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August 14

Cardinality of Topology

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

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

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

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

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

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

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

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

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

Rings, subfields and integral domains

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

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

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

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

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

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

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

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

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

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

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

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

August 15

Math Child Prodigies of 21st Century

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Partial fractions

Hi,

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

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

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

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

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

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

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

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

We can write the function as:

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

with

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

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

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

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

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

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

So, we can write:

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

And

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

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

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

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

Rearranging a formula - getting confused!

Hi folks,

How can you get from:

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

to here:

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

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

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

and similarly:

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

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

Should have noted that I've tried the following:

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

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

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

Did you get as far as

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

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

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

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

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

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

This equality actually comes from the equality

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

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

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

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

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

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

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

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

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

Another attempt... still a bit stuck

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

The goal

Prove that:

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

is the same as:

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

Proof

OK, so the formula here is:

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

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

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

Then you expand the numerator:

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

Which is the same as:

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

Which is:

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

This cancels a term...

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

Leading to the simplified numerator...

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

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

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

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

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

Which is:

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

This leads to:

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

or also:

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

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

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

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

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

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

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

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

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

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

Pivoting

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

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

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

August 16

trying to compare homicides

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

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

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

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

How many in England & Wales?

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

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

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

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

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

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

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

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

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

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

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

SO(n) vs Spin group

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

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

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

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

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

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

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

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

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

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

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

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

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

Some notions in topology

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

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

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

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

correlation conundrum

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

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

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

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

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

Where did this list of 20 bad locations come from? You might be better off going back to the starting point and using whatever metric was used to compile the list rather than using the list itself. --Tango (talk) 08:18, 18 August 2010 (UTC)

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

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

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

Identifying Multiples of low primes

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

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

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

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

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

August 17

Starting values for Root-finding algorithms (polynomials)

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

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

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

integral of sinx cosx

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

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

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

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

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

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

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

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

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

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

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

Can't somebody mention the trigonometric identity that says

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

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

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

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

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

Simple math problem

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

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

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

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

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

August 18

Optics: 35mm focal length equivalent

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

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

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

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

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

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

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

Asymptotic order

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

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

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

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

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

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

Be careful,

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

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

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

You are correct. I stated it backwards. I do have a related question... Given that ${\frac {x}{y}}$ and ${\frac {\lg {x}}{\lg {y}}}$ are not always equal, is it valid to use that reduction here? If x is asymptotically greater than y, won't lg x be asymptotically greater than lg y? -- kainaw™ 15:43, 19 August 2010 (UTC)

Yes. Assuming everything is positive, ${\frac {x}{y}}<1\iff x<y\iff \lg x<\lg y\iff {\frac {\lg x}{\lg y}}<1$. However, as Invrnc explains, this will not work if you are talking about big O. -- Meni Rosenfeld (talk) 18:08, 19 August 2010 (UTC)

I believe that I found an example where it doesn't work. Proving that lg(f/g) is bounded from above did work. Thanks. -- kainaw™ 19:06, 19 August 2010 (UTC)

Space Curves

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

I'm not sure if the question is well posed as infinitesimally the box will be moving along the line as well as spinning. You also need to specify how fast the box is moving along the curve, you could say its a unit speed curve.

Those caverts taken into account you can consider rotation given by

${\frac {d\mathbf {v} }{dt}}={\begin{pmatrix}0&\kappa &0\\-\kappa &0&\tau \\0&-\tau &0\end{pmatrix}}\mathbf {v}$

The axis of this rotation is $(\tau ,0,-\kappa )\,$. To find the angular frequency take a vector perpendicular to the axis, say $(0,1,0)$ its tangental velocity is just $(\kappa ,0,-\tau )\,$, so the angular frequency is just $\omega =|(\kappa ,0,-\tau )|={\sqrt {\kappa ^{2}+\tau ^{2}}}$. --Salix (talk): 09:30, 19 August 2010 (UTC)

