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.
I'm not exactly asking this because I can't do it, but because someone here has the data to do it more quickly, and s/he may as well have the opportunity to see the answer first.
What is the side-length of the regular octahedron composed of all anthropogenic Carbon currently in the atmosphere transformed into the diamond structure?
You should be made aware that not all functions have a derivative and that differentiability is a local characteristic (i.e., in this case may be differentiable only at zero if the problem has a solution). The problem's statement is not very good, but it does have a solution nevertheless. Use the basic definition of the derivative (at x=0) and see what you get.Julzes (talk) 10:22, 8 January 2010 (UTC)[reply]
You don't have anything to prove. Just by definition of derivative |f(x)|≤x2 means in particular that f is differentiable at 0 and that f'(0)=0.--pma13:16, 8 January 2010 (UTC)[reply]
Of course there is something to prove. Just because it takes about 2 lines, one of which is just stating a definition, doesn't mean it isn't a proof. --Tango (talk) 16:20, 8 January 2010 (UTC)[reply]
Well, there is always a proof, sure. To be precise you just have to recognize " |f(x)|≤x2 " as a particular case (i.e. x0=l=0) of the definition of derivative, that is f(x0+x)=f(x0)+lx+o(x) as x→0. To do this you don't have anything to write. Here the only thing to prove, if anything, is that x→0 as x→0. --pma17:56, 8 January 2010 (UTC)[reply]
I'm not convinced. While what you say is obviously true, I think there is still something to prove in there - I would write down something about A+Bx+o(x)<x2 (as x->0) implying A=B=0, I don't think that would be immediately obvious to someone taking a course at the level where this question would come up. I think an elementary proof directly from the definition (as the limit of a quotient) is preferable. --Tango (talk) 18:52, 8 January 2010 (UTC)[reply]
I'd say it depends on what is the definition of derivative you have given. I personally believe that the right definition of derivative (at every level, even at the level you mention) is not by means of a limit of a quotient, but it is via the first order expansion. There are several reasons of fundamental importance to do that, both from theory and from practice. Therefore, if you have introduced the derivative properly, and with a reasonable comment on the definition, then the OP's statement is really self-evident. --pma23:55, 8 January 2010 (UTC)[reply]
Informally, you can see that f(x) is squeezed between x2 and −x2. If you want a pathological example that is differentiable only at x=0, consider f(x) = x2 if x is rational; −x2 if x is irrational. Gandalf61 (talk) 17:02, 8 January 2010 (UTC)[reply]
So what is ƒ(0) if |ƒ(x)| ≤ x2 ? After you've answered that, you can simplify the numerator of the fraction, and then further simplify the fraction after that, and finally use the inequality |ƒ(x)| ≤ x2 one more time. Michael Hardy (talk) 17:37, 8 January 2010 (UTC)[reply]
Number of partitions of a set in classes of given cardinalities.
Suppose there were 150 people in a building with six rooms, and only 25 people were allowed in each room. How many possible combinations can those people be divided into the rooms? jciindyysgvxc (my contributions) 11:22, 8 January 2010 (UTC)[reply]
See combinations. How many ways are there of picking 25 people from the 150 to go into the first room? How many ways of picking 25 from the remaining 125 to go into the second room? What should be done with the two numbers obtained? Proceed likewise to get the answer.→86.160.55.155 (talk) 12:21, 8 January 2010 (UTC)[reply]
Alternatively, how many of the 150! permutations of the 150 people leave each person in the same room? Then you have to divide 150! by this number. The general problem is the same. I put a more specific header.--pma13:06, 8 January 2010 (UTC)[reply]
If you regard the rooms as distinguishable, then you'll get a bigger number than if you regard them as indistinguishable. It would be 6! = 720 times as big. More later.... Michael Hardy (talk) 17:28, 8 January 2010 (UTC)[reply]
You'll find in some very elementary books the statement that the number above is the number of distinguishable "words" that can be made from 25 indistinguishable "A"s, 25 indistinguishable "B"s, etc. through the sixth letter. Now let the letters A–F correspond to the six rooms. Michael Hardy (talk) 17:33, 8 January 2010 (UTC)[reply]
Letter grid problem
Here's a simple example of a more general problem which I'm considering without success - suppose that distinct letters are at the corners of a square, and that 4-letter words are formed by traversing each possible route linking all corners without repetition. There are two cases
(a) only horizontal and vertical connections are possible, giving 8 words
(b) diagonal connections too are possible, giving 24 words.
What I want is the number of possible complete words from the mn letters put on an mXn grid (m,n>1), for linking rules (a) and (b). Is this a standard result?→86.160.55.155 (talk) 12:12, 8 January 2010 (UTC)[reply]
I don't know the answer to your question, but I can phrase it in different terms that might lead you somewhere. In case (a), you're looking for the number of (directed) Hamiltonian paths in a square grid graph. Case (b) is similar, but the graph includes diagonal edges too. —Bkell (talk) 13:16, 8 January 2010 (UTC)[reply]
As a further example, consider a grid of 2 rows and 4 columns, with the points in the top row A, B, C and D consecutively from L to R and in the bottom row E, F, G and H consecutively from L to R. Then AEFBCDHG would be a valid word in the case of either linking rule, but AEFBCHDG would be valid only with the permitted diagonal linking of case (b). If x(m,n) is the number of words with m rows and n columns without diagonal linking, it's easy to show, I think, that x(n,m)=x(m,n), x(1,n)=2, x(2,2)=8, x(2,3)=16, x(2,4)=28 and x(3,3)=56. I'm guessing that a recurrence relation might give general x(m,n), but I can't reliably count the number of words in larger grids so as to determine its form. Hence my question as to whether this is a standard result. If y(m,n) is the corresponding number of words with diagonal linking possible, y(n,m)=y(m,n), y(1,n)=2, y(m,n)>x(m,n) for m,n>1. Again, is there a known result?
The stimulus for this query was a common kind of letter puzzle, to find the 9-letter word in a 3X3 grid with diagonal linking possible, where I wondered how many possibilities there were. A particular one had the letters FUN, LET and GEC in consecutive rows (not all letters different), where you'll find GENUFLECT without too much trouble.→86.160.55.155 (talk) 16:10, 8 January 2010 (UTC)[reply]
I still don't know the answer, but I have a hunch that a general recurrence relation, if one exists, is not going to be pretty. Why do I say this? Well, if you start with a Hamiltonian path for a grid graph, and you slice off the last column (say), usually you will not get a Hamiltonian path for the smaller grid graph, but rather a collection of paths that together cover all the vertices. The "interaction" of such paths may be complicated, and whether such a collection of paths can be extended to a Hamiltonian path for the original grid graph may depend on what happens in the very first column of the grid.
Now, having said that, it is clear that the 1×n case is trivial, and you may be able to work out recurrence relations for the 2×n and 3×n cases, for example, but as the shorter dimension increases my intuition tells me that the number of special cases you will have to consider will grow quickly. Of course, I could very well be wrong. —Bkell (talk) 21:54, 8 January 2010 (UTC)[reply]
2xn seems easy (isn't that a Hamiltonian path is determined by its endpoints?). Anyway, it seems the general problem has been studied. If you state the question as you suggested, "number of Hamiltonian paths in a rectangular grid" and plug it into Google, you get a number of interesting results and articles; there is also something in the On Line Encyclopedia of Integes Sequences: the case nxn is [1] --pma23:18, 8 January 2010 (UTC)[reply]
Limits
Is there a number such that
exists? If so, find the value of a and the limit.
