Wikipedia:Reference desk/Mathematics

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January 1

Probability Formula

I am looking for formula that would solve this question and ones like it:

If I have a deck of 52 standard cards (26 black, 26 red). If I draw 20 cards, what is the chance that 15, or more, are red? --155.144.251.120 23:50, 1 January 2007 (UTC)

I don't think there's a simple formula (although there would be simplifications available in certain special cases) but the first step is calculating the probability of getting exactly 15 red, which you do as follows: the total number of ways to draw 20 cards is ₅₂C₂₀. The number of ways to draw 15 red (and hence 5 black) is ₂₆C₁₅ * ₂₆C₅ (i.e. the number of ways to draw 15 from the group of red AND 5 from the group of black), so the probability is ₂₆C₁₅ * ₂₆C₅ / ₅₂C₂₀ = 0.00403 according to both Windows and Google calcuators. Then you just have to do that for drawing 16 red, 17 red, ... 20 red, and add them up. There may be an easier way but sadly I can't think of it. Confusing Manifestation 08:48, 2 January 2007 (UTC)

(I have added subscripting.) The notation “₅₂C₂₀” denotes the number of combinations of 52 distinct items taken 20 at a time, with numeric value given as a binomial coefficient,

${}_{n}C_{k}=C(n,k)=n?k={n \choose k}={\frac {n!}{k!(n-k)!}}.$

Here n! denotes, as usual, the factorial,

$n!=n\times (n-1)\times (n-2)\times \cdots \times 2\times 1.\,\!$

For example, C(26,5) is 65780. We might be able to use Vandermonde’s convolution,

$\sum _{k\in \mathbb {Z} }{r \choose {m+k}}{s \choose {n-k}}={{r+s} \choose {m+n}},$

to simplify the proposed sum. (See Graham, Knuth, Patashnik, Concrete Mathematics.) --KSmrq^T 03:20, 3 January 2007 (UTC)

January 2

Nagorno-Karabakh

Hi, Dear Wikipedia!

I am from Azerbaijan. You have indicated the Nagorno-Karabakh (an integral and historical part of Azerbaijan)as an independent country in the list of World Countries. It is very erronous. It is not an independent country, you have made mistake. The Armenia republic invaded our lands.

Please make amenment to this prestiged encyclopedia.

Azerbaijan, Baku

Mr Qurban Rzayev

P.S.

Please forward this letter to the Editorial Board of Wikipedia I couldn't find their e-mail — Preceding unsigned comment added by 62.217.141.48 (talk • contribs)

Mr Rzayev - Wikipedia is written by anyone who wishes to contribute, so there is no Editorial Board - to find out more about how Wikipedia works, you might like to read Wikipedia:About and Wikipedia:Who writes Wikipedia. With regard to Nagorno-Karabakh, our article on that region says "the Nagorno-Karabakh Republic (NKR) ... remains unrecognized by any international organization or country", and Nagorno-Karabakh appears on our list of unrecognized countries. There is a note at the top of the list of countries which explains that this list includes some "countries" that are not internationally recognised. This list page also notes that Nagorno-Karabakh is a de facto (i.e. self declared) independent state which is unrecognised by any United Nations member state. Gandalf61 10:15, 2 January 2007 (UTC)

(Somehow I have the feeling that this is not a mathematics topic.) Additionally, the list of countries (which is what I presume you mean by "the list of World Countries") mentions clearly in its first sentence that the list includes states that are generally unrecognized.  --Lambiam^Talk 11:52, 2 January 2007 (UTC)

Sure it's a math question, we are debating whether Nagorno-Karabakh is a proper or improper fraction of Azerbaijan. :-) StuRat 13:29, 2 January 2007 (UTC)

Also note that Nagorno-Karabakh is an integral. -- Meni Rosenfeld (talk) 14:22, 2 January 2007 (UTC)

I would have thought that Nagorno-Karabakh culture was derivative of Azerbaijani culture. :-) StuRat 00:38, 3 January 2007 (UTC)

Radian

Ok, I feel awkward for not understanding this but I'll throw away my egoism and ask the following:

A circles length is 0.4π we roll it so it rotates up until it creates an angle of 80rad. What will be the distance that has covered? Please, give me a brief explanation along with the result, as there is more homework like this. --87.203.55.12 20:21, 2 January 2007 (UTC)

Is that really how the question is worded? —Bkell (talk) 20:30, 2 January 2007 (UTC)

Of course not, just my english are bad. Let's try again... It's a really long exercise in physics, whose mathematics part goes: on the one side of a pulley (r=0,2, so c=0,4π) we have a movable with a mass M and we let it fall. After the pulley has rotated for 80rad, how far has the movable gone from his original location? --87.203.55.12 20:37, 2 January 2007 (UTC)

A radian is the name for a length of a circle's circumference equal to the circle's radius, or for the angle defined by that measurement. If one radian of circumference has a length of one radius, then how many radiuses are there in 80 radians of circumference? --Carnildo 20:35, 2 January 2007 (UTC)

Thanks everyone for your will to help, even though you didn't understand the question. I have just called a friend and found out. Have a nice day and a happy new year! --87.203.55.12 20:56, 2 January 2007 (UTC)

January 3

All the King's marbles

You are the royal statistician for the King of a small country. Recently, the King’s spies captured a machine from the enemy that manufactures orange and green marbles, along with an instruction manual. The instructions say that the machine can only produce one marble per day, and that the color of the marble will be determined at random. The manual further states that a factory preset determines the fraction of orange marbles that the machine will produce on average. Unfortunately, the spies could not find any evidence to indicate the value of the factory-preset for the machine they captured.

