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

June 3

How Many Triangles?

Problems like this are relatively easy to solve with a little planning and some brute force. I've been working on a similar problem, but with the medians drawn in for each of the sixteen smallest triangles. (Sorry, no image.) With so many triangles, it's much harder to develop a strategy that avoids double counting. Any thoughts about how I should approach this? Is there a general solution or strategy for problems like this? For what it's worth, I've found 482 so far. 24.1.231.240 (talk) 06:08, 3 June 2009 (UTC)

You could number the vertices v₁ to v_n and then for each vertex v_i, find all verticies v_j connected by a straight path where j > i, and then for each such pair, find any v_k which connects via a straight path to both v_i and v_j, where k > j, and where v_k are not colinear. This brute force method will give you all triangles and no duplicates. This approach is probably not practical as a manual method.

Another approach (one you may have already taken) would be to attempt to identify each uniquely sized and shapes triangle that could possibly exist in the structure, including rotations and reflections, and then for each one, pick a vertex and find all possible positions for that vertex within the structure. This more closely matches that approach typically used in the original problem and easier to perform manually. -- Tcncv (talk) 06:39, 3 June 2009 (UTC)

The first problem (for all size n of the triangle) is contained in the OEIS : [1]. Maybe your variant is there also: try introducing the number of triangles for small values of n. In any case a closed formula seems easy to obtain (but why should one do it? leave it where it is). --pma (talk) 17:58, 4 June 2009 (UTC)

Modelling "value" for used cars

I'm in the market for a used car, and I've been trying to figure out a simple score that (in a handwavey sense) captures something approximating "value". It's just a first approximation (naturally condition and history will affect the real value), but as the dealers are all rather far away (in vexingly diverse directions) this should hopefully allow me at least to reject the overly ambitious. Given the same model (and ignoring age) leaves mileage and asking price.

I've been calculating the difference between the new price of the model (it's still made) minus the asking price; call that the saving. A large saving is good, a small one bad. Then I'm simply dividing the saving by the mileage (as a big mileage is bad, a small one good). Assuming that car prices vary linearly with mileage (they don't, it's of a gentle exponential) does the logic of my calculation hold? That is, if V is (original-asking)/mileage, is it true to say that a higher V is (practical things aside) a better "value"? 87.115.17.103 (talk) 13:14, 3 June 2009 (UTC)

Mileage and price are nonnegative variables. So the approximate linear relationship may be between the logarithms of the mileage and price. Bo Jacoby (talk) 17:00, 3 June 2009 (UTC).

If the mileage is small then it doesn't matter. The OP did mention that the linear relationship is an approximation to the slightly better approximation of the value decaying exponentially with mileage. -- Meni Rosenfeld (talk) 20:14, 3 June 2009 (UTC)

I'm not sure this is valid. Let's say the original price is 10,000$ and it depreciates at 1$/mile. If you are offered a car at 9,999$ with negligible mileage, then by your method it has infinite value - even though all you are saving is 1$. On the other hand, a 9,000$, 500 mile car will have a value of only 2, even though you are saving 500$ compared to the true value.

Your calculation would work if you were buying many cars and limited by the total mileage you can buy. But you're only buying one car, and you want to gain as much as possible from this purchase. To do this you need to determine how much each mile decreases the value of the car to you, and simply maximize the difference ValueToYou - AskingPrice. -- Meni Rosenfeld (talk) 20:14, 3 June 2009 (UTC)

The minute somebody buys a new car and drives it off the dealer lot, it becomes a used car. I've always heard that the value drops about $1000 at that instant. After that, there's a slower decline in value (you could get a used car price guide and plot a curve through it to see the shape). Also, brand new cars often have manufacturing defects, which get sorted out under warranty during the car's first year. Cars more than 3 years old have usually developed some problems from normal wear and tear. So there are some folks who claim that the most problem-free cars you can buy are used cars between 1 and 3 years old, AND you get the savings resulting from the instant depreciation that the first owner incurred.

That all said, this is probably about as good a time as it gets to buy a new car (dealers are desperate to sell). After a series of used cars I bought my first new one several years ago (I'm still driving it) and although I knew it that decision wasn't optimal in pure financial terms, I found it to be a much more satisfying experience than buying a used car, due to freedom from uncertainty about undisclosed problems with the car, etc. 207.241.239.70 (talk) 03:12, 4 June 2009 (UTC)

u dv

Okay, I haven't learned how to do this. I keep getting "parse error" or something similar when I try to make it appear.

In my calculus book, the first integral on the list of integrals is (integral sign) u dv = uv - (integral sign) v du. List of integrals on Wikipedia doesn't have it, so it must be somewhere else.

I was looking to see how Wikipedia explained this, just out of curiosity. I know how my calculus book does.Vchimpanzee · talk · contributions · 21:44, 3 June 2009 (UTC)

Integration by parts, no? Angus Lepper^{(T, C)} 21:45, 3 June 2009 (UTC)

Yep, that's integration by parts all right (it's easy enough to derive, just integrate the product rule for differentiation and rearrange it a bit). To write in on Wikipedia you type this: <math>\int u\,\mathrm{d}v=uv-\int v\,\mathrm{d}u</math> which displays as: $\int u\,\mathrm {d} v=uv-\int v\,\mathrm {d} u$. --Tango (talk) 21:56, 3 June 2009 (UTC)

"just integrate the product rule for differentiation and rearrange." Although slightly pedantic, it is important to understand that the functions involved in "integration by parts" must be continuously differentiable. Otherwise, the obvious proof is incorrect. However, this is just a side comment to Tango's post above. --PST 01:00, 4 June 2009 (UTC)

