Lie algebra of vector fields on a manifold, she could not read it because she did not know what "cohomology" was, what a "Lie algebra" was, what a "vector field" was, or what a "manifold" was!

Harvard wants to make sure that this does not happen to students. Not everyone who does a PhD at Harvard goes on to research multiple areas of mathematics, of course, but there are a select few students who do benefit from these qualifying exams. It is also true that intuitions from other branches of mathematics are sometimes very helpful when one does research - of course, as Fly by Night says, Harvard primarily tests whether you know the definitions. (For example, if you know what an Artinian ring is, how can one not be able to prove that it has finitely many maximal ideals?!)

I agree with you Fly by Night - I have known some very, very specialized researchers who have done excellent work in their fields while having only a very narrow knowledge. But there are some areas which are highly connected to the qualifying exam topics. The topics that one is required to know for the general Harvard qualifying exams are algebra, algebraic geometry, algebraic topology, complex analysis, differential geometry, and real analysis. (More information can be found here.) In actual fact, if one researches number theory, algebraic geometry, or PDE's one may need each and every one of these topics at one's fingertips. And many people tend to go into one of these three areas. (Although many do not as well.)

Also bear in mind that what goes on in Harvard is "secret" to the "outside world". One does not need to "directly" pass these qualifying exams. For example, if you are able to pass a course in say, differential geometry, and obtain a satisfactory grade, you may be excused from sitting the qualifying exam in differential geometry. And of course, this is Harvard! They expect hard work from you in your first two years of graduate study (this is time when one prepares for the qualifying exams), but even harder work from you after this during your original research. But on the whole I agree with you Fly by Night - ultimately most people see these qualifying exams as an annoyance and unless one actually wants to dish out the books and study them on one's own, one will never be able to get an appreciation of other branches of mathematics. Qualifying exams force one to learn - but that is no where near as effective as learning out of one's own interest. PST 23:09, 23 August 2010 (UTC)

PST, when you said "There is no need to panic just yet if you have recently completed your undergraduate education. In fact, typically the first two years in graduate school is spent to prepare for these qualifying exams so you will have plenty of time to learn the necessary background material" Do you mean you take the quals after 3years in grad school? I thought these quals are meant for undergrads trying to apply for a phd position. Please tell me I thought wrong... I don't understand how life works after undergraduate can someone explain it to me?Money is tight (talk) 02:51, 24 August 2010 (UTC)

What I meant was that a student is typically expected to pass the qualifying exams within the first two years of graduate school. (I should have been more clear.) This is, of course, for a top graduate schools such as Harvard or Chicago. Other schools may give additional time to a student to pass the qualifying exams.

If you are an undergraduate student, in which year are you currently? There are so many pathways one can take after undergraduate studies so it is difficult to give a "hard and fast rule" as to how life works after undergraduate. If you plan to go to graduate school, you typically would have to apply to several graduate schools. Once you get into a good graduate school (which hopefully you should!) you would then have to pass certain requirements to do a PhD. One requirement is, of course, the qualifying exam - as I said, if you have a broad background in mathematics already, you should usually be allowed to sit the qualifying exams as soon as you enter graduate schools. (And you should not worry about passing - there is usually no stigma associated to taking the qualifying exams several times.) Another requirement is that you should be able to read, to a reasonable extent, mathematics in one or two of several other languages. (German, French, Russian and Italian, for example, are the standard choices.) There are other requirements as well - for example, teaching requirements. Perhaps you should read about it yourself since I could go on for several pages about this ... One link for Harvard's requirements is this. But you should research other schools as well.

One point that I should elaborate on is your application to graduate schools. There are so many points one could talk about, but the most important factors are: good grades, the GRE (Graduate Records Examination) mathematics subject test, the general GRE, letters of recommendation, a strong statement of intent etc. (Some of these factors are more of a formality that a practical requirement if you ask me. You can do quality research in mathematics without being able to get a strong score on the general GRE. But rules are rules - I do not make them; I state them!) The most important thing (or, one of the most important things) that graduate schools want to see is that you have attempted to take "hard mathematics courses". Usually applicants to top graduate schools are serious - they have everything: good scores, good grades etc. The best way to separate the "good from the bad" is to see what courses they have taken. E.g., if you take a 4th year course while still in your second year, that would be very useful for your application. (Provided you get a good grade!) Ideally, you should be taking graduate level courses in your 4th year at the latest. Also try to take a variety of courses from several different areas of mathematics - that shows graduate schools that you have a "back-up plan" in case something went wrong with doing the area of mathematics you originally wanted to do.

Hope that helps! PST 03:47, 24 August 2010 (UTC)

It's worth noting that this is how the American system works. It's not the same in the UK. I imagine it's different in the rest of Europe too. Not to mention Russia and Asia; although I don't really know about the last two. — Fly by Night (talk) 19:16, 25 August 2010 (UTC)

Convex subnetwork?

Let "network" mean "undirected weighted graph with positive edge weights". Given such a network N, is there a specific term for a subnetwork S in N such that for any two vertices in S, the distance between them in S is equal to the distance between them in N? This of course meaning that it is not beneficial to leave S when trying to travel between the vertices, it seems both reasonable and quite natural to call S a "convex subnetwork", but I haven't found any references. Maybe I'm searching for the wrong thing, or maybe the concept is so unfruitful that it's not in use? 85.226.207.22 (talk) 15:24, 23 August 2010 (UTC)

"Isometric subgraph" seems to be fairly standard. Algebraist 17:26, 23 August 2010 (UTC)

Ah, thanks. To be sure, I mostly find uses related to unweighted graphs, but the concept above certainly is a natural generalization of that (an unweighted graph is just a weighted graphs with all weights 1, right?), so it seems fine to use the same term. 85.226.206.72 (talk) 05:46, 26 August 2010 (UTC)

Magicicada lovers...

Ok, I'm not entirely sure how to pose this question: I'm writing a science fantasy story - it's a strange love triangle: three characters, two are in love, the third wants to murder the man so he can have the woman. But this is the thing: they're populations of magicicadas, or creatures inspired by them - usually living beneath the ground, rising at set intervals, at which times they can interact. What I guess I would like from you guys is a set of three numbers - the periods at which the three characters rise from the ground - that brings about an interesting interaction between the three characters. So let's say the hero male rises after the passage of every M number of years, the female after F number of years, the villain every V years - it'd be nice if V works out so it touches M before it touches F, because the villain wants to kill the male before he gets to the female; but maybe the male and the female have worked it out so they can join forces to destroy the villain at some point....

Sorry - floundering here - told you I was having trouble posing the question. I'm hoping someone might find the question fun enough to play with...

Thanks for your forebearance.

