The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
If a sporting league consisted of n teams, which each played g games, how many games would there be in total? —Preceding unsigned comment added by 76.69.241.74 (talk) 04:12, 28 May 2008 (UTC)[reply]
Rather than giving you the answer, let's look at it this way... Major League Baseball has 30 teams, each of which are scheduled to play 162 games. According to Major League Baseball season, that comes out to 2430 scheduled games per season. How does one get that figure? Alternatively, try to break it down into a scenario such as this: you have 6 teams, each of which has to play the other teams exactly once each. How many situations can you form based on this? I hope this leads you in the right direction. --Kinut/c 04:32, 28 May 2008 (UTC)[reply]
Hint: You'll need to consider how many teams take part in each game. --Tango (talk) 10:13, 28 May 2008 (UTC)[reply]
I initially though n*g/2, but someone told me, and he seems quite knowledgable in sports, that there are 1335 games in an NHL season (30 teams and 82 games apiece). So is he wrong (predicted value=1230)? —Preceding unsigned comment added by 76.69.241.74 (talk) 03:04, 29 May 2008 (UTC)[reply]
The regular season does hold 1230 contests, as the formula (which you are correct on) dictates. I'm assuming the difference (105 games) your friend is including is the maximum number of playoff games (7 game series) that could be played in the postseason: 8 first round series, 4 conference semifinal series, 2 conference final series, and 1 Stanley Cup final series; 7*(8+4+2+1)=105. --Kinut/c 04:22, 29 May 2008 (UTC)[reply]
Mappings
.
What are the number of mappings from ?
My book gives the straight answer as but I was wondering how to arrive at that conclusion, I assume by induction. If there are n mappings from 1 to S, and n mappings from 2 to S, and n mappings from 3 to S, and so on then there are n * n * n e.t.c mappings from S to S, but this doesn't convince me. Any thoughts? Damien Karras (talk) 14:08, 28 May 2008 (UTC)[reply]
There's no need to use induction: your reasoning seems perfectly sound to me. To reword it only slightly, there are n choices for what 1 maps to, n choices for 2, etc, and these choices are independent, so the total number is n*n. Try doing it explicitly for n = 2 or 3 and it might become clearer. AndrewWTaylor (talk) 14:31, 28 May 2008 (UTC)[reply]
Actually, any function from S to itself can be considered as an ordered n-tuple (the i-th entry in the tuple tells you what element i gets mapped to). Count the number of such n-tuples. –King Bee(τ • γ) 16:23, 28 May 2008 (UTC)[reply]
I'm pretty sure Andrew meant , with n*n being either a typo or part of a larger description which was cut short due to lack of space. -- Meni Rosenfeld (talk) 16:46, 28 May 2008 (UTC)[reply]
Oops, you're quite right Meni, I did indeed mean . Thanks for the correction. (I've amended my previous comment.) AndrewWTaylor (talk) 19:14, 28 May 2008 (UTC)[reply]
Diagonalizing a matrix.
Regarding diagonalizing matrices, I know that a matrix P whose columns are linearly independent eigenvectors will always diagonalize a matrix by taking P-1AP, and that a matrix is diagonalizable iff it has a full set of linearly independent eigenvectors. What I can’t figure out is if those are the only such matrices. Specifically, if at least one of the columns of P is not in an eigenspace of an eigenvalue of A, is it possible that P-1AP is a diagonal matrix?
More generally, if one has a linear transformation of vector spaces , does the corresponding result hold for infinite dimensional vector spaces? GromXXVII (talk) 16:55, 28 May 2008 (UTC)[reply]
The columns of P must be eigenvectors of A. Let . For every i, denote the ith column of P, and . Then . Therefore , and .
I don't know about the infinite dimensional case. -- Meni Rosenfeld (talk) 17:31, 28 May 2008 (UTC)[reply]
Set Notation
I have a set of classes, , and a set of documents, . I want to define a relation, , that says which documents belong to which classes.
I then want to sum over all documents of a particular class, .
I think the relation can be defined either with
or
My question is, how can I express the subset of documents belonging to class ?
I don't think you want the "there exists c" bit - that means any c, I think you mean a specific c, so you just want {d|(d,c) in B}, or, a shorter notation: {d|dBc} (using B in the same way we use =, <, > or whatever). --Tango (talk) 22:08, 28 May 2008 (UTC)[reply]
Ah yes, that makes lots of sense. It's surprising how simple it is when you see it. Proberts2003 (talk) 22:26, 28 May 2008 (UTC) [Same user as 80.175.114.193 above][reply]
Also, the statement "" is identical to just "", so you could also simply have . It makes it less clear that you are summing over different ds, though. -- Meni Rosenfeld (talk) 22:51, 28 May 2008 (UTC)[reply]
I believe the usual notation for that would be —Ilmari Karonen (talk) 22:57, 28 May 2008 (UTC)[reply]
(ec) Or you could even define D(c) = {d|(d,c) ∈ B} and then just write your sum over d ∈ D(c). You could even use that notation exclusively, starting with a definition of D(c) as a function of c and writing d ∈ D(c) instead of (d,c) ∈ B or dBc. Of course, if you also need the sets of classes for a particular document, C(d) = {c|(d,c) ∈ B}, then going with the relational notation as suggested by Tango may be simpler. —Ilmari Karonen (talk) 22:57, 28 May 2008 (UTC)[reply]
Or, better yet, since c is supposed to be a class, why not simply define it as the set of documents it contains? Then you could simply write d ∈ c. —Ilmari Karonen (talk) 23:00, 28 May 2008 (UTC)[reply]
Try y = ax + b, and try to come up with values of a and b that fit. (Hint: b should be the difference between consecutive values of y. Also, when x = 1, y = a + b.) —Ilmari Karonen (talk) 01:23, 29 May 2008 (UTC)[reply]
I have no idea how to write this as a function, but its the difference between x and 3. 3-1=2, 3-2=1, 3-3=0, 3-4=-1. well, i guess -x+3=y? --Xtothe3rd (talk) 01:58, 29 May 2008 (UTC)[reply]
Very good. That is correct. is the function. In order to check this, if you have a TI calculator, hit the "y=" button and type in the corresponding x part of the equation. Then hit the 2nd key and then graph. This should display a table with the corresponding y and x values. By the way, this is a linear function. Ζρς ι'β'¡hábleme! 07:26, 29 May 2008 (UTC)[reply]
This looks like homework, so the only appropriate answer is really the first. The rest of you shouldn't go this far. 130.95.106.128 (talk) 11:10, 29 May 2008 (UTC)[reply]
Oops. It did look like homework to me as well. However, I gave my answer at 3:00 am local time, so I overlooked the fact that the OP and Xtothe3rd are not the same person. That was my bad. Ζρς ι'β'¡hábleme! 17:08, 29 May 2008 (UTC)[reply]
Different differences
How many different differences can be obtained by taking only 2 numbers at a time from 3,5,2,10 and 15?