Yeah, I knew where it came from, but I just forgot what it was called and how it would appear in Wikipedia. I could have opened the calculus book up and found that term. Thanks.Vchimpanzee · talk · contributions · 13:25, 4 June 2009 (UTC)

Indeed, my parenthetical asides do sometimes lack rigour! --Tango (talk) 16:37, 4 June 2009 (UTC)

June 4

Cirumference of a circle

Pi is unending as I understand it, there is no final digit after the decimal point (there is an internet site giving it to 200 billion decimal places - http://ja0hxv.calico.jp/pai/epivalue.html . Since the circumference of a circle is Pi times the diameter, how can the circumference be finite if the value of pi goes on to infinity? I know I’m missing something obvious but I don’t know what it is. Thanks - Adrian Pingstone (talk) 16:54, 4 June 2009 (UTC)

• Having a decimal expansion in which infinitely many digits are nonzero is not the same as being infinite. Otherwise, even simple numbers like 1/3 = 0.333333... (repeat ad infinitum) would be infinite and arithmetic would be pointless. —David Eppstein (talk) 17:02, 4 June 2009 (UTC)
• (2x edit conflict) Actually, pi does not "go to infinity". It is lesser than 3.2, for instance. So if d is the diameter of a given circle, the circonference is certainly lesser than 3.2 * d. The fact that (a particular way of writing) pi has infinitely many digits has nothing to do with its size. Hope this helps, Goochelaar (talk) 17:04, 4 June 2009 (UTC)
• It might help if you consider this analogy. Let's say that you must walk a mile to your destination, but you can only walk half the distance every time before stopping. That is, the first time, you walk 1/2 mile before stopping. Next, you walk half of that distance before stopping, which is 1/4 mile. Then 1/8 mile, 1/16 mile, and so on ad infinitum. It takes an infinite amount of sequences to reach the 1 mile mark, even though you're only walking a finite distance.

If that is too theoretical for you, simply consider any infinitely-repeating decimal, like 1/3 (which is .3 + .03 + .003 + .0003 + ...). Hope this helps, MuZemike 17:09, 4 June 2009 (UTC)

• (ec, and I even have the same example (but a nicer formula)) The value of pi is not going to infinity, only the length of its decimal representation is infinite. This can be seen as a variant of Xeno's paradox. The important insight is that the sum over an infinite number of summands is not infinite. Consider the simpler case of 1/3, which is 0.333.... It can be written as 0.3+0.03+0.003+0.0003+..., or as $\sum _{i=1}^{\infty }{}3\times 10^{-i}$, but still will always be smaller than 0.4, just like pi is smaller than 3.2 and hence the circles circumference is bounded. --Stephan Schulz (talk) 17:11, 4 June 2009 (UTC)

Many thanks for your answers, now I see my error - Adrian Pingstone (talk) 18:44, 4 June 2009 (UTC)

It's not really your error so much as not having it presented to you as the examples above. It's difficult to understand any mathematical concepts poorly presented. Glad the the folks here were able to help. hydnjo (talk) 20:30, 4 June 2009 (UTC)

That's the basis of mathematical analysis. MuZemike 23:05, 4 June 2009 (UTC)

Let a_n = ${\frac {1}{n^{2}}}$. It may be interesting to note that the sum of all a_n (where n is a natural number), converges; that is, this sum equals to π²/6. This maybe non-trivial but it illustrates how the sum of infinitely many numbers may actually be finite. See Convergence (mathematics) for a simple detailed analysis of this. --PST 04:03, 5 June 2009 (UTC)

You couldn't have used an example that doesn't involve pi, an "infinite" number? —Tamfang (talk) 15:39, 27 June 2009 (UTC)

Adrian is not the first philosopher to argue that infinitely many contributions must produce an infinite total amount. See Zeno's paradoxes. Bo Jacoby (talk) 07:28, 6 June 2009 (UTC).

As we often see with threads on the RD ;) pma (talk) 10:17, 7 June 2009 (UTC)

June 5

Topology: Density of rational functions and Euclidean Balls

Hi there:

a while back I asked a question on Euclidean balls, [2] - but I wasn't sure I completely followed the solution: specifically I was looking for how to find the 'k' I mentioned in the question for a ball of radius r slightly > 3 (3.001) containing unit balls, such that the number of balls in R^n is an exponent of (1+k) - and also how to find whether the same exists for a ball of radius slightly >2 (2.001) - I'd be very surprised if the latter did exist since it seems a bit pointless asking a question about '3.001' rather than 3 and then dropping down to just over 2: but how can I show it?

Also, what would be the cleanest way to show that the rational functions $\subset C[0,1]:\mathbb {Q} \mapsto \mathbb {Q}$ are dense in C[0,1] under the metric $d(f,g)=max_{x\in [0,1]}|f(x)-g(x)|$?

I could use a good explanation for my first question on the balls if anyone's willing to give one (which I'd very much appreciate!) - the second question I could probably do fine if someone just pointed me in the right direction with the best way to show that its completion is C[0,1]: I'm happy to show any space is dense in its completion, the starting off just seems so obvious in a way I can't see where to go with it!

Thanks very much guys!