Adambrowne666 (talk) 20:25, 23 August 2010 (UTC)

The cicadas choose (or evolve) a large prime number so that they don't interact with their predators, of course. You presumably want regular interactions, so why not choose the first three prime numbers: say M = 2 ; F = 3 ; V = 5. This will give an interaction between M & F after 6 years for the romance, between M & V after 10 years for possible killing, M & F again after 12 years if M survives, V & F after 15 years for V to claim F, but leaves a twist after 18 years when a surviving M can meet F again, but another chance for V to kill M after 20 years, another romantic reunion between M & F after 24 years, and a final denouement after 30 years when they will all meet. How complex is your plot? Dbfirs 21:32, 23 August 2010 (UTC)

Good answer well expressed, thank you. Although primes are great, and I hate to seem an ingrate, they've been a bit overdone in sf - does anything else recommend itself? Still, I like the plot you've set out; it's a nice satisfying backbone - but it could go more baroque - longer, weirder - I'm open to imaginary numbers, square root of negative-one, that sort of thing... Adambrowne666 (talk) 11:37, 24 August 2010 (UTC)

Coprimes would be slightly less obvious than primes. 93.95.251.162 (talk) 15:34, 24 August 2010 (UTC) Martin.

Well you could use Quaternions, but they are not used for counting years in the real world! Dbfirs 07:41, 25 August 2010 (UTC)

Ooh. I like quaternions. I don't understand the first thing about the first thing about them. How confident are you with them? Could you give me a sequence like the one you did for primes? Or on second thoughts, maybe not... Reading articles like this one make me realise how little I know about maths. Stuff they were doing 300 years ago is completely opaque to me. Completely. How about just imaginary numbers then? Could you maybe help me work out a sequence? I could credit you if/when the story's published (just about everything I write ends up being published in semi-pro magazines).... Thanks for the coprimes suggestion too, Martin. Adambrowne666 (talk) 05:29, 26 August 2010 (UTC)

We do have an article on Imaginary time, but my imagination fails me on how to bring this into a story that uses real time as perceived by normal people in the real world. I can't see how to map time that flows in two (or more) dimensions into the real world. Does anyone else have any ideas? Perhaps we should ask Stephen Hawking!

Ah, but remember, this isn't 'real time as perceived by normal people in the real world' - this is an ultraweird sf story. If you can come up with a sequence as you did with the primes, I reckon I can get it to work in the context, which only touches but lightly on the real. I say go for it! Hit me with something bizarre!

Puzzles

Do all puzzles need logic for their solution? Is there some subset of puzzles depending on some other method for solution? The question arises from the WP Category:Puzzles that lists a subcategory of Logic puzzles, which by implication suggests that there are puzzles solved by some other means. 196.30.31.182 (talk) 20:47, 23 August 2010 (UTC)

Puzzles that can be solved by the systematic application of an algorithm are often called logical puzzles. Other puzzles rely mainly on prior knowledge for their solution, and yet others require "lateral thinking". Many puzzles require a combination of methods. Dbfirs 20:55, 23 August 2010 (UTC)

Could you give examples of each? —Preceding unsigned comment added by Androstachys (talk • contribs)

Hi anon 196.30.31.182, it happens that there was a mildly interesting chat related to this at Category talk:Logic puzzles, Talk:String girdling Earth and Category Talk:Puzzles, where the previous contributor (Androstachys) insists that the string puzzle belongs in the category Logic Puzzles because matematics, and therefore logic is needed to solve it, and beacause mathematics and logic "are virtually synonymous", whereas to me it is clear that this is a mathematics puzzle or a geometry puzzle at most, and by no means a logic puzzle. DVdm (talk) 06:59, 24 August 2010 (UTC)

If you read my contributions with more care you'll see that I feel that no puzzle can be solved without using logic, which is why I think that the category Logic puzzles is belabouring the point. Subcategorising Puzzles into Logic puzzles I think "is looking for a bright-line distinction that simply does not exist". I was hoping to keep you out of this discussion in order to get some unbiased opinions about the matter. Androstachys (talk) 12:58, 24 August 2010 (UTC)

Ha, you were hoping to keep me out of this discussion. I thought this was a genuine question by an anonymous user. Sorry for having blown your cover, and of course for having disturbed the 'discussion' with my biased opinions. DVdm (talk) 13:30, 24 August 2010 (UTC)

It seems to me that 196.30.31.182/Androstachys is looking for a bright-line distinction that simply does not exist. In its most restrictive sense, the term logic puzzle refers to a partricular type of puzzle in which logical deductions are made from a set of given statements. Other types of puzzle, such as sudoku, can also be solved using logic and so could be loosely described as "logic puzzles" too - this seems to be the sense in which the term is used in Category:Logic puzzles. However, a crossword is clearly not a logic puzzle (in either sense), as it requires lateral thinking ratrher than logic; crosswords and similar puzzles are grouped in the category Category:Word puzzles. Gandalf61 (talk) 09:19, 24 August 2010 (UTC)

I think most crossword puzzlers would take issue with the idea that they do not use logic in their solutions. Perhaps I have a broader view of logic that feels that lateral thinking, like any other form of thinking, decidedly involves logic. Androstachys (talk) 13:03, 24 August 2010 (UTC)

Nonsense. To take a famous example, there is no process of logical deduction (in the accepted sense of the word) that can take you from the clue "Gegs (9,4)" to its solution. The majority of cryptic crossword clues are language-dependent, using puns, homophones, anagrams and abbreviations. Logic puzzles, on the other hand, are largely language-independent and are easily translated form one language to another. Your definition of "logic" as "any form of thinking" is absurd - for example, where is the logic behind the thought "I like bananas" ? In particular, our article on lateral thinking defines its as "reasoning that is not immediately obvious and about ideas that may not be obtainable by using only traditional step-by-step logic.". Gandalf61 (talk) 13:30, 24 August 2010 (UTC)

It seems to me that the problem is that Androstachys interprets the term "logic puzzle" as a puzzle whose solution, among other things, also needs logic, whereas the usual meaning is that it is a puzzle which can be solved using only logic. This is not helped by the fact that typically this "logic" needs to be interpreted in a very broad sense encompassing at least combinatorics, hence a "combinatorial puzzle" or "mathematical puzzle" would be a more appropriate description.—Emil J. 13:51, 24 August 2010 (UTC)

That is why I feel that the word "only" should be added to the category description of [[Category:Logic puzzles]] again. See Category talk:Logic puzzles. Thanks for this 3rd opinion. DVdm (talk) 14:00, 24 August 2010 (UTC)

