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September 14

chess variables and states

Is it more proper to think of the game of chess as having 32 variables with 65 states (64 in-play states and 1 out-of-play state) or as having 64 variables with 33 states (32 occupied and 1 unoccupied state)? -- Taxa (talk) 02:19, 14 September 2009 (UTC)

There isn't one objectively proper way to think about it. It's just a question of what the most useful scheme is for what you happen to be doing with the information. Is there a particular problem you're trying to solve? Rckrone (talk) 02:31, 14 September 2009 (UTC)

I'm just trying to figure the best way to think of it. I mean if you do the same for tic-tac-toe then it would seem the positions (squares) always remain while the players vary so positions as variables seems better. -- Taxa (talk) 03:04, 14 September 2009 (UTC)

Okay, if you are not going to buy that then how about the idea that there are programs that can reduce many-valued logic equations to minimum form. If you had a computer big and fast enough to assign it the task of reducing all known winning games to minimum form then you might come up with a way to win at chess all the time. -- Taxa (talk) 03:33, 14 September 2009 (UTC)

Assuming one can win all the time. A "best" strategy, when used by both players, may always result in a draw. I don't know. But, what constitutes a minimum form? One uses different parameterizations to solve different problems. And sometimes the most concise form isn't the easiest to solve. As for chess, well, I don't know, and I'm pretty sure nobody does.--Leon (talk) 08:49, 14 September 2009 (UTC)

Most serious chess programs these days use bitboards. 67.122.211.205 (talk) 04:30, 14 September 2009 (UTC)utnot

The reason seems to be speed and memory due to the compact nature of the data format rather than the accomodation of a superior method of play. -- Taxa (talk) 05:28, 14 September 2009 (UTC)

Neither of your methods are proper, because neither actually work. You seem to be ignoring promotions, the details of the rules of castling, draw by repetition, and the fifty move rule. Not to mention whose turn it is. Algebraist 13:56, 14 September 2009 (UTC)

To expand on Algebraist's point, you can use:

• 65 variables with 13 possible states each (since there are only 12 distinct chesspieces)
• 48 variables, corresponding to the 32 starting pieces and upto 16 (?) promoted queens, each with upto 65 possible states.
• Maintain a variable length list of live pieces and their position on the board.
• Use bitboards

Of course, there are many more possibilities. All of them carry exactly the same information, so none are inately "more proper" than the other. However depending upon the specific application, hardware architecture, performance metric of interest etc one may be preferred over the other. Abecedare (talk) 15:57, 14 September 2009 (UTC)

This can be analysed from a computer science perspective: 32 variables with 65 states is ugly since one needs to check for the out-of-play state everytime the value is used. Also, this representation provides too many ways to represent a single position, since the n pawns, etc can be interchanges in factorial(n) ways. So, I prefer the 64 variables with 33 states representation, but I would change it to 7 states: (king, queen, rook, bishop, knight, pawn, empty), to minimize redundancy. Thought this representation also has redundancy (in the sense that there cannot be more than one king), this is more elegant from a programming perspective. Also, you need to store ability to en-passant, castle king-side, castle queen-side, and maybe even time controls, etc. --Masatran (talk) 07:55, 17 September 2009 (UTC)

Also note that some pieces can occupy any square (the kings, queens, rooks, and knights), but others can only occupy a subset of the squares (32 possible squares for each bishop and 48 for each pawn). StuRat (talk) 03:08, 20 September 2009 (UTC)

winning games

Is there a list of winning games for tic-tac-toe? -- Taxa (talk) 05:29, 14 September 2009 (UTC)

If you mean winning strategies then no, the best strategy on both sides leads to a draw (see tic-tac-toe). If you mean configurations of X's and O's where there are three in a row, why would someone make or need such a list?--RDBury (talk) 06:48, 14 September 2009 (UTC)

Maybe the question is about winning sequences, e.g. O in 5, X in 2, O in 7, X in 3, O in 1, X in 4, O in 9 to win. It would be easy enough to list these, particularly if reflections etc are removed→86.148.186.231 (talk) 09:09, 14 September 2009 (UTC)

Easy, but pretty pointless. --Tango (talk) 13:47, 14 September 2009 (UTC)

I remember making such a list during a boring history high-school lesson once. The game mentioned above is the only "interesting" one: all others either result in a draw or require the loosing party to make an obvious blunder.195.128.251.10 (talk) 22:46, 14 September 2009 (UTC)

Yes, I've done similar things during boring lessons! As you say, most of the games are not interesting, that is why making the list is pointless. --Tango (talk) 23:22, 14 September 2009 (UTC)

Why so many 20ps

I always have a mixture of loose change in my purse, but at any time, about half of them are 20p pieces, even though there is a mixture of all eight common UK coins. Is there a mathematical reason why I always end up with so many 20ps?

I doubt it. I can think of two explanations which are quite likely - you might regularly buy things that result in you getting 20p coins as change, or you might have a tendency not to spend 20p coins. You might well choose to make 60p by 50p+10p rather than 3x20p, for example. There might be something in the most efficient way to make up various common sums not involving 20ps, but that would be different to work out since we would need to know the distribution of prices. --Tango (talk) 13:55, 14 September 2009 (UTC)

Let's assume prices are uniformly distributed between whole pound amounts, (do people typically use a bill for 1 pound or a coin?) and that people always make change by starting with the highest denomination they can (that's how it's usually done in the US at least). Then on average you'll get 0.4 1ps, 0.8 2ps, 0.5 5ps, 0.4 10ps, 0.8 20ps and 0.5 50ps each time you get change. By that measure 2p and 20p coins would be a bit more frequent than the rest. If the UK has sales tax that gets added on top of the listed price, and you tend to be making purchases of over a few pounds, then these might be reasonable assumptions. If not, then there might be some weird patterns in the distribution. Rckrone (talk) 16:01, 14 September 2009 (UTC)

In addition, Gresham's law may also be at play. For example, if you are a typical consumer you probably don't like to be weighted down by too many 1-2 penny coins. Thus if you purchase something worth (say) 32 pence, you probably hand a pound note and 2 pennies, and accept back 70 pence ( 2 coins) in change, rather than take 68 pence (minimum of 6 coins) back. Such behaviour will similarly skew the distribution of coins in your purse. Abecedare (talk) 16:11, 14 September 2009 (UTC)

Just to correct you both on a cultural, rather than mathematical note, there are no pound notes in England and Wales, only pound coins (and we wouldn't call them "bills" if there were ;)). I say England and Wales, because the Royal Bank of Scotland still issues pound notes, although I believe they are increasingly rare.

Also, the UK has value added tax, rather than straight sales tax, and retailers are generally required to display the "with VAT" price as the main price, so reverse engineer that to be a round number - £x.00, £x.50, £x.99, £x.95 etc. On the other hand, adding up those 99s and 95s tends to produce all sorts of prices for a whole basket of goods anyway. - IMSoP (talk) 23:50, 18 September 2009 (UTC)

Geometry question

I have a triangle ABC and a smaller one DEF inside it. ADF, CFE and BED are straight lines. If it is known that CF = xFE, BE = yED and AD = zDF and the area of triangle is a m², is there a quick way to find the area of triangle ABC in general? Professor M. Fiendish, Esq. 13:20, 14 September 2009 (UTC)

Yes. Here's a hint: To find the area of triangle ADB in terms of the area of DEF, utilize the formula Δ = (1/2)ab*sinC and remember that sinθ = sin(180° - θ). Rckrone (talk) 15:32, 14 September 2009 (UTC)

integral approximation name

The name of this method escapes me and searching Wikipedia and Googling are not helping. What is the name of the process of using little rectangles under a function to estimate the integral? -- kainaw™ 14:42, 14 September 2009 (UTC)

Wikipedia calls it the rectangle method. Gandalf61 (talk) 14:49, 14 September 2009 (UTC)

Thanks. -- kainaw™ 15:54, 14 September 2009 (UTC)

