mm-2509 === Subject: Benchmark Bias Has anybody studied the benchmark bias effect? You have a program that you try to improve, and each change gets tested against a large set of benchmark runs. If the program typically stumbles into linear-e solutions of exponential-e problems, changes tend to random-walk the log of the run e of individual benchmarks. As a result, every change tends to raise the average run e (.5 + 2)/2 = 1.25 not 1.0 The question to look at is: how do you take this bias out, so you can judge the effectiveness of a change. The benchmarks you tend to accumulate are ones that work with the old program (not infinite runners, for instance). If you had done the change backwards, you'd have a set of benchmarks that favor the now-changed program and the former program would be measured as slower. -- === Subject: Re: JSH your ship has come in!!!! message Maybe the only point is that I fear James being overwhelmed by > evil. > Hmmm. > I have to ask myself, Why should I care? James may be the > reincarnation > of Gauss, but is it really any skin off my nose if he goes > unrecognized? > I've been worrying about the guy for months (ever since I > realized > that > he > was not, in fact, a crank, but a genius) and defending him on > this > newsgroup. > My reward? Laughter and bile from the peanut gallery. And not > a > word > from > James himself. A word of advice to Prof. Connes: Don't waste > your > e > on > . If he loves being the solitary genius so much, > let > him > fight > his own damn battles. He's not worth losing sleep over. Well damn it, I'm losing sleep at least partly because of your > scary > dream!!! Good writing their Ferry, and I have to give you credit for > that, > but hey, how many years were you ridiculing me, and now you expect > me > to just go, hey, pal? Show me you're serious and wade in and respond to some of these > ant-mathematicians in the current battle. Prove your sincerity, and um, keep posting any interesting dreams > you > have, as I found it interesting puzzling over that one. Exactly! This Ferry now thinks you are a genius, but has he > defended > you in your current battle? No! I suspect that he is not *fully* committed to your view of > mathematics. > That > is just a feeling, and I suppose I could be wrong. The primary focus is the odd definition error in core. Once that's accepted for what it is, and most importantly FIXED, then > it's not about committing to my view of mathematics, but about showing > your commitment to mathematics itself. There is ONE mathematics. I, however, have been trying to find out more about your Object > Ring, > but I > feel you are pushing me away. I have spent some effort guiding you along at the Mega Foundation > discussion area. Yes, indeed. But, I expect you remember that I did ask you a question that I was > unable > to answer. I was using the notation [a,b,c] to represent an ordered triple of > complex > numbers. And I was wondering if the ordered triple [1,2,8] was an > element of > the Object Ring. You replied: > Quit being lazy!!! You have the definition, figure it out for > yourself!!! > What amazes me is how often people are willing to ask someone else to > do > their work for them. > If you're smart enough, answer your own question. > I'm curious to see if you can. > I've given the definition for the object ring, so no excuses. Well, I am ashamed to say I still cannot figure it out. > Sounds like a ploy. > You have to understand that your record Clive Tooth is rather long and > involves some rather...sleazy behavior...like that attack webpage, and > quite a few negative postings over a period of YEARS. > That was just... just... boyish high spirits, James. And you joined in the > fun by threatening to sue me for libel!! Ah... happy days... Well, it was effective writing on your part as even when you took that webpage down that other guy copied it for his own webpage, and then some other, um, person ran that robot program. And *someone* out there was at one point trying a somewhat meek denial of service attack on one of my old websites, or, weirdly enough, they were trying to convince me I had thousands of hits per day on my webpages!!! That's one of the reasons I was happy with MSN as I never even knew how many hits my pages were actually getting with them, but there was no worry about them charging me extra! > You don't get the benefit of the doubt from me but have to make the > extra effort yourself, so quit being lazy!!! > Oh... James... You have to admit that I did help you out, on the Mega board, > with sqrt(i) which you didn't realize was a complex number. Can't you help > me out just a little with this one? That was a silly error. Like I there though I was reaching as I really *wanted* to believe that the object ring isn't a subset of complex numbers, but didn't have the proof, so I guess I did what I've done many es before and told myself what I wanted to hear. As for your question, again, quit being lazy!!! Figure it out yourself, or even go get help from someone else besides me, as I've spent enough e with you already, and given your history, it's not sensible to spend any more. > And I noticed that you that Ferry is on the team! How can I get > on the team if you won't help me when I am struggling? By the way, is > anybody else on the team? I think that all the team-members should have > well-defined roles for the up-coming battle. So far the team as I call it are people who recognize that my work is indeed correct, and so far seem to only be very high IQ people. I guess that's not too surprising. > Could you help me please? > It sounds to me like you think you have some angle for even more > negativity and I've given enough e explaining. > Remember mathematics is a continuing process. The object ring is fascinating in and of itself, so you can't expect > me to know all the answers just because I'm a discoverer. After all, if it were that easy, then math research would have ended > long ago with the first mathematician explaining it all. Your friend. Clive e will tell. > Yes, as I , e will tell. > Very true. Yup. === Subject: Re: polysigned numbers An interesting point about reduced form is that it so far has not been absolutely necessary except for the dimensional analysis. That is to say that a number like: ( - 1 + 2 * 3 # 4 ) can be worked with even though it is not in its reduced form. So far none of the math, including product and graphical analysis fails to work with non-reduced numbers. The reduced form of the above number is: ( + 1 * 2 # 3 ). Is this is the reduction you are speaking of? The neat thing is that and so is coherent. I believe that the same will happen with the tetrahedral. The origin is at the center of a tetrahedron. the poles go out from the origin through the points that form the tetrahedron. This is the symmetrical mapping of four-signed math in cartesian space. Any point in RxRxR can be uniquely defined in four-signed math. Since I don't have the math I can't say to have proven this. I can see it though. Making this assumption and putting the #-pole in the i direction (as in i,j,k) we get the following partial transform for a four-signed value x: a = n(x) - ( 1/3 )( m(x) + p(x) + s(x) ) where a is the i component of the three dimensional vector ai + bj + ck. n(x) is the # (number) component, m the - (minus), p the + (plus), and s the * (star). Now putting the minus pole in he ij plane and going in a left-handed direction we see that the one third component yields an angle of: pi - arccos( 1 / 3 ) where arccos is the inverse cosine. This should be the angle from the center to any two corners of the tetrahedron. Please note that I am not proving this. I'd like to find that angle in a book somewhere or sharpen up my triginometry skills so that I could verify this. If this is true then there should be no problem with a clean RxRxR transformation for four-signed math. The same concept should work upward beyond our sight to five signed(pi-arccos(1/4)) and beyond. Please keep in mind the simple sum rule: - x + x * x # x = 0. > I've done some work on the four-signed math this weekend. It looks like > the key to making it work is carefully defining a reduced form. Under > certain definitions of reduced from, n-signed math exists in R^(n-1), > Under other definitions, n-signed math is isomorphic to C. I need to > finish some more details, then I'll let you know what I come up with. I > think that thinking in terms of n-tuples will make things easier when > dealing with general n-signed math. Otherwise you may need to switch to > subscripted signs. I will try to work with your representation, but I am pretty happy with the symbolics that are used here since it is more arithmetic. At some point I'd like to go to code. Do you know C or C++? I think four signed is as far as I'd like to go for now in symbolic format. I've written n-signed code for C++ and performed the Mandelbrot mapping for four-signed in the simplest planes. Some look like the usual Mandelbrot and some are simple solids. It's quite likely that there are some bugs in my code so don't take my results too seriously. I can't get too excited about the Mandelbrot mapping anyhow. I was very disappointed when my three-signed code spat out the usual mandelbrot shape. I was really hoping for something new. What does the Mandelbrot test mean anyways? It has become apparent to me in writing code that the symbols used for sign should really include a zero sign below the minus sign. This zero sign is identical to the highest sign ( # for four-signed, * for three-signed, + for two-signed). Since the reals are symbollically faulty in this way I will continue on with the symbols I have chosen for now. I'm looking forward to your results but I fear that you are straying from the zero sum rule. Golden === Subject: Re: Help is needed with a Bayes Rule exercise > In a certain region of the country it is known from past experience > that the probability of selecting an adult over 40 years of age with > cancer is 0.05. If the probability of a doctor correctly diagnosing a > person with cancer as having the disease is 0.78 and the probability > of incorrectly diagnosing a person without cancer as having the > disease is 0.06, what is the probability that a person is diagnosed as > having cancer? See http://www.acad.sunytccc.edu/instruct/sbrown/stat/falsepos.htm for a useful look at problems like this, although the approach does not use Bayes theorem. === Subject: Re: The ... spacee; answer to critic. > I know it is a bit stupid, but > 1> how do you prove that a discrete topology is metrizable? Write down a metric for it :-) The simplest possible metric should give the discrete topology ... > 2> X is an set of all positive integers and > T={{},{1,2,3,4...},{2,3,4...},{3,4...},{4...},...} Why is (X,T) not > metrizable? Is it Hausdorff? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 es) === Subject: Re: Ces.88ro-convergence - analysis question >Hi all, >Regular convergence implies Ces.88ro-convergence. The inverse statement is >not generally true. Can one give conditions on a sequence to guarantee >that when you have a sequence that is Ces.88ro-converging it is also true >that it is regularly converging? Yes. The fact that convergence implies Cesaroo convergence is an abelian theorem, and the fact that Cesaro convergence plus some extra condition implies convergence is a tauberian theorem, where the extra condition is a tauberian condition. It's fairly easy to show that if the sequence s_n is Cesaro convergent and also (s_{n+1} - s_n)*n -> 0 then the sequence s_n is convergent - if I recall correctly this is the world's first tauberian theorem, proved by Tauber. It's true but not nearly as easy to show that you only need to assume that (s_{n+1} - s_n)*n is bounded - this is due to Littlewood I think. (I'm assuming here you really want to talk about convergence of sequences - of course there are corresponding results for sums of series...) At least I've read that Littlewood's theorem is not so easy, and the proofs I've seen are not quite elementary. Thinking about it just now it seems that one can give a very simple proof of Littlewood's theorem. Maybe there's a stupid error below, maybe the theorem is really not supposed to be that hard, or maybe this is a huge event in the history of mathematics: Assume that (i) (s_1 + ... + s_n) / n -> 0 and (ii) |s_{n+1} - s_n| <= c/n for all n. We want to show that s_n -> 0. Suppose not; then there exists d > 0 such that (iii) |s_N| >= d. for infinitely many N. So we may as well assume (iii') s_N >= d for infinitely many N. Suppose N satisfies (iii'). Now for n > N we have s_n >= d - c sum_N^n 1/j ~ d - c log(n/N) >= d/2 as long as c log(n/N) < d/2. This condition is the same as n < N * exp(d/(2c)) = N * K = M; note that K > 1. Hence (iv) (s_N + ... + s_M) / M >= (d/2)(M - N)/M = (d/2)(K - 1)/K = C > 0. But (iv) shows that |s_1 + ... + s_M|/M does not tend to zero, since the difference between this and the left side of (iv) is smaller than |s_1 + ... + s_{N-1}| / M, which _does_ tend to zero, since M > N. ************************ === Subject: Re: Chess/Go/etc: Continuous Game-Boards? > I was just wondering today if there has been anything interesting > games such as Chess, Go, etc, where the game-boards are without > any subdivision? > For example, each piece's base's shape/size would be significant. The > pieces would move (or be placed on the gameboard or removed from > it) under certain rules, but the exact distance/direction could be > within a continuous interval. > For instance, in Chess, the pieces (except knight) would not be > allowed to touch other pieces while moving. (So, either this game > would be played on a computer so as to be precise in moving, or the > game would take on aspects of Operation if the pieces are moved by > hand {no touching pieces to each other!}.) > I think, with Chess, what is considered a capture might be ambiguous. > Go, on the other hand, might be very interesting if played > continuously, especially if the pieces are different-sized circles. > (Each player would have the same number of pieces as their opponent > of any given radius, since bigger pieces would be more valuable > than smaller pieces.) > Any other games that might be interesting if played this way, > either on a computer or by hand? Modifying the basic bridging game (Twixt, Hex, etc.) to be played on a continuous board makes a fun game, but would require a computer implementation to play seriously (too much room for argument about size of circles, thickness of lines, etc.). Somebody should write one. The basic idea is that you have a square or rectangular board. Players take turns drawing circles, and trying to make a chain of their own circles from one end to the other. If two circles overlap, the first one drawn gets the overlapping space, but the other one can control whatever space it has that is distinct from earlier circles. Player one goes from top to bottom and player two goes across. If the board is square, each player draws two circles per turn (except the very first turn). If rectangular, it is (ideally) carefully calibrated so that the extra distance required of the first player neutralizes the advantage of going first. Jonathan Dushoff === Subject: Re: =?ISO-8859-1?Q?Ces=E0ro-convergence_-_analysis_question?= > Hi all, > Regular convergence implies Ces.88ro-convergence. The inverse statement is > not generally true. Can one give conditions on a sequence to guarantee > that when you have a sequence that is Ces.88ro-converging it is also true > that it is regularly converging? Such extra conditions are called Tauberian conditions after the first theorem of this kind, due to Tauber. === Subject: Re: 0! = 1 >Did someone know a simple demonstration of >0! = 1 ? > n! is equal to the product of j as j varies from 1 to n by steps of 1. > 0! is therefore an empty product, and the empty product evaluates to 1. > David McAnally Yes, and why does the empty product evaluate to 1? Because that is the multiplicative identity. Likewise, the empty sum evaluates to 0. You shouldn't change the value of a number by not adding anything to it. Spencer === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana. >>The only minimal counter-example to the FCT is K5! >> No, K5 is NOT a counterexample to the Four Color Theorem, because the >> 4 color theorem states that any ->planar<- graph can be colored with >> at most 4 colors in such a way that no two adjacent vertices share the >> same color. >>The conjecture that there exists a 5-chroma graph may be recolored to >>4-chroma is false. >> There is no such conjecture. >>Let H be any subgraph of G, where G has n vertices and H has n-1 >>vertices. Then, the description of H seems to imply that the deletion >>of 'any' vertex from G will make chi(H)<=4. >> This is true if G is a minimal counterexample for the 4 Color Theorem. >>But this interpretation is generally false and is valid only for >>n=5!!! >> The triple exclamation points make you look like a raving loon. So >> start by removing them. >Point taken, thank you. Could you explain why? Can I explain why the triple exclamation points make you look like a raving loon? Because they do. It makes it seem like you are jumping up and down, yelling, spitting, and foaming at the mouth. That's the mental image they conjure up. >> Then note that the original argument started by ->assuming<- that the >> FCT is ->false<-, from which we deduce that if this is the case, then >> among them, there is one with the least number of vertices. Call n the >> number of vertices of this HYPOTHETICAL counterexample. Then, by the >> definition of n, any graph with fewer than n vertices must be >> 4-colorable. In particular, if you took this HYPOTHETICAL example G, >> and removed one vertex, then the resulting graph would be 4-colorable. >> What exactly are you having trouble understanding about the above >> argument? Try to answer without using a ->single<- exclamation point. >I understand the argument perfectly. Then why did you think somebody was claiming the Four Color Theorem was ->false<-? >I have given the problem some >thought and I have concluded that HYPOTHETICAL G is impossible. Good for you. > No >graph meets all three criteria; ie, G is 5-chroma, G is planar, H is >4-chroma. Good for you. But your argument seems to be no such G can exist, because then G would be K5, and that does not even begin to make sense. G cannot be K5 if it is assumed to be planar. Indeed, the proof of the Four Color Theorem rests on showing that there does not exist any graph G which requires 5 colors, is planar, and such that the removal of any vertex results in a graph which can be colored with only 4 colors. But you have not given any coherent argument to establish this proposition that I can see anywhere. All you have done is yell like a loon that G would be K5!, which is nonsense. === Subject: Re: Minimal Graph, Four Color Theorem |I understand the argument perfectly. I have given the problem some |thought and I have concluded that HYPOTHETICAL G is impossible. No |graph meets all three criteria; ie, G is 5-chroma, G is planar, H is |4-chroma. I assume by this last clause you mean that all the graphs one obtains by deleting a single vertex from G are 4-chromatic. This conclusion is equivalent to the four color theorem. If the four color theorem is true, then no planar graph has chromatic number 5. On the other hand, on the assumption that your conclusion above is correct, the four color theorem follows by induction on the number of vertices. Once we've shown it's true for planar graphs of up to n vertices, then it also must be true for planar graphs of n+1 vertices, since whatever graph H is, it's already been shown to have chromatic number <=4 (and its chromatic number differs from that of G by at most 1). So the only way you can reach such a conclusion is by an argument which is at most one short paragraph shorter than a proof of the four color theorem. I would assume that you're just relying upon the existing proof, except that it wouldn't usually take some thought to conclude that a 5-chromatic planar graph having a certain kind of subgraph doesn't exist, given that no 5-chromatic planar graph exists at all. I just hate to see someone go away still confused, so I hope your clarity on the argument has reached the point of recognizing that this conclusion you state above is very far from trivial, without taking the proof of the four color theorem for granted. If there's some simple way to show such a G doesn't exist, a number of smart people have failed to see it despite working hard on it for a long e. === Subject: Re: Numeric one-way hash function > I need to find an algorithm that can produce a unique non-predictable > 12 > digit (0-9) number for any given 12 digit number. This is to be used to > create a unique barcode on a ticket that cannot be predicted. It is not > required that the original seed number be computed from the resulting > barcode, so some form of one-way hashing function would be acceptable. > Any help in this problem would be appreciated. > The simplest way is to encrypt the first number using AES or >> 3DES. You will have to convert the result from binary to decimal. >> Any extra digits can be thrown away. Change the key variable >> regularly and keep the old ones secret. >> But shouldn't the bar codes be unique? >> Your procedure can generate duplicate bar codes. > With high grade ciphers like AES there will be very few > duplicates. I assume that you throw away extra digits by computing the block cipher output modulo 10^12. When the block cipher is cryptographically strong this should give a random map, in other words you can regard the output as a random number with range [0..10^12). So when you take the entire set of outputs for all 10^12 inputs this set should behave like a set of 10^12 random numbers that are independently drawn from a uniform distribution. Now let's compute the probability that one specific number is not present in this output set, this probability is (1 - 1/n)^n and when n is large this is approximately equal to 1/e = 0.368 . So about 36% of the possible outputs never occur, this means that at least 36% of the inputs will lead to a duplicate output value. > The simplest way to ensure that the bar codes are > unique is to add a prime number to the previous > value. Lap round when you get to the top (or a prime > number near the top). Start at a weird value. Could you be more precise, I am afraid that I don't understand your algorithm. Anyhow, I can't see that your algorithm would be unbiased, or how it would prevent duplicates. greetings, Ernst Lippe === Subject: Re: Help is needed with a Bayes Rule exercise >In a certain region of the country it is known from past experience >that the probability of selecting an adult over 40 years of age with >cancer is 0.05. If the probability of a doctor correctly diagnosing a >person with cancer as having the disease is 0.78 and the probability >of incorrectly diagnosing a person without cancer as having the >disease is 0.06, what is the probability that a person is diagnosed as >having cancer? A = person has cancer => P(A) = 0.05 Ac = person is healthy => P(Ac) = 0.95 B = diagnosis is positive Bc = diagnosis is negative P(B | A) = ???, P(B | Ac) = ??? Now, P(B | A) = P(A | B) * P(B) / P(A) => P(B) = P(A) * P(A | B) Bayes' formula gives us: P(A | B) = P(A)*P(B | A) / (P(A)*P(B | A) + P(Ac)*P(B | Ac)) so the final result can be calculated by plugging the second equation into the first one. === Subject: Re: Help is needed with a Bayes Rule exercise > Hello good people, > I am taking a course in probability and for some reason found the > Bayes' Rule topic not so easy to understand. > I need your help with showing me the right way to solve a problem. I > don't need only the solution as this is not a test. I would really > appreciate it if someone could show me the different steps involved in > the solution with some explanations. > I have to solve the following problem: > In a certain region of the country it is known from past experience > that the probability of selecting an adult over 40 years of age with > cancer is 0.05. If the probability of a doctor correctly diagnosing a > person with cancer as having the disease is 0.78 and the probability > of incorrectly diagnosing a person without cancer as having the > disease is 0.06, what is the probability that a person is diagnosed as > having cancer? > Again, I'd love to see the steps and way to the solution so I can do > similar exercises by myself. > Dotan. Here's another web page if you need it : http://faculty.vassar.edu/lowry/bayes.html === Subject: Re: Boolean Algebra - Arithmetic Relationship > ... Can arithmetic be further simplified into boolean logic?? Yes. Computers - or rather those who design computers, are quite good at it. ;-) === Subject: Re: 0! = 1 Spencer Mortensen scribbled the following: >>Did someone know a simple demonstration of >>0! = 1 ? >> n! is equal to the product of j as j varies from 1 to n by steps of 1. >> 0! is therefore an empty product, and the empty product evaluates to 1. > Yes, and why does the empty product evaluate to 1? Because that is the > multiplicative identity. > Likewise, the empty sum evaluates to 0. You shouldn't change the value of a > number by not adding anything to it. If I know my algebra, then an identity for an operation (let's call it #) is such an i, that for all x, x#i=x. But there is another concept too: such a j, that for all x, x#j=j. For multiplication, this is 0. For addition, there doesn't seem to be one. What is this concept called? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ As we all know, the hardware for the PC is great, but the software sucks. - Petro Tyschtschenko === Subject: Re: Properties of f(x) given the graph of f'(x) You can tell where its max/min are. You can also tell where f(x) is increasing and decreasing. All functions f(x) will be the same with the exception that they will differ by a constant. Lurch > What conclusions about function f(x) can be drawn given the graph of > its derivate, function f'(x)? > That is where the ordinate is f'(x) and abscissa is x. === Subject: Re: irreducible polynomial ? I asked: > Is the polynomial 2*cos(2^k*arccos(x/2)) irreducible for each > positive integer k? If so, how is it proved? I received email from Joe Silverman answering this question and giving a good deal of additional information on the factorization of Chebeshev polynomials. --------------------------------------------------------------- Comments from Joe Silverman: The n'th Chebeshev polynomial (up to factors of 2 used by different authors) is the polynomial F_n(x) with the property that F_n(2*cos(t)) = 2*cos(n*t). So if we put t = arccos(x/2), then we get F_n(x) = 2*cos(n*arccos(x/2)). This is your formula. If we write cos(t) = (e^{it} + e^{-it})/2, and for ease of notation, let z = e^{it}, then we get F_n(z+z^{-1}) = z^n + z^{-n}. This is another characterization of F_n. The roots of F_n(x), from your formula or from this alternative, are x = 2*cos(pi*(j+1/2)/n) for j = 0,1,2,...,n. Notice this can also be written as x = e^{it} + e^{-it} with t = (pi*(j+1/2))/n. it's easy to see that the splitting field of F_n(x) is the real subfield of the 4n'th cyclotomic field. In other words, the roots of F_n(x) generate the field K_{4n} = Q(e^{pi*i/2*n}+e^{-pi*i/2*n}). The degree of this field is [K_n:Q] = phi(4*n)/2, where phi(n) is Euler's phi function. On the other hand, the degree of F_n is n. So the conclusion is the following: Proposition: F_n(x) is irreducible over Q if and only if phi(4*n) = 2*n. The case you're asking about is n = 2^k, and indeed phi(4*2^k) = phi(2^{k+2}) = 2^{k+1} = 2*2^k, so your polynomials are irreducible. Further, I think this is probably the only case that phi(4*n) = 2*n, so the only case that F_n is irreducible. Hope this is of some help. Feel free to post this, if you want. ------------------------------------------------------ -- Buddenhagen To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE === Subject: Re: Is it mass; or is it weight >Take any object; body or mass of material substance: It's a misnomer to call >it a weight; because it's really not: >which is the mutual thrust that they exert on each other when they obstruct >each other from simultaneously occupying, and/or passing through the exact >same place; because they can't: This is the phenomena of impenetrability. >It's how matter interacts, and aggregates into masses. >So any object or body of material substance _is a mass_, and will exert >force on any other object; body or mass of material substance when any two >or more of them resist each other's passage through the same place: The >force they exert will either reverse, stop, slow or change each other's >direction of motion. >All masses of material substance resting on Earth's surface are constantly >being restrained from free falling further toward Earth's center by a mutual >force called weight; which we all feel as heaviness, and which can be >measured on weight-scales: Where weight-scales are calibrated by standard >masses; whose weight, divided by the acceleration at which they will free >fall is a constant: >Thus a slug is a standard _unit_ of mass; A slug is a little used 20th century invention which exists only in one specialized subsystem of mechanical units--and a unit which doesn't even have an official definition. >as is a gram and a kilogram. These >units of mass and subdivisions thereof, are often simply called weights; >which they aren't: They are masses, whose inertia is a ratio of the thrust >exerted on, and/or by them, to the acceleration - and/or deceleration - that >it causes: >Now it's e for all of us to realize that a customary pound is a unit of >force, You know better, Donald. Pounds everywhere have always been units of mass. Pounds force are such a recent bastardization that they are uniquely identified by that name. Back in 1959, the national standards laboratories of the United States of America, Canada, the United Kingdom of Great Britain and Northern Ireland, the Union of South Africa, Australia, and New Zealand got together and agreed on a common definition of the most commonly used pound, the avoirdupois pound. They defined it as a unit of mass exactly equal to 0.45359237 kg. Of course, you already knew that, Dishonest Don. This is for the benefit of anybody else who hadn't been paying attention. > and a slug or a subdivision thereof is a mass: Always. But they only exist in one specialized subsystem of mechanical units, a system which includes none of those pints you are so fond of, not U.S. liquid pints, not U.S. dry pints, not imperial pints, and not those ill-defined Dense Donny Pints either. So is a hyl, another obsolete unit of mass which like the slug was only used in calculations. Look it up; it isn't so hard to find with internet search engines. What is the base unit of force in the system in which the hyl (aka the TME from a German acronym, or the mug, or the metric slug) is the derived unit of mass? > Just as a metric >newton is a unit of force, And so is the poundal in English units. >and a kilogram or a subdivision thereof is a mass! Kilograms force were quite proper and legiate units before the introduction of the International System of Units in 1960, and we still see many vestiges of their use. For example, I have a torque wrench in meter kilograms and they are still readily available in those units. Good grief, those kilograms force even have another name, the kilopond, a name more common in Europe than it ever was in America. Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ Gentlemen of the jury, Chicolini here may look like an idiot, and sound like an idiot, but don't let that fool you: He really is an idiot. Groucho Marx === Subject: Re: 0! = 1 > If I know my algebra, then an identity for an operation (let's call > it #) is such an i, that for all x, x#i=x. But there is another concept > too: such a j, that for all x, x#j=j. For multiplication, this is 0. > For addition, there doesn't seem to be one. What is this concept called? It's called a zero. === Subject: Re: What to tell students in a 10-15 minute talk @vixen.cso.uiuc.edu: >> So if you think of the three hands as rotating unit vectors, >> their sum will never be zero. So here's another problem. Assign >> angular velocities to the three unit vectors so that the mimimum >> length of their sum is as large as possible. > There are various answers depending on what additional restrictions > you apply. For j = 1,2,3, let Case j allow up to j hands to have the > same angular velocity. Define subcase (a) by requiring all hands to > meet at some e, and subcase (b) by not requiring this. > Case 1a: maxmin length is sqrt((47 - 14 sqrt(7))/27) ~= 0.607346. > The second hand moves three es as fast as the minute hand in the > frame of reference of the hour hand. Yeah, that's the interesting case, and that's the right answer. We can restate it as trying to evaluate: sup over {a_1 That was just... just... boyish high spirits, James. And you joined in the > fun by threatening to sue me for libel!! Ah... happy days... > Well, it was effective writing on your part as even when you took that > webpage down that other guy copied it for his own webpage, and then > some other, um, person ran that robot program. > And *someone* out there was at one point trying a somewhat meek denial > of service attack on one of my old websites, or, weirdly enough, they > were trying to convince me I had thousands of hits per day on my > webpages!!! > That's one of the reasons I was happy with MSN as I never even knew > how many hits my pages were actually getting with them, but there was > no worry about them charging me extra! > You don't get the benefit of the doubt from me but have to make the > extra effort yourself, so quit being lazy!!! > Oh... James... You have to admit that I did help you out, on the Mega board, > with sqrt(i) which you didn't realize was a complex number. Can't you help > me out just a little with this one? > That was a silly error. Like I there though I was reaching as I > really *wanted* to believe that the object ring isn't a subset of > complex numbers, but didn't have the proof, so I guess I did what I've > done many es before and told myself what I wanted to hear. > As for your question, again, quit being lazy!!! > Figure it out yourself, or even go get help from someone else besides > me, as I've spent enough e with you already, and given your > history, it's not sensible to spend any more. I am not sure who to turn to, James. I would guess that neither Arturo nor Nora would be able to help me with this. They just don't seem to have grasped the basic concepts of the Object Ring. > And I noticed that you that Ferry is on the team! How can I get > on the team if you won't help me when I am struggling? By the way, is > anybody else on the team? I think that all the team-members should have > well-defined roles for the up-coming battle. > So far the team as I call it are people who recognize that my work > is indeed correct, and so far seem to only be very high IQ people. James, I know you have been in the Army. Perhaps we could organise the Team along Army lines... with ranks! You could be the General, of course... and... um... I could be a colonel, say. And that Ferry could be a private. Talking about Ferry, I wonder if he could help me with my question about the ordered triple of complex numbers. I don't know where to turn... -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Can you help me? >prove the following: > lim c^1/n = 1 where c>0 >n -> inf. >Do you have any pointers? > Can't you take | nth-root(c) - 1 | and es it by (nth-root(c) + 1)/(nth-root(c) + 1) ? Work with that. I think that will work, but I am not positive. Lurch === Subject: Re: Properties of f(x) given the graph of f'(x) > What conclusions about function f(x) can be drawn given the graph of > its derivate, function f'(x)? > That is where the ordinate is f'(x) and abscissa is x. This question becomes reasonably interesting if it is read as implying simply that one is given the graph of f' in the way that graphs are usually given, namely, as rather fuzzy drawings. (After all, if the real intention of the question was to ask what conclusions about f(x) can be drawn given f'(x)?, why bring graphs in at all?) When one is given a fuzzy graph, it might not be at all clear where (if it all all) it touches, or crosses, the (equally fuzzy) x-axis, nor where (if at all) the function it represents is increasing or decreasing. Here's a precise version of the question: what conclusions about a continuously differentiable function f on a closed interval [a,b] can be drawn, given a continuous function g on [a,b], an epsilon > 0, and the knowledge that |f'(x)-g(x)| is at most epsilon for x in [a,b]? === Subject: Re: Is this newsgroup useless? >> But my other interest is homebrewing, > What's your favorite style? I like to brew IPA's. But, so far, my best > brew is a lagered Bohemian Pilsner. Smooooth. I'm having a little > problem with mashing ... poor yield. Math + Homebrew = FUN. I probably to an IPA every other batch, because they're ready early and I seem to run out of beer so fast. I often make the Sister Star variety because I like heavily hopped versions. But my favorites are double bocks, even though they have to age forever. The thing that increased my mash efficiency the most was doing a mash-out. (In a 10-gallon batch) just before sparging I drain off 3 gallons of wort, bring to almost boil, and then add it back to the mash. This brings the temp up into the 165 range and the sparging rinses out much more fermentables. (But you gotta' be careful not to go higher than 170.) But maybe you do that already. The other thing that helped me a lot was paying attention to pH and adding the right amount of calcium carbonate to the mash water. Gotta keep those enzymes happy. rec.crafts.brewing is a great group. I've nicknamed my liver Kenny, by the way... Bart === Subject: Re: JSH your ship has come in!!!! > As for your question, again, quit being lazy!!! > Figure it out yourself, or even go get help from someone else besides > me, as I've spent enough e with you already, and given your > history, it's not sensible to spend any more. Translation: You haven't the slightest idea how to answer the question, so you're stalling and hoping someone else will do it for you. Of course, you'd still be quite happy to take the credit for the answer. -- rs, Silverlock === Subject: Re: Measure extension proof > Well, it happens a lot that the _existence_ of the extension is > used to define measures, by first defining them on a (non-sigma) > field... when you do that you need the uniqueness to know that > you've defined _a_ measure. >In other words, is understanding of it >important in terms of understanding new material down the road? > Other people may have different opinions: If you're just learning > measure theory my advice would be to skim through this part as > quickly as possible and concentrate on the stuff coming up that > gets used in applications of measures, as opposed to constructions > of measures - if it turns out you get into something where this is > important there will be plenty of e to go back to the details. >Your >help is always appreciated. >> ************************ >> === Subject: Re: 0! = 1 A N Niel scribbled the following: >> If I know my algebra, then an identity for an operation (let's call >> it #) is such an i, that for all x, x#i=x. But there is another concept >> too: such a j, that for all x, x#j=j. For multiplication, this is 0. >> For addition, there doesn't seem to be one. What is this concept called? > It's called a zero. So intuitively, if #' is a repetitive #, which means that x#'i is x#x performed (i-1) es in succession, then the zero of #' is the identity of #? For example multiplication is repetitive addition. 0 is the additive identity and the multiplicative zero. If we switch the definition of a zero around so that it means that for all x, j#x=j, then we get that exponentation is repetitive multiplication, and that 1 is the multiplicative identity and the exponentative zero. Is this making any sense? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ My absolute aspect is probably... - Mato Valtonen === Subject: All masses have inertia The way I see it guys, is that all masses have inertia! That is it requires a net impulse - the product of a net force [f] and its duration [t] - to change a body's present velocity [vi], to some other velocity [vt]; during which period of e, the body is displaced a distance [s]: The ratio of this impulse [ft], to the e rate of displacement [s/t = (vt-vi)] that it causes is a constant: That is ft/(s/t) = ft/(vt-vi); which can be written more concisely as f/tî/s = ftî/ (vt-vi)! Isn't _that_ inertia? === Subject: Re: irreducible polynomial ? > I asked: > Is the polynomial 2*cos(2^k*arccos(x/2)) irreducible for each > positive integer k? If so, how is it proved? > I received email from Joe Silverman answering this question > and giving a good deal of additional information on the > factorization of Chebeshev polynomials. > --------------------------------------------------------------- > Comments from Joe Silverman: > The n'th Chebeshev polynomial (up to factors of 2 used by different authors) > is the polynomial F_n(x) with the property that > F_n(2*cos(t)) = 2*cos(n*t). > So if we put t = arccos(x/2), then we get > F_n(x) = 2*cos(n*arccos(x/2)). > This is your formula. > If we write cos(t) = (e^{it} + e^{-it})/2, and for ease of notation, > let z = e^{it}, then we get > F_n(z+z^{-1}) = z^n + z^{-n}. > This is another characterization of F_n. > The roots of F_n(x), from your formula or from this alternative, are > x = 2*cos(pi*(j+1/2)/n) for j = 0,1,2,...,n. > Notice this can also be written as > x = e^{it} + e^{-it} with t = (pi*(j+1/2))/n. > it's easy to see that the splitting field of F_n(x) is the real subfield > of the 4n'th cyclotomic field. In other words, the roots of F_n(x) > generate the field > K_{4n} = Q(e^{pi*i/2*n}+e^{-pi*i/2*n}). > The degree of this field is [K_n:Q] = phi(4*n)/2, where phi(n) is Euler's > phi function. On the other hand, the degree of F_n is n. So the conclusion > is the following: > Proposition: F_n(x) is irreducible over Q if and only if phi(4*n) = 2*n. > The case you're asking about is n = 2^k, and indeed > phi(4*2^k) = phi(2^{k+2}) = 2^{k+1} = 2*2^k, > so your polynomials are irreducible. Further, I think this is probably the > only case that phi(4*n) = 2*n, so the only case that F_n is irreducible. > Hope this is of some help. Feel free to post this, if you want. by essentially the same argument you see that the odd chebyshev polynomials of prime order are irreducible if you eliminate the trivial factor x (zero x=0) sincerely Klaus > ------------------------------------------------------ > -- Buddenhagen > To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE === Subject: Re: What to tell students in a 10-15 minute talk > sup over {a_1 Where the a_k and t are real. > Let f(n) be the maxmin length for n hands. Then f(1)=1, f(2)=0, > and f(3)=sqrt((47 - 14 sqrt(7))/27). > Some questions: > 1. Is f(n) always attainable by specific values of the a_i? > (And if so, is it attainable by integer values as in the case > of f(3)?) > 2. Is f(n) monotonically increasing? > 3. What are some other values for f(n)? (4- or 5-handed clock.) > 4. f(n) has a trivial upper bound of sqrt(n+1), but this is pretty > gross. How about a substantial improvement? > I have a particular interest in question 1. Some comments: * It seems like a good idea to normalize a_1 = 0 (i.e., replace a_k with a_k - a_1 for all k). * The hands all meet at t=0, z=1. Define T to be the minimum positive value of t such that all values of exp(a_k*t*i) are identical. With the normalization a_1 = 0, this is equivalent to requiring that each a_k*t is an integer multiple of 2 pi. This minimum can fail to exist in two ways: (1) because no t exists to make the values identical, in which case let T = infinity, and (2) because a_k = 0 for all k, in which case let T = 0. Let's assume that n > 1 so that the latter case never occurs (because the a_k are unequal). * If T = infinity, the function |sum_{k=1}^{n} exp(ak*t*i)| attains all values between 0 and n, so this case can be ignored. (Exercise for the reader.) * Now we now apply a further normalization (for finite, positive T): let T = 2 pi. I.e., replace each a_k by (2 pi/T) * a_k (making the complete normalization to replace each a_k by (2 pi/T) * (a_k - a_1)). This makes the a_k's co-prime integers (meaning that the D of all of them together is 1, not that they're pairwise co-prime). * So the candidate a_k's are n-tuples like (0,3,7,8,11): i.e., n-tuples of increasing integers beginning with 0. I would now apply one final normalization. Replacing each a_k by a_n - a_k results in the same solution (i.e., frame of reference is now the fastest hand, and we look at the clock in the mirror). In the example, this yields (0,3,4,8,11). The final normalization is to take the smaller of the two (in lexicographical order), which is (0,3,4,8,11) in this case. * Now the key to proving things about this problem is to obtain a bound on the minimal sum, showing that for big (a_k) it's small, leaving only a few cases to check. In particular, the answer to 1 should be yes, because the maximal minimum sum will be obtained for small (a_k), not approached by a sequence of large (a_k). * Okay, that's a lot of hot air, and no real proofs, but I it's how I'd approach the problem if I wanted to prove things. -- | Ferry | Center for Simulation | +------------------------------------+ of Advanced Rockets | | http://www.uiuc.edu/ph/www/jferry/ +------------------------+ | jferry@[delete_this]uiuc.edu | University of Illinois | === Subject: Nice summation puzzle Starting from arcsin(x)=x+1/2/3*x^3+... and using an integration operation show well-known sum: oo 1 S ------------------- = (pi^2)/6 n=1 n^2 === Subject: Re: Properties of f(x) given the graph of f'(x) Visiting Assistant Professor at the University of Montana. >You can tell where its max/min are. You can also tell where f(x) is >increasing and decreasing. And also where f is concave up, concave down, and where its points of inflection are. f is increasing where f' is positive; f is decreasing where f' is negative. f has critical points where f' does not exist or is equal to zero, which can usually be figured out from the graph of f'. Also, f is concave up where f' is ->increasing<-, and f is concave down where f' is ->decreasing<-. The inflection points of f are where f' has relative extremes. === Subject: Re: How to calculate the total coverage area of a few circles? > ...Randy Poe...: > In my simulation, N circles with the same radius r are randomly > placed. Let P_i denote the center of circle i. For any i, p_i lies > within the coverage range of at least one other clicle, i.e. at least > one other circle contains p_i. How to calculate the total coverage > area of the N overlaped circle? The method should be easy to be > implemented by programming for simulation. ... > http://mathworld.wolfram.com/Circle-CircleIntersection.html > However, your problem might involve a large number of these, > and in addition you need to calculate overlaps of 3 circles, > 4 circles, etc. > Edelsbrunner has inclusion-exclusion formulas that depend only on > overlaps of at most three circles: The union of balls and its dual > shape, http://portal.acm.org/citation.cfm?id=161139 > [...] The number of circles is about 10. I think that the calculation > e of Monte Carlo integration might be too long. Is it possible to > use numerical integration [...] need to consider the sunk parts while > integrating the area [...] The sunk parts presumably are accounted for by the inclusion-exclusion formulae of Edelsbrunner. Here is an alternative plane-sweep method that would be ok for ~ 10 circles (its complexity is probably O(N^3) as stated): Make a list of the N*(N-1) circle intersection points and the 2N points p_i with x +/- r that are not in the interior of another circle, and sort into ascending order by x coordinate, then sum the areas of vertical zones bounded by these critical points. There are no arc intersections within a zone. A zone may contain disjoint segments but each segment is bounded above and below by an arc of a circle, and left and right by straight lines. For example, if we have 3 circles of radius 5 and centers at (5,12), (8,9), and (9,5), the event-points list for the plane sweep is: 0.0 12.0 3.3 7.3 4.0 5.0 4.1 5.9 9.7 13.7 12.9 8.1 13.0 9.0 14.0 5.0 The following edges or intersections are interior and not relevant: 3.0 9.0 4.4 7.0 9.6 10.0 10.0 12.0 -jiw === Subject: Re: Mass is a quantity of matter >> I'm sure that people would stand in line for blocks to get a signed >copy!! >As a matter of fact RJ, I have already written a couple, and can't even give >'em away! As long as the gravy train keeps running, nobody wants to rock the >gravy boat. > Your metaphors are as mixed as your understanding of elementary > physics. I think his metaphors are funny! But the idea is that THE ESTABLISHMENT (shiver, shiver) controls the TROOTH (tm) and suppresses publication of HERESY, especially because they're making lotsa money is certainly neither novel nor humorous. You'd think that he could at least GIVE them away to social friends who (Nudge,nudge, wink, wink) Bob Pease === Subject: Re: Numeric one-way hash function > .... [snipped] ... >... [I think Ruby-Lackov can >tolerate a small amount of bias in f. If not, I'm sure someone will post >another suggestion.] > Depends what you mean by tolerate. The security > proofs for Luby-Rackov certainly don't hold up if > there's any knowledge at all of the f() functions. I was thinking that, with sample size limited to 4, the slightly biased set of functions from which the four particular functions used are drawn could not be distinguished from the unbiased set of functions from which the four function should have been drawn. Given hash output words w distributed uniformly in 0..2**32-1, one way the code could avoid bias is by using w%1000000 when w>=967296 (which is 2**32%1000000) and repeating with a new w value otherwise. The probability of going on to a second w value is about 1/4440. The probability of exhausting all four words available from an MD5 hash even once in 10**12 barcodes 1/338. Code that just unconditionally used the first w%1000000 would, on average, in 1000000 trials, produce 999775 output values uniformly distributed in 0..999999 and 225 output values uniformly distributed in 0..967295. I conjecture that the number of plaintext-ciphertext pairs that an attacker would need to take advantage of that bias would exceed the number of barcodes the OP intends to print. In any event, I agree that qualms about the bias can be removed by using the uniform-in-[0..N-1] algorithms you mention. I think Benjamin posted good code for one. > While on that subject, I'll also point out that > the proofs require four *independent* f() > functions, not one that is reused. You can > simulate this with f_n = hash(key, n, data) for n = 1..4. My mistake. > That , for such a small data input, you'd > probably be safe ignoring those two nits. Problems > are more likely to surface somewhere else. >>You would have to be careful in the selection of your hash function. >>All standard hash functions have 2**n different outputs, and >>I don't know any hash function that produces 10**6 outputs. >>An example of a bad hash function would be to take the first 20 bits >>mod 1000000 of a standard cryptographical hash, because this hash >>function is extremely biased (some values occur twice as often as the >>others). When you use a larger number of bits the bias is reduced. > Yes, there's a regular subject here about how to > get unbiased uniform random in [0..N-1] given a > pseudorandom bitstream (such as generated by a > hash function), avoiding this bias that > creeps in when you least expect it. > Greg. --Mike Amling === Subject: Re: combination > Suppose we take x numbers out of y numbers in a decreasing sequence. > Say, take 2 numbers from {1,2,3} and arrange as {2,1}. What is the > number of possible combinations ? > y choose x (assuming the y numbers are distinct). === Subject: Re: JSH your ship has come in!!!! >And I noticed that you that Ferry is on the team! How can I get >on the team if you won't help me when I am struggling? By the way, is >anybody else on the team? I think that all the team-members should have >well-defined roles for the up-coming battle. Sarcasm. Sigh. >>So far the team as I call it are people who recognize that my work >>is indeed correct, and so far seem to only be very high IQ people. I must reiterate: I don't recognize that your work is indeed correct. (Maybe my IQ is not high enough...) My position is that it's not absurd to think it could be correct rather than to assume, arrogantly, that if I don't understand it, it must be wrong. I'm sickened by this carnival of dogs driven wild by raw meat, and I'm outraged that James is being treated like raw meat. James Harris is a child of God. We are all children of God. Where did it all go so wrong? I am supporting James because if he is correct (and I'm entirely unable to assign a number to the probability of that), it will have a profound impact on society. For the better, despite short-term chaos. > James, I know you have been in the Army. Perhaps we could organise the Team > along Army lines... with ranks! You could be the General, of course... > and... um... I could be a colonel, say. And that Ferry could be a > private. I'd prefer that when you dream about James's privates, my name not spring to mind. James would probably prefer people not to think about his privates at all, especially people named Tooth. Or Ferry, for that matter. But enough silliness. I don't take orders from James, nor does James give them at all, as far as I know. At first I didn't realize what my role in all this was to be, but it's becoming clear to me now. James needs a plan, and for reasons not clear to me, I have been receiving a plan. It keeps me awake at night. It distracts me from my work. It is vast and beautiful, and I worry that I'll never get it typed up, or worded right. I also worry that James will simply reject it out of hand because I've been such an asshole all these years. In fact, why would his plan even come to me? Why wouldn't it just come to him? That would seem more direct. Maybe it's too much for one person to do the math and receive the plan. It is, after all, a vast plan. Already it's wearing me out, and I've only typed up some notes, sketching the broad themes. I want to just chuck it all -- I don't owe James anything, not really -- but there's a reason I can't, a reason that has nothing to do with James, or his math. I'd sort of like to say, but I can imagine the kind of reaction I'd get on *this* newsgroup. Anyway, better to save it for the final product. Here's to the revolution! > Talking about Ferry, I wonder if he could help me with my question about > the ordered triple of complex numbers. I don't know where to turn... Oh poor Clive. It's really burning you up, isn't it? As James , figure it out yourself, or just forget about it. Stop pretending to be James's buddy. Colonel Clive? Colonel of Sarcasm! Colonel of Chuckles! Colonel of, I don't know, figure something else out, wise guy. Sorry, that was uncalled for. Clive Tooth, too, is a child of God. It's just that I wish you wouldn't bother James with such nonsense at a e like this. His day approaches rapidly. -- | Ferry | Center for Simulation | +------------------------------------+ of Advanced Rockets | | http://www.uiuc.edu/ph/www/jferry/ +------------------------+ | jferry@[delete_this]uiuc.edu | University of Illinois | === Subject: Re: All masses have inertia > The way I see it guys, is that all masses have inertia! That is it requires > a net impulse - the product of a net force [f] and its duration [t] - to > change a body's present velocity [vi], to some other velocity [vt]; during > which period of e, the body is displaced a distance [s]: The ratio of > this impulse [ft], to the e rate of displacement [s/t = (vt-vi)] that it > causes is a constant: That is ft/(s/t) = ft/(vt-vi); which can be written > more concisely as f/tî/s = ftî/ (vt-vi)! Isn't _that_ inertia? Why are you polluting this newsgroup with your topic. This is the third thread you started with almost the same topic, isn't it? -- Andreas For replying, remove the fruit from my address. === Subject: (job) Scientific Programmer - New York, NY Scientific Programmer - New York, NY This client is a technology leader specializing in software solutions for life science research and development. They are seeking computational scientists to create the next generation of computer-aided drug design methods, algorithms, and software. Two Ligand-based drug design positions are available for individuals with a background in 3D pharmacophore development (focusing on algorithm and methodology development, not strict modeling) in the New York City office. Responsibilities include: Participate in scientific and algorithmic development as well as software design and implementation. Extend and maintain large software packages written by others, as well as develop new applications. Required qualifications include: Ph.D. in computational chemistry, biochemistry, or biology. Post-doctoral research and/or relevant experience in the commercial sector. Proven skill and experience in ligand-based drug design with specialization in development of 3D pharmacophore models. Experience in methodology development for conformational analysis, pharmacophore feature definition and 3D pharmacophore model identification is preferred. Experience in pharmacophore modeling, 3D ligand-based drug design, and structure-based design experience is a plus. Strong algorithm development and methodology development skills. Strong programming experience. Good knowledge of C/C++ and object-oriented design is a plus. Experience writing and maintaining large (100,000-line-plus) software packages is preferred. Peer-reviewed publications in computational chemistry, bioinformatics, cheminformatics, or computational medicinal chemistry. For consideration, please send CV (including publications, thesis topic, academic scores and references) to resumes@workwondersstaffing.net. Benefits include medical, dental, 401(k), flexible spending account, 3+ weeks vacation, and tuition reimbursement. === Subject: (job) Computational Chemist - NY Computational Chemist - NY A great company in New York is seeking computational chemists (junior, senior and management level). Here is more information: Candidates should have world-class credentials in computational chemistry, biology, or physics, or in a relevant area of computer science or applied mathematics, and must have unusually strong research and software engineering skills. Relevant areas of experience might include the computation of protein-ligand binding free energies, molecular dynamics and/or Monte Carlo simulations of biomolecular systems, application of statistical mechanics to biomolecular systems, free energy perturbation methods, and methods for speeding up -- but specific knowledge of any of these areas is less critical than exceptional intellectual ability and a demonstrated track record of achievement. Current areas of activity for the firm include structure- and ligand-based drug design, protein structure determination through homology modeling, molecular mechanics force field development, de novo drug design algorithms, and the development of special-purpose hardware to accelerate computational chemistry simulations. This client will offer above-market compensation to candidates of truly exceptional ability. Please send your resume (including list of publications, thesis topic, and advisor, if applicable), along with a history of academic performance (including GPAs as well as SAT, GRE, and other standardized test scores) to resumes@workwondersstaffing.net. === Subject: ARCSIN function, single precision floating point. -- Example routine needed? I'm trying to write ATAN2 function for a small basic language that has IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are availible in the language. I've tried a few methods I've found but the results are way off due to low precision, rounding, etc. Are there any repositorys of old fortran routines or algorithms that I could use to get a good accuracy single precision routine. Speed or space aren't as important as reasonably good accuracy. Jon === Subject: Re: (job) Scientific Programmer - New York, NY Who does the composition on these things? I feel overqualified. -- On the internet, nobody knows you're a jerk. === Subject: Re: Properties of f(x) given the graph of f'(x) <> a .8ecrit s le message de > And also where f is concave up, concave down, and where its points of > inflection are. > [...] > Also, f is concave up where f' is ->increasing<-, and f is concave > down where f' is ->decreasing<-. Is a function with f'' > 0 called concav up in English? Is convex a synonym for that? I'm just wondering, because in German, such functions are called konvex (f''>0) and konkav (f''<0). -- Andreas For replying, remove the fruit from my address. === Subject: Re: How to calculate the total coverage area of a few circles? > about 10. I think that the calculation e of Monte Carlo integration might > be too long. Is it possible to use numerical intergration, i.e calculating > the area enclosed by the envelope of those circles? But how can I get the > expression of this envelop easily in my simulation? I also need to consider > the sunk parts while integrating the area, right? The calculation is related > Leng Supeng quoted, but it sounds like it contains an efficient algorithm to do exactly what you want to do. Note that it's from a computer graphics conference. Monte Carlo integration is a quick way of esating the integral you're talking about doing explicitly. On my Solaris machine here's the result of a quick run with 10 circles and a half million points. That took 9 seconds of real e, 6.6 seconds of CPU e, and converged to 4 decimal places. By the way, another method that occurs to me is to actually render the circles in some pixelated medium and then count colored pixels. ------------------------------------------ Circle centers C = 0.1942 0.1138 0.0846 0.9897 0.9635 0.5098 0.4557 0.0639 0.6524 0.0272 0.0005 0.0413 0.3786 0.4947 0.0858 0.8082 0.5010 0.4129 0.3872 0.9048 Circle radii ans = Columns 1 through 8 0.9391 0.4621 0.9122 0.2243 0.6262 0.2088 0.4072 0.7326 Columns 9 through 10 0.3542 0.2420 N A(est) clocke CPUe 20000 4.791736 0.3326 0.2700 40000 4.768794 0.6383 0.5400 60000 4.775305 0.9956 0.8100 80000 4.774684 1.3191 1.0800 100000 4.772452 1.6082 1.3500 120000 4.772566 2.0246 1.6300 140000 4.770211 2.4158 1.9000 160000 4.768678 2.9425 2.1700 180000 4.769965 3.2977 2.4400 200000 4.769445 3.5792 2.7000 220000 4.769358 3.9876 2.9700 240000 4.767554 4.4163 3.2400 260000 4.769772 4.7735 3.4900 280000 4.769901 5.1083 3.7600 300000 4.769807 5.4628 4.0100 320000 4.768561 5.8629 4.2600 340000 4.766387 6.1869 4.5300 360000 4.768725 6.5527 4.7900 380000 4.769789 7.0027 5.0500 400000 4.768623 7.3632 5.3100 420000 4.767096 7.7723 5.5800 440000 4.767117 8.0896 5.8400 460000 4.765990 8.4463 6.1000 480000 4.764828 8.8470 6.3600 500000 4.765148 9.3062 6.6200 === Subject: Re: Is it mass; or is it weight Cut< > A slug is a little used 20th century invention which exists only in > one specialized subsystem of mechanical units--and a unit which > doesn't even have an official definition. Cut > Gentlemen of the jury, Chicolini here may look like an idiot, > and sound like an idiot, but don't let that fool you: He > really is an idiot. > Groucho Marx Jeepers Gene, I think all of your knowledge about _so many_ different systems of weights and measures must have driven you buggy as Chicolini: According to Webster's online, definition #7 for Slug is : the gravitational unit of mass in the foot-pound-second system to which a pound force can impart an acceleration of one foot per second per second and which is equal to the mass of an object weighing 32 pounds. === Subject: Can this be proved? Has anyone ever seen a proof of the following; Let X(n) be a function defined for all positive integers n, X(n) = 0, n even X(n) = 0, if the prime factors of n include repeated primes X(n) = 1, if the prime factors of n do not include repeated primes example; x(15) = x(3*5) = 1, x(45) = x(3*3*5) = 0 Prove that (1/n)*integral( X(n) ) converges with increasing n. Bob Adams === Subject: Re: 0! = 1 >Did someone know a simple demonstration of >0! = 1 ? >PS Sorry about my english Refer to the implication from http://www.wikipedia.org/wiki/Factorial n!=n(n-1)! Let n=1; you know that 1!=1 n! becomes when n=1, 1*(1-1)!=1 Divide both sides by 1: (1-1)!=1/1=1 Since (1-1)=0, 0!=1 G C === Subject: Re: 0! = 1 > A N Niel scribbled the following: >> If I know my algebra, then an identity for an operation (let's call >> it #) is such an i, that for all x, x#i=x. But there is another >> concept too: such a j, that for all x, x#j=j. For multiplication, this >> is 0. >> For addition, there doesn't seem to be one. What is this concept >> called? > It's called a zero. As to what the _concept_ is called, my answer is absorption. And in a general context, instead of zero, one might use the term absorptive element. > So intuitively, if #' is a repetitive #, which means that x#'i is x#x > performed (i-1) es in succession, then the zero of #' is the > identity of #? For example multiplication is repetitive addition. 0 is > the additive identity and the multiplicative zero. If we switch the > definition of a zero around so that it means that for all x, j#x=j, then > we get that exponentation is repetitive multiplication, and that 1 is > the multiplicative identity and the exponentative zero. > Is this making any sense? Yes. But exponentiation isn't commutative. Perhaps one might say that 1 is left-absorptive, and that exponentiation has no right-absorptive element. David === Subject: Re: Can this be proved? > Has anyone ever seen a proof of the following; > Let X(n) be a function defined for all positive integers n, > X(n) = 0, n even > X(n) = 0, if the prime factors of n include repeated primes > X(n) = 1, if the prime factors of n do not include repeated primes > example; x(15) = x(3*5) = 1, x(45) = x(3*3*5) = 0 > Prove that (1/n)*integral( X(n) ) converges with increasing n. ? integral( X(n) ) ?? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 es) === Subject: Homeomorphism and boundaries I've read that homeomorphisms are open and closed bijections. Does that mean that if f:X->Y is a homeomorphism, then f(boundary(A)) = boundary(f(A)) for all sets A subset X? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ I will never display my bum in public again. - Homer Simpson === Subject: Re: All masses have inertia > The way I see it guys, is that all masses have inertia! That is it > requires > a net impulse - the product of a net force [f] and its duration [t] - > to > change a body's present velocity [vi], to some other velocity [vt]; > during > which period of e, the body is displaced a distance [s]: The ratio > of > this impulse [ft], to the e rate of displacement [s/t = (vt-vi)] > that it > causes is a constant: That is ft/(s/t) = ft/(vt-vi); which can be > written > more concisely as f/tî/s = ftî/ (vt-vi)! Isn't _that_ inertia? > Why are you polluting this newsgroup with your topic. This is the third > thread you started with almost the same topic, isn't it? > -- > Andreas > For replying, remove the fruit from my address. You've got to learn; like it or not: Fruit for your address; very appropriate(:^) === Subject: Re: a baseball odds calculation problem >here's the problem, specifically: there are three divisions in the american league. every season, each one > of >these divisions will be clinched by one team on a certain date. this is >mathematical certainty. past baseball history tells us that dates before september are extremely >rare for teams clinching. in fact, the later days of september, roughly > the >15th through the 28th, are where the likelihood of teams clinching is the >greatest. You need to specify the probability distribution of the date of > clinching for each division. For example, assuming the the three > divisions are independent and have the same distribution, completely > supported (a model of your approximation) by {15, 16,..., 28}, then the > probability is sum(k=15..28, p(k)^3), where p(k) is the probability of > clinching on the date k. So if you are saying that each division is > equally likely to be clinched on each day and is independ of the others, > the probability is 14*(1/14)^3 = 1/196 = .0051 (0.51%), approximately. > Or if you assume identical triangular distributions, p(k) = > min(k-14,29-k)/56, the probability (still assuming independence) is > 2/56^3 sum(k=1..7, k^3) = 1/112 = .0089 (0.89%), approximately. > Another model would be a discretized normal over the whole season. > Perhaps you want to make an empirical esate of the distribution; you > may then do the calculation yourself. > baseball seasons. in those seasons, there have been 126 different clinches > (4 clinches per year from 1975-1993 excluding 1981 strike year and then 6 > and i was able to find 96 of those dates (they're not very easy to find). i > did an esate based on this here's the distribution of clinches i found: [...data omitted...] It seems you are not satisfied with the quality of the answers you got so far and want a more accurate answer. But I suspect you collected the wrong data. It would be far more informative to know how many days from the end of the season the clinch occurred. That is because the last (scheduled) day of a season is a Sunday, whose date obviously varies from year to year. === Subject: Re: Mass is a quantity of matter >But the idea is that THE ESTABLISHMENT (shiver, shiver) >controls the TROOTH (tm) and suppresses publication of HERESY, especially >because they're making lotsa money is certainly neither novel nor humorous. Well that's the standard crackpot conspiracy theory, but somehow it's even more ludicrous when adapted to the scientific community. >You'd think that he could at least GIVE them away to social friends who >(Nudge,nudge, wink, wink) The trouble is, most people have learned elementary physics in school so it narrows down his target audience a bit. === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed? > I'm trying to write ATAN2 function for a small basic language that has > IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are > availible in the language. > I've tried a few methods I've found but the results are way off due to > low precision, rounding, etc. > Are there any repositorys of old fortran routines or algorithms that I > could use to get a good accuracy single precision routine. Speed or > space aren't as important as reasonably good accuracy. > Jon Two suggestions: (i) See The Standard C Library by P J Plauger, published by Prentice Hall. You will find two short pages in C. Do you need to code in Fortran? (ii) Google for CORDIC; with it you can calculate arctan and much else. -- G.C. === Subject: 2nd Principle of Finite Induction I'm hoping to get help in how to approach the following problem on number theory: Use the 2nd prinjicple of finite induction to establish that a^n - 1 = (a - 1)(a^(n - 1) + a^(n - 2) + a^(n - 3) + ... + a + 1) for all n geq 1 the trouble i'm having is there's a hint:a^(n + 1) - 1 = (a + 1)(a^n - 1) - a(a^(n - 1) - 1) and i am clueless as to what to do with it. === Subject: Re: Properties of f(x) given the graph of f'(x) Visiting Assistant Professor at the University of Montana. > <> a .8ecrit s le message de >> And also where f is concave up, concave down, and where its points of >> inflection are. >> [...] >> Also, f is concave up where f' is ->increasing<-, and f is concave >> down where f' is ->decreasing<-. >Is a function with f'' > 0 called concav up in English? Is convex a >synonym for that? >I'm just wondering, because in German, such functions are called >konvex (f''>0) and konkav (f''<0). Yes, it is; in fact, I learned it convex and 'concave', but most books now seem to use concave up and concave down instead, as being, I assume (a) easier to remember and (b) more evocative. Of course, you still have students who don't remember if concave up means opens up or the round part points up, so I'm not sure it accomplished that much. (Except that f''>0 means f' goes up means concave up may be easier). === Subject: Re: All masses have inertia > You've got to learn; like it or not: What do I have to learn? I'm a happy SI user who never gets confused distinguishing between mass and force, so I don't have a problem with that topic. No one asked you to start three threads. It seems that you're the only one here having problems. > Fruit for your address; very appropriate(:^) Yeah, I like apples! -- Andreas For replying, remove the fruit from my address. === Subject: Re: Homeomorphism and boundaries Visiting Assistant Professor at the University of Montana. >I've read that homeomorphisms are open and closed bijections. Does that >mean that if f:X->Y is a homeomorphism, then f(boundary(A)) = >boundary(f(A)) for all sets A subset X? I was going to say depends on how you define boundary, but then I remembered we already had ->that<- conversation. Looking back through boundary(A) = cl(A) - int(A). A homeomorphism is a continuous bijection that sends open sets to open sets (equivalently, a bijection which sends open to open and closed to closed). Thus, if f:X->Y is a homeomorphism, it follows that for all A, f(int(A)) = f( union{B : B is open, B contained in A}) = union{ f(B): B is open, B contained in A} (since f is bijective) = union{ f(B): f(B) is open, f(B) contained in f(A)} = int(f(A)) and likewise f(clos(A)) = f( intersection{B: B closed, B contains A}) = intersection {f(B): B closed, B contains A} = intersection {f(B): f(B) closed, f(B) contains f(A)} = clos(f(A)). So f(boundary(A) ) = f(cl(A) - int(A)) = f(cl(A)) - f(int(A)) = cl(f(A)) - int(f(A)) = boundary(f(A)). === Subject: Re: 2nd Principle of Finite Induction >I'm hoping to get help in how to approach the following problem on number >theory: >Use the 2nd prinjicple of finite induction to establish that >a^n - 1 = (a - 1)(a^(n - 1) + a^(n - 2) + a^(n - 3) + ... + a + 1) for all >n geq 1 >the trouble i'm having is there's a hint:a^(n + 1) - 1 = (a + 1)(a^n - 1) - >a(a^(n - 1) - 1) >and i am clueless as to what to do with it. The 2nd Principle of Induction (applied to the natural numbers) is the statement that if a property P of natural numbers, if [for all n, (for all kp(n) ] then for all n, P(n). That is, you assume the property is true for all natural numbers less than n, and from that you prove it is true for n. If you can do that for each n, then you have proven it for all n. In this case, your induction hypothesis is that for all k<(n+1), you have a^k - 1 = (a-1) (a^{k-1} + a^{k-2} + ... + a + 1). and you want to prove that a^(n+1) - 1 = (a-1)(a^n + a^{n-1} + ... + a + 1). So, use the hint and the induction hypothesis. Caveat: your hint presupposes that n>1; so you would need to do the case n=1 separately; this is called a special case of your argument. === Subject: Re: factoring to satisfiability but lets just restate this in a more interesting way that immediately leads to an interesting research direction, along the lines of exploring structural properties of SAT factoring instances Ive been advocating: what is the smallest x-SAT formula that can be used to factor an arbitrary sized number? by electrical engineering principles a half adder and full adder can be implemented in 5-SAT, so as I recall (working from memory on my old SAT generator) 5-SAT is sufficient. but is it necessary? RE has a reason that if 3-SAT is sufficient then we have some reason to believe that factoring might be in P. if it is not, then we have a demonstration of the power of nonuniform versus uniform computation. if factoring can be done with 3-SAT but is not in P, then factoring is in nonuniform P, but not in uniform P. actually, now that I think about it, this is interesting because for the same reason 5-SAT has C(n,5) term possibilities. so we may already have some kind of implication here for nonuniform vs uniform power of factoring. we can find the C(n,5) clauses that create the factoring problem, but not solve them. I guess factoring is in the C(n,5) clause class and Ive refuted my own claim that the clauses go up exponentially. hmmmmmm note: uniform= computations done by TMs uniform= computations done by arbitrary circuit families > Why should all the clauses in this formula have at most three literals? > You can't necessarily represent a function in N variables using a 3-CNF > formula. The number of functions that can be represented that way is very > small by a counting argument: there are 2^(2^N) functions in N variables, > but only 2^(O(N^3)) possible 3-CNF formulas. What are the odds that > factoring is one of those lucky functions for interesting values of N? === Subject: Re: Homeomorphism and boundaries <> scribbled the following: >>I've read that homeomorphisms are open and closed bijections. Does that >>mean that if f:X->Y is a homeomorphism, then f(boundary(A)) = >>boundary(f(A)) for all sets A subset X? > I was going to say depends on how you define boundary, but then I > remembered we already had ->that<- conversation. Looking back through > boundary(A) = cl(A) - int(A). (snip) > So > f(boundary(A) ) = f(cl(A) - int(A)) = f(cl(A)) - f(int(A)) > = cl(f(A)) - int(f(A)) > = boundary(f(A)). -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ Roses are red, violets are blue, I'm a schitzophrenic and so am I. - Bob Wiley === Subject: Re: factoring to satisfiability 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > what is the smallest x-SAT > formula that can be used to factor an arbitrary sized number? > by electrical engineering principles a half adder > and full adder can be implemented in 5-SAT, so > as I recall (working from memory on my old SAT generator) 5-SAT is > sufficient. but is it necessary? 5-SAT can be reduced to 3-SAT with only a constant factor blowup in number of clauses. More specifically, any 5-clause (t1 + t2 + t3 + t4 + t5) can be replaced by three 3-clauses (t1 + t2 + x)(-x + t3 + y)(-y + t4 + t5) where x and y are new variables. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Chess/Go/etc: Continuous Game-Boards? Nick Wedd scribbled the following on sci.math: > For Go, I assume: > The difference between Japanese (territory) rules and Chinese (area) > rules would be very important in this game, unlike in normal Go where it > rarely matters. For one thing, under Chinese rules, you get captured > pieces back and can use them again. What's the precise difference between these rules? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ === Subject: Re: factoring to satisfiability hi all. conversion of factoring to SAT does NOT (*always* or *usually*) give easy instances as was stated in another msg on this thread. it MAY or MAY NOT give easy instances. if you want to factor a prime number then the resulting SAT clause must be easy based on the new agrawal proof. and it would be extremely interesting to try to find a SAT algorithm that runs in P e on prime number factoring instances... I think it is virtually guaranteed that the basic DPLL etc algorithms do NOT.. and so why not?? and in fact if you look at a SAT algorithm that does give P e performance for factoring prime numbers.. it might give more insight into the ASK algorithm, and possibly new ways to opize it. this all falls into a basic strategy & general research program of analyzing computational problems via their boolean circuits, which imho is more fundamental and will ulately lead to more/most insight into the inherent problem complexities. I think this will soon be shown for some number theory problem(s). soon gauged in research e which is glacial prior to a tipping point. 