User:NorwegianBlue/refdesk

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[edit] Questions to the reference desk

Contents 1 Questions to the reference desk 1.1 See also: Some questions I have answered 2 The area of a square on the surface of a sphere. 3 Static electricity 4 Cubic equations with three real, irrational roots 5 Algebra stumper 6 Is it possible to see clearly underwater by wearing spectacles instead of goggles? 7 Statistical process control 8 Looking for decent maps throughout (modern) history 9 Norwegian name pronunciation 10 R licence question 11 Spanish pronunciation of "s" as "h" 12 Once upon a time, in Spanish 13 Maths resources 14 Generating permutations 14.1 Algorithm to generate permutations 15 GIMP question 16 Statistics & Propagation of error with least square fits 17 Integration in Maxima 18 Sample size vs. extreme observations 19 Creating duplicate hard links, then making them independent 20 Magnetic Resonance Imaging 21 Catalan and Castellano, Mutual intelligibility 22 Catalan y castellano 23 Consultas, castellano y catalan 24 Consulta español-catalán 25 Linear regression problem 26 No. of chromosomes in Neanderthals? 27 Mexico City 28 Radicals 29 Arcsin sqrt(x) transformation 30 AVI files that can't fast foward 31 CD28 gif 3d rotating structure 32 Solving cubic equations 33 Best calculus textbook 34 Balancing a bicycle

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[edit] See also: Some questions I have answered

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[edit] The area of a square on the surface of a sphere.

Moved to a separate subpage.

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[edit] Static electricity

When you get a shock of static electricity, simply stated, the reason is presumably that your body has acquired a surplus or deficit of electrons compared to the environment. How would one go about to estimate the number of electrons that is transferred in such a shock? Does somebody have an idea of the approximate number? When walking on synthetic carpets etc., do we usually acquire a positive or negative charge? --NorwegianBlue 12:24, 21 May 2006 (UTC)

No ,when you get a shock, its because charge has passed through your body causing the muscles to contract. THe amount of current passing depends upon the voltage, the resistance of the static generator, and the body's internal reistance (which is about 700R).

As to charge build up, I would try the Faradays ice pail experiment with the victim standing in the pail. The pail needs to be insulated from earth. THe victim would not actually get a shock because current would not pass thro the body (hopefully). Then by knowing the capacitance of the pail and its final voltage after charge transfer, one can calcualte the amount of charge passed.8-)--Light current 12:36, 21 May 2006 (UTC)

There are some calculations done by Mr Static here. The answer in his case was a positive charge of about 3 x 10^-8 coulombs, which is about 2 x 10¹¹ electrons. --Heron 13:39, 21 May 2006 (UTC)

Thanks a lot, Heron, that was exactly what I was looking for. The value of 3 x 10^-8 coulombs appears to be per step, the graph might suggest a charge buildup about 20 times larger. --NorwegianBlue 13:58, 21 May 2006 (UTC)

[edit] Cubic equations with three real, irrational roots

(Copied from the reference desk, maths section).

The equation 8x³ - 6x + 1 = 0 has the solutions sin 10º, sin 50º and -sin 70º, and is solved readily using trigonometric methods. However, attempting to solve the equation using Cardano's method (see Cubic equation) yields some rather nasty expressions, such as

Question 1: Is it possible to reduce this to an expression involving radicals and non-complex rational numbers only?
Question 2: Can anybody give an example of a cubic equation with three different real irrational roots, where the roots can be expressed using radicals and non-complex rational numbers only? --NorwegianBlue 16:58, 21 May 2006 (UTC)

I don't know about the first question, but my guess is that no. About the second, the equation

Satisfies your conditions. If you want the coefficients to be integers as well, I guess this is equivalent to the first question. -- Meni Rosenfeld (talk) 17:50, 21 May 2006 (UTC)

Yes, I did want the coefficients to be integers. And I suspect you are right that the answer to both questions is no. If so, is anybody aware of a proof? --NorwegianBlue 17:56, 21 May 2006 (UTC)

I think that the answer to Question 1 in general is "no", because of the result refered to here:

"One of the great algebraists of the 20th century, B.L. van der Waerden observes in his book Algebra I, that the Casus Irreducibilis is unavoidable. There will never be an algebraic improvement of the cubic formula, which avoids the usage of complex numbers." [1]

However, it may still be possible that for this particular equation, such a formula exist. Perhaps that reference is enough to get you started if you are really interested. You could also try asking User:Gene Ward Smith. -- Jitse Niesen (talk) 03:05, 22 May 2006 (UTC)

Thank you, Jitse, as far as I can see, the reference answers both questions. And I had no idea that it was this very problem that initiated the study of complex numbers! --NorwegianBlue 16:57, 22 May 2006 (UTC)

• Solving the Casus Irreducibilis, the case of three real roots, of the cubic equation requires the use of complex numbers if you insist on doing it with radicals; this was a strong factor in the adoption of complex numbers by European mathematicians, which may be the first clear example of modern European mathematics getting somewhere which the rest of the world hadn't gotten to earlier and better. The article cubic equations goes into great deal on how you can solve the cubic equation algebraically, not just in terms of transcendental functions, without using complex numbers. In that case, the algebraic funcion C_1/3(x) can be used. While this can be computed in terms of trig functions, that does not make it a transcedental function, any more than the fact that P_1/3(x) = x^1/3 can be computed in terms of logs and exponentials makes it a transcendental function. In general, solving a solvable algebraic equation of degree not a power of two, all of whose roots are real, requires complex numbers if you use radicals but no complex numbers if you use Chebychev radicals C_1/n. The cubic equation article is long--would an article on the Casus Irreducibilis help, I wonder? Gene Ward Smith 19:38, 24 May 2006 (UTC)

Thank you Gene, for commenting my question. I do realize that sin 10º, sin 50º and -sin 70º are algebraic numbers, being solutions of the equation 8x³ - 6x + 1 = 0, but I did insist on expressing the solution with radicals. I'm grateful for the response that it cannot be done without using complex numbers. As to whether there should be a separate article on the Casus Irreducibilis, I think that would be overkill, but it wouldn't hurt if, at the end of the section on Cardano's method, you mentioned the fact that, when applied to an equation with three real, irrational roots, it will always give a solution which includes a sum of two conjugate complex numbers, such that the imaginary parts cancel out. --NorwegianBlue 21:47, 25 May 2006 (UTC)

Second thought, maybe for some special classes, it can be done by denesting Nested radicals? (See the reference section, through you may have to use an algebra system) --Lemontea 02:37, 24 May 2006 (UTC)

Well, the Cardano solution is expressed in nested radicals involving complex numbers. I did try to construct cubic equations using nested radicals of real numbers only, but always ended up with irrational coefficients. It is quite some time since this problem nearly drove me nuts (33 years, to be exact), and I didn't have access to a computer algebra system. --NorwegianBlue 14:25, 24 May 2006 (UTC)

Anyway, at least now I know that it is impossible for your example above - see Exact trigonometric constants, which said "No finite radical expressions involving real numbers for these triangle edge ratios are possible because of Casus Irreducibilis. 9×2X-sided 70°-20°-90° triangle - enneagon (9-sided) 80°-10°-90° triangle - octakaidecagon (18-sided)" PS:my fault, the reference I cited actually duels with nested square root more, and for general radicals, not all are denestable. PPS:If question 2 doesn't require there to be three real roots, then x³ + 6x + 2 = 0 satisfy the other requirement. --Lemontea 14:50, 24 May 2006 (UTC)

Thanks. The point was that there be three real roots. The real solution to your example () would have satisfied the conditions if only there were two more real, irrational roots that could be expressed in a similar manner. However, the example is only a combination of the equations
y³ = 2 and z³ = 4.
Subtract the equations, i.e. y³ − z³ = − 2,
factorize the l.h.s., substitute with x = y − z, observe that yz = 2 and that
x² + 6 = (y² + yz + z²), and you have your equation. --NorwegianBlue 19:10, 25 May 2006 (UTC)

[edit] Algebra stumper

Years ago, a calculus professor gave me this one. The problem is to either solve the following set of equations by giving x, y and z; or to prove no such solution exists (among the complex numbers)

—The preceding unsigned comment was added by Plf515 (talk • contribs) 03:50, 2006 November 24.

Suppose a is one of x, y, and z. Then (a − x)(a − y)(a − z) = 0, so

We already know one of the coefficients, and we can find the other two:

So, x, y, and z are the roots of the cubic polynomial a³ − a² − (1 / 2)a − 1 / 6, which are about 1.43, 0.215 + 0.265i, and 0.215 - 0.265i. —Keenan Pepper 05:30, 24 November 2006 (UTC)

We might explicitly point out that all three equations lead to symmetric polynomials in x, y, and z. This has a double relevance. First, as shown, it assists in finding one solution. Second, it tells us that any permutation of one solution is also a solution.

A modern tool, that may not have been available when the problem was first posed, is the computation of a Gröbner basis for the three polynomials using Buchberger's algorithm. One such basis here is

The cubic in z should look familiar. --KSmrq^T 06:52, 24 November 2006 (UTC)

Expressed algebraically, the real root of the 3rd degree polynomial equals:

This was found by first using the substitution z: = 1 / ζ and subjecting the result to Cardano's third degree technique of interrogation.  --Lambiam^Talk 09:14, 24 November 2006 (UTC)

Thanks~ Plf515 10:56, 24 November 2006 (UTC)plf515

[edit] Is it possible to see clearly underwater by wearing spectacles instead of goggles?