Spamalert101 (talk) 07:16, 5 June 2009 (UTC)

The quickest way to answer your second question is to note that it's an immediate consequence of the Stone–Weierstrass theorem. I don't know if you want something lower-technology than that. Algebraist 11:17, 5 June 2009 (UTC)

I think part of the first question is answered in Wikipedia:Reference desk/Archives/Mathematics/2009_May_21#Spheres_in_an_n-dimensional_sphere The change from a linear to exponential numbers happens at diameter 1+sqrt(2). Dmcq (talk) 12:29, 5 June 2009 (UTC)

I'm not sure about what you mean by "the rational functions $\subset C[0,1]:\mathbb {Q} \mapsto \mathbb {Q}$"? --pma (talk) 13:26, 5 June 2009 (UTC)

I thought the OP just meant the normal rational functions mapping the closed interval [0,1] to itself and I don't know why Algebraist deleted his/her/its/their contribution. Dmcq (talk)

In fact my doubt was about the meaning of $\mathbb {Q} \mapsto \mathbb {Q}$. Maybe he wants rationals functions mapping Q into Q, i.e. with rational coefficients? its: you mean he/she could be just a program?? I also had this suspect as he/she has always perfect answers... instead, we humans happen to say silly things sometimes :) --pma (talk) 16:49, 5 June 2009 (UTC)

If I did in fact have always perfect answers, this would prove my nonmachinehood. Algebraist 17:14, 5 June 2009 (UTC)

I try not to make assumptions. Statements like 'If I don't go to bed I'll not be able to get up in the morning' are what I aspire to Dmcq (talk) 17:55, 5 June 2009 (UTC)

In fact the Approximations section of that article states the version the OP wants with rational coefficents. However it doesn't give a proof which is what the OP wanted. Dmcq (talk) 15:22, 5 June 2009 (UTC)

Which article is that? As far as I can see, none of the articles linked above has an "Approximations" section. Anyway, Stone–Weierstrass theorem#Weierstrass approximation theorem refers to Bernstein polynomial#Approximating continuous functions which is followed by a proof, though it relies on the weak law of large numbers so it is rather an unusual one. — Emil J. 15:41, 5 June 2009 (UTC)

Sorry I meant Stone–Weierstrass theorem#Applications. Don't know how that became approximations. Dmcq (talk) 07:02, 7 June 2009 (UTC)

On reflection, I think the OP wants to show that the continuous functions on [0,1] which take rational values at rational points are dense in the whole space. In that case, my original remark stands, since the set in question is obviously dense in its R-linear span, which is dense in the whole space by the theorem. Algebraist 17:14, 5 June 2009 (UTC)

volume of torus

How is the volume of a torus (donut) calculated? is it the area of the cross section multiplied by the circumfrence at the center of the cross section (assuming it is a circular cross section and the torus itself is circular?) —Preceding unsigned comment added by 65.121.141.34 (talk) 14:10, 5 June 2009 (UTC)

Yes, Torus#Geometry says so. PrimeHunter (talk) 14:16, 5 June 2009 (UTC)

And the reason is Pappus's centroid theorem. It thus does not really matter whether the cross section is circular. — Emil J. 14:19, 5 June 2009 (UTC)

June 6

Separated/connected equivalence

If A and B are disjoint from each other's closure, can they be separated by open disjoint sets? The converse is easy, but going from disjoint closure to open sets seems impossible. I don't need the answer, only a confirmation of whether it's right or wrong. ThanksStandard Oil (talk) 06:30, 6 June 2009 (UTC)

If two sets satisfy the hypothesis satisfied by A and B above, they are called "separated sets." A topological space is said to be completely normal, if any two separated sets in the topological space, can be separated by open sets (according to the Wikipedia article). Equivalently, a topological space is completely normal if every subspace of X is normal. Given the high probability that the context of which you speak above is that of the metric spaces, the statement which you assert is correct. In particular, metric spaces are completely normal since a subspace of a metric space is a metric space in its own right. However, note that this property is not satisfied by topological spaces in general. --PST 09:30, 6 June 2009 (UTC)

Thanks, I figured a while later that non-normal spaces don't satisfy the property that I couldn't prove, but didn't know it had to be completely normal for the property to be true. I was aware that metric spaces are normal (and thus completely normal like you said), but I needed this for all topological spaces...Standard Oil (talk) 17:15, 6 June 2009 (UTC)

Spherical wedge

Hello. I recently created Spherical wedge. An image would really brighten and clarify it, I think. For the pleasure of it, and so I know where to go for next time, could someone forward me to an application I might use to construct a sphere and perform operations such as colouring and labelling on it? I'm grateful for any response. —Anonymous Dissident^Talk 10:19, 6 June 2009 (UTC)

POV-Ray works exceptionally well for these purposes. It's a small, free and easy to learn application. But beware, you'll have to write the scene down. There's no graphical interface. — Kieff | Talk 19:24, 6 June 2009 (UTC)

Slightly off topic, but the term wedge to me does not inherently mean a wedge meade with two great circles. Is there some historic reason why wedge here applies to that condition? What does one call a wedge made with (a) Two equal sized non-great intersecting circles and (b) any two intersecting circles? -- SGBailey (talk) 19:54, 6 June 2009 (UTC)

That's a good question. All I know is that all my sources, old and new, have always mentioned the intersection of two halfplanes. The real definition for a spherical wedge is a solid of revolution obtained by revolving a semicircle through only x degrees, rather than the full 360 to produce a sphere. I'll look into it some more... —Anonymous Dissident^Talk 22:36, 6 June 2009 (UTC)

Another thing: I'm fairly sure the exterior lune is a defining feature, so you couldn't have a wedge defined within the sphere, as far as I know. —Anonymous Dissident^Talk 22:56, 6 June 2009 (UTC)

June 7

Hypothesis testing, backwards?

So far, I have learnt how to carry out a hypothesis test using the binomial expansion in the following way:

A company believes that 80% of the lightbulbs it makes are not faulty. They test a sample of 15, and find that six are faulty. Can their initial claim be justified?