Since the nation’s GNMO (Gross National Marble Output) is critical to national security, the King has asked you to provide a daily report stating the current best estimate of the factory preset and to give an indication of the uncertainty of the estimate. The first report is due immediately, before the machine has produced any marbles. What do you tell the King? After 1 day, the machine produces an orange marble. What do you tell the King? The next day another orange marble is produced. What do you tell the King? What do you report on day 1000, by which time the sample consists of 1000 orange marbles and 0 green marbles?

I’ll attempt to answer. I will use the equation f = x/n, where f is my estimate of the factory preset fraction of orange marbles, x is the number of orange marbles in my sample, and n is the total number of marbles in my sample. I am currently unaware of what equation to use to express the uncertainty in the estimate.

In the first report, I would state that no estimate is possible because no data is available on which to base an estimate. The uncertainty is 100%.

In the second report, I have a sample of 1 orange marble, and I would state that the best estimate of the factory preset is 1 (f = x/n = 1/1 = 1), but that the uncertainty of the estimate is large. What I currently can’t figure out is how to express the uncertainty numerically.

In the third report, I have a sample of 2 orange marbles, and I would still give an estimate of 1 for the factory preset (f = x/n = 2/2 = 1), but now, my uncertainty is less (but how much less)?

In the 1001st report, I have a sample of 1000 orange marbles. I am now very certain that the factory preset is 1 or at least very close to 1 (f = x/n = 1000/1000 = 1). My uncertainty is far lower than it was in the first two reports, but by how much?

dryguy 14:14, 3 January 2007 (UTC)

A relatively acceptable way of measuring the uncertainty would be to see how much difference pulling a green next time would make. So the first time, you'd have an uncertainty of 100% (1/1), the second time, 50% (1/2), the third time 33% (2/3), and the last time, 0.1%. I'm not sure this is the right way to go about it, but I seem to remember having done something like this (a long time ago). yandman 14:21, 3 January 2007 (UTC)

I would look at the number of marbles produced as the sample size. I believe 1100 marbles works out to about a 3% margin of error over a 95% confidence interval (meaning 95% of the time the setting would be between 97% and 100% orange marbles). For the no marbles case, I would estimate a 50% setting with a 50% margin of error. (If you knew all the possible settings, say 50%, 60%, 70%, 80%, 90%, or 100% orange, then you could make a more intelligent guess by averaging all the possible settings to get 75% orange marbles, on average.)

Margin of error can also be calculated using a given confidence interval:

            99% confidence 95% confidence 90% confidence   
Sample size $\approx 1.29/{\sqrt {n}}\,...\approx 0.98/{\sqrt {n}}\,...\approx 0.82/{\sqrt {n}}\,$
----------- -------------- -------------- --------------
     2                                                   \
  1000                                                    > Margin of error
  1100           3.9%            3.0%          2.5%      /

Those formulae don't work well for small samples, however, even giving over a 100% margin of error for a sample size of 1. I guess I'd say for a sample of 1 the margin of error equals the confidence interval. Thus, there is a 99% margin of error within a 99% confidence interval. So, adding the info from the above chart, I get:

Sample size Estimate 99% confidence 95% confidence 90% confidence   
----------- -------- -------------- -------------- --------------
     0         50%       50.0%           50.0%         50.0%      \
     1        100%       99.0%           95.0%         90.0%       \
     2        100%                                                  > Margin of error
  1000        100%                                                 /
  1100        100%        3.9%            3.0%          2.5%      /

You can fill in the missing numbers. StuRat 16:21, 3 January 2007 (UTC)

This is a textbook example of Bayesian inference. Let x be the fraction of orange marbles. Then, given x, the probability of drawing x orange marbles is xⁿ. That means that the likelihood for x is xⁿ times the prior probability of x. If you initially make the not unreasonable assumption that each fraction x between 0 to 1 is equally likely (i.e. a uniform prior) then after 0 samples, the mean and median values of x are 1/2. After n samples, the probability distribution for x is given by the normalized likelihood: (n+1)xⁿ. Thus, the most likely value of x is 1 whereas the estimated median value of x is 2^-1/n and the mean value is (n+1)/(n+2) (computing this is a simple integral). The variance about this mean is (n+1)/(n+2)(n+3) (the standard deviation is the square root of this). –Joke 16:45, 3 January 2007 (UTC)

