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August 9

Stone-Cech Compactification

Hi, Does anyone have an idea how to calculate the power of the Stone-Cech compactification of a given space X with a given power? Is it 2^X? And if so, how do I show it? Thanks! —Preceding unsigned comment added by Topologia clalit (talk • contribs) 06:17, 9 August 2010 (UTC)

Certainly not in general. For example βN is the set of ultrafilters on N (with the appropriate topology), so it has cardinality $2^{2^{\aleph _{0}}}$. On the other hand, if X is already compact, then βX is just X (I think). So I don't believe there's a general formula independent of the topology on X. --Trovatore (talk) 06:22, 9 August 2010

(UTC)

Thanks. But can you explain why βN has cardinality $2^{2^{\aleph _{0}}}$? Why isn't it just $2^{\aleph _{0}}$? Topologia clalit (talk) 06:37, 9 August 2010 (UTC)

Well, on the face of it, an ultrafilter on N is a set of subsets of N, not a subset of N, so you'd kind of expect there to be $2^{2^{\aleph _{0}}}$ of them. Of course not every set of subsets of N is an ultrafilter, so this isn't a proof. Nevertheless it's true. It's a theorem of some dude named Pospisil or something like that, with diactritical marks in various places. You can look it up in Jech, Set Theory. --Trovatore (talk) 06:43, 9 August 2010 (UTC)

Ah, turns out someone had asked about this a long time ago on sci.math, and I wrote up Pospíšil's proof and posted it. You can find it at http://www.math.niu.edu/~rusin/known-math/00_incoming/filters . --Trovatore (talk) 06:48, 9 August 2010 (UTC)

OK Thanks! I'll have a look and try to figure it out. Topologia clalit (talk) 07:12, 9 August 2010 (UTC)

And of course since you can build the Stone–Čech compactification out of ultrafilters of zero-sets, β of any Tychonoff space X has cardinality at most $2^{2^{|X|}}$. Algebraist 09:52, 9 August 2010 (UTC)

For specific spaces, one can often get a better bound using the same method. For example, there are only $2^{\aleph _{0}}=|\mathbb {R} |$ zero sets (or even closed sets) in $\mathbb {R}$, hence $|\beta \mathbb {R} |\leq 2^{2^{\aleph _{0}}}=2^{|\mathbb {R} |}$. (In fact, it's easy to see that one can embed βN in βR, hence the cardinality of βR is exactly $2^{2^{\aleph _{0}}}$.)—Emil J. 11:41, 9 August 2010 (UTC)

Thanks a lot Guys. I can see now that the cardinality of the Stone–Čech compactification can not be more then $2^{2^{|X|}}$. Even just because filters are sets of subsets of X. But, why can one say that there are only $2^{\aleph _{0}}=|\mathbb {R} |$ zero sets or closed sets in $\mathbb {R}$? Also, if you conclude from the fact that one can embed βN in βR, that the cardinality of βR is exactly $2^{2^{\aleph _{0}}}$, does that mean that we know that $|\beta \mathbb {N} |=2^{2^{\aleph _{0}}}$? I mean, if we take N with the discrete topology then I guess that every subset of N is clopen and therefore can be a zero set? How can you be sure that $|\beta \mathbb {N} |<2^{2^{\aleph _{0}}}$? less then $2^{2^{\aleph _{0}}}$ is not the case? Topologia clalit (talk) 08:50, 10 August 2010 (UTC)

Every closed set corresponds to exactly one open set (its complement). To see there are only $2^{\aleph _{0}}$ open subsets of R, note that every open set is the union of open intervals of the form (r,s), where r and s are both rational. There are only countably many intervals of that form, so there are only $2^{\aleph _{0}}$ possible unions of them. The answer to your other question is the Pospíšil result again. --Trovatore (talk) 08:55, 10 August 2010 (UTC)

You are right, I should take a second look at the Pospíšil result and proof. But still I miss something here. Why can't we have the interval $({\sqrt {2}},{\sqrt {5}})$ as an open set? —Preceding unsigned comment added by Topologia clalit (talk • contribs) 09:03, 10 August 2010 (UTC)

It is an open set. Who said it wasn't? --Trovatore (talk) 09:10, 10 August 2010 (UTC)

I uderstood from what you wrote above that I have to "note that every open set is the union of open intervals of the form (r,s), where r and s are both rational.." I don't get it.. Topologia clalit (talk) 09:18, 10 August 2010 (UTC)

Every open subset of R is indeed such a union. Including $({\sqrt {2}},{\sqrt {5}})$. --Trovatore (talk) 09:17, 10 August 2010 (UTC)

You are defenetly right, sorry for the confusion, somehow i had in mind that you ment that the power of set of open intervals is countable while you actually said that it has the power of continuity... Topologia clalit (talk) —Preceding undated comment added 09:22, 10 August 2010 (UTC).

I keep on thinking about what you have all said above and I wonder, can one characterize, in which cases $|\beta X|=2^{|P(X)|}$? Topologia clalit (talk) 10:25, 10 August 2010 (UTC)

I doubt there's any very enlightening characterization. For example, consider X to be the disjoint union of the closed unit interval [0,1] and N (disjoint union so that e.g. the 0 of the closed interval has nothing to do with the 0 of N). Then I expect, though I haven't proved it, that βX will be homeomorphic to β[0,1] union βN. Since [0,1] is compact, β[0,1] is just [0,1] and has cardinality $2^{\aleph _{0}}$. βN as before has cardinality $2^{2^{\aleph _{0}}}$. So βX has cardinality $2^{2^{\aleph _{0}}}$, and easily X has cardinality $2^{\aleph _{0}}$, so X is one of the spaces you're trying to characterize. But it's thus for a completely silly reason! The part of X that makes its cardinality large is the part that plays no role in making the cardinality of βX large. --Trovatore (talk) 18:24, 10 August 2010 (UTC)

Oh, now I realize I was answering the wrong question — I thought you wanted $|\beta X|=2^{|X|}$. --Trovatore (talk) 18:26, 10 August 2010 (UTC)