The claim that logic is required to solve all puzzles is flawed. Consider a jigsaw puzzle. You can randomly try to place the jigsaw pieces on a table until you see the proper picture. No logic was required. Logic would help, but it is not required. Consider a common knot-puzzle (where you have to undo some knot of string/metal). You can just wiggle the thing around until, eventually, they come apart. No logic needed, but logic would help. Consider a crossword puzzle. You can write random words into the boxes. Then, see if those words match the clues. If not, erase and try again. No logic required. When it comes to a logic puzzle, it is possible to give an answer without using logic, but logic is required to check the answer. In the jigsaw puzzle, checking the solution simply requires looking at the picture. In the knot-puzzle, count how many separate parts there are. If there are two, you are done. In the crossword puzzle, you must check every answer in the puzzle against every clue. If the answer doesn't match the clue, try again. This is language, not logic. For a logic puzzle, you must form a logical proof to show your answer is correct. -- kainaw™ 14:02, 24 August 2010 (UTC)

I absolutely reject the counter-argument: Everything in existence may be defined with math and all of math may be defined with logic, so everything is logic, so all puzzles are logic puzzles. -- kainaw™ 14:03, 24 August 2010 (UTC)

Solving a jigsaw puzzle is not a job that you could easily give to a robot. Fitting the pieces together calls for judgement and reasoning. If the robot worked randomly it would - eventually - come up with the solution, just as a computer putting words together at random would eventually come up with "War and Peace" and not know when it had done so. Any reasoning or thinking of necessity involves logic or a rational approach. Androstachys (talk) 16:31, 24 August 2010 (UTC)

If it were pure logic, Eternity II puzzle would have been solved. Kittybrewster ☎ 16:50, 24 August 2010 (UTC)

What sort of logic is that? All logical problems or puzzles capable of solution would have been solved by now? There are very many unsolved and insoluble problems in maths and physics. Ouch, my stomach hurts...

My Chambers' defines logic as "the science and art of reasoning correctly" and the Oxford "...the application of science or the art of reasoning...". Some Wikipedians obviously have different views on the meaning of logic. Androstachys (talk) 19:01, 24 August 2010 (UTC)

Either I wasn't clear or you ignored my point... A logic puzzle requires logic to check the solution. What logic is required to look at a jigsaw puzzle and say, "Yep, that looks right"? -- kainaw™ 23:59, 24 August 2010 (UTC)

The same logic that will tell you that the Eternity puzzle has been solved - the jigsaw puzzle is just a simpler version. Androstachys (talk) 07:06, 25 August 2010 (UTC)

Yes, perhaps I should have written (in my original reply) that a "logic puzzle" is one that can be solved by the systematic application of an algorithm that is not just "trial and error". Perhaps there is no clear distinction? Alfred Jules Ayer and Edward de Bono were friends, but stopped discussing anything related to logic because they couldn't even begin to agree. Ayre insisted on digging deeper and deeper, whilst de Bono kept looking for new places to dig. Dbfirs 07:20, 26 August 2010 (UTC)

August 24

A term

What is a word for "non-arbitrarily close" as in "We can find the equation of a function f if we know its degree to be n and n+1 "non-arbitrarily close" points on f"? 68.76.159.51 (talk) 00:12, 24 August 2010 (UTC)

Discrete, maybe? As an aside, finitely many points are always discrete. Invrnc (talk) 03:01, 24 August 2010 (UTC)

Whoops, I mean finitely many distinct points. Invrnc (talk) 03:02, 24 August 2010 (UTC)

I'm not sure what you are describing is meaningful, or at least I don't really understand what it could mean. Any two points in R² have some positive distance between them unless they're equal. You could say the points are distinct as Invrnc mentioned to describe that no two of them are equal. If you want some at least some minimum distance d between any pair of points you could say they are separated by at least a distance d. Rckrone (talk) 05:22, 24 August 2010 (UTC)

I, too, don't understand exactly what you want. The suggestions above are good if you meant what the respondents think you may have. If OTOH you meant "not necessarily close, just any old points" then arbitrary is the usual term.—msh210℠ 16:20, 24 August 2010 (UTC)

Certainly there is a unique nth degree polynomial that fits prescribed values at n + 1 distinct points in the domain. I suspect that is what is meant. I don't know what "non-arbitrarily close" means, unless maybe it's a clumsy way of saying "distinct". Michael Hardy (talk) 01:52, 25 August 2010 (UTC)

Set formula conversion

I was looking at Dice's coefficient and Jaccard index. In some papers, it is claimed that Dice's coefficient is twice the Jaccard index. In others, it is claimed that Dice's coefficient cannot be translated to Jaccard index. In the articles here, it claims that the relationship between Dice's coefficient (D) and Jaccard index (J) is D=2J/(1+J). It appears that it is converting |X|+|Y| = |X∪Y|+|X∩Y|. Is any of this correct? If I could translate Dice's coefficient directly to Jaccard index, it would be helpful. -- kainaw™ 12:08, 24 August 2010 (UTC)

I am not familiar about this particular topic, but |X| + |Y| = |X ∪ Y| + |X ∩ Y| is correct, which implies that the conversions D = 2J/(1 + J) and J = D/(2 − D) are also correct, given the definitions in the two articles. Therefore the other papers you mention either use different definitions or are in error.—Emil J. 12:48, 24 August 2010 (UTC)

Actually, now I noticed that the Dice's coefficient article contradicts itself. It says that he coefficient is defined as twice the shared information (intersection) over the combined set (union), and then it gives the expression 2|X ∩ Y|/(|X| + |Y|). However, twice the intersection over the union is actually 2|X ∩ Y|/|X ∪ Y|, which is exactly twice the Jaccard index. Go figure.—Emil J. 12:53, 24 August 2010 (UTC)

Because these come from botany, it is very possible that there are multiple formulas running around with the same name. I will try to hunt down Dice's original paper and see what his formula was. -- kainaw™ 13:14, 24 August 2010 (UTC)

Quaternion question

If I have a quaternion $\Omega$, which I know is equal to $2{\dot {q}}q^{\dagger }$, where the overdot denoted differentiation with respect to time, how is it possible to find $q$? I know $q$ is a unit quaternion if that helps.--Leon (talk) 18:13, 24 August 2010 (UTC)

Could you explain your notation a little, please. What does the $\scriptstyle \dagger$ represent? Is that a footnote marker that you have copied verbatim from a text, or does it have some other notation meaning? — Fly by Night (talk) 19:32, 24 August 2010 (UTC)

The dagger refers to quaternion conjugation. ALSO, I forgot to explicitly mention that the quaternion varies as a (differentiable) function of time but is always a unit quaternion.--Leon (talk) 19:57, 24 August 2010 (UTC)

What you have is a nonlinear differential equation. You can solve it numerically, but it is non-trivial. You have a relation that defines q multiplied by dq/dt - in otherwords, not a linear combination of q and its derivatives. Your constraint to be unit-quaternion is not a proper boundary-condition, but it can serve as a regularization condition for a numerical solver. Nimur (talk) 20:42, 24 August 2010 (UTC)