That is one of the two most convincing answers anyone has given me so far. The other suggestion came from someone at Craigslist who said to use Wilson's interval. I will be mulling this over for a while, but would love to hear any more comments. My thanks to everyone who contributed! dryguy 17:29, 3 January 2007 (UTC)

Vertex operator algebra and vertex algebras - question

In wikifying an article on the mathematician, James Lepowsky, there is this statement: "His current research is in the areas of infinite-dimensional Lie algebras and vertex algebras." Is Vertex operator algebra the same area as vertex algebras? That is, could I use the working link as a substitute for the red link? Sincerely, Mattisse 14:33, 3 January 2007 (UTC)

Yes. –Joke 16:48, 3 January 2007 (UTC)

Thanks for answering. I realised it was a stupid question to as on closer inspection I could see they were the same. Sincerely, Mattisse 00:05, 4 January 2007 (UTC)

product

Can someone direct me to a page dealing with the product

Product (from x=0 to x=N) of (n+x)/(x+1) or similar eg 1/n!{n(n+1)(n+2)...(n+N)}

Thanks.87.102.4.89 16:39, 3 January 2007 (UTC)

I don't think we have an article on these product per se, but as they both ultimately come down to factorials, maybe that article will help. Can you specify what do you want to know about these? -- Meni Rosenfeld (talk) 17:43, 3 January 2007 (UTC)

See binomial coefficient or combination:

$\prod _{i=0}^{N}{\frac {i+n}{i+1}}={\frac {(N+n)!}{(n-1)!(N+1)!}}={N+n \choose N+1}$

Moreover, by Stirling's approximation

$\log {N+n \choose N+1}\sim N\log {\frac {N+n}{N+1}}+n\log {\frac {N+n}{n-1}}+{\frac {1}{2}}\log {\frac {(N-n)(n-1)}{2\pi (N+1)^{3}}}$

for large values of N+n. –Joke 20:43, 3 January 2007 (UTC)

Thanks is the above brakets the same as c(N+x,N+1) ie a combination - I honestly hadn't spotted that it was an exmaplpl of a combination Cr. Thanks.87.102.4.89 21:24, 3 January 2007 (UTC)

I wanted to know if there was a 'name' or symbol as per combinations and permutations (Cr Pr), since I've been finding this equation cropping up a lot (distinct ways of arranging n 'things' over x 'holes', sum of integers, number of balls in a tetrahedral pyramid, derivations sums of powers eg 1⁴+2⁴+3⁴ etcto name a few I know). It seems common in counting problems and thought there might be a conventionally used symbol etc.

I'd also bee interested in other properties/uses of it if any common ones are known, and also any history of it's usage. Thanks.87.102.4.89 21:18, 3 January 2007 (UTC)

Please read: combination, binomial coefficient, combinatorics. –Joke 21:44, 3 January 2007 (UTC)

1 plus 1

Whats 1 plus 1? —The preceding unsigned comment was added by 74.102.217.142 (talk) 22:20, 3 January 2007 (UTC).

It's -1 in $\mathbb {Z} _{3}$, 0 in $\mathbb {Z} _{2}$, 1 in boolean algebra, 2 in the integers, 10 in binary and 11 in reflective binary code (about which I could not find an article). Pick your favorite. -- Meni Rosenfeld (talk) 22:47, 3 January 2007 (UTC)

I added a redirect for reflective binary code -> Gray code. --Ornil 20:07, 4 January 2007 (UTC)

One plus one is two, and 1 + 1 is 2, but 1 plus 1 is an unholy chimera, so let's say it's undefined and pretend it didn't come up. Melchoir 23:33, 3 January 2007 (UTC)

As 1 red apple is a red apple, 1 plus 1 can be interpreted as a plus 1, i.e. a single occurrence of a unary plus preceding the integer 1.86.132.238.241 19:51, 4 January 2007 (UTC)

January 4

Nanosecands

How many nanosecands are there in a 986 billion years?

I assume you mean nanoseconds? That depends on the length of your year: the common year of 365 days, the Julian year of 365.25 days, the Gregorian year of 365.2425 days, the sidereal year of 365.256 363 051 days, or some other year entirely. --Carnildo 06:39, 4 January 2007 (UTC)

You can find this out with Google calculator. Try Googling for 986 billion years in nanoseconds and your answer will be right there. Note that Google appears to use a Gregorian year. For more information, check out the Year article, which lists all the different year lengths in days. Once you've chosen a year, multiply its length in days by 986 billion (for the number of years) and then multiply that by 864 x 10¹³, which is the number of nanoseconds in a day. Maelin (Talk | Contribs) 07:14, 4 January 2007 (UTC)

Why 986 billion years? That's an awfully long time. Where does the number come from? The universe may not last nearly that long; see Big Crunch and Big Rip. --Lambiam^Talk 09:40, 4 January 2007 (UTC)