Specifically, it's that version of the rectangle method where you calculate a lower Darboux sum. Algebraist 17:57, 14 September 2009 (UTC)

Probabilities of unobserved observations

I'm writing a hidden Markov model to calculate the probability of utterances (the probability of the given fragment out of all fragments with the same number of words). To do this, I need to assign each word in the utterance a probability (the probability that a word taken at random from English text is this word). The naive way is to say that if the word appears k times in my n-word corpus, then the probability is k/n, but this ignores the fact that my corpus is just a sample of English text (and thus rare words may not appear in it at all, etc.). Is there a more accurate way to work out this probability? --196.210.182.23 (talk) 15:46, 14 September 2009 (UTC)

Your sample is the only window you have into what English text looks like, so there's not anyway for your program to tell that "normalizable" is a word and "asljdasjksgs" isn't if neither appear in the sample. In stuff I've done with Markov models, the probability is taken to be (k+1)/n to reflect the fact that things you haven't seen are still possible. I don't know how standard that is though. Rckrone (talk) 16:22, 14 September 2009 (UTC)

Shouldn't honest-to-god probabilities sum to 1? Also, do you really assign probability 1/n to "asljdasjksgs" on the grounds of its having k = 0? — Emil J. 16:33, 14 September 2009 (UTC)

On the first point, you'd normalize them, so I guess it would really be (k+1)/(n+a) where a is the number of possible outcomes. On the second point, you're right this method doesn't work at all for what the OP is doing. The case where I used it modeled the probability of the next character, so a was only about 35 and a<<n. But there are too many words. Maybe someone else has a better method... Rckrone (talk) 17:19, 14 September 2009 (UTC)

That is additive smoothing and I think it's very common in practice, even if not necessarily optimal. 70.90.174.101 (talk) 00:07, 15 September 2009 (UTC)

I think this should be viewed in a Bayesian context. Assign a prior distribution to the frequency of each "word" (beta with parameters which depend on the complexity of the word works nicely) and update it based on your empirical data (corpus). The dependency between the frequencies can probably be neglected. -- Meni Rosenfeld (talk) 19:59, 14 September 2009 (UTC)

Of course, the same result can be achieved without considering this mechanism. Just replace "+1" with "+ a term which depends on the complexity". -- Meni Rosenfeld (talk) 20:16, 14 September 2009 (UTC)

That is additive smoothing and I think it's very common in practice, even if not necessarily optimal. 70.90.174.101 (talk) 00:07, 15 September 2009 (UTC)

Assuming Zipf's law you can do a maximum likelihood estimate which would adjust the very low probability ones a little, though you'll still have to assume a maximum number of words, and I'm not sure what it would come out as. I'd be a bit surprised if this really is the sort of thing you want. Dmcq (talk) 22:20, 14 September 2009 (UTC)

The Zipf's law article shows the frequencies dropping after 10000 words to something near 1/x^2 so you do get a proper density function for a prior distribution even with an infinite number of words including things like Zipfian. I guess the Zipf-Mandelbrot law it refers to is about getting that extra accuracy. The whole phrases should probably also follow a Zipf law, it would be interesting to see how one could ensure that and keep the Zipf law for the individual words. Dmcq (talk) 09:37, 15 September 2009 (UTC)

If you do not observe something then your unbiased estimator of its frequency is 0, which sometimes isn't very helpful, as you're realizing. Can you incorporate 95% confidence intervals into your model? There is a law we refer to in medicine which is that if you observe n patients and event X doesn't happen, the 95% confidence interval for the frequency of X is 0 to 3/n. Perhaps you can extend the statistical approach to calculate 95% confidence intervals through your Markov model. AFAIK the 3/n rule was first explained to medical audiences in [1][2], although it's pedestrian non parametric statistics. An article that is free to access is here: [3] I hope this helps. RupertMillard (Talk) 10:21, 15 September 2009 (UTC)

The 3/n rule arises from the apparent fact that, as n increases, so (0.05)^(1/n) approaches 1 - 3/n. Can someone explain how this limit arises?→81.131.163.227 (talk) 20:57, 15 September 2009 (UTC)

$0.05^{1/n}=e^{\ln 0.05/n}\approx e^{-3/n}\approx 1-{\frac {3}{n}}$

Where the last estimate follows from the Taylor series of $e^{x}$. -- Meni Rosenfeld (talk) 05:01, 16 September 2009 (UTC)

The computation of the mean value ± standard deviation of a sample, based on data of the population, is called deduction. The computation of the mean value ± standard deviation of the population, based on data of a sample, is called induction. Using the J (programming language), the deduction formula is this:

   deduc =. (* (% +/)) ([ ,: %:@*) (*/@}: % {:)@(-.@(([ , ] ,: 1:) % +/@]))

Using this formula, the estimated mean values ± standard deviations of a sample of 1 ticket taken from a population of 1 red and 1 green ticket is

   1 deduc 1 1
0.5 0.5
0.5 0.5

showing that the sample contains 0.5±0.5 red and 0.5±0.5 green tickets.

A sample of 6 tickets taken from a population of 6 red and 12 green and 18 blue tickets contains 1±0.845 red and 2±1.07 green and 3±1.13 blue tickets:

   6 deduc 6 12 18
    1    2    3
0.845 1.07 1.13

The induction problem is more useful, because usually an uninteresting sample is known, but the interesting population is unknown.

The induction formula, using the deduction formula above, is:

   induc =. *&_1 1@(+&1 0)@(-@>:@[ deduc~ -@(+ #)~)

The simplest example is taking no sample from a population containing one ticket which is either red or green

   0 0 induc 1
0.5 0.5
0.5 0.5

showing that the population contains 0.5±0.5 red and 0.5±0.5 green tickets.

Taking no sample from a population containing one ticket which is either red or green or blue:

   0 0 0 induc 1
0.333 0.333 0.333
0.471 0.471 0.471

showing that the population contains 0.333±0.471 red and 0.333±0.471 green and 0.333±0.471 blue tickets.

The formula gives what you would expect in the simplest cases. A more complicated example which completely answers the question: 'Is there a more accurate way to work out this probability? '

   corpus =. 55 40 29 18 1 0 0
   # corpus NB. number of different words in the corpus
7
   +/corpus NB. number of words in the corpus
143
   (%~ corpus&induc)100000 NB. probabilities of unobserved observations, ± standard deviation
 0.37  0.27   0.2  0.13  0.013 0.0067 0.0067
0.039 0.036 0.033 0.027 0.0093 0.0066 0.0066

See also [[4]].

Bo Jacoby (talk) 20:38, 16 September 2009 (UTC).

Parabola in projective space

Hi. I'm not formally studying mathematics, but am trying to get my head round projective geometry. As it's a raster image, I was trying to redraw File:Parabola in projective space.png as an SVG using gnuplot. My best effort so far is File:Parabola in projective space.svg. Please can you experts let me know what you think? Is it mathematically correct, and could someone alter the image description page so it precisely says what I have done? Does anyone have a tip for making the image reflect that the plane is at a tangent to the sphere? Should the cubic line on the ball go all the way round the sphere as it does in the first image? I didn't understand how that was a projection.Many thanks. RupertMillard (Talk) 22:07, 14 September 2009 (UTC)

You don't really have a description of the mapping you're trying to apply so it's hard to tell if what you're doing is correct or not. It looks like you're trying to map the parabolas onto the Riemann sphere which isn't really projective space at all. You can map project space onto a sphere but its a 1-2 mapping. Also keep in mind that all conics are basically the same curve in projective since the equations can all be transformed under a projective transformation to x²+y²=z², which just happens to be the equation of the cone in space. Does that help?--RDBury (talk) 10:24, 15 September 2009 (UTC)