One might thing the maximum number of differences would be 2 out of 5 (5*4), and that the number of different ones must be less than this. (At least I would).
Yet the answer is 1440.
Anyone?
83.167.112.140 (talk) 15:32, 29 May 2008 (UTC)john[reply]
Where did you hear it should be 1440? Your reasoning is correct for what I understand the problem is. Either the problem should be interpreted in some unusual manner, you haven't stated the problem correctly, or your source is dead wrong. -- Meni Rosenfeld (talk) 16:08, 29 May 2008 (UTC)[reply]
This was an on-line gre practise question. I recited the question verabatim, so I guess it must be a mistake. Thanks. 83.167.112.140 (talk) 16:27, 29 May 2008 (UTC)[reply]
Perhaps the differences need not come from the original set. For instance, take 10-2, and then repeat the process on the numbers 3, 5, 2, 10, 15, 7? That’s not how I would interpret it, but could be what the author of the question meant. GromXXVII (talk) 17:02, 29 May 2008 (UTC)[reply]
10-2=8 That interpretation either gives infinitely many possibilities (if differences can be negative) or 15 (if they cannot). I can't see any way to get 1440. In any case I would be very unhappy with an exam that included questions this loosely worded. Algebraist 18:12, 29 May 2008 (UTC)[reply]
Another stupid question; graphs of functions
Hello, nice refdesk people. Another stupid question I've been puzzling over:
For each of these functions, decide whether the function is always increasing, always decreasing, or sometimes increasing and decreasing:
y=x3 - 3x + 1
y=x3 + 3x + 1
y=1- 3x - x3
I don't actually understand the question: By the 'function is always increasing', do they mean it always has a positive gradient? Or what? :/ *confused once again* naerii - talk 18:01, 29 May 2008 (UTC)[reply]
A function y=f(x) is increasing if f(a)<f(b) whenever a<b. That's not quite the same as having positive gradient, but certainly a function with positive gradient everywhere is increasing. Algebraist 18:08, 29 May 2008 (UTC)[reply]
Thanks for making it more clear Algebraist! :) naerii - talk 21:09, 29 May 2008 (UTC)[reply]
The word "always" in the question apparently means "everywhere", that is, everywhere on the domain of the function. --Lambiam 04:41, 30 May 2008 (UTC)[reply]
You need to find the gradient of each equation, as these will give you the values:
. So increasing when x > 1, decreasing when x < -1.
. Since it contains a squared term, which cannot be negative, the function is always increasing.
. Since it contains a squared term, which cannot be negative, then multiplied by a negative term, the resulting solution will always be negative, ie always decreasing. 86.153.37.241 (talk) 23:28, 1 June 2008 (UTC)[reply]
No, the first example is increasing in both and and decreasing in . Algebraist 23:49, 1 June 2008 (UTC)[reply]
But when differentiated the equation is a parabola, and it can be easily seen when the equation is increasing and when it is decreasing. 86.153.37.241 (talk) 00:18, 2 June 2008 (UTC)[reply]
Yes indeed, and it's easily seen to be what I said and not what you said. Algebraist 00:20, 2 June 2008 (UTC)[reply]
That's a matter of personal opinion. 86.153.37.241 (talk) 00:22, 2 June 2008 (UTC)[reply]
It's a matter of personal opinion because it depends entirely on what people prefer to see as the given solution and how to get the correct one. 86.153.37.241 (talk) 00:35, 2 June 2008 (UTC)[reply]
People usually prefer to be given the correct solution... It doesn't really matter how you get to that solution, but any method that results in an incorrect solution is, by definition, incorrect. --Tango (talk) 00:58, 2 June 2008 (UTC)[reply]
(ec) Yes, it can, and it's what Algebraist said. It's a parabola that just dips below the x-axis between -1 and 1, and it above it everywhere else. --Tango (talk) 00:30, 2 June 2008 (UTC)[reply]
Probability
First, yes, this is a school question. A question on a test I took today said, if you're dealt a 5 card hand from a deck of 52 cards, what is the probability of getting exactly one King? I worked it, and I came up with
But, according to whoever made up this test, the actual answer is . I'm stumped, and so is my teacher. So, can you please enlighten me on what the right answer is, and how to get it? Digger3000 (talk) 19:34, 29 May 2008 (UTC)[reply]
This is a hypergeometric distribution. Define a "trial" as drawing a card. Define a "success" as the drawn card being a king. In the deck, there are 52 possible trials and 4 possible successes. You are going to run 5 trials and you seek 1 success. Plug in the values to the hypergeometric distribution and you get 0.299. Wikiant (talk) 19:47, 29 May 2008 (UTC)[reply]
So, what you calculated originally, was, in fact, probability that the first card would be the only king. But you have to calculate the probability that any one of the cards is a king, meaning that you have to multiply the result by 5 - the probability that the second (third, fourth, fifth) card will be the only king will be the same 0.0599. And that gives you 0.299. --Martynas Patasius (talk) 20:10, 29 May 2008 (UTC)[reply]
Is this a homework question? ;) Is 'm' here the mass? Is it moving at constant velocity? naerii - talk 21:08, 29 May 2008 (UTC)[reply]
Linear algebra without vector spaces
Hi. I'm working through Algebra by Michael Artin, and I'm stuck on one of the miscellaneous problems at the end of chapter 1. The problem (which has a star by it) is:
Consider a general system AX=B of m linear equations in n unknowns. If the coefficient matrix A has a left inverse A', a matrix such that A'A=I, then we may try to solve the system as follows:
AX = B
A'AX = A'B
X = A'B
But when we try to check our work by running the solution backward, we get into trouble:
X = A'B
AX = AA'B
AX = B ???