Apparently because of the news that came out last year saying the Universe could be 986 billion years old. Why nanoseconds, then? I think it was just one of those "lol science" or "lol big numbers" question. — Kieff 09:54, 4 January 2007 (UTC)

LOL. If you read that article carefully it says "986 billion years older than physicists thought", because somone had come up with a multi-big-bang theory in which the universe is at least 1 trillion (10¹²) years old, and 986 billion is the difference between 1 trillion and the "orthodox" age of 14 billion years. —The preceding unsigned comment was added by Gandalf61 (talk • contribs) 15:50, 4 January 2007 (UTC).Who came back to add his sig and found some darn bot had beaten him to it. Gandalf61 15:54, 4 January 2007 (UTC)

Heh. – b_jonas 19:24, 5 January 2007 (UTC)

When addition maps A×B->A but not A×A->A

In physical applications, if one is pedantic there is a difference between duration and time, and between absolute and relative temperature. For example, the difference between 400K and 300K is −100K even though a temperature of −100K is meaningless. In mathematical terms, it looks like this:

addition $:(A\times B)\cup (B\times A)\to A$ (E.g., the sum of a position and a displacement is a position.)

subtraction $:(A\times A)\cup (B\times B)\to B$ (E.g., the difference of positions is a displacement, as is the difference of displacements.)

Is there mathematics that describes this? It appears that relative position, relative temperature, etc. is a vector space but absolute position is not. —Ben FrantzDale 15:34, 4 January 2007 (UTC)

A quick study suggests affine space. — Arthur Rubin | (talk) 15:52, 4 January 2007 (UTC)

That looks right. Thanks! —Ben FrantzDale 16:13, 4 January 2007 (UTC)

See also torsor. Dave Rosoff 00:11, 5 January 2007 (UTC)

A related question

In physics, the vector space of displacements (measured in units of distance) is distinct from the vector space of velocities (measured in units of speed), however we can define an angle between a velocity vector and a distance vector. Is there mathematical language to describe this relationship? Thanks again. —Ben FrantzDale 18:23, 4 January 2007 (UTC)

Thinking simply - at a given instance - both are just directions - ie vectors in 3 dimensional space - so just use the formala that gives the angle between two lines eg the dot product#Geometric interpretation87.102.8.102 19:45, 4 January 2007 (UTC)

Your 'vector space of velocities' is really just a 'vector space of distance' where the distances are functions of time (an external variable). So you don't really have two 'vector spaces' at all. 87.102.8.102 20:15, 4 January 2007 (UTC)

Since d(distance)/d(time) = velocity you could say that the 'mathematical language to describe this relationship' is one of differentiation and integration.87.102.8.102 20:20, 4 January 2007 (UTC)

I guess {displacements} × {durations} forms a 4-D vector space, but there seems to be some additional structure allowing for division by duration. It seems that velocity is either in a different vector space from displacement in that you can't add them, but they obviously have some relation to one another since you can take angles between them. (That is, I feel like group theory has something to say about this.) —Ben FrantzDale 20:37, 4 January 2007 (UTC)

Good observation; we cheat. (But not much.) Suppose we have an ice hockey rink marked off in cartesian coordinates x and y, measured in metres. The displacement of a puck on the surface would be a vector such as (3 m,4 m), where the units of displacement are metres. If the puck traverses that distance in 0.5 seconds, its velocity is a different vector, (6 m/s,8 m/s), where the units of velocity are metres per second. Physics training tells us to pay attention to the difference between units, for "we must not mix apples and oranges". But the velocity has a direction which is a pure displacement, and the angle between two nonzero vectors does not depend on the length of the vectors. So we quietly convert from velocity to displacment.

Such quiet conversions are everywhere, even in pure mathematics. (Computer programming languages do it so often they have a formal term to describe it in reference manuals: "coercion".) For example, when we write 3 + 0.1415 we appear to be adding an integer and a decimal fraction. In fact, we quietly use the canonical injection of the integers into the rationals (or the reals) to make both numbers rational (or real) before we add. --KSmrq^T 02:12, 5 January 2007 (UTC)

The thing is, I'm sure this can be done without this sort of hand waving. I just haven't seen it done. My intuition tells me that velocity is in a separate vector space from displacement but they are somehow closely linked in that, like the anonymous person above said, velocity is in the space of time derivatives. I just haven't seen rigorous math to flesh out that intuition. —Ben FrantzDale 14:26, 5 January 2007 (UTC)

1 2 3...?

what number is after three? —The preceding unsigned comment was added by 74.102.217.142 (talk • contribs) 4 January 2007.