No idea how to approach this problem; although I guess solving for would be a start, if I could set it up for that. Any ideas? (ಠ_ಠ) - 17:26, 8 January 2010 (UTC)
What is the value of the denominator at that point? If we assume the limit exists, what does that say about the numerator? --Tardis (talk) 17:54, 8 January 2010 (UTC)[reply]
You want to be able to factor x+2 out of the numerator, in order to cancel with x+2 in the denominator. That is, you want the numerator to evaluate to 0 when x=-2. --COVIZAPIBETEFOKY (talk) 17:55, 8 January 2010 (UTC)[reply]
Okay, that's fine then - , which gives . Although I don't quite understand the motivation. As by Michael Hardy's post below, why won't the limit exist if the numerator is not 0 when x = -2?(ಠ_ಠ) - 18:20, 8 January 2010 (UTC) —Preceding unsigned comment added by Damien Karras (talk • contribs)
Again, what is the value of the denominator there? What does any fraction with such a denominator do at that point if its numerator remains non-zero? --Tardis (talk) 18:35, 8 January 2010 (UTC)[reply]
Well, around the point, I assume it would shoot off to infinity. I'm not sure. There are many functions that are just discontinuous at a single point. (ಠ_ಠ) - 19:01, 8 January 2010 (UTC)
The denominator f(x) = x2 + x - 2 is a continuous function, so not only is f(x) = 0 when x = -2, but as x gets near -2, f(x) gets near zero. When the denominator gets small that means the fraction gets large. When x approaches -2, f(x) gets arbitrarily small, which mean the fraction gets arbitrarily large (i.e. approaches infinity), unless the numerator does something to counter that growth. You're looking to find the value of a that let's the numerator do that. Rckrone (talk) 19:50, 8 January 2010 (UTC)[reply]
Solving WHAT for a??
The limit won't exist if the numerator is not 0 when x = −2. So plug −2 in for x in the numerator, set that equal to 0, and solve that for a. Whether the limit exists then is a question of what happens when the resulting polynomial is divided by (x + 2). Michael Hardy (talk) 18:04, 8 January 2010 (UTC)[reply]
No, that's not correct. In a more general case somewhat like this, whether a limit exists at x=-2 depends on whether x=-2 is a multiple root. If not, there is a limit regardless of the denominator. Generally, there is a limit if the multiplicity of the root in the numerator is not lower than the multiplicity in the denominator.Julzes (talk) 23:12, 8 January 2010 (UTC)[reply]
No, sorry, it is correct, and it is the efficient way to solve the given problem (nobody is speaking of more general or different cases: we are speaking of this case).--pma23:28, 8 January 2010 (UTC)[reply]
That was a response to Michael Hardy's last sentence. In the specific case, once you divide out all of the factors of (x+2) from the denominator, you can be certain there is a limit.Julzes (talk) 00:10, 9 January 2010 (UTC)[reply]
Although, we should check that the denominator doesn't have x=-2 as a repeated root, since if it does the numerator would have to be equal to the denominator (up to multiplication by a constant) for there to be a limit. That doesn't take long, though, since the denominator is quadratic so the only way it could have a repeated root is if it is a perfect square, which it clearly isn't. --Tango (talk) 00:01, 9 January 2010 (UTC)[reply]
Anyway I'd follow MH's directions (find the proper value of a that makes the numerator vanish; consider the fraction only for that value of a --the fraction should be 3 + 12/(x-1); compute the limit); there's nothing to check. --pma00:36, 9 January 2010 (UTC)[reply]
January 9
Arrow's impossibility theorem and real elections
Arrow's impossibility theorem says that given three or more candidates, we can't convert a list of individual rankings into an aggregate ranking that satisfies certain criteria. But in real elections, we don't need a complete aggregate ranking, just a choice of the most preferred candidate (we don't have to rank the losers against each other). How much of the theorem still applies? NeonMerlin06:31, 9 January 2010 (UTC)[reply]
Interestingly enough, Gibbard's version linked voting schemes with social preferences, and applied Arrow's Impossibility theorem to derive his result. Pallida Mors19:09, 9 January 2010 (UTC)[reply]
It is also noteworthy the fact that normal voting procedures you are mentioning are just one set of possible voting schemes: the outcome they produce is formed using as informational input just the tops of individual preferences; since more general schemes are manipulable, that is of course true especially of traditional majority voting, under which many individuals are tempted to vote (i.e. declare the top of their preferences) for a second choice whenever their first choice is believed to have no real electoral chance. Pallida Mors02:08, 10 January 2010 (UTC)[reply]
Actually, I'm also talking about alternative systems that take a complete ranking of preferences from each voter but output only a first choice. NeonMerlin08:24, 10 January 2010 (UTC)[reply]
Is there a known counterexample to the conjecture that 1001 is the only product of consecutive primes that is one more than a perfect power?
If not, is there a heuristic estimate of the probability that the conjecture is true? (I assume it has not been proven, though someone may set me straight on this.)Julzes (talk) 06:48, 9 January 2010 (UTC)[reply]
I have noticed a worrying trend of this sort of behaviour on Wikipedia reference desks. There are often many ways in which one may "misinterpret" a question in order to respond, but it is also important to note that which the OP desires. In this case, you have answered the question to some extent which is appreciated, but you should bear in mind that such answers will have little effect on the questioner unless he/she has not even thought about his/her own question! Of course, your contributions are much appreciated, but you should let others answer a question if you cannot respond with a non-trivial answer and know that your answer is trivial (if you truly thought that your answer shed light on the question, then there is not a problem; if you knew that your answer did not, but included it anyway, then you should have let others respond instead of frustrate the OP with trivial answers). This is merely my opinion, however, and you may ignore it if you wish. --PST14:23, 9 January 2010 (UTC)[reply]
It's not always perfectly crystal clear what exactly it is that an OP intends to ask, or how much effort he has already put into the question. Sometimes an OP himself is confused about what he is after. Pointing out trivial cases can help an OP focus the question, which benefits other responders as well.
I agree with that myself, but in my case isn't 'consecutive' implying something pretty clearly. Is anybody going to say what's known? Let me Google it to see if I've made a booboo.Julzes (talk) 17:37, 9 January 2010 (UTC)[reply]
But the question, in this case, was crystal clear! The phrase "product of consecutive primes" is pretty clear. Consecutive means to follow, to be in succession[2], so the empty product and the "silly counterexamples" given by Bkell above clearly don't match the conditions given by the OP. The only problem I have is as to how many consecutive primes we allow. Is it the product of two, three, four, …, or arbitrarily many consecutive prime numbers? ~~ Dr Dec(Talk) ~~18:25, 9 January 2010 (UTC)[reply]
I posted my first comment after just a little bit of thinking about the problem, thinking perhaps that it was the only kind of trivial counterexample, with the idea that often trivial counterexamples are easy to overlook but can shed light on the original problem. Then shortly after I realized that there were lots of silly counterexamples in which only one prime is "multiplied"; since I had already posted one silly counterexample, I figured I should mention the others as well. I was and am fully aware that this is not what Julzes was asking. I was attempting to post my thoughts, the idea being that other contributors could join in and we could make some progress on the question by working together (as I was also doing in my answer to the letter grid problem). I was not aware that the mathematics reference desk is only to be used for complete, final answers that spring fully formed from the minds of great mathematicians. I apologize for my problem-solving technique, in which sometimes I investigate silly cases in an attempt to understand how the problem behaves. It won't happen again. I'm sure one of you would have posted a complete "non-trivial answer" to this question by now if I hadn't screwed everything up by posting a couple of innocuous, true, but trivial facts. —Bkell (talk) 20:05, 9 January 2010 (UTC)[reply]
Bkell: I sincerely apologize if I have upset you with my comment; it was a suggestion that people could consider, rather than a criticizm of your contributions. I firmly believe that you are an excellent contributor to the reference desks; it was not my intention to represent you otherwise. In fact, I will be the first to admit that I posted nothing in the direction of the question, so of course I should be criticized if you are. However, please do note that my comment was merely a suggestion that you need not follow, although it does have a purpose. In some cases, when people observe that a question has already been answered, they feel that there is no furthur need to respond (some merely glance over questions and answers, therefore, without carefully examining the content). Therefore, it may well be that no one responds furthur to Julzes' question, thinking that your answer is complete without examining it thoroughly. Secondly, I just feel that, given the sorts of questions Julzes has asked in the past, it is unlikely that he had not even thought about his question (in fact, the wording of his question suggests that he has thought about the trivial cases already). However, I did not intend to fuel a heated debate regarding your post, and for that I shall apologize. --PST03:08, 10 January 2010 (UTC)[reply]