Could the expression $2{\dot {q}}q^{\dagger }$ have emerged from a chain rule? Did you mean that q varies or that Ω varies? Are you saying Ω is a known unit-quaternion-valued function of time, or maybe that Ω is a fixed (constant) and known unit quaternion? Michael Hardy (talk) 01:29, 25 August 2010 (UTC)

$\Omega$ is a function of time, as is q. I know $\Omega$ and I know that q is a unit quaternion. I want to find q. I also know that $\Omega$ is a pure (complex) quaternion, which can be demonstrated from the above expression, knowing q is a unit quaternion for all time.--Leon (talk) 07:23, 25 August 2010 (UTC)

If

$q=a+bi+cj+dk\,$

where a, b, c, d are real, then q is a unit quaternion precisely if

$1=qq^{\dagger }=a^{2}+b^{2}+c^{2}+d^{2},\,$

and the derivative of that with respect to time is

$0={d \over dt}1={d \over dt}(a^{2}+b^{2}+c^{2}+d^{2})=2a{\dot {a}}+2b{\dot {b}}+2c{\dot {c}}+2d{\dot {d}}.$

If I look at $2{\dot {q}}q^{\dagger }\,$ and expand it, I get

$2a{\dot {a}}+2b{\dot {b}}+2c{\dot {c}}+2d{\dot {d}}+(a{\dot {b}}-b{\dot {a}}+c{\dot {d}}-d{\dot {c}})i+\cdots \,$

and the sum of the first four terms vanishes as above. Now $a{\dot {b}}-b{\dot {a}}\,$ is the numerator in an application of the quotient rule, as is $c{\dot {d}}-d{\dot {c}}\,$, etc. I don't know where this will take us. So just some preliminary scratchwork. Michael Hardy (talk) 02:02, 25 August 2010 (UTC)

The equation is equivalent to $\Omega =2{\dot {q}}q^{-1}$ on account of |q| = 1. If it were complex numbers rather than quaternions, then the solutions to this equation would be $\textstyle q(t)=\exp \left({\frac {1}{2}}\int _{t_{0}}^{t}\Omega (u)\,du+c\right)$, where c is any constant (with real part 0 to ensure |q| = 1). I don't know whether one can do such thing with quaternions, but since the complex numbers are a special case (where the j and k components of q and Ω are 0), the solution will have to involve some quaternion generalization of the exponential.—Emil J. 11:43, 25 August 2010 (UTC)

Well, actually, would it work to just take this expression as is, and interpret it in quaternions using the exponential here? The only problem is the chain rule. The definition ensures that $\exp(z)\in \mathbb {R} (z)\simeq \mathbb {C}$, hence exp(z) commutes with z for any z; is this enough to make (exp(f(t)))' = f'(t)exp(f(t))?—Emil J. 12:10, 25 August 2010 (UTC)

I don't think it is, though I might be wrong. I know that doesn't hold for general matrices, but it might hold in some sense for quaternions. —Preceding unsigned comment added by Star trooper man (talk • contribs) 12:15, 25 August 2010 (UTC)

I guess you are right, I'd need f(t) to commute with f'(t). The solution above is valid if q(t) commutes with its derivative, or Ω(t) commutes with $\textstyle \int _{t_{0}}^{t}\Omega (u)\,du+c$. IIUIC this condition is equivalent to the values of q(t) being confined to a fixed subfield isomorphic to C, so it is rather a degenerate case.—Emil J. 13:15, 25 August 2010 (UTC)

August 25

Help on ACT/SAT math

Hey guys. I'm taking the ACT and SAT in the near future. The concepts tested in the math section are pretty elementary, in my opinion (but I take AP calc :|). BUT! I still get one or two wrong in easy places when doing practice tests due to arithmetic errors. SAT math is ~25 questions per section and ACT math is 60 questions, and I generally have 5-10 minutes left at the end. I don't usually get the "hard" one wrong because to me they're not really hard to me. So does anyone have tips on how to reduce these, and how to check over my work? I'm most interested in the latter; I can't go over all of the questions in the time left over, but I also can't really identify problems I might have done wrong because of an arithmetic error. 76.230.231.191 (talk) 00:11, 25 August 2010 (UTC)

The most important point is that if you are scoring above 700 (for the mathematics section of the SAT), and there are still a few weeks to go before the test, then you are doing fine. What score are you achieving on an average? Unfortunately, the mathematics section of the SAT is all too easy these days and so if you make just one mistake in the whole test, you might get 720. I know you probably do not want to be hearing this, but I think that these points are important to keep in mind. (And for some universities such as Caltech, they might reject your application if you have scored below 760 on the mathematics section without even bothering to read the rest of your application - this is also unfortunate.)

As for your question, the best advice I can give you is to write all your computations without ommitting a single detail - not one. It does not matter if you are moving from $(5+4)-(2+3)$ to $4$ or something more complex - expand the brackets carefully in at least 3 steps! I know this sounds boring, but it is the only way I know to avoid silly errors. There is ample space to work out your questions so you should use this as much as you can. And once you have worked out a question in many steps, it is very likely that you will not make a mistake. However, always double check at the end of the test.

Also confirm that you are bubbling everything correctly - double, no triple check this! One mark is "gold" on the higher end of the scale and so you should not be losing marks because of incorrect bubbling.

You are probably thinking that if you do all these things, then you will run out of time. I think that taking tests at home is the key here. Try different strategies, decide how to manage your time on the questions, and you should be fine. The first 15 questions should take 5 minutes at the most especially if you have done AP calculus. The last 10 questions should take 10 minutes at the most. And this should give you ample time to check. But I strongly believe that you should try to do everything carefully the first time around - then, there will be no need to check! (But still check!) In particular, while this strategy is possible, you should also try other strategies that enable you to spend more time on the easier questions. If you have 10 minutes left over, then you are doing the test too fast. Try to divide that extra 10 minutes among the easier questions. Hope that helps! (Disclaimer: I am no expert on these matters.) PST 02:37, 25 August 2010 (UTC)

When you check an answer, it's useful to try to solve the question in a different way. If you have an error you will be less likely to repeat it. -- Meni Rosenfeld (talk) 09:01, 25 August 2010 (UTC)

I have one possibly weird piece of advice. A lot of the multiple-choice questions on those tests are of the form "what number x satisfies [some condition]". These can be further split into two types. It's usually obvious at a glance what type a question is:

• "Which number gives 15 if you double it and then add 3?" (type 1)
• "Which number is 2 less than the square of a prime number?" (type 2)

Type 1 means you can set up an equation with a unique solution (2x+3=15 in the example), then solve for x, then look at the multiple choices and find the one equal to x and fill in the bubble. Type 2 doesn't have a unique answer so you have to examine the 5 choices and see which one satisfies the condition. Type 1 is more common than type 2, or at least it was on the SAT that I took, back in the day.

