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May 10[edit]

Suggestion for a good book on axiomatic set theory[edit]

What is a good book on axiomatic set theory? I am not familiar iwth first-order predicate calculus so I want the book to have some discussion on that. I also want the book to be at graduate level. I know sets but don't know the axioms for them. I also don't know the continuum hypothesis well. Thanks for help. —Preceding unsigned comment added by 58.161.138.117 (talk) 02:03, 10 May 2009 (UTC)[reply]

Paul Halmos' "Naive Set Theory" gives quite a good treatment of the subject. There is very little first-order predicate calculus in it (in fact, I don't think there's any) - I found that to make it a good, intuitive introduction to the subject, as someone who also was not familiar with FOPC. (Don't be misled by the name - the book IS about axiomatic set theory; the 'naive' in the title is just referring to the intuitive treatment of the axioms.) Icthyos (talk) 07:54, 10 May 2009 (UTC)[reply]

I think the standard graduate texts on set theory are Kunen's and Jech's. I'm not at that level so I haven't done more than flip through them for a minute. "Naive set theory" is more of an undergraduate book. At the undergraduate level, Enderton's is a bit more modern. The class I took in college used it. If you're not familiar with first-order predicate calculus, you're likely to find any serious graduate set theory text to be absolutely impenetrable. I would start with Enderton's "Introduction to Mathematical Logic" or something comparable. Finally, Wikipedia has some very enthusiastic and knowledgeable editors writing about set theory and logic, so I've learned a lot from the articles here. 67.122.209.126 (talk) 11:49, 10 May 2009 (UTC)[reply]

I attempted to read Jech's book a few months ago, but found it far too terse when it came to explaining the FOPC - he just explained what the symbols he was using signified, then launched straight into long strings of them. I've yet to come across a book that sets it out at a nice, even pace - I might give the other ones mentioned above a look. Icthyos (talk) 16:15, 10 May 2009 (UTC)[reply]

The standard introductory graduate-level books for general set theory are indeed Jech's book and Kunen's book. Jech's is more thorough and much longer; Kunen's book spends more time being pedagogical but is not as good as a reference. In general I would recommend Kunen's book to start with for a graduate-level text. If you make it through that, you will have learned a significant chunk of basic set theory. You are unlikely to find a set theory book that spends a long time on first order logic; you will want a more general-purpose mathematical logic text for that. — Carl (CBM · talk) 04:35, 11 May 2009 (UTC)[reply]

Numerical Methods for Solving PDEs[edit]

Consider the PDE $u_{t}+au_{x}=f$ and consider the scheme

${\frac {11v_{m}^{n+1}-18v_{m}^{n}+9v_{m}^{n-1}-2v_{m}^{n-2}}{6k}}+a\left(1+{\frac {h^{2}}{6}}\delta ^{2}\right)\delta _{0}v_{m}^{n+1}=f_{m}^{n+1}.$

We are supposed to show that this scheme is accurate of order (3,4) and it is UNSTABLE for all values of $\lambda =k/h$ where h is the space step size and k is the time step size. I have several questions about this scheme.

1.Is this scheme correct the way it is? I think that it is supposed to be

${\frac {11v_{m}^{n+1}-18v_{m}^{n}+9v_{m}^{n-1}-2v_{m}^{n-2}}{6k}}+a\left(1+{\frac {h^{2}}{6}}\delta ^{2}\right)^{-1}\delta _{0}v_{m}^{n+1}=f_{m}^{n+1}.$

The inverse is missing, right? Can anyone verify that?

2.Assuming the scheme is correct the way it is, for the stability, I perform von Neumann analysis and my amplification factor comes out to

$\phi _{3}(z)=(11+\alpha i)z^{3}-18z^{2}+9z-2,\,\,\alpha =6a\lambda \sin(\theta )\left(1-{\frac {2}{3}}\sin ^{2}\left({\frac {\theta }{2}}\right)\right)$

and I want to see under what conditions is this polynomial a simple von Neumann polynomial. So I got all the way to $\phi _{1}(z)$ and I get that $|\alpha |\geq 3{\sqrt {15}}$ which then tells me that $|a\lambda |\geq {\sqrt {30{\sqrt {3}}-45}}$ . My questions is what does this mean? Is this a condition on stability? Or does this somehow imply that the scheme is unconditionally UNSTABLE? How is this implied? This is kind of fishy because everything I have worked with so far, the stability condition has always been less than or equal to something. But this time I have greater then something so I am not sure. By the way, the same thing happens if I assume the missing inverse on the scheme so this scheme appears to be unconditionally unstable in either case.

