User talk:Ozob

Thank you...

...for your improvements to the problem of Apollonius! :)

Welcome to Wikipedia, by the way! I hope that you like it here, and if I can help you, I'll do my best. I like your name; perhaps it's Bozo spelled backwards? ;) A friendly hello from Willow (talk) 23:22, 14 May 2008 (UTC)

....for helping me understand partitions and riemman integrals... —Preceding unsigned comment added by 69.116.88.176 (talk) 23:40, 14 July 2009 (UTC)

Empty product

I appreciate the compromise at empty product. The difficulty is that the material about 0^0 isn't particularly relevant to the article on empty products, except for the case where 0^0 is viewed as an empty product. That's why all the material you added is already in the exponentiation article, with a pointer from the empty product article. I hope you'll look through the talk archives of those pages to see that Bo Jacoby has made the same arguments before, but didn't find consensus. The previous round of discussion about the topic is what led to the consolidation in the exponentiation article. — Carl (CBM · talk) 22:06, 12 June 2008 (UTC)

Hi Carl,

Yes, I agree that the material's not relevant. I did read part of the talk page, and I agree that there's no consensus. And I, too, would prefer that the material be on the exponentiation page. But I saw a war starting, and I was hoping I could pacify both sides. I actually like the article better the way you've left it (because, as you say, the article is about empty products and not about 0^0); but I thought that by including the other side's arguments, I would make everyone happy. I just want the discussion to be peaceful. Ozob (talk) 17:49, 13 June 2008 (UTC)

If you feel a discussion is more appropriate somewhere else, the usual thing to do is to direct people there and copy material over. People leave messages to bring matters to community attention so at least Trovatore's request should not have been removed. You can't expect people to watch this page religiously. I almost missed seeing the request because of your removal. In general, you should not remove other people's comments unless they violate some serious policies. Even then, it may be preferable to archive them (see Wikipedia:Talk_page_guidelines). --C S (talk) 23:54, 12 June 2008 (UTC)

Hi C S,

I'm puzzled. Just as you suggested, I left a notice [1] redirecting people to Talk:Empty_product#Moved from WikiProject talk where I had moved the discussion [2]. Trovatore's original comment didn't provide any context except for a mention of Bo Jacoby (I had to look at the article history to know what was going on), so I thought my notice was no worse. So I'm wondering how I might have done better. I see that you would have preferred if I had left Trovatore's original comment, and now that I think about it, that would have been the right thing to do. Do you have any other ideas on better ways to move a discussion? Thanks. Ozob (talk) 17:49, 13 June 2008 (UTC)

Well, as you've figured out, without knowing the context, it's more prudent just to leave things the way they were said. Trovatore could have phrased it better but this is like the umpteenth time, so he probably was too exasperated to do so. To learn about Bo Jacoby you have to go through a lot more than one article history (you can try searching the WP:Math archives). Except for the removal of the original request, moving the other comments wasn't a bad idea. But it's unusual. It's not uncommon for the discussion to sometimes become a little article specific before someone advocates moving the focus of discussion elsewhere. Even then, the comments themselves are not expunged. I personally think it's just best to leave things as they are. The bot will archive them anyway. Also, sometimes people want to make a comment to the general community and not get involved on the article discussion page (which they may not even watch). --C S (talk) 21:01, 13 June 2008 (UTC)

Oh, I've seen Bo Jacoby at work before, and I realize that he's part of the problem. (But I also see User:Michael Hardy, User:JRSpriggs, and User:Kusma on his side.) Counting from the first reply (which was also the first to mention the issue at hand rather than plea for help) at 15:07 UTC to 21:02 UTC when I moved it, it generated ten posts totaling about 7,000 bytes, all of which simply repeated arguments that could be found on the talk page. My real concern was that the thread was pure noise, no signal, and a lot of noise at that (for that page, at least). I felt entirely justified in moving it away, "to a better place", as I put in my edit summary, and I meant that not only in a literal but also a more figurative way. I know that it's unusual, but I strongly feel that it was (and still is) the right thing to do. Ozob (talk) 00:28, 14 June 2008 (UTC)

Apollonius

Hi Ozob, I’ve replied to your points at: Talk:Problem_of_Apollonius#Oldest_significant_result_in_enumerative_geometry.3F

Sorry for the delay!

Nbarth (email) (talk) 11:49, 22 June 2008 (UTC)

With thanks

Mmmm, gratitude...

I'd like to thank you for your contributions to Emmy Noether. As a total mathematics moron, I feel infinitely indebted to number-smart folks like you and WillowW. I appreciate your support and the many edits you've made. Have a donut. – Scartol • Tok 00:03, 23 June 2008 (UTC)

Emmy Noether

I saw that you have almost entirely reverted an IP's edit, which I'm not that happy about. The person obviously knew what s/he was talking about. Missing citations is bad, but one cannot make everything perfect in one go. Please reconsider your revert. Thank you, Randomblue (talk) 19:13, 30 June 2008 (UTC).

I have discussed this privately with the user over email, and I hope we are both satisfied with the situation. I'm actually looking forward to their contributions, since (as you said) they seem to know what they're talking about. If there's something you'd like to rescue from the edit in question, please go ahead and change the article. I'm not infallible, after all, and as I said initially, I'm probably being a little overprotective. Ozob (talk) 22:31, 30 June 2008 (UTC)

The footnote to Einstein's letter to the New York Times links to a Web page that claims Einstein's letter appeared on May 5, 1935. However, one can retrieve the actual letter from the New York Times archive, and see that the letter appeared in the Times on May 4, 1935. Therefore, the Web page cited in the footnote has the date wrong. I corrected the date recently, but just discovered that my correction was reverted on the grounds that my change was incorrect. I invite you to check for yourself that the letter appeared on May 4, 1935. —Preceding unsigned comment added by 66.108.13.221 (talk) 00:23, 4 July 2009 (UTC)

expert point of view

Hi Ozob. I just wanted to point out that Moon Duchin, one of the sources in the Noether article, has kindly accepted to review the article. I see she has left a message on the talk page today discussing various issues. There is still room for improvement! Best, Randomblue (talk) 10:22, 9 July 2008 (UTC).

RFC at St. Petersburg paradox

As you have contributed to an earlier related discussion at Wikipedia talk:Manual of Style (mathematics)#Punctuation of block-displayed formulae, you may be interested in Talk:St. Petersburg paradox#Request for comments: punctuation after displayed formula.  --Lambiam 18:18, 8 August 2008 (UTC)

Thanks!

Yes, \textstyle{} is much better. Thanks for fixing it up! siℓℓy rabbit (talk) 15:18, 9 August 2008 (UTC)

Common interests, clearly

You're right, we've been chasing same articles all over the place for the past weeks. I just ended up reverting your IPA guide to étale (see the talk page). I accidentally got more involved in fixing articles related to algebraic geometry (just back from vacation so I really should not have time for this!), so it's very good to have some company doing this. I think we've had some good progress on many topics. A lot remains to be done to AG-related topics — one example is just adding the maths rating template to many (majority of) articles on important topics. With things getting busier at work by the day, I'm afraid my contributions will slow down from now on, but let's see. Cheers, Stca74 (talk) 09:13, 10 August 2008 (UTC)

Ozob's proposed deletion of "Non-Newtonian calculus"

The article "Non-Newtonian calculus" provides a brief description about a subject of interest to scientists, engineers, and mathematicians. The omission of this subject from Wikipedia would be a huge disservice to those people. The article is coherent, meaningful, and unbiased. Exactly what do you object to? Shouldn't an encyclopedia contain as much pertinent knowledge as possible? Please reconsider your decision. Thank you.

Sincerely, Michael Grossman —Preceding unsigned comment added by Smithpith (talk • contribs) 19:48, 12 September 2008 (UTC)

Citation, reviews, and comments re "Non-Newtonian Calculus"

"Non-Newtonian Calculus" is cited by Professor Ivor Grattan-Guinness in his book "The Rainbow of Mathematics: A History of the Mathematical Sciences" (ISBN 0393320308). Please see pages 332 and 774.

"Non-Newtonian Calculus" has received many favorable reviews and comments:

The [books] on non-Newtonian calculus ... appear to be very useful and innovative.

                       Professor Kenneth J. Arrow, Nobel-Laureate 
                       Stanford University, USA 

Your ideas [in Non-Newtonian Calculus] seem quite ingenious.

                       Professor Dirk J. Struik 
                       Massachusetts Institute of Technology, USA 

There is enough here [in Non-Newtonian Calculus] to indicate that non-Newtonian calculi ... have considerable potential as alternative approaches to traditional problems. This very original piece of mathematics will surely expose a number of missed opportunities in the history of the subject.

                       Professor Ivor Grattan-Guinness 
                       Middlesex University, England 

The possibilities opened up by the [non-Newtonian] calculi seem to be immense.

                       Professor H. Gollmann
                       Graz, Austria 

This [Non-Newtonian Calculus] is an exciting little book. ... The greatest value of these non-Newtonian calculi may prove to be their ability to yield simpler physical laws than the Newtonian calculus. Throughout, this book exhibits a clarity of vision characteristic of important mathematical creations. ... The authors have written this book for engineers and scientists, as well as for mathematicians. ... The writing is clear, concise, and very readable. No more than a working knowledge of [classical] calculus is assumed.

                       Professor David Pearce MacAdam
                       Cape Cod Community College, USA

... It seems plausible that people who need to study functions from this point of view might well be able to formulate problems more clearly by using [bigeometric] calculus instead of [classical] calculus.

                       Professor Ralph P. Boas, Jr.
                       Northwestern University, USA

We think that multiplicative calculus can especially be useful as a mathematical tool for economics and finance ... .

                       Professor Agamirza E. Bashirov
                       Eastern Mediterranean University, Cyprus/
                       Professor Emine Misirli Kurpinar
                       Ege University, Turkey/
                       Professor Ali Ozyapici
                       Ege University, Turkey

Non-Newtonian Calculus, by Michael Grossman and Robert Katz is a fascinating and (potentially) extremely important piece of mathematical theory. That a whole family of differential and integral calculi, parallel to but nonlinear with respect to ordinary Newtonian (or Leibnizian) calculus, should have remained undiscovered (or uninvented) for so long is astonishing -- but true. Every mathematician and worker with mathematics owes it to himself to look into the discoveries of Grossman and Katz.

                       Professor James R. Meginniss
                       Claremont Graduate School and Harvey Mudd College, USA

Note 3. The comments by Professors Grattan-Guinness, Gollmann, and MacAdam are excerpts from their reviews of the book Non-Newtonian Calculus in Middlesex Math Notes, Internationale Mathematische Nachrichten, and Journal of the Optical Society of America, respectively. The comment by Professor Boas is an excerpt from his review of the book Bigeometric Calculus: A System with a Scale-Free Derivative in Mathematical Reviews.

Thank you.

Sincerely, Michael Grossman —Preceding unsigned comment added by Smithpith (talk • contribs) 01:03, 13 September 2008 (UTC)

Dab page

The dab page is at Janko; why did you redirect the other dab page (Janko_group_(disambiguation)) to the aricle Janko group instead of the correct dab page ? SandyGeorgia (Talk) 00:02, 14 September 2008 (UTC)

"Janko group" on its own is ambiguous. Shouldn't Janko group (disambiguation) disambiguate Janko group rather than Janko?

I agree that the present Janko group article is not a proper disambiguation page, but I'll fix that. Ozob (talk) 00:08, 14 September 2008 (UTC)

Janko group is now a "proper" article, as it should be. Which page is going to be the dab page, and the other (duplicate) dab page should direct to it. SandyGeorgia (Talk) 00:10, 14 September 2008 (UTC)

Janko group is incapable of containing actual content since it could refer to any of four separate objects. I've made it a disambiguation page. Janko group (disambiguation) points there. It's worth pointing out that I created Janko group (disambiguation) because Template:Group navbox pointed to Janko group but clearly intended to refer to all four groups. Ozob (talk) 00:18, 14 September 2008 (UTC)

This was an article. OK, you'll need to sort this with the folks at the Group Math FAC, as the article will now show as incorrectly linked. I did what I could to try to help. SandyGeorgia (Talk) 00:21, 14 September 2008 (UTC)

I have this feeling that we're both trying to do the right thing, and somehow we're not communicating.