Thanks Salix. I think I mentioned that the curve was parametrised by arc length. Wouldn't that make it unit speed? You said that the axis of rotation is (τ,0,–κ) = τT – κB. Isn't there a sign error? I read that the axis of rotation is τT + κB. If you take a vector v = aT + bN + cB and solve dv/ds = 0 then you get, assuming v ≠ 0, that a = τ, b = 0 and c = κ; meaning that v = τT + κB. I've also seen it calculated by working out a vector perpendicular to dT/ds, dN/ds and dB/ds, i.e. perpendicular to κN, τB – κT and –τN. If κ and τ aren't both zero then you get τT + κB. Also, what did you mean by the "tangential velocity" of (0,1,0)? Do you mean the tangential velocity of N? And if so, what is it's tangential velocity? (It seems to be –dN/ds from your answer). I would be very pleased if you could fill in the details and answer some of the whys as well as the hows. Thanks in advance. — Fly by Night (talk) 12:35, 19 August 2010 (UTC)

Looks like brain not fully working this morning. Yes being parametrised by arc length does make it unit speed. Yes it is a sign error axis should be (τ,0,κ) = τT + κB. Tangential velocity comes from the angular frequency article, I'm meaning work out component of the velocity of the point (0,1,0) = N which is tangental a circle lying in a plane normal to the axis. The velocity is dN/ds = (–κ,0,τ) (another sign error on my part). Note that this has no component in the N direction and no component along the axis, so it is already the tangental velocity. The radius is 1 so the angular frequency is simply |dN/ds|. --Salix (talk): 18:57, 19 August 2010 (UTC)

Thanks again Salix! I appreciate your help. Let me check that I understand what I need to do. If the axis of symmetry is spanned by the vector v then I need to choose a unit vector, say w, at a right angle to v and then work out || dw / ds || to give the angular frequency? (Like you said, this all depends on the parametrisation of the curve, but once it's unit speed we can relax.) — Fly by Night (talk) 19:44, 19 August 2010 (UTC)

Yep. You may find the Angular velocity article more helpful as it has a few more details than angular frequency.--Salix (talk): 20:32, 19 August 2010 (UTC)

August 19

Best Mathematicians of 2010

Does anyone know who won the fields medal? I can't seem to find this information anywhere! ... If it hasn't been released, when will it be? I heard that it would be released on 19 August,2010 but I don't want to be waiting till midnight for the declaration of the world's best mathematicians! Please guys, put me out of my misery and tell me how many more hourse I need to wait? I've waited for 4 years already ... Thanks! Signed:THE DUDE. (BTW, I fixed my typsetting bug!) —Preceding unsigned comment added by 110.20.26.228 (talk) 08:37, 19 August 2010 (UTC)

I think I know now ... Damn, why do applied mathematicians always get the awards? Where are the algebraists, topologists and geometers (and number theorists for that matter)? And why on Earth hasn't anyone received it for logic and set theory yet? (FOrget the the first question in above paragraph, I think I know who won it.) Thanks guys ... —Preceding unsigned comment added by 110.20.26.228 (talk) 08:37, 19 August 2010 (UTC)

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

Meni Rosenfeld, I just don't want to be known to the public. I want to be anonymous. You have a name, Meni Rosenfeld. If THE DUDE is taken, can I have another name. I never vandalized signatures. I just didn't want to be made public. Isn't that a simple rquest? Or are the people here at Wikipedia so serious. What's a change in signature between friends? —Preceding unsigned comment added by 110.20.26.228 (talk) 08:37, 19 August 2010 (UTC)

Dmcq has advised you to create an account if you want to remain anonymous. You ignored his advice, lied about your motivation, and now you are playing dumb. We have no reason to believe your intentions are in any way friendly. -- Meni Rosenfeld (talk) 09:44, 19 August 2010 (UTC)

OK, I fixed up all signatures? Now can you please answer my question Meni Rosenfeld? —Preceding unsigned comment added by 114.72.209.200 (talk) 11:34, 19 August 2010 (UTC)

I don't know the answer, but I'm sure that now those who do will be much happier to assist you. (Your next step is to sign your posts by typing 4 tildes, ~~~~ ). -- Meni Rosenfeld (talk) 11:45, 19 August 2010 (UTC)

114.72.244.249 (talk) 08:35, 20 August 2010 (UTC) Is this right? 114.72.244.249 (talk) 08:35, 20 August 2010 (UTC) How did you type 4 tildas without a signature coming up?