Arbitrarily many. That makes for less likelihood that the conjecture is correct, but I doubt it makes that much difference except for the estimate I was seeking. There is nothing in Unsolved Problems in Number Theory (2nd ed.), and I couldn't find anything on Google. I'm deletingI've deleted what's below this as out of sequence and irrelevant.Julzes (talk) 19:27, 9 January 2010 (UTC)[reply]
Out of interest, I searched for small (less than 1020) instances of power + 1 that are the product of distinct primes close together. Apart from 1001, closest I could find are:
Those factorisations don't even contain a pair of consecutive primes, so this suggests that there aren't any small counterexamples. Gandalf61 (talk) 10:06, 10 January 2010 (UTC)[reply]
Interesting. Less than 1020? And these 8-digit numbers have the closest factors? That's acording to their relative orderings, I assume. Anyway, that's what I would choose. My general intuition is and has been that this is an essentially unprovable fact, but now you've actually told me something that strikes me as peculiar. It seems certain to me that whatever measure of closeness you chose, this is a little strange. Hmmm. Thanks for filling me in. Now I'm more curious about this basically useless subject.Julzes (talk) 10:36, 10 January 2010 (UTC)[reply]
Oops, missed out an even smaller example with close but non-consecutive prime factors:
Also found some examples of power+1 with close (but again non-consecutive) pairs of prime factors, although the separate pairs are not close to each other:
Of course, I knew about 1729. So your measure is then absolute ordering. I'm going to have to look at this question of close-but-no-cigar myself some time. There is one rather strange thing about your whole collection, as you've related it. I can't say off the top of my head about the factors of 86350889, but the rest all look like pairs of almost-consecutive primes. Well, enjoy the strange task I set you on. I'm going to sleep.Julzes (talk) 11:41, 10 January 2010 (UTC)[reply]
I have a related very strong conjecture now: Among all n>1, the ratio of the largest and smallest prime factors of np+1 for a fixed prime p>3 is smallest for n<5 except for an isolated exception at p=43 and n=7.Julzes (talk) 13:29, 11 January 2010 (UTC)[reply]
I'll also make the conjecture that the record-setters for the ratio between largest and smallest prime factors (excluding the cases where they are equal) of squares plus one and cubes plus one always have two and three distinct prime factors, respectively, and that there is not a last record-setter in either case.Julzes (talk) 15:53, 11 January 2010 (UTC)[reply]
Here's more base-ten stuff: The record-setters for the square case appear to always be round numbers with the exception of one case after a short start in which it's not true. The single biggest example I have is Julzes (talk) 12:39, 12 January 2010 (UTC)[reply]
PrimeHunter has given an explanation that n=50m^2 is the form that produces a lot of the record-setters (at his talk page). Getting an approximate probability that this is always true will be a task for me at some point in the next couple of weeks.Julzes (talk) 16:04, 12 January 2010 (UTC)[reply]
I pointed out that (50m2)2+1 = (50m2-10m+1) × (50m2+10m+1) gives a low ratio (50m2+10m+1)/(50m2-10m+1) when both factors are prime, and that probably happens infinitely many times. It happens for m = 152129×10500 with two 1013-digit primes found and proved by PrimeForm/GW. The ratio is approximately 1 + 2.6×10-506. PrimeHunter (talk) 17:23, 12 January 2010 (UTC)[reply]
The record-setters and the cases where the two factors are prime coincide after a short run, unless there is a semiprime with very close factors by chance for a different type of n. It has only been remarked on in an edit summary, so I'll repeat it here the last case is likely to be one giving the factorization 1973*1993^2.Julzes (talk) 17:50, 12 January 2010 (UTC)[reply]
I did a little work on this on paper. What I came up with is:
If the largest prime factor of 2s2+2s+1, p, and its cofactor, k, satisfy 2s+k2+1<p, and if p-2*(2s-k+1) is prime; then n=p-(2s+1) outperforms all smaller n of the form 50m2.
The question's formulation seems to me to require improvement (randomness from a countable set), but it is clear what is meant. I doubt that the result I got can be simplified (unlike the question with just two numbers), but it can be computed. What one should do to get what I have is consider divisibility by each prime as independent from divisibility by each other prime. Then you should be able to get what I have. Unless someone beats me to it, I'll get back to you with a numerical value (which I'm certain is already in print somewhere and is easy to generate).Julzes (talk) 11:20, 9 January 2010 (UTC)[reply]
Assuming I have interpreted the question correctly, the probability equals the reciprocal of Apéry's constant. In general, the probability of k randomly chosen integers being coprime is where is the Riemann zeta function. I quote a relevant discussion from one of our articles below:
Given two randomly chosen integers and , it is reasonable to ask how likely it is that and are coprime. In this determination, it is convenient to use the characterization that and are coprime if and only if no prime number divides both of them (see Fundamental theorem of arithmetic).
Intuitively, the probability that any number is divisible by a prime (or any integer), is (for example, every 7th integer is divisible by 7.) Hence the probability that two numbers are both divisible by this prime is , and the probability that at least one of them is not is . Now, for distinct primes, these divisibility events are mutually independent. (This would not, in general, be true if they were not prime.) (For the case of two events: A number is divisible by p and q if and only if it is divisible by pq; the latter has probability 1/pq.) Thus the probability that two numbers are coprime is given by a product over all primes,
≈ 0.607927102 ≈ 61%.
Here ζ refers to the Riemann zeta function, the identity relating the product over primes to ζ(2) is an example of an Euler product, and the evaluation of ζ(2) as π2/6 is the Basel problem, solved by Leonhard Euler in 1735. In general, the probability of randomly chosen integers being coprime is .
Unfortunately, the exact value of the probability of which you inquire is not known; an approximate numerical value is which is approximately 83.2%. --PST11:39, 9 January 2010 (UTC)[reply]
What you've given looks like the probability that there will be no prime dividing all three, while the questioner asked about pairwise coprimality (no prime divides more than one). I can't TeXify it yet, but that's the product of .Julzes (talk) 11:58, 9 January 2010 (UTC)[reply]
You are right; I did not read the OP's question carefully enough (I missed the "pairwise" bit). Thanks. --PST12:04, 9 January 2010 (UTC)[reply]
The probability of picking three integers between 1 and N at random being pairwise coprime goes to 0 as N increases toward infinity. Proof?Julzes (talk) 12:20, 9 January 2010 (UTC)[reply]
I'd have thought it would be more like the cube of the chance of two numbers being coprime considering the 3 possibilities of being coprime, of course that leaves a lot out! Dmcq (talk) 12:57, 9 January 2010 (UTC)[reply]
If the density (=limit probability) of the set of triples of pairwise coprime natural numbers is what you suggested above, that is that I trust very much, then it can't be 0. An infinite product of non-vanishing terms of the form with always converges to a non-zero limit. --pma15:33, 9 January 2010 (UTC)[reply]
And, since for all prime it is true that , Julzes' formula is also consistent with Dmcq's bound, for --pma16:20, 9 January 2010 (UTC)[reply]
Okay, I figured out what I was expecting from PARI/GP wasn't working and, more or less why. The result I got from the first 500000 primes using rounding at twenty digits is .Julzes (talk) 17:31, 9 January 2010 (UTC)[reply]
I agree with your product formula (the finite probabilities indeed do converge to the infinite product by a dominated convergence argument). The Plouffe's inverter [3] seems to have nothing special about the value 0.286747.., but Google gives [4] and this [5] (pag.11). But there should be for sure some more classical result. --pma18:27, 9 January 2010 (UTC)[reply]
So, there is (almost certainly) a closed form given at the bottom of that first Google result. I imagine the other--later--source says more (but I can't get it right now).Julzes (talk) 18:55, 9 January 2010 (UTC)[reply]
But the value of that expression is 0.286747419846(..), possibly found by an inverse calculator; it's close but seems different from the value you computed. An interesting remark there, is that the product formula easily generalizes to the case of n numbers. The other paper has no relevant information.--pma20:37, 9 January 2010 (UTC)[reply]
I'm a little skeptical of both my precision and the idea that a nice coincidence of something looking reasonable times Apery's constant was found by inverse calculator in this instance. I'm going to explore the question more precisely.Julzes (talk) 03:54, 10 January 2010 (UTC)[reply]
Oh, I don't know. It looks to be moving solidly toward a higher degree of precision with more factors included, but the source already said it was close and it doesn't look likely to be true on paper. I'm inclined to reverse any skepticism I had that they were merely close. It probably was done by reverse calculator. It's simple, but doesn't look simple enough.Julzes (talk) 06:27, 10 January 2010 (UTC)[reply]
The same counting argument that gives Julzes' product formula applies; as the first link above remarks, for (n+1) pairwise coprime numbers it's (note: We should speak of natural density rather than probability, for there is no uniform probability on ).--pma13:04, 10 January 2010 (UTC)[reply]
I would like to know whether value of pi could be calculated in this way.