I'm not sure what part of the MoS Janko group now violates; it does not give extended definitions as you stated in your edit summary, but only the simplest possible fact which could be used to distinguish the groups (besides their names), namely their order. I'll ask at the Group FAC and we'll sort it out there. Ozob (talk) 00:34, 14 September 2008 (UTC)

I'll wait for you all to sort it; the article was fine and did the job, now it's back to being a dab trying to be an article, and we have Group (mathematics) pointing to a dab, that easily could have been (was) an article. SandyGeorgia (Talk) 00:36, 14 September 2008 (UTC)

Janko

Hi Ozob,

before we are all going mad!!! (<- notice the complete anti-MOS-hly markup), I have replaced the Janko group link in the navbox by all four groups (which is, I believe, worse than having the link to J.gr. itself, but at least compliant to MOS). (I'm so tired by these nitpicking comments at FAC whose sole objective it seems to be to follow the guidelines mm per mm) Jakob.scholbach (talk) 10:06, 14 September 2008 (UTC)

Easy as pi?: Making mathematics articles more accessible to a general readership

The discussion, to which you contributed, has been archived, with very much additional commentary,
at Wikipedia:Village pump (proposals)/Archive 35#Easy as pi? (subsectioned and sub-subsectioned).
A related discussion is at
(Temporary link) Talk:Mathematics#Making mathematics articles more accessible to a general readership and
(Permanent link) Talk:Mathematics (Section "Making mathematics articles more accessible to a general readership"). Another related discussion is at
(Temporary link) Wikipedia talk:WikiProject Mathematics#Making mathematics articles more accessible to a general readership and
(Permanent link) Wikipedia talk:WikiProject Mathematics (Section "Making mathematics articles more accessible to a general readership").
-- Wavelength (talk) 01:38, 29 September 2008 (UTC)

Derivative with respect to a vector

Hi Ozob. On the talk page for Euclidean vector, you wrote that there is a section in the article called:

• Derivatives with respect to a vector (wrongly labeled the "derivative of a vector")

I don't want to sidetrack the discussion on that page, so I'm asking you here. Why do you believe it is incorrectly titled? Isn't

${\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}$

the derivative of the vector v with respect to the scalar t? MarcusMaximus (talk) 06:15, 5 October 2008 (UTC)

It should be "Derivative of a vector-valued function" or "Derivative with respect to a vector". I believe that the section is incorrectly titled because vectors are not functions, and therefore the idea of differentiation is meaningless. Ozob (talk) 01:23, 6 October 2008 (UTC)

What about the concepts of relative velocity as the derivative of displacement, and acceleration as the derivative of velocity? Is the displacement between two points not a vector, but rather a vector function? Is it worth making such a semantic distinction? The text says that the vector is a function of a scalar.

Also, I think it would be incorrect to title it "derivative with respect to a vector", because that is the phrasing typically used to refer to the quantity that appears in the "denominator" part when using the d/dt notation. Certainly there is nothing in there about taking the derivative of something with respect to a vector, in the sense of measuring the rate of change of a dependent function as an independent vector changes. MarcusMaximus (talk) 02:32, 6 October 2008 (UTC)

Displacement between two points is a vector. If one (or both) of those points is variable, then you get a function: Each choice of points determines a unique displacement vector. Functions have derivatives, so this makes sense. But to talk about the derivative of a vector—just a vector, not a vector-valued function—is meaningless. Derivatives are only defined for functions; even when we differentiate a constant (as in (d/dx)(1) = 0) we are really differentiating the constant function.

I think "derivative with respect to a vector" is an appropriate description of a directional derivative. It seems to me that it would be consistent with the usual notation to write the directional derivative in the direction v as d/dv, even though that's not usually done.

I have a question for you: You write d/dt where I would write d/dt. Surely you learned this notation somewhere. Where? The only place I've ever seen an upright d is on Wikipedia. Ozob (talk) 14:53, 6 October 2008 (UTC)

I understand your point about vector functions. In fact, the reference I was using refers to them as vector functions, so I agree that we should change the title to Derivative of a vector function. I've made the change.

The section we are talking about doesn't discuss the directional derivative, only the derivative a of vector with respect to scalars. Regardless, I don't believe that d/dv would be the correct notation for the directional derivative. The directional derivative in the direction of v is defined using the gradient function as

$\nabla _{\mathbf {v} }{f}(\mathbf {x} )=\nabla f(\mathbf {x} )\cdot \mathbf {v} .$

In contrast, the notation you are suggesting indicates that there is some vector-valued function that is dependent on v and we want to differentiate it with respect to v; in other words, find the rate of change of the function as the vector v changes. I don't think that's what the directional derivative does; often v is a constant vector. It doesn't make sense to take the derivative with respect to a constant.

You could argue that the gradient of f at x is equal to df/dx

$\nabla {f}(\mathbf {x} )={\frac {df(\mathbf {x} )}{d\mathbf {x} }}$

but I haven't given it enough thought to decide if that idea is problematic.

I actually agree with you on the d/dt notation, but several places on Wikipedia people have gone through and changed my italic d's to upright, so I assumed that was some sort of style guide convention. MarcusMaximus (talk) 16:24, 6 October 2008 (UTC)

When I wrote d/dv, I was thinking in terms of a connection, specifically in its manifestation as a covariant derivative. From that perspective, d/dt is differentiation with respect to the a vector which points in the direction of the positive t axis and has length 1.

I think it's reasonable to write the gradient as df/dx. But this sort of notation is never used, nor is d/dv, and I don't know why. I'd guess that in practice, it's messier than the usual notation.

Currently the WP MoS specifies that the d in d/dx can be either italic or upright. It also says not to change it from one to the other except for consistency within the article (or if you're rewriting from scratch). So if someone changes your italic d's to upright, I think you ought to revert! Ozob (talk) 21:57, 6 October 2008 (UTC)

As an engineer I'm not well educated in the more abstract principles of connection and contravariance. It sounds like you can call a scalar variable a vector in its own right? That seems trivial to me, but I'm not here to judge the value of mathematical concepts.

After thinking about it more, it seems to me the reason the gradient is not generally written df/dx is because the function f(x₁,x₂,x₃) of which we are taking the gradient is not truly a function of the vector x in general. It is a scalar function of an arbitrary number of scalars. If you wrote it out, you generally wouldn't have any occurrences of the vector x made up of components in the standard basis. However, each scalar has its own axis and can be thought of as a scalar component of a vector x = x₁i + x₂j + x₃k, but the vector x does not actually appear in the function f itself, unless you wanted to recast the function so that every instance of (x₁,x₂,x₃) was a dot product of x with i,j,k. MarcusMaximus (talk) 05:01, 7 October 2008 (UTC)

Yes, I am calling a scalar a vector. I agree that it's trivial: A one-dimensional vector space is just the space of scalars. But if you think in those terms then you can see an analogy between the notation d/dt and the notation d/dv.

I disagree that f is not a function of x. To every x, you get a unique output f(x); this is the definition of a function. The way it's written is unimportant. As you say, you could write dot products every time you want to take a component. Doing so demonstrates that f is a function of x. But you could also add and subtract the same number, for example, f(x) = (x·i + 1) - 1. This formula specifies an algorithm to compute f(x) which goes like: Take the dot product of x and i, add one, then subtract one. As a function, this is the same as f(x) = x·i, even though the latter specifies a different algorithm, namely, take the dot product of x and i. In the same way, the distinction between f(x) = x·i and f(x) = x₁ is only a matter of the algorithm specified by the formula, not the function.

I suppose that if one wanted to be very pedantic, then one could argue that even though x is an element of a three dimensional vector space with a fixed basis, the underlying set of that vector space might not be the set of all triples (x₁, x₂, x₃); instead it might be something like "all polynomials of degree less than or equal to two with basis 1, x, x²". But vector spaces with chosen basis are naturally isomorphic to arrays of numbers, so I don't think it's an important distinction. Ozob (talk) 22:39, 7 October 2008 (UTC)

I do see your analogy now. Here's my theory on d/dv question.

Usually people expect to the arguments to appear explicitly in the formula for the function. With that in mind, f can be cast as a function of x; however, in general, it is not cast that way.

To use your example f(x) = x₁, there is no connection between f and x unless you use a separate equation to define the relationship between x and x₁, such as x = x₁i + x₂j + x₃k. In that case, most people would infer that x is a function of x₁ rather than the inverse (even though both are true). In that case it's counterintuitive to say that f is a function of the vector x; most people would say that both f and x are functions of x₁.

The alternative is to define x₁ = x•i, which does create the necessary intuitive order. But even in this instance you don't get the explicitness most people expect and desire. If you write

f(x) = x₁ where x₁ = x•i,

even though it is strictly true, it makes about as much intuitive sense as writing g(z) = y. People wonder what the heck you're talking about, until you tell them, oh, by the way, y = h(z). Then they wonder why the heck you wrote g(z) on the left side instead of g(y) but wrote the right side in terms of y rather than explicitly in terms of z.

So back to the main point, I think it is clear that we should really be talking about the gradient, not the directional derivative. In my opinion the main reason the gradient is not commonly written as d/dv is that it often requires counterintuition. I'm not arguing that you can't write the gradient as d/dv, but it only makes sense sometimes and it's more general to just use $\nabla$.

It's also not obvious to me (without knowing a priori how to take the gradient) what exactly you're supposed to do when you see df(x)/dx. You're actually taking the derivative of a scalar function f with respect to the sum of three vectors x₁i, x₂j, x₃k, which doesn't make a lot of sense to me. It appears that you have to take the formula on the right hand side of f(x), which is a scalar algebraic expression containing x₁,x₂,x₃, and implicitly differentiate it with respect to x using the chain rule. Then you have an algebraic expression in terms of x₁, x₂, x₃, dx₁/dx, dx₂/dx, and dx₃/dx. Next you have to find expressions for dx₁/dx, dx₂/dx, and dx₃/dx. I'm not sure what those would be, since there is no such thing as vector division that I know of. Maybe they are i,j,k, respectively? MarcusMaximus (talk) 08:54, 8 October 2008 (UTC)

That's a good point that while f(x) = x₁ is counterintuitive. I was only paying attention to the formal logic, but you're right.

If we continue to think of d/dx as the gradient, then dx₁/dx would be the gradient of x₁, hence i, and similarly you'd get j and k for x₂ and x₃. And no, there is no such thing as vector division in general; division is a very, very restrictive condition.

I agree that the "right thing" to think about in a lot of cases is the gradient, not the directional derivative. A better thing to think about, it turns out, is differential forms and exterior derivatives. But in order to make sense of them and their relationship to the gradient, you need to distinguish between the tangent space and cotangent space, and this is more effort than most people are willing to make. Ozob (talk) 17:42, 8 October 2008 (UTC)

Forgive my ignorance, but could you be more explicit? You said, "If we continue to think of d/dx as the gradient, then dx₁/dx would be the gradient of x₁, hence i, and similarly you'd get j and k for x₂ and x₃."

This becomes circular, because I'm trying to find an expression for dx₁/dx in order to prove that d/dx is the gradient. I just need to show that dx₁/dx is i, but how do I get there? MarcusMaximus (talk) 23:26, 10 October 2008 (UTC)

Well, after a little thought,

$x_{1}=\mathbf {x} \cdot \mathbf {i} ,$

therefore

${\frac {dx_{1}}{d\mathbf {x} }}={\frac {d\mathbf {x} }{d\mathbf {x} }}\cdot \mathbf {i} +\mathbf {x} \cdot {\frac {d\mathbf {i} }{d\mathbf {x} }}.$

Since di/dx is the zero vector, we are left with

${\frac {dx_{1}}{d\mathbf {x} }}={\frac {d\mathbf {x} }{d\mathbf {x} }}\cdot \mathbf {i}$

If you substitute the derivative of x the equation becomes a triviality, collapsing to dx₁/dx = dx₁/dx. It seems that the value of dx/dx must be an entity that dot multiplies with the vector i and leaves it unchanged. The only thing I know of that does that is the unit dyadic (ii + jj + kk), but I have no idea how that would come into play. I'm still stumped. MarcusMaximus (talk) 04:04, 11 October 2008 (UTC)

Hmm. If I understand you correctly, you're looking for a way to manipulate the symbol d/dx (using the usual rules) that makes the expression for the gradient pop out. I'm not sure whether or not one can do this. At some point one has to define what d/dx means; I was intended to define it to be the gradient because that seemed to be the only way to make it consistent. You seem to be looking to define it by certain properties that it should satisfy (the product rule, chain rule, etc.), but I don't think that's enough to get a unique expression out. (My reasoning comes from Riemannian metrics; the gradient is what one gets by taking the exterior derivative of f and contracting with the metric, so if one changes the metric one gets a different gradient. If you normalize d/dx by choosing the values of dx₁/dx and similar expressions in the other variables, you should have enough information to determine the gradient. But without that there are too many possible metrics.)