This is very helpful, thank you. I've never read about Riemann spheres before. (Isn't this illustration stunning? [5]) What I was really trying to do was just redraw the figure File:Parabola in projective space.png in the context of Algebraic_geometry#Projective_space. I thought it looked like the curves are projected onto a unit sphere through its centre, where the sphere touches the origin of the graph, but I can see this is all wrong now. As you say, this would yield a 1-2 mapping. I guess I shouldn't even be trying to draw figures that I don't understand. (I don't understand exactly what you mean about the conics being basically the same curve - in the original diagram, the parabola goes from an unclosed figure, to a closed one. Is this related to one of the foci of a parabola being at infinity, which is about where my conics left off.) RupertMillard (Talk) 10:52, 15 September 2009 (UTC)

Try the conic sections paragraph in Projective geometry#description. Really the most natural setting for algebraic curves is complex projective space. But any attempt to visualize that make my head hurt.--RDBury (talk) 12:55, 15 September 2009 (UTC)

Just to be clear, the Riemann sphere represents one-dimensional complex projective space CP¹ which is topologically a sphere. We can think of wrapping the complex plane up and then completing the sphere by adding a single point at the top, which is called the "point at infinity". This is different from what you were dealing with when making your diagram, which is two-dimensional real projective space RP². RP² isn't topologically a sphere, but there is a 1-2 mapping from RP² to the sphere. Each point on the sphere is identified with the point on the direct opposite side of the sphere. In this case we're mapping the x-y plain onto just the top half of the sphere and then there's a whole "line at infinity" which is the equator, and then the bottom half is just an identical copy of the top half.

I didn't check the math, but your diagram certainly looks correct. It would probably be more complete to show both sets of points on the sphere, although to be fair the original diagram you were reproducing failed to do so for the y=x² graph. Rckrone (talk) 21:06, 15 September 2009 (UTC)

Ah thanks. This all makes a lot more sense now. I've added the second halves, and described it as "y=x² & y=x³ projected on RP² mapped onto a sphere". But I think I might animate this or redraw it in POV-Ray because it's not looking very clear to me. (Ironically, this would bring it back to a raster format of some description.) RupertMillard (Talk) 23:07, 15 September 2009 (UTC)

The key idea of projective space is that the infinite can become finite (and visa versa). Take a piece of paper and cut yourself a rectangle. Put the thin end against the thin end to form a loop. (The join formed by the thin end of the rectange meeting thr other thin end will be called the join.) We will draw some circles in relation to the join and these will give us conics. If the circle misses the join, then when we open the strip of paper we will have an elipse. If the circle is tangent to the join then when we open then strip we will have a parabola. If the circle crosses the join then we get a hyperbola. ~~ Dr Dec (Talk) ~~ 21:29, 15 September 2009 (UTC)

visa versa? --131.114.73.84 (talk) 09:56, 16 September 2009 (UTC)

September 15

Box and Whisker Plot

Ok I understand the general idea behind this graph method. The very top hash represents the maximum value in the sample, the fatter part of each line represents the data between the 1st and 3rd quartile (interquartile range I assume?), The bolder line within the fat box is the sample median, and the bottom hash represents the lowest value in the set. Does that all sound right? OK...

So, in experiment # 3 above How is it that the lowest value also represents the firs quartile? Or is it possible for that to happen?

I am currently looking a published article where the fat part of the box starts at 0 and ends at 3 with no median line and no bottom range line, but the fat box does end at 3 and the top of the range is indicated normally. So does this mean that the lowest value in the data set, the median, and the first quartile are all 0?

Much thanks in advance. Let me know if you need clarification

Mrdeath5493 (talk) 00:13, 15 September 2009 (UTC)

Lowest value can also be first quartile as in 1 1 2 2 3 3 4, for example. And the median can be the same as the first quartile, and the lowest value, eg. 1 1 1 1 5 5 5. Can you link to the published article? Without seeing it, I would imagine that the bottom range line is obscured by the box because the lowest value is also the median, as in your example of #3 above. RupertMillard (Talk) 00:31, 15 September 2009 (UTC)

I think you answered my question well enough. I'm not sure exactly which article the graph is from, but I bet it isn't licensed under GNU or whatever. It was patient reported data measuring pain from 0-10 after receiving 30 minutes of a IV pain reliever. So, I bet The low value, 1st quartile, and median were all 0 since they were all on pain meds. That makes sense :) Thank you for the help. Mrdeath5493 (talk) 00:46, 15 September 2009 (UTC)

I don't know if the first and last bar mean what you said they mean. I am in a stat class right now and my book calls your description a "skeletal box plot". The skeletal box plot has only minimum, first quartile, median, third quartile, and max. The thing is, your pictures have things other than that. Notice there are also a few little circles. That matches a "box plot" in my book, and also in the article box plot. The difference is the top and bottom lines are no longer max and min. They are upper adjusted value and lower adjusted value (that could be wrong terms, but it is UAV and LAV... wiki article does not have those terms). But, the definition of UAV and LAV I remember, and it agrees with wiki article. You look at Q_1 - 1.5 IQR. LAV is the smallest measurement that is bigger than that number. Then you consider Q_3 + 1.5 IQR. UAV is the biggest measurement which is smaller than that. Then, anything that does not fall in that range, from Q_1 - 1.5 IQR to Q_3 + 1.5 IQR is called an outlier. And, those 3 little circles represent outliers. See the Wikipedia article for more detail. StatisticsMan (talk) 01:13, 15 September 2009 (UTC)

You wrote:

The very top hash represents the maximum value in the sample,

That is incorrect. You see some outlying values in some of these plots—beyond the top or bottom hash marks. You'd better pay attention to those.

You glaringly omitted the sample sizes. Think about where the quartile boundaries are in a sample of only 3. That explains how the minimum and first quartile can coincide. Michael Hardy (talk) 20:30, 15 September 2009 (UTC)

I'm not sure what the sample size was, it wasn't indicated by my professor. I was looking at this plot out of the context of the study. I never took stats, but am now basically assumed to have mastered it in a class I'm currently taking. While I guess it is possible that there were only 3 people in the data set I was looking at, I'm pretty sure there were just a lot of 0's in the data set I was referring to. Since negative values were impossible and since the median was zero, this necessitated that the min value and first quartile would be as well. I have learned that the extra dots (outliers) on the graph even come in two flavors, hollow and filled, referring to the degree of outlierness they possess. Again, thank you for your help, even as my quest for knowledge was "glaringly" misguided. —Preceding unsigned comment added by 67.167.150.120 (talk) 03:37, 16 September 2009 (UTC)

Brain Teaser

Hey. I recently ran into a brain teaser that I havn't been quite able to figure out. Maybe someone here could help? I read this is the place to ask knowledge questions. OK, here goes:
There are 5 values, whose sum is 60 and the sum of whose squares is 765. I can't quite get past this part though. Help? Thanx Lucretion (talk) 02:15, 15 September 2009 (UTC)

I assume you want them integer. But how many are odd, and how many are even, of your 5 numbers if the sum is even and the sum of squares is odd? --pma (talk) 03:38, 15 September 2009 (UTC)

This is the intersection of a sphere with a hyperplane, so it's a sphere of smaller dimension. Maybe more later.......... Michael Hardy (talk) 20:33, 15 September 2009 (UTC)

A quick program search shows that there aren't any solutions to this problems for integers in the range -27 thru +27. FWIW, the same program rapidly finds a solution for sum = 60 with sum of squares = 766. -- SGBailey (talk) 19:32, 16 September 2009 (UTC)

As pma has hinted, you don't need a program search to know that there are no integer solutions. -- Meni Rosenfeld (talk) 19:44, 16 September 2009 (UTC)

Zeros of Derivatives

Okay, here it goes. My deficiencies in analysis have forced to me to ask for help from all of you again. I am doing this proof in numerical analysis regarding some error analysis and this is the information given to me. The domain given is [-1,1] and G(x) is a continuous given function. We know that G(-1)=G(1)=0 and also there is another third real distinct zero of G(x), denoted t so G(t)=0 for some t in (-1,1) and we have no idea what t actually is. In addition, we also know that $G'(0)=G''(0)=0$ as well. Using all this, how can I deduce that $G''''(x)$, the fourth derivative, has a zero in (-1,1)? I have a proof but it is ugly and complicated. I hate it and there are a billion subcases. I am sure that there is a nice concise proof. Thanks!-Looking for Wisdom and Insight! (talk) 02:37, 15 September 2009 (UTC)

The way to do these questions is always the same: apply Rolle's theorem a lot of times. It tells you (under some hypotheses...) that between every two zeros of f, there's a zero of f'. Use this, and the other zeros you were given, enough and the result follows with no subcases.Silverfish70 (talk) 09:02, 15 September 2009 (UTC)

These simple notions should help to make it short. Let f(x) be a smooth function on (a,b) and a<x₀<b.