Creating an account is the only way to hide you IP address. No matter what edit you do to this page its and easy matter just to look at the history to find your IP. You can create an account at Special:UserLogin/signup --Salix (talk): 08:46, 19 August 2010 (UTC)

Is the argument of a given unary function a singleton?

The argument of a given binary function is an (ordered) pair, so the argument of a given unary function is a singleton, isn't it? Eliko (talk) 08:45, 19 August 2010 (UTC)

Yes--Shahab (talk) 10:38, 19 August 2010 (UTC)

Unfortunately, it's mentioned nowhere in Wikipedia, as far as I know. Eliko (talk) 10:47, 19 August 2010 (UTC)

Well, technically, it's an individual rather than a singleton. But you could just as easily take it to be a singleton, and nothing important would change. --Trovatore (talk) 10:47, 19 August 2010 (UTC)

I'm looking for the formal definition. "technically" (as you've put it), a binary function doesn't have two arguments each of which is an individual, but rather has an argument being an (ordered) pair of individuals (because the order of both individuals must be preserved), so I would expect that, "technically", a unary function won't have one argument which is an individual, but rather will have an argument being a singleton of an individual. Eliko (talk) 12:33, 19 August 2010 (UTC)

There isn't necessarily a "the" formal definition. You're talking about details at the "implementation" level, which may vary from formalization to formalization, but don't change the conceptual content for most purposes.

I would say that a unary function takes an individual, and a binary function, at the most fundamental level, takes two individuals (distinguishing between them, as you note). For many purposes it's convenient to package those two individuals into a single individual, that we then call an ordered pair. But at the very fine level you seem to be examining this, that technically makes it a unary function that takes as its sole argument an ordered pair, rather than a binary function with two arguments. --Trovatore (talk) 22:11, 19 August 2010 (UTC)

You seem to distinguish between "the most fundamental level" and the "very fine level". So would you agree with the following? "At the most fundamental level", a binary real function takes two arguments each of which is a real number, while a unary real function takes one argument which is a real number. On the other hand, i.e. "at the very fine level", every real function is unary, the following way: a function that would be called "binary" at "the most fundamental level", is in fact (i.e. "at the very fine level") a unary function that "takes as its sole argument an ordered pair" of real numbers, while a function that would be called "unary" at "the most fundamental level", is in fact (i.e. "at the very fine level") a unary function that "takes as its sole argument a singleton" of a real number. Eliko (talk) 07:15, 20 August 2010 (UTC)

Monoliths in 2001:A Space Odyssey

In 2001: A Space Odyssey, the Monoliths are 1:4:9. That would be 1 by 4 by 9. But 1 what by 4 what by 9 what? Centimetres? Metres? Yards? Feet? Inches? Can somebody help? --Editor510 ^{drop us a line, mate} 08:58, 19 August 2010 (UTC)

PS: I know this is about an entertainment film but it is a mathematical question, so please don't ask me to move the question.

The author carefully avoided that question by using the smaller side of the monolith as unit of measurement, and then noticing that the larger side was 9 units and the intermediate side was 4 units. Bo Jacoby (talk) 10:56, 19 August 2010 (UTC).

