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June 3

Laurent Series

I have a complex analysis exam coming soon, and I'm not very confident with Laurent series. Suppose I have a complex function f(z) which has a finite number of poles. Is there a general method for finding its Laurent expansion? Admittedly, my knowledge of Laurent series is somewhat limited. I know what they are, but not really how they work. Can anyone help? Maelin 03:12, 3 June 2006 (UTC)

Have you read Laurent series? Conscious 06:56, 3 June 2006 (UTC)

Yes, but it's an infinite series from -infinity to infinity. If I'm asked to find the Laurent series of a function, where do I start? The examples in the article don't show where the expressions come from, it's just "consider this function. Now, abracadabra! Here are some Laurent expansions for it depending on where you're interested!"

The second formula of the article gives a general formula for calculating the coefficients in a Laurent series, so it's not magic, just calculation. Admittedly, that's not usually the easiest way to calculate. -lethe ^{talk +} 07:40, 3 June 2006 (UTC)

You can always use the general formula and well-known series for some functions (see the example for $e^{-{1}/{x^{2}}}$ in the article). Conscious 07:49, 3 June 2006 (UTC)

Okay, I've done some more work with them. In our lecture note examples, we generally have a function of the form f(z) = 1 / g(z) where g(z) is some polynomial in z (usually conveniently factorised into linear terms). Then we reduce it to some form of geometric series and find the Laurent series. The problem is that I have no idea how we end up with a Laurent series that is valid for the particular region we're interested in. We just seem to head off reducing it in one particular way and then voila, we have the right one. In the article, we have an example just like this, and then three different Laurent expansions miraculously appear, correct for each region. Clearly, whoever made the example didn't do an infinite number of integrals around γ and then build a series out of them, so what is going on here? Apologies if I sound terse, this is very frustrating. Maelin 05:19, 4 June 2006 (UTC)

Well, doing infinitely many integrations for the coefficients is not as bad as it sounds, since you can often tell at a glance what a contour integral will be. Anyway, like I said, that's not usually the easiest way. The example in the article goes like this. Split that thing up by partial fractions. When |z| < 1 and 2, you can use the geometric series with the z – 1 and the z – 2i terms. When z is between the two roots, you can use the geometric series with the z –2i term, but for the other term, you have to use like z/(1 – 1/z). And for the last region, you have to invert z for both terms. Basically, you can only invoke the geometric series when the ration has modulus less than 1, and this fixes how you find the Laurent series in each region. -lethe ^{talk +} 20:58, 4 June 2006 (UTC)

voronoi diagram

I would like the definition and applications of a voronoid diagram (scatter application). Please put a definition on your web site.

Voronoi diagram Melchoir 03:19, 3 June 2006 (UTC)

qauadratic formula

what is the quadratic formula — Preceding unsigned comment added by 04:03, 2006 June 3 (talk • contribs) 69.231.27.28

Please read and follow the first bullet point at the top of this page. (And sign your posts with four tildes, ~~~~.) Thank you. --KSmrq^T 05:33, 3 June 2006 (UTC)

See quadratic formula. Conscious 06:54, 3 June 2006 (UTC)

Firefox: how to search edit areas?

How do I search edit areas in Firefox? For example, when I edit a long article and want to find where it links to a specific category, the text entered into the search field seems to be searched for only outside the wikicode. Is there any way to get it straight? Conscious 06:07, 3 June 2006 (UTC)

I don't know of a way to do that. It would be very useful. However, if you use the "Show preview" button, you'll be able to search the edited version (though not the edit box itself), and that makes it a lot easier to find the required text, by counting paragraphs.... TheMadBaron 19:28, 3 June 2006 (UTC)

Mathematica to Wikipedia

I have Mathematica 5.0. What is the best way to transform the *.nb to a format that Wikipedia understands.

You don't. You write text for an article explaining what you're doing. Dysprosia 04:14, 4 June 2006 (UTC)

The HTMLSave function comes to mind. Realistically, the text will mostly need to be copied as plain text, perhaps with some UTF-8 characters; and each formula will need to be rewritten in either wiki syntax or TeX format. The TeXForm facility may help with the latter, though compatibility with MediaWiki's limited version is not guaranteed. Come BlahTeX, the MathMLForm version may be of interest, though it would make editing obnoxious. Note that it would be inappropriate to import Mathematica style sheets to override Wikipedia's own. --KSmrq^T 04:27, 4 June 2006 (UTC)

Upper bound on n!

Given an integer n, what are the tightest bounds on factorial n? More specifically, I want to calculate the number of binary digits required to represent n! for a given n. -- Sundar ^{\talk \contribs} 10:37, 3 June 2006 (UTC)

The article on factorial gives the following approximation based on Stirling's approximation :

$\ln(n!)\approx n\ln(n)-n+{\frac {\ln(n)}{2}}+{\frac {\ln(2\pi )}{2}}.$

Can this be taken as an upper bound on $\ln(n!)$? If not what would be an upper bound? -- Sundar ^{\talk \contribs} 10:45, 3 June 2006 (UTC)

The article on Stirling's approximation says that the error is the same sign and size as the first omitted term. The next term in the series after the one you've written is negative, so your series is an upper bound. -lethe ^{talk +} 11:30, 3 June 2006 (UTC)

Oh whoops. My comment above will be true if you also include the 1/12n term. So do that. -lethe ^{talk +} 11:32, 3 June 2006 (UTC)

Oh thanks, Lethe. -- Sundar ^{\talk \contribs} 11:44, 3 June 2006 (UTC)

Upper bound on $\ln(n!)$: $\ln(n!)\approx n\ln(n)-n+{\frac {\ln(n)}{2}}+{\frac {\ln(2\pi )}{2}}+{\frac {1}{12n}}$

Sposta be a 12. -lethe ^{talk +} 11:48, 3 June 2006 (UTC)

Thanks, fixed it. By the way, what would be the equivalent one for log to the base 2 (my original question)? (Excuse my laziness.) -- Sundar ^{\talk \contribs} 12:06, 3 June 2006 (UTC)

Well, log₂ x = log x/log 2. So divide both sides of the equation by ln 2, and you're good to go. -lethe ^{talk +} 12:10, 3 June 2006 (UTC)

I now realise. It wasn't laziness, but naivety. -- Sundar ^{\talk \contribs} 12:16, 3 June 2006 (UTC)

Well considering that summation is cheap for computer, for relatively small n you can get a very good estimate by using

$\ln n!=\sum _{k=1}^{n}\ln k$

Igny 12:48, 3 June 2006 (UTC)

Thanks. But, summation was not the main concern, logarithm was. -- Sundar ^{\talk \contribs} 06:29, 5 June 2006 (UTC)

i cut a little piece out of a torus (donut), what is the remaining space

Consider a torus, cut a tiny CLOSED piece out on the side. So the piece I cut out is homeomorphic to $\mathbb {D} ^{2}$ .

What is the remaining space? My professor tells me I can only see it when i start stretching that hole open 'until my fingers touch on the other side'. However I was born with no 3D mind, I simply do not see it.

It should be an 'easy space involving cilinder(s)' I would like to understand this completely for the understanding of homotopy and homology groups.

Evilbu 11:35, 3 June 2006 (UTC)

Well, a torus is a sphere with a handle, and a sphere with a disc removed is a disc, so a torus with a disk removed is a disc with a handle. I don't know if that's the answer you're looking for though. -lethe ^{talk +} 11:53, 3 June 2006 (UTC)

Well uhm, what is a handle, my syllabus says every $\mathbb {D} ^{k}\times \mathbb {D} ^{l}$ is a handle? The article http://en.wikipedia.org/wiki/Handle_%28mathematics%29 doesn't really help me out right now. Evilbu 12:16, 3 June 2006 (UTC)

A handle is a cylinder attached at its two end circles. It is not the product of two balls, which is trivial. -lethe ^{talk +} 12:29, 3 June 2006 (UTC)

$T\setminus \mathbb {D} ^{2}$ is $\mathbb {S} ^{1}\vee \mathbb {S} ^{1}$. To see this, represent $T$ as $\mathbb {D} ^{2}$ with side identifications; remove a chunk around the corner; shrink the remainder. Alternatively consider $\mathbb {S} ^{1}\vee \mathbb {S} ^{1}$ as the skeleton of a torus and imagine growing a patch of skin around the torus from that skeleton; the remaining hole is $\mathbb {D} ^{2}$-shaped. EdC 12:39, 3 June 2006 (UTC)

Is that supposed to be a disjoint union? The torus with a disc removed is certainly a connected space. Perhaps you mean the wedge sum of two circles instead? But that's not right either, as the latter is 1 dimensional, while the former is two dimensional. They are certainly homotopy equivalent though. -lethe ^{talk +} 12:55, 3 June 2006 (UTC)

Yeah, I meant wedge sum, sorry. As for dimension - if you cross each with a $\partial \mathbb {D} ^{1}$, so that they join at a $\partial \mathbb {D} ^{2}$, then you get something like two cylinders tangent at right angles. EdC 17:09, 3 June 2006 (UTC)

isn't the boundary of a 1-ball just a disjoint pair of points? If you go crossing anything with that, you'll get a disconnected space again. Surely not what you want. -lethe ^{talk +} 20:48, 3 June 2006 (UTC)

$T\setminus D^{2}$ is not the same as $S^{1}\vee S^{1}$. $S^{1}\vee S^{1}$, the deformation of the space in question, is a 1-manifold except at one point, whereas $T\setminus D^{2}$ is a 2-manifold with boundary. $T\setminus D^{2}$ looks like a thickened copy of $S^{1}\vee S^{1}$---though it's not just a figure-8 drawn with a really fat-tipped marker. (The boundary of a this figure-8 has three components, where the boundary of $T\setminus D^{2}$ has only one by the construction.) To visualize $T\setminus D^{2}$, take a big fat blocky plus sign (like the Red Cross logo), then glue the top and bottom ends together in the back, and the left and right ends together in the front. Tesseran 01:43, 6 June 2006 (UTC)