1. Say as usual that x₀ is a zero of f(x) with multiplicity m (at least) iff f vanishes at x₀ together with its derivatives of all orders up to m-1 included. In what follows, we count zeros with multiplicity.
2. Fact: if f has n zeros in the interval I (an interval of any kind: open, closed..), then f' has at least n-1 zeros in I (then also, of course, f^(j) has at least n-j zeros in I &c).
3. Fact: (Generalized Rolle's theorem). If f(a)=f(b), and f(x)≠0 in (a,b), then f' has an odd number of zeros in (a,b), or infinitely many.
4. I assume your G is four times differentiable. Case I : G(0)=0. Then, 0 is a zero of G of multiplicity 3; with 1 and -1 this makes at least a multiplicity 5, so G⁽⁴⁾ has at least a zero. Case II : G(0)≠0, so e.g. t>0. Then by the Rolle's thm G' has a zero in (t,1), and, by the generalization, an odd number of zeros in [-1,t] (precisely, in the largest interval I around 0 where G(x)≠0) , thus at least 3, as 0 is a zero of G' of multiplicity at least 2. Hence G' has at least 4 zeros, and again G⁽⁴⁾ has a zero.

Note: since you wanted the zero of G⁽⁴⁾ to be in (-1,1), some small details are to be added. Finally, a great reference for these elementary yet subtle facts is Pólya - Szegő : "Problems & Theorems in Analysis".--pma (talk) 16:07, 15 September 2009 (UTC)

September 16

What is the correct pronunciation of George Pólya's last name?

What is the correct pronunciation of George Pólya's last name? (I consider the way he pronounced his name to be the correct pronunciation.) —Preceding unsigned comment added by 98.114.146.57 (talk) 04:20, 16 September 2009 (UTC)

here is it [6]--84.220.118.109 (talk) 06:08, 16 September 2009 (UTC)

Hungarian pronunciation: [ˈpoːjɒ ˈɟ͡ʝørɟ͡ʝ]. — Emil J. 10:35, 16 September 2009 (UTC)

How many combinations are there?

We have a lock box on our house similar to this: http://www.buyasafe.net/S6-Supra-Pushbutton-Lockbox-Surface-Mount./M/B000M7OXPY.htm?traffic_src=froogle

My question is, how many possible combinations are there?

Here are a few things to consider: 1) There are 10 possible digits (numbers 0 through 9). 2) The order the numbers are pressed doesn't matter. 3) We chose a 4-digit combination; however, we could have selected anywhere from a 0 - 10 digit combination.

To break the 4 digit combination, the odds would be 10 x 9 x 8 x 7 = 5,040, right? However, when you consider we could have selected any number of digits, how does this affect the total number of possible combinations?

Thanks in advance for your help! —Preceding unsigned comment added by 209.206.158.57 (talk) 17:02, 16 September 2009 (UTC)

Your 5,040 isn't right, because that distinguishes different orders. There are 5,040 4-number permutations using 1-10 w/out repetition. You want to divide that by 4! = 24, meaning there are only 210 4-number combinations.

Now, if you add up the number of 0-number combinations (1) + the number of 1-number combinations (10) + the number of 2-number combinations (45) + . . . + the number of 10-number combinations (1), you'll get 2^10 = 1,024.

One way to see why it's 2^10 is that each possible code either uses or doesn't use each of the 10 numbers. Thus, you either use '1' in your code, or you don't - 2 options. You either use '2' or you don't - 2 options. . . You either use '10' or you don't - 2 options. Multiply all those 2's together, boom.

Does that answer your question? -GTBacchus^(talk) 17:19, 16 September 2009 (UTC)

Sorry if I'm being a bit slow, but I don't see how that's right. Surely a code using a '5' say, could use it more than once? As a disproof (I think!), imagine a similar lock using two digits, which can either be 1,0 or not used. That leaves 6 combinations, which isn't 2^2.--Leon (talk) 17:33, 16 September 2009 (UTC)

No, I've used locks like this. When you press a button, it stays pressed, and having all n buttons down at once, with no others down, is how it opens. Thus, with two buttons, labelled '1' and '2', the 4 possible codes are {}, {1}, {2}, and {1,2}. We're really just looking at the power set of {1,...,n}. -GTBacchus^(talk) 18:45, 16 September 2009 (UTC)

I'm sure GTBacchus is right. But for an edit conflict, I would have posted a similar answer. The person who asked the question stipulated that the order the numbers are pressed in doesn't matter. This implies a simple mechanism which doesn't care if a button is pressed once or many times. If you have two digits, you have two 1-digit combinations, not six. RupertMillard (Talk) 17:53, 16 September 2009 (UTC)

Apologies: I neglected to check the link, and I presumed that a number could be selected multiple times!--Leon (talk) 18:53, 16 September 2009 (UTC)

Here's a puzzle. How many distinct, valid combinations are years since 1900? Harder - how many are dates? (eg. 0916) How many combinations does this total for you to try to crack the safe if you knew the combination was a date or a year? I think the answer is the days of our age, but I'm not certain. RupertMillard (Talk) 18:55, 16 September 2009 (UTC)

To your second question... combinations representing dates can involve 2, 3, or 4 buttons (1/10, 1/23 and 12/03 being the first occurrence of each). I count 82 combinations that can represent 149 different dates. If a thief can try a combination in 5 seconds, that means 410 seconds, which is just under 7 minutes to run through all date combinations. I would note that 0916 is not actually an option, because there's no zero button. If there were a 0 instead of a 10, then only 49 combinations represent dates, which is just over 4 minutes of burgling. -GTBacchus^(talk) 20:10, 16 September 2009 (UTC)

Interpretation help

Hi. I have a small problem with some homework but it's only in understanding what the question means.

"Let $g(z)={\frac {z-i}{z+i}}$ where z is a complex number. Writing $g^{2}(z)$ for $g(g(z))$ and $g^{-1}(z)$ for the inverse function, what is the image of the real line under the successive maps $g$, $g^{2}$, $g^{3}$? What is the image of this set of images under these maps?"

Letting z=x+iy, does the first part just mean determine $g(x)$, $g^{2}(x)$, $g^{3}(x)$? And I don't have a clue for the second part. Any ideas? Thanks. 92.2.21.125 (talk) 19:56, 16 September 2009 (UTC)

They want you to determine the images of R under these linear fractional transformations, that is, the sets A_n:=gⁿ(R)={gⁿ(x): x in R}, for n=1,2, and 3. The other question is not clearly stated indeed; probably they just want gⁿ(A_k), which is trivially A_n+k; or they mean the image of the union of the sets A_k under the maps gⁿ. --84.221.68.30 (talk) 20:19, 16 September 2009 (UTC)

(edit conflict) Finding the "image of the real line" means that, if you plug in every point on the real line to g, and then look at where they all end up, what shape on the complex plane does that give you? For example, the point 0 gets sent to -1, the point 1 gets sent to g(1) = (1-i)/(1+i) = -i, etc. Try plugging in some more real numbers, and see if you can make a sketch of where the real line is going, point by point. Once you have an idea, try and prove your conjecture.