(EC) Monolith (Space Odyssey)#Appearance and capabilities discusses this, the monoliths are not all the same size.81.98.38.48 (talk) 10:58, 19 August 2010 (UTC)

Ratios don't require units. HiLo48 (talk) 11:09, 19 August 2010 (UTC)

Good estimation of number of divisors d(n)

Hello! Let d(n) be a number of positive divisors of n, for example, d(6)=4. In 1907 S. Wigert proved that for any $\epsilon >0$ holds two statements:
1) for infinitely many n $d(n)>2^{(1-\epsilon ){\frac {\ln n}{\ln \ln n}}}$;
2) for all sufficiently large n $d(n)<2^{(1+\epsilon ){\frac {\ln n}{\ln \ln n}}}$;

Does anybody know more precise estimations than these?
Actually I am able to prove better estimation than 1), but I don't have any information about better results.
Thank you!
RaitisMath (talk) 12:52, 19 August 2010 (UTC)

Nowhere dense subsets

Hi, I have recently encountered this proposition that seems somewhat vague to me: Let X be a topological space without isolated points having countable $\pi$-weight and such that every nowhere dense subset in it is closed. Then it is a Pytkeev space.

The thing which is not clear to me is this, if every nowhere dense subset is closed, doesn't that means that it has to be discrete? and doesn't discrete means that every point is isolated? So, is the condition given in this proposition is that, there arn't any nowhere dense subsets in X? Which means that every subset of X is dense somewhere? which doesn't make sense.. I mean, what about subsets of X that contain one point for instance? Thanks! Topologia clalit (talk) 21:55, 19 August 2010 (UTC)

Consider an infinite set with the cofinite topology. This space is not discrete, but every nowhere dense set is closed: if a set is infinite, its closure is the entire space, so the only nowhere dense sets are finite. (This space also has no isolated points, and, if the underlying set is countable, has countable $\pi$-weight.) —Preceding unsigned comment added by 130.195.5.7 (talk) 00:09, 20 August 2010 (UTC)

Alternatively, you can start with any topology and make a finer one by defining all the nowhere dense sets to be closed (and then taking appropriate unions). This new space will have isolated points precisely if the old space did, will have no greater $\pi$-weight (maybe the same?), and all nowhere dense sets will be closed. —Preceding unsigned comment added by 130.195.5.7 (talk) 00:35, 20 August 2010 (UTC)

August 20

How do you write this augmented matrix in reduced row echelon form?

${\begin{bmatrix}1&2&0&1\\0&1&1&2\\0&0&0&1\end{bmatrix}}.$
115.178.29.142 (talk) 03:25, 20 August 2010 (UTC)

Subtract twice the second row from the first row

${\begin{bmatrix}1&0&-2&-3\\0&1&1&2\\0&0&0&1\end{bmatrix}}.$

Subtract multiples of the third row from the other rows

${\begin{bmatrix}1&0&-2&0\\0&1&1&0\\0&0&0&1\end{bmatrix}}.$

This matrix is in reduced row echelon form. Bo Jacoby (talk) 06:42, 20 August 2010 (UTC).

Comprehensive Linear Algebra Book With Relations To Other Branches of Mathematics

I've always wanted to find "THE" Linear algebra book. A book that would serve me for my entire math life. But I never could find it. Hence I ask the gurus at this reference desk.

My ideal linear algebra book would: (a) Cover all the fundamentals of linear algebra such as what's covered in Herstein's Topics in Algebra in a nice and elegant way (canonical forms, bilinear forms, hermitian matrices, you know advanced topics + the basics) (b) Be significantly more advanced than this and cover topics in linear algebra that pop up deeply in other branches of math (say tensor algebra) (c) Be a PURE MATH book. This means applied mathematicians: OUT!.

So is there such a book? I like the style of Herstein's topics in algebra so a book with this style would be desirable. My problem with Herstein's topics in algebra is: it's great and solid but I want a more advanced book. I like my book to give some calculations but also lots of nice theory in a elegant way. + the book should cover as MANY advanced topics as possible. This is important. It should be comprehensive. If a series of texts is required, that's fine. I don't mind if my book requires math background in general algebra, in fact I want a book like that. Thanks guys ... Oh and I forget to add: this book should be primarily on FINITE DIMENSIONAL VECTOR SPACES. It's alright if other vector spaces are covered, but I want the emphasis on FINITE. Thanks guys again. .. —Preceding unsigned comment added by 114.72.248.27 (talk) 07:29, 20 August 2010 (UTC)