Difficult question - could you give some more contextual information.87.102.8.102 20:16, 4 January 2007 (UTC)

See above question (1 plus 1) by same anon user, who appears to be trolling. -- NorwegianBlue ^talk 20:27, 4 January 2007 (UTC)

Also, take a look at the user's contributions. — Kieff 20:41, 4 January 2007 (UTC)

The sequence you have described seems to be F₂ to F₄ of the Fibonacci series. Therefore your answer is F₅, which is 5. Readro 20:35, 4 January 2007 (UTC)

The fourth number can be any number n that you care to chose, if the numbers are the first four values of the sequence given by $f(x)={\frac {(x-2)(x-3)(x-4)}{-6}}+(x-1)(x-3)(x-4)+{\frac {3(x-1)(x-2)(x-4)}{-2}}+{\frac {(x-1)(x-2)(x-3)n}{6}}$ This function can be adapted to generate any arbitary finite sequence that you care to suggest, and is fundamentally uninteresting. -- AJR | Talk 00:56, 5 January 2007 (UTC)

For these questions, the right resource is the OEIS. If you search for 1, 2, 3, you'll find that the next number is most likely one between 1 and 6 inclusive: the first few hits give 5, 5, 6, 3, 1, 2, 2, 4, 5, 4, 2, 6, 2, 4, 4, 4, 4, 4, 4, 4, 4, 2, 3, 3, 3, 3, 3, 3, 4, ... as the next single number after 1, 2, 3 —The preceding unsigned comment was added by B jonas (talk • contribs) 19:21, 5 January 2007 (UTC).

Philosophical question

This question may be more suitable for the humanities page but here goes..

I notice that my world experience (outside imagination) can be mostly explained by simple mathematical concepts - euclidean space, mechanics, statistics, fractals etc. (ie basic science and engineering type maths)

My question for any more advanced mathematicians is - does knowledge and understanding of other more 'obscure' mathematical ideas cause an increase in the level of experience of the physical world - eg does an understanding of hyperbolic geometry, greater than 3 dimensional space, number theory etc enhance your experience of the world? (I'm working on the principle that 'inner knowledge' (in the 'brain') is reflected in 'outer experience' (in the world) that so far has seemed to be true.)

I hope someone can understand what I'm asking and I would really appreciate any replies you can give. Thanks.87.102.8.102 20:43, 4 January 2007 (UTC)

To turn the question on its head, consider if it is possible for any accurate knowledge to degrade or detract from your experience of the physical world. The answer there is that all knowledge, if true, at worst is irrelevant to your set of outside world experiences, in which case there would be no impact positive or negative. If knowledge is relevant, though, it can not help but to enhance the level of an experience. The situations where increased knowledge fails to enhance experience are those where knowledge is faulty or incomplete or misleading, all of which can lead to a correspondingly flawed experience or flawed understanding.

A beauty of mathematics, then, is that it possesses a particularly high degree of accuracy within it, such that knowledge based on mathematical principles has a correspondingly high probability of enhancing understanding. At worst, an understood mathematical principle will have no practical application to the outside world, but when it does possess relevancy it can not help but to enhance the personal understanding of the situation. (Not sure if that relates to your question, but just a thought.) Dugwiki 21:51, 4 January 2007 (UTC)

I do think that knowing obscure mathematics can improve your way of perceiving the world, but not always in the direct sense of understanding it more, but of getting more awe out of it, of seeing more beauty or interesting things on it. By knowing about the homeomorphic equivalence between a donut and a coffee mug, you can have a nice feeling of surprise during breakfast. That sort of thing. — Kieff 22:01, 4 January 2007 (UTC)

Just choosing one area of higher math as an example, I find that abstract algebra is reflected in my day-to-day experience, both on a literal level and on a metaphorical one. I certainly find groups in various places, particularly anytime I'm dealing with permutations. I can't arrange the billiard balls in the rack for 8-ball without thinking about permutation groups, and decomposition into disjoint cycles. (Well, maybe I can, but I prefer not to!)

On a different level, I remember finding it tremendously satisfying to learn the Fundamental theorem on homomorphisms for groups, and to apply it to human interactions. The theorem basically says that, to have a nice mapping from one group to another, they have to have certain correspondences in their internal structures (namely, the first group has to have a normal subgroup that induces a quotient group isomorphic to some subgroup of the second group). I like to think of that in terms of people - in order to interact with another human, you have to have something inside you that's common with something in that person, and that can form the basis for a relationship. A bit whimsical perhaps, but that's the kind of stuff I think about. -GTBacchus^(talk) 23:26, 4 January 2007 (UTC)

Yes thanks for all those answers - I have similar experiences. But I'd really be interested in the point of view of someone who understands say hyperbolic geometry or 4d space or something else I don't know much about. (Though I probably wouldn't understand what they were saying) - and if they can find examples of that in the world we inhabit. Thanks to you above though.87.102.23.224 23:45, 4 January 2007 (UTC)

When I studied Lobachevsky, I remember a professor making an analogy between hyperbolic geometry and love. Parallel lines, in a space with negative curvature, don't remain the same distance apart, but continually approach one another, becoming more and more aligned the closer they get. This professor used it as a metaphor for marriage, which I thought was sweet.

