Wikipedia:Reference desk/Archives/Mathematics/March 2006

= March 1 =

Third order covariance matrix
Can anyone give me a clear definition of a third order covariance matrix? The only thing I can find on the net is at mathworld but it is not clear to me what is meant by the superscript notation mn --Will


 * The superscript on the letter V is just for notational convenience. The m and n on the right side are exponents though. Whenever m + n = 3 (and presumably m and n are positive) you have your "third order" covariance matrix. From this definition, there are 4 different "third order" matrices found by setting (m,n) = (0,3),(1,2),(2,1),(3,0)...in some books, only (1,2) and (2,1) are considered as "third order covariance matrices". --HappyCamper 19:08, 1 March 2006 (UTC)

vector planes?
= March 2 =

how do i make a combination of eighteen numbers in groups of six

 * I'm afraid you'll have to suitly emphazi your question. —Keenan Pepper 03:24, 2 March 2006 (UTC)
 * I'll make a wild guess and assume you meant "How many possibilities are there to select 6 elements of a set of 18 elements, without repetition and without order being important". In that case, the answer is
 * $${18 \choose 6} = \frac{18!}{6! 12!} = 18564$$
 * -- Meni Rosenfeld (talk) 07:29, 2 March 2006 (UTC)

convert 2'-9" in inches

 * Can you really not do this in your head? If not, there's always Google Calculator. —Keenan Pepper 05:48, 2 March 2006 (UTC)


 * In case it's the notation that's giving you problems, see our article prime (symbol). In the context of length, ′ denotes a foot, ′′ an inch. ×Meegs 06:01, 2 March 2006 (UTC)


 * The notation is certainly giving me problems. Does the questioner want two feet plus nine inches, or two feet minus nine inches?  —Blotwell 13:02, 2 March 2006 (UTC)
 * Or perhaps cm to in? DuctapeDaredevil 20:36, 6 March 2006 (UTC)

Wikipedia growth prediction question (third-degree polynomial equation)
If I know that Wikipedia reached 500,000 articles on March 17 2005, 666,666 articles on August 4 2005, and 1,000,000 articles on March 2, 2006, and I know that Wikipedia "currently" has x articles, then how can I make this into a third-degree polynomial approximating Wikipedia's growth and solve it to get the time Wikipedia will reach a given milestone? J I P | Talk 08:17, 2 March 2006 (UTC)
 * Okay, first you need to convert the dates to explicit amount of days. I'll leave it to you to work that out, but let's assume you get March 2005 = 90 days, August 2005 = 270 days, March 2006 = 450 days. Now you write:
 * $$f(x) = \frac{500000(x-270)(x-450)}{(90-270)(90-450)} + \frac{666666(x-90)(x-450)}{(270-90)(270-450)} + \frac{1000000(x-90)(x-270)}{(450-90)(450-270)}$$

Simplify, and solve using Newton's method. -- Meni Rosenfeld (talk) 08:31, 2 March 2006 (UTC)

Wait, you got me mixed up. With 3 points of data you only get a quadratic polynomial. So you can solve it directly. -- Meni Rosenfeld (talk) 08:34, 2 March 2006 (UTC)

Or perhaps you meant there's an additional point, the current time x0 and the current articles y0? Stick it to the formula and you get
 * $$f(x) = \frac{500000(x-270)(x-450)(x-x_0)}{(90-270)(90-450)(90-x_0)} + \frac{666666(x-90)(x-450)(x-x_0)}{(270-90)(270-450)(270-x_0)} + \frac{1000000(x-90)(x-270)(x-x_0)}{(450-90)(450-270)(450-x_0)} + \frac{y_0(x-90)(x-270)(x-450)}{(x_0-90)(x_0-270)(x_0-450)}$$

And see also Lagrange polynomial. -- Meni Rosenfeld (talk) 08:40, 2 March 2006 (UTC)

This isn't my best day. You can also solve a third-degree polynomial directly, but I guess Newton's method is simpler. -- Meni Rosenfeld (talk) 08:42, 2 March 2006 (UTC)


 * Beware, though, that an exactly-fit polynomial will probably not make such an accurate a model of WP's growth, and could actually be astonishingly bad. Depending on what that fourth data point is, the cubic could easily predict negative growth in the future (as well absurd things back between your data points).  There is some discussion about modeling the growth exponentially at Modelling Wikipedia's growth that might interest you.  ×Meegs 09:11, 2 March 2006 (UTC)

Fourier transform of a positive semidefinite function
Under which additional conditions (on the function f(x)) is the Fourier transform of a real, positive semidefinite function f(x), defined for the whole real axis, again real and positive semidefinite? --CA

(Please sign your questions.) The Fourier transform is a change of basis. Contemplate the definition of positive semidefinite. --KSmrqT 13:06, 2 March 2006 (UTC)

(I have signed and made slightly more specific the question.) I do not understand your answer. If you want to insinuate that the Fourier transform F(k) of every real, positive semidefinite function f(x) is again real and positive semidefinite, then this is not correct. First of all, f(x) has to obey f(x) = f(-x) in order to have F(k) real. Within this class of functions, a simple counterexample is the symmetric, rectangular pulse. Indeed, if f(x) is  f(x) = 1    for   -a <= x <= a   f(x) = 0    for   |x| > a (here a is a positive, real constant and <= means "less or equal") then its Fourier transform F(k) is (up to a factor) F(k) = sin(ak)/k which certainly is not positive semidefinite. --CA

Area bounded by a hyperbola
I need to prove that the area bounded by a unit hyperbola and a line from the origin to the hyperbola is equal to twice the angle of that line.



I know that the point where the line and the hyperbola meet is (cosht, sinht). Using that, I was able to find the equation of the line: y = xtanht. Im running into trouble with the integral. I think I did it right. I wish I could show it better, but I can't get the syntax of the math language to work: I figured I have to integrate along the y axis, from sinht to 0, because sinht is the y value of the intercection. Assuming I did it right, I got an integral of sqrt(y^2+1) - y/tanht. Please let me know if I am on the right track, and if you can help me out with the math formatting, that would help too. Thanks a lot. --Chris 17:07, 2 March 2006 (UTC)
 * Looks okay to me. Now all you need is to look at List of integrals of irrational functions to solve the integral
 * $$\int_0^{sinh(t)} \left(\sqrt{y^2+1}-\frac{y}{tanh(t)}\right)dy$$
 * Btw, an efficient way to do math formatting is to seek an equation that uses the symbols you need, copying the code and altering it to match your expression. -- Meni Rosenfeld (talk) 18:05, 2 March 2006 (UTC)
 * Hey, thanks for helping out. Just one question though: is it supposed to be the difference of the two integrals, not the sum?
 * You're right, I fixed it. -- Meni Rosenfeld (talk) 19:10, 2 March 2006 (UTC)

Riemann hypothesis, who cares?
I've been told that the Riemann hypothesis has many broad implications, if only it were proven, but what are those implications? And why should I, a person who has an interest in math but doesn't spend all day looking at numbers (though I read quite a bit about philosophy, science and computers) care about it?

(Also, couldn't we construct a statement along the lines of "X is true iff the Riemann hypothesis is true." And then see if X is true or not, and use that information to either prove or disprove the Riemann hypothesis?) -86.138.233.25 20:09, 2 March 2006 (UTC)
 * I'm definitely no expert about the Riemann hypothesis, but what I can say is this: If you haven't already, you can read about several of the implications of it at Riemann hypothesis. If none of what is written there is of particular interest to you, than I guess you really shouldn't care about it.
 * About your second question: Of course we can, there are many statements that are equivalent to the Riemann hypothesis. But these are just as difficult to prove\disprove as the hypothesis itself. It is possible indeed that if\when the hypothesis is proved\disproved, it will be using an alternative statement and not with the standard formulation. -- Meni Rosenfeld (talk) 20:18, 2 March 2006 (UTC)
 * I suggest you to read the proof of Prime number theorem which involves complex analysis. While it doesn't require Riemann hypothesis, it shows why the roots of zeta-function are important. If Riemann's hypothesis were true, prime number theorem would be MUCH easier to prove.  Grue   20:47, 2 March 2006 (UTC)
 * If the proof results in a clarification of the Hilbert-Pólya conjecture, then it will have wide-ranging effects on quantum chaos (the quantum mechanics of chaotic systems) and as a side-effect will spill over to string theory. (Curiously, the nuclear physics people care as well, as the spectra of excited nuclei are "random", and are modelled by the Gaussian unitary ensemble, which also models the zeros of the zeta function.) linas 05:36, 6 March 2006 (UTC)

= March 3 =

= March 4 =

Nabla
Suppose you were to define an arbitrary operator, Û, in 3-space, using cartesian coordinates, as
 * $$\hat{U} = lim_{n \to \infty} \nabla ^n$$

Would it reduce all linear functions it was applied to, to 0?--64.12.116.11 01:43, 4 March 2006 (UTC)


 * Well, of course it would reduce all linear functions to zero, because the gradient of a linear function is zero, and so are all the elements of the sequence after that. —Keenan Pepper 21:28, 3 March 2006 (UTC)


 * This is sort of interesting...what inspired you to ask this question? --HappyCamper 04:54, 4 March 2006 (UTC)


 * I believe it would reduce all polynomial functions to zero as well. I don't think it would be well defined on a transcendental, e.g. trigonometric function though. moink 11:33, 4 March 2006 (UTC)


 * You can't take iterated gradients, so I assume we're taking iterated Laplacians. In that case, U is either zero or infinity on various Fourier modes, and the dividing line depends pretty arbitrarily on the units you use. I guess it's best to say that the limit doesn't make sense, since every time you take the Laplacian you get two inverse powers of the spacial dimension, so the terms of the sequence don't naturally belong to the same function space anyway. Melchoir 23:24, 5 March 2006 (UTC)

Double Contour Integral
What command should I use in TeX to get double contour integral?
 * Is this for on- or off-Wikipedia use? For off-Wikipedia use, check out The Comprehensive LaTeX Symbol List (PDF): both the "mathabx" and the "wasysym" package provide "\oiint" for this. On Wikipedia, I don't know if there is a way to get this symbol. Kusma (討論) 17:22, 4 March 2006 (UTC)


 * The Unicode symbol is " ∯ " (U0222F, [SURFACE INTEGRAL] ), which should work with MathML and perhaps eventually BlahTeX. You probably already know to check Help:Formula for wiki symbols; however, it shows nothing for this one. Nor do we have a comprehensive table or other description of what our pseudo-TeX accepts. --KSmrqT 18:58, 4 March 2006 (UTC)


 * I don't have those packages. I can download the .zip file from CTAN but I don't know how to proceed with installation. How can I add the mathabx package?


 * Hello?... Could someone please respond to this? How do you set up mathabx, given you only have the .sty files, etc. but have received the them in a .zip file as opposed to a .cab file.


 * It can be done by typing \mathop{{\int\!\!\!\!\!\int}\mkern-21mu\bigcirc}{} but it is not really elegant. Rudolf Saathof (talk) 15:12, 4 August 2009 (UTC)

= March 5 =

= March 6 =

Napier and Briggs' imaginary math-world
So, I'm looking up information on logarithms, and it says that Briggs and Napier decided that the log of 10 shuld be one, and the log of 1 should be zero, and that at first they thought log 10 should be 10^10 to avoid fractions. Was there some mathematical basis for this that someone would be willing to explain to me, or were they just making it all up for fun? DuctapeDaredevil 17:06, 6 March 2006 (UTC)
 * Well, I think they were definitely not making it up for fun. I'm not sure where you read that information - I didn't find it in the logarithm article, the closest I could find was that in Napier's system log 10^7 = 0. Napier used a system which is somewhat different than modern logarithms, but essentially the difference is in the naming of things and not in the underlying logic. So there is nothing "imaginary" about his system. The point is that he probably defined his system in a way which he believed would be the most efficient or convenient for calculations and\or tables. Only later systems which are mathematically less arbitary were created. -- Meni Rosenfeld (talk) 18:14, 6 March 2006 (UTC)
 * I didn't say it was imaginary, that was a joke based on the fact that I couldn't find any mathematical basis for their decision that log 1=0 and log 10=1. (Though I apologize, jokes don't always work over teh Internets.) It seems like they were just 'poking' numbers, and that they weren't actually basing it off anything. It says that Napier "...proposed a table using log 1=0 and log 10=10^10 (to avoid fractions).  The two men finally concluded that the logarithm of one should be zero and that the logarithm of ten should be one."  I'm looking at A History of Mathematics by Carl Boyer.  And the logs in this table are modern logarithms.  They're 'Briggisan', or common logs, so it's not something that was made up later.  DuctapeDaredevil 19:54, 6 March 2006 (UTC)
 * I figured the tone was humorous, but still appeared to me very critical of their definitions. Anyway, what you quoted sounds peculiar - If anything, I'd expect log (10^10) to be 10, and not the other way around. Nor do I see what does this have to do with fractions. And you say you don't find a mathematical basis for log 1=0 and log 10=1, but I don't see any problem with that - That is necessary in base-10 logarithms; And since we, as it happens, use a decimal number system, it is very natural to use such logarithms. So basically, I don't really understand what is it that you are trying to ask. In any case, whatever definitions they chose, they almost certainly did it for some kind of practical purpose. -- Meni Rosenfeld (talk) 20:11, 6 March 2006 (UTC)
 * The log 10^10 being 0 was the definition they decided against, leading me to believe that they were just making it up as they went along. I'm not saying that they were just playing a joke, just that it looked more like a decision than something they proved mathematically.  So to pose my question more simply:  Is there any mathematical proof that this has to be true, other than the fact that they said it? How did they determine that log 10 would be 1?  DuctapeDaredevil 20:31, 6 March 2006 (UTC)
 * And by fractions, they meant decimals. They didn't want to be endlessly working with point something - they wanted whole numbers as often as possible. DuctapeDaredevil 20:34, 6 March 2006 (UTC)
 * I'm still not sure I understand. Did you mean "why did they decide to use base-10 logarithms and not some other base"? (After all, log 10=1 follows immediately from the definition of such logarithms). In that case, the reason is that we are using a decimal number system, so this is the most convenient for calculations - If you have a table that says log 3.69754 = 0.567913, you can tell immediately that, for example, log 3697540 = 6.567913 without the need for an additional entry. That's what makes calculations with logarithm tables so simple. -- Meni Rosenfeld (talk) 10:30, 7 March 2006 (UTC)
 * Nevermind, I got my answer. Thank you for your help, though.  DuctapeDaredevil 19:52, 7 March 2006 (UTC)

Using the + sign in Arab countries
Hello, I have already seen in the article about plus and minus signs that the Jewish usually do not use the plus sign. However, they use an inverted T. I am interested if Arab countries use the usual symbol or another (as + may resemble a Christian cross). Thank you. --84.21.200.224 18:38, 6 March 2006 (UTC)
 * A correction is in order here: Jews, in Israel at least, probably use the symbol you have mentioned only in elementary schools. I don't think it ever appears in any serious work. This makes me doubt that arabic countries do use any non-standard symbol, but can't say for sure. -- Meni Rosenfeld (talk) 18:57, 6 March 2006 (UTC)

= March 7 =

Std Dev on the fly
I am calculating average values in a database of a few million records. It takes a very long time to scan each record and decide if it should be included in the running average. So, I'd like to do standard deviation on the fly rather than go through all the records once to get the mean and then go through them all to get the standard deviation. Is that possible? --Kainaw (talk) 16:00, 7 March 2006 (UTC)


 * AFAIK, standard deviation is defined in terms of the mean, so I don't see how you could calculate the former without the latter. What about calculating the std dev of a random sample of the population? --LarryMac 16:43, 7 March 2006 (UTC)


 * You might at least read the standard deviation article before asking. When working with floating point numbers, do take seriously the caution there about increased roundoff error when using the on-the-fly calculation, especially when the standard deviation is large. --KSmrqT 16:49, 7 March 2006 (UTC)

If there are only a few million of them, why not make one pass in which you decide whether a record should be included and cache the value on which you want the statistics? Then make a second pass through those values and get your s.d. It should all fit in physical memory so it will be very fast, and you won't have to worry about the roundoff-error problem. --Trovatore 18:23, 7 March 2006 (UTC)


 * I've done database query programming, with efficiency in mind, before. You should try to formulate a query which includes your criteria and does the average for you, something like:

SELECT ave(SALARY) FROM EMPLOYEE_TABLE WHERE STATUS != "TERMINATED" AND SALARY >          0.00 AND SALARY < 99999999999.99;


 * I am assuming you have something like SQL to do queries. You might even have a function which does standard deviations, in which case the query would look like this:

SELECT ave(SALARY),stdev(SALARY) FROM EMPLOYEE_TABLE WHERE STATUS != "TERMINATED" AND SALARY >          0.00 AND SALARY < 99999999999.99;


 * In short, anything that can be done through a query is bound to be much faster than something you do manually or with a program, as they have spend thousands of man-years perfecting query methods. StuRat 21:39, 13 March 2006 (UTC)

= March 8 =

CDF of Log-normal distribution
This question shouldn't be too hard, which is good, since I need this done tonight. I have managed to be pretty ignorant in probability to this point in my life, which is very sad. I should be able to get this from the article, but it's very technical and I'm having trouble reading it.

How do I get the cumulative distribution function for the log-normal distribution? Let's say I have a silly program which will give me the CDF of the normal distribution, but not the lognormal. I know the values $$\mu$$, which isn't a mean but kinda looks like one, and $$\sigma$$.

So if z is a random variable with the lognormal distribution with that $$\mu$$ and $$\sigma$$, what exactly is the probability that $$z<X$$ where X is some number I also have? moink 10:12, 8 March 2006 (UTC)


 * Wait, I think I figured it out. $$\mu$$ is the mean of the natural log of z, and $$\sigma$$ is its standard deviation.  So if I just plug in ln(X) into the value for the normal distribution, with $$\mu$$ and $$\sigma$$ as mean and stddev, then I get the right answer?  Could someone just confirm this, cuz I'm pretty unsure of myself here.  moink 10:19, 8 March 2006 (UTC)


 * I'm not an expert either, but that sounds right. Fortunately for both of us, the formulas are in the boxes on the right-hand sides of Normal distribution and Log-normal distribution! Melchoir 10:26, 8 March 2006 (UTC)
 * [via edit conflict]According to log-normal distribution, the cdf of this distribution, which I shall call cdflog, is:
 * $$cdflog(x)=\frac{1}{2}+\frac{1}{2} erf\left[\frac{\ln(x)-\mu}{\sigma\sqrt{2}}\right]$$
 * The error function is a scaled and translated version of the cdf of the standard normal distribution:
 * $$erf(x) = 2cdfnorm(x\sqrt{2})-1$$
 * So what you have is:
 * $$cdflog(x)=\frac{1}{2}+\frac{1}{2} (2cdfnorm\left[\frac{\ln(x)-\mu}{\sigma\sqrt{2}}\sqrt{2}\right]-1)=cdfnorm\left[\frac{\ln(x)-\mu}{\sigma}\right]$$
 * Another way to obtain the same result: If z is distributed Log-normally, then ln(z) is distributed normally, so:
 * $$P(z<X) = P(\ln(z)<\ln(X)) = cdfnorm_{\mu,\sigma}(\ln(x)) = cdfnorm\left[\frac{\ln(x)-\mu}{\sigma}\right]$$
 * -- Meni Rosenfeld (talk) 10:30, 8 March 2006 (UTC)


 * Thank you both for being so quick! I hadn't even noticed the box on the RHS of the article.  Anyway, that last line of Meni Rosenfeld's is what I decided was true, so I'm very glad to have it confirmed for me.  moink 10:44, 8 March 2006 (UTC)

exponential distance or linear distance
hi, please i would like to know the difference between exponential distance and linear distance? i would also like to know if the impact of distance on the visibility of an object is either exponential or linear...thanks


 * In what context did you see those phrases? I suppose it might mean something like this: The linear distance between 10 and 100 is 100 − 10 = 90, but the exponential distance between them is 100/10 = 10 = 1 order of magnitude.


 * I think the impact of distance on visibility would be linear, because the angle an object makes on your visual sphere is tan(s/d), where s is the size of the object and d is its distance from you, and by small angle approximation tan(s/d) is about s/d. In other words, if an object is twice as far away from you it appears about twice as small. —Keenan Pepper 15:33, 8 March 2006 (UTC)


 * What about a point source of light, for which a star is a good approximation? The light spreads spherically… --KSmrqT 15:53, 8 March 2006 (UTC)


 * You can also take into account the fact that some of the light is absorbed in the atmopshere, and the amount of light that manages to pass through a certain distance is exponential in that distance. Over long distances, this dominates the polynomial issues discussed by Keenan Pepper and KSmrq, so ultimately I believe it's safe to say that visibility is exponential. -- Meni Rosenfeld (talk) 15:58, 8 March 2006 (UTC)


 * Agreed. Specifically, visibility thru a vacuum decreases in proportion to the square of the distance between the object and observer.  In the real world, particles in the intervening distance which can absorb light will cause visibility to decrease at an even faster rate. StuRat 21:27, 13 March 2006 (UTC)


 * If you're nearsighted, visibility decreases considerably faster than that. Black Carrot 20:43, 14 March 2006 (UTC)

Math Help
Three cake boxes are stacked. Find the surface area of the entire stack if all the prism bases are squares. Top Box: L=12 W=12  H=6 Middle Box: L=26 W=26  H=6 Bottom Box: L=36 W=36  H=6




 * Which part of the problem is giving you trouble? —Keenan Pepper 15:12, 8 March 2006 (UTC)


 * To the original questioner: As a checking of concepts ask these yourself: 1)Do you know how to calculate surface area of prism? 2)Do you know the area for which the stacks are touching each others? If you know both question's answer, I see no reason you can't do this question. On the other hand, you may search the formula for surface area of prism using the searchbox on the left if you need to. --Lemontea 14:26, 9 March 2006 (UTC)

It seems that some assumptions must be made. These would be my assumptions:

1) The area each box has in contact with another box is not counted as surface area for either box.

2) The area the bottom box has in contact with the table is not counted.

3) The boxes are stacked such that the largest possible area is in contact with the table or box underneath it. That is, none are lying on their sides or hanging over the edge of the table or box underneath. StuRat 21:21, 13 March 2006 (UTC)

We do not need to know the table area.

Division by zero
How did 'Renee' prove that '1=0'? ( If she ever existed and what I read isn't just part of a novel).--Cosmic girl 19:20, 8 March 2006 (UTC)
 * See division by zero. —Ilmari Karonen (talk) 19:29, 8 March 2006 (UTC)

thanks :D.--Cosmic girl 20:54, 8 March 2006 (UTC)


 * You might like the proof that horses have an infinite number of legs:


 * "In the rear, horses have two legs, while in the front, they have forelegs. That makes six legs total.  Six is even and also certainly an odd number of legs for a horse to have.  The only number which is both odd and even is infinity.  Therefore, horses must have an infinite number of legs." StuRat 21:12, 13 March 2006 (UTC)

do my homework
do my homework--Gefploxer 21:56, 8 March 2006 (UTC)
 * &mdash; Arthur Rubin | (talk) 22:10, 8 March 2006 (UTC)
 * Actually, I'm feeling particularly generous today, so I've done your homework. Here are the answers:
 * 1. False
 * 2. Naproxen sodium
 * III. Roman numerals
 * 4. All of the above, except for years divisible by 400
 * -- E lkman - (talk) 23:23, 8 March 2006 (UTC)

infinity/infinity
What is infinity divided by infinity? &mdash;The preceding unsigned comment was added by Biofireball (talk • contribs).


 * There is no general way to give a useful meaning to this expression, see Indeterminate form. Kusma (討論) 22:35, 8 March 2006 (UTC)


 * Division and multiplication require numbers. Infinity is not a number but more like a philosophical concept. JackofOz 22:56, 8 March 2006 (UTC)


 * I disagree with that. For example, infinity times infinity is definitely infinity, which is to say that if sequence an increases without bound and sequence bn increases without bound, then the sequence whose terms are the products anbn also increases without bound. On the other hand, infinity divided by infinity is truly indeterminate, because the behavior is different for different pairs of sequences. —Keenan Pepper 23:17, 8 March 2006 (UTC)


 * I knew we'd have this sort of debate. Last time I studied maths, there were a multitude of infinities (aleph null, alpha 1 etc), some infinitely larger than others.  So if you're going to be (hypothetically) multiplying them together, you have to specify which ones you're talking about because the result will vary, rather a lot actually.  But you can't do that, not even hypothetically.  Infinity times infinity is an absurd concept because it depends on the existence of a bounded number, which is the very thing that infinity is not.  The whole basis of the 6 standard mathematical operations depends on the elements being known or knowable quantities.  Infinity does not fit this bill, although mathematicians can pretend to manipulate it using mathematical tools.  That's why I said it's "more like" a philosophical concept. JackofOz 00:33, 9 March 2006 (UTC)


 * It's not really a philosophical concept. In the extended reals, infinity times infinity = infinity, and infinity plus infinity = infinity.  In the projective reals or complex numbers, infinity times infinity = infinity, but infinity plus infinity is undetermined.  infinity / infinity is undetermined in all those systems. &mdash; Arthur Rubin | (talk) 01:04, 9 March 2006 (UTC)


 * The IEEE floating-point standard includes signed infinities, and specifies precisely what the result of such a division must be. Fun to know, but that's only one choice among many. Absent further context, no general answer is possible. --KSmrqT 00:12, 9 March 2006 (UTC)

Isn't infinity/infinity = 0 or 1?.--Cosmic girl 23:55, 8 March 2006 (UTC)


 * No: $$\lim_{n\to\infty}\frac{n^2}{n}=\infty$$ and $$\lim_{n\to\infty}\frac{\pi n}n=\pi$$, to name just two trivial examples. Kusma (討論) 00:09, 9 March 2006 (UTC)


 * Perhaps a few points should be emphasized:


 * The value of things like infinity/infinity depends on the context and the definitions we use, but given certain definitions, it can be derived explicitly from them. The same applies to everyday numbers, too - 1/2 is indeterminate in integers, but equal to 0.5 in reals. So it really can't be said that it's more philosophical.