Hence, X~B (15, p) and H₀: p = 0.8, where p is P (Not faulty). $1-{\frac {6}{15}}=1-0.4=0.6<0.8\therefore$ H₁: p < 0.8.

Or something along those lines. Anyway, then you use cumulative binomial tables to figure out whether H₀ can be accepted or rejected at a given percentage tail. That's all well and good, but how would one go about doing this backwards, as it were? Id est, if one began by carrying out that test and finding that p = 0.6, how would one find out the range of H₀s that one could reasonably fob off on people (without changing the result of the test, of course)? It Is Me Here ^{t / c} 09:09, 7 June 2009 (UTC)

The binomial distribution, that out of n bulbs you find i faulty ones, is ${\binom {n}{i}}P^{i}(1-P)^{n-i}$ where P is the probability that any bulb is faulty. The very same expression describes an (unnormalized) beta distribution for the likelihood density of P, knowing n and i. The formula for the mean value ± the standard deviation is $P\approx {\frac {i+1}{n+2}}\pm {\sqrt {{\frac {1}{n+3}}\cdot {\frac {i+1}{n+2}}\cdot \left(1-{\frac {i+1}{n+2}}\right)}}.$ When i = 6 and n = 15 you get $P\approx {\frac {7}{17}}\pm {\sqrt {{\frac {1}{18}}\cdot {\frac {7}{17}}\cdot {\frac {10}{17}}}}=0.412\pm 0.116.$ The believed probability P = 1−0.8 = 0.2 that a bulb is faulty, is (0.412−0.2)/0.116 = 1.8 standard deviations away from the mean value. Based on this the initial claim (P = 0.2) cannot be rejected. (See [[3]]). Bo Jacoby (talk) 16:26, 7 June 2009 (UTC).

What the original poster asks for is a confidence interval for p. Certainly there are reasonable approximate confidence intervals based on the normal approximation to the binomial distribution. I'm not sure if we have an account of exact confidence intervals for this situation somewhere on Wikipedia. More later.... Michael Hardy (talk) 03:34, 8 June 2009 (UTC)

I cannot tell if the above method is original research, so I did not include it in Wikipedia, even if it solves a commonly occuring problem. A confidence interval, 0 < a < P < b < 1, with confidence level 0 < L < 1, satisfies ${\frac {\int _{a}^{b}P^{i}(1-P)^{n-i}dP}{\int _{0}^{1}P^{i}(1-P)^{n-i}dP}}\geq L.$ This is computationally more difficult than computing the exact mean value and standard deviation of the beta distribution by the above formula, and then using the Chebyshev inequality to estimate an exact confidence interval pessimistically, or the normal distribution to estimate an approximate confidence interval more realistically. Bo Jacoby (talk) 07:03, 8 June 2009 (UTC).

Connected sets

Following my question on June 6, there's a bigger problem I need help with. I'm trying to find a space that contains two non-empty closed subsets A and B such that they are disjoint and A is disconnected, and their union is connected. I have a possible model:

Let C1 be the circle with radius 1 centered at the origin, C2 be the circle of radius 1 centered at x=1.5 y=0 and C3 be the circle with radius 1 centered at x=3 y=0.

Let X=union of these circles, and its topology be generated by the 3 circles. Let A=complement of C2, B=complement of ((C1)U(C3))

Clearly they are disjoint, and closed because they are complements of open sets, and also A is disconnected because it can be dissected by C1 and C3. But is AUB connected? It seems impossible to check with all the possible open sets... I really need an answer... And please say it's correct, I've spent 2 days working on this problem now...Standard Oil (talk) 09:55, 7 June 2009 (UTC)

But there is no such a space, because A and B would be closed also in their union wrto the relative topology. So, just by definition, their union can't be connected. --pma (talk) 10:08, 7 June 2009 (UTC)

Are you talking about disjoint closed sets can be separated by open sets? Yes that's exactly what I'm trying get around because there are non-normal spaces... Can you give me two disjoint open sets that dissect AUB?Standard Oil (talk) 10:14, 7 June 2009 (UTC)

Yes: A and B. --pma (talk) 10:19, 7 June 2009 (UTC)

But how can you get A and B from just performing union and intersection on the circles? Maybe I just don't see it... Can you write the process to attain A and B? —Preceding unsigned comment added by Standard Oil (talk • contribs) 10:24, 7 June 2009 (UTC)

It is simpler than that: if A and B are disjoint closed subset of any space X, then they are both open and closed in AUB since they are the complement of each other in AUB. In general, it can be useful for you to reflect on useful equivalent forms of the definition of a connected topological space X, e.g.:

• X is not union of 2 nonempty disjoint open sets;
• X is not union of 2 nonempty disjoint closed sets;
• The only clopen (i.e. both closed and open) subsets of X are the empty set and X itself;
• The only subsets of X with empy boundary are the empty set and X itself.

--pma (talk) 10:19, 7 June 2009 (UTC)

Ohhhhh I see what you mean... There's a misunderstanding between us here, my mistake. Your talking about connected space. I should have said connected set, meaning a set A such that you can't find two open disjoint sets V1 V2 in the embedding space that dissects A (V1 intersection A and V2 intersection A both non-empty, and A is a subset of (V1)U(V2)). What I'm asking here is, is it possible to find two disjoint open sets in X that dissects AUB, so I'm not asking whether AUB is a connected space in its own right, but rather is AUB a connected set in X.Standard Oil (talk) 10:39, 7 June 2009 (UTC)

OK. So what you want are two nonempty closed disjoint subsets A and B of a topological space X, that are not separated. Briefly, (X,A,B) is a counterexample to normality. Moreover, you wanted A disconnected (say, in the stronger sense that it is union of two separated closed subset of X).