The second part of the question seems to be saying, do the same thing, but instead of just the real line as your starting set, use the set $\mathbb {R} \cup g(\mathbb {R} )\cup g^{2}(\mathbb {R} )\cup \cdots$. Does that help at all? -GTBacchus^(talk) 20:29, 16 September 2009 (UTC)

OK I'm with the first part of the question now - just determine the sets (/regions of the complex plane) produced by the results of $g(x)$, $g^{2}(x)$, $g^{3}(x)$. That was mainly what I was thinking before, it's just that the wording '... the successive maps...' seemed to imply you might do $g^{3}(g^{2}(g(x)))$ but clearly not. You both seem to be going for the $g^{n}(\mathbb {R} \cup g(\mathbb {R} )\cup g^{2}(\mathbb {R} )\cup \cdots )$ idea for the second part so I'll try that. Thank you both! 92.2.21.125 (talk) 20:43, 16 September 2009 (UTC)

But notice that what is the image of the real line means: what set is g(R) --you should find that it's a circle. So you do not have to find what is the image of every single real point g(x). Also, recall that the composition of linear fractional transformations is easily computed multiplying the associated 2x2 matrices of coefficients. Here g³ sounds like a sort of well known map ;-) --84.221.68.30 (talk) 21:00, 16 September 2009 (UTC)

Of course nobody's going to try every single real point. We're mortal. I think he's supposed to figure out whether it's a circle; that's why it's homework. Let's not give away so much, eh? -GTBacchus^(talk) 21:18, 16 September 2009 (UTC)

For extra credit (and the pleasure of a Eureka moment) visualise and describe the action of g(z) on the Riemann sphere. Gandalf61 (talk) 09:42, 17 September 2009 (UTC)

Ooh, nice! Thank you. :) -GTBacchus^(talk) 14:44, 17 September 2009 (UTC)

I've just worked through g(R) and, once simplified, have $g(x)={\frac {x^{2}-1}{x^{2}+1}}-{\frac {2ix}{x^{2}+1}}$. Now I can show that the modulus of this is 1 but when I graph this, I most certainly do not get a circle. Where am I going wrong? 92.2.18.79 (talk) 14:24, 17 September 2009 (UTC)

Right, for $g^{2}(x)$ I have $g^{2}(x)=(1-i){\frac {x+1}{x-1}}$ but don't know where to go from here. Taking the modulus still leaves it in terms of x, which doesn't seem helpful. Any suggestions? Thanks 92.2.18.79 (talk) 14:48, 17 September 2009 (UTC)

Yeah, double-check your work. That's not g^2(x). -GTBacchus^(talk) 14:55, 17 September 2009 (UTC)

My mistake. $g^{2}(x)=-i{\frac {x+1}{x-1}}$. So does this just mean that every point on the real line, excluding x=1, is mapped onto a point on the imaginary line? 92.2.18.79 (talk) 15:06, 17 September 2009 (UTC)

Yes, you've just shown that $g^{2}(\mathbb {R} )\subset i\mathbb {R}$. If you can show inclusion in the other direction, then you'll have $g^{2}(\mathbb {R} )$! (That's an exclamation point at the end, and not a factorial, of course.) -GTBacchus^(talk) 15:24, 17 September 2009 (UTC)

OK, $g^{3}(x)=x$ and so every point on the real line is mapped onto itself. Now for the final part of the question, I just need to check the circle. I managed to get $g(x)={\frac {x^{2}-1}{x^{2}+1}}-{\frac {2ix}{x^{2}+1}}$ and that its modulus is one. So is the equation of this circle given by $(x-1)^{2}+y^{2}=1$? 92.2.18.79 (talk) 18:31, 17 September 2009 (UTC)

If the modulus of every point on the circle is 1, then where should the center of the circle be? -GTBacchus^(talk) 19:04, 17 September 2009 (UTC)

Hmm, I sense you mean the origin. This is the first time I've ever considered the modulus of the points on the circle, rather than the distance of the points o the circle from the centre, to determine the centre. Thanks GTBacchus. 92.2.18.79 (talk) 19:11, 17 September 2009 (UTC)

You're quite welcome. Complex analysis is fun; enjoy! -GTBacchus^(talk) 19:15, 17 September 2009 (UTC)

Expectation

Resolved

So, I have emailed my TA a question once and his response indicated he did not understand my question at all. I just got back from asking him other questions in person and he again did not understand what I was even asking, though I explained it several times. The professor does not have any office hours. So, I come here.

I'm doing a simple problem and I need help in understanding. The problem says to show if you have random variables X and Y then

E(max(X, Y)) = EX + EY - E(min(X, y))

They even give a hint to show that X + Y = max(X, Y) + min(X, Y). Well, this is a problem that is intuitively simple but I don't know how to explain it. I know if I am dealing with 2 real numbers x and y, then x + y = max(x, y) + min(x, y) and the proof is easy. So, it seems obvious to me from this that the same equation with random variables must be true. But, just saying it is obvious does not prove it. This is my only difficulty with this problem. Clearly, if I prove the equation with random variables, I can just take the Expectaion of both sides and use linearity and then subtract the min from both sides.

One thought is to let A = X + Y and B = max + min, new random variables. Then, P(A = a) = P(X + Y = a) = P(X = a - Y)... I don't know. We have only talked about functions of one random variable, but not functions of multiple random variables.StatisticsMan (talk) 20:48, 16 September 2009 (UTC)

According to Random variable#Formal definition X and Y are both functions. Taemyr (talk) 21:15, 16 September 2009 (UTC)

I know they are functions. What does that have to do with my question? StatisticsMan (talk) 21:22, 16 September 2009 (UTC)

ecThis: you already have all you need. Recall that X and Y are in particular functions Ω →R, so at any ω in Ω, X(ω) and Y(ω) are certain real numbers, and for them you do know the equality:

X(ω) + Y(ω) = max( X(ω), Y(ω)) + min( X(ω), Y(ω) );

this equality for all ω in Ω just means the equality between functions

X + Y = max(X, Y) + min(X, Y);

so you conclude as you said by linearity of the expectation. --pma (talk) 21:24, 16 September 2009 (UTC)

Well, I guess it has everything to do with my question! And, I believe I understand this. Thanks a lot! StatisticsMan (talk) 02:41, 17 September 2009 (UTC)

September 17

Formula for placing things equidistant along a circle circumference

As the subject implies, i need to find a formula to place items an equal distance apart on the circumference of a circle. Specifically i need 24 equidistant locations. I dont know quite yet how i will define the circle, probably with parametrics? But ive looked on google for an answer to this and there doesnt seem to be any. Please help?

Thanks! 137.81.115.58 (talk) 01:26, 17 September 2009 (UTC)

You can define a circle with radius R in terms of the angle θ around the origin with the parametric equations x = Rcosθ, y = Rsinθ. Evenly spaced points along the circle are at evenly spaced angles, so we divide the full 2π angle of the circle up into 24ths. The points with coordinates (Rcos(2πn/24), Rsin(2πn/24)) with n ranging from 1 to 24 satisfy the conditions you're looking for. Rckrone (talk) 02:44, 17 September 2009 (UTC)

See Root of unity. Bo Jacoby (talk) 08:58, 17 September 2009 (UTC).

Note that 24 = 6x2x2 - you can easily divide the circle into 6 arcs, and then half those arcs twice. When you can define how you want to describe the circle it will be easy to give a method for the process.87.102.94.154 (talk) 11:12, 17 September 2009 (UTC)

help me ...it will safe my future..??

sir/mam....which extra book i adopt ?? i am the student of commerce stream..i choose maths because i want to become good C.A.

you are good to go rambharose..best of luck

You are going to need to be more specific. --Tango (talk) 17:56, 17 September 2009 (UTC)

Quite so :o) ~~ Dr Dec (Talk) ~~ 18:30, 17 September 2009 (UTC)

You seem like a nice person, though I'm not sure who 'rambharose' is. I'm going to guess that you need an introductory maths book for a commerse coarse, possibly including accountancy and economic maths? (and maybe statistics?) Maybe someone else would be good enough to give a suggestion?83.100.251.196 (talk) 19:12, 17 September 2009 (UTC)

Rambharose means in trust of Rama(God) in Hindi:)--Shahab (talk) 18:42, 19 September 2009 (UTC)

Writing Roman Numerals

Hello

Hope someone can answer this as I can't find it in the system. I understand how roman numerals work but get stuck at large numbers, ie one million is M with a bar, but how do you write 2 million or 10 million etc ?