3.Furthermore, when I try to determine the order of accuracy, I need to take a smooth function, call in $\phi$ and then do its taylor expansion. The problem is that some terms (when I expand the delta expressions) I need to do taylor expansion in two variables simultaneously and I think that I need to do it up to $O(h^{5})$ and $O(k^{5})$ but it gets very cumbersome and complicated. First I didn't even know how to write the expansion but then the page here talks about multi-index notation but then I can't simplify using addition, subtraction, and division. Am I doing something wrong? Is it suppose to be much easier than this? This is actually preparation for the final in this class and I have spent the entire day today on this with no progress. Any help would be appreciated. Thanks! -Looking for Wisdom and Insight! (talk) 06:29, 10 May 2009 (UTC)[reply]

Means[edit]

I am collecting ice data and I got my hands on sea-ice area data for each 10 degrees of longitude for Antarctica by month. I would like to remove the longitude dimension in order to have only one average value per month. What would be the natural method to do this? Arithmetic mean, geometric, harmonic, any other? Thank you. 213.204.122.116 (talk) 08:05, 10 May 2009 (UTC)[reply]

It depends on the distribution of the (random) variable, x. If x has a normal distribution, the arithmetic mean is natural. If x has a natural zero point and 0<x, transform into y=log(x) before taking the arithmetic mean and transforming back. (This is using the geometric mean). If x has both a natural zero and a natural unit and 0<x<1, then use the transformation y=log(x/(1-x)). Bo Jacoby (talk) 10:09, 10 May 2009 (UTC).[reply]

Thank you for your answer, but what do you mean by "natural"? I meant it as the "best" or "mostly used", but you seem to use it as a technical statistical term which I'm unfamiliar with. 213.204.119.133 (talk) 13:37, 10 May 2009 (UTC)[reply]

"Natural" means "occurring in nature". In this context, it means the value automatically comes up when you consider the source/meaning of the data. For example, if you were looking at the thickness of the ice, there would be natural zero point (no ice at all). The idea of negative thickness is meaningless, so you don't want the fact that there are no negative numbers in your data to affect your results. If you were looking at the ratio of two different isotopes of oxygen dissolved in the ice, there would be natural zero (all one isotope) and a natural one (all the other isotope). The idea of a value less than 0 or more than 1 would be meaningless in this case, so you don't want the lack of such values to affect your results. --Tango (talk) 13:59, 10 May 2009 (UTC)[reply]

Seems to me the obvious thing to do, if you want the total amount of ice, is to add up the areas in the wedges. Or are you given the proportion of ice to open water in each wedge? In that case you need to weight them by the amount of water (since Antarctica is not a circle). —Tamfang (talk) 05:28, 11 May 2009 (UTC)[reply]

Maybe you could answer my question on the science desk about the total volume of ice in the world please. 84.13.171.69 (talk) 15:00, 10 May 2009 (UTC)[reply]

Top mathematical journals[edit]

Is there any impartial neutral agency that ranks the mathematical journals? Is there any pointer for this list? Thank you in advance. twma 09:23, 10 May 2009 (UTC)[reply]

as for your second question, no there is no point to having such a list. 94.27.208.52 (talk) 09:56, 10 May 2009 (UTC)[reply]

The ISI Science Citation Index (accessible through the Web of Science) is most commonly used in practice, even though the impartiality and neutrality of this organization as well as the rationale behind the algorithm it uses to compute its citation index are rather doubtful. — Emil J. 14:35, 10 May 2009 (UTC)[reply]

There is no "master list" that ranks all journals in order. However, because some sort of ranking is needed when universities evaluate faculty performance, there are many systems that are used by individual schools. Some use a measure of the number of citations to the journal that appear in print; the usefulness of this measure for mathematics is not as clear as its usefulness for lab science, but at least it's quantitative and objective. Other research universities simply present a list of journals broken down into groups, and expect faculty to publish in the higher groups and not the lower ones.

Typically, mathematicians who work in a particular subfield will have a sense of which journals are more prestigious and which are less so. If you have a particular field in mind, someone here might be able to give you a rough sense of the order. — Carl (CBM · talk) 14:58, 10 May 2009 (UTC)[reply]

Thanks for the general mathematical education. Based on Science Citation Index, mathematical Charlie Brown may be considered as good mathematics. Next question: How to determine the impact factor? Also based on Science Citation Index? Thank you in advance. twma 23:12, 10 May 2009 (UTC)[reply]

See Impact factor. — Carl (CBM · talk) 04:29, 11 May 2009 (UTC)[reply]

Thank you. In a hypothetical situation, a group of idiots making a lot of noise about a topic get their professorship while a clever person puts it into final definitive form but could never get a job because there is no more need to say anything about the topic. In a hypothetical situation, it could happen in scientific world. Thank you for the entertaining information about the reality. twma 16:39, 12 May 2009 (UTC)[reply]

Simple vector equation: most general solution[edit]

Hi there guys, was just wondering what the most general solution of x+y=2a, x.y=c is in vector form? It's not half as tough as my usual questions but I'm feeling quite ill & my brain has switched off today! Thanks a lot, Spamalert101 (talk) 15:47, 10 May 2009 (UTC)[reply]

Well, if you wanted to solve only for x, or only for y, then you could express the equation as

x·(2a − x) = c.

Then:

x·x −2a·x = −c.

Then complete the square:

x·x −2a·x + a·a = a·a − c.

|x − a|² = |a|² − c,

and you've got a sphere centered at a.