First let us assume, there are pi for all shapes like, regular triangle (3_pi), square (4_pi), pentagon (5_pi), hexagon (6_pi), .. n-gon (n_pi) and, these values can be calculated from their perimeter,
2 * 3_pi * r = 3 * a
2 * 4_pi * r = 4 * a
2 * 5_pi * r = 5 * a
2 * 6_pi * r = 6 * a
2 * 7_pi * r = 7 * a
2 * 8_pi * r = 8 * a
2 * 9_pi * r = 9 * a
....................
....................
....................
2 * n_pi * r = n * a
Let,
r be the radius of circle you draw through all the corners of the regular polygons mentioned above
a be one side of a regular polygon (this can be ignored, it is equal to 1, for simplicity)
As we know values of a, and r, we can easily calculate corresponding 'n_pi' values.
Now by considering a circle as a regular polygon, with infinitly big number of sides, Is it possible to extrapolate the data what we got from first few polygons (3_pi, 4_pi, etc.), and calculate n_pi ?!
I think the problem is you'll just end up proving π == π. I.e. the formula for the boundary is N tan (π / N) for a regular polygon with N sides around a circle radius 1. As N -> ∞, π / N -> 0 and tan (π / N) -> π / N. So the length of the boundary tends to π. --JohnBlackburnewordsdeeds16:03, 9 January 2010 (UTC)[reply]
Rather than a limiting process, he found bounds by inscribing and circumscribing a polygon.
Rather than considering a general n (which, as JohnBlackburne points out, leads nowhere), he considered , starting with known values and applying the ancient equivalent of the half-angle trigonometric formulae. Once the geometric argument is extracted into pure arithmetic, we end up with a simple calculation involving square roots.
For 4 numbers, is it ALWAYS true that the 2nd largest number lies in 1 standard deviation? Mathematically, for 4 numbers a, b, c and d where a>b>c>d, is b≤μ+σ ALWAYS true? —Preceding unsigned comment added by 219.79.198.73 (talk) 15:43, 9 January 2010 (UTC)[reply]
No, I think the OP has it right. Your inequality is obviously false in general, consider the set {1,5,5,9}. --Tango (talk) 17:07, 9 January 2010 (UTC) You changed it while I typing! --Tango (talk) 17:09, 9 January 2010 (UTC)[reply]
To me the phrase "2nd largest number lies in 1 standard deviation" looks as if it means c ≤ μ + σ. That's a somewhat subtler question. It's obvious that b and c cannot both lie outside of the interval [μ − σ, μ + σ], since then the average of the squares of deviations would be smaller than all four of the squares of deviations—everyone would be above average. So at least one of the middle numbers lies within the interval. The question is whether both of them must. Michael Hardy (talk) 03:10, 13 January 2010 (UTC)[reply]
No, b is the second largest number, and it must have , by the proof I outlined. c is smaller so obviously . By negating the numbers you get . So . -- Meni Rosenfeld (talk) 09:30, 13 January 2010 (UTC)[reply]
The target audience here is unclear. If you understand what "without loss of generality" means, you don't need to have the proof spelled out this way. -- Meni Rosenfeld (talk) 11:46, 13 January 2010 (UTC)[reply]
I just thought if Michael Hardy was asking a question about your hint then it wasn't obvious and deserved a full answer. And it isn't actually immediately obvious how to go about it, I guess I did put in too many steps, it takes more time and effort to make things shorter and simpler and it seemed adequate. Dmcq (talk) 13:04, 13 January 2010 (UTC)[reply]
The mean of a and b is more than 2σ from the mean of c and d
Therefore the total variance var(a,b)+var(c,d)+4((mean(a,b)−mean(c,d))/2)2 is greater than 4σ2, which is a contradiction. QED or whatever Dmcq (talk) 13:23, 13 January 2010 (UTC)[reply]
I worked this out by a brute-force method involving trivial but opaque algebra and thought there must also be an intelligent way to do it. Now I know what that is and it seems as if I should have seen this right away. Maybe this is what Dmcq had in mind. Split {a,b,c,d} into an upper part {a,b} and a lower part {c,d}. The average of the upper part is
r = (a + b)/2
and the average of the lower part is
s = (c + d)/2.
The mean μ of {r,s} is the same as the mean of {a,b,c,d}. But the standard deviation τ of {r,s} is smaller than the standard deviation σ of {a,b,c,d}, since the variance of {r,s} doesn't take into account the variability within the upper part or within the lower part (see law of total variance). Now observe that the interval whose endpoints are μ ± τ is just the interval [s, r], and that interval contains the two middle values b and c. Therefore the longer interval whose endpoints are μ ± σ must also contain those two middle values. Michael Hardy (talk) 04:38, 15 January 2010 (UTC)[reply]
So by "all possible products" you mean products of two terms a, b where 1 ≤ a, b ≤ n and ab and ba are counted separately if a ≠ b. That looks like (1 + 2 + 3 + ... + n)2 to me. Gandalf61 (talk) 16:18, 10 January 2010 (UTC)[reply]
No. Imagine the following grid is made up of blocks one unit thick:
-+---+-------+-----------+---------------+
^| | | | |
|| | | | |
|| | | | |
4| O | M | K | J |
|| | | | |
|| | | | |
V| | | | |
-+---+-------+-----------+---------------+
^| | | | |
|| | | | |
3| H | F | E | P |
|| | | | |
V| | | | |
-+---+-------+-----------+---------------+
^| | | | |
2| C | B | I | N |
V| | | | |
-+---+-------+-----------+---------------+
1| A | D | G | L |
-+---+-------+-----------+---------------+
|<1>|<- 2 ->|<--- 3 --->|<----- 4 ----->|
The volume of the whole thing is . On the other hand, A is a 1×1×1 cube; using B as one layer, and C & D put together as the second layer, we can make a 2×2×2 cube; using E, F & G, and H & I we can form a 3×3×3 cube; and from J, K & L, M & N, and O & P we can form a 4×4×4 cube. So the total volume is also . (My officemate Derek Boeckner and I decided one day to figure out geometrically why the sum of the first n cubes is the square of the sum of the first n integers, and this is the picture we came up with.) —Bkell (talk) 18:32, 10 January 2010 (UTC)[reply]
Another construction for the sum of the first n squares is presented as a marginal note in Infinitesimal Calculus by Henle and Kleinberg, ISBN 0486428869. Part of the construction is used as the illustration on the front cover. It packs six copies of the first n squares together to form a rectangular prism measuring . —Bkell (talk) 20:25, 10 January 2010 (UTC)[reply]
I guess the construction in Infinitesimal Calculus is pretty similar to what you described. For some reason I thought you had used the formula for the volume of a pyramid, but that was a different discussion. —Bkell (talk) 20:48, 10 January 2010 (UTC)[reply]
That page doesn't give a true closed form. The closest thing to that is the integral, which doesn't exactly count, since calculating the integral requires manually adding up all the terms 1/1+1/2+1/3+...+1/n. --75.41.186.123 (talk) 16:49, 10 January 2010 (UTC)[reply]
The sum doesn't have a closed form in terms of elementary functions.