5-10 years would be soon > As primality testing has been shown to be in P would not a reduction of > SAT to factoring be a proof that NSAT in in P? What does this say about > P=NP? === Subject: Re: Help is needed with a Bayes Rule exercise Your explanations and links were of great value for me. Dotan. === Subject: Re: What to tell students in a 10-15 minute talk > * If T = infinity, the function |sum_{k=1}^{n} exp(ak*t*i)| attains all > values between 0 and n, so this case can be ignored. (Exercise for the > reader.) Oops. Not correct. I was thinking of the case where each pair (a_j,a_k) has an irrational ratio. The T = infinity case should never pertain, but Justifying this is more complicated than I thought. -- | Ferry | Center for Simulation | +------------------------------------+ of Advanced Rockets | | http://www.uiuc.edu/ph/www/jferry/ +------------------------+ | jferry@[delete_this]uiuc.edu | University of Illinois | === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed? > I'm trying to write ATAN2 function for a small basic language that has > IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are > availible in the language. > I've tried a few methods I've found but the results are way off due to > low precision, rounding, etc. A Basic language with Sin, Cos, and Tan, should also have at least an ArcTan ? I mention that because a division result sent to an ArcTan and combined with some simple quadrant rules should arrive at an ArcTan2. === Subject: Re: factoring to satisfiability another obvious point. a proof that C(n,3) clauses are not possible to specify factoring instances (or arbitrary other problems for that matter) would be quite interesting. it probably would not be too difficult to construct. just look for some bit relationships in factoring that are more complex than the C(n,3) relationships; these could easily be identified empirically and then possibly generalized into a proof without too much trouble/hassle/effort. also, we are referring to a natural complexity class that has been studied in the literature somewhat. bounded width circuits I believe they are called, or bounded input circuits. note that bounded width measures are apparently more fine-grained than P vs NP. the various NP complete problems may have different bounded-width complexities. not too much is known in this area so far. another theory-edge. > Why should all the clauses in this formula have at most three literals? > You can't necessarily represent a function in N variables using a 3-CNF > formula. The number of functions that can be represented that way is very > small by a counting argument: there are 2^(2^N) functions in N variables, > but only 2^(O(N^3)) possible 3-CNF formulas. What are the odds that > factoring is one of those lucky functions for interesting values of N? === Subject: Re: Chess/Go/etc: Continuous Game-Boards? In message , Joona I Palaste >Nick Wedd scribbled the following >on sci.math: >> For Go, I assume: >> The difference between Japanese (territory) rules and Chinese (area) >> rules would be very important in this game, unlike in normal Go where it >> rarely matters. For one thing, under Chinese rules, you get captured >> pieces back and can use them again. >What's the precise difference between these rules? The basic difference is that with Japanese rules, you count points for territory surrounded and lose points for pieces you have lost, while with Chinese rules, you count points for territory surrounded _or_ occupied, and ignore pieces you have lost. Usually this makes no difference, there is a simple proof of this depending on the fact that each player has made the same number of moves. Other differences (which have little practical effect): With Japanese rules you don't count points inside a seki; with Chinese rules you do. With Japanese rules bent four in the corner is rules to be dead; with Chinese rules, you play it out, and it almost always turns out to be dead anyway. Japanese put handicap stones on the marked hoshi points; Chinese allow the recipient of the handicap to choose where to put them. There are more details at http://www.britgo.org/rules/compare.html Nick -- Nick Wedd nick@maproom.co.uk === Subject: Re: Chess/Go/etc: Continuous Game-Boards? Visiting Assistant Professor at the University of Montana. >Nick Wedd scribbled the following >on sci.math: >> For Go, I assume: >> The difference between Japanese (territory) rules and Chinese (area) >> rules would be very important in this game, unlike in normal Go where it >> rarely matters. For one thing, under Chinese rules, you get captured >> pieces back and can use them again. >What's the precise difference between these rules? Don't know if they are all of the differences, but: In Chinese rules, you get your captured pieces back and you return to your opponent his captured pieces. At the end of the game, you total the number of positions ->occupied<- by your pieces, and add them to the number of empty positions ->controlled<- by your pieces. You compare your total to your opponents (plus or minus the handicap), and highest total wins. In Japanese rules, you keep the pieces you capture from your opponent, he/she keeps the pieces he captures from you. At the end of the game, you use your prisoners to occupy territory he controls, and he uses the prisoners to occupy territory you control. At the end, you count the number of empty positions controlled by your pieces, he counts the number of empty positions he controls, and after adding the appropriate handicap the highest total wins. If you have more prisoners than your opponent has empty but controlled territory, he has a negative score. === Subject: Re: All masses have inertia > You've got to learn; like it or not: Plonk Him.. I'm even wasting bandwith with this advice. Bob PEase === Subject: Re: Properties of f(x) given the graph of f'(x) <> a .8ecrit s le message de >Is a function with f'' > 0 called concav up in English? Is convex a >synonym for that? >I'm just wondering, because in German, such functions are called >konvex (f''>0) and konkav (f''<0). > Yes, it is; in fact, I learned it convex and 'concave', but most > books now seem to use concave up and concave down instead, as > being, I assume (a) easier to remember and (b) more evocative. Of > course, you still have students who don't remember if concave up > means opens up or the round part points up, so I'm not sure it > accomplished that much. (Except that f''>0 means f' goes up means > concave up may be easier). -- Andreas For replying, remove the fruit from my address. === Subject: Re: Chess/Go/etc: Continuous Game-Boards? Visiting Assistant Professor at the University of Montana. >In message , Joona I Palaste >>Nick Wedd scribbled the following >>on sci.math: > For Go, I assume: > The difference between Japanese (territory) rules and Chinese (area) > rules would be very important in this game, unlike in normal Go where it > rarely matters. For one thing, under Chinese rules, you get captured > pieces back and can use them again. >>What's the precise difference between these rules? >The basic difference is that with Japanese rules, you count points for >territory surrounded and lose points for pieces you have lost, while >with Chinese rules, you count points for territory surrounded _or_ >occupied, and ignore pieces you have lost. Usually this makes no >difference, there is a simple proof of this depending on the fact that >each player has made the same number of moves. Which is certainly not the case under Japanese rules. At least the way I learned them: a player is allowed to pass, and then the other player may either pass (terminating the game), or play. Is this different under Chinese rules? === Subject: Re: Numeric one-way hash function > I need to find an algorithm that can produce a unique non-predictable 12 > digit (0-9) number for any given 12 digit number. This is to be used to > create a unique barcode on a ticket that cannot be predicted. It is not > required that the original seed number be computed from the resulting > barcode, so some form of one-way hashing function would be acceptable. > Any help in this problem would be appreciated. > Mark. 1. Create a 10 element array with digits 0-9 in linear order (0 in 0, 1 in 1, etc.) 2. For each digit of the 12-digit number: 2a. Shuffle the 10-element array. 2b. Using the original decimal digit as an offset into the array, look up the replacement digit value. === Subject: a puzzle related to artinian group cards are dealt to them. There is no asumption >on the number of cards a player receive. In each round, all players with 2 or more cards pass >one card to the left and one card to the right. Prove that eventually, all players but one have >exactly one card. I do not have the solution, so any hint would be highly appreciated. Please >restrain from posting messages saying this is obvious, and For fixed n this is, of course, a cellular automaton. You can allow any positive integer states for each of the n cells. The transition rule is such that if a cell has state x >=2 then the state never exceeds x and if it has state x < 2 then it either stays the same or increase by 1 or 2. If N is the maximum value of a cell state, then the maximum value of the states in the next generation can be at most N, unless N = 2 and then a maximum of 3 is possible in the next generation. It follows that there are there are only a finite number of possible global states, so eventually it is periodic. One question might be to figure out which of Wolfram's four classes this cellular automaton lies in. These remarks don't answer your question, but may be of some interest. --Edwin === Subject: Re: Chess/Go/etc: Continuous Game-Boards? In message , >>In message , Joona I Palaste >Nick Wedd scribbled the following >on sci.math: >> For Go, I assume: >> The difference between Japanese (territory) rules and Chinese (area) >> rules would be very important in this game, unlike in normal Go where it >> rarely matters. For one thing, under Chinese rules, you get captured >> pieces back and can use them again. >What's the precise difference between these rules? >>The basic difference is that with Japanese rules, you count points for >>territory surrounded and lose points for pieces you have lost, while >>with Chinese rules, you count points for territory surrounded _or_ >>occupied, and ignore pieces you have lost. Usually this makes no >>difference, there is a simple proof of this depending on the fact that >>each player has made the same number of moves. >Which is certainly not the case under Japanese rules. At least the way >I learned them: a player is allowed to pass, and then the other player >may either pass (terminating the game), or play. >Is this different under Chinese rules? It is the same under Chinese rules. A player may always pass - two consecutive passes end the game. So, unless there are passes before the final two passes, each player has played the same number of stones; except that Black may have played one more point than White. So you are right - it can make one point of difference. The North American rules are a compromise between Chinese and Japanese rules, tending to be closer to the Chinese rules. They deal with this one-point problem by requiring White to make the last move. Nick -- Nick Wedd nick@maproom.co.uk === Subject: Irrationality of the sqrt(2) by unique factorization I studying a different way of proving that the sqrt(2) is irrational, or n th roots in general. It proceeds as follows: Assume the sqrt(2) is rational. Thus, m/n = sqrt(2). Which then implies 2^1 * n^2 = 2^0 * m^2 Now, by unique factorization, a perfect square will have all even exponents in its prime factorization. Since, the LHS of the above equation has an odd exponent, namely 2^1, can I then conclude that the sqrt(2) is irrational? It seems to me that some step is missing? Shouldn't there be something about the RHS being even and the LHS being odd ? How does one arrive at the conclusion that if a square isn't perfect, then it is irrational? I understand intuitively that if a sqrt() is not perfect, then it is irrational, but I don't see how one can logically conclude that from the above equation. Lurch Lurch === Subject: is there a name for this operation? Let the members of set Ai be ai1, ai2, ... aiNi. input, returns as an answer { [a11, a21, a31, ...], [a12, a21, a31, ...], [a13, a21, a31, ...], ..., [a1N1, a21, a31, ...], [a11, a22, a31, ...], [a11, a23, a31, ...], ..., [a11, a2N2, a31, ...], ..., [a11, a21, ..., aM1], [a11, a21, ..., aM2], ..., [a11, a21, ..., aMNM] } In other words, the answer is a subset of the Cartesian product, containing only those sets that differ from [a11, a21, a31, ...] in at most one place. I'm providing this operation as one of a number of possibilities in a test-case generation program. So far I'm just calling it perturbation which is unsatisfying. Does this operator have an accepted name? --JMike === Subject: Exercise in Projective Linguistics I am teaching a class this term in which students are wrestling with projective planes. I gave them an exercise to identify (i.e., coordinatize) the plane over the field of 4 elements. It can be described as an assembly of 21 points, some subsets of 5 of which are called lines. When I posed the problem, I just assigned names A, B, C, ... to the points and gave the lines as 5-tuples. Later, it occurred to me I might have been able to find a different assignment of the points into the alphabet in such a way that each line was ( a permutation of ) the letters in a 5-letter word. So here is the challenge: how well can you do ? Can you embed these letters back into the alphabet so that all, or most, of these 5-tuples become English words? (Interpret word as you wish.) (I would be willing to take a solution which works only for an affine plane inside here, if necessary.) By the way, I hereby claim my title as inventor of a new field. Google finds no match for projective linguistics except for random-word generators. The 21 lines are: {A, B, C, D, Q} {A, F, K, P, R} {A, G, L, N, T} {A, H, J, O, S} {A, I, E, M, U} {B, E, L, O, R} {B, F, J, N, U} {B, H, K, M, T} {B, I, P, G, S} {C, E, J, P, T} {C, F, L, M, S} {C, G, K, O, U} {C, I, R, H, N} {D, E, K, N, S} {D, F, I, O, T} {D, G, J, M, R} {D, H, L, P, U} {E, F, G, H, Q} {I, J, K, L, Q} {M, N, O, P, Q} {Q, R, S, T, U} === Subject: Re: Express As Single Fraction > How do I do this? Express the following as a single fraction: 4/3ab - 5/6bc (1) find a common denominator. The least common denominator is a > good one to use. > What, for example in the first one, is the lowest common denominator? Is it > (3ab)(6bc)? If it is, do I then multiply that out? If not, what is it? 6abc = (3ab)(2c) = (6bc)(a), so 6abc is a common multiple of 3ab and 6bc. Since (2c) and (a) have no common factor greater than 1, nothing can be left out of 6abc and still have a common multiple of 3ab and 6bc, so 6abc is the least common multiple. === Subject: Re: Irrationality of the sqrt(2) by unique factorization >I studying a different way of proving that the sqrt(2) is irrational, or n >th roots in general. It proceeds as follows: >Assume the sqrt(2) is rational. Thus, m/n = sqrt(2). Which then implies >2^1 * n^2 = 2^0 * m^2 >Now, by unique factorization, a perfect square will have all even exponents >in its prime factorization. Since, the LHS of the above equation has an odd >exponent, namely 2^1, can I then conclude that the sqrt(2) is irrational? Yes. >It seems to me that some step is missing? Shouldn't there be something >about the RHS being even and the LHS being odd ? Consider the factors of 2 that appear in either n or m or both; there is an even number of factors of 2 in n^2, and an even number of factors of 2 in m^2. So the RHS has, in total, an even number of factors of 2 (0 + the number of factors in m^2), while the LHS has an odd number of factors of 2 (1 + the number of factors in 2). This is impossible. This would be the long-winded explanation. What you have noticed is simply that all the primes on the right hand side appear raised to an even exponent, while all but ONE of the primes on the left hand side appear raised to an even exponent, the exception being raised to an odd exponent. That's impossible by unique factorization. > How does one arrive at the >conclusion that if a square isn't perfect, then it is irrational? No such conclusion is reached. Or do you mean, that an integer which is not a perfect square has an irrational square root? > I >understand intuitively that if a sqrt() is not perfect, then it is >irrational, but I don't see how one can logically conclude that from the >above equation. The above equation only proves it for 2. A similar argument yields the result for sqrt(p) for any prime p; then it is not hard to prove it by a similar argument for any product of the form p_1*...*p_r where p_1,...,p_r are distinct primes. Then simply note that positive integer greater than 1 can be written uniquely as n=k*s, where s is a perfect square, and k is a product of distinct primes. So then sqrt(n) = sqrt(k)*sqrt(s), sqrt(s) is an integer, and sqrt(k) is irrational. === Subject: absolute moments of the normal distribution Hi to everyone!! Some of you know where I can find the formula of the k-th absolute moment for the normal distribution? My own calculations are different from the Mathematica output and I am not sure about what to do... Massimiliano Bultrini. === Subject: Re: hw help -- continuity === > Subject: Re: hw help -- continuity >> 1)prove that if f,g continuous, then so are max(f,g) and min(f,g) >> Any ideas how to approach this? >Proving a function continuous at any point >proves it continuous at every point. > f(x) = 0 when x <= 0 > = 1 when 0 < x > is continuous at 1. Thus it's continuous at 0? But 1 is not any point, it is a specific point. To restate less ambiguously: If you prove f continuous at an arbitrary point, then it is continuous at every point And to prove it at an arbitrary point, you only need show that for an arbitrary x, g(x) = lim_{y -> x} g(y) === Subject: Re: Antidiagonal, Infinity >What I propose is that given any >rational that the value greater than it and less than any other >greater is irrational, There is no such number, as several different people have shown you. In non-standard analysis, there might be, however. > See Alain Robert's book about NSA. Rather than being > irrational, it would be non-standard, though. > I have yet to see any standard or non-standard model of the reals in > which there is a smallest positive number. > Who says that that is meant? > In nsa, there can be a nonstandard number that can be to be > 'greater than (a given standard rational) and less than any other (standard) > greater' > That it is not unique, who cares? Ross wants to use non-standard numbers to confirm his hypothesis that there is a next real after any real, and that the irrationals and rationals alternate on the real line or on some non-standard real line. So he cares. On the other hand, his hypothesis is way out in left field, where he has been stuck for months, if not years. === Subject: Re: quantum echo > --I could have baked both cakes yesterday. > --I could have baked both cakes today. > --I could have baked this one yesterday > and that one today. > --I could have baked that one yesterday and > this one today. Your example doesn't work! > How am I going to distinguish 'one cake' from 'the other cake'! Since they now are identical-with-somethings, if they shared > all the same properties they would be one cake rather than > two. THerefore, since they are two cakes, they don't share > all their properties, and I can distinguish them by means > of any property that one has and the other doesn't. > you mean argument by GPS or by just determining their global positioning > with sattellites we have a difference even if there are no other clear > attributes of different classes besides location? That is a call which no logician qua logician has any particular > comnce > to make. Are you saying it is possible for something to exist at location A and > location B at the same e? I thought it was inference by the rule of > non-contradiction. > I have no idea whether this is possible or not. > My suggestion that quanta may satisfy '~(x=x)' is meant to translate > into logical terms the claim by some physicists that quanta > 'lack individuality' (Heller suggests that quanta 'lack haecceities'). > The scenario with cakes-to-be suggests that the behaviour of other > entities without individuality resembles that of quanta in the > relevant respect. Heller's description of that behaviour follows > my sig. Metaphysical Background, Thomas McTighe asserted that the quiddity of a thing is nothing other than unity itself. Hence, by virtue of its positive content, the sun differs not at all from the moon or any other particular thing. The diversity which is exhibited by the natural world is merely the product of accidental differences; no object possesses any specific form which interposes itself between a particular existing thing and the source of their being e.g. the Absolute.15 All individual entities are nothing more than differing contractions of the whole devoid of any being of their own. ...because the restricted quiddity of a thing is the thing itself. http://www.crvp.org/book/Series01/I-10/chapter_ii.htm Aristotle and Aquinas and Scotus and Bonaventura all believed that human minds can conceive and express the intelligibilities or quiddities of things and their properties, intelligibilities that are not simply mind-dependent. We can capture in thought and language the actual natures of things, spelling out their genera and specific differences. Definition brackets or delimits for us as knowers just what it is we attempt to understand and nothing else. The mind-independent thing-substance or the characteristics that we are attempting to define measure the epistemic correctness of a definition. Such real (as opposed to nominal) definition relies on the intelligible and perceptible characteristics thing-substances exhibit to perception and thought for understanding what they are and for picking out individuals of a type. In this way the epistemological realism of the definition corresponds to an ontological realism of actual formal features in mind-independent entities. http://www.sunysb.edu/philosophy/faculty/lmiller/Delinonaliud.htm After all, the three tenets that largely define Nicholas's 'metaphysic of contraction' seem altogether remote from Anselm's Scholasticism. For Anselm has no use for the triad of notions (1) that there is an infinite disproportion between the Creator and His creatures, (2) that, therefore, finite minds can never positively know what God is, given the alleged ground (3) that He is the Coincidence of opposites, i. e., is undifferentiated 'Being' itself, which, with respect to its Quiddity, can never be conceived by anyone except itself. http://www.cla.umn.edu/jhopkins/CusaAnselm.pdf > --John JJ 1 I. The Problem, and the Problem with the Problem, > 2 of Identical Articles and Quantum Statistics > 3 > 4 Suppose we have a box with two qualitatively > 6 bouncing around inside. We think of the box > 7 as having a left (*l*) and a right (*r*) side. > 9 at random without interacting, so that their > 10 motions are independent; in particular we > 12 that we may neglect collisions. What are the > 13 chances for finding one or both on one side > 14 or the other? > 15 > 16 Many find the following reasoning persuasive. > 19 in *r*, and 2 in *l* and 1 in *r*. These should > 20 be equally likely, so that each has a probability > 21 of 1/4, or a probability of 1/4 for two in *l*, > 22 1/4 for two in *r*, and 1/2 for one on each side. > 23 > 24 This stylized example is a simple mock-up for a > 25 kind of situation that can occur with quantum entities > 26 and properties. For many of these situations the > 27 probabilities are in fact found to be 1/3 for each > 28 of the three cases: two in *l*, two in *r* and one > 29 on each side. Many interpreters have found this > 30 fact utterly astonishing. > 31 > 32 But on the face of it, there is a very simple > 33 resolution of the puzzle: give up supposing > 34 that there are two qualitatively identical but > 36 there are two *quanta*, as I'll put it, to which > 37 the notion of being numerically distinct does not > 38 apply. . . (p. 114) === Subject: Re: Exponentative closure >A further question along those lines is what is is the set, E, of >numbers generated by repeated sums of products of rational powers of >rationals? It is a subset of the algebraics, but does it form a >field? > I believe so: if x_1 is a member of E, then so are its conjugates x_2, > x_3, ..., x_n. > Note that x_1 x_2 ... x_n is rational (being the constant term of a > monic polynomial over the rationals whose roots are x_1, x_2, ..., x_n). > And then 1/x_1 = x_2 ... x_n / (x_1 x_2 ... x_n) is in E. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 If we say that a set, P, is exponentially closed if for all p,q in P, p^q is P, then can the complex numbers be defined as the field that is exponentially closed? Or even more simply, the set that is exponentially closed and contains -2 and 0? If not, what are the properties of the smallest field, S, that is exponentially closed (cardinality, algebraic completeness, convergence of Cauchy-Sequences)? Note: that I am only considering fields with characteristic 0. === Subject: Re: Can this be proved? How do you propose to integrate X(n) since it is defined only over the integers? Do you mean integrate as a Stieltje's integral? You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana. [.snip.] >I understand the argument perfectly. I have given the problem some >thought and I have concluded that HYPOTHETICAL G is impossible. No >graph meets all three criteria; ie, G is 5-chroma, G is planar, H is >4-chroma. As has been noted, if by H you mean any graph obtained by removing a single vertex from G, then your statement is exactly equivalent to the 4 Color Theorem. Reading back through the thread, I ->think<- I'm beginning to understand what exactly it is that you are finding problematic. You are arguing that a G which requires 5 colors, and with the property that removing any vertex results in a planar graph that requires just 4 colors, and with a minimal number of vertices among all graphs that satisfy that condition, must be equal to K5. This is true, but note that we have dropped the key property of planar from the assumptions of G. Rather, we are expected to take a planar G which requires 5 colors, and with the property that removing any vertex results in a graph that requires just 4 colors, and with the minimal number of vertices among all graphs that satisfy that condition. Clearly, any graph G that satisfies the conditions of (a) Being planar; (b) Requiring 5 colors; (c) Removing any vertex results in a planar graph that requires only 4 colors; also satisfies just (b) and (c); so a graph which satisfies (a)-(c) and has a minimal number of vertices among all graphs satisfying this condition would necessarily have 5 or more vertices (since a graph that satisfies just (b) and (c) and has a minimal number of vertices among all graphs satisfying (b) and (c) has exactly 5 vertices). But the only minimal graph that satisfies (b) & (c), namely K5, does not satisfy (a). So the number of vertices of this hypothetical G will be strictly greater than 5. But after that, you seem to be arguing that in fact this hypothetical G would necessarily have no more than 5 vertices. And I think that the reason you are making this argument is that you point out that in order to reduce G to a 4-colorable graph, you must remove all vertices of some color, which means that the vertex you removed was the only vertex of the given color; which in turn means that each vertex is the only vertex of its color, which means G has 5 vertices, which means G is K5, which is a contradiction. Or something like that. Now, if that is not what you are arguing, then you may ignore this post and everything that follows. So, assuming I got the gist of your argument correct, the error is in the step that goes from the vertex you removed was the only vertex of the given color to each vertex is the only vertex of its color. The fact that G-{v} can be 4 colored but G cannot means that for each vertex v, there exists a coloring C(v), which ->depends on v<-, with the property that v is the only vertex of its color. However, if v and w are two distinct vertices, there is no guarantee that the coloring of G-{v} is compatible with the coloring of G-{w}; that is, there is a coloring C(v) which depends on v and in which v is the only vertex of its color, and there is a different coloring C(w) which depends on w in which w is the only vertex of its color, but there is no reason to assume that w is the only vertex of its color under the coloring C(v), and there is no reason to assume that v is the only vertex of its color under the coloring C(w). This was mentioned by Erick Wong in Again, consider the 5-cycle, that is the graph consisting of 5 vertices, {1,2,3,4,5}, with adjacencies 1-2-3-4-5-1 (so each n is adjacent to n+1 (mod 5) ). Removing any vertex results in a 2-chromatic graph; the graph itself, however, is not 2-chromatic, it requires 3 colors. For ->each<- vertex, there is a coloring of G in which that vertex is the only vertex of its color. However, there is no 3 coloring of the graph in which ->each<- vertex is the only vertex of its color, and there is no reason to assume that this is the case from the fact that there is a coloring for each vertex. That is, we are encountering a typical fallacious exchange of quantifiers. We have: (1) For every vertex v, there exists a coloring C such that v is the only vertex of its color; and you seem to be interpreting this as being equivalent to (2) There exists a coloring C such that for every vertex v, v is the only vertex of its color. The two statements are not equivalent; (2) implies (1), but (1) does not imply (2). === Subject: Re: Exponentative closure >Can the reals be defined using repeated exponentative closure? >By the exponentative closure F, I define F/x as the set of all the >zeroes of all the polynomial functions with coeffeicients AND >exponents in F. For example, the the algebraics are the exponentative >closure of the integers. Thus, it can be written A=Z/x. Does >C=A/x? If not what does A/x equal? Can C be generated by >repeatedly exponentatively closing the integers a finite number of >es? If so, how many? A countable number of es? An uncountable >number of es? The set of all that is obtained is countable. This is because each polynomial has only a finite number of coefficients and exponents, and returns only a finite number of results. This is assuming that a clear definition exists of x^y if x is negative and takes the value |x|^y; more can be done. Doing this more than omega (the order type of the integers) es yields nothing new, because there are only a finite number of arguments for each extension. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. === Subject: Re: Find Integral >Dont know how to start with calculation of integral >Triple Integral from 0 to Inf . >Integrant: 1/(1+x^a+y^b+z^c)dxdydz >Calculate the integral and find a,b,c for which integral convergent. >(Should be solved using a Gamma functions I think) You might start by showing that int_0^infinity (A+x^a)^(-q) dx = A^(1/a-q) Gamma(q - 1/a) Gamma(1/a)/(b Gamma(q)) under suitable assumptions on A, a, q. This is related to the Beta function integral. === Subject: Re: Chess/Go/etc: Continuous Game-Boards? >I was just wondering today if there has been anything interesting >as Chess, Go, etc, where the game-boards are without any subdivision? On a slightly related basis, I've been thinking about something similar in the past, but I was rather day-dreaming about quantum-like delocalized pieces, or continuous distribuitions of pieces. Needless to say I had no actual idea of one such game... Rather than chess something like draughts would be more suitable IMHO for such an anti-discretization, but in any case it would be hard to define what a move could be, or how to eat an opponent's piece (ehm, a component of the opponent's distribuition of pieces). Funny to think of, having such a game to be played on a computer, at some point a (much finer) discretization should be introduced anyway... Just my 2 (Euro)cents, Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === Subject: Re: polysigned numbers Note: I'm posting through Google since my regular news-feed seems to be screwed up. > An interesting point about reduced form is that it so far has not been > absolutely necessary except for the dimensional analysis. That is to > say that a number like: > ( - 1 + 2 * 3 # 4 ) > can be worked with even though it is not in its reduced form. So far > none of the math, including product and graphical analysis fails to > work with non-reduced numbers. The reduced form of the above number > is: > ( + 1 * 2 # 3 ). > Is this is the reduction you are speaking of? This is one of two possible reductions. This one corresponds to the tetrahedral interpretation, RxRxR. There is a second possible reduction: ( -1 +2 *3 #4) = (*2 #2) which directly corresponds with ( i -2 -3i 4) = (-2i 2) so is isomorphic (again) with C. > The neat thing is that > and so is coherent. I believe that the same will happen with the > tetrahedral. The origin is at the center of a tetrahedron. the poles > go out from the origin through the points that form the tetrahedron. > This is the symmetrical mapping of four-signed math in cartesian > space. Any point in RxRxR can be uniquely defined in four-signed math. > Since I don't have the math I can't say to have proven this. You can fairly easily use trig to compute the (x,y,z) from the (-,+,*,#) coordinates. > I can see > it though. Making this assumption and putting the #-pole in the i > direction (as in i,j,k) we get the following partial transform for a > four-signed value x: > a = n(x) - ( 1/3 )( m(x) + p(x) + s(x) ) > where a is the i component of the three dimensional vector ai + bj + > ck. > n(x) is the # (number) component, m the - (minus), p the + (plus), and > s the * (star). > Now putting the minus pole in he ij plane and going in a left-handed > direction we see that the one third component yields an angle of: > pi - arccos( 1 / 3 ) > where arccos is the inverse cosine. > This should be the angle from the center to any two corners of the > tetrahedron. > Please note that I am not proving this. I'd like to find that angle in > a book somewhere or sharpen up my triginometry skills so that I could > verify this. > If this is true then there should be no problem with a clean RxRxR > transformation for four-signed math. The same concept should work > upward beyond our sight to five signed(pi-arccos(1/4)) and beyond. > Please keep in mind the simple sum rule: > - x + x * x # x = 0. Notice that *both* simplifications obey the sum rule, but represent very different structures. This will be important if you develope this out to 5-signed math. > I've done some work on the four-signed math this weekend. It looks like > the key to making it work is carefully defining a reduced form. Under > certain definitions of reduced from, n-signed math exists in R^(n-1), > Under other definitions, n-signed math is isomorphic to C. I need to > finish some more details, then I'll let you know what I come up with. I > think that thinking in terms of n-tuples will make things easier when > dealing with general n-signed math. Otherwise you may need to switch to > subscripted signs. > I will try to work with your representation, but I am pretty happy > with the symbolics that are used here since it is more arithmetic. At > some point I'd like to go to code. Do you know C or C++? > I think four signed is as far as I'd like to go for now in symbolic > format. I'm reasonably comfortable with both. I'm working on C++ right now. > I've written n-signed code for C++ and performed the Mandelbrot > mapping for four-signed in the simplest planes. Some look like the > usual Mandelbrot and some are simple solids. It's quite likely that > there are some bugs in my code so don't take my results too seriously. > I can't get too excited about the Mandelbrot mapping anyhow. I was > very disappointed when my three-signed code spat out the usual > mandelbrot shape. I was really hoping for something new. What does the > Mandelbrot test mean anyways? I believe it will determine if your interpretation corresponds with C or not. the 3-signed is equivalent to C, so Mandelbrot looked the same. Since 4-signed as you are doing it is *not* equivalent to C, Mandelbrot calculations will produce different results. > It has become apparent to me in writing code that the symbols used for > sign should really include a zero sign below the minus sign. This zero > sign is identical to the highest sign ( # for four-signed, * for > three-signed, + for two-signed). Since the reals are symbollically > faulty in this way I will continue on with the symbols I have chosen > for now. I wouldn't worry about it. > I'm looking forward to your results but I fear that you are straying > from the zero sum rule. I'm not, I just want to make sure the tetrahedral model works in a consistent way. Will Twentyman === Subject: Re: Exponentative closure Visiting Assistant Professor at the University of Montana. >>A further question along those lines is what is is the set, E, of >>numbers generated by repeated sums of products of rational powers of >>rationals? It is a subset of the algebraics, but does it form a >>field? >> I believe so: if x_1 is a member of E, then so are its conjugates x_2, >> x_3, ..., x_n. >> Note that x_1 x_2 ... x_n is rational (being the constant term of a >> monic polynomial over the rationals whose roots are x_1, x_2, ..., x_n). >> And then 1/x_1 = x_2 ... x_n / (x_1 x_2 ... x_n) is in E. >> Robert Israel israel@math.ubc.ca >> Department of Mathematics http://www.math.ubc.ca/~israel >> University of British Columbia >> Vancouver, BC, Canada V6T 1Z2 >If we say that a set, P, is exponentially closed if for all p,q in P, >p^q is P, then can the complex numbers be defined as the field that is >exponentially closed? Or even more simply, the set that is >exponentially closed and contains -2 and 0? >Note: that I am only considering fields with characteristic 0. No; either set is too small. I assume that your operation ^:PxP -> P is just a partial operation, not defined at (0,0); so just redefine it to give ^(0,0) = 0. The smallest field of characteristic zero that is exponentially closed would be the closure of Q under the operation ^. Since ^ is a finitary operation (takes only a finite number of arguments), the closure of Q is obtained as follows: S_0 = Q; T_{n} = S_n cup ^(S_n,S_n) where ^(S_n,S_n) is the image of (S_n,S_n) in C; that is, all numbers of the form p^q with p,q in S_n, under that definition; S_{n+1} = T_{n} cup +(T_{n},T_{n}) cup -(T_n) cup *(T_n,T_n) cup ^{-1}(T_n-{0}) where +(T_{n},T_{n}) is the sum of any two elements of T_n, -(T_n) is the additive inverse of any element of T_n, *(T_n,T_n) is the product of any two elements of T_n, and ^{-1}(T_n-{0}) is the multiplicative inverse of any nonzero element of T_n Then the closure of Q is S_{omega} = Union_{nIf not, what are the >properties of the smallest field, S, that is exponentially closed >(cardinality, algebraic completeness, convergence of >Cauchy-Sequences)? So S=S_{omega}. Cardinality is clearly countable, by the argument above (a standard argument of General Algebra). You would not get convergence of Cauchy sequences: since this is countable, there is a real number which is not in the set, and of course that real number is the limit of a cauchy sequence of rationals, which are all in S. I suspect algebraic completeness will also fail: if r is an algebraic number such that Q(r) is not solvable by radicals, how would you get r in S? Of course, since you also have non-algebraic numbers in S, this is hardly a 'proof', more of a 'I would look at these kinds of numbers to try to settle the question in the negative'. === Subject: Re: Can you help me? >>
I am stuck on a proof.  I have gotten a series down to where I need 
to
>>prove the following:
>> lim     c^1/n = 1  where c>0
>>n -> inf.
>>Do you have any pointers?