--Bruceglasgow (talk) 18:42, 17 September 2009 (UTC)

Two million derives from one million just as two derives from one. If one million is M, two million is easy to generate: MM. Alternately, parentheses may be used to represent "times 1000", such that two million is (MM). Ten million requires some extrapolation from our Roman numerals article, but might be fairly represented as (X), or a longer form. — Lomn 19:21, 17 September 2009 (UTC)

Equation of oscillator - flat section

A question arises from another persons question about oscillations of things travelling through a spherical body exerting and inverse square force law Wikipedia:Reference_desk/Science#Gravity_at_the_center_of_a_body.

My question begins: If a spherical centre part is removed then within this part the field force is zero. (as per Shell theorem), assuming a body is dropped into the whole hollow sphere (assumming a negligable cylindrical hole for it to travel in or whatever) - then it seems to me that the body will accelerate towards the centre, then when it reaches the empty part no field is experienced and the body will move at constant velocity, until it exits the spherical hole at which point the deacceleration occurs, and so on. Thus it oscillates.. but with a part at which the acceleration is zero...

So the equation of motion (eg acceleration vs time) I would expect to be expressable as a fourier series (ie a series of powers of sines or similar).. Since the function is periodic. But at the times at which the particle is inside the spherical void the acceleration is 0, as are the derivatives of the accelerations.. This seems to contradict the taylor-maclaurin type theorems (ie that a function at any point can be constructed from the function at a given time and all it's derivatives.

My questions are:

• Can the function be expressed as a series of sines, ie a fourier series
  • If not what sort of function
• What goes wrong with the taylor-maclaurin series here? (or what is my error?)83.100.251.196 (talk) 19:22, 17 September 2009 (UTC)

Taylor-Maclaurin-type results apply to analytic functions. This function is not analytic. The Fourier series of position as a function of time would converge to that function. This is one of the difference between Fourier series and power series. Michael Hardy (talk) 19:57, 17 September 2009 (UTC)

The page analytic function seems to define itself as "A function is analytic if and only if it is equal to its Taylor series in some neighborhood of every point." (which is somewhat circular in respect to your statement, but I'm not arguing this).

The actual function of this oscillation can be expressed as a fourier series, and so also as a power series (is there an exception here?). I imagine that the square wave is a simple(r) example to discuss.

Is there another way of showing that a given power series (or fourier series) will not be analytic (other than trying the taylor maclaurin series).

Does the fact that eventually a differential of the function includes a discontinuity mean that it will be non-analytic ? - it seems to me that this will be the case, but can't imagine a proof.83.100.251.196 (talk)

You can't have a power series that's not analytic, since a power series is analytic by definition. You're right that a function has to be smooth in order to be analytic. To prove it, any power series is smooth (assuming it converges), so if a function isn't smooth, it's not equal to any power series. The Taylor series article mentions that the converse isn't true: there are smooth functions that aren't analytic.

As for the Fourier series, sine functions are analytic, and so any finite sum of sine functions is analytic, but that doesn't necessarily extend to an infinite sum. So the partial sums of the Fourier series are analytic, but they may converge to something that's not analytic, which is the case with the function being discussed. Rckrone (talk) 21:41, 17 September 2009 (UTC)

A power series is analytic by definition: not quite so; to be precise, you need to prove that at any point within the convergence disk of your power series, the function represented by the series is also representable by a power series centered at that point: not difficult, but still a computation. --84.221.68.30 (talk) 22:34, 17 September 2009 (UTC)

ok thanks so far, For the square wave the 'x¹' term would be of the form 1+1/3+1/5+1/7... etc that relates to pi , (I haven't checked the other terms x^3 etc) , but it looks like it can be expressed as a infinite power series (is this right?).

I'm not very familiar with the radius of convergence - though at first sight it looks like the square wave will converge for all values of x .. ah not for the differentials.

I'm fairly familiar with smooth functions though.

It would be great if a few points could be clarified - specifically about the square wave (I'm confused because I think it is non-analytic, yet expressible as a power series):

• Can the square wave be expressed in the form a₀ + a₁x¹ + a₂x² + a₃x³ + etc all the way to infinity. (as I suspect) - but this form does not converge - but oscillates and diverges does this relate to it being non-analytic?
  • Is this form a power series, or is this the definition of a power series more complex.
• The radius of convergence of the square wave - infinite ? but only for the fourier series not for the differentials, starting to partially understand.
• Apart from lacking smoothness (ignoring smooth non-analytic functions for now) is there another test for lack of analyticalisity in a function? (apart from the obvious taylor maclaurin method)? The test doesn't have to be absolute - just able to reject a set of equations.. Thanks/83.100.251.196 (talk) 23:39, 17 September 2009 (UTC)

I think you might be confusing approximating the square wave with a sequence of analytic functions (such as with partial Fourier series), and a single power series that converges to the square wave. There's no power series that converges to the square wave, which is to say that it's not analytic. We know this because it's not smooth, and all analytic functions are smooth. We can however approximate the square wave with analytic functions, for example with the Fourier series. The partial sums of the Fourier series give us a sequence of different analytic functions that approach the square wave (at most points).

Besides smoothness I don't know what other criteria there are to decide if a function is analytic besides just comparing it to its Taylor series. Maybe someone else who knows more about this has something. Rckrone (talk) 01:23, 18 September 2009 (UTC)

Yes exactly that, I know see why attempting to make the corresponding power series for the fourier series fails - because the terms for x are all infinite (or non converging) beyond x¹.

I get the principle now thanks.

Resolved

83.100.251.196 (talk) 11:25, 18 September 2009 (UTC)

The taylor series for g(z) defines the fourier series for f(x) = g(e^ix) . Bo Jacoby (talk) 10:04, 18 September 2009 (UTC).

I'll have a look at that too, thanks.83.100.251.196 (talk) 11:25, 18 September 2009 (UTC)

This corresponds to the rather unusual case when the nth (exponential) Fourier coefficient of f is zero for every negative n. In general, the Laurent series for g defines the Fourier series for f. However, in practice the Laurent series corresponding to the Fourier series of a typical periodic function will have both radii of convergence equal to 1, meaning that it's degenerate (its annulus of convergence is empty). — Emil J. 16:34, 18 September 2009 (UTC)

September 18

Substitution in the integral

Suppose we have the integral of the form $\int _{0}^{h(x)}f(g(t))dt$ and then we do a substitution of the form $g(t)=u$. what will happen for the integral limits, 0 and $h(x)$ ? —Preceding unsigned comment added by Re444 (talk • contribs) 15:32, 18 September 2009 (UTC)

I think of it by imagining a little "t=" next to each limit. You want those to turn into "u=", so figure out what u is when t=0, and similarly when t=h(x). Does that do it? -GTBacchus^(talk) 15:36, 18 September 2009 (UTC)

Oh thanks, I really need that! It solved all my problems. Re444 (talk) 18:53, 18 September 2009 (UTC)

A really brainy sum !!!!!

My friend sent me this question via mail and i tried various options for getting this sum.Even after 3 hrs of futile trying I failed pls help.The question as follows

A farmer was dividing his property among his children. This was what the farmer told to his first son “Take as many number of cattle as you can care for and your wife may take one-ninth of the remaining number.” To his second son he said “Take one more than what the first son took and your wife will have one ninth of the cattle remaining after you have taken.” This applied to the remaining sons too i.e. each son would take one more than the next oldest brother and their wives would have one ninth of the remaining after their husbands have got their share.