With a type 1 question, it's usually faster and less error-prone to solve the equation first and look at the choices afterwards. It's faster because you do just one calculation instead of trying out several guesses, and less error-prone because if you make a mistake, you'll probably get an x that doesn't match any of the choices, so you know something is wrong and can go back and fix it. With type 2, you just check the choices one by one.

It is possible to hit a type 2 question that looks like it could be a type 1 question, causing you to burn a lot of time if you're trying to stick to the above strategy rigorously (I did that and probably lost a few points because of it). Just remember that all the questions are basically easy, so if you're spending more than a few moments figuring out a type 1 approach, it's possible that there isn't one or that you're missing it somehow, so just switch to type 2 and get on with it. 67.117.146.38 (talk) 07:34, 26 August 2010 (UTC)

here's the non-answer, what is the question?

Resolved

There is a certain unsolved problem in combinatorics (Ramsey theory I guess) that seeks to determine the value of a certain number n. The situation is something like this: just about everyone believes n=6. It's obvious that n>4, and it can be proven with a little work that n>5. There is a ton of experimental evidence pointing to n=6, though it might not come as a complete shock if it turns out that n=7 or even n=8.

However, the best lower upper bound anyone has been able to rigorously prove on n is something like n = the 10 millionth Ackermann number.

Anyone know what I'm referring to? Thanks.

67.117.146.38 (talk) 21:48, 25 August 2010 (UTC)

You're thinking of Graham's problem. The upper bound you're thinking of is probably Graham's number (which article also states the problem), though that's not the best bound known. Even the best known upper bound is absurdly large, however: much bigger than the 10 millionth Ackermann number. For a long time the solution to the problem was believed to be 6, but it is now known to be at least 11. Algebraist 22:09, 25 August 2010 (UTC)

Thanks! I guess the deal with the number is less clear than I thought. I remembered there was a general belief that the actual value of the number was quite small, not that it was somewhere unknown within a vast interval. 67.117.146.38 (talk) 22:34, 25 August 2010 (UTC)

Line integrals

Hey guys. We're studying line integrals in class, and I've got three homework assignments about them. The trouble is we don't have our books yet and I left my notes on campus :( The problems are pretty basic (two involve linear bottom function things--forgot the word, sorry!--and the third is an ellipse). Since it would be simple to solve these with basic calculus or even geometry, we're getting graded on the work more so than the answer. So i guess what I'm asking is, can someone demonstrate how you would do the line integral (being sure not to skip steps please), with any function you think is best or just a circle centered at the origin and with radius 1 (That's not one of the problems, I just want to see how it's done)? thanks. 68.249.1.8 (talk) 23:39, 25 August 2010 (UTC)

All right, let's do the easy thing and integrate $f(x,y)=1$ along the unit circle, using the natural parametrization ($x=\cos t,y=\sin t$). By the first definition in Line integral, we then have:

$\int _{C}1\,ds=\int _{0}^{2\pi }1\,|(-\sin t,\cos t)|\,dt=\int _{0}^{2\pi }{\sqrt {(-\sin t)^{2}+(\cos t)^{2}}}\,dt=\int _{0}^{2\pi }{\sqrt {1}}\,dt=\int _{0}^{2\pi }dt=2\pi .$ 85.226.206.72 (talk) 06:29, 26 August 2010 (UTC)

The crucial step was deciding the best parameterization for a circle. In the case of a circle, it is easy to map two-dimensions (x,y) to a single parameter (t) (distance along the circle). From basic geometry/trigonometry, we know that the relationship {x=sin(r), y=cos(r)} is a suitable description of a circle. In general, you need to find the "simplest" parameterized representation of your line path - preferably as a one-dimensional function in the form of easily-integrable parts. Nimur (talk) 21:46, 26 August 2010 (UTC)

August 26

Volume of revolution

How do you find the volume of revolution generated by the area enclosed between $y=x^{n}$, $x=0$, $y=a^{n}$, $y=b^{n}$ rotated around the x axis?--115.178.29.142 (talk) 03:24, 26 August 2010 (UTC)

It would be easier to help if you show us how much you've already done. The obvious first step here is to draw a sketch, however crude, of the curves and find out what species the region in question belongs to. Have you done that? 85.226.206.72 (talk) 06:18, 26 August 2010 (UTC)

... and tell us if you don't understand the methods given in the article: Solid of revolution. Dbfirs 06:50, 26 August 2010 (UTC)

Inclined sit-up

OK, tonight I did 3 sets of 5 inclined sit-ups. The incline was at approximately 25-30 degrees. I placed a 25 lb weigh on my chest, and sat straight up. I also tried to maintain myself parallel to the ground for a few seconds, which was surprisingly difficult.

Can you help me come up with a formula, based on my body weight of 185 lbs, for how many calories I actually burned tonight. I do not know the breakdown of how weigh is distributed in the body (sorry). I would like to have an accurate representation, so if we have to add just a little bit for calories wasted, that's fine with me (I imagine most exercises have this type of fudge factor, though I could be wrong). Magog the Ogre (talk) 04:14, 26 August 2010 (UTC)

The amount of calories burned is pretty small, but you are building muscle tone. If you are trying to lose weight through mathematics, you might like the downloadable book The Hacker's Diet. 67.117.146.38 (talk) 08:41, 26 August 2010 (UTC)

I did a very rough calculation and it turned out about 5 kcal (it could easily be 10 or something).

Assuming that by "formula" you meant just that rather than a number, then I'd say the energy burned per sit-up is

${\frac {mgh(\sin(\alpha _{2})-\sin(\alpha _{1}))}{\eta }}$

Where

• m is the moving mass (not including, say, your legs)
• g is the acceleration due to gravity
• h is the average (weighted by mass) distance of the moving mass from the fulcrum
• $\alpha _{1},\ \alpha _{2}$ are the angles at the start and finish (negative angle for the mass being below the fulcrum)
• $\eta$ is the efficiency of the muscles involved.

If you're using metric units you'll get a result in Joules; you'll need to convert them using $1kcal=4184J$.

-- Meni Rosenfeld (talk) 15:10, 27 August 2010 (UTC)

Optimal decision making under uncertainty.