Inspired by the preceding question, as well as this one and this one: The reason we do not see clearly underwater is that the refractive index of the cornea is almost equal to that of the surrounding water, so that we lose the refraction at the interface between air and cornea. Is it possible to correct this by wearing spectacles underwater, (i.e. with water between the lenses and the eyes)? It would be kinda nice, goggles tend to get foggy... And if it indeed is possible, what lens strength would you need? --NorwegianBlue 21:07, 23 May 2006 (UTC)

http://www.liquivision.ca/fluidgogglesfeatures.html —Keenan Pepper 21:12, 23 May 2006 (UTC)

Thanks, gotta get one of those... --NorwegianBlue 21:49, 23 May 2006 (UTC)

Thanks a trrrrrrrrrrrrrrrrrrrrrrrrrrrillion, that was the EXACT answer I was looking for 206.172.66.172 00:10, 24 May 2006 (UTC)

Or you could just get some defog for your SCUBA mask. There are commercial preparations, or you can use a dish soap solution, or your own saliva. They all work. And you can even get a prescription in your mask. --Ginkgo100 23:59, 24 May 2006 (UTC)

[edit] Statistical process control

In statistical process control using control charts, I have noticed that presenters often recommend calculating the standard deviation in a, so to speak, nonstandard way. The recommended procedure is to calculate a mean moving range, i.e. , using a relatively small dataset, and then divide the mean moving range by the magic number 1.128. If you google for "(1.128 and calculate)" and are feeling lucky today, you will find a such a presentation. The number 1.128 is often represented by the symbol d₂. Does anybody know the maths behind this non-standard estimator of the standard deviation? --NorwegianBlue 19:00, 1 June 2006 (UTC)

If the SD (innate variability about the current mean) was estimated from all the values, there would be an over-estimate in the presence of a trend (shift of mean), whether upward, downward or cyclic. Using the difference between successive observations removes this effect. If the sum of the squares of these differences is used, it has to be divided by 2(n-1) and square-rooted to estimate SD. Your formula uses absolute differences, without a "2" in the denominator. But for a Normal distribution, Mean Absolute Deviation is SD*root(2/pi).

Combining these corrections, 1.128 is root (4/pi).

86.132.237.108 19:49, 2 September 2006 (UTC)

[edit] Looking for decent maps throughout (modern) history

Hello,

does anyone know a site (or an article here?) where you can either type in a year (like 1925) and find a world map, or at least find many maps categorized by time?

I am asking because yesterday I was watching a documentary about de Gaulle and his relations with other leaders, and when planning a campaign in Africa, he was looking at a completely different map. So I understand now how weird it is to look at my world map when trying to understand history.

If for instance, I were to to understand international politics in 1963 or whatever, a world map of that year would be extremely useful.

So if anyone has any useful links or whatever, I will be very interested.

Evilbu 19:50, 8 August 2006 (UTC)

And we could have pages here like "Political world in 1925", &c. Very nice request (no ideas, though). --DLL ^{.. T} 19:55, 8 August 2006 (UTC)

Try Ancient world maps and Library of Congress Map collections. There are a lot more links on Maps that will probably get you what you need. Nowimnthing 20:22, 8 August 2006 (UTC)

My all-time favorite online map collection is the Perry-Castañeda Library Map Collection from UTexas. It's a truly amazing resource - maps of all types from all parts of the worlds throughout history - and well organized, too. For instance,here is their section on historical maps of Africa. They also have an excellent links section at the bottom of all the map lists. Lots of other good sites. --Bmk 21:29, 8 August 2006 (UTC)

About seven years ago I saw a professor used some commercial software which would allow you to view political maps of the world (or just of Europe?) for any point in time (you could run it like a movie). It was pretty neat stuff. Searching around, I think it was Centennia Historical Atlas Software, which is limited only to Europe and the Middle East (and not cheap, and from the screenshot doesn't look like it has been worked on lately), if you're interested. --Fastfission 22:55, 8 August 2006 (UTC)

[edit] Norwegian name pronunciation

How is the Norwegian name Trygve Lie (the first Secretary General of the United Nations) pronounced?

Is the last name a long "i" sound (as in a falsehood), or is it like "lee", or is it "lee-eh", or something else?

What are the vowel sounds in the first name? One or two syllables, and which gets emphasis?

This is my understanding, but I'm not a speaker of Norwegian, so I'd be happy to be corrected:

/?t?yg?? 'li?/

Rather /?t?yg?? 'li:?/, anyway... (Roughly "lee-eh"). By the way, he's a popular fellow in Norwegian crosswords... ;) ?? ???? 14:08, 20 November 2006 (UTC)

I don't know how to represent the pitch accent. Informally, the last name is like "lee-eh" but the second syllable is very weak, like the first "a" in "around". The first name has two syllables, of which the first gets the stress. The "r" is not trilled, but a single flap, almost like a "d". The "y" is like French "u", German "ü". The "v" is soft, like a "w" but also a bit like initial "r" in English. The final "e" is again like "a" in "around".  --Lambiam^Talk 08:31, 20 November 2006 (UTC)

Here is the pronunciation by a native: Image:Trygve lie.ogg. --NorwegianBlue ^talk 12:58, 20 November 2006 (UTC)

And here's an audio link: Norwegian pronunciation (help·info) --Kjoonlee 14:26, 20 November 2006 (UTC)

[edit] R licence question

R (programming language) is distributed under the GPL Version 2, June 1991. If I develop an application that (among various other things) writes R scripts, in order to use R as a graphing engine, do I have to distribute this application as open source, under the GPL 2.0, or can I distribute it commercially under a different licence? --NorwegianBlue ^talk 08:04, 16 November 2006 (UTC)

The GNU GPL requires that programs based on the source code of another program that is GPLed must be also licensed under the GNU GPL. It does not say the output (in your case, the program) must be licensed under the GNU GPL. In short, your program can be distributed commercially. --wj32 ^{talk | contribs} 08:32, 16 November 2006 (UTC)

See [2], especially the section on "Combining work with code released under the GPL". There may be situations where your program will (according to the FSF) become covered by the GPL, such as linking with GPL'd libraries (calling runtime functions may be harmful) or subclassing GPL'd classes. As far as I know FSF's assertions on this subject have not been tested in court. Consider consulting a lawyer, or getting a programming language that has no such worries. Weregerbil 10:37, 16 November 2006 (UTC)

Languages are not distributed (how would you distribute German?); compilers and interpreters and libraries are distributed. Similarly, languages are not licensed. As such, merely writing an R script can have no impact on anything; it's just text that you output. (An exception would be if there were trademark considerations in your output or so, but typically interoperability has trumped trademark law. See the Sega case.) The trick is actually if/when you run an R interpreter or so; consult both the FSF and a lawyer, as the details of how you invoke R programs, how your program would behave in the absense of R support, and how closely your program works with the graphing code may affect the issue. --Tardis 16:22, 16 November 2006 (UTC)

• Let's see if I get this right... The graphics is an important component of the application that I am considering writing. If the program writes the R code, and R is started separately to produce the pdf's, it would be OK. However, if the program calls R in batch mode via a system call, it would be a violation of the GPL. Is this a correct interpretation? --NorwegianBlue ^talk 16:54, 16 November 2006 (UTC)

If your program never interacts with any other programs (e.g., R interpreters), it's irrelevant what you produce. If you do interact with an R program (via system() or otherwise), you may count as associated. That's when you have to talk to someone in a legal capacity and/or discuss it with the FSF (although I'm sure they'd be none too happy to hear you were trying to avoid writing free software). --Tardis 17:38, 16 November 2006 (UTC)

Thanks. Actually, I'm not trying to avoid writing free software, but I may not be in a position to make the decision myself. Furthermore, the application in question would need to interact more or less tightly with closed-source commercial software, so if we were to release it under the GPL, the same kind of considerations would apply to that interaction. --NorwegianBlue ^talk 10:33, 18 November 2006 (UTC)

[edit] Spanish pronunciation of "s" as "h"

I have noticed that some Spanish speakers pronounce "s" at the end of unstressed syllables as "h", "las ciudades" might be pronounced as "lah siuthatheh". An example is found here Vanguardia de la ciencia - liquenes , in the item on lichens. The first time I noticed this was in an interview of Augusto Pinochet. Is this feature of certain dialects (which ones?) or sociolects, or what? --NorwegianBlue ^talk 19:11, 14 November 2006 (UTC)

At Spanish dialects and varieties#Evolution, we are told, "The realization of syllable-final /s/ as a barely audible [h] or simply nothing is rather noticeable in many dialects, including the Argentine ones. In the Castilian variety, this tendency exists but is less marked." The brief discussion at Spanish phonology (under /s/) also seems to imply that the phenomenon can occur in Madrid Spanish too. Wareh 19:48, 14 November 2006 (UTC)

The person with the "s as h" pronunciation in the audio file that I linked to is Ana Crespo, who works in Madrid as a botanist at the Universidad Complutense, but I don't know if she actually is from Madrid. She also pronounces "c" as "s" (seseo). Can anyone pinpoint her dialect from the podcast? --NorwegianBlue ^talk 21:04, 14 November 2006 (UTC)

Anyone with a ceceo in Madrid would not likely be from Madrid. If she's Spanish, she'd be from Andalusia or the Canary Islands. I can never make audio files work on Wikipedia so I won't try.

The weakening of /s/ to [h] or elision is very widespread in Spanish. It is typical of what is sometimes called the lowland dialects, including most of Southern Spain, the Canaries, the Caribbean, and most of coastal Ibero-America except Pacific Mexico, Peru, and perhaps parts of Central America (I'm not sure about that). Because of the large-scale migration of Andalusians to Barcelona and Madrid, the weakening can be found there too, particularly in the working classes. Someone, not me, oughta clean up the Spanish phonology pages. Argentine Spanish has that feature, but it is far more widespread. I did ceceo, and that was enough for now. mnewmanqc 01:55, 15 November 2006 (UTC)

• Thank you! --NorwegianBlue ^talk 17:33, 15 November 2006 (UTC)

[edit] Once upon a time, in Spanish

Spanish fairy tales often begin with the phrase "Érase una vez", which I assume means "Once upon a time". I'm a bit puzzled by the first word, "Érase". It's not in my dictionary, and appears to be a reflexive use of the verb "ser". Is this correct? "Lavarse", "irse" etc are in my dictionary, but not "serse". If it indeed is a reflexive use of "ser", is it used in any other context, or any other tense? --NorwegianBlue ^talk 18:42, 14 December 2006 (UTC)

"Érase una vez... " o "érase que se era... ", son formas del verbo "ser", o mejor dicho serían del verbo pronominal "serse", pero este verbo de hecho no existe salvo estas 2 frases hechas, equivalentes al inglés "once upon a time". Skarioffszky 19:05, 14 December 2006 (UTC)

Muchas gracias. I was wondering, could this construct be used in the future tense, "Se sera un dia, cuando ..."? Google gives some hits indicating that this indeed may be the case, but it is difficult for a non-native speaker to understand how the use of the reflexive pronoun is modifying the intended meaning of the sentence. Couldn't it just have been omitted, with no loss of meaning or clarity? --NorwegianBlue ^talk 21:30, 14 December 2006 (UTC)