The page does give a good way to calculate it approximately,
A trapezoid(Assuming that you mean the shape with two parallel sides, called a trapezoid in the USA) If you know the lengths of the parallel sides, the height, and the two angles at A and B, then the perimeter is just a + b + h(cosec + cosecB). As Bkell says, for other possibilities, you can often use "Pythagoras" or Trigonometry if you have enough information. (Trivially, if you know the lengths of all four sides, just add them together!) Dbfirs10:02, 11 January 2010 (UTC)[reply]
Smoothness Scale In Number Theory
Have there been any reasonably good attempts to refine the idea of relative smoothness of numbers? I'm looking for something like a scale that defines powers of 2 as 1 and primes as 0 (or the reverse), something that gives a pretty good--and increasingly better with size--relative smoothness by some reasonable consideration of it. Is the best out there what's essentially already in the smooth number article?Julzes (talk) 14:37, 11 January 2010 (UTC)[reply]
January 12
Normal Closures Qual. Exam
I'm stuck on the problem: Given a non-normal separable extension [E: F] = 4, bound the degree [K: F] of the
normal closure of E. [Bergman] from a grad. qual. exam http://www.math.harvard.edu/graduate/quals/topics/galois.pdf. Please help! I think that [K: F] is bounded by 8 by testing a couple of examples like F = Q, E = Q[a] where a4 = 2 where the minimal polynomial of a over Q has a splitting field of degree 8 and other examples. Obviously [K: F] is bounded by 24, but I'm not sure that's what the question is after (I don't even know whether 24 is a strict bound, that you can get a normal closure of degree 24 in the problem). I haven't much experience with normal closures, but I don't believe I require it for this question. The other qual. exam problems in the link are dead easy and that's why this one question is bugging me. Thanks. —Preceding unsigned comment added by 122.109.42.56 (talk) 06:25, 12 January 2010 (UTC)[reply]
Can you show me how [E: F] = 4 can exist with [K: F] = 24? Thanks for your answer but I can't think of an explicit example of how the degree could get as high as 24. Why I'm unsure that the degree is 24, is because of the specificity of the problem ([E: F] = 4, and not [E: F] = n for any n), and that it's a qual. problem.—Preceding unsigned comment added by 122.109.42.56 (talk) 11:15, 12 January 2010 (UTC)[reply]
Let (the field of rational functions in 4 variables, that is, the xi are algebraically independent). Any induces an automorphism of K by permuting the corresponding variables. Let F be the fixed field of all these automorphisms, and let E = F(x0). The minimal polynomial of x0 over F is , hence [E : F] = 4. The normal closure of E (i.e., the splitting field of f over F) is K, and the construction ensures Gal(K/F) = Sym4, hence [K : F] = 24. Of course, the same argument works for any n, there's nothing special about 4. — EmilJ.11:30, 12 January 2010 (UTC)[reply]
Maybe I should also stress that this F can be easily described explicitly, namely it is the field generated by the elementary symmetric polynomials:
Thanks muchly so F can be chosen as the fixed field of that Galois group. Just to investigate furthur suppose F = Q. Is it then true that the bound of 8 is optimal? I've been basically looking at the problem for F = Q so that's probably why I wasn't thinking clearly but in the case F = Q my bound seems to work. Thanks.—Preceding unsigned comment added by 122.109.42.56 (talk) 11:58, 12 January 2010 (UTC)[reply]
Please, sign your posts, and don't remove signatures supplied by others without adequate replacement. This is a basic rule of conduct here.
No, you can get degree 24 even for F = Q. The problem is equivalent to asking for a normal extension K of Q such that Gal(K/Q) = Sym4. This is a particular instance of the inverse Galois problem. While the general problem for arbitrary finite groups remains unsolved, many special cases are known, and in particular, one can always find such an extension for any Symn (the argument is described at Inverse Galois problem#Symmetric and alternating groups; your polynomial is x4 − t(x + 1) for a suitably chosen rational number t). — EmilJ.13:35, 12 January 2010 (UTC)[reply]
If you're concerned about your IP address being visible, don't worry (or worry more, depending on how you look at it). Your IP address is publicly available in the history page (and this is common knowledge, so don't accuse me of further compromising your privacy). -- Meni Rosenfeld (talk) 15:33, 12 January 2010 (UTC)[reply]
Sorry about that, I was just worried that people may find me out. I didn't know you were adding the signature, so I thought it was alright to remove it. Thanks for the help. —Preceding unsigned comment added by 122.109.42.151 (talk) 02:15, 13 January 2010 (UTC)[reply]
solution of differential equation of motion of a quarter car
The only thing you can look at by only looking at a quarter of a car is the suspension of one wheel, i.e. disregarding all effects (lateral, turning, anti-roll bar, drag etc.) which require knowledge of the other wheels or the body. The page Vehicle dynamics links to Tuned mass damper which may be what you need.--JohnBlackburnewordsdeeds18:45, 12 January 2010 (UTC)[reply]
I want to show that for , and , f is locally invertible around each point of C except (0,0) and , i.e. show that there are open sets U and V about any in C (but for the aforementioned 2 points) such that f maps bijectively: UV. Could anyone give me a hand? I've tried saying f(a,b)=f(c,d), so a=c, and then if b is not equal to d, we get 3a=, and I'm not sure where to go from here or whether I am indeed going in the right direction. I haven't actually used the fact that we're local to points where the second coordinate of f(x,y) is 0, and I can't really see what makes the above 2 points special either. Could anyone give me a hand finishing this off?
I also want to find the derivative of the inverse function (locally) - what would be the neatest way to find this?
Jacobian matrix and determinant might be a good place to start. Basically the Jacobian gives you the linear transformation that your function looks like locally. The two points you mentioned are the points on C where the Jacobian determinant is zero. The derivative of the inverse function at a point will be the inverse of the Jacobian there. Note that won't work when the determinant is zero. Rckrone (talk) 21:58, 12 January 2010 (UTC)[reply]
Just want to add, if the Jacobian determinant is zero that doesn't necessarily imply a problem, but those are the places where things can potentially go wrong. Rckrone (talk) 00:16, 13 January 2010 (UTC)[reply]
To be precise, if the Jacobian is zero at a point, the map is not locally invertible there in the C1 sense, that is with a C1 local inverse. So, to give a complete answer to the question as you stated it, now you just have to look more closely at the two points where the Jacobian determinant vanishes (as Rckrone sais, the Jacobian determinant being zero doesn't necessarily imply that the map is not locally bijective there). For instance, the map is not locally bijective at (0,0), because e.g. for any 0 < ε < 1 there are at least two distinct solutions (x1,y1) and (x2,y2) of f(x,y) = (ε,0) close to (0,0), namely with x1 = x2 = ε and 0 < y1 < ε1/2 < y2 < 2ε1/4 (just look at the sign of y3 - 3εy + ε3). Can you see how to do the other point? (It is also possible a more detailed geometric description of the local structure of C at these points, of course) --pma08:24, 13 January 2010 (UTC)[reply]
For instance 2.68:1 ratio, if I want to round to nearst 0-5-0 should 2.68 round up to 3 or it should round to 2.5 if I'm douing halfway rounding scale. For 3.04:1 to estimating sentence is it could be cite as more than 3 times xx and vice versa, or 3.04 is usually said like about 3 times the xx vice versa since 3.04:1 is only 4 cents past 3?--209.129.85.4 (talk) 20:26, 12 January 2010 (UTC)[reply]
It's unclear to me what you are asking; maybe you will find something helpful at rounding. If I were to choose between rounding 2.68 to either 2.5 or to 3, I would likely choose to round to the nearer number. As for your second question, it is correct to say that 3.04 is more than 3 and it is also correct to say that 3.04 is about 3, so either sentence is correct. Eric. 131.215.159.171 (talk) 09:14, 13 January 2010 (UTC)[reply]
I was also unclear about what you were asking. It is not usual to round to the nearest half, but if you really need to do this, then any value more than .00 and less than .25 gets rounded down, and so does any value more than .50 and less than .75 For values from .25 up to .50, and values from .75 up to the next whole number, the convention is to round up. Dbfirs20:01, 13 January 2010 (UTC)[reply]
January 13
Cube root function on the space Mn of matrices
Hi all:
I've been asked by my supervisor to investigate the possibility of the existence of a continuous square root function in some open ball about the identity matrix in ; that is, a continuous function g, with for all matrices A in this open ball about I. Now I've used the inverse function theorem and the fact that is continuously differentiable about I to prove that the inverse of f exists and is continuous in some open ball about I (an I-ball, as it were - oh dear...), but I also managed to prove that you can't extend such a function to the whole space , because a matrix like, for example,
[0,1
0,0]
has no square root (I don't know how to LaTeX a matrix, sorry!). However, I couldn't find any such matrix which has no cube root: does there exist then a continuous 'cube root' function which extends to the whole of ? (I presume that every matrix does have a cube root, and so a discontinuous function would be fairly trivial by defining it pointwise, right?) My intuition tells me that no such function should exist - but how would I go about proving it? If not, why am I wrong?