β is left adjoint to the inclusion functor, and hence preserves colimits. So as you suspected, β of a disjoint union is the disjoint union of the βs. Algebraist 18:33, 10 August 2010 (UTC)

(When the union is finite, that is. Infinite coproducts in the category of compact Hausdorff spaces are more fancy than disjoint unions.)—Emil J. 18:41, 10 August 2010 (UTC)

I'll have to think about it for a while.. Thanks! 77.125.113.164 (talk) 09:41, 11 August 2010 (UTC)

Total differential

Good day. As we know, a total differential for

$z=x+y$

is

${\operatorname {d} z}={\frac {\partial z}{\partial x}}\operatorname {d} x+{\frac {\partial z}{\partial y}}\operatorname {d} y$

but what is the total differential for

$z=f(x,y)+g(t)$

? --Merry Rabbit (talk) 14:05, 9 August 2010 (UTC)

If x, y and t are independent then it is

$dz={\frac {\partial f}{\partial x}}dx+{\frac {\partial f}{\partial y}}dy+{\frac {dg}{dt}}dt$

On the other hand, if x and y depend on t then it is:

$dz=\left({\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}+{\frac {dg}{dt}}\right)dt$

Gandalf61 (talk) 14:54, 9 August 2010 (UTC)

what's wrong with this proof?

What's wrong with this proof? This is not homework. 85.181.49.221 (talk) 21:13, 9 August 2010 (UTC)

There are a couple of claimed flaws in the blog comments here (comments #28 and #55, and possibly others). -- BenRG (talk) 02:45, 10 August 2010 (UTC)

And a summary of possible issues, including those two, here. It will be a few days/weeks/months before anyone is sure of anything. -- BenRG (talk) 03:36, 10 August 2010 (UTC)

P versus NP problem for dummies

OK, I want to understand how this works but I don't actually understand what polynomial time is or all those other bits are. So here's how I think it works and you guys can tell me if I'm right or wrong on the level that I'm working at, which is very low.

Some things are easy to figure out using computers and some things are very hard. Some things are easy to figure out an answer to and easy to prove that an answer is correct for. This is P. On the other hand, some things are easy to prove an answer for but very difficult to figure out an answer to: for example, it's easy to prove that two prime numbers multiply into another number, but it's very tricky to take that number again and figure out the two prime numbers that we started with. These are NP problems. But what the P versus NP problem is is proving that P problems and NP problems are ACTUALLY different and not that we're just stupid and haven't figured out easy ways to do these NP problems yet. Do I have this right? —Preceding unsigned comment added by 88.108.242.37 (talk) 21:50, 9 August 2010 (UTC)

Almost. The thing you're missing is that all P problems are automatically NP. NP doesn't say that it's hard to figure out the answer; it only says that it's easy to check the answer. NP-complete says it's hard to figure out the answer (or at least as hard as it can be for an NP problem). --Trovatore (talk) 22:16, 9 August 2010 (UTC)

OK, sweet. 88.108.242.37 (talk) 22:23, 9 August 2010 (UTC)

Also, there are (probably) more than two levels of difficulty here. Factoring, for example, is likely not in P but it's also thought to be easier than the most difficult problems in NP (the NP-complete problems). A proof that P ≠ NP would tell you that the NP-complete problems are not in P, but wouldn't necessarily tell you anything about the difficulty of factoring. And one can argue with the idea that problems in P are "easy". If the best known solution for a certain problem takes n¹⁰⁰ steps for a case of size n, then that problem is in P, even though the algorithm is unusably slow in practice. Likewise an algorithm that takes Kn² steps, where K is Graham's number or something equally ridiculous. It's very rare for the best known algorithm for a problem to have a run time like that, but I think there are a few examples (?). -- BenRG (talk) 03:02, 10 August 2010 (UTC)

Conversely, not being in P doesn't make it slow. 1.00000000000001ⁿ isn't polynomial, but it's still very quick for any size of n you are likely to come across in a real-world problem. --Tango (talk) 03:59, 10 August 2010 (UTC)

I feel that you shouldn't use "hard" and "easy". You should use "time-consuming" and "quick". Factoring, which BenRG used, is an example of why "hard" and "easy" are terribly confusing. Factoring is very easy. Want to factor 28475938273? See if divides evenly by 2, then 3, then 4, then 5, then 6... That is easy, but time consuming. Checking it is easy also. If I said the factors were 17 and 295, you just multiply the two and see if you get 28475938273 as an answer. It is easy and quick. So, you can see that factoring is not "hard". It is time-consuming. To date, there is no quick way to factor large numbers. So, it doesn't have a polynomial-time solution. You can read "polynomial-time" to mean "quick". However, checking the solution is very quick - so checking it is polynomial-time. -- kainaw™ 03:14, 10 August 2010 (UTC)

A good indication of why, when you want to be fully accurate about such things, the best approach is to learn the technical language and then use it correctly. --Trovatore (talk) 03:55, 10 August 2010 (UTC)

I wouldn't use "quick" either, for the reason given by Ben above. Just say "solvable in polynomial-time" and be done with it. Anything else is likely to be wrong in some way. --Tango (talk) 03:59, 10 August 2010 (UTC)

The problem is defining "polynomial time" for "dummies" (as the question states). In order to do so, you need to discuss upper bounds, big O, and things like that. So, giving a "dummy" a word like "quick" allows the dummy to latch on to a word that has meaning. Giving the dummy "polynomial time" makes as much sense as saying NP problems are a duck while P problems are a goose. If you can define what polynomial time means without getting into more complicated topics like big O, please do. -- kainaw™ 04:21, 10 August 2010 (UTC)

If you want to actually know what's going on, you have to learn about upper bounds and things like that. At a level short of that, "hard" and "easy" are about as good as anything else. --Trovatore (talk) 06:01, 10 August 2010 (UTC)

Some things just can't be understood by dummies. P=NP is a very technical problem and require technical knowledge to understand it. We just have to accept that. --Tango (talk) 13:17, 10 August 2010 (UTC)