I asked a well-known Spanish linguist, who assures me that it is simply the reflexive in the archaic word order for that tense verb clitic. Se es is used to this day. Here's a citation from the Spanish writer, Javier Marías's, blog: [[3]] mnewmanqc 15:59, 15 December 2006 (UTC)

Thank you very much! --NorwegianBlue ^talk 10:23, 16 December 2006 (UTC)

[edit] Maths resources

I am delighted by your question. It takes uncommon wisdom and humility to acknowledge a need to learn more and to seek assistance. In this case, your need is a shared one, and there are quality sources on the web. One place to start is The Math Forum at Drexel, where you will find a page of resources. For online mathematics texts, the AMS has links, though many are on more advanced topics. Still, you can find jewels like Calculus, by Gilbert Strang, a Professor of Mathematics at MIT. You might also browse topics at The Mathematical Atlas. If I have a chance, I'll see if I can dig up a good elementary discussion of the binomial theorem. Meanwhile, these sources should keep you in good reading. --KSmrq^T 22:07, 29 December 2006 (UTC)

[edit] Generating permutations

I refer to this section of the Permutation article:

BEGIN

[edit] Algorithm to generate permutations

For every number k () this following algorithm generates the corresponding permutation of the initial sequence :

 function permutation(k, s) {
     var int factorial:= 1;
     for j = 2  to length(s) {
        factorial := factorial* (j-1);
        swap( s[j - ((k / factorial) mod j)], s[j]);
     }
     return s;
 }

Notation

• k / j denotes integer division of k by j without rest, and
• k mod j is the remaining rest of the integer division of k by j.

END

The algorithm is supposed to generate every different permutation of the integers 1 to n, but when I coded it there were repetitions - could someone else check this, please?

Also, the bit following "Notation" seems to be written very clumsily, omitting the generally-understood word "remainder".81.153.219.51 16:42, 9 January 2007 (UTC)

I don't claim I understand why the algorithm does what it does, but I've checked all n! outputs for sequence length n up to and including 9, and it is working just fine for me. Are you perhaps using the permuted s as input to a next call?

As to the clumsiness, I'd say: {{sofixit}}. If you don't feel comfortable doing that (but why wouldn't you?), consider leaving a comment on the article's talk page. --Lambiam^Talk 21:47, 9 January 2007 (UTC)

Yes, I was using successive sequences - when I reset to 123...n order before each use, everything worked fine, thanks. With this apparently not working I searched for alternative algorithms - this one seems by far the shortest. It's a pity there is no attribution of source.

Re. the text, yes I'll change it - I hadn't fully appreciated how things were here.

With a dynamic IP address it may not look as if I'm the same person as before, but I am.81.153.220.80 18:27, 10 January 2007 (UTC)

[edit] GIMP question

In the GIMP, is it possible to have a layer with a transparent background, and add a fully opaque layer (for example, a JPG photograph) on top of that so that the transparent pixels stay transparent, but the non-transparent pixels get their colours from the new layer? Or is it possible to "substract" a layer with a transparent background from a layer with an opaque background, so that the "substracted" pixels would become transparent? JIP | Talk 18:28, 9 January 2007 (UTC)

Yes I'm sure it's possible as I've done it before, but I haven't used it in ages and haven't the slightest idea of how to do it now --froth^{T C} 20:05, 9 January 2007 (UTC)

If I understood your intention correctly, this is something that can be done with masks. Go to the layer with the transparent background and choose Layer → Mask → Add mask... and then select Layer alpha and click OK. Now activate the opaque layer, choose Layer → Mask → Add mask... and proceed with whatever settings there are. Go to the transparent layer's mask (by clicking it in the layer list), Ctrl+A Ctrl+C, then go to the opaque layer's mask, Ctrl+V. Anchor the mask, and you're done. –mysid☎ 21:01, 9 January 2007 (UTC)

Yes, this works. Thank you! JIP | Talk 10:11, 10 January 2007 (UTC)

[edit] Statistics & Propagation of error with least square fits

I originally came across this question my freshman year (at college), and although I am now a senior, I have yet to get a concrete answer. Basically, in my program (chemistry), we must do a lot of propagation of error, and there is one particular point that confuses me. Basically, if I plot a set of data that already has an associated error with it, how can that error be incorporated into the slope and intercept of a best-fit line? For example, if I plot absorbance vs. concentration of some solution, each point will have an error in the X direction (concentration, due to errors in preparation) and an error in the Y direction (absorbance, error and variablity due to instrument). If I calculate a best fit line, how can I incorporate these errors with the normal errors generated with a best fit? Let me know if you don't understand and I will dig up an example. As a similar question, if I take a set of points with error and try to take an average, how can these be incorporated into the error (and/or standard deviation) of the average? Again if you need an example, let me know. I would be happy with just a name of an equation or point me to a page. I would be very grateful :) Thanks a lot --Bennybp 02:39, 11 January 2007 (UTC)

Well I sort of found an answer in Bayesian linear regression, although it looks much more complicated than I wanted it to be. Anything simpler? --Bennybp 03:06, 11 January 2007 (UTC)

If your data sets are humungous, incorporating the individual point errors will have only a minimal effect. Otherwise, one method you can use is to replace each data point in your data set by K copies, each of which is randomly perturbed by X and Y amounts having the known error distribution for that point. Then compute the mean and variance of the miraculously multiplied data set. The mean should be the same, but the variance should be larger. If N was the size of the original set and you would have used the bias correction for the variance estimator of replacing N by N−1 in the divisor, then now replace KN by KN−K. The same method can be used for least square fits; in fact, computing an average is a form of least square fit in one dimension. Unless you use a fixed pseudo-random sequence you should get slightly different results each time; if the differences are too large, then N was not large enough. --Lambiam^Talk 05:54, 11 January 2007 (UTC)

Depending on how variable the errors in the X variable are relative to the errors in Y, the standard least squares "best-fit" line may be a very poor fit. Our regression dilution article gives a good overview of the issues, with good references. You might also be interested in errors-in-variables models. -- Avenue 15:39, 11 January 2007 (UTC)

The average is easy: the variance of the sum (of independent variables) is the sum of the variances, and the standard deviation scales linearly with what it describes. Therefore the standard deviation of the average is just as the average is the sum divided by N; . I separated the two factors of to emphasize that the standard deviation of the average is the RMS of the individual standard deviations, divided by . It is that division which makes repeating a measurement worthwhile; the resulting error is better than the "average" errors of the individual measurements.

Handling something like least-squares is more complicated, but it can be handled by general error propagation methods — see the first equation for Δf in that article. You have two functions f for the slope and intercept of your line; if you do the fit with only symbolic values (this gets complicated for large N), you can then take the required partial derivatives and evaluate the total error. As a trivial example, consider just two points: obviously the line passes through both so the slope is just and the intercept is b = y₁ − mx₁. The partial derivatives for m are

we can use those (with the chain rule) to easily evaluate the ones for b:

The total errors (using Greek letters for uncertainties for brevity, with η = Δy) are then

.

Throwing in numbers I get . The much higher uncertainty for b comes from the distance the points are from the y-axis; perhaps a more useful measure would be the uncertainty at the midpoint, which is . (I doubt it's a coincidence that with two points the uncertainty at the middle is just the distance to the middle times the uncertainty in the slope.) Obviously this algebraic approach would become tedious for N any bigger, but one can probably make it somewhat easier using the linear algebra approach to LSF and a CAS. Does this help? --Tardis 17:06, 11 January 2007 (UTC)

That all seems like a big help to me. I didn't think of applying propagation of uncertainty to the linear regression equation. And I usually get everything mixed up anyway (error due to measurements, standard deviation, etc.). Thanks a lot :) --Bennybp 00:47, 12 January 2007 (UTC)

[edit] Integration in Maxima

I'm experimenting with open-source computer algebra systems, and am trying out Maxima. I've encountered a problem when doing symbolic integration. The value returned from integrate, although displayed as an algebraic expression, behaves differently from the same expression entered by hand. Here is an example of what is going on:

• f(x):=45^(-x^2);

• F(x):=integrate(f(x),x);

• F(x);

// Seems reasonable...

// Now I'd like to evaluate F(x) for a specific value of x

• F(5);

Attempt to integrate wrt a number: 5#0: F(x=5)

-- an error. To debug this try debugmode(true);

// Hmm, why doesn't it simply substitute x=5?

// Ok, lets try defining G(x) = F(x) from scratch, and see what happens...

• G(x):=sqrt(%pi)*erf(sqrt(log(5)+2*log(3))*x)/(2*sqrt(log(5)+2*log(3)));

• G(5);

• float(G(5));

• F(t)/G(t);

• F(5)/G(5);

Attempt to integrate wrt a number: 5#0: F(x=5)

-- an error. To debug this try debugmode(true);

My question is this: How do I make F(x) behave as though it had been entered by hand. --NorwegianBlue ^talk 19:19, 14 January 2007 (UTC)

When you wrote F(x):=integrate(f(x),x) you intended the right side to be evaluated before the function was defined, but that is not what you actually said. So what is happening is that F(5) means integrate(f(5),5) and the error is exactly what you would expect. Try the quote-quote input:

• F(x):=''(integrate(f(x),x));

This forces the integral to be evaluated before the function is defined. --KSmrq^T 23:36, 14 January 2007 (UTC)

Yes, that fixed it. Thanks! --NorwegianBlue ^talk 20:59, 15 January 2007 (UTC)

[edit] Sample size vs. extreme observations

For a one-variable statistic with an infinite, standard-normally-distributed population, what is the relationship between the sample size and the expected highest value observed? NeonMerlin 23:49, 15 January 2007 (UTC)

Given a single measurement of a random variable with a cumulative distribution function F, the probability of getting a result that does not exceed s is F(s). For N measurements, the probability of not exceeding s in any of the measurements is F(s)^N. The derivative of F(s)^N with respect to s will thus tell you the probability of getting no result larger than s+ds in any of them, but a result of at least s in at least one of the measurements. That is thus the probability that your highest observed value was s. In other words, F(s)^N would be the cumulative distribution function of the highest observed value. I made that up from scratch, but a quick googling confirms it, see for example the abstract of [4] --mglg_(talk) 00:51, 16 January 2007 (UTC)