I also think that if f is a vector-valued function, then d/dx should mean (by definition again) the total derivative. This is consistent with the use of d/dx to mean "derivative with respect to the variable x"; if x happens to be a one-dimensional vector (so that it's okay to write x = x), then d/dx is equal to d/dx just like it should be.

Does this work? I think I've dodged circularity this time by making a definition. Ozob (talk) 23:07, 11 October 2008 (UTC)

That is rather unsatisfying. It just seems to me (with no particular reason) that we should be able to prove that d/dx is the gradient. We already know how to take derivatives according to the definition based on the limit of the slope of the secant, and we know what a vector is. What we don't know, I guess, is what it means to take the derivative with respect to a vector, but I was hoping to be able to derive it. MarcusMaximus (talk) 09:14, 12 October 2008 (UTC)

Hey, I just got an idea! Let's go back to the definition. I'm going to take this as the definition of the derivative in one variable:

$\lim _{h\to 0}{\frac {|f(x+h)-f(x)-f'(x)h|}{|h|}}=0.$

(That is to say, f'(x) is the unique number which makes the above equation hold.) OK, now I make everything a vector:

$\lim _{\mathbf {h} \to 0}{\frac {\lVert f(\mathbf {x} +\mathbf {h} )-f(\mathbf {x} )-f'(\mathbf {x} )\mathbf {h} \rVert }{\lVert \mathbf {h} \rVert }}=0.$

Now, would you agree that df/dx ought to satisfy this equation? But this is exactly the definition of the total derivative.

When f is a real-valued function, this is not exactly the same as the gradient; instead it's a linear transformation R³ → R. If one fixes a basis of R³ (which is the same as fixing a Riemannian metric at the point we're differentiating at) then one can identify R³ with its dual space and convert the linear transformation into a vector in R³. That vector will be the gradient. Ozob (talk) 20:19, 12 October 2008 (UTC)

Excellent. So ƒ’(x) (ƒ prime) is dƒ/dx, the gradient of ƒ(x). Then ƒ’(x)h is the juxtaposition of two vectors...a dyadic? Or does there need to be a dot product in there, ƒ’(x)•h, because the other two terms in the numerator are scalars? Even if ƒ(x) is a vector function you have the sum of two vectors with a dyadic. MarcusMaximus (talk) 07:07, 13 October 2008 (UTC)

No, $f(\mathbf {x} )$ is a linear transformation. If f is a real-valued function, then $f(\mathbf {x} )$ is a linear functional: It takes a vector, in this case h, and returns a scalar. When f is a vector valued function, then $f(\mathbf {x} )$ can be written as a matrix. (Sorry for the TeX, but I'm having strange formatting issues.)

I think it's worth pointing out that a dyadic tensor is the same thing as a linear transformation (see the last paragraph of the dyadic tensor article as well as dyadic product). In that interpretation, f'(x)h is the application of h to the dyadic tensor f'(x); in the case when f is real-valued, however, it's a dyadic tensor in the basis vectors ii, ij, ik (and no others). Ozob (talk) 15:00, 13 October 2008 (UTC)

So is this definition operational? Can I plug in real expressions and do some algebra and calculus to get a real answer? Starting with ƒ(x) = x•i = x₁,

${\begin{aligned}0&=\lim _{\mathbf {h} \to 0}{\frac {\lVert f(\mathbf {x} +\mathbf {h} )-f(\mathbf {x} )-f'(\mathbf {x} )\mathbf {h} \rVert }{\lVert \mathbf {h} \rVert }}\\&=\lim _{\mathbf {h} \to 0}{\frac {\lVert (x_{1}+h_{1})-(x_{1})-f'(\mathbf {x} )(h_{1}\mathbf {i} +h_{2}\mathbf {j} +h_{3}\mathbf {k} )\rVert }{\sqrt {h_{1}^{2}+h_{2}^{2}+h_{3}^{2}}}}\\&=\lim _{\mathbf {h} \to 0}{\frac {\lVert h_{1}-f'(\mathbf {x} )(h_{1}\mathbf {i} +h_{2}\mathbf {j} +h_{3}\mathbf {k} )\rVert }{\sqrt {h_{1}^{2}+h_{2}^{2}+h_{3}^{2}}}}\\&=?\end{aligned}}$

MarcusMaximus (talk) 03:21, 17 October 2008 (UTC)

I don't think so. Usually one proves that for a continuously differentiable function, the total derivative equals the matrix of partial derivatives; then one computes the total derivative using partial derivatives. But I don't know how else to do it. Ozob (talk) 18:35, 17 October 2008 (UTC)

I suppose it is useful in mathematics to prove something is true after you have the correct answer. However, for an engineer using applied mathematics, it is important that definitions be operational. I'll keep looking. MarcusMaximus (talk) 23:32, 18 October 2008 (UTC)

The point of such a theorem is that d/dx can (under mild hypotheses) be computed easily. Taking partial derivatives is easy, and putting them in a matrix is even easier. So this theorem tells you that the total derivative, while a priori hard to compute, is actually easy to compute. I agree that the computation can't be done easily from the definition itself, but that's no different from computing any complicated derivative. Think about differentiating a function where you need to use the product and chain rules in combination several times; directly from the definition, it's a huge and nearly impossible mess, but with the product and chain rules it becomes easy. Computing the total derivative is analogous: An hard definition (the one above) which one proves to be the same as an easy rule (take partial derivatives and put in a matrix). Ozob (talk) 01:00, 19 October 2008 (UTC)

Vector spaces

Hi Ozob,

I have asked for a GA review at the round table, but people are busy/dizzy with LateX formatting and icon questions ;) I thought you might be interested in having a look at vector spaces and giving it a GA review? This is the page. Thanks a lot. Jakob.scholbach (talk) 14:13, 1 December 2008 (UTC)

AfD nomination of Bishop–Keisler controversy

An article that you have been involved in editing, Bishop–Keisler controversy, has been listed for deletion. If you are interested in the deletion discussion, please participate by adding your comments at Wikipedia:Articles for deletion/Bishop–Keisler controversy. Thank you. Mathsci (talk) 05:42, 14 December 2008 (UTC)

Radius of convergence

Hi, your tweak does help, thanks. Do you happen to know what the coefficients of x^n would be in standard form?Regards, Rich (talk) 02:08, 16 December 2008 (UTC)

It'd be something to do with the round down of the base 2 logarithm of n. Looks like it'd be

$a_{n}={\frac {(-1)^{\lfloor \lg n\rfloor }}{2^{\lfloor \lg n\rfloor +1}\cdot (\lfloor \lg n\rfloor +1)}}$

Ozob (talk) 02:53, 16 December 2008 (UTC)

Table on trig identities

[3] I disagree. Provide reasoning as to why it's clearer. It looks to me as though there are more functions than there are. The parenthesis makes it clear what are the abbreviations. —Anonymous Dissident^Talk 05:12, 20 December 2008 (UTC)

Flagged Revisions Trial

Thanks for all your work on this difficult proposal. I am enjoying that a great deal of the effort into ensuring that any trial would be well-run and give us the right information is coming from editors on the oppose side of things. :) Lot 49a^talk 22:31, 6 January 2009 (UTC)

Flagged Revs

Hi,

I noticed you voted oppose in the flag revs straw pole and would like to ask if you would mind adding User:Promethean/No to your user or talk page to make your position clear to people who visit your page :) - Thanks to Neurolysis for the template   «l| Ψrometheăn ™|l»  (talk) 07:22, 8 January 2009 (UTC)

About math formulae

Thanks for addressing 800x600 window size of a math formula. See explanation at User_talk:Wikid77#Confusion over math formulas. -Wikid77 (talk) 12:35, 24 February 2009 (UTC)

Calculus lead

Hi Ozob, I really don't think that "a course of study" is a good descriptor there, both on the substance and in terms of compatibility with the rest of the lead. For one thing, it's too narrow: calculus, as a mathematical discipline, consists of a collection of techniques and some proofs that go along with them. Now, there are some courses that purport to teach these techniques … (Techniques: fundamental; how/whether they are taught: secondary). Do you see what I mean? Further down, the lead talks about how calculus is widely applicable. Obviously, it's the principles and techniques of calculus that are applied, not "the course of study". I don't insist on the word "discipline", but calculus is a fairly well defined part of mathematics, and it's a lot more than the name of a course or a popular textbook title. I've made a few other changes (might as well …), but this is what triggered my involvement. Good job, otherwise! Cheers, Arcfrk (talk) 04:16, 23 April 2009 (UTC)

Having read your argument, I agree with you: "course of study" isn't as good as I thought it was. I'm going to put "discipline" back in. Thanks for the feedback! Ozob (talk) 19:19, 23 April 2009 (UTC)

Calculus

Hi Ozob, I removed places that for my view have confusing grammar. however if you find some more places with poor grammar i will appreciate your help. Aleks kleyn (talk) 02:27, 26 April 2009 (UTC)

List still up to date?

Hallo Ozob, is this list Talk:Problem_of_Apollonius#Old_sources still up to date? Is the help still needed? Have you found some of the sources? --DrJunge (talk) 17:59, 1 May 2009 (UTC)

Derivative of a Vector Function

Hi Ozob.

I like your improvement of the intro to this section, except that I disagree with this statement:

A different choice of reference frame will produce the same derivative function, but it will usually have a different formula because of the different choice of reference frame.

In general, the derivative of a vector function is different in every reference frame. Your statement seems to confuse a reference frame with set of basis vectors. If we were only talking about multiple different bases fixed in the same reference frame, the statement would be true.

The concept of relative velocity is a good counterexample, I think. If you consider that the velocity (as the derivative of displacement or position) of a pilot sitting in an airplane, it should be apparent that you don't get the same function by differentiating in different reference frames. The velocity of the pilot relative to the Earth may be very large and time-varying, but his velocity relative to the airplane is identically zero. It is not merely switching bases, but it is an entirely different function.

Do I misunderstand you? MarcusMaximus (talk) 00:50, 1 August 2009 (UTC)

No, I think you do understand me, but we have a difference in viewpoint. I think I am still right; what's confusing is that in a certain sense I think you are also right. I am considering the reference frame as a coordinate chart on a manifold, i.e., we have functions x,y,z : M → R, where M is the space under consideration (which just so happens to be Euclidean space) and the map (x,y,z) : M → R³ is a diffeomorphism. The velocity function is a function on M, so it and its derivative are absolute in some sense. I.e, assuming a choice of connection, there is only one derivative of a function, namely the one given by the connection. But the derivative can be coordinatized in many different ways, and a choice of x,y,z leads to a choice of a derivative in your sense. Your sense, as far as I understand it, is that there is no space M, but only what I would call coordinate charts and their automorphisms. From this viewpoint, one always defines things in coordinates and then checks that it behaves under coordinate chart changes; that is, that it transforms tensorially. And since there is no analog of M, there is no sense in which there is one function with many possible formulas. Instead you get a different function in every reference frame.

There is another complication, which is that you seem to always use the standard Euclidean connection in each reference frame. This explains your airplane example: "Velocity relative to Earth" is one connection, and in the reference frame in which Earth is fixed, it is the standard Euclidean connection; "Velocity relative to the plane" is a different connection, given in the reference frame of the plane by the standard Euclidean connection. Of course, when you transform one reference frame into another, you don't transform the two connections into each other!

Does that make sense? I'm not sure what to do about the article, but am I at least understanding your viewpoint? Ozob (talk) 19:02, 1 August 2009 (UTC)

No, I actually have no idea what that means. I apologize, but I'm not trained in the mathematical theory you are talking about. I'm only trying to say that for a Euclidean vector, as they are called in the article, the right hand sides of the derivatives taken in two different reference frames are not merely the same function expressed in different bases, in which case you could perform a series of substitutions and get the same formula. They are literally different functions, related by the second formula shown here defining the derivative of a vector function is one reference frame as a function of the derivative taken in another reference frame.