As for higher dimensional spaces, those are easy to apply to situations if we don't assume that all dimensions have to be spatial dimensions. A lot of things can be described by more than 3 mutually independent variables, so they could be thought of as inhabiting points in an n-dimensional space, where n is just about any number you like. One very concrete example is the phase space of a particle, like Planck used in his development of the idea of entropy. One particle is described by six numbers - three for position and three for velocity. The air in a room can be described as a point in (6*n)-dimensional space, where n is the number of molecules of air in the room. This is a useful tool for understanding how some aspects of thermodynamics work, and can be extended to many other areas, literally or by analogy. -GTBacchus^(talk) 23:56, 4 January 2007 (UTC)

Thanks again - it seems that hyperbolic geometry (like love) will remain in the realms of the mind for now.87.102.19.164 01:12, 5 January 2007 (UTC)

Ha, ha. Lobachevsky's "Theory of Parallels" - his work where he develops hyperbolic geometry - is quite accessible, actually. If you follow along with some paper and a pencil, you can get quite a ways into it. Good luck to you. -GTBacchus^(talk) 01:17, 5 January 2007 (UTC)

We could get into theories of mind and explore philosophy and recall the history of mathematics and physics. But perhaps one or two examples will be more informative and inspiring. Euclidean geometry, by rejecting the parallel postulate split into three choices: Euclidean, spherical, and hyperbolic. All of these have one thing in common, namely constant curvature (zero, positive, and negative, respectively). Although this mathematics began as an abstract inquiry, surfaces with these properties exist tangibly. Then Gauss and Riemann went one step further, and began the study of differential geometry, where the curvature could vary depending on location. This more accurately describes most surfaces. Einstein realized that a synthesis of varying curvature with higher dimensions provided exactly the tools he needed to extend special relativity to include gravity. He, and all modern physicists, could "see" curved higher-dimensional space in the physical universe.

Mathematics is sometimes described as the formal description of patterns, and the more patterns we know about the more we "see". If you would like to experience this yourself, read "Caustic (mathematics)" and then "Caustic (optics)". Working in reverse, computer graphics generates images simulating the real world out of pure mathematics. When trained mathematician and Academy-award winner Alvy Ray Smith looks at a plant, what he "sees" is surely shaped by his study of L-systems. Visit the pages of Przemyslaw Prusinkiewicz for beautiful examples of what that vision can accomplish. --KSmrq^T 02:46, 5 January 2007 (UTC)

Balls in a bag

If i have a bag of balls, containing 3 red balls, 2 blue balls, and 2x105 orange, what is the chance of picking a red ball?

For a single draw, it's the number of red balls divided by the total number of balls. StuRat 02:51, 5 January 2007 (UTC)

it is 7 draws —The preceding unsigned comment was added by 81.132.66.73 (talk) 17:17, 5 January 2007 (UTC).

Elementary questions in a plane

I am interested in the two dimensional plane only. Because my questions are elementary at the second year undergraduate level, perhaps somebody could offer pointers of solutions from the existing literatures.

A curve with endpoints is the continuous image of a closed interval. It is easy to prove that a simple curve C with distinct endpoints is homeomorphic to the closed unit interval [0,1]. Question#1: Is there any homeomorphic automorphism on the plane such that C is carried onto [0,1]?

Let A,B be nonempty closed bounded sets in a plane. The distance from a point x to A is denoted by d(x,A). It is easy to prove that the set C of points x with d(x,A)=d(x,B) is a closed set without any interior point. Question#2: Is it a smooth curve? Thank you in advance. Twma 00:53, 5 January 2007 (UTC)

Regarding Q2: If I understand you correctly, it is easy to come up with examples where C is a curve that is not smooth, and other examples where C is not a single curve. For example, consider small (closed) circles centered at the corners of a square. If A is the union of two circles at opposite corners, and B is one circle at one of the other corners, C will become two semi-infinite lines that meet at a right angle at the center of the square. This is a non-smooth curve. If, for the same A, B instead is the union of both the remaining circles, C becomes the union of two perpendicular lines that meet at the center of the square. If A is a small circle centered at (0,0) and B is the union of two small circles centered at (0,1) and (0,-1), C becomes two parallel lines. Am I missing something? Did you intend for A and B to be connected as well as closed? Even for that case, it is easy to make up examples that yield non-smooth curves (in the first example above, just connect the two parts of A with a loop that avoids getting closer to B than either of the circles, and A becomes connected without changing C). -- mglg_(talk) 02:21, 5 January 2007 (UTC)

Actually, without further assumption, it is not true that the set C of points x with d(x,A)=d(x,B) is a closed set without any interior point. Consider two overlapping disks, for instance; the set of points with d(x,A)=d(x,B) includes the entire region of overlap (where both distances are zero). Did you intend to require A and B to be disjoint from each other? —David Eppstein 02:23, 5 January 2007 (UTC)

Consider A = B = { P }, where P is a single point on the plane......... --CiaPan 14:39, 5 January 2007 (UTC)

Suppose that A,B must be disjoint. Is it possible to have any interior points? Thank you in advance.