100 cups are placed on a desk. Underneath one of them is $100, and the rest contain nothing. At any time, you may pick a cup and claim the contents of it. You may only do this once. You may also pick a cup and identify the contents of it, but doing so costs $1. You may do this as many times as you like. What is the optimal strategy to maximise the expected profit?--Alphador (talk) 09:04, 26 August 2010 (UTC)

Keep spending $1 until you find the $100. On average you'll have looked at ~50 cups before you find it and expect to earn ~$50. Dragons flight (talk) 09:26, 26 August 2010 (UTC)

With the obvious, very minor caveat that if you investigate 99 and don't find it, take the 100th without investigating. —Preceding unsigned comment added by 203.97.79.114 (talk) 10:55, 26 August 2010 (UTC)

This would be much harder if, as you wait idly trying to make up your mind, other players are allowed to participate in the same game with the same set of cups. The mathematics for deciding when to act are described in Nash equilibrium (to maximize your own profit), and the mathematics to maximize the total profit for all participants (i.e., somebody gets the $100, and the least amount of money is spent) is described in Pareto optimality. Nimur (talk) 21:41, 26 August 2010 (UTC)

It's customary to say "thank you" when people answer one's question, along with any followup questions. I don't know about others, but for me, knowing that my answer was helpful gives me closure which allows me to stop thinking about the question. -- Meni Rosenfeld (talk) 14:50, 27 August 2010 (UTC)

Assigning probabilities to curve fittings

Let's suppose you're given the following data points: (0,1) (1,2) (2,4) (3,8) (4,16) (5,32) (6,64) (7,128) (8,256) (9,512)
There are infinitely many curves that will fit these data points, however if you saw this as a natural phenomenon (for instance, the population of bacteria with respect to time), intuition would tell you that the most likely equation that would fit these is $y=2^{x}$. Is there any mathematical justification for believing that $y=2^{x}$ is the equation which most probably fits these data points out of the infinitely many that could potentially fit them?--Alphador (talk) 11:10, 25 August 2010 (UTC)

The general principle for this is Occam's razor. One concretization for it is that your prior probability for a sequence (or anything else) should be smaller the higher its Kolmogorov complexity. Because the sequence $y=2^{x}$ has such a short description, it will have a relatively high prior probability, and its posterior probability will be overwhelming after observing these data points. -- Meni Rosenfeld (talk) 11:01, 26 August 2010 (UTC)

The exponential function assumes fractional values while the population at any time is an integer. So the population is not an exponential function.

The exponential function grows without limit while the population is limited. So the population is not an exponential function.

So the best you can get is an approximate fit in a finite interval of time.

Consider the function g(t)=f(t)−2^t. Then g(t)=0 for t=0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Consider only a finite number, N, of points of time in the finite interval, because you cannot measure with unlimited frequency. So this is like a lottery with N tickets, and K of these tickets are marked 'zero', and N−K of these tickets are marked 'nonzero', and you have opened n=10 tickets, out of which k=10 are all marked 'zero' and n−k=0 are marked 'nonzero'. The likelihood function L(K)=$\scriptstyle {\binom {K}{k}}{\binom {N-K}{n-k}}$=$\scriptstyle {\binom {K}{10}}$ for 0≤K≤N. The maximum likelihood estimate is K=N. So YES, there is a mathematical justification for believing that y = 2^x is the equation which most probably fits these data points out of the infinitely many that could potentially fit them. Bo Jacoby (talk) 18:30, 25 August 2010 (UTC). The question and the answer has been removed, and the question reinstalled without the answer. Don't know why. Bo Jacoby (talk) 11:07, 26 August 2010 (UTC).

Sorry, the question was removed several times, I restored the question as it appeared the first time while I should have looked for the last time. -- Meni Rosenfeld (talk) 11:22, 26 August 2010 (UTC)

how to integrate 1/(1+cosx)?

tried making denominator (1^2+(sqrt(cosx)^2)) but stuck after that. the answer supposedly has a tan(0.5x). —Preceding unsigned comment added by 218.186.8.247 (talk) 18:34, 26 August 2010 (UTC)

Divide and multiple by 1-cos x -Shahab (talk) 19:11, 26 August 2010 (UTC)

You're right that the answer is tan(½x). Why not differentiate tan(½x) and then try to retrace the steps to prove the integration? For example:

${\frac {\text{d}}{{\text{d}}x}}\tan \!\left({\frac {x}{2}}\right)={\frac {\text{d}}{{\text{d}}x}}\left({\frac {\sin \left({\frac {x}{2}}\right)}{\cos \left({\frac {x}{2}}\right)}}\right)={\frac {{\frac {1}{2}}\cos ^{2}\!\left({\frac {x}{2}}\right)+{\frac {1}{2}}\sin ^{2}\!\left({\frac {x}{2}}\right)}{\cos ^{2}\!\left({\frac {x}{2}}\right)}}={\frac {1}{2}}\left({\frac {\cos ^{2}\!\left({\frac {x}{2}}\right)+\sin ^{2}\!\left({\frac {x}{2}}\right)}{\cos ^{2}\!\left({\frac {x}{2}}\right)}}\right)={\frac {1}{2\cos ^{2}\!\left({\frac {x}{2}}\right)}}.$

If you use the cosine angle formulas, which say that cos(α ± β) = cos(α)cos(β) ∓ sin(α)sin(β), then you will be able to show that 2⋅cos^{2(½x) = 1 + cos(x).} Notice that cos(x) = cos(½x + ½x) = cos^{2(½x) − sin2(½x),} and so 1 + cos(x) = 1 + cos(½x + ½x) = 1 + cos^{2(½x) − sin2(½x).} Since cos^{2(½x) + sin2(½x) = 1} it follows that 1 − sin^{2(½x) = cos2(½x),} and so 1 + cos(x) = 2⋅cos^2(½x) as required. Thus:

${\frac {\text{d}}{{\text{d}}x}}\tan \!\left({\frac {x}{2}}\right)={\frac {1}{1+\cos x}}\ .$

If you run these trigonometric identities backwards then you'll have the proof. Don't forget the constant of integration! — Fly by Night (talk) 19:38, 26 August 2010 (UTC)

There's some material on this in the article titled tangent half-angle formula. Michael Hardy (talk) 20:43, 26 August 2010 (UTC)

differential equation

80(dx/dt) = 1 + k/(x+1) where k is a constant.

i can see that if k/(x+1) is combined with the dx/dt term, can integrate by ln. also tried multiplying whole equation by x+1 but then the 1 will become an x+1 and I can't transfer the x over. also tried making RHS (x+1+k)/(x+1) but after transferring reciprocal over to LHS, cant think of a formula for (x+1)/(x+1+k). —Preceding unsigned comment added by 218.186.8.247 (talk) 18:40, 26 August 2010 (UTC)

I'm going out on a limb, but you might have to solve this approximately or numerically - I'm pretty sure it's nonlinear. So: taylor series! k/(x+1) is exactly k(x+1)^-1, which can be expanded and approximated as k*(1-x). This linearizes the equation for small x. For large x, dx/dt approximately equals 1/80. So, you have a suitable first-order approximation. The function diverges, so if this were a real-world problem, it would be of little value to solve it exactly or approximate it very accurately anyway. Maybe a pure mathematician can come through and clean up my messy engineer's work. Nimur (talk) 21:34, 26 August 2010 (UTC)

Separation of variables is the way to go, try writing (x+1)/(x+1+k)=1-k/(x+1+k).--RDBury (talk) 23:25, 26 August 2010 (UTC)

Not nonlinear! Just ugly. Nimur (talk) 23:51, 26 August 2010 (UTC)

Separate variables:

$80{dx \over dt}={x+1+k \over x+1}$

and then

${80(x+1)\,dx \over x+1+k}=dt.$

Then integrate both sides. Michael Hardy (talk) 00:58, 27 August 2010 (UTC)

Substitute t=80s and get rid of the common factor 80:

${x+1 \over x+1+k}dx-ds=0.$

Substitute x=y-1-k:

$dy-k{dy \over y}-ds=0.$

Integrate:

$y-{k\log y}-s\,$ is constant.