Perhaps you could explain how what I see as two different functions, i.e. identically zero velocity vs. a time-varying velocity, are the same function? (that is a serious question, not rhetorical) MarcusMaximus (talk) 23:48, 1 August 2009 (UTC)

OK, I'll try again. In more elementary language, what I'm saying is that your notion of derivative changes when you change reference frames. Not only is the identically zero function not the same as a non-identically zero function; they are not the same kind of derivative! Abstractly, this is described by a connection. A connection is just an abstraction of the differentiation operator. When you take velocity relative to the Earth, you're measuring a rate of change with respect to the Earth, and this is a totally different rate of change than rate of change with respect to the plane. But once you have chosen something to measure against, the rate of change is independent of the reference frame.

Let me put some formulas to things. Suppose that x(t) = t is the position of the plane relative to the Earth. If we want to find the velocity of the plane with respect to the Earth, we take the derivative and get 1 unit per unit time. Suppose now we measure the position of the plane in its own reference frame and we get y(t) = 0. To measure the velocity of the plane with respect to itself, we take the derivative and get zero. OK so far. But what if we are in the reference frame of the Earth and we want to know the velocity of the plane with respect to itself? We can't take the derivative because we are in the wrong reference frame to do that. What we would like to do is convert to the reference frame of the plane, take the derivative there, then convert back to the reference frame of the Earth. That three-step operation satisfies all the formal properties of the derivative, e.g., it has a Leibniz rule. Therefore it's a connection. But it's not the usual connection (because that's the derivative, and the derivative in the reference frame of the Earth is velocity with respect to the Earth). Instead it lets you measure things relative to the plane while you're in the reference frame of the Earth; e.g., the Earth itself is found to be moving at velocity -1 relative to the plane. And that velocity is independent of the reference frame: The Earth is always moving at velocity -1 with respect to the plane, no matter what reference frame you're in!

Does that make more sense? Connections turn up in general relativity and in the Yang-Mills equations, so you might be able to find a better physically-motivated description of them in stuff on those. Ozob (talk) 20:00, 2 August 2009 (UTC)

Yes, I think we are in agreement on the basic phenomenon,

When you take velocity relative to the Earth, you're measuring a rate of change with respect to the Earth, and this is a totally different rate of change than rate of change with respect to the plane.

which I believe is the end of the discussion. However, this part really baffles me:

But what if we are in the reference frame of the Earth and we want to know the velocity of the plane with respect to itself? We can't take the derivative because we are in the wrong reference frame to do that. What we would like to do is convert to the reference frame of the plane, take the derivative there, then convert back to the reference frame of the Earth.

With respect to the bold text, why does it matter what reference frame "we" are "in" when we take this derivative? "We" don't appear anywhere in the equations that define the position of the pilot, the airplane, or the Earth. The equation for the derivative only takes account of the reference frame in which the derivative is being taken. There appears to be no other dependency; we are only talking about the relative motion of points with respect to each other. What I mean is, the velocity of the pilot relative to his plane is merely the velocity of the pilot relative to his plane—the velocity of a pilot-fixed point moving through a field of infinitely many airplane-fixed points. It matters not how any other objects or persons in the universe are moving. Am I mistaken? MarcusMaximus (talk) 06:47, 3 August 2009 (UTC)

I think we've just about worked it out, but that part you quoted I phrased badly. Let me try yet again. This time I'll start by asking, physically, what is the derivative? It is the rate of change in a reference frame; in the reference frame of the Earth, it's the rate of change with respect to the Earth, and in the reference frame of the plane, it's the rate of change with respect to the plane. Suppose that we want to measure the velocity of a bird relative to the plane. We observe that in the reference frame of the Earth, the plane has position x(t) = t and the bird has position b(t) = 0.1t. How do we measure the velocity of the bird relative to the plane? We can't do this by taking the derivative of the position function of the bird. This is what I meant when I said that we were in the wrong reference frame to do that: "Take the derivative" means "Find the velocity with respect to the present reference frame, i.e., with respect to the Earth"--which of course is not what we want. That's where the three-step procedure (and the connection it determines) come in. In the reference frame of the plane, the bird moves like -0.9t. The derivative in this reference frame is -0.9. Finally, that -0.9 was measured in the reference frame of the plane, so to be scrupulous and pedantic we must convert back to the reference frame of the Earth, because that's where we got our measurements from. The velocity doesn't change when do this, though; this last step isn't actually necessary. (But I'm including it anyway, because if you were trying to determine a position in the reference frame of the Earth by converting to the reference frame of the plane, then this step would be important.)

Are we agreed? As I said before, I dont't know what to do about the article; what I wrote has inspired so much discussion above that there must be a better way of saying it. Ozob (talk) 03:52, 4 August 2009 (UTC)

I still don't think I agree. While the velocities of the bird and the airplane relative to Earth are interesting, they are not necessary. The velocity of the bird in the reference frame of the airplane is the derivative of the position vector connecting a point in the bird to any point fixed in the airplane (since in the airplane frame, by definition, all points fixed in the airplane have zero velocity). There is no need to worry about the Earth or any measurements taken from it--it's a problem involving only two points and one reference frame.

If, as you are suggesting, we are constrained to take measurements from a different reference frame, that is a subjective limitation, not one inherent to kinematics. MarcusMaximus (talk) 04:22, 4 August 2009 (UTC)

Such limitations can be real. What if instead of a plane and a bird we have two elementary particles or two stars? We currently have no way of taking measurements in those frames directly.

Besides your comment about necessities of measurement, I couldn't see any specific objections in what you wrote. Ozob (talk) 04:30, 4 August 2009 (UTC)

Now I am interested to know what this statement means:

There are infinitely many possible reference frames because there are many nonequivalent bases for a vector space, even after rescaling.

I think different reference frames are more than just different bases in a vector space, aren't they? MarcusMaximus (talk) 07:34, 4 August 2009 (UTC)

And by the way, let me say that I really appreciate the discussions we have had on this page. Thank you for your patience and insight. MarcusMaximus (talk) 07:48, 4 August 2009 (UTC)

The discussions have been good for me, too. It's clarified my own understanding.

Or at least, it has to the point of realizing that I'm clueless. I've decided that I don't know what a frame of reference is. What I thought at the beginning was that it was the same as choice of an orthonormal basis, what I would call a frame. That would be nice, right? Frame and frame of reference? That's what I was thinking when I made that edit. But of course you can't represent curvilinear coordinates by taking linear combinations of basis vectors. So then I decided that it was a choice of coordinates, as I said above. In that case, the statement you quoted above is obviously wrong. But a mere choice of coordinates isn't enough to distinguish an inertial frame of reference from a non-inertial one. You can't tell whether there are any fictitious forces solely on the basis of your coordinates; you can use cartesian or polar coordinates for both inertial and non-inertial reference frames.

What I think a reference frame might be now is a choice of a connection. This might sound kind of goofy, because a connection is something that lets you take derivatives. It makes some sort of sense to me because an acceleration is a second derivative, so if you get fictitious forces, then your second derivatives are bad. If I'm right, an inertial reference frame is a flat connection, and a non-inertial reference frame is a non-flat connection. This is speculation, though; someone at WP:WPPhys might actually know. Ozob (talk) 00:32, 5 August 2009 (UTC)

Ok, well, I can't comment on connections. But I can say that to confuse vector bases, coordinate systems, and reference frames is one of the most common errors in kinematics, and there is no shame in it. In my experience even most college professors don't seem to understand it clearly. Many people blend the concepts together by talking about a hybrid entity called a "coordinate frame", which I and my colleagues regard as just asking for trouble. There is actually a hierarchy among them. First you need a reference frame. Then you can fix in that reference frame a set of mutually orthogonal basis unit vectors, which are free vectors that have a direction only. Tnen you pick any point in the reference frame as the origin and establish a coordinate system by defining coordinate axes parallel to the basis vectors. Reference frames are still quite useful even without coordinate systems. MarcusMaximus (talk) 01:09, 5 August 2009 (UTC)

When you pick a reference frame, what does that mean mathematically? How does the choice of reference frame affect a calculation? Ozob (talk) 14:00, 5 August 2009 (UTC)

I'm afraid I can't give you a rigorous mathematical definition of a reference frame in terms of manifolds or connections all that stuff. But a reference frame is fundamentally a set of points whose relative positions are constant with time. Therefore, if you have a door on a hinge with a vector embedded in the door (imagine an arrow painted on the door representing the vector), there are two reference frames that are immediately relevant: a reference frame consisting of the door and all points fixed in the door, and all "imaginary" points in the universe extending in all directions that move with the door as it swings; and a reference frame fixed relative to the doorframe/house/Earth. Now imagine swinging the door open and closed--the vector you painted on the door obviously remains unchanged in the reference frame of the door, always pointing exactly parallel to the door. But relative to the Earth, that vector is swinging back and forth, pointing in a different direction relative to the house at every instant.

The calculation of the derivative is different in each reference frame, for example, because when you express the vector using a formula in terms of a basis fixed in the door and take the derivative in the reference frame of the door, the derivatives of all the door-fixed basis vectors are zero, so the second term in the product rule goes to zero and you have a derivative like this one.

However, using exactly the same formula for the vector (using basis vectors fixed in the door) and taking the derivative in the Earth-fixed frame, the basis vectors fixed in the door have nonzero derivatives as the door swings. This is shown here.

If you compare the formula at the first link to the first formula at the second link, you'll see that the only difference is the second part of the product rule for differentiation is zero in the first case.

You can also use a linear transformation (like a direction-cosine matrix) to express the vector in terms of vectors fixed in the Earth frame at any step along the way and get a different formula. However, a linear transformation cannot get you from one reference frame to the other, because they are fundamentally different functions.

This concept took me a while to wrap my mind around. MarcusMaximus (talk) 16:43, 5 August 2009 (UTC)

By the way, I just took a look at the article on frames of reference. It is terrible. It is self-contradictory. In the opening sentence it says a reference frame is a coordinate system. Then it quotes 5 or 6 guys who say the concepts of reference frame and coordinate system are distinct. No wonder nobody knows what they are talking about! MarcusMaximus (talk) 04:00, 6 August 2009 (UTC)

Well, I tried looking in the two books on mathematical physics that I happened to have handy, namely Frankel's Geometry of Physics and Dubrovin–Fomenko–Novikov's Modern Geometry, Part 1 (which isn't really on physics, but it talks about physics a bit). The first wasn't helpful at all; I didn't see any mention of moving reference frames. The second was only slightly helpful; at one point they talk about reference frames moving with respect to each other in terms of Lagrangians. I've never studied enough physics to really figure out Lagrangians, so this left me unenlightened. All I got out of it was that it may be possible to describe the rotation (or lack of rotation) of a reference frame by putting certain terms in the Lagrangian.

But now I've gotten pretty confident that a reference frame is really a connection like I was thinking before. I'm going to write this all down here, mostly for my own benefit, but maybe you'll get something out of it too.

Here's the setup I want to use: Let the manifold M be the real numbers R. Choose a coordinate t on M. Let E be the trivial vector bundle on M of rank three (for simplicity). In other words, E is just one copy of R³ for each point of M; since M is just R, E is really R¹ × R³ together with the information that the first copy of R is special and corresponds to M. A (smooth) section of E is by definition a (smooth) function s from M to E of the form t → (t, s¹(t), s²(t), s³(t)); that is, for each time t, s chooses a vector in R³, i.e., s is really a parameterized curve in R³ written in a funny way.