That gives me, for the standard normal distribution, a CDF of Φ(x)ⁿ = and a PDF of d/dx that. I imagine someone who knows integral calculus will be able to simplify these, but how do I turn either of them into the expected value? NeonMerlin 01:40, 16 January 2007 (UTC)

In general, given a PDF you can calculate the expectation value as Integral[x PDF(x) dx]. I don't think you can get any analytic answer in this case. There is probably a useful approximation for large N, though, but I don't know it. By the way, (essentially) your integral above has a name: the Error function. --mglg_(talk) 03:40, 16 January 2007 (UTC)

Let Z denote the highest of n independent random variables having the standard normal distribution. It is possible to give the following upper lower bound on the expected value of Z: E(Z) > Φ⁻¹(n/(n+1)). This can be seen as follows. Φ is a monotonic functions, so it commutes with max. So Φ(Z) is the highest of the Φ-values of n random variables whose (cumulative) distribution function is Φ. But these Φ-values are then simply uniformly distributed on the interval (0,1), so Φ(Z) has distribution function F(x) = xⁿ (as per above), and E(Φ(Z)) is now easily found to be n/(n+1). Since function Φ⁻¹ is concave convex in the area of interest (for n > 1), E(Z) = E(Φ⁻¹(Φ(Z))) > Φ⁻¹(E(Φ(Z))) = Φ⁻¹(n/(n+1)). I have not tried to estimate how tight this is, but I wouldn't be too surprised if E(Z) = o(Φ⁻¹(n/(n+1))). --Lambiam^Talk 05:21, 16 January 2007 (UTC)

Nice, Lambiam. But shouldn't it be "convex" and "lower bound", i.e. E(Z) > Φ⁻¹(n/(n+1)) ? --mglg_(talk) 19:12, 16 January 2007 (UTC)

You're right. In the meantime I have done some calculations, which suggest that the difference is not a runaway one; putting E(Z) = Φ⁻¹(n/(n+δ_n)), the quantity δ_n appears to decrease very slowly as n increases:

      n  delta_n
  -----  -------
      1  1.00000
      2  0.80231
      4  0.71501
      8  0.67000
     16  0.64408
     32  0.62783
     64  0.61688
    128  0.60909
    256  0.60326
    512  0.59874
   1024  0.59511
   2048  0.59213
   4096  0.58965
   8192  0.58755
  16384  0.58578

It looks like δ_n will remain above 0.5. --Lambiam^Talk 21:51, 16 January 2007 (UTC)

[edit] Creating duplicate hard links, then making them independent

Under Windows XP, I have a large library of PDFs that I need to edit down by deletion, and I want to keep both the edited and unedited versions on my hard drive. There will be room for both, but there isn't room to store the unedited version a second time. So what I want to do is make a copy of the folder with an identical tree, but where each file is a second hardlink to the same file in the original folder. I know this can be done with

fsutil hardlink create C:\oldfolder\subfolder\file.pdf C:\newfolder\subfolder\file.pdf

The problem is that fsutil only handles one file at a time, and the destination folder must already exist. How can I write a batch file that will find all the subfolders and files of the original folder, create corresponding subfolders in the duplicate folder and then fsutil hardlink all the files? NeonMerlin 00:35, 23 January 2007 (UTC)

Why would you store the unedited versions a second time? And why would you want to use multiple hardlinks, they're almost never useful. --froth^T 01:01, 23 January 2007 (UTC)

Well, can you think of a better way to make two copies of a folder, with the ability to delete files from one and keep them in the other, but without taking up twice as much space as a single copy? I don't intend to edit the files themselves, only to delete some and keep others (and maybe rename and move a few). NeonMerlin 01:10, 23 January 2007 (UTC)

Nope. How many directories are we talking about? I'm not sure if you'd be able to preserve directory structure, but maybe you could change the file handler from acrobat reader or whatever to fsutil, and mass-open all of the pdfs. It would be a lot of Ctrl-A and Enter, but it might work. --froth^T 01:15, 23 January 2007 (UTC)

From a command prompt, do "dir /s /b >file.txt" then use a text editor to convert each line in file.txt into two lines of the form "md C:\newfolder\subfolder\file.pdf\.." and then your " fsutil hardlink create C:\oldfolder\subfolder\file.pdf C:\newfolder\subfolder\file.pdf" (yes it appears possible to make a new folder structure by simply adding \.. to a file which doesn't yet exist). Tested on Win2000. -- SGBailey 11:18, 23 January 2007 (UTC)

[edit] Magnetic Resonance Imaging

In trying to understand how MRI's work, I discovered that the wikipedia article on the subject is a bit dense. I would like to try and rewrite the physics aspects of the article so that it might be a bit more accessible to non-physicists. But, I still don't understand the physics myself. I posted the following message to the MRI talk page, but it doesn't seem to get much traffic. The following is a disection of Wikipedia's mri article section called 'Principal'. Any and all feedback greatly appreciated:

Medical MRI most frequently relies on the relaxation properties of excited hydrogen nuclei in water and lipids.

Does the MRI work only with hydrogen atoms? Does it work only with hydrogen atoms in water and lipids? Are there any hydrogen atoms in a human (or other things commonly scanned) which are not contained in water or lipids? (i.e., could this be changed to state simply it relies on relaxation properties of hydrogen nuclei (or perhaps any nuclei with net non-zero spin)?

When the object to be imaged is placed in a powerful, uniform magnetic field, the spins of atomic nuclei with a resulting non-zero spin have to arrange in a particular manner with the applied magnetic field according to quantum mechanics. Hydrogen atoms (= protons) have a simple spin 1/2 and therefore align either parallel or antiparallel to the magnetic field.

My understanding of spin is a little rusty; "with resulting non-zero spin" - what does the word 'resulting' mean here? The spin is created as a result of applying the magentic field? Any given proton may obtain either spin 0, 1/2 or -1/2? Is the direction of alignment (parallel/antiparallel) arbitrary or a result of some other property of the proton which will precisely determine the direction (+ or - spin)?

Common magnetic field strengths range from 0.3 to 3 T, although field strengths as high as 9.4 T are used in research scanners [5] and research instruments for animals or only small test tubes range as high as 20 T. Commercial suppliers are investing in 7 T platforms. For comparison, the Earth's magnetic field averages around 50 μT, less than 1/100,000 times the field strength of a typical MRI.

Is this paragraph appropriate/needed in the principal section?

The spin polarization determines the basic MRI signal strength. For protons, it refers to the population difference of the two energy states that are associated with the parallel and antiparallel alignment of the proton spins in the magnetic field and governed Boltzmann's statistics.

Is this saying that spin polarization is the integral difference in protons in one state versus the other? So, if I have 500,001 protons that are parallel and 500,000 protons that are antiparallel then my spin polarization is 1? What is Boltzmann's statistics and what does that have to do with the calculation of spin polarization? What is the cause for the descrepancy; is it arbitrary or is it a result of some fundemental property of protons?

In a 1.5 T magnetic field (at room temperature) this difference refers to only about one in a million nuclei since the thermal energy far exceeds the energy difference between the parallel and antiparallel states. Yet the vast quantity of nuclei in a small volume sum to produce a detectable change in field.

The thermal energy causes protons to switch between parallel and antiparallel? (which seems to argue that the state is arbitrary) If the selection of parallel and antiparallel is arbitrary than each "small volume" would statistically cancel out each other "small volume" (since that would not be helpful; seems to argue that the selection isnt arbitrary and must favor either parallel or antiparallel statistically. If so, why?) '...detectable change in magnetic field...'?

Most basic explanations of MRI will say that the nuclei align parallel or anti-parallel with the static magnetic field though, because of quantum mechanical reasons, the individual nuclei are actually set off at an angle from the direction of the static magnetic field. The bulk collection of nuclei can be partitioned into a set whose sum spin are aligned parallel and a set whose sum spin are anti-parallel.

This is describing the heisenberg effect? Each proton could never be precisely in a specific alignment, but simply has a probablity cloud of alignments which is centered along the magnetic field? If thats the case then there is a certain probablity that a given proton is precisely aligned with the magnetic field and the probablity cloud is statistically parallel with the magnetic field? Can be partitioned theoretically or is partitioned in practice during the scan?

The magnetic dipole moment of the nuclei then precesses around the axial field. While the proportion is nearly equal, slightly more are oriented at the low energy angle. The frequency with which the dipole moments precess is called the Larmor frequency.

Ok, starting to get really lost... Is this saying that similar to -- electron orbitals around a nuclei--, the proton's axis has multiple quantized "orbitals" which effect the probablity cloud of its angle of incidence with the magnetic field? Ok, all my previous theories start to breakdown when adding that the proton precesses at a specific rate. And the angle of incidence with the magnetic field is a function of energy level. Perhaps the higher level orbitals are donuts centered around the magnetic field which allows for a specific frequency, but then it seems that the 0 level orbital should still have no frequency (and should still include the magnetic axis itself). ...the proportion is nearly equal... Proportion of 0 level to 1 level? Why would it be nearly equal? Why are the protons in the level 1 state at all? Aren't there states above level 1? Larmor frequency is specific to the level 1 state or all states precess at the same frequency?

The tissue is then briefly exposed to pulses of electromagnetic energy (RF pulses) in a plane perpendicular to the magnetic field, causing some of the magnetically aligned hydrogen nuclei to assume a temporary non-aligned high-energy state. Or in other words, the steady-state equilibrium established in the static magnetic field becomes perturbed and the population difference of the two energy levels is altered. The frequency of the pulses is governed by the Larmor equation to match the required energy difference between the two spin states.

Is the EM applied from a single direction or radially? The EM waves hit the protons and move them from level 0 to level 1? How are the states measured; photons created when the proton relaxes and released in a random direction as opposed to the original direction of the source beam? What does that information tell one about the material being scanned? What is really being measured; proportion of hyrdrogen atoms at every given location? Different regions are detected by measuring intensity of photons detected bouncing randomly? I have no idea how far off from reality I am at this point, but if that's the case how does measurement work in 3D?