 * The Aleph's JackofOz mentioned are just one kind of types of infinities, called (transfinite) cardinal numbers. There are other classes, such as ordinals, and those appearing in non-standard analaysis.


 * In all the structures mentioned, infinity*infinity=infinity, but that is not always true; In ordinals, for example, if by "infinity" we mean &omega;, then &omega;*&omega;=&omega;2, which is different than &omega;.


 * -- Meni Rosenfeld (talk) 11:29, 9 March 2006 (UTC)


 * Aren't indeterminate forms with limits a special case though, when L'Hôpital's rule can be applied and reduce it to a determinate form? --BluePlatypus 14:17, 9 March 2006 (UTC)


 * I guess you could call it that way. If you're calculating the limit of the ratio of specific functions, each of which approaches infinity, you can use L'Hopital's to do that. But the point is that if you write just infinity/infinity, you don't know which functions are used, so you can't give a specific result. -- Meni Rosenfeld (talk) 14:56, 9 March 2006 (UTC)


 * I see. Cancer is not a single disease but a collective term for a number of diseases, and the treatment will depend on which particular disease it is.  Talking about infinity/infinity is like talking about "the cure for cancer".  Is that a good analogy?  JackofOz 02:19, 13 March 2006 (UTC)


 * I think the current study of cancer is seriously flawed. They look at what "causes" it, when just about everything causes cancer (such as sunlight).  Instead, they should focus on why certain people's immune systems are unable to destroy the cancer before it spreads.  In my model, everyone has cancer all the time, in a few cells here and there, which are wiped out before we ever notice anything.  Only those with suppressed immune systems develop a noticeable disease. StuRat 21:05, 13 March 2006 (UTC)


 * In other words, yes. Except that one of them has already been discovered. And the other one kills people. Black Carrot 20:40, 14 March 2006 (UTC)

= March 9 =

Fair cake cutting problem
Long time ago I remember reading a problem and solution which I can no longer recall. Problem is to cut and distribute a completely homogeneous cake into n pieces among n people such that each person perceives that he received fair share of cake. Note that it is not important whether it is fair share or not, but all must believe that it is. For n=2, solution is one person cuts the cake and another choses his piece. Thus person cutting the cake will know that both pieces are equal since he cut it (and took atmost care on his behalf) and person picking will get to pick whichever he thinks is bigger share. I remember that solution exists for n=3 or more, but I don't know what. Can someone help me here? Again, cake is plain, no fancy icing, and all participants have no preference over piece except its size/volume. Thanks.


 * I saw this in a Chinese textbook. It wasn't cake, but the concept was the same.  For 3 people, one guy cuts it into what he thinks are 3 equal slices.  Then, the other 2 decide which slice is smallest and give it to him.  They take both remaining slices.  One of them cuts both slices in equal halves.  For each new pair, the third guy takes what he thinks is the largest.  It works in theory, but not in practice.  You will quickly end up making paper-thin slices of cake. --Kainaw (talk) 01:50, 9 March 2006 (UTC)


 * Fair division? Melchoir 02:06, 9 March 2006 (UTC)


 * In theory, there is no difference between theory and practice. But in practice, there is (Jan L.A. van de Snepscheut). JackofOz 03:11, 9 March 2006 (UTC)


 * Kainaw, in your solution what if there are differences between two remaining persons in deciding which piece is smallest (step 2)? Otherwise it is optimal. I am pretty much looking for theoritical solution so it's not a problem. Can someone think of n=4?

Actually this page has some information, including solution for n=3 which is different & much complex then one proposed above. And it says that n=4 or more is just too difficult. So I take back my question. By the way, this reference desk is great place for fun!


 * I don't really see the issue. As long as the person who cuts the cake gets the last piece left after the others have chosen, it's in their interest to cut it as evenly as possible so they won't be left with a smaller piece. StuRat 20:55, 13 March 2006 (UTC)


 * What if they're getting kickbacks from one of the other partygoers? Black Carrot 20:37, 14 March 2006 (UTC)


 * LOL, shall we call it "Cakegate" ? StuRat 01:41, 15 March 2006 (UTC)

= March 10 =

How does 1.54 come out?
Suppose X has a normal distribution with expected value 0 and variance 1. Let Y = -X if -C<=X<=C or Y=X if |X|>c, where c is a positive number to be specified below. If c is very small, then the correlation corr(X, Y) is near 1; if c is very large, then corr(X, Y) is near -1. Since the correlation is a continuous function of c, the intermediate value theorem implies there is some particular value of c that makes the correlation 0. That value is approximately 1.54. In that case, X and Y are uncorrelated, but they are clearly not independent, since X completely determines Y.
 * I guess all of those things are true, but it's not very clear to me what is the question here. -- Meni Rosenfeld (talk) 12:26, 10 March 2006 (UTC)
 * I think the questioner is asking for an exact expression for this value "1.54" and possibly for a proof of its correctness. —Blotwell 03:33, 11 March 2006 (UTC)
 * In that case, the correlation can be shown to be equal to:
 * $$\int_{-\infty}^{\infty}x y f(x) dx = -2\left(\int_0^c\frac{e^{-t^2/2}}{\sqrt{2\pi}}t^2dt - \int_c^{\infty}\frac{e^{-t^2/2}}{\sqrt{2\pi}}t^2dt\right) = 1-4\int_0^c\frac{e^{-t^2/2}}{\sqrt{2\pi}}t^2dt$$
 * So finding the value for which the correlation is 0 amounts to finding a positive value of c such that
 * $$\int_0^c\frac{e^{-t^2/2}}{\sqrt{2\pi}}t^2dt=\frac{1}{4}$$
 * This, as far as I know, can't be done explicitly with elementary functions, but it can be solved numerically to get
 * $$c \approx 1.53817225445505233$$
 * -- Meni Rosenfeld (talk) 08:36, 11 March 2006 (UTC)


 * The question could be, "How can X and Y be uncorrelated despite X completely determining Y?" Two reasons. The first is that correlation and causation are different things, and one doesn't neccessarily imply the other. The second is that correlation tests for exactly one thing: the strenth of a linear relationship between X and Y. Here, there is a strong positive linear relationship when C is small, a strong negative linear relationship when C is large, and no linear relationship when C is medium. This is only an effective test when you expect the relationship to be vaguely linear in the first place, and would be just as bad an idea were Y to be defined as X2. Black Carrot 20:34, 14 March 2006 (UTC)

sqrt(−1)
So, I've been told that the square root of -1 is i. I've looked at the square root article but it doesn't say anything about this. I understand sqrt(-1) has no integer number, so why i? Any help would be greatly appreciated :) -Benbread 20:08, 10 March 2006 (UTC)
 * See imaginary unit and complex number. ☢  Ҡ i∊ ff   ⌇  ↯  20:10, 10 March 2006 (UTC)


 * Note that although i is not an ordinary integer, it is a Gaussian integer. —Keenan Pepper 21:51, 10 March 2006 (UTC)
 * By the way, the square root article does have a section about complex numbers, but it assumes the reader is already familiar with the concept. Also, understanding where i comes from can be illustrated pictorially like this (this isn't how things really happened, but that's the general idea): Long ago, mathematicians didn't use negative numbers. One day they asked, "how come we can't calculate 1-2? This isn't right. Let's invent a new number, call it -1, and define it as -1 = 1 - 2." Later they asked, "how come we can't calculate sqrt(-1)? This isn't right. Let's invent a new number, call it i, and define it as i = sqrt(-1)". See the links given by Kieff for more details. -- Meni Rosenfeld (talk) 07:19, 11 March 2006 (UTC)
 * How come we can't calculate ln(0)? (Igny 21:53, 12 March 2006 (UTC))
 * Invent a new number, call it ɰ define it as ɰ = ln(0). -64.24.35.16 01:31, 13 March 2006 (UTC)
 * Well, $$\lim_{x \to 0} \log x = -\infty$$. Doesn't do you any good. This is not like i. You can't just invent a new number whenever you can't find a solution for something, because in most cases, the solution is just undefined (like division by zero, 0^0 and log(0)), and a new math doesn't really change anything. Oh yeah, while you're on this, check quaternion to see how you can expand complex numbers even more. ☢  Ҡ i∊ ff   ⌇  ↯  21:06, 13 March 2006 (UTC)
 * Can't say I agree with kieff here. First, you can just invent a new number whenever you can't find a solution for something - Whether you'll get useful results or not is the question. $$\lim_{(x,y) \to (0,0)}x^y$$ "almost always" converges to 1, so it makes sense to define 0^0=1, and it often is defined this way. Division by zero is defined in the real projective line (which is derived from the reals by "inventing" a new number $$\infty$$). Defining $$\log{0} = -\infty$$ in the extended real numbers doesn't raise any problems I'm aware of. In short, you can always define something, but you'll often have to make sacrifices. -- Meni Rosenfeld (talk) 20:18, 14 March 2006 (UTC)
 * On a railroad track, a locomotive can pull the train in one direction; call that positive. For quite some time positive numbers were comfortable. A locomotive can also push the train; call that negative. So now we have quantity (how far along the track) but also a sense of direction (pull or push). To obtain a number that squares to −1 we must free ourselves from the track and turn 90° (either right or left, it doesn't matter). Our number line (the track) becomes a number plane (complex numbers presented as an Argand diagram).
 * Another approach uses 2×2 matrices. For the real number r use the matrix rI,
 * $$r \mapsto r{\mathbf I} = \begin{bmatrix}r&0\\0&r\end{bmatrix}$$
 * We can easily verify that −1, or rather its 2×2 matrix version, can be obtained by squaring the matrix J,
 * $${\mathbf i} \mapsto {\mathbf J} = \begin{bmatrix}0&-1\\1&0\end{bmatrix}$$
 * This J matrix also happens to represent a planar 90° rotation, consistent with the turning off the track metaphor. --KSmrqT 03:26, 15 March 2006 (UTC)

= March 11 = arch z = -i cos^-1 z = -i arccos z. But |z| < 1. So let's say M * arch 0,11 = 1000 * -i * 1,46057328 = -1460i. Another question please. --DLL 11:23, 11 March 2006 (UTC)


 * Huh? —Keenan Pepper 18:28, 11 March 2006 (UTC)
 * I agree with Keenan. -- Meni Rosenfeld (talk) 18:34, 11 March 2006 (UTC)
 * I'm not sure what your question is, but it looks like you got your formula wrong. According to the site you link, it should be i arccos z. Black Carrot 21:45, 14 March 2006 (UTC)

where's that one rupee?
the question comes along a story one person in SriLanka rented out a room for 3 persons,each one must pay him ten rupees as key money and he will get totally 30rupees.When he collected 30rupees and returned home his wife said that the 10 rupees for each one is too much and return them 5 rupees,so owner called his servant and gave him a 5 rupee coin and ordered him to handover it to the new boardes,servant took it and walked .While the way he think to buy a toffee for him and bought one spending 2 rupees.Finally he  return only three rps to boarders.Only now the question begins,when the servant returned the 3rps each boarder gets 1 rupee.So actually each one have spent only  9rps for the room.Finally if we see through the account, the boarders spent money=9+9+9=27rps.the servant spent for toffee=2rps.so the total money spent in this story is 29rps.But initially we got 30 rupees,so where is that one rupee? --222.165.169.196 12:09, 11 March 2006 (UTC)


 * It doesn't exist. I'm sure this old problem has an article somewhere, but the quick answer is something like this: The 2rps the servant kept shouldn't be added to the money spent by the people in the room. The correct equation is 3*7 spent on the room, 1*3 given back, 2 kept by servant, adds to 30 in total. Confusing Manifestation 12:59, 11 March 2006 (UTC)
 * What ConMan wrote doesn't add up to 30, so he probably mistyped something. I'll give my version: of the 30 rupees paid by the men, 3 were returned, 2 were paid for the coffee, and 25 are in the landlord's pocket. 25+3+2=30. Another way: Each man ultimately spent 9, totaling 27, minus 2 which is the amount paid for the coffee, is 25, the amount in the landlord's pocket. 27-2=25. -- Meni Rosenfeld (talk) 15:06, 11 March 2006 (UTC)


 * Did he have coffee with his toffee ? That might be better than coffee with biscotti, I'll suggest it to Starbucks. 20:27, 13 March 2006 (UTC)


 * Missing dollar paradox? Melchoir 22:22, 11 March 2006 (UTC)


 * It's not missing. It's merely a trick the words play on us. Three men pay $30. The landlord gives back $5. These 5 dollars consist of a)$2 for a toffee and b)$3 returned. Therefore, the total they payed should have been $25. But the servant cheats them by taking $2 for himself; the total they payed is now $(30 - (5 - 2)) = $27. It all adds up. The problem here is that people don't see the logic and add the $2 to the sum of $27. In effect, they're saying that the servant bought two toffees (or [30 - (5 - 2 - 2)]) and not one. Igor the Lion  (Roar!)  21:26, 13 March 2006 (UTC)

A Clock Conundrum
Earlier this week while discussing a question regarding the angles between clock hands with some of my friends, I posed them a question of my own: assuming a clock with second, minute and hour hands, in which all hands move continuously (ie. the minute hand doesn't wait until the second hand reaches 12 to move), at what time(s) will the angle between each pair of hand be 120 degrees? I figured, while it might take some figuring out, it wouldn't be that difficult.

Not knowing the answer myself, I've spent the week trying to figure it out to no avail. I've tried doing some series of equations to express the angle of each in terms of hour, minutes and seconds, but that hasn't worked. I've tried doing it simply in seconds, but dealing with the number of complete rotations of the second hand has proved troublesome. My friends are equally stumped. Any suggestions on how to go about solving such a problem would be much appreciated.

--Elzaban 16:22, 11 March 2006 (UTC)
 * I don't see why this shouldn't be 8:00:20 (8 hours, 0 minutes, 20 seconds). If that's not what you meant, you will have to better explain yourself. -- Meni Rosenfeld (talk) 16:31, 11 March 2006 (UTC)


 * It's not 8:00:20 because all the hands are moving continuously, so at 8:00:20, the minute and hour hands will have moved slightly from where they were at 8:00:00 when they are seperated by 120° (the minute hand will have moved 2° and the hour hand will have moved 10', so the gaps between the hands will be 121°50' (hour hand to minute hand), 118° (minute hand to second hand), and 120°10' (second hand to hour hand), not 120°.) The way I would go about finding what times, if any, meet this criteria, would be to first list all the times when the hour & minute hands are 120° apart and then check the angle to the second hand at each of those times.  (Actually I would probably write a computer program to do all the calculations and give me the answers because that's more interesting than doing numerical grunt-work, but that's the strategy my program would use.) -- AJR | Talk 17:23, 11 March 2006 (UTC)
 * Ah, you're right, I forgot about that thing. I'll give it some more thought. -- Meni Rosenfeld (talk) 17:28, 11 March 2006 (UTC)


 * I don't think it ever happens exactly, but I can't do it in my head. Let me think....
 * By the same argument I gave in Golden Times on a clock face, the hour and minute hand have a 120 degree angle at $$\frac{12}{11}\left(n\pm \frac{1}{3}\right)$$ hours since midnight or noon, for some integer n. Similarly, the minute and second hand have a 120 degree angle at $$\frac{60}{59}\left(m\pm \frac{1}{3}\right)$$ minutes after the hour, for some integer m.  This isn't going to align.  &mdash; Arthur Rubin | (talk) 17:32, 11 March 2006 (UTC)


 * Ah, well that would certainly help explain the lack of sensible answers when I tried solving. Thanks for the help on that. I'll have to go and try to come up with a solveable clock puzzle now to make up for it. --Elzaban 22:09, 11 March 2006 (UTC)


 * You could also look at it from a probability perspective. There are a small number of finite cases where the hour hand and minute hand are 120 degrees apart, and each of these will have an exact location for the second hand, as well.  If there were only 60 positions the second hand could occupy, each of these would have a 1/60 chance of being at the proper location.  However, since you specified continuous movement, there are now an infinite number of positions the second hand can occupy.  Any finite number of positions divided by an infinite number of possible positions leaves approximately zero probability that the second hand will ever be in the exact position needed to be 120 degrees from both the hour and minute hand.  Thus you could assess your chances of finding a solution before doing any heavy math.StuRat 20:13, 13 March 2006 (UTC)


 * On the other hand, you could look for points where the deviations are as small as possible. Preliminary results suggest that the time should be $$5:49:09+\epsilon$$ or $$6:10:51-\epsilon$$, where &epsilon; depends on the precise measure of the deviation;


 * $$\epsilon=\frac{69}{719}$$ if the figure of merit is the sum of the deviations or maximum deviation, and $$\epsilon=\frac{62823}{599173}$$ if the figure of merit is the RMS of the 3 deviations.


 * For $$5:49:09\frac{69}{719}$$, the hands are at $$174\frac{414}{719},294\frac{654}{719},54\frac{414}{719}$$ degrees, with the minute hand being only $$\frac{240}{719}$$ degrees out of position.
 * &mdash; Arthur Rubin | (talk) 22:39, 13 March 2006 (UTC)

= March 12 =

Solution of equations
I have these two equations:

$$ \int _{0}^\!\sec \left( \theta \left( x \right) \right) {dx}={\it Le} $$

$$ \int _{0}^{{\it Xmax}-d\sin \left( \theta \left( {\it Xmax} \right) \right) }\!\sec \left( \theta \left( x \right) \right) {dx}={\it Lc} $$

The two unknowns are $$ \theta \left( x \right) $$ and Xmax. How can I solve this. I'm not even sure if I have sufficient information to solve, since one of the unknowns is a function. Help would be highly appreciated. deeptrivia (talk) 02:15, 12 March 2006 (UTC)


 * What's the d sin on the second upper limit supposed to mean? Melchoir 05:24, 12 March 2006 (UTC)


 * d is a known constant. d times sine of theta(Xmax). deeptrivia (talk) 05:38, 12 March 2006 (UTC)


 * Oh! In that case, couldn't you design &theta; such that sec &theta;(x) is a pair of delta functions in x, one between 0 and Xmax &minus; d sin (&theta;(Xmax)) and one between Xmax &minus; d sin (&theta;(Xmax)) and Xmax, and adjust the amplitudes to get Le and Lc? Melchoir 06:05, 12 March 2006 (UTC)

Thanks! I found a missing equation. I still don't know how to solve it, but I haven't tried much. It's not a design problem, and there has to be a unique solution. Thanks for your help :) deeptrivia (talk) 18:16, 13 March 2006 (UTC)
 * Oh... okay! Melchoir 20:48, 13 March 2006 (UTC)

Maximisation Question
Yes, this is homework, but I have done most of the work. We have been given the function y=exp(-x2). We are asked: What are the dimensions and area of the largest rectangle that fits into the curve and the x axis, as shown on the below:

Let x equal the distance between one of the vertical sides of the rectangle and the y-axis. The area of the rectangle is therefore 2x(exp(-x2)-x) (length times breadth). I call this function A(x), and it has a maximum when I graph it on my calculator. To find the maximum, I use calculus, letting the derivative of A(x) equal zero:

$$A'(x)=2 (e^{-x^2}-x)+2x(-2x e^{-x^2}-1)$$

$$0=(1-2x^2) e^{-x^2}-2x$$

How on earth would I solve the last equation for x? The closest I can get to is:

$$e^{-x^2}=2x(1-2x^2)^{-1}$$

Taking natural logs will get rid of the exponential expression on the left, but leave a log on the right. I'm sure I haven't been taught the tools to deal with equations like this. (Year 12 Specialist Mathematics, Victoria, Australia).

Did I miss an easier way? Can the above be solved using high-school mathematics? --Alexs letterbox 04:58, 12 March 2006 (UTC)


 * Probably I'm missing something. Why is the breadth (exp(-x2)-x) ? Why not exp(-x2)? deeptrivia (talk) 05:05, 12 March 2006 (UTC)


 * PS: If breadth = exp(-x2) as I think, then you'll find it's very easy to see that $$x = 0.707 = {\sqrt{2} \over 2}$$ for optimum. deeptrivia (talk) 05:09, 12 March 2006 (UTC)


 * Oops, I just saw the x on my diagram, and did the subtraction. It really should be exp(-x2)-0 for the breadth. Thanks. --203.129.42.40 07:19, 12 March 2006 (UTC)

I changed the format of the first two equations and another, just for legibility, in case anyone else had trouble reading it. D. F. Schmidt 02:39, 18 March 2006 (UTC)

Simultaneous Equation
I ran into a problem recently while trying to solve a pair of equations that appeared to be fairly easy, but had me stumped when I actually tried to do them. It was part of my maths work at school, and the thing that had me most puzzled was that it was only the lead-in part to the harder part of the question which was on some basic integration.

The specific question asked for the points of intersection of $$y = \sin x \,$$ and $$y = {2x \over \pi}$$. I couldn't do it at all, so I'd appreciate it if anyone could demonstrate a reasonably simple method, because it should be doable with fairly basic maths. I'm fine with manipulating them around and so forth, it's just I can't work towards something with less than two unknowns, for instance.

I can solve most simultaneous equations but when it came to this I wasn't sure how to communicate a decent proof on paper. In connection with solving this I decided to see if I could solve $$x = \sin x \,$$ which I can easily see answers to in my head. I couldn't, however, prove my answers elegantly on paper. So it would appear that my problem lies in not being able to get down to one unknown when I'm restricted by the sine function, and if anyone could offer any general help on that it'd be most welcome. 81.157.152.22 19:44, 12 March 2006 (UTC)
 * Surely you realize that the interesting part of the equations you mention is equivalent to $$\frac{\pi}{2}\sin{x} = x$$. It is easy to see that a solution must satisfy $$|x| \le \frac{\pi}{2}$$, otherwise $$\frac{\pi}{2} < |x| = \frac{\pi}{2}|\sin{x}| < \frac{\pi}{2}$$. Within this range, you can, using basic trigonometry, show that -&pi;/2, 0 and +&pi;/2 are solutions. Then, either graphically or with some basic analysis, you can show that these are the only solutions. Then your equations are immediately solved. -- Meni Rosenfeld (talk) 19:58, 12 March 2006 (UTC)


 * Both equations for y pass through the origin, so (x,y) = (0,0) is a solution. Both right-hand sides are odd functions of x, so if x > 0 is a solution then so is −x. And sin x cannot exceed 1, so x cannot exceed &pi;&frasl;2. For a visual thinker, this suggests plotting both right-hand sides for x from 0 to &pi;&frasl;2. We see one nonzero intersection, which appears to be exactly at (x,y) = (&pi;&frasl;2,1); this solution is easily confirmed by substitution. It remains to confirm that no other solutions exist over this interval, if we don't trust our plot as proof; but under the circumstances, that's probably excessively paranoid. --KSmrqT 21:27, 12 March 2006 (UTC)


 * I think I see it now, thanks for the help. I never thought of it in terms of that inequality which simplifies it quite a bit, and there was also an unreasonable unwillingness to rely on a graphical proof (maybe partially down to the fact that my attempting to draw a plot in my jotter at school normally results in a large-scale mess of ink).  81.157.152.22 16:11, 13 March 2006 (UTC)


 * Ink ? Yuk !  Try a graphing calc like the one at:


 * http://www.calculator.com/calcs/GCalc.html


 * You skip the "y=" part when entering equations in this graphing calc, so just enter "six(x)" and then "2x/pi", without the quotation marks, of course. StuRat 17:00, 13 March 2006 (UTC)


 * Indeed I would, but as I say this was at school. Probably in more mature schools people would be allowed to go up and try out things on a computer or something but in my case I was sitting scribbling away feverishly on a jotter, and my whole proof had to be on there, so it would have been an ink-oriented attempt (or at best, pencil, which would probably result in rubber droppings all over the place, kinked paper and graphite smears).  Granted I should probably have grabbed a graphing calculator, but there was always that distrust of graphical proofs.