After the cattle were divided, the farmer proceeded to divide the gold bars he had among the children. Each gold bar was valued at 3.5 times each cow's value. The gold bars were divided such that each couple had equally valued inheritance. What are the number of cows, gold bars and the number of sons the farmer had?

My deductions :

1. Obviously the wife of last son will not get any cows as there should be no cow left.
2. The number of cows of each couple consists of three terms.All terms except the constant term follow a particular pattern except the integer.
3. The difference between the number of cows(assuming the cost of a cow is 1 unit ) of each couple should be equal to a multiple of 7.

so pls help...3 to 4 hrs of working on this didnt help much :-(164.100.170.4 (talk) 16:14, 18 September 2009 (UTC)

Does the wife round down or up?83.100.251.196 (talk) 16:56, 18 September 2009 (UTC)

The wives don't round. That's part of the problem, is that there's always a multiple of nine left when the wife chooses.

Nice problem. This'll take a minute... -GTBacchus^(talk) 19:37, 18 September 2009 (UTC)

what does that mean???

Simplest non-trivial solution: 2 sons, 135 cows, 2 bars of gold. The first son takes 63 cows, leaving 72. His wife takes 8. The second son gets the remaining 64 cows, and the second wife, none. That's 71 cows for the first couple, and only 64 for the second. The 2 bars of gold make it even.

Of course, if there's only 1 son, it's trivial, and (2 sons, 16 cows, no gold) works, but isn't very interesting. There are other solutions, with 2 sons. Not sure about 3 or more. -GTBacchus^(talk) 20:13, 18 September 2009 (UTC)

(after ec) The problem has several possible solutions. For example:

• Number of sons = 1; Number of cows = arbitrary; number of gold bars = arbitrary. The first son takes all the cows, and his wife gets none.
• Number of sons = 2; Number of cows = 16; Number of goldbars = any even number, say 2p. First son takes 7 cows, his wife gets (16-7)/9 = 1; second son gets 7+1=8, and his wife gets none. Both couples have 8 cows each, so the gold bars are divided equally.

I am confident that there are other solutions with more sons, but I haven't worked them out yet. Note though, that the number of gold bars will always be non-unique, because for any solution with n sons, one can always add np to the number of gold bars while meeting all the conditions of the problem. Abecedare (talk) 20:15, 18 September 2009 (UTC)

Ah, good point about the gold. You can always throw more gold around, as long as the extra is divisible by the number of couples. For more unique solutions, one would add the condition that the couple with the most cows gets no gold. -GTBacchus^(talk) 20:18, 18 September 2009 (UTC)

There are no solutions with 3 sons. The number of cows the first son takes would have to be congruent to both 6 and 0, modulo 64, which is impossible. -GTBacchus^(talk) 20:24, 18 September 2009 (UTC)

I can post the general solution for 2 sons, but I don't want to ruin any more of the fun for others... feel free to post to my talk page for it. -GTBacchus^(talk) 20:28, 18 September 2009 (UTC)

Hmmm, I found a few solutions with three sons!

• Number of cows = 24; Goldbars = 3p; First son takes 6 cows, and his wife takes (24-6)/9=2; Second son takes 7 cows and his wife takes (24-6-7-2)/9=1; Third son takes the remaining 8 cows. Each couple has 8 cows, and can divide the gold evenly.
• Number of cows = 1543; Gold bars = 3p+52. First son takes 454 cows, and his wife gets (1543-454)/9 =121 cows. Second son takes 455 cows and his wife gets (1543-454-455-121)/9 =57 cows. Third son takes the remaining 456 cows. So first couple has 575 cows, second couple 512 cows and third couple has 456 cows. So first couple gets p goldbars, second couple p+18 goldbars, and third couple p+34 goldbars, to even the inheritence.

I am guessing there are solutions with even more number of progeny, but I'm donw with this family :-) Abecedare (talk) 20:40, 18 September 2009 (UTC)

Huh, what did I do wrong? Runs off to double-check congruences... -GTBacchus^(talk) 20:42, 18 September 2009 (UTC)

Aaargh. How silly of me. Thanks for the down-a-peg; those are good for the soul. -GTBacchus^(talk) 20:44, 18 September 2009 (UTC)

(ec) It's easier to run it in reverse. Start with no cows on the pile. First wife adds no cows, first son (last really) adds some number of cows that has to be a multiple of 8, call it 8k, so that the second wife can add 1/8 of the existing pile. The second son adds 8k-1, and the new pile (with 17k-1) now has to be a multiple of eight for the third wife, etc. The only exception is that the last son (first really) doesn't need to make the pile a multiple of 8. For each additional son that there is in the family, there are stricter constraints, but always an infinite number of solutions. For example with 2 son there are no constraints on k. With 3 sons, k = 1(mod 8). With 4 sons k = 1(mod 64). With n sons, k = 1(mod 8^n-2) I think (it's definitely a sufficient condition, but it might be looser than that). Then there's the gold bar thing which adds some more constraints. 1 son puts no constraints on k. With 2 sons, k = 1(mod 7). I'm less sure about how this one grows with the number of sons, although I think requiring k = 1(mod 7^n-1) is always safe. So combining these, for n sons, requiring k = 1(mod 7*56^n-2) should always get you solutions I think, but I'm not totally sure.

Although there are infinite solutions for any given number of sons, an easy solution that pops out is k=1. Each couple gets 8 cows, and no gold is required. That would work for any number of sons, except that the cows each son takes has to be non-negative, so it fails for more than 9 sons. Rckrone (talk) 21:19, 18 September 2009 (UTC)

That is easier. I made my variable the number of cows that the first son takes, which took a bit more calculating. -GTBacchus^(talk) 21:26, 18 September 2009 (UTC)

With no rounding I came to the conclusion that there could be any number of sons - though clearly as 'sons' increase the number of cattle increases vaguely 'factorially' (to satisfy the modular relationships). The gold bars don't seem to affect it at all (at any number of sons I can multiply all cattle by 7 to get a integral relationship with number of gold bars..) Did anyone find a reason not to suggest any number of sons?

the issue I have is that there doesn't seem to be a proper answer 83.100.251.196 (talk) 22:00, 18 September 2009 (UTC)

"at any number of sons I can multiply all cattle by 7" : That's not true since the condition that each son recieves one more cow than his elder brother would then not be satisfied.

The analysis by Rcrone above is right on the point: we need to simultaneously satisfy congruency relations with respect to some power of 8 and some of power of 7, and since this is achievable for any power n, (infinitely many) solutions exist for any given number of sons. Abecedare (talk) 22:10, 18 September 2009 (UTC)

Yes, I just came back to correct that.83.100.251.196 (talk) 22:34, 18 September 2009 (UTC)

I think with last wife taking no cows the method fails after 2 sons, (though probably made a mistake) I did. Doing sons wrong way round.83.100.251.196 (talk) 22:50, 18 September 2009 (UTC)

cows=16 sons=2

first son = 7 , wife = 1

second son 8 , wife =0

cow difference = 0, thus no gold

No solution for 3 sons I can see.(reason:algebraic (5k-1)/8 must equal integer for integer k - impossible)

Solutions also exist for cows = 16+34n , fs=16n+7,fw=2n+1,ss=16n+8,sw=0, cow difference=2n = 7n gold.

Ignore 1 son solution since it seems irrelevent to the question. Is that ok?83.100.251.196 (talk) 23:30, 18 September 2009 (UTC)

Doh, me dumbo see correct solution below.83.100.251.196 (talk) 00:09, 19 September 2009 (UTC)

(5k-1)/8 is an integer if k=5+8n for any integer n. In that last solution, remember, ss=fs+1.

General solution for 2 sons: let k be any non-negative integer.

Cows: 16+119k.

Gold: 2k.

Son 1 takes: 7+56k cows.