The second question is completely answered. Thanks. For Q1, by homeomorphic automorphism, I mean a homeomorphism from the plane onto itself. Twma 19:33, 5 January 2007 (UTC)

symmetry operation in 3 dimensional space

I'm trying to find out what symmetry operation (not an inversion) operates about a point in 3 dimensional space. I've added the table below to clarify what I'm asking for. (Note in the 4d case I've 'assume(d) that all dimensions have to be spatial dimensions.' and orthogonal).

Dimensions (d)           1(number line)   2(plane)       3 (3d space)        4 (?)
.
Inversion in a..          point            point          point            point
.
Reflection in a..         point (d=0)      line  (d=1)    plane (d=2)      'volume' (d=3)
Rotation around a..       impossible       point (d=0)    line  (d=1)      (plane   (d=2)???)
Operation X               impossible       impossible     point (d=0)       ...

Am I missing something obvious (or is it impossible) (Note I've bracketed rotation about a plane in 4 dimensions since I haven't worked out the maths of it..) 87.102.19.164 01:25, 5 January 2007 (UTC)

This is not a complete answer, since I am not qualified to give one, but here goes anyway: There are reflections in points (i.e. inversions), in lines, in planes, etc, up to dimension n-1, where n is the dimension of your space. There are also rotations around objects of dimension n-2, as you correctly guessed, though one might want to think of them as rotations within a 2D plane rather than around anything. The math of rotations in higher dimensions is simply to apply a 2x2 rotation matrix containing sines and cosines to the 2 coordinates the rotation acts on, after choosing an appropriate coordinate system. By combining such 2D rotations in different planes you can define rotations within k-dimensional subspaces. I don't see an obvious candidate for your operation X, and after adding the different kinds of reflections, and incorporating inversions as one of the reflections, the table structure may no longer suggest (as strongly?) the existence of an operation X. -- mglg_(talk) 02:21, 5 January 2007 (UTC)

Could "Operation X" be two rotations about orthoginal lines intersecting at the given point ? StuRat 02:43, 5 January 2007 (UTC)

See point group. All such symmetries can be constructed by composing reflections, where a reflection essentially negates one coordinate in a suitably chosen coordinate system with origin at the fixed point. For example, the composition of two reflections is a rotation. The composition of three can be an inversion (if they are mutually perpendicular) or a rotary inversion (otherwise). In 4D, the composition of four reflections again gives a rotation, but one that simultaneously rotates two planar subspaces. If you like matrices, try these:

${\begin{aligned}{\begin{bmatrix}-1\end{bmatrix}}&\qquad {\text{reflection of line}}\\{\begin{bmatrix}-1&0\\0&-1\end{bmatrix}}&\qquad 180^{\circ }{\text{ rotation of plane}}\\{\begin{bmatrix}0&-1&0\\1&0&0\\0&0&-1\end{bmatrix}}&\qquad 90^{\circ }{\text{ rotary inversion of space}}\\{\begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}}&\qquad {\text{(double) rotation of 4-space}}\end{aligned}}$

Each of these is, and any point symmetry corresponds to, an orthogonal matrix. --KSmrq^T 05:55, 5 January 2007 (UTC)

Qualifying examination preparation

Hi everyone, I'm in the midst of studying for one of my PhD qualifying examinations, and I've hit a snag on a problem on one of the previous qualifiers; I'm kind of lost as to where to begin. Anyway, here's the statement:

Let G denote the group of all $n\times n$ nonsingular complex matrices (under multiplication). Let H be a finite group and $f:H\rightarrow G$ a group homomorphism. Prove that $\forall h\in H,\ f(h)$ is a diagonalizable matrix.

I don't want you to just give me the answer, I'd like a hint or something. Thanks in advance! –King Bee ^{(talk • contribs)} 04:08, 5 January 2007 (UTC)

I haven't worked it out, but it's got to be enough to know that some power of h is the identity. It's really all the information you're given. From there it should be pure linear algebra. Melchoir 04:34, 5 January 2007 (UTC)

Of course! Then $f(h)$ is a involutary matrix whose Jordan Canonical Form must be diagonal for all h, since every Jordan block $J_{\ell }$ must have the property that $J_{\ell }^{k}=I_{\ell }$. and is thus diagonalizable (Jordan Canonical Form argument). Thank you so much! Maybe you can help me with this one:

Let G be a group with trivial center. Prove that Aut(G) has trivial center.