>>
There is a proof of this in (among many others) What is Mathematics by Courant and Robbins. It starts with Bernoulii's inequality (1+x)^n >= 1+n*x if n is a positive integer and x >= 0. Rewrite this as (1+y)^(1/n) <= 1 + y/n and choose an appropriate y. === Subject: Re: Chess/Go/etc: Continuous Game-Boards? Visiting Assistant Professor at the University of Montana. >In message , [.snip.] >The basic difference is that with Japanese rules, you count points for >territory surrounded and lose points for pieces you have lost, while >with Chinese rules, you count points for territory surrounded _or_ >occupied, and ignore pieces you have lost. Usually this makes no >difference, there is a simple proof of this depending on the fact that >each player has made the same number of moves. >>Which is certainly not the case under Japanese rules. At least the way >>I learned them: a player is allowed to pass, and then the other player >>may either pass (terminating the game), or play. >>Is this different under Chinese rules? >It is the same under Chinese rules. A player may always pass - two >consecutive passes end the game. >So, unless there are passes before the final two passes, each player has >played the same number of stones; except that Black may have played one >more point than White. So you are right - it can make one point of >difference. Well, no, it can make an arbitrarily large difference (limited to the number of positions on the board, of course). You play, I pass. Then you play, I pass. Then we alternate plays until the final two passes. Then it makes 2 points difference, does it not? And since I may pass an arbitrary number of es, would that not allow a large difference? The difference would equal the absolute value of the difference in the number of passes each player made before the final two passes. === Subject: Re: Can you help me? >I am stuck on a proof. I have gotten a series down to where I need to >prove the following: > lim c^1/n = 1 where c>0 >n -> inf. >Do you have any pointers? The limit of a discrete series is the same as the limit of a smooth function with those values. Why not say: lim[n->oo, c^(1/n)] = lim[x->oo,c^(1/x)] = lim[x->0,c^x] = c^0 = 1 (since c =/= 0). === Subject: Re: real analysis: construct this set ... >Can you construct a set E in [0, 1] s.t. for every open interval I in [0,1] m(I intersect E) > 0 & m(I intersect E^c) > 0 >m is lebesgue measure >E^c is the complement of E >This is so tricky! I was thinking something with the generalized Cantor set but everything I'm trying isn't working. Any suggestions? Ideas? I have previously posted an example of such a set, which is Borel measurable. Use the Cantor set idea, but with different sized parts removed from the intervals still in, and added to the intervals already removed. It can be done in terms of the digits to any base. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. === Subject: Corrected inflation dark energy/matter equations Technical typo correction on the math formula below for the exotic vacuum zero point stress-energy density residual inflation field whose local control could be applied to the Doomsday WMD described by Sir Rees in Chapter 9 of his book Our Final Hour. tuv(x)vac = (c^4/8piG*)/(x)zpfguv(x) Lp*^2 = hG*/c^3 = Lp^4/3(c/Ho)^2/3 (t'Hooft-Susskind) ~ (1 fermi)^2 ~ (10^-13 cm)^2 Ho is from cosmology (Hubble parameter), c/Ho ~ 10^28 cm Lp^2 = hG(Newton)/c^3 ~ (10^-33 cm)^2 /zpf(x) = Lp*^-2[1 - Lp*^3|Higgs Field(x)|^2] is the unified dark energy/matter zero point vacuum field. Giant vacuum superfluid local inflation wave from BCS spontaneous U(1)em symmetry breaking pair instability in the -mc^2 Fermi sphere band of the Dirac electron micro-quantum vacuum is PSI(vac) = (0|e+(x)e-(x)|0) = |Higgs Field(x)|e^i(Goldstone Phase(x) - Integral TaA^au(x)dx^u) Ta are the generalized charges of standard model of lepto-quarks and gauge forces. [D^uDu + V(|Higgs Field)] PSI(vac) = 0 V(|Higgs Field|) in large scale FRW limit of wavelet transform is the potential of inflationary cosmology is the BIT FROM IT Diff(4) covariant Landau-Ginzburg nonlinear local field equation Du is the Diff(4) covariant derivative guv(x) = nuv + (1/2)[du(x),v + dv(x),u] is Einstein's gravity c-number local geometrodynamic field where nuv is the flat constant Minkowski metric, and the elastic world crystal lattice distortion field is du(x) = Lp*^2 (Goldstone Phase(x) - Integral TaA^av(x)dx^v),u in 4D where ,u is the ordinary partial derivative is the IT FROM BIT Bohm guice constraint analogous to v = (h/m)GradPhase - (e/mc)A in the micro-quantum Bohm-Aharonov --> macro-quantum Josephson effect === Subject: Re: Can you help me? >I am stuck on a proof. I have gotten a series down to where I need to >prove the following: lim c^1/n = 1 where c>0 >n -> inf. Do you have any pointers? > The limit of a discrete series is the same as the limit of a smooth function > with those values. Why not say: > lim[n->oo, c^(1/n)] = lim[x->oo,c^(1/x)] = lim[x->0,c^x] = c^0 = 1 (since c > =/= 0). Well, at least if the function has a limit, then the series has one too. Somees the function might not even have a limit. In this case, f(x) = c^(1/x) does have a nice limit as x->oo. === Subject: Re: real analysis: construct this set ... >> Can you construct a set E in [0, 1] s.t. for every open interval I in >[0,1] m(I intersect E) > 0 & m(I intersect E^c) > 0 >> m is lebesgue measure >> E^c is the complement of E >A doubt. You say the Lebesgue measure. Are you implying that the >constructed set must be Lebesgue measurable? Or the problem somehow >gives us a notion that, even if the wording Lebesgue measure is >replaced by Lebesgue outer measure,it is impossible to find a >Lebesgue immeasurable set satisfying the condition as the problem >indicates? By saying m(...) where m is Lebesgue measure, it's clearly asking for measurable sets. As far as outer measure m^* is concerned, you can do a lot better: there is a (nonmeasurable) set E such that for every open interval I, m^*(E intersect I) = m(I) and m^*(E^c intersect I) = m(I). === Subject: lopital's rule? could someone please explain what lopital's rule is? my teacher likes to say things like, now, we could use lopital's rule, but that'd be too easy and you don't know it, so naturally it piqued my interest. sorry to ask such a broad question, but i really don't know anything at all about it to possibly narrow the question down. === Subject: Re: History of blackboard bold? >> ... The example BBB in >> the TeXBook (my edition anyway) is pretty obviously a CMR I and R >> run together. (I think it looks better than any currently available >> doublestroke font and have considered adapting cmr to produce >> something along those lines.) >You mean like the bbm.sty and fonts do? No, I don't mean like that. I'm pretty sure I've checked all currently available (free) doublestroke fonts. There's a blackboard.ps by Olaf Kummer available on CTAN that lists them (or at least those that were known to the author). And I usually check out any new ones announced here. As to bbm: compare the example R in the TeXBook (in index under blackboard bold) to the bbm R. Not at all alike. -- Luecking Department of Mathematical Sciences University of Arkansas Fayetteville, Arkansas 72701 luecking at uark dot edu === Subject: Re: lopital's rule? > could someone please explain what lopital's rule is? my teacher likes > to say things like, now, we could use lopital's rule, but that'd be > too easy and you don't know it, so naturally it piqued my interest. It's L'Hopital's rule (with a circumflex over the o). It's the principle that if f(x) -> 0 and g(x) -> 0 as x -> a but if f'(x)/g'(x) -> L as x -> a then f(x)/g(x) -> L as x -> a. many polemics from me on how evil and useless this rule is and how it should never be mentioned to impressionable undergraduates :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 es) === Subject: Re: lopital's rule? L'Hospital's Rule (pronounced lopital in French) can be found in any book on calculus of a single variable (look it up!). Briefly, the Rule allows you to calculate the limit of a function where the normal method would give you 0/0 - ie is indeterminate. Here is an example: say you want to find out the value of (3x - sin x)/x as x approaches zero if you take the limit as x approaches zero, both numerator and denominator approach zero, so you get 0/0 the Rule says that the limit as x approaches a of f(x)/g(x) = f''(a)/g'(a), where f'is d/dx of f(x), etc so, in our example above, f(x) is (3x - sin x), g(x) is x and a is zero the limit of (3x - sin x)/x as x approaches zero is, by the Rule: (3 - cos x)/1 as x approaches zero, which = 2 hth Guy > could someone please explain what lopital's rule is? my teacher likes > to say things like, now, we could use lopital's rule, but that'd be > too easy and you don't know it, so naturally it piqued my interest. > sorry to ask such a broad question, but i really don't know anything > at all about it to possibly narrow the question down. === Subject: Re: lopital's rule? Guy Corrigall scribbled the following: > L'Hospital's Rule (pronounced lopital in French) can be found in any book > on calculus of a single variable (look it up!). L'Hospital or L'H.99pital. The circumflex (the ^ thingy) means that the following s is understood implicitly. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ It's e, it's e, it's e to dump the slime! - Dr. te === Subject: help !!! I need help giving (p->q) xor (p<->q) an equivalence less than 12 === Subject: Re: lopital's rule? > Guy Corrigall scribbled the following: > L'Hospital's Rule (pronounced lopital in French) can be found in any book > on calculus of a single variable (look it up!). > L'Hospital or L'H.99pital. The circumflex (the ^ thingy) means that the > following s is understood implicitly. Is the s pronounced? -- G.C. === Subject: Re: Ces.88ro-convergence - analysis question >>Hi all, >>Regular convergence implies Ces.88ro-convergence. The inverse statement is >>not generally true. Can one give conditions on a sequence to guarantee >>that when you have a sequence that is Ces.88ro-converging it is also true >>that it is regularly converging? >Yes. The fact that convergence implies Cesaroo convergence is an >abelian theorem, and the fact that Cesaro convergence plus some >extra condition implies convergence is a tauberian theorem, where >the extra condition is a tauberian condition. >It's fairly easy to show that if the sequence s_n is Cesaro convergent >and also (s_{n+1} - s_n)*n -> 0 then the sequence s_n is convergent - >if I recall correctly this is the world's first tauberian theorem, >proved by Tauber. I realized later that I didn't recall correctly, sorry. Tauber's tauberian theorem was about convergence versus Abel summability, not Cesaro summability. (Which means that Tauber was not the sort of idiot it seemed this morning he must have been...) >It's true but not nearly as easy to show that you >only need to assume that (s_{n+1} - s_n)*n is bounded - this is due >to Littlewood I think. Again, the theorem of Littlewood I had in mind was about Abel summability - the simple theorem I prove below is actually due to Hardy. ************************ === Subject: Re: lopital's rule? > L'Hospital's Rule (pronounced lopital in French) can be found in any book > on calculus of a single variable (look it up!). Briefly, the Rule allows you > to calculate the limit of a function where the normal method would give you > 0/0 - ie is indeterminate. > Here is an example: > say you want to find out the value of (3x - sin x)/x as x approaches zero > if you take the limit as x approaches zero, both numerator and denominator > approach zero, so you get 0/0 > the Rule says that the limit as x approaches a of f(x)/g(x) = f''(a)/g'(a), > where f'is d/dx of f(x), etc Apart from the typo (f'' for f'), that's not right. lim_x->a f(x)/g(x) = f'(a)/g'(a) _if_ f'(a)/g'(a) _exists_. > so, in our example above, f(x) is (3x - sin x), g(x) is x and a is zero > the limit of (3x - sin x)/x as x approaches zero is, by the Rule: > (3 - cos x)/1 as x approaches zero, which = 2 > hth > Guy > could someone please explain what lopital's rule is? my teacher likes > to say things like, now, we could use lopital's rule, but that'd be > too easy and you don't know it, so naturally it piqued my interest. > sorry to ask such a broad question, but i really don't know anything > at all about it to possibly narrow the question down. -- G.C. === Subject: Re: help !!! Visiting Assistant Professor at the University of Montana. >I need help giving (p->q) xor (p<->q) an equivalence less than 12 p->q is true in every case except when p is true and q is false. p<->q is true in all cases when p and q are both equal. xor is true when one is true but the other not. In other words, xor is the opposite of <->. So, you want to know when p->q and p<->q are different. If p=0, then p->q is 1, so you would need q=1 for p<->q to be 0. If q=1, then p->q is 1, so you would need p=0 for p<->q to be 0. So p=0 if and only if q=1 for the xor to be true. if p=1, q=0, then p->q is 0, so for the xor to be true you would need p<->q to be true, but that does not happen. So the xor is true exactly when p=1 and q=0; it is false in all other cases. Can you come up with a propositional formula involving p, q, and, or, not, implies, and if and only if which uses less than 12 symbols and is equivalent to saying true when p=1 and q=0, false in all other cases? === Subject: Re: lopital's rule? > could someone please explain what lopital's rule is? my teacher likes > to say things like, now, we could use lopital's rule, but that'd be > too easy and you don't know it, so naturally it piqued my interest. > It's L'Hopital's rule (with a circumflex over the o). > It's the principle that if f(x) -> 0 and g(x) -> 0 > as x -> a but if f'(x)/g'(x) -> L as x -> a then > f(x)/g(x) -> L as x -> a. > many polemics from me on how evil and useless this rule > is and how it should never be mentioned to impressionable undergraduates :-) Why's that then? It seems to me to be a mildly interesting theorem. I can see that it might not be a good idea to prove it and then give students lots of 0/0 limit problems for them to solve using L'Hospital's when doing so is just pedagogically useless manipulation. But that is no reason for not proving the theorem. We could have a little contest: readers are invited to pose a 0/0 limit problem that is _best_ solved using L'Hospital's rule. (And Robin is required to post not only a non-L'Hospital solution, but a solution that is better than the L'Hospital one. Don't ask me what better means--having lit the blue touch paper, I shall retire:-) -- G.C. === Subject: Re: Homeomorphism and boundaries > I've read that homeomorphisms are open and closed bijections. Does that > mean that if f:X->Y is a homeomorphism, then f(boundary(A)) = > boundary(f(A)) for all sets A subset X? Boundary is a property of a set *and* the space in which it is embedded. Of two homeomorphic sets, one can have a big fat boundary, and the other can have empty boundary (because, for example, it is the whole space). === Subject: Re: help !!! >I need help giving (p->q) xor (p<->q) an equivalence less than 12 My approach is to translate everything into terms of and, or, and not. Then, use the basic rules for manipulating logical expressions to reduce the translated expression into a simpler expression. For example, p->q translates into not p or q. I got a result the exact opposite of what got. Of course, my method is more complex than just evaluating a truth table for the original expression. [cut] > So the xor is true exactly when p=1 and q=0; it is false in all other > cases. [cut] -- Bill Hale === Subject: Re: Minimal Graph, Four Color Theorem >>The only minimal counter-example to the FCT is K5! No, K5 is NOT a counterexample to the Four Color Theorem, because the >> 4 color theorem states that any ->planar<- graph can be colored with >> at most 4 colors in such a way that no two adjacent vertices share the >> same color. The conjecture that there exists a 5-chroma graph may be recolored to >>4-chroma is false. There is no such conjecture. >>Let H be any subgraph of G, where G has n vertices and H has n-1 >>vertices. Then, the description of H seems to imply that the deletion >>of 'any' vertex from G will make chi(H)<=4. This is true if G is a minimal counterexample for the 4 Color Theorem. But this interpretation is generally false and is valid only for >>n=5!!! The triple exclamation points make you look like a raving loon. So >> start by removing them. >Point taken, thank you. Could you explain why? > Can I explain why the triple exclamation points make you look like a > raving loon? Because they do. It makes it seem like you are jumping up > and down, yelling, spitting, and foaming at the mouth. That's the > mental image they conjure up. >> Then note that the original argument started by ->assuming<- that the >> FCT is ->false<-, from which we deduce that if this is the case, then >> among them, there is one with the least number of vertices. Call n the >> number of vertices of this HYPOTHETICAL counterexample. Then, by the >> definition of n, any graph with fewer than n vertices must be >> 4-colorable. In particular, if you took this HYPOTHETICAL example G, >> and removed one vertex, then the resulting graph would be 4-colorable. What exactly are you having trouble understanding about the above >> argument? Try to answer without using a ->single<- exclamation point. >> >I understand the argument perfectly. > Then why did you think somebody was claiming the Four Color Theorem > was ->false<-? >I have given the problem some >thought and I have concluded that HYPOTHETICAL G is impossible. > Good for you. > No >graph meets all three criteria; ie, G is 5-chroma, G is planar, H is >4-chroma. > Good for you. But your argument seems to be no such G can exist, > because then G would be K5, and that does not even begin to make > sense. G cannot be K5 if it is assumed to be planar. Assume that G is 5-chroma, then show that no 5-chroma graph can be planar. This seems to be a make sense. > Indeed, the proof of the Four Color Theorem rests on showing that > there does not exist any graph G which requires 5 colors, is planar, > and such that the removal of any vertex results in a graph which can > be colored with only 4 colors. But you have not given any coherent > argument to establish this proposition that I can see anywhere. All > you have done is yell like a loon that G would be K5!, which is > nonsense. You have misinterpreted my use of the triple explanation marks. I am suprised that you don't have me 'foaming at the mouth' if I use just one. I agree that it is nonsense to 'yell like a loon'; so I don't. Although, I would be in good company, ie, Archimedes. I think of G as a 5-chroma graph that might be planar; while you think of it as a planar graph that is or could be 5-chroma. I am afraid that I overlooked our differing points of view. By the way, are there any other punctuation marks that you consider signs of emotional unbalance === Subject: Re: 0! = 1 > scribbled the following: >>Did someone know a simple demonstration of >>0! = 1 ? > Well this is actually the _definition_ of 0!, so it doesn't > need to be demonstrated. > But one can explain why this definition is a good idea: > For example with this definition the equation > (n+1)! = (n+1)*n! > remains true for n = 0. With this definition the formula > for binomial coefficients C(n,k) in terms of factorials > gives the right answer for k = 0 and k = n. With this > definition the traditional expression for the Taylor > series of a function comes out right. Etc. > David Ullrich explained the mathematical side. I can show you a > philosophical reason why it's a good idea. > The number x! is intuitively understood as the number of *different* > orderings you can put x different items into. > Suppose x=0. Meaning you don't have any items. For better visualisation, > think of someone holding an empty basket and asking: How many > *different* orderings can I put all the balls in this basket in? > Well, there aren't any balls in that basket, so you could move balls > around to your heart's content, and the ordering of balls inside the > basket would never change. Therefore there is *ONE* possible ordering > of no balls: the one that the basket already contains. > Thus 0! = 1. so you can arrange 1 item in 1 way {x} and you can arrange 0 items in 1 way {}. Herc === Subject: Re: 0! = 1 > Is this making any sense? to which David W. Cantrell replied > Yes. But exponentiation isn't commutative. Perhaps one might say that > 1 is left-absorptive, and that exponentiation has no right-absorptive > element. You sure its making sense? Or you can't tell? Herc === Subject: Re: Factorial/Exponential Identity, Infinity I got to thinking about sum 2^n and how it was equal to 2^(x+1) - 1. I wonder: does it work for 3? The answer is not quite, no. For example sum 3^0 = 1, 3^1-1= 2* sum 3^0, 3-1=2*1, sum 3^1 = 4, 4*2=10-2. Anyways here is a short list of values of sum(3^n): 1 = 3^0 4 = 3^0 + 3^1 13 = 3^0 + 3^1 + 3^2 40 = 3^0 + 3^1 + 3^2 + 3^3 121 = 3^0 + 3^1 + 3^2 + 3^3 + 3^4 and values of 3^n: 3 = 2*1+1 9 = 2*4+1 27 = 2*13+1 81 = 2*40+1 Thus it appears that the sum of 3^i for i from zero to n is (3^(n+1)-1) / 2 For sum 4^n: 1 5 21 85 341 4^n: 1 4 16 64 256 1024 It appears that sum ((4^n) = 4^(n+1) - 1 ) / 3. I would thus surmise that sum(x^n) = x^(n+1)-1)/(x-1), the sum of x^i for i=1 to n equal (x^(n+1)-1)/(x-1). Where should I look to find a proof of this, so I don't have to write one, and that I can read the surrounding material to which it refers to gain understanding of its concept? This doesn't help me solve sum(n^x), because that is the sum of i^x for i=1 to n, the closed form of that quantity is determined as has been noted by the Bernoulli polynomials, sum of powers. I guess after that would be sum(n^n). Anyways in the limit n! = sqrt( sum(n)^n - sum(n^n) ), another in the long list of approximations of n!. Ross === Subject: Re: lopital's rule? >> L'Hospital's Rule (pronounced lopital in French) can be found in any book >> on calculus of a single variable (look it up!). Briefly, the Rule allows you >> to calculate the limit of a function where the normal method would give you >> 0/0 - ie is indeterminate. >> Here is an example: >> say you want to find out the value of (3x - sin x)/x as x approaches zero >> if you take the limit as x approaches zero, both numerator and denominator >> approach zero, so you get 0/0 >> the Rule says that the limit as x approaches a of f(x)/g(x) = f''(a)/g'(a), >> where f'is d/dx of f(x), etc >Apart from the typo (f'' for f'), that's not right. lim_x->a f(x)/g(x) >= f'(a)/g'(a) _if_ f'(a)/g'(a) _exists_. No, that's not right. lim_x->a f(x)/g(x) = lim_x->a f'(x)/g'(x) if lim_x->a f'(x)/g'(x) exists. (The existence of that second limit does not follow from the existence of f'(a)/g'(a), nor does it imply the existence of f'(a)/g'(a).) >> so, in our example above, f(x) is (3x - sin x), g(x) is x and a is zero >> the limit of (3x - sin x)/x as x approaches zero is, by the Rule: >> (3 - cos x)/1 as x approaches zero, which = 2 >> hth >> Guy >> could someone please explain what lopital's rule is? my teacher likes >> to say things like, now, we could use lopital's rule, but that'd be >> too easy and you don't know it, so naturally it piqued my interest. >> sorry to ask such a broad question, but i really don't know anything >> at all about it to possibly narrow the question down. ************************ === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana. [.snip.] >> No >>graph meets all three criteria; ie, G is 5-chroma, G is planar, H is >>4-chroma. >> Good for you. But your argument seems to be no such G can exist, >> because then G would be K5, and that does not even begin to make >> sense. G cannot be K5 if it is assumed to be planar. >Assume that G is 5-chroma, then show that no 5-chroma graph can be >planar. This seems to be a make sense. Then show that no 5-chroma graph can be planar ->IS<- the 4 Color Theorem. What you are saying is that to prove the 4-color map theorem, you should just prove the 4 color map theorem. Well, duh. >> Indeed, the proof of the Four Color Theorem rests on showing that >> there does not exist any graph G which requires 5 colors, is planar, >> and such that the removal of any vertex results in a graph which can >> be colored with only 4 colors. But you have not given any coherent >> argument to establish this proposition that I can see anywhere. All >> you have done is yell like a loon that G would be K5!, which is >> nonsense. >You have misinterpreted my use of the triple explanation marks. I didn't say it ->made<- you into a raving loon. I it made you ->look<- like a raving loon. > I am >suprised that you don't have me 'foaming at the mouth' if I use just >one. One exclamation mark, if not overused, indicates emphasis, surprise, any number of things. A triple exclamation mark reads like yelling. Continual use of multiple exclamation marks, joined with (apparently) not reading the responses to your questions, is what makes you look like you are foaming at the mouth. >I agree that it is nonsense to 'yell like a loon'; so I don't. >Although, I would be in good company, ie, Archimedes. Yes, and some people were laughed at and turned out to be geniuses. On the other hand, most of the people who are laughed at are clowns. >I think of G as a 5-chroma graph that might be planar; while you think >of it as a planar graph that is or could be 5-chroma. I am afraid that >I overlooked our differing points of view. If you think of G as a graph which has chi(G)=5 and may or may not be planar, then you completely misunderstood the paragraph you quoted when this began, and then that is at least part of the reason for your arguments/misunderstandings/confusion. The paragraph you quoted started as part of a proof by contradiction, by assuming that there was a graph G which was planar, and which had chi(G)=5, and which had a minimal number of vertices from among all graphs that satisfied those properties: being planar, AND having chi(G)=5. If you thought that such a graph might be planar, then you missed the point entirely. The assumption is that it ->is<- planar. I do not think of this hypothetical planar G as a graph that could have or fail to have chi(G)=5. If it is part of a proof by contradiction, then I ->must<- assume that the graph is planar AND that it satisfies chi(G)=5. Now say you are trying to prove The 4 Color Theorem by induction on the number of vertices, as Keith Ramsey suggested. You have proven that a planar graph with fewer than 5 vertices is 4 colorable. Now as an induction hypothesis, we assume that a planar graph with fewer than n vertices is necessarily 4-colorable, and consider a graph G which is planar and has n vertices. At this point, G is a graph that could, indeed, have chi(G)>=5 or chi(G)<=4. Since for any given vertex v we have, by the induction hypothesis, that G-{v} is 4-colorable, that shows that G is certainly 5-colorable, so chi(G)<=5. Thus, at this stage, we have a graph which is planar and which may or may not satisfy chi(G)=5. Perhaps that is what you meant. Note that in this situation, since the assumption is that ALL planar graphs with fewer than n vertices are 4-colorable, that would mean that if chi(G)=5, then G is a minimal counterexample to the conjecture. However, I fail to see how assuming that the graph G has n vertices and chi(G)=5, and being unsure as to whether G is or is not planar, would help you in figuring out the situation. Removing a vertex is not enough to be able to apply the induction hypothesis, since the result may not be planar. You would not be able to say that G is a minimal counterexample. Or rather, assuming that G is a graph with the smallest number of vertices among all graphs with chi(G)=5 is NOT the correct assumption to make; the induction hypothesis does not guarantee that chi(G) is minimal with this property, it only guarantees that any proper subgraph of G ->which is planar<- is 4-colorable. So, for example, for all you know the graph properly contains K5. There is no minimality property you could apply to this G, or rather, the minimality hypothesis here is just plain ->wrong<-. >By the way, are there any other punctuation marks that you consider >signs of emotional unbalance Sigh. It was the entirety of your interaction. You started by asking a reasonable question. When you encountered replies, your immediate reaction was to ->argue<- about those replies, using multiple exclamation marks. So you were being ->very<- emphatic, at the very least. The more exclamation marks you put, the more emphasis/volume one is expected to read into the statement. Surprise! is not read the same way as Surprise!! or as Surprise!!! or as Surprise!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Frankly, I felt like you were yelling in my face, and making little sense to boot. In addition, your statements made it seem like you were either not reading, or not understanding, what people were writing. As such, we have someone who asks a question, and starts yelling to everyone who replies, apparently without listening to their answers. ->That<- makes you look like a raving loon. === Subject: Quantum Gravity Part 2 Comments by Jack Sarfatti on excerpts from: Excerpts are from: The three perspectives on the quantum-gravity problem and their implications for the fate of Lorentz symmetry1 Dipart. Fisica Univ. La Sapienza and Sez. Roma1 INFN P.le Moro 2, I-00185 Roma, Italy ñthere is still not a single measured number whose interpretation requires advocating ïquantum gravityÍ. ñThe third possibility is a condensed-matter perspective (see, e.g., the research programs of Refs. [13] and [14]) on the quantum-gravity problem, in which some of the familiar properties of spacee are only emergent. Condensed-matter theorists are used to describe some of the degrees of freedom that are measured in the laboratory as collective excitations within a theoretical framework whose primary description is given in terms of much different, and often practically unaccessible, fundamental degrees of freedom. Close to a critical point some symmetries arise for the collective-excitations theory, which do not carry the significance of fundamental symmetries, and are in fact lost as soon as the theory is probed somewhat away from the critical point. Notably, some familiar systems are known to exhibit special-relativistic invariance in certain limits, even though, at a more fundamental level, they are described in terms of a nonrelativistic theory. For a rather general class of fermionic systems one finds [13] that at low energies, as a Fermi point is approached, fermions gradually become chiral Weyl fermions, while bosonic collective modes of the vacuum transform into gauge fields and gravity. Clearly from the (relatively new) condensed-matter perspective on the quantum gravity problem it is natural to see the familiar classical continuous Lorentz symmetry only as an approximate (emergent) symmetry. [Revised Comment #3: This notion is very compatible with the Bohm-Hiley-Vigier pilot BIT wave landscape IT hidden variable system point flow on the landscape picture of quantum theory. Orthodox micro-quantum theory is the limit of An Valentini's sub-quanta heat death with signal locality in which the quantum potential is fragile (Bohm-Hiley) with a pilot BIT wave (nonlocal in configuration space for entangled systems) that has no sources (Bohm-Hiley). This means that the IT hidden variable system point flowing on the BIT orders (J.A. Wheeler) from BIT but not vice versa. This is action without reaction! Here reaction means back-action where the BIT pilot wave landscape has sources so that IT is no longer a passive test landscape it is flowing on according to vs = (h/m)Grad(Phase) - (q/c)A or in the 4D elastic world crystal lattice (Hagen Kleinert) model distortion field du(x) we have the IT FROM BIT constraint equation showing how the IT geometrodynamic field gets its ñmarching ordersî (WheelerÍs term in a different but related context) from the MACRO-QUANTUM BIT vacuum inflation field. du(x) = Lp*^2 (Goldstone Phase),u - TaA^a,u Lp*^2 = hG*/c^3 = Lp^4/3(c/Ho)^2/3 in t'Hooft-Susskind world hologram model for renormalized Planck scale Lp* ~ 1 fermi determined by Newtonian Planck scale Lp = 10^-33 cm and size 10^28 cm of our contingent local Level I (Max Tegmark) parallel IT universe in an infinity of parallel IT universes on a single 3D spatially flat American). Ta are the generators of the Lie algebra of the standard model in globally flat Minkowski spacee with fiber connections A^a,u for parallel transport. a is internal space and u is spacee. This is normalized such that TaA^a,u is a length. ,u is the ordinary partial derivative operator relative to x^u, u = 0, 1,2,3. The attraction between virtual electrons and positrons near the -mc^2 band in the Dirac vacuum spectrum Fermi sphere is a pair instability that creates the virtual giant local vacuum wave BEC <0|e+(x)e-(x)|0> a complex scalar inflation field whose phase is the Goldstone phase and whose amplitude is the Higgs field <0|e+(x)e-(x)|0> = |Higgs field(x)|e^i(Goldstone phase(x)) = PSI(vac) such that the unified dark energy/matter zero point stress-energy density Diff(4) tensor field is : tuv(x)zpf = (c^4/8piG*)/zpf(x)guv(x) where /zpf(x) = Lp*^-2[1 - Lp*^3|Higgs field(x)|^2] /zpf(x) > 0 is strongly scale-dependent anti-gravitating dark energy exotic vacuum with negative pressure and positive zero point energy density. /zpf(x) < 0 is strongly scale-dependent gravitating dark matter exotic vacuum with positive pressure and negative zero point energy density. All quantum fields of all spins contribute to the net residual /zpf(x) LOCAL FIELD in the emergent More is different collective mode c-number background spacee with both Diff(4) base space and local Lorentz group tangent space emergent symmetries at least near a critical point of the renormalization group flow sense. Einstein's gravity c-number background field is guv(x) = nuv(Minkowski) + (1/2)[du(x),v + dv(x),u] with anholonomic torsion tensor field (string Goldstone phase singularities) suv(x) = (1/2)[du(x),v - dv(x),u] and the back-action nonlinear LOCAL virtual BEC Landau-Ginzburg Diff(4) symmetric BIT FROM IT ñback-actionî equation: {D^uDu + V(|<0|e+(x)e-(x)|0>|)}<0|e+(x)e-(x)|0> = 0 Where V(|<0|e+(x)e-(x)|0>|) limits to the effective spontaneous broken vacuum symmetry potential of chaotic inflation cosmology in the large-scale limit of the isotropic homogeneous FRW metric with zero point energy density induced cosmological constant for ñdark energyî using an adaptive windowed wavelet transform version of the Wigner phase space density with ODLRO rather than a rigid windowed Fourier transform. Shorter scales have longer bandwidth and longer scales have smaller bandwidth with total area in phase space per conjugate pairs of incompatible dynamical variables constant ~ h (in phase space) or Lp*^2 (in spacelike slice of spacee) to preserve HeisenbergÍs uncertainty principle and the Bekenstein-tÍHooft-Susskind ñworld hologramî reinterpretation of generalized ñblack hole thermodynamicsî. See also the somewhat related 3D spacelike quantized area operators in Penrose spin networks of the loop quantum gravity approach. Note that Lp^2 = hG(Newton)/c^3 Maps the discrete structure of quantum gravity foam in phase space to a discrete ñareaî structure in spacelike slices of spacee consistent with the world hologram Ansatz. Du is the Diff(4) operator so that D^uDu is the GR wave propagation operator on a complex numbered scalar field PSI(vac).] Results obtained over the last few years (which are partly reviewed later in these notes) allow us to formulate a similar expectation from the general-relativity perspective. Loop quantum gravity and other discretized-spacee quantum-gravity approaches appear to require some departures, governed by the Planck scale, from the familiar (continuous) Lorentz symmetry. And in the study of noncommutative spacees some Planck-scale departures from Lorentz symmetry might be inevitable, since (at least in a large majority of noncommutative spacees) a Lie algebra is not even the appropriate language for the description of the symmetries of a noncommutative spacee (one must resort to the richer structure of Hopf algebras). reason to renounce to exact Lorentz symmetry. Minkowski classical spacee is an admissible background spacee, and in classical Minkowski there cannot be any a priori obstruction for classical Lorentz symmetry. Still, a breakup of Lorentz symmetry, in the sense of spontaneous symmetry breaking, is of course possible. This possibility has been studied extensively [10, 15] over the last few years, particularly in String Theory, which is the most mature quantum-gravity approach that emerged from the perspective. 1.2 What do we know about quantum-gravity? The theory debate clearly is a confrontation between very different perspectives on the quantum-gravity problem. If we had any robust information on quantum gravity certainly at least some of these ideas would have been proven to fail. But after more than 70 years [16] of work on the quantum-gravity problem there is still not a single measured number whose interpretation requires advocating quantum gravity. I have so far mentioned the quantum-gravity problem as if it was a well-established and familiar concept, but it is perhaps useful to give here an intuitive characterization of this problem. The quantum-gravity problem is somees described as a sort of human discomfort, as a problem pertaining to the achievement of a more satisfactory philosophical worldview. For example, as motivation for research in quantum gravity it is somees stated that quantum theory (in an appropriate generalized sense) has turned out to be relevant for the description of measurement results in all other branches of fundamental physics, and we therefore must assume that it will eventually be relevant also for spacee/gravity physics. Analogously (and amounting to the same thing), it is somees stated that it is unsatisfactory to have on one side our present unified quantum-field-theory description of electromagnetic, weak and strong forces and on the other side gravity which is still described in a very different way. These human discomforts do not of course define a scientific problem, but actually there is, as emphasized by some, a well-defined scientific problem which can be naturally called quantum-gravity problem. The scientific problem that can be reasonably called quantum-gravity problem is actually the problem of producing numbers (predictions), in a logically-consistent quantum-field theory effects cannot be neglected. For example, although we are presently (and for the foreseeable future) unable to set up and observe collisions between two electrons each with energy of, say, 10^50 eV, our present theories provide no obstruction for the analysis of such high-energy collisions, but are unable to produce a logically consistent number for, say, the probability that such a collision would result in two muons with certain energies and momenta. The problem, as I shall try to point out later in these notes, resides in the fact that quantum field theory implicitly assumes that gravity effects can be neglected. When the gravity effects are so large that (from the field theory perspective) space geometry evolves significantly on very short e scales, field theory cannot be consistently appliedb. Similarly, field theory runs into trouble when gravity effects are strong enough to admit the emergence of spacee singularities (e.g. black holes). We are able to get numbers out of quantum field theory in contexts in which there is a curved static (or slowly-varying) nonsingular space, but fast-varying and/or singular space geometries are untreatable.