Exponentiate:

$y^{-k}e^{y-s}\,$ is constant.

Substitute back:

$(x+1+k)^{-k}e^{x+1+k-t/80}\,$ is constant.

Discard the constant factor:

$(x+1+k)^{-k}e^{x-t/80}\,$ is constant.

Bo Jacoby (talk) 20:34, 27 August 2010 (UTC).

August 27

Semicircles problem

This was a question in math league a while ago, and I did not know how to do the problem.

Givens: There are 3 semi circles, one with radius 3, one with radius 2, and one with radius 1. They are inlaid as shown in the diagram. There is a circle tangent to the sides of each of the semicircles. What is $x$, the radius of that circle?

What I attempted to do before getting confused: I made a triangle (not drawn in the diagram), connecting the center of the cicle with the center of the semicircle radius=1, connecting the circle center with the center of semicircle radius=2, and the final line connecting the two centers of the semicircles r=1 and r=2. So now I got a triangle with sides of 3, 1+x, and 2+x.

Was making a triangle helpful in solving the problem? What is the next step I should take? Spencer^T♦C 01:36, 27 August 2010 (UTC)

I seem to recall this very same question being asked here maybe between six and twelve monts ago. I'm not sure if I can find it though. Michael Hardy (talk) 01:42, 27 August 2010 (UTC)

That's when it was the math league problem, so it's quite possible someone else asked how to do it. Spencer^T♦C 01:45, 27 August 2010 (UTC)

Found: Wikipedia:Reference_desk/Archives/Mathematics/2010_March_25#Circle. Can someone explain how the Descartes' theorem applies? Spencer^T♦C 01:46, 27 August 2010 (UTC)

Wikipedia's article on Descartes' theorem says it's a relationship among the four curvatures. Three of the curvatures you've given are 1, 1/2, and 1/3. The problem is to find the fourth. So plug them in. Michael Hardy (talk) 02:07, 27 August 2010 (UTC)

...and it turns out the quadratic equation you get works out very neatly, since its solutions are rational. Michael Hardy (talk) 02:13, 27 August 2010 (UTC)

...but remember that the largest circle surrounds the other three circles, so you have to take its curvature as -1/3, which leads to only one solution (there are two kissing circles, but they are the same size). In general, given circles with radii a, b and a+b, the term under the square root is 0 because

${\frac {1}{ab}}-{\frac {1}{a(a+b)}}-{\frac {1}{b(a+b)}}=0$

and the radius of the kissing circles is

$\left({\frac {1}{a}}+{\frac {1}{b}}-{\frac {1}{a+b}}\right)^{-1}={\frac {ab(a+b)}{a^{2}+ab+b^{2}}}$

Gandalf61 (talk) 10:47, 27 August 2010 (UTC)

Break the side of length 3 into two parts: from the center of the circle of radius 2 to the center of the circle of radius 3, of length 1, and from there to the center of the circle of radius 1, of length 2. Draw a line from the center of the big circle (of radius 3) to the center of the circle of unknown radius, and keep going until that line hits the point where the big circle and the unkown circle touch each other. (Notice that if a line passes through the centers of two circles in the same plane that touch each other, then it must pass through the point where they touch each other.) The portion of that line that lies within your triangle has length 3 − x. Keep going from there. Michael Hardy (talk) 02:01, 27 August 2010 (UTC)

Apollonian gasket has some relevance here.--RDBury (talk) 14:50, 27 August 2010 (UTC)

Oldest open problem

What is the oldest open problem in mathematics? --84.61.172.89 (talk) 07:53, 27 August 2010 (UTC)

There must be countless old forgotten open problems in mathematics, thus there is no telling what is the oldest. However, among well-known problems, I believe that the problem of the existence of odd perfect numbers goes back to Euclid, making it fairly old.—Emil J. 10:33, 27 August 2010 (UTC)

sin(ln(z)) in annulus around z=0

Although ln(z) is multivalued and discontinuous near 0, i'm thinking sin(ln(z)) collapses the multivalues into a single valued function and is continuous. Am i correct? If so, what's the Laurent series? Thanks, 24.7.28.186 (talk) 09:57, 27 August 2010 (UTC)

No, you are not correct. For example, the values of ln 1 are 2kπi for any integer k, and plugging it into the definition $\sin z={\tfrac {1}{2i}}(e^{iz}-e^{-iz})$ gives $\sin \ln 1={\tfrac {i}{2}}(e^{2k\pi }-e^{-2k\pi })$. However, sin(i ln z) is single-valued, it is a meromorphic function with a simple pole at 0, and it equals ${\tfrac {i}{2}}(z-z^{-1})$.—Emil J. 10:27, 27 August 2010 (UTC)

(To provide more context: this question appears to be a followup of Wikipedia:Reference desk/Archives/Mathematics/2010 August 16#Laurent Series for ln(z) in annulus around z=0?.—Emil J. 10:54, 27 August 2010 (UTC))

Right you are, thanks.-Rich Peterson199.33.32.40 (talk) 22:40, 27 August 2010 (UTC)

Roots of unity for non-integer exponents

Does $x^{n}=1$ have multiple solutions for $x$ when $n$ is not an integer, and if so, how many?--Alphador (talk) 11:23, 27 August 2010 (UTC)

First, how do you interpret the equation? If n is not an integer, then xⁿ is multi-valued. Do you call x a solution if some value of xⁿ is 1, or if all values of xⁿ are 1 (i.e., if 1 is the only value)?—Emil J. 11:30, 27 August 2010 (UTC)

Have you studied root of unity? Bo Jacoby (talk) 13:14, 27 August 2010 (UTC).

Why do you ask me? Anyway, I know the article, I even contributed to it, and as far as I am aware it does not contain a single word on the topic of roots with non-integer exponents.—Emil J. 13:25, 27 August 2010 (UTC)

Bo was probably just misusing indentation. -- Meni Rosenfeld (talk) 14:43, 27 August 2010 (UTC)

Yes. Sorry. Bo Jacoby (talk) 22:12, 27 August 2010 (UTC).

