The business cycles has a varying period and amplitude. Also cycles are not always that clear when you get 'double dips' and the such like. Has anybody ever studied business cycle data with FFT-based spectral > techniques? ==== I think that the idea would then be to simply examine the spectral function and see if it obeys a power law. Has that ever been observed in the data? Power law dependence would mean that it is aperiodic, that basis function expansions are a waste of time (not counting wavelets) and that active control wouldn't work. Passive control might still work, (ie. setting interest rates to neutral and forgetting about them) but I wouldn't bet on that either. I'm not up on the math, but a few years ago there was a breakthrough in aerodynamics, where engineers realized when a fighter jet went into a tumble (chaotic trajectory) the pilot was causing it by fighting the stick. If he let go of the stick, the plane would spontaneously come out of the tumble. A sloppy person might be inclined to simply try the same solution for stabilizing the business cycle, but that would be a bad idea. It might be that there is no stable limit cycle in the available range of parameters except the one where all activity stops dead. This would be the same as if the time constant for recovery were so long that everybody starved before the economy came back. In the absence of actual state equations for the process model, I think you'd have to assume that this is the way things are and that the business cycle is the best solution allowed by the system. > Yes, a long time ago. The business cycles has a varying period and amplitude. Also cycles > are not always that clear when you get 'double dips' and the such > like. ==== power law ... would you mind to give a little bit more than you already said? I think that the idea would then be to simply examine the spectral function > and see if it obeys a power law. Has that ever been observed in the data? Power law dependence would mean that it is aperiodic, that basis function > expansions are a waste of time (not counting wavelets) and that active > control wouldn't work. Passive control might still work, (ie. setting interest rates to neutral > and forgetting about them) but I wouldn't bet on that either. I'm not up > on the math, but a few years ago there was a breakthrough in aerodynamics, > where engineers realized when a fighter jet went into a tumble (chaotic > trajectory) the pilot was causing it by fighting the stick. If he let go > of the stick, the plane would spontaneously come out of the tumble. A sloppy person might be inclined to simply try the same solution for > stabilizing the business cycle, but that would be a bad idea. It might be > that there is no stable limit cycle in the available range of parameters > except the one where all activity stops dead. This would be the same as if the time constant for recovery were so long > that everybody starved before the economy came back. In the absence of > actual state equations for the process model, I think you'd have to assume > that this is the way things are and that the business cycle is the best > solution allowed by the system. > Yes, a long time ago. The business cycles has a varying period and amplitude. Also cycles > are not always that clear when you get 'double dips' and the such > like. ==== If you do not insist in FFT you may find a lot on time series analysis + econometrics (i never had the patients to work through J. Hamilton) and even in TA it is used. The problem - as Rod says - is data are disturbed and not clearly follow one rule (you need 'windowing'). A concurrent approach is to set up a process model and estimate parameters from market data. Your FFT will be translated to characteristic functions (take care to get not lost in mathematical physics). For the time varying aspect you may dig for wavelets + econometrics. A third may be adaptive low pass filter (being slow i prefer that). Anyway you need lots of data. And may think about ARMA / GARCH according to your intention. Just my 2 Euro cents ... Yes, a long time ago. The business cycles has a varying period and amplitude. Also cycles are not > always that clear when you get 'double dips' and the such like. > Has anybody ever studied business cycle data with FFT-based spectral > techniques? ==== understand what I am missing. The problem is as follows: The Feige-Fiat-Shamir identification scheme is modified s.t.: - Peggy sends each of y_1,...,y_k individually, where every y_i = r*(s_i)^(b_i) - The verification in Step 4 is replaced with checking (y_i)^2 * (v_i)^(b_i) = r^2 mod n. As before, there are t rounds of the above protocol, so we can use the notation y_(i,j) for the y_i of round j, and similarily, b_(i,j) is the b_i round of j. Of course, Victor can find out k-1 of the s_i's if t=1, and can find out all of the s_i's if t > 1 by a judicious choice of the b_(i,j)'s, but we assume Victor is not interested in doing so, so that each b_(i,j) is random with equal probability of being 1 or 0, and then the b_(i,j)'s are independent of each other. In this case, what is the probability that Eve, who is recording all the t rounds between Peggy and Victor, can obtain s_1? This is my line of thinking: The probablity we can obtain s_1 is based on two things: 1. b_1 = 1, i.e. y_1 = r*s_1 2. b_k = 0, i.e, k != 1 i.e some y_i = r P(1) = 1/2 P(2) = (2^(k-1)-1)/(2^(k-1)) P(pass in a single round) = P(1) * P(2) = (1/2) - (1/2^k) P(pass in t rounds) = P(1-fail in t rounds) = 1-(1-((1/2) - (1/2^k))) = 1-((1/2) + 1/2^k)^t This logic is somehow flawed. I have the posted solution of 1 - 2^(-t) - 2^(t-kt) + 2^(-kt) without a good explination, which varies from my answer by one term. Can anyone help me find the error in here? -Chu ==== > > But how do we know (from theoretical considerations) that the batteries > aren't using up their chemical energy fighting each other to keep a current > from flowing??? They aren't. The space charge in the diode is doing that. There is a static field caused by the accumlation of charge that creates the space charge. Basically you form a capacitor in the diode. ==== > There are instances in physics where we encounter transcendental > equations of the form: exp(-constant*x) = x-r1 where r1 is a constant Has anyone encountered real-life cases of: exp(-constant*x) = (x-r1)*(x-r2) where r1 and r2 are real numbers? > Well, I don't think that's exactly the kind of example you would like to get, but once I had an experince almost like the one you mentioned. I work on the planning of the expansion of electric systems and I was trying to adjust, by the least square method, a curve that represented the marginal cost C of a thermal plant as a function of it's factor of utilization, f. Through simulation programs I got a table relating the values of C to several values of f. The curve I found was divided into 2 branches, the fist comprehending the interval [0.2 , k] and the other the interval [k, 0.96]. In the first branch the curve was given by a 3rd degree polynomial and in the second by a negative exponencial. Since the curve was supposed to be differentiable, at the point k we had an equation almost like yours. I used an Excel spreadsheet to find the the value of k, the values of the coefficients of the polynomial the constants of the exponential curve. Artur ==== > I bet you already knew that cos pi = -1 and such. But I was shocked to find, > just now, that there appears to be a _vegetable_ in the domain of cos x. > (Yes, you read that correctly. Obviously this is a silly pseudo-mathematical > post, but then you could tell that from the thread's title.) What is that vegetable? What is the value of cos x there? (I guess I should wait for other people to try to answer first. But this is > far too silly to be drawn out that way, and so I've given the answers below.) David The vegetable is lettuce. According to my dictionary: cos lettuce = romaine You're wrong, lettuce is NOT in the domain of cos. It would be if it was cos(lettuce) = romaine, which is not the case. Amanda ==== I have never reviewed material for exams when I took the >course. I like to say that a test which can be studied for is not worth > giving (or taking). Been to a medical doctor lately? I'm glad mine studied for their exams. What is the graduate with the lowest GPA in a Medical School class > called? Doctor. What do you call the person with the least achievement to graduate from Stanford Chem with a Ph.D.? Doctor Schwartz > A license to practice medicine isn't meaningful until the doctor does > years of clinical work supervised by nurses (re green lieutenants and > seargents) and senior doctors. An MD is not a license to practice medicine. What you are trying to discuss above is called a Residency. After MD, before license. Without it, you don't see patients unsupervised. ==== > An MD is not a license to practice medicine. What you are trying to > discuss above is called a Residency. After MD, before license. > Without it, you don't see patients unsupervised. Here's something not to do: When lying on the emergency room table and the ER Dr. is stitching up your hand (where there are a LOT of nerve endings) and he tries to distract you from the fact that the pain- killing shot hasn't really taken effect, by making small talk with you, by saying: So, what do you do for a living? DO NOT SAY I'm a math prof. Because odds are that Calculus almost kept the poor bastard out of med school and he may inform you of that fact while he continues to ram that needle in and out of your hand. Indeed, sewing may become throwing stitches. Case in point, my ER Dr. put stitches through my thumbnail. In one side and up through the other. I'd have paid extra not to have that done. What the heck was wrong with adhesive tape. Or duct tape. I didn't give him his stupid C-, but I sure took the hit for it. The ONLY saving grace of the entire experience was that it wasn't a groin injury. ==== What are your credentials Niles? Are you only capable of whining and bitching? ==== > Anyone know why Grad Schools will not accept your application unless > you have a Bachelor's? Is it not possible to have the required > knowledge and no degree? Is it so very hard to verify that one's claim > of knowledge. some reasons: 1. it invalidates the whole college education schtick. > 2. lack of institutional admissions exams. > 3. the general population lacks the discipline to pull off such > thing as self-education. now to address your latter question. it wouln't be very hard to verify knowledge, but it is politically > incorrect and somewhat costly and time-consuming. Which is why grad schools can't be bothered unless you indicate that you are > special. You don't get special consideration unless you're a special > person. well, it seems that our advise seeker is reporting that the typical graduate programme does NOT offer such special consideration, even to special persons. > Jon Miller ==== Suppose we take the chess rules and add a new one: A1: You have the option of moving two pieces simultaneously, as long as each move considered separately was legal in 'single move' chess. The two pieces can't land on the same square, obviously. So you can't move a pawn, then the bishop behind it which was previously blocked by the moved pawn. Or, if your king is checked, you move your king, and move another piece to kill the check. I'm just wondering how drastic would a change be? Would a typical chess game have half the number of turns for each player? Would the games be faster? Would there be some opening trick that white can play that will always guarantee a win in a few turns? Now, if we were to look at the entire decision tree for double-move chess, it would be bigger than single-move chess, because double-move chess would have all the moves of single-move chess too. But now, let's look at the decision tree for those double-move chess games where two pieces were always moved if legal to do so. Would the tree be roughly twice as wide but half as deep as the single-move chess decision tree? I know silly questions... just curious to know the answer. ==== > Suppose we take the chess rules and add a new one: A1: You have the option of moving two pieces simultaneously, > as long as each move considered separately was legal in > 'single move' chess. The two pieces can't land on the > same square, obviously. http://homepages.stayfree.co.uk/gpj/gvc.htm and scroll down to Two-Move Chess. Marco -- ==== > Suppose we take the chess rules and add a new one: A1: You have the option of moving two pieces simultaneously, > as long as each move considered separately was legal in > 'single move' chess. The two pieces can't land on the > same square, obviously. > So you can't move a pawn, then the bishop behind it > which was previously blocked by the moved pawn. Or, > if your king is checked, you move your king, > and move another piece to kill the check. > I'm just wondering how drastic would a change be? > Would a typical chess game have half the number of > turns for each player? Would the games be faster? Would there be some opening trick that white can play > that will always guarantee a win in a few turns? > It would probably go much faster. One thing that might help is to give white one move, then black gets 2 and they alternate 2 moves from there on. Two moves in a row is such a huge advantage that in Monster Chess (white gets Ke1, Pc2d2e2f2 and two moves per turn, black gets the normal setup and one move per turn) is pretty much a forced win for white. --Harold Buck I used to rock and roll all night, and party every day. Then it was every other day. . . . -Homer J. Simpson ==== In Harold Buck typed: > Two moves in a row is such a huge advantage that in Monster Chess > (white gets Ke1, Pc2d2e2f2 and two moves per turn, black gets the > normal setup and one move per turn) is pretty much a forced win for > white. On the contrary, with careful play, Black should win. It isn't easy, and if you never seen it before, Black is very hard to play, but nevertheless it's better for Black. -- Ken Blake Please reply to the newsgroup ==== > In > Harold Buck typed: Two moves in a row is such a huge advantage that in Monster > Chess > (white gets Ke1, Pc2d2e2f2 and two moves per turn, black gets > the > normal setup and one move per turn) is pretty much a forced win > for > white. > On the contrary, with careful play, Black should win. It isn't > easy, and if you never seen it before, Black is very hard to > play, but nevertheless it's better for Black. Really? I thought I'd read that it was hugely in white's favor. In any case, the fact that it's even playable shows how important those extra moves are. --Harold Buck I used to rock and roll all night, and party every day. Then it was every other day. . . . -Homer J. Simpson ==== In Harold Buck typed: > In >> Harold Buck typed: >> Two moves in a row is such a huge advantage that in Monster >> Chess >> (white gets Ke1, Pc2d2e2f2 and two moves per turn, black gets >> the >> normal setup and one move per turn) is pretty much a forced win for >> white. >> On the contrary, with careful play, Black should win. It isn't >> easy, and if you never seen it before, Black is very hard to >> play, but nevertheless it's better for Black. > Really? I thought I'd read that it was hugely in white's favor. In any > case, the fact that it's even playable shows how important those extra > moves are. Yes, I agree. -- Ken Blake Please reply to the newsgroup ==== Find the natural numbers k, as many as possible, so that all the > solutions of the scalar equation x'' = x^k can be extended > infinitely far? Do you mean you want there to be a solution x(t) with a domain of the form (a, infty) ? > d(x^(k+1))/dt = (k+1)(x^k)x' > = (k+1)(x'')(x') > = ((k+1)/2)(d((x')^2)/dt) This tells you that x^(k+1) - (k+1)/2 (x')^2 is a constant, C. Thus dx/dt = sqrt( 2/(k+1) x^(k+1) - C ). You can solve this by separation of variables: t will be an antiderivative of 1 / sqrt( 2/(k+1) x^(k+1) - C ). More accurately, if (a0, b0) and (a1, b1) are two points on the graph of x(t), then a1 - a0 = int_b0^b1 dx / sqrt( 2/(k+1) x^(k+1) - C ). Since you want the domain to extend to infinity, you need the integral to grow unbounded as b1 increases. But your integrand is big-O of x^{-(k+1)/2}, and so the integral stays bounded if (k+1)/2 > 1, i.e. if k > 1. For k=1, of course, all solutions can be defined across the whole line. dave ==== Find the natural numbers k, as many as possible, so that all the > solutions of the scalar equation x'' = x^k can be extended > infinitely far? d(x^(k+1))/dt = (k+1)(x^k)x' > = (k+1)(x'')(x') > = ((k+1)/2)(d((x')^2)/dt) How could I do this? -- What do you mean by extended infinitely far? Lurch ==== >> >> Also, can we have things such as e^(quat) or ln(quat) where quat is a >> quaternion? Has anybody ever figured out how to do this? Also, what >> about using lie algebras which work on quaternions? >> Any quaternion q can be written a+bu, where a,b are real and u^2 = -1. >So the set of {x+yu:x,y real} for this fixed u is isomorphic >to the complex numbers. Compute e^q and ln(q) in it just as in the >complex numbers. Generalizing functions of two variables is where problems come, if you >use two quaternions that don't commute. -- >G. A. Edgar http://www.math.ohio-state.edu/~edgar/ This proves that representing quaternions as 3x3 matrices is not exact, since they can at most have 2 eigenvalues, whereas 3x3 matrices usually obey a cubic. ==== >Problem (Calculus of variations) > b > / >minimise s= | (1+u^2)^0.5 dx for u(x) > / > a where s>0 and b>a, subject to Did you mean to write u > 0? b > / > |u dx - c = 0 c>0 > / > a and b > / du > | --- dx - m = 0 m>0 > / dx > a >when m = 0, the solution is trivial, namely u=constant. I am interested in the case m>0. > You can use the standard variation formula to show that the value of the functional for any non-constant u(x) can be decreased by changing u near two points where the value of u differs, in such a way as to leave the integral of u equal to c (and leaving the end points fixed). This shows that, in general, there is no continuous minimizing function. A generalized solution is a function u(x) that equals c/(b-a) everywhere except at x = b, where the value is chosen to satisfy the last condition. John Mitchell ==== >I have a partial difference equation whose sum over >> a certain range can be used to count prime numbers. 2) Presumably you are going to justify why that's true at some point > Who cares if Mr. Harris can justify it? Are you really interested in his explanation? The interesting questions about his formula are: (1) Over what range does the method work? and (2) Is it faster than other methods on this range? Simple experiments should answer both questions. rich ==== > >I have a partial difference equation whose sum over > a certain range can be used to count prime numbers. >>2) Presumably you are going to justify why that's true at some point >> Who cares if Mr. Harris can justify it? Are you really interested in his > explanation? The interesting questions about his formula are: (1) Over what > range does the method work? and (2) Is it faster than other methods on this > range? Simple experiments should answer both questions. rich I care. Sufficiently many people whose opinions I respect have decided it will count primes, in a manner similar to Legendre's method. I would like to see James, who has consistently denied it has anything to do with Legendre (in fact for a while he even refused to admit Legendre's formula could be used to count primes, so how he knew it was different from his is a mystery), prove he understands what's going on and can justify his claims that it is an all new and powerful method that no one else 'in the history of mankind' has discovered. And just proving a range in which it works is a start. Unlike all the rest of his maths that he's shown recently, I think he may well have something here. I just don't see that it is new, exciting, or useful. He has also in the currently running debate on his crankhood repeatedly said he wants a point by point critique by a mathematician, without insults, so that he can rebut them all. So there were some observations by someone who's not a number theorist. If anyone who does number theory seriously wants to point out where he or I am misunderstanding something (which possibly I am too) please tell me. As it is the thread currently gives the lie to James assertion that he would respond to any fair criticisms. I'm not remotely interested in insulting him, or debating his mental health; it seems counterproductive. I am interested in people learning more mathematics, and that includes me. ==== >Dear Group Members, I ran into a problem where I would need to solve for a system (specifically >2) of quadratic equations. My search on the net lead me to systems of >polynomial equations. Unfortunately my math knowledge does not yet allow me >Could anyone please point me to a text that could be understood by a >interested technician? For completeness I formulate my problem, and what I know so far: 1)The equations: >transpose (x-a) * A * (x-a) = r0 >transpose (x-b) * B * (x-b) = r1 where x is the (column) matrix of unknowns, a and b are column matrices and >A,B are symmetric matrices. r0 and r1 are scalars. 