We seem to want A' to be a right inverse: AA'=I, which isn't what was given. Explain. (Hint: Work out some examples.)
Ok, so I know what's going on in this problem, and it comes down to this. If A is m x n with m<n, then it can't have a left inverse, because it represents a transformation from Rn → Rm, and no transformation back from Rm → Rn can be surjective.
On the other hand, if m>n, then A' can have a left inverse (depending on the rank of A), and if it does, then AA', while not being an identity matrix, is a square matrix for which B is an eigenvector with a corresponding eigenvalue of 1; in other words: A'AB = B.
Here's my problem. I'm in chapter 1 of this book, and there's nothing about vector spaces, or surjective maps, or eigenvectors, or any of that yet. All the book has presented so far are matrices as systems of equations, square matrices as products of elementary matrices, a simple version of the invertible matrix theorem, the definition of determinants, and Cramer's rule. I don't know how to use only that to establish that the coefficient matrix of an underdetermined system has no left inverse.
So... does this question make any sense, and does anyone have any idea how to help me? -GTBacchus(talk) 21:57, 29 May 2008 (UTC)[reply]
Why should B be an eigenvector of AA' with eval 1? Suppose we have two equations in 1 unknown, x=a and 0=b. Then our 2x1 matrix A has a left inverse A', and AA' is the 2x2 matrix with rows 1,0 and 0,0. Thus B is only an evec if b=0, and indeed, if b is not zero, then there's no solution to our system of equations, even though A had a left inverse. We conclude that while A having a left inverse guarantees uniqueness of solutions, it's the right inverse that guarantees existence. Algebraist 22:08, 29 May 2008 (UTC)[reply]
Put differently: if, then, but not the other way around. -- Meni Rosenfeld (talk) 22:33, 29 May 2008 (UTC)[reply]
Applying a function (e.g., a linear transformation) to two equal quantities (an equation) is guaranteed to preserve any difference between them (create no new solutions) if and only if the function is injective. (It is guaranteed not to destroy solutions iff its domain includes the range of each quantity.) For multiplying on the left by a matrix, this means that the matrix has full column rank. If A has only a left inverse, then it is "tall" () and has full column rank. Its left inverse, however, is "wide" and cannot have full column rank. The problem is that A is not surjective, so there are bs for which there is no solution. The equation is interesting nonetheless: the matrix is the projection into A's column space and arranges for a solution to exist. Projection matrices have eigenvalues of 0 and 1; it is precisely when b is in the "1" eigenspace that it is in A's column space, so it is an eigenvector as you describe. --Tardis (talk) 22:50, 29 May 2008 (UTC)[reply]
You guys rock. Thank you all for those very helpful explanations. I had to chew on it for a couple of days, but I think I've got it now. Let me try to summarize:
Since the matrix A has a left inverse, then the transformation it represents () in injective - injectiveness is a necessary condition for having a left inverse in the category of sets and mappings, so we don't need to appeal to vector spaces for that. The mapping T is injective, but we don't know that it's surjective, so B (or B's columns - we can deal with them one at a time) might not be in the range of T.
If B is in the range of T, then A', being a left inverse for A, will map B back to its preimage, namely A'B. In that case, we're all good, and even though A' may not be a right inverse for A, we know that AA' will be a square matrix that maps B to B.
On the other hand, if B is not in the range of T, then there is no solution. In that case, A' is certainly not injective, so applying its transformation to the equation AX=B creates an extraneous solution. While A'B is a solution to the equation A'AX=A'B, it is not a solution to AX=B. The point A'B in the domain of A is mapped to AA'B, which is not the same as B, although A' takes them both to A'B.
You've about got it, but I would refine a couple of points: first, is never injective when it but not exists — it obviously doesn't depend on B being in T's range. Also, in the same case, it should be pointed out that there are infinitely many such left inverses: we can add to any row of a any vector that is orthogonal to all of A's columns (which is possible because it has fewer columns than the dimension of its columns) and not affect the product . --Tardis (talk) 13:46, 4 June 2008 (UTC)[reply]
May 30
Mathematicians objecting Cantor
This quote can be found in the web:
"something on the order of 90% or so of working mathematicians accept Cantorian set theory both in theory and in practice, to some extent. (Source: The Mathematical Experience, Davis/Hersh)"
Although it's not clear what is meant by "Cantorian set theory" the 10% against set theory seems a rather high percentage. So my question is: are there notable working mathematicians not accepting traditionally accepted belief or proof techniques involving modern set theory?--Pokipsy76 (talk) 06:34, 30 May 2008 (UTC)[reply]
Particularly amusing considering his other work, Edward Nelson is another. He doubts the consistency of Peano arithmetic; cf his site and his Predicative Arithmetic. The question is not too clear.John Z (talk) 09:23, 31 May 2008 (UTC)[reply]
n > 3, n = Number Of Dimensions
I totally understand the 0th, 1st, 2nd, and 3rd dimension. I have ABSOLUTELY no idea how any other dimension above the 3rd works. I mean, let's take the simple cube 3rd dimension analagoe into the 4th dimension: the hypercube. So the net is the same as a cube's net, but instead of squares, it's cubes. So when a hypercube rotates, it looks like it's volumes are warping. How is this possible? and then, the "sides" of the hypercubes don't even look like cubes, they look have trapzoidial sides (you know what I mean?)68.148.164.166 (talk) 07:21, 30 May 2008 (UTC)[reply]
You can project a cube into a plane (in a photo or a drawing), when you do so the squares do not seems squares anymore and when it rotates the area and of the 2D picture changes. The same happens when you project a hypercube into a 3D space: cubes do not appear as cubes and when it rotates the volume of the solid picture changes.--Pokipsy76 (talk) 07:45, 30 May 2008 (UTC)[reply]
I still do not understand. Is it because my brain is to tiny? Why can't I understand?68.148.164.166 (talk) 10:16, 30 May 2008 (UTC)[reply]
It's hard for us, as creatures who are more or less three dimensional, to envision what a fourth dimension would be like. But then, as in Flatland, consider how hard it would be if you were two-dimensional to envision a third dimension. The easiest way to envision the 4th dimension is to just consider out 3D world as a horizontal 2D plane, and the 4th dimension being the vertical. This is what you see in all those pictures of space being warped by gravity. -mattbuck (Talk) 18:13, 30 May 2008 (UTC)[reply]
Hey, thanks for the link to Flatland, I really understand it a lot more now. But I think why I don't understand is because matter is 3 dimensional. In Flatland, that wouldn't be possible because the polygons wouldn't exist, because matter can't exist 2 dimensionally.68.148.164.166 (talk) 00:57, 31 May 2008 (UTC)[reply]
Actually, matter probably isn't three dimensional - I believe that most current theories postulate that the universe is 17-dimensional or something like that. However, there's no real reason you shouldn't have 2D matter as far as I know.