I am very confused now. What do you mean, a cilinder attached at its two end circle, you mean take a cilinder, then attach the upper and lower circle? Wouldn't that be a torus? Why would the product of two balls be trivial? What is the union of twice $\mathbb {S} ^{1}$ Evilbu 12:44, 3 June 2006 (UTC)

A coffee cup has a handle. The handle attaches to the mug in two different places, once at each end. -lethe ^{talk +} 12:58, 3 June 2006 (UTC)

Seems like I know much less than I thought. Is there a precise definition of handle (I am familiar with the language of quotient spaces) In order to make sure we understand each other, I will say a couple of things, and please tell me when you disagree : $\mathbb {D} ^{1}=[0,1]$ $\mathbb {D} ^{2}$ is the closed disk in two dimensions $\mathbb {S} ^{1}$ is the unit circle (it is one dimensional $\$\mathbb {T} ^{2}$ , the (empty,thus 2d)torus, is homeomorphic with $\mathbb {S} ^{1}\times \mathbb {S} ^{1}$ the full,thus 3d torus is homeomorphic with $\mathbb {S} ^{1}\times \mathbb {D} 2$

I disagree that a 3d torus is equal to a circle times a ball. Rather, such a thing is called a genus 1 handlebody. A 3d torus is the cartesian product of 3 circles. If you'd like a technical definition of a handle, you can take it to be a torus with a disk removed, though that won't help you visualize what it is. Hence the coffee cup description. -lethe ^{talk +} 20:46, 3 June 2006 (UTC)

Maybe it would be relevant to say why I want this. I want to find the torus' homology groups, especially the $H_{2}(\mathbb {T} ^{2})$ group Now my professor told us to do a Mayer Vietoris trick on the torus, by cutting out a little piece (a disk) . The two spaces I get then, have a 'relatively easy' intersection in my Mayer Vietoris sequence, it is a cilinder, homotopic with a circle, and thus completely known. But what is the other space?Evilbu 15:31, 3 June 2006 (UTC)

It's two cylinders kissing, homotopic to a bouquet of two circles. EdC 19:26, 3 June 2006 (UTC)

I suppose if you've previously worked out the homology of the more complicated piece this makes sense. (It also hints at a general construction involving a 2n-gon and edge gluing.) But otherwise, wouldn't it be more natural to use two cylinders (retracting to circles), with overlap deformation retracting to two disjoint circles? Also, this space is simple enough that you could take a rectangle split diagonally into two triangles, and glue the outer edges of the triangles together in such a way as to give a simplicial decomposition for an explicit computation. And if you've gotten as far as Mayer-Vietoris sequences you're probably not far from the Künneth theorem, which makes this T² = S¹×S¹ product a trivial computation. You can check your work by these other means, and also by comparing the known Euler characteristic of the torus, namely zero, to that given by its Betti numbers summed with alternating signs.--KSmrq^T 20:58, 3 June 2006 (UTC)

Also, isn't H_n(X,Z) always isomorphic to Z when X is a connected manifold? Generated by the fundamental class of X. -lethe ^{talk +} 03:22, 4 June 2006 (UTC)

How best to respond to a learning exercise? I'm assuming that the purpose is to get comfortable with computations, here using the Mayer-Vietoris sequence, rather than to get the answer. It's not as if computing the homology of a torus is much of a burden. Also, Evilbu doesn't tell us if the coefficients are Z, Z₂, R, or something else. (Has the class covered the universal coefficient theorem yet?) Nor do we know if this is singular homology, though that seems a good assumption, and relatively unimportant. In short, I'm trying hard not to say "The answer is …!"  :-D --KSmrq^T 05:03, 4 June 2006 (UTC)

To (sort of) answer lethe's question, consider the homology groups of Rⁿ; these are the same no matter what n may be. For, Rⁿ is homotopy equivalent to a punctured n-sphere, and deformation retracts to a point. (It is a contractible space.) So, what is H₂(R²,Z)? Remember, homology measures cycles modulo boundaries under homotopy equivalence; to get a Z there must be a "hole" to catch a cycle, preventing repetitions from collapsing. --KSmrq^T 19:20, 5 June 2006 (UTC)

Well, I'm a little concerned by your phrase "homology measures cycles modulo boundaries under homotopy equivalence". It's possible to have homologous paths that are not homotopic, I think. But your point is well-taken, there's something wrong with my assertion that H_n(X,Z) is always Z for an n-dimensional manifold, since, as you rightly point out, for Rⁿ, which is trivial. Suitably chastened by your correction (and Blotwell's above), I will put forth that perhaps it's true for closed manifolds? -lethe ^{talk +} 08:54, 6 June 2006 (UTC)

Yes, there's a "problem" with non-compact manifolds which you can fix by taking compactly supported homology. This gives you H_n (Rⁿ, Z) = Z, but at the price that (by the above argument) it can't be a homotopy invariant. But there's one other problem with your claim, which is that if your manifold isn't orientable then the integer top-dimensional homology is 0. You can fix that by taking coëfficients in Z₂. —Blotwell 18:05, 7 June 2006 (UTC)

Uhm, no I do not know anything of universal coefficient theorem. If it is relevant, we always consider these groups as$\mathbb {Z}$ modules, thus abelian groups. I know I cold do a Mayer Vietoris trick by using two cilinders, who intersection are two disjoint cilinders, then I find everything except $H_{2}(\mathbb {T} ^{2})$ Basically I was hoping by doing this cutting out of a little sphere, I would be able to find it in another way.Evilbu 09:47, 4 June 2006 (UTC)

June 4

cost management accounting - indirect labor unit cost.

Please help a new small manufacturing company by locating "indirect labor unit cost." We need this for our financial portion of the business plan as required by Small Business Administration. Our leaders, Maasters of Science in Healthcare, for some reason did not include this information. If anyone out there can help, it will greatly appreciated. Our company sews dresses and suits for premature infants and low birth weight infants. The SBA booklet indicates that this "cost" is needed to be included with total production costs.

This question may not be a typical for Wikipedia, but we have really been trying to find this and to date have not been successful.

"Labor unit cost" is the cost of labor per unit produced. So if in a month your labor cost is $100,000.00, and you produce in that time 5000 garments, the labor unit cost is $100,000.00 divided by 5000 equals $20. "Indirect labor cost" is that portion of labor cost that can be ascribed to activities not directly contributing to production (e.g. administration, salespeople, advertising). And the unit cost is as before. So if in that month your indirect labor cost is $40,000.00, then the indirect labor unit cost is $8. See http://www.uwm.edu/Course/IE360-Saxena/two.pdf section 2.5, elements of cost. --Lambiam^Talk 13:51, 4 June 2006 (UTC)

Formula used in civil engineering

Can someone give me a simple mathematical forumla sude in civil engineering? It's for a math homework project. Thanks in advance! --Wizrdwarts 01:38, 4 June 2006 (UTC)

Somewhere not far from where you live is a firm that does civil engineering. Contact them. Chances are excellent that they would be delighted to spend a few minutes talking to you about what they do, and offer a sample calculation or two. It should be much more fun and educational than asking a bunch of mathematicians on the web. (Though we can be fun, too, in our own way.) --KSmrq^T 05:19, 4 June 2006 (UTC)

Does Hooke's law count? Dysprosia 09:20, 4 June 2006 (UTC)

Catenary has a cool formula for suspension bridges... (and that is TYPICAL engineering right?) Oh by the way, you should write formula. Evilbu 10:48, 4 June 2006 (UTC)

Minor pedantic correction - the curve of a free-hanging uniform chain is a catenary, but the curve of the cables of a suspension bridge, in the ideal case where we assume that the weight of the cables is neglible compared to the weight of the horizontal road deck, is a parabola. Gandalf61 08:59, 5 June 2006 (UTC)

random drawing of numbers 1–100

Suppose you had the values of 1 to 100. Then, you randomly organized the 100 numbers into 10 groups of 10. One group might contain the numbers 7, 13, 17, 38, 41, 52, 59, 71, 90 and 95 for example. Then, by group, the highest number would be given a corresponding value of 10, the second highest a corresponding value of 9, and so on. Then, the process is repeated, and any value earned is added to the last value earned. For example, the number 100 will always be given a value of 10 (because it is always the highest number out of 100), so after 3 random "draws," its corresponding value would be 30. Likewise, the number 1 would have a value of 3 after 3 "draws." My question is: how many random "draws" would it take so that all the numbers were in numerical order based on their values. Likewise, how many random "draws" would it take so that less than 10 numbers were not in correct placement when organized by values. I understand that this question is confusing (and it is quite hard to word), so if you have any questions as to what I mean, then please ask and I will update and clarify accordingly. Thank you in advance for all of your help. - Zepheus 02:55, 4 June 2006 (UTC)

Fun challenge, though not a practical way to sort. But you cannot use random draws and ask those precise questions. For example, although it is statistically unlikely, fifty random draws could be exactly the same. Or did you know that? Anyway, perhaps you should think about why you expect that the cumulative "10-ranking" scores will converge to a correct 100-ranking order. That will be essential to answering the first question. For simplicity, consider four numbers in groups of two. Write out all possible draws and consider how they combine. Notice that after one draw the values will be ⟨1,1,2,2⟩, with only two distinct quantities; and after two draws the cumulative values will include 1+1 and 2+2, and typically 1+2 as well, but nothing else. So it is impossible to sort properly with only one or two draws. More generally, after n draws the largest cumulative value will be n times the group size and the smallest cumulative value will be n. Randomness aside, the difference of these must be large enough to permit a distinct value for each number in the full set. Since 11×(10−1) = 99 = 100−1, clearly at least 11 draws are necessary for a full sort. Of course, this necessary condition may not be sufficient; nor does it address the likelihood of a correct sort. --KSmrq^T 06:02, 4 June 2006 (UTC)

11 draws are sufficient:

 (1-10)           (11-20)    (21-30)    ... (91-100)
 (1,11,21,31,...) (2,12,...) (3,13,...) ... (9,19,...)
 ... (10 times) ...
 (1,11,21,31,...) (2,12,...) (3,13,...) ... (9,19,...)