2) I know about how to transform the quadratic forms into their >eigensystems, but then still can see no systematic way of how I could >eliminate one of the unknowns. 3) I guess the order of my (specific) problem is at most of degree 4. Any pointers appreciated. Roland BTW.: Of course I would be interested how to solve for more than two >equations too. First expand both equations and rewrite them as M1 v = v1, M2 v = v2, where in x has dimension n, v has dimension equal to all quadratic, plus all linear, plus one constant, terms. Now if M1 is invertable, M2 (M1^-1) v1 = v2. So it is possible you have no solution at all. ==== > (x,y)--> sqrt{x^2+y^2} cos (arctg y/x) if x=/=0 > comes from thinking about (r,t)-->r cos(t) > it isn't linear time to review polar coordinates ==== Mystery remains as journal withdraws paper JOHN WHITFIELD [LONDON] A mathematics journal has withdrawn a paper that claimed to crack one of the discipline's great mysteries after reviewing and accepting the work and publishing it online. On 18 November, Nonlinear Analysis published a paper by Elin Oxenhielm - a postgraduate student in mathematics at the University of Stockholm, Sweden - sixteenth problem, one of a set of challenges laid out by German If a solution were validated, mathematicians agree, it would be a significant step towards a complete solution to the problem. Oxenhielm predicts just that: We could find one in a year or so, if we're lucky, she says. The work was described in a 24 November press release from Oxenhielm and covered in several media outlets including the BBC. But the paper immediately came under fire from mathematicians. It's completely inadequate - I can't imagine who would have thought it was a proof, says John Mather of Princeton University, New Jersey. Critics include Oxenhielm's supervisor, Yishao Zhou, who put a statement on her website saying: The paper is incomplete and contains serious mistakes. research. Solving any one of them is almost guaranteed to make a mathematician's name, and by 2000 all but three had been solved. The sixteenth, the problem of the topology of algebraic curves and surfaces, deals with the territory where geometry meets algebra. Its second part involves showing that the number of periodic solutions to a differential equation is finite. Such periodic solutions are also known as limit cycles - stable, oscillating trajectories to which a system will return if perturbed. Limit cycles are common in nature, and a proof of the second part could lead to a better understanding of heartbeats, animal movements and the kind of runaway vibrations that can shake a structure to bits. Oxenhielm formulated her proof using 'describing functions' - which can predict roughly the presence of limit cycles in nonlinear equations. A few minutes' scrutiny is enough to show that her reasoning is false, says mathematician Grigori Rozenblioum of Chalmers University of Technology in Gothenburg, Sweden. The approximate solutions studied by Oxenhielm cannot provide the exact answers demanded in a proof, he says, and some of her equations contain exact terms where approximate ones should be used. The work should never have been published, Rozenblioum says: It's impossible to understand the behaviour of the journal, which is one of the leaders in its field. Nonlinear Analysis pulled the paper on 4 December. Publication has been halted until a thorough investigation into the matter has been handled, says editor-in-chief V. Lakshmikantham, a mathematician at the Florida Institute of Technology in Melbourne. Originally approved by one reviewer, the paper has now been sent to two more mathematicians for further round of review, along with a defence by Oxenhielm, who says that the critics do not understand her methods. She refuses to comment further. Nonlinear Analysis' editors have evaluated the paper, they accepted it for publication and they have the copyright of its contents - and thus they are responsible for its correctness, she told Norwegian newspaper Aftenposten. ---------------------------------------------------------------------------- ---- ==== > [The author] refuses to comment further. Nonlinear Analysis' > editors have evaluated the paper, they accepted it for publication > and they have the copyright of its contents - and thus they are > responsible for its correctness, she told Norwegian newspaper > Aftenposten. Remarkable. Watch out, James. Another eminently .siggable quote factory. But can she maintain this standard? -- Run mathematicians, RUN!!! I'm coming for you. It may take a few months, but I'll get [computer verification of my proof] and then your lives will be ended as you previously knew it. -- JSH meets PVS ==== >That is the theorem...but the converse is: If a triangle is inscribed in a semicircle with the angle opposite to the >hypoteneuse as 90¡ then the hypoteneuse is the diameter... How would you prove that ? I tried it many times and I get stuck > Well that is why I put a question mark at my statement of the converse. I wondered if you asked what you meant to ask. Here is what you asked in the original post: >How do you prove the converse of this theorem? >If a right triangle is inscribed in a semicircle then the hypoteneuse is the >diameter of the circle So you meant to ask how to prove this theorem, not its converse. --Lynn ==== >> Dear Nutjob, >> Stay out of alt atheism with your insane claims. >> There I said something else to you. What's my full legal name? >> 100,000 witnesses is a well founded claim, just examine the evidence >> someone to guess the names, its all in google. >You are the one making the claim you can tell a persons name by what they >said to you. >Your answer is meaningless. > 6 witnesses admit I'm the Truman, all in google. I've got evidence and >> witnesses, you're just a heathen. >Sadly people like you need help but probably don't qualify under 5150 H&S. >However I bet that if you live in California you are familiar with the >section. > He lives in Townsville, Queensland. A population of about 140,000 > people, businesses, and major tourism. Strangely though, he seems > unable to find employment, or a woman. That should tell you something. Under 5150 if you are a danger to yourself or others or can't take care of yourself you can be committed for 72 hours. A doctor can over ride this. It is a good and powerful law that will probably be overturned one day because of it's abuse. I once had some cops that wanted me to commit a woman because she didn't agree with her boy friend. The cops (I used to be one) were not smart enough to know they could do it. This guy is sick and I should not have responded to him. He is plonked. ==== > is a good and powerful law that will probably be overturned one day because > of it's abuse. I once had some cops that wanted me to commit a woman because > she didn't agree with her boy friend. The cops (I used to be one) were not > smart enough to know they could do it. at least you admit you're not smart and corrupt This guy is sick and I should not have responded to him. > He is plonked. You can't silence the truth, but you can PLONK it! after he actually contacted his famiily in townsville, i assume because they confirmed my story this was his last post in aus.tv If what you say is true, wouldn't I be able to confirm what you say by 'hearing' your thoughts or watching a broadcast of your life on television? I have satellite cable, why can't I get the Herc show? My brother in law and his wife life in Townsville, and, amazingly enough, they can't seem to find this show on their television either. Nor have they ever heard of you or your demented claims. My partner's sister who lives in Rockhampton seems surprisingly unfamiliar with The 24 / 7 Herc Show!, starring everyone's favourite Why not use the imprison without charge law on all applicants to skeptic companies? They're all insane right, cut out the middle man of formal applicaitons, test procedure, just put up a big million dollar prize and send all the names and addresses to the mental lock up! Pigs are so entrenched with the skeptic companies already wave a INC at a pig and you can lock up any unemployed person you want. Noone examining my claim of 3 extraordinary posts on the 1 day, just group abuse from 5 newsgroups I've spent numerous months at each not one person to give me any benefit of doubt for a few minutes investigation into my claim, does the 3 question test have an answer, are replies to herc's authors guessable? Very well, if YOU wont cooperate with fellow man then you can't be ruled with religion, you will be ruled with force like you are now under a real nutcase Bush who floods my life with noise from a spy satellite, believe me you would have preferred the former. Herc EVIDENCE B Rust 3 attacks metal http://tinyurl.com/nd56 E Rich Shewmaker 1 rich showmaker James Randi http://tinyurl.com/nd52 H See You In Hell My Friend. 2 It all depends http://tinyurl.com/nd53 predictions that paranormal will be demonstrated on 02 02. http://tinyurl.com/yo1b one year before http://tinyurl.com/nw1r one day before ----------------------------- Then I have 100,000 witnesses that I'm paranormal, here they admit that I am the TRUEman http://tinyurl.com/iky5 http://tinyurl.com/iky8 http://tinyurl.com/iky9 http://tinyurl.com/iky4 http://tinyurl.com/rv5f http://tinyurl.com/v1yf Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! We're sick of you You rule Truman. >Is the truman is living in Townsville? I've been hearing stuff, yeah. and watch here live in Google as I predict back in 2000 Jim Carrey's next romance costar Laurie Holden with http://tinyurl.com/fuf8 she looks exactly like Laurie Holden of the x-files with bright green eyes. Just like any Truman could have predicted, Majestic hey? Adam and Eve ==== Let X be compact Hausdorff, let {A_i} be a collection of closed connected subsets of X that is simply ordered by proper inclusion. Prove Y = {intersection of A_i over all i} is non-empty and connected. How can I prove this? The way I want to start is to suppose U,V are disjoint non-empty open subsets of X. Then I want to analyze A_i - (U union V) over all i....can I get anywhere with this? How can I complete the proof? Steven ==== ==> some K1,..Kn with K1 /../ Kn subset U Now assume some open disjoint U,V with /C subset U/V and use the exercise. I don't think Hausdorff is needed. / = cap = intersection ---- ==== >Let X be compact Hausdorff, let {A_i} be a collection of closed connected >subsets of X that is simply ordered by proper inclusion. Prove Y = >{intersection of A_i over all i} is non-empty and connected. >How can I prove this? >The way I want to start is to suppose U,V are disjoint non-empty open >subsets of X. Then I want to analyze A_i - (U union V) over all i....can >I get anywhere with this? How can I complete the proof? Suppose that U union V contain the intersection of the A_i. Show that it must contain some A_i. Compactness is helpful. --Dan Grubb ==== > Let X be compact Hausdorff, let {A_i} be a collection of closed connected > subsets of X that is simply ordered by proper inclusion. Prove Y = > {intersection of A_i over all i} is non-empty and connected. How can I prove this? The way I want to start is to suppose U,V are disjoint non-empty open > subsets of X. Then I want to analyze A_i - (U union V) over all i....can > I get anywhere with this? How can I complete the proof? Consider this: let U be an open subset of X such that cap A_i subset U. Prove that A_i subset U for some i (hence, for all sufficiently large i--in the sense of the order). Why? Because if not, A_i cap U^c is nonempty, compact and has the finite intersection property, hence has empty intersection. --Ron Bruck ==== Suppose that the random variables X_i are independent and P(X_i = 2^k) = 2^(-k) for all k >= 1. Show that (X_1 + X_2 + ... + X_n)/(n*log(n)) converges in probability to log(2). Prove or disprove that limsup (X_1 + ... + X_n)/(n*log(n)) = infinity with probability 1. Can anyone prove this problem? Sincerely, Rob ==== Suppose that the random variables X_i are independent and P(X_i = 2^k) = > 2^(-k) for all k >= 1. Show that (X_1 + X_2 + ... + X_n)/(n*log(n)) converges in probability to > log(2). > Prove or disprove that limsup (X_1 + ... + X_n)/(n*log(n)) = infinity > with probability 1. Can anyone prove this problem? Well, Feller can. For the first part, see Feller's An Introduction to Probability Theory and Its Applications: Volume 1. In my 3rd edition (1968), it's in section X.4 starting on p. 251. Feller also gave a very complete answer to the second problem in a 1946 paper (A limit theorem for random variables with infinite moments. Amer J. Math. 68, 257-262), but that result is much more general than you need. You should be able to prove/disprove the second part fairly directly using Borel-Cantelli. Think about the events {X_n > n log n}. Do they occur infinitely or finitely often? If the former, you should be able to easily show that the limsup is infinite. If the latter, you can't quite prove that the limsup is finite, but you can get close. If you replace n log n with something smaller such that the events *still* happen only finitely often, you ought to be able to do it. Does that help? -- Kevin ==== Suppose that the random variables X_i are independent and P(X_i = 2^k) = > 2^(-k) for all k >= 1. Show that (X_1 + X_2 + ... + X_n)/(n*log(n)) converges in probability to > log(2). > Prove or disprove that limsup (X_1 + ... + X_n)/(n*log(n)) = infinity > with probability 1. Can anyone prove this problem? Well, Feller can. For the first part, see Feller's An Introduction to Probability > Theory and Its Applications: Volume 1. In my 3rd edition (1968), > it's in section X.4 starting on p. 251. > Where can I see your 3rd edition? What is your textbook called? (more text below) > Feller also gave a very complete answer to the second problem in a > 1946 paper (A limit theorem for random variables with infinite > moments. Amer J. Math. 68, 257-262), but that result is much more > general than you need. You should be able to prove/disprove the second part fairly directly > using Borel-Cantelli. Think about the events {X_n > n log n}. Do > they occur infinitely or finitely often? If the former, you should be > able to easily show that the limsup is infinite. If the latter, you > can't quite prove that the limsup is finite, but you can get close. > If you replace n log n with something smaller such that the events > *still* happen only finitely often, you ought to be able to do it. > I don't see how you would do this. How about this approach : What we want to show is P(limsup (X_1 + X_2 + .... + X_n)/nlog(n) = infinity) = 1 where log is base e. (correct?) Now, is the following proposition true? For any sequence of independent random variables {Z_k}, if Summation (from k = 1 to infinity) of P( |Z_k| > epsilon) = infinity for all epsilon > 0, then Z_k converges almost surely to infinity. (Maybe I can prove this with borel-cantelli lemma of independent events?) If true, then I'd like to use Markov's inequality, by saying that P((X_1 + X_2 + .... + X_n)/nlog(n) > epsilon) <= (1/epsilon) * E((X_1 + X_2 + .... + X_n)/nlog(n))) or some variation of markov's inequality, and then sum up these values to possibly get infinity and our proof would be done. If this doesn't work, how can I view the events {X_n > nlog(n)}? I agree this would be helpful. I don't see how I can tell if they occur infinitely often. Furthermore, if this were true, how would this imply that the limsup is infinite? And which Borel-Cantelli lemma would you use? (a bit lost) > Does that help? -- > Kevin ==== >> For the first part, see Feller's An Introduction to Probability >> Theory and Its Applications: Volume 1. In my 3rd edition (1968), >> it's in section X.4 starting on p. 251. Where can I see your 3rd edition? What is your textbook called? > (more text below) Sorry, I meant my 3rd edition copy of Feller's An Introduction to Probability Theory and Its Applications: Volume 1. > Now, is the following proposition true? For any sequence of independent random variables {Z_k}, if > Summation (from k = 1 to infinity) of P( |Z_k| > epsilon) = infinity for > all epsilon > 0, > then Z_k converges almost surely to infinity. (Maybe I can prove this with > borel-cantelli lemma of independent events?) No, it's not true. Constant random variables (i.e., random variables that are always equal to a specific constant) are independent of anything, so any deterministic sequence is a sequence of independent random variables, too, but your proposition fails for Z_k=-k (where the limit is -infinity instead of infinity) or Z_k=k for k even and =0 for k odd (where there is no limit). It *is* true that a sequence of nonnegative, independent random variables {Z_k} satisfying sum_{k=1}^infinity P(Z_k > epsilon) = infinity for all epsilon > 0 will have limsup Z_k = infinity, and this can be proved by Borel-Cantelli. (Note that we have added the hypothesis of nonnegativity and changed the conclusion to specify the limsup rather than the limit.) > If true, then I'd like to use Markov's inequality, by saying that > P((X_1 + X_2 + .... + X_n)/nlog(n) > epsilon) <= (1/epsilon) * E((X_1 + > X_2 + .... + X_n)/nlog(n))) or some variation > of markov's inequality, and then sum up these values to possibly get > infinity and our proof would be done. The problem is that the inequality goes the wrong way. To apply the above result, you need to show that the sum of the probabilities is infinite, but the inequality will only let you conclude that this sum is less than or equal to something, and that won't help. > If this doesn't work, how can I view the events {X_n > nlog(n)}? I agree > this would be helpful. Well, if n=6, say, then nlog(n) is about 10.75. So, how would you go about calculating P(X_n > nlog(n)) = P(X_6 > 10.75)? You should be able to generalize this to an arbitrary n. > I don't see how I can tell if they occur infinitely often. > Furthermore, if this were true, how would this imply that the limsup > is infinite? Well, okay, I spoke too glibly. It still wouldn't be *quite* enough, but it would tell you that, infinitely often (i.e., whenever X_n > nlog(n)), you'd have to have: (X_1 + ... + X_n)/nlog(n) > X_1/nlog(n) > 1 Now, if you could do the same thing with {X_n > a n log(n)} for every a > 0 (instead of just for a=1), *then* that would be enough. Do you see why? > And which Borel-Cantelli lemma would you use? Which do you have? The two I know about are the one that applies when the sum of the probabilities is infinite and the other that applies when the sum of the probabilities is finite. So, once you've determined whether the sum of the probabilities is infinite or finite, you'll know. -- Kevin ==== > If this doesn't work, how can I view the events {X_n > nlog(n)}? I agree > this would be helpful. I don't see how I can tell > if they occur infinitely often. Furthermore, if this were true, how would > this imply that the limsup is infinite? And which Borel-Cantelli lemma > would you use? > Consider the events A_n = {X_n > a n log n} where a is a constant. These are independent events. Show that sum P(A_n) = infty (for any choice of a). ==== > If this doesn't work, how can I view the events {X_n > nlog(n)}? I agree > this would be helpful. I don't see how I can tell > if they occur infinitely often. Furthermore, if this were true, how would > this imply that the limsup is infinite? And which Borel-Cantelli lemma > would you use? Consider the events A_n = {X_n > a n log n} where a is a constant. > My apologies. These are discrete random variables. Now it appears that sum P(A_n) = infty, as I plug numbers into my calculator, but I need to come > These are independent events. Show that sum P(A_n) = infty > (for any choice of a). Cancel-Lock: sha1:fo+GBypctlzmhtte3qiDdDA9jI4= ==== My apologies. These are discrete random variables. Now it appears > that sum P(A_n) = infty, as I plug numbers into my calculator, but > I need to come up with a rigorous proof of this. Yes, and calculators can be deceiving. Rumour has it that log(log(n)) goes to infinity as n does, but just try proving that with a calculator. -- Kevin ==== > If this doesn't work, how can I view the events {X_n > nlog(n)}? I agree > this would be helpful. I don't see how I can tell > if they occur infinitely often. Furthermore, if this were true, how would > this imply that the limsup is infinite? And which Borel-Cantelli lemma > would you use? Consider the events A_n = {X_n > a n log n} where a is a constant. > This is where my problem is....how do I calculate P(A_n)? I understand they are independent events and then by Borel-Cantelli I am golden. But all I have is that P(X_i = 2^k) = 2^(-k)...how can I calculate P(X_n > an*log(n))? I really think this must be a stupid question I am asking but I am just forgetting something from my elementary probability...isn't P(X_i = 2^k) = 2^(-k) my density function? Then do I just integrate from -infinity to an*log(n)? but I must change the density function to P(x) = 2^{-log(x)/log(2)} since I need a density function for P(X_i = x), not P(X_i = 2^k)... I think I am just getting more and more confused... > These are independent events. Show that sum P(A_n) = infty > (for any choice of a). ==== : : > : Anyway, my point had nothing