You don't need to explain big O, since "O(polynomial)" is the same as "solvable in polynomially many steps", where "steps" are more or less what you'd expect—"add x and y and call the result z", "if z < w then answer Yes, else go to step 10", etc. The way I used "easy" and "hard" is standard, and I think it's reasonable enough, since the time complexity of the best algorithm does tell you something about the intrinsic difficulty of the problem. If "easy" meant "mindlessly/mechanically solvable, given unlimited time" then practically anything would be easy. -- BenRG (talk) 22:01, 11 August 2010 (UTC)

For dummies: A problem may be solved in a slow way and a fast way. Adding natural numbers can be done by counting: 518+357 = 519+356 = 520+355 = ...= 874+1 = 875+0 = 875, or by computing: 518+357 = (5·100+1·10+8)+(3·100+5·10+7) = (5·100+3·100)+(1·10+5·10)+(8+7) = (5+3)·100+(1+5)·10+(8+7) = 8·100+6·10+15 = 875. The time used by some method to solve a problem depends on the amount of data in the problem. Bigger problems take more time. The degree of some function f is lim_x→∞ log|f(x)|/log(x). For polynomials this use of the word degree matches the standard use of the word. Exponential functions have infinite degree, and logarithms have degree 0. The time used for addition by counting has infinite degree, while addition by computing has degree 1. The standard multiplication algorithm has degree 2, Karatsuba multiplication has degree 1.585, and the Schönhage–Strassen algorithm has degree 1. All known methods for integer factorization has infinite degree, but checking that some integer factorization is correct has finite degree. The N=NP hypothesis says that if checking a solution has finite degree, then solving the problem has also finite degree. Bo Jacoby (talk) 08:28, 10 August 2010 (UTC).

Not all problems can be solved in a slow way and a fast way, that's the point (if all problems could be solved in polynomial-time then P=NP would be trivial). I'm not sure what a list of multiplication algorithms was supposed to add to the explanation, either... --Tango (talk) 13:17, 10 August 2010 (UTC)

And what are the dummies supposed to do with "lim_x→∞ log|f(x)|/log(x)"? -- kainaw™ 18:58, 10 August 2010 (UTC)

No, not all problems can be solved both in a slow way and a fast way, and that's why I wrote that it may be solved in a slow way and a fast way. I wrote later that 'all known methods for integer factorization has infinite degree', so that problem apparently cannot be solved in a fast way. The list of multiplication algorithms exemplifies that the problem of multiplication can be solved in ways of different speed. Dummies need examples. The definition degree(f)=lim_x→∞ log|f(x)|/log(x) is elementary compared to complexity theory, and the dummy can verify that if f is the polynomial, f(x)=Σ_i=0^Na_ixⁱ where a_N≠0, then degree(f)=N. I should add that polynomial time means finite degree. Thanks for your comments. Bo Jacoby (talk) 07:33, 11 August 2010 (UTC).

But you haven't actually defined this function f. To do so, you need to explain about upper and lower bounds and all that complicated stuff. Working it terms of degrees doesn't help at all. Once you've defined the function, you can just say "if it's a polynomial or something that grows slower than a polynomial (such as a logarithm) then we call it polynomial-time". Mentioning degrees doesn't add anything. --Tango (talk) 12:43, 11 August 2010 (UTC)

The function f is the computing time as function of the size of the input. I do not need to explain the complicated stuff, and so working in terms of degrees does help. Bo Jacoby (talk) 20:44, 11 August 2010 (UTC).

Either way, "polynomial" is easier to understand than "of finite degree", so why not just say "polynomial"? --Tango (talk) 22:19, 11 August 2010 (UTC)

Feel free to use what you find easier to understand, of course. To me "polynomial", meaning literally "many terms", is a little confusing, because the important thing is not that it has many terms. Mathematicians say "algebraic" where computer scientists say "polynomial". That x log x is "polynomial" is not nice in mine ears, while the definition shows that it has degree 1. It is a matter of taste. Bo Jacoby (talk) 04:52, 12 August 2010 (UTC).

August 10

Different groups with the same order signature

Resolved

For a finite group G, let g_n be the number of elements of G with order n; call this sequence its "signature". For example, the signature of S₃, with trailing zeroes omitted, is [1, 3, 2], and the signature of Z₈ is [1, 1, 0, 2, 0, 0, 0, 4]. Isomorphic groups have the same signature, but the converse is not necessarily so. What are the smallest distinct groups with the same signature? —Mark Dominus (talk) 05:13, 10 August 2010 (UTC)

This first occurs at |G|=16. $\mathbb {Z} _{4}^{2}$ and $Q_{8}\times \mathbb {Z} _{2}$ each have signature [1, 3, 0, 12]. Algebraist 17:52, 10 August 2010 (UTC)

Thank you very much! —Mark Dominus (talk)

E-books

Hello does anyone know a place where I can find e-books? I'm at a university so I can access things like springer journals and other stuff that require you to login via some institution. I'm looking for the book An introduction to homological algebra by Weibel and my library doesn't have it. I never buy books because of the insane price. For example Hatcher and J P May have online versions of their books. Thanks Money is tight (talk) 06:04, 10 August 2010 (UTC)

If you want to know what e-books you can access through your institution and you can't find the information on the institution's website I'd suggest asking one of the institution's librarians. Qwfp (talk) 06:18, 10 August 2010 (UTC)

Have you thought about getting Saunders Maclane's Homology? (See this.) While there are many other excellent books on homological algebra, the novel feature of this book is the way in which the author has managed to provide applications of the theory to numerous branches of mathematics. (My copy is not with me at the moment but if I remember correctly, the book discusses homological algebra in the context of algebraic topology (especially; for example, a treatment of spectral sequences in this context is given), group theory (especially in relation to group cohomology which is an important tool in number theory and group theory itself (e.g., the Schur-Zassenhaus theorem)), commutative algebra and algebraic geometry.) The book is also accessible if you "know undergraduate mathematics" and at $48, it is definitely worth the money (at least in my opinion). (If you so desire, you can borrow it from your library and see whether you like it but I have little doubt that you will!) PST 08:15, 10 August 2010 (UTC)