It's fairly obvious that this schema assigns each number n the value n+10. EdC 14:23, 4 June 2006 (UTC)

Additionally, this (with permutations) is the only way to get a full sort in only 11 draws.

You're going to have to clarify how many random "draws" would it take so that all the numbers were in numerical order based on their values. As KSmrq pointed out, the numbers can stay unsorted through an indefinite number of draws. One possible question is how many draws are needed such that P(sorted | n draws) exceeds some value (say ½). I wouldn't think that P(sorted | n draws) has a nice form, though. — Preceding unsigned comment added by EdC (talk • contribs)

Thank you for all your help so far, and it's been a long time since I've had a math class so it's tricky to write in mathematical ways. I knew originally that every draw could be the same, but it is statistically improbable. Also, I figured that when the number of draws approached infinity, the number of errors (or numbers not in correct placement when sorted by value) would reach zero. I was just wondering how many draws would probably be sufficient. I think my question has pretty much been answered, unless EdC has more to say on the matter. - Zepheus 16:56, 4 June 2006 (UTC)

I'm currently running a simulation of the problem, and it seems that the average amount of draws required is around 2045 - If that is of any use. Keep in mind, though, that I have not double-checked my program's correctness, and it is very inefficient, so it could take a while until an accurate result is obtained. -- Meni Rosenfeld (talk) 17:28, 4 June 2006 (UTC)

Cdf of the number of draws needed

I've tried such a simulation too, and I've got a similar result: average 2069 draws needed from a sample of 160 tries. The same warnings apply as above. Btw, see birthday paradox as for why you need so many draws. – b_jonas 18:35, 4 June 2006 (UTC)

Update^2: run on a faster SMP machine (from 13919 iterations) gives average 2058 draws, quartiles of number of draws are 1521, 1911, 2443. – b_jonas 19:19, 4 June 2006 (UTC)

Some thoughts on the convergence of the order. Consider a random variable $X^{k}$, the rank of number k in a random draw. Denote $Y^{k}=X^{k+1}-X^{k}$. Denote by $A^{k}$ the event that the numbers k and k+1 are in different groups, its probability is $P(A^{k})=9/10$. Given $A^{k}$, $X^{k}$ has the same distribution as $X^{k+1}$, and $Y^{k}$ is symmetric with respect to 0 ($P(Y^{k}=x|A^{k})=P(Y^{k}=-x|A^{k})$ for x>0), thus $E(Y^{k}|A^{k})=0$. With probability 1/10 the numbers k and k+1 are in the same group ($A^{kc}$, complementary to $A^{k}$) and $X^{k+1}=X^{k}+1$ and $P(Y^{k}=1|A^{kc})=1$. That is,

$P(X^{k+1}=x)=P(X^{k}=x|A^{k})P(A^{k})+P(X^{k}=x-1|A^{kc})P(A^{kc})$

Besides, we have

$E(Y^{k})=E(Y^{k}|A^{k})P(A^{k})+E(Y^{k}|A^{kc})P(A^{kc})=1/10$.

After realization of n draws, the numbers 1,...100 have the correct order if

${\bar {Y}}^{k}={\frac {1}{n}}\sum _{i=1}^{n}Y_{i}^{k}>0$ for all k=1,...99

It is known than ${\bar {Y}}^{k}\to 1/10$ in distribution so the convergence to the correct order should be expected. The questions arise whether the r.v. $Y^{k}$ are independent of each other, and for which smallest n the event $\inf _{k}{\bar {Y}}^{k}>0$ occurs the first time, and what is the distribution of such n. (Igny 22:46, 4 June 2006 (UTC))

These mathematical functions are getting crazy. I wish I could decipher them. I'll definitely archive this page. One more question, the first answer I receive was that 11 draws would be sufficient. The next answer was that roughly 2,050 draws would be needed. How are these related? Also, what is the rough estimate for the number of draws needed for less than, say, 10 mistakes. - Zepheus 19:09, 5 June 2006 (UTC)

No, the first answer was that 11 draws are necessary. That is, with less than 11 draws you have zero chance of getting it right. No finite amount of draws is sufficient (in the sense of having a probability of 1 of winning). Roughly 2047 (obtained after over 55,000 experiments) is the average number (expectation) of draws until a success is obtained. Assuming the distribution is roughly symmetric, this also means that 2047 draws will give you a 50% chance of success. -- Meni Rosenfeld (talk) 19:27, 5 June 2006 (UTC)

Well, according to the numbers I gave above, you have about 50% success after less draws than that: about 1911 draws. – b_jonas 20:47, 5 June 2006 (UTC)

Okay. I understand now. Thanks for the update, and all of your hard work. - Zepheus 21:16, 5 June 2006 (UTC)

Above is a graph of the cdf of the distribution of the number of draws needed I've made from the output of my simulation. – b_jonas 21:26, 5 June 2006 (UTC)

This graph is awesome thanks. - Zepheus 17:29, 6 June 2006 (UTC)

Loudspeakers

Can anyone recommend me loudspeakers for an integrated sound card (asus a7v8x-x) with a good quality/price ratio? thanks.

The web site for NewEgg lists numerous speakers along with specs and customer ratings. That should help you narrow down your interests to price range, number of channels, optional subwoofer, wattage, and sensitivity. A quick web search suggests that the integrated ADI 1980 sound chip is not exceptionally good, so if you do not intend to some day add a separate card, it may not be worth investing in top-quality speakers. It does provide 6 channels of output, so you may be interested in a 5.1 speaker setup, consisting of front stereo, rear stereo, center, and subwoofer. Finally, listening habits and tastes vary considerably, so it matters whether you are interested in gaming, hip-hop, classical, and so on. Again, a little reading will be quite helpful in narrowing your options. --KSmrq^T 19:22, 4 June 2006 (UTC)

free group (abelian?) <->free product<->coproduct confusion

Hi,

I have yet again a topology inspired question. First of all though I would like to express my gratitude for the many people who have helped me here.

Right now I mostly receive from Wikipedia being a student in exams, but I have and I will again give to the community myself:)

I am confused about http://en.wikipedia.org/wiki/Free_product_with_amalgamation

Suppose I take a free product of the groups$\mathbb {Z}$ and $\mathbb {Z}$ the article states it should give me a free group on two generators. Now the article on free groups it links to says that free groups and free abelian groups are not the same. There goes my hope that it would be $\mathbb {Z} \oplus \mathbb {Z}$

but wait! , later that article says it is a coproduct of two groups in the categorical sense. But I was thought in my algebra class that for Rmodules, and thus also abelian groups (as they are the same as \mathbb{Z} modules ), simply taking the outer direct sum of two modules should do just fine to give you a categorical coproduct.

So what is going on, can anyone point out the difference between coproduct and free product. What am I doing wrong?

This confusion has led me to believe than 'eight' or an 'infinity symbol' has fundamental (homopoty) group$\mathbb {Z} \oplus \mathbb {Z}$}.

Thanks, Evilbu 14:53, 4 June 2006 (UTC)

The coproduct of $\mathbb {Z}$ and $\mathbb {Z}$ in the category of $\mathbb {Z}$-modules (i.e. abelian groups) is indeed $\mathbb {Z} \oplus \mathbb {Z}$. The fundamental group of $S^{1}\vee S^{1}$ is the coproduct of $\mathbb {Z}$ and $\mathbb {Z}$ in the category of groups, that is, the free product (which, for example, has uncountably many elements, so is clearly not $\mathbb {Z} \oplus \mathbb {Z}$). —Blotwell 15:04, 4 June 2006 (UTC)

But that is bad for me! So you are saying  : free product of two groups is NOT the same as coproduct?Evilbu 17:05, 4 June 2006 (UTC)

Coproducts look different in every category. In the category of groups, the coproduct is the free product. In the category of abelian groups and the category of modules, it is the direct sum. In the category of topological spaces, it is the disjoint union (also the category of sets). In the category of pointed spaces, it is the wedge sum. Z is a set, a space, a group, an abelian group, and a module. So there are many different coproducts you can make out of Z. The fundamental group is a functor from the category of pointed spaces into the category of groups, and the coproduct that you have to use is the free product (the coproduct in Grp), not the direct sum (the coproduct in Ab). So π1 takes coproducts of pointed spaces to coproducts of groups. It is a continuous functor. -lethe ^{talk +} 20:39, 4 June 2006 (UTC)

π₁ isn't a continuous functor (one which preserves limits) because this would contradict the long exact sequence of a fibration. But more importantly, it isn't cocontinuous (preserving colimits) which is what I imagine you meant. For example it doesn't preserve the colimit of the diagram ${\begin{matrix}I&\rightarrow &S^{1}\\\downarrow &&\\S^{1}&&\\\end{matrix}}$ where both arrows take the line segment to a circle by identifying the two endpoints. (Hint: the colimit is again S¹.) —Blotwell 01:37, 6 June 2006 (UTC)

Firstly, you're right, of course I meant cocontinuous. I don't understand the bit about the long exact sequence of a fibration. I was about to complain that your claim contradicts the Seifert-van Kampfen theorem, but then I saw the edit you made to (my addition to) the article fundamental group. π1 preserves pushouts along injections, but not every pushout. Your counterexample of course also helps. But doesn't π1 have an adjoint? -lethe ^{talk +} 08:44, 6 June 2006 (UTC)

The classifying space functor? It's a "homotopy adjoint": the homotopy equivalence classes of maps biject, $[X,BG]\simeq {\mathrm {Hom} }(\pi _{1}X,G)$, but you can't make this into an actual categorical adjunction. Correspondingly, the neatest statement to make about π₁ is that it takes homotopy colimits to colimits. Coproducts, and more generally limits of diagrams of cofibrations, are examples of homotopy colimits: my counterexample above is not. —Blotwell 16:38, 6 June 2006 (UTC)

So classifying spaces are only defined up to homotopy equivalence? I was under the impression that they were defined up to homeomorphism, but I'm not so comfortable with the whole business, so I could be out for a six. Anyway, it sounds like we will be able to see that π₁ is cocontinuous as a functor from the homotopy category of topological spaces then, no? Only, I don't know what the colimits look like in that category. -lethe ^{talk +} 02:20, 7 June 2006 (UTC)