 * The main problem was that I could vaguely see the plots in my head but wasn't sure how to communicate a nice proof, and didn't realise that a graphical one would probably be acceptable. 81.157.152.22 17:22, 13 March 2006 (UTC)


 * We're not your instructors, nor your graders, and cannot speak for them as to what's acceptable. But based on what you told us these intersections were peripheral to the thrust of the bigger problem. If we want to be a little paranoid, let's see what the plots suggest in the way of a "proof". At x = 0 the slope of sin x is 1, while at x = &pi;&frasl;2 the slope is 0. Compare these to the slope of the line, which is a constant 2&frasl;&pi;. So at 0, the sinusoid is rising faster, pulling away from the line; and at &pi;&frasl;2 the line rises from below to meet it. Since the shape of the quarter sinusoid is convex, this simple argument should give us confidence that the only intersections are at the ends. --KSmrqT 22:24, 14 March 2006 (UTC)

March 12/ Googol's
I got this question as a homework assignment and i just can't figure it out. Here's the question: What is the least value of N such that N! is greater than a googol?
 * I suggest using Stirling's approximation to get an estimate; if you need to narrow it down after that, you can try direct computation on a computer. Melchoir 22:21, 12 March 2006 (UTC)
 * Actually, on second thought, you can figure it out by trial-and-error just by computing factorials on a TI-89 in approx mode, which automatically converts them to scientific notation. But that's cheating, of course. Melchoir 22:33, 12 March 2006 (UTC)
 * ...and actually, the answer is already at Googol. Oh well! Melchoir 22:37, 12 March 2006 (UTC)

= March 13 =

Help with solving an equation with elliptic integrals
I have to solve the equations:


 * $$ L = p [ F (p, \theta_{c}/2) - F (p,\pi/4)]/k \,\!$$

and


 * $$ sin( \theta_{c}/2) = ( 1/p^2 - e^2 k^2/4)^{1/2} \,\!$$

for $$ \theta_{c} $$. Here, F is the incomplete elliptic integral of the first kind. e, k and L are known.

I am doing this in MATLAB as follows:

opt = optimset('Display','notify','TolFun',1e-8,'TolX',1e-8,'MaxFunEvals',10000); pfunc=inline('L - (mfun(EllipticF,sqrt(1/p^2-e^2*k^2/4),p)-mfun(EllipticF,sin(pi/4),p))*p/k','p','e','k','L'); [p,xval,exitflag] = fsolve( pfunc,0.707,opt,e,k,L) thetaC = 2*asin(sqrt(1/p^2-e^2*k^2/4));

I know that for sufficiently large values of e, I should get $$ \theta_{c} > \pi $$. A trial value for such case is e = 137, k=0.01, L=100. The numerical computation for finding p fails to converge after values of e that should result in $$ \theta_{c} > \pi $$. What's the reason? Does it have to do with the way asin and sin are defined in matlab? What's the way out? I'm doing this to calculate elastica, and I know that after a particular value of e, it will start making a loop, forcing $$ \theta_{c} > \pi $$. How can I do this calculation? deeptrivia (talk) 18:37, 13 March 2006 (UTC)


 * I dont know Matlab that well, but am I right to assume that mfun(EllipticF,sin(th),p) computes $$F(p,th)$$? In this case you forgot to divide theta by 2.(Igny 20:04, 13 March 2006 (UTC))


 * Oops. No I missed the division in the equation above, so my code is correct, but doesn't converge. deeptrivia (talk) 21:31, 13 March 2006 (UTC)


 * I can't test your example (I don't recognize mfun, so I suppose that mfun is a function you wrote yourself for evaluating the elliptic integrals?). However, solving a system of nonlinear equations is hard, and the algorithm that fsolve use does not always terminate. Without seeing the output of fsolve, I can't tell exactly why it does not terminate. Usually, giving a better initial guess for the solution instead of 0.707 does the trick.
 * In this case you have an alternative. The function fzero is meant for the solution of a single nonlinear equation. This is a simpler problem, and fzero is almost always succesful. So, I'd advice you to use fzero instead of fsolve. -- Jitse Niesen (talk) 10:38, 14 March 2006 (UTC)
 * If you denote $$\sigma=\sin(\theta/2)$$, $$F_0(p,\sigma)=F(p,\theta/2)$$, then $$p^2=(\sigma^2+e^2k^2/4)^{-1}$$, and you probably can solve the equation
 * $$\frac{p}{k}(F_0(p,\sigma)-F_0(p,\sqrt{2}/2))=L$$
 * for $$\sigma$$. Then $$\sin(\theta/2)=\sigma$$ will have infinitely many solutions for $$\theta$$ (Igny 17:28, 14 March 2006 (UTC))


 * Thanks a ton everyone! Actually, I simplified my problem, so I don't need elliptical integrals now! deeptrivia (talk) 21:42, 15 March 2006 (UTC)


 * What a shame. Elliptic integrals are so interesting; I'm sorry you didn't need them. linas 06:09, 19 March 2006 (UTC)

Euclid
Is there any easy explanation (one that I can understand) of how Euclid's parallel postulate is false?.--Cosmic girl 18:39, 13 March 2006 (UTC)
 * It's not that it is false, but that when you consider it false, a new form of geometry arises. This is the non-euclidian geometry. The first image on that article pretty much explains everything, I think. If not, let us know.☢  Ҡ i∊ ff   ⌇  ↯  20:44, 13 March 2006 (UTC)
 * Also, the actual universe is most likely not Euclidean at a very large scale, in roughly the same way the surface of the Earth is not a Euclidean plane but a sphere. Septentrionalis 22:08, 13 March 2006 (UTC)


 * I remember something about an experiment to measure the sum of the angles in a triangle formed by three mountain peaks, using laser interferometry. Anyone know what I'm talking about? Do we have an article about it? —Keenan Pepper 23:40, 13 March 2006 (UTC)


 * The parallel postulate was introduced in Euclid's Elements because it was necessary in the development of standard geometry, but apparently could not be derived from previous axioms. It was not clear then, nor for many centuries, whether it actually was a consequence of the other axioms, or exactly what its status was. (David Hilbert eventually showed that a solid foundation for standard geometry required a bit more than Euclid included, but that's a separate story.) One radical possibility was that perhaps geometry was not singular (Euclidean geometry) but plural (geometries), depending on the choice of this postulate. It's easy to see how hyperbolic geometry might have been overlooked, but somehow spherical geometry (the simplest elliptical geometry) was missed as well.
 * One of the consequences of the postulate is that the interior angles of any triangle sum to exactly 180°. For spherical geometry, a "point" is a point on the surface, while a "line" is a great circle. Taking one vertex at the North Pole and two others on the Equator, we can easily construct a spherical triangle with three right angles, so the sum of this triangle's interior angles is 270°! The more area a spherical triangle encloses, the more the angle sum exceeds 180°; in fact, the excess is equal to the area. --KSmrqT 09:45, 14 March 2006 (UTC)
 * (this is covered in our spherical trigonometry article.) ☢  Ҡ i∊ ff   ⌇  ↯  10:48, 14 March 2006 (UTC)


 * To be fair, no axiom is derived from previous ones. If it could be, it wouldn't be an axiom. Superm401 - Talk 02:05, 16 March 2006 (UTC)

Wow! :O ...thanks! I understood but I'll read the articles too.--Cosmic girl 20:29, 14 March 2006 (UTC)

It's an axiom

 * Properly speaking the statement that co-planar parallel lines never intersect is an axiom of Euclidean geometry. As others have said in this thread, by making other statements about how they do intersect, we can create other, non-Euclidean, geometries.  In fact the efforts to prove the hypothesis (to make the statement a theorem rather than an axiom) by reductio ad absurdum led to the creation/discovery of non-Euclidean geometries. This led, more generally, to the realization that any axiomatic system is only one among many possible other axiomatic systems which is one of the pivotal discoveries of mathematics.  JimD 02:22, 20 March 2006 (UTC)

= March 14 = This is Pi Day. But when, in history, was the best approximation to pi equal to 3,14 ?
 * Have you checked out approximations of pi? 3.14 was apparently suggested by the Chinese as "good enough" (though they had what they believed a more correct value) in  AD 263. &mdash; Lomn Talk 16:18, 14 March 2006 (UTC)
 * This is also Steak and Blowjob Day. Happy S&BJ Day! Black Carrot 19:52, 14 March 2006 (UTC)


 * How cool that Pi Day just happens to be Albert Einstein's birthday! JackofOz 05:52, 15 March 2006 (UTC)


 * Hmm. I'm pretty sure AD 263 is wrong. I think its AD 1592. About 6 in the morning, 53 minutes and 58 seconds and 979 milliseconds after the hour. linas 06:18, 19 March 2006 (UTC)

SA of a sphere
How was the forumla for the surface area of a sphere derived? Are there any good sites with information about this? 64.198.112.210 16:41, 14 March 2006 (UTC)


 * It's a simple calculus problem. I'll give my own intuitive explanation, which you may or may not like: The volume of a sphere is $$V = {4\over3}\pi r^3$$, and when the radius changes by dr the volume changes by dr times the surface area. (You can visualize this as adding or removing an infinitesimally thin spherical shell.) The surface area is then $${dV\over dr} = 4\pi r^2$$. —Keenan Pepper 18:13, 14 March 2006 (UTC)


 * The question is based on a faulty premise; the surface area of a sphere has been calculated many times in many ways. A proper question is "how did so-and-so do it at such-and-such a time", or "how might we do it". One of the more ancient and intriguing methods first demonstrates that we can project the sphere out onto a cylinder wrapped tightly around it at the equator without affecting the area. Now we need only calculate the area of a cylindrical band with circumference 2&pi;r and height 2r; and this, of course, is simply 4&pi;r2. Still, this depends on being able to show that the projection leaves area unchanged. A simple argument for area preservation is that tilt of the sphere surface towards the central axis gives the projection onto the cylinder less area (by foreshortening), but expansion of the surface out to the cylinder gives more area, and these two factors exactly cancel everywhere on the sphere. Archimedes found this relationship of sphere to cylinder so appealing that he is said to have asked for it to be drawn on his tombstone. --KSmrqT 19:43, 14 March 2006 (UTC)

= March 15 =

A number theory problem
Find all n such that (20n+2) divides evenly into (2003n+2002)

L33th4x0r 05:46, 15 March 2006 (UTC)


 * Shouldn't you be doing your own homework? The way to do it is to define a variable x as the quotient of the division of (20n+2) into (2003n+2002). Then derive n as a function of x, so that every whole number value of x = 1,2,3... will give the values of n you need. --Canley 07:46, 15 March 2006 (UTC)


 * Found it. It's not The Answer to Life, the Universe, and Everything - quite the opposite, in fact. Gandalf61 17:23, 15 March 2006 (UTC)

About 2 D creatures traveling on a 2 D plane seen from a 3 D point of view
I've got this strange theory that I conjured out of thin air som 15 years ago, have there been anybody else than me thinking about this stuff.

I postulate hereby that C a 2D creature's fastest way (like in shortest/less resource expenditure) of traveling (in an XY plane) from point A to point B would be to travel in one direction (X or Y) (not distance AB) until it would be at the distance AB - (A - (X or Y)) = B and then the distance B.

This way the creature C would only have to change direction of travel (90 degree change) once.

Try to think about this and and tell me how much further would 2 D creature have to travel compared to a 3D (capable) creature, my answer is squareroot 2 (appr. : 1.4) times more.

Reason: while the 2D creature could change direction of travel (X or Y) for every second or mm (or whatnot) that it went, this would take even more time/resources ..stop/change direction etc.... than it would to just travl the distance in the X direction and then after that in the Y direction....one start/stop/start/stop

Relate this to a 3D creature like us in a 4D world....and the answer would be 0.5 times the value of pi.....and bring in the double slith photon/electron experiments and other relativistic points... anybody got any thoughts about this...?

Could this be elaborated a bit more? [Azalin1999]


 * Uh. Perhaps you should phrase your question better. Are A and B points or lines? You use them as both. What's X and Y? And if it travels from A to B in a straight line, it doesn't ever need to change its direction. --BluePlatypus 18:05, 15 March 2006 (UTC)


 * I think maybe I see what he's getting at, but it doesn't really have anything to do with 2 vs. 3 dimensions. It sounds to me like he's describing the Manhattan distance.  At least in the first part where he's talking about 2 dimensions.  If you compare the Manhattan distance (let's call it m) with the ordinary, Euclidean distance (d), between any two points in a Euclidean plane, it's not too difficult to show that $$1 \le \frac{m}{d} \le \sqrt{2}$$.  But I have no idea where he's getting the $$\frac{\pi}{2}$$ factor from for three dimensions.  In three dimensions, you have $$1 \le \frac{m}{d} \le \sqrt{3}$$.  More generally, in n dimensions, $$1 \le \frac{m}{d} \le \sqrt{n}$$. Nor do I see what it has to do with quantum mechanics or relativity. Chuck 18:40, 15 March 2006 (UTC)

Hi again I'm the one who phrased the question. A little note about myself. Never did score high in maths and equations but I'm really curious about this 'theory'. Much thanks for the link to the 'manhattan distance' article, I think I remember reading about it some years ago. My native tongue is Danish so that's my excuse for the inaccurate phrasing.

I originally made some drawings on a XY crossed paper to illustrate the theory. And I had a really hard time to explain my arguments for those I presented it to (mostly indifferent friends). And I still have difficulties atm. because the thing I would like to verify is that it would be possible to apply something like this 'manhattan distance' to distances in a 3D world seen from the perspective of a fourth dimension. Which should be possible. Basically like --BluePlatypus went about distances in an euclidian plane...

The distance 1/2 Pi was derived from projecting a imaginary path (distance traveled) through 4D space compared to what a 3D creature would have go by not using the 4th dimenson...like a 2D creature can't (see, thus) travel the path that a 3D creature can. In regards to the 'Manhattan Distance' this would be akin to a race between a cab an a helicopter going over the cityblocks.


 * Too bad your friends are indifferent; this stuff is fun! As Chuck wrote above, yes, you can define Manhattan distance in 3D, and you don't even need a 4D perspective to do it. I'm not sure where the pi is coming from.
 * As for bringing in modern physics like quantum mechanics and relativity, let's just say that they're not really compatible with a Manhattan metric. If you lived in a world where the physics was affected by a Manhattan metric, you would easily be able to experimentally determine the coordinate axes you lived on, whether you're a 2D or a 3D creature. But we've never observed any such special directions in space. Melchoir 19:30, 15 March 2006 (UTC)
 * That's an understatement! :) If all directions weren't equivalent, you wouldn't have conservation of angular momentum, and that'd really yank out the carpet from under physics-as-we-know-it. --BluePlatypus 20:08, 15 March 2006 (UTC)


 * OK, I kind of see where you're going here. First of all, it's important to remember that the $$\sqrt{2}$$ factor in 2 dimensions is a maximum; it doesn't hold for all pairs of points.  For example, if you're travelling from (0,0) to (5,5), the person constrained to taxicab geometry has to travel 10 units, but the person who can travel in any direction only has to travel $$5\sqrt{2}$$ units, and the person limited to taxicab geometry has to travel $$\sqrt{2}$$ times farther in that particular case.  However, if you're traveling from (0,0) to (7,24), the Euclidean traveler has to travel 25 units, but the taxicab traveler only needs to travel 31 units--just 1.24 times further than the Euclidean traveler, which is considerably less than $$\sqrt{2}$$.  And of course, if you're going from (0,0) to (13,0), both people travel the same distance.


 * The analagous situation in three dimensions, it seems to me, is someone who can travel in any one of six orthogonal directions (which we might call "north," "south," "east," "west," "up," and "down," but it's important to remember here that these are idealized directions where all north-south lines are parallel with each other and do not meet, as are the east-west lines, and also the up-down lines; and not the directions as defined on the earth, where the "north" lines all meet at the north pole, the "south" lines all meet at the south pole, and the "down" lines meet at the center of the earth).


 * In this case the maximum factor between the distance of the taxicab traveler and the distance of the Euclidean traveler would be $$\sqrt{3}$$, for example going from (0,0,0) to (5,5,5) the taxicab traveler has to move 15 units, as opposed to the Euclidean traveler's $$5\sqrt{3}$$ units. But as before, that's a maximum: in going from (0,0,0) to (3,4,12) the taxicab traveler goes 19 units to the Euclidean traveler's 13, but $$\frac{19}{13}<\sqrt{3}$$.  It's possible to choose a pair of points where the taxicab traveler's distance is exactly $$\frac{\pi}{2}$$ times as long as the Euclidean traveler's, but that's only because such a pair of points can be found for any number x such that $$1 \le x \le \sqrt{3}$$, which $$\frac{\pi}{2}$$ happens to satisfy.


 * The "car vs. helicopter" analogy is an interesting way to think about it, but at the same time it's misleading because you don't actually have to move in the 3rd dimension for this to happen. You can have one traveler who can move in any direction, and one traveler who is limited to moving in a set of orthogonal directions, and still have both of them confined to a plane, so it's not really an issue of one person being constrained to two dimensions while another can move in three (or three vs. four, or four vs. five, etc.) Chuck 20:33, 15 March 2006 (UTC)

You're Chuck making some good points. But what if we assume that in our (unlimited) reference 3D space we are looking down on the 2 plane (which could be from any arbitrary reference point/perspective) let's say head on, straight down.

Let's assume some more - Our little 2D creature can only observe (look) through either the X or the Y, and thus only observe it's destination when its in line with either the x axis of the destination or th Y axis, whichever comes first, depending on which route it took.

From the 3D POV this would satisfy my squareroot 2 clause about a 3D being would be able to go the shorter distance right? Because it can sort of observe the direct line of vision...X & Y combined.

Now if the 2D creature were able to in some way observe the destination it could of course just travel the direct route. But this would only happen after it had moved all the way along either the X or Y axis to be in line with the destination. If this 2D thing were sentinent the perceived distance the 2D being traveled would be? the squared X+Y which is less than the distance it really took if a 3D being measured...also there are 2 routes that are the same distance but starting with etiher going out of the X or the Y axis...

Relate this to a 3D world...(damn crunching)hmm non-euclidian geometry..........Spherical trigonometry


 * That's all true--but the "2D" and "3D" specifications are red herrings. What is important is that you have one creature which can only travel parallel to the X and Y axes, and another that can travel any direction within the XY-plane.  The first creature, who can only travel parallel to the X and Y axes, will have to travel at least 10 units to go from (0,0) to (5,5).  It doesn't matter if he's two dimensional, or if he can perceive three, or four, or eighteen dimensions, as long as his movement is limited to directions parallel to the X and Y axes, he has to travel at least 10 units.  Similarly, for the creature that can travel any direction in the XY-plane, he can get from (0,0) to (5,5) by moving only $$5\sqrt{2}$$ units, regardless of whether he can travel, or even see, in any additional dimensions or not.  (For this second creature, imagine not a helicopter, nor a car confined to a grid of city streets, but instead a car on a wide open dry lakebed.)


 * In other words, it's true that in your example with a 2D-creature confined to moving parallel to the X and Y axes, and a 3D-creature which can move in any direction, everything you say is true. But your choice of "2D" and "3D" are arbitrary, and there is nothing special about 2-dimensional or 3-dimensional creatures which makes your statements true; I could just as easily talk about a dog which could only move parallel to the X and Y axes, and a cat which could move in any direction, and the statements would all still be true. Chuck 22:44, 15 March 2006 (UTC)


 * Just googled and found this again - I'm the OP - well now I'm interested in two things. One to see if anybody else but me will notice that I've been here 5 years later, and two : Is there someone out there who can explain in laymans terms why this
 * Quote "In this case the maximum factor between the distance of the taxicab traveler and the distance of the Euclidean traveler would be $$\sqrt{3}$$, for example going from (0,0,0) to (5,5,5) the taxicab traveler has to move 15 units, as opposed to the Euclidean traveler's $$5\sqrt{3}$$ units. But as before, that's a maximum: in going from (0,0,0) to (3,4,12) the taxicab traveler goes 19 units to the Euclidean traveler's 13, but $$\frac{19}{13}<\sqrt{3}$$.  It's possible to choose a pair of points where the taxicab traveler's distance is exactly $$\frac{\pi}{2}$$ times as long as the Euclidean traveler's, but that's only because such a pair of points can be found for any number x such that $$1 \le x \le \sqrt{3}$$, which $$\frac{\pi}{2}$$ happens to satisfy. " Unquote is right?

As I see it - Going from A to B in 3D is just (always) a straight line. But this C cannot see that so it has to travel from the perspective of the 3D creature in X + Y ie. at the point where it needs to change direction is where the 2D creature actually has traveled the least distance in the X distance (or Y) and then changes direction 90 degrees to be 'the fastest' without changing directions every x/Y mm (or whatever).

So the 3D viewer would, if seeing the 2D creature move along the xy points, percieve the path at some fractional degree from the perfect - Squareroot 2 path - which means the 3D viewer would be able to see the 2D creature's energy expenditure adhere to my postulation.... 2D travelers always expend square root 2 times more energy when the 3D viewer multiply this with hmmm is it cosinus or sinus ? (Perfect angle view that makes a straight line away from the XY plane) That's the hint for ½ PI value. To elaborate : A photon is emitted from somewhere..... this photon is measured at some point and only there you can make the mark that you measured a photon (in whereever space) how it got there (Cab or Taxi) you don't know - you can just measure the time difference between you emtting the photon - and collecting information about when it hit 'point B' 85.81.121.107 (talk) 19:59, 7 October 2011 (UTC)

Graph Gradients
Hi, I'm working with gradients at the moment but can't for the life of me rember if you count the squares (i.e. measure) or use the unit values for the graph. To see an example of what I mean: Click here is the gradient 3 over 6 (as attained by counting squares) or 15 over 12 (as attained by using the unit values). If it helps I'm dealing with "real life" data here. Thanks! --Christopher Denman 17:50, 15 March 2006 (UTC)
 * I guess that would depend on the context, but since the scales of the x- and y- axes are different, it's more plausible to discuss the units and not the squares. So the slope is 15/12 = 5/4. -- Meni Rosenfeld (talk) 18:01, 15 March 2006 (UTC)


 * You need to use the units, otherwise you'll be off by a factor 5/2, since you presumably want the slope given in "y units per x unit", and not in "5 y units per 2 x units" the latter being the equivalent to "y lines per x line". --BluePlatypus 18:11, 15 March 2006 (UTC)


 * Thanks very much both of you for your answers!--Christopher Denman 18:34, 15 March 2006 (UTC)


 * When we talk about a road having a steep gradient, the term "gradient" has a different meaning than is usual in mathematics, where the common term would be "slope". To compute slope, a dimensionless number, we divide the vertical displacement by the horizontal displacement, being careful to use the same measurement units for both. Your graph is therefore an invitation for trouble, misleading both the eye and the calculation by displaying as a square what is actually a rectangle 5 high and 2 wide. See the slope article for a more extended discusssion. --KSmrqT 21:18, 15 March 2006 (UTC)

Making things appear 3D without axes
On right is a plot of x-, y-, and z- coordinates of two ropes on which some forces are acting. I made this using the plot3 command in matlab. Now I want to remove the axis labels, and marks. In that case, this will look like 2d. How can I make it look 3d? Some kind of shading/lighting/perspective tricks? I need to do this in MATLAB. Thanks! deeptrivia (talk) 21:39, 15 March 2006 (UTC)
 * It depends on what you want to emphasize, but I guess that's not too helpful... how about conical arrowheads? I have no idea about MATLAB. Melchoir 23:21, 15 March 2006 (UTC)


 * Do a little reading on depth cues. Two of the easiest for you might be motion and shadows/occlusion. If the curves were rendered as ropes, the variations in texture size would be a natural cue. For simplicity, let MATLAB provide data points along the curve (or spline coefficients), and use a free rendering package like POV-Ray or Blender to produce the fancy images. This would allow fog (aerial perspective), for example. --KSmrqT 23:48, 15 March 2006 (UTC)

Thanks for your answers. I tried shadows and it helps a bit. I have no choice but to use matlab because this will be a part of a GUI I'm making in Matlab. I think I can manage if I get a solution to the problem regarding cylindrical surfaces below. Cheers, and I appreciate it highly. deeptrivia (talk) 01:30, 16 March 2006 (UTC)


 * You could also use color, with, say, red increasing in the X direction, blue in the Y direction, and green in the Z direction. This method is particularly useful for surface plots.  Showing the graph from different views would be another option. StuRat 03:48, 16 March 2006 (UTC)


 * Or use perspective, can u make thickness of the lines varying? (Igny 18:04, 20 March 2006 (UTC))

definition of manipulatives
I need the definition in mathematical terms for manipulatives


 * Aren't those those little wooden blocks that come in sticks of ten, plates of a hundred, and cubes of a thousand? Those things were great. —Keenan Pepper 01:01, 16 March 2006 (UTC)


 * Cuisenaire rods are one of the early tools for "hands-on" mathematics teaching. Today there are many others. --KSmrqT 03:28, 16 March 2006 (UTC)

= March 16 =

Equation of cylindrical surface
This is a follow up question from above (3D effects without axes). Suppose we have a spatial curve defined by x = f(s), y = g(s) and z = h(s). What equations will describe a cylindrical surface of radius r which has this curve as its axis? I can do real neat 3D things in matlab if I plot this surface. Will post the results here, of course. deeptrivia (talk) 01:28, 16 March 2006 (UTC)


 * I'm not sure what you mean. Are you talking about an envelope of spheres centered on the curve? —Keenan Pepper


 * Well, I'm not 100% sure what would an envelope of spheres exactly mean, but I think that's what I mean. Think of a tube/pipe (eg a water hose) whose axis is twisted to coincide this curve. I need the equation of the surface of the tube. deeptrivia (talk) 02:39, 16 March 2006 (UTC)


 * Assuming the curvature never vanishes, you can use
 * $$r'(s,\theta)=r(s)+N(s)\cos\theta+B(s)\sin\theta,$$
 * where N and B are defined at the article on the Frenet-Serret formulas. Melchoir 02:52, 16 March 2006 (UTC)


 * Thanks for the answer. Since this is a 3D curve, we need r, $$ \theta $$ and $$ \phi $$ to describe the surface. deeptrivia (talk) 03:01, 16 March 2006 (UTC)


 * Oh...I think I better understand your answer now. Let me try. Thanks ! deeptrivia (talk) 03:04, 16 March 2006 (UTC)

For everyone's benefit. I got a readymade solution based on Melchoir's answer. See deeptrivia (talk) 03:08, 16 March 2006 (UTC)
 * Yeah, I should have figured that it's been done. Well, it doesn't hurt to rederive the wheel every now and then, eh? Melchoir 03:11, 16 March 2006 (UTC)

Surface Area of a Rotated Curve
I am writing this as an AP Calculus BC student. Assume f(x) describes a continuous, differentiable curve, such as
 * $$f(x)=\sqrt{x}$$

I want to find the surface area of the solid that results from a rotation of part of the curve, say 0 to 4. For this example, let's assume I'm rotating about the x axis. I know the correct method to solve this problem (surface areas of cone frustrums). However, why can't it be found as a summation of the surface areas of right circular cylinders (excluding the bases)? Thus, my presumed integral would be
 * $$\int_{0}^{4} 2 \pi r\, dx =$$
 * $$\int_{0}^{4} 2 \pi \sqrt{x}\, dx$$

Obviously, it is wrong. However, why is it wrong? My textbook helpfully includes the formula above (though I found it myself before seeing this). It then says it can not be used because it has "no predictive value and almost never gives results consistent with other calculations." However, I find this somewhat disappointing, as there is no explanation of why the formula is flawed. Please do not simply show me the right way to do it, and explain that this is therefore the wrong way; that is unhelpful Superm401 - Talk 01:58, 16 March 2006 (UTC)
 * Try visualizing the geometric interpretations of the two formulas really close to the origin, where the slope of f is infinite. The correct formula accounts for the "part" of the surface area facing perpendicular to the x-axis; the incorrect formula cannot. Whatever you do, don't try to visualize the difference where the slope of f is small. Melchoir 02:21, 16 March 2006 (UTC)
 * Thanks. That's a helpful distinction.  I can begin to see the problem now, though a more formal explanation would help too. Superm401 - Talk 02:46, 16 March 2006 (UTC)

When calculating the area under a curve, one can approximate either by rectangles, or by trapezoids bounded by some "diagonal segments" on top. I think the second approximation is called the trapezoid method, and it's better than the rectangle method, but both work in the limit. Both converge to the Riemann integral.