Consider X':=X U {*}, with * a clopen point, and A':=AU{*} and B. Is it what you want?

Remark: In any case by "connected set" of a topological space people universally mean the same thing as "connected subspace" that is as a topological space with the subspace topology. The same convention is standard with any quality of topological spaces (so a compact subset, paracompact subset, metrizable subset etc all refer to the induced topology). --pma (talk) 11:38, 7 June 2009 (UTC)

I don't understand exactly what you mean by clopen point, you mean like the singleton is clopen? Furthermore, having A and B not separated doesn't mean their union is connected (in my sense), because we can well have 2 disjoint open sets both containing parts of A and parts B. Is my example right though? ThanksStandard Oil (talk) 12:05, 7 June 2009 (UTC)

Oops. --pma (talk) 12:13, 7 June 2009 (UTC)

Thanks for your help, maybe I'm just too confusing (a beginner can't be too lucid). Let's say that for a set A, you can find two open disjoint sets V1,V2 such that V1 intersection A and V2 intersection A are both not empty, and furthermore A is a subset of (V1)U(V2). There are 4 conditions here, and we'll just call A dissectable. So, what I want is two non-empty disjoint closed sets A and B satisfying: A is dissectable, AUB is NOT dissectable.Standard Oil (talk) 12:20, 7 June 2009 (UTC)

A particular quotient space.

I'm having difficulty trying to show that the unit square, quotiented out by the relation that identifies $(x,y)$ with $(1-x,1-y)$, is homeomorphic to S². I've had no luck simply trying to stretch the space around and gluing the relevant points together (I can't find a way to carry this out in a finite number of steps, there are always points left unglued) - the closest I came was to put a half twist in, then join one pair of opposing edges together (like a Moebius strip), but that just gets me into a mess. I've noticed the square (with identifications) has half-turn symmetry, but I can't see how to use that fact. It seems at first that any line drawn on the square, passing through the centre, is a great circle on the sphere (since the ends are identified), which seems like a good avenue to head down, but then you have to perform subsequent gluings to pairs of points on the great circle, which gets into a bit of a mess.

I've tried thinking about it purely in terms of open sets of each space and the like, but I'm never really sure how to do that generally. Any help would be much appreciated, thanks! Icthyos (talk) 12:48, 7 June 2009 (UTC)

I assume your unit square has edges identified so that it has the topology of a torus (without some such identification of edges, your square has a boundary so cannot be homeomorphic to S²). Cut the torus in half and work out how to sew up the cut edges so that $(x,y)\sim (1-x,1-y)$. Gandalf61 (talk) 16:07, 7 June 2009 (UTC)

Ahh, yes, the exercise sneaked in at the begin that it was a further identification, on top of the torus, which I missed. I think I can see why the two are homeomorphic - if you cut the torus in half in such a way that it corresponds to cutting along the diagonal of the square, you get it so that the two halfs are identified with each other, so all that's left to do is clamp, say, the top half shut, with two 'hinges' on opposite sides of the boundary (one being on the longitudinal (...meridional?) circle where we originally glued to get the torus), which gives us a surface which we can easily deform into a sphere! Is that correct? I'm not terribly confident at this. It doesn't help that it's so difficult and hand-wavey to explain just what you're doing in your head! Thanks a lot, Icthyos (talk) 18:16, 7 June 2009 (UTC)

Yes, that is correct. Parameterise your torus with parameters (θ,φ) with θ and φ taking values between 0 and 2π. Cut it in half say along the circles φ = 0 and φ = π. Then join each point (θ,0) to (2π - θ, 0), so your "hinges" are at (0,0) and (π,0). Do the same along the circle φ = π, with "hinges" at (0,π) and (π,π) and you are done. Gandalf61 (talk) 22:40, 7 June 2009 (UTC)

I'm not following this so far (maybe I will after I've thought it through....). If all this is correct, there should be a two-to-one continuous mapping from the torus to the sphere that's locally one-to-one everywhere and whose local inverse is continuous. As if you had a torus-shaped map of the earth on which every location appears twice. Can you specify that mapping? Michael Hardy (talk) 20:53, 8 June 2009 (UTC)

OK, maybe I am seeing it. For any point on the surface of the earth, there are two points on the torus where the tangent plane is parallel to the tangent plane to the earth at that point and the side of the plane that's in the interior of the torus is the same side as the side of the tangent plane to the sphere at the point in question. Someone who's good at computer graphics should be able to start with an ordinary globe and create a picture of this thing. Michael Hardy (talk) 20:57, 8 June 2009 (UTC)

...but the point (1/2, 1/2) corresponds only to itself, so that won't work. And also, the way I describe it, more than two points on the torus would correspond to the north pole (if you oriented things that way). OK, back to the drawing board. I suppose I should try to do this systematically instead of looking at it like this. Michael Hardy (talk) 21:02, 8 June 2009 (UTC)

See torus - "The 2-torus double-covers the 2-sphere, with 4 ramification points". The ramification points are the images of Icthyos's "hinges". They only have one pre-image on the torus; every other point on the sphere has two pre-images. On the original unit square the ramification points are at its corners (only one point on the torus); mid-points of its sides (two points on the torus); and at its center. There is a nice diagram on p194 of Peter Cromwell's Knots and Links (that page is, unfortunately, not accessible in Google books). Gandalf61 (talk) 22:49, 8 June 2009 (UTC)

Thank you, Gandalf.