Both Homology and An introduction to homological algebra are on gigapedia.com -Shahab (talk) 08:20, 10 August 2010 (UTC)

(Edit Conflict) Not to say that you should not get Weibel; this is an excellent book as well and does resemble Maclane in some ways. However, Maclane can be found in most libraries (at least it can be found in my library, but I believe that it can be found in other libraries as well), and so could be an option if no other decent book can be found. PST 08:23, 10 August 2010 (UTC)

Also, most university libraries do Inter-Library Loan. It can't hurt to ask. Septentrionalis PMAnderson 00:10, 16 August 2010 (UTC)

Name of function

Is there a name for this function: f(n) = Number of connected graphs possible on n vertices. Is a formula for this function. Thanks -Shahab (talk) 12:28, 10 August 2010 (UTC)

See Graph enumeration and (sequence A001187 in the OEIS) Dmcq (talk) 12:41, 10 August 2010 (UTC)

Transforming a principal stress tensor to maximum shear stress tensor?

Hi,

I'm feeling a little stupid here, because I thought I understood stress tensors, but here we go... All notation is MatLab, but it should be fairly easy to understand.

I have a stress tensor, A, stored as a 3x3 matrix on MatLab. It's a standard stress tensor, 6 independent components.

I do:

[V,D] = eig(A)

to obtain the eigenvalues, i.e. principal stress tensor, D and eigenvectors, i.e. direction cosines, V.

Such that performing:

inv(V)*D*V = V\D*V returns the original stress tensor A

I thought (and the wikipedia page on stress confirms this) that the maximum shear stress plane lies at 45 degrees to the principal stress plane.

So I tried:

V = V + cos( acos(V) + pi/4 ) and then V\A*V again to give the maximum shear tensor, but that didn't seem to work?

or, alternatively:

DIRCOSINES = zeros(3,3)

DIRCOSINES = cos(pi/4)

DIRCOSINES\D*DIRCOSINES

which should also work if I understand it correctly, but it doesn't.

I know how to extract the maximum shear stress (0.5*(sigma1-sigma3)), but what I really need are the direction cosines to go from A to that tensor where that maximum shear stress exists.

Any help would be really really appreciated.

Thank you! —Preceding unsigned comment added by 89.204.153.131 (talk) 16:27, 10 August 2010

Edit: Well, I've sorted it now, so if anyone else ever has to do this...

Firstly, my MatLab notation above was off, because it was from memory and I haven't been using it for long. It's now been corrected.

Secondly, a\D*a works, were a is a standard rotation matrix about an axis, i.e. for the 1-3 plane (if sigma1 was the highest principal stress and sigma3 was the lowest), it would be:

a = [0.7071 0 -0.7071; 0 1 0; 0.7071 0 0.7071]

(where one of those is a negative due to the -sin(angle) term, though I might have written the wrong one down there by mistake, but it's just the standard rotation matrix..) —Preceding unsigned comment added by 89.204.137.172 (talk) 15:46, 11 August 2010

Irreducible polynomial over the rationals

Hi there,

I want to show that x^3 + x^2 - 2x + 1 is irreducible over the rationals, or at the very least justify why it should be - I tried using Eisenstein's criterion with some sort of trick like the cyclotomic polynomials where you set X = Y-1, but I couldn't find any sort of substitution which gave the right coefficients. Suggestions, anyone?

Thankyou! :) 86.30.204.236 (talk) 17:51, 10 August 2010 (UTC)

If it were reducible, it would have to have a linear factor, i.e., a rational root. Since it is a monic integer polynomial, its roots are algebraic integers, and the only rational algebraic integers are true integers. It's easy to see that all roots of the polynomial have to have absolute value at most 4 or so, and then it's a matter of checking that no integer of absolute value at most 4 is a root of the polynomial.—Emil J. 18:00, 10 August 2010 (UTC)

You can make that slightly easier by, instead of thinking about absolute values, noting that any root must be a factor of 1 (the constant term), and so must be ±1. This is the idea of the rational root test. Algebraist 18:05, 10 August 2010 (UTC)

Ahh, you are a star, thanks :D 86.30.204.236 (talk) 18:04, 10 August 2010 (UTC)

Divergence Theorem

I've been battling with the following question all day and, after a wide variety of different answers, seem have to achieved the same answers enough times to confirm that I'm wrong.

"Let $\mathbf {F} (\mathbf {r} )=(x^{3}+3y+z^{2},y^{3},x^{2}+y^{2}+3z^{2})$ and let S be the open surface $1-z=x^{2}+y^{2}$, 0≤z≤1. Use the divergence theorem (and cylindrical polar coordinates) to evaluate $\iint \limits _{S}\mathbf {F} \cdot d\mathbf {S} .$ Verify your result by calculating the integral directly."

So, using the divergence theorem, I calculate the divergence of F integrated over V and get 4.5π. I then close the surface with the unit disc in the z=0 plane, define the unit normal as (0,0,-1), integrate over the unit disc and get -π/2, giving me an answer for the original surface integral of 5π. But when I calculate the integral directly, using cylindrical polars, I get 2π. These all seem like plausible answers so it's very difficult to tell which is wrong. Can someone help me out? Thanks asyndeton talk 18:46, 10 August 2010 (UTC)

I get 3π/2 for the first one and the same answers as you for the other two. -- BenRG (talk) 06:15, 11 August 2010 (UTC)

OK time for a stupid question. Is div dependent on the coordinate system? I attempted to calculate the first integral by finding the divergence and then changing to cylindrical polars and then integrating but it appears changing variables first and then taking the divergence gives a different answer. asyndeton talk 13:27, 11 August 2010 (UTC)