to do with path categories specifically This almost looks like a mis-attribution; despite the fact that my name is the last one occurring before it, the above line was written by the other arguer, as was the following one: : > : it was entirely about your refusal to accept the general phenomenon that : > : different ways of looking at the same mathematical objects can lead to : > : better insights about those objects. : > : > All I can say to that, frankly, is that you are starting to deserve : > insults at the same level at which you are dishing them out. : : What insult? This insult: your refusal to accept the general phenomenon that different ways of looking at the same mathematical objects can lead to better insights about those objects. I have NOT so refused and I am not so refusing now. Or, as I said it better the first time, : > I DO NOT refuse to accept the general phenomenon that : > different ways of looking at the same mathematical objects can : > lead to better insights about those objects. ... : > In projective geometry, points : > and lines are REALLY dual. : Perhaps your understanding of projective duality is different from mine. Duality is one thing. The fact that it actually occurs in projective geometry is another. But it does NOT occur (not between points and lines anyway) in affine geometry. It is a DIFFERENCE between affine and projective geometries that points&lines ARE dual in projective geometry, but ARE NOT dual in affine geometry. Unfortunately (terminology-wise), it is also the case that for any given affine plane, there are some fairly simple things you can do to it to turn it into a projective one. This transition is SO small and simple that sometimes people are tempted to say that the affine plane that you started with, and the projective one that you get to (from applying this simple transformation to it), are THEMselves dual to each other. But THAT is a BAD use of dual. Or at least a distracting one, in the context of the more real duality, which is between the projective plane's points and the projective plane's lines. : In my understanding, the dual of a configuration of points and lines : consists of the same objects, but the objects that were points now get : called lines and the objects that were lines now get called points. You can always perform this rename/exchange but you canNOT always CALL it the dual. In projective geometry, points and lines ARE dual, which means that that re-calling works in the relevant sense. But where it doesn't work, there simply IS NO duality. : A projective geometry is a system of objects with incidence relations : satisfying certain axioms and when you rename the objects they still : satisfy the appropriate axioms. Right, i.e., when you exchange points with lines, the same statements all keep the same truth-values. AFTER that happens, THEN you can say that points and lies are dual, withIN that axiomatic framework. But affine geometry is governed by DIFFERENT axioms and under THEM, you canNOT perform this renaming without changing truth-values. : The only difference is in the human : terminology for the objects, not in their mathematical behavior or : identity. There is no difference in projective geometry, yes; that by definition is why you CALL it projective. But we were talking about the affine case. : To get back to an earlier point, one possible way of forming objects : satisfying the set of projective geometry axioms is to have points that : are just points (atomic from the point of view of the geometry) while : the lines are sets of points and the point-line incidence relation is : set membership. Projective duality tells us that we could instead view : the lines as being atoms, and the points as being sets of lines through : them, with incidence again being equivalent to set membership. Of course. : As someone who has personally gained insight into geometry problems by : applying projective duality, Oh, SHUT UP! The AFFINE plane IS NOT PROJECTIVE and DOES NOT EXHIBIT projective duality! In order to apply projective duality, you would FIRST need to APPLY SOME TRANSFORMATION TO the affine plane so that the result WOULD IN FACT BE projective! ONLY AFTER that do you get to exploit duality! : despite duality's inability to change the : underlying mathematical objects involved, I find your claim You will : not gain any insight into the nature of a point by thinking about it as : an infinite collection of lines to be counterfactual and strange. No, you don't. I was talking about the affine case. In the projective case, obviously, points and lines are dual. AND I SAID THAT. : Why are you trying to tell me not to apply a technique that I know from : personal experience has worked? SHUT the FUCK up, bitch. THAT is, AGAIN, the sort of insult up with which I need not put. -- --- It's difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America ==== > : Why are you trying to tell me not to apply a technique that I know from > : personal experience has worked? SHUT the FUCK up, bitch. > THAT is, AGAIN, the sort of insult up with which I need not put. You keep using that word insult. I do not think it means what you think it means. I also do not think you are making a positive contribution to this newsgroup. *plonk* -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science ==== David Eppstein insultingly asks, : > : Why do you want to blinker us? I politely replied, : > I don't. : Give it a break George. Go back to the hell that spawned you, mitch. : You know what you are doing. Of course I do, but you don't, so why are you presuming to pontificate about it? : > The general thrust of this thread is about : > representation theorems, not blinkering. : : Now who is lying? You are, fool. I FOUNDED this thread. I POSTED several messages in it about representation theorems. : You bring ridiculous concepts like burden of proof concerning metaphysical : positions like foundationalism WHERE? QUOTE ME OR SHUT THE *FUCK* UP, BITCH. : to mathematical discussions I STARTED this discussion, IDIOT. IF I WANT to discuss foundations, THEN I CAN DO THAT, both on sci.logic and sci.math, AND THERE ISN'T *SHIT* that *YOU* can do about it. : and aren't honest about : it to the people with whom you are in discussion. About what, specifically, am I being dishonest? Will you please QUOTE SOMETHING I SAID and then EXPLAIN why it is false? THEN you will become entitled to say what you just said. UNTIL you can do that, you are perfectly welcome to eat shit and die. ==== > David Eppstein insultingly asks, : > : Why do you want to blinker us? I politely replied, : > I don't. : Give it a break George. Go back to the hell that spawned you, mitch. : You know what you are doing. Of course I do, but you don't, so why > are you presuming to pontificate about it? > I'm not. If after having changed the topic in this thread you had redirected it under sci.logic alone (as you have repeatedly stated that you do because you do not want to be on math threads) I would have left my simple lol response without further comment, if even that. But you are in the business of a separate debate having nothing to do with mathematics proper. My presumption comes from a year of experience with your posts on sci.logic. It is inappropriate to engage people with a specific disinterest in the background motivation for your questions (forcing ontological commitment), blinker them, and then claim that is not what you are doing. As for the hell that spawned me--it was a course on intermediate logic in a philosophy department. I received an 'A' and a recommendation for graduate school. As for why I have been on sci.logic... that is because it is an appropriate unmoderated newsgroup for posting ideas about the foundations of mathematics (given the general disinterest in these matters among mathematicians as you yourself have noted). Look George. I am sorry for coming down on you like I did. But, there is a difference between opinions concerning foundations and formal philosophic proposals. Likewise, there is a difference between a discussion and an interrogation. :-) mitch Hell hath no bounds, Nor is circumscribed in any one self place. For where we are is hell, And where hell is, There must we ever be. And to conclude, When all the world dissolves And every creature shall be purified, All places shall be hell that is not heaven. --Mephistopheles The Tragedy of Doctor Faustus by Christopher Marlowe It is, perhaps, true. Ignorance is bliss. One thing I find admirable about people of faith is that they begin with the principle that they cannot know the essential truths of the universe and then use that as a foundation for ethical guidance. ==== : > : > : You know what you are doing. I replied, > Of course I do, but you don't, so why > are you presuming to pontificate about it? : I'm not. Yes, you are. I will QUOTE you doing so, in a second. : If after having changed the topic in this thread you had redirected it : under sci.logic alone (as you have repeatedly stated that you do because you do not : want to be on math threads) I would have left my simple lol response without : further comment, if even that. This is idiotic. I already had an exchange with Tim Chow where I EXPLAINED WHY this thread is still on sci.math. And you OF ALL PEOPLE are NOT qualified to judge what does or doesn't belong on sci. math. : But you are in the business of a separate debate having nothing to do with : mathematics proper. That is an idiotic lie and MORE OF the same pontification-from-ignorance about what I am doing that you just falsely alleged about yourself that you don't engage in. This debate is not separate from math. It does MATTER what you think about concreteness and it does MATTER what pedagogically useful prototypical categories might be. : My presumption What EXACTLY are you presuming? Do you mean presumption as in presuming a logical premise, or presumption as in daring to DO something you are not entitled to do? : comes from a year of experience with your posts on sci.logic. Which, tragically, has taught you nothing. : It is inappropriate to engage people with a specific disinterest in : the background motivation for your questions A motivation which you are personally both intellectually and psychologically completely UNABLE to compute : (forcing ontological commitment), and that was a really shitty guess at it, since I am NOT trying to force any kind of ontological commitment. I already posted a reply to James Dolan in which I completely agreed with him that this was about noticing that tons of viewpoints are ontologically relevant, whether anyone wants them to be or not, and that far from committing to some of them, we were rather about recognizing isomorphisms and representation-theorems that would allow us to think in terms of 1 representative viewpoint. In this thread, I am about electing representatives, NOT ontological commitment. But more to the point, They can ALWAYS JUST IGNORE IT. I ADVERTISED that I was gear-shifting the discussion. I did not lure anybody into this under false pretenses, or engage anyone against their will. Insinuations and pontifications to the effect that I *have* done that are just bullshit and should be flushed ALONG with the fool who was fool enough (or hateful enough) to assert them. : blinker them, I have not blinkered anybody. I can't help it if Dave Eppstein thinks you can have a vertical line in an affine geometry (I suppose next he will be telling us that you can have an origin). I can't help it if the fact that there are easy transformations from affine to projective geometry has him wrongly speaking of a point/line duality in the affine case, when in fact such only obtains in the projective case. I am NOT blinkering anybody. If you are going to publicly claim that I am then you are just going to make yourself part of the problem. : and then claim that is not what you are doing. Well, I RE-claim, IT'S NOT. : As for the hell that spawned me--it was a course on intermediate logic in a : philosophy department. Just because you got it THEN doesn't mean you get it NOW. In your own opinion, you have grown up a lot since then, and you now feel free to embrace perspectives that were NOT taught in THAT course. Not that that course would even be EXPECTED to give a shit about philosophy of math in ANY case. : I received an 'A' and a recommendation for graduate school. Well, that makes two of us. : Look George. I am sorry for coming down on you like I did. You would do well to just never apologize to me again, period. It won't matter. : But, there is a difference between opinions concerning foundations : and formal philosophic proposals. Yeah, but since I'm the one with the philosophy degree here, maybe you should ASK me instead of TELLING me what that difference is. : Likewise, there is a difference between a discussion and an interrogation. No, not likewise, not at all. And interrogation is going to remain appropriate as long as you keep trying to evade basic questions. I mean, you don't see *me* whining about getting interrogated. ==== Download a free beta version of my new proof checking software, DC Proof 1.0 at http://www.dcproof.com/ (for MS Windows 95 and later) It is a learning aid to teach the fundamentals of logic and proof. After entering a premise or assumption, the user can select rules of inference for Logic, Set Theory and Number Theory from easy-to-use, drop-down menus. It should be impossible to write an invalid proof in this system! Feedback is immediate with each new line. A tutorial with examples and exercises demonstrating the main features of DC Proof is included along with the user manual. (Click the Help** Button.) DC Proof is interesting from both pedagogical and theoretical perspectives. The axioms for set theory, for example, are a simplified version of the standard ZF model. Yet they are robust enough to avoid the known contradictions of naive set theory, and, I believe, to develop much of modern-day mathematical theory. Please let me know what you think of my new system. Dan Christensen Toronto, Canada ==== > Download a free beta version of my new proof checking software, DC Proof 1.0 > at http://www.dcproof.com/ (for MS Windows 95 and later) It is a learning aid to teach the fundamentals of logic and proof. After > entering a premise or assumption, the user can select rules of inference for > Logic, Set Theory and Number Theory from easy-to-use, drop-down menus. It > should be impossible to write an invalid proof in this system! Feedback is > immediate with each new line. A tutorial with examples and exercises demonstrating the main features of DC > Proof is included along with the user manual. (Click the Help** Button.) DC Proof is interesting from both pedagogical and theoretical perspectives. > The axioms for set theory, for example, are a simplified version of the > standard ZF model. Yet they are robust enough to avoid the known > contradictions of naive set theory, and, I believe, to develop much of > modern-day mathematical theory. Please let me know what you think of my new system. Dan Christensen Toronto, Canada Neither the Help menu nor the Help** button produced any help. Here is what got installed... C:Program FilesDC Proofasycfilt.dll C:Program FilesDC Proofcomcat.dll C:Program FilesDC ProofCOMDLG32.OCX C:Program FilesDC ProofDC Proof.chm C:Program FilesDC ProofDC Proof.exe C:Program FilesDC ProofDefaultDir.bin C:Program FilesDC Proofe.gif C:Program FilesDC Proofhhctrl.ocx C:Program FilesDC Proofmsvbvm60.dll C:Program FilesDC Proofoleaut32.dll C:Program FilesDC Proofolepro32.dll C:Program FilesDC ProofRecentFiles.bin C:Program FilesDC ProofRICHTX32.OCX C:Program FilesDC Proofstdole2.tlb C:Program FilesDC Proofunins000.dat C:Program FilesDC Proofunins000.exe -- Clive Tooth http://www.clivetooth.dk ==== Download a free beta version of my new proof checking software, DC Proof > 1.0 > at http://www.dcproof.com/ (for MS Windows 95 and later) It is a learning aid to teach the fundamentals of logic and proof. After > entering a premise or assumption, the user can select rules of inference > for > Logic, Set Theory and Number Theory from easy-to-use, drop-down menus. It > should be impossible to write an invalid proof in this system! Feedback is > immediate with each new line. A tutorial with examples and exercises demonstrating the main features of > DC > Proof is included along with the user manual. (Click the Help** Button.) DC Proof is interesting from both pedagogical and theoretical > perspectives. > The axioms for set theory, for example, are a simplified version of the > standard ZF model. Yet they are robust enough to avoid the known > contradictions of naive set theory, and, I believe, to develop much of > modern-day mathematical theory. Please let me know what you think of my new system. Dan Christensen Toronto, Canada Neither the Help menu nor the Help** button produced any help. Here is what > got installed... C:Program FilesDC Proofasycfilt.dll > C:Program FilesDC Proofcomcat.dll > C:Program FilesDC ProofCOMDLG32.OCX > C:Program FilesDC ProofDC Proof.chm > C:Program FilesDC ProofDC Proof.exe > C:Program FilesDC ProofDefaultDir.bin > C:Program FilesDC Proofe.gif > C:Program FilesDC Proofhhctrl.ocx > C:Program FilesDC Proofmsvbvm60.dll > C:Program FilesDC Proofoleaut32.dll > C:Program FilesDC Proofolepro32.dll > C:Program FilesDC ProofRecentFiles.bin > C:Program FilesDC ProofRICHTX32.OCX > C:Program FilesDC Proofstdole2.tlb > C:Program FilesDC Proofunins000.dat > C:Program FilesDC Proofunins000.exe -- > Clive Tooth > http://www.clivetooth.dk Somehow the HTML Help viewer isn't being made available to my program as it is downloaded. I will investigate and report back. In the mean time, the download is temporarily not available. Sorry for the inconvenience. Dan ==== Download a free beta version of my new proof checking software, DC Proof > 1.0 > at http://www.dcproof.com/ (for MS Windows 95 and later) It is a learning aid to teach the fundamentals of logic and proof. After > entering a premise or assumption, the user can select rules of inference > for > Logic, Set Theory and Number Theory from easy-to-use, drop-down menus. It > should be impossible to write an invalid proof in this system! Feedback is > immediate with each new line. A tutorial with examples and exercises demonstrating the main features of > DC > Proof is included along with the user manual. (Click the Help** Button.) DC Proof is interesting from both pedagogical and theoretical > perspectives. > The axioms for set theory, for example, are a simplified version of the > standard ZF model. Yet they are robust enough to avoid the known > contradictions of naive set theory, and, I believe, to develop much of > modern-day mathematical theory. Please let me know what you think of my new system. Dan Christensen Toronto, Canada Neither the Help menu nor the Help** button produced any help. Here is what > got installed... C:Program FilesDC Proofasycfilt.dll > C:Program FilesDC Proofcomcat.dll > C:Program FilesDC ProofCOMDLG32.OCX > C:Program FilesDC ProofDC Proof.chm > C:Program FilesDC ProofDC Proof.exe > C:Program FilesDC ProofDefaultDir.bin > C:Program FilesDC Proofe.gif > C:Program FilesDC Proofhhctrl.ocx > C:Program FilesDC Proofmsvbvm60.dll > C:Program FilesDC Proofoleaut32.dll > C:Program FilesDC Proofolepro32.dll > C:Program FilesDC ProofRecentFiles.bin > C:Program FilesDC ProofRICHTX32.OCX > C:Program FilesDC Proofstdole2.tlb > C:Program FilesDC Proofunins000.dat > C:Program FilesDC Proofunins000.exe -- > Clive Tooth > http://www.clivetooth.dk > Clive, Perhaps you do not have the Windows HTML Help viewer. For the latest version of it, go to http://msdn.microsoft.com/library/default.asp?url=/library/en-us/htmlhelp/ht ml/hwHTMLHelpFrequentlyAskedQuestions.asp Click on Downloads. Read the notes carefully. Different versions of Windows may require different solutions. If this does not seem applicaple, see the FAQ's at this link. Please let me know if you continue to have problems. Dan ==== > Clive, Perhaps you do not have the Windows HTML Help viewer. For the latest version > of it, go to http://msdn.microsoft.com/library/default.asp?url=/library/en-us/htmlhelp/htm l/hwHTMLHelpFrequentlyAskedQuestions.asp Click on Downloads. Read the notes carefully. Different versions of > Windows may require different solutions. If this does not seem applicaple, see the FAQ's at this link. Please let me > know if you continue to have problems. Dan Haven't had a look at the software yet, but you've usually got to write the path to the required help file in the registry.... Does this registry entry exist? Bruce. ==== within my diploma thesis I read chapter 2 in the book NumberTheory in Function Fields from Michael Rosen. I followed the advice to translate some results for $Z$ into the context of the polynomial ring $A=F_{p}[x]$. I did this also for the following function: Let $m=p_{1}^{e_{1}} cdot p_{2}^{e_{2}} cdot ldots cdot p_{t}^{e_{t}}$ where $p_{i}$ prime. Then define $g(m)=1$ if $m=1$ and $g(m)=max(e_{1},e_{2},ldots,e_{t})$ if $m>1$. If one calculates the average value of $g$, one gets 1.705221, the constant of Niven. I translated the function to the ring $A$ in the following way: Let $f=P_{1}^{e_{1}} cdot P_{2}^{e_{2}} cdot ldots cdot P_{t}^{e_{t}}$ where $P_{i}$ irreducible