On the other hand, when calculating the length of that same curve, you have to approximate the curve by diagonal chords. If you approximate the curve with rectangles, the sum of the rectangles will always add up to b–a (where the domain of the curve is the interval [a,b]).

The technical reason behind this is that the form which gives you the length is not a linear differential form. In other words, the Pythagorean equation is not linear, and that's needed to calculate length. The form that gives you the area is linear, and therefore any sum will do. Stated even more simply, the triangle inequality tells you that the hypotenuse of a triangle is greater than the leg. Rectangles get you the leg of the triangle, and never approximate the length, though they do approximate the area. Calculating surface area is just calculating length and throwing in a 2πr, so the same issues apply to volume versus surface area. -lethe talk [ +] 03:31, 16 March 2006 Th(UTC)
 * Thank you. That's very helpful. Superm401 - Talk 04:18, 16 March 2006 (UTC)

To illustrate the problem somewhat differently, consider a cone obtained by rotation of "y=ax" around the x-axis. The area and the volume are given by integrals
 * $$S=\int_0^h 2\pi r(x) \frac{dl}{dx}dx= \int_0^h 2\pi ax \sqrt{1+a^2}dx=\pi h^2 a\sqrt{1+a^2}$$
 * $$V=\int_0^h \pi r^2(x) dx= \int_0^h \pi (ax)^2 dx=\frac{1}{3}\pi h^3 a^2=\frac 13 \pi h r^2(h)$$

You can see that omitting dl/dx will lead to a wrong result. For the curve $$y=\sqrt{x}$$, dl/dx varies with x. Another example is the surface area of a unit circle
 * $$S=\int_{-1}^1 2\pi r(x) \frac{dl}{dx}dx=\int_{-1}^1 2\pi r(x)\sqrt{1+\left(\frac{dr}{dx}\right)^2}dx=

\int_{-1}^1 2\pi \sqrt{1-x^2}\frac{1}{\sqrt{1-x^2}} dx=4\pi$$ (Igny 13:04, 16 March 2006 (UTC))
 * And how do you know that result is wrong? Because it has "no predictive value and almost never gives results consistent with other calculations"? -lethe talk [ +] 17:13, 16 March 2006 (UTC)

Volume of cone
can anyone give the derivation for this?
 * A cone is a special type of pyramid, and the derivation is of the volume formula is therefore the same. It involves finding the area of a "step" pyramid, and then making the steps finer and finer. You can take a look at this site. Sjakkalle (Check!)  13:59, 16 March 2006 (UTC)
 * Or alternatively:
 * $$V = \int_0^H\pi\left(\frac{hR}{H}\right)^2dh = \frac{\pi R^2H}{3}$$
 * -- Meni Rosenfeld (talk) 18:11, 16 March 2006 (UTC)


 * It's very interesting that the volume of a pyramid (including a cone) is equal to one-third of the volume of a prism with the same base area and height. :P —OneofThem 19:10, 16 March 2006 (UTC)
 * I guess so, but it's just a special case of the fact that the volume of an nth dimensional pyramid is 1/n the volume of an nth dimensional prism - which itself is a consequence of the equality
 * $$\int_0^1t^{n-1}dt=\frac1n$$
 * -- Meni Rosenfeld (talk) 19:35, 16 March 2006 (UTC)

determinat of a parabola
how do you find the determinant of a parabola?? please help
 * Did you mean, "discriminant"? The discriminant of the parabola $$y = ax^2 + bx + c$$ is $$\Delta = b^2 - 4ac$$. -- Meni Rosenfeld (talk) 18:13, 16 March 2006 (UTC)

= March 17 =

discontinuous linear functional on test spaces
On page 74, L. Schwartz, Mathematics for the Physical Sciences,Hermann

Quote: The existence of linear functionals which are discontinuous on $\script D$ may be demonstrated mathematically using the axiom of choice. UN-Quote.

Question: Where can I actually find a proof for the existence of a discontinuous linear functional on test spaces?

Be more grateful if you could send your answer directly to me by visiting http://www.maths.uwa.edu.au/~twma/mathematics/index.htm Thank you in advance. twma
 * See discontinuous linear functional. -lethe talk [ +] 04:12, 17 March 2006 (UTC)

Dear Lethe, Thank you for your quick response. Because you did me a favor to visit my page above, I was alert immediately without having to wait until tomorrow.

On page 73 of the same book above, a sequence of test functions f_n tends to 0 if

(a) the supports of f_n are contained in the same bounded set and

(b) for each fixed m, the m-th derivative of f_n tends to 0 uniformly.

\sin n^2x/n could not satisfy the above conditions.

I multiplied \sin n^2x/n with a test function g satisfying g(0)=1. Could not get any further. STILL NEED HELP. twma


 * You have not broken any of our rules, this kind of thing is what this page is here for. Welcome.  Now, as for your question, I guess you're talking about the topology of smooth functions of compact support.  This is the topology used to define the space of distributions (the space of distributions is the continuous dual space of the space of smooth functions of compact support with the topology you mention: uniform convergence on compact sets of all derivatives).  Now what exactly is it that you want to do?  The sequence you mention does not converge; the function itself converges uniformly, but none of its derivatives do.  But what task are you trying to accomplish?  I do not have this text by Schwartz. -lethe talk [ +] 06:13, 17 March 2006 (UTC)
 * You have not broken any of our rules, this kind of thing is what this page is here for. Welcome.  Now, as for your question, I guess you're talking about the topology of smooth functions of compact support.  This is the topology used to define the space of distributions (the space of distributions is the continuous dual space of the space of smooth functions of compact support with the topology you mention: uniform convergence on compact sets of all derivatives).  Now what exactly is it that you want to do?  The sequence you mention does not converge; the function itself converges uniformly, but none of its derivatives do.  But what task are you trying to accomplish?  I do not have this text by Schwartz. -lethe talk [ +] 06:13, 17 March 2006 (UTC)

To find, explicitly by displaying formulas or implicitly by axiom of choice, a counter example that a linear functional T is not continuous, we need to find a sequence of test functions f_n satisfying both conditions (a) and (b) and in addition T(f_n) fails tending to 0 as n tends to infinity. The sequence \sin n^2x/n does not satisfy (a) and not (b).

Distribution theory is well established nowadays. There are many good books on the subject. For example,

Page 222, F. Treves, Topological vector spaces, distributions and Kernels, Academic.

Pages 28, 26, M.A. Al-Gwaiz, Theory of distributions, Monographs and textbooks in pure and applied mathematics, vol 159.

I quote L. Schwartz because the book was written by the creator. I even looked up the French version of the same book with disappointment. My next possible step is to look up the original paper. Twma 18:52, 17 March 2006 (UTC)twma

You people have something like --- 17 March 2006 (UTC) but I do not have it. Am I overlooking some rules and regulations of this site? If so, apologize and please tell me what to do in order to CONFORM.


 * See top of the page, which explains that you just have to type four tildes to sign (gives IP + date when not logged). --DLL 06:18, 17 March 2006 (UTC)
 * When editing, just above the window is a row of buttons. Hovering the cursor above them should popup text describing their function. The antepenultimate one, the scrawl between the red no-wiki and the dash, adds a signature with one button click. For those who are typing challenged, this may be easier. Also note that the contents of a signature can be changed by using the "my preferences" link at the top of the page while logged in. This is how some of us include a link to our talk page. --KSmrqT 16:31, 17 March 2006 (UTC)

I wrongly thought that four tildes are examples of letters of a signature. I am able to conform now. Thank you. My identity is twma in lower case but this site changed it by default to Twma in order to conform.

To save the room, I shall delete part of this section if I have no objection in 24 hours. Twma 18:52, 17 March 2006 (UTC)twma
 * Which room? What do you want to delete and why? -- Meni Rosenfeld (talk) 19:14, 17 March 2006 (UTC)
 * The buttons row is java script, does every browser allow it ?
 * Everything is interesting in our discussions. Errors are amongst the most because everyone may learn from them. There is enough space (room) left on the page, thank you. --DLL 20:50, 17 March 2006 (UTC)

All right! People want the complete record even full of irrelevant junk such as part of mine. Good to consult first and take action later. No action in this case. Twma 22:57, 18 March 2006 (UTC)

OK, twma, now I think I see what you were trying to do. You want a sequence of functions fn and a linear functional T such that fn → 0 but T(fn) does not → 0. But you chose a sequence of functions which does not converge, so clearly that is not going to show anything (your sequence doesn't satisfy your (a) and (b)). Now, the article discontinuous linear functional teaches us that on a complete vector space, all examples are nonconstructive, so you can't just look for a sequence and a functional, you have to invoke the axiom of choice. The vector space you have chosen is complete (though not in the uniform norm). The examples look like the one you find in that article. Let fn be a sequence of linearly independent functions. Denote by ||g||m,K the seminorm supK Dm|g|. If {Ki}i is any countable collection of compact sets which cover the real line, then ||*||Ki,m is a countable sequence of seminorms. Create a directed countable family of seminorms, and denote the kth seminorm by ||*||k. Then define a linear functional T such that T(fn) = n||fn||n. Then we have for any k, I can choose an fn so that |T(fn)| is greater than any ||fn||m (or even any finite linear combination thereof). Thus T is unbounded. Now use the axiom of choice to find a basis of the space which contains the fn, and define T to be 0 on the rest of the vectors. Any unbounded linear map is discontinuous.

Now, this example is the same in essence as the one at discontinuous linear functional, except that I had to get involved with families of seminorms, since the vector space in question is not a Banach space. This makes the details of the proof a little more involved, but doesn't change the idea behind it. Nevertheless, I found it quite confusing dealing with those seminorms, I wonder if you'll find it just as confusing when trying to read it as I did trying to write it. -lethe talk [ +] 21:39, 17 March 2006 (UTC)

I do NOT know how to produce nice looking mathematical stuff on this page. Be grateful if SOMEBODY could REPLACE this and next paragraph with something equivalent. It compiles well before I pasted here. Thank you in advance. PS: Somebody has done this. Thanks from twma.

twma's example
Here is an example of a discontinuous linear form on the test space D(R) on the real line R. For every compact subset K of R, D(K) is the vector space of all test functions with support contained in K. The topology of D(K) defined by the seminorms $$\|\varphi\|_\alpha=\sup\{|(d/dx)^\alpha\varphi(x)|:x\in K\}$$ for all integers α ≥ 0 coincides with the subspace topology induced by the test space D(R) equipped with the locally convex inductive topology induced by D(K) for all compact subsets K of R.

Let ρ be a test function with ρ(0) = 1 and ρ(x) = 0 for all |x|>1. Choose 0 α, we have ||fn||α ≤ 1/n → 0. Now fn → 0 in D(K) and hence also in D(R). Next because all intervals [an,bn] are disjoint, the sequence {fn} is linearly independent and hence it can be extended to an algebraic basis B of D(R). Define T(φ)=1 for all φ in B and extend T to a required linear form on D''(R).

Thanks to the enthusiasm of people of this group that provides encouragement to me in sharp contrast to DingoBabyAffair. Twma 08:24, 19 March 2006 (UTC)


 * I'm trying to read through your example, but your ρ function bothers me. It is discontinuous at x=0, whereas we usually choose test functions to be smooth.  For starters, none of the seminorms is defined for this function (except the 0th one).  Well, anyway, examples are not hard to come by.  Just find any linearly independent sequence and make an unbounded operator on it.  PS I'm glad you find this place hospitible; feel free to hang around or come back any time.  But who or what is DingoBabyAffair? -lethe talk [ +] 08:54, 19 March 2006 (UTC)

Let ρ be a test function with ρ(0) = 1 and ρ(x) = 0 for all |x|>1 (typo corrected). Movie: Lost in translation. Hope it should be all right now (waiting for your approval). In order to make life easier for myself, I only look at the positive side but rarely at the ugly part of the world until I HAVE TO face it NOW. I agree that based on one example, we can find many more without any difficulty but we DO need one example for a copycat. ONCE AGAIN, THANKS TO THE HELPING HANDS OF THIS GROUP. DingoBabyAffair appeared on the front page of many Australian newspapers for several years in early seventies. A baby had been severely hurt continuously but nobody wanted to help while the politicians argued whether the surrounding dingoes or the mother should be responsible. In SHARP CONTRAST, this group helps without any argument. After more than 30 years of protracted war, this baby grew up as a lost child of cold war. Now he/she is rising like a fabulous phoenix from the ash, new and fresh and young. Saving the Private Ben happened only in the movie while the politicians argued once again whether it did happen recently in reality. Perhaps I should claim copyright for another movie. Am I violating the rules and regulations of this site by comparison in my thank-you-note and by answering query about something irrelevant to Mathematics? Any way, I do not think it will drag on. Twma 00:51, 20 March 2006 (UTC)
 * OK, with the typo fixed (and the fix was obvious, I should have been able to figure it out), I think it's a fine example. In the end, you have lim fn=0, but lim T(fn) = 1.  I chose to make my operator unbounded, while you chose to make the discontinuity explicit.  We can also check that your operator is unbounded: for any α, 1/||fn||α outgrows any bound.  You also explicitly constructed a convergent independent sequence, which was nice.  I like it.  About the DingoBabyAffair, now I realize you're talking about the Azaria Chamberlain disappearance.  I think I saw a made-for-TV movie about that when I was a little kid.  And then there was some famous scene in Seinfeld about it.  So you're comparing the response you got on other internet sites to the public response to the dingo affair?  Well OK, whatever.  Anyway, you're not violating our rules.  You came here with a good math question.  It's OK to chat about other things on the side while you're doing math, I think. -lethe talk [ +] 15:42, 20 March 2006 (UTC)
 * PS I've copied your example to talk:discontinuous linear map, for the possibility of inclusion in our article. Hope you don't mind! -lethe talk [ +] 17:39, 20 March 2006 (UTC)


 * I am NOT talking about different internet sites but the mentality of different ENVIRONMENT.
 * In the academic world, we do not make money. If we are lucky, we make a name. In this case, your name should be with the example too. Do whatever people want. I do not claim copyright and take no responsibility if any mistake is found. It appears that THIS CASE IS CLOSED with a happy ending. Thanks again. Perhaps I should help the others whenever I have one or two days free. I got this site through some search engine. Twma 23:47, 20 March 2006 (UTC)

Simple Equation
How would one solve the equation 3^x = x^2? Is there a rule which specfies which explains the solvability of equations? --Alexs letterbox 05:35, 17 March 2006 (UTC)


 * Take the square root of both sides, and then use Newton's method to try to find the fixed point. There might be more than one fixed point for the different signs, so consider positive and negative signs first. --James S. 05:44, 17 March 2006 (UTC)


 * As a total cheater, I graphed it on my TI-83 and got x = -0.6860276 --DevastatorIIC 07:41, 17 March 2006 (UTC)


 * I asked something similar a few weeks ago (but was 2^x = x^2). Apparently there's a tool for this, called Lambert's W Function ☢  Ҡ i∊ ff   ⌇  ↯  08:06, 17 March 2006 (UTC)
 * Yes, this equation can't be solved using elementary functions. It can be solved numerically using your favorite numerical method, or with Lambert's W function, in terms of which the solutions are:
 * $$-\frac{2W(\pm\frac{\ln{3}}{2})}{\ln{3}}$$
 * One of them is real, and is roughly -0.686026724536251319. -- Meni Rosenfeld (talk) 08:38, 17 March 2006 (UTC)

Hello: Angle questions
(Title was changed to actually say something useful. StuRat 16:01, 17 March 2006 (UTC))

I know this might be a simple question but can someone please explain to me what Perpendicular mean? And the angle of Depression and Elevation mean ? Thankz


 * Is there something wrong with the article Perpendicular? —Keenan Pepper 13:09, 17 March 2006 (UTC)
 * About the angles of depression and elevation, I guess that would depend on the context. -- Meni Rosenfeld (talk) 15:41, 17 March 2006 (UTC)
 * Perpendicular is used to describe the convergance of two lines so that they form a ninety degree angle. If a line is given, then naturally a perpendicular line is at a ninety degree angle to that line.65.95.44.127 17:12, 17 March 2006 (UTC)


 * Generally, angles of depression or elevation are the angles you have to look down or up (respectively) to see something. I found a concise explanation for you. Superm401 - Talk 20:32, 18 March 2006 (UTC)
 * An angle of depression is the angle between the horizontal and the line looking down towards something. The angle of elevation looks from the horizontal up towards something. Most of the time, these are opposite angles on a transversal, and therefore congruent. --Geoffrey 03:17, 22 March 2006 (UTC)

Primorials
I need to know the growth rate of Primorials(for factorials it is X!=O(e^(x*ln(X)) using Big O notation.) I also need to know how fast you can calculate primorials, is it polylogarithmic?  For factorials it is not.  Ozone 23:12, 17 March 2006 (UTC)
 * You're wrong about the growth rate of factorials, but you're close. Stirling's approximation is
 * $$n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$

As for primorials,
 * $$\pi(x) \sim \int_2^x \frac{\mathrm{d}t}{\ln(t)}$$

suggests, (using some manipulations I am unable to generate here, because I can't figure out how to get the "#" symbol within a $$$$ equation)
 * n# =O(e^n)

&mdash; Arthur Rubin | (talk) 22:44, 18 March 2006 (UTC)


 * How about $$n\# \sim e^n$$? —Ilmari Karonen (talk) 00:20, 19 March 2006 (UTC)


 * Pretty! ☢  Ҡ i∊ ff   ⌇  ↯  03:35, 19 March 2006 (UTC)


 * Actually, that formula is wrong, because of the variability of prime gaps. How about (at least as a conjecture),
 * $$n\# = O \left( e^n \right)$$
 * &mdash; Arthur Rubin | (talk) 17:28, 19 March 2006 (UTC)
 * Just testing my understanding here: What you mean is that the ratio $$\frac{n\#}{e^n}$$ is bounded, but does not converge to a limit? -- Meni Rosenfeld (talk) 17:43, 19 March 2006 (UTC)
 * I think it's bounded, but if it converged to a limit (other than 0), then the prime gaps would be bounded. &mdash; Arthur Rubin | (talk) 17:46, 19 March 2006 (UTC)
 * I'm almost certain that $$\ln n\# = n+o \left( n^\alpha \right)$$ for &alpha; = 1, and possibly (assuming the Riemann hypotheses) for &alpha; = 0.5. The conjecture that $$\frac{n\#}{e^n}$$ is separate.  I stand by it as a conjecture, but I don't have an idea of a proof. &mdash; Arthur Rubin | (talk) 18:24, 19 March 2006 (UTC)

= March 18 =

covariant hom functor is left exact
http://en.wikipedia.org/wiki/Exact_functor

this article defines left exact functor

Suppose 0→A→B→C is an exact sequence

I am having difficulty to see that the hom(M,-) is a left exact covariant functor. Why is it left exact? Somehow it seems to me that that means proving that if Y is a submodule of X, and alpha is a module morphism of Y to Z, it can be extended to the full X?

Thanks,

Evilbu 16:48, 18 March 2006 (UTC)


 * No, the statement you have given is the statement of right exactness of contravariant hom (i.e., hom(-,X)), and as you apparently have noted, this does not hold. The corresponding statement in the context you want says that if you can map M into A, and A is a submodule of B, then the composite map M into B uniquely determines the original map.  Perhaps that's a little more believable?  —Blotwell 17:13, 18 March 2006 (UTC)

Solving an equation
How would you solve this equation:


 * $$L = \sqrt{J^2-P^2} + \sqrt{K^2-P^2}$$

for P? (So that P is only on one side.)

Thanks for any help.