Anyway, I'm sure I'll have more questions and if this goes well I'd like to try and tackle the k-space section also. Aepryus 18:02, 25 January 2007 (UTC)

Hmm - that's a lot of questions. MRI is a pretty involved topic, and I don't think you'll get far without reading a lot of more basic material that makes up the technique. I'll try to answer some of the questions and reference you to pages that you could read to understand nuclear magnetic resonance better. First of all, as you can see from the MRI article, there are many different variations on MRI technology which are used to image different aspects of a sample, but I'll go over the basics. The first article you should read is quantum spin, which should clear up many of your question. You could also take a look at a somewhat poor article on spin 1/2, and at another poor article on the Bloch sphere, which is an essential theoretical tool for understanding NMR. Speaking of which, Nuclear Magnetic Resonance is the basic theory, and the article is decent - you should read the "Theory of nuclear magnetic resonance" section.

Ok. Here's the basic idea. Protons are spin 1/2 particles, meaning they are a two state quantum system. That is, you can find only two orthogonal eigenstates, and the proton will be in some superposition of the two eigenstates. For instance, / sqrt2( | z + > + | z − > ) is an equal superposition of the eigenstate in the positive z direction and the negative z direction in the basis of the z-direction eigenstates, and | x + > represents a proton only in the x+ direction eigenstate. The idea behind the Bloch sphere representation is that any superposition of eigenstates can be represented by a vector in space which represents the direction of the magnetic moment of the spin (each spin produces a magnetic field like a little bar magnet). In the absence of an external magnetic field, your choice of basis is arbitrary - there is no eigenstate with more energy than any other eigenstate. However, when you apply a magnetic field (say, in the z direction), the z+ eigenstate gains a higher energy than the z- eigenstate. It is the peculiar property of a 2-state quantum system that when it is put in a superposition of the energy eigenstates, it precesses. I am amazed that we don't have an article on spin precession - perhaps I'll write one. In the Bloch sphere geometrical representation, that means if you point the spin off axis, it will spin about the magnetic field axis at the larmor frequency. Now: the idea behind NMR. If you have a bunch of these proton spins in a magnetic field, then you zap them with a circularly polarized RF pulse at the larmor frequency, then an interesting thing happens. Imagine a small magnetic field component (the RF pulse) rotating with the precessing spins at the larmor frequency. The spin sees a small constant magnetic field perpendicular to it (the RF field in it's rotating frame), and precesses about it. The result is a spiral down the Bloch sphere which, depending on the duration of the RF pulse, can put the spin at any latitude on the sphere. Once the spin is precessing at a nonzero angle to the z-axis, it acts like a little antenna, and can be detected by placing an induction coil around the sample. Thus, you can zap the sample, then detect a response.

How do you use this for imaging? That is another very long story, and you'll have to do some reading before you'll understand the fourier space idea. -bmk

[edit] Catalan and Castellano, Mutual intelligibility

I noticed on Catalan television last summer in a talk show that one of the participants spoke Castellano, while the others spoke Catalan. Evidently, the Castellano speaker understood Catalan without trouble. Just about everybody in Catalunya speeks Castellano fluently (although I have noticed that people in their twenties often have a characteristic accent that the parent generation lacks). I doubt, however, that the converse holds. I have three questions:

1. To what extent do children in the Castellano-speaking parts of Spain learn Catalan, and the others Spanish languages?
2. Approximately what proportion of native inhabitants of Madrid understand Catalan?
3. Has anyone else here observed this phenomenon - conversations where one participant speaks Catalan and another speaks Castellano? In a conversation between a Norwegian and a Swede, each uses his own language because they are mutually intelligible, but (to me, at least) the "distance" between Castellano and Catalan is much greater than the "distance" between Norwegian and Swedish. --NorwegianBlue ^talk 22:24, 25 January 2007 (UTC)

PS: Is there a technical term for "distance" as used in question 3? --NorwegianBlue ^talk 22:24, 25 January 2007 (UTC)

"Distance" is a term widely used to discuss the concept, although I just saw a presentation that used the word "dialectometry". Measuring the distance between dialects and languages is not easy. The only rigorous, objective effort I've heard of is this one, which uses conditional entropy and Levenshtein distance to try to measure information loss between dialects.

The main problem is that where two language communities coexist for a long time, people become multilingual. For example, in Switzerland, I've heard conversations where one person spoke German and the other spoke French, each speaking their own language and understanding the other just fine. Yet, no one would ask if French and German are similar enough for people to understand each other automatically. Catalan/Spanish bilingualism is a bit like Swedish/Norwegian bilingualism in that the two language are in many ways similar, but it's also like French/German bilingualism in Switzerland in that many people just speak and understand both, so you can use whichever. There's something of a continuum between the two kinds of multilingualism.

I would expect that teenagers in Barcelona might well be highly bilingual, but that teenagers in Madrid would have no easier a time understanding Catalan than they would Italian. --Diderot 10:00, 26 January 2007 (UTC)

Thank you Diderot, for an insightful comment, and for a very interesting link! I posed this question also on the Spanish equivalent of the reference desk, and the answer somewhat surprised me:

(From Usuario Camimo, any errors in translation are of course mine!):

In lack of surveys on the subject, and leaving aside the fact that the concept of "understanding a language" implies a gradation, is evident that any Spanish-speaking adult is moderately able to understand Catalan without too much difficulty, always better if the conditions are optimal: clear articulation and separation between words, etc. As with Galician, the linguistical similarities between the languages are so strong that this is logical.

In no community of Spain where the only official language is Castellano, does the obligatory curriculum of a pre-college student include the language of other communities (there are particular cases like, for example, the possibility of voluntarily attending such classes in some centers of Asturias Gallego...).

Interestingly, the Spanish page on mutual intelligibility lists Castellano and Catalan as mutually inteligible, as well as Castellano and Italian! --NorwegianBlue ^talk 09:14, 28 January 2007 (UTC)

When a Czech person and a Slovak person meet, the Czech speaks Czech and the Slovak speaks Slovak. (Of course, if a Slovak person is living in the Czech Republic and interacts with Czechs every day he/she would learn and use Czech with his/her colleagues.) When Czechoslovakia was around, there was a "federal" TV channel that alternated between the two languages, even on the same newscast. So Czechs were used to hearing Slovak. Interestingly, Czech kids today can't understand Slovak like their parents can because they haven't been exposed to it as much. When I was living in Prague and would visit Bratislava, I would speak in Czech, and the waiters/bartenders/ticket sellers would answer in Slovak. I had to explain that I could barely understand Czech, let alone Slovak. -- Mwalcoff 23:45, 27 January 2007 (UTC)

That's interesting! We have a similar situation in the eastern parts of Norway. When I grew up, Norway had only one TV channel, but the eastern parts of the country could receive Swedish television as well. Therefore, I and my generation understand Swedish much better than our children do. --NorwegianBlue ^talk 09:29, 28 January 2007 (UTC)

Holds true for Swedish and Danish as well. I grew up in the southern part of Sweden, with convenient access to Danish TV. As a result, I understand Danish without a problem, and can even speak (a little). When I meet people who live in Stockholm however, they tell me that it's just impossible to understand what a Dane says. TERdON 18:17, 29 January 2007 (UTC)

[edit] Catalan y castellano

Observé en un "talk show" (entrevista) en la televisión catalana el verano pasado, que uno de los participantes habló castellano, mientras que los otros hablaron catalán. Evidentemente, el hablante de castellano entendía catalán sin problemas. Yo sé que todo el mundo en Cataluña domina el castellano, y sé tambien que el revés no es verdad.

Tengo cuatro preguntas relacionadas con esto:

1. ¿Aproximadamente qué proporción de madrileños nativos entienden el catalán?
2. ¿En cataluña, es comun que un participante en una conversación habla catalán y otro habla Castellano?
3. ¿Los niños en las partes de España donde se habla castellano, aprenden catalán, y los otros idiomas españolas en colegio?
4. ¿Existe un continuo dialectal entre las regiones donde se hablan el catalan y el castellano?

Gracias, --Noruego azul] 22:33 27 ene 2007 (CET)

A falta de localizar datos precisos al respecto en forma de encuestas, le puedo comentar lo siguiente:

Respecto de su primera pregunta, y dejando a un lado que el concepto "entender una lengua" implica una gradación, es evidente que cualquier hispanohablante adulto está moderadamente capacitado para entender sin demasiadas dificultades el catalán, siempre mejor si las condiciones son las más propicias: articulación diáfana y espaciosa, etc. Al igual que ocurre con el gallego, las similitudes lingüísticas entre ambas lenguas son tan fuertes que es algo lógico.

En cuanto a su segunda cuestión, existe también un problema con el concepto de "común" pues depende de una cuestión meramente aleatoria que dos hablantes de castellano y catalán queden frente a frente; en cualquier caso, e independientemente de lo que puedan decir las posibles encuestas que haya al respecto, e intuyendo hacia dónde dirige Ud. su pregunta, lo que sí se puede afirmar es que en la actualidad esa situación no llevaría a que ninguno de los dos participantes cambiase de código o recriminase al otro por usar la otra lengua, pues el grado de normalización del uso de ambas lenguas es elevadísimo.

Por lo que respecta a su tercer pregunta, en ninguna comunidad de España donde no haya más lengua oficial que el español, forma parte del currículo educativo pre-universitario la enseñanza obligatoria de otra u otras lenguas de otras comunidades (hay casos particulares como, por ejemplo, la posibilidad de cursar voluntariamente en algunos centros de Asturias el gallego...).

Por último: por supuesto, como en cualquier otro lugar del mundo, el continuo dialectal existe en las zonas limítrofes entre castellano y catalán, por más que esas hablas no tengan reconocimiento oficial y por más que sea difícil establecer si pertenecen al catalán, aragonés, castellano, francés, valenciano, etc.