General solution for 3 sons: let k be any non-negative integer.

Cows: 24+1519k.

Gold: 52k.

Son 1 takes: 6+448k cows.

General solution for 4 sons: let k be any non-negative integer.

Cows: 32+17225k.

Gold: 902k.

Son 1 takes: 5+3584k cows.

-GTBacchus^(talk) 23:51, 18 September 2009 (UTC)

Can anyone write it down for n sons? I'm working on 5... -GTBacchus^(talk) 23:56, 18 September 2009 (UTC)

Empirically observed pattern: first son's take when there are n sons = (9-n) + (8^(n-1))*7k. One can trace back the total number of cows from this, but I'm too lazy to do that at the moment. My conjecture is that this pattern will hold for n<9, and possibly for n=9. Abecedare (talk) 03:36, 19 September 2009 (UTC)

September 19

Field extensions

Hi. I'm working through a p-adic analysis book, and there's a quick review of fields at the beginning of chapter 3. I read the following there:

A field extension K of F is an F-vector space; if it is finite-dimensional, it must be an algebraic extension, and its dimension is called the degree [K:F]. If $\alpha \in K$ has the property that every element of K can be written as a rational expression in $\alpha$, we write $K=F(\alpha )$ and say that K is the extension obtained by adjoining $\alpha$ to F.

Two paragraphs later:

If F is a perfect field, then any finite extension K of F is of the form $K=F(\alpha )$ for some $\alpha \in K$. $\alpha$ is called a primitive element. Knowing a primitive element $\alpha$ of a field extension K makes it easier to study K, since it means that everything in K is a polynomial in $\alpha$ of degree <n.

Super. So... the first paragraph there says that everything in K is a rational expression in $\alpha$. The second says that everything in K is a polynomial in $\alpha$. Those aren't equivalent, are they? What polynomial in $\alpha$ is the same as ${\frac {1}{1-\alpha }}$, for example? What am I missing? Is the polynomial thing something extra that we get when F is perfect? -GTBacchus^(talk) 00:10, 19 September 2009 (UTC)

They're equivalent as long as α is algebraic over K. Then any rational expression in α over K can be rewritten as a polynomial, using the extended Euclidean algorithm. For example, if α is a root of the polynomial x²-x+1, then 1/(1-α)=α. Algebraist 00:59, 19 September 2009 (UTC)

The algorithm at that link seems to apply in a finite field. What if we're talking about an extension of $\mathbb {Q}$, for example? That's a perfect field, because is has characteristic 0, right? -GTBacchus^(talk) 01:02, 19 September 2009 (UTC)

The algorithm works whenever α is algebraic over the ground field (perfection is not relevant, and neither is finiteness). Algebraist 01:09, 19 September 2009 (UTC)

Ok, I'll study that. Thank you. -GTBacchus^(talk) 01:29, 19 September 2009 (UTC)

calculations over three years

I learned one porcelain bowl from the estate of Barbara Hutton was auctioned and sold by Christie's Hong Kong for $22,240,000 in 2006. When I tried to use the inflation calculator provided by the United States Department of Labor to figure out what that sum of money would be today, I ran into a problem. The original sum has to be at least $10,000,000 or less. So if anyone could improvise and help me figure out what $22,240,000 in 2006 would be in today's money, that would be great. Thank you.69.203.157.50 (talk) 05:43, 19 September 2009 (UTC)

Try the inflation calculator on $22,240 and multiply the result by 1,000. Bo Jacoby (talk) 06:14, 19 September 2009 (UTC).

Method to generate symmetries of a point lattice

I am interested in algorithmically generating the symmetries of a 2D or 3D point lattice, given reduced lattice basis vectors. I know that there are a finite number of crystallographic point groups in these dimensions, but I am wondering how those tables of symmetries were generated. Is there an automatic way of discovering lattice symmetries? Victor Liu (talk) 19:39, 19 September 2009 (UTC)

One way you could go about it is to begin at the origin and generate all the lattice points near the origin, expanding your search outward until you have in your collection 3 points that are linearly independent (for a 3D lattice). Then divide all the points in your collection into sets of points that are equidistant from the origin. Any symmetry has to map the points in each set into one another, and any isometry that successfully maps all the points in your collection to other points in the collection is a symmetry of the lattice. That should narrow down the search for what isometries work to something finite. There might be something more clever though. Are you looking to implement this on a computer? Rckrone (talk) 06:56, 20 September 2009 (UTC)

Yes, I plan to implement this on a computer. I thought it would be interesting to be able to generate the standard tables of crystallographic point groups in 2D and 3D, and then see if I can go to higher dimensions. That's the mathematically curious reason. The practical reason is to implement very generally, among other possible applications. Your proposal sounds good. I'm think to use the set of equidistant nearest neighbors from the origin, and trying rotations and flips on them. I still need to convince myself this works. 76.102.161.22 (talk) 08:36, 20 September 2009 (UTC)

September 20

Collection of six related primes

The numbers 931981, 941981, 961981, 971981, 991981 and 1001981 are all prime. I have verified that no other number of one to four digits can be put in place of 1981 for this to remain true, and crude estimates of how many five- and six-digit numbers there are for which the six implied numbers will all be primes are 'around two' and 'around eight-and-a-half'. (The formula used for the first approximation is 12000*[3.75/log(9700000)]^6, where 12000=90000/7.5 comes from congruence modulo 30 considerations and the square brackets encloses a value derived from the prime number theorem and consideration of the first three primes.) If anybody can spare the time to generate by program all of the five- and six-digit values one can substitute for 1981, I would appreciate it. I imagine there is a database listing of primes that one can simply search.Julzes (talk) 03:52, 20 September 2009 (UTC)

Did the sequence "93, 94, 96, 97, 99, 100" come from anywhere in particular? It's not surprising that no other 4-digit numbers can be prefixed by elements of that sequence to make primes, more interesting would be some kind of how many such numbers work for various sequences. Is there any sequence of 6 consecutive prefixes that has a 4-digit number that gives all primes, for example? --Tango (talk) 04:02, 20 September 2009 (UTC)

I would say I just took a kind of numerological curiosity in all of the values beginning with 11981 and worked upward. You could say I just stumbled upon it. Obviously the answer to your other question is no by consideration modulo 3. I'm in the midst of formulating another nice question beginning at 100 and working backward, but one could come up with all sorts of related questions. You can sort of figure out what the question I am trying to formulate is if I say that 9 is the first value in sequence, and 1981 is the sixth. Julzes (talk) 04:21, 20 September 2009 (UTC)

The first six members in the sequence would be 9, 67, 67, 787, 1981 and 1981 (1009 is prime; 10067, 9967 and 9767 are prime; 100787, 99787, 97787 and 96787 are prime; and then the above is the first that allows 94 as a prefix and it also allows 93. The sequence of prefixes continues 91, 90, 87, 85, 84, 82, 79, 78, 76, 72, 69, etc., with modulo 7, 11 and 13 considerations first necessarily coming in at 95 (taking 88 and 74 off), 92 (81 and 70) and 88 (75), respectively. How would we expect the sequence to grow, rather than asking what its precise values are (the next few terms would be nice, but I expect the growth to be pretty fast)? What does the sequence starting with prefixes of 0 (no prefix), 1, 3, 4, 6, 7, 9, 10, 13, 15, 16, etc., look like? This is probably a more natural question. Is there a good base-independent related question is also something I wonder about.Julzes (talk) 06:17, 20 September 2009 (UTC)

For 5 digits, the only answer is starting 93,94, etc. is

9346501,9446501,9646501,9746501,9946501,10046501.

For 6 digits, there are 4 solutions:

93132667,94132667,96132667,97132667,99132667,100132667

93338149,94338149,96338149,97338149,99338149,100338149

93484393,94484393,96484393,97484393,99484393,100484393

93740131,94740131,96740131,97740131,99740131,100740131

70.90.174.101 (talk) 09:01, 20 September 2009 (UTC)