65.31.80.100 19:12, 18 March 2006 (UTC)
 * You need to raise both sides of the equation to the second power, move the remaining square root to its own side, and then once more raise to the second power. From there it's quite straightforward - it's just a biquadratic equation in P. -- Meni Rosenfeld (talk) 19:32, 18 March 2006 (UTC)
 * I don't think that would work. By squaring the right side, you would not remove the roots. The end result would look something like this:
 * $$L^2 = J^2-2P^2 + 2\sqrt{J^2K^2 - P^2J^2 - P^2K^2 + P^4} + K^2$$
 * or something along those lines. The point is that you do not remove the roots. - APower
 * You didn't read the whole thing. You square both sides, removing one of the roots. Then you isolate the root and square again.  Superm401 - Talk 20:37, 18 March 2006 (UTC)
 * In other words, we produce
 * $$4 L^2 P^2 = -J^4 + 2 K^2 J^2 + 2 L^2 J^2 - K^4 - L^4 + 2 K^2 L^2 \,\!$$
 * To solve this, we must assume that L is non-zero. --KSmrqT 06:31, 19 March 2006 (UTC)
 * And of course, if L=0, then the solution is J=K=P. -- Meni Rosenfeld (talk) 14:36, 19 March 2006 (UTC)
 * Be careful; that's too restrictive. Although we assume the (non-negative) principal square root, so each square root must separately vanish in the original formula, that still leaves the signs of J, K, and P to vary freely. And to be thorough, we should note that since we have no control over J and K, there may be no solution. --KSmrqT 19:54, 19 March 2006 (UTC)
 * You're right, that's what I meant, I guess I didn't write it accurately enough. -- Meni Rosenfeld (talk) 10:44, 20 March 2006 (UTC)
 * The signs of J, K and P will be arbitrary anyway, since only their squares appear in the equation. So one might as well just restrict the problem to non-negative values and note that other symmetric solutions also exist.  —Ilmari Karonen (talk) 16:22, 20 March 2006 (UTC)


 * Multiplication by the conjugate ($$\sqrt{J^2-P^2} - \sqrt{K^2-P^2}$$)and some algebraic manipulations result in a system of equations
 * $$\sqrt{J^2-P^2} + \sqrt{K^2-P^2}=L$$
 * $$\sqrt{J^2-P^2} - \sqrt{K^2-P^2}=\frac{J^2-K^2}{L}$$
 * Which we can solve to get
 * $$P^2=J^2-\left(\frac{L^2+(J^2-K^2)}{2L}\right)^2

=K^2-\left(\frac{L^2-(J^2-K^2)}{2L}\right)^2$$
 * Interestingly enough, unless I made a mistake with signs I get
 * $$P^2=\frac{(K+J-L)(K+L-J)(L+J-K)(L+J+K)}{4L^2}$$
 * (Igny 18:42, 20 March 2006 (UTC))
 * The signs are ok, but it should be L^2 in the denominator and not L^4. In any case, I'm not sure your method is really easier. -- Meni Rosenfeld (talk) 19:34, 20 March 2006 (UTC)
 * Thanks, I corrected this. The methods (Viète formula and quadratic formula) are essentially equivalent. (Igny 19:43, 20 March 2006 (UTC))

= March 19 =

Integral of a Rational Function
I know all about integration by parts, but is there any one integral formula that can find the integral of (a^nxn + a^(n−1)x(n−1) + ... + a2x^2 + a1x + a0) / (b^nxn + b^(n−1)x(n−1) + ... + b2x^2 + b1x + b0). This is clearly the classic definition of a polynomial divided by a polynomial. Is there an integral formula that uses similar notation to account for all possible terms? --Chris 03:24, 19 March 2006 (UTC)
 * I doubt there is. Any such formula would have to be capable of spontaneously generating factors of pi; for an example that's recent in my memory see the integral in Proof that 22 over 7 exceeds π. So you can't expect a simple function of the coefficients and endpoints. Melchoir 03:46, 19 March 2006 (UTC)
 * Recall that transcendental functions can be the antiderivatives of rational functions. -lethe talk [ +] 05:08, 19 March 2006 (UTC)
 * Yes, but I'm trying to point out that the problem in question isn't going to be easy. The integral of a polynomial is another polynomial, and the coefficients are simply related. For a rational function, there's no hope of a formula using "similar notation". Melchoir 05:14, 19 March 2006 (UTC)
 * Well, I think your argument about π doesn't seem relevant. -lethe talk [ +] 05:31, 19 March 2006 (UTC)
 * It proves by counterexample that the definite integral of a rational function is not itself a rational function with rational coefficients of the original coefficients and endpoints. I could've made that point with a simpler (and less interesting) example like 1/x, but it's still worthwhile to point out, and I suspect that it speaks to the question. Melchoir 06:31, 19 March 2006 (UTC)
 * I suppose you mean indefinite integral. The definite integral of any integrable function is just a number.  I guess if your only point was that the antiderivative of a rational function need not be a rational function, then of course I agree with you.  But that doesn't meaan that we can't write down a formula for the antiderivative of rational functions.  On the contrary, we can. -lethe talk [ +] 07:03, 19 March 2006 (UTC)
 * I meant the definite integral, to be considered as a function of all the numerical inputs to the problem, which I leave indeterminate. Chris is a college student, and he probably has access to the method of partial fractions. He asks for "one integral formula" using "similar notation", and neither of us has such a thing. The review of partial fractions below is valuable, as is the Risch algorithm. But to start with, there is nothing wrong with explaining why some things can't be done. Melchoir 07:27, 19 March 2006 (UTC)
 * Um, so you refer to indefinite integrals as "definite integrals with indeterminate bounds of integration"? Weird, but I guess that's your prerogative.  In that terminology, is there even any distinction between definite and indefinite integrals?  Do you even need two terms?  Anyway, I'm still not clear on what it is that can't be done.  Is it that the notation can't be kept similar?  Or is it that the relation between the coefficients can't be kept "simple", for some definition of simple? -lethe talk [ +] 08:07, 19 March 2006 (UTC)
 * Ugh, no, I meant something else, but that's not important. As for how simple a formula one might concoct, we can only speculate, and to that end, it is helpful to consider what feats that formula is expected to perform. That's enough. Melchoir 08:25, 19 March 2006 (UTC)

Chebyshev's theorem says that
 * $$x^p(a+bx^r)^q$$

has an elementary antiderivative if and only if one of (p+1)/r, q, or (p+1)/r + q is an integer. But I guess q is always an integer for rational functions, so this doesn't answer your question. Probably the answer is: perform synthetic division, factor the denumerator, and integrate by partial fractions. I think you can write a general formula this way. -lethe talk [ +] 04:40, 19 March 2006 (UTC)

So like, assume you're working over the complexes, so that your polynomial can be factored. And assume that the denominator has greater degree than the numerator; you can always make it so by synthetic division. Then you're left with an integral of the form
 * $$\int \frac{x^{n-1}+a_{n-2}x^{n-2}+\dotsb+a_0}{(x-r_1)^{m_1}\dotsb (x-r_q)^{m_q}}\,dx,$$

with
 * $$m_1+\dotsb+m_q=n,$$

so there are q distinct roots, and the ith root has multiplicity mi. You find the integral of this bad boy by partial fractions. The partial fraction method gives you a general formula for the coefficients. The article on partial fractions gives you this formula for the case that all the roots are distinct (have multiplicity 1). I don't know about the general case. Then you perform the integration. For the distinct root case, you're left with
 * $$\sum c_i\log |x-r_i|.$$

For the general case, once you have the coefficients, the integrals are equally easy.

Note that you can't write the entire thing in one formula in terms of the coefficients of the polynomials, because by a result of Abel, there is no formula (in terms of finitely many radicals) for the roots of a polynomial. -lethe talk [ +] 05:05, 19 March 2006 (UTC)


 * Try reading about the famous Risch algorithm, and its relatives. --KSmrqT 06:39, 19 March 2006 (UTC)

Probability
There's something basic I don't understand about probability. It isn't the manipulation of the numbers (which has never been a problem for me), it's why the whole thing's possible. What about a natural series of events causes them to fall, given a large sample size, so perfectly into a particular arrangement like 50/50 or 16.7/16.7/16.7/16.7/16.7/16.7? This might be a better question for the science desk, but it seems to fit here too, and most people work both. Black Carrot 22:05, 19 March 2006 (UTC)


 * Have you read Law of large numbers? —Keenan Pepper 22:28, 19 March 2006 (UTC)


 * Depending on the level of sophistication you wish for the answer, these explanations may or may not help you. Suppose you throw a perfectly symmetrical coin 1,000,000 times, and calculate the h, the proportion of heads. Then:


 * The probablity that you will get, say, h<0.49 or h>0.51, can be calculated and shown to be extremely small. This is a kind of a circular argument, explaining probability by using probability - But it shows the the concept is internally consistent.


 * Suppose you get h=0.6. That would mean that the coin "prefers" to land on heads. But why would it do that? We assumed it was perfectly symmetrical. The same would happen if you got h=0.3. Since the coin has no preference, the only "logical" result is h=0.5, with equal proportions to heads and tails. Of course, deviations can occur. But if you threw the coin only 10 times, deviations can be attributed to "coincidence". If you throw it a million times, you can no longer attribute a significant deviation to a "coincidence", but only to a real preference of the coin - Which, again, it shouldn't have.


 * -- Meni Rosenfeld (talk) 10:38, 20 March 2006 (UTC)

If you accept that the chances of a combination of random (equally weighted) events occurring is 1/c, where c is the total possible combinations of events, then you could demonstrate it yourself. This is because the more tosses of a coin, say, you do, the more combinations of all tosses will end up in any given range about the predicted average, say the 0.4 - 0.6 range. However, the chances of getting exactly 0.5 will actually go down (and of course that probability is zero for an odd number of tosses). As the number of tosses of a coin goes up, the chances of having all heads or all tails goes down. In the case of a single toss, you get either all heads or all tails (H or T) 100% of the time, while with two tosses, you get all heads or all tails (HH or TT) only 50% of the time, and with 3 tosses (HHH or TTT), only 25% of the time. The probability of getting a head/tail ratio somewhere in the middle goes up as a result, particularly as you get closer to half heads and half tails, because a larger percentage of combinations works out to be near half heads and half tails when you have more tosses. You can see this trend with Pascal's triangle (which models coin toss behavior as well as many other things), where the numbers in the center of each row continue to get larger relative to the edges of the row as you move further down the rows. StuRat 16:36, 20 March 2006 (UTC)


 * Cogent replies. I'm familiar with the law of large numbers (though I as yet don't understand enough of the symbols to grasp that article), and I understand how to manipulate an already determined probability (such as 1/c) into its necessary consequences. I hadn't really thought about using a coin's symmetry to prove the likely result, that's a good idea. I'm glad you all said what you did, because I think I can pinpoint what I don't get about it. I don't get how the unordered and unconnected interactions of the universe (the infinitely complex causes that shape a random event), which in themselves seem to show no interest in leaning one way or another as a matter of course, wind up so reliably filling in probability curves the way they're supposed to. I don't quite mean the average as sample size goes to infinity, which as you say can be reasoned directly from an assumption that each event is uniformly weighted (or whatever weights are convenient). I mean, why do the individual trials scatter so perfectly along that original curve in the first place? Or do they? Black Carrot 02:03, 22 March 2006 (UTC)


 * Let me try explaining with a two-dice example. Here are all 36 possible rolls listed by their totals, and the odds of rolling each total:

Total Rolls                     Odds ===== =======================   ====  2 -> 1:1                        1/36  3 -> 1:2,2:1                    2/36  4 -> 1:3,3:1,2:2                3/36  5 -> 1:4,4:1,2:3,3:2            4/36  6 -> 1:5,5:1,2:4,4:2,3:3        5/36  7 -> 1:6,6:1,2:5,5:2,3:4,4:3    6/36  8 -> 2:6,6:2,3:5,5:3,4:4        5/36  9 -> 3:6,6:3,4:5,5:4            4/36 10 -> 4:6,6:4,5:5                3/36 11 -> 5:6,6:5                    2/36 12 -> 6:6                        1/36


 * This chart says that you are 6 times more likely to roll a 7 than a 12. So, for this not to work out to be the case after a large number of trials, the dice would have to be weighted towards rolling 6:6.  Does this help any ? StuRat 03:58, 22 March 2006 (UTC)
 * To emphasize, each experiment you can imagine can usually be represented as a union of equivalent cases, neither of which should be preferred over the other. In the simple example of throwing 2 dice, getting a sum of "3" is really identical to getting 1:2 or 2:1. Since the dice have no preference, when doing the experiment many times, each result should appear an equal number of times. Only 2 of the 36 possible & equivalent results will give a sum of 3. So 1/18 of the throws should give 3. Any other proportion would indicate a preference one way or the other.
 * What is true in such simple cases, is also true in any other case. Suppose you measure the lifetime of an unstable particle. The experiment can (not exactly but with arbitary precision), be broken down to discrete and equivalent results, and we can count (or mathematically calculate) the proportion of results that will give a specific range of lifetimes. By checking many particles, we expect that the proportion of particles with lifetimes in the given range, will match the proportion of suitable cases - Anything else will, again, indicate a preference to specific cases, which we assumed to be all equivalent. -- Meni Rosenfeld (talk) 13:06, 22 March 2006 (UTC)

'''I mean, why do the individual trials scatter so perfectly along that original curve in the first place? Or do they?''' You are trying to over-interpret probabilities. Probability theory (PT) provides a mathematically rigorous logical framework to fit a rule on the occurrence of some phenomenon. It cannot and does not deal with explaining the phenomenon. Say in the case of a biased dice---on the basis of given observations---all that PT tells you is the preference of the dice towards some of the possibilites (for example, say p(1),p(3),p(4)=0.2 and p(2),p(5),p(6)=0.4/3). PT never claims to explain the reason for that bias (you can analyze the reason of that bias by going in details of the geometric shape of the dice etc., but you are not concerned with such details, when you are simpling fitting a rule over the observations). The same reasoning applies reverse: when the events scatter along the proability distribution, in no single trial, the dice is making a conscious or deterministic choice, and even if say, it is making a conscious choice---All in all, PT provided you was the aggregate characteristic or rule about how trials would turn out on an average (in expectation). This is "the" subtlity of probability theory, and many greats have erred when they tried to over-interpret probabilities. The famous debate between Bohr and Eeinstein about correctness of Quantum mechanics is, in fact, centered around this understanding. Of course, both these greats understood PT perfectly and their argument was mainly concerned about whether it is worth to go in additional details,which---at least right now---seems impossible to verify, when PT explains pretty much everything about the world that we could verify experimentally. Check out the article "Probability Theory as Logic" by late Edwin Jaynes at. --35.9.136.15 18:49, 23 March 2006 (UTC)Keyur Desai

Black Carrot, perhaps you are also making a logic error, in thinking that if heads comes up the first time a fair coin is flipped, that makes it more likely that tails will come up the next time "to even out the probabilities". This is not the case. It's has a 50:50 probability each time it's flipped, regardless of past flips. StuRat 19:18, 23 March 2006 (UTC)
 * Perhaps you can be clearer about what you mean by "that original curve", and why shouldn't the trials scatter perfectly along it? StuRat and I gave examples why should the trials scatter along a piecewise-linear curve and an exponential one. Which curve do you have in mind? -- Meni Rosenfeld (talk) 19:24, 23 March 2006 (UTC)

Hi all, sorry, but if I understand Black Carrot's original question, that user is asking something much more simple: why is it that when you throw a die a bunch of times, it lands on each of the 6 sides about 1/6th of the time? Am I right? If I'm wrong, sorry.

If so, the short answer is, of course, Why not? The longer answer is that we are using, as a mathematical model of the physical situation, the assumption that the die does this. This would be true for a "fair" die. This, of course, would not be true for a loaded one. One could build a die so that it was much more likely to land as a 1 than anything else. Then, of course, this 1/6 model wouldn't be a good model. The probability of landing on different sides would change because of this, and you'd have to build that into your model. Of course, mathematicians have worked this out as well: see Bernoulli trials for a start.


 * No, I recognize that unconnected events don't influence each others' probabilities. Although they can, as it happens, influence the actual outcome, just in a wildly unpredictable (aka random) way. Do you think the way a coin comes up one time doesn't effect how the person moves the next time?
 * To answer this: This is, of course, true. It's possible that seeing a heads will put you in a different psychological frame of mind than seeing a tails, and this could conceivably cause you to flip the next coin in a different way.  But the modelling hypothesis we would use there is this:  the eventual outcome of the coin-flip depends so strongly on microscopic changes in how the coin is flipped so that the user can't really affect it.  I mean, certainly it's clear that a coin flipped or dice-roller can't consciously control what the outcome is; if so, casinos would be out of business (or more into the business of taking "lucky" people out back).  But, you know, maybe you're right, and this could be a scientific experiment:  get 10,000 people to flip a coin 10,000 times each, and measure the correlations.  On the other hand, the model of probability we're using assumes that the trials are independent, so if it turned out that this thing was true for people, you've not shown that there is a problem in probability theory, you've shown that the error is in modelling a particular real-world situation by a particular theory.  Hope this helps! --Deville (Talk) 16:55, 25 March 2006 (UTC)
 * Probability Theory as Logic- I could only find the abstract and table of contents. Did I do something wrong?
 * Original Curve- I meant the curve from which others were reasoned. For the shrinking normal curve of coin flipping, it'd be the evenly weighted binary curve of a single flip. Of course, that could also be broken up, by tracing it back to the motion of the thumb or even further, but then there'd just be a new first one. I meant, how do we work with and explain the original one, which has to be directly defined in terms of natural laws? From there, of course, everything can be inferred. The simplest seems to me to be a flat line across the graph from 0 to 1, but I can't think of a situation that describes.
 * Preference/Symmetry- This is a tricky one. I'll get to it when I know what to say. Black Carrot 16:33, 25 March 2006 (UTC)
 * I'm still not sure I understand. In any case, I think symmetry is a key issue here. If you take symmetry as a premise, I think you can derive anything else. Are you asking, why would it be symmetrical in the first place? That's a tricky one, but you can relate the symmetry of the result of the coin toss to symmetries in the physical process of tossing the coin. Why would those be symmetrical? My guess is, because the fundumental laws of the universe are symmetrical. But ultimately, I don't believe that it can be "proved" that probability works, just that is internally consistent and backed up by experimental evidence. -- Meni Rosenfeld (talk) 17:07, 25 March 2006 (UTC)
 * More specifically, if you take an actual, say, US quarter, and flip it, it turns out that the probability of it turning up heads is different than the probability of it turning up tails. I'm not sure which one is greater, but the experiment is easily done by a machine and they can figure these things out.  Also, in fact, there is some non-zero probability that the coin actually lands on its edge.  In short, for a real coin there isn't symmetry, and the 50/50 assumption is invalid.  However, if one postulates a truly fair coin, i.e. a short fat cylinder of metal which is completely homogeneous throughout and completely symmetric, well then, of course, the probability is equal that it lands on one side or the other.  As for the proof of this:  Let us assume that the probability of landing on side 1 is p, and the probability of landing on side 2 is q.  Now, consider any coin flip where you take into account the exact position of the flipper's fingers, the air currents, everything.  The coin will land on whichever side it lands on.  Now, take the same initial condition, except imagine exchanging the two sides of the coin right before it leaves the flippers' hand.  Then it will obviously land on the opposite side.  Since our model doesn't distinguish with how the flipper holds the coin (and in fact, if the coin is truly symmetric he wouldn't even know which side is up), this means that p must equal q.  If we further assume that $$p+q=1$$ (no landing on sides), then we get 50/50.  Again, however, in the real world with a real coin, the probabilities are NOT equal. --Deville (Talk) 17:22, 25 March 2006 (UTC)

= March 20 =

Line integrals and path independence
Let's say you're taking the line integral (path integral) along four straight lines C1, C2, C3, and C4, where C1 is y=-2, C2 is x=2, C3 is y=2, and C4 is x=-2, so you're integrating along a closed curve C, where C is a 4x4 square centered at the origin. (The lines C1-C4 are segmented so that they only go from x=-2 to 2 and y=-2 to 2; the lines are oriented counterclockwise, therefore positive.) Then, let's say you're taking the integral of -ydx/(x^2+y^2)+ xdy/(x^2+y^2) along the curve.

Now, dP/dy equals dQ/dx, so the differential is exact, so the integral is path-independent. Because the curve is closed, this would make the integral equal to zero, because you can choose any path you want, and you could choose a path that remains at the starting point and not go anywhere. All path-independent integrals over closed curves are equal to zero, right? (Right?)

But then my textbook (this is an example from my textbook) shows how one can use Green's theorem here. Generally, one wouldn't be able to use Green's theorem because the denominator in the integral makes the integral undefined at (0,0). But one could draw a circle C', x^2 + y^2 = 1, around the hole, and the textbook shows how the line integral along C (the square) is equivalent to the line integral along C' (the circle). Evaluating this integral quickly gets the answer of 2pi.

My question is this: aren't all path-independent integrals over closed curves 0? What's different here? I'm not so concerned with the specific example - it doesn't have to be that initial curve C. When there's a hole in the region bounded by C, and C is path-independent (b/c the differential is exact), why does a discrepancy arise between a) assuming the integral is 0 because, being closed, the start point is the endpoint, and b) encircling the hole and using Green's? zafiroblue05 | Talk 19:31, 20 March 2006 (UTC)
 * If I remember correctly my multivariate calculus, I think the problem is with the singularity at (0, 0). True, the integral does not depend on the specifics of the path, but it does depend on the number of revolutions around the singularity(ies). Each revolution (in the positive direction) adds a constant term to the integral (in your case, 2&pi;). You can think of your integral as going between 2 different points on a Riemann surface. -- Meni Rosenfeld (talk) 19:46, 20 March 2006 (UTC)
 * Meni is exactly right. Poincare's lemma only holds in contractible spaces.  Your function is defined on a space with a hole in the middle, therefore Poincare's lemma doesn't hold, the differential is not exact, and the integral is not zero. -lethe talk [ +] 19:58, 20 March 2006 (UTC)
 * Viewed another way, the source of confusion is with the exactness of the differential form &omega; = P dx + Q dy = −y/(x2+y2) dx + x/(x2+y2) dy. Naively, it appears that ∂P/∂y |x and ∂Q/∂x |y are equal for all x and y, which should ensure that &omega; = dα for some differential form α. However, there is a problem at x = y = 0. Also note that Wikipedia's statement of Green's theorem insists that P and Q have continuous partial derivatives over the region in question. Any way we look at it, the vanishing of the denominator, x2+y2, cannot be ignored. --KSmrqT 21:00, 20 March 2006 (UTC)

Thank you all, that explains it perfectly. :) zafiroblue05 | Talk 22:32, 20 March 2006 (UTC)
 * Sorry if I'm being superfluous, since everyone has answered this question in different ways, but let me tie it together. If you take any closed curve in the plane, then deform the curve smoothly so that you do not pass through a singularity of the vector field, then the line integrals do not change.  (One can see this by the same argument made above for the two contours C and C'.)  As long as the vector field doesn't have any singularities, you can simply deform this curve out of existence; make it smaller and smaller and then simply disappear.  However, when there is a singularity inside your curve, you're only allowed to move the curve to a small circle around the singular point.  Thus, one can see that the line integral is simply going to depend only on which singular points it contains.  One can compute that this is simply $$2\pi$$times the residue at that point; using complex analysis makes this calculation quite nice.  Hope this helps!  --Deville (Talk) 19:27, 23 March 2006 (UTC)

Coplanarity test
This one should be easy for you guys. Suppose I have a set of 100 points defined by P[i] = X[i]i + Y[i]j + Z[i]k, i = 1..100. How can I test whether these points lie in a plane or not? I can think of choosing any three points and finding the normal vector of the plane defined by them, and then taking the dot product of all P[i] - P[0] with that normal and checking if it is zero. Is this the best way of going about it, or is there something more elegant? How can I find the normal vector of the plane defined by 3 points. Sorry for asking such elementary questions, but I'm kinda retarded on this. deeptrivia (talk) 21:19, 20 March 2006 (UTC)
 * There are a number of independent ways of determining whether a set of points is coplanar. The simplest that I can think of is that the 100 &times; 4 matrix described below has rank 3 (or less, if the points are collinear or coincident):


 * $$\begin{bmatrix}X_1 & Y_1 & Z_1 & 1 \\ X_2 & Y_2 & Z_2 & 1 \\ \vdots & \vdots & \vdots & \vdots \\ X_{100} & Y_{100} & Z_{100} & 1\end{bmatrix}$$


 * If the rank is 3, the normal vector is the first three components of the (unique, up to scale factor) vector in the (right) Null space of the matrix. &mdash; Arthur Rubin | (talk) 21:42, 20 March 2006 (UTC)

Okay, there's a slight problem with this method in my case. Since these arrays X[i] etc are floating point numbers, the points can be very slightly out of plane (deviations such as 1e-10.) I want the algorithm to somehow ignore these small deviations. How do I manage that? The rank of the above matrix might come out to be 4 even if practically the points are coplanar. How do I increase the tolerance value for calculation of rank. Thanks! --deeptrivia (talk)


 * A floating point calculation is qualitatively different from a precise calculation with exact real numbers. What you likely want is a regression analysis, typically a linear regression, to find a best-fit plane, followed by an examination of the residuals. Then you must make a judgement call, based on your circumstances, about how much deviation from planarity is acceptable. Some of the deviation will be a result of limited floating point precision, some of it will come from measurement error, and some of it may be honest non-planarity. --KSmrqT 23:25, 20 March 2006 (UTC)


 * Thanks for your comment. I'm getting these floating point numbers not from a measurement, but from some functions. If I'm allowed to slightly change my question, suppose you have closed form functions X(s), Y(s) and Z(s). You have to find whether the curve defined by X(s)i + Y(s)j + Z(s)k is a plane curve or not. How can you do that? Regards, deeptrivia (talk) 23:41, 20 March 2006 (UTC)


 * A function can be "closed form" and yet horrible to contemplate. The ideal goal would be a test for linear dependence. Suppose N = ai+bj+ck is a normal to the plane (if it exists), and that P = xi+yj+zk is a point in the plane. Then for any value of s it must be the case that
 * $$0 = N \cdot (C(s) - P) ,\,\!$$
 * where C(s) = X(s)i+Y(s)j+Z(s)k. Or, in components,
 * $$0 = a(X(s)-x) + b(Y(s)-y) + c(Z(s)-z) .\,\!$$
 * Choose P however is convenient, such as letting x = X(0), y = Y(0), z = Z(0). If such a normal N exists, it can be found exactly from three independent points. And if the closed forms are not too complicated, then it may be possible to decide if the weighted sum is identically zero for all s.
 * It may be helpful to experiment with degree 2 polynomials; the curves are guaranteed planar. For example, try
 * $$X(s) = 1 - s^2 \,\!$$
 * $$Y(s) = 2s \,\!$$
 * $$Z(s) = 1 + s^2 \,\!$$
 * which includes the point P = C(0) = (1,0,1). Subtracting this gives (-s2,2s,s2), so we can clearly use N = (1,0,1). --KSmrqT 00:59, 21 March 2006 (UTC)
 * Alternative conditions are
 * $$0=N\cdot (C(s)-C(s_0))$$ for some $$N$$, as mentioned above
 * $$0=N\cdot C'(s)$$, for some N, tangential vectors lie in one plane.
 * $$\alpha N=(C(s)-C(s_0))\times(C(s_1)-C(s_2))$$, for some constant N and scalar $$\alpha(s;s_0,s_1,s_2)$$
 * $$\alpha(s,s_0) N=C'(s)\times C'(s_0)$$, for some constant N and scalar $$\alpha$$
 * By the way, in the example of Ksmrq above, $$C'(s)\times C'(s_0)=-4(s-s_0)(\mathbf i+\mathbf k)$$
 * (Igny 20:36, 21 March 2006 (UTC))
 * A more elegant way to do this is to use the Frenet-Serret formulas formula. It the curve has zero torsion than in must lie in a plane. Essentially equivilent to the above. --Salix alba (talk) 20:37, 25 March 2006 (UTC)

What's the name of this little thing
A high-math question:

a * x + b --- z + y * x

This reads "a * x + b over z + y * x". But what is the English name of this little line betveen the numerator and denominator? It is not "over", not "fraction line". Sign of division? Sounds funny. Can't find anything in relevant Wikis, not in Math books. In Math you can go without mentioning it, if you do some typesetting, you have to name it somehow :-)

Miklos Somogyi


 * A typesetter would call the line a horizontal rule. In fact, HTML markup for the web uses the tag &lt;HR /&gt; for Horizontal Rule. The MathML discussion of fractions calls it a "bar", a "rule", and a "line", all within a few paragraphs. However, what we might use as a fraction bar here can be used with an entirely different meaning in, say, mathematical logic or elementary arithmetic. --KSmrqT 23:54, 20 March 2006 (UTC)

Thank you, I've got a few good-sounding names. It was not more difficult than Fredholm integral equations with a double-layer Cauchy-Kolmogorov kernel of the Third Kind, after all :-) Thanks, MS

I think this is also called a Vinculum. J. LaRue 01:58, 22 March 2006 (UTC)

This one is not right. Vinculum is on top of things (plural), to show that they are in some kind of a group. MS


 * Vinculum (meaning chain, binding or cord in Latin) usually means a horizontal line placed above an expression, but it can also be used as a term for the horizontal rule in a fractional expression, as it binds the terms in the numerator and denominator together - see and . Gandalf61 10:23, 23 March 2006 (UTC)

Yes, I may use the term "vinculum". If I want to join the Borg Collective, I even must. I suspect that mathematicians would suspect what I am talking about re vinculum but actually, do they USE the term themselves? Thanks, MS


 * I, like you, resist the use of overly complex terms for simple things (although, perhaps, "resistance is futile"). I simply call it "the line" as in "the numerator is the portion of a fraction above the line, while the denominator is that portion below the line". StuRat 18:03, 23 March 2006 (UTC)


 * The short answer is that I don't think a single mathematician would even use the term vinculum. However, everyone should, and I will now.  ;)--Deville (Talk) 19:14, 23 March 2006 (UTC)

= March 21 =

the kernels' chicken?
If kernel sanders really makes real mathematically accurate chicken, shouldn't the kernel's chicken have no calories? hence be = to the 0 vector? serious responses only please-Kolin farrellovsky 18:04, 21 March 2006 (UTC)
 * Well, if the chicken falls into the kernel of the "calories of" operator, then the chicken has no calories, but you can't assume that the chicken is 0, since the kernel can be a large space. This would be equivalent to kernel sanders himself falling into the kernel of the "calories of chicken made by" operator.