The claim that every matrix has a cube root sounds highly unlikely, and in fact I think that
is a counterexample. Let us assume that B3 = A, and write B = P−1CP, where P is invertible, and C is in Jordan normal form. Note that A = P−1C3P. If J is a Jordan block of C with eigenvalue λ, then J3 appears as a block of C3, and it is a triangular matrix with λ3 on the diagonal, hence λ3 is an eigenvalue of A. However, the only eigenvalue of A is 0, hence λ = 0. Thus C is a triangular matrix with 0 on the diagonal, hence C3 is a triangular matrix whose main diagonal as well as two adjacent shifted diagonals are zero. As it is a 3-by-3 matrix, it is in fact the zero matrix, thus A = P−1C3P = 0, a contradiction. — EmilJ.14:48, 13 January 2010 (UTC)[reply]
Actually, now that I think about it, your matrix should have no cube root either, by pretty much the same argument. — EmilJ.14:55, 13 January 2010 (UTC)[reply]
The way I see it is: if N is any nilpotent nxn matrix, then Nn=0 (because of the normal form of nilpotent matrices, which is just their Jordan form. Check also PST's hint to a close question below). This fact implies a necessary condition on the existence of a p-th root of N. Indeed if Xp=N then X is nilpotent too, hence Xn=0, therefore N is more nilpotent than X: Nk=0 as soon as kp≥n. In particular a nilpotent N such that Nn-1≠0 (for instance the above mentioned shift matrix) has no root of any order p>1. (not even if you allow complex coefficients) --pma15:34, 13 January 2010 (UTC)[reply]
Also recall that an analytic p-th root in the unit ball around the identity is provided by the binomial series, that works in any Banach algebra A. You may also like to investigate the set of all square roots of the identity: it is an analytic submanifold of A (in general not connected); the tangent space at x is given by the closed linear subspace V of all v that anti-commute with x, that is vx=-xv. Note that A splits as A=V⊕W where W is the closed linear subspace of all w that commute with x. --pma16:16, 13 January 2010 (UTC)[reply]
If ever you forget how to LaTeX a particular mathematical notation, go to the corresponding Wikipedia article, click the "edit page" tab, and observe how the LaTeX is done. By repeatedly following this procedure, you will quickly learn how to LaTeX the given notation. For instance, to LaTeX a matrix, go to Matrix (mathematics), and notice that some examples of matrices are given at the beginning of the article. Now, click the "edit page" tab at the top, and you can observe the exact LaTeX for the particular matrix shown. At this point, it should not be difficult to duplicate the LaTeX for other matrices. Of course, you can also see how the above posters have written matrices in LaTeX, in this instance, by simply editing this section and viewing the underlying <math></math>. --PST03:36, 14 January 2010 (UTC)[reply]
In fact, in this case, I can give you an example of how to LaTeX a particular matrix:
It is simple if you try it a couple of times, and become accustomed to the LaTeX, but you are of course not obliged to do so (but just in case you wish to learn how to do so, I have included these tips). Hope this helps. --PST03:43, 14 January 2010 (UTC)[reply]
Firstly thanks for any help and if wiki has an article on this topic already then can you please point me in the right direction?
Anyway my problem is this i have a set of (x,y) co-ordinates randomly distributed inside a circle of radius R and i'd like to find out where the centre of the circle is. At best if we assume all points are evenly distributed over the circle then i guess calculating the mean of the (x,y)'s would yield the centre, but at worst, say all points clustered in one quadrant taking the straight mean would give a wrong answer. Is there a solution to this problem?--86.27.192.94 (talk) 17:02, 13 January 2010 (UTC)[reply]
Without knowing the distribution, you cannot definitively state where the center is. By omitting points in one area of the circle, you have lost information and cannot recreate it. -- kainaw™17:24, 13 January 2010 (UTC)[reply]
No. If they are randomly distributed then the best you can do if find an expected value for the centre (whether that is the mean of the coordinates or not will depend on the details of the distribution). It would be unlikely that the points will be significantly skewed, so that expected centre should be close to the actual centre, but it is entirely possible for it to be a long way off. --Tango (talk) 17:27, 13 January 2010 (UTC)[reply]
If you want something non-randomdefined as a function of the set of points, then there is a unique closed disk of minimum radius containing your bounded set of points; it depends continuously on the set wrto the Hausdorff distance; it's indeed Lipschitz continuous of constant 1 --this is true in a Hilbert space too. --pma17:48, 13 January 2010 (UTC)[reply]
That is really an answer to the OP's question, although it may be the best the we can do. The OP's circle has a fixed radius, though, so finding the smallest circle won't necessarily help. --Tango (talk) 18:54, 13 January 2010 (UTC)[reply]
Sorry, I'm not sure I got what you meant: did you mean "That isn't "? Anyway. the OP question is interesting but clearly ill-posed. It's not even clear if R is known or not. So something has to be clarified and possibly changed. If the point is, to have a unique non-random solution, then I suggested to modify the question as I wrote. --pma22:26, 13 January 2010 (UTC)[reply]
Assuming the distribution is uniformly random, i.e. each point in the circle has an equal chance of being sampled, can we give good estimators of center and radius? I guess in that case the best estimator for the center would indeed be the mean (and it should be unbiased, I think). If we have the center, the distance to the farthest point is a lower bound for the radius. As the sample size goes towards infinity, it should also converge against the true radius. However, for any finite set it's bound to underestimate the true radius. I suspect we can do better... --Stephan Schulz (talk) 18:55, 13 January 2010 (UTC)[reply]
The OP says the radius is R, which I interpreted as meaning R is known, so we only need to worry about the centre. Assuming a uniform distribution (which actually means any region of the circle has a probability of being sampled equal to the ratio of its area to the area of the whole circle - if you look at individual points you just get the probability for every point is zero, which doesn't tell you much) then I agree that the mean should be unbiased. --Tango (talk) 19:29, 13 January 2010 (UTC)[reply]
I'd still like to know if we can estimate the radius as well...
I can imagine some sets of points where the mean is not going to be the best estimate. For example if I have the points (0,0), (1,0), (2,0) and (100,0), I don't think this provides any different information than if I had only the points (0,0) and (100,0). Wouldn't you also need some prior distribution on what the radii of the circles tend to be?67.100.146.151 (talk) 21:14, 13 January 2010 (UTC)[reply]
Sure, in non-trivial cases you can always find a sample that will mis-estimate the distribution. The charm of an unbiased estimator is that for a random sequence it will stolastically converge to the true value. --Stephan Schulz (talk) 21:40, 13 January 2010 (UTC)[reply]
(the IP 67... is me.) Actually I was wrong. The two cases give different information by merit of the fact that they have a different number of points. But the idea still holds. A simple average is not the best you can do.
I expect one could readily prove that on average the "simple average" performs considerably better than the center of the smallest disk enclosing the data. Michael Hardy (talk) 03:20, 14 January 2010 (UTC)[reply]
We can find the region SR formed by the centers of all possible circles of radius R that enclose the points in the initial set (it'll be the intersection of the disks of radius R centered at each point in the set). I'm pretty sure there's an even distribution of where the center is likely to be in SR. We can think about if we pick a circle and then choose points at random, the chance of the points being clustered in one region of the circle is the same as it is in any other. So the average location of the center over all those possible circles should be the centroid of SR. I don't know if there's a straight forward way to calculate that though. Rckrone (talk) 22:50, 13 January 2010 (UTC)[reply]
If I understand what you said, SR is the disk with the same center as the minimal disk, and radius R-r (r being the radius of the minimal disk). Correct?--pma23:14, 13 January 2010 (UTC)[reply]
SR contains that disk of radius R-r, but I think it will generally be larger since a disk of radius R that contains the points in the set doesn't have to contain the minimal disk. Rckrone (talk) 23:48, 13 January 2010 (UTC)[reply]
Tango writes of an expected value of the center. The average location of observed points is the expected value only of the discrete distribution of the set of points actually observed; it's not the expected value of the center. Only if there's a prior distribution, and hence a posterior distribution, can you speak of an expected value of the center.
pma proposed unique smallest disk enclosing the observed points is not "non-random" if those observed points are random.