polynomials in A. Then I defined $g(f)=1$ if $deg(f)=0$ or $deg(f)=1$ and $g(f)=max(e_{1},e_{2},ldots,e_{t})$ if $deg(f)>1$. I calculated $frac{1}{p^{n}} cdot sumlimits g(f)$ (summing over all monic $f$ of degree $n$) for different $p$ and increasing $n$. I got a lot of numerical values, which led me to the following statement: $lim_{n to infty} frac{1}{p^{n}} cdot sumlimits g(f) = 1 + frac{1}{p-1}$. Now I`d like to know if one of you out there knows this result and if so, could tell me, where I can find it. I`d like to compare my prove to the known one. Yours sincerly Stefan Wagenbrenner ==== >I've never posted before, but as it is, it's 4:48am and for some >>stupid reason, I've been trying to solve a double integral for almost >>3 hours now. >>It's just personal now (not sure of notation here so i'm going to >>display it twice): >>/ 1 / arccos y >>| | exp (sin(x)) dx dy >>/ 0 / 0 >>or >>integrate relative to y {0..1} >>integrate relative to x {0..arccos(y)} >>the equation: exp(sin(x)) dx dy >>My trusty calculator can give me an answer, but for some reason >>(fatigue for example) I can't figure out how to come up with the >>answer. >>any help? >> >Change the order of integration: x from 0 to pi/2 and y from 0 to cos x. KP That's the wrong area, y goes from cos x to 1, which still leaves a problem. --Lynn ==== No, y goes from 0 to cos x. If 0 < =x <= arccos y, then upon taking cosines (a decreasing function when 0 <= x <= pi/2) 1 >= cos x >= y or y <= cos x <= 1. This is NOT the same as cos x <= y <= 1! cos x <= 1 is assured in the real domain. leaving y <= cos x. The outer limits in the original integral imply y >= 0, so we end up with 0 <= y <= cos x. --OL ==== >No, y goes from 0 to cos x. If 0 < =x <= arccos y, then upon taking >cosines (a decreasing function when 0 <= x <= pi/2) 1 >= cos x >= y or y ><= cos x <= 1. This is NOT the same as cos x <= y <= 1! cos x <= 1 is assured in the real domain. leaving y <= cos x. The outer >limits in the original integral imply y >= 0, so we end up with 0 <= y ><= cos x. --OL Oops, you're correct of course. Guess I was a little hasty on the keyboard. --Lynn ==== : (What I presented was an incidence matrix for the trivial affine plane because : it was easy to cut-and-paste from recent material I had written. Like a good : mathematician, George visualized the affine geometry from the presentation. My : words explaining otherwise made no difference.) That is a lie about what actually happened, not that any of you should care. If in fact the incidence matrix posted was the incidence matrix for the smallest affine plane, then no words EVER COULD POSSIBLY explain otherwise, for the sole and simple reason that the TRUTH is NOT otherwise. If mitch wanted to say something about this matrix OVER AND ABOVE, IN ADDITION TO (as OPPOSED to otherwise than) its being the one for this little plane, well, I read it, and it is NOT the case that it made no difference to me. It is, however, the case that what he was saying about it was ridiculous. -- --- It's difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America ==== > .... > the axis are the eigenvector of B, and the 'diameters' are > d^2= 1/ eigenvalue > ( is it correct? I think it is) > .... Not quite. You've missed a factor of 2 or 4. The semi-axis (_half_ the diameter you're looking at) is 1/sqrt(eigenvalue). There's a handy little reference written by a statistician, giving lots of facts about such things: M.G. Kendall, A course in the what you want. Ken Pledger. ==== >and sorry for the silly q., but i'm not a mathematician preamble( I hope this is correct) >let > A= >[a11 a12 0; >a21 a22 0; >0 0 -1] A is an ellipse >a11 x^2 + a22 y^2 +2 a12 x y =1 >if, let > B= >[a11 a12 ; >a21 a22] - B is symmetric >- B is positive definite the axis are the eigenvector of B, and the 'diameters' are >d^2= 1/ eigenvalue >( is it correct? I think it is) I wish to know if I can generalize in Rn ; A= [a11 .... a1n 0; > ..... >an1 an2 ... ann 0; >0 0 0 .... 0 -1] B= [a11 .... a1n ; >.... >an1 an2 ... ann] >What about the eigenvalues and eigenvector of B? Are they related to the axes of the ' hyperellipsoid', like in R2 ?? > Yes. You can see this by using an eigenbasis of B as a coordinate system. In that (orthonormal) coordinate system, the equation is e1*x1^2 + e2*x2^2 + ... + en*xn^2, where e1, e2, etc. are the eigenvalues and x1, x2, etc. are the coordinates (with respect to the eigenbasis). This describes an ellipsoid with the coordinate directions as axes and with diameters as you indicated. John Mitchell ==== > and sorry for the silly q., but i'm not a mathematician preamble( I hope this is correct) > let > A= > [a11 a12 0; > a21 a22 0; > 0 0 -1] A is an ellipse > a11 x^2 + a22 y^2 +2 a12 x y =1 > if, let > B= > [a11 a12 ; > a21 a22] - B is symmetric > - B is positive definite the axis are the eigenvector of B, and the 'diameters' are > d^2= 1/ eigenvalue > ( is it correct? I think it is) Yes. I wish to know if I can generalize in Rn ; A= [a11 .... a1n 0; > ..... > an1 an2 ... ann 0; > 0 0 0 .... 0 -1] B= [a11 .... a1n ; > .... > an1 an2 ... ann] > What about the eigenvalues and eigenvector of B? Are they related to the axes of the ' hyperellipsoid', like in R2 ?? You can use exactly the same formulas. Perhaps you may want to look up the term quadratic form over R. This has to do with the so-called spectral theorems. -- Phyics is much too hard for physicists. ==== > One thing I don't get, is a big picture of how well I'm > doing. My investment dollars are spread across multiple > IRA/401K/SEP/Roth accounts, as well as six different > non-retirement accounts at 2 brokers. (don't ask -- this > can't currently be changed). Anyway, aggregating the > data from these various sources is easy, analysing the > data, in aggregate, is *not*. > The software 'KBH Investor Accounting' outputs a year- to-date gain of realized and un-realized positions combined together...for each date of portfolio activity. It calls this a mark-to-market accounting. The application also shows a year-to-date portfolio percentage gain relative the net of deposits and withdrawals...and it calls this a profit margin. Finally, the application shows a year-to-date portfolio percentage gain relative to a time-weighted net of deposits and withdrawals and it calls this a time-weighted profit margin. Note that this last feature is an annualization or an IRR within the yearly period. Obviously, a consolidated accounting should be input. I would recommend a consolidated accounting for the taxable portfolio and a consolidated accounting for the tax-deferred portfolio. Here is a user link to the application: http://pages.prodigy.net/halsteadinvest/kbh.htm ==== > This may be better placed in one of the investing newsgroups, > but I find that most of the posters there can't handle basic > arithmetic, let alone what I'm trying to do.... You could find it worth while to follow some of the links from Ken Pledger. ==== http://online.itp.ucsb.edu/online/cmb02/caldwell/ http://qedcorp.com/APS/EmergentGravity.pdf http://qedcorp.com/APS/StarGate1.mov Caldwell does not seem to realize that the attractive dark matter can also be a zero point energy /zpf effect in addition to the repulsive dark energy on different scales. Latest data is dark energy/dark matter split is 0.7 to 0.3 in spatially flat sum Omega = 1 of FRW cosmology metric with uncertainty 0.04. Ordinary matter (stars, planets, radiation, dust clouds etc are inside that 0.04 uncertainty). My key new idea here is that dark matter detectors will not click with the that is on mass shell in terms of quantum field theory Feynman propagators, whizzing around in space like neutrinos passing through our bodies. That is wrong IMHO. Dark matter is, like dark energy a virtual, i.e. off mass shell exotic vacuum effect with w = (pressure)/(energy density) = -1 because the total local zero point stress-energy density vacuum Diff(4) symmetry group tensor is tuv(Vacuum) = (String Tension)/zpf(x)guv(Curved Space-Time) where /zpf (x) = Lp*^-2[Lp*^3/2|Vacuum Coherence(x)|^2 - 1] Lp*^2 = hG*/c^3 hc/Lp*^2 = String Tension G* is the variable scale-dependent effective gravity coupling coefficient. Something like G*/G(Newton) = e^ ? The NON-EXOTIC i.e. non-gravitating vacuum region of space-time is when /zpf = 0 When Vacuum Coherence = 0, like in the inside of a vibrating string of pure energy (Brian Greene NOVA Elegant Universe) we have a strongly attractive dark energy core /zpf(core) = -1/Lp*^2 that, for example, explains the self-cohesiveness of the spatially extended electron, which is a micro-geon Kerr-Newman ring singularity of size ~ e^2/mc^2 ~ 1 fermi surrounded by a virtual plasma of electron-positron pairs and virtual photon extending out to ~ h/mc ~ 10^-11 cm. This whole spatially extended complex looks more and more like a Bohmian hidden variable point enormous space-warping since G*m^2/hc ~ 1 where m ~ 10^-27 gms. realism is consistent with J.P. Vigier's idea for a non-nuclear release of atomic energy from tight atomic states like the experiments in Beograd, Serbia were looking for. ==== http://online.itp.ucsb.edu/online/cmb02/caldwell/ > http://qedcorp.com/APS/EmergentGravity.pdf http://qedcorp.com/APS/StarGate1.mov Caldwell does not seem to realize that the attractive dark matter can > also be > a zero point energy /zpf effect in addition to the repulsive dark energy on > different scales. Latest data is dark energy/dark matter split is 0.7 to 0.3 in spatially > flat sum > Omega = 1 of FRW cosmology metric with uncertainty 0.04. Ordinary > matter (stars, planets, radiation, dust clouds etc are inside that 0.04 > uncertainty). > My key new idea here is that dark matter detectors will not click > with the > that is on mass shell in terms of quantum field theory Feynman > propagators, whizzing > around in space like neutrinos passing through our bodies. That is wrong > IMHO. > Dark matter is, like dark energy a virtual, i.e. off mass shell > exotic vacuum effect > with w = (pressure)/(energy density) = -1 because the total local zero > point stress-energy density > vacuum Diff(4) symmetry group tensor is tuv(Vacuum) = (String Tension)/zpf(x)guv(Curved Space-Time) where /zpf (x) = Lp*^-2[Lp*^3/2|Vacuum Coherence(x)|^2 - 1] Lp*^2 = hG*/c^3 hc/Lp*^2 = String Tension G* is the variable scale-dependent effective gravity coupling coefficient. Something like G*/G(Newton) = e^ ? The NON-EXOTIC i.e. non-gravitating vacuum region of space-time is when /zpf = 0 When Vacuum Coherence = 0, like in the inside of a vibrating string of > pure energy (Brian Greene NOVA Elegant Universe) > we have a strongly attractive dark energy core /zpf(core) = -1/Lp*^2 > that, for example, explains the self-cohesiveness of the > spatially extended electron, which is a micro-geon Kerr-Newman ring > singularity of size ~ e^2/mc^2 ~ 1 fermi surrounded by > a virtual plasma of electron-positron pairs and virtual photon extending > out to ~ h/mc ~ 10^-11 cm. This whole spatially extended > complex looks more and more like a Bohmian hidden variable point > enormous space-warping since G*m^2/hc ~ 1 where m ~ 10^-27 gms. realism is consistent with J.P. Vigier's idea for a non-nuclear release > of atomic energy from tight atomic states like the experiments in > Beograd, Serbia were looking for. I love it when Sarfatti uses IMHO. It means just the opposite. ==== Here is the question: > Find the distance between the lines L(1) = (1, 1, 1) + s (1, 2, 3) and L(2) = (0, 1, 1) + t (1, -1, -1). I would really appreciate a full solution because many people have > given me tips on another message board and I have been unable to solve > this. I know that is not the style of this newsgroup but I would really > appreciate it as I have an exam tomorrow morning and I am sure a > question almost identical to this will be on it. Zack Using vector notation , with P*Q as dot product of vectors P and Q > and P^Q as cross product of P and Q (see below), let A = (1,1,1), > U = (1,2,3), B = (0,1,1), V = (1,-1,1-) then U is parallel to line > L(1) and V is parallel to line L(2) so that C = (A^B) is > perpendicular to both lines. Then (A - B)*C/sqrt(C*C) is the projection of A-B parallel to C ( > and therefore perpendicular to both lines) and is its absolute value > is desired perpendicular distance between the lines. > Note, dot product is given by > (a,b,c)*(c,d,e) = ac + bd + cf > and cross product by > (a,b,c)^(d,e,f) = (bf - ce, cd - af, ae - bd) Urq. It has always seemed to me that this kind of problem can be solved with a lot less machinery and a lot more clarity by simply considering the distance (squared) between a point on line L(1) and a point on L(2): f(s,t) = || (1+s,1+2s,1+3s) - (t,1-t,1-t)||^2 = (1+s-t)^2 + (2s+t)^2 + (3s+t)^2. This function is a minimum when its partial derivatives are both zero. (And it does take a minimum, because f is convex and --> +infty as s^2 + t^2 --> +infty.) That leads to simultaneous equations 2(1+s-t) + 4(2s+t) + 2(3s+t) = 0 i.e. 28s + 8t = -2, -2(1+s-t) + 2(2s+t) + 2(3s+t) = 0, i.e. 8s + 6t = 2. These are easily solved to give unique solution s = -7/26, t = 9/13 (hence, there is a unique pair of closest points). We have f(-7/26, 9/13) = 1/26, and therefore the distance between the lines is 1/sqrt(26). I would MUCH prefer that students be able to use a seat-of-the-pants, common-sense method like this instead of the cookbook procedure taught in section 8.2 of chapter 4, on page 167. Besides, your solution fails as soon as you go to R^4. --Ron Bruck ==== > Here is the question: >> Find the distance between the lines >> >> L(1) = (1, 1, 1) + s (1, 2, 3) and L(2) = (0, 1, 1) + t (1, -1, -1). >> >> I would really appreciate a full solution because many people have >> given me tips on another message board and I have been unable to solve >> this. I know that is not the style of this newsgroup but I would really >> appreciate it as I have an exam tomorrow morning and I am sure a >> question almost identical to this will be on it. >> >> Zack Using vector notation , with P*Q as dot product of vectors P and Q >and P^Q as cross product of P and Q (see below), let A = (1,1,1), >U = (1,2,3), B = (0,1,1), V = (1,-1,1-) then U is parallel to line >L(1) and V is parallel to line L(2) so that C = (A^B) is >perpendicular to both lines. > Of course, you mean C = U^V, not A^B. --Lynn ==== > > Here is the question: >> Find the distance between the lines >> >> L(1) = (1, 1, 1) + s (1, 2, 3) and L(2) = (0, 1, 1) + t (1, -1, -1). >> >> I would really appreciate a full solution because many people have >> given me tips on another message board and I have been unable to solve >> this. I know that is not the style of this newsgroup but I would really >> appreciate it as I have an exam tomorrow morning and I am sure a >> question almost identical to this will be on it. >> >> Zack Using vector notation , with P*Q as dot product of vectors P and Q >and P^Q as cross product of P and Q (see below), let A = (1,1,1), >U = (1,2,3), B = (0,1,1), V = (1,-1,1-) then U is parallel to line >L(1) and V is parallel to line L(2) so that C = (A^B) is >perpendicular to both lines. > Of course, you mean C = U^V, not A^B. --Lynn Right! Proof reading is such a bore! ==== > Huh? How in the world does it suggest that? None of the replies > have agreed with you... I felt there might be a bit of defensive feeling behind some of the posts. Is the geometrical approach the best one for anyone who just wants to study functions? Then as a technique analytical continuation is only practical with the aid of a computer program. If cuts are allowed I can treat some algebraic functions (I do not claim all) without using it. ==== >The fact that my original post has to date attracted 19 replies >suggests that there are others who do not see Riemann surfaces as an >entirely finished subject. >> Huh? How in the world does it suggest that? None of the replies >> have agreed with you... I felt there might be a bit of defensive feeling behind some of the >posts. Huh? >Is the geometrical approach the best one for anyone who just >wants to study functions? Huh? What does this have to do with anything that's been discussed here? The question was whether there are necessarily arbitrary choices involved in constructing the Riemann surface for a function... >Then as a technique analytical continuation >is only practical with the aid of a computer program. Huh???????? >If cuts are >allowed I can treat some algebraic functions (I do not claim all) >without using it. Huh? Nobody's said anything about cuts not being allowed. The question was [see above]. ************************ David C. Ullrich ==== large n (say n > 100 digits). Since n is that large, it's not very efficient to store the lookup table. I know it's not easy to find a polynomial that works. I am looking for a deterministic algorithm (likely to be recursive) probably with some finite lookup table storage. Would studying the distribution of primes help? As I know given a particular prime, it is possible to find the next m primes -- is that true? in which I need to map an integer identity (ID) to a > prime number. I want h( ): Z_n -> Z-m (m > n) s.t. for every integers in > Z_n, I can easily find a prime in Z_m that is distinct from > the image of another integer. i.e. h(ID_i)=h(ID_j) => ID_i = ID_j. Please note that I don't want a lookup table for the mapping --- > that's what I meant efficient. What's not efficient about table lookup? Do you want to do this for one particular n? or do you want something > that works for all n? If the latter, you're looking for a simple (efficient) formula > f(a) that returns a different prime for each positive integer a. > You're unlikely to find anything to your liking. Certainly there > is no polynomial that works. If n < 40 then you can use Euler's polynomial, n^2 + n + 41 or > something like that. -- ==== > large n (say n > 100 digits). Whoa! You want a function that gives you more than 10^100 primes? and only primes? You're dreaming. > As I know given a particular prime, it is possible to > find the next m primes -- is that true? Yes, but only by testing each number after your prime until you find m more. Suppose you know some 100-digit prime, p, and you want to find the next 10 primes. Well, roughly one number in each 230 is prime up there, so, to be on the safe side, maybe you want to look at p + k for k from 1 to 5000. Now you can sieve out a lot of those k by deleting multiples of 2, of 3, of 5, etc, and then subject the remaining numbers to primality tests. Exactly how much sieving to do, I'll leave to the experts (who may know a better approach than what I'm outlining, but I don't see how one can do fantastically better). -- ==== > If I have 2 - sqrt(12)i how do I find the polar form? So far I've worked out that r = 4 and the argument is 5pi/6. The next step I'm not sure about. Do I put it in the form r(cos(theta) + > isin(theta))? Making it 4(cos(5pi/6) + isin(5pi/6))? Then what do I do? BTW - what is the correct notation on here for angle theta? TIA > You've done it perfectly Sorry I should have elaborated. How do I get it into polar form with cis > (whatever that's called)? cis(theta) is merely an abreviation for cos(theta)+i*sin(theta), and is equivalent to exp(i*theta) and e^(i*theta) Thus given x = r*cos(theta) and y = r*sin(theta), one can write x + i*y = r*cis(theta) = r*exp(i*theta) = r*e^(i*theta) ==== In sci.math, Michael : > If I have 2 - sqrt(12)i how do I find the polar form? So far I've worked out that r = 4 and the argument is 5pi/6. (Note: the mathematical symbol for theta looks a bit like an O with a horizontal hyphen through it.) If z = a + bi, then r = sqrt(a^2 + b^2) and theta = atan2(b,a), where atan2() is a fairly widespread function on computer systems. More conventionally, theta = atan(b/a) if a > 0, atan(b/a) + pi if a < 0, pi/2 if a = 0 and b > 0, -pi/2 if a = 0 and b < 0. (For a = b = 0, theta is conventionally taken to be 0. Since r is 0 it doesn't matter all that much in that case.) So r = sqrt(4 + 12) = 4 and theta = atan(-sqrt(12)/2) = atan(-sqrt(3)) = -pi/3. The next step I'm not sure about. Do I put it in the form r(cos(theta) + > isin(theta))? Making it 4(cos(5pi/6) + isin(5pi/6))? Then what do I do? Some use the notation (r,theta) but that's obviously slightly confusing if accidentally taken out of context. I think most simply state r = value, theta = value. Converting back to Cartesian isn't difficult: z = r*cos(theta) + i*r*sin(theta). On occasion one can also use the identity z = r*exp(i*theta) or your form z = r*(cos(theta) + i*sin(theta)). BTW - what is the correct notation on here for angle theta? By convention, just theta. TIA > -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > ... stuff deleted ... > ... given a map f : X --> Y between > sufficiently nice spaces (homotopy type of a CW complex should be nice > enough), the path-space construction: P_f = {(x,p) in X x map(I,X) | p(0) = x } yields a space homotopy-equivalent to X ... stuff deleted ... > ... and the map j > P_f -------> Y (x,p) |--> p(1) is a fibration. Further, the diagram f > X --------------->_Y > /| > / > / > i /j > / > / > / > _| / > P_f is commutative. In other words, every map is a fibration, up to > homotopy. Well, just as long as the spaces are > sufficiently well-behaved to permit the above > construction, and it's been long enough for me > since I worked with that construction that I > don't recall just what's needed. Dale > Just to tie up this loose end, I consulted Spanier (Algebraic Topology, now published by Springer-Verlag). He claims it's true for any spaces X,Y, and any (continuous) map f. I would be surprised if some further technical condition (e.g., normality) weren't necessary, but Spanier doesn't mention it. The introduction to the text does say that, absent any explicit restriction, the term space means any topological space, so it could be that this construction holds for arbitrary topological spaces X,Y. I would, of course, welcome any correction on this matter. Dale. ==== I need help on the following problem. that Kerv={h in H| v(h)=0}={0} and v(H)is dense in H but does not coincide with it. Is it true that v(H)=H or there can exist such an operator?? I am looking forward to receiving your answers . Greetings Mladen ==== >I need help on the following problem. >that >Kerv={h in H| v(h)=0}={0} and v(H)is dense in H but does not coincide >with it. Is it true that v(H)=H or there can exist such an operator?? >I am looking forward to receiving your answers . >Greetings Mladen basis |n>, n = 1, 2, 3, ..., and define the linear operator v by v |n> = 1/n |n> for n = 1, 2, 3, .... Then v is continuous, Ker(v) = 0, and the range of v is dense, but the element sum_{n=1}^infty 1/n |n> Hope this helps. David McAnally -------------- ==== I need help on the following problem. >that >Kerv={h in H| v(h)=0}={0} and v(H)is dense in H but does not coincide >with it. Is it true that v(H)=H or there can exist such an operator?? Certainly there can exist such an operator. For example say H is L^2(R) and v f(x) = f(x)/(1+x^2). >I am looking forward to receiving your answers . Greetings Mladen ************************ David C. Ullrich ==== > I need help on the following problem. > that > Kerv={h in H| v(h)=0}={0} and v(H)is dense in H but does not coincide > with it. Is it true that v(H)=H or there can exist such an operator?? square summable sequences. ==== > >[...] Well this morning I was pissed I didn't have a solution to B6 - so >you can imagine how I felt when you recorded my pissed-off >non-solution there following an actual solution. So I thought >about it. It's easy if int f = 0, which seems like it should be >the hard case, for example it's trivial if f >= 0. This morning >I was going to say just as a joke that you could get a solution >in general by interpolating between these two cases. But >it turns out that you can do exactly that. The part you want >to record for posterity starts here: Aargh. What's below is wrong. Wrong, wrong wrong. I still think it's curious that the case int f = 0 is easy, > and it still seems that the general case should follow, > since if f has a lot of cancellation that should make it > harder. But what's below is nonsense. > >It's clear if f >= 0. Suppose that int f = 0. Choose a function >g so that |g| = 1 and gf >= 0. Then int int |f(x) + f(y)| >= int (f(x) + f(y)) g(x) = int f(x) g(x) = int |f(x)|. You can do the general case by interpolating between >these special cases: Suppose that int f > 0. Then >f = g + h, where int g = 0, h >= 0, and g and h have >disjoint support. Say g is supported on A and h is >supported on B. Then int int |f(x) + f(y)| >= int_A int_A |f(x) + f(y)| + int_B int_B |f(x) + f(y)| > Let me try this again. int int |f(x) + f(y)| = int_A int_A |g(x)+g(y)| +int_B int_B (h(x)+h(y)) + 2 int_A int_B |g(x)+h(y)| >= int_A int_A |g(x)+g(y)| + int_B int_B (h(x)+h(y)) + 2 int_{x in A} int_{y in B} (g(x)+h(y)) Now David's argument shows that the first integral on the extreme right is >= int_A |g(x)|. The second is equal to 2m(B) int_B h, and the third is equal to 2m(A) int_B h. This yields int int |f(x) + f(y)| >= int_A |g| + 2m(B) int_B h + 2m(A) int_B h = int_A |g| + 2int_B h >= int_A |g| + int_B h = int |f|. -- A. ==== > >>[...] >>Well this morning I was pissed I didn't have a solution to B6 - so >>you can imagine how I felt when you recorded my pissed-off >>non-solution there following an actual solution. So I thought >>about it. It's easy if int f = 0, which seems like it should be >>the hard case, for example it's trivial if f >= 0. This morning >>I was going to say just as a joke that you could get a solution >>in general by interpolating between these two cases. But >>it turns out that you can do exactly that. The part you want >>to record for posterity starts here: >> >> Aargh. What's below is wrong. Wrong, wrong wrong. >> >> I still think it's curious that the case int f = 0 is easy, >> and it still seems that the general case should follow, >> since if f has a lot of cancellation that should make it >> harder. But what's below is nonsense. >> >>It's clear if f >= 0. Suppose that int f = 0. Choose a function >>g so that |g| = 1 and gf >= 0. Then >> int int |f(x) + f(y)| >= int (f(x) + f(y)) g(x) >> = int f(x) g(x) = int |f(x)|. >>You can do the general case by interpolating between >>these special cases: Suppose that int f > 0. Then >>f = g + h, where int g = 0, h >= 0, and g and h have >>disjoint support. Say g is supported on A and h is >>supported on B. Then >> int int |f(x) + f(y)| >> >= int_A int_A |f(x) + f(y)| + int_B int_B |f(x) + f(y)| Let me try this again. int int |f(x) + f(y)| = int_A int_A |g(x)+g(y)| +int_B int_B (h(x)+h(y)) > + 2 int_A int_B |g(x)+h(y)| >= int_A int_A |g(x)+g(y)| + int_B int_B (h(x)+h(y)) > + 2 int_{x in A} int_{y in B} (g(x)+h(y)) Now David's argument shows that the first integral on the extreme right >is >= int_A |g(x)|. Does it? It looks more like >= m(A) int_A |g(x)| to me (although considering my record here what it looks like to me is of little interest...) >The second is equal to 2m(B) int_B h, and the third >is equal to 2m(A) int_B h. This yields int int |f(x) + f(y)| >= int_A |g| + 2m(B) int_B h + 2m(A) int_B h = int_A |g| + 2int_B h >= int_A |g| + int_B h = int |f|. ************************ David C. Ullrich ==== > > >>[...] >>Well this morning I was pissed I didn't have a solution to B6 - so >>you can imagine how I felt when you recorded my pissed-off >>non-solution there following an actual solution. So I thought >>about it. It's easy if int f = 0, which seems like it should be >>the hard case, for example it's trivial if f >= 0. This morning >>I was going to say just as a joke that you could get a solution >>in general by interpolating between these two cases. But >>it turns out that you can do exactly that. The part you want >>to record for posterity starts here: >> >> Aargh. What's below is wrong. Wrong, wrong wrong. >> >> I still think it's curious that the case int f = 0 is easy, >> and it still seems that the general case should follow, >> since if f has a lot of cancellation that should make it >> harder. But what's below is nonsense. >> >>It's clear if f >= 0. Suppose that int f = 0. Choose a function >>g so that |g| = 1 and gf >= 0. Then >> int int |f(x) + f(y)| >= int (f(x) + f(y)) g(x) >> = int f(x) g(x) = int |f(x)|. >>You can do the general case by interpolating between >>these special cases: Suppose that int f > 0. Then >>f = g + h, where int g = 0, h >= 0, and g and h have >>disjoint support. Say g is supported on A and h is >>supported on B. Then >> int int |f(x) + f(y)| >> >= int_A int_A |f(x) + f(y)| + int_B int_B |f(x) + f(y)| Let me try this again. int int |f(x) + f(y)| = int_A int_A |g(x)+g(y)| +int_B int_B (h(x)+h(y)) > + 2 int_A int_B |g(x)+h(y)| >= int_A int_A |g(x)+g(y)| + int_B int_B (h(x)+h(y)) > + 2 int_{x in A} int_{y in B} (g(x)+h(y)) Now David's argument shows that the first integral on the extreme right >is >= int_A |g(x)|. Does it? It looks more like >= m(A) int_A |g(x)| to me (although > considering my record here what it looks like to me is of little > interest...) You and me both. Uncle. -- A. ==== >[...] >>Now David's argument shows that the first integral on the extreme right >>is >= int_A |g(x)|. >> >> Does it? It looks more like >= m(A) int_A |g(x)| to me (although >> considering my record here what it looks like to me is of little >> interest...) You and me both. Uncle. Yeah, but there _must_ be a one or two line proof along these lines. I mean, it's very simple if int f = 0. And int f = 0 says f has a lot of cancellation, which would make int int |f(x) + f(y)| small, so the case int f > 0 must follow, by a similar argument. (Shoulda just said that the case int f > 0 followed and seem whether anyone wanted to admit they didn't see how...) ************************ David C. Ullrich ==== >Rather, continue as: > = int_A int_A |g(x) + g(y)| + int_B int_B (h(x) + h(y)) > >= int_A int_A |g(x) + g(y)| + int_B h(x) Nuh-uh, this step is still wrong. You pick up a factor of 2 meas(B) on the right-hand side. I suspect that no solution which ignores the terms coming from the cross-regions A x B and B x A can be made to work for functions f with int f <> 0: if f is large when it's positive and small when it's negative, without the contributions |f(x) + f(y)| when f(x) is positive and f(y) is negative, you won't get the inequality that you want. Dave ==== > >Rather, continue as: = int_A int_A |g(x) + g(y)| + int_B int_B (h(x) + h(y)) > >= int_A int_A |g(x) + g(y)| + int_B h(x) Nuh-uh, this step is still wrong. You pick up a factor of 2 meas(B) on > the right-hand side. Ack! > I suspect that no solution which ignores the terms > coming from the cross-regions A x B and B x A can be made to work for > functions f with int f <> 0: if f is large when it's positive and small > when it's negative, without the contributions |f(x) + f(y)| when f(x) is > positive and f(y) is negative, you won't get the inequality that you want. -- A. ==== >> > > One of their favorite labels is crank, where the idea is that you > have some person who is unreasonable, who has ideas that don't work, > who refuses to acknowledge the truth. > > Let's see now... > > ...some person who is unreasonable, who has ideas that don't work, > who refuses to acknowledge the truth. > > Sounds just like you, James! >>Your opinion is noted Alec McKenzie. >>I do admit that if enough people think that I'm just some nut, then I >>have to wonder. >> >> _If_ enough people thought that? What reason do you have to >> think that _anyone_, with only one exception that I can think of >> right now, doesn't think of you as just some kind of nut? >> Note that I made a *specific* challenge in my previous post, which has >so far been ignored. I'll talk about it more a little later. And note that you're not answering the specific question I asked, in regard to a specific thing you said. >Now then, I'm reasonable, as I recognize that hey maybe I am nutty, >but then again, I also know enough math to realize that mathematicians >haven't made a logical case for total dismissal of my prime counting >discovery, and I know enough about human nature to be very wary when >people rely on insults, like here David Ullrich is imputing that >everyone with one exception thinks I'm a nut!!! Again: What reason do you have to think that anyone thinks otherwise? ************************ David C. Ullrich ==== > >> > > One of their favorite labels is crank, where the idea is that you > have some person who is unreasonable, who has ideas that don't work, > who refuses to acknowledge the truth. > > Let's see now... > > ...some person who is unreasonable, who has ideas that don't work, > who refuses to acknowledge the truth. > > Sounds just like you, James! >>Your opinion is noted Alec McKenzie. >>I do admit that if enough people think that I'm just some nut, then I >>have to wonder. >> >> _If_ enough people thought that? What reason do you have to >> think that _anyone_, with only one exception that I can think of >> right now, doesn't think of you as just some kind of nut? >> Note that I made a *specific* challenge in my previous post, which has >so far been ignored. I'll talk about it more a little later. And note that you're not answering the specific question I asked, > in regard to a specific thing you said. That's a rude question. My family doesn't think I'm insane you sick turd. So far you've shown just how despicable of a human being you can be, without any regard for common decency or caring, or ethics. If you have an issue with my prime counting discovery then why can't you keep to the mathematics? I have a *specific* challenge to you David Ullrich, I dare you to give an outlined dispute of the value of my prime counting discovery providing *all* the information necessary for readers to evaluate your claim, without trying to rely on links or insults. What kind of math professor are you if you can't handle a *math* challenge but need to insult instead? If you are a math expert, prove it. Give a point-by-point exposition to support your case without insults, innuendo, or any other attempts at sly or back-handed insults, but with *mathematics* and logic. I don't think you have the capacity to engage in an honest and open debate without trying to use tricks or insults David Ullrich. I think you lack the capacity to have a *reasoned* discussion, but instead rely on childish behavior. The challenge is to you now, show that you can reason and discuss mathematics reasonably, if you can. James Harris ==== [snip] >Note that I made a *specific* challenge in my previous post, which has >so far been ignored. I'll talk about it more a little later. The specific challenge you mention above has been ignored because it represents a non-issue. In your previous post you invite readers to attack a particular point when there is *already* a specific attack on a particular point on the table. You have so far *refused* to either acknowledge or defend the challenge to your claim that the so-called 'partial differential equation' solves the prime counting problem -- a claim you made repeatedly. C'mon, James Harris. It is *you* who claimed that the difference equation you posted does things which no other method does do or can do. Name them! What things? Where is the proof that the so-called 'partial differential equation' solves the prime counting problem? Where is your evidence? Where is your data? Note that no one is fooled or tricked by your childish attempt to divert attention from the *real* issues to some side show by inviting certain challenges. You ARE ALREADY CHALLENGED ON YOUR OWN CLAIMS. Put up or SHUT UP. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== >> > >> >> One of their favorite labels is crank, where the idea is that you >> have some person who is unreasonable, who has ideas that don't work, >> who refuses to acknowledge the truth. >> >> Let's see now... >> >> ...some person who is unreasonable, who has ideas that don't work, >> who refuses to acknowledge the truth. >> >> Sounds just like you, James! Your opinion is noted Alec McKenzie. I do admit that if enough people think that I'm just some nut, then I >have to wonder. > > _If_ enough people thought that? What reason do you have to > think that _anyone_, with only one exception that I can think of > right now, doesn't think of you as just some kind of nut? > >>Note that I made a *specific* challenge in my previous post, which has >>so far been ignored. I'll talk about it more a little later. >> >> And note that you're not answering the specific question I asked, >> in regard to a specific thing you said. That's a rude question. I was rude to _you_? Golly. >My family doesn't think I'm insane Right. Actually I was talking about your correspondents on usenet and elsewhere in the mathematical community. And I asked whether you had any reason to think that any of those people did not think of you as some kind of nut - there's a significant difference in strength between some kind of nut and insane. Look. _You_ said if enough people think that I'm just some nut, then I have to wonder. This makes the question of how many people would be enough, and why you think there are any exceptions, a perfectly natural question. > you sick turd. Uh, the expression you're looking for is fucking dogshit. I like it when you complain about people being rude to you... >So far you've shown just how despicable of a human being you can be, >without any regard for common decency or caring, or ethics. If you have an issue with my prime counting discovery then why can't >you keep to the mathematics? I have a *specific* challenge to you David Ullrich, I dare you to give >an outlined dispute of the value of my prime counting discovery >providing *all* the information necessary for readers to evaluate your >claim, I and many other people have done this, many times. > without trying to rely on links or insults. What kind of math professor are you if you can't handle a *math* >challenge but need to insult instead? If you are a math expert, prove >it. Give a point-by-point exposition to support your case without insults, >innuendo, or any other attempts at sly or back-handed insults, but >with *mathematics* and logic. I don't think you have the capacity to engage in an honest and open >debate without trying to use tricks or insults David Ullrich. I think >you lack the capacity to have a *reasoned* discussion, but instead >rely on childish behavior. The challenge is to you now, show that you can reason and discuss >mathematics reasonably, if you can. >James Harris ************************ David C. Ullrich ==== First, I'll directly address the title Common sense: If crank label were believed. Some already believe the label of crank applies, so dialogue present, it appears that the majority believe you are a crank to some degree. If you wish to reverse that image, your best option is to conduct yourself in a professional manner, not insult anyone for any reason, and fully respond to all questions and challenges about your claims. > Some of you may have noticed these free-wheeling mathematical > discussions I've been having with all these people, and wondered. Maybe mostly you were wondering what they were doing on your > newsgroup. Well I've been someone who has dabbled in mathematical research in > full knowledge that I'm NOT a mathematician, but partly because it's > fun, and partly because I have confidence that people can miss things. > Well, I'm not a mathematician either. > Mathematicians don't believe they could miss anything simple enough > that a non-mathematician could find that's important. And I'm not > just saying that as I've read it more than once from mathematicians. > Thats just wrong. I have corrected mathematics professors on several occasions and they thanked me. I have yet to find any mistake in a published paper by a mathematician, though when I find one, I do not expect to be belittled. > Now as I've talked about my various ideas and research, math people > have talked a bit about the math, but mostly they've retaliated by > insulting me and saying nasty things about me on webpages. > The post archive on google tells a different story. I looked at the early posts, and there were no insults. > Yes, it's odd but true. One of their favorite labels is crank, where the idea is that you > have some person who is unreasonable, who has ideas that don't work, > who refuses to acknowledge the truth. Only problem is that I enjoy tracing out the steps of my mathematical > arguments, and I can do things like give you a Java program that > works. > I have seen your mathematical arguments. Your tracing out the steps is actually rather poor. In your Advanced Factorization/Core Error thread, you only put in seven steps when I could see as many as thirty. You should try programming in C. Its much better for heavy mathematical computation. > The truth is that I've dabbled in a lot of mathematical areas, poking > around looking for something dramatic, and pissed off some math > people. Does that make me a crank? > What you need to address is what pissed off the other sci.math users. > When you contact mathematicians about specifics, what are the responses? > There's some guy who was going around quoting from me all the time in > his posts! Now then, use your common sense. Is it simpler that I'm really some crank, or that I've just upset a > lot of oversensitive people who think that negative labeling and > insults are the way to go? Now the problem is bad lessons learned on Usenet. I'm sure lots of > dabblers who had some ideas here or there were rapidly chased off of > the sci.math newsgroup when they were ganged up on and insulted. So > posters learned that they could control people by insults! They were > happy!!! They could insult people and get rid of them! But I came > along and decided that I didn't want to let people control my speech > by insults. So because I refused to be insulted into silence you have a weird > religion that's developed in the math community which is based on > obsessively insulting me! > You CAN stop the exchange of insults by not responding to them. > And I wanted others on other newsgroups to see that oddity. It's rather freaky, eh? But then again, who really thought math > people were normal? And don't worry, this post is very unlikely to > change anything. Yup, believe it or not, it will all just keep going. I'll talk about > math, there will be some who will argue with the math, while most will > just lob insults. > Stick to just the math and ignore the insults. When the insulters