--Camima 00:11 28 ene 2007 (CET)

[edit] Consultas, castellano y catalan

¡Muchas gracias por su respuesta! Mi impresion era que la diferencia entre el catalan y el castellano fuera mucho mas grande que la diferencia entre el castellano y el gallego o el portugués. ¿No es asi? En la wikipedia ingles, la pagina Mutual intelligibility dice que el castellano y el portugés tiene un alto nivel de inteligibilidad entre sí, pero no menciona el hecho que catalan y castellano son mutuamente inteligibles. Sin embargo, despues de leer su respuesta, he leido la página correspondiente aquí, y esta dice que sí son mutuamente intelegibles. Para mi sorpresa, también indica que castellano y italiano son mutuamente inteligibles! --azul Noruego azul] 03:09 28 ene 2007 (CET)

Solo quisiera notificarle que he citado su respuesta aqui en el "Reference desk" de la Wikipedia ingles. --Noruego azul] 10:21 28 ene 2007 (CET)

¡Gracias! --Noruego azul 11:44 28 ene 2007 (CET)

[edit] Consulta español-catalán

Piensa, primero, que todas ellas (gallego, portugués, catalán, español...) son lenguas_románicas; segundo, que han compartido durante siglos un mismo espacio cultural que ha implicado, inevitablemente, numerosísimos contactos; tercero que, en algunos casos, cuestiones políticas han ido matizando sus relaciones: en la Edad Media, el gallego y el portugués eran la misma lengua, el llamado gallegoportugués; tras la independencia de Portugal, la variante correspondiente, el portugués, se fue diferenciando cada vez más del gallego, el cual, sin embargo, estuvo durante siglos viviendo únicamente como lengua oral, pues la lengua de cultura pasó a ser, tras la edad media, el castellano. Por el otro lado, el catalán siempre ha tenido más vitalidad por razones geográficas y su "personalidad" se vio aderezada, en la Edad Media, con muchos toques del provenzal, lo que le han conferido un pequeño matiz de distinción respecto de otras lengua peninsulares. Además, su proceso de normativización (de establecimiento de unas claras reglas de uso) comenzó ya a comienzos del siglo XX, mientras que en el caso del gallego ese proceso ha sido mucho más tardío (después de 1975).

Así las cosas, para un hispanohablante resultará siempre más próximo el gallego (que de una forma u otra se ha mantenido siempre mucho más cerca del español) que el catalán o el portugués. Sin embargo, y como ya te comentaba, las similitudes son tan grandes que bajo unas determinadas condiciones de conversación el entendimiento mutuo es relativamente fácil.

--Camima 11:34 28 ene 2007 (CET)

[edit] Linear regression problem

I have some data that I tried analysing using a linear model with R (programming language). The dependent variable is Y. There is an explanatory variable X1. The data consists of two populations, represented by another variable X2. I did a regression analysis Y~X1 in each subpopulation (X2=1 and X2=2), as well as a regression Y~X1 for the combined data set. My problem is that the regression lines, when plotted, are quite different from what I would have expected from the scatterplot. I've put the data, plots and R code on an external page, and would be extremely grateful for feedback on whether I am making a silly mistake in the regression analysis or plotting, or if I might need to consult an opthalmologist.

The first figure shows the results of the regression analysis for each subpopulation (red and blue), as well as the result for the whole data set (green). The second figure repeats the regression line for the combined data set (green), and shows where I had expected to find the regression line(orange). Thanks in advance for any feedback. --NorwegianBlue ^talk 13:46, 28 January 2007 (UTC)

1.

Different lines can be obtained by using different weights to off mean points eg |x_n-x_av| , (x_n-x_av)² it depends on how you want to emphasise the shape of the scatter plot. You could do an analysis of the scatter plot to see if the spread of values seems to match any known distribution.87.102.33.144 14:23, 28 January 2007 (UTC)

2.

The slope of the line of the combined data set didn't match what you expected - here's what I would do.

There are n points. Take a pair of points and find the slope and second variable (constant) for this pair of points eg y=ax+c, y2-y1=(x2-x1)a. Repeat this for each combination of points - there are n(n-1)/2 pairs of points..

Now you should have n(n-1)/2 values for 'a' (the slope) and 'c' (the constant). You can analyse this data set for and average 'a' and 'c', and also find the variance of both.

This should give you not only average slope but a range (amount of accuracy) of values in which 'a' can be expected to lay based on your data set.

Question - does this range of values include your expected (orange) line.. Hope that helps. I'm suggesting you find the 'amount of error' in the line.87.102.33.144 14:40, 28 January 2007 (UTC)

Thanks! Re 1: I was thinking plain old least squares linear regression. My problem isn't that I had some preconceived expectations about the regression lines, but that they are less steep than what my eyes are telling me that they should be when I look at the scatter plot. I have plotted the residuals against quantiles of the normal distribution (qqnorm in R). Except for two outliers at the lower end, and four at the upper end, the residuals appeared to be reasonably normally distributed. And with approx. 1500 data points, I wouldn't expect 6 outliers to have a tremendous effect.

Re 2. That's interesting, I'll try to do what you suggest. Should the average slope and intercept using this procedure be identical to those obtained using least squares linear regression? --NorwegianBlue ^talk 15:11, 28 January 2007 (UTC)

As I understand it - not identical - method of least squares emphasises 'erratic' points more than a magnitude (absolute value) method. So there will be a minor difference (except of course when all the points are exactly on a line - not the case here).

I've just 'thought' of something else - is your method using - vertical offsets (my guess is it is - as it is simpler and more common) - if so you could try perpendicular offsets (much better) see http://mathworld.wolfram.com/LeastSquaresFitting.html (second set of diagrams down) - I think the picture speaks for itself. If your not using perpendicular offsets you really should try this - it does give better results (especially when the slope of the line is high) - you might find it gives a result closer to the one you expected - the eye is usually a good measurer..87.102.33.144 16:19, 28 January 2007 (UTC)

You also might want to look at http://www.tufts.edu/~gdallal/slr.htm which gives "confidence bands from the regression line " - I haven't checked the maths on this - what you need to do is find "the confidence bands from the regression line" - a google search might help or you could ask here - I'd be very suprised if your orange line didn't lie within the confidence bands.

The confidence bands effectively give a measure (statistical) of where the actual line can be expected to lie (within certain probabilities) - like a broad thick line that the actual line must lie within.87.102.33.144 16:19, 28 January 2007 (UTC)

I think you are right in that the lm() function of R minimizes the squared vertical offsets. Browsing the help files, I found that R has a procedure called line() which does robust line fitting. The line thus obtained was slightly "better" than the green line obtained using lm(), but still far from my orange line. --NorwegianBlue ^talk 18:48, 28 January 2007 (UTC)

Looking at your picture, the orange line is more or less the major axis of an ellipse containing most of the points of the joint scatter plot. In general, the line obtained by linear regression has a slope that is less steep. Let U and V be two independent normally distributed (μ = 0, σ = 1) random variables. If X = aU + b, Y = mX + c + dV, for sufficiently large samples the least squares linear regression fit will approximate the line y = mx + d. The lines of equal density in a scatter plot are then ellipses of the form u² + v² = r², for various values of r, where u = (x−b)/a and v = (y-(mx+c))/d. Expressed in x-y coordinates: ((x−b)/a)² + ((y-(mx+c))/d)² = r². In general, the slope of the major axis of these ellipses is not equal to m. As you change the scale of Y, the value of m changes accordingly. But the major/minor axes of an ellipse are not projective invariants. In particular, they are perpendicular, but linear scale transformations do not preserve angles and consequently do not preserve these axes. --Lambiam^Talk 12:36, 29 January 2007 (UTC)

Thanks, Lambiam. It's correct that I expected the regression line to more or less follow the major axis of that ellipse, and I didn't know that a less steep line was to be expected. Can I infer from your comment, then, that there is nothing obviously wrong with the regression lines as plotted? --NorwegianBlue ^talk 18:58, 29 January 2007 (UTC)

You can infer from my comment that I see nothing obviously wrong with these lines. For a simple visual check, if you have a nicely ellipse-shaped cluster, take the vertical tangents at the left and right sides of the ellipse. The regression line should pass through the points where these lines touch the ellipse. For each point on the regression line, the vertical extents above and below should be (roughly) equal. As far as I can see without a visit to the ophthalmologist, your red, blue and green lines pass this visual check; the orange line does not. --Lambiam^Talk 13:36, 30 January 2007 (UTC)

Thanks a lot! --NorwegianBlue ^talk 22:11, 30 January 2007 (UTC)

[edit] No. of chromosomes in Neanderthals?

I know that it is not a well researched topic. But it'd be great if anyone could give me an idea about the number of chromosomes in the genome of Neanderthals? Thanks.nids(♂) 22:19, 28 January 2007 (UTC)

Neanderthals are now widely believed to have bred with homo sapiens, therefore they must have had a corresponding number of chromosomes, 46. Vespine 00:42, 29 January 2007 (UTC)

Widely believed? I thought this was merely a theory. Do you have some references? Clarityfiend 03:03, 29 January 2007 (UTC)

Isn't it truer to say that, if Neanderthals bred with homo sapiens then they must have had 46 chromosomes and therefore must have been human beings and therefore the distinction between Neanderthals and homo sapiens breaks down; or, if they didn't have 46 chromosomes then they were not humans and could never have bred with homo sapiens? JackofOz 03:09, 29 January 2007 (UTC)

Recent genetic work (e.g. the work of Krings [6], Bryan Sykes, and others) strongly argues that the humans and Neanderthals did not interbreed. Samples of mitochondrial DNA from well preserved Neanderthal fossils are too distinct from modern humans for the gene pools to have been mixing. Given that the two populations almost certainly had opportunities to interact, a lack of interbreeding would likely indicate that they were genetically incompatible, and one plausible (but by no means certain) explanation for that would be that Neanderthals had a different number of chromosomes from modern homo sapiens. Notably, homo sapiens have 46 chromosomes, while living great apes have 48, and that transition could have occurred after the Neanderthal population split from the pre-homo sapien line. Dragons flight 05:42, 29 January 2007 (UTC)

No it isn't, number of chromosomes isn't a unique species identifier, a domestic cat and a domestic pig has 38, doesn't mean they are the same species, nor that they can interbreed. But animals that DO interbreed do need to have the same number of chromosomes. Neanderthals had 46 chromosomes but that doesn't mean they were the same as humans, but they were probably close, it doesn't 'prove' that they bred. And I think recently it went beyond being just a theory, there was evidence found that Neanderthals could have actually just been bred into the population, not died out due to homo sapiens as previously thought. I'll try to find the sources. Vespine 05:15, 29 January 2007 (UTC)

We can have both the scenarios. i.e. Neanderthals had 48 chromosomes and they interbred with humans to produce Hybrids (they would be mostly sterile though). Also, Neanderthals may be having 46 chromosomes but still unable to interbreed with Humans. So, even if they interbred with humans, this does not prove that they had 46 chromosomes. But have you read somewhere that Neanderthals had 46 chromosomes. I am not asking for a source, just confirming if you had read that in a reliable source.nids(♂) 10:30, 29 January 2007 (UTC)