One should be able to find confidence regions for the center; these would be large if the number of points is small. I'd have to work out details though.
I think people (at least I was) were working under the assumption that the prior distribution of the center is an even distribution on R2 (or on some really large closed disk if that makes more sense. As long as the points in the set aren't near the edge it doesn't matter). I'm not really sure confidence regions are really useful here, unless I'm misunderstanding the problem. There's a well defined region where the center could be, and the posterior probability is evenly distributed across it. Outside that region the probability is zero. Rckrone (talk) 05:14, 14 January 2010 (UTC)[reply]
"posterior probability is evenly distributed across it" That is WRONG. I didn't even notice this comment the first time I commented. The posterior probability, with any reasonable prior, would be highly concentrated near the mean of the data points, and get less concentrated as you moved further away, and the distribution would have unbounded support unless the prior (i.e. pre-data) distribution had bounded support. Your comments are really seriously confused. Michael Hardy (talk) 05:04, 15 January 2010 (UTC)[reply]
That's why I said "(or on some really large closed disk if that makes more sense. As long as the points in the set aren't near the edge it doesn't matter)". Obviously you can't really have an even distribution on the whole plane, but an even distribution on any large bounded region doesn't effect the outcome of the problem as long as the points are not within 2R of the edge of the region. We can essentially ignore its specific properties. Rckrone (talk) 05:33, 15 January 2010 (UTC)[reply]
You seem confused. The uniform distribution on the whole plane is not a proper probability distribution; it assigns infinite measure to the whole space. I doubt that's what people assumed. I think they were assuming the DATA are uniformly distributed, not within the whole plane, but within the interior of the circle. That is NOT a prior distribution of the center of the circle. Given the data, the center of the circle COULD be anywhere in the plane, even quite remote from a disk-like cloud of data point, but that's improbable---it means by some accident all the points landed close together, far from the center. At any rate, "prior" and "posterior" refer to the probability distribution of the center, not of the data. Using frequentist methods, one wouldn't usually have prior and posterior distributions, but only a distribution of the data. Michael Hardy (talk) 04:54, 15 January 2010 (UTC)[reply]
The only explanation I have for what you're saying is that you're thinking about a quite different problem than I am. Here is the problem as I understood it: You have some large large bounded region of the plane and you choose a point P at random according to an even distribution on the region. Then you draw a circle of radius R centered at P and pick some number of points at random from inside the circle according to a even distribution on that disk. Given that set of points and the radius R, describe the probability of where P lies.
In the context of that problem, P cannot lie farther than R from any of the points in the set. It CAN'T be anywhere in the plane, rather it's restricted to the region SR that I described above. Moreover, the posterior distribution is an even distribution on SR.
Maybe you are assuming that R is not given? In that case (a) you need to know a prior distribution of the values the radius can take in order to answer the question (as I said and then crossed out when I modified my interpretation of the question) and (b) the prior distribution for the location of P that I described makes less sense since the edge effects will always matter for any set of data points, unless the prior for the radius is bounded. Rckrone (talk) 05:33, 15 January 2010 (UTC)[reply]
OP here, thanks for the help so far if it helps i'll describe the actual situation, i initially said "taking point (A,B) get all data within a circle of radius R, now some time later (after losing track of the individual (A,B)), i have all the data points and would like to reconstruct the original (A,B), or even just the best guess at what it would be--86.27.192.94 (talk) 08:37, 14 January 2010 (UTC)[reply]
Well that gives a whole lot more information. You have a whole lot of points which weren't within radius R, so this circle of radius R must be somewhere in the gap between these two sets of points. For instance if you were choosing all towns of more than 10000 inhabitants within 10 miles of a point you might only get three towns and the towns you didn't choose might be as good or better at determining the point.
Where the problem given the points inside determines a convex polygon with curved sides of radius R within which the center must lie, excluding points means you have sides that bend the other way and the figure isn't convex. The corners where the curves intersect still form a convex figure though as far as I can see so it should be a fairly reasonable problem for a computer program. Dmcq (talk) 09:31, 14 January 2010 (UTC)[reply]
Sigh. OK, I wrote some comments neglecting the fact that R is known. But some of what's said about makes me think some people are confusing the terms "prior" and "posterior" with the distribution of the DATA given the location of the center. Michael Hardy (talk) 05:20, 15 January 2010 (UTC)[reply]
I think I'll sigh even deeper. If you consider the points which aren't included the possible positions of the centre with a fixed R needn't even be connected never mind anywhere near convex. If you have three points excluded at the corners of an equilateral triangle and one included point at the centre then with R = half the length of a side or a bit more you get four completely separate areas where the centre can be. Dmcq (talk) 09:29, 15 January 2010 (UTC)[reply]
By the way using Bo Jacoby's very simple version of an interval of length 1 the midpoint of the interval containing the samples is a better estimate of the center of the interval than the mean - see Uniform_distribution_(continuous)#Estimation of midpoint. Generalizing to the circle case and ignoring any points outside I'd be pretty certain the center of the minimum circle would be a better estimate of the center than the mean though of course working out SR as above would give a better estimate. In fact just getting the midpoint of the minimum and maximum x and y values would probably be a good quick and dirty way of estimating the center. Dmcq (talk) 13:20, 15 January 2010 (UTC)[reply]
Of course Im(A) is included in ker(A) (so e.g. the quotient H=ker(A)/im(A) is well defined); further information may be seen via the normal form above. Now that you say it, I have a vague memory that somebody in some book defines a "module with differentiation" to be a module M with an endomorphism A such that A2=0, as an elementary example of chain complex. I think I saw it in Jacobson's "Basic Algebra II" (typical of it). --pma18:08, 13 January 2010 (UTC)[reply]
The way in which these questions were phrased suggests that they constitute homework. As pma has already confirmed that Im(A) is contained in Ker(A), the OP might like to consider the following exercise. If V is an n-dimensional vector space over a field K, and A is a linear transformation of V over K with the property that Ap = 0 for some p, show that An = 0 (n is the dimension of V over K). A hint is hidden below, but make a solid attempt at the problem before viewing the hint to attain a satisfactory intuition of nilpotent transformations.
(Click the "show" button at the right to see the hint or the "hide" button to hide it.)
Prove that is an ascending chain of subspaces (that each inclusion is indeed correct).
Hope this helps but please note that in order to learn anything, you must show us that you have made an attempt at the problem. Simply firing questions at us in quick succession will not aid your intuition of nilpotent transformations. --PST03:29, 14 January 2010 (UTC)[reply]
If you trust the above normal form, note that the number of 2x2 blocks in it is r, and that that matrix differs from the one you wrote only by a permutation of the base. If you want an independent proof, start with PST's hint. --pma20:30, 14 January 2010 (UTC)[reply]
Your most recent question suggests that you have not yet attempted to visualize the linear operator A. Visualization can prove an important tool in determining canonical forms, and thus it may increase your chances of solving the questions you pose. Personally, I feel that your most recent question should not require a formal proof in that it is very intuitive; you need only formally pinpoint the intuition (which is done in my hint above). Since you are unable to solve it, either you do not have any intuition of A at all, or you do not know how to construct a mathematical proof. If the former, please read my hint and pma's responses carefully. If the latter, we cannot help you. If neither, it is very difficult to decide exactly that which you are asking, especially with the wording "How to prove...". I do not intend to discourage you from asking questions here but please provide context (rather than merely fire questions at us). --PST03:26, 15 January 2010 (UTC)[reply]
Since the plate is moving 35 degrees east of north (that is, it's moving NNE), and moving at a rate of 67 mm per year, you can find how fast the plate is moving in both the East-West and North-South direction by drawing a right triangle (with the legs lined up along the North-South and East-West directions, and a Hypotenuse of 67 mm/yr) and using trigonometry to determine the lengths of the legs. The plate is moving somewhat faster to the north than you guessed, and significantly faster to the east. Does someone know a good way to get a diagram in here? Buddy431 (talk) 02:00, 14 January 2010 (UTC)[reply]
Simplistic solution is that northwrds component of velocity is 67 cos(35 degrees) = 54.9 mm/yr and eastwards component is 67 sin(35 degrees) = 38.4 mm/yr. However, as the figure given for the plate's velocity is an average of small movements over a large geographic area and over a long period time, this notional calculation of northwards and eastwards components is probably over-simplifying a much more complex pattern of movement. Different places on the plate boundary may be moving at different rates. Gandalf61 (talk) 10:58, 14 January 2010 (UTC)[reply]
January 14
An elementary homotopy equivalence
Continuing with my attempts to grasp the basics of quotient spaces and homotopy, I find myself quickly running into trouble when trying to show that two simple spaces are homotopically equivalent. Consider, for example, the problem of showing that the two spaces "S2 with a straight line joining north and south poles" and "wedge sum of S2 with a circle" are homotopically equivalent.