It seems unlikely that div is different for different coordinate systems so if I show my working, perhaps someone can tell me where I've gone wrong. In Cartesians, div F is $3x^{2}+3y^{2}+6z$. Convert this to cylindrical polars to get $3r^{2}+6z$. Multiply by the Jacobian to give $3r^{3}+6rz$. This has no term in theta so the theta integration just gives 2π$(3r^{3}+6rz)$. Integrating between one and zero for r gives $2\pi ({\frac {3}{4}}+3z)$. Integrating again between one and zero for z gives ${\frac {9\pi }{2}}$, which is not what I want. Where's it going wrong? asyndeton talk 13:47, 11 August 2010 (UTC)

I think the problem is with your integration limits. The region you integrate over is a cylinder. You don't want to integrate all the way to $r=1$ for each $z$ value. Martlet1215 (talk) 17:31, 11 August 2010 (UTC)

I don't understand why I don't want 0≤r≤1. In Cartesians $1-z=x^{2}+y^{2}$ so in Cylindrical polars $1-z=r^{2}$ so when z=1, r=0 and when z=0, r=1 and since z is continuous, r must take every value in between. Where have I gone wrong and what limits do I actually want then? asyndeton talk 18:08, 11 August 2010 (UTC)

I think Martlet1215 meant z=1, not r=1. You need to integrate z from 0 to 1−r² first and then integrate r from 0 to 1, or else integrate r from 0 to ${\sqrt {1-z}}$ first and then integrate z from 0 to 1.

The divergence formula does depend on the coordinate system. In cylindrical coordinates it's $\nabla \cdot F={\frac {1}{r}}{\frac {\partial }{\partial r}}(rF_{r})+{\frac {1}{r}}{\frac {\partial }{\partial \theta }}F_{\theta }+{\frac {\partial }{\partial z}}F_{z}$ (copied from Griffiths, Introduction to Electrodynamics). -- BenRG (talk) 19:42, 11 August 2010 (UTC)

Ah, I'm with you on the limits now. I'm just so used to the limits, one transformed into cylindrical or spherical polars, being constants that I didn't even think they might depend on the other variables. Thank you for that. And as for the divergence, I know that there's a different formula for working them out in different coordinate systems but what I meant was will your answer be different if you take the divergence and then change variables rather than change variables and then take the divergence? Thanks asyndeton talk 20:09, 11 August 2010 (UTC)

Oh, I see. It will be the same either way (as I think you already know). -- BenRG (talk) 21:55, 11 August 2010 (UTC)

Yeah, just as I expected. Thanks for all your help. It's much appreciated. asyndeton talk 22:49, 11 August 2010 (UTC)

Just to clarify my earlier statement. I admit there are other choices but I meant what I said: I was using the latter of BenRG's limits, integrating $r=0$ to $r={\sqrt {1-z}}$ and then $z$ from 0 to 1. So $r$ needn't go all the way to 1 for each $z$. Martlet1215 (talk) 23:29, 11 August 2010 (UTC)

Hyperbolas

Hello. For some hyperbola, suppose a is the length of the semi-transverse axis, b is the distance from the centre of the hyperbola to a point on the perpendicular bisector of the transverse axis, and c is the distance from the centre to one focus. Why is $a^{2}+b^{2}=c^{2}$? I don’t see a right triangle anywhere. Thanks in advance. --Mayfare (talk) 20:09, 10 August 2010 (UTC)

I think you're misstating the question. $a$ and $c$ are fixed for a given hyperbola, while $b$ will change depending on the point on the bisector. So in general, $a^{2}+b^{2}\neq c^{2}$. —Preceding unsigned comment added by 203.97.79.114 (talk) 10:23, 11 August 2010 (UTC)

I think if the equation of the hyperbola is

${x^{2} \over a^{2}}-{y^{2} \over b^{2}}=1$

and c is the distance from the center to either of the two foci, then

$a^{2}+b^{2}=c^{2}.\,$

One should be able to give a nice simple geometric explanation. I'll see what I can do. Michael Hardy (talk) 17:16, 11 August 2010 (UTC)

In the case of an ellipse there is a nice geometric interpretation for the analogous equation. I don't know what it would be for the hyperbola unless you start using imaginary values for a, b, and c.--RDBury (talk) 02:28, 12 August 2010 (UTC)

August 11

Tricky complex number / infinite series problem

Given $z={\frac {1}{2}}(\cos \theta +i\sin \theta )$ show that the imaginary part of the infinite series $1+z+z^{2}+z^{3}+...$is ${\frac {2\sin \theta }{5-4\cos \theta }}$
Being an infinite geometric series of common ration ${\frac {e^{i\theta }}{2}}$ I expect I should evaluate $\Im ({\frac {1}{1-{\frac {e^{i\theta }}{2}}}})$ but no avail so far.--220.253.98.30 (talk) 08:52, 11 August 2010 (UTC)

Here's a start:

${\frac {1}{1-{\frac {e^{i\theta }}{2}}}}={\frac {2}{2-e^{i\theta }}}={\frac {2(2-e^{-i\theta })}{(2-e^{i\theta })(2-e^{-i\theta })}}={\frac {4-2e^{-i\theta }}{5-2(e^{i\theta }+e^{-i\theta })}}$

I'll let you take it from there. Gandalf61 (talk) 09:01, 11 August 2010 (UTC)

It is simpler to keep z as ${\frac {1}{2}}(\cos \theta +i\sin \theta )$. Apply ${\frac {1}{x}}={\frac {\bar {x}}{|x|^{2}}}$ to ${\frac {1}{1-z}}$.