Skull suggests human-Neanderthal link. Anchoress 05:55, 29 January 2007 (UTC)

This skull, by itself, suggests nothing. For all we know, he could have been just a sterile hybrid.nids(♂) 10:11, 29 January 2007 (UTC)

Just another note. It is not required that species that breed have the same number of chromosomes (see Donkey, Horse, Mule). If the homo sapien population exploded and interacted with Neanderthals, cross breeding (and the subsequent sterile offpsring) would have wiped Neanderthals off the planet quickly and with no genetic trace. Imagine 10,000 horses and 100 donkeys. Females are almost constantly pregnant and the large majority of female donkeys would be carrying sterile offspring. Homo Sapien/Neanderthal offspring wouldn't be sterile for the same reason as mules but they may be for other reasons. (This is my own pet theory, no pun intended :) ). --Tbeatty 06:25, 29 January 2007 (UTC)

The Cell paper from Svante Pääbo's group that was linked to, studied mitochondrial DNA. Humans are but a medium in which mitochondria propagate themselves. The fact that Neanderthal mtDNA is not found in contemporary populations, only shows that there is no unbroken maternal line between contemporary humans and neanderthals. If the carriers of particular mitochondrion variants had even the slightest disadvantage compared to others, they would be less likely to have children, and those mitochondria would die out. This does not mean that the people that carried these mitochondria have no ancestors today, only that there is no unbroken chain of maternal ancestors. Consider also, that the fitness of mitochondria reasonably might depend on climate, since they are the cells' "power plants". And the neanderthals lived in a climate that was very different from today's. Therefore, to me it proves absolutely nothing that neanderthal mitochondria are extinct. I also disagree with the statement made in the Cell paper, that the observation that the time of divergence between neanderthal mitochondria and modern human mitochondria, being much longer ago than than the estimated divergence time of modern human mitochondria, argues against interbreeding. The fact that all contemporary human mitochondria stem from the mitochondria that lived in a woman some 200,000 years ago only proves that mitochondria are subjected to evolutionary pressure. Studies of the Major histocompatibility complex of primates, show that humans, chimps, orangutans and gorillas share many polymorphisms, particularly in the class II region. This argues strongly against very narrow bottlenecks in the size of the human (and ape) populations, and also shows that speciation is not an event that happens in a single individual, but in large groups which, while diverging, gradually lose the ability to interbreed. Jan Klein wrote a very interesting review on this way back in 1987 (Origin of major histocompatibility complex polymorphism: the trans-species hypothesis. Hum Immunol. 1987 Jul;19(3):155-62.). Note, incidentally, that this also shows that a difference in the number of chromosomes cannot be an absolute obstacle to interbreeding. --NorwegianBlue ^talk 20:22, 29 January 2007 (UTC)

[edit] Mexico City

Hi, guys. I will be leaving at the end of this week for Mexico City, my first visit there, where I will be spending the best part of February. I would be grateful for any inside information any of you may be able to supply-where are the hot spots, cold spots, places to go and places to avoid? Some information on good restaurants and the night life would be a help. I will be staying in the Zona Rosa, if that's any help, though will obviously go wherever necessary. Also, I will be travelling out from a wintery London, and was hoping to get away with my summer clothes, but I understand the Mexican nights can be quite chilly. Should I take a coat? All advice gratefully received. Clio the Muse 01:59, 29 January 2007 (UTC)

I've had some delightful vacations in Mexico City in February. It is cool at night and in the morning. Not as cool, though, as London in February. (Typically, around 8 C at the first light of dawn, but between 10 C and 15 C by the time most tourists venture out.) When I have gone, I have brought a spring/autumn jacket, and that has been sufficient. You might also bring a sweater so that you can double up (sweater plus jacket) if it is unusually cold and/or for your trip to and from Heathrow (or Gatwick). On the other hand, it is often quite warm, or even hot (around 25 C) during the afternoon, and the sun is typically scorching because of the altitude and latitude. Unless you have very dark skin, bring sun screen, as it may not be available or may be overpriced in Mexico .

I assume that you have a guidebook and know not to carry valuables, especially not on the subway/underground, and to sit in cars that are full but not too full, with your purse or backpack firmly closed or tied shut and in front of you where you can see it. Also, avoid the ubiquitous VW bug taxis, some of which are driven by armed robbers. If you want to take a cab, call a (more expensive but safer) radio taxi, or hire a cab at one of the cab stands for the regulated and safer cabs. A Lonely Planet Guide will give the latest tips on safety.

For transport, I had good experiences with both the metro/subway/underground and the peseros, or minibuses, which are a good way to reach places not near a metro stop. If you speak a little Spanish and have a sense of directions, just stand on the side of the major street running in the direction you want to go at a pesero stop and ask people how to get where you want to go. People will tell you which one to take, and if you need to transfer.

I strongly recommend the National Museum of Archaeology. Also, don't miss Diego Rivera's murals in the Palacio Nacional. To me, the most enjoyable neighborhood is Coyoacán, particularly Plaza Hidalgo, with its street musicians, Frida Kahlo museum, people watching, good restaurants, and generally wonderful atmosphere.

Also, you absolutely must make the excursion to Teotihuacán. It is really spectacular, particularly the climb to the top of the Temple of the Sun. I recommend taking the public bus from the main Norte bus station (accessible by metro) instead of the overpriced turista coaches that charge ten times as much for the same trip and insulate you from interacting with locals.

¡Buen viaje! Marco polo 02:32, 29 January 2007 (UTC)

Also, if you expect an area to be rather slummy, by all means avoid it. Mexico's drug trade is getting dangerous, so watch out, and have a fun time. ;-) The velociraptor 03:52, 29 January 2007 (UTC)

Great response, Marco Polo. You make me want to go there! Anchoress 06:36, 29 January 2007 (UTC)

Here's WikiTravel's article on Mexico City[7]. Vranak

www.tripadvisor.com is a very good site for people's recommendations of places to stay, places to go and attractions etc (http://www.tripadvisor.com/Tourism-g150768-Mexico-Vacations.html) is a Mexico page result. ny156uk 17:15, 29 January 2007 (UTC)

Thank you, one and all. This is very much appreciated. Clio the Muse 19:07, 29 January 2007 (UTC)

A couple other points. I'm sorry that I did not mention restaurants or night spots. My last trip there was three years ago, such places tend not to be "in" for very long, and to be honest, I don't remember names of places I went. But I found them in a guidebook, and I remember that several of the meals were wonderful. Also, I have one more tip, regarding Xochimilco. The standard thing for tourists to do in Xochimilco is to take a trajinera (covered boat) from the Xochimilco village landing, typically with a mariachi band on board, and travel a very short distance through a canal filled with other trajineras and tourists and mariachi bands. You will embark briefly on an island filled with stands selling cheap souvenirs. This excursion has some campy, festive appeal, I suppose, but when I visit a foreign city, I want to experience an environment that does not revolve around tourists. I was much more satisfied with an excursion in a trajinera through the Parque Ecológico de Xochimilco, 1-2 km to the north of Xochimilco village along the Anillo Periferico Oriente. Here is a link. Some sources say that these trajineras travel only on weekends, but we went on a weekday and had a trajinera and the whole park to ourselves. You will see how chinampas, the basis of Aztec agriculture, are built and maintained, and you will travel along beautiful and very peaceful waterways that are a welcome break from the noise and frenzy that is Mexico City. Marco polo 19:23, 29 January 2007 (UTC)

[edit] Radicals

Just a quick question: is it correct to have a radical inside of another radical, or should I attempt to work the second out of the first radical? I guess "correct" isn't the best word. I guess I'm asking is it more formal format to have it out? 75.18.9.71 03:48, 30 January 2007 (UTC)

It depends on the context, but I would certainly say that there is nothing wrong with having square roots inside other square roots. Do you mean something like ? –King Bee ^{(T • C)} 04:05, 30 January 2007 (UTC)

Yes, that's exactly what I mean. Thanks :-) -75.18.9.71 04:11, 30 January 2007 (UTC)

That's perfectly ok. In fact, there are some numbers defined by an infinite amount of nested radicals. Check out that article. --Ķĩřβȳ♥ŤįɱéØ 07:13, 30 January 2007 (UTC)

If there is an unnested version, it will be considered simpler, so by the unwritten rule that (in the absence of other rules to the contrary) the simpler expression is preferred, it is the better form. For example, rather than

use

--Lambiam^Talk 14:13, 30 January 2007 (UTC)

How do you go about to simplify such an expression? I tried the expression in Maxima (which I've only just begun to get acquainted with). It confirmed, of course, that both expressions evaluate to 0.6180339887499. But I was unable to make it simplify the expression (tried procedures ratsimp, radcan, trigsimp and trigreduce, i.e. all the buttons that had the words "simplify" or "reduce" on them). It was no more successful at simplifying

.

How does one approach the problem "by hand"? --NorwegianBlue ^talk 22:08, 30 January 2007 (UTC)

One possible approach, which can have varying degrees of success, it to conjecture that will be of the form with a and b rational. This gives , or which gives a³ + 15ab² = − 2 and 3a²b + 5b³ = 1. Solving this (the substitution helps) should give the result. -- Meni Rosenfeld (talk) 22:26, 30 January 2007 (UTC)

Thanks for the quick reply! --NorwegianBlue ^talk 22:45, 30 January 2007 (UTC)

• From Exact_trigonometric_constants

Main article: Nested radicals

In general nested radicals cannot be reduced.