A connection (mathematics) can be defined in many ways—I like connection (vector bundle)—but for the present purposes it's good to think of it as a covariant derivative. A covariant derivative ∇ with coefficients in E is a function that takes two things, a vector field and a section of the vector bundle E, and determines another section of E called the derivative of the original section with respect to the vector field. It satisfies a bunch of good properties which I won't list here. The important thing for the present purposes is that a covariant derivative can always be expressed as follows: Suppose that the section is written (using the Einstein convention) as sⁱe_i, where e₁, e₂, e₃ are the standard basis sections of E (i.e., for each time t, e_i(t) is the ith standard basis vector of R³ at time t). Suppose also that d/dt represents the tangent vector field on M which points to the right with length one at all times. Then we always have the following formula:

$\nabla _{d/dt}(s^{i}\mathbf {e} _{i})={\frac {ds^{i}}{dt}}\mathbf {e} _{i}+s^{i}\nabla _{d/dt}(\mathbf {e} _{i}).$

This is the Leibniz rule for connections. $\nabla _{d/dt}(\mathbf {e} _{i})$ is a section of E by definition, so it must have an expression in terms of the basis sections, too. We set:

$\nabla _{d/dt}(\mathbf {e} _{i})=\omega _{i}^{k}\mathbf {e} _{k}.$

The ω^k_i are called the coefficients of the connection; it's fair to interpret them as the curvilinear partial derivative of e_i in the kth coordinate direction. Note that they have only two indices instead of the usual three because M has only one coordinate and hence there's no point in indexing over that one coordinate. In terms of these coefficients, we get, after reindexing:

$\nabla _{d/dt}(s^{i}\mathbf {e} _{i})={\frac {ds^{i}}{dt}}\mathbf {e} _{i}+s^{j}\omega _{j}^{i}\mathbf {e} _{i}=\left({\frac {ds^{i}}{dt}}+s^{j}\omega _{j}^{i}\right)\mathbf {e} _{i}.$

This should be looking familiar. If the connection coefficients are zero, then we're in the case of a non-moving reference frame. If not, then we should be able to use the connection coefficients to describe how the reference frame's movement affects the derivative of s. That is, we need to replicate the formula:

${\frac {{}^{\mathrm {N} }d\mathbf {a} }{dt}}=\sum _{i=1}^{3}{\frac {da_{i}}{dt}}\mathbf {e} _{i}+\sum _{i=1}^{3}a_{i}{\frac {{}^{\mathrm {N} }d\mathbf {e} _{i}}{dt}}$

somehow. This is easy. We let ωⁱ_j equal the ith component of ${}^{\mathrm {N} }d\mathbf {e} _{j}/dt$. Tada! There we go. Ozob (talk) 22:49, 6 August 2009 (UTC)

Just out of curiosity, does this definition of angular velocity agree with your analysis?

${}^{\mathrm {N} }\omega ^{\mathrm {E} }=\mathbf {e} _{1}({\frac {{}^{\mathrm {N} }d\mathbf {e} _{2}}{dt}}\cdot \mathbf {e} _{3})+\mathbf {e} _{2}({\frac {{}^{\mathrm {N} }d\mathbf {e} _{3}}{dt}}\cdot \mathbf {e} _{1})+\mathbf {e} _{3}({\frac {{}^{\mathrm {N} }d\mathbf {e} _{1}}{dt}}\cdot \mathbf {e} _{2})$

Or in simpler notation, where the overdot replaces ${}^{\mathrm {N} }d/dt$,

${}^{\mathrm {N} }\omega ^{\mathrm {E} }=\mathbf {e} _{1}{\dot {\mathbf {e} }}_{2}\cdot \mathbf {e} _{3}+\mathbf {e} _{2}{\dot {\mathbf {e} }}_{3}\cdot \mathbf {e} _{1}+\mathbf {e} _{3}{\dot {\mathbf {e} }}_{1}\cdot \mathbf {e} _{2}$

MarcusMaximus (talk) 03:50, 7 August 2009 (UTC)

Well, let's find out if I can fit it in somehow. I don't know how one would define "angular velocity" in an intrinsic way except by trying to use the curvature of the connection, so I'm going to try that first.

I'm going to describe connections in the way that connection (vector bundle) does. That is, now ∇ is a vector bundle homomorphism E → E ⊗ Ω that satisfies the Leibniz rule; here Ω is the cotangent bundle of M. In coordinates, this means that ∇ looks like this:

$\nabla (\mathbf {e} _{i})=\omega _{i}^{k}\mathbf {e} _{k}\otimes dt.$

The curvature is ∇² : E → E ⊗ Λ²Ω. But Λ²Ω is the zero bundle; in local coordinates, the unique coordinate dt on Ω determines the coordinate dt ∧ dt, which is zero.

OK, that didn't work. But one thing that's been bothering me since yesterday is that the approach I'm taking really doesn't unify space and time; it treats them as separate (time is the manifold, space is the vector bundle). So I'm going to try something else.

Let M be R⁴. Give M the standard Lorentzian metric, that is, we define the dot product of two vectors (x₀ = t, x₁, x₂, x₃) and (y₀ = u, y₁, y₂, y₃) to be tu − x₁y₁ − x₂y₂ − x₃y₃. (Note c = 1.)

The tangent bundle of M is TM = M × R⁴. A connection on the tangent bundle of M is a vector bundle homomorphism ∇ : TM → TM ⊗ Ω satisfying the Leibniz rule; in coordinates, it sends

$\nabla (\partial _{i})=\Gamma _{ji}^{k}\partial _{k}\otimes dx^{j}.$

(This is really the same as the definition of a connection I gave above, but rephrased.) The Γs are the ωs from before, but now a lot more of them are zero. If we have a tangent vector field $s^{i}\partial _{i}$, then using the Leibniz rule we get:

$\nabla (s^{i}\partial _{i})=\partial _{i}\otimes ds^{i}+s^{i}\nabla (\partial _{i})=\partial _{i}\otimes ds^{i}+s^{i}\Gamma _{ji}^{k}\partial _{k}\otimes dx^{j}.$

To replicate the results of my last post, we just let all the coefficients $\Gamma _{ji}^{k}$, where j is not zero, equal zero. Sort of; to handle a curve a(t), what we ought to do is pullback the tangent bundle of M along a and use the pullback connection. Unfortunately then we're in the same situation as above, where the curvature is zero for dimension reasons. I'm going to press on with the computations in the hope that this reveals something to me.

Anyway, there's a standard formula for the curvature of a connection which can be found at Riemann curvature tensor:

$R_{jkl}^{i}=\partial _{k}\Gamma _{lj}^{i}-\partial _{l}\Gamma _{kj}^{i}+\Gamma _{km}^{i}\Gamma _{lj}^{m}-\Gamma _{lm}^{i}\Gamma _{kj}^{m}.$

Look at the last two terms, the ones with only Γs and no partial derivatives. Recall that $\Gamma _{ji}^{k}$ is zero when j is not zero; so unless both l and k are zero, those last two terms vanish; and if they're both zero, then upon substituting zero we see again that the last two terms also vanish. So we get:

$R_{jkl}^{i}=\partial _{k}\Gamma _{lj}^{i}-\partial _{l}\Gamma _{kj}^{i}.$

By similar reasoning, we get:

$R_{jkl}^{i}=0,$ when k and l are both not zero,

$R_{jk0}^{i}=\partial _{k}\Gamma _{0j}^{i},$ when k is not zero,

$R_{j0l}^{i}=-\partial _{l}\Gamma _{0j}^{i},$ when l is not zero,

$R_{j00}^{i}=\partial _{0}\Gamma _{0j}^{i}-\partial _{0}\Gamma _{0j}^{i}=0.$

This looks like it can't replicate the angular velocity that you have above. There are no partials in the time direction, i.e., nowhere do you get a $\partial _{0}$. I guess this is consistent with the observation that it didn't work before.

For the moment, I'm out of ideas as to what the angular velocity represents in a strictly mathematical sense. Maybe this is obvious, but could you tell me where that formula comes from? It's pretty opaque to me. Ozob (talk) 00:36, 8 August 2009 (UTC)

It came from a very good dynamics textbook, Dynamics: Theory and Applications by Kane and Levinson. I work with Levinson on a daily basis and he taught me most of what I know about kinematics and dynamics. You can download the book for free here. The definition of angular velocity is presented at the top of page 16 of the text, which is page 36 of the PDF. I'm not sure if the definition of angular velocity is derived from anything, but it's a very intuitive definition--at least it makes sense physically in my head. I'll try to think of a way to explain it in text.

I was hoping this equation

$\nabla _{d/dt}(s^{i}\mathbf {e} _{i})={\frac {ds^{i}}{dt}}\mathbf {e} _{i}+s^{j}\omega _{j}^{i}\mathbf {e} _{i}=\left({\frac {ds^{i}}{dt}}+s^{j}\omega _{j}^{i}\right)\mathbf {e} _{i}.$

would lead logically to this one

${\frac {{}^{\mathrm {N} }d\mathbf {a} }{dt}}=\sum _{i=1}^{3}{\frac {da_{i}}{dt}}\mathbf {e} _{i}+\sum _{i=1}^{3}a_{i}{\frac {{}^{\mathrm {N} }d\mathbf {e} _{i}}{dt}}={\frac {{}^{\mathrm {E} }d\mathbf {a} }{dt}}+{}^{\mathrm {N} }\mathbf {\omega } ^{\mathrm {E} }\times \mathbf {a}$

because somehow the indices on the $s^{j}\omega _{j}^{i}\mathbf {e} _{i}$ would work cyclically to form ${}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {a} .$

If you start with the statement you made, can you make any progress from there?

We let ωⁱ_j equal the ith component of ${}^{\mathrm {N} }d\mathbf {e} _{j}/dt$.

I don't really know what ranges those indices are supposed to cover, or I'd do it myself.

MarcusMaximus (talk) 06:41, 8 August 2009 (UTC)

The indices are supposed to go from 1 to 3; x₁, x₂, and x₃ are the x, y, and z coordinates, respectively. (The reason for using this notation rather than using x, y, and z is just that it's convenient for making computations.)

The first part of the first equation that you quoted leads to the first part of the second equation that you quoted. We have:

$\nabla _{d/dt}(s^{i}\mathbf {e} _{i})={\frac {{}^{\mathrm {N} }d\mathbf {a} }{dt}},$

${\frac {ds^{i}}{dt}}\mathbf {e} _{i}+s^{j}\omega _{j}^{i}\mathbf {e} _{i}=\sum _{i=1}^{3}{\frac {da_{i}}{dt}}\mathbf {e} _{i}+\sum _{i=1}^{3}a_{i}{\frac {{}^{\mathrm {N} }d\mathbf {e} _{i}}{dt}}.$

I should reindex the left-hand side of the first equation by swapping the is and js in the second term. If I also stop using the Einstein summation convention on the left-hand side, then I get:

$\sum _{i=1}^{3}{\frac {ds^{i}}{dt}}\mathbf {e} _{i}+\sum _{i=1}^{3}\sum _{j=1}^{3}s^{i}\omega _{i}^{j}\mathbf {e} _{j}=\sum _{i=1}^{3}{\frac {da_{i}}{dt}}\mathbf {e} _{i}+\sum _{i=1}^{3}a_{i}{\frac {{}^{\mathrm {N} }d\mathbf {e} _{i}}{dt}},$

which I think makes the equality a little more transparent: a_i equals sⁱ, and ω^j_i is the component of ^Nde_i/dt in the direction of the jth basis vector.

The one thing which I haven't yet worked out is where the cross product comes in. So I looked at the book you referenced. It looks very good! Unfortunately, as you say, it doesn't justify the definition of angular velocity, it simply states it. It's still helpful, though: It's convincing me that angular velocity behaves in a good way, and that the definition above isn't arbitrary but really comes from something.