I think the answer depends on the exact nature of your non-integer n; specifically if it's rational, and if so whether the numerator is odd or even in simplest form. If we extend the general technique we use for the roots of unity then we can rewrite the equation as:

$\textstyle x^{n}=e^{2\pi ki}$ (where k is an integer) which rearranges to:

$\textstyle x=e^{2\pi i{k \over n}}$

giving us $\textstyle x=1$ whenever $\textstyle k=0$ OR $\textstyle k/n$ is a non-zero whole number (which can happen easily enough for rational-but-non-integer n)

but we can also get $\textstyle x=-1$ when $\textstyle k/n=m+1/2$ for m an arbitray integer. Unless I'm missing something, this can happen when n is rational and has a factor of 2 in the numerator in its simplest form (trivial example: $\textstyle n=2/3$ gives $\textstyle k/n=3/2$ when $\textstyle k=1$, and $\textstyle e^{2\pi i{2 \over 3}}=e^{3\pi i}=-1$).

For irrational n, though, we only have the $\textstyle k=0$ case giving us $\textstyle x=1$, and then an infinite number of complex roots. (I think it's safe to say that we're sort of precessing around the Argand diagram in steps of $\textstyle k/n$, and I'm sort of tempted to say that $\textstyle x=-1$ when $\textstyle m=\infty$, but that seems like wishful thinking.)

Summarising the conclusions:

If n is rational with an even numerator, there are two real roots, -1 and 1, and some number of complex roots..

If n is rational with an odd numerator, there is one real root, 1, and some number of complex roots.

If n is irrational, we have 1 and an infinite number of complex roots.

I've probably missed a subtlety somewhere in all that, though. (Using n for an explicitly non-integer number is messing with my head; too much FORTRAN in my youth, perhaps.) --217.41.233.67 (talk) 15:52, 27 August 2010 (UTC)

(Replying to myself, sorry) An easier way to think about the requirement that n has an even numerator to get -1 as a root is to write $\textstyle n=2a/b$, giving $\textstyle x^{2a/b}=1$ which gives $x^{2}=1^{b/a}$. For $\textstyle x=-1$ this gives the obviously valid $\textstyle (-1)^{2}=1^{b/a}$ which works for any a and b.

Also, I think the 2/3 example I gave is something of a special case, as AFAICS there are no complex roots there at all -- each time k increases by 1, we're back to either an odd or even multiple of $\textstyle \pi$ in the exponential form. --81.158.2.129 (talk) 17:04, 27 August 2010 (UTC)

Non atomic poset

Resolved

I have the following definition in my book:

Definition: If a poset T has a smallest element 0, then any cover of 0 is called an atom or point of T. A poset with 0 is atomic if every nonzero element contains an atom.

Now can someone give me an example of a non-atomic poset. Since 0 is the smallest element all nonzero elements contain it, and so either are atoms or contain atoms. So isnt any poset with 0 an atomic poset?-Shahab (talk) 11:46, 27 August 2010 (UTC)

The rational or real interval [0,1] with the usual ordering has no atoms.—Emil J. 12:29, 27 August 2010 (UTC)

Oh, of course. Dont know why I was thinking only finite. Thanks-Shahab (talk) 16:29, 27 August 2010 (UTC)

partial derivative

find the wxtreme values of function $f(x,y)=x^{3}y^{2}(1-x-y)$ —Preceding unsigned comment added by Himanshu.napster (talk • contribs) 19:23, 27 August 2010 (UTC)

I have written the definition of your function in LaTeX so it is easily intelligible by volunteers at this reference desk. I hope you do not mind. PST 06:19, 28 August 2010 (UTC)

Partial derivative should help you with your homework. But then I guess you knew that since you made it the title of your question. --81.158.2.129 (talk) 20:50, 27 August 2010 (UTC)

On what region do you wish to work out the extreme values of the function f? If you wish to work out its extreme values on the entire plane, do the standard routine. Work out the critical points, compute the determinants of the corresponding Hessians, and use the relevant extrema tests. (All of this should be possible to find in whatever textbook you are using.) PST 06:19, 28 August 2010 (UTC)

Calculus on science desk: atmospheric carbon scenario difference projections

There is a calc problem that I don't understand at Wikipedia:Reference desk/Science#What is the wind-water-solar climate change mitigation scenario atmospheric carbon projection? Why Other (talk) 22:22, 27 August 2010 (UTC)

mgf of the square of a standard normal random variable

How can I get the moment generating function of the square of a standard normal random variable? Do I just replace $x$ with $x^{2}$ in the integral? I know it's meant to come out the same as a gamma mgf but I can't get it to work. —Preceding unsigned comment added by 118.208.51.232 (talk) 23:08, 27 August 2010 (UTC)

The law of the unconscious statistician applies. (I finished my Ph.D. without ever hearing that name for it, although I used it all the time. However, apparently some respectable authors call it that.) The moment-generating function of a continuous random variable with density ƒ is given by

$m_{X}(t)=E\left(e^{tX}\right)=\int _{-\infty }^{\infty }e^{tx}f(x)\,dx.$

The moment-generating function of the square is

$m_{X^{2}}(t)=E\left(e^{tX^{2}}\right)=\int _{-\infty }^{\infty }e^{tx^{2}}f(x)\,dx.$

To be continued..... Michael Hardy (talk) 02:01, 28 August 2010 (UTC)

Michael, I have known that rule/law/theorem since I began high school but to this day I never knew it had a name! Thank you for noting that! This is exactly the reason Wikipedia is so great. (I do not (really) work in probability theory by the way, so I probably have not heard of many results in the field and I guess I can be excused for that ;), but I feel awkward knowing that one of the very first (and easiest) results I learnt in the field has a name, and I never knew it for so many years ...) Or perhaps the name was invented recently. Was it? If not, do you know when it was invented? PST 06:11, 28 August 2010 (UTC)

Let me try doing this...

$\int _{-\infty }^{\infty }e^{tx^{2}}f(x)\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{tx^{2}}e^{\frac {-x^{2}}{2}}\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{{\frac {-1}{2}}(x^{2}-2tx^{2})}\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{{\frac {-x^{2}}{2}}(1-2t)}\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{\frac {-x^{2}}{2}}e^{(1-2t)}\,dx.$

$=e^{(1-2t)}\int _{-\infty }^{\infty }{\frac {1}{\sqrt {2\pi }}}e^{\frac {-x^{2}}{2}}\,dx.$

$=e^{(1-2t)}$

Alas! I'm sure the answer is supposed to be $(1-2t)^{\frac {1}{2}}$. What am I doing wrong? —Preceding unsigned comment added by 130.102.158.15 (talk) 06:02, 28 August 2010 (UTC)