I know I'm missing something really obvious here. Again, hint over a solution is preferred. –King Bee ^{(talk • contribs)} 04:44, 5 January 2007 (UTC)

Uh... possibly the representation map helps? This time I'm afraid I also get the feeling that I'm missing the easy solution! Melchoir 04:51, 5 January 2007 (UTC)

Well, I really feel like it should have something to do with the fact that G is isomorphic to Inn(G), but I'm just not clever enough to make it work. –King Bee ^{(talk • contribs)} 05:06, 5 January 2007 (UTC)

Okay, this problem was bothering me, so I found a brute-force solution: given a non-identity automorphism, you can construct a group element -- in hindsight, the obvious one -- whose conjugation doesn't commute with the automorphism at a certain point. I don't feel any cleverer for slogging through this method, though, so I hope someone finds a solution using some theoretical machinery. Melchoir 07:03, 5 January 2007 (UTC)

I see what you're saying. It's kind of annoying when the solution is like that. I'm going to query a few more people today, and if I come up with something else, I'll let you know. Thanks again for your help, I really appreciate it. –King Bee ^{(talk • contribs)} 13:25, 5 January 2007 (UTC)

Great to hear, good luck! Melchoir 18:45, 5 January 2007 (UTC)

Proving the Viète formula

In the article for the Viète formula a proof is stated which assumes the following identity:

${{\sin(2^{n}x)} \over {2^{n}\sin(x)}}=\prod _{i=0}^{n-1}\cos(2^{i}x)$

The article also states that the identity can be derived from the double angle identity. How can this be done?--Fiber B 06:15, 5 January 2007 (UTC)

Eh? The article says use

$\sin(2x)=2\sin(x)\cos(x),\,\!$

which is fairly explicit, and suggests induction over n. This is the n = 1 base case. Try n = 2.

${\begin{aligned}\sin(2^{2}x)&{}=2\sin(2x)\cos(2x)\\&{}=2\left(2\sin(x)\cos(x)\right)\cos(2x)\\&{}=2^{2}\sin(x)\cos(x)\cos(2x)\end{aligned}}$

I will leave the general induction step in your able hands. --KSmrq^T 06:53, 5 January 2007 (UTC)

exponents

when and how were exponents first used? —The preceding unsigned comment was added by 192.133.105.19 (talk) 17:18, 5 January 2007 (UTC).

Casting of nines

What properties, behavior and/or history is known about casting of nines. Is the term I know it as proper? I originally learned the trick in gradeschool in Madrid Spain. Very few know of this in the United States. It is an error prone method for checking multiplication.

Example:

2537865 × 4589002 = 11646267560730

Add up the individual term digits until you get a single digit answer, whenever you get an answer of 9 just continue any number +9 "simplifies" to 9, so for example 9 + 5 = 14 = 1 + 4 = 5, you don't have to do this neccessarily, I just think its where the name "casting of 9s" comes from

So you add up the digits of the multiplicand: 2+5+3+7+8+6+5 = 36 = 3+6 = 9 Add up the multiplier: 4+5+8+9+0+0+2 = 28 = 2+8 = 10 = 1+0 = 1

Then you multiply 9 × 1 = 9

If the answers adds up to 9 it is most likely correct.

1+1+6+4+6+2+6+7+5+6+0+7+3+0 = 54 = 5+4 = 9

It is most definetly not foolproof, but for a quick check for doing the multiplication by hand... Haloway13 17:56, 5 January 2007 (UTC)

See Casting out nines. --mglg_(talk) 17:55, 5 January 2007 (UTC)

In looking at that article I noticed it kind of glosses over proving that the method works. I put a note about that in the article's talk page, but if anybody happens to have a good published reference that they can cite and include in the article that includes a more formal proof the method works, feel free to include it. One useful proof to include would be that for an integer x, x mod 9 is equal to the sum of its digits mod 9. For example, 587 mod 9 = (5+8+7) mod 9. It can be proven using pretty simple math and would be a nice addition to the article. Dugwiki 19:02, 5 January 2007 (UTC)

Approximation of a simple curve by polygonal lines

A polygonal line is obtained by joining consecutive line segments. Let B(e) denote the open disk with center at the origin and radius e>0. Given a SIMPLE curve with dinstinct end points A,B, is it possible to find a SIMPLE polygonal line P with end points A,B such that P is contained in the tube C+B(e)={c+b:c in C, b in B(e)}? The approximation follows easily from uniform continuity but being SIMPLE (injective) is difficult for me to handle. I am interested in plane curves only. Because it is so basic and foundamental, I expect that it is a well known result. Be grateful if somebody could offer me a pointer. Thank you in advance. Twma 19:44, 5 January 2007 (UTC)

Value of money in 1800's`

I am trying to find out what the value of 150,000 dollars would have been in the mid 1800's in contrast to what it is worth today. Can you help me? Barb —The preceding unsigned comment was added by 70.176.52.116 (talk) 19:59, 5 January 2007 (UTC).

Equation of a line

The line has coordinates (-5,6) and (6,-7)

What is he equation and how can it be worked out? THanks in advance81.132.66.73 20:37, 5 January 2007 (UTC)