I'm going to try to reason my way backwards this time, by equating the two terms that I don't understand. For each i, we can suppose that a(t) = s(t) = e_i, and then we ought to have:

$\omega _{i}^{j}\mathbf {e} _{j}={}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {e} _{i}.$

Equations (9) and (10) on page 17 of the textbook tell me that the right-hand side should equal ${\dot {\mathbf {e} }}_{i}.$ But that's just the statement that we can write things in terms of ^Nde_i/dt as above, so that's not really helpful. What I think I'd like to do is to see what components ^Nω^E should have. Let's write its kth component as ^Nω^E_k. If we write out all the terms of that equation above, we get:

$\omega _{1}^{1}\mathbf {e} _{1}+\omega _{1}^{2}\mathbf {e} _{2}+\omega _{1}^{3}\mathbf {e} _{3}=\omega _{1}^{i}\mathbf {e} _{i}={}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {e} _{1}={}^{\mathrm {N} }\omega _{3}^{\mathrm {E} }\mathbf {e} _{2}-{}^{\mathrm {N} }\omega _{2}^{\mathrm {E} }\mathbf {e} _{3},$

$\omega _{2}^{1}\mathbf {e} _{1}+\omega _{2}^{2}\mathbf {e} _{2}+\omega _{2}^{3}\mathbf {e} _{3}=\omega _{2}^{i}\mathbf {e} _{i}={}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {e} _{2}={}^{\mathrm {N} }\omega _{1}^{\mathrm {E} }\mathbf {e} _{3}-{}^{\mathrm {N} }\omega _{3}^{\mathrm {E} }\mathbf {e} _{1},$

$\omega _{3}^{1}\mathbf {e} _{1}+\omega _{3}^{2}\mathbf {e} _{2}+\omega _{3}^{3}\mathbf {e} _{3}=\omega _{3}^{i}\mathbf {e} _{i}={}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {e} _{3}={}^{\mathrm {N} }\omega _{2}^{\mathrm {E} }\mathbf {e} _{1}-{}^{\mathrm {N} }\omega _{1}^{\mathrm {E} }\mathbf {e} _{2}.$

So we deduce that:

${\begin{array}{lll}\omega _{1}^{1}=0,&\omega _{1}^{2}={}^{\mathrm {N} }\omega _{3}^{\mathrm {E} },&\omega _{1}^{3}=-{}^{\mathrm {N} }\omega _{2}^{\mathrm {E} },\\\omega _{2}^{1}=-{}^{\mathrm {N} }\omega _{3}^{\mathrm {E} },&\omega _{2}^{2}=0,&\omega _{2}^{3}={}^{\mathrm {N} }\omega _{1}^{\mathrm {E} },\\\omega _{3}^{1}={}^{\mathrm {N} }\omega _{2}^{\mathrm {E} },&\omega _{3}^{2}=-{}^{\mathrm {N} }\omega _{1}^{\mathrm {E} }&\omega _{3}^{3}=0.\end{array}}$

OK, this is looking promising! Apparently there really is some sort of good relation between the connection coefficients and the angular velocity. But I'm not really sure how to express it abstractly: All I see at the moment is "plug in the numbers in a nice-looking way". This is some sort of low-dimensional phenomenon relating three-dimensional vector spaces and three-dimensional matrices; you can't do this in two or four and higher dimensions because you don't have the right number of matrix entries. And at the moment I can't see what the relation is; I'm sure I can look it up somewhere, though. Ozob (talk) 16:09, 8 August 2009 (UTC)

Oh, wait, this is obvious. There is a linear transformation of vector spaces R³ → R³ determined by "send v to its cross product with ^Nω^E". ω^j_i is just the matrix of that linear transformation. The skew-symmetry of the matrix corresponds to the symmetry you see in the formula for a cross product (which ultimately comes from the skew-symmetry of the wedge product).

In a sense, this answers the question of the relationship between ^Nω^E and ω^j_i. But in another sense it doesn't: For a linear transformation to have the form "cross product with a vector" is very, very special. In three dimensions (which we're in), this happens exactly when the matrix is skew-symmetric. As it turns out, skew-symmetry of the matrix of connection coefficients happens exactly when the connection is "compatible with the metric", i.e., that it satisfy a certain condition relating the connection and inner products. So that explains why angular velocity is a vector and why the interesting part of the connection can be represented as a cross product: It's because the connection is compatible with the metric and we're in three dimensions.

So here is the overall picture as I see it at the moment:

• Spacetime, in classical physics, is R¹ × R³.
• A reference frame is a connection on the tangent bundle of spacetime which is:
  1. Compatible with the metric, and
  2. Zero in all the spacelike directions.
• The derivative in a certain reference frame is just an application of the connection.
• The angular velocity of a reference frame with respect to an inertial frame of reference is the connection form.
• Because the connection is zero in all the spacelike directions, the interesting part of the connection form is a 3 × 3 matrix.
• Because the connection is compatible with the metric, the 3 × 3 matrix is skew-symmetric.
• Skew-symmetric 3 × 3 matrices are determined by vectors. The vector corresponding to the connection form is the angular velocity.
• A connection determines an inertial frame of reference if and only if its torsion tensor vanishes.
• All reference frames correspond to a flat spacetime, meaning that the Riemann curvature tensor vanishes.

OK, that's a lot more complicated than I thought it would be way back when we started this discussion. But I think I've got it; do you have any comments? I hope I haven't lost you! Ozob (talk) 20:51, 8 August 2009 (UTC)

I just moved, so I've been without internet access for a while. Unfortunately a lot of what you just said is over my head, but I do know about the relationship between skew-symmetric matrices and the cross product.

The only comment I have is that you don't have to specify that one of the two reference frames is an inertial reference frame; this relationship of derivatives works for any two reference frames. This is a purely kinematical (which roughly means "geometric and mathematical") relationship, while inertia and "inertiality" only is important when you're dealing with dynamics and kinetics (the interaction of bodies and forces).

Thanks for the fantastic discussion. I'm glad to see that all of this makes sense at a fundamental level, even though my knowledge only starts somewhere in the middle of the hierarchy. MarcusMaximus (talk) 18:42, 20 August 2009 (UTC)

Hmm. Well, even if neither of the reference frames is assumed inertial, we should still be able to figure out the angular velocity somehow. I'm going to guess that all one does is compute the angular velocities of the two reference frames with respect to an inertial frame, then takes their difference.

Come to think of it, I bet there's a direct way of doing this. There was a step above where I said:

$\omega _{i}^{j}\mathbf {e} _{j}={}^{\mathrm {N} }\omega ^{\mathrm {E} }\times \mathbf {e} _{i}.$

Now, while it doesn't look like it, there's an inertial frame of reference hidden here on the right hand side. The cross product is usually defined as a goofy looking differential operator with some weird stuff happening in the indices; in fact, it's not, it's very natural and comes right out of the wedge product and the Hodge dual. The Hodge dual is something to do with the ordinary differential d on Euclidean space, that is, it's related to the Euclidean derivative. I wouldn't be surprised if there's some way to define a curved Hodge dual with respect to a connection; and that using this you could define a curved cross product. If you stuck in the curved cross product on the right hand side, then you'd eliminate the implicit use of an inertial reference frame. That may not be particularly useful; after all, it should just work out to be what you'd get by finding the angular velocities of N and E with respect to an inertial frame and taking their difference. But it would be satisfying if it were true, because then all of this stuff would be kinematic like it ought to be.

Discussing these things with you is always enlightening. If you have any more questions, just ask. I'm sure I'll learn something. Ozob (talk) 22:04, 20 August 2009 (UTC)

Did your talk page stuff get deleted some how?

First, I don't think I deleted any of your comments on quaternion talk. If I did I am terribly sorry, and am wondering if it might have somehow been an edit conflict, where we were both typing at the same time.

Second yes, this question that I am wondering about right now, I have been editing a bit, because the way I first typed it did not really sound as clear as I could make it after I went back and read it some more. Plus I wanted to add links to it.

Right now I am reading Jasper Jolly's 1905 Manual of Quaternions. Jolly if you don't know him well, was a royal astronomer of Ireland, a post that Hamilton had once held, and edited the 1898 version of Elements of Quaternions that Hamilton died before completing, based on the earlier 1865 version.

I will try and be more diligent in avoiding deleting comments, not sure if I actually did that, but if I did I am sorry and will not do it again.

Thanks for all your interesting comments and contributions. I always find it very interesting reading about what you have ask for. TeamQuaternion (talk) 01:34, 8 August 2009 (UTC)

Jacobson radical

Although this is relatively minor, I happened to notice the addition of, "If R is unital and is not the trivial ring {0}, the Jacobson radical is always distinct from R.", to the article. Are there any interesting non-trivial consequences of this property? In many interesting cases of the theory, it is desirable to study rings with J(R) = R; for instance "If R is nil, is the polynomial ring over R Jacobson radical?", is an equivalent form of the (open) Köthe conjecture. Apart from the fact that the quotient R/J(R) will be non-trivial for a unital ring, I do not know of any properties which can be deduced from this. Could you please tell me what you had in mind? Thanks, --PST 00:46, 20 August 2009 (UTC)

Oh, all I had in mind there was to not lose the fact that you'd removed (because it had been incorrectly stated). I noticed that the fact, properly stated, was a trivial consequence of Zorn's lemma, and while I really don't know anything about these sorts of rings, it seemed to me that J(R) equaling R or not would be pretty important. You seem to know more about these things; if it's actually not an interesting fact, then feel free to remove it. Ozob (talk) 21:44, 20 August 2009 (UTC)

Thanks, I have kept the fact but added a sentence which covers more general cases that I mentioned above. --PST 04:22, 25 August 2009 (UTC)

Mathematic typing style

Dear Ozob, thanks for your advise on User talk:129.97.227.25. The edit was mine. I do not understand, why dx should be written as dx as this is to my view clearly against any common sense: variables are generally written in italics and are composed out of one (1) letter, every other use would start confusion (this notations is according to AMS style). This "d" here is not to be taken as a variable multiblied by x.

Otherwise, what is the value of $\int _{0}^{\pi }\sin(t)d\rho dt$

I wonder, whether it is $\pi \sin(t)dt\,$ or $2d\rho \,$

Well, as it seems, you define this with spaces and so on, ok. but I disagree. However it will be like it says.

Cheers, Saippuakauppias ⇄ 01:13, 5 December 2009 (UTC)

I agree that variables are often written in italics and are only one letter long. However, dx can be considered as an infinitesimal variable, in which case it does not make sense to separate the d from the x. One can also consider d to be a function, the exterior derivative, and functions are also often written in italics. There are many ways of interpreting dx, and it often makes sense to italicize the d. I think that I have always seen the d italicized in AMS publications.

I would like to know where you learned to write an upright d. I have only ever seen it done on Wikipedia, but you seem to have learned it somewhere else. Where? Ozob (talk) 13:44, 5 December 2009 (UTC)

Thank you

I very much appreciate your work here. :) --Moonriddengirl ^(talk) 14:52, 20 December 2009 (UTC)

Speedy?

The lead section of arithmetic variety didn't seem to be copied from the source, although the section on Kazhdan's theorem was. I removed the speedy tag and stubbed the article. If you disagree, then please restore the tag. Also, I have been tagging the articles with {{copyvio}} rather than {{db-copyvio}}. Is the latter preferred in some more obvious cases? Sławomir Biały (talk) 13:09, 21 December 2009 (UTC)

Yes, I think I may have been a bit too hasty with the speedy tag on that article. I've been using {{db-copyvio}} when I felt that nothing substantial would be left. In the case of arithmetic variety, we would be left with two not obviously equivalent definitions and no context; I suppose that's something, so {{copyvio}} looks more appropriate. But the {{db-copyvio}} template I just added to noncommutative resolution is okay, for example, because the article had a single sentence, and that sentence was a copyvio. Ozob (talk) 17:38, 21 December 2009 (UTC)

Another potential approach

Hi. A contributor to the clean-up at the CCI asks whether material can be presumptively deleted. It can, in accordance with policy. There's more about this at Wikipedia talk:WikiProject Mathematics#Copyright concerns related to your project, including a template that may prove helpful should you wish to take this approach. --Moonriddengirl ^(talk) 18:16, 21 December 2009 (UTC)

Speedy deletion declined: Albert–Brauer–Hasse–Noether theorem

Hello Ozob, and thanks for your work patrolling new changes. I am just informing you that I declined the speedy deletion of Albert–Brauer–Hasse–Noether theorem - a page you tagged - because: Not an unambiguous copyright infringement, or there is other content to save. Please review the criteria for speedy deletion before tagging further pages. If you have any questions or problems, please let me know. JohnCD (talk) 18:50, 21 December 2009 (UTC)

Apologies - I meant to come back sooner and add to the rather curt notice that the CSDHelper script produces. My reasoning was that the copyvio passage could be removed leaving a valid stub article. I didn't realise what a swamp I was stepping into, see conversation with Moonridengirl here. Regards, JohnCD (talk) 20:39, 21 December 2009 (UTC)

That's okay. But the first sentence may also be a copyvio: See the first sentence of the source he cites for it. Given that and this user's history (and that the template lets you fill in only one url as far as I'm aware) I thought it was better to just nuke the page. Failing that I think it should be blanked and rewritten; I'm sure I could use the sources he cites to come up with appropriate content. Ozob (talk) 04:22, 22 December 2009 (UTC)