To solve for both variables, looking at (x, y) space rather than just at x space, might involve more than just that terse conclusion; I'm not sure. Michael Hardy (talk) 17:03, 10 May 2009 (UTC)[reply]

Excuse me, what do you mean? Haven't you solved completely the problem in both variables? You said x is a point in the sphere, and the corresponding y is 2a - x ; what should one do better than that? --pma (talk) 18:59, 10 May 2009 (UTC)[reply]

I agree, he's solved the problem completely. There are two variables and two equations (which are neither contradictory or equivalent), so there is a unique solution. --Tango (talk) 19:53, 10 May 2009 (UTC)[reply]

¿unique solution? indeed he proves that the set (x,y) of solutions is affinely isomorphic to an Euclidean sphere of the x space, and it is also implicit in his solution that there is a unique solution iff |a|² = c , and no solutions if |a|² < c. --pma (talk) 20:36, 10 May 2009 (UTC)[reply]

Oops! The second equation is a scalar equation, not a vector one, so what I said was nonsense (even without that, it was still not quite right - the solution set would be discrete, it wouldn't necessarily be a single point). In terms of scalars, you have 2n variables and n+1 equations, so the solution set will be (n-1)-dimensional (which the sphere obviously is). I apologise for my idiocy! --Tango (talk) 20:49, 10 May 2009 (UTC)[reply]

The set of values of x satisfying this equation is a sphere. The set of y values is likewise a sphere. The equation gives a one-to-one correspondence between the two spheres. So how does one geometrically describe the set of pairs (x, y)? Topologically it's just another sphere of the same dimension as either of the two aforementioned spheres. Michael Hardy (talk) 20:59, 10 May 2009 (UTC)[reply]

Wonderful, I'm appalled at quite how slow I was being but things are back on track now, thanks! Spamalert101 (talk) 22:01, 10 May 2009 (UTC)[reply]

We may just rephrase your solution formula saying that a solution (x, y) is exactly an ordered pair of antipodal points on the sphere of center a and radius (|a|² - c)^1/2 --always assuming |a|² > c --pma (talk) 06:47, 11 May 2009 (UTC)[reply]

Areas of circles[edit]

Can someone help me with this problem? [1]I need to find the shaded region's area. Yes, it is a homework problem if you're wondering, I'm completely stumped by it. All the the middle circle is tangent to the rest of them, and the diameter of the large circle is split into 4 congruent parts by the centers of the circles. 24.6.46.177 (talk) 20:47, 10 May 2009 (UTC)[reply]

You are probably already aware, that the grey area can be expressed by sums and differences of the individual circle areas in the figure. (If you want to know more about that, see Constructive solid geometry). The core of the problem is to find the radius for every circle. One must be given in any case, two are easy. For the radius of the small center circle, you need to establish a relation (=equation) to other sizes in the figure which you already know and solve it.

There are probably many ways; my first idea was this: connect the center of one medium sized circle to the center of the small circle and also draw a vertical line through the center of the small circle across the whole figure. If you cannot see it immediately, try to label all resulting line segments and start collecting every bit of information you have about their relative lengths. You might need to use some ancient Greek mathematics in doing so. —Preceding unsigned comment added by 84.187.75.89 (talk) 01:24, 11 May 2009 (UTC)[reply]

I'd solve it as above since it is homework, but when you're finished you might be interested that it is an instance of Soddy circles, he produced an execrable poem with the solution which is in Problem of Apollonius Dmcq (talk) 11:46, 11 May 2009 (UTC)[reply]

bicycling paradox[edit]

I just read something on the internet that seems like a "paradox." It was an article saying that a bicyclist riding against traffic (left lane in the US) on average passes more cars in that lane than a bicyclist riding with traffic (right lane in the US) does. It makes a sort of intuitive sense: if there's a steady stream of cars, your higher relative velocity in the left lane means that you pass more cars per minute and thus more cars on your whole trip. But realistically come on. if there are 5 people driving down that road to the park and back during the hour you're riding, they will all pass you, whether you're against traffic or with traffic. Do we have an article on something like this? —Preceding unsigned comment added by 71.176.161.131 (talk) 22:41, 10 May 2009 (UTC)[reply]

5 people will not be a continuous steam of traffic, though. If we assume all the cars go at the same speed and with the same distance between them, regardless of what direction they are going, we can characterise the number of cars that will pass us in terms of the length of time during which they can start their journey and pass us. Let's assume we go at 10mph and the cars go at 60mph and it is a 10 mile journey. If we travel with the traffic, a car cannot start before we do and they have to start early enough that they can reach the end of the journey before we do. It will take us an hour to do the journey and a car would take 10 minutes, so there is a 50 minute window in which they could leave. If we are travelling against the traffic then they can't leave after we finish our journey or more than 10 minutes before we start it. That gives them a 70 minute window in which to leave. That's 20 minutes longer so allows more cars to pass us. --Tango (talk) 23:28, 10 May 2009 (UTC)[reply]