Huh, how why? I REALLY don't understand that.68.148.164.166 (talk) 04:22, 31 May 2008 (UTC)[reply]
Physicists sometimes work with toy models of physics that have fewer than three dimensions of space, and there are real models of physics (like Kaluza-Klein theory and string theory) that have more than three dimensions for various reasons. But the extra dimensions, if they exist, are nothing like the ones we see. They don't extend off to infinity but rather curl around on themselves with a very small radius. -- BenRG (talk) 05:15, 31 May 2008 (UTC)[reply]
Rather than to try to envision four dimensions, make an algebraic attempt. The coordinates of the two end points of a one-dimensional line segment are 0 and 1. The four corners of a two-dimensional square are (0,0), (0,1), (1,0), and (1,1). The eight corners of a three-dimensional cube are (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), and (1,1,1). Now generalize. Each point in four dimensions have four coordinates. The sixteen corners of a four-dimensional hypercube are (0,0,0,0), (0,0,0,1), (0,0,1,0), (0,0,1,1), (0,1,0,0), (0,1,0,1), (0,1,1,0), (0,1,1,1), (1,0,0,0), (1,0,0,1), (1,0,1,0), (1,0,1,1), (1,1,0,0), (1,1,0,1), (1,1,1,0), and (1,1,1,1). That was not too difficult. Bo Jacoby (talk) 20:45, 30 May 2008 (UTC).[reply]
You can't intuitively visualize what something with more then 3 spatial dimensions looks like. You can get indications of what that might look like, but since our basic visual vocabulary is based on three spatial dimensions, it's just not going to ever work out. Don't fret about it. You're not the only one who has problems visualizing it intuitively. There are little ways to try to get around this—a 4D object would have a 3D shadow, ooooh—but in the end it's just not visualizable in any sort of obvious way. A hypercube is a good example of that—the volumes do look like they are warping, and that's how we'd perceive something that was in 4D, as horribly warping and being an impossible object, because those 4D would be still only available to use in the 3D (the hypercube example actually is a little problematic because it is basically modeling a 4D object in only 2D, which makes things even worse). --98.217.8.46 (talk) 00:56, 31 May 2008 (UTC)[reply]
Why are our basic visual vocabularies based on three spatial dimensions?68.148.164.166 (talk) 04:35, 31 May 2008 (UTC)[reply]
We evolved an ability to visualize objects in three dimensions, but no more than that. Even professional mathematicians can't visualize four-dimensional objects directly. They have to visualize a three-dimensional projection or slice or analogy of the object, or else analyze it in non-visual ways. -- BenRG (talk) 05:15, 31 May 2008 (UTC)[reply]
Why have we evolved an ability to visualize objects in three dimensions, but no more than that. If that was the case, then that means 3 dimensions has more importance than 2 dimensions as well as more importance than 4 dimensions or more.68.148.164.166 (talk) 17:40, 1 June 2008 (UTC)[reply]
I've met mathematicians that claim to be able to visualise higher dimensions - I've no idea how they do it, or even if they are really doing what they claim (and not just visualising projections, like you say), but they certainly claim to be able to do it. One suggested doing it in a pitch black room so there's no 3D world to confused you. --Tango (talk) 12:09, 31 May 2008 (UTC)[reply]
We all do a similar trick all the time without even thinking about it, when we infer three-dimensional properties of objects from the two-dimensional projection that we see in our visual field. We know that the faces of a cube are squares, and the fact that they change appearance as we turn the cube and don't look like squares from most directions doesn't worry us at all. In fact, this interpretation is so ingrained that most graphic artists require a degree of practice and training to supress it in order to draw an accurate representation of what they see, instead of what they think they see. Mathematicians like the late Donald Coxeter can, through practice, develop a similar intuitive feel for inferring four-dimensional qualities by visualising a three-dimensional projection (well, really by visualising a two-dimensional projection, so there is another level of interpretation involved). The human brain is truly amazing. Gandalf61 (talk) 10:13, 31 May 2008 (UTC)[reply]
May 31
A Decreasing Sequence
While working on a problem, I came across this intermediate problem. Let . I need to show that these functions form a strictly decreasing sequence for x strictly positive. In other words, for I need to show that for all . A couple of us have worked on this problem and we have tried all sorts of techniques (including the standard tests and treating the sequence as a continuous function of n and using the derivative test) but we either get inconclusive results or the inequality comes out the wrong way. The question is correct (there is no typo) as one can easily verify the statement by graphing a few terms. Any tips would be welcomed. Thanks guys! A Real Kaiser (talk) 00:04, 31 May 2008 (UTC)[reply]
Well, the result is obviously true for sufficiently large x. This is the route I'm thinking would work:
For the sequence to be decreasing, we require the nth term to be less than the (n-1)th term, ie . We want to simplify what we need to show to be negative, so multiply out the fractions to get . It's quite clear the denominator is positive, so we proceed by looking solely at the numerator. Now, . I suspect that you can equate this in terms of powers of k by . Those terms which are negative are obviously fine, and if a term is positive, it ought to have modulus less than the previous term (working from k=0 upwards). I think that should work. -mattbuck (Talk) 01:22, 31 May 2008 (UTC)[reply]
Here is an outline for proving the above:
Define .