Of course, the claim is true when θ is real, but not in general. -- Meni Rosenfeld (talk) 09:10, 11 August 2010 (UTC)

Duality (order theory)

Resolved

In the article Duality (order theory) it says that in the dual poset, meet becomes join and join becomes meet. While this seems intutively okay, what is a mathematical proof of this fact. Thanks-Shahab (talk) 16:50, 11 August 2010 (UTC)

Take the definition of a meet expressed in terms of ≤, replace all ≤ with ≥, and see what you get.—Emil J. 17:26, 11 August 2010 (UTC)

Ah! Its obvious after you told me.-Shahab (talk) 18:02, 11 August 2010 (UTC)

Puzzle Cube solutions

Resolved

How many possible combinations are there of the pieces of a Happy Cube to form a six-piece cube? I'm guessing that there are 122,880 (20*16*12*8*4), as there are 20 places to put the second piece, 16 places to put the third, etc. If this is correct, what is a better way to express this? I've never been very good with permutations/combinations. Other questions I'm thinking about include how the answer changes if you consider each side of a piece as distinct and how to tell how many solutions exist. Mannerisky (talk) 20:56, 11 August 2010 (UTC)

Assuming that the outside vs inside face of each piece is already known, that appears correct. If the sense of the faces is not initially known, the there would be 122,880 × 2⁶ = 2 × 40 × 32 × 24 × 16 × 8 = 7,864,320 possible combinantions. That assumes that the position and rotation of the first piece can be ignored, but the inward/outward sense of the first and all remaining pieces is relevant. -- Tom N (tcncv) talk/contrib 03:44, 12 August 2010 (UTC)

Thanks Tom! That's what I was looking for. Mannerisky (talk) 18:18, 12 August 2010 (UTC)

August 12

Efficiency of a factorization algorithm

Assuming a function Prime(x) that ordinally generates every prime for {x|x>0,x∈ℤ} and generates the primes in polynomial time, what would the efficiency (in layman's terms) be of an algorithm that:

1. is given a number N.
2. exhaustively divides N by Prime(x) until N modulo Prime(x) does not equal zero.
3. returns the result of each division as in number 2, each result being a non-distinct prime factor of N.

--Melab±1 ☎ 04:48, 12 August 2010 (UTC)

In layman's terms, it would be grossly inefficient. It's pretty much the same as other variants of trial division.—Emil J. 10:04, 12 August 2010 (UTC)

It's time function is basically the prime-counting function, which is asymptotic to x/log(x). However, I think the only way you could have your function Prime() run in polynomial-time is if it is simply looking up prime numbers in a pre-existing list (the only way to know what the xth prime is to generate that list and I'm pretty sure you can't generate that list in polynomial-time). --Tango (talk) 18:16, 12 August 2010 (UTC)

When I say assume Prime(x) runs in polynomial time, I also mean that I know there is no efficient prime generation formula. --Melab±1 ☎ 21:42, 12 August 2010 (UTC)

Spaces of Countable tightness - Is the following example OK?

Hi, I'm trying to understand the notion of countable tightness (http://en.wikipedia.org/wiki/Countable_tightness). So I have tried to find a simple example for a space which is not countably tight and came with te following example: Let X be the space of functions from R to {0,1}. Take the set A to be the set of all functions such that f(x) = 1 for a counable (which includes finite) set of x's and 0 otherwise. Then X is not a space of countable tightness since A is not closed in X but the intersection of A with any countable set of functions U is closed in U. Is this example OK? Also, I am trying to figure out, is the space of ultrafilters on R or on N for example is of countable tightness? Thanks! —Preceding unsigned comment added by Topologia clalit (talk • contribs) 08:50, 12 August 2010 (UTC)

Forgot to mention, I am using the point open topology here... Topologia clalit (talk) 08:54, 12 August 2010 (UTC)

Yes, this is an example. I prefer to think of Countable Tightness as saying that accumulation points are limits of countable sequences. Then since A is dense, the constant 1 function is an accumulation point of A, but it is not a limit of any countable sequence from A. —Preceding unsigned comment added by 203.97.79.114 (talk) 10:22, 12 August 2010 (UTC)

Yes, but, isn't saying that accumulation points are limits of countable sequences, is the definition of a sequential space (http://en.wikipedia.org/wiki/Sequential_space)? I mean, if we use your definition then what is the difference between a sequential space to a space with countable tightness? Topologia clalit (talk) 11:01, 12 August 2010 (UTC)

You're right, I misspoke. I should have said "are accumulation points of countable sets". For an example of the distinction, let $x$ be a non-principle ultrafilter on $\mathbb {N}$. Then $x$ is an accumulation point of $\mathbb {N}$, but for any subsequence of $\mathbb {N}$, choose $A\in x$ which omits infinitely much of the subsequence. Then said subsequence is never eventually in $[A]$.

However, your example is still good. It's not hard to see that the constant 1 function is not an accumulation point of any countable subset of A. —Preceding unsigned comment added by 203.97.79.114 (talk) 12:33, 12 August 2010 (UTC)

Ok Thanks. So, these two are examples for non-countable tughtness spaces which helps me. Does anyone have an idea for a space which is of countable tightness but is not sequential? I'm still not sure I got the exact difference between these two definitions.. Thanks! Topologia clalit (talk) 13:18, 12 August 2010 (UTC)

Wait, I didn't say that $\beta \mathbb {N}$ has non-countable tightness; I'm not sure if it does or not.

As for an example which has countable tightness but is not sequential, the Arens-Fort space does it. As a countable space, it automatically has countable tightness. On the other hand, (0,0) is in the closure of the rest of the space, but isn't the limit of any sequence: if the sequence hits a column infinitely often, the open set which omits that column is never eventually entered by the sequence; if it doesn't, the complement of the sequence is an open set containing (0,0). —Preceding unsigned comment added by 203.97.79.114 (talk) 13:59, 12 August 2010 (UTC)

Thanks!!! Some of it is much clearer. But I think still miss some things here.. I mean, The Arens-Fort space is countable, then can't we make a sequense out of the whole space? Is the problem that our whole space sequence will "get in and out" an infinite number of times for any open neighborhood of (0,0)? I'm not quite sure I understand what does a column mean.. If our pairs is of the form (n,m), then, a column means something like all the pairs of the form (n,5) for example? Topologia clalit (talk) 15:12, 12 August 2010 (UTC)

The issue is that the definition of sequence convergence requires that for every open neighborhood of (0,0), the sequence must at some point enter that neighborhood and never leave. So yes, the problem is that for some open neighborhoods, the sequence is going in and out an infinite number of times.