Sometimes a page that seems not accessible by google books becomes accessible by cleverly altering the search terms. But you might have to think of some words that occur together only on that one page. Any ideas along those lines? Michael Hardy (talk) 05:00, 9 June 2009 (UTC)

Could it turn out that no axiomatic system is consistent?

Could it turn out to be the case that no axiomatic system of mathematics is consistent? If that happens, would Mathematics as a discipline then turn into a branch of Physics, or what? 94.27.225.206 (talk) 15:57, 7 June 2009 (UTC)

No. Presburger arithmetic is consistent. Taemyr (talk) 20:51, 7 June 2009 (UTC)

I don't think we exactly say with the kind of certainty being requested that Presburger arithmetic is consistent, I mean, sure, we can prove its consistency from Peano arithmetic, but if Peano arithmetic is inconsistent, then it can prove anything. For that matter, there are self-verifying theories that prove their own consistency, but again, who knows if they are really consistent?

We also have proofs that Peano arithmetic is consistent (Gödel's incompleteness theorem only proves that systems like PA can't prove their own consistency, not that their consistency can't be proven from "outside"). The first one, Gentzen's consistency proof, uses an induction principle on trees that turns out to be equivalent to PA but is still "intuitive"; another, in Gödel's Dialectica interpretation, uses higher-order functions, subject to the same problem.

Presberger arithmetic and Peano arithmetic both contain a "successor axiom" that says that every integer has a successor, i.e. there are infinitely many integers. There are actually a few mathematicians (see ultrafinitism) who reject that axiom and say there is a largest integer, which means all proofs relying on the successor axiom are wrong. Doron Zeilberger has written some interesting articles about this. Zeilberger is of the view that we are approaching an era when mathematics indeed is becoming like physics, where we have a lot of observations that we can experimentally seem to confirm, but which are beyond our capability to prove. Gregory Chaitin says similar things but from a different philosophical standpoint. 67.122.209.126 (talk) 03:12, 8 June 2009 (UTC)

Your first two paragraphs seemed to have some valid points, though anyone who doesn't accept the consistency of both Presburger and Peano arithmetic is probably just being a bit too anal. The last paragraph, however, I have to say seems to be confusing the issue between consistency and truth of an axiomatic system. The question had nothing to do with whether the system is "true" in any sense. A better approach, for the purpose of answering the OP's question, would be: are there mathematicians who will claim that those basic number theoretic systems mentioned are somehow inconsistent? Note: I mean that they will claim that the system absolutely is inconsistent, not that inconsistency is a possibility, as you have pointed out already. --COVIZAPIBETEFOKY (talk) 14:31, 8 June 2009 (UTC)

I have never heard of a serious mathematician saying "such and such theory is inconsistent, but I just haven't found the inconsistency yet; be patient".

However Edward Nelson has put effort into trying to find a contradiction in the extremely weak theory Q (Robinson arithmetic).

Also there was somebody whose name I can't remember for certain — I want to say Wette but the only remotely close hit I get is for an Elisabeth Wette, and I thought this was a man — who was claiming actually to have found a contradiction in, I think, Peano Arithmetic. As I remember it no one could find the error, because no one could actually follow the argument. --Trovatore (talk) 18:11, 8 June 2009 (UTC)

(Update: I did a little more searching, and I think it was Edward Wette. This is based on second-hand info, like Usenet posts. I can't find out very much about him in more traceable sources. He does appear to have a paper in a symposium for Goedel's sixtieth birthday.) --Trovatore (talk) 18:30, 8 June 2009 (UTC)

Pairwise intersecting sets of bounded size

For every natural number k there is a natural number G such that if each member of a set C is a set of size at most k and any two members of C have a non-empty intersection, then there exists a set A of size at most G such that the intersection of A and any two members of C is still non-empty.

This statement is not hard to prove. What I'd like to know is whether it is known, and where it appears in literature. I'd be interested about anything similar too.

The draft of the proof is this. Given C, construct a set A_p by induction on p such that for every element X of C either X and A_p share at least p elements or X intersects every member of C even in A_p. Once you have A_p, choose one element X of C for each possible intersection of X with A_p, and then choose A_p+1 as the union of these sets. The size of these A_p can be bounded by a number depending only on k and p.

This also proves a slightly stronger variant of the above statement, namely that you can have X and Y and A intersect for any member X of C and any set Y that intersects all members of C.

– b_jonas 20:13, 7 June 2009 (UTC)

June 8

statistics

can you explain more on validity and reliability of statistics. the truthfulness of statistics —Preceding unsigned comment added by 81.199.149.148 (talk) 09:58, 8 June 2009 (UTC)

See Statistics#Misuse. Bo Jacoby (talk) 20:54, 8 June 2009 (UTC).

You may also be interested in a (radio/podcast/online) series from the BBC called 'Go Figure'. A Google search for 'BBC Go Figure' turns up various pages (like this one) from the online magazine feature - the dropdown box in the top right lets you access others in the series. It's more light hearted, and written for a general audience, but may be of some use. Angus Lepper^{(T, C)} 23:31, 8 June 2009 (UTC)

Why doesn't 10 divided by zero equal infinity?

This seems like a really plausible result since division is nothing more than repeated subtraction. When I divide 10 by 1 then I in essence only count the number of times I can subtract 1 from the result of the previous subtraction, i.e. 10-1=9, 9-1=8... etc. until I am left with a value of zero or a remainder between zero and 1.