Consider .
Show that this implies
Demonstrate by series expansion of the left-hand side that this equation has no solutions for x > 0.
Explain why knowing that is sufficient to solve the problem.
series expansion of the logarithms, convergent for |2y|<1 (or n>2|x|)
reinsert x:
For fixed x, for sufficiently big values of n, the series is convergent and the omitted terms are neglegible, and so the sequence fn(x) is strictly decreasing. It remains to be shown that this is also the case for smaller values of n. Bo Jacoby (talk) 07:47, 31 May 2008 (UTC).[reply]
Wow guys, a couple of really good answers. But Dragons flight, in your step three shouldn't it be
.
This is how I did it all. I will show all the steps. Maybe I made a stupid mistake somewhere.
Let .
Let .
.
And then, from here how can we conclude that this can't have a solution for x>0? I expanded the series for log(1+y) and plugged in x/(n+x).
But how does this show me that this expression cannot equal to for any positive x. Even if I perform partial fraction decomposition (treating n as a variable and x as a constant), I get that
but I still don't see how that helps us. Should I equate the coefficients? I can equate the fraction with the denominator n+x but what about the one with n+2x? There is nothing on the other side with that same denominator. Now assuming that, I can show this, I even understand the last step The last one is that basically our function's derivative is never zero. Therefore (as a function of n) the derivative is either always strictly positive or negative. We plug in n=1 and see that the derivative
is always negative for x>0. Which means that the derivative is always negative for all n which means that our original function is a strictly decreasing function of n which means that the first function is also a strictly decreasing function of n because the exponential is a continuous increasing function and hence it preserves the inequality.A Real Kaiser (talk) 01:49, 1 June 2008 (UTC)[reply]
shortest distance in a circle
If you have to move in a circle (i.e. go from degree 300 to 0 or 45 to 90), how you can calculate the shortest distance? Or should you just discard every result above 180 and go to the other one. Is there a formula that always gives the shortest distance? —Preceding unsigned comment added by 217.168.1.77 (talk) 00:30, 31 May 2008 (UTC)[reply]
Well, do you mean the direct distance (as the crow flies)? If so, use the cosine rule. For movement by degrees, yes discard anything over 180 and go the other direction. For movement around circumference measured in units or whatever, multiply the circumference by degrees moved and divide by 360. -mattbuck (Talk) 00:49, 31 May 2008 (UTC)[reply]
If you're looking for a single expression to give you the shortest distance in the range -180 to +180, you can use modulo operation. Something like: ((angle2 - angle1 + 180) mod 360) - 180. Take the absolute value if you only want the magnitude. -- Tcncv (talk) 03:30, 31 May 2008 (UTC)[reply]
Assuming that angle mod 360 returns a value in the range from 0 to 360−, if you only want the magnitude, you can also use
min((angle1 - angle2) mod 360, (angle2 - angle1) mod 360).
I've looked at your article on the Mordell Curve, which doesn't tell me what I want. I want the parametric form of the Mordell Curve (fixed k = 1), ie y^2 = x^3 + 1. This would help me
a lot in anti-differentiating tough functions.
In other words, I want something in the form y = f(t) and x = g(t) that describes y^2 = x^3 + 1.
Is this impossible? It says Euler found the only three integer solutions to the equation, but is there a general solution in terms of a parameter?
I suspect that the solution f(t) = t, g(t) = (t2 − 1)1/3 will not satisfy you. --Lambiam 07:15, 31 May 2008 (UTC)[reply]
The equivalent quadratic equation
(which is, of course, the equation of an hyperbola) is parameterised by the hyperbolic functions
Notice that f(t) and g(t) satisfies the differential equations
For the cubic case you can do the same trick if you have a g(t) that satisfies
The functions that parameterise elliptic curves in this way are called Weierstrass's elliptic functions. They are similar to the trigonometric and hyperbolic functions, except that they have two periods, not one. Gandalf61 (talk) 09:55, 31 May 2008 (UTC)[reply]
Thanks Gandalf. The only problem is, trying to find (g')^2 = g^3 + 1 leads me to attempting to find the integral of 1/√(g^3+1) with respect to g, which I was trying to do in the first place. I assume, then, that my integral can be expressed in terms of this Weierstrass function?
can we conclude that u is a constant function on ?