And yes, that's what's meant by a column. (I would have said pairs of the form (5,m), but it doesn't matter which version you go with, as long as you pick one and stick with it.) —Preceding unsigned comment added by 130.195.253.1 (talk) 23:11, 12 August 2010 (UTC)

Got it. Thanks! Topologia clalit (talk) 06:21, 13 August 2010 (UTC)

Pearson product-moment correlation coefficient

According to Pearson product-moment correlation coefficient, the formula is:

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

However, according to the following University website, the formula is:

$r={\sum {(X-\mu _{X})(Y-\mu _{Y})} \over N\sigma _{X}\sigma _{Y}}$

How can this be? - 114.76.235.170 (talk) 12:23, 12 August 2010 (UTC)

They're the same, since $\sigma _{X}={\sqrt {\frac {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}{n}}}$. Algebraist 12:28, 12 August 2010 (UTC)

[ec] It's the same thing. The second version uses a somewhat odd notation, what they really mean is $\sum _{i=1}^{n}{(X_{i}-\mu _{X})(Y_{i}-\mu _{Y}) \over N\sigma _{X}\sigma _{Y}}$. -- Meni Rosenfeld (talk) 12:31, 12 August 2010 (UTC)

Oh! Doh... silly me... thanks folks :-) P.S. thanks Meni, very helpful note on the notation. 114.76.235.170 (talk) 12:34, 12 August 2010 (UTC)

Er, btw, I stuffed up the math syntax... do you mean:

$r={\sum _{i=1}^{n}{(X_{i}-\mu _{X})(Y_{i}-\mu _{Y})} \over N\sigma _{X}\sigma _{Y}}$

I'm quite new to the math syntax... - 114.76.235.170 (talk) 12:37, 12 August 2010 (UTC)

These, too, are the same thing - since $N\sigma _{X}\sigma _{Y}$ doesn't depend on i, you have

$\sum _{i=1}^{n}{(X_{i}-\mu _{X})(Y_{i}-\mu _{Y}) \over N\sigma _{X}\sigma _{Y}}={\sum _{i=1}^{n}{(X_{i}-\mu _{X})(Y_{i}-\mu _{Y})} \over N\sigma _{X}\sigma _{Y}}$.

I used the original version you wrote. -- Meni Rosenfeld (talk) 12:43, 12 August 2010 (UTC)

Oh brother, that's the second time I've done that... I asked almost exactly the same questions about a week ago about the sum of operator, I must have promptly forgotten/not noticed! Thanks though :-) 114.76.235.170 (talk) 14:05, 12 August 2010 (UTC)

Hold the phone!

I think we have a problem here. The definition above uses r, but that's meant to be used for the sample correlation coefficient! That formula above is for the population correlation coefficient. Shouldn't this be using ρ, instead of r? - 114.76.235.170 (talk) 14:41, 12 August 2010 (UTC)

In fact, my textbook says that it is:

$r={n(\sum {xy})-(\sum {x})(\sum {y}) \over {\sqrt {[n(\sum {x}^{2})-(\sum {x})^{2}]\times [n(\sum {y}^{2})-(\sum {y})^{2}]}}}$

What am I missing? - 114.76.235.170 (talk) 14:54, 12 August 2010 (UTC)

These formula work for either the sample coefficient (in which case r is appropriate) or for the population coefficient when the population is finite. I don't think the distinction is very important in practice, so it's possible some texts don't specify which they are talking about.

The formula in your textbook is - you've guessed it - the same as the other two formulas. Showing that this is so is a nice exercise. -- Meni Rosenfeld (talk) 15:44, 12 August 2010 (UTC)

Someone's not being entirely careful about the sample-versus-population issue. I'd use the Greek letters for the population means, not for sample means. Michael Hardy (talk) 17:04, 12 August 2010 (UTC)

I don't understand. This point has already been raised by the OP, and as far as I can see, all mentioned formula are for a sample mean (assuming the sums are over the sample), so our use of the Latin r is appropriate. -- Meni Rosenfeld (talk) 17:55, 12 August 2010 (UTC)

Quadrilateral proof

Imagine a random quadrilateral (not parallelogram, not kite etc.). Prove that the sum of 3 sides is longer than the remaining side.--Mikespedia is on Wikipedia! 17:32, 12 August 2010 (UTC)

It's meaningless to speak of random quadrilaterals until you specify the random distribution they are drawn from, there is no canonical choice here. Fortunately, it does not matter, since the property you mention holds for all quadrilaterals, and you can prove it using the triangle inequality.—Emil J. 17:43, 12 August 2010 (UTC)

Let the vertices be A, B, C, and D, connected by lines in that order. If the quadrilateral is convex, than you can always divide it in two triangles in two ways: either you can consider ABC and ACD, or you can consider ABD and BCD. If the quadrilateral is not convex, you can only construct one of these pairs. Hence, in any case, you can divide the quadrilateral in two triangeles in at least one way. Assume that you can divide it into ABC and ACD. By the triangle inequality, AB + BC > AC, and AC + CD > AD. Add these inequalities to obtain AB + BC + AC + CD > AC + AD, or AB + BC + CD > AD, Q.E.D. --Andreas Rejbrand (talk) 17:52, 12 August 2010 (UTC)

I think that's what the OP meant to say, without knowing the terminology - that it holds for an arbitrary quadrilateral and not only for some specific type. -- Meni Rosenfeld (talk) 17:57, 12 August 2010 (UTC)

As "they" say, "A straight line is the shortest distance between two points." That "remaining side" is a straight line. So there. Michael Hardy (talk) 23:34, 13 August 2010 (UTC)

Proof that P != NP

Could someone explain the proof in less-advanced terms? I'm not a TCS expert and the Wikipedia article only has two sentences about the proof. --70.134.48.188 (talk) 19:29, 12 August 2010 (UTC)

There is a question about this earlier on this page. In addition, there is StackOverflow. --Andreas Rejbrand (talk) 19:31, 12 August 2010 (UTC)