î [Comment: An adaptive windowed ñzoom in/outî wavelet transform reformulation may help here.] ñOne might argue that 10^50 eV electrons should be the least of our concerns, since we are never going to be able to produce and/or observe them, but first of all in cosmology there are some numbers we should produce that depend on very early es after the high energies were abunt), and, secondly, the fact that our theories fail to produce numbers in some contexts which those same theories describe as accessible (in principle) makes us concerned in general about the robustness of these theories. Since we know that new elements would have to be introduced in our theories for the description of collisions between 10^50 eV electrons (or for a justification of an in-principle exclusion of such collisions from the list of processes that can occur in Nature), it is natural then to wonder whether those new elements can affect also some of the contexts in which our present theories do provide us an apparently acceptable prediction. In some cases the issues we encounter in analyzing, say, collisions among 10^50 eV electrons might bring to the surface some issues that could also modify more ordinary (but still untested) predictions produced by our theories. b Here the reader should keep in mind that general relativity governs self-consistently the spacee dynamics in terms of (and together with) asymptotically, in the S-matrix sense, in quantum field theory.î [Comment: Lenny Susskind tried to fix this at Cornell in 1964 when we were students together. He failed because the wavelet transform method was not known back then.] ñDuring a collision process the, say, electrons involved are not following any trajectories. We can associate to them some (however fuzzy) trajectories only asymptotically, much before and much after the collision.î [Comment: The Bohm picture would help here as well.] ñIf one tries to apply general relativity to the formally-classical trajectories that appear in the path integral formulation of quantum mechanics, the problem becomes anyway ill defined (and affected by enough to induce significant geometrodynamics. There is a very natural explanation for our lack of insight on this quantum-gravity problem. One of the few (perhaps the only) robust hint we have about quantum gravity quantum-field-theory description starts to appear inadequate is the Planck scale Ep ~ 1028^eV.î [Comment: That may not be true. In my theory Lp* ~ 10^-13 cm, i.e. energy scale is 20 powers of ten lower than the na.95ve Newtonian esate for quantum gravity. This explains alphaÍ = 1/(1 Gev)^2 for hadronic resonant parallel Regge trajectories, it explains why the lepto-quarks are ñmicro-geonî Kerr-Newmann spatially extended yet stable and why they appear more and more point-like when probed to smaller scales at larger momentum scattering transfers. Furthermore, there is seamless integration with the large scale dark energy/matter precision chaotic spatially flat inflationary observations (WMAP, type 1a supernovae, gravity lensing, gamma ray bursts).] unsatisfactory. And usually the scale that sets the break point of an effective low-energy theory is also the scale that sets the magnitude of the new effects to be expected going beyond the effective low-energy theory. It is therefore reasonable to expect that ñquantum-gravity correctionsî to our low-energy predictions would be very small, with their magnitude set by some power of the ratio between the Planck length (Lp ~ 10-35m, which is the inverse the Planck scale Ep ~ 1028eV ) and So we have good reasons to suspect that the quantum-gravity effects would be very small (and actually they must be typically small, since we have not managed to see them yet).î [Comment: I disagree here. To the contrary we are seeing quantum gravity on scale of 1 Mev and 1 Gev but we have not properly understood what we are seeing.] to be continued === Subject: Re: lopital's rule? >> L'Hospital's Rule (pronounced lopital in French) can be found in any book >> on calculus of a single variable (look it up!). Briefly, the Rule allows you >> to calculate the limit of a function where the normal method would give you >> 0/0 - ie is indeterminate. >> Here is an example: >> say you want to find out the value of (3x - sin x)/x as x approaches zero >> if you take the limit as x approaches zero, both numerator and denominator >> approach zero, so you get 0/0 >> the Rule says that the limit as x approaches a of f(x)/g(x) = f''(a)/g'(a), >> where f'is d/dx of f(x), etc >Apart from the typo (f'' for f'), that's not right. lim_x->a f(x)/g(x) >= f'(a)/g'(a) _if_ f'(a)/g'(a) _exists_. > No, that's not right. lim_x->a f(x)/g(x) = lim_x->a f'(x)/g'(x) > if lim_x->a f'(x)/g'(x) exists. (The existence of that second > limit does not follow from the existence of f'(a)/g'(a), nor > does it imply the existence of f'(a)/g'(a).) Yes, I'm deeply ashamed of myself. I could see that Guy's limit as x approaches a of f(x)/g(x) = f'(a)/g'(a) wouldn't work and I was too hasty to criticize. I am reminded of a principle in software (an area that I used to work in) debugging: if you find a bug, always look around the same area for another bug--especially the one that you're just about to introduce with your fix! I shall go and boil my head. -- G.C. === Subject: Re: infinitary DEs With my humilty point of view, I believe that I can give you an idea: there is an analogy between diferential equations and diference equations, in the sense that there are a correspondence between an diferential equation and its diference equation. So why not we see what we will try with a infinitary difference equation, and then when we know the solution we pass to its diferential equation?. It's only an idea. Sorry for no develope this idea, but I have oxided some of the concepts that you manage!. Xan. > >Here's a creature we all studied as undergraduates: > a_0(x)y + a_1(x)y' + ... + a_n(x)y^(n) = b(x). >It has only just (25 years later!) occurred to me to wonder why it stops >at n. Is there anything interesting to be about > a_0(x)y + a_1(x)y' + ... = b(x) >in which, for each n there is a derivative of y of order n or >greater? I'm sure there is. Consider the case where the a_j are constants. > Write your equation as f(D) y = b(x), where > f(q) = sum_{j=0}^infinity a_j q^j. Suppose the radius of convergence > of this series is infinite, so f is an entire function. > Note that f(D) exp(p x) = f(p) exp(p x). In particular, if f(p) = 0 > then exp(p x) is a solution of the homogeneous equation. > Formally, the Fourier transform of a solution is y^(k) = b^(k)/f(ik), > and under certain conditions this will make sense. E.g. if b is in L^2 > and |f(ik)| > epsilon > 0 for all real k, then this produces a solution > in L^2. > the case a_j = constant. I don't know what to make of the more general > case. === Subject: Hard (unsolved?) search problem revisited BASIS ----- A two dimensional table have integers distributed this way (in this examle, only odd numbers): 3,11,13,21,27,... 7,13,17,29,35,... 9,15,19,33,41,... . . . Assume that this table is huge, so a fast algorithm is needed - especially since many tables of this form must be searched many es. So the values always increase when going from left to right, top to bottom, no matter where you start. That is, t[i,j] < t[i+1,j] and t[i,j] < t[i,j+1] PROBLEM ------- Given such a table, t, can you locate a given integer N in polynomial e - or - can you determine that N is not in the table in polynomial e (or better, as in constant :-)? If yes, what does that algorithm look like? STATUS ------ I tried the following approach: start in the middle, t[i_max / 2, j_max /2] and see if it's less than N. If so, the top/left quadrant of the table can be eliminated because all values in that quadrant must be smaller than N. Apply algorithm recursively to the remaining quadrants. Of course, this is not a fast algorithm at all... Any insight would be highly appreciated. === Subject: Re: limit points of Q Sure the question can be asked, it just that the larger field isn't R, it's the p-adic rationals Q_p. And indeed, Q is dense in Q_p with respect to the p-adic absolute value. Actually, if you define R to be the (unique up to isomorphism) completion of Q with respect to the usual absolute value, then it's pretty immediate that Q is dense in R, and ditto Q being dense in Q_p. More generally, if K is any field and |x| is an absolute value of K, then there is a unique (up to isomorphism) smallest field L containing K with the properties that |x| extends to an absolute value on L and L is complete wrt this absolute value. Then one can prove that K is dense in L. An interesting example if K = R(X), the field of rational functions with real coefficients, and the absolute value is |f(x)| = e^{-deg(f)}. The completion of K is often denoted L = R((X)), it's the field of formal (i.e., not necessarily convergent) Laurant series. > It actually depends on the valuation you are using. If your valuation > is the normal absolute value, then you are correct. But know that for > each prime there exists a valuation called the p-adic valuatian, and > in these cases your question can't even be asked. === Subject: Re: factoring to satisfiability > hi all. conversion of factoring to SAT does NOT (*always* > or *usually*) give easy > instances as was stated in another msg on this thread. > it MAY or MAY NOT give easy instances. if you want to > factor a prime number then the resulting SAT clause must > be easy based on the new agrawal proof. > and it would > be extremely interesting to try to find a SAT algorithm > that runs in P e on prime number factoring instances... > I think it is virtually guaranteed that the basic DPLL > etc algorithms do NOT.. and so why not?? I have been investigating methods that work on UNSAT problems better than on SAT problems. There are simple algorithms that generate ALL solutions to a SAT problem. I had looked at such methods some e ago and abandoned them because they generate enormous numbers of intermediate solutions. Recently, I determined that problems with few or no solutions should not generate as many intermediate solutions. Most hard SAT problems have few solutions. I now have a working program that generates all solutions for the first M clauses of a 3-SAT problem. The results are interesting. This program works best on UNSAT problems. I just ran it on a series of 10 UNSAT problems with 50 variables and 218 clauses. The largest number of intermediate solutions was 301. This is much smaller than the 3^M intermediate solutions that might be expected. The largest number of clauses required to show a problem was unsolvable was 40 (out of 218). A method like this might work well on UNSAT problems based on factoring primes. - 2 many 2 count === Subject: Why Inflation? I do not have the e to wade through those papers. If you can show me one thing I would be more motivated. Show me in e-mail how you derive Einstein's field equation Guv = (8piG/c^4)Tuv from your idea of the wave structure of matter. I claim I have done exactly that where the wave is a macro-quantum vacuum coherent local inflation field wave from a QED BCS instability. Up until now there has been no micro-quantum dynamical explanation why such a large scale inflation field should even exist. According to Michael Turner that is one of the current mysteries as to why inflation should be there at all. It is introduced by Linde ad hoc - indeed Linde admits it was a sudden idea out of the blue when he was trying to get a Green Card to emigrate to America from Russia. Show me how you do it. Also show me how you explain the following facts: 1. The spatially flat expanding accelerating universe is at least 73% Note I claim it is at least 96% i.e. dark matter is also exotic vacuum IMHO - not on mass shell axions, not neutralinos etc. 2. Explain the stability of the electron. 3. Explain the Regge trajectories of hadronic resonances. 4. Explain why lepto-quarks look more and more point-like in deep inelastic electron scattering as the scattering momentum transfer gets larger and larger. Hi Milo and Jack, The URLs are below as links. www.QuantumMatter.com www.SpaceandMotion.com Hope this helps. Geoff Haselhurst ----- Original Message ----- Cc: Geoff Haselhurst === Subject: Re: The Wave Structure of Matter You need to send the proper URL with http://... I am not able to find these websites unless you format them in proper syntax so I can click on it to get to it. Google has nothing for them. Hi Jack, I can feel your intense interest to find the mechanism of gravity and objects but are instead wave structures in a quantum space. Our perception of their properties was 'schaumkommen' of the wave structures. (appearances.) Now, it has been worked out. see QuantumMatter.com and SpaceandMotion.com The results are amazing. 1) All the natural laws are found as properties of the wave structure of the electron. 2) Everything grows out of only two principles which are properties of one thing - space. Awesome. Gravity is the simplest piece of cake. Take a look. I would love to have your thoughts. I do not know what you mean. I mean it is necessary to study ( take a long look) at the above two web sites and the references given in them. Have you derived the equations for general relativity from the information wave? That is precisely what I have done for the giant vacuum pilot wave along with the unified dark energy/matter local field. YES. All the natural laws flow mathematically and logically out of the Wave Structure of the Electron and other matter. GTR included of course. as Schroedinger and Clifford proposed. Any new proposal must be couched in mathematical language and must in suitable limiting cases yield the battle tested equation of theoretical physics such as Guv = (8piG/c^4)Tuv Maxwell's equations etc. YES. There is complete agreement with all the experimental observations of the natural laws. Otherwise it is not legiate physics IMHO. Take a look. The Wave Structure of Matter makes it all quite simple but there is lot there to learn. It will take you many hours of study. But it will be well worth it! You are an expert on the forefront of the problems of physics. I will really appreciate your serious considered opinion after your study. Also there must be contact with experimental observations both in terms of prediction and explanation as nicely presented in David Deutsch's book The Fabric of Reality for example in the chapters on proper methodology in theoretical physics. Milo You can always tell a pioneer by the arrows in his back === Subject: Re: Boolean Algebra - Arithmetic Relationship > ... Can arithmetic be further simplified into boolean logic?? > Yes. Computers - or rather those who design computers, are quite good at it. > ;-) Or rather, no. Computers only ever handle a finite number of finitely-representable numbers, so pi for example is beyond them. -- G.C. === Subject: Re: factoring to satisfiability oops, my factoring -> SAT generator does indeed only use 5-clauses or less, but introduces auxiliary variables. I got mixed up the same way the other poster did. so its still an open question, what is the smallest clause size possible using no auxillary variables? this seems to be closely related to the gate-width complexity of the problem. > More specifically, any 5-clause (t1 + t2 + t3 + t4 + t5) can be replaced > by three 3-clauses (t1 + t2 + x)(-x + t3 + y)(-y + t4 + t5) where x and > y are new variables. === Subject: Re: Boolean Algebra - Arithmetic Relationship > To all, > Forgive me if I sound naive, I have an appreciative but highly > abstracted perception of mathematics. My Question: > I have been told that most of mathematics (geometry, algebra, etc.) > is derived, or at least reducible, to arithmetic. Can arithmetic be > further simplified into boolean logic?? Which I understand is > equivalent or corresponds highly to propositional logic (correct me if > I'm wrong here). I'm not sure what you mean by boolean logic, but let's assume that it's the same as propositional logic. Then the answer is no, quantification is needed. Now, are first order quantifiers sufficient or are second order (at east) needed? If set theory counts as logic, then first order quantification suffices. If it doesn't then second order arithmetic will do for analysis. > Is this what Bertand was attempting to accomplish in Principia > Mathematica?? Or did he mean logic in a different sense?? He meant logic in a _much_ broader sense than propositional logic. P.S. It may be that by boolean logic you mean computer hardware gates. A physically realized (or realizable) logic is certainly inadequate--only a finite number of the infinity of numbers could be handled, and only finite approximations of numbers like pi could be handled. -- G.C. === Subject: Re: Minimal Graph, Four Color Theorem > |I understand the argument perfectly. I have given the problem some > |thought and I have concluded that HYPOTHETICAL G is impossible. No > |graph meets all three criteria; ie, G is 5-chroma, G is planar, H is > |4-chroma. > I assume by this last clause you mean that all the graphs one obtains by > deleting a single vertex from G are 4-chromatic. > This conclusion is equivalent to the four color theorem. If the four color > theorem is true, then no planar graph has chromatic number 5. On the other > hand, on the assumption that your conclusion above is correct, the four > color theorem follows by induction on the number of vertices. Once we've > shown it's true for planar graphs of up to n vertices, then it also must > be true for planar graphs of n+1 vertices, since whatever graph H is, it's > already been shown to have chromatic number <=4 (and its chromatic number > differs from that of G by at most 1). > So the only way you can reach such a conclusion is by an argument which is > at most one short paragraph shorter than a proof of the four color theorem. > I would assume that you're just relying upon the existing proof, except that > it wouldn't usually take some thought to conclude that a 5-chromatic > planar graph having a certain kind of subgraph doesn't exist, given that > no 5-chromatic planar graph exists at all. > I just hate to see someone go away still confused, so I hope your clarity > on the argument has reached the point of recognizing that this conclusion > you state above is very far from trivial, without taking the proof of the > four color theorem for granted. If there's some simple way to show > such a G doesn't exist, a number of smart people have failed to > see it despite working hard on it for a long e. > I hope you will be gracious and respond to my previous posting Re: Four Color Theorem Simplified. I notice that you have not responded to any of my previous posting re the FCT. May I inquire as to why you chose to respond to this one? === Subject: Re: hw help -- continuity > Folks, > I have a couple questions. This is homework, so please post a nudge, > not a solution. > 1)prove that if f,g continuous, then so are max(f,g) and min(f,g) > After drawing some graphs, this seems pretty obvious for the single > point a0 -- max(f,g) has 2 cases: it equals to f or g. Either is > continuous. However, this question implies continuous on R, not just > at a single point. Any ideas how to approach this? I'm confused. If you can show that max(f,g) is continuous at each point, then you have shown that max(f,g) is continuous. However, you still have to prove that max(f,g) is continuous at a point given that f and g are both continutous at that point. Remember, the ideal of continuity is that one is making a statement about how control over the independent variable allows control over the dependent variable. For example, if h(t) represented the height of a balloon (h) as a function of e (t), then to say that h is continous is to say that if at a given e t0, if I confine my attention to es sufficiently close to t0, then the height of the ballon is guaranteed to be sufficiently close to h(t0). Formally, the two occurances of sufficiently are replaced by delta and epsilon, respectively. Thus, if I want h is a continutous function of t, suppose I want a guarantee that the balloon will now have change height by more than 1000ft. You might respond that this will be true if I confine my attention to a e period of 60seconds. So you have told me that if |t-t0|<60, then |h(t)-h(t0)|<1000. Suppose I want better--I want a guarantee that the balloon has not changed altitude by more than 100 ft. Well, you might tell me that now I must confine my attention to 20 seconds. You have told me that if |t-t0|<20, then |h(t)-h(t0)| < 100. Now, the above is the basic idea of continuity. In the first case, I demanded what to do of an epsilon=1000, and you told me that a delta=60 would suffice. Then I asked what I would need to get a guarantee for epsilon=100, and you told me delta=20. Note, but the way, that if delta=20 works for epsilon=100, then delta=10 also works for epsilon=100. That is, for each epsilon you have to produce a delta small enough so that changes in the independent variable of less than delta result in a change of less than epsilon in the dependent variable. For checking continuity, one need not provide an opal or largest delta--this is part of the abstraction. Of course, if you told me that for epsilon=100ft I would need a delta of 0.0000000001sec, then this might not be useful for practical purposes; a continuous function can still jump around a lot. But for the abstract idea of continuity, you must merely argue that for at an arbitrary point, for a given epsilon>0, there exists a delta (perhaps a very small delta, of no real use) such that a change in the independent variable of less than delta results in a change of the dependent variable that is *logically guaranteed* to be less than epsilon. Now back to the problem at at. We want to guarantee that max(f,g) doesn't change too quickly. We know that f and g don't change too quikly, so this is looking reasonable. Now we follow our nose and try the standard openning Let a0 be a point, and epsilon>0. [Ok, what now. Well we know that f and g are continuous. Let's write down what that means] So, for the particular epsilon listed about, there exists delta_1 such that |x-a0| 2)Let f be a function with the property that every point of > discontinuity (ie the lim (x->a) f(x) exists, but is not equal to > f(x)) is a removeable discontinuity. This implies lim (y->x)f(y) > exists for all x, but f may be discontinuous at some (even infinitely > many) numbers x. > Define g(x) = lim (y->x) f(x). Prove g is continuous. Try the one above again--if you still have problems, write back. -MIke > --I don't even know where to start with this one. > -earl- === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed? > I'm trying to write ATAN2 function for a small basic language that has > IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are > availible in the language. > I've tried a few methods I've found but the results are way off due to > low precision, rounding, etc. > A Basic language with Sin, Cos, and Tan, should also have at least an ArcTan > ? I mention that because a division result sent to an ArcTan and combined > with some simple quadrant rules should arrive at an ArcTan2. I made a type (reused my paragraph).. It should have ARCSIN() function in the body of the text. The BASIC compiler doesn't have much for intrinsic's as its for the PIC uProcessors and they don't have much memory. Does anyone have a softcopy of the ARCSIN() single precision routine from the C library? Or any other library like Fortran? Thansks, Jon === Subject: Re: Exercise in Projective Linguistics >When I posed the problem, I just assigned names A, B, C, ... to the >points and gave the lines as 5-tuples. Later, it occurred to me >I might have been able to find a different assignment of the >points into the alphabet in such a way that each line was >( a permutation of ) the letters in a 5-letter word. >So here is the challenge: how well can you do ? Can you embed >these letters back into the alphabet so that all, or most, of >these 5-tuples become English words? (Interpret word as you wish.) >(I would be willing to take a solution which works only for >an affine plane inside here, if necessary.) This might be a good one to send to rec.puzzles. I doubt that you can get all 21. The best I've been able to come up with so far is 13, with AB...U = DYCAOEBTJRMLSNKUGFWIV generating the words cadgy, fumed, blind, folky, nervy, misty, curio, clews, woman, barfs, vault, begot, gunks Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Boolean Algebra - Arithmetic Relationship > To all, > Forgive me if I sound naive, I have an appreciative but highly > abstracted perception of mathematics. My Question: > I have been told that most of mathematics (geometry, algebra, etc.) > is derived, or at least reducible, to arithmetic. Can arithmetic be > further simplified into boolean logic?? Which I understand is > equivalent or corresponds highly to propositional logic (correct me if > I'm wrong here). > Is this what Bertand was attempting to accomplish in Principia > Mathematica?? Or did he mean logic in a different sense?? Few will recommend that you read PM to find out, but you might like to read 's Introduction to Mathematical Philosophy, Dover, ISBN 0486277240. If you want to see a logical (whatever that means) foundation (whatever that means) of arithmetic, try set theory, Zermelo-Fraenkle perhaps, and Suppes' Axiomatic Set Theory, Dover, ISBN 0486616304 maybe. -- G.C. === Subject: Re: Chess/Go/etc: Continuous Game-Boards? > I was just wondering today if there has been anything interesting > as Chess, Go, etc, where the game-boards are without any subdivision? http://chessvariants.com/other.dir/continuouschess.html === Subject: Re: Factorial/Exponential Identity, Infinity > I got to thinking about sum 2^n and how it was equal to 2^(x+1) - 1. > I wonder: does it work for 3? The answer is not quite, no. Look up geometric series. If you must reinvent the wheel, you need not do in so publicly. === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed? >> I'm trying to write ATAN2 function for a small basic language that has >> IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are >> availible in the language. >> I've tried a few methods I've found but the results are way off due to >> low precision, rounding, etc. >> A Basic language with Sin, Cos, and Tan, should also have at least an ArcTan >> ? I mention that because a division result sent to an ArcTan and combined >> with some simple quadrant rules should arrive at an ArcTan2. > I made a type (reused my paragraph).. It should have ARCSIN() > function in the body of the text. The BASIC compiler doesn't have > much for intrinsic's as its for the PIC uProcessors and they don't > have much memory. > Does anyone have a softcopy of the ARCSIN() single precision routine > from the C library? Or any other library like Fortran? You didn't say whether your library includes the ARCTAN() function. If so, you can define ARCSIN(x) = ARCTAN(x/SQRT(1-x^2)). -- Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Hard (unsolved?) search problem revisited > BASIS > ----- > A two dimensional table have integers distributed this way (in this > examle, only odd numbers): > 3,11,13,21,27,... > 7,13,17,29,35,... > 9,15,19,33,41,... > Assume that this table is huge, so a fast algorithm is needed - > especially since many tables of this form must be searched many es. > So the values always increase when going from left to right, top to > bottom, no matter where you start. > That is, t[i,j] < t[i+1,j] and t[i,j] < t[i,j+1] > PROBLEM > ------- > Given such a table, t, can you locate a given integer N in polynomial > e - or - can you determine that N is not in the table in polynomial > e (or better, as in constant :-)? > If yes, what does that algorithm look like? > STATUS > ------ > I tried the following approach: start in the middle, t[i_max / 2, > j_max /2] and see if it's less than N. If so, the top/left quadrant of > the table can be eliminated because all values in that quadrant must > be smaller than N. Apply algorithm recursively to the remaining > quadrants. Of course, this is not a fast algorithm at all... > Any insight would be highly appreciated. Your quartering method is not as slow as you seem to think. For reasons similar to the analysis of a binary search of linearly ordered data, this is about the best you can expect, and is about O(log(max(R,C))), which is better than polynomial. === Subject: Dedekind Cuts I have heard that the reals can be defined by Dedekind cuts. What is this definition, exactly? === Subject: Re: a puzzle related to artinian group > cards are dealt to them. > There is no asumption >on the number of cards a player receive. In each > round, all players with 2 or more cards pass >one card to the left and > one card to the right. Prove that eventually, all players but one have >exactly one card. > I do not have the solution, so any hint would be highly appreciated. > Please >restrain from posting messages saying this is obvious, and > For fixed n this is, of course, a cellular automaton. You can allow any > positive integer states for each of the n cells. The transition rule is > such that if a cell has state x >=2 then the state never exceeds x and if > it has state x < 2 then it either stays the same or increase by 1 or 2. > If N is the maximum value of a cell state, then the maximum value of the > states in the next generation can be at most N, unless N = 2 and then a > maximum of 3 is possible in the next generation. It follows that there are > there are only a finite number of possible global states, so eventually > it is periodic. > One question might be to figure out which of Wolfram's four classes this > cellular automaton lies in. > These remarks don't answer your question, but may be of some interest. > --Edwin For what it is worth, I have just verified that the game always terminates for n <=9 players no matter how the n-1 cards are dealt. I did it using brute force with Maple. --Edwin === Subject: Re: Vedctor Calculus Question >>A single equation, such as f2(x,y,z)=c2, can describe in a 3d space >>a surface, possibly a plane, but not a line. >> I could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a line, >> last e I looked. >> >Note that z is unrestricted. Oh, I do note that, I do, I do. >It might not hurt to look again. You go first! === Subject: Re: Dedekind Cuts > I have heard that the reals can be defined by Dedekind cuts. What is > this definition, exactly? There's a crazy little thing called Google you might like to try. === Subject: Re: Numeric one-way hash function > The simplest way to ensure that the bar codes are > unique is to add a prime number to the previous > value. Lap round when you get to the top (or a prime > number near the top). Start at a weird value. > Could you be more precise, I am afraid that I > don't understand your algorithm. > Anyhow, I can't see that your algorithm would > be unbiased, or how it would prevent duplicates. Where C, Max are prime constants and C != Max Initialise N to a valid random number loop N = N + C if N > Max then N = N - Max Print N end loop Andrew Swallow === Subject: Re: JSH: About e > So it's about e, as I wait, and wonder, how many of you can handle > the truth. Pure megalomania. Has anyone written any interesting psych papers about the JSH phenom yet? I mean, there's enough material for at least a nice masters' thesis. V. === Subject: Re: 0! = 1 > Did someone know a simple demonstration of > 0! = 1 ? Proof by C: main () { if (0!=1) printf(truen); else printf(falsen); } (I wish this was original but it isn't.) . === Subject: Re: 0! = 1 >> Did someone know a simple demonstration of >> 0! = 1 ? >Proof by C: >main () > if (0!=1) printf(truen); > else printf(falsen); >(I wish this was original but it isn't.) >. There's no way to prove or demonstrate a definition. -- John E. Prussing University of Illinois at Urbana-Champaign Department of Aerospace Engineering http://www.uiuc.edu/~prussing === Subject: Re: ab... = (a*b*...)^n ? > In Maple: > dp:=proc(i, B) if i < B then i > else (i mod B)*dp((i - (i mod B))/B, B) fi:end: > B:=3:k:=4: # for example > for i from B to B^k-1 do pr:=dp(i,B):if pr>1 then > n:=round(evalf(log(i)/log(pr))):if i=pr^n then lprint(i,pr,n):fi:fi:od: You get a lot further by examining only those numbers whose prime factors are less than B. I did some of this and got a number of examples. Some of the longer ones: Base 5: 212323421441324231_5 = 1761205026816 = 1327104^2 Base 6: 411412_6 = 32768 = 32^3 Base 7: 523251_7 = 90000 = 300^2 Base 11: 118239_11 = 186624 = 432^2 Base 16: b73351_16 = 12006225 = 3465^2 Base 17: 1cce112423bf2_17 = 1019744590233600 = 31933440^2 Base 24: 122n2g_24 = 8667136 = 2944^2 Still nothing in base 10 up to 10^200. But if we allow rational exponents, there are more possibilities. Base 3: 1122221122_3 = 32768 = 64^5/2 Base 5: 224_5 = 64 = 16^3/2 Base 6: 4424_6 = 1024 = 128^10/7 Base 9: 12212_9 = 8192 = 8^13/3 24424_9 = 16384 = 256^7/4 48848_9 = 32768 = 8192^15/13 Base 10: 128_10 = 128 = 16^7/4 Base 12: 1228_12 = 2048 = 32^11/5 Base 14: 288_14 = 512 = 128^9/7 Base 15: 2585_15 = 8000 = 400^3/2 319c9_15 = 157464 = 2916^3/2 Base 18: 777_18 = 2401 = 343^4/3 Base 19: g26c_19 = 110592 = 2304^3/2 Base 22: 16c8_22 = 13824 = 576^3/2 6l1e_22 = 74088 = 1764^3/2 Base 26: 3993_26 = 59049 = 729^5/3 The example in base 3 is instructive--I believe the proposition that every power of 2 greater than 2^15 has zeroes in its base-3 representation is an open conjecture. Hoey haoyuep@aol.com === Subject: Re: Dedekind Cuts > I have heard that the reals can be defined by Dedekind cuts. What is > this definition, exactly? This is a division of the rationals into a pair of sets, which we will call Left and Right. Each element of Left is less than all elements of Right. Each element of Right is greater than all elements of Left. It follows that Left intersect Right is empty. We also require that a rational number either be in Left or Right so that Left union Right = the set of rational numbers. In the case that Right has a lest element we put that element in Left. There are two case. Left has no greatest element and Right has no least element. If so the division or cut Left,Right corresponds to an irrational number. Or Left has a greatest element and Rigt does not. This sup of Left is the number (a rational) represented by the cut. There are rules of addition, subtraction, multiplication and division defined for cuts and it is shown they form a field. For a good account of Dedikind Cuts set -A Course in Pure Mathematics- by G.H.Hardy. Bob Kolker === Subject: Re: How to calculate the total coverage area of a few circles? > about 10. [...] ... > Monte Carlo integration is a quick way of esating the > integral you're talking about doing explicitly. On my Solaris > machine here's the result of a quick run with 10 circles > and a half million points. That took 9 seconds of real e, > 6.6 seconds of CPU e, and converged to 4 decimal places. > By the way, another method that occurs to me is to actually > render the circles in some pixelated medium and then count > colored pixels. ... > Circle centers > C = > 0.1942 0.1138 > 0.0846 0.9897 > 0.9635 0.5098 > 0.4557 0.0639 > 0.6524 0.0272 > 0.0005 0.0413 > 0.3786 0.4947 > 0.0858 0.8082 > 0.5010 0.4129 > 0.3872 0.9048 > Circle radii > ans = > Columns 1 through 8 > 0.9391 0.4621 0.9122 0.2243 0.6262 0.2088 0.4072 0.7326 > Columns 9 through 10 > 0.3542 0.2420 > N A(est) clocke CPUe > 20000 4.791736 0.3326 0.2700 > 40000 4.768794 0.6383 0.5400 [...] > 500000 4.765148 9.3062 6.6200 If I correctly understand your output, you used different radii for different circles, rather than going with Leng Supeng's N circles with the same radius r formulation. Probably a minor point, except that it makes it difficult to compare your results to single-radius code. :) Anyhow, for the 10 centers you gave, if there were a common radius of 0.4, there would be only 12 critical points for the method I briefly outlined earlier. They would be: -0.3995 0.0413 -0.3154 0.9897 -0.3142 0.8082 -0.3044 0.8964 0.2328 -0.2843 0.2542 -0.2817 0.3353 1.3014 0.4830 -0.3352 0.7872 0.9048 1.0421 0.1176 1.0524 0.0272 1.3635 0.5098 which took about 1 millisecond to compute*, so a precise integration probably could be completed in about 2 ms (on a 450 MHz Athlon). Your render the circles in some pixelated medium and then count pixels might be easier to properly implement, of course, and probably only a few milliseconds slower. -jiw * http://pat7.com/jp/circles-area.c computes event points but not area at the moment === Subject: Re: lopital's rule? > could someone please explain what lopital's rule is? my teacher likes > to say things like, now, we could use lopital's rule, but that'd be > too easy and you don't know it, so naturally it piqued my interest. > sorry to ask such a broad question, but i really don't know anything > at all about it to possibly narrow the question down. It is spelled L'Hospital's rule. Use Google to look it up. Bob Kolker === Subject: Re: Numeric one-way hash function > The simplest way to ensure that the bar codes are > unique is to add a prime number to the previous > value. Lap round when you get to the top (or a prime > number near the top). Start at a weird value. > Could you be more precise, I am afraid that I > don't understand your algorithm. > Anyhow, I can't see that your algorithm would > be unbiased, or how it would prevent duplicates. > Where C, Max are prime constants and C != Max > Initialise N to a valid random number > loop > N = N + C > if N > Max then N = N - Max > Print N > end loop > Andrew Swallow p.s. If you want to analyse the algorithm for bias or duplicates just set C equal to 1. Or in a simple case setting C = 3, Max = 5 and initialise N to 2 5 3 1 4 2 You have to stop after producing Max numbers. Andrew Swallow === Subject: Re: All masses have inertia >> The way I see it guys, is that all masses have inertia! That is it >requires >> a net impulse - the product of a net force [f] and its duration [t] - >> change a body's present velocity [vi], to some other velocity [vt]; >during >> which period of e, the body is displaced a distance [s]: The ratio >> this impulse [ft], to the e rate of displacement [s/t = (vt-vi)] >that it >> causes is a constant: That is ft/(s/t) = ft/(vt-vi); which can be >written >> more concisely as f/tî/s = ftî/ (vt-vi)! Isn't _that_ inertia? >Why are you polluting this newsgroup with your topic. This is the third >thread you started with almost the same topic, isn't it? Hardly. You'd be much closer if you guessed 3333rd--and you'd still probably be significantly short. Not all in these two newsgroups, not just in the last week or so; he's been regurgitating the same vomit for six and a half years. Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: Re: Brainteaser: answer correctly to win the admiration and adulation of all. The formula is incomplete. You are lacking the name of the supermarket. Give me more information and will solve your problem. Notice, if its Wal-Mart Stuper Center, they don't give a flip. They will open anything, anywhere. > A supermarket chain has 3 outlets in the suburbs of a large town. The > locations of these outlets relative to the town centre (in miles) are as > follows: > Outlet North South East West > 1 5 5 > 2 6 3 > 3 4 3 > The supermarket chain wish to site a warehouse so that it is centrally > located relative to the positions of the outlets. Advise the supermarket > chain on the opum location of the warehouse relative to the town centre. === Subject: Re: 0! = 1 > In <1g2k59v.1onstxx1i0ew6aN%bdm@cs.anu.edu.au> bdm@cs.anu.edu.au ( >> Did someone know a simple demonstration of >> 0! = 1 ? >Proof by C: >main () >{ > if (0!