--89.97.102.209 (talk) 16:34, 31 May 2008 (UTC)[reply]
My gut feeling is yes, u must be constant (in fact, u must be identically zero because any other constant has unbounded integral), but I can't think of a nice proof yet. The maximum principle will surely be useful. (BTW, in English it's spelled harmonic, not armonic.) —Keenan Pepper 17:28, 31 May 2008 (UTC)[reply]
Huh? Any constant is possible, since any constant function has zero gradient. Algebraist 18:25, 31 May 2008 (UTC)[reply]
Well, any constant will surely satisfy this but that doesn't mean that ONLY a constant will satisfy this inequality. I mean, what if we had the modulus come out to be a decaying exponential which will have a finite volume in all of R^3?A Real Kaiser (talk) 19:49, 31 May 2008 (UTC)[reply]
Sorry, forgot about that gradient operator for a second. —Keenan Pepper 19:59, 31 May 2008 (UTC)[reply]
Ah, I see a proof now. If u(x,y,z) is a harmonic scalar field (), then is a vector field with zero divergence (). It also has zero curl, because the curl of a gradient is zero. Now simply consider the Helmholtz decomposition of . —Keenan Pepper 20:09, 31 May 2008 (UTC)[reply]
game theory question about tennis
I have a question about tennis that's fundamentally a game theory or math problem. It could be generalized to other games with similar scoring structures. The question is, should a player pursue a different strategy during game, set, or match points (either about to lose or about to win) than he would during other points? This is ignoring psychological factors such as surprising the opponent by changing one's strategy. It seems that there is a natural tendency to want to play more conservatively during a match point. But intuitively, it also seems that the rational thing is to stay with the strategy you had used and considered optimal during the rest of the game. However, on the other hand, the outcomes of match points have greater variance than the outcomes of entire matches because the sample sizes are smaller. So players with superior opponents will do better during match points than during entire matches. --MagneticFlux (talk) 20:14, 31 May 2008 (UTC)[reply]
You might get better answers if you clarify your question for those of us not familiar with tennis scoring. If I understand Tennis score correctly, games have no time limit, and thus the points are played independently of each other. This means that there are only two possible outcomes for each point - "win" and "lose", and so the only parameter is the probability of winning the point. If we ignore factors such as psychology and fatigue, the optimal strategy (which maximizes this probability) does not depend on "the big picture". However, it certainly can depend on who the opponent is. -- Meni Rosenfeld (talk) 20:46, 31 May 2008 (UTC)[reply]
I would argue that it is rational to play more conservatively for your opponent's match-point than for an ordinary point. If you lose your opponent's match-point, you lose the match but if you lose one of the other points, the match goes on. So it's not about surprising your opponent but about playing less risky shots to allow for the heavier downside of losing match-points. I'm not sure if this is the psychological factor you're considering ignoring but I wouldn't ignore it. Zain Ebrahim (talk) 22:08, 31 May 2008 (UTC)[reply]
No (again under the assumption that I understand the rules and the points are independent). For a match point, if you win it, you win the match. If you lose it, you lose the match the match goes on and you might lose it. So, you want to win the point, and since this is a boolean random variable, you want to maximize the probability of winning. If it was any other point, you would want to... maximize the probability of winning. There is no difference. -- Meni Rosenfeld (talk) 22:17, 31 May 2008 (UTC)[reply]
Not quite - if you lose a match point, the match carries on. The way tennis rules work, it's impossible for both players to be able to win on a single point (you have to win by at least two points). --Tango (talk) 22:21, 31 May 2008 (UTC)[reply]
Yeah, I don't know why I wrote that. The conclusion holds, though. -- Meni Rosenfeld (talk) 22:44, 31 May 2008 (UTC)[reply]
(edit conflict) I was thinking that, but when I considered it more, it doesn't actually hold up. As Meni says, there are only two possible outcomes, you win the point, or you lose. The optimal strategy is, therefore, to maximise your chances of winning, and that's going to be the same for any point. The idea of playing conservatively is to avoid significant loss at the expense of also avoiding significant gain, but that requires there to be something inbetween winning and losing, which there isn't in a tennis point. --Tango (talk) 22:21, 31 May 2008 (UTC)[reply]
It may be that I'm not ignoring the psychological aspect (and I don't think that one should). I was referring to the opponent's match-point (where there is significant downside but the usual upside). Compare this to an ordinary point where the upside and downside are equivalent. With the former, you'd be more interested in minimising the probability of losing the point (is this a different strategy?) where with the latter, you'd want to maximise the probability of winning. Surely, a strategy that minimises the probability of losing would be more conservative than one that maximises the probability of winning? Or have I totally missed something? Zain Ebrahim (talk) 22:41, 31 May 2008 (UTC)[reply]
[3ec] Exactly. In chess for example, there is a possibility for a tie, so if a tie or better is needed to win the match, the player should play defensively (or whatever it is that minimizes probability of losing, without caring about the division between a win and a tie). But in tennis the possibilities are more limited, and the phenomenon does not exist. -- Meni Rosenfeld (talk) 22:44, 31 May 2008 (UTC)[reply]
You either win a tennis point or you lose it. Thus the two strategies you describe are identical. Algebraist 22:43, 31 May 2008 (UTC)[reply]
Yes, if w is the probability of winning, and l is the probability of losing, then and thus maximizing the former is identical to minimizing the latter. -- Meni Rosenfeld (talk) 22:47, 31 May 2008 (UTC)[reply]
To clarify about what "match point" means, it could be either the point on which you are one point away from winning, or the point on which your opponent is one point away from winning. --MagneticFlux (talk) 20:16, 1 June 2008 (UTC)[reply]
Permutations
Hi, I've got a question which i'm a bit stuck with.
The question gives a permutation , then asks to find a permutation in with or to prove cannot exist.
I've had a little attempt, but i don't think i understand the question properly, so a little explanation would be greatly appreciated. Thanks for any help.
212.140.139.225 (talk) 22:39, 31 May 2008 (UTC)[reply]
For starters, you'll want to multiply out your expression for π. Algebraist 22:45, 31 May 2008 (UTC)[reply]
Oh sorry, i meant to put the disjoint cycle, , is this what you mean by multiply out? Thanks.
Yes, or that. I didn't notice the easier route because I was thinking of general values of 4. Algebraist 23:22, 31 May 2008 (UTC)[reply]
Ok thanks for your help. I'm not sure i understand what you mean by parity, but here goes. I know is even because it has 8, an even number of inversions so , but how does this help.