That was asking to explain P and NP. I know what P and NP are, but specifically, I need to understand the proof that P != NP. --70.134.48.188 (talk) 19:34, 12 August 2010 (UTC)

There is no proof yet. At least not one that's accepted widely. You've probably heard of the new paper by Vinay Deolalikar. See Vinay_Deolalikar#Status_of_proof for some words about it. Current consensus seems to be that there might be some interesting new ideas in the paper, but they probably do not constitute a valid proof. Typically it takes the experts several months at least to confirm or refute a proof like this, so we'll have to wait to be sure. Staecker (talk) 00:20, 13 August 2010 (UTC)

The best place to follow developments with this proof is Dick Lipton's blog, http://rjlipton.wordpress.com . IMHO the proof is basically toast at this point. Yes it takes months to confirm a proof like this, but refuting can be much faster than confirming, as we're seeing. 67.122.209.167 (talk) 09:43, 13 August 2010 (UTC)

My brother wrote an explanation that I found helpful. Paul (Stansifer) 05:44, 14 August 2010 (UTC)

Strong Convergence of Operators

Ok, so a couple of us are studying for an upcoming analysis test and came upon a question which we couldn't answer so we appeal to higher authorities here. The question asks for an example of a Hilbert space H and a sequence of operators $(S_{n})$ on H such that

i) $||S_{n}||\leq 1$ for all n=1,2,3,...

ii) the operators $V_{N}=\sum _{n=1}^{N}{\frac {S_{n}}{n}}$ converge strongly as $N\rightarrow \infty$

iii) The strong limit of the operators $V_{N}$ is not compact.

Any help/hints would be greatly appreciated. Thanks! 174.29.63.159 (talk) 19:54, 12 August 2010 (UTC)

What's wrong with letting S₁ be the identity and all the other S_i be zero? Algebraist 21:38, 12 August 2010 (UTC)

Wow, I don't think I have ever felt dumber than this before.174.29.63.159 (talk) 01:39, 13 August 2010 (UTC)

Wait, sorry! The example you gave is perfect but I asked the wrong question. I wanted to ask for a sequence of COMPACT operators $(S_{n})$ each with norm less than or equal to one such that their partial sums (properly scaled) converge strongly to something not compact. Thanks! 174.29.63.159 (talk) 04:53, 13 August 2010 (UTC)

What does strong convergence mean, then? If it means convergence in the operator norm, then this is impossible (a limit of compact operators is compact). If it just means pointwise convergence, then you can get a noncompact limit by exploiting the divergence of the harmonic series:

Let H be ℓ², and let P_i be projection onto the ith basis vector. Let S₁=P₁, then let enough of the S_i be P₂ for the sum to contain a P₂-component larger than 1: that is, since 1/2+1/3<1<1/2+1/3+1/4, we let S₂=S₃=S₄=P₂. Then let S₅ through S₁₂ all be P₃ (since 1/5+1/6+1/7+1/8+1/9+1/10+1/11<1<1/5+1/6+1/7+1/8+1/9+1/10+1/11+1/12), and so on. The divergence of the harmonic series means we can do this for ever, and it's easy to see the resulting series converges pointwise. The limit has every basis vector as an eigenvector with value≥1, so cannot be compact. Algebraist 10:02, 13 August 2010 (UTC)

family of hyperplanes

What is the topology of the "space" of p-planes through the origin in n-space?

This question is motivated by the problem of choosing a basis for orthographic projection from n-space into 2– or 3-space; I want to avoid some parts of the solution-space, and to do that I need to know what the solution-space is! —Tamfang (talk) 21:46, 12 August 2010 (UTC)

This space is normally called a Grassmannian. That article mentions several ways of thinking about it. Algebraist 22:02, 12 August 2010 (UTC)

I read it (insofar as I can) but feel no wiser. Oh well. —Tamfang (talk) 05:44, 13 August 2010 (UTC)

Basically, your intuition of when two planes are close to each other is correct; with lines, for example, the smaller the angle between them, the closer they are. —Preceding unsigned comment added by 130.195.253.1 (talk) 23:20, 12 August 2010 (UTC)

Uh, yeah, I was pretty confident in that. —Tamfang (talk) 05:44, 13 August 2010 (UTC)

There is a definition for planes that's very similar to the one for lines. Given planes A and B, we would like to define an "angle" between them. The way to do this is as follows. For every nonzero vector x in A, consider the least angle α_x formed by x and an arbitrary nonzero vector y of B. Now take the maximum of α_x over all nonzero x in A. This defines a metric on the space of all planes. The underlying topology is what you want. Tkuvho (talk) 10:59, 13 August 2010 (UTC)

An engineer, a physicist, and a mathematician are discussing how to visualize four dimensions:

Engineer: I never really get it

Physicist: Oh it's really easy, just imagine three dimensional space over a time- that adds your fourth dimension.

Mathematician: No, it's way easier than that; just imagine $\mathbb {R} ^{n}$ then set n equal to 4. PST 08:53, 13 August 2010 (UTC)

Okay, then here's a technical answer: Consider the space of injective linear maps from $V^{p}$ to $V^{n}$. Topologize this with the compact-open topology. Identify two maps if they have the same image, and quotient out by this identification. —Preceding unsigned comment added by 203.97.79.114 (talk) 09:41, 13 August 2010 (UTC)

August 13

Mathematicians and Math (What's the deal?)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

permutations and combination

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

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

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

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

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

Unsolved problems

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

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

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

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

Topological 'closeness'

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Cardinality of Topology

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

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

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

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

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

August 14

Rings, subfields and integral domains

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

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

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

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

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

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

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

August 15

Math Child Prodigies of 21st Century

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

The future of math is in the hands of child prodigies and we need to start appreciating that IMO.So who's the greatest child prodigy right now?BTW,Typsetting Bug!And can we document any unsuccessful child prodigies?I'd like to hear examples. —Preceding unsigned comment added by 119.20.19.24 (talk) 01:43, 15 August 2010 (UTC)