 * On the other hand, if kernel sanders falls into the kernel of the "chicken made by" operator, then his chicken is equal to 0, but you can make that conclusion without examining the calories. Melchoir 18:29, 21 March 2006 (UTC)


 * Er, to clarify, the second situation implies the first, but the first does not imply the second. So the moral of the story is that the meaning of "kernel" depends on which operator you're talking about. Melchoir 18:34, 21 March 2006 (UTC)


 * There's a kernel of truth in there somewhere. StuRat 18:46, 21 March 2006 (UTC)


 * The "calories operator" has a range which contains negative numbers, or at least I've heard that celery has a negative caloric content, as it causes you to burn more calories digesting it than it contains nutritionally. Since it is negative, then choose any food that actually tastes good, and some linear combination of that, and celery, will have zero calories.  As will any scalar multiple of that meal.  Thus there exist arbitrarily large meals with zero calories!
 * However, I guess it is not entirely accurate to talk of the kernel of the calories operator, because food doesn't form a group (or a vector space, for that matter). It's at best a module:  you can't eat negative 2 potatoes.  ;)--Deville (Talk) 19:12, 23 March 2006 (UTC)


 * It, does, and you can, if you pass to the Grothendieck group! If you're competing with someone else to lose (or gain) weight and all you care about is your relative food intake, then eating &minus;2 potatoes is the same as forcing 2 potatoes down your opponent's throat. Of couse, this technique is probably felony assault, so our budding mathematicians shouldn't try it at home. Melchoir 22:48, 23 March 2006 (UTC)

Where am I?
If I travel 10km North, then 10km East and then 10km South and find myself back where I started where am I?
 * Given that this is such an old and hoary one, a) it's almost certainly your homework, and b) elementary Google skills should give you the answer anyway (try using quote marks to enclose some relevant phrases and excluding the specific distances as they can be changed without affecting the basic characteristics of the problem). However, I'll give you a hint anyway: think about which areas on the Earth's surface are special in relation to the cardinal directions.  --Bth 18:14, 21 March 2006 (UTC)


 * There are actually two answers, the obvious one, and one that's almost on the opposite side of the world. StuRat 18:42, 21 March 2006 (UTC)

I prefer this version:

If you leave home, then travel 1000 km south, then travel 1000 km east, then see a bear, then travel 1000 km north, and end up back home, what color was the bear ?

StuRat 18:37, 21 March 2006 (UTC)


 * Actually, aren't there infinitely many circles which solve the problem? The "other" answer you mention is a latitude from which travelling 10km north leads you to a place, where travelling 10km east doesn't get you anywhere. But the latter is satisfied by any circle with circumference of 10km/n, where n is a positive integer. -- Meni Rosenfeld (talk) 19:14, 21 March 2006 (UTC)


 * Yes, that's right, but I am calling all of those circles "one answer" as they all are very close together (relative to the obvious answer on the other side of the world) and use a similar solution method. StuRat 20:13, 21 March 2006 (UTC)


 * I suggest we stick with the longstanding convention that 1 = 1, and 1 < 1 + n, where n is any positive number.71.224.204.167 04:24, 16 May 2006 (UTC)

= March 22 =

how math is used in a beautician job
How is math used in a beautician job, or better yet, what kind of math do beauticians use?


 * Well, if nothing else they have to figure out how much to charge their customers. Financial math pervades most businesses. Also, mathematical concepts like symmetry and the golden ratio are widely believed to be associated with beauty... Keenan Pepper 03:27, 22 March 2006 (UTC)


 * If you can work out the surface area of someone's lips, or someone's face, you can estimate the amount of cosmetic you may use to cover these. Dysprosia 04:13, 22 March 2006 (UTC)


 * Apart from any specific job, mathematics is useful with regard to many transactions. Gratuities of 15% or 20% require understanding multiplication and maybe addition. Percent off savings on groceries or other items, likewise. Interest on savings, or on loans, can be hugely important to understand. Budgets and taxes require mathematical calculations. And throughout life, mathematical knowledge can enrich experiences. A surprising number of people get through life without being able to read or write in their native language, but most would admit it limits them. People may defend lack of mathematics skills, yet never understand why a state lottery consistently takes their money and pays out far less. Which is the more dangerous way to travel from Atlanta to Chicago: automobile, train, or private plane? Perhaps an understanding of statistics would help. When ordering a pizza, is a 14 inch pizza for $9 a better deal than two 10 inches pizzas at $5 each? An understanding of geometry is relevant. Ever lock the keys in the car? That's an introduction to the algebra of noncommutative operators (key in door, key in ignition) the hard way. Not sure if a politician's arguments are sound? Perhaps a class in logic would help. This message is stored in bits and bytes, but what does that mean? The mathematics of information theory provides a simple and beautiful explanation. The beautician need never listen to any music either, whether metal or pop or country or classic; is a utilitarian argument necessary to realize that something is worthwhile? --KSmrqT 06:30, 22 March 2006 (UTC)

= March 23 =

Zeta constants series
Here's an interesting function:
 * $$f(x) = \sum_{n=2}^\infty \zeta(n) x^{-n} = \sum_{n=1}^\infty \frac{1}{(xn)^2-xn}$$

With:
 * $$f(2) = \log 2.\,$$

Anyone seen it before, and does it have a closed form? Fredrik Johansson 17:26, 23 March 2006 (UTC)
 * Well, Mathematica says that:
 * $$f(x) = -\frac{\gamma + \psi\left(\frac{x-1}{x}\right)}{x}$$
 * Where &gamma; is the Euler-Mascheroni constant, and &psi; is the Polygamma function. I guess you can't simplify this any further. -- Meni Rosenfeld (talk) 17:49, 23 March 2006 (UTC)
 * Right; it's basically the Taylor series for the digamma function (and a rational zeta series). Thanks. Fredrik Johansson 18:05, 23 March 2006 (UTC)

Factoring Random Integers
Choose a random integer with n decimal digits. I want to be reasonably certain that I can factor it using publically available software on a modern PC within, say, 24 hours. How large can n be? -Alecmconroy 22:32, 23 March 2006 (UTC)


 * Pretty darn big. I would think the largest integer which could be fully represented could be processed in 24 hours.  A 4 byte integer is up to 2^32 and an 8 byte integer is up to 2^64 (each is half as big if negatives are allowed).  So, I would say 2^64.  At that size, you would need to check a little over 2 billion possible factors.  On a computer that can do, say a billion operations per second (1 Ghz), that should only take a few seconds.  If you can find a factoring program that will handle 16 byte integers, that would allow integers up to 2^128.  Unfortunately, those would require several centuries to process. StuRat 22:53, 23 March 2006 (UTC)


 * With GGNFS, a bit over 100. Fredrik Johansson 22:46, 23 March 2006 (UTC)


 * How is that possible ? By my calcs, an integer of size 1x10^100 must be checked against all prime numbers up to it's square root.  If the square root is 1x10^50 and we figure around 7% of all integers are prime, that would give us 7x10^48 numbers to check.  Going with a 3 GHz computer, this would take a minimum of 2.33x10^39 seconds, or 7.4x10^31 years.  This isn't including the time to find the primes.  Do you have some vastly more efficient algorithm ? StuRat 23:06, 23 March 2006 (UTC)


 * See general number field sieve. Fredrik Johansson 23:11, 23 March 2006 (UTC)


 * I can't say I understand a word of that article, other than the number of operations calculations, which does work out to 9.25 hours, and with a little padding for overhead that would get us to 24 hours. So that part seems to work. StuRat 23:23, 23 March 2006 (UTC)


 * I don't understand it either. The source code for the above-mentioned implementation of the GNFS is over 1 MB (!) in size. There are also simpler algorithms that are better than trial division (the article integer factorization has a list), but none that is simple and a reasonable choice for factoring a random 100-digit integer. Fredrik Johansson 23:40, 23 March 2006 (UTC)


 * Another program is MSieve: "On a 2GHz Opteron, a 95 digit factorization takes 4.5 hours, and a 100-digit factorization takes around 14.5 hours (possibly more)". Fredrik Johansson 23:46, 23 March 2006 (UTC)

The OP said "Choose a random integer...". If this is really what was meant, then factorisation is probably much easier than these worst case examples. There is a better than 6 in 7 chance that a random integer is divisible by a prime under 50. Of course, this doesn't help if the problem is really related to cryptography, for example, where key numbers are definitely not random, but are deliberately chosen to be hard to factor. Gandalf61 09:56, 24 March 2006 (UTC)


 * That only helps part of the way. Sure, a random integer is likely to contain a few small factors, but you want to find all of them, so in the end you'll still likely have to factor a 80-digit integer into a 30-digit and a 50-digit piece. I'm sure a good estimate for the sizes of the two largest factors of a random n-digit integer exists; does anyone know? Fredrik Johansson 13:21, 24 March 2006 (UTC)


 * The answer lies in Dickman's function (we need an article). Knuth discusses it in The Art of Computer Programming vol. 2. One result is that the probability that a random positive integer $$\le x$$ has a prime factor $$ > \sqrt{x} $$ is about 69 percent. There is also an interesting algorithm by Erich Bach that allows you to generate an uniformly random integer including its factorization efficiently ( or for a slightly slower, but much simpler algorithm). Rasmus (talk) 13:42, 24 March 2006 (UTC)


 * But a positive integer $$\le x$$ cannot have a prime factor $$ > \sqrt{x} $$, can it? -- Jitse Niesen (talk) 15:18, 24 March 2006 (UTC)


 * Yes it can. Only one, though.  —Ilmari Karonen (talk) 15:40, 24 March 2006 (UTC)
 * For example, 291 = 97 * 3, where 97 > &radic;291. -- Meni Rosenfeld (talk) 15:43, 24 March 2006 (UTC)
 * ...or, for that matter, 6 = 3 * 2, where 3 > &radic;6. —Ilmari Karonen (talk) 16:40, 24 March 2006 (UTC)
 * Of course. I guess it's time for me to go to bed. Just one remark to rescue my face: you don't have to search for the greatest prime factor; you can calculate it if you know the other prime factors. -- Jitse Niesen (talk) 15:52, 24 March 2006 (UTC)

= March 24 =

balanced product of two modules
let R be a ring, M a right R module, N a left R module

a balanced product (P,f) consists of an abelian group, a map f from $$M\times N$$ to P, such that f(u+v,w)=f(u,w)+f(v,w) f(u,v+w)=f(u,v)+f(u,w) f(va, w)=f(v,a w)

Is this a general definition? It is used in my course before introducing tensor products? Why can't I find balanced products anywhere else, wikipedia and the rest of the internet included? Evilbu 15:16, 24 March 2006 (UTC)
 * This is what Rotman calls an R-biadditive function (not bilinear because you can't pull the a out front). The purpose of the definition is that these are the morphisms that fill in the universal property for the tensor product, which sounds like this: for all abelian groups P and biadditive functions f: M×N → P, there exists a unique group homomorphism M⊗RN→P such that the obvious diagram commutes.  R-biadditivity is what replaces bilinearity which is used in the maps in the case of tensor products of vector spaces (or indeed, modules over commutative rings).  But I'm a little confused by the way your definition is phrased.  I think this should be a property for maps to have, the phrase "balance product" makes it sound like you're doing some operation to M and N, but for starters, your balanced product is not unique (even up to isomorphism), so it can't be viewed that way.  Anyway, if our tensor product article weren't so crappy, it would explain all this.  -lethe talk [ +] 17:01, 24 March 2006 (UTC)
 * OK, actually we have something at Tensor product of modules, these maps are called bilinear maps there. -lethe talk [ +] 17:25, 24 March 2006 (UTC)

Thanks, after we define balanced product, we define morphisms between two of them (P1,f1) and (P2,f2) it has to be a group morphism g of P1 to P2 and for all m, n in M and N g(f1(m,n))=f2(m,n)

a tensor product is then a balanced product such that there is a unique morphism to each balanced product

Now it is supposed to be obvious (without even proving existence of one) that the elements f(m,n) if (P,f) is a tensor product, generate the abelian group P now why is that? Evilbu 17:56, 24 March 2006 (UTC)
 * Well, I'm not sure, but lemme take a stab. Let G be the range of f in P.  The idea is that G also satisfies the universal property, and so is isomorphic, so the range is all of P, so those elements generate P.  Explicitly, then if i: G → P is the inclusion map, I have the a map q: M×N→G such that iq=f.  Then let r be the unique map P→G which the universal property promises us.  So now I have a map qr: P→P, which satisfies fqr=f.  But the identity map also satisfies that equation, and so by uniqueness, that map is the identity.  I guess we do a similar argument for rq, and then we're done.  This seems pretty standard for proofs involving universal properties, but I've never described universal properties as “obvious”, so maybe that‘s not the proof we‘re looking for?  I dunno. -lethe talk [ +] 18:22, 24 March 2006 (UTC)

thanks but as i understand you used universal property what actually did you use then as i understand, you do not think this is such a trivial question( my syllabus works that way :balanced product, then morphisms between them, then tensor product, then this exercise)
 * I'm sorry, I don't understand what you're asking here. Are you asking whether I used the universal property?  The answer is: yes, of course I did.  The universal property characterizes the tensor product, if you don't use it, you've written a wrong proof.  Shall I explain the step where I used the universal property again?  It's like this, the universal property guarantees me a morphism from the tensor product to any group.  The image of f is such a group, so the universal property guarantees a map to it.  But the equations that hold for all these maps, along with the uniqueness, make the image isomorphic to the entire tensor product. -lethe talk [ +] 19:45, 24 March 2006 (UTC)
 * You know, given that you're not supposed to use use the explicit construction of the tensor product (I think that's what you meant by "without even proving existence of one"), the only way to proceed is with the universal property. I think it's probably just a case of your professor using the word "obvious" too liberally. -lethe talk [ +] 20:12, 24 March 2006 (UTC)

forgive me for insisting but I am quite interested allow me to continue for a moment with my not so general terminology if (Z,f) is a tensor product, let P be the subgroup of Z generated by the image of f ( in general, you can't say the image is a group right?) now (P,f) will be a balanced product as well so according to my definition of tensor product there is a unique morphisms p from Z to P such that p(f(x,y))=f(x,y)

this is all correct right? so how exactly should I proceed now?
 * Yes, and I guess you're right about the image not being a subgroup, that's my mistake, oops.  So since P is a subgroup of Z, there is an inclusion map i:P→Z and there is a morphism q: M×N→P such that iq=f.  You've chosen to use the same symbol for f and for q, but I prefer to use different symbols; they are different morphisms, since they have different codomains.  This morphism q is just f restricted to its range.  OK, so ip is a morphism from Z to itself, and we see that it satisfies ipf=iq by the universal property, which is equal to f by the property of subgroups.  Thus ipf=f.  Now since Z is itself a balanced product, then there exists a unique morphism id from Z to itself such that idf=f.  So we have ipf=f and idf=f.  But the universal property says that the morphism which makes the diagram commute must be unique, so we must have ip=id.  We have to do something similar to show that also pi=id, and then we know that P and Z are isomorphic. -lethe talk [ +] 06:03, 25 March 2006 (UTC)

Far too hard.
If 45^x = 1 (mod 56), what does x equal? Is it ?
 * 1) 56186
 * 2) 29629586
 * 3) 29629608
 * 4) 29629633

Please, how do I go about this question? Thanks Reffies. --Dangherous 17:58, 24 March 2006 (UTC)


 * Simply trying each value with the aid of a computer shows that the answer is 29629608. Are you looking to solve it by hand? Fredrik Johansson 18:19, 24 March 2006 (UTC)
 * There's probably a number theory theorem which makes this easier, but a possible way is this: Calculate small powers of 45 modulu 56: For x=0, 1, 2... You'll get 45^x = 1, 45, 9, 13, 25, 5, 1, ... . So 45^6 = 1, therefore 45^(6n)=1. So x should be divisible by 6. The answer is 29629608. -- Meni Rosenfeld (talk) 18:22, 24 March 2006 (UTC)


 * This isn't an answer to the problem (Meni essentially nailed this for our purposes), but this strikes me as very reminiscent of Fermat's Little Theorem or Euler's Totient Theorem, maybe something there would help.
 * Yes, it seems Euler's theorem tells us that 45^24 = 1 (mod 56). At worst, this narrows down our search - In this particular case, it can immediately give a solution since 29629608 is divisible by 24. -- Meni Rosenfeld (talk) 19:15, 24 March 2006 (UTC)

i think the point is that you have to Chinese remainder theorem 45^x=1 mod 56 is equivalent with the set of equations

{45^x=1 mod 8, 45^x=1 mod 7} or {5^x=1 mod 8,3^x=1 mod 7} now you see immediately this can only happen if x is true without calculations that require computers
 * My calculations didn't require computers. But I agree that the method you present may be easier. -- Meni Rosenfeld (talk) 19:33, 24 March 2006 (UTC)


 * Using the Euclidean algorithm we can quickly determine that the greatest common divisor of 45 and 56 is 1, so these two numbers are coprime.
 * GCD(45,56) → GCD(11,45) → GCD(1,11) → 1
 * We can also easily find the prime factorization of 56 as 237. Thus we can easily compute the Carmichael function,
 * &lambda;(56) = lcm(23−2,71−1(7−1)) = lcm(2,6) = 6.
 * Now we know a very small power of 45 is congruent to 1 modulo 56, either 2 or 3 or 6; and we can quickly eliminate 2. This tells us the correct exponent must be a multiple of 3 (or 6), so we can use the ancient trick of casting out nines:
 * 5+6+1+8+6 → 2+6 → 8
 * 2+9+6+2+9+5+8+6 → 4+7 → 1+1 → 2
 * 2+9+6+2+9+6+0+8 → 4+2 → 6
 * 2+9+6+2+9+6+3+3 → 4+0 → 4
 * The only number divisible by 3 is 29629608, so that must be the correct exponent.
 * Warning: A multiple choice question like this surely only appears in an educational context, where the person taking the test is assumed to have absorbed some relevant instruction to support answering this quickly. We have no way of knowing what you are expected to know. For example, perhaps you should know Euler's theorem. ;-) --KSmrqT 19:58, 24 March 2006 (UTC)

conference graph (spectrum of strongly regular graphs)
let G be a strongly regular graph. $$srg(v,k,\lambda,\mu)$$ assume nontriviality : do not allow it to be disconnected or to have a disconnected complemenent it has k as eigenvalue once then, and two other eigenvalues

when $$2 k\neq (v-1)(\mu-\lambda)$$ all eigenvalues are integers

apparently the conference graphs, those are graphs of this form

$$srg(2 k+1,k,\frac{k}{2}-1,\frac{k}{2})$$, are the only ones that don't have integer eigenvalues

but how to see this, essentially there are four parameters here that are restricted by two equations

$$2 k\neq (v-1)(\mu-\lambda)$$ and $$(k-1-\lambda) k=(v-1-k)\mu$$ (this holds for all strongly regular graphs)

So where does the third restriction come from, as I see only one parameter in a conference graph?

I already requested a conference graph page on wikipedia The problem is that on the internet, google etc. think you talk about conferences ON graphs :)

Evilbu 23:37, 24 March 2006 (UTC)

= March 25 =

Circle
Lets say a circle, with diameter D was placed on the corner of a wall. Obviously, there would be a gap between the wall and the circle. The question is what is the diameter of the largest possible circle that can fit in this gap. Using Pythagorean Theorem, I found that the length of the gap would be $$ (D^2 - D)/2 $$. However, I have to contend with the gap between this hypothetical circle and its gap between the wall. The length of this gap I called x. I wrote the diameter of the second circle as $$ [(D^2 - D)/2] - x $$. However, I cannot solve for x. I realised that the new gap would follow the same rule ($$ (D^2 - D)/2 $$). So the new equation would be:
 * $$\frac{(\frac{D^2 - D}{2} - x)^2 - (\frac{D^2 - D}{2} - x)}{2}=\frac{D^2 - D}{2}$$

With that, I was hoping to bring all the x's to one side. However, this proved troublesome. Can you help. Thanks.


 * I get a different result for the initial gap: (sqrt(2)-1)D/2 = 0.2071D, approximately. Check your work and then see if you can proceed using this value. StuRat 17:23, 25 March 2006 (UTC)


 * Yes, the gap should be linear in D, i.e. not involve D2. Also, I can't figure out what you mean by "the gap between this hypothetical circle and its gap between the wall". How many circles are there? —Keenan Pepper 17:37, 25 March 2006 (UTC)


 * Maybe this crude diagram will help:

********************************************************** *   *           *                                 + *  *               *                       + * *                 *                 + *                                 + **                   *         + *                           + **                   *   + *                      + * *                 *+ *  *               * *    *           * *          *   + *            + *           + *          *         + *        *       + *     *     *     +
 * StuRat 18:16, 25 March 2006 (UTC)
 * Yes, I forgot to square root. Sloppy mistake. Thanks.