dont get a response, they will eventually stop. > Fascinating. > James Harris ==== > Granted, some might think I do the same, but he comes to me. That is, > David Ullrich makes sure to come to my threads and reply to my posts, > and then whines when I tell him to go away. He's a tag-along that just won't go away when he's told that he's not > wanted. This is usenet, a public service. You don't get to choose who participates in your threads. ==== James Harris, barely-literate yard-ape. Yard ape, curtain climber, carpet shark, ankle biter, nose miner, > sprog, linoleum lizard, loin monkey, fartling, crotch fruit, > shriekling, flesh loaf, womb dropping, little sticky person, crotch > trophy, howling shit-machine. But give James Harris credit: he fought off the rusty coat hanger to > evenutally emerge as a twat lugie. Hey stooopid loud troll James > Harris, put up or shut up, > LOL! Git 'im! ==== I *dare* any posters who disagree to give just a point-by-point attack > on my prime counting function, numbering out each point, and providing > all the information in their post rather than just posting links. > That's just silly. Your prime counting function is correct, as has been demonstrated several times. As I pointed out in an earlier post (quoting) # Thus we see that the proposed prime counting method # is indeed correct, though that wasn't in doubt. # # Is this Legendre's method? That depends on one's point of view. # It's certainly not an exact copy, but it is not difficult to # see that Legendre's prime counting formula may be derived from # this method and conversely, so in that sense the two techniques # are equivalent. If one were inclined to be unchartiable, one # could call this obfuscated Legendre; otherwise, it could be # designated as streamlined Legendre. # # Is it better than Legendre's method? Well, it does not require # knowing the primes, since they're computed on the fly. # Unfortunately, the resulting method is several orders of # magnitude slower than some existing methods, at least when # implemented as written. # # Does it have any applicability other than to produce a slow # method of counting primes? I don't see any, though it is # tidy and compact. It might have some use as a pedagogical # device, perhaps as an exercise in a text, but I have to admit # that I don't see any use beyond that. # # Is it publishable? Perhaps as a classroom note, but the journal # would have to be selected with care. It would face the same # difficulty as an improvement on an algorithm like bubble sort-- # the response by a great number of reviewers would be, I suspect, # This is mildly interesting, but it hardly advances the state # of the art. Rick ==== A geometry can be expressed by a form w: T(M^n) -> C^n. We can define invariant vector fields by, w_i ( v_j) = d_ij A * operator by, *(w) e = sum(i1 .. in = 0 .. n - 1) sign_of_permutation(i1 ... in) e(v_i1, ..., v_ip) w_i(p+1) ^ ... ^ w_in / (p! (n-p)!) Some special types of geometries obey, ( l1 *(w) d *(w) d + l2 d *(w) d *(w) ) w = 0. For example, smooth groups have l1 = 0. A value of the parameters might approximate gravitation. l1 = l2 seems likely. Other values of the parameters might be expressible as an algebraic relation. The rules of physics might. A generalization of the equation is, ( l0 + sum(i = 1, infinity) ( l_(2i - 1) ( *(w) d *(w) d )^i + l_(2i) ( d *(w) d *(w) )^i ) ) w = 0. mihai ==== let us assume, we have an compact topological space X, where the topology comes from an absolut value |.| on X. let us assume further we have an surjective continous function f : X^n -> X^m and let us measure X^i by the ith product of the haar measure comming from X. my question is: if N has measure 0 in X^m, does f^(-1)(N) then also have measure 0? is there any literature dealing with these problems? thank you for any answer... tanja ==== >let us assume, we have an compact topological space X, where the >topology comes from an absolut value |.| on X. let us assume further >we have an surjective continous function f : X^n -> X^m and let us >measure X^i by the ith product of the haar measure comming from X. my >question is: if N has measure 0 in X^m, does f^(-1)(N) then also have >measure 0? is there any literature dealing with these problems? thank >you for any answer... There are many things wrong with your set-up. First of all, a general topological space doesn't have an absolute value. Do you want X to be asubset of the complex numbers? Second, Haar measure is something on a topological *group*, in fact, a locally compact one. I am going to make a wild guess about what you mean. Please correct me if I am wrong. I am assuming that X is a subset of the complexes and that Haar measure is just Lebesgue measure restricted to X. Even then, the answer to your question is trivially no. Just consider any function f:X->X which is constant on a set of positive measure and is still surjective. Then the inverse image of a point will be a set of positive measure. I doubt there is much literature on a problem so broad as this. --Dan Grubb ==== let us assume, we have an compact topological space X, where the >topology comes from an absolut value |.| on X. let us assume further >we have an surjective continous function f : X^n -> X^m and let us >measure X^i by the ith product of the haar measure comming from X. my >question is: if N has measure 0 in X^m, does f^(-1)(N) then also have >measure 0? Well, I have no idea what an absolute value on a compact topological space is, nor what haar measure on a compact topological space is. But whatever you mean, the answer to your question is almost surely _no_ - inverses of null sets under continuous functions are not going to be null sets, except under very special conditions. >is there any literature dealing with these problems? thank >you for any answer... tanja ************************ David C. Ullrich ==== > I need a suggestion. I took linear algebra this semester up here in Montana > (MSU-Billings). I thought that I could handle it, but I'm really > struggling. In fact, I'm having to take an incomplete in the course because > it's just not making any sense to me. I've always been strong in math, but > matrices and vectors spaces have got me entirely stumped. Are there any good resources (either in print or online) out there that > anyone knows of that would help me out. Something like Linear Algebra for > Complete Morons would be nice . . . :) Okay, this is pretty simplistic, but you asked for it: http://ceee.rice.edu/Books/LA/contents.html It's intended for highschool students as a precalculus course. Doesn't assume any background. (I used it to teach myself.) -- - Laurel * * * http://amberdine.com ==== no, trust me, very basic is very good. i don't know why i'm so lost with this, but i feel like a bumbling idiot. and to everyone: thank you very much for all the responses, both here in -- matt > I need a suggestion. I took linear algebra this semester up here in Montana > (MSU-Billings). I thought that I could handle it, but I'm really > struggling. In fact, I'm having to take an incomplete in the course because > it's just not making any sense to me. I've always been strong in math, but > matrices and vectors spaces have got me entirely stumped. Are there any good resources (either in print or online) out there that > anyone knows of that would help me out. Something like Linear Algebra for > Complete Morons would be nice . . . :) Okay, this is pretty simplistic, but you asked for it: http://ceee.rice.edu/Books/LA/contents.html It's intended for highschool students as a precalculus course. Doesn't > assume any background. (I used it to teach myself.) > -- > - Laurel * * * http://amberdine.com > ==== >no, trust me, very basic is very good. i don't know why i'm so lost with >this, but i feel like a bumbling idiot. and to everyone: thank you very much for all the responses, both here in I'm curious. Are you having problems with matrix arithmetic? I also hit a very large brick wall when my book went from talking about arithmetic to vectors and I still don't seem to be able to get past the section where the unit vector is defined. I haven't spent much time on figuring out why yet but it has something to do with the notation |v|. It is possible that my brain is hardwired to map the two vertical glyphs to absolute value, which also causes me enormous problems. /BAH ==== My reply is interwoven within original post below. > Here is an unoriginal game (inspired by Dots and Boxes{name?}, Go, > and by Scrabble somewhat). > But this post is not just a game-idea, but a question about strategy. > The game can be played with any number(m) of players, m>=2, > where each player has a black pencil/pen and a colored pencil/pen of a > color different than the player's opponents' colors. > The game is played on an n-by-n grid, where n^2 is divisible by m. > I would suggest an n of at least 6 or 8 for 2 players. The players take turns each writing (in black) 1, 2, 3, ...(n^2/m), each > integer being written into any still-unoccupied grid-square. > (with his/her colored pencil) all grid-squares within any rectangle in > the grid, IF, and only if, each grid-square in the rectangle has an integer written in it, the SUM of every integer in the rectangle is any PRIME number, and all the squares in the rectangle are not yet colored in. > After each player has moved (ie written an integer and perhaps colored in > a rectangle) > (n^2/m) times, the winner is the player with the most grid-squares of > that player's pencil-color. > Variations: Players can only fill in rectangles which contain the square they just > I thought this would be the best at first. But now I believe that being able to color in rectangles not necessarily including the last integer written might be better. For then the players can, for one thing, leave prime-rectangles for later, hoping their opponent(s) do not fill them in. Anyway, either way would affect strategy differently. > Players can write their integers into the grid in any order they wish. > (I like this variation, but it requires remembering which integers have > already been played.) Players draw at-random their integers from a deck of cards. > (Either from 4 decks of 1 through (n^2/m), or from one master deck of 1 > through n^2.) > In this game, with or without any of the variations (or with any > variations of your own), what would be a good strategy??? > One strategy might involve putting down consecutive integers in adjacent squares, so as to ensure that the line of integers is composite (when length of line is >= 3). Players can ensure that certain rectangles are composite until a given point in the game, where the rectangle is finished with an integer which makes it prime. But the stategy depends upon what is known about an opponent's unplayed integers (if integers are not necessarily played sequencially). (If you do not know what I ma talking about above, then okay. I do not know exactly myself...) > By the way, ideally, the game board will look aesthetically interesting > when the game is done, especially with the right m and n. > (Any suggestions for m and n from this standpoint?) > Actually what really matters, probably, regarding the beauty of the final board are the colors used, rather than the number of colours used. And I bet most players would use x's and o's or something else as such in-practice, instead of filling the squares with colored pencil, since then they only need one pencil. > Leroy Quet ==== > Here is Eucleides's proof that there are infinitely many Pythagorean > triplets. > If n is a natural, then n^2 - (n-1)^2 = (n+n-1)(n-n+1) = 2n-1. This > means that every odd natural (>1) is the difference between two > squares. > Some of these odd naturals are squares themselves. There are (at least) > as many Pythagorean triplets as there are odd squares. And because there > are infinitely many odd numbers, and an odd number's square is odd, then > there are infinitely many Pythagorean triplets. > Do I remember this proof correctly? Just for grins, here is a way to get every odd number as the smallest term of a Pythagorean triplet. I noticed this when I hadn't yet had any algebra in school, so I found the proof very difficult. Now of course it is extremely easy. We were always given (3, 4, 5) and (5, 12, 13) as Pythagorean triples in homework problems. What I noticed was that 4 + 5 = 9 = 3^2 and that 12 + 13 = 25 = 5^2, and of course that 4 and 5 as well as 12 and 13 differed by 1. After trying to extend this pattern to (7, 24, 25), (9, 40, 41) and perhaps a few more, I saw that it always worked. I asked my big brother to prove it for me and the rat made me do it for myself. I had a hard time figuring out that the pattern was (2n+1, 2n^2+2n, 2n^2+2n+1), but once that was done, it was easy enough to calculate that (2n^2+2n+1)^2 = (2n^2+2n)^2 + 2(2n^2+2n) + 1 = (2n^2+2n)^2 + (4n^2+4n+1) = (2n^2+2n)^2 + (2n+1)^2 as desired. Now it is actually easy to see that this triple is primitive, for clearly the last two terms differ by 1 and thus cannot have a common factor, anything dividing the first two terms would also divide 2n^2+n, hence n, hence 1, so they are relatively prime, and anything dividing the first and third terms would have to divide 2n^2+1, hence n+1, hence n, hence 1, and so they are also relatively prime and the triplet is thus primitive. I just figured that out now, not way back when. It would not surprise me if this result dates back to the ancient Greeks. Does anyone know? Achava ==== >> >> Here's Euclid's proof that there are infinitely many Pythagorean triples. >> If n is a natural, then n^2 - (n-1)^2 = (n+n-1)(n-n+1) = 2n-1. This >> means that every odd natural (>1) is the difference between two squares. >> Some of these odd naturals are squares themselves. There are (at least) >> as many Pythagorean triplets as there are odd squares. And because there >> are infinitely many odd numbers, and an odd number's square is odd, then >> there are infinitely many Pythagorean triplets. >> Do I remember this proof correctly? Just for grins, here is a way to get every odd number as the smallest > term of a Pythagorean triplet. I noticed this when I hadn't yet had > any algebra in school, so I found the proof very difficult. Now of > course it is extremely easy. We were always given (3, 4, 5) and (5, 12, 13) as Pythagorean triples > in homework problems. What I noticed was that 4 + 5 = 9 = 3^2 and > that 12 + 13 = 25 = 5^2, and of course that 4 and 5 as well as 12 and > 13 differed by 1. After trying to extend this pattern to (7, 24, 25), > (9, 40, 41) and perhaps a few more, I saw that it always worked. I > asked my big brother to prove it for me and the rat made me do it for > myself. I had a hard time figuring out that the pattern was (2n+1, > 2n^2+2n, 2n^2+2n+1), but once that was done, it was easy enough to > calculate that (2n^2+2n+1)^2 = (2n^2+2n)^2 + 2(2n^2+2n) + 1 = > (2n^2+2n)^2 + (4n^2+4n+1) = (2n^2+2n)^2 + (2n+1)^2 as desired. It's very easy to derive directly, requiring only trivial algebra and some parity: n^2 odd => n odd (<=> even^2 = even ); to wit: If (a,b,c) is a Pythagorean triple with c = b+1 (so primitive) then (b+1)^2 = b^2 + a^2 => 2b+1 = a^2 odd => a odd, say a = 2n+1 => 2b+1 = (2n+1)^2 = 4n(n+1)+1 => b = 2n(n+1), a = 2n+1, c = b+1 Fibonacci did it by summing 2n+1 = (n+1)^2 - n^2, as is described in his entry at the MacTutor math history archive: Liber Quadratorum, written in 1225, is Fibonacci's most impressive piece of work, although not the work for which he is most famous. The book's name means the book of squares and it is a number theory book which, among other things, examines methods to find Pythagorean triples. Fibonacci first notes that square numbers can be constructed as sums of odd numbers, essentially describing an inductive I thought about the origin of all square numbers and discovered that they arose from the regular ascent of odd numbers. For unity is a square and from it is produced the first square, namely 1; adding 3 to this makes the second square, namely 4, whose root is 2; if to this sum is added a third odd number, namely 5, the third square will be produced, namely 9, whose root is 3; and so the sequence and series of square numbers always rise through the regular addition of odd numbers. Thus when I wish to find two square numbers whose addition produces a square number, I take any odd square number as one of the two square numbers and I find the other square number by the addition of all the odd numbers from unity up to but excluding the odd square number. For example, I take 9 as one of the two squares mentioned; the remaining square will be obtained by the addition of all the odd numbers below 9, namely 1, 3, 5, 7, whose sum is 16, a square number, which when added to 9 gives 25, a square number. -Bill Dubuque ==== Bruce Harvey scribbled the following: >> Here is Eucleides's proof that there are infinitely many Pythagorean >> triplets. >> If n is a natural, then n^2 - (n-1)^2 = (n+n-1)(n-n+1) = 2n-1. > Apart from the algbra you have it. Although the three expressions are indeed > equal the middle one should be > n^2 -(n^2-2n+1) 12 year old's algebra!!!!!! Why should it be that? I was using the rule a^2 - b^2 = (a+b)(a-b), with b being equal to (n-1). -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ That's no raisin - it's an ALIEN! - Tourist in MTV's Oddities ==== > Bruce Harvey scribbled the following: >> Here is Eucleides's proof that there are infinitely many Pythagorean >> triplets. >> If n is a natural, then n^2 - (n-1)^2 = (n+n-1)(n-n+1) = 2n-1. Apart from the algbra you have it. Although the three expressions are indeed > equal the middle one should be n^2 -(n^2-2n+1) 12 year old's algebra!!!!!! Why should it be that? I was using the rule a^2 - b^2 = > (a+b)(a-b), with b being equal to (n-1). > but you did say proof and to my simple mind as a maths teacher, a proof is set out line by line stating any other theorems invoked. > -- > /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- > -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ > That's no raisin - it's an ALIEN! > - Tourist in MTV's Oddities ==== > Bruce Harvey scribbled the following: >> Here is Eucleides's proof that there are infinitely many Pythagorean >> triplets. >> If n is a natural, then n^2 - (n-1)^2 = (n+n-1)(n-n+1) = 2n-1. Apart from the algbra you have it. Although the three expressions are indeed > equal the middle one should be n^2 -(n^2-2n+1) 12 year old's algebra!!!!!! Why should it be that? I was using the rule a^2 - b^2 = > (a+b)(a-b), with b being equal to (n-1). example google? I asked the first original reference. Nobody replied. I'm interested in math history. Here is a partial copy of my post (equations) dated above: For example: 1) For ODD a , a general solution exists: p^2 - (p-1)^2 = q^2 => p^2=(p-1)^2+q^2, because p^2-(p^2-2p+1)=> 2p-1 =q^2 Examples: 5^2-4^2= 3^2 , where 3 is odd in Eq. in the title. 13^2-12^2= 5^2 25^2-24^2= 7^2 41^2-40^2= 9^2 .... etc. infinitely for (odd)^2 on rhs. 2) For EVEN b , a general parametric solution exists: s^2-(s-2)^2=t^2 => s^2-(s^2-4s+4) =4(s-1) =t^2 Examples: 5^2 -3^2=4^2, where 4 is even in Eq.in the title. 