Also, this is where i feel i dont understand the question, i thought , so as the order of is 12. But i've tried this and it doesn't work so i think theres a fault in my logic rather than assuming doesn't exist. Thanks again 212.140.139.225 (talk) 23:48, 31 May 2008 (UTC)[reply]
I don't follow your reasoning for π being even; what eight transpositions is it the product of? And . Algebraist 23:52, 31 May 2008 (UTC)[reply]
8 Inversions i have a feeling that is completely unrelated. I think there are 5 transpositions so the cycle is odd.(?)
I probably confused myself with the second point. But since would it be as simple to say and since the exponent has to be an integer there is no ? Thanks again 212.140.139.225 (talk) 00:13, 1 June 2008 (UTC)[reply]
It's simpler than that. is odd, therefore so is . is either even, or odd. Try each case and see what happens. --Tango (talk) 00:18, 1 June 2008 (UTC)[reply]
(ec)Yes, π is odd. What is the parity of σ4? The argument you suggest does not work. Firstly, you have not considered the possibility (for example) that (since π=π25), but this can be dealt with. More importantly, all this shows is that σ cannot be a power of π. This does not mean σ cannot exist: for example, taking π=(12)(34), we have π=σ2 (where σ=(1324)), even though π has order two. Algebraist 00:24, 1 June 2008 (UTC)[reply]
Thanks for the replies, as is odd is odd, i don't know if this applies but with numbers if is odd then is odd. Is this at all helpful, i feel theres something blidingly obvious i'm missing, I'm new to the idea of cycles so please explain if you feel i'm not understanding it. Thanks again 212.140.139.225 (talk) 00:45, 1 June 2008 (UTC)[reply]
The key thing with parity (that is, evenness and oddness) is that it doesn't depend on how you write it. You can write the same permutation using numerous different products of transpositions and it will either always be even or always be odd (for example (12)=(12)(34)(34), one has 1 transposition, the other 3, both are odd numbers). That means you don't have to worry about simplifying to get a minimal number of transpositions, you can just write it as a long sequence of them. If has n transpositions, how many will have before any simplification? --Tango (talk) 11:49, 1 June 2008 (UTC)[reply]
If a permutation is decomposed into a product of disjoint cycles, then a power of that permutation is the product of the corresponding powers of these cycles. For example, ((1 3 4)(2 7 5 6))5 = (1 3 4)5(2 7 5 6)5. A power γn of a cyclic permutation γ is a product of disjoint cycles, all having the same order. For example, (1 5 9 3 2 8 7 4 6)3 = (1 3 7)(2 4 5)(6 9 8). The number of these component cycles and their order depends solely on n and the order of γ. For what orders of γ does γ4 have component cycles of order 4? --Lambiam 08:41, 1 June 2008 (UTC)[reply]
That seems overly confusing... It's actually a really simple problem once you spot how to do it. --Tango (talk) 11:49, 1 June 2008 (UTC)[reply]
I'd say this approach is more instructive, actually, since it (without too much difficulty) solves all problems of this form, not just the ones that happen to allow parity tricks. Algebraist 00:17, 2 June 2008 (UTC)[reply]
Is there a way of knowing which powers of γ will have how many factors?
Clearly the order of those factors has to divide the order of γ. The prime order, and order 2^m are pretty simple, but for instance is there a way of knowing that (1 5 9 3 2 8 7 4 6)3 was going to be a product of 3 cycles without working it out?
I think a cycle γ of length k can only break into smaller cycles when n divides k, because of the order of γn. But does it always? If so how do you know what lengths of cycles it breaks into? GromXXVII (talk) 11:06, 2 June 2008 (UTC)[reply]
A k-cycle, raised to the power n, becomes hcf(n,k) (k/hcf(n,k))-cycles. Algebraist 11:57, 2 June 2008 (UTC)[reply]
[ec] It may help to consider that a cycle and its powers form a cyclic group. The length of cycles in a power is the order of the element, which is , and the number of cycles is (thus breaking occurs not only when n divides k, but when n and k have a common factor). -- Meni Rosenfeld (talk) 12:01, 2 June 2008 (UTC)[reply]
could you please explain why division in matrices is not defined Kasiraoj (talk) 07:14, 1 June 2008 (UTC)[reply]
(Fixed leading space causing prefomatted text section.) Well, if we tried to define matrix division as multiplication by an inverse, we'd run into a couple of problems. First, many matrices don't have inverses. Second, matrix multiplication does not commute, so it would be important to distinguish whether you did it as left multiplication or right multiplication by the inverse. I don't see any particular benefit to defining such an operation, but go for it if you want; you'll just have to explain it where you use it since it isn't a commonly used notation. --Prestidigitator (talk) 07:59, 1 June 2008 (UTC)[reply]
As long as you stick to square matrices, I don't think the lack of commutativity is a problem. A non-singular square matrix has a unique inverse - it's both a left-inverse and the right-inverse. If you have non-square matrices, it's obviously a problem - the left and right inverses have to be different sizes, so they can't be the same. The main problem is that not all matrices have inverses (that is, ones with determinant zero). If you restrict yourself to non-singular, square matrices, then division is perfectly well defined - that's why is a group (the general linear group). --Tango (talk) 11:55, 1 June 2008 (UTC)[reply]
A group doesn't have "division" - it has multiplication and inverses, which invertible matrices surely do. But what would "A divided by B" be - the matrix X such that BX = A or that XB = A? Put differently, will it be or ? -- Meni Rosenfeld (talk) 11:58, 1 June 2008 (UTC)[reply]
Division is just the name we give to multiplication by an inverse. In a non-commutative situation, you have to specify "on the left" or "on the right", just as you do with multiplication. Using your logic, matrix multiplication isn't defined either, since "the product of A and B" could be either AB or BA - we don't consider that a problem, we just need to be more precise in our notation and terminology. There are notations that work for dividing on different sides, for example you could use / and \, so that A divided by B on the left would be B\A, A divided by B on the right would be A/B. It's generally easier to write B-1A and AB-1, though. --Tango (talk) 12:41, 1 June 2008 (UTC)[reply]