 * I'm not sure I follow the problem statement or any of the reasoning here. Assume a unit circle for simplicity. A diagonal line from the corner to the circle center has length &radic;2, so the corner of the square is &radic;2 − 1 away from from the nearest point on the circle, where the diagonal crosses. However, the diagram makes clear that the sought center cannot be halfway between.
 * Instead, the sought center C will be at a point where a horizontal (or vertical) line to C is the same length as the distance from C to the circle. Let C be at (r,r), which places the circle contact at r+r1&frasl;&radic;2 in x and y. But we already know the coordinates of the circle contact, namely (&radic;2 − 1)1&frasl;&radic;2 in x and y. Thus the equation to solve is
 * $$r (1 + 1/\sqrt{2}) = \frac{\sqrt{2} - 1}{\sqrt{2}}, $$
 * which gives
 * $$ r = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}, $$
 * or 3−2&radic;2 (approximately 0.17). In general this would be scaled by the radius of the given circle. --KSmrqT 22:10, 25 March 2006 (UTC)


 * That's the answer I got, too, although by a different method:


 * As stated previously, the gap distance from the large circle to the corner is approximately 0.2071D. The total of the large circle diameter and the large gap is therefore approximately 1.2071D.  This makes the ratio of the large gap distance to this total approximately 0.2071D/1.2071D = 0.1716.  If we assume the same ratio for the small gap to the small circle, then the size of the small gap is approximately 0.1716(0.2071D) = 0.0355D.  So, if we subtract the small gap from the large gap, we get approximately 0.2071D - 0.0355D = 0.1716D.  This is the diameter of the small circle in terms of the diameter of the large circle. StuRat 23:53, 25 March 2006 (UTC)

HOW MUCH SHOULD I PUT INTO A SAVINGS ACCOUNT TO HAVE A BANK MAKE AUTO MATIC WITHDRAWALS
IF I AM PAYING A MORTGAGE OF $225.00 PER MONTH FOR 15 YRS ,,HOW MUCH MONEY SHOULD I PUT IN THE BANK TO HAVE THE BANK PAY OUT THE PAYMENTS?ASSUMING I'M GETTING 3% INTEREST WHILE MY MONEY IS IN THE BANK,,,AT THE END OF 15 YEARS MY ACCOUNT SHOULD BE ZERO..

CAN YOU CALCULATE THIS???

THANK YOU,,,JAMES
 * If I understand the question correctly, it's something like 32232$. But if this is a real-world problem, you should discuss it with a financial consultant instead. -- Meni Rosenfeld (talk) 18:10, 25 March 2006 (UTC)


 * What? Give me the name of your bank if you have a mortgage of only $225 a month and you get 3% interest on savings... You must have a mortgage on a doghouse! KWH 04:43, 27 March 2006 (UTC)

F = $32,580 dollars


 * I = 0.03 / 12 = 0.0025 (or 0.25% per month)
 * T = 12 * 15 = 180 months
 * Y = $225 dollars payment per month

Ohanian 09:53, 27 March 2006 (UTC)

= March 26 =

Subject: naming points on graphs (house numbers on twisty roads)
Let's say that:

>> Lane 256 Alley 34 House 7 means approximately: >> we travel 256 units and turn right on to Lane 265; then >> we travel 34 units and turn right on to Alley 34; then >> we travel  7 units and there is the house, on the left. >> >> Telephone pole 22 L 33 R 6 means >> we travel to pole 22, follow the left fork wire 33 poles, then fork >> right for 6 poles and arrive at our goal pole. >> >> ( jidanni.org/geo/house_numbering/ ) P> What is your question? Do you want to assign house numbers? Do you P> want to build a cross references from those numbers to spatial P> location? Do you want to compute the distance from one to another? P> Do you want to uniformly store house numbers like that in a software P> data structure?

I guess I just want to find which part of graph theory deals with naming points and segments on graphs the closest.

e.g., http://en.wikipedia.org/wiki/Glossary_of_graph_theory doesn't ever mention (systematic) ways to name each segment and point... That's what I want: what's the branch of math that deals with names for points in twisty graphs... maybe they have a better way of naming the two houses on the image on jidanni.org/geo/house_numbering/mountain_en.html

I want to know if there are even smarter ways of naming such points, or how folks go about naming points on graphs.

Yes, we would use nice grid coordinates if we were in a flat city, but we have twisty hilly roads.

--Dan Jacobson, jidanni.org


 * Looks like you copied the above discussion from another bulletin board, email exchange, or something like that. It would still be possible to number each location by the XY cartesian coordinates.  That type of label would not be useful in describing the path you take to find the location, however.  The general term for describing the distance along a curve is it's parametric coordinate.  I'm not sure what you would call it when you chain multiple parametric coordinates together, though, perhaps a parametric graph ?  Parametric coords in 2D use parallel curves on a surface, however, which doesn't sound like what you have here.  (I think the article here might be called curvilinear coordinates instead of parametric coordinates, but it 's written in such an inaccessible way (no pics !), that I find it hard to understand.)  StuRat 00:19, 26 March 2006 (UTC)

topology
prove or disprove:A and B are path connected subsets of a space X and intersection of A and B closure is non-empty.A union B is path connected.
 * check out the topologist's sine curve. -lethe talk [ +] 11:16, 26 March 2006 (UTC)

= March 27 =

l2 tail
It's damn obvious, but I can't remember how to prove that the tail of an l2 sequence goes to zero.

That is, given $$x = {\left ( x_i \right )}_{i=1}^\infty x_i \isin \mathbb{C}$$ where $$\sum_{i=1}^\infty{|x_i^2|} < \infty$$, to show that $$\lim_{n \rightarrow \infty}{\sum_{i=n}^\infty{|x_i^2|}} = 0$$. Confusing Manifestation 11:04, 27 March 2006 (UTC)


 * It follow almost by definition: If the first limit exists, per definition $$\forall \epsilon > 0 :\exists n \in N: \forall m>n: |\sum_{i=1}^m{|x_i^2|}-\sum_{i=1}^\infty{|x_i^2|}| < \epsilon$$. But $$|\sum_{i=1}^m{|x_i^2|}-\sum_{i=1}^\infty{|x_i^2|}|=|\sum_{i=m+1}^\infty{|x_i^2|}|$$ so we have (n'=n+1) $$\forall \epsilon > 0 :\exists n' \in N: \forall m>n' |\sum_{i=m}^\infty{|x_i^2|}-0| < \epsilon$$, which gives you your other limit. Rasmus (talk) 11:49, 27 March 2006 (UTC)


 * What Rasmus wrote is true, of course, and answers your question. But it might be helpful to point out that there's nothing special about $$l^2$$; this is just a statement about convergence of infinite sums.  The following more general theorem is true:  If $$a_i >0$$, then
 * $$\sum_{i=1}^\infty a_i = L < \infty$$ implies $$\lim_{n\to\infty} \sum_{i=n}^\infty a_i = 0$$. As you see, this gives you the result you need without using anything about $$ l^2$$.


 * The proof is pretty straightforward. If we define $$S_N = \sum_{i=1}^N a_i$$, then the assumption of the theorem is that $$\lim_{N\to\infty} S_N = L$$.  But then it is easy to see that $$\lim_{N\to\infty} |L - S_N| = 0$$, and $$L - S_N = \sum_{i={N+1}}^\infty a_i$$, so we are done. --Deville (Talk) 18:48, 27 March 2006 (UTC)

Thanks for both answers. I knew it was so obvious as to practically follow from the definition, but for some reason the actual steps eluded me. Confusing Manifestation 00:12, 28 March 2006 (UTC)

Height of Parallelogram
I am working on producing a spreadsheet and have hit a road block in some geometry. I am trying to upload my diagram but something isn't working right so I'll need to explain it verbally first.

I have a parallelogram, with theta in the lower left hand corner, by starting at theta and traveling clockwise I have B and then the horizontal leg A. Theta and A are my inputs. I can also even determine F, which is perpendicular to B and terminates at angle BA, but I don't really see a way that it helps me out. On lower leg A, if we go to it's far right end and draw a verticle line we get a right triangle with leg d, B, and H. I had a hunch that the quantity (A-d) relates to theta. So in AutoCAD I set A = 1 and varied theta from 0-90 in increments of 10 degrees, converted into radians and plotted the values against (A-d). Found using linear regression with a third polynomial equation, [-0.0698(theta)^3 + 0.3564(theta)^2 - 0.031(theta) + 0.44 = (A-d) ] resulted in a R^2 = 1, so someone please set me straight if I am way off target here but I assumed I since the R^2 = 1 everything was fine and I would be able to multiply the equation by left hand said of the equation by A and I would be able to get (A-d) do some geometry and have the Height pop out. (This part may get confusing and I can set up a junk hotmail account, post it, and if anyone is interested in seeing how I did this in excel email me and I'll send it to you). But I took two arbitrary values of A varied theta, got more equations by from trend lines, divided each term by A and compared the terms to the first equation and they came out reasonably the same......I thought I was good to go but when I do it in AutoCAD....it doesn't really work......don't know if I'm just spinning my wheels or if there really is a way to determine the height of a parallelogram from by given input. But doesn't there just seem a way that theta can be a direct function of (A-d)???


 * You fail to state a problem. Name your inputs. And what is each: lengths, angles, line segments? Likewise, what is your desired output? Stream-of-consciousness babble doesn't help. Also, please sign your post using four tildes, which the wiki software automagically converts to your id and a time stamp. --KSmrqT 18:44, 27 March 2006 (UTC)


 * See that with the same imput data : an angle and a segment lenghth only, you have different parallelogram shapes. Make sure you said everything; type in a simple drawing for us like those :

#####       A       #   #     ########## A#  #     ########## #  #     ##########   θ#####    θ##########


 * Is what you're looking for H=Bsin(theta)? TimBentley (talk) 00:48, 28 March 2006 (UTC)
 * Nevermind, I think there's no way to determine the area from your inputs. TimBentley (talk) 00:54, 28 March 2006 (UTC)

Multivariable Max/Min Problem
I do Independent Study math, so it's not graded - I make up my own assignments. Anyway, here's a problem I can't quite figure out:

Find the local max/min values and saddle points of the function:


 * $$f(x,y)=(x^3)y+12x^2-8y \,\!$$

I take the first partial derivatives and equate them to 0 (I can do that part ;):


 * $$F_x=3(x^2)y+24x=0 \,\!$$ $$F_y=x^3-8=0 \,\!$$

Then I can find one set of points, (2, −4): x=2 (from $$F_y$$) → y=−4 (plugged into the other partial derivative)... but then I'm stuck. It's a tiny roadblock but I can't seem to figure it out -_-

Thanks. – ugen64 22:44, 27 March 2006 (UTC)
 * Well, you seem to have solved most of the problem correctly to me. What's left to do?  Deciding whether this critical point is a maximum, minimum or critical point, I guess?  For that, use the Hessian.  If the matrix is positive definite, it's a minimum, negative definate, it's a maximum, or indefinite, saddle point.  One criterion for 2x2 matrices is that they're definite if their determinant is positive and indefinite if their determinant is negative.  So it goes like this:
 * $$H=

\begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy}\\ \end{pmatrix} = \begin{pmatrix} 6xy+24 & 3x^2\\ 3x^2 & 0\\ \end{pmatrix} $$
 * from which I observe that the determinant is –9x2, which is –144 at x=2. This is negative, so I conclude that f(2,–4) = 112 is a saddle point (and there are no extrema since this function is linear in y).  Hope that helps. -lethe talk [ +] 23:12, 27 March 2006 (UTC)


 * See, the back of the book (answer) says that there is a 2nd critical point... but I can't figure out how to find it. Guess I could have made that clearer :) Thanks for that part of the answer, though. – ugen64 01:15, 28 March 2006 (UTC)
 * Well, uh... maybe it's supposed to be a complex function? There are three roots over the complexes...  that doesn't seem right though.  And you're expecting only 1 more point, not 2.  Maybe there's a mistake in the text.  I'll bet the function isn't supposed to be linear in y. -lethe talk [ +] 02:16, 28 March 2006 (UTC)


 * One guess is that the second term should use y, not x; namely,
 * $$f(x,y)=x^3 y+12y^2-8y \,\!$$
 * This would have real extrema at (0,1&frasl;3) and (2,0), and the two Hessian signatures differ. --KSmrqT 03:07, 28 March 2006 (UTC)

= March 28 =

China's dollar reserves
(Not sure where to ask Macroeconomics question, so I will try here.) I read in The Economist that China has >$800 billion in reserves. Why doesn't China spend this money on sorely needed infrastructure projects, rather than saving it? I.e., what marginal benefit exists for such savings? Lokiloki 02:23, 28 March 2006 (UTC)
 * The benefit isn't in having the savings, but in buying and selling it. Buying and selling dollar reserves allows them to manipulate the exchange rate of the currency. By buying dollars (i.e. selling yuan), they can decrease the value of the yuan.  This helps the economy by making it cheaper for other nations to buy goods from China.  If the yuan gets too cheap (making it overly expensive for China to buy US [and other foreign] goods), they can sell the reserves, driving up the value again. For such transactions to have any noticeable effect, the quantities must be massive. Superm401 - Talk 03:36, 28 March 2006 (UTC)
 * Thanks very much for your answer... but since China artificially sets the exchange rate at a fixed rate, how does the buying and selling play a role? In that there is a "ghost" exchange rate that is unfixed? Lokiloki 05:22, 28 March 2006 (UTC)
 * The reserve system is how they peg the exchange rate. If they just set rates and ignored the market, people would resort to the black market. I found a helpful Forbes article. Superm401 - Talk 21:27, 28 March 2006 (UTC)


 * India has the same problem. Lots of foreign exchange reserves, and a sore need to develop infrastructure. It doesn't spend the money for infrastructure, in part because suddenly introducing so much money in the economy will lead to hyperinflation, and the poor will find it difficult to afford things. deeptrivia (talk) 20:16, 31 March 2006 (UTC)

On Construction of Ramanujan Graphs (LPS Constructions)
Optimal expander graphs are known as Ramanujan Graphs. Explicit constructions for them is given in the Ramanujan Graphs by LPS. I have a doubt in the construction of them. While using PSL or PGL, how they are different from SL and GL. It has been said that PSL is quotient group with out zero transform. Can any one explain me more about how to construct them? Any full example available with any one? --Guru 03:45, 28 March 2006 (UTC)

Cube
A giant lamington cake in the shape of a cube.How many small lamingtons have 0,1,2 or 3 sides iced if the cake was cut into 64 and 125 cubes?
 * They all have four sides. Isn't this obvious? It is like asking: "I have 64 cubes. How many of the cubes have 0, 1, 2, or 3 sides?" -- [[Image:Flag_of_the_United_States.svg|17px]] Mac Davis ] ⌇☢ &#xE0D;&#x19B;. 10:17, 28 March 2006 (UTC)
 * ahem "iced sides" (judging by our lamington article, this is a cake with icing on all sides including the bottom and hence a good candidate for a maths question to do with vertices, sides and so on; given the geographical specialisation, I also deduce that our questioner is from Oz or New Zealand). On the assumption that this is homework (but fun homework that I enjoyed overgeneralising), I shall give the solution for a cut into n3 smaller cubes (n>=2; for n=1 we trivially have one six-sides-iced piece) and then our questioner will at least have to do a bit of algebra to get the answers for his problem:
 * 3 sides iced: always 8
 * 2 sides iced: 12(n-2)
 * 1 side iced: 6(n-2)2
 * 0 sides iced: (n-2)3
 * The proof that these add up to n3 is left to the reader. --Bth 10:49, 28 March 2006 (UTC)
 * A quick note: your math seems to assume that the bottom of the cake is iced, which is rare (to say the least) in my experience. On the other hand, I have no idea what a "lamington" cake is vis a vis other cakes.  &mdash; Lomn Talk 15:33, 28 March 2006 (UTC)
 * Appears to be one of Australia's national treasures. Yep, all six sides are iced. --jpgordon&#8711;&#8710;&#8711;&#8710; 22:45, 28 March 2006 (UTC)
 * Indeed so, but it's an assumption I explicitly stated with a link to the article about lamington cakes here on Wikipedia to justify it. --Bth 07:33, 29 March 2006 (UTC)
 * You wretched creatures who know not the Glorious Lamington are all invited to my place next weekend for the first annual International Wikipedians Lamington Festival. Seriously though, I promise you they taste much, much better than Vegemite. (That's not a slur on Vegemite - I love it - but I know Americans just don't get it.) JackofOz 14:53, 29 March 2006 (UTC)

Cost of Iraq War per person
A recent study indicated the total cost of the war is 1 to 2 trillion dollars depending on how long it lasts. Going with the 1 trillion figure and figuring that the US population is close to 300 milllion--how much would the cost be per person?


 * The answer is 2.50, but not necessarily in current US dollars. Sheesh. --KSmrqT 07:48, 28 March 2006 (UTC)


 * Answering the maths question first of all ... a trillion (in this context) is a million million, so you want to divide a million million dollars by 300 million. Answer is $3,333 and some cents - say $3,000 in round numbers.


 * Now ask yourself a few questions. Is the "total cost" in this estimate all paid by the US, or is it an international cost (other countries are paying financial and human costs for the Iraq occupation too) ? Is it an estimate of the cost so far, or of the "final" cost when everything is finished (and if so, how does the estimator define "finished") ? And is it useful to divide by the total US population, or would it be more meaningful to divide by the number of US tax payers, which is more like 100 million ?


 * Anyway, whichever way you cut the numbers, it is clear that the Iraq war and occupation have cost far more than originally envisaged, both financially and in human lives. But financial considerations are unlikely to sway the opinion of those who take a moral stance on the issue. Those who believe that the war was a justifiable way of achieving its objectives will still believe that this is true at almost any cost. Those who believe the war was unjustified will not agree it would have been the right thing to do if only it had cost less. Gandalf61 09:12, 28 March 2006 (UTC)


 * You guys are monsters. The cost of human life is priceless. Have you no dignity? :P -- [[Image:Flag_of_the_United_States.svg|17px]] Mac Davis ] ⌇☢ &#xE0D;&#x19B;. 10:12, 28 March 2006 (UTC)


 * It also might be important to note that all of this is not taken from current tax income, but a huge chunk is in loans (considering that there is a sizable deficit in the US right now). You could argue that that would mean that the taxpayers would pay even more than that, just over a significantly longer time. Oskar 20:58, 28 March 2006 (UTC)
 * Gandalf61, those seem like extreme positions but they're not even entirely separate. The war turns out to be the right thing to have done in both positions, and the human and monetary costs are reduced to mere quibbles.  Sorry, but you don't speak for me.  JackofOz 13:54, 29 March 2006 (UTC)


 * JackofOz - what a strange thing to say ! Of course I wasn't trying to speak for you - whatever gave you that impression ? The point I was trying to make (speaking only for myself) was that someone who takes an absolute moral position on the Iraq war one way or the other is unlikely to be influenced by estimates of its cost. Gandalf61 14:26, 29 March 2006 (UTC)


 * Then I have inaccurately divined your meaning. for which my apologies. I should explain, I've re-read your post, and I think your logic baffles me.  A will believe X, and B will not believe Y.  Which is confusing enough.  By "B will not believe Y", I inferred you were implying B should believe Y, hence my indignation.  Then Y turns out to be modified form of X anyway. The message I keep on getting is that there is only one serious attitude to the Iraq war worth having, that of support for it.  I accept that was not what you were intending to say, but that's what comes thru.  To me.  (Maybe it's just me.  I'm having a funny week.)  :-)) JackofOz 14:47, 29 March 2006 (UTC)


 * The atomic bomb dropped on Hiroshima killed 66,000 people. So far over 30,000 Iraqis have been killed, over 2,300 American soldiers have died, over 17,000 American soldiers have been injured, and the human and financial costs will continue to mount until at least the end of Bush's administration without an unconditional surrender from the Iraqi insurgents.

Looking at this from a logical perspective, it is obvious that the human and financial costs would have been lower if America had dropped an atomic bomb in 2003. It is just that the word "nuclear bomb" makes everyone shout and become emotional. These people are essentially saying that killing people a few at time is preferable to massive deaths even if the number of massive deaths can be fewer. --Patchouli 01:15, 3 April 2006 (UTC)

Kudos to Harry Truman for saving American and ultimately Japanese lives.--Patchouli 01:18, 3 April 2006 (UTC)


 * Where would you drop the bomb ? Such an idea wouldn't work in Iraq, since, unlike Japan (which honored the surrender agreement at the urging of their god-emperor, Hirohito), there is no single leader in Iraq.  It should be thought of as a collection of many different factions, not a nation at all. StuRat 04:14, 5 April 2006 (UTC)


 * We have created the factions ourselves to make sure they don't unite against us. There are essentially (1) Shiites, (2) Sunnis, (3)Kurds, and (4) a negligible number of secularists.  We drop one bomb, and give them a chance to stop.  After the second bomb, the Iraqis themselves will stop sympathizing with the insurgents like now.  The sympathizers are inside the police and military and this is how the insurgents have the endless supply of weapons without ever running out of IEDs and other bombs.  Why is it that in the United States it is now impossible for anyone to get his hands on a bombs?  The same measures need to be implemented in Iraq.Patchouli 22:49, 5 April 2006 (UTC)

Factorization
Hi,I am a new person to this website and i am getting a lot of information from you guys.

I am an IGCSE student that studies in Britain and am studying at grade 10.

I have 4 question to ask you about from the IGCSE.

First question is that you give me brief information about the sine and cosine rule. Secondly, What is factoization of expression, and also it's definition. Thirdly, The simplification of long expressions with exponents. Last question is that can you give me information about the sets.

Thank you very much for your help and i hope that it will come back to you one day.


 * We have some nice little articles on the sine and cosine rules, which are basic identities in trigonometry.
 * I'm not quite sure what you mean by "factorization of expression", but am guessing it refers to, eg, writing down expressions like $$n^2+5n+4 = (n+1)(n+4)$$, which is useful for solving polynomial equations where such factorisations exist (in general it can't always be done). We have an article on that too.  "The simplification of long expressions with exponents" I think just refers to using powers notation to avoid hand cramp, eg $$n \times n \times n \times n \times n \times n \times n$$ can just be written $$n^7$$.  Sets are a big topic that in many ways underpin large chunks of mathematics itself, so you'll need to be a bit more specific there. --Bth 11:52, 28 March 2006 (UTC)


 * The significance of the law of sines and the law of cosines is that you can use them to solve a triangle if you know two sides and an angle (law of cosines) or two angles and a side (law of sines). &#x2013; b_jonas 12:13, 28 March 2006 (UTC)

Help
How would I figure this out 1−1. I can't remember how to figure the answer if if the power is negative. Thank you for the help. I Lov  E Plankton 16:36, 28 March 2006 (UTC)


 * $$a^{-b}=\frac{1}{a^b}$$, so e.g. 2−1=1/2. Kusma (討論) 16:47, 28 March 2006 (UTC)
 * So 2−2 would be 1/4? I  Lov  E Plankton 17:00, 28 March 2006 (UTC)
 * Precisely. You got it. Kusma (討論) 17:09, 28 March 2006 (UTC)


 * Remember that addition in the exponents equals multiplication in the base, so xaxb = xa+b. In the case where b = −a, the exponent simplifies to 0 and the answer to 1; therefore additive inverse (negative) in the exponent implies multiplicative inverse (reciprocal) in the base. As always, reciprocation is only meaningful if the base (here, x) is nonzero. --KSmrqT 20:04, 28 March 2006 (UTC)

I have been wondering if everyone has the concept of order in coombinations backwards.
I did a little experiment. I tested the Combination and Permutation functions.

P(n, r) = n! / (n - r)!

C(n, r) = n! / ((n - r)!r!)

Let's say that for each one I'm using 4 (n) items in set of 3 (r).

P(4, 3) = 4! / 1! = 4! = 24

C(4, 3) = 4! / (1! * 3!) = 4

Well everywhere I go, it says that the Combination returns the number of combinations of items in which the order does not matter. Well tell this to the following sets.

(1 2 3), (1 2 4), (1 3 4), (2 3 4)

Notice that it only has 4 items (n), sets of 3 items (r), 4 sets (C), and it IS ordered.

Thats not all though. They always say that Permutation returns the number of ordered pairs, but I found that I can get the same number of unordered sets with all the same variables.

(1 2 3), (1 2 4), (1 3 2), (1 3 4), (1 4 2), (1 4 3), (2 1 3), (2 1 4), (2 3 1), (2 3 4), (2 4 1), (2 4 3), (3 1 2), (3 1 4), (3 2 1), (3 2 4), (3 4 1), (3 4 2), (4 1 2), (4 1 3), (4 2 1), (4 2 3), (4 3 1), (4 3 2)

Again notice that it only has 4 items (n), sets of 3 items (r), 24 sets (C), and it IS NOT ordered.

You think it stops there? Wrong. It is the same for the functions with repetition. I'm not going to go into detail about them though. I have already talked long enough.

I was just wondering about this, and though somewhat might look into it.