Email

Hi Ozob, I've sent you email. Paul August ☎ 18:21, 29 December 2009 (UTC)

Thank you

		The Copyright Cleanup Barnstar
		Your work on this contributor copyright investigation is very much appreciated. Moonriddengirl (talk) 17:14, 12 January 2010 (UTC)

Thanks. :) --Moonriddengirl ^(talk) 17:14, 12 January 2010 (UTC)

MoS markup messup

What process was used to create this edit to the Manual of Style? Was it some external editor? It generated some real howlers in the resulting page. Eubulides (talk) 17:08, 8 February 2010 (UTC)

:-O I had no idea that happened. No, all I did was use the usual edit link—but the old edit box has been replaced by a new weird and broken one. (You can see a notice about its brokenness on your watchlist.) It looks from this like it mangles ampersands, too. I'm going to submit a bug report. Ozob (talk) 18:01, 8 February 2010 (UTC)

It's now bug 22435. Ozob (talk) 18:22, 8 February 2010 (UTC)

I'm afraid it happened again, later, with this edit. I suggest disabling this new edit box, and sticking with the old edit box, until the bugs are fixed. Also, have you recently edited any other pages that might be affected by the bug? Eubulides (talk) 00:14, 9 February 2010 (UTC)

Blast. Of the articles I've edited, it seems that only the MoS is affected by this; it's weird. The bug has been fixed in the current Mediawiki source, but I don't think that fix has been pushed out to Wikipedia's servers yet. Thanks for cleaning up again. I hope this bug died a painful death. Ozob (talk) 04:47, 9 February 2010 (UTC)

DYK nomination of Lefschetz theorem on (1,1)-classes

Hello! Your submission of Lefschetz theorem on (1,1)-classes at the Did You Know nominations page has been reviewed, and there still are some issues that may need to be clarified. Please review the comment(s) underneath your nomination's entry and respond there as soon as possible. Thank you for contributing to Did You Know! Marylanderz (talk) 01:42, 18 February 2010 (UTC)

DYK for Lefschetz theorem on (1,1)-classes

On February 21, 2010, Did you know? was updated with a fact from the article Lefschetz theorem on (1,1)-classes, which you created or substantially expanded. You are welcome to check how many hits your article got while on the front page (here's how, quick check ) and add it to DYKSTATS if it got over 5,000. If you know of another interesting fact from a recently created article, then please suggest it on the Did you know? talk page.

Ucucha 18:10, 21 February 2010 (UTC)

List of algebraic geometry topics

I've added dévissage and generic flatness to the list of algebraic geometry topics. If you know of other articles that should be listed there and are not, could you add those too? Michael Hardy (talk) 04:42, 2 March 2010 (UTC)

Four Egyptian geometry formulas may qualify. In the Rhind Mathematical Papyrus problems 41, 42, and 43 report the area of a circle as A = [(8/9](D)]^2 cubit^2 (formula 1.0), where pi =256/81 and radius R = semi-diameter D/2. MMP 10 also used formula 1.0 to compute the area of a semi-circle/. Gillings suggested C = (pi)D was involved in formula 1.0. Skipping over the area debate, scribal algebra added height (H) in two cases V = (H)[8/9)(D)]^2 cubit^3 (formula 2.0) and V = (3/2)(H)[(8/9)(D)]^2 khar (formula 3.0). An interesting algebraic geometry devives from V = (2/3)(H)[(4/3)(D)]^2 khar (formula 4.0), reported in RMP 43 and the Kahun Mathematical Papyrus. Did two scribes modify formula 3.0 by applying these algebraic steps:

1. considering V = (3/2)(H)(8/9)(8/9)(D)D)

multiplying both sides by 3/2 such that

2. (3/2)V =(3/2)(3/2)(H)(8/9)(8/9)(D)(D) khar = (4/3)(4/3)(D)(D) khar

and multiplying both sides by 2/3, such that

3. V = (2/3)(H)[(4/3)(D)]^2 khar (formula 4)?

as cited on the math forum: http://mathforum.org/kb/thread.jspa?threadID=2109309&tstart=0

Anneka Bart may wish to comment on this topic for an added reason. Ahmes in RMP 41, 42 and 43 divided the khar unit by 20, and found a 100-hekat unit (*reported by Peet and Clagett). A single hekat total was reported by multiplying the 100-hekat value by 100, a step that Ahmes did not clearly report. The scaled scribal hekat context shows that Ahmes scaled a khar to 5 hekat, a conclusion that Dr. Bart disagrees. She suggests without writing out mathematical statements associated with scholar references that a khar properly contained 20 hehat. Scribal algebraic geometry discussions may offer conflicting points of view. Let the raw data and the scholars openly debate the history of math geometry formulas and the unit values contained therein. Wikipedia entries that contain controversial topics should be noted, thereby avoiding needless Wiki-debates and Wiki-wars.

Best Regards, Milogardner (talk) 14:53, 14 September 2010 (UTC)

I would say that the material from the Rhind Mathematical Papyrus is not algebraic geometry at all. The rest of the comments about disagreements is completely besides the point and not relevant to the discussion here. --AnnekeBart (talk) 19:08, 15 September 2010 (UTC)

Thanks

Thank you very much for brokering the WQA and ANI reports on my behalf. Things seem to be moving in a more productive direction at Gravitational potential now, although they are no less frustrating. Even though no action came out of the incident, RHB at least seems to be a little less confrontational now. Best wishes, Sławomir Biały (talk) 01:37, 18 March 2010 (UTC)

Which vs that

In the US, it is incorrect to use 'which' in a restrictive clause, according to the Chicago Manual of Style and every other reference I know. It is not incorrect in the UK, but 'that' in a restrictive clause is also not wrong...so following the US rule creates something that is correct in both countries. Thanks. —Preceding unsigned comment added by 75.0.176.7 (talk) 04:37, 2 April 2010 (UTC)

Reference

I have a question about your edit. I wanna know that ... is there any policy about referencing/citation in Wikipedia? (It's just a question, not quarrel) -- Modamoda (talk) 16:14, 19 April 2010 (UTC)

Yes, there's WP:CITE. For footnotes, see WP:FOOT. Neither of these specify whether we should group adjacent footnotes together or not. I prefer not to; that's what I'm used to, and that's what I usually see others do. (You can observe WP:FOOT not group them under "Ref tags and punctuation".) I agree that the cluster of footnotes at the start of integral domain looks odd, but it's there because of a quarrel on the talk page. Ozob (talk) 23:45, 20 April 2010 (UTC)

I see, thanks anyway -- Modamoda (talk) 10:01, 21 April 2010 (UTC)

Good faith

You wrote in Wikipedia talk:Words to watch#Question for Philip as a parting shot "I believe therefore that it is fair to characterize your views as primarily a content objection. Your process objections are red herrings intended to slow us down."

If I did not assume good faith which you do not seem to be extending to me I could argue that "Your process objections are red herrings intended to speed up the process to sneak in changes and then game the system by arguing that it needs consensus to change them back". But as you are clearly a honest person I am sure that you would not sink to such depths and I would appreciate it if you would extend good faith to my motive. If so you will strike out the comment as it does not help us reach a consensus. -- PBS (talk) 02:05, 26 April 2010 (UTC)

I will not strike out my comment. Please answer the question I posed you. Ozob (talk) 02:17, 26 April 2010 (UTC)

Spectral sequences

Hi Ozob,

I was the anonymous user that changed the discussion of the differentials in spectral sequences.... I'm a noob here, so I hope this is the right place to discuss.....

I think it's actually "up or down" at the zero level, and "left or right" at the one level - according to J. McCleary's User's Guide, the bidegree is (-r,r-1) for homological type and (r,1-r) for cohomological type.

Cheers, Mathjd

Whoops. You're right; I completely botched that one. Thanks. Ozob (talk) 19:49, 2 May 2010 (UTC)

Substing Welcome Templates

Just a quick note, can you make sure you subst welcome templates when you add them to a users talk page? Thanks =] ·Add§hore· ^{Talk To Me!} 18:59, 18 May 2010 (UTC)

Whoops. Thanks for catching that. Ozob (talk) 02:34, 19 May 2010 (UTC)

I have marked you as a reviewer

I have added the "reviewers" property to your user account. This property is related to the Pending changes system that is currently being tried. This system loosens page protection by allowing anonymous users to make "pending" changes which don't become "live" until they're "reviewed". However, logged-in users always see the very latest version of each page with no delay. A good explanation of the system is given in this image. The system is only being used for pages that would otherwise be protected from editing.

If there are "pending" (unreviewed) edits for a page, they will be apparent in a page's history screen; you do not have to go looking for them. There is, however, a list of all articles with changes awaiting review at Special:OldReviewedPages. Because there are so few pages in the trial so far, the latter list is almost always empty. The list of all pages in the pending review system is at Special:StablePages.

To use the system, you can simply edit the page as you normally would, but you should also mark the latest revision as "reviewed" if you have looked at it to ensure it isn't problematic. Edits should generally be accepted if you wouldn't undo them in normal editing: they don't have obvious vandalism, personal attacks, etc. If an edit is problematic, you can fix it by editing or undoing it, just like normal. You are permitted to mark your own changes as reviewed.

The "reviewers" property does not obligate you to do any additional work, and if you like you can simply ignore it. The expectation is that many users will have this property, so that they can review pending revisions in the course of normal editing. However, if you explicitly want to decline the "reviewer" property, you may ask any administrator to remove it for you at any time. — Carl (CBM · talk) 12:33, 18 June 2010 (UTC) — Carl (CBM · talk) 13:34, 18 June 2010 (UTC)

Thanks! Ozob (talk) 02:29, 21 June 2010 (UTC)

Talkback

Hello, Ozob. You have new messages at Wikipedia talk:Manual of Style (words to watch).
You can remove this notice at any time by removing the {{Talkback}} or {{Tb}} template.

Weaponbb7 (talk) 02:43, 26 June 2010 (UTC)

Andre - Quillen vs.Harrison (co)homology ?

Dear colleague, I do not have Andre's and Quillen's papers. From what You have written it seems to me that Andre-Quillen is the same as Harrison's ones. I guess the relation is very well-known, would You be so kind to comment on it. —Preceding unsigned comment added by Alexander Chervov (talk • contribs) 17:48, 22 July 2010 (UTC)

Quillen discusses this in his paper. Try [4]. Ozob (talk) 00:13, 27 July 2010 (UTC)

Thank You very much. So it seems Harrison was the first to define this cohomology, while Quillen generalizes to the "relative case" A->B, while Harrison treats just "B" meaning k->A , "k" is basic field. I think in some future it would be better to create a page "Homology of commutative rings" with redirect from Andre-Quillen and Harrison cohomology to such page. Harrison's approach is downtoearth while Quillen's is most not downtoearch... 17:14, 28 July 2010 (UTC) —Preceding unsigned comment added by Alexander Chervov (talk • contribs)

Your revert on Template:Group-like structures

https://secure.wikimedia.org/wikipedia/en/w/index.php?title=Template%3AGroup-like_structures&action=historysubmit&diff=385478429&oldid=385423672

Actually, Magmas are called groupoids sometimes. This is actually on the first line of the article. It is also in several books that I've been reading.

I think your change is not a good idea. The additional information makes it less confusing for readers, like me, who don't know much about the subjects. Tony (talk) 03:11, 19 September 2010 (UTC)

Hmm. As I said, I've never seen "groupoid" used this way, only in the way it's used in the groupoid article. I suspect that "groupoid" for "magma" is obsolete. But you say that it's in books you're reading, so maybe I'm just out of touch. I think the right thing to do is for you to bring this up on the article's talk page. Maybe we can get some additional opinions as to the right way forward here. Ozob (talk) 12:57, 19 September 2010 (UTC)

Barnstar!