The status of division is different from that of multiplication. Given two matrices A and B, there are just two ways you can multiply them - AB and BA, but four ways to divide them - . Put differently, multiplication is a binary operation on (that is, a function from to ). You need to specify the first operand and the second operand, and that's it. Of course the outcome can change if you switch the operands. "Division" is actually two distinct binary operations - left division and right division. For each of them you need, again, to specify the two operands. In short, there is an operation called "matrix multiplication", but no operation called "matrix division". -- Meni Rosenfeld (talk) 13:07, 1 June 2008 (UTC)[reply]
Ok, so there is no single operation called "matrix division", there is however such a thing is dividing matrices, it just requires a little more information than for multiplying them. --Tango (talk) 14:20, 1 June 2008 (UTC)[reply]
It could be noted that the two division operations for matrices are only formally different in that they provide an isomorphic structure. I mean, every result for left division has a symmetrical result for right division. In more general settings left and right division might act completely differently. GromXXVII (talk) 23:16, 1 June 2008 (UTC)[reply]
Sine Operations
I was wondering: if i was lost in the dessert without a calculator, and my life depended on finding the side opposite to angle b, using the trigonometric functions of sine cosine and tangent, how would i do that? Or, how do you do sine without a calculator? --Xtothe3rd (talk) 17:24, 1 June 2008 (UTC)[reply]
How do you differentiate the factorial function? Also how do you integrate it? I suspect that I am too much of an amateur in maths to understand the second answer because integrals.com couldn't do it but I'll ask anyway. Thanks 92.4.5.56 (talk) 18:29, 1 June 2008 (UTC)[reply]
The factorial function, strictly speaking, is defined only on the natural numbers, so you can't differentiate it. But you can differentiate the gamma function -- start there. --Trovatore (talk) 18:33, 1 June 2008 (UTC)[reply]
June 2
Quasiconvex subgroups of a finitely generated group
I'm trying to show that if G is a finitely generated group, and H is a quasiconvex subgroup of G (that is, there's some natural k such that for any a, b in G, any geodesic γ from a to b in the Cayley graph of G lies within the k-neighbourhood of H), then H is itself finitely generated. I'm having trouble because I don't really know how to show that a subgroup is finitely generated short of coming up with an actual generating set for it, but H seems too arbitrary for an explicit generating set formulation. Google has found me a couple of people asserting the result, but no proofs. Can anybody suggest any approaches to take? Thanks. Maelin (Talk | Contribs) 11:44, 2 June 2008 (UTC)[reply]
Aha! Figured it out with a friend. Thanks to anybody who spent some time thinking about it! Maelin (Talk | Contribs) 03:25, 3 June 2008 (UTC)[reply]
Complex cube root
Real and imaginary part of complex square roots can be expressed as radicals:
We're looking for the real solutions for x and y for given reals a and b so that:
Solution:
Is there a similar representation for the cube root? I derived an equation of degree 12 which looks like it's maybe not impossible to transform it to a quartic equation of x3 or a cubic equation of x4:
It is certainly possible. The derivation is not nice though. Let x be the required root, and z the initial value. Then obviously x^3=z. This expression in polar form after applying demoivre's theorem yields rx^3(cos(3θx)+isin(3θx))=rz(cos(θz)+isin(θz)), where rx is the radius of x, θx is its angle, and respectively rz and θz are the radius and angle of z. Without loss of generality, we can make both rz and rx one. If |z| does not equal 1, then let z'=z/|z| and x'=x/|x|. |x| can easily be solved from |z| and x can be solved from x' and |x|. So, we can find x from x' and |z|. With this, the statement is simply (cos(3θx)+isin(3θx))=(cos(θz)+isin(θz)). Using trig identities, 4cos^3(θx)-3cos(θx)+3isin(θx)-4isin^3(θx)=cos(θz)+isin(θz). Seperating real and imaginary parts, this yields
4cos^3(θx)-3cos(θx)=cos(θz)
3sin(θx)-4sin^3(θx)=sin(θz).
These two expressions are equivalent, so we choose the first one. Let a=cos(θx) and C=cos(θz). We get the expression
4a^3-3a=C, which is solveable through the cubic equation. By backtracking through our process with the found value of a, you can find x', and subsquently x. Indeed123 (talk) 16:24, 2 June 2008 (UTC)[reply]
Thank you very much - that seems to be a much easier way. Icek (talk) 19:53, 2 June 2008 (UTC)[reply]
An easier method would be to just convert the a+ib into for and . Then, where and . Polynomials are overrated. -mattbuck (Talk) 20:07, 2 June 2008 (UTC)[reply]
Well, that's beside the point, I knew that way. Icek (talk) 21:21, 2 June 2008 (UTC)[reply]
A series
Is there a closed formula for ? Thanks, --Dr Zimbu (talk) 15:32, 2 June 2008 (UTC)[reply]
Yes, assuming you meant , it is equal to . -- Meni Rosenfeld (talk) 15:49, 2 June 2008 (UTC)[reply]
3-dimensional geometry
If a pyramid is built with its apex on top of the center of a unit square, what is the relationship between the height of the pyramid and the angle subtended by each side of the square at the apex? --Masatran (talk) 18:38, 2 June 2008 (UTC)[reply]
Call the length of the sides of the base b. Call the apex point A. From A drop a perpendicular to the base and call the intersection point B and the length of the segment h (I assume that's the height you are talking about). Now from B draw a perpendicular to one of the edges of the base and call the point of intersection C (it will be the midpoint of the edge since the base is a square). The distance BC is b/2. ABC forms a right triangle, and we are trying to find the hypotenuse AC. By the pythagorean theorem it is . --Prestidigitator (talk) 01:08, 3 June 2008 (UTC)[reply]
Is there any example of a non-measurable set in the plane whose boundary is a continuous closed curve?--Pokipsy76 (talk) 07:09, 3 June 2008 (UTC)[reply]
Yes. You can choose a Vitali set which is dense in [0,1]. Cross this with the interval, and we get a non-measurable subset of [0,1]x[0,1] whose boundary is the whole square. But [0,1]x[0,1] is [[peano curve|a continuous closed curve]. Algebraist 10:25, 3 June 2008 (UTC)[reply]
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