Therefore when I subtract zero from 10 I get 10 but the number of times I can do this subtraction and do it again and again and again is infinite. -- Taxa (talk) 20:58, 8 June 2009 (UTC)

Because it could be positive infinity or it could be negative infinity. 10/x gets bigger and bigger as x approaches zero from above, but it gets more and more negative if x approaches zero from below. For a limit to exist, it has to exist and be equal in both directions. There are some number systems in which positive and negative infinity are considered to be the same thing, in those systems 10/0 really does equal infinity. See real projective line. --Tango (talk) 21:03, 8 June 2009 (UTC)

See also Division_by_zero. Friday (talk) 21:04, 8 June 2009 (UTC)

10 / 0 = x is equivalent to saying what solves 0 x = 10, yes? But any value of x will not solve this equation. 10 / 0 isn't really meaningful. —Preceding unsigned comment added by 210.11.75.201 (talk) 02:54, 9 June 2009 (UTC)

Yes, but 0*infinity is indeterminate, which means it could be anything, including 10. That's why you can get meaningful and useful mathematics by defining 10/0 to be infinity (as long as you identify positive and negative infinity). --Tango (talk) 03:20, 9 June 2009 (UTC)

What you are telling is the limit (that too on the positive side) as the denominator approaches zero. 10/0 is simply not defined. Division by zero is not defined, which means there isn't any specified value for it. You can define it yourself, saying 10/0 is 10 from now onwards, but there wouldn't be much point in that as it wouldn't help you to solve any problem. Now the logic you are telling is the limit, which has already been clearly explained by others that it does not exist, as it can be both infinity or minus infinity. See One sided limit. Rkr1991 (talk) 03:59, 9 June 2009 (UTC)

For some purposes it makes sense to add just one "infinity" to both ends of the real line and say 10/0 = infinity. This serves the purposes of trigonometry, projective geometry, and complex analysis (where the whole plane has just one extra "point at infinity" tacked onto it. Michael Hardy (talk) 04:53, 9 June 2009 (UTC)

Plotting contour lines from discrete points

Suppose I have a number of discrete points with values attached to them. I want to now draw contour lines (isopleths) around these points. As a real example, plotting isobars on data from pressure data from weather stations.

Naturally, the plots such as these are performed by meteorological organizations are automated somehow.

What are the standard algorithms used to plot these points? Any online references I can look at? —Preceding unsigned comment added by 210.11.75.201 (talk) 23:24, 8 June 2009 (UTC)

June 9

Can a limit be infinite?

I suddenly got this doubt, and am a bit confused. Consider the function 1/|x|. As x tends to zero, should i say limit does not exist, or should i say, limit exists but is infinity? Better yet, consider y=x, as x tends to infinity. Rkr1991 (talk) 07:02, 9 June 2009 (UTC)

Yes, you'd usually consider those limits to be infinity. Evaluating a limit or some other expression to infinity is usually just fine. It's when you start treating infinity as a number that you get into trouble. --Pykk (talk) 08:19, 9 June 2009 (UTC)

In a casual context, saying the limit "goes to infinity" or "is infinity" will be understood by other people correctly. Saying the limit "exists" is sloppy at best but is probably ok.

Formally, you need to be clear about what set (more precisely, metric space or topological space) one is taking the limit in. If you are working in the real numbers, then the function 1/|x| has no limit as x goes to zero (i.e., the limit does not exist). If you are working in the extended real numbers, which are like the real numbers together with plus and minus "infinity", then the function 1/|x| has a limit of positive infinity as x goes to zero (i.e., the limit exists and is positive infinity). Eric. 131.215.159.106 (talk) 08:47, 9 June 2009 (UTC)

the function 1/|x| has a limit of positive infinity as x goes to zero — not quite: there are two limits, one approaching 0 from above which is positive infinity, and the other which approaches from below, which is negative infinity. — Charles Stewart (talk) 10:15, 9 June 2009 (UTC)

Absolute value? -- Meni Rosenfeld (talk) 11:11, 9 June 2009 (UTC)

Ah, yes. I managed to ignore those vertical lines. Quibble withdrawn. — Charles Stewart (talk) 11:37, 9 June 2009 (UTC)

I think it's common to treat limits involving infinity as complete pieces of notation. That is, "$\lim _{x\to a}f(x)=+\infty$" is defined to mean "for every $M\in \mathbb {R}$ there is a $\delta >0$ such that for every $0<|x-a|<\delta$ we have $f(x)>M$", "$\lim _{x\to +\infty }f(x)=+\infty$" means "for every $M\in \mathbb {R}$ there is an $N\in \mathbb {R}$ such that for every $x>N$ we have $f(x)>M$", and so on. You don't have to explicitly consider a topological space in which there is an element "$+\infty$". -- Meni Rosenfeld (talk) 11:11, 9 June 2009 (UTC)

While you don't need to be explicit about what you are doing, you shouldn't really use that kind of notation without knowing how to make it explicit, otherwise you risk making assumptions about how those limits behave that aren't true. For example, you might a limit that you've shown exists (with this broad definition of existence), so you call it L. You then prove that L+a=L+b for some real variables a and b, so you then deduce that a=b. This would be fallacious since, if L is infinite, you cannot subtract it from both sides. --Tango (talk) 16:36, 9 June 2009 (UTC)

I generally understand those written infinite limits as being little more than an abuse of notation. Under the standard epsilon-delta definition of a limit, those limits do not exist. However, a limit can fail to exist without the limit "being infinity". So saying that the limit is infinity can be thought of as being more specific than saying the limit doesn't exist. --COVIZAPIBETEFOKY (talk) 12:24, 9 June 2009 (UTC)