I can intuitively see some vague ways of deforming these spaces into each other; for example, one can take the second space and imagine the circle standing inside S2 and joined at the north pole, and then quotient half the circle into the side of the sphere; the result should by rights be homeomorphic to the first space. But when it comes to actually writing down explicit maps or homotopies, I'm lost; the text (linked above) is rather light on examples, so it's not even clear what a correct solution would look like. How do topologists approach problems like this? — merge12:30, 14 January 2010 (UTC)[reply]
This is example 0.8 in Hatcher's Algebraic Topology (relevant chapter here). He does it by collapsing a curve in the sphere from N to S to a point, using some general technical lemmas to make this precise. Algebraist13:17, 14 January 2010 (UTC)[reply]
Hmm. Thanks for the reference, but it doesn't seem of much use for this problem; in Bredon's book, it's posed in Chapter 1 just after introducing the most basic definitions and properties of homotopy. There are certainly no CW-complexes in sight, and no propositions about homotopy equivalence of quotient spaces (at least that I can identify as such). — merge16:23, 14 January 2010 (UTC)[reply]
Here's an approach that might help. Imagine the sphere as a (hollow) cube and line as a semicircular "handle" above two points near the centre of one face of the cube. Now apply a homotopy that moves the two points together while preserving the boundary of the cube - it should be reasonably easy to define an explicit formula to do this. Extend to the whole cube using the identity map on the other faces, and you're done (messy details left as an exercise..) (Disclaimer - it's been a long time since I did this sort of thing) AndrewWTaylor (talk) 17:16, 14 January 2010 (UTC)[reply]
Just because the approach of using CW complexes and a few technical lemmas isn't the intended approach of your text doesn't mean it can't be the best approach to such problems. Hatcher clearly thinks it is. Algebraist17:28, 14 January 2010 (UTC)[reply]
To AndrewWTaylor: Your homotopy takes place outside of both spaces, so with the same reasoning the OP's first space is homotopic to the sphere, which is impossible because one has a non-trivial fundamental group and the other doesn't. Correct me if I'm wrong I'm a novice. Money is tight (talk) 17:34, 14 January 2010 (UTC)[reply]
I recall being messily taught this example in an introductory algebraic topology course - the lecturer used a (non-obvious) deformation retraction of both spaces onto a common space...but I can't fully recall what that was. Not that that matters really, since Brenon seems to want it done using basic definitions. Icthyos (talk) 19:19, 14 January 2010 (UTC)[reply]
Heh. This section does contain definitions for deformation retracts and some homotopy properties of mapping cylinders, for what it's worth. :) If this problem is bad, the next two problems on the page must be much worse... — merge19:31, 14 January 2010 (UTC)[reply]
How about this: If X is the one-point union of the sphere and the circle at point x0 and Y is the other space, let f:Y → X map the points on the closed left side of the Y sphere to x0 and the rest of the Y sphere to the X sphere minus x0, and then the line gets mapped to the circle minus x0 in the way you'd expect. Then let g:X → Y map the X sphere to the Y sphere identically with the point x0 at the north pole and then map half of the circle to the line connecting the poles and the other half to a path along the left side of the sphere connecting the poles. I'm pretty sure fg and gf are homotopic to 1 but it might be annoying to show. Rckrone (talk) 22:10, 14 January 2010 (UTC)[reply]
I've come up with a couple of ideas along these lines, but it always ends up seeming far from obvious how to even show the maps in question are continuous. — merge23:35, 14 January 2010 (UTC)[reply]
Yeah, that's fairly similar to the attempt at a solution I provided - I was told it wasn't clear that the compositions of f and g were homotopic to the identity maps, then we moved on to some hand-waving. Icthyos (talk) 00:21, 15 January 2010 (UTC)[reply]
Consecutive integers with same sum of proper divisors
At one point, I was having trouble remembering the definition of Ruth-Aaron pairs, and that got me into some other interesting questions.
Are there pairs of consecutive integers with the same sum of all their divisors? Yes. (14,15) and (206,207) are the first two such pairs; this is (sequence A002961 in the OEIS).
What if we look only at proper divisors? (2,3) works and there are no others under 5,000. Note that this sum is always the same parity as the number itself, unless the number is a square or twice a square, so if we want two consecutive ones to be the same, then one of them must be of that type. Before I spend a lot of time programming this to at least start things off, it would be useful to know if there is already a result in the literature.
I'd say that such a set is ordered by hence well ordered because it is explicitly required by ZF; it's an ordinal. --pma09:20, 15 January 2010 (UTC)[reply]
No, it doesn't have to be an ordinal (nor do all ordinals qualify). For example Vω has the desired property, but 1 (understood as ) does not; its union is just . It would have to be a transitive set, but not all transitive sets work (e.g. 1 again). I don't know of a specific name for the property. --Trovatore (talk) 09:26, 15 January 2010 (UTC)[reply]
ach I realized it and was just on the point of correcting it but you are too fast :-) --pma 09:31, 15 January 2010 (UTC) So we may rephrase the assumption on such a set X saying it's a transitive set without "maximal element" wrto the relation (into commas because needn't to be transitive as a relation on X; X has no maximal element wrto in the sense that for any there is such that ) --pma09:38, 15 January 2010 (UTC)[reply]
Is a set that contains only itself hereditary?
In the process of copyediting hereditary set, I found myself writing the sentence
It then occurred to me not only that this may or may not be true, but that it might not even be a meaningful statement. Consider the set E = {E}. By the definition of hereditary sets, if E is hereditary, then {E} is hereditary, which merely restates the initial premise. If E isn't hereditary, then E isn't hereditary, again restating the inital premise. I can't see how to get a better handle on this problem. Can anyone help? -- The Anome (talk) 14:59, 15 January 2010 (UTC)[reply]
The usual way to unambiguously phrase such definitions in non-well-founded set theories is to define that A is a hereditary xxx iff every object in the transitive closure of {A} is a xxx (note that this is equivalent to the inductive definition if the universe is well-founded). Your E is thus indeed a hereditary set. — EmilJ.15:09, 15 January 2010 (UTC)[reply]
Resolved
![{\displaystyle 4n^{2}+1=[2n+(2s+1)]*[2n-(2s+1)+2k]}](https://wikimedia.org/enwiki/api/rest_v1/media/math/render/svg/8de703c4b20c639ad34b6d0fba66fedaf7049f5e)
![{\displaystyle b,c\in [\mu -\sigma ,\mu +\sigma ]}](https://wikimedia.org/enwiki/api/rest_v1/media/math/render/svg/d6661bbe16894dc686f68fe2ef47bb51ceee95ea)
![{\displaystyle [n(n+1)/2]^{2}}](https://wikimedia.org/enwiki/api/rest_v1/media/math/render/svg/afd6ad7cef74ee5d17b5736487d984b0af3d5bd1)
![{\displaystyle [g(A)]^{2}=A}](https://wikimedia.org/enwiki/api/rest_v1/media/math/render/svg/a57f5c8787ef3cf1f125bb8dec3e27284273b8a9)