10^2-8^2=6^2 17^2-15^2=8^2 ....... etc. infinitely for (even)^2 on rhs My point was: Because any square, even or odd, is the difference of two squares, then there are infinite many Pythagorean triplets so that any even or odd integer is a member in some Pythagorean triplet. As there are infinite many primes, then there are infinite many different non-trivial Pythagorean triplets. But who invented that first. Was it really Euclides? Somebody else? Tapio -- > /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- > -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ > That's no raisin - it's an ALIEN! > - Tourist in MTV's Oddities ==== Bruce Harvey scribbled the following: >> Here is Eucleides's proof that there are infinitely many Pythagorean >> triplets. >> If n is a natural, then n^2 - (n-1)^2 = (n+n-1)(n-n+1) = 2n-1. Apart from the algbra you have it. Although the three expressions are > indeed > equal the middle one should be n^2 -(n^2-2n+1) 12 year old's algebra!!!!!! Why should it be that? I was using the rule a^2 - b^2 = > (a+b)(a-b), with b being equal to (n-1). example google? > I asked the first original reference. Nobody replied. I'm interested in math > history. Here is a partial copy of my post (equations) dated above: For example: > 1) For ODD a , a general solution exists: > p^2 - (p-1)^2 = q^2 => p^2=(p-1)^2+q^2, because p^2-(p^2-2p+1)=> 2p-1 =q^2 Examples: 5^2-4^2= 3^2 , where 3 is odd in Eq. in the title. > 13^2-12^2= 5^2 > 25^2-24^2= 7^2 > 41^2-40^2= 9^2 > .... > etc. infinitely for (odd)^2 on rhs. 2) For EVEN b , a general parametric solution exists: > s^2-(s-2)^2=t^2 => s^2-(s^2-4s+4) =4(s-1) =t^2 Examples: 5^2 -3^2=4^2, where 4 is even in Eq.in the title. > 10^2-8^2=6^2 > 17^2-15^2=8^2 > ....... > etc. infinitely for (even)^2 on rhs My point was: Because any square, even or odd, is the difference of two > squares, then there are infinite many Pythagorean triplets so that any even > or odd integer is a member in some Pythagorean triplet. As there are > infinite many primes, then there are infinite many different non-trivial > Pythagorean triplets. > But who invented that first. Was it really Euclides? Somebody else? Tapio > I am not happy about the even numbers because infinity is a difficult concept: with x^2 as x aproaches infinity being of a different order than x as x aproaches infinity. Or even (x^x)! as x aproaches infinity. They are different concepts of infinity. So in your series >13^2-12^2= 5^2 > 25^2-24^2= 7^2 > 41^2-40^2= 9^2 12, 24, 40 ...... is increasing faster than 5, 7, 9 ....... so its a different order of infinity As a concept, infinity needs pinning to a definite conceptual entity such as a number line or a grid of unit cubes formed by the grid lines of a cartesian co-ordinate system sapnning infinite space. -- Bruce Harvey bruce@bearsoft.co.uk The Alternative Physics Site http://users.powernet.co.uk/bearsoft -- > /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- > -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ > That's no raisin - it's an ALIEN! > - Tourist in MTV's Oddities ==== Michigan State University grad student Michael Shafer has succeeded in identifying the largest known prime number to date, using a distributed computer network of more than 200,000 computers located around the world. The new number is 6,320,430 digits long and is only the 40th Mersenne prime to have ever been discovered (Mersenne primes are an especially rare breed that take the form of 2-to-the-power-of-P, where P is also a prime number). ==== > Michigan State University grad student Michael Shafer has succeeded in > identifying the largest known prime number to date, using a distributed > computer network of more than 200,000 computers located around the world. > The new number is 6,320,430 digits long and is only the 40th Mersenne prime > to have ever been discovered (Mersenne primes are an especially rare breed > that take the form of 2-to-the-power-of-P, where P is also a prime number). Sounds like a waste of processor power (and electricity) to me. There is no reason in discovering these large numbers (even if you want to use them for encryption. I believe the encryption process would need veeeery much time ;-}} ). Karl ==== > Michigan State University grad student Michael Shafer has succeeded in > identifying the largest known prime number to date, using a distributed > computer network of more than 200,000 computers located around the world. > The new number is 6,320,430 digits long and is only the 40th Mersenne > prime > to have ever been discovered (Mersenne primes are an especially rare breed > that take the form of 2-to-the-power-of-P, where P is also a prime > number). Sounds like a waste of processor power (and electricity) to me. There is no > reason > in discovering these large numbers (even if you want to use them for > encryption. > I believe the encryption process would need veeeery much time ;-}} ). I don't think Mersenne primes would be very useful for encryption. For example in the RSA method, what you need is the product of two large primes with the idea that nobody can factor that product. Give me any product of two primes of which one is a Mersenne prime, and I can factor it easily. ==== Mike Parker > Michigan State University grad student Michael Shafer has succeeded in > identifying the largest known prime number to date, using a distributed > computer network of more than 200,000 computers located around the world. > The new number is 6,320,430 digits long and is only the 40th Mersenne prime > to have ever been discovered (Mersenne primes are an especially rare breed > that take the form of 2-to-the-power-of-P, where P is also a prime number). My mother phoned me today to tell me that she had seen something on CNN about a breakthrough in mathematics. As soon as I got home I scoured CNN's website, finding nothing. Turns out it's just a 6-million digit prime. What a letdown. LH ==== Larry Hammick scribbled the following: > Mike Parker >> Michigan State University grad student Michael Shafer has succeeded in >> identifying the largest known prime number to date, using a distributed >> computer network of more than 200,000 computers located around the world. >> The new number is 6,320,430 digits long and is only the 40th Mersenne > prime >> to have ever been discovered (Mersenne primes are an especially rare breed >> that take the form of 2-to-the-power-of-P, where P is also a prime > number). > My mother phoned me today to tell me that she had seen something on CNN > about a breakthrough in mathematics. As soon as I got home I scoured CNN's > website, finding nothing. Turns out it's just a 6-million digit prime. What > a letdown. Finding new Mersenne primes is just a question on spending more and more time and processor power on distributed computer networks. Now a real mathematical breakthrough would be for example solving a famous conjecture, like Goldbach's conjecture, either proving or disproving it. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ 'It can be easily shown that' means 'I saw a proof of this once (which I didn't understand) which I can no longer remember'. - A maths teacher ==== Isn't there a power of 2 that is in fact prime? --OL ==== >Isn't there a power of 2 that is in fact prime? --OL > 2^1 ==== Chergarj schrieb: >>Isn't there a power of 2 that is in fact prime? >> 2^1 Fascinating! Two of the most fundamental mathematical constants in a simple formula... giving a prime... ==== <28085-3FD7D447-106@storefull-2315.public.lawson.webtv.net>, > Isn't there a power of 2 that is in fact prime? --OL > Yes, but it is not a large power, and was discovered to be prime long ago. ==== ><28085-3FD7D447-106@storefull-2315.public.lawson.webtv.net>, > Isn't there a power of 2 that is in fact prime? >> >Yes, but it is not a large power, and was discovered to be prime >long ago. You mean we won't be seeing this in the papers then? Newsgroup participant discovers smallest prime. Ever. ==== <28085-3FD7D447-106@storefull-2315.public.lawson.webtv.net>, Isn't there a power of 2 that is in fact prime? Yes, but it is not a large power, and was discovered to be prime > long ago. I wonder if we could generalize the result. There may be others of these, hmmm, not-large powers that turn out to be prime. Jim Burns ==== <28085-3FD7D447-106@storefull-2315.public.lawson.webtv.net>, > Isn't there a power of 2 that is in fact prime? Yes, but it is not a large power, and was discovered to be prime > long ago. I wonder if we could generalize the result. There may be > others of these, hmmm, not-large powers that turn out to > be prime. I would be interested in a list of all known primes that are powers of Mersenne primes. I bet there are not many of them, probably not more than fourty at the moment. ==== > <28085-3FD7D447-106@storefull-2315.public.lawson.webtv.net>, Isn't there a power of 2 that is in fact prime? Yes, but it is not a large power, and was discovered to be prime > long ago. I wonder if we could generalize the result. There may be > others of these, hmmm, not-large powers that turn out to > be prime. I would be interested in a list of all known primes that are powers of > Mersenne primes. I bet there are not many of them, probably not more > than fourty at the moment. How about zero? As in There are no prime numbers which are powers of Mersenne primes. None at all. Ain't gonna happen. End of story. Next. ==== > <28085-3FD7D447-106@storefull-2315.public.lawson.webtv.net>, Isn't there a power of 2 that is in fact prime? Yes, but it is not a large power, and was discovered to be prime > long ago. I wonder if we could generalize the result. There may be > others of these, hmmm, not-large powers that turn out to > be prime. I would be interested in a list of all known primes that are powers of > Mersenne primes. I bet there are not many of them, probably not more > than fourty at the moment. How about zero? As in There are no prime numbers which are > powers of Mersenne primes. None at all. Ain't gonna happen. > End of story. Next. p^n for p = 1. ==== Mike Parker scribbled the following: >> <28085-3FD7D447-106@storefull-2315.public.lawson.webtv.net>, >> Isn't there a power of 2 that is in fact prime? >> Yes, but it is not a large power, and was discovered to be prime >> long ago. >> I wonder if we could generalize the result. There may be >> others of these, hmmm, not-large powers that turn out to >> be prime. >> I would be interested in a list of all known primes that are powers of >> Mersenne primes. I bet there are not many of them, probably not more >> than fourty at the moment. > How about zero? As in There are no prime numbers which are > powers of Mersenne primes. None at all. Ain't gonna happen. > End of story. Next. Depends on your definition of power. If p is a Mersenne prime, then p^1 sure as heck is a prime. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ It's time, it's time, it's time to dump the slime! - Dr. Dante ==== > Michigan State University grad student Michael Shafer has succeeded in > identifying the largest known prime number to date, using a distributed > computer network of more than 200,000 computers located around the world. > The new number is 6,320,430 digits long and is only the 40th Mersenne prime > to have ever been discovered (Mersenne primes are an especially rare breed > that take the form of 2-to-the-power-of-P, where P is also a prime number). When I read it the first time I found the most surprising not that he had discovered a large power of 2 but that he had discovered one that is prime. There must be something wrong with the core of mathematics. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== >Message-id: Michigan State University grad student Michael Shafer has succeeded in > identifying the largest known prime number to date, using a distributed > computer network of more than 200,000 computers located around the world. > The new number is 6,320,430 digits long and is only the 40th Mersenne >prime > to have ever been discovered (Mersenne primes are an especially rare breed > that take the form of 2-to-the-power-of-P, where P is also a prime >number). When I read it the first time I found the most surprising not that he >had discovered a large power of 2 but that he had discovered one that >is prime. There must be something wrong with the core of mathematics. Yup. >-- >dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, >+31205924131 >home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ -- Mensanator Ace of Clubs ==== > Dear all, It is easy to prove that if the first number in a row of Pascal's > triangle is a prime then all other numbers (except 1) in that row are > multiples of that prime. However, several sources say that this is also a sufficient condition > for this phenomenon, ie if the first number n is composite, then there > exists a binomial coefficient in that row not a multiple of n. I've tried looking for a proof online, but with no success. I would > greatly appreciate it if you could help me out. Apologies if this is rather trivial. Red If we have n!/(k!(n-k)!) = binomial(n,k), then binomial(n,k) = (n/k) binomial(n-1,k-1), for k and n = positive integers. So, if GCD(n,k) = j, and n = n' j, k = k' j, then binomial(n,k) = (n' j)/(k' j) binomial(n-1,k-1) = multiple of n' / k'. And so, for n and k = positive integers, binomial(n,k) is divisible by n'/k'. But k' does not divide n', so k' divides binomial(n-1,k-1), since binomial(n,k) is an integer. So, n' = n/GCD(n,k) always divides binomial(n,k), for n and k = positive integers. Specifically, if n and k are relatively prime, then n always divides binomial(n,k). Yet n MAY still sometimes divide the binomial if it is not coprime with k. (What is the first example??? Is there an example ??) thanks, Leroy Quet ==== > Yet n MAY still sometimes divide the binomial if it is not coprime with k. > (What is the first example??? Is there an example ??) binomial(10,4) = 210 -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W ==== > Dear all, It is easy to prove that if the first number in a row of Pascal's > triangle is a prime then all other numbers (except 1) in that row are > multiples of that prime. However, several sources say that this is also a sufficient condition > for this phenomenon, ie if the first number n is composite, then there > exists a binomial coefficient in that row not a multiple of n. I've tried looking for a proof online, but with no success. I would > greatly appreciate it if you could help me out. Apologies if this is rather trivial. Red Here's something I posted several years ago: ============================================================================ === |> |> Is it true that GCD of {mC1,mC2,...,mC[m-1]} is p if m=p^n with p prime and |> 1 otherwise? |> Yes. |> Here mCn is m combination n, or the nth entry in the mth row of Pascal's |> triangle. |> |> If so, is there a really obvious proof? More important: can anybody name a |> reference (textbook, paper etc..)? |> Using a result due to Kummer which is nicely presented in [1], it's not _too_ hard to prove. Let me change your notation a bit: it'll be easier to follow the argument if m combination n is denoted by C(m,n). That said, Kummer's result is as follows: given n, k > 0, let c_p(n,k) = the number of carries generated when adding n & k using base p. Then Theorem(Kummer). If p is a prime, then p^{c_p(n,k)} is the highest power of p which divides C(n+k,k). If you grant me this, then I can argue as follows. Since C(m,1) = m, the gcd is always a divisor of m. So, if m=p^n, the gcd of the set is p^r for some r >= 0. But, given n and k (0 < n,k < p^n) such that n+k = p^n, there _has_ to be at least 1 carry generated when we add the base p versions of n and k -- that's the only way to generate the '1' that's the coefficient of p^n in the result. Hence, every element of our set is divisible by (at least) p. On the other hand, if 0 < k < p, then Kummer's result shows that C(p^n,kp^{n-1}) is _not_ divisible by p^2. Hence, the gcd is p in this case. For the other case, suppose that m is not a power of a prime, and let p be any prime which divides m. In the base p representation of m, there is either a term rp^u (with 1 < r < p) or a pair of terms rp^u + sp^v.(There are probaly a *bunch* of other terms, as well -- I don't care about those). In the first case, C(m,(r-1)p^u) is _not_ divisible by p; in the second case, C(m,rp^u) is not divisible by p. Since, for every prime p dividing m, there's some C(m,n) in the set that is not divisible by p, it follows that the gcd of the set is 1. 1. Peitgen,H.-O. _et_al._, Chaos and Fractals/New Frontiers of Science, Springer-Verlag, 1992. ISBN 0-387-97903-4 |> Flo. |> |> |> -- |> ---------------------------------------------------------- |> Florian Breuer at University of Stellenbosch, South Africa |> reply to: s9520155@itu3.sun.ac.za |> ---------------------------------------------------------- ============================================================================ == The result that you're wanting to prove follows easily -- unless n is a prime, some C(n.k) with 1 < k < n is not a multiple of n. ==== Dear all, Trying to figure out the 1-dimensional representations of D_n, the dihedral group of order 2n. Should be easy... Let r be rotation, s be reflection. Then r^n = 1, s^2 = 1, rs = sr^-1... Now let F be the representation from D_n to k* , F : D_n --> k* where k = complex numbers. Then F(r)F(s) = F(s)F(r^-1) <==> F(r) = F(r^-1) since these numbers commute , so F(r^2) = 1 <===> F(r) = 1 or -1....likewise since s^2 = 1, F(s) = 1 or -1... So there are 4 1-dimensional representations of D_n for all n? Rob ==== > Dear all, Trying to figure out the 1-dimensional representations of D_n, the > dihedral group of order 2n. > Should be easy... Let r be rotation, s be reflection. Then r^n = 1, s^2 = 1, rs = sr^-1... Now let F be the representation from D_n to k* , F : D_n --> k* > where k = complex numbers. Then F(r)F(s) = F(s)F(r^-1) <==> F(r) = F(r^-1) since these numbers > commute , so F(r^2) = 1 <===> F(r) = 1 or -1....likewise since s^2 = 1, > F(s) = 1 or -1... So there are 4 1-dimensional representations of D_n for all n? Are there four 1-dimensional reps of D_3? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== Dear all, Trying to figure out the 1-dimensional representations of D_n, the > dihedral group of order 2n. > Should be easy... Let r be rotation, s be reflection. Then r^n = 1, s^2 = 1, rs = sr^-1... Now let F be the representation from D_n to k* , F : D_n --> k* > where k = complex numbers. Then F(r)F(s) = F(s)F(r^-1) <==> F(r) = F(r^-1) since these numbers > commute , so F(r^2) = 1 <===> F(r) = 1 or -1....likewise since s^2 = 1, > F(s) = 1 or -1... So there are 4 1-dimensional representations of D_n for all n? Are there four 1-dimensional reps of D_3? Ouch, you are right. There are only 2 1-dimensional reps of D_3. Then how can I generalize to D_n? -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) ==== > >> Dear all, >> Trying to figure out the 1-dimensional representations of D_n, the >> dihedral group of order 2n. >> Should be easy... >> Let r be rotation, s be reflection. >> Then r^n = 1, s^2 = 1, rs = sr^-1... >> Now let F be the representation from D_n to k* , F : D_n --> k* >> where k = complex numbers. >> Then F(r)F(s) = F(s)F(r^-1) <==> F(r) = F(r^-1) since these numbers >> commute , so F(r^2) = 1 <===> F(r) = 1 or -1....likewise since s^2 = >> 1, F(s) = 1 or -1... >> So there are 4 1-dimensional representations of D_n for all n? >> Are there four 1-dimensional reps of D_3? Ouch, you are right. There are only 2 1-dimensional reps of D_3. Then > how can I generalize to D_n? So far you have F(r) = +- 1 and F(s) = +-1. Are there any more restrictions you can find on these? Have you used all the relations in the group? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== >> Dear all, >> Trying to figure out the 1-dimensional representations of D_n, the >> dihedral group of order 2n. >> Should be easy... >> Let r be rotation, s be reflection. >> Then r^n = 1, s^2 = 1, rs = sr^-1... >> Now let F be the representation from D_n to k* , F : D_n --> k* >> where k = complex numbers. >> Then F(r)F(s) = F(s)F(r^-1) <==> F(r) = F(r^-1) since these numbers >> commute , so F(r^2) = 1 <===> F(r) = 1 or -1....likewise since s^2 = >> 1, F(s) = 1 or -1... >> So there are 4 1-dimensional representations of D_n for all n? >> Are there four 1-dimensional reps of D_3? Ouch, you are right. There are only 2 1-dimensional reps of D_3. Then > how can I generalize to D_n? So far you have F(r) = +- 1 and F(s) = +-1. Are there any more > restrictions you can find on these? Have you used all the relations > in the group? I guess the last thing I would have to do is analyze the conjugacy classes of D_n and that is all. (I can do this on my own) -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) ==== > >> Dear all, >> Trying to figure out the 1-dimensional representations of D_n, the >> dihedral group of order 2n. >> Should be easy... >> Let r be rotation, s be reflection. >> Then r^n = 1, s^2 = 1, rs = sr^-1... >> Now let F be the representation from D_n to k* , F : D_n --> k* >> where k = complex numbers. >> Then F(r)F(s) = F(s)F(r^-1) <==> F(r) = F(r^-1) since these numbers >> commute , so F(r^2) = 1 <===> F(r) = 1 or -1....likewise since s^2 = 1, >> F(s) = 1 or -1... >> So there are 4 1-dimensional representations of D_n for all n? >> Are there four 1-dimensional reps of D_3? Ouch, you are right. There are only 2 1-dimensional reps of D_3. Then how > can I generalize to D_n? > >> -- >> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >> Needless to say, I had the last laugh. >> Alan Partridge, _Bouncing Back_ (14 times) 1, time consuming: by learning some more representation theory or 2. lazy option: reading chapter 18 in james and liebeck: there is an obvious 2-d rep. show, that there are in fact many more such. you need to consider n odd and even separately. that odd and even explanantion will tell you why there are 2 1-d reps n odd, 4 n even. you're on the right track. if you want i could write out the proof, but i'd rather not have to. oh, the n odd and even thing means there are different 'types' of conjugacy classes. if you can read the book cos it's much more clearly laid out than i'll manage to make it on here. ==== > I know guys like Abel and Galois did their best work early and then died, > but can people give me some examples of mathematicians who did their best > work in their early 20s (say), then continued to work for a long time but I would like to see an *objective* study on this. First, a group of unknown mathematicians' work would be evaluated for greatness (whatever 'greatness' means...), without those doing the evaluating knowing any biographical info about the mathematicians at first. And *then* the mathematicians' ages (and sexes and race and education level and economic status and...) would be revealed. Perhaps even the evaluators' own biases can be studied as well, where they each fill out a survey (after evaluating the math work, but before learning anything personal about the mathematicians) as to what they GUESS what the mathematicians' biographical info might be. ... So...by looking over MY past work posted here to sci.math over the last few years, how old do you all *think* I might be???..... ;) (And I know I have posted my age here on occasion. So no fair peeking at that info!) thanks, Leroy Quet ==== >> I know guys like Abel and Galois did their best work early and then died, >> but can people give me some examples of mathematicians who did their best >> work in their early 20s (say), then continued to work for a long time but > > > I would like to see an *objective* study on this. > [...] > So...by looking over MY past work posted here to sci.math over the > last few years, how old do you all *think* I might be???..... > ;) (And I know I have posted my age here on occasion. So no fair peeking > at that info!) We don't have to peek. You react negatively to the suggestion that mathematical genius peaks at an early age. Conclusion: you're old. Me too. At least I'm old enough to be past that peak. Of course, I first realized I wasn't going to hit the greatness for which I was destined when, at twenty-four, I learned that Robert Johnson died at age twenty-four[1]. Meanwhile, I had never had a hit blues record like I expected. Of course, I didn't play an instrument or sing either, but evidently it was too late to learn. Or to quote Tom Lehrer, When Mozart was my age, he had been dead for two years. Footnotes: [1] Evidently, I was misinformed. He died at twenty-seven, but I'm past that, too. -- So, at this time, I'd like to assure you that I am not interested in I'll have prosecutors knocking on your doors. I have no problem with ==== >message >> What with the war on terror, and Canada's puny military and corrupt >> police force, the only way to insure peace and safety for that >> country, is for it to be annexed by America. >> It is the only way foward for Canada aka Canuckistan. >> --------------------------------- >> Well, another solution is the liberation of U.