You'd think, wouldn't you? But I'm not convinced that the homotopy category is cocomplete and I can prove that colimits in the homotopy category are not the same as homotopy colimits: ${\begin{matrix}S^{n}&\rightarrow &*\\\downarrow &&\\\,*\,&&\\\end{matrix}}$ has homotopy colimit Sⁿ⁺¹. Homotopy colimits are generally the Right Thing and colimits in homotopy categories don't exist in general, but I can't actually think of a counterexample to your statement. I would say classifying spaces are only defined up to homotopy equivalence, though of course the bar construction picks a canonical representative for you. —Blotwell 16:54, 7 June 2006 (UTC)

Oh yes, I see, I think I have made a serious mistake in assuming something. I was seeing the abelian groups as a subcategory of the group category. I cannot take some abelian groups, take the free product (defined in the categorical sense) and assume it will still be that in the bigger category right? Evilbu 20:58, 4 June 2006 (UTC)

There is nothing wrong with thinking of Ab as a subcategory of Grp. But you're right, the coproduct of groups does not restrict to the coproduct of Abelian groups on this subcategory. Stated more explicitly, the coproduct of two groups which are abelian in the category of groups (this coproduct is a free product; always nonabelian) is not the same as the coproduct of two abelian groups in the category of abelian groups (this coproduct is a direct sum; always abelian). -lethe ^{talk +} 21:02, 4 June 2006 (UTC)

More generally, it is a mistake to think that any operation must restrict to a suboperation on a subset. Just because Ab and Grp both have coproducts does not mean the coproduct in Grp restricts to the coproduct in Ab on that subcategory. Similarly, the Killing form of a Lie algebra need not restrict to the Killing form of a subalgebra (this will happen for the Cartan subalgebra in the semisimple case, but need not in general). The covariant derivative of a vector in a submanifold of a Riemannian manifold need not equal the induced covariant derivative of that vector. Subcategories are, by definition, closed under composition of morphisms. This does not imply that they must be closed under every operation, like coproducts -lethe ^{talk +} 21:28, 4 June 2006 (UTC)

Bike problem

I know how easy it is to fall off a bike while not moving - I also know that it's harder to turn the wheel when I am in motion. I've had a look at angular momentum and related topics and I can't figure why a rotating bike wheel is harder to turn from its line of motion than a stationery one. Any help please? Anand 18:54, 4 June 2006 (UTC)

Although your second question should be answered at Gyroscope, it is not generally considered the correct reason for bicycle stability. See Bicycle#Balance, including the link to Bicycle physics. Walt 19:48, 4 June 2006 (UTC)

June 5

Circular number

I just started this article. If anyone believes they can add anything more to it, even one more sentence, go right ahead. — BRIAN0918 • 2006-06-05 05:52

I added my comments on the talk page. --vibo56 09:05, 5 June 2006 (UTC)

What is the difference with automorphic numbers? Can you give an example of an automorphic number that is not circular? --Lambiam^Talk 14:51, 5 June 2006 (UTC)

If the definition in circular number is interpreted as "there exists a power such that...", than it's the other way around - 4 is circular (4^3 = 64) but not automorphic (4^2 = 16). -- Meni Rosenfeld (talk) 15:09, 5 June 2006 (UTC)

Discussion on the talk page has reached the conclusion that circular number is equivalent to automorphic number. Gandalf61 16:00, 5 June 2006 (UTC)

Complex Functions & Poles

A question about poles. Consider a function of this form:
$f(z)={\frac {(z-a)(z-b)}{(z-a)(z-c)}}$
Will it have a pole at z = a? A removeable singularity? Or some other form of oddness? Maelin 06:26, 5 June 2006 (UTC)

Removable discontinuity. No pole. -lethe ^{talk +} 06:41, 5 June 2006 (UTC)

f is in fact differentiable at z = a -- there's no oddness. Cancel the z-a factor. Dysprosia 06:55, 5 June 2006 (UTC)

It's a removable singularity, and as one professor said, "At this point we assume all removable singularities are removed." In other words, while f(z) is technically undefined at z = a, since it is identical to ${\frac {z-b}{z-c}}$ elsewhere and can be analytically continued through it, you can essentially work with the analytic continuation instead of the function as you defined it. More interesting is a function like ${\frac {\sin z}{z}}$, which has no obvious cancellation, but because of its removable singularity is basically treated as though f(0) was automatically defined as 1. Confusing Manifestation 09:19, 5 June 2006 (UTC)

Am I missing something (Analysis has never been my strong suit)? Why is f(z) undefined at z = a -- surely f(a) = (a-b)/(a-c)? Dysprosia 09:31, 5 June 2006 (UTC)

Simply put, you can only cancel out nonzero terms. When z = a, you cannot cancel. The function is undefined there, though as others point out, it can easily be extended. -lethe ^{talk +} 10:07, 5 June 2006 (UTC)

Of course. Excuse the diversion. Dysprosia 10:59, 5 June 2006 (UTC)

Having studied them some more, I can now answer. The reason is that at z = a, you get z - a = 0, and to cancel that you must divide by zero and that's not defined. Everywhere else, z - a is some finite nonzero quantity, and the terms will cancel normally, but zero terms do not cancel in that way.

As an example, (3x) / (3y) = x / y everywhere because 3 is never equal to zero, but (0x) / (0y) is indeterminate. -Maelin 10:00, 5 June 2006 (UTC)

There could still be a pole at z = a, in the special case that c = a and b ≠ a. --Lambiam^Talk 14:57, 5 June 2006 (UTC)

Measurement unit on measuring tapes.

Several of my measuring tapes have a measure mark that is a small black diamond shape. There are five of them for every eight feet or 19 1/5 inches for each mark. What is this measure? Sincerely

Albert J. Hoch

Its purpose is to allow carpenters to divide 8 feet exactly in five. See this page and this page (scroll down to diamond) for documentation. There has been a dispute on the Wikipedia where a user claimed it to be an "English cubit", but this is apparently not correct. --vibo56 09:45, 5 June 2006 (UTC)

BTW, is there a method to divide a line in five equal parts ? (ruler and compass only). Thanks. --DLL 21:29, 5 June 2006 (UTC)

Yes. You draw two parallel lines extending from the endpoints of the line, then mark off five equal intervals on each of the parallel lines. Lines drawn through corresponding marks will divide the original line into five equal parts. --Serie 00:12, 6 June 2006 (UTC)

So the black diamond is a carpenter's mark! Very interesting, I'd been supposing it was some foriegn unit of measurement. I was guessing Chinese!

Thanks very much. Albert J. hoch Jr.

Integral of dirac delta over step function

What does the following integral evaluate to:

$\int _{0^{-}}^{0^{+}}{\frac {\delta (x)}{H(x)}}dx$

where $\delta (x)$ is the Dirac delta function and H(x) is the Heaviside step function.

What about the following:

$\int _{0^{-}}^{0^{+}}{\frac {\delta (x)}{1+cH(x)^{2}}}dx$

$\int _{0^{-}}^{0^{+}}{\frac {\delta (x)^{2}H(x)}{1+cH(x)^{2}}}dx$

$\int _{0^{-}}^{0^{+}}{\frac {\delta (x)H(x)}{1+cH(x)^{2}}}dx$

$\int _{0^{-}}^{0^{+}}{\frac {\delta (x)^{2}(1+c_{1}H(x))}{1+c2H(x)}}dx$

If these do not exist, what are reasonable approximations to them I can make? deeptrivia (talk) 18:27, 5 June 2006 (UTC)

As a distribution, the delta function is really only supposed to be integrated against certain smooth functions, which the Heaviside function is not. Strictly speaking, the integral isn't defined. As for whether a meaningful approximation can be made, I do not know. -lethe ^{talk +} 21:32, 5 June 2006 (UTC)

Thanks lethe. Is it terribly unsafe to assume, for engineering purposes, that H(0) = 0.5, while integrating these? deeptrivia (talk) 21:42, 5 June 2006 (UTC)

That's probably perfectly safe. For the purposes of integration, we only care about the equivalence class of functions which differ almost everywhere. You can do anything you want on a set of measure zero without affecting the integration. So you can take H(0) to be anything you want, and 1/2 is a sensible value. -lethe ^{talk +} 23:12, 5 June 2006 (UTC)

NO! Integration with respect to Lebesgue is independent of sets of measure zero, but integration with respect to the Dirac measure, which is what you are doing, cares about pointwise values. You can only integrate functions that are continuous at zero with respect to Dirac. You cannot do this integral, you need to use an approximation for either Dirac or Heaviside. The problem is, the answer depends on which approximation, which is bad, and the reason the integral isn't defined. But, maybe your problem really involved some function that is being approximated by a Dirac or Heaviside. You should use that function instead. (Cj67 15:05, 6 June 2006 (UTC))

So basically, I guess you want to do these integrals by substituting 0 in as an argument for H. I expect that this bit is unsafe, though I'm not sure exactly how unsafe it is. -lethe ^{talk +} 23:14, 5 June 2006 (UTC)

There is a problem here: the symbol

$\int _{A}\delta (x)f(x)dx$

means "take the integral of the function $f$ on the set $A$ w.r.t. the measure $\delta (x)$". While it is perfectly true that you can change the values of a function on a $0$-measure set without changing its integral, you must not miss that it is not the Lebesgue measure we are integrating against, but the Dirac delta, which assignes a weight of $1$ to $\{0\}$, hence you cannot change the value of $H(0)$. The integral you are asking, actually, indeed equals $H(0)$ (which is undefined, according to Heaviside step function). Cthulhu.mythos 09:00, 6 June 2006 (UTC)

Okay, if I ask maple to evaluate:

$\int _{-\epsilon }^{\epsilon }\!{\frac {{\it {Dirac}}\left(x\right)}{1+c\left({\it {Heaviside}}\left(x\right)\right)^{2}}}{dx}$

it gives:

${\it {Heaviside}}\left(\epsilon \right)-{\it {Heaviside}}\left(-\epsilon \right)$