		The Barnstar of Diligence
		Hi Ozob – just noticed your lil’ edit to Riemann integral, and recognized you (from Problem of Apollonius). Thanks for your specific edit (yeah, jargon should really be avoided on elementary pages – oops), and for your consistent pattern of making Wikipedia (Math) just that much better and more polished – thanks! —Nils von Barth (nbarth) (talk) 06:09, 20 September 2010 (UTC)

RfC closing

Letting you know that there is opposition to your closing and interpretation of the results about italic titles at Wikipedia talk:Article titles#Again. _Xeworlebi ^(talk) 21:31, 27 September 2010 (UTC)

RFC of Italics closing

I read the RFCs on this. Simple, plain response is that i wholeheartedly dispute both you closing it for being involved and for your assessment. Your own tally does not support what you put forward as consensus. I waited for a response as Xeworlebi advised you of this 10 days ago. Today i reverted the policy to what it was before you changed it. Surely someone will not like that but hopefully someone notices why i changed it. The problem is that italic titles have now been implemented site wide in all relevant infobox template, which are all protected. It will be a pain to undo all of it. I read all of it and i see you having added the "limited use support" into the "full support" to claim that sufficient people support it to implement it. Thing is those same people who support limited use would also oppose full use (or else they would have voted for full use). Hence the consensus was to not implement italics beyond the limited use and you misrepresented things. I call upon you to revise your close or to re-open the RFC, and to have italics removed from all infoboxes where it has been added subsequent to your close of the RFC on 19 September, and to then have someone who truly is not involved close the RFC. Other than this i am not too sure what to do short of an RFC on your closing of the RFC and even i know that sounds somewhat silly just to write. But i am serious. delirious & lost ☯ ^~hugs~ 17:29, 7 October 2010 (UTC)

Go for it on the RFC on my RFC. If there is someone who actually wants to read all the arguments, tally the votes, and come up with a coherent policy statement, then I'd prefer that they do it. I did it only because nobody else was going to (look at the timestamps). I tried to be as objective as I could, but as I think was clear from what I wrote, I'm not a fully objective party. Ozob (talk) 21:03, 7 October 2010 (UTC)

Since you've posted essentially the same objection at WT:AT, I have made essentially the same reply more publicly there. Let's continue the discussion there instead. Ozob (talk) 21:23, 7 October 2010 (UTC)

Just for a record, i posted there some time ago now and you didn't respond. So today i added a 2nd notice of my contesting the closing by reverted the policy change. That i think is what got someone's attention. However the consensus is that it is too late to contest since it is now in full widespread use and would require a new consensus to undo what you claimed. :S
I would prefer the formal RFC on the RFC but someone beat me to it with another subsection simply contesting the close based on my stated concerns and their own. That you just simply leave it closed and invite someone else to re-open it... really, who is going to do that? Noöne.
As my issue is first and foremost with you closing it despite being about as involved as anyone there is, and as my secondary issue is with being completely baffled by your conclusion that results in changing the policy, and as my attempt at addressing this on the policy talk page went pretty much unnoticed i would rather discuss this here.
At the most basic level i do not see where you find the support. To simplify it, there is a Yes, No, & Kinda. Yes and No are about even. Kinda is conveniently about the same and thus the majority could swing a few ways depending on what you want.
Those who like the close you did surely will not be objecting to it.
Those not liking the close you did are either not in the mood to bother or believe it to be futile to object, or they don't know it was closed.
Then there is the flaw in the entire set-up. I hate consensus ruling (not a secret) but i do know that it will surely fail if there are more than 2 options. That part of consensus on WP is to allow as many options as people care to put forward is why things get stalemated, such as the removal of admin rights. This RFC had 3 major options with a couple of subsections depending on your choice of major. By design that will result in indecision and massive vote splitting. That is what happened and yet you found a 'clear winner'.
If someone had simply told me of this or if i had stumbled upon it before its implementation flooded my watchlist then yes i would have closed it. I read it and my conclusions do not agree with yours. Now i too am too involved to close it and if i were to well it would not go as smooth as your closing did, and i never actually voted :P I had no idea the template existed let alone the long ongoing discussion about forcing its use. delirious & lost ☯ ^~hugs~ 22:11, 7 October 2010 (UTC)

Since you say, "I would prefer the formal RFC on the RFC", please do that. If you don't want to do that, I suggest raising the issue at WP:VPP. I would prefer either of those to another meandering debate on WT:AT. Ozob (talk) 22:28, 7 October 2010 (UTC)

Shell (mathematics)

How do you know its a hoax? Do you mean a hoax, that there is no such thing, or just erroneous? I have removed the speedy, and suggest you take it to AfD, not prod, for a community decision. It would be well to notify the mathematics wikiproject to get some informed comment. DGG ( talk ) 23:28, 25 October 2010 (UTC)

Composite Number Factoring Theorem

I see you tagged this article for speedy deletion. That was rejected. I've posted a proper AfD. The article's AfD page can be found here. Thanks. — Fly by Night (talk) 17:51, 30 November 2010 (UTC)

Email

Hello, Ozob. Please check your email; you've got mail!
It may take a few minutes from the time the email is sent for it to show up in your inbox. You can at any time by removing the {{You've got mail}} or {{ygm}} template.

Access

I hope you're enjoying this discussion about access. It is something our project needed to address. --Anthonyhcole (talk) 12:29, 21 January 2011 (UTC)

Dashes and music-album lists

Hi, I think this is what it's referring to. Tony (talk) 12:25, 31 January 2011 (UTC)

WP:WPM interview

Update: Thanks for participating in the interview. Just a heads up that section editor Mabeenot, has move the publication date to this coming Monday, 21 February. The final draft has now been posted. Please go through it to check for any inaccuracies, etc. Thanks again. – SMasters (talk) 23:48, 16 February 2011 (UTC)

The Signpost

Hi, could you review how I've presented the successful nomination of the two animations in "Featured pictures"? If there's a way of presenting what they're about to the intelligent non-mathematician, that would be good. link. Tony (talk) 10:29, 20 February 2011 (UTC)

Sorry, I was away all weekend, so this is too late to help you. But the description is fine. Ozob (talk) 12:21, 23 February 2011 (UTC)

Derivative

Hi Ozob, I see that you've reverted my modification in the article Derivative, section Derivatives of elementary functions, from $f'(x)=0.25x^{-0.75}\,\!$ back to $f'(x)=(1/2)x^{-1/2}\,$. I wonder why the latter seems superior to you? I do see a couple of reasons myself why mine might be somewhat clearer. Iamthedeus (talk) 22:15, 25 February 2011 (UTC)

Well, my reasoning is mainly that 1/2 is a simpler number to think about than 0.25. I also prefer to use fractions rather than decimals because too many students identify numbers with their decimal expansions, and I think that using fractions fights that. So it's nothing deep. It sounds like you have a reason for preferring 0.25, so please say why. Ozob (talk) 02:05, 26 February 2011 (UTC)

Well in the case of using 1/2 or 0.5 as the value of r in $f(x)=x^{r},\,$ one finds created in the resulting derivative a visual symmetry that is meaningless and thus potentially misleading: the number one-half appears both as the coefficient and, in negative form, as the exponent. Of course this is an anomaly that occurs for only one of the infinite possible values for r, and thus is meaningless, and yet it is a rather salient feature of the example, all the more so for readers with less of a grasp on math. This seemed to me reason enough to change the value of r in the example to some other number, such as one-quarter, in order to avoid any unnecessary confusion as to the true rationale for the values of the coefficient and exponent in the derivative function, while preserving the other important features of the example (such as the derivative being undefined for negative values of x).

This, too, is nothing deep, and surely most readers would not be thus confused upon encountering the one-half version, but I see no harm in opting to avoid that possibility entirely.

As for my choice of decimal over fraction, it was merely that the "0.25" strikes in me a slightly sharper conception of the number than does "1/4"—I suppose I myself am one of those who at heart identify numbers with their decimal expansions—but I certainly see the value of discouraging that, so I do not dispute your preference for using fractions.

Also, I noticed that you reverted a number of modifications I had made to various formulas on the page; specifically, I inserted "\,\!", forcing them to be rendered as TeX PNGs. I did this because they had been rendered in HTML (for me, at least—even if they contained just "\,") and as such happened to have some readability issues that were fully avoided by rendering them as images. Is there any reason I should know of for which "\,\!" would be undesirable in these cases?

Iamthedeus (talk) 01:31, 3 March 2011 (UTC)

I agree with your reasoning, so I've changed the article to use 1/4 instead of 1/2. To me, 0.25 seems much less precise than 1/4: The former literally means "25 parts out of one hundred", and I have a hard time visualizing one hundred parts; whereas the latter means "one part out of four", which I can easily visualize.

\, should always force a formula to be rendered as HTML, unless perhaps you've changed your math rendering preferences. Wikipedia:Manual of Style (mathematics) explicitly suggests \, as a way to force PNG rendering. I figured that your use of \,\! was an attempt to prevent the formulas from having extra space at their ends. This can't happen in practice because of how they're rendered, so I thought the extra characters amounted only to clutter. But if they have a real effect then you should start a thread about this at the math MoS, as a lot of articles may have to be changed. Ozob (talk) 01:57, 3 March 2011 (UTC)

I'm not sure what the default math rendering preference is, but mine is set to "HTML if possible or else PNG", which would explain why \, was insufficient. In any case I was guided by Help:Displaying_a_formula#Forced_PNG_rendering which seemed to indicate that the practice of using \,\! is acceptable. I suppose, though, that the rendering option "Recommended for modern browsers" would be more ideal for someone like me with a modern browser, so I'll switch to that. Iamthedeus (talk) 04:59, 3 March 2011 (UTC)

The italics issue

Hi. Just a note to say that I have raised the italics issue here (with a mention here). I am broadly in favour of re-opening the discussion — as a proper centralized discussion — though obviously without wanting to take on a huge workload personally. Regards. --Kleinzach 04:23, 28 March 2011 (UTC)

Salebot

Hi Ozob; "Boubaker polynomials" are pretty infamous on frwiki, and Salebot is trained to automatically revert edits containing the term (in French), ignoring 1RR for this specific case. You're on Salebot's whitelist now. --gribeco (talk) 00:32, 4 May 2011 (UTC)

Hi. Just for your information, Rirunmot was blocked (global account) in May 2009 on it: as sockpuppet of Softer, main account of the "boubaker team" on it:. Rhadamante (talk) 04:33, 4 May 2011 (UTC)

Wikipedia talk:WikiProject Mathematics/Straw poll regarding lists of mathematics articles

In light of your participation in the discussion(s) regarding the treatment of disambiguation pages on the "Lists of mathematics articles" pages, please indicate your preference in the straw poll at Wikipedia talk:WikiProject Mathematics/Straw poll regarding lists of mathematics articles. Cheers! bd2412 T 18:58, 23 May 2011 (UTC)

Good advice at WT:WPM

I shouldn't have gotten heated. I'll try to count to ten in the future. In hind sight I can see that if I don't understand the words it is a bit silly to take offense to them. Thenub314 (talk) 21:29, 25 May 2011 (UTC)

Spacing of en dashes

I just wanted you to know that I've since loosened up my thinking about the rules on the spacing of certain en dashes (the "von Keipert" thing) – and I think I took a rather too hard line against your proposals at MoS, last year, was it? Tony (talk) 04:08, 18 June 2011 (UTC)

(Sorry for the delay, I've been on vacation.) Yes, you did take a rather hard line. I did too. I've also softened on this issue, and I don't object to spaces as much as I used to anymore. I'm sure that we can come to some consensus! Ozob (talk) 12:41, 20 June 2011 (UTC)

pentagram map questions

Hi Ozob, I tried to answer your questions on the pentagram map discussion page. I didn't do a perfect job, but I hope it helps. RichardEvanSchwartz (talk) 05:46, 30 June 2011 (UTC)

Riemann Integral

Thank you for your edits...I made a lot of stupid typos despite my efforts with the "Show Preview" button.

Fraqtive42 (talk) 18:54, 22 July 2011 (UTC)

Infinite Numbers

Thank you for your mathematical insights. I could follow almost all of them. Indeed Infinite reals can be straightforwardly added and subtracted in a convergent manner, but they CANNOT be multiplied convergently. You may not even multiply a real by an infinite real by your argument. Then again, who says that numbers have to be multiplyable, maybe we can be SATISFIED with mere addition and subtraction. Anyway I had fun concocting this and fun reading your eloquent treatise. Thank you Sir! — Preceding unsigned comment added by BenHeideveld (talk • contribs) 21:04, 14 August 2011 (UTC)

Hey ozob

Thanks for showing me what was wrong, I'm just writing here to let you know I read your response. I won't make those mistakes again, but I definitely need to do something about that derivatives page. It is way too hard for learners to understand.

Anyway, thanks for your message!