$.A by Canadians and the >> return of democracy to the Americans peoples. >> I am sure Canadians will be welcome with (french) kisses and flowers >when >> they will topple the Little Dictator and his gang of corporate crooks. >> Patriot act will be scraped; free press will be restaured, Health Care >> will be available to all citizens, Yeah, like it is in Canada. If you wait. And wait. And wait. And... oops, >died already. Or you can join an American HMO where they deny you care because they don't want to pay out the money. ooopps one dead wife...mine! Capitalist medicine is an abomonation and a human rights violation. THOM But at least the bigtime politicians and bureaucrats get their healthcare! >(By paying cash in Yankee hospitals...) -- >zimriel sbc dot > at global net >. >http://pages.sbcglobal.net/zimriel/blog/zimblog.html >because everyone else is doing it ==== > What with the war on terror, and Canada's puny military and corrupt >> police force, the only way to insure peace and safety for that >> country, is for it to be annexed by America. >> It is the only way foward for Canada aka Canuckistan. You seem to forget that the Canadians kicked our ass in two conflicts.... > Really? Which two? ==== >> What with the war on terror, and Canada's puny military and corrupt >> police force, the only way to insure peace and safety for that >> country, is for it to be annexed by America. >> It is the only way foward for Canada aka Canuckistan. You seem to forget that the Canadians kicked our ass in two conflicts.... Really? Which two? > 1775 http://www.civilization.ca/cwm/chrono/1000invasion_e.html 1812 http://www.civilization.ca/cwm/chrono/1774invasion_repelled_e.html ==== What with the war on terror, and Canada's puny military and corrupt > police force, the only way to insure peace and safety for that > country, is for it to be annexed by America. It is the only way foward for Canada aka Canuckistan. >>You seem to forget that the Canadians kicked our ass in two conflicts.... >Really? Which two? he'se probably talking about ice hockey THOM > ==== > What with the war on terror, and Canada's puny military and corrupt > police force, the only way to insure peace and safety for that > country, is for it to be annexed by America. It is the only way foward for Canada aka Canuckistan. >>You seem to forget that the Canadians kicked our ass in two conflicts.... >Really? Which two? he'se probably talking about ice hockey Those ignorant of history are bound to repeat it. ==== Let c(1,k) = 1, for all positive integers k. For all m >= 2, Let c(m,k) = --- - ln(k) ------- > c(j,k) ln(m) / --- j|m j 1, f(x) = oo --- --- > ( > c(k,m/k) ) /m^x / / --- --- m=1 k|m (which in linear mode is...) f(x) = sum{m=1 to oo} (sum{k|m} c(k,m/k) ) /m^x. (Inner-sum is over all positive divisors, k, of m) Puzzle: What is a closed-form (in terms of known functions) for f(x)?? thanks, Leroy Quet ==== > If you fucking morons think that I will let you get away with not > giving me credit for my fucking math discoveries then you have another > fucking thing coming. As far as I know, you don't get paid for mathematical discoveries. Certainly you don't get paid by announcing them on a newsgroup. Consider - who's going to pay you, and for what? There exist a few prizes in mathematics that I've heard of, but you have to submit your work to these judges and so forth. To my knowledge, these prizes aren't altogether spectacular. And you can forget about a Nobel Prize in Mathematics. There isn't one, as I'm sure you know. I have never heard of any corporation paying for a mathematical discovery, with the exception perhaps of those made in the R&D departments. Same with the government, which probably wouldn't be able to even judge a discovery anyway (they'd need a committee, and it probably would be composed of politicians, not mathematicians). I can understand you being upset if you think you'd be denied money due to you, or if you think someone else would be able to steal it from you by plagiarizing your work. But as I said - who's going to pay *anyone* for a mathematical discovery? Be well. Baruch ==== synaptic energy redistribution (audio) filter http://apollonicon.com/music/gold.wav http://apollonicon.com/music/black.wav http://apollonicon.com/music/gold.mp3 http://apollonicon.com/music/black.mp3 google video filter? http://apollonicon.com/series/pierre_21.jpg ==== a good score. I know the word good is fairly arbitrary, but if we consider the three cases: 1) Want to get into a top 10 school 2) Want to get into a top 20 school 3) Want to get into an upper Tier 2 school Then for each of the above, what would be the subject test score/percentile that one would recommend? ==== >Then for each of the above, what would be the subject test >score/percentile that one would recommend? I can tell you that a lot of mathematics graduate programs won't even look at your score; just that you took it. Doug ==== > a good score. I know the word good is fairly arbitrary, but if we > consider the three cases: 1) Want to get into a top 10 school > 2) Want to get into a top 20 school > 3) Want to get into an upper Tier 2 school > Then for each of the above, what would be the subject test > score/percentile that one would recommend? Get a 900, and I think that you can call that good. Otherwise, I would say at least a 700. Most universities want the top 1/3 of test takers. That's what I have found anyway. Lurch ==== > a good score. I know the word good is fairly arbitrary, but if we > consider the three cases: 1) Want to get into a top 10 school > 2) Want to get into a top 20 school > 3) Want to get into an upper Tier 2 school > Then for each of the above, what would be the subject test > score/percentile that one would recommend? Get good recommendations from your profs. If you did especially well in a course taught by a famous prof, that's the best. ==== a good score. I know the word good is fairly arbitrary, but if we > consider the three cases: 1) Want to get into a top 10 school > 2) Want to get into a top 20 school > 3) Want to get into an upper Tier 2 school > Then for each of the above, what would be the subject test > score/percentile that one would recommend? Get good recommendations from your profs. If you did especially well in > a course taught by a famous prof, that's the best. I realize that, but for the purposes of this message I am concerned only about GRE scores. Why are people so resistant to answering questions about GRE scores? Of all the times I've asked this question, I've never once gotten a number back as an answer. Even my advisor was like Don't worry so much about GRE scores. Just worry about doing your best. Well if GRE scores weren't important, then why does practically every university in the country require them to get into their PhD program? Clearly at least one person at each graduate school must know what GRE scores are good and which are not, otherwise nobody would ever take GRE scores into consideration. That's all I'm asking for here. I know GRE scores aren't the final word on everything, but I hope that for once someone can finally give me a straight answer, and hopefully respond with 3 numbers/intervals corresponding to the above 3 ranges of schools. ==== The Prophet Nobody known to the wise as toolshed37@yahoo.com, opened the Book of Words, and read unto the people: >Why are people so resistant to answering questions about GRE scores? >Of all the times I've asked this question, I've never once gotten a >number back as an answer. Even my advisor was like Don't worry so >much about GRE scores. Just worry about doing your best. Well if People don't answer it because, frankly, it's an ill-formed question. GRE scores _don't_ get you into good universities. Good recommendations/research history gets you into good universities. You need at least a good GRE to get into a mediocre university, but even a great GRE won't get you into a great university. The subject GRE covers competence, not excellence; a GRE score above a certain level guarantees you know a good survey of undergrad math -- knowing more above that simply means you have a good memory. The general GRE may honestly be of more value to math programs, since the main thing is not so much what you know as how you think, and the general is honestly a better predictor of that. But, y'know, don't let GRE scores bother you. If you think they're important, study hard. Apply everywhere you'd apply anyways. I don't see how knowing the relationship between GREs and acceptance (one which is actually tenuous, as I say above) helps you in any way. >know GRE scores aren't the final word on everything, but I hope that >for once someone can finally give me a straight answer, and hopefully >respond with 3 numbers/intervals corresponding to the above 3 ranges >of schools. The numbers/intervals you're asking for don't exist. If it would make you happy, I could say '600-1000' for all three (I'm guessing at the lower bound at which 'competence' is assessed). +-------------------------------------------------------------+ | D. Jacob Wildstrom -- Math monkey and freelance thinker | | Graduate Student, University of California at San Diego | | A mathematician is a device for turning coffee into | | theorems. -Alfred Renyi | +-------------------------------------------------------------+ The opinions expressed herein are not necessarily endorsed by the University of California or math department thereof. ==== > The Prophet Nobody known to the wise as toolshed37@yahoo.com, opened the Book of Words, and read unto the people: >Why are people so resistant to answering questions about GRE scores? >Of all the times I've asked this question, I've never once gotten a >number back as an answer. Even my advisor was like Don't worry so >much about GRE scores. Just worry about doing your best. Well if People don't answer it because, frankly, it's an ill-formed > question. GRE scores _don't_ get you into good universities. Good > recommendations/research history gets you into good universities. You > need at least a good GRE to get into a mediocre university, but even a > great GRE won't get you into a great university. The subject GRE > covers competence, not excellence; a GRE score above a certain level > guarantees you know a good survey of undergrad math -- knowing more > above that simply means you have a good memory. The general GRE may > honestly be of more value to math programs, since the main thing is > not so much what you know as how you think, and the general is > honestly a better predictor of that. But, y'know, don't let GRE scores bother you. If you think they're > important, study hard. Apply everywhere you'd apply anyways. I don't > see how knowing the relationship between GREs and acceptance (one > which is actually tenuous, as I say above) helps you in any way. > >know GRE scores aren't the final word on everything, but I hope that >for once someone can finally give me a straight answer, and hopefully >respond with 3 numbers/intervals corresponding to the above 3 ranges >of schools. The numbers/intervals you're asking for don't exist. If it would make > you happy, I could say '600-1000' for all three (I'm guessing at the > lower bound at which 'competence' is assessed). So with a GRE score of 600 you'd get into MIT? Wrong. What about with a GRE score of 700? Wrong. People seem to confuse my question with what scores will get you into a good university? However, nowhere did I ask that question. I said what scores are good if you want to go to such and such university? Hence a score which would cause the reviewer to throw away your application would not be considered good. So the correct response should be an interval which cause the reviewer to probably not throw away your application. These intervals do exist, as a few people have pointed out that. For example, if you want to go to a top 10 university your GRE score should be in the top 95 percentile or so. Yes, I understand that doesn't mean anything other than getting in the top 95 percentile will cause them to not eliminate you. But after all, that's what I was asking about. ==== Do negative numebrs exist in the real world or are they just a purely symbolical and mathematical construct? -- nethlek ==== > Do negative numebrs exist in the real world or are they just a purely > symbolical and mathematical construct? Do even positive numbers exist? To answer that, you need to define exist I think. Something for philosophy rather than mathematics. Here is a start: Realism in Mathematics by Penelope Maddy Paperback: 224 pages Oxford Univ Pr ISBN: 019824035X -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== >Do negative numebrs exist in the real world Negative numebrs are the opposite of positive numebrs, so if you can prove the existence of positive numebrs, then you immediately see negative numebrs. Doug ==== >Do negative numebrs exist in the real world or are they just a purely >symbolical and mathematical construct? What are the respective charges of the electron and proton? ==== > Do negative numebrs exist in the real world or are they just a purely > symbolical and mathematical construct? Do positive numebrs exist in the real world or are they just a purely symbolical and mathematical construct? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== > Do negative numebrs exist in the real world or are they just a purely > symbolical and mathematical construct? -- > nethlek i think it just abstract. consider 3 apples. does it really three? is the size of the apples is the same? ==== > Do negative numebrs exist in the real world or are they just a purely > symbolical and mathematical construct? > When you run a credit balance on a credit card or your phone bill, in other words when you pay more than you owe, they send you a bill for a negative amount. If the banks and credit card companies believe in negative numbers, who are the rest of us to disagree? ==== > Do negative numebrs exist in the real world or are they just a purely > symbolical and mathematical construct? Numbers (even couting numbers) don't exist in a physical sense. You can't go out to the shop and buy yourself a box of 7s for example... But if you pay for a box containing 8 apples and you notice that there is only 7 in there you've applied this abstract notion of a counting number to a real world phenomena. The same thing goes for negative numbers... They are useful for example in determining direction. e.g The stock market is going down today! Numbers provide a conceptual framework that has proven to be useful. Bruce. ==== >Numbers (even couting numbers) don't exist in a physical sense. You can't >go out to the shop and buy yourself a box of 7s for example... But I live at 7777 Seventy-Seventh street. How am I supposed to ==== > >>Numbers (even couting numbers) don't exist in a physical sense. You can't >>go out to the shop and buy yourself a box of 7s for example... But I live at 7777 Seventy-Seventh street. How am I supposed to ==== Do negative numebrs exist in the real world or are they just a purely > symbolical and mathematical construct? Numbers (even couting numbers) don't exist in a physical sense. You can't > go out to the shop and buy yourself a box of 7s for example... As a teenager I used to buy boxes of Number 6s! ==== > Do negative numebrs exist in the real world or are >>they just a purely >> symbolical and mathematical construct? >> Numbers (even couting numbers) don't exist in a physical >>sense. You can't >> go out to the shop and buy yourself a box of 7s for >>example... >As a teenager I used to buy boxes of Number 6s! > At the beginning of every school year, I buy my kids a box of No. 2s. haha hehe. OK, I'll stop now. ==== Do negative numebrs exist in the real world or are they just a purely > symbolical and mathematical construct? Numbers (even couting numbers) don't exist in a physical sense. You can't > go out to the shop and buy yourself a box of 7s for example... But if you pay for a box containing 8 apples and you notice that there is > only 7 in there you've applied this abstract notion of a counting number > to a real world phenomena. The same thing goes for negative numbers... > They are useful for example in determining direction. e.g The stock > market is going down today! > Yes but you can have a box with five apples in it. You can't have a box with negative five apples in it. I think that's what the original poster was getting at. ==== > Do negative numebrs exist in the real world or are they just a purely > symbolical and mathematical construct? Numbers (even couting numbers) don't exist in a physical sense. You can't > go out to the shop and buy yourself a box of 7s for example... But if you pay for a box containing 8 apples and you notice that there is > only 7 in there you've applied this abstract notion of a counting number > to a real world phenomena. The same thing goes for negative numbers... > They are useful for example in determining direction. e.g The stock > market is going down today! > Yes but you can have a box with five apples in it. You can't have a box > with negative five apples in it. I think that's what the original > poster was getting at. Yes, but if I own a warehouse I can! This is very easy to apply to real life situations. Hypothetically ---- I own a warehouse that stocks among other things x, y and z widgets. Due to shortages experienced by my supplier my supply of z widgets is z = -230 units. In other words my orders to date for z widgets outstrips my z widget stock by 230 units. I am sure stock status reports' for many manufacturers or distributors deal with negative numbers all the time. Also in many areas of finance negative numbers are the norm. Gee that reminds me, I have to get on my suppliers' ass because I am loosing $ on these z widget sales. ;-) Dan ==== > Yes but you can have a box with five apples in it. You can't have a box > with negative five apples in it. I think that's what the original > poster was getting at. I can't recall who it was first pointed it out, but there is nothing inherently natural about integer arithmetic. Yeah, integer arithmetic works for apples. But for clouds? If you add one cloud to one cloud, you don't get two clouds. The individuation and the physics of clouds are different than for apples. It is a matter of experimentally determined fact that integer arithmetic works for apples, and not some automatic thing. This means that positive integers are not somehow the simple counting numbers--they are the simple *apple* counting numbers. Or for counting other discrete objects that work like apples do. Other things that don't work like apples are crumbs. If you cut an apple in half, you get two half-apples. But if you break a crumb in two, you don't get two half-crumbs, you get two crumbs. So crumbs add like apples do, but they don't divide like apples do. So nature uses all different kinds of numerical entities, and only some kinds of counting use positive numbers. If we are counting, for example, generations, then it is perfectly natural to count descendent generations with positive numbers and ancestor generations with negative numbers (or vice versa). Or, if you are counting electrons, then you can have -5 electrons: when you have 5 positrons. Indeed, you can do that with apples too, which makes the quoted observation above wrong. A little physics, and it turns out that apples do *not* behave such that natural numbers are the right counting tools, but rather it turns out, integers! Because (however rare), there is such a beastie as an anti-apple. Thomas ====