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March 26[edit]

winding[edit]

Do we have an article on the problem of the size of wound-up like cable or toilet paper? You know, the more you wind the greater the circumference so the more it takes to increase the diameter.. .froth. (talk) 05:18, 26 March 2009 (UTC)[reply]

To a very close approximation ignoring that the paper isn't a exact circle or that the thickness may depend on the curvature the length will be proportional to the area. In a sort of way putting thin sheets together was the start of integral calculus, here each thin sheet round in a circle has length 2πr and integrating gives the area. Dmcq (talk) 08:49, 26 March 2009 (UTC)[reply]

See Clackson scroll formula (WHAAOE). Gandalf61 (talk) 10:47, 26 March 2009 (UTC)[reply]

By the way instead of measuring the inner and outer diameters you can just lay a ruler across the outside of the inner cylinder (the 'former') and measure the diameter of the outside of the roll at that point - so one easy measurement instead of two harder one. Left as an exercise to see why :) Dmcq (talk) 13:20, 26 March 2009 (UTC)[reply]

For some cable, wire, and, of course, string, you get more than the width of one thickness on a spool. And it's not as simple as taking the width of the wire, dividing the width of the spool by this, and figuring that that many times as much wire would store as on a single-wire-width spool. The reason is that the packing isn't always ideal, sometimes you get a nice hexagonal packing when viewed in wire cross-section, but not always. It very much depends on how the wire or string is wound on the spool. So, for such cases it would be best to measure the actual amount of wire per sample spool, than to try to apply formulae, at least initially. After enough samples are taken, one could find a suitable "fudge factor" to apply to the formula to account for non-ideal packing, bearing in mind that this factor is subject to change if the spool winding method changes. StuRat (talk) 15:00, 26 March 2009 (UTC)[reply]

See [1] (OR). —Tamfang (talk) 10:25, 28 March 2009 (UTC)[reply]

What's that, Tamfang ? It appears to be a page with lots of illustrations of fractals and such. How does that relate to this Q ? StuRat (talk) 13:25, 28 March 2009 (UTC)[reply]

If your browser fails to take you to the section of the page specified by the #-tag, search for "bead". There's a formula for a parametric curve which is relevant to the question. —Tamfang (talk) 16:24, 30 March 2009 (UTC)[reply]

On second thought, I may as well give the formula here rather than try to pump my stats.

r = √(t/π)

θ = u-atan(u) where u = √(4πt-1)

This curve wraps around the origin in the way that seems to be sought by the OP. —Tamfang (talk) 06:35, 1 April 2009 (UTC)[reply]

Matrix notations (quick lighthearted question)[edit]

I've just had a linear assignment handed back marked by a postgradute student on behalf of a lecturer, and it was unfortunately full of unhelpful and cheeky remarks, but one I found rather interesting was the disgust expressed by the marker over my use of square brackets for matrices. The PG wrote use round brackets for matrices, square brackets have a specialized meaning in other contexts. I use square brackets because the I first learned about matrices from a book which used the square bracket notation, and have always stuck with it. I tried looking for these other contexts on WP but have only found that it appears to be a case of simple aesthetic preference of the author; square brackets and parentheses seem to be used fairly interchangably in elementary matrix algebra. Is there any difference in the meaning of an array surrounded by parentheses versus by square brackets? 131.227.167.206 (talk) 15:34, 26 March 2009 (UTC)[reply]

I've seen square brackets used for matrices plenty of times (more often in print than handwriting). As long as it is clear from context what you are doing, it really doesn't matter what notation you use. Vertical lines on either side (with nothing at the top and bottom) denotes the determinant, so make sure your square brackets have clear horizontal marks at the tops and bottoms. Other than that, I wouldn't worry. (Obviously, if you are working in one of those specialised contexts where it make a difference [projective geometry, maybe], then you need to distinguish them, but there are plenty of mathematical notations that mean different things in different contexts.) --Tango (talk) 15:51, 26 March 2009 (UTC)[reply]

FWIW, matrix uses box brackets throughout, but says that parentheses are an acceptable alternative notation. Gandalf61 (talk) 15:52, 26 March 2009 (UTC)[reply]

I doubt there's any symbol that doesn't "have a specialized meaning in other contexts." I've only ever seen matrices with square brackets, though. Black Carrot (talk) 22:09, 26 March 2009 (UTC)[reply]