=1) printf(truen); > else printf(falsen); >} >(I wish this was original but it isn't.) >. > There's no way to prove or demonstrate a definition. I think the original post was sarcastic exploring 0 != 1 vs. 0! = 1 ambiguity. === Subject: Compact Hausdorff Spaces This problem is killing me and I know I am pretty close, but I can't make that last leap. Show that if X is compact Hausdorff under both T and T ', then either T and T ' are equal or they are not comparable. Suppose without loss of generality that T ' is finer (or equal) to T. Then, pick an open set U in T '. Then, X U is closed in T '. Thus, since closed subspaces of compact spaces are compact, X U is compact. Now, since X U is compact in T ', it is also compact in T. This is where I am stuck. Where do I use Hausdorffness? Am I going about this the wrong way? Steve === Subject: Re: combination Suppose we take x numbers out of y numbers in a decreasing sequence. > Say, take 2 numbers from {1,2,3} and arrange as {2,1}. What is the > number of possible combinations ? y choose x (assuming the y numbers are distinct). C(x + y - 1, x). Which is the number of non-negative integer solutions to X_1 + X_2 + ... + X_y = x. Using your example of 2 numbers from {1, 2, 3} you want the solutions to X_1 + X_2 + X_3 = 2 where X_1 counts the number of 1's, X_2 the 2's and X_3 the 3's. There are C(3 + 2 - 1, 2) = C(4, 2) = 6 possibilities. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Ces.88ro-convergence - analysis question > Hi all, > Regular convergence implies Ces.88ro-convergence. The inverse statement is > not generally true. Can one give conditions on a sequence to guarantee > that when you have a sequence that is Ces.88ro-converging it is also true > that it is regularly converging? A word of caution: I assume by regular convergence you mean the usual notion of convergence. But you should be aware that regular convergence has a precise technical definition in summability theory. Let A = (a_{ij} : i = 1, 2, ..., j = 1, 2, ...) be an infinite matrix. A is to be REGULAR if its entries are non-negative, its row sums are 1 and, for each fixed row i, lim_j a_{ij} = 0. (Somees it's only required that the LIMIT of the row sums be 1. Same difference.) A sequence {x_n} is to be A-summable to x provided the sequence {sum_j a_{nj}x_j}, for n = 1, 2, ..., converges to x. {x_n} is to be REGULARLY summable to x if it is A-summable to x for every regular matrix A. Cesaro convergence IS a regular convergence method (the n-th row consists of 1/n (n es) followed by all zeros). A regular matrix is to be STRONGLY regular provided the row variations, sum_j |a_{ij} - a_{i,j+1}|, converge to 0 as i --> infinity. Obviously Cesaro summability is a strongly regular method, too. The most important result along these lines is that a sequence {x_n} is A-convergent to x for every strongly regular matrix A if and only if LIM_n x_n = x for every Banach limit; or, equivalently [and remarkably] provided (x_{p+1} + x_{p+2} + ... + x_{p+n})/n --> x as n --> infinity, UNIFORMLY in the shift p. This is a very famous result due to, hmm, what's-his-name, hmmm, another senior moment... [This is also called ALMOST convergence. And x_n - x_{n+1} --> 0 **is** a Tauberian condition for almost-convergence. Alas, almost convergence has nothing whatsoever to do with almost everywhere convergence.] -- === Subject: Re: Express As Single Fraction > Express the following as a single fraction: 4/3ab - 5/6bc > (m^2 + 2)/(m^2 + m) - (m - 2)/m You do it the same way you do it for fractions in arithematic. > The general formula is derived thus > a/b + r/s = as/bs + br/bs = (as + br)/bs > Yep, I understand the basic priniciple, but I just don't know how to put it > in practise with these types of fractions. Show us how you make 1/5 + 1/3 and 3/4 + 1/6 into a single fractions. Show us what you've done trying to make 4/3ab - 5/6bc (m^2 + 2)/(m^2 + m) - (m - 2)/m into a single fraction. What have you tried === Subject: Re: Express As Single Fraction alt.math deleted because it is not recognized by my newsreader. alt.algebra.help added because of all the helpful people there. >>How do I do this? >>Express the following as a single fraction: >>4/3ab - 5/6bc >> Multiply the first term by 2c/2c and the second by a/a. >I'm obviously doing it wrong but that appears to leave two different >denominators. I thought we were trying to get a common one? Multiplying the first denominator (3ab) by (2c) gives 6abc. Multiplying the second denominator (6bc) by a gives 6abc. What values did you get? This is only a little more abstract than doing common denominators with numbers. You look for the factors of both terms, and a common denominator has to contain all those factors. 3ab has a factor of 3, a and b. 6a has a factor of 2, a factor of 3, and factor of a. For the numeric part, 6 will serve as common denominator for both (since it's a multiple of 3). So you just need to include all the other factors: a, b and c. Hence: 6abc. You can always multiply the terms together to get *A* common denominator, just not the lowest one. (3ab)(6bc) = 18abc is a common denominator that you can use for both fractions. === Subject: Re: How to calculate the total coverage area of a few circles? >If I correctly understand your output, you used different radii >for different circles, rather than going with Leng Supeng's >N circles with the same radius r formulation. Probably a >minor point, except that it makes it difficult to compare your >results to single-radius code. :) Ah, yes, I must have missed that in the problem statement. Also my centers and radii were drawn from uniform distributions with no attempt to guarantee overlap. Some of the circles may well be disjoint. I just wanted to experiment with some quicky code to see how well it did. >Anyhow, for the 10 centers you gave, if there were a common >radius of 0.4, there would be only 12 critical points for the >method I briefly outlined earlier. They would be: > -0.3995 0.0413 > -0.3154 0.9897 > -0.3142 0.8082 > -0.3044 0.8964 > 0.2328 -0.2843 > 0.2542 -0.2817 > 0.3353 1.3014 > 0.4830 -0.3352 > 0.7872 0.9048 > 1.0421 0.1176 > 1.0524 0.0272 > 1.3635 0.5098 >which took about 1 millisecond to compute*, so a precise >integration probably could be completed in about 2 ms (on >a 450 MHz Athlon). I have an Ultra something Sparc station (Ultra 60?) I think your 2 milliseconds beats my 6600 milliseconds. So I guess your algorithm is a wee bit more efficient. To the OP: Um... what he . Ignore my approach. >Your render the circles in some pixelated medium and then >count pixels might be easier to properly implement, of course, >and probably only a few milliseconds slower. Yeah, I was thinking that could be horrendously slow in C code, but extremely fast with assembly code reading graphics memory directly. I remember doing very fast pixel-level stuff centuries ago in EGA mode on 10 MHz machines carved from stone and powered by gophers. === Subject: Re: Vedctor Calculus Question >>A single equation, such as f2(x,y,z)=c2, can describe in a 3d space >>a surface, possibly a plane, but not a line. >> I could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a line, >> last e I looked. >> >Note that z is unrestricted. It might not hurt to look again. (psst: the points in that set have the property x=0, y=0, z any real. That's a line. Think about it.) === Subject: Re: a baseball odds calculation problem >here's the problem, specifically: there are three divisions in the american league. every season, each one > of >these divisions will be clinched by one team on a certain date. this is > a >mathematical certainty. past baseball history tells us that dates before september are extremely >rare for teams clinching. in fact, the later days of september, roughly > the >15th through the 28th, are where the likelihood of teams clinching is the >greatest. > You need to specify the probability distribution of the date of > clinching for each division. For example, assuming the the three > divisions are independent and have the same distribution, completely > supported (a model of your approximation) by {15, 16,..., 28}, then the > probability is sum(k=15..28, p(k)^3), where p(k) is the probability of > clinching on the date k. So if you are saying that each division is > equally likely to be clinched on each day and is independ of the others, > the probability is 14*(1/14)^3 = 1/196 = .0051 (0.51%), approximately. > Or if you assume identical triangular distributions, p(k) = > min(k-14,29-k)/56, the probability (still assuming independence) is > 2/56^3 sum(k=1..7, k^3) = 1/112 = .0089 (0.89%), approximately. Another model would be a discretized normal over the whole season. > Perhaps you want to make an empirical esate of the distribution; you > may then do the calculation yourself. full > baseball seasons. in those seasons, there have been 126 different clinches > (4 clinches per year from 1975-1993 excluding 1981 strike year and then 6 divisions) > and i was able to find 96 of those dates (they're not very easy to find). i > did an esate based on this here's the distribution of clinches i found: > [...data omitted...] > It seems you are not satisfied with the quality of the answers you got so far > and want a more accurate answer. i'm fine with it, others over in teh baseball ng i referenced are pointing out that this isn't a completely random and independent probability problem since the probability is weighted heavily toward the end of the season. they're hoping this gets taken into account in the calculation. But I suspect you collected the wrong data. > It would be far more informative to know how many days from the end > of the season the clinch occurred. That is because the last (scheduled) day > of a season is a Sunday, whose date obviously varies from year to year. this is very true. also worth noting that baseball used to schedule games well into october andhasn't done so for a couple years now, which makes the chance of clinching on october 3 these days almost nil whereas in years past obviously there was a good chance of clinching on that date. instead of calling it month/day X ishould be referring to it as day Y which gives us a much more accurate reading from season to season. === Subject: Re: Irrationality of the sqrt(2) by unique factorization >I studying a different way of proving that the sqrt(2) is irrational, or n >th roots in general. It proceeds as follows: >Assume the sqrt(2) is rational. Thus, m/n = sqrt(2). Which then implies >2^1 * n^2 = 2^0 * m^2 >Now, by unique factorization, a perfect square will have all even exponents >in its prime factorization. Since, the LHS of the above equation has an odd >exponent, namely 2^1, can I then conclude that the sqrt(2) is irrational? > Yes. >It seems to me that some step is missing? Shouldn't there be something >about the RHS being even and the LHS being odd ? > Consider the factors of 2 that appear in either n or m or both; there > is an even number of factors of 2 in n^2, and an even number of > factors of 2 in m^2. So the RHS has, in total, an even number of > factors of 2 (0 + the number of factors in m^2), while the LHS has an > odd number of factors of 2 (1 + the number of factors in 2). > This is impossible. > This would be the long-winded explanation. What you have noticed is > simply that all the primes on the right hand side appear raised to an > even exponent, while all but ONE of the primes on the left hand side > appear raised to an even exponent, the exception being raised to an > odd exponent. That's impossible by unique factorization. > How does one arrive at the >conclusion that if a square isn't perfect, then it is irrational? > No such conclusion is reached. Or do you mean, that an integer which > is not a perfect square has an irrational square root? > I >understand intuitively that if a sqrt() is not perfect, then it is >irrational, but I don't see how one can logically conclude that from the >above equation. > The above equation only proves it for 2. A similar argument yields the > result for sqrt(p) for any prime p; then it is not hard to prove it by > a similar argument for any product of the form p_1*...*p_r where > p_1,...,p_r are distinct primes. > Then simply note that positive integer greater than 1 can be written > uniquely as n=k*s, where s is a perfect square, and k is a product of > distinct primes. So then sqrt(n) = sqrt(k)*sqrt(s), sqrt(s) is an > integer, and sqrt(k) is irrational. > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) perfect sense now. Lurch === Subject: Mathematics I need to show that the edge connectivity (the lease number of edges if removed will disconnect the graph) of a complete graph is equal to n-1 Here is my proof by induction: Let G be a complete graph on n vertices, n>1 Let n = 2 then e(G)= 1 = n - 1 is true. Let n = 3 then e(G) = 2 = n - 1 is true. Let n = 4 then e(G) = 3 = n - 1 is true. Assume e(G) = n - 1 is true for every n, G is a complete graph on n vertices. We need to show that e(G) = n when G is on n+1 vertices. Since e(G) = n .9a1, then for a graph on n + 1 vertices, then e(G) = (n+1) -1 = n is true since n equals to the number of vertices of G minus 1 . Thus e(G) = n - 1 for every n when G is a complete graph. Is it right and complete or missing something? Please let me know. === Subject: Re: Compact Hausdorff Spaces > This problem is killing me and I know I am pretty close, but I can't make > that last leap. > Show that if X is compact Hausdorff under both T and T ', then either T and > T ' are equal or they are not comparable. > Suppose without loss of generality that T ' is finer (or equal) to T. Then, > pick an open set U in T '. Then, X U is closed in T '. Thus, since > closed subspaces of compact spaces are compact, X U is compact. Now, > since X U is compact in T ', it is also compact in T. This is where I am > stuck. Where do I use Hausdorffness? Am I going about this the wrong way? Consider the identity map id: (X,T') --> (X,T). It's continuous (because T' is finer than T), 1-1 and onto. Because X is compact Hausdorff, id is therefore a homeomorphism. And thus T = T'. When you recall the proof that id is a homemorphism, i.e. that id^{-1} is continuous, it runs as follows: let K be closed relative to T'. Since (X,T') is compact, therefore K is compact relative to T'. A fortiori, K is compact relative to T. Since (X,T) is Hausdorff, therefore K is closed relative to T. In other words, K closed relative to T' implies K is closed relative to T. In other words, T' subset T. -- === Subject: Re: Express As Single Fraction >>How do I do this? >>Express the following as a single fraction: >>4/3ab - 5/6bc >> Multiply the first term by 2c/2c and the second by a/a. >I'm obviously doing it wrong but that appears to leave two different >denominators. I thought we were trying to get a common one? What did you get after performing the multiplication? <> === Subject: Zeroes, extrema, inflection points, ..., etc? If you consider the graphs of y=f(x), y=f'(x), and y=f''(x), the values of y are called the value, the slope, the concavity, respectively, and the points at which y=0 are called the zeroes, the vertices, and the inflection points. My question is, does f'''(x) have a name? If so, what is it? Do any other f'(n)'(x) have names? Note: f'(4)'=f'''' and f'(n)' is f with n primes. === Subject: Re: All masses have inertia In sci.math, Bob Pease : >> You've got to learn; like it or not: > Plonk Him.. > I'm even wasting bandwith with this advice. > Bob PEase I'll admit to some surprise as to why he comes over here in sci.math. He used to hang around sci.physics, but we kinda drove him off. :-) As it is, what he's apparently proposing is nothing horribly new; the main problem with SI as he sees it is that the newton != the kg, when the kg is placed in a 1-g gravity field. However, the pound-force == pound-mass in 1-g gravity field, numerically speaking. (Personally, I think that leads to confusion. Besides, in SI g = 9.805 m/s/s anyway, which is almost 10 -- good enough for back of the napkin calcs.) There are a number of problems with this approach, not the least of which is the variance of g (around 0.4%) as one wanders about the globe. And he has a magic number: the number of pounds in a gallon of water. SI has an artifact (a 1 kg iridium block somewhere in France, IIRC) and a few magic numbers relating to e and length using the speed of light. Of the two, I think the artifact is a little easier to handle. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: JSH your ship has come in!!!! >So far the team as I call it are people who recognize that my work >is indeed correct, and so far seem to only be very high IQ people. Hey James, did I ever tell you about how the high IQ people fared when I challenged them in their annual Quiz Bowl? Undefeated in a double elimination tournament, culminating in two consectutive victories over the reigning champions! High IQ my ass. -- Mensanator Slayer of the Mensa === Subject: Re: Compact Hausdorff Spaces > Show that if X is compact Hausdorff under both T and T ', then either T and > T ' are equal or they are not comparable. > Suppose without loss of generality that T ' is finer (or equal) to T. Then, > pick an open set U in T '. Then, X U is closed in T '. Thus, since > closed subspaces of compact spaces are compact, X U is compact. Now, > since X U is compact in T ', it is also compact in T. This is where I am > stuck. Where do I use Hausdorffness? Am I going about this the wrong way? Compact sets of Hausdorff spaces are closed, that's where. === Subject: Re: lopital's rule? following: >> Guy Corrigall scribbled the following: >> L'Hospital's Rule (pronounced lopital in French) can be found in any book >> on calculus of a single variable (look it up!). >> L'Hospital or L'H.99pital. The circumflex (the ^ thingy) means that the >> following s is understood implicitly. > Is the s pronounced? No. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ 'I' is the most beautiful word in the world. - John Nordberg === Subject: Re: Algebra proof Let a, b, r and s be integers with r>1, s>1 and d(r,s)=1. Prove >> that if a=b(mod r) and a=b(mod s) then a=b(mod rs). How do I do this???? a = b + m*r for some m, and a = b + n*s for some n, right? That means >> that m*r = n*s. Now use the fact that d(r,s) = 1 to conclude something >> about m (or n). I guess since r and s are relatively prime, then r must divide n. so > r*x=n. and a=b+r*x*s, so a=b(mod rs), right? > Exactly. Is there a name to this theorem? That is, the theorem that says that if d(a,b)=1, and a*c=b*d, then a must divide d and b must divide c? === Subject: Re: a puzzle related to artinian group > Suppose n people sit around a table and n-1 cards are dealt to them. > There is no asumption on the number of cards a player receive. In each > round, all players with 2 or more cards pass one card to the left and > one card to the right. Prove that eventually, all players but one have > exactly one card. R. Anderson, L. Lovasz, P. Shor, J. Spencer, E. Tardos, S. Winograd, ``Disks, balls, and walls: analysis of a combinatorial game'', Amer. Math. Monthly, 6, 96, pp. 481--493, 1989. It talks about the same problem (of course along a line, not around a table). Their discussion is thrilling. Best, === Subject: Re: lopital's rule? > so, in our example above, f(x) is (3x - sin x), g(x) is x and a is zero > the limit of (3x - sin x)/x as x approaches zero is, by the Rule: > (3 - cos x)/1 as x approaches zero, which = 2 This is the kind of example that gets some of us upset over the use (or overuse, or misuse) of l'Hopital. To use l'Hopital for this one, you must know the derivative of sin x. To compute the derivative of sin x (from the Newton quotient), you must know the limit, as x goes to zero, of (sin x) / x. Thus the use of l'Hopital on this circular function involves circular reasoning. -- === Subject: Re: =?ISO-8859-1?Q?Ces=E0ro-convergence_-_analysis_question?= > for infinitely many N. Suppose N satisfies (iii'). Now for > n > N we have > s_n >= d - c sum_N^n 1/j > ~ d - c log(n/N) > >= d/2 David, I have a question about the above step you take in your proof. I don't understand why d - c log(n/N) >= d/2 log(n/N) goes to infinity as n goes to infinity if N stays constant. ~ Chris === Subject: Re: Joe Uptaught (Was Re: David Ullrich on Identity) A propos, here are cites from Pierre Bourdieu's > _Language & Symbolic Power_ (the titles are mine). > Enjoy! Censorship There Oughta Be A Law Cool-Hand Luke > The Social Conditions for the Effectiveness of Ritual Discourse Heretical Discourse The *Skeptron* Symbolic Power & the Symbolism of Power There is no symbolic power without the symbolism of power. Symbolic attributes--as is well illustrated in the paradigmatic case of the *skeptron* and the sanctions against the improper wearing of uniforms--are a public display and thereby an officialization of the contract of delegation: the ermine and the robe declare that the judge or the doctor is recognized as having just cause (in the collective recognition) for declaring himself judge or doctor, that his imposture--in the sense of the pretension expressed by his appearance--is legiate. The comnce that is specifically linguistic--the Latin once spoken by doctors or the eloquence of the spokesperson--is also one of the manifestations of comnce in the sense of right to speech and to power through speech. There is a whole dimension of authorized language, its rhetoric, syntax, vocabulary and even pronunciation, which exists purely to underline the authority of its author and the trust he demands. In this respect, style is an element of the *mechanism*, in the Pascalian sense, through which language aims to produce and impose the representation of its own importance and thereby help to ensure its own credibility.[14] The symbolic efficacy of the discourse of authority always depends, in part, on the linguistic comnce of the person who utters it. This is more true, of course, when the authority of the speaker is less clearly institutionalized. It follows that the exercise of symbolic power is accompanied by work on the *form* of discourse which, as is clearly demonstrated in the discourse of poets in archaic societies, has the purpose of demonstrating the orator's mastery and gaining him the recognition of the group. (This logic is also found in the popular rhetoric of insults which seeks, by flagrant overstatement and the regulated deformation of ritual formulas, to produce the expressive accomplishment which allows one to 'get those laughing on one's side'. Notes 14. The two senses of comnce come together if one sees that, according to Percy Ernst Schramm, just as the crown of the medieval king designates both the thing itself and the set of rights which constitute royal dignity (as in the term 'crown property'), so too linguistic comnce is a symbolic attribute of the authority which *designates* a socially recognized status as a set of rights, beginning with the right to speak and the corresponding technical capacity. (Pierre Bourdieu, _Language and Symbolic Power_ pp. 75-76) === Subject: Re: help !!! === Subject: help !!! >I need help giving (p->q) xor (p<->q) an equivalence less than 12 Prove the following are a sequence of equivalent statements. (p->q) xor (p<->q) (p->q)&~(p<->q) or ~(p->q)&(p<->q) (p->q) & ~(p<->q) (p->q) & (~(p->q) or ~(q->p)) (p->q) & ~(q->p) (p->q) & q & ~p q & ~p If instead you just verify that (p->q) xor (p<->q) and q & ~p have the same truth tables, you've learned nothing. ---- === Subject: Re: Mathematics === Subject: Mathematics Graph theory would have been a more welcomed and informative title. >I need to show that the edge connectivity (the lease number of edges >if removed will disconnect the graph) of a complete graph is equal >to n-1 Pick any vertex V from K_n the complete graph on n vertices. V has n-1 edges coming from it. Removal of these disconnects V from K_nV. Thus the edge connectivity is at most n-1. Now remove n-2 edges from K_n. Call that graph G. Case 1: some vertex V has no edges removed G is connected as every vertex in G has an edge going to V Case 2: every vertex has an edge removed pick any vertex V; V has degree at least one the remaining vertices are K_(n-1) with at most n-3 edges removed by induction hypothesis, the edge connectivity of K_(n-1) is n-2 thus the remaining vertices are connected and V connects to them >Here is my proof by induction: >Let G be a complete graph on n vertices, n>1 >Let n = 2 then e(G)= 1 = n - 1 is true. >Let n = 3 then e(G) = 2 = n - 1 is true. >Let n = 4 then e(G) = 3 = n - 1 is true. >Assume e(G) = n - 1 is true for every n, G is a complete graph on n >vertices. We need to show that e(G) = n when G is on n+1 vertices. >Since e(G) = n^1, then for a graph on n + 1 vertices, then What's that strange stuff for e(G) = ???. >e(G) = (n+1) -1 = n is true since n equals to the number of vertices >of G minus 1 . >Thus e(G) = n - 1 for every n when G is a complete graph. >Is it right and complete or missing something? Please let me know. A badly written circular argument. To clarify terminology, consider K_(n+1) and for some subgraph K_n, apply e(K_n) = n-1. ---- === Subject: Re: Nice summation puzzle === Subject: Nice summation puzzle >Starting from arcsin(x)=x+1/2/3*x^3+... What's 1/2/3 ? The next term is 1/2/3/4/5 x^5 ? >and using an integration operation show well-known sum: >oo 1 >S ------------------- = (pi^2)/6 >n=1 n^2 integral(0,1) arcsin x dx = x arcsin x + sqr(1-x^2) |_0^1 = arcsin 1 - 1 = pi/2 - 1 /= (pi^2)/6 ---- === Subject: PV Integration Help! I'm having problems visualising integration of pressure with respect to volume.... If I integrate velocity wrt e I get distance. Practically this means that I sum the measured velocity over a monotonically increasing varible (e), this I can see. But with pressure volume loops the data is cyclic and if I integrate delta Pressure wrt Volume what do I get?? I cant see what I get from this process or how I go about doing the math in Matlab.... Mark. === Subject: Re: a puzzle related to artinian group > Suppose n people sit around a table and n-1 cards are dealt to them. > There is no asumption on the number of cards a player receive. In each > round, all players with 2 or more cards pass one card to the left and > one card to the right. Prove that eventually, all players but one have > exactly one card. Construct a function f that has as its domain the state of the table at any given round and has as its range the nonnegative integers. Define f as the sum of the number of cards the each of the n players have BEYOND 1. I will give a brief example in case my definition of f is unclear. Let n = 3, so there are 2 cards. If the first and second players each have a card, then f = 0, since no player has any more than 1 card. However, if the first player has both cards, then f = 1 since the first player has 1 more than 1 card. Similarly, if n = 5, and both the first and the second players have 2 cards, then f = 2, since f is a sum. Notice that f cannot increase. Each of the n people can have at most n-1 cards at any e. This is finite, so eventually, the state of the table will become periodic. We will now only consider the states after the states become periodic. We know that f is invariant. Find any player who has no cards. If he were to ever get a card, then f would decrease. So we know that if a player has no cards, he will never get any cards. That means that any player next to a player who has no cards will never get more than 1 card. Doing so would force him to, on the next turn, give a card to his neighbor who has no cards. That means that any player next to a player who is next to a player who has no cards will never get more than 1 card. Doing so would force him to, on the next turn, give a card to his neighbor who has a neighbor has no cards. I'm sure you can see where I'm going. Any player who has a path of neighbors to someone who has no cards cannot have more than 1 card. It follows that the only periodic behavior shown can be of n-1 people each having one card. ~ Chris === Subject: Pluecker coordinates of symmetric powers Let V be a finite dimensional vector space with a given ordered basis e_1, ..., e_n . Let W be a p-dimensional linear subspace. If x_1, ..., x_p is a basis of W, express the x_i as linear combinations in the e_j. The coefficients give a p x n-matrix. The maximal minors of this matrix are known as the *Pluecker coordinates* of the subspace W . Up to a scalar multiple they are independent of the basis x_1, ..., x_p and, considered as homogeneous coordinates, characterize W uniquely. For any k, consider the k-th symmetric power S^kV of V. The ordered basis e_1, ..., e_n determines an ordered basis of S^kV, and the subspace W of V determines a subspace S^kW of S^kV. So the Pluecker coordinates of S^kW as a subspace of S^kV are determined by the Pluecker coordinates of W in V and, in fact, should be polynomials Boudewijn === Subject: Re: Vedctor Calculus Question > >A single equation, such as f2(x,y,z)=c2, can describe in a 3d space >a surface, possibly a plane, but not a line. I could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a line, > last e I looked. > Note that z is unrestricted. It might not hurt to look again. My bad My incredible, incredible, bad === Subject: Re: lopital's rule? > Why's that then? It seems to me to be a mildly interesting theorem. I > can see that it might not be a good idea to prove it and then give > students lots of 0/0 limit problems for them to solve using > L'Hospital's when doing so is just pedagogically useless manipulation. > But that is no reason for not proving the theorem. Who anything about proving anything. My experience is that it is taught as a method, not as a theorem, and when taught it, students will use it *exclusively* for all limit problems thereafter (it's a formula isn't it) no matter how inappropriate it is for the problem at hand. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 es) === Subject: Re: a puzzle related to artinian group Nevermind. My proof is horridly flawed. === Subject: Re: a puzzle related to artinian group > Suppose n people sit around a table and n-1 cards are dealt to them. > There is no asumption on the number of cards a player receive. In each > round, all players with 2 or more cards pass one card to the left and > one card to the right. Prove that eventually, all players but one have > exactly one card. > Construct a function f that has as its domain the state of the table > at any given round and has as its range the nonnegative integers. > Define f as the sum of the number of cards the each of the n players > have BEYOND 1. [snipped example] > Notice that f cannot increase. Why is that: ....121.... -> ....202...., so f may increase by one??? > Each of the n people can have at most n-1 cards at any e. This is > finite, so eventually, the state of the table will become periodic. > We will now only consider the states after the states become periodic. This is true and applies without the rectriction that the number of cards is limited to precisely n-1. > We know that f is invariant. If it were non-increasing, we could deduce that. However, see above. > Find any player who has no cards. If he were to ever get a card, then > f would decrease. So we know that if a player has no cards, he will > never get any cards. Again, the situation is more subtle than that. Somehow you MUST use the fact that the total number of cards is n-1. Otherwise you can have, e.g. the cycle 02020202...02 (even number of players) alternating with 202020...20. Notice that the function f is constant here. > That means that any player next to a player who has no cards will > never get more than 1 card. Doing so would force him to, on the next > turn, give a card to his neighbor who has no cards. [snipped a natural recursive reasoning] I was thinking along the similar lines, so IMHO your scheme for a proof is a very natural one. In other words, it is natural to study the periodic situations. However, the devil is in the detail. Back to the drawing board! Jyrki Lahtonen, Turku, Finland === Subject: Re: absolute moments of the normal distribution > Hi to everyone!! > Some of you know where I can find the formula of the k-th absolute moment > for the normal distribution? I presume you mean a_k = integral_{-infinity}^infinity (2pi)^{-1/2} |x|^k exp(-x^2/2) dx. This is 2(2pi)^(-1/2) integral_0^infinity x^k exp(-x^2/2) dx. We convert this into a gamma integral by making the substitution y = x^2/2. We get dy = x dx and so a_k = 2(2pi)^(-1/2) integal_0^infinity (2y)^{(k-1)/2} exp(-y) dy = 2^{k/2} pi^(-1/2) Gamma((k+1)/2). To simplify this further, recall that for integers r >= 0, Gamma(r+1) = r! and Gamma(r + 1/2) = pi^(1/2)2^{-2r}(2r)!/r!. Hence for even k = 2r, a_k = 2^{-r}(2r)!/r! and for odd k = 2r+1 a_k = 2^{r + 1/2} pi^{-1/2} r!. > My own calculations are different from the Mathematica AAAARRRGGHHHH! > output and I am not > sure about what to do... -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 es) === Subject: Question on Hilbert & Godel What did Hilbert ask and claim concerning Foundations of Mathematics (sets, predicate calculus), metamathematics, Logic, Incompleteness, etc? Something about the Continuum Hypothesis? Completeness or Incompleteness of Logic? Was he contradicted by kindergarten logic from Godel? How would we exactly formally represent Hilbert's questions and claims using current day notation? Formal or programmed derivations? Charlie Volkstorf Cambridge, MA === Subject: Re: factoring to satisfiability > (10-year-old) conjecture: factoring -> SAT will eventually > give greater magnification > of the exact structure of the problem that will be invisible > to straight factoring algorithms, ***if studied carefully***, > leading to asymptotically equivalent or superior algorithms. > that seems to be exactly the same motivation/inspiration for Below is a 3x2 bit multiplication circuit. I can use this circuit to make assertions about the factors of any number. I'm curious if this has been studied. 3x2 Input ECA * DB = ZYXWV V = A1B1 W = A1B0D1 + A1C0D1 + A0B1C1 + B1C1D0 X = A0C1D1E0 + B0C1D1 + A1B1E1 + B1C0E1 + B1D0E1 Y = A1B1C1D1E0 + B0D1E1 + C0D1E1 Z = B1C1D1E1 Using the circuit above I can make the following assertions: V1 -> A1B1 W1V1 -> A1B1C0D1 + A1B1C1D0 X1W1V1 -> A1B1C0D1E0F1 + A1B1C0D1E1F0 + A1B1C1D0E0F1 + A1B1C1D0E1F0 (this last assertion is based on a 3x3 bit multiplier.) Let z = x * y The first assertion shows that if the low order bit of z is 1 then both x and y must have 1 for their low order bit. This is fairly obvious. If z is odd then both x and y must be odd. But, the multiplication circuit allows me to extend these assertions. If the low order bit of z is 1 then both x and y must have 1 as their low order bit. z = ?1 -> (?1 * ?1) If the bottom two bits of z are 11 then one factor must end in 01 and the other factor must end in 11. z = ?11 -> (?01 * ?11) or (?11 * ?01) If the last three bits of z are 111 then there are only two possible combinations for the last three bits of x and y: z = ?111 -> (?111 * ?001) or (?011 * ?101) or (?101 * ?011) or (?001 * ?111) These assertions can be extended with larger multiplication circuits. I am also curious if assertions like these can be used to show that prime numbers must end with certain bit patterns. - Zeno was right. Motion is impossible. === Subject: Re: Deep Thoughts # 1: A new limitation to the human mind > 1. Mathematics is the science in which we make something out of > nothing. Wrong. Mathematics is built on the 13 Axioms, which are not nothing. > And where did those 13 axioms come from? What is needed to develop > them? They are definitions. That is however irrelevant. Mathematics is not the the science in which we make something out of nothing. Are you trying to imply that because 'something' requires 'nothing' to be created, it is 'nothing'? By that logic, everything would ulately come from 'nothing'. I guess you got your terms mixed up. > 2. All of man-made Mathematics consists of an abstraction from > physical processes. Wrong again. Mathematics consists of theorems derivated by following a > set of formal rules. > And when is this not just an abstraction from physical processes > (example, please)? Group theory? Banach's Fixpoint theorem? most of mathematics? > 3. Since Mathematics in general needs nothing to be created, then the > human mind seems to be limited in that it can only consider > possibilities that are analogous to physical processes. Well, I think first of all mathematics exist all by themselves, as a > logical system. > And so needs nothing to be created. > However one may argue that for humans this exists in > its precise own way, because the human mind works the way it works. > That's right. Exactly. And in particular, the human mind seems to > only discover the mathematics that is an abstraction from physical > processes. Sorry no. You are building on false premises. > Thus mathematics actually may help delineate the limits of the human > mind. > That is exactly what I am postulating. No, what you postulated is the human mind seems to be limited in that it can only consider possibilities that are analogous to physical processes, which I think is not only wrong because you derive it from false premises but wrong in it own terms. The human mind can consider the possibility of, lets say contradiction, or poetic justice which are not analogous to a physical processes. You will agree with me that I did not state in which way the human mind is limited. === Subject: Re: Quantum Gravity ........ ...However, as there is a never a two without the third. Quantum Gravity, definitely, would be as should be the third after a Relativity and Quantum Mechanics. Whether, the later has been and remains, the guide as a controler of the first, the third would be as should be formulated Mathematically as a guiding as a controler of the both. Therefore, in that moment, a specific as a crucial event would be as should be, as it is, that it would appear a fastidious transition allowing the three turning over a closed a Mathematical circle. Therefore, over all that matter, a Space-e would definetely apply, definitely as a matter a fact!!!!!!!!!!!........... ... -- Ahmed Ouahi, Architect Jack Sarfatti kirjoitti viestiss.8a > Part 2 > Comments by Jack Sarfatti on excerpts from: > Excerpts are from: > The three perspectives on the quantum-gravity problem > and their implications for the fate of Lorentz symmetry1 > Dipart. Fisica Univ. La Sapienza and Sez. Roma1 INFN > P.le Moro 2, I-00185 Roma, Italy > .8bthere is still not a single measured number whose interpretation > requires advocating ëquantum gravity.89. > .8bThe third possibility is a condensed-matter perspective (see, e.g., the > research programs > of Refs. [13] and [14]) on the quantum-gravity problem, in which some of the > familiar properties of spacee are only emergent. Condensed-matter > theorists are > used to describe some of the degrees of freedom that are measured in the > laboratory as > collective excitations within a theoretical framework whose primary > description is given > in terms of much different, and often practically unaccessible, > fundamental degrees of > freedom. Close to a critical point some symmetries arise for the > collective-excitations > theory, which do not carry the significance of fundamental symmetries, > and are in fact > lost as soon as the theory is probed somewhat away from the critical > point. Notably, > some familiar systems are known to exhibit special-relativistic > invariance in certain > limits, even though, at a more fundamental level, they are described in > terms of a nonrelativistic theory. > For a rather general class of fermionic systems one finds [13] that at > low energies, as a Fermi point > is approached, fermions gradually become chiral Weyl > fermions, while bosonic collective modes of the vacuum transform into > gauge fields and > gravity. > Clearly from the (relatively new) condensed-matter perspective on the > quantum gravity > problem it is natural to see the familiar classical continuous Lorentz > symmetry > only as an approximate (emergent) symmetry. > [Revised Comment #3: This notion is very compatible with the > Bohm-Hiley-Vigier pilot BIT wave landscape IT hidden variable system > point flow on the landscape picture of quantum theory. Orthodox > micro-quantum theory is the limit of An Valentini's sub-quanta heat > death with signal locality in which the quantum potential is > fragile (Bohm-Hiley) with a pilot BIT wave (nonlocal in configuration > space for entangled systems) that has no sources (Bohm-Hiley). This > means that the IT hidden variable system point flowing on the BIT > orders (J.A. Wheeler) from BIT but not vice versa. This is action > without reaction! Here reaction means back-action where the BIT > pilot wave landscape has sources so that IT is no longer a passive test > landscape it is flowing on according to > vs = (h/m)Grad(Phase) - (q/c)A > or in the 4D elastic world crystal lattice (Hagen Kleinert) model > distortion field du(x) > we have the IT FROM BIT constraint equation showing how the IT > geometrodynamic field gets its .8bmarching orders.8a (Wheeler.89s term in a > different but related context) from the MACRO-QUANTUM BIT vacuum > inflation field. > du(x) = Lp*^2 (Goldstone Phase),u - TaA^a,u > Lp*^2 = hG*/c^3 = Lp^4/3(c/Ho)^2/3 in t'Hooft-Susskind world hologram > model for renormalized Planck scale Lp* ~ 1 fermi determined by > Newtonian Planck scale Lp = 10^-33 cm and size 10^28 cm of our > contingent local Level I (Max Tegmark) parallel IT universe in an > infinity of parallel IT universes on a single 3D spatially flat > American). > Ta are the generators of the Lie algebra of the standard > model in globally flat Minkowski spacee with fiber connections A^a,u > for parallel transport. a is internal space and u is spacee. This is > normalized such that TaA^a,u is a length. ,u is the ordinary partial > derivative operator relative to x^u, u = 0, 1,2,3. > The attraction between virtual electrons and positrons near the -mc^2 > band in the Dirac vacuum spectrum Fermi sphere is a pair instability > that creates the virtual giant local vacuum wave BEC <0|e+(x)e-(x)|0> a > complex scalar inflation field whose phase is the Goldstone phase and > whose amplitude is the Higgs field > <0|e+(x)e-(x)|0> = |Higgs field(x)|e^i(Goldstone phase(x)) = PSI(vac) > such that the unified dark energy/matter zero point stress-energy > density Diff(4) tensor field is : > tuv(x)zpf = (c^4/8piG*)/zpf(x)guv(x) > where > /zpf(x) = Lp*^-2[1 - Lp*^3|Higgs field(x)|^2] > /zpf(x) > 0 is strongly scale-dependent anti-gravitating dark energy > exotic vacuum with negative pressure and positive zero point energy density. > /zpf(x) < 0 is strongly scale-dependent gravitating dark matter > exotic vacuum with positive pressure and negative zero point energy density. > All quantum fields of all spins contribute to the net residual /zpf(x) > LOCAL FIELD in the emergent More is different collective mode c-number > background spacee with both Diff(4) base space and local Lorentz > group tangent space emergent symmetries at least near a critical point > of the renormalization group flow sense. > Einstein's gravity c-number background field is > guv(x) = nuv(Minkowski) + (1/2)[du(x),v + dv(x),u] > with anholonomic torsion tensor field (string Goldstone phase singularities) > suv(x) = (1/2)[du(x),v - dv(x),u] > and the back-action nonlinear LOCAL virtual BEC Landau-Ginzburg Diff(4) > symmetric BIT FROM IT .8bback-action.8a equation: > {D^uDu + V(|<0|e+(x)e-(x)|0>|)}<0|e+(x)e-(x)|0> = 0 > Where V(|<0|e+(x)e-(x)|0>|) limits to the effective spontaneous broken > vacuum symmetry potential of chaotic inflation cosmology in the > large-scale limit of the isotropic homogeneous FRW metric with zero > point energy density induced cosmological constant for .8bdark energy.8a > using an adaptive windowed wavelet transform version of the Wigner phase > space density with ODLRO rather than a rigid windowed Fourier transform. > Shorter scales have longer bandwidth and longer scales have smaller > bandwidth with total area in phase space per conjugate pairs of > incompatible dynamical variables constant ~ h (in phase space) or Lp*^2 > (in spacelike slice of spacee) to preserve Heisenberg.89s uncertainty > principle and the Bekenstein-t.89Hooft-Susskind .8bworld hologram.8a > reinterpretation of generalized .8bblack hole thermodynamics.8a. See also > the somewhat related 3D spacelike quantized area operators in Penrose > spin networks of the loop quantum gravity approach. > Note that > Lp^2 = hG(Newton)/c^3 > Maps the discrete structure of quantum gravity foam in phase space to a > discrete .8barea.8a structure in spacelike slices of spacee consistent > with the world hologram Ansatz. > Du is the Diff(4) operator so that D^uDu is the GR wave propagation > operator on a complex numbered scalar field PSI(vac).] > Results obtained over the last few years (which are partly reviewed > later in these notes) allow us to formulate a similar expectation from > the general-relativity perspective. Loop quantum gravity and other > discretized-spacee quantum-gravity approaches appear to require some > departures, governed by the Planck scale, from the familiar (continuous) > Lorentz symmetry. And in the study of noncommutative spacees some > Planck-scale departures from Lorentz symmetry might be inevitable, since > (at least in a large majority of noncommutative spacees) a Lie > algebra is not even the appropriate language for the description of the > symmetries of a noncommutative spacee (one must resort to the richer > structure of Hopf algebras). > reason to renounce > to exact Lorentz symmetry. Minkowski classical spacee is an > admissible background > spacee, and in classical Minkowski there cannot be any a priori > obstruction for > classical Lorentz symmetry. Still, a breakup of Lorentz symmetry, in the > sense of > spontaneous symmetry breaking, is of course possible. This possibility > has been studied > extensively [10, 15] over the last few years, particularly in String > Theory, which is > the most mature quantum-gravity approach that emerged from the > perspective. > 1.2 What do we know about quantum-gravity? > The theory debate clearly is a confrontation between very different > perspectives on > the quantum-gravity problem. If we had any robust information on quantum > gravity > certainly at least some of these ideas would have been proven to fail. > But after more > than 70 years [16] of work on the quantum-gravity problem there is > still not a single > measured number whose interpretation requires advocating quantum gravity. > I have so far mentioned the quantum-gravity problem as if it was a > well-established > and familiar concept, but it is perhaps useful to give here an intuitive > characterization > of this problem. The quantum-gravity problem is somees described as a > sort of > human discomfort, as a problem pertaining to the achievement of a more > satisfactory > philosophical worldview. For example, as motivation for research in > quantum gravity > it is somees stated that quantum theory (in an appropriate > generalized sense) > has turned out to be relevant for the description of measurement results > in all other > branches of fundamental physics, and we therefore must assume that it > will eventually > be relevant also for spacee/gravity physics. Analogously (and > amounting to the > same thing), it is somees stated that it is unsatisfactory to have on > one side our > present unified quantum-field-theory description of electromagnetic, > weak and strong > forces and on the other side gravity which is still described in a very > different way. These > human discomforts do not of course define a scientific problem, but > actually there is, > as emphasized by some, a well-defined scientific problem which can be > naturally called > quantum-gravity problem. > The scientific problem that can be reasonably called quantum-gravity > problem > is actually the problem of producing numbers (predictions), in a > logically-consistent > quantum-field theory effects cannot be neglected. For example, although > we are presently (and for > the foreseeable future) unable to set up and observe collisions between > two electrons > each with energy of, say, 10^50 eV, our present theories provide no > obstruction for the > analysis of such high-energy collisions, but are unable to produce a > logically consistent > number for, say, the probability that such a collision would result in > two muons with > certain energies and momenta. The problem, as I shall try to point out > later in these > notes, resides in the fact that quantum field theory implicitly assumes > that gravity > effects can be neglected. When the gravity effects are so large that > (from the field theory > perspective) space geometry evolves significantly on very short e > scales, field > theory cannot be consistently appliedb. Similarly, field theory runs > into trouble when > gravity effects are strong enough to admit the emergence of spacee > singularities (e.g. > black holes). We are able to get numbers out of quantum field theory > in contexts in > which there is a curved static (or slowly-varying) nonsingular space, > but fast-varying > and/or singular space geometries are untreatable..8a > [Comment: An adaptive windowed .8bzoom in/out.8a wavelet transform > reformulation may help here.] > .8bOne might argue that 10^50 eV electrons should be the least of our > concerns, since we > are never going to be able to produce and/or observe them, but first of > all in cosmology > there are some numbers we should produce that depend on very early es > after the > high energies > were abunt), and, secondly, the fact that our theories fail to > produce numbers in > some contexts which those same theories describe as accessible (in > principle) makes us > concerned in general about the robustness of these theories. Since we > know that new > elements would have to be introduced in our theories for the description > of collisions > between 10^50 eV electrons (or for a justification of an in-principle > exclusion of such > collisions from the list of processes that can occur in Nature), it is > natural then to > wonder whether those new elements can affect also some of the contexts > in which our > present theories do provide us an apparently acceptable prediction. In > some cases the > issues we encounter in analyzing, say, collisions among 10^50 eV > electrons might bring > to the surface some issues that could also modify more ordinary (but > still untested) > predictions produced by our theories. > b Here the reader should keep in mind that general relativity governs > self-consistently the spacee dynamics in terms of (and together with) > asymptotically, in the S-matrix sense, in quantum field theory..8a > [Comment: Lenny Susskind tried to fix this at Cornell in 1964 when we > were students together. He failed because the wavelet transform method > was not known back then.] > .8bDuring a collision process the, say, electrons involved are not > following any trajectories. We can associate to them some (however > fuzzy) trajectories only asymptotically, much before and much after the > collision..8a > [Comment: The Bohm picture would help here as well.] > .8bIf one tries to apply general relativity to the formally-classical > trajectories that appear in the path integral formulation of quantum > mechanics, the problem becomes anyway ill defined (and affected by > enough to induce significant geometrodynamics. There is a very natural > explanation for our lack of insight on this quantum-gravity problem. One > of the few (perhaps the only) robust hint we have about quantum gravity > quantum-field-theory description starts to appear inadequate is the > Planck scale Ep ~ 1028^eV..8a > [Comment: That may not be true. In my theory Lp* ~ 10^-13 cm, i.e. > energy scale is 20 powers of ten lower than the na.95ve Newtonian esate > for quantum gravity. This explains alpha.89 = 1/(1 Gev)^2 for hadronic > resonant parallel Regge trajectories, it explains why the lepto-quarks > are .8bmicro-geon.8a Kerr-Newmann spatially extended yet stable and why they > appear more and more point-like when probed to smaller scales at larger > momentum scattering transfers. Furthermore, there is seamless > integration with the large scale dark energy/matter precision chaotic > spatially flat inflationary observations (WMAP, type 1a supernovae, > gravity lensing, gamma ray bursts).] > unsatisfactory. And usually the scale that sets the break point of an > effective low-energy theory is also the scale that sets the magnitude of > the new effects to be expected going beyond the effective low-energy > theory. It is therefore reasonable to expect that .8bquantum-gravity > corrections.8a to our low-energy predictions would be very small, with > their magnitude set by some power of the ratio between the Planck length > (Lp ~ 10-35m, which is the inverse the Planck scale Ep ~ 1028eV ) and > So we have good reasons to suspect that the quantum-gravity effects > would be very small (and actually they must be typically small, since we > have not managed to see them yet)..8a > [Comment: I disagree here. To the contrary we are seeing quantum gravity > on scale of 1 Mev and 1 Gev but we have not properly understood what we > are seeing.] > to be continued === Subject: Re: Can this be proved? Oops; I meant (1/n)*sum(X(n)) over all n > 1, not the integral. Bob Adams > How do you propose to integrate > X(n) since it is defined only over the > integers? Do you mean integrate as > a Stieltje's integral? > You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: Ces?ro-convergence - analysis question >> for infinitely many N. Suppose N satisfies (iii'). Now for >> n > N we have >> s_n >= d - c sum_N^n 1/j >> ~ d - c log(n/N) >> >= d/2 >David, > I have a question about the above step you take in your proof. I >don't understand why > d - c log(n/N) >= d/2 >log(n/N) goes to infinity as n goes to infinity if N stays constant. If you look closely you see that there's no . at the end of the part you cited. This means you're looking at only part of a sentence. Often part of a sentence doesn't make sense by itself. What I actually was this: Now for n > N we have s_n >= d - c sum_N^n 1/j ~ d - c log(n/N) >= d/2 as long as c log(n/N) < d/2. (where that as long as... means for values of n > N such that...) > ~ Chris ************************ === Subject: Re: lopital's rule? >>[...] that's not right. >>[...] >> No, that's not right. [...] >Yes, I'm deeply ashamed of myself. [...] >I shall go and boil my head. That seems a little extreme. But if you find it helps with the math let us know. ************************ === Subject: Re: Can this be proved? Liz top-posted: > Oops; I meant (1/n)*sum(X(n)) over all n > 1, not the integral. X(n) = 0 (n even) X(n) = |mu(n)| (n odd). Your question still doesn't make sense, but I essay that it means, does N^{-1} sum_{n=1}^N X(N) converge to a limit as N -> infinity. If so, the answer is yes. Replacing X(n) by |mu(n)| one gets a well-known problem: with answer 1/zeta(2) = 6/pi^2. Exactly the same method applies here: with answer (1 - 1/2)(1/zeta(2)) = 3/pi^2. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 es) === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed? >> I'm trying to write ATAN2 function for a small basic language that has >> IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are >> availible in the language. >> I've tried a few methods I've found but the results are way off due to >> low precision, rounding, etc. >> A Basic language with Sin, Cos, and Tan, should also have at least an ArcTan >> ? I mention that because a division result sent to an ArcTan and combined >> with some simple quadrant rules should arrive at an ArcTan2. > I made a type (reused my paragraph).. It should have ARCSIN() > function in the body of the text. The BASIC compiler doesn't have > much for intrinsic's as its for the PIC uProcessors and they don't > have much memory. > Does anyone have a softcopy of the ARCSIN() single precision routine > from the C library? Or any other library like Fortran? > You didn't say whether your library includes the ARCTAN() function. > If so, you can define ARCSIN(x) = ARCTAN(x/SQRT(1-x^2)). Jon === Subject: Re: All masses have inertia > You've got to learn; like it or not: > What do I have to learn? I'm a happy SI user who never gets confused > distinguishing between mass and force, so I don't have a problem with > that topic. You _do_ have a problem; you just don't know it! SI is fatally flawed in that it uses mass as a fundamental concept; not as the mathematically derived _ratio_ of force to acceleration; that it really is. > No one asked you to start three threads. It seems that you're the only > one here having problems. Oh posh: I've started dozens of threads; all about this same flaw, and nobody's getting the message: But despite the many false starts - I'm only learning the lesson too - don't expect me to throw in the sponge. Not 'til we all learn the lesson, and learn it well! SI seems like a boone to international commerce, with it's standard kilogram of the archives; which is almost a kilogram; but it's the worst thing that ever happened to physics: Which consists of only three fundamental measures: One for Length; one for Force, and one for e. These three combine mathematically: As a ratio of the net impulse [ft], divided by the e rate of displacement [(vt-vi)/t = s/t] so that the inertia of any object or body of material substance is mathematically defined as ftî/s, and is equal to its gravitational inertia [2w/g]! === Subject: Re: is there a name for this operation? I'll take twenty hours of silence as a no? --JMike > Let the members of set Ai be ai1, ai2, ... aiNi. > input, returns as an answer > { [a11, a21, a31, ...], > [a12, a21, a31, ...], [a13, a21, a31, ...], ..., [a1N1, a21, a31, > ...], > [a11, a22, a31, ...], [a11, a23, a31, ...], ..., [a11, a2N2, a31, > ...], > ..., > [a11, a21, ..., aM1], [a11, a21, ..., aM2], ..., [a11, a21, ..., > aMNM] } > In other words, the answer is a subset of the Cartesian product, > containing only those sets that differ from [a11, a21, a31, ...] in at > most one place. > I'm providing this operation as one of a number of possibilities in a > test-case generation program. So far I'm just calling it > perturbation which is unsatisfying. Does this operator have an > accepted name? > --JMike === Subject: Re: All masses have inertia > You _do_ have a problem; you just don't know it! SI is fatally flawed in > that it uses mass as a fundamental concept; not as the mathematically > derived _ratio_ of force to acceleration; that it really is. What is the mass of an object on which 0 net force is acting? What is the mass of an object which is in unaccelerated rectalinear motion? Bob Kolker === Subject: Re: lopital's rule? > It's L'Hopital's rule (with a circumflex over the o). or: L'Hospital's with S and no circumflex. I saw once a letter the man orthography has changed in the last 200 years... -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Nice summation puzzle >Starting from arcsin(x)=x+1/2/3*x^3+... > What's 1/2/3 ? in computer languages where computation goes from left to right, we write 1/2/3*x^3 instead of (1/(2*3))*x^3 because it saves all those parentheses. > The next term is 1/2/3/4/5 x^5 ? Hard to say, the original poster is confused about the arcsine series, I guess. === Subject: Re: Question on Hilbert & Godel > What did Hilbert ask and claim concerning Foundations of Mathematics > (sets, predicate calculus), metamathematics, Logic, Incompleteness, > etc? In a speech in 1900, he *asked* himself and his audience a number of interesting mathematical questions (see http://aleph0.clarku.edu/~djoyce/hilbert/problems.html) of which he rightfully assumed that many of them would be solved in the upcoming century. Asking meaningful questions is a non-trivial task: the questions should be non-trivial, interesting, and still have a perspective of being solvable. As a mathematician, he did not *claim* things he could not prove. > Something about the Continuum Hypothesis? This was the first of these problems: stated as an open problem with no claim. > Completeness or > Incompleteness of Logic? Was he contradicted by kindergarten logic > from Godel? The way we talk about completeness today depends on a formalisation of logic with its modern notions of axioms. Hilbert was among those who contributed most to this formalisation. He had probably the goal of developing complete axiom systems (Hilbert's programme, see http://www.rbjones.com/rbjpub/logic/jrh0104.htm), and he did not reach this goal which was shown by G.9adel to be unreachable. One cannot say he was contradicted; every serious mathematician has goals that prove unreachable - only claiming to have reached such goals can be contradicted. > How would we exactly formally represent Hilbert's > questions and claims using current day notation? To a large extent, current day notation *is* Hilbert's work. Helmut Richter === Subject: Group generated by a and b ... I have a group G generated by 2 elements a and b such that ba=a(b^k) for some integer k. Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m and n ? -- Ce message a ete poste via la plateforme Web club-Internet.fr This message has been posted by the Web platform club-Internet.fr http://forums.club-internet.fr/ === Subject: Re: polysigned numbers The four-signed isomorphism to C is not legiate in my opinion. There is a symmetry that exists between each sign that is destroyed when allowing: - a + b * a # b = 0. I understand the mapping but it is not a one to one map since + 2 is the same as + 3 # 1. I do see your thinking in terms of the reduction. You can claim that -1 is the same as -2 + 1 * 1 # 1, and so the symmetrical reduction is not one to one. The symmetry involved comes from a purely arithmetic means. This is partly why thinking in vectors for this particular math is less desirable. They abstract the fundamental concept beyond where it ought to go. The math should be able to go back to a numberline concept. The elements of the multisigned values are positions on a branched numberline(or just branch). The general math is an accumulation of these fundamental elments, like moving form T to Y for three-signed(see the three-signed thread). The graphical representation then follows. I believe the the symmetric reduction is acceptable and interesting. It follows along with a general principle of accumulation. I suppose that the isomorphisms to C for higher dimensions could be of interest. If you turn up something good that's great. Perhaps the following gets close to the problem that I have with them: - a + b * a # b = 0. - a + a * a # a = 0. a = b ? Golden === Subject: Re: Question on Hilbert & Godel > To a large extent, current day notation *is* Hilbert's work. It's to a much larger extent based on Peano's and 's work. I don't think Hilbert contributed heavily to notation at all, but I'd be happy if someone could provide information to prove me wrong. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Group generated by a and b ... > I have a group G generated by 2 elements a and b such that ba=a(b^k) for some > integer k. > Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m > and n ? Yes, when the order of a is finite. === Subject: Re: What to tell students in a 10-15 minute talk > * If T = infinity, the function |sum_{k=1}^{n} exp(ak*t*i)| attains > all > values between 0 and n, so this case can be ignored. (Exercise for > the reader.) I can prove that it suffices to consider only integer values for the a_i. So if a_1=0, the others are positive integers, and so the ratios of the a_k/a_j are rational, and there's no need to consider T=infinity. Indeed, at this point f(n) can be redefined as f(n)=sup_{0=a1 * So the candidate a_k's are n-tuples like (0,3,7,8,11): i.e., > n-tuples of increasing integers beginning with 0. The winning tuple for f(4) is (0,2,3,4) and f(4)=0.7524.... which is the decimal expansion for an algebraic number which takes a page to write out. I can't prove what f(5) is, but the best tuple I can find is (0,1,2,6,9). Does it give a horrible algebraic number? No, it's maxmin value is exactly 1. I think f(5)=1. The tuple (0,6,9,10,17,24) proves that f(6)>= 1.13.... The growth of f(n) in these small cases is more like c ln(n), but I've never been able to improve on the gross bound of sqrt(n+1). Bart === Subject: Re: a puzzle related to artinian group > R. Anderson, L. Lovasz, P. Shor, J. Spencer, E. Tardos, S. Winograd, > ``Disks, balls, and walls: analysis of a combinatorial game'', Amer. > Math. Monthly, 6, 96, pp. 481--493, 1989. > It talks about the same problem (of course along a line, not around a > table). Their discussion is thrilling. > Best, several days' worth of spare e thinking about this. I came up with an argument for termination of the linear variant, but the cyclic one seems quite a bit harder to me. A quick sketch of my argument for the linear case is that you number cards and players and arrange the play so that the cards are always in order from left to right and so that each player holds a contiguous set of card numbers. You then show that the centre of gravity of the cards is fixed, that the variance of the positions of the cards relative to some point to the left of all of them is strictly increasing whenever a round of play takes place, and that the distance between the first and last cards is never more than C + k, where C is the distance between them as initially dealt and k is the number of cards. The last condition means the variance is bounded above so play must terminate with the k cards distributed amongst C + k + 1 players each holding at most one. As for the cyclic version? Rob. === Subject: Re: Zeroes, extrema, inflection points, ..., etc? > If you consider the graphs of y=f(x), y=f'(x), and y=f''(x), the > values of y are called the value, the slope, the concavity, > respectively, and the points at which y=0 are called the zeroes, the > vertices, and the inflection points. My question is, does f'''(x) > have a name? In physics when the x-variable represents e and f(x) represents position, the first three derivatives are called velocity, acceleration, and jerk. But in general I don't think there's a name (and calling f'' the concavity is new to me as well). > If so, what is it? Do any other f'(n)'(x) have names? The n-th derivative with respect to x > Note: f'(4)'=f'''' and f'(n)' is f with n primes. A more common notation is just to have (n) as a superscript. In ASCII I guess you could suggest that with f^(n)(x). === Subject: convex hull Given three points p1,p2,p3 in 2d and given point g also in 2d how can i decide whether the point is inside the convex hull that p1p2p3 create or not. === Subject: Re: Dedekind Cuts > I have heard that the reals can be defined by Dedekind cuts. What is > this definition, exactly? > This is a division of the rationals into a pair of sets, which we will > call Left and Right. Each element of Left is less than all elements of > Right. Each element of Right is greater than all elements of Left. It > follows that Left intersect Right is empty. We also require that a > rational number either be in Left or Right so that Left union Right = > the set of rational numbers. > In the case that Right has a lest element we put that element in Left. > There are two case. Left has no greatest element and Right has no least > element. If so the division or cut Left,Right corresponds to an > irrational number. Or Left has a greatest element and Rigt does not. > This sup of Left is the number (a rational) represented by the cut. > There are rules of addition, subtraction, multiplication and division > defined for cuts and it is shown they form a field. > For a good account of Dedikind Cuts set -A Course in Pure Mathematics- > by G.H.Hardy. I understand what a Dedekind cut is, but what what is the definition that goes the set of reals is some collection of Dedekind cuts? Is it simply the set of of all Dedekind cuts of the rationals? Or is it the set of numbers such that all Dedekind cut of the reals the reals yields a greatest element for Left? Or something else? > Bob Kolker === Subject: Re: JSH: About e > So it's about e, as I wait, and wonder, how many of you can handle > the truth. > Pure megalomania. > Has anyone written any interesting psych papers about the JSH phenom > yet? I mean, there's enough material for at least a nice masters' > thesis. > V. The argument at http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 is not just some guy talking maliciously. === Subject: Re: Numeric one-way hash function >> Because you snipped my proof that your original procedure generates a lot of duplicates, I assume that you could not find any errors in the proof. >> The simplest way to ensure that the bar codes are >> unique is to add a prime number to the previous >> value. Lap round when you get to the top (or a prime >> number near the top). Start at a weird value. >> Could you be more precise, I am afraid that I >> don't understand your algorithm. >> Anyhow, I can't see that your algorithm would >> be unbiased, or how it would prevent duplicates. >> Where C, Max are prime constants and C != Max >> Initialise N to a valid random number >> loop >> N = N + C >> if N > Max then N = N - Max >> Print N >> end loop > p.s. If you want to analyse the algorithm > for bias or duplicates just set C equal to 1. > Or in a simple case setting C = 3, Max = 5 > and initialise N to 2 > 5 3 1 4 2 > You have to stop after producing Max numbers. But this is a simple linear congruential PRNG, that can be easily broken. (Also, Max is not a prime, this is not a real problem when C is relative prime to Max) greetings, Ernst Lippe === Subject: Dirt Simple Proof Re: Symmetric Groups. Hi all. Can anyone think of a dirt simple proof that S_6, the symmetric group of order 6, is *not* generated by (1 2 3 4) and (3 4 5 6)? By dirt simple I mean something that can be explained to someone who knows very little math. In essence, consider the puzzle: A - D - E | | | B - C - F Suppose your two legal moves are (A B C D) and (C D E F), and you want to show that you can't switch A and B without disturbing anything else. How might you explain this to someone who knows as much math (none) as is required to describe the puzzle? Justin === Subject: Re: Numeric one-way hash function >> I need to find an algorithm that can produce a unique non-predictable 12 >> digit (0-9) number for any given 12 digit number. This is to be used to >> create a unique barcode on a ticket that cannot be predicted. It is not >> required that the original seed number be computed from the resulting >> barcode, so some form of one-way hashing function would be acceptable. >> Any help in this problem would be appreciated. >> Mark. > 1. Create a 10 element array with digits 0-9 in linear order (0 in 0, > 1 in 1, etc.) > 2. For each digit of the 12-digit number: > 2a. Shuffle the 10-element array. > 2b. Using the original decimal digit as an offset into the array, look > up the replacement digit value. You did not specify how the array should be shuffled. How do you prevent duplicates? It seems that you'll have to store a full list of all generated numbers in order to verify a specific bar-code, this is very awkward in most situations. greetings, Ernst Lippe === Subject: Re: Dedekind Cuts >> I have heard that the reals can be defined by Dedekind cuts. What is >> this definition, exactly? >> This is a division of the rationals into a pair of sets, which we will >> call Left and Right. Each element of Left is less than all elements of >> Right. Each element of Right is greater than all elements of Left. It >> follows that Left intersect Right is empty. We also require that a >> rational number either be in Left or Right so that Left union Right = >> the set of rational numbers. >> In the case that Right has a lest element we put that element in Left. >> There are two case. Left has no greatest element and Right has no least >> element. If so the division or cut Left,Right corresponds to an >> irrational number. Or Left has a greatest element and Rigt does not. >> This sup of Left is the number (a rational) represented by the cut. >> There are rules of addition, subtraction, multiplication and division >> defined for cuts and it is shown they form a field. >> For a good account of Dedikind Cuts set -A Course in Pure Mathematics- >> by G.H.Hardy. > I understand what a Dedekind cut is, but what what is the definition > that goes the set of reals is some collection of Dedekind cuts? If a real number is defined to be a Dedekind cut, then it follows that the set of real numbers is the set of all Dedekind cuts. > Is it simply the set of of all Dedekind cuts of the rationals? Or is > it the set of numbers such that all Dedekind cut of the reals the > reals yields a greatest element for Left? Or something else? Dedekind cuts are defined on the rationals. You could carry out a similar construction on the reals, but it wouldn't be a Dedekind cut. -- Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Dedekind Cuts > Dedekind cuts are defined on the rationals. You could carry out a > similar construction on the reals, but it wouldn't be a Dedekind cut. If you did you just would get the real numbers. No new numbers would be added by using D.C.-s Bob Kolker === Subject: Re: Dedekind Cuts > I understand what a Dedekind cut is, but what what is the definition > that goes the set of reals is some collection of Dedekind cuts? > Is it simply the set of of all Dedekind cuts of the rationals? Or is > it the set of numbers such that all Dedekind cut of the reals the > reals yields a greatest element for Left? Or something else? Dedikind Cuts are a way of extending the rational numbers. It is the metric completion of the rationals. The rational numbers do not contain all limit points for cauchey sequences of rationals, so the limit points must be added to make a complete space. Dedikind cuts are one way of doing this. Bob Kolker === Subject: Re: lopital's rule? > or: L'Hospital's with S and no circumflex. I saw once a letter the man > orthography has changed in the last 200 years... In freshman calculus we called this rule, the hosptial rule. Bob Kolker === Subject: Re: All masses have inertia > You _do_ have a problem; you just don't know it! SI is fatally flawed in > that it uses mass as a fundamental concept; not as the mathematically > derived _ratio_ of force to acceleration; that it really is. > What is the mass of an object on which 0 net force is acting? What is > the mass of an object which is in unaccelerated rectalinear motion? > Bob Kolker In order to determine the mass of an object; body, or any mass, it requires _being measured_: Like putting it on a weight-scale - resting on Earth's or some similar planet's terra firma - or exerting a measured force - with a spring scale - and determining the acceleration [a], and or deceleration [g], and mathematically dividing that net force [f = F-uw] by the acceleration; to get the inertia of the mass: That inertia is the measure of its mass. Near the sun it seems that we'd find it impossible to measure the inertia of any mass. === Subject: Re: JSH: About e >> So it's about e, as I wait, and wonder, how many of you can handle >> the truth. >> Pure megalomania. >> Has anyone written any interesting psych papers about the JSH phenom >> yet? I mean, there's enough material for at least a nice masters' >> thesis. >> V. >The argument at >http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 >is not just some guy talking maliciously. Huh? Nobody the argument there was some guy talking maliciously. The argument there is not some guy talking maliciously, and it's also not correct. Otoh the statement above, about your wondering how many of us can handle the truth, is pure megalomania. > ************************ === Subject: Re: All masses have inertia > In order to determine the mass of an object; body, or any mass, But, but, but relativity shows the effective mass increases with velocity. As a result your definition of mass is meaningless. If you apply a constant force to a body, it relativistic mass increase so the acceleration decreases. The ratio F/a tends to infinity. So at what velocity shall we measure the mass? If we get different answers at different velocities we have achieved bogusity-1 on the bogus scale. In short you have given a bogus definition of mass, which is not surprising. Just about everything you say is bogus. Why don't you learn physics instead of constantly demonstrating what an ingnoramus you are. Bob Kolker === Subject: Re: JSH: About e > not just some guy talking maliciously. I megalomania, not malice. Btw, the link you want is http://www.geocities.com/jrstrader2000/Incomnt.htm V. === Subject: Re: Dirt Simple Proof Re: Symmetric Groups. > Hi all. > Can anyone think of a dirt simple proof that S_6, the symmetric group > of order 6, is *not* generated by (1 2 3 4) and (3 4 5 6)? They both fix a syntheme --- but are synthemes dirt simple? How about this: draw a complete graph with vertices labelled 1, ..., 6. Colour the edges as follows red: 1-3, 2-4, 5-6 blue: 1-4, 2-5, 3-6 green: 1-5, 2-6, 3-4 yellow: 1-6, 2-3, 4-5 purple: 1-2, 3-5, 4-6. Now applying each of those permutations takes each set of three coloured edges, to another set of three coloured edges. Thus so does each permutation in the group they generate --- but that's not true for all permutations in S_6. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 es) === Subject: Re: matrix differential equation. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h987JD930241; >hello group. >i'm extremely confused as to how to change the system of ode's into a >matrix version. >let's say you have the the lorenz system. >{ x'[t] = - a*y[t] - a*z[t], > y'[t] = r*x[t] + y[t] - x[t]*z[t] , > z'[t] = x[t]*y[t]- b*z[t]} >where a, r, b, are constants, >how do i represent the above as a matrix differential equations >system? >and then find the eigenvalues and so on. >any help is most appreciated. >john Hi John, if your system of differential equations is nonlinear (like the one above), then it only makes sense to consider a matrix version of the system locally. Consider the case your ODE system is given by y' = f(y) , y(t0) = y0 (*) where y = (y1,...,yn), y' = (y1',...,yn') and f = (f1,...,fn). Then, for the given point y(t0) = (y1(t0),...,yn(t0)), you may linearize (*) by considering the derived system z'(t) = f(y0) + df/dy(y0)*(z-y0), z(t0) = y0 and look for the properties of the (constant) matrix df/dy(y0). This will give you hints about the behaviour of the solution of (*) in a neigbourhood of y0 (stiff and nonstiff solution components etc.). Best wishes Torsten. === Subject: Re: Chessboard knight metric? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h98Bw2g14972; >>Take a chessboard (with or without infinetely many squares) let the >>distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >>chessboard be defined as the minimum number of moves a knight takes >>to reach y from x. >>Is d a metric? >With this distance, the triangular equation is obviously true, and that >makes it a metric. >> Not trying to suggest that this is some new >>question that hasn't been asked/answered before. Is there a general >>formula for calculating d? More generally, the same question may be >>asked for the other pieces (queen, king, knight, biship)? Actually, I >>asked myself this question a few years ago. If I remember back to the >>notes I took, I had something like (x_1-y_1, x_2-y_2)= (even number, >>even number), then d(x,y) = even number. If (x_1-y_1, x_2-y_2)= (even >>number, odd number), then d(x,y) = odd number. Finally, if (x_1-y_1, >>x_2-y_2)= (odd number, odd number), then d(x,y) = even number. In other >>words, the same rules for adding natural numbers... >> C.Dement >With this distance, the triangular equation is obviously true, and that >makes it a metric. >I don`t (yet) see the triangle inequality as so inherently >obvious that it is not worth further thought. Take the >points x=(1,1), y=(77,79), and z=(163,163). It is still feasable >that d(x,y)+d(y,z) is less than d(x,z) (especially as an >explicit formula for d doesn`t exist). The fact that I would >be somewhat surprised if that wasn`t the case shouldn't be used >as part of the proof. C.Dement === Subject: Re: matrix differential equation. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h98Fu4g31804; > hello group. > i'm extremely confused as to how to change the system of ode's > into a matrix version. >let's say you have the the lorenz system. >{ x'[t] = - a*y[t] - a*z[t], > y'[t] = r*x[t] + y[t] - x[t]*z[t] , > z'[t] = x[t]*y[t]- b*z[t]} > where a, b, r, are constants, re-write your system as x'[t] = - a y[t] - a z[t] y'[t] = r x[t] + y[t] - x[t] z[t] z'[t] = - b z[t] + x[t] y[t] > how do i represent the above as a matrix differential > equations system? x[t] .89 0 - a - a x[t] y[t] = r 1 0 y[t] z[t] 0 0 - b z[t] 0 0 0 x[t] + 0 0 - x[t] y[t] 0 x[t] 0 z[t] > and then find the eigenvalues and so on. since this system is non-linear, formally, you find eigenvalues by linearising. for hints and reference see page 330 of _ordinary differential equations_ by wolfgang walter > any help is most appreciated. === Subject: Re: who to invert 3*3 singular matrix by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h98J7J613882; Who indeed?! phil === Subject: Re: real analysis: construct this set ... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h98JjfM16569; >Hint: Suppose {Bn} is a countable basis for the topology of (0,1). > B2 (K1 U J1) take away fat disjoint Cantor sets K2 and J2. > Continue, and set E = ... (If this is a homework problem and you >use this hint, be sure to give credit to sci.math.) :D Yes yes, a challenge homework problem. The whole class is wracking our brains trying to work something out. I will bring this to our groups & of course no secrets where we got the hint from :) === Subject: Pi formula finished? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h98Lpl825658; Last year my science teacher told me and some friends that she saw on the news that the pi formula had been completed. Is this true? === Subject: Super LOW price for all softwares - All you will receive is the actual software and your own unique registration codeMwGpmGgRWP by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h99CEWo18092 by support2.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.6 secondary) with SMTP id h99CEBk15623