I would add: ...does not exist in R (whereas it exists as limit in the extended real line). In general, the epsilon-delta definition applies to metric spaces and needs a distance. The compactification of R is metrizable and the epsilon-delta definition applies perfectly. Notice that two elementary compactifications of R are commonly used: the one point (or Alexandrov) compactification , which is homeomorphic to S¹, and where the added point is usually denoted $\infty$ (without sign); and the order compactification, which is homeomorphic to the interval [-1,+1], that has two added points (positive and negative infinity) denoted $+\infty$ and $-\infty$. --pma (talk) 13:16, 9 June 2009 (UTC)

Wedge again

Can anyone with adequate knowledge provide some insight about this? —Anonymous Dissident^Talk 10:55, 9 June 2009 (UTC)

Look-and-say sequence

This is maybe rather a question concerning language than math: When Conway introduced this sequence in his 1986 article, he directly explained the rule and stated: "I note that more usually one is given a sequence such as 55555 ; 55 ; 25 ; 1215 ; 11121115 and asked to guess the generating rule or next term." Can we conclude from that whether he invented the system himself or not? If not - is there a way to ask him (there is no public contact address)? --KnightMove (talk) 12:20, 9 June 2009 (UTC)

Conway is current associated with Princeton University, so can almost certainly be contacted there. Our article says (in the Origins section): "It was introduced and analyzed by John Conway in his paper "The Weird and Wonderful Chemistry of Audioactive Decay" published in Eureka 46, 5-18 in 1986." So that would imply he invented it or, at least, was the first to publish something about it. I haven't read the paper, does it not make it clear where the sequence came from? If it doesn't say he got it from someone it probably means it was his own, since academics are generally very careful to correctly attribute things. --Tango (talk) 16:44, 9 June 2009 (UTC)

The next term is 31123115, and then 132112132115. Fun. I think it's an old game, that predates Conway, but I'm not sure. -GTBacchus^(talk) 16:49, 9 June 2009 (UTC)

Harmonic function

Hi. I'm studying for a complex analysis qualifying exam, and I'm working on a question about harmonic functions:

Find a harmonic function h on the upper half-plane such that $h=1$ on the positive real axis and $h=-1$ on the negative real axis.

So, I figure that, given $w=u+iv$, the function $u(w)=Rew$ is harmonic, and it equals 1 on the line Re w = 1, and -1 on the line Re w = -1. All I need is a function that takes the upper half-plane to that strip, sending the appropriate parts of the boundary to the right places. I found a function that pretty much does that. I start with the upper half-plane, apply the principal logarithm to get the strip $\{x+iy|0<y<\pi \}$. Then we just multiply by ${\frac {2}{i\pi }}$, and then subtract 1, and finally multiply by -1. This ought to send the upper half-plane to the strip between x=-1 and x=1. The whole function, composed together, comes out to: $w=1-{\frac {2}{i\pi }}Log(z)$. Our function h should be the real part of w, and we'll let z = x + iy.

Thus, $h(x,y)={\frac {-2}{\pi }}tan^{-1}({\frac {x}{y}})$. This function is harmonic on the upper half-plane (I checked its derivatives), but it's not defined on the real axis. Thus, do I want a function defined piecewise, that is h(x,y) when y>0; -1 when y=0, x<0; and 1 when y=0, x>0? Is that legit? Is it still a harmonic function, thus extended? -GTBacchus^(talk) 16:46, 9 June 2009 (UTC)

I think everything is fine, but perhaps you have a sign error? You just want the piecewise function to be continuous at the boundary (real axis other than the origin). Any sequence approaching (x,0) for x>0 must have all but finitely many x_n > x/2 > 0, so you get h(x,0)=-2/pi*arctan(+infinity) = -1, and for x < 0, you have x_n < x/2 < 0, so you get h(x,0)=-2/pi*arctan(-infinity) = 1. For harmonic functions, sometimes "boundary values" can be assigned even when the extension is not continuous, but you function is nice enough not to have to fool with this. JackSchmidt (talk) 17:40, 9 June 2009 (UTC)

Set of coordinates in R^2 with exactly 1 rational point - connected?

Hi there,

I've just shown why the set of points in $\mathbb {R} ^{2}$ with 1 or 2 rational coordinates is path connected - but I was wondering, what happens if we remove $\mathbb {Q} \times \mathbb {Q}$ though? Is the space still connected? Are there even any non-trivial open sets in this space? It seems like any open set would contain a rational point if I'm not being stupid, which would force the indiscrete topology and thus connectedness on this set, right?

Thanks, 131.111.8.97 (talk) 17:13, 9 June 2009 (UTC)

Take any point $\langle a,b\rangle \notin \mathbb {Q} \times \mathbb {Q}$. Assume for instance $a\notin \mathbb {Q}$ (the other case is symmetric). Using two line segments going through $\langle a,{\sqrt {2}}\rangle$, the point is connected by a path to the point $\langle {\sqrt {2}},{\sqrt {2}}\rangle$. We can do this for any point, hence $\mathbb {R} ^{2}-\mathbb {Q} ^{2}$ is path-connected. You seem to be confused about the definition of the subspace topology: if U is any open subset of $\mathbb {R} ^{2}$, then $U-\mathbb {Q} ^{2}$ is an open set in $\mathbb {R} ^{2}-\mathbb {Q} ^{2}$. — Emil J. 17:33, 9 June 2009 (UTC)

Sorry - to clarify, I meant we start with points without 2 irrational coordinates, so at least 1 coordinate is irrational, and then remove also points with 2 rational coordinates, so only points with exactly 1 rational coordinate are allowed... 131.111.8.97 (talk) 17:42, 9 June 2009 (UTC)