Might this have to do with different national customs? In Italy I have only seen parentheses, and have seen my first brackets in some English-language text. What about other languages/countries? May I ask where the original poster is from? Goochelaar (talk) 22:27, 26 March 2009 (UTC)[reply]

If we're fishing for OR: as a product of the British education system I was taught with parentheses at both school and university without any mention of alternative notation. Obviously variation exists in the literature but I guess the pragmatist's approach is to explain the notation as you introduce it if you think there could be any ambiguity. (Also, I'd adopt whatever notation your examiner/marker says, it tends to stay on their good side.) 86.140.160.93 (talk) 22:42, 26 March 2009 (UTC)[reply]

Both round and square brackets are commonly used. As indicated above, straight lines usually refer to the determinant, but to avoid any ambiguity, I recommend using det(A) for the determinant of the matrix A. Parenthesis (which to me means the curly brackets {}) are so widely used to indicate the object under consideration is a set, that I think they should be reserved for that purpose. Of course, anyone reading mathematics at the level of linear algebra (second year university, say) should know enough to infer from the context the intended meaning of any symbol. The final remark in the comment above is good practical advice.Aliotra (talk) 18:07, 27 March 2009 (UTC)[reply]

Note that a "parenthesis" is ( or ), not { or }. See Bracket. —JAO • T • C 18:33, 27 March 2009 (UTC)[reply]

Indeed. "{}" are "braces" (or just "curly brackets"). --Tango (talk) 18:36, 27 March 2009 (UTC)[reply]

I was introduced to matrices in my second year of A-levels (year before uni). I would be surprised if anyone can get to the second year of a maths degree without having met them. (Although my first formal course in linear algebra was 2nd year - my dept is moving it to 1st year, though.) --Tango (talk) 18:36, 27 March 2009 (UTC)[reply]

Square is the form of bracket that is easiest to extend to encase a 2-D matrix of any size. This can be managed even when using a limited fixed-size font in ASCII art. Square brackets are used for matrix (called array) references in C programming language. Parentheses ( and ) seem called for only when the matrix is small or when other brackets are unavailable such as on many typewriters or BASIC programming language. IMHO braces {} are unnecessarily ornamental (I never wear them). Cuddlyable3 (talk) 20:02, 27 March 2009 (UTC)[reply]

In the way of deriving the 1st kind christoffel symbol[edit]

Hi, I'm wondering how the 1.4.1 formula in the following, could be derived? when I write down the combination, I don't know how to get to the result? Any suggestion and hint?

File:Christoffel symbol 1st.jpg —Preceding unsigned comment added by Re444 (talk • contribs) 15:47, 26 March 2009 (UTC)[reply]

It follows almost immediately from plugging your first equation into your definition or $(\alpha ,\beta ,\gamma )$ and carrying out the derivative on each of the three terms. Maybe you could show what working you have done to show where you got stuck? Sorry, I misread your problem! Ignore me! 163.1.176.253 (talk) 17:11, 26 March 2009 (UTC)[reply]

You need some kind of additional information to get 1.4.1 from the previous formula. For example, if we know

g_{ab}{\frac {\partial ^{2}x^{a}}{\partial {\overline {x}}^{\alpha }\partial {\overline {x}}^{\gamma }}}=g_{bc}{\frac {\partial ^{2}x^{c}}{\partial {\overline {x}}^{\alpha }\partial {\overline {x}}^{\gamma }}}

(and the two analogous relationships) then the result follows. I'm afraid I'm not familiar with any of this notation so I don't have an intuition for if that assumption is reasonable. (Judging from the article it seems that there's some kind of coordinate transformation going on here?) Eric. 18:49, 26 March 2009 (UTC) —Preceding unsigned comment added by 131.215.158.238 (talk)

Your assumption is fine. In summation convention the contracted index is a dummy index and so your assumption just relies on the symmetry of the metric,

g_{ab}=g_{ba}

, which is true by defintion (see Metric tensor). 86.140.160.93 (talk) 22:34, 26 March 2009 (UTC)[reply]

Ahhhhh, I see now... never did feel comfortable with Einstein notation.... Let's hope our OP understands? Eric. 131.215.158.238 (talk) 02:01, 27 March 2009 (UTC)[reply]

Oh!! Thanks so much. I was looking for the solution almost a long time, while it was pretty straightforward! Thanks again everybody! —Preceding unsigned comment added by Re444 (talk • contribs) 10:03, 27 March 2009 (UTC)[reply]