But if for ,

is rational,

and both

and are rational,

with the appropriate choice of the four signs,

then

Example:

[edit] Arcsin sqrt(x) transformation

My officemate had a reviewer on a paper write "data should be transformed by arcsin sqrt(x)." I recall (I think) that this is a normalizing transformation for some kind of data set (binomial/n, I think), but neither she nor I can find a reference on it. Any ideas? --TeaDrinker 00:30, 20 January 2007 (UTC)

Google gave an immediate reference: http://www.tina-vision.net/tina-knoppix/tina-memo/2002-007.pdf

The point seems to be that if the data are binomially-distributed, the transformation gives a variance independent of the mean. I question, however, the arrogance of the reviewer in saying what should be done, without explanation or even considering that another approach could be valid.86.132.163.126 12:25, 27 January 2007 (UTC)

[edit] AVI files that can't fast foward

I've noticed that certain video files me and my friends have encountered cannot be fast forwarded, you can't skip to any other point in the clip. The only thing you can do is just play it forward, and to get to a certain point you have to sit and watch everything preceding it. Is there any way that this can be avoided? I try taking it into Windows Movie Maker and it's unable to import. What's going on here? NIRVANA2764 21:36, 31 January 2007 (UTC)

These files are lacking indexes, which tell a program which position in the file corresponds to what time in the movie. virtualdub may be able to help. Droud 01:44, 1 February 2007 (UTC)

Try it in VLC. I haven't heard of problems playing avi files, but if there are still problems seeking, then use ffmpeg to re-encode the file's container. This guide explains fixing flv files using ffmpeg - it is the same procedure for fixing (broken) avi to (fixed) avi. --h2g2bob 02:06, 1 February 2007 (UTC)

Indexes? That would be a huge waste of storage space and computation. Think about it: a 32-bit pointer for each frame. --wj32 ^{talk | contribs} 06:30, 1 February 2007 (UTC)

Even with the 64b timestamps and 64b pointers, it still comes out to less than 4MB to index every frame in a 2 hour movie. In any case, most AVIs can only be played from keyframes, so the index would only contain those. Droud 12:48, 1 February 2007 (UTC)

[edit] CD28 gif 3d rotating structure

I really enjoyed the rotating 3d image of CD28 on your CD28 page.My question is simply that I would like to know if I could use your .gif format software (code) to portray the 3d structure of another protein molecule for which I have the .pdb file (3d coordinates)on my website?Can you please help me with how to do this if it is indeed legal?Thanks again for another terrific Wikipedia page, as always!

Don Kaiser <Rm email to reduce spam>

• If you click on the image, you will find yourself on a page that indicates who created it...ask him how he did it. I've used the free Jmol program to make 3D molecular models from PDB files. I think it can export animated gif images, but not sure. DMacks 20:48, 2 February 2007 (UTC)

[edit] Solving cubic equations

In school, we learnt how to solve cubic equations of the form ax³ + bx² + cx + d.

To factorise it into (x+r)(x+s)(x+t), we have to find a value for r by trial and error (with the factor theorem), then solve a quadratic equation to get the other two roots, s and t.

In the exams, the value of r will usually be 3 or less, so finding r by trial and error will not take a long time.

However, if r is large, finding r by trial and error will take a long time. Are there any faster, more systematic ways of finding r, besides trial and error? —The preceding unsigned comment was added by 218.186.9.3 (talk • contribs) 09:18, 2007 February 13.

(Please sign your posts with four tildes, "~~~~".) The roots of a cubic polynomial can be expressed in closed form, so in theory no searching is required. However, the technique is more difficult than using the quadratic formula. When the coefficients, a, b, c, and d, are real numbers (with a nonzero), the cubic is guaranteed to have a real root; but it may have only one. For example,

has only one real root.

But we can say much more. The coefficients are normalized so that the x³ term has coefficient 1. If we list them all, (1,−3,2,−6), the −6 has the largest absolute value. The Cauchy bound says that all real roots must lie between −m and +m, where m is one plus the maximum absolute value. In this example, the roots will be between −7 and 7.

We can do better. The positive coefficients before the −6 are 1 and 2, summing to 3; and before the −3 we have just 1. For an upper bound we can choose the maximum of ⁶⁄₃ and ³⁄₁, which will be 3, and again add 1. That is, the maximum real root is 4. Negating every other term (so we are evaluating the polynomial with −x), we can similarly find a lower bound, which here will be −1.

There is also a method known as Descartes' rule of signs that may help us restrict the number of positive or negative real roots. In this example it is no help with the positive roots (we will have three or one), but it tells us we cannot have a negative root.

This example polynomial has integer coefficients, and may have integer roots. If r is such a root, then it must divide the constant term, so the only possibilities here are 1, 2, 3. We have used the known bounds; and we know that 0 cannot be a root unless the constant term is 0.

We also have ways to bracket roots. If the polynomial evaluates to positive for x = a and negative for x = b, then there must be at least one root between a and b. Using a Sturm sequence, we can learn even more.

This does not exhaust our inventory of tools, but perhaps this is enough for now. Computer programs for solving cubics typically use closed form solutions. Above degree four, however, a famous theorem states that we have no closed form solutions. --KSmrq^T 11:48, 13 February 2007 (UTC)

You might want to read cubic equation. – b_jonas 14:17, 13 February 2007 (UTC)

[edit] Best calculus textbook

I've heard very often of Tom Apostle's Calculus I and Calculus II written in the 40s and 50s, which are very proof-heavy, well, entirely proof-based. What are your opinions on the best calculus textbooks? [Mαc Δαvιs] X (How's my driving?) ❖ 08:28, 19 February 2007 (UTC)

I can also recomend Leithold's book. Specially for beginners it is a good choice. Mr.K. (talk) 13:48, 19 February 2007 (UTC)

More information about your expectations would help. Do you want a graphic-rich easy introduction? Well-grounded proofs? Higher mathematics insight? Emphasis on engineering/physics applications? A modern revision? One of the non-standard analysis presentations?

Most texts I've seen do not impress me. Richard Courant and Fritz John do. Try a variety of books online, including those listed by Stef and at AMS. --KSmrq^T 04:27, 20 February 2007 (UTC)

[edit] Balancing a bicycle

Why is it (relatively)easy to 'balance' on a moving bicycle and the same, impossible on a standing cycle?

We all know this.... i just want to know the physics behind this everyday action

How can we easily balance ourselves (stay upright) on a moving bicycle (or bike) but the same thing is impossible when a try on a stationary two-wheeler? _________________

i did some re-searching and it looks like its not entirely 'angular momentum' as in the case of a top and also not entirely the gyro effect either... http://en.wikipedia.org/wiki/Bicycle_and_motorcycle_dynamics

all the facts are here... i can almost see the answer forming but not quite there... help me out - somebody pls put this phenomenon in nice understandable sentences

Thanks a lot

Seethahere 21:21, 22 February 2007 (UTC)

Contrary to a common belief, angular momentum (the gyroscope effect) has (almost) nothing to do with it. The main point is that you have an effective means of controlling your balance when moving, but not when standing still. It works like this: suppose your bike is starting to lean over slightly to the left (i.e. your center of mass is to the left of the line between the wheels' contact points with the ground, which is the axis you would pivot around if you were to fall over). Thus you are (your center of mass is) starting to fall over towards the left. When you notice this, your learned bicycling skill makes you make a nearly unconscious correction, by turning the handle bar slightly to the left. This causes your front wheel to move to the left under you, until the line between the wheels' contact points is once again vertically under your center of mass. In other words, you have regained balance. This correction doesn't work if you are standing still, because turning the handlebar doesn't then cause the front wheel to move sideways. --mglg_(talk) 21:36, 22 February 2007 (UTC)

Ok, that is one of the most complex articles ever, for something so simple. Bikes have self-stability. That means if you just hold a bike still, you can make it steer left or right by leaning it. You can ride a bike without hands, and steer just by leaning. It's a bit like a skateboard. --Zeizmic 23:27, 22 February 2007 (UTC)

Sorry I don't believe it, a well "ghostied" bicycle will go forward without a rider for a long time without falling over, which obviously has nothing to do with slight, almost unconscious steering inputs from the rider. I've ridden both bicycles and motorcycles a lot in the past and definitely there is more to it then 'corrections' made by the rider especially on a motorcycle that's going at speed, so much so in fact that when you TRY to turn the motorcycle and then release the pressure the motorbike will return on its own to travelling in a straight line. Can you source your assumption of gyroscope effects not having much to do with it? Vespine 23:48, 22 February 2007 (UTC)

http://www2.eng.cam.ac.uk/~hemh/gyrobike.htm Skittle 10:43, 23 February 2007 (UTC)

Ah, ah. No complaining, unless you made a pretext of reading that long, long, article! :) (which covers this particular turning thing somewhere..) --Zeizmic 01:01, 23 February 2007 (UTC)

The principle is correct, but there's no unconscious inputs involved - the stability comes from the bike turning itself. The horizontal "trail" (as pictured) is the key. Explaining the details of it is beyond me, but the gist of it is that when the bike starts to lean, the trail causes the wheel to steer into the turn, straightening it up and preventing the fall. It only works above a certain minimum speed, but it's why you can let go of the handlebars and not fall over. There's no special skill involved; the bike is genuinely driving itself.

Apparently a lot of time and money goes into getting the trail right, to ensure stability. I saw a really good explanation of this very recently, but I can't for the life of me remember where. If it springs to mind, I'll post the link. Spiral Wave 01:08, 23 February 2007 (UTC)

Might it have been in New Scientist, in the "The Last Word" section? I've linked to a bit of it, although I don't know how well this will work for those without a subscription. It also provides an answer for those who don't believe how little effect the gyroscopic effect has: a bike designed to cancel out all gyroscopic effects which people could ride with ease. Skittle 10:42, 23 February 2007 (UTC)

Turns out there is also a (very) brief mention of the effect in the trail article linked in the caption. I do remember that it's why bikes are built with trail. If the steering column was vertical, the result would be highly unstable. Spiral Wave 01:31, 23 February 2007 (UTC)

You can see from the diagram that if the bike is stationary - and leans to one side, there is a gravitational force acting through the frame of the bike and a ground reaction from the point where the front wheel touches the ground. This causes a rotational moment through the steering tube that causes the front wheel to try to turn 'into' the direction the bike is leaning. If you stand next to a bicycle and just lean it to one side, you can see that happening. But when you are moving rapidly, you have forward momentum and it takes a significant amount of force to cause the bike to change direction. The reaction to that force is pushing the wheel back into a straight line - so there is a net upward force countering gravity and making the bike stand up straight. It's hard to explain without a white-board and some force vector diagrams! SteveBaker 13:50, 23 February 2007 (UTC)

All of this is true, but please realize that the self-correction induced by the rake is only there to help you make roughly the right correction. The rider still does the fine tweaking to perfect the balance (and yes, this tweaking can be done either without hands, by leaning the frame so that the rake induces the front wheel to turn, or more simply by turning the handlebar by hand). The bike is NOT fully self-balanced, as you can establish by letting your rider-less bike roll down a hill by itself and seeing what happens (I don't recommend that test with an expensive bike...). --mglg_(talk) 18:31, 23 February 2007 (UTC)