The limit of this expression as epsilon --> 0 is undefined. However, as an engineering approximation, we can assume $\epsilon$ to be something small, like 1e-24, and then ${\it {Heaviside}}\left(\epsilon \right)=1$ and ${\it {Heaviside}}\left(-\epsilon \right)=0$, and so the integral evaluates to 1. Is there any flaw in this reasoning? I'm asking because there's a significant thing happening between -1e-24 and 1e-24, which will be ignored by this assumption. Another thing is, if I were doing this integration from, say -1 to 1, then, say you pointed out I won't have cared about values at finite points. But here, the integration has to be done in a range that encloses 0 and is as small as can be imagined, so the value at 0 might have a significant effect. Regards, deeptrivia (talk) 01:14, 6 June 2006 (UTC)

You say that the limit of the expression is undefined, but it appears well-defined to me. The right=hand limit of Heaviside is 1, the left-hand limit 0, and so the difference between the two is 1. Your subsequent comments bear out this limit, and by the way, it does not matter how small an interval you integrate over, the result is always the same. What I fail to understand is how Maple arrived at the expression you quote. I don't know how to arrive at the integral that Maple has given you, so I'm not sure how bulletproof it is, but anyway, your reasoning about the value is correct: it is 1, no matter the interval. -lethe ^{talk +} 01:57, 6 June 2006 (UTC)

But how can this be? Shouldn't the c from the integrand return in the result? Does Maple assume that Heaviside(0) = 0? After all, you expect the answer 1/(1+c(Heaviside(0))²).  --Lambiam^Talk 02:18, 6 June 2006 (UTC)

The rule is

$\int _{-\epsilon }^{\epsilon }\delta (x)\,f(x)\,dx=f(0),$

if f is nice enough. My guess is that "nice enough" in this case is "continuous". In any case, since H(0) is ill-defined, I guess that none of the integrals are defined.

If I am pressed to give a value to the integral

$\int _{-\epsilon }^{\epsilon }{\frac {\delta (x)}{1+c(H(x))^{2}}}\,dx,$

then I'd use that the delta function is a derivative of the Heaviside function, and hence

$\int _{-\epsilon }^{\epsilon }{\frac {\delta (x)}{1+c(H(x))^{2}}}\,dx=\left.{\frac {1}{\sqrt {c}}}\arctan({\sqrt {c}}\,H(x))\right|_{x=-\epsilon }^{\epsilon }={\frac {1}{\sqrt {c}}}\arctan({\sqrt {c}}).$

I have grave doubts about this reasoning, but perhaps it can be made rigorous.

If your integral arises from an engineering application, I would take a closer look to the limiting process that you are using. -- Jitse Niesen (talk) 03:11, 6 June 2006 (UTC)

The limit of

${\it {Heaviside}}\left(\epsilon \right)-{\it {Heaviside}}\left(-\epsilon \right)$

as epsilon --> 0 is undefined according to Maple itself (using the 'limit' function). The first approach I followed was the one proposed by Jitse Niesen, but I can't remember any more why I gave up on it. Anyway, these equation arise from a nonlinear treatment of point loads and moments at various points on a flexible beam with discontinuities in cross sectional area (like a stepped beam.) I am using heaviside functions to model steps in areas, and dirac functions to model point loads and moments. Any suggestions appropriate to this situation? Thanks. deeptrivia (talk) 04:18, 6 June 2006 (UTC)

If you use the definition of the Riemann-Stieltjes integral, then the first integral is equivalent to

$\int _{0^{-}}^{0^{+}}{\frac {1}{H(x)}}dH(x)=ln{\frac {H(0+)}{H(0-)}}=\infty$ (this is because ${\frac {dH(x)}{dx}}=\delta (x)$)

On the other hand, $\int _{0^{-}}^{0^{+}}\delta (x)H(x)dx=1/2$. Conscious 05:57, 6 June 2006 (UTC)

The integral no. 4 seems to be equal to ${\frac {ln(1+c)}{2c}}$, and no.3 and no.5 are infinite (because of δ²). No.2 was evaluated by Jitse. I'd say all these integrals are undefined as Riemann integrals, but well-defined as Riemann-Stieltjes integrals. (And since you start getting infinite results, you might need to tweak your physical model, as results seem to be dependent on how abrupt the edges are and how pointy loads are). Conscious 06:45, 6 June 2006 (UTC)

I tweaked my model a bit. It's now working atleast for small values of parameters. Hopefully, won't have problems with large values. Thanks for your help. deeptrivia (talk) 18:04, 8 June 2006 (UTC)

two loops on a cylinder , are in same homology class??

Hi,

consider $\mathbb {S} ^{1}\times \mathbb {R} =X$ a cylinder thus

I am still working on that torus, and I thought, this would be handy :

suppose I have a loop ,so let's say a path

$p:[0,1]\rightarrow X:t\rightarrow (e^{2\pi ti},a)\,\!$

and another , a path

$q:[0,1]\rightarrow X:t\rightarrow (e^{2\pi ti},b).\,\!$

Are these two loops in the same homology class, I mean, is their difference, in $S_{1}(X)$ an element of the image of $\partial _{2}$

--Evilbu 20:01, 5 June 2006 (UTC)

If each loop goes around the cylinder the same number of times, then there is a homotopy taking one into the other: they are homotopy equivalent. By definition, they then are in the same cycle class. Consequently, yes, they are also in the same homology class (cycles modulo boundaries). The Z of H₁ comes from that fact that a once-around cycle can be added or subtracted with itself any number of times to give homotopically different cycles (twice-around, once-around-reversed, and so on). Viewed at a slightly higher level, a deformation retraction of the cylinder produces a circle, S¹; therefore these spaces have the same homology. Furthermore, this is true more generally: X×Rⁿ has the same homology as X. --KSmrq^T 21:10, 5 June 2006 (UTC)

"If each loop goes around the cylinder the same number of times, then there is a homotopy taking one into the other: they are homotopy equivalent." You mean homotopic. Tesseran 22:46, 9 June 2006 (UTC)

A nit about your notation. What you've written are not paths. Your paths should have domain [0,1] and codomain X. You should better write something like

$[0,1]\ni t\mapsto (e^{2\pi ti},a)$

to indicate that the number t in the unit interval is mapped to a point in the path on the cylinder. -lethe ^{talk +} 21:36, 5 June 2006 (UTC)

Thanks, yet I'm sorry but I don't completely get it. Please be very clear in what you mean : homotopy between paths, or homotopy between continuous maps in general. Here was my idea : a 'push up u' of (b-a) is a continuous map from the cilinder to the cilinder, homotopic with the identity this means $H_{1}(id)=H_{1}(u)$ and thus $H_{1}(u)([p])=[q]$ A little weird I think. Why would a homotopy between those two points suffice? And what kind of homotopy do you speak, usually they mean with 'homotopy between two paths' : the homotopy fixes begin and end point all the time, which cannot be the case here as both are even disjoint. Evilbu 21:42, 5 June 2006 (UTC)

It's quite easy to see that your two paths are homologous: the boundary of the finite cylinder segment bounded by p and q is p – q. Thus they differ by the boundary of a 2 chain, so they are homologous. As for homotopy, you often consider homotopies with fixed endpoints, but you don't have to. The point is that given a homotopy between two curves with fixed endpoints, the two curves are the boundary of the image of the unit square under the homotopy. This also works on the cylinder for a homotopy without fixed endpoint, because the sides of the square are not in the boundary of the image. -lethe ^{talk +} 23:03, 5 June 2006 (UTC)

To amplify on what lethe has said, in this example we have two options to consider. The definition of homotopy applied to paths says that the ends can move. A path is a map, f, from the unit interval [0,1] to the space X. Given two such maps, f and g, a homotopy continuously deforms one path into the other. Before we start to compute homology groups, we want to take our huge number of cycles and reduce them to classes by homotopy equivalence. A closed loop on the cylinder is a 1-cycle and also a path for which the start and end points coincide. If we have two such loops a homotopy will necessarily deform loop to loop, but it need not leave any point fixed.

This not quite the same as computing the fundamental group, π₁(X,x₀), which requires a relative homotopy leaving point x₀ fixed. (Of course, the homotopy group π₁ is independent of the choice of x₀ if the space X is a path-connected space.)

A formally different option is the definition of homology groups as cycles modulo boundaries. Even without the reduction by homotopy equivalence this can cause two cycles to be identified in a homology group.

The definitions and implications in algebraic topology take time and exercise to grok. It will come; and besides, (modern) algebraic geometry is worse. I had the strange experience that algebraic topology seemed to have more geometric appeal than algebraic geometry! --KSmrq^T 23:58, 5 June 2006 (UTC)

Thanks everyone, it's a bit hard to understand that all completely. But I surely would like to know this : lethe, you wrote that p-q is the boundary of a segment. But I was taught that you need linear combinations (over the integers) of 2-simplices (those are maps from a triangle in the plane to your space). How would you proceed? Evilbu 08:15, 6 June 2006 (UTC)

I'm not sure what you're asking. Two 1-cycles are homologous if they form the boundary of a 2-chain, which can be thought of as a particular type of linear combination of 2-simplices. So a cylinder segment is a 2-chain, and its boundary is the two circles. Thus the two circles are homologous. -lethe ^{talk +} 08:39, 6 June 2006 (UTC)

The earlier reference to Mayer-Vietoris suggested Evilbu already had a solid grounding in some of the basics, but maybe we'd do better to take more explicit steps.

So, consider a cylindrical strip, a circle swept perpendicular to its plane. Topologically, the circle is S¹, the sweeping is some interval such as I = [0,1], and the cylinder is the product space S¹×I. (We could just as well use an infinite cylinder, S¹×R.)

Now suppose we consider a strip in the midsection, in the interval [a,b], with 0 ≤ a,b ≤ 1. If we slice it open parallel to the sweep, flattening gives a rectangle. Split the rectangle into two triangles. Each triangle is a 2-simplex. In fact, we can consider these triangles "on the surface". That is, we have a map from plane triangles to cylinder triangles.

To compute explicitly, we'll need some names. Call the edges of the rectangle T (top), B (bottom), L (left), R (right), and call the diagonal edge from top-left to bottom-right D. Call the vertices tl, tr, bl, br. This gives us a 2-simplex, σ₁, with edges D, R, T (in that order); and another, σ₂, with edges L, B, D (in that order).

tl			T			tr
	●		⟵		●
L	↓		↘		↑	R
	●		⟶		●
bl			B			br

But we need to be a little more careful, because each edge is a 1-simplex with its own vertex order. Thus

T = (tr, tl)	B = (bl, br)	L = (tl, bl)	R = (br, tr)	D = (tl, br)
∂T = tl − tr	∂B = br − bl	∂L = bl − tl	∂R = tr − br	∂D = br − tl

∂σ1 = D + R + T	∂σ2 = L + B − D
∂∂σ1 = (br−tl) + (tr−br) + (tl−tr) = 0	∂∂σ2 = (bl−tl) + (br−bl) − (br−tl) = 0

So far, so good. Each 2-simplex has a boundary that is a chain of 1-simplexes, and the boundary of each boundary is 0. Thus, automatically, a boundary is a cycle. Now we want to add our two 2-simplexes to form a chain, σ₁+σ₂. But before we do, we should glue the left and right edges of our rectangle together, which is what happens on the cylinder. Being careful with orientation, we declare that R = −L. This implies

∂σ1 = D − L + T	∂σ2 = L + B − D
∂(σ1+σ2) = T + B

We ordered the vertices of T right-to-left, and those of B left-to-right. (If we do many of these calculations we need to adopt a consistent ordering convention, and there is one that works well automatically.) Taking that into account, notice that we have verified what lethe asserted, that a loop around the cylinder at height a is homologous to a loop (in the same direction) at height b, because their difference is the boundary of a 2-chain, σ₁+σ₂.

I apologize for not including a good picture. (Anyone?) It would make most of this easier to see. --KSmrq^T 23:17, 6 June 2006 (UTC)

Addendum: Since we've come this far, we might as well relate the rectangle to the torus. Simply identify the top and bottom edges with proper orientation, T = −B, and we're done. To remove a disc for Mayer-Vietoris, cut out σ₂. Or, identify generator cycles and compute homology directly.

Notice that when we identify top and bottom edges, the chain σ₁+σ₂ becomes a cycle: its boundary is zero, since T+B = −B+B. Notice that we have also necessarily identified all four vertices. Claim: For the torus, L, B, and D are 1-cycles that are not boundaries. (Verify!) Are any of them equivalent? Claim: For the torus, D is homotopic (and homologous) to L+B. (Verify!) --KSmrq^T 07:03, 7 June 2006 (UTC)

Sine-rule like formula for radians

There is a highly useful formula which looks like the sine rule, ie something/something = something/something = something/something but I've completely forgot it. I believe terms like arc length, area, theta etc were included in it but I can't remember the other ones, nor can I remember the order. Thanks, Matt. — Preceding unsigned comment added by 80.229.237.12 (talk) 19:39, 5 June 2006 (UTC)

Have you read the radian article? Radian is the ratio between the arc length and the radius, i.e. $\theta ={\frac {s}{r}}$, where s is arc length and r is radius. Notice that since circumference of a complete circle is $2\pi r$, it follows that(from the radian article)

$2\pi {\mbox{ rad}}=360^{\circ }$

$1{\mbox{ rad}}={\frac {360^{\circ }}{2\pi }}={\frac {180^{\circ }}{\pi }}\approx 57.29577951^{\circ }$

or:

$360^{\circ }=2\pi {\mbox{ rad}}$

$1^{\circ }={\frac {2\pi }{360}}{\mbox{ rad}}={\frac {\pi }{180}}{\mbox{ rad}}\approx 0.01745329{\mbox{ rad}}$

What the article didn't mention(can anyone expand the radian article?) is however the formula of sector area in radian terms. Since the area ratio between the sector and the full circle is the same as the ratio between their radian measures, we have

${\frac {A}{\pi r^{2}}}={\frac {\theta }{2\pi }}$

$A={\frac {r^{2}\theta }{2}}$

or:

$A={\frac {rs}{2}}$

where A is the area. In anycase, I don't think a sine-like formula will be very useful, really, the only things you need to know is that radian is simply the ratio between the arc length and its radius, and everything else follows logically. --Lemontea 02:48, 6 June 2006 (UTC)

Hi, thanks for the help. I've worked out with the help of those two that the one I was looking for is:

${\frac {\theta }{2\pi }}={\frac {arclength}{\pi d}}={\frac {area}{\pi r^{2}}}$

I find this very useful for solving radian problems. Is this a well-known formula? — Preceding unsigned comment added by 80.229.237.12 (talk)

Of course. In fact, this is an accurate restatement of the ideas presented by Lemontea. -- Meni Rosenfeld (talk) 15:57, 6 June 2006 (UTC)

Multiplying 2 16-bit numbers with 32-bit registers?

I'm trying to implement IDEA in assembly language on a 386, and I'm having trouble because it was optimized for 16-bit processors. I want to be able to concatenate two 16-bit numbers to multiply mod $2^{16}+1$ and then separate them later, but I'm having trouble with it. I've tried using $\left(2^{16}x_{1}+x_{4}\right)\left(2^{16}k_{1}+k_{4}\right)$ and $\left(\left(2^{16}+1\right)x_{1}+x_{4}\right)\left(\left(2^{16}+1\right)k_{1}+k_{4}\right)$ but neither lets me extract just the multiplications I want and it's all mixed up. --Zemyla^t 21:05, 5 June 2006 (UTC)

Using $2^{16}a+b\equiv b-a\mod (2^{16}+1)$, we have:

$\left(2^{16}x_{1}+x_{4}\right)\left(2^{16}k_{1}+k_{4}\right)\equiv (x_{4}-x_{1})(k_{4}-k_{1})\mod (2^{16}+1).$

So perform the r.h.s. multiplication in 32 bits, giving $2^{16}u+v\equiv v-u\mod (2^{16}+1).$  --Lambiam^Talk 02:43, 6 June 2006 (UTC)

Geometry question -- truncated icosahedron

I assure you this is not homework, just a question from someone who hasn't taken geometry since high school and did almost nothing with three dimensional shapes, at that.

Let's say I have a truncated icosahedron that should fit into a sphere an inner diameter of 150 cm. How long should each of the vertices be? I'm sure this is probably easy for someone to calculate given all of those wonderful symbols on the icosahedron page but I'm totally baffled by them.

Many thanks. --Fastfission 22:48, 5 June 2006 (UTC)

A vertex is a single point; it has no length. So, do you mean how long should the edges be? Or do you want coordinates for each of the 60 vertices? If the latter, the section Canonical coordinates has all the data, assuming a sphere of radius r, where r² = 9φ + 10. The associated edge length is not given, but can be computed if so desired. --KSmrq^T 23:08, 5 June 2006 (UTC)

Length of edges, sorry. I probably picked up the habit of calling the lines between vertices as themselves being vertices from computer graphics or sometihng like that. I don't need the coordinates, just the edge lengths. I can't computer them myself, because I don't understand the formulation and am not really interested in learning it from scratch just for this one question. :-) --Fastfission 02:33, 6 June 2006 (UTC)

No matter where you picked up the habit, get rid of it; it's wrong. As for edge lengths, the article links to MathWorld, where the ratio of radius to edge length is given as

${\frac {1}{4}}{\sqrt {58+18{\sqrt {5}}}},\,\!$

or approximately 2.478. For a radius of 150 cm this implies an edge length of approximately 60.53 cm. --KSmrq^T 03:05, 6 June 2006 (UTC)

Thanks. --Fastfission 03:11, 6 June 2006 (UTC)

splicing WAV files

I have to do some cutting and splicing of audio files in WAV format, but I don't seem to have suitable software handy. Does anyone know of any freeware which might do the job and run under Windows XP? — Preceding unsigned comment added by Physchim62 (talk • contribs) 19:00, 6 June 2006 (UTC)

There is a list of free audio software at Free audio software. One I'd recommend is Audacity. Harryboyles 10:23, 6 June 2006 (UTC)

The number 9

After i have played around with the number 9 i noticed that 9 will always end up as 9.

For example:

9x9 = 81 (8 and 1) 8+1=9

9x15 = 135 (1 and 3 and 5) 1+3+5=9

9x265 = 2385 (2 and 3 and 8 and 5) 2+3+8+5=18 (1 and 8) 1+8=9

9x996633 = 8969697 (8 and 9 and 6 and 9 and 6 and 9 and 7) 8+9+6+9+6+9+7=54 (5 and 4) 5+4=9

This you can do with any random number. 9x? = 9

So my question is: When was this noticed the first time and who noticed it?

-Randi Hermansen, Denmark

See digital root and casting out nines. —Ilmari Karonen (talk) 12:22, 6 June 2006 (UTC)

Indeed, this is the simplest test for working out if a number is divisible by 9 (add digits together, if that sum is divisible by 9, then the number is divisible by 9). A similar test exists to check if 3 is a factor (add all the digits together, and if that sum is divisible by 3, then the number is divisible by 3). Sjakkalle (Check!) 13:42, 6 June 2006 (UTC)

I can't answer the bolded question, but you'd probably be interested in Divisibility rule and the general way to determine these properties. Walt 14:14, 6 June 2006 (UTC)

There was a somewhat related question on the science desk awhile back. Y'all should really keep a better eye on that page–some of us a apt to write foolish answers when confronted by anything more complicated than addition. EricR 18:53, 6 June 2006 (UTC)

Manual Calculation

Does any one have any information on how the vast tables of logarithms were calculated. Clearly, it is not a simple task or tables would not have been necessary. 68.6.85.167 21:52, 29 May 2006 (UTC)

It's simple, it just takes a long time and it's prone to error, so it's useful to make a handy book. I could do it using power series, but I'm sure there are more efficient methods. —Keenan Pepper 14:38, 6 June 2006 (UTC)

I'm not sure what method the people at those times used, but the power series are(natural log)

$\ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}\cdots$, provided the absolute value of x is smaller than 1.

or $\ln x=2\left[\left({\frac {x-1}{x+1}}\right)+{\frac {1}{3}}\left({\frac {x-1}{x+1}}\right)^{3}+{\frac {1}{5}}\left({\frac {x-1}{x+1}}\right)^{5}\cdots \right]$, which converge for all positive real numbers.

However, last time I tried, the second one seems to converge dead slowly for some numbers, so I think it's still easier to use the first power series, and use the identity $\ln(ab)=\ln(a)+\ln(b)$ to break down the number until it's small enough to be within the radius of convergence. (reference though on that page, it should be $x={\frac {y-1}{y+1}}$ rather than $x={\frac {y+1}{y-1}}$) --Lemontea 04:24, 7 June 2006 (UTC)

• I think this was done by hand (and yes, that was a laborious task). You might be interested in reading the biographies of Henry Briggs and John Napier. You can also take a look here. Sjakkalle (Check!) 14:40, 6 June 2006 (UTC)

In the years 1614 (Napier) and 1624 (Briggs), necessarily all computations were manual. Both men were clever inventors and calculators. Logarithms themselves were a brilliant labor-saving device. Curiously, there seems to be a parallel between the methods Briggs described and modern methods for hardware, described here. --KSmrq^T 23:56, 7 June 2006 (UTC)

There must be more fancy power series expansions for the logarithm which converge faster. Anyone know of a neat one to share with the RD? --HappyCamper 19:16, 8 June 2006 (UTC)

Thanks

I just wanted to say thanks to all of the contributers here. The people here are really knowledgable and have help satiate my intellectual curosity on different occasions. Mayor Westfall 20:21, 6 June 2006 (UTC)

Thank you also M. Westfall. --DLL 20:32, 6 June 2006 (UTC)

Recurrence Relations and Logarithm

I'm learning recurrence relations in school and I frequently come across problems involving finding the number of generations needed for the recurrence relation to have an answer twice its original value. So basically we are asked to find n when:

$u_{n}=2u_{0},u_{n}=au_{n-1}\,$

The method that is taught in school is to go through each generation by iteration and eventually finding n when you find that u_n has grown to twice the size of u₀. However, the obviously easier method is to use the equation:

$n=\log _{a}2\,$

Is there a similar shortcut when the recurrence relation is instead defined as this?:

$u_{n}=au_{n-1}+b\,$

I have tried working it out by firstly expanding each iteration:

$u_{1}=au_{0}+b\,$

$u_{2}=au_{1}+b=a(au_{0}+b)+b=a^{2}u_{0}+ab+b\,$

$u_{3}=au_{2}+b=a(a^{2}u_{0}+ab+b)+b=a^{3}u_{0}+a^{2}b+ab+b\,$

From this the whole series can be expressed as:

$2u_{0}=u_{n}=a^{n}u_{0}+(a^{n-1}b+a^{n-2}b+...+a^{2}b+ab+b)\,$

We can see the bracketed part of the equation is actually a geometric series, with the starting value b, ratio a and number of terms n (not n−1 as the starting value counts as one term). Also, the number of terms appearing in the geometric series seems to equal the number of iterations in the recurrence relation. Therefore we can use the equation for the sum of the geometric series:

$\sum _{k=0}^{n}ar^{k}={\frac {s(r^{n}-1)}{r-1}}\,$

Where s is the starting term, r is the ratio and n is the number of terms. Thus, substituting into the recurrence relation:

$2u_{0}=u_{n}=a^{n}u_{0}+{\frac {b(a^{n}-1)}{a-1}}\,$

Unfortunately at this point I hit a dead end because I'm unable to change the subject to n. Is this along the right lines or is there a completely different method? ----★Ukdragon37★^talk 20:27, 6 June 2006 (UTC)

You're nearly there. Solve for $a^{n}$:

$a^{n}={\frac {2(a-1)u_{0}+b}{(a-1)u_{0}+b}}$

Take logarithms:

$n\log a=\log {\frac {2(a-1)u_{0}+b}{(a-1)u_{0}+b}}$

Divide by $\log a$ (assuming $a\neq 1$):

$n=\log _{a}\left({\frac {2(a-1)u_{0}+b}{(a-1)u_{0}+b}}\right)$ EdC 21:36, 6 June 2006 (UTC)

Thank you so much! That question has been bugging me for weeks! ----★Ukdragon37★^talk 22:39, 6 June 2006 (UTC)

June 7

a fat delta function?

hello, i'm a student of physics taking an engineering course. can anyone give some intuition on the following:

the fourier transform of a sine wave of frequency w extracts the frequency w. but a fourier transform of a delta function (which has no width (in the time domain) and so no duration) extracts all frequencies.

my question is, how can a delta function accommodate all frequencies?

thanks -crj

Probably because the Fourier transform of a delta function is constant for all frequencies and the Fourier transform of sin(wt) is zero for all the frequencies but w.(Igny 03:11, 7 June 2006 (UTC))

Consider the family of functions A sech(A t√^π⁄₂), whose Fourier transforms are sech(^{1⁄_{A ω√π⁄2). Each of these has the shape of a hump centered at the origin, and as A gets larger the hump gets higher and narrower. The delta function is the limit of just such a concentration, with the area remaining constant. Observe that as the function gets narrower, its transform gets broader. Intuition says that the more abrupt a transition, the higher the frequencies required to produce it. In fact, the limit of the transforms of functions in this family as A goes to infinity is 1, meaning that all frequencies are present equally. The precise shape of the family is not important in arriving at the delta limit. For example, the Gaussian "bell", A exp(−(A t)2/2), is a nice family to try. --KSmrqT 04:17, 7 June 2006 (UTC)}}

ha! it makes sense that higher frequencies are needed to produce more abrupt transitions. thank you all for taking the time to answer. (incidentally, not sure how the answer sech(^{1⁄_{A ω√π⁄2) was obtained. i searched through my signal analysis books, an engineering mathematical handbook, and a book on mathematical methods of physics but could not find the result. i even tried to get this result using Maple, but all i get in return is my original input!)}}

I first ran into sech so long ago I no longer remember the original source. One reason I like to use it as an example is precisely because it is not well-known. Most people learn about Gaussians and impulse trains transforming to versions of themselves, but the typical transform pair involves two different kinds of functions. That's not important for the delta function limit, but it's nice to have something different for a change. Recent versions of Mathematica have a FourierTransform function you might like to try. Better still, maybe someone reading this would like to give a simple argument for correctness. --KSmrq^T 10:23, 8 June 2006 (UTC)

Oooh...that sech example is a nice example... --HappyCamper 19:12, 8 June 2006 (UTC)

Macs and PCs

Apple has released Boot Camp public Beta, which will be included in Mac OS X v10.5.
Question: Will Windows Vista also be able to run in a Mac usinq Boot Camp? --Alexignatiou 08:49, 7 June 2006 (UTC)

If not in Boot Camp, then surely using Parallels Workstation. Two indications that it will be possible in Boot Camp are this report that some hackers have already done it, and this Cringely column stating "One reason why Microsoft isn't surprised by Boot Camp is because Microsoft has been working with Apple to make sure that Windows Vista runs well on IntelMacs." --KSmrq^T 09:14, 7 June 2006 (UTC)

Is every function in Cp also locally integrable?

Hello,

I am quite unsure about something my professor told me.

Let $\Omega$ be open, and let f be a function in $C_{p}(\Omega )$ , unless I am really mistaking, that just means that f is defined in $\Omega$ , and that is can be continously derived p times

Now he told me if I just define an extension by making it zero outside \Omega , I get an$L_{1}^{loc}$ function, locally integrable thus : integrable over every compactum.

Either I got my definitions incorrect, or this is incorrect : what about $f={\frac {1}{1-x^{2}}}$ in $]-1,1[$  ?

Evilbu 10:14, 7 June 2006 (UTC)

If I am not mistaken, one also requires the function to be zero near $\partial \Omega$, but I am not sure. Observe also that in order for the extension of $f$ to be locally integrable you just need $f$ to be $L^{1}$ (which is nothing surprising). Hope this helps. Cthulhu.mythos 15:39, 7 June 2006 (UTC)

Hm, I don't understand, is that true? What about $\Omega =]0,1[$ and $f={\frac {1}{x^{2}}}$ That function can be extended to a $L_{1}^{loc}$ fuction, but it certainly will never be integrable?? So what is your definition in your opinion? I am guessing you would go for this then :

$f\in C_{p}(\Omega )\Leftrightarrow f:\mathbb {R} ^{n}\rightarrow \mathbb {C}$ and $f(x)=0\forall x\notin \Omega$ and f can be derived continuously p times

Evilbu 17:10, 7 June 2006 (UTC)

It is true that the function is in $L_{1}^{loc}(\Omega )$, and you don't need the function to be zero at the boundary of $\Omega$ or extend it. Remember, we are talking about $L_{1}^{loc}(\Omega )$, so we are talking about the integral being finite over sets that are compactly contained in $\Omega$. It is not true that the extension by zero is in $L_{1}^{loc}(R)$, as you illustrate. (Cj67 19:30, 7 June 2006 (UTC))

Thanks, but then what is your definition exactly of $C_{p}(\Omega )$ ? Evilbu 19:52, 7 June 2006 (UTC)

$C^{p}(\Omega )=\{f\in C(\Omega ):D^{k}f\in C(\Omega ),k\leq p\}$. Functions from $C({\rm {cl}}\,\Omega )$ can be extended by 0 to $L_{1}^{loc}(R^{n})$ (Igny 22:01, 7 June 2006 (UTC))

AMD64 With 64-bit OS

If I install Ubuntu (64-bit linux) on my computer with AMD64 3200+ should I expect an increase in performance? --Username132 (talk) 13:22, 7 June 2006 (UTC)

As compared to what? Compared to a 32-bit Ubuntu system (compiled for ix86), yes. – b_jonas 20:39, 7 June 2006 (UTC)

It should run faster, but depending on what you're doing the difference may not be all that noticeable. You also give up a certain amount of flexibility since you can't as easily run 32-bit applications. For example, your web browser will not be able to use some plugins, such as Flash. (There are ways around this, involving setting up a 32-bit chroot jail, which are probably quite easy to do but I haven't bothered).-gadfium 01:40, 8 June 2006 (UTC)

Sorry to butt in - can I ask how soon such problems (eg with Flash) are likely to be resolved? I'm weighing up the same proposition. Thanks --The Gold Miner 06:34, 8 June 2006 (UTC)

You may find this forum discussion to be of interest. Apparently the latest versions of K/Ubuntu don't require chroot to run 32-bit applications. My information was out of date.-gadfium 08:23, 8 June 2006 (UTC)

Do I Really Need Expensive HD?

I don't want to pay for hardware capable of more than I need. I want to buy a new system disk and the way I see it, I've a few options; a) buy two WD raptors and put into RAID-0 configuration b) buy two budget HDs and put into RAID-0 c) buy one raptor d) buy one budget drive

When loading the OS for example, is the speed of the HD a bottleneck for an AMD64 3200+ system with Corsair value RAM? And what games really benefit from 300 Mb/s data transfer to and from your HD? I would have thought the graphics card would be the bottleneck in any system with an old ATA-100 - I mean most important, immediately required information will be in the RAM, wont it? —The preceding unsigned comment was added by Username132 (talk • contribs) .

Assuming you have enough RAM, your assumption is correct and a faster drive will only decrease startup times. It won't have an important effect on performance once everything is loaded into RAM. —Keenan Pepper 23:39, 7 June 2006 (UTC)

You may need faster HD to do audio/video/picture editing, as well as data mining. Copying, archiving, compiling, and obviously defragging take less time with faster HD. (Igny 00:01, 8 June 2006 (UTC))

June 8

Statistics questions... related to dice in an indie role-playing game called Donjon

If I roll three six-sided dice (3d6) and only take the median roll (not the mean or average), what kind of bell curve or distribution odds would it give me for the result 1 to 6?

If I roll two 20-sided dice (2d20) and only look at the higher result of the two, what is the average result I will get? What about 3d20 and only look at the highest die? 4d20 etc. up to 16d20...?--Sonjaaa 11:43, 8 June 2006 (UTC)

1. The probabilities are :

1 : 2/27    2 : 5/27    3 : 13/54    4 : 13/54    5 : 5/27    6 : 2/27

2. I'm too lazy to find analytical solutions right now, so I'll give you approximate numerical solutions instead : The averages are, repsectively: 13.83, 15.49, 16.48, 17.15, 17.62, 17.97, 18.24, 18.46, 18.64, 18.79, 18.91, 19.02, 19.11, 19.19, 19.26. Write again if you need something more accurate.

-- Meni Rosenfeld (talk) 15:34, 8 June 2006 (UTC)

For the second question, the probability of getting x as the highest result when rolling n 20-sided dice is ${\frac {x^{n}-(x-1)^{n}}{20^{n}}}$. The average value is $20-{\frac {\sum _{i=1}^{19}i^{n}}{20^{n}}}$. Chuck 14:18, 9 June 2006 (UTC)

JavaScript date object

It is quite straightforward to use JavaScript to return today's date, but how would you go about getting it to return tomorrow's date, or the date in ten days' time? — Gareth Hughes 14:18, 8 June 2006 (UTC)

Get today's date, then add a day (86400 seconds, or whatever the conversion is). — Lomn Talk 15:03, 8 June 2006 (UTC)

But how do you add 86,400 s to a date object (should I not use milliseconds?)? Could someone give an example? — Gareth Hughes 15:29, 8 June 2006 (UTC)

Try this site. It appears you use setDate(getDate() + n) — Lomn Talk 17:58, 8 June 2006 (UTC)

Ah, that's how you do it! THank you. — Gareth Hughes 23:20, 8 June 2006 (UTC)

Microsoft C++

Not long ago I purchased an "Introductory" copy of Microsoft's Visual C++ on eBay. Since I have not programed in C since 1984 I was really surprized at how far it looked like C had come. However the "Introductory" version will not compile programs that worked great back in 1984 and even when a console program is written that compiles with no errors it stops and says that because it is an "Introductory" version that I can't make an execute file. I need another compiler but would like to avoid giving away any more of my hard earned money to Microsoft. What C++ compiler do you recommend? ...IMHO (Talk) 17:12, 8 June 2006 (UTC)

Borland has a great C++ command-line compiler which can be combined with Spetniks C++ Compiler Shell if you want a visual environment. —Mets501^talk 17:46, 8 June 2006 (UTC)

The GNU compilers are an excellent free option for nearly any language/platform combination. — Lomn Talk 18:00, 8 June 2006 (UTC)

If programs that worked in 1984 won't compile, it might be because they were written in K&R C, while the compiler expects Ansi-C, or because they are using libraries or header files which lack in the Microsoft compiler. You can find a list of free C compilers here. --vibo56 ^talk 19:41, 8 June 2006 (UTC)

BTW, I thought Microsoft was givning the introductory version of Visual C++ away for free. I find it difficult to believe that the compiler refuses to create executables. Is the problem only related to command-line programs? --vibo56 ^talk 19:48, 8 June 2006 (UTC)

visual studio 2005 has microsoft's latest visual c++, but it compliles to CLI (you need the .net framework 2.0 to run the executables). visual studio express edition is free. If you're just poking around for personal programming pleasure, I recommend it; it's very good. --Froth 03:11, 9 June 2006 (UTC)

Reinstallation of .net framewoek 2.0 did not solve the problem. Actually I'm thinking of getting back into assembler since it was the first compiler that brought to an end the need to program in machine code (binary). ...IMHO (Talk) 14:54, 9 June 2006 (UTC)

Screenshots

Intro was sold out of Seattle on eBay. I imagine that someone got a bunch of copies for free and then put them on eBay.

File:MS C++ v6 intro.PNG

File:MS C++ v6 intro error screen.PNG

...IMHO (Talk) 20:42, 8 June 2006 (UTC)

---

I can believe it. Only running in interpreter mode would allow them to demonstrate most of their capabilities but still make you want to pay them money to get the ability to create compiled executables. Bill Gates didn't create his Evil Empire by being stupid, after all. StuRat 20:24, 8 June 2006 (UTC)

Doesn't that message just say that the redistribution of executables is not allowed? It doesn't say anything about the introductory version not being able to create executables. In fact it explicitly acknowledges that executables can be created. —Bkell (talk) 21:49, 8 June 2006 (UTC)

Yes, you are right. I misinterpreted it when I saw the next screen with the error flag. How is no redistribution enforced and more importantly what is the cause of the error? Thanks ...IMHO (Talk) 23:27, 8 June 2006 (UTC)

It's probably not enforced through technological means; if you make an .exe file with that compiler, then I can't think of any way they could prevent you from giving it to someone else. But by using the software you agreed to some licensing agreement, which is a legal contract, and part of that contract probably said that you will not redistribute the executables you make. (The reason is probably that they don't want software companies buying the cheaper introductory version in place of the full version.) As for the cause of the dialog, I would guess that it will come up every time you compile something, just to remind you that you are not allowed to redistribute the executables. But that's only a guess; I've never used that particular software myself. —Bkell (talk) 02:15, 9 June 2006 (UTC)

Matlab's bvp4c

I have a very strange problem with Matlab's bvp4c. Hopefully, someone here would have some idea what's going wrong. I am solving a 30-variable boundary value problem, and I know that for certain inputs, a particular variable u3 must be 0 everywhere. bvp4c returns the values of variables y(x), as well as their derivatives yp(x). The derivative of u3 in the result is always zero (sth like 1e-17 to be precise, and x ranges from 0 to 1), so that looks good. But the value of u3 is varying a lot (instead of staying 0, it goes smoothly, but not linearly to 0.9). Doesn't it clearly mean there's a bug in matlab's bvp4c? What else could be the problem? deeptrivia (talk) 18:10, 8 June 2006 (UTC)

Matrix question

Suppose we have a large symmetric matrix which can be partitioned into smaller blocks matricies. These smaller blocks happen to be symmetric too. Suppose further, that we are given all the eigenvalues of each of these blocks. Is there a way to infer the eigenvalues of the entire original matrix easily? --HappyCamper 19:09, 8 June 2006 (UTC)

Depending on what you mean by "partition", it's likely that the eigenvalues are the same. Meaning that the eigenvalues don't depend on the basis. If you find a basis in which the matrix is block diagonal, bully for you. Find the eigenvalues of the blocks, you have also found the eigenvalues of the original matrix. Edit: after reading KSmrq's followup below, I realize that my reply is only useful for block diagonal matrices, something you didn't specify in the question, so this answer may be entirely useless, in which case, my apologies. -lethe ^{talk +} 01:28, 9 June 2006 (UTC)

Was the intent that the blocks be on the diagonal? --KSmrq^T 03:03, 9 June 2006 (UTC)

In case you are not restricting yourself to block diagonal matrices, then knowing the eigenvalues of the blocks does not give enough information to get the eigenvalues of the big matrix. Example: consider

${\begin{bmatrix}1&0&2&0\\0&0&0&0\\2&0&0&0\\0&0&0&0\end{bmatrix}}\quad {\mbox{and}}\quad {\begin{bmatrix}1&0&0&0\\0&0&0&2\\0&0&0&0\\0&2&0&0\end{bmatrix}}.$

The 2-by-2 blocks have the same eigenvalues, but the matrices themselves do not. -- Jitse Niesen (talk) 13:36, 9 June 2006 (UTC)

Nice simple example. – b_jonas 14:53, 9 June 2006 (UTC)

I guess that was a bit of wishful thinking, huh? Thanks guys. --HappyCamper 16:49, 9 June 2006 (UTC)

June 9

3D taxicab world

Has there been any research done into extending taxicab geometry into 3D space? --Tuvwxyz 21:11, 9 June 2006 (UTC)

In that article, it's defined for any dimension, only restricting to the plane after "For example, in the plane..." Is there some question you wanted answered? Melchoir 22:03, 9 June 2006 (UTC)