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=== Subject: Partial d.8erivates by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99CGUs18307; I try to solve a patial d.8erivates equation with two variables but I could not under math.8ematica, my system is : (a1+b1.exp(-x^2/c1))dE(x, t)/dx + a2.E(x, t).(d^2E(x, t)/dx^2).dE(x, t)/dt+ a3.dE(x, t)/dx.d^2E(x, t)/dt^2+ (a4+b4.exp(-x^2/c1))E(x, t).d^2E(x, t)/dx^2+ a5+b5.exp(-x^2/c1))dE(x, t)/dx dE(x, t)/dt+ a6.exp(-x^2/c1))E(x, t)=0 such as, a1, b1, c1, a2, a3, a4, b4, a5, b5, a6 ; constants. I thank you for giving me at least an idea for the solution of this very complex equation. na.95ma === Subject: Re: Do Physicists Understand Their Own Peer-Reviewed Literature? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99CGd718325; It is not in the best interests of physicists to understand physics. Understanding empties their rice bowl. A rice bowl kept to overflowing by the taxpayer. Proof at http://www.thewebspert.com/cresswell/ Diagram 9-1. So much for the 'Conservation of Energy'. Shall I hold my breath waiting to hear from ya'll. === Subject: Re: prime numbers factoring by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h99CGZC18317; The repunits R(311), R(509), R(557), R(617), R(647) and R(991) are apparently known to be composite (I have not seen this anywhere) but no factor has been found yet for those numbers. In fact, that last repunit R(P) completely factorised is R(733). It is true that (relatively) small prime factors can be found for some of these repunits such as R(359) and R(659), but how can they be sure that R(311) and R(991) are not prime? How can they even know where possible factors might be?? === Subject: Re: Pi formula finished? > Last year my science teacher told me and some friends that she saw on the > news that the pi formula had been completed. Is this true? There was some April Fool press release to the effect that the final nonzero digit of pi had been reached by a supercomputer. And a few months later an Australian newspaper printed it without noticing the date. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: a puzzle related to artinian group quick computation could be wrong (and generalizing shamelessly from small-n computer results), circular table: n players all eventually have <= 1 cards if you deal n-1 cards max (quoted puzzle) n players all eventually have <= 2 cards if you deal n + (n mod 2) cards max n players all eventually have <= X cards if you deal 2n + X - 2 cards max, X >= 3 == n players all eventually have >= 1 cards if you deal 2n - (n mod 2) cards min n players all eventually have >= 2 cards if you deal 2n + 1 cards min no way to guarantee >= 3 cards these are all maximally tight limits. -- On the internet, nobody knows you're a jerk. === Subject: Re: convex hull > Given three points p1,p2,p3 in 2d > and given point g also in 2d > how can i decide whether the point is inside the convex hull that p1p2p3 > create or not. See http://mathworld.wolfram.com/TriangleInterior.html and http://mathforum.org/library/drmath/view/54505.html === Subject: Re: Boolean Algebra - Arithmetic Relationship Originator: hack@watson.ibm.com (hack) >P.S. It may be that by boolean logic you mean computer hardware >gates. A physically realized (or realizable) logic is certainly >inadequate--only a finite number of the infinity of numbers could be >handled, and only finite approximations of numbers like pi could be >handled. Replace approximations with descriptions and you'd be ok. The finite number of gates could be programmed for symbolic math, in which case a mathematically exact pi could be handled in formulae or theorems up to a cretain size or complexity (bounded indeed by some function of the number of gates). (The gate count includes those needed to implemement memory, including storage for the program, of course. All is finite here. So clearly we're talking about a finite subset of recursive functions -- but one that is arbitrarily large if you have enough gates.) Michel. === Subject: Re: help !!! Visiting Assistant Professor at the University of Montana. >>I need help giving (p->q) xor (p<->q) an equivalence less than 12 >My approach is to translate everything into terms of and, or, >and not. Then, use the basic rules for manipulating logical >expressions to reduce the translated expression into a simpler >expression. >For example, p->q translates into not p or q. >I got a result the exact opposite of what got. Actually, I just made a mistake in the line you did not delete; had you seen the rest of the post, you would have seen we got the same thing. >[cut] >> So the xor is true exactly when p=1 and q=0; it is false in all other >> cases. >[cut] Here's what preceded this mistake: -- Begin insert -- So p=0 if and only if q=1 for the xor to be true. if p=1, q=0, then p->q is 0, so for the xor to be true you would need p<->q to be true, but that does not happen. -- End Insert --- so I clearly have p=0 and q=1; I just miscopied. === Subject: Re: A polynomial problem > I'm trying to prove the following theorem: Let P be a polynomial with real coefficients such that P(x) >=0 for > every real x. Then, there are polynomials R and S such that P(x) = > R^2(x) + S^2(x) for every complex x. > Show P factors over the reals as a product of irreducible quadratics > es a product of squares of linear polynomials. > Show that an irreducible quadratic is a sum of 2 squares. > Show that a product of two sums of two squares is a sum of two squares. Sure. I almost got there! All I had to do to complete my proof was show that a product of two sums of two squares is a sum of two squares. This is simple: (a^2+b^2)*(c^2+d^2) = (ac)^2 + (ad)^2 + (bc)^2 + (bd)^2 = (ac)^2 + 2abcd+ (bd)^2 + (ad)^2 -2abcd + (bc)^2 = (ac + bd)^2 + (ad - bc)^2. Thak you all for the help Amanda === Subject: Re: Fundamental Reason for High Achievements of Jews === >Subject: Re: Fundamental Reason for High Achievements of Jews >> >> :> : Most historians believe that Jews avoid pork, >> :> : because the ancient Jews associated pigs with leprosy, >> :> : and pigs and people with leprosy were unclean. >> :> >> :> Name one historian who believes that, and give a citation to the >> :> place where he says it. >> : >> : As I recall, this was in Tacitus' Histories >> : which was written in the first century A.D. >> If that's the best you can do, I think that we can safely ignore your >theory. >It's soooo easy . Someone in your camp >should have the information at his fingertips. >GOOGLE: tacitus histories ~13,600 hits > tacitus histories jews ~3,950 hits > tacitus histories jews leprosy ~154 hits This is all irrelevant. Quoting from Tom Potter: 2. This is the computer age. Almost everyone has access to the first hand historical accounts, and can do wild card searches on the source material. It is STUPID to provide detailed cites, as these focus ONLY on the POINTS trying to be emphasized by the writer. It also STUPID to use second hand, accounts which have a racial, religious, national, or personal spin on them, rather than using the FIRST HAND historical accounts. Anything written by Tacitus would NOT be a FIRST HAND account, so only a STUPID person would allow himself to be brainwashed by Tacitus' racial, religious, national, personal spin on history. >Other possible search terms: pork, trichinosis, Visa, Master Card.... >The New York Library may have something...I might >have the e to look for that which you should've. > Mark (My institution consists of three tents > pitched on the desert sands....) -- Mensanator Ace of Clubs === Subject: Re: Benchmark Bias > Has anybody studied the benchmark bias effect? You have a program > that you try to improve, and each change gets tested against a > large set of benchmark runs. > [ ... ] Was that a rhetorical question, designed to introduce some musing, or was it trying to elicit a *specific* sort of critique? - I'm afraid that I did not track the specificity very well. Back when I read computer magazines, the one benchmark area where I remember folks concerned about happenstance-baises, or another kind, was 'opizing compilers'. < benchmark opizing compilers bias > returns 240 hits. The ones at the top seem to be potentially on the topic. Perhaps they can help you refine your search. Hope this helps. -- Rich Ulrich, wpilib@pitt.edu http://www.pitt.edu/~wpilib/index.html Taxes are the price we pay for civilization. === Subject: Name of construction? What is the name of the construction for proving that a countable union of countable unions of As is a countable union of As? The construction goes like this: Let X = biup_{i in I} biup_{j in J} A_{i_j} and then arrange X as an infinite-sized quarter plane like so: A_1_1 A_1_2 A_1_3 A_1_4 ... A_2_1 A_2_2 A_2_3 A_2_4 ... A_3_1 A_3_2 A_3_3 A_3_4 ... A_4_1 A_4_2 A_4_3 A_4_4 ... : : : : and then enumerate all the As in a zigzag pattern starting from the top left corner: A_1_1, A_1_2, A_2_1, A_1_3, A_2_2, A_3_1, A_1_4, A_2_3, A_3_2, ... What is the name of this construction? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ He : 'I'm not Elvis'. Who else but Elvis could have that? - ALF === Subject: Re: who to invert 3*3 singular matrix > Who indeed?! A linear algebraicist? -- P.A.C. Smith 'If the Apocalypse comes, beep me.' <*> http://www.srcf.ucam.org/~pas51 === Subject: Re: Is it mass; or is it weight >>Now it's e for all of us to realize that a customary pound is a unit of >>force, > You know better, Donald. Pounds everywhere have always been units of mass. > Pounds force are such a recent bastardization that they are uniquely > identified by that name. > Back in 1959, the national standards laboratories of the United States of > America, Canada, the United Kingdom of Great Britain and Northern Ireland, the > Union of South Africa, Australia, and New Zealand got together and agreed on a > common definition of the most commonly used pound, the avoirdupois pound. > They defined it as a unit of mass exactly equal to 0.45359237 kg. > Of course, you already knew that, Dishonest Don. This is for the benefit of > anybody else who hadn't been paying attention. So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net which stretches an additional 4.4-ft before thrusting the man back, would the potential energy of the net at the instant it plunges the man back be U = mgh; U = 220 * (36 + 4.4). -- Ayaz Ahmed Khan Yours Forever in, Cyberspace. === Subject: Re: Pi formula finished? > Last year my science teacher told me and some friends that she saw on the news that the pi formula had been completed. Is this true? I think you omitted maybe the funniest part of the story: did your science teacher believe the news ?? ... === Subject: multiplication negs Hopefully, I'm in the right place here.... I've been making my way throught the Principia, taking my e, working along. One day I was talking to my young neice, helping her out with some simple algebra. We were discussing mulitiplication and division of negative numbers and my neice brought up a question and for some reason I have been unable so far to reason my way through the answer: exactly why is it that when one multiplies 2 negative numbers one ends up with a positive? It's disturbing me that I cannot come up with a solid answer, and certainly 'We've been taught that that is the case. will not suffice. Any help would be greatly appreciated. TIA === Subject: Re: All masses have inertia > In order to determine the mass of an object; body, or any mass, > But, but, but relativity shows the effective mass increases with > velocity. As a result your definition of mass is meaningless. If you > apply a constant force to a body, it relativistic mass increase so the > acceleration decreases. The ratio F/a tends to infinity. > So at what velocity shall we measure the mass? If we get different > answers at different velocities we have achieved bogusity-1 on the bogus > scale. > In short you have given a bogus definition of mass, which is not > surprising. Just about everything you say is bogus. Why don't you learn > physics instead of constantly demonstrating what an ingnoramus you are. > Bob Kolker Bobby: You are whistling loudly as you walk past the cemetary. It won't help the gobblins will get you anyway(;^) Only at speeds close to that of light rays will the formula f/a = w/g even it nears the sun. All your learning of physics - as well as my _more extensive_ learning of it - tells us both, unequivically that accelerating a measurable object; body or mass of material substance to anywhere near where its inertia will change without adding to its matter or burning it, is a practical impossibility: You're just grasping at straws; from a strawman to try and discredit the truth! === Subject: Re: multiplication negs Visiting Assistant Professor at the University of Montana. >Hopefully, I'm in the right place here.... >I've been making my way throught the Principia, taking my e, working >along. One day I was talking to my young neice, helping her out with some >simple algebra. We were discussing mulitiplication and division of negative >numbers and my neice brought up a question and for some reason I have been >unable so far to reason my way through the answer: exactly why is it that >when one multiplies 2 negative numbers one ends up with a positive? It's >disturbing me that I cannot come up with a solid answer, and certainly >'We've been taught that that is the case. will not suffice. >Any help would be greatly appreciated. There are a number of answers. For integers, we can envision multiplication as repeated addition. Multiplying by a negative number means do the repeated addition, then change the sign. So if n and m are integers, with n positive, then n*m = add n copies of m; and (-n)*m = change the sign of n*m. If m is negative, adding n copies of m yields a negative number, and changing its sign yields a positive number. But that only sort of works for integers. Otherwise, note that you can reduce the problem to showing that (-1)*(-1)=1, and that -a = (-1)*a, because using commutativity and associativity of multiplication, we have: (-a)*(-b) = [(-1)*a]*[(-1)*b] = [(-1)*(-1)]*[a*b] So, how do we prove that? To see that (-1)*a = -a, note that -a is the one and only number that added to a gives 0. So we take (-1)*a, add it to a, and see if we get 0. If we do, then (-1)*a must be equal to -a, yes? [(-1)*a] + a = [(-1)*a] + [1*a] = [(-1)+1]*a (distributivity) = 0*a = 0. Aha, so (-1)*a must be equal to -a. As for (-1)*(-1), we have the same thing. This is equal to 1 if and only if, when we add it to -1 we get zero. [(-1)*(-1)] + (-1) = [(-1)*(-1)] + [1*(-1)] = [(-1) + 1] * (-1) = 0*(-1) = 0. So, why is the product of two negative numbers a positive number? So that the rules of addition and multiplication hold. === Subject: Re: factoring to satisfiability if you systematically generalize this to larger numbers, then yes, afaik, it has not been studied. but a 3x2 circuit is too trivial to lead to much insight. one interesting approach would be to take the circuit graph for multiplication and look at the way that SAT algorithms visit various nodes to reverse engineer the solution. in other words, if x*y=z, and z is specified, then the information for 'z' percolates backwards into the circuit. it seems that some kind of proof may be possible that an opum algorithm must visit the nodes in reverse order. in other words if it visits nodes in the middle of the circuit 1st, that is (possibly) provably nonopal. afaik studying the graph nature/properties of multiplication has not been done by almost anyone. > Below is a 3x2 bit multiplication circuit. > I can use this circuit to make assertions > about the factors of any number. > I'm curious if this has been studied. === Subject: Re: multiplication negs > Hopefully, I'm in the right place here.... > I've been making my way throught the Principia, taking my e, working > along. One day I was talking to my young neice, helping her out with some > simple algebra. We were discussing mulitiplication and division of negative > numbers and my neice brought up a question and for some reason I have been > unable so far to reason my way through the answer: exactly why is it that > when one multiplies 2 negative numbers one ends up with a positive? It's > disturbing me that I cannot come up with a solid answer, and certainly > 'We've been taught that that is the case. will not suffice. > Any help would be greatly appreciated. Here's an example I like: I have been giving away five dollars each minute. Currently (at t = 0), I have zero dollars, but I'm continuing to give away five dollars each minute in the form of IOU's. Thus, the equation for the number of dollars I have at e t is: D = -5t, where D represents dollars. How many dollars did I have four minutes ago? -- Bob Day http://www.bob.day.name === Subject: Re: ab... = (a*b*...)^n ? > You get a lot further by examining only those > numbers whose prime factors are less than B. > Still nothing in base 10 up to 10^200. Would you describe further how you did that? (Obviously, you didn't test each number to see whether its prime factors were less than 10, since to finish within a week the testing alone would average > 10^194 tests/second.) Using the brute force method with a C program running in backgound for a day or so, I managed to confirm the negative result only up to a paltry 3 10^10. === Subject: Joint distribution of distances and angles in hyperspheres Suppose p(x) is a rotationally symmetric distribution in R^n. Denote by U and V i.i.d. random variables with this distribution. Denote by d=|U-V| the distance bewteen the two points and by $alpha$ the angle between them. The distribution of the distances between points, p(d), has been determined by Hammerseley and Lord. I am interested in finding out the joint distribution p(d,alpha) as a function of the space dimension, n, and the distribution p(x). -- Dario === Subject: Re: prime numbers factoring See: http://www.cerias.purdue.edu/homes/ssw/cun/ You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: multiplication negs See.. that's where I thought you were going to go... and I've been wrapping my head around this and basically.. we simply have to take it for granted ( because I cannot recall any axiom) that when: (-n)*m = change the sign of n*m. ?? This seems rather odd, even arbitrary. In reducing multiplication to a series of successive additions I fail to see how we end up, when working with negative integers, with a negative product. Am I missing something here?? (I've always had problems with the a priori. :-) ) === Subject: 0! = 1 (episode 2) 0! = sqr(pi) eh eh eh === Subject: Re: multiplication negs Visiting Assistant Professor at the University of Montana. >See.. that's where I thought you were going to go... and I've been wrapping >my head around this and basically.. we simply have to take it for granted >( because I cannot recall any axiom) that when: (-n)*m = change the sign of >n*m. ?? This seems rather odd, even arbitrary. In reducing multiplication >to a series of successive additions I fail to see how we end up, when >working with negative integers, with a negative product. Did you read the rest of the reply, or just that line? >Am I missing something here?? >(I've always had problems with the a priori. :-) ) What would it mean to add something minus 2 es? Adding it twice means: start with zero, then add m, then add m. So adding something minus 2 es should mean: start with zero, then ->subtract<- m, then ->subtract<- m again. Which reduces to -m -m = -(m+m) so again, you are down to proving that (-1)*m = -m, which amounts to showing that if you take (-1)*m and you add it to m, you should get 0. === Subject: Re: Is it mass; or is it weight >Now it's e for all of us to realize that a customary pound is a unit of >force, >> You know better, Donald. Pounds everywhere have always been units of mass. >> Pounds force are such a recent bastardization that they are uniquely >> identified by that name. >> Back in 1959, the national standards laboratories of the United States of >> America, Canada, the United Kingdom of Great Britain and Northern Ireland, the >> Union of South Africa, Australia, and New Zealand got together and agreed on a >> common definition of the most commonly used pound, the avoirdupois pound. >> They defined it as a unit of mass exactly equal to 0.45359237 kg. >> Of course, you already knew that, Dishonest Don. This is for the benefit of >> anybody else who hadn't been paying attention. >So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net >which stretches an additional 4.4-ft before thrusting the man back, would >the potential energy of the net at the instant it plunges the man back be >U = mgh; >U = 220 * (36 + 4.4). No, because U is not a dimensionless quantity. Did you have particular units in mind? What's your point, anyway? Sure, it might be, if you choose your units right, and your local g is close enough to the standard value used to define one of those units. Then, of course, the magnitude of your omitted multiplicand will be close to 1 in the units chosen for it. It might also be, depending where you are (limiting yourself to real rooftops will limit the possible range), something like these U = (220 lb)(32.06 ft/sî)(36 ft + 4.4 ft) or U = (220 lb)(32.24 ft/sî)(36 ft + 4.4 ft) Now, if you carry out those operations, what are the units of the final result? Or, it might also be, knowing that in the definition most often used today for a slug (which doesn't have an official definition), 1 slug = 32.1740 lb, so 220 lb = 6.84 slug U = (6.84 slug)(32.06 ft/sî)(36 ft + 4.4 ft) or U = (6.84 slug)(32.24 ft/sî)(36 ft + 4.4 ft) Now what are the units of your final result? Now, have you learned anything today? Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: Re: Is it mass; or is it weight > So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net > which stretches an additional 4.4-ft before thrusting the man back, would > the potential energy of the net at the instant it plunges the man back be > U = mgh; > U = 220 * (36 + 4.4). You left off the factor of g. Which either means that you forgot about it or that you are using a system of units within which the For instance, if you measure m in pounds mass, h in feet and energy in jump site with standard g from the ratio of pounds force to pounds mass. Gene Nygaard would be correct to point out that there is no single standard g that is officially sanctioned for this purpose. John Briggs === Subject: topological terminology So... homotope or homotop, which is the verb? === Subject: Re: multiplication negs > Hopefully, I'm in the right place here.... > exactly why is it that > when one multiplies 2 negative numbers one ends up with a positive? We have certain rules of arithmetic that we like: commutative, associative, distributive laws, for example. If we want these laws to continue to hold for negative numbers, we are forced to the conclusion you mention. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Group generated by a and b ... >> I have a group G generated by 2 elements a and b such that ba=a(b^k) for some >> integer k. >> Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m >> and n ? >Yes, when the order of a is finite. But in general, no. As a simple example, let G be the multiplicative matrix group over the rational generated by a = [2 0] and b = [1 0] [0 1] [1 1] Then ba = ab^2, but ba^-1 = [1/2 0] [1/2 1] whereas all products (a^m)(b^n) have integral entries. Derek Holt. === Subject: Re: topological terminology > So... homotope or homotop, which is the verb? > Use it in a sentence? The space X homotoped into Y? === Subject: Re: 0! = 1 (episode 2) correction Correction, i mean (-1/2)! = sqr(pi) === Subject: Re: lopital's rule? > Why's that then? It seems to me to be a mildly interesting theorem. I > can see that it might not be a good idea to prove it and then give > students lots of 0/0 limit problems for them to solve using > L'Hospital's when doing so is just pedagogically useless manipulation. > But that is no reason for not proving the theorem. > Who anything about proving anything. My experience > is that it is taught as a method, not as a theorem, and when I'm probably being dim, but a method is only useful if it works. Hence one needs a proof that the method works. Such a proof is a proof of a theorem. > taught it, students will use it *exclusively* for all limit > problems thereafter (it's a formula isn't it) no matter > how inappropriate it is for the problem at hand. So, tell the students that they will be marked down for inappropriate methods just as they will be for wrong answers. Alternately tell them to Prove/find ... without appeal to L'Hospital's theorem. -- G.C. === Subject: Re: multiplication negs boundary=----=_NextPart_000_004B_01C38E65.85C09DE0 --------------------------------------------------------------------- OK, what you say makes sense but... (and I really don't think this is just semantics): we're not talking here of adding something 'minus 2 es'; rather, we're talking about adding negative something x amount of es. So, in -(m)*-(n), where m=2, and n=3... we have: (-2) + (-2) + (-2) = -6 .... right? or are you saying it would be: -[ (-2) + (-2) + (-2) ] = 6.... (which, obviously, is the correct answer, but I fail to see the move here.. even though I can figure it out.) P.S.: I must ask you to forgive my ignorance, as I am but a lowly student of Philosophy, with little math skills. === Subject: Re: multiplication negs Aha! That, is as much as I suspected. Nevertheless, I still like to understand the reasoning behind the movements... TY === Subject: Re: Boolean Algebra - Arithmetic Relationship >P.S. It may be that by boolean logic you mean computer hardware >gates. A physically realized (or realizable) logic is certainly >inadequate--only a finite number of the infinity of numbers could be >handled, and only finite approximations of numbers like pi could be >handled. > Replace approximations with descriptions and you'd be ok. The finite > number of gates could be programmed for symbolic math, in which case a Indeed. The numeral for pi might be the approximation 3.141592653 or the precise pi. But still you can't encompass mathematics in a finite number of gates. > mathematically exact pi could be handled in formulae or theorems up to a > cretain size or complexity (bounded indeed by some function of the number > of gates). (The gate count includes those needed to implemement memory, > including storage for the program, of course. All is finite here. So > clearly we're talking about a finite subset of recursive functions -- but There are functions which aren't recursive, in fact more that aren't than are. > one that is arbitrarily large if you have enough gates.) > Michel. -- G.C. === Subject: Re: Dirt Simple Proof Re: Symmetric Groups. : They both fix a syntheme --- but are synthemes dirt simple? Sweet explanation! That's perfect for my needs, thank you indeed! Justin === Subject: Re: Is it mass; or is it weight >>Now it's e for all of us to realize that a customary pound is a unit of >>force, > You know better, Donald. Pounds everywhere have always been units of mass. > Pounds force are such a recent bastardization that they are uniquely > identified by that name. > Back in 1959, the national standards laboratories of the United States of > America, Canada, the United Kingdom of Great Britain and Northern Ireland, the > Union of South Africa, Australia, and New Zealand got together and agreed on a > common definition of the most commonly used pound, the avoirdupois pound. > They defined it as a unit of mass exactly equal to 0.45359237 kg. > Of course, you already knew that, Dishonest Don. This is for the benefit of > anybody else who hadn't been paying attention. > So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net > which stretches an additional 4.4-ft before thrusting the man back, would > the potential energy of the net at the instant it plunges the man back be > U = mgh; > U = 220 * (36 + 4.4). > -- > Ayaz Ahmed Khan > Yours Forever in, > Cyberspace. This problem could be answered correctly if the sbf system is used (Stones, barleycorns and fortnights) RJ Pease === Subject: Re: multiplication negs > Hopefully, I'm in the right place here.... > I've been making my way throught the Principia, taking my e, working > along. One day I was talking to my young neice, helping her out with some > simple algebra. We were discussing mulitiplication and division of negative > numbers and my neice brought up a question and for some reason I have been > unable so far to reason my way through the answer: exactly why is it that > when one multiplies 2 negative numbers one ends up with a positive? It's > disturbing me that I cannot come up with a solid answer, and certainly > 'We've been taught that that is the case. will not suffice. > Any help would be greatly appreciated. > TIA One answer is that we want the distributive property of multiplication over addition, as well as certain other properties, to contintue to work with negatives as well as with positives. This also justifies why negative es negative should come out positive. One form of the distributive property says that for al x,y and z x*(y+z) = (x*y) + (x*z). Then, for positive x and y, x*(y + (-y)) = (x*y) + (x*(-y)) but y + (-y) = 0, and x*0 = 0 , so (x*y) + (x*(-y)) = 0, and z + (-z) = 0 for any z, so x*y + (-(x*y)) = 0, so x*(-y) = -(x*y). === Subject: Re: Is it mass; or is it weight >Now it's e for all of us to realize that a customary pound is a unit >force, >> You know better, Donald. Pounds everywhere have always been units of >mass. >> Pounds force are such a recent bastardization that they are uniquely >> identified by that name. >> Back in 1959, the national standards laboratories of the United States >> America, Canada, the United Kingdom of Great Britain and Northern >Ireland, the >> Union of South Africa, Australia, and New Zealand got together and >agreed on a >> common definition of the most commonly used pound, the avoirdupois >pound. >> They defined it as a unit of mass exactly equal to 0.45359237 kg. >> Of course, you already knew that, Dishonest Don. This is for the >benefit of >> anybody else who hadn't been paying attention. >> So, if a 220-lb man jumps off a roof-top down 36-ft to a stretchable net >> which stretches an additional 4.4-ft before thrusting the man back, would >> the potential energy of the net at the instant it plunges the man back be >> U = mgh; >> U = 220 * (36 + 4.4). >> -- >> Ayaz Ahmed Khan >> Yours Forever in, >> Cyberspace. >This problem could be answered correctly if the sbf system is used >(Stones, barleycorns and fortnights) Sure. But can you help me figure out what the units for m, g, and h would have to be, if I want to get the energy U in barn yard atmospheres? Gene Nygaard http://ourworld.compuserve.com/homepages/Gene_Nygaard/ === Subject: Re: Question on Hilbert & Godel > What did Hilbert ask and claim concerning Foundations of Mathematics > (sets, predicate calculus), metamathematics, Logic, Incompleteness, > etc? > Something about the Continuum Hypothesis? Completeness or > Incompleteness of Logic? Was he contradicted by kindergarten logic > from Godel? How would we exactly formally represent Hilbert's > questions and claims using current day notation? You're in luck, if you read Hilbert [1,2,3] you'll find that it's in today's notation. [1] Hilbert & Ackerman Principals of Mathematical Logical [2] Hilbert & Bernays Grundlagen der Mathematik, I, II (Only available in German?) The last item puts it all in an historical context. -- G.C. === Subject: Re: Antidiagonal, Infinity What I propose is that given any >rational that the value greater than it and less than any other >greater is irrational, There is no such number, as several different people have shown you. In non-standard analysis, there might be, however. > See Alain Robert's book about NSA. Rather than being > irrational, it would be non-standard, though. I have yet to see any standard or non-standard model of the reals in > which there is a smallest positive number. Who says that that is meant? > In nsa, there can be a nonstandard number that can be to be > 'greater than (a given standard rational) and less than any other (standard) > greater' > That it is not unique, who cares? > Ross wants to use non-standard numbers to confirm his hypothesis > that there is a next real after any real, and that the irrationals > and rationals alternate on the real line or on some non-standard > real line. So he cares. On the other hand, his hypothesis is way out > in left field, where he has been stuck for months, if not years. Actually, I have not been claiming to be using the nonstandard numbers. In light of the issues that we cover and mutually understand to some extent, it might be the case that of the set of reals that for each that it has a next element of opposite rationality that that would be in neither the nonstandard nor classical model. Its elements would still be the elements of the set of real or hyperreals. About equating density with measure in the unit interval, that means equating density in the unit interval to measure in the unit interval. In this model, say alternating analysis, or a rationally alternating model of the continuous reals, the alternating measure of the rationals in [0,1) is 1/2, as is that of the irrationals, as is their density in the unit interval and the density of the rationals in the reals. This doesn't disagree with the measure of the reals in the unit interval being equal to one. The results of classical analysis would hold true. Anything that didn't hold true would be suspect. What's a proper subset of the rationals or irrationals that is dense in the reals, where its complement is infinite? The rationals and irrationals are each dense in the reals, their union, the reals are continuous. Ross === Subject: Re: 0! = 1 (episode 2) correction > Correction, i mean > (-1/2)! = sqr(pi) But for not integer arguments, generally is used Gamma(x + 1) -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Dedekind Cuts >> I have heard that the reals can be defined by Dedekind cuts. What is >> this definition, exactly? >> This is a division of the rationals into a pair of sets, which we will >> call Left and Right. Each element of Left is less than all elements of >> Right. Each element of Right is greater than all elements of Left. It >> follows that Left intersect Right is empty. We also require that a >> rational number either be in Left or Right so that Left union Right = >> the set of rational numbers. >> In the case that Right has a lest element we put that element in Left. >> There are two case. Left has no greatest element and Right has no least >> element. If so the division or cut Left,Right corresponds to an >> irrational number. Or Left has a greatest element and Rigt does not. >> This sup of Left is the number (a rational) represented by the cut. >> There are rules of addition, subtraction, multiplication and division >> defined for cuts and it is shown they form a field. >> For a good account of Dedikind Cuts set -A Course in Pure Mathematics- >> by G.H.Hardy. > I understand what a Dedekind cut is, but what what is the definition > that goes the set of reals is some collection of Dedekind cuts? > If a real number is defined to be a Dedekind cut, then it follows that > the set of real numbers is the set of all Dedekind cuts. > Is it simply the set of of all Dedekind cuts of the rationals? Or is > it the set of numbers such that all Dedekind cut of the reals the > reals yields a greatest element for Left? Or something else? > Dedekind cuts are defined on the rationals. You could carry out a > similar construction on the reals, but it wouldn't be a Dedekind cut. The purpose of the Dedekind cut construction, or the equivalence classes of Cauchy sequaences construction of the real fraom the rationals, is to create a complete ordered field, in which for every non-empty bounded-above set there is a least upper bound. The rationals do not have this property, since, for example, the set of rationaTs S = {q: q^2 < 2, q in Q} has no rational least upper bound. A similar construction based on cuts of the set of reals has been done, but the result is isomorphic as an ordered field to the standard Dedekind construction real field. === Subject: Re: Group generated by a and b ... > I have a group G generated by 2 elements a and b such that ba=a(b^k) for some > integer k. > Is it possible to express b(a^(-1)