Matt DeKok


 * I think you've reversed "ordered" and "unordered". I don't see a problem.  &mdash; Arthur Rubin | (talk) 18:10, 28 March 2006 (UTC)
 * Yeah, it's just confusion with how you're reading the terms. For combinations, (1 2 3) is not order dependent.  All of the order-dependent permutations of that combination [(1 2 3), (2 1 3), (3 2 1), etc] reduce to order-independent (1 2 3).
 * You could also write the combinations as (1 3 2), (4 1 2), (4 1 3), (4 3 2) and it would be the same set -- standard ordering just seems to make the most sense. &mdash; Lomn Talk 19:05, 28 March 2006 (UTC)


 * Anytime you find yourself wondering if "everyone has &lang;whatever&rang; backward", stop and use a little common sense. Almost certainly you are the one who has it backward. --KSmrqT 20:09, 28 March 2006 (UTC)


 * KSmrq, this is good advice for anyone in any context, but especially in math. Let me say that I could have heeded this advice many times in my younger days and obtained more optimal outcomes :-)--Deville (Talk) 13:34, 29 March 2006 (UTC)


 * Let me see if I can rephrase this. "Ordered" means that you count different orders as different things. "Unordered" means that you count different orders as merely rearrangments of the same thing. I think what you're seeing is that "Unordered" only includes one particular order listed, but it actually includes all orders since they're the same thing. Similarly, "Ordered" includes all possible orders because each order is different. Think unordered=committee (all spots are equally powerful) and ordered=hierarchy (different offices have different powers), and ask yourself how many committees vs. how many arrangements into offices. --Geoffrey 05:46, 30 March 2006 (UTC)


 * Here's a way to remember: When we shuffle a deck of cards, we want to randomly choose a permutation, preferably with each of the 52! = 52×51×50×⋯×3×2×1 orderings equally likely. But when we look at the cards dealt to us, we only care about the combination, not the order in which the cards come to us.
 * Now for a little fun. Assuming a simple mathematical model of a riffle shuffle, how many shuffles are required to have confidence that the deck is well-shuffled? One way to analyze this is using a Fourier transform on the group of permutations. It can be shocking to learn how poorly shuffled a deck is after as many as four shuffles. Of course, all bets are off (literally) if a magician is shuffling the deck—especially if that magician happens to be Persi Diaconis, author of Group representations in probability and statistics. --KSmrqT 10:15, 30 March 2006 (UTC)

Inequality
Does the inequality: (x^(t))(y^(1-t))<=(t)(x)+(1-t)(x) hold? with 0<=t<=1 and x,y > 1 Also any proof? Oh and by "<=" I mean less than or equal to. Thanks Qeee1 17:24, 28 March 2006 (UTC)


 * I'll describe a hint from which you should be able to get the answer. But the real reason it works is that the function $$f(t) = x^t y^{1-t}$$ is convex.


 * Fix any $$x>1,y>1$$. Define the functions $$f(t) = x^t y^{1-t}$$ and $$g(t)=tx + (1-t)y$$ (I'm assuming that's what you meant above).  Compute $$f(0), g(0)$$ and compare, compute $$f(1),g(1)$$ and compare.  Finally, compute $$f''(t)$$, and what does this tell you?--Deville (Talk) 18:11, 28 March 2006 (UTC)


 * Take the logarithm on both sides and use convexity of logarithm. Kusma (討論) 18:17, 28 March 2006 (UTC)
 * Concavity, of course. Kusma (討論) 18:42, 28 March 2006 (UTC)
 * You mean
 * $$x^t y^{1-t} \leq t x + (1-t)\color{Red} y$$?
 * Yes. See, for example inequality of arithmetic and geometric means.  (written before I saw Deville's and Kusma's answers.  Grumble.) &mdash; Arthur Rubin | (talk) 18:24, 28 March 2006 (UTC)


 * Thanks, I used Kusma's method. (I have problems differentiating, and I had already checked AM-GM generalisations) But also thanks to both of you too. I'm impressed at the speed of the response.Qeee1 20:09, 28 March 2006 (UTC)
 * Oh and yes I meant Y, just to be clear.Qeee1 20:10, 28 March 2006 (UTC)

Graphing
Hi there,

I am looking for a good freeware software to plot graphs. I looked through the Graph-Plotting Software Page but was very unsure...does anyone of you know a good programme. I do need it for school calculus, so it should be able to do soem analysis.

Thank you --165.165.228.18 19:16, 28 March 2006 (UTC)


 * It may be more than you need, but the freeware Maxima computer algebra software has a variety of graphing options, and much more. --KSmrqT 20:18, 28 March 2006 (UTC)


 * Octave, a clone of matlab, has a nice graphing system. It's essentially a frontend over gnuplot which is a graphing software usable by itself too.  &#x2013; b_jonas 20:52, 28 March 2006 (UTC)

I have no experience to use any packages on computer graphic. Only Maple and Matlab are available to me paid by my University but I never try them. My simple question is how do I produce a JPG-file, say sample.jpg of the function joining (-1,0) to (0,1), from (0,1) to (1,0) and set all other values to be zero. Can I use formula such as \sin nx/(\pi x) for n=1,5,10 on the same output? I can start Maple on MacIntosh with a double click, then what? How do I draw a circle inscribed in a triangle? Do I have to study a lot from their users' manual (RTFM) before I can use Maple or Matlab? If successful, I can use TEX-graphicx to paste this file as part of a PDF-file. Thank you in advance. Twma 02:03, 29 March 2006 (UTC)


 * For geometric constructions of the inscribed circle variety, a specialized package like C.a.R. (which is written in cross-platform Java) is likely to be simpler than learning Maple or MATLAB. For more general illustrations, including plots, Maple or gnuplot are powerful with only a modest learning curve. For those comfortable around TeX, MetaPost may be your cup of tea. For 3D illustrations, the task itself tends to be challenging, regardless of the software used. --KSmrqT 07:25, 29 March 2006 (UTC)


 * Microsoft PowerToys Graphing Calculator is pretty good at basic graphing, easy to use, and technically freeware, but it's Windows XP only. And no analysis. If you can figure out Maxima or Octave, those are useful too. --Geoffrey 05:42, 30 March 2006 (UTC)


 * Not exactly a graphing calculator, but a quality cross-platform tool for graphical mathematics, is GrafEq. Its specialty is implicit functions, so the results are raster images. --KSmrqT 07:07, 30 March 2006 (UTC)


 * Actually I think trying to use the POVRay renderer might be a good idea, of course you shall have to invest yourself quite a lot but you'll then be able to produce amazingly beautiful 2D and 3D graphs. Sure worth the time. Xedi 13:26, 30 March 2006 (UTC)


 * Yes, raytracing can produce lovely mathematical illustrations. Yet I repeat my caution that 3D illustrations are a significant challenge, much harder than 2D. --KSmrqT 01:10, 31 March 2006 (UTC)

trigonometry
a radius of a circle is 15cm. find the length of the arc of the circl intercepted by a central angle of 3pie/4radians. (leave in terms of pie)


 * mmm... pie.... -lethe talk [ +] 22:10, 28 March 2006 (UTC)


 * Pie is a perfectly valid mathematical constant...it's the ratio of the amount of pie left unattended to the amount of pie remaining after one hour. :-) StuRat 21:13, 1 April 2006 (UTC)
 * That wouldn't work in most cases, Stu. Division by zero is undefined. :--) JackofOz 21:50, 1 April 2006 (UTC)


 * It says at the top of the page to DO YOUR OWN HOMEWORK! But in spite of that I decide to give you the formula:

$${x^\circ\over360^\circ}={y\over 2 \pi r}$$ $$x^\circ$$=degree measure of the arc

$$y \,\!$$=arc leangth

there is an algebraic way, but I find this way easier. (updated by myself schyler 00:01, 29 March 2006 (UTC))
 * I'm guessing the "algebraic way" is just:
 * $$\frac{x}{2\pi}=\frac{y}{2 \pi r}=$$
 * $$x=\frac{y}{r}=$$
 * $$xr=y \,\!$$


 * $$x\,\!$$-radian measure of arc
 * $$y\,\!$$-arc length


 * The two equations are equivalent. Superm401 - Talk 01:16, 30 March 2006 (UTC)

Pi and the Universe
On a disk the same radius as the observable universe what is the minimum number of digits to describe an arc the length of a ... pick something interesting. The width of a finger, a human hair, a hydrogen atom. I was just reading the Pi article and thought this might be good to add.Trieste 14:57, 29 March 2006 (UTC)
 * Well, you don't use π to calculate just any arc length, rather you use it to calculate the entire perimeter of a circle or semicircle. But you want some upper bound on how many digits of π one might need to talk about even cosmological distances, right?  So let's say you wanted to calculate the surface area of our Hubble volume.  I think our Hubble volume is 80 billion lightyears across.  That's 13.4 billion years old, times the speed of light, adjusted for the cost of expansion.  About 10^27 meters?  So in order to get the surface area accurate to, say, a meter squared (absurd, given the error bars on the Hubble volume, which might be 50%), then you need maybe 52 digits of π (double 27, since we're squaring).  This is a rather silly calculation, doesn't have much physical relevance; nevertheless, I think it's a nice example to show how completely physically irrelevant knowing a million digits of π is.  Doing so is mostly for testing computer algorithms or hardware, or for testing mathematical notions about the distribution of digits, and not for knowing the perimeter of a circle more accurately. -lethe talk [ +] 15:01, 29 March 2006 (UTC)
 * Yes about the relevance. That is what I was hoping to illustrate e.g. one only needs x digits of π to accurately calculate the area of this period . in relation to a sphere of radius the mean distance  of the earth from the sun. Or how about what you can calculate with 5,10,20,30 etc digits of precision of π
 * About thirty digits. It's good to know that you have to fold a sheet of paper in two about a hundred times to make it as thick as the radius of the observable universe, and 2100 is approximately 1030. &#x2013; b_jonas 11:06, 3 April 2006 (UTC)

Functions with latex
Hi guys,

is there any packet for LaTeX to draw and display functions in a LaTeX Document, such that u enter a formula and parameters and it will draw it into the document.

Thank you --Da legend 15:54, 29 March 2006 (UTC)


 * If I understand you correctly, you want to draw graphs and display them in a LaTeX document. Probably the easiest way to do that is to draw them with a package like gnuplot, export them as embedded postscript, and embed those in the LaTeX document. --04:12, 30 March 2006 (UTC)


 * I seem to recal there is some picture drawing capabilities in TeX/LaTeX, (might be a add on) and the syntax was horible. Irrelavant for wikipedia as its not supported. --Salix alba (talk) 08:06, 30 March 2006 (UTC)


 * Haha I remember people trying to do analytical geometry with LaTeX, they must have spent week to do a triangle !


 * What if "Functions with latex" was a crossword definition ... --DLL 21:06, 30 March 2006 (UTC)

Dow Jones Average
I have been trying to learn "how the $11,000+" amount is calculated. I have looked up the 30 stocks and added the stock price but doesn't come close to $11,000. How is this number derived on a daily basis?

Thanks, Bill


 * It is a "price-weighted" ratio. See the article Dow Jones Industrial Average. --LarryMac 22:39, 29 March 2006 (UTC)


 * You add up the stock prices and then divide the total by a term called the divisor, which is currently about 0.125. Whenever there is a change in the constituent stocks that make up the index, the value of the divisor is adjusted so that the index value immediately after the change is the same as its value immediately before the change. This avoids artificial jumps in the index value and ensures that the index has continuity. For more details see . Gandalf61 11:42, 30 March 2006 (UTC)

= March 30 =

Function for making 1,2,3,4,5,6 then any number
When I was studying the problem of induction at University the lecturer wrote '1,2,3,4,5,6' on the board, and asked what number came next. The general consensus was '7'. He then wrote out a simple function that, when we worked out the next number, it turned out to be something in the hundreds. Depending on a certain variable in the function, it would give a series of successive integers followed by any number you specify. Does anyone know what this function is? Purely out of interest, no hurry. Phileas 04:40, 30 March 2006 (UTC)
 * Here's one easy way: The function $$f(x)=x+(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-k)$$ will always return 1-6 for inputs 1-6, but for any other value, the output will depend on k. (The function basically works by adding y=x to an expression with roots at 1 through 6.) In particular, to get any given output at 7, use k=(847-f(7))/120. Is this kinda what you saw? --Geoffrey 05:40, 30 March 2006 (UTC)
 * That looks pretty familiar, yeah! I'm not certain that that's the same one for sure, but it definitely answers my question. Thanks! Phileas 05:54, 30 March 2006 (UTC)

I'd be more impressed if someone could write a recursive sequence whose first 6 terms were one through six and whose seventh was anything. OEIS has nothing of the like. -lethe talk [ +] 06:03, 30 March 2006 (UTC)
 * Try
 * $$f(1) = 1; f(n) = 1+f(n-1)+\left \lfloor \frac{f(n-1)}{6} \right \rfloor (A-7)$$
 * where A is anything. Replace the floor expression with the polynomial with roots 1-5 and adjust constants accordingly for a slightly hairier but "continuous" expression. Fredrik Johansson 10:34, 30 March 2006 (UTC)

Uncle Scrooge's wealth
In the 1952 story Back to the Klondike, Uncle Scrooge claims that he has 3 cubic acres of cash in his Money Bin. "Cubic acres" is an unusual measurement for volumes so I need to ask: How much is 3 cubic acres in cubic metres?

Also, the depth gauge in the bin shows a depth of 99 feet. Is that reasonable? Thuresson 16:27, 30 March 2006 (UTC)


 * Well, it doesn't quite make sense on first glance. Since "acre" is a measurement of area, it's not clear what a "cubic acre" should be in the first place.  But, for example, if we make the following definition:  a cubic acre is a cube each of whose face has an acre of area, it then makes sense to talk about the volume of a cubic acre.  According to Acre, an acre is 4046.8564224 m2.  A square which has an acre in area would have to have side length a square root of that, roughly 63.615 m.  A cube whose side lengths are each 63.615 m would have a volume of 257 440 m3 and three of these would have a volume of 772 321 3.  That's a lot of cubic yards.


 * As for depth, if the height is 99 feet, that's roughly 33 m which is about half of 63.615 m. Thus a cubic acre with depth 99 ft would have to have a base area of about 2 acres.  Big bin.--Deville (Talk) 17:15, 30 March 2006 (UTC)


 * Cubic acres make perfect sense—as a measurement of six-dimensional hypervolumes. I would take the comic-book usage as a joke, not intended to have any precise meaning. --Trovatore 17:30, 30 March 2006 (UTC)


 * Thanks for the answer. The weight of all this cash (mostly coins?) must be massive. Thuresson 17:36, 30 March 2006 (UTC)


 * Here's a trivial but cute-sounding exercise that occurred to me: How many square gallons in a cubic acre? --Trovatore 23:27, 30 March 2006 (UTC)
 * Interestingly, even Google can do it. Superm401 - Talk 00:37, 31 March 2006 (UTC)
 * Hehe, that made my day. SanderJK 13:55, 31 March 2006 (UTC)

How manys suns would fill the sky.
How many suns would it take to fill the sky,I know the size of the sun is about 1/2degree, and the sky is basically half a sphere.But what is the formula to work it out step by step. Looking at the sky I tryed to visalize how suns it would take,the answer must be quiet large,but it is the maths that I have find hard.
 * Well, the question could use a more-precise formulation. Do you have to be able to see a solar surface in every direction, with no gaps in between? If so, then some of the suns are going to have to overlap.
 * If we ignore this point, though, we should be able to get a quick-and-dirty approximation easily enough. Half a degree is about 0.01 radian, so the area of the solar disc should be about 0.0001 R2, where R is the distance to the Sun. (Aside—can that be right? Seems too much.) That's taking the Sun to be a square; you'd multiply by &pi;/4 to get the area of a circle, so we'll remember that the answer might be off by 30% or so.
 * Now the area of the entire hemisphere is 2&pi;R2, so you'd need about 60,000 Suns if they were squares. Closer to 80,000 if you could distribute circular area to fill the gaps, which of course you can't, so maybe guesstimate 100,000 allowing overlaps. --Trovatore 19:32, 30 March 2006 (UTC)
 * I just want to add that How manys suns would fill the sky. would be an excellent first line for a poem. Could you be channeling E. E. Cummings? --Trovatore 19:38, 30 March 2006 (UTC)
 * And yes, the average diameter of the sun is (roughly) 0.0093 radians, which is 0.53 degrees. -- Meni Rosenfeld (talk) 08:20, 31 March 2006 (UTC)


 * I see Trovatore used 2&pi;R2. That is for the whole hemisphere, not the visible amount of sky. An interesting way to find the circumference of the visible sky would be looking at Image:Anticrepuscularpano.jpg and comparing the size of the sun with the length of the photograph. Assuming the photograph is full-circle. -- [[Image:Flag_of_the_United_States.svg|17px]] Mac Davis ] ⌇☢ &#xE0D;&#x19B;. 08:52, 31 March 2006 (UTC)

= March 31 =

Approximation by polynomials
It is well known that for every m-times continuously differentiable real function f on R^n, every compact subset K of R^n, and every e>0, there is a polynomial p on R^n such that for every multi-index $$\alpha$$ with $$|\alpha|<m+1$$ and every x in K, we have $$|\partial^\alpha f(x)-\partial^\alpha p(x)|<e$$.

On page 155 of Topological vector spaces, distributions and kernels by F. Treves, Academic 1967, there is a proof using extension of f to a complex entire function on C^n.

Most of the standard undergraduate textbooks in numerical analysis AVAILABLE to me do not deal with this specialized topic.

Question: Is there any proof which is internal to real analysis? Any references?

Thank you in advance. Twma 07:18, 31 March 2006 (UTC)


 * For the case m=0 this can be done with Bernstein polynomials; I think no results from outside real analysis are used. A concise discussion online is Shadrin's lecture notes on Approximation Theory; you need Lectures 2 and 3. I think this is also treated in M.J.D. Powell, Approximation Theory and Methods, Cambridge University Press, 1981. My guess is that extension to general m is quite possible. -- Jitse Niesen (talk) 08:56, 31 March 2006 (UTC)
 * The m=0 case goes by the name of the Stone-Weierstrass theorem. -lethe talk [ +] 14:13, 31 March 2006 (UTC)

I visited Shadrin's notes and checked out Powell's book (waiting for 4 days from our Library). It appears that we do not have any alternative proof in our archives. Thanks. Twma 07:39, 5 April 2006 (UTC)

Probability Question
Suppose an event e has two equally likely outcomes, x or y. Suppose e has occurred k=5 times in succession, each with outcome x. What is that probability that the 6th outcome of event e will be x? --Simian1k 18:37, 31 March 2006 (UTC)


 * It's still 50:50. Truly random events have no memory. The common belief that a Y will be more likely after a run of Xs comes from a misinterpretation of the law of large numbers. —Keenan Pepper 18:53, 31 March 2006 (UTC)


 * Be careful not to overstate the facts. It is possible for "truly random" events to be correlated. We explicitly assume that, say, coin flips are independent of each other. A Markov process, however, is also random. --KSmrqT 22:45, 31 March 2006 (UTC)


 * Good point. —Keenan Pepper 23:32, 31 March 2006 (UTC)


 * See also Tom Stoppard's wonderful Rosencrantz & Guildenstern Are Dead (in which a flipped coin lands as "heads" eighty-five consecutive times, engendering an interesting discussion from the main characters). Joe 00:02, 1 April 2006 (UTC)


 * This is not one coin flipped repeatedly; it's a bag of individual coins each flipped once. When the play opens, a run of 69 coins has landed "heads"; by the time it ends, the number is much, much higher. The title characters are executed in Hamlet; here they don't know their fate, but GUIL ominously observes:
 * The equanimity of your average tosser of coins depends upon a law, or rather a tendency, or let us say a probability, or at any rate a mathematically calculable chance, which ensures that he will not upset himself by losing too much nor upset his opponent by winning too often. This made for a kind of harmony and a kind of confidence. It related the fortuitous and the ordained into a reassuring union which we recognized as nature. The sun came up about as often as it went down, in the long run, and a coin showed heads about as often as it showed tails. Then a messenger arrived. We had been sent for. Nothing else happened. Ninety-two coins spun consecutively have come down heads ninety-two consecutive times ...
 * Rather a nice line, about "the fortuitous and the ordained". --KSmrqT 21:01, 1 April 2006 (UTC)


 * I believe the word that is missing from the problem statement is that they must be independent random events.StuRat 20:44, 1 April 2006 (UTC)


 * If the outcomes are equally likely they will still be after the event happened 5 times. What you need to know if the chances of the outcome are indeed equal and if there's not some hidden thing affecting the results. - Mgm|(talk) 11:04, 4 April 2006 (UTC)


 * That's not true in the case of dependent events. For example, if a card is drawn from a deck and not replaced, the odds of drawing the same color card again most definitely goes down. StuRat 04:06, 5 April 2006 (UTC)


 * That's true, but it doesn't really fit the description. He states quite clearly that event e, each time it occurs, has two equally likely outcomes x and y. Even though he doesn't say it outright, this is a description of independent events. Unless it's a Markov something-or-other, which I've never heard of. Black Carrot 04:43, 6 April 2006 (UTC)

Infinite Sum
What is the value of the series: $$s=\sum_{i=0}^{\infty} (-1)^i * a = a-a+a-a+\ldots$$? Thanks in advance, Mickey 195.93.60.7 19:30, 31 March 2006 (UTC)
 * The traditional definition of an infinite sum says that this does not have a sum. It is, however, summable by the methods of Cesaro and Borel, and they give answers of a/2. -lethe talk [ +] 19:41, 31 March 2006 (UTC)


 * More precisely, see for example pages 256-8, Distributions in the Physical and Engineering Sciences: Volume 1: Distributional and Fractal Calculus, Integral Transforms and Wave, ISBN:0817639241 Twma 02:46, 3 April 2006 (UTC)


 * Someone asked a closely related question over at AskMetafilter some time ago, and I'll just point you to the many interesting comments there. Chuck 23:17, 31 March 2006 (UTC)


 * Thank you! Mickey 195.93.60.7 22:09, 1 April 2006 (UTC)

Brachystochrone and Snell's Law
In the article on the brachystochrone Snell's law is given as sin theta divided by velocity equals Cste. What is Cste ? --204.69.190.14 19:38, 31 March 2006 (UTC)
 * Snell's law says that n sin θ is a constant. So sin θ/v is a constant.  I expect that "constant" is what is meant by "Cste".  Looking at the history of the article, I see that that section was copied from the french, which might explain its funny terminology.  I don't know how they would say "constant" in french, or how they would abbreviate it, so that's a bit of speculation on my part.  Based on this assumption, I'm going to fix the article. -lethe talk [ +] 19:55, 31 March 2006 (UTC)
 * "A constant" is "une constante" in French. I've no idea how they abbreviate it, but I'm pretty sure that your assumption is correct. -- Jitse Niesen (talk) 02:06, 3 April 2006 (UTC)

A cruel set of differential equations
I have to solve the following set of differential equations numerically:

$$ \psi'' = f_1 (\psi, \phi, \theta, \psi', \phi', \theta',s) $$ $$ \phi'' = f_2 (\psi, \phi, \theta, \psi', \phi', \theta',s) $$ $$ \theta'' = f_3 (\psi, \phi, \theta, \psi', \phi', \theta',s) $$

Boundary conditions:

$$ \theta (0) = \pi/2, \phi(0) = 0, \psi (0) = 0 $$ $$ \theta'(L) \sin\phi(L) - \psi'(L) \sin\theta(L) \cos\phi(L) = M_1/EI_x $$ $$ \theta'(L) \cos\phi(L) - \psi'(L) \sin\theta(L) \sin\phi(L) = M_2/EI_y $$ $$ \phi'(L) + \psi'(L) \cos\theta(L) = M_3/GJ $$

I have three 2nd order equations, and 6 boundary conditions, so I should be able to solve this. The derivatives are with respect to s. Functions f1, f2, f3 are a bit complicated. How can I find solutions to this problem numerically to find ψ(s), θ(s) and φ(s)? I have Matlab at my disposal. Any help will be greatly appreciated. deeptrivia (talk) 19:55, 31 March 2006 (UTC)


 * Aha, the boundary-value problem. I'm guessing you're doing something in elasticity or the like?  In any case, the most straightforward way to solve a system like this is to use a "shooting" method.  The idea is this:  if instead of setting the BC at 0 and L as you do, what if you specified six initial  conditions at 0, namely $$\phi(0),\psi(0),\theta(0),\phi'(0),\psi'(0),\theta'(0)$$?  Then you could integrate the problem from 0 to L like a standard initial value problem, and see what you get when you get to L.  You'll probably not solve the BC at x=L, so you need to pick different initial velocities.
 * But now think of it like this. For any choice $$A,B,C$$, define $$f(A,B,C)$$ as follows.  First, integrate your ODEs with initial conditions $$\phi(0)=0,\psi(0)=0,\theta(0)=\pi/2,\phi'(0)=A,\psi'(0)=B,\theta'(0)=C$$ until x=L.  Then let $$f(A,B,C)$$ be the difference between what you get when you evaluate the BC on your numerical solution with what you want to get.  Then $$f:R^3\to R$$ is just a function you want to find the roots of, and then you use Newton's method or whatever to find the roots.
 * A much more detailed explanation is given in the Numerical Recipes series. For example, check here, Chapters 16 and 17.  Even if you don't want to do it in C, they describe the algorithms in words quite well and you should be able to get started from there.
 * If you want more details, let me know here or my talk page.--Deville (Talk) 00:27, 1 April 2006 (UTC)


 * Thanks a lot Deville. I hope this will solve the problem. It looks very promising. Cheers :) deeptrivia (talk) 15:14, 1 April 2006 (UTC)


 * In matlab, the function bvp4c solves boundary value problems by a different method (collocation). Both this method and the shooting method often need a good initial guess for the solution; this can sometimes be a big challenge. -- Jitse Niesen (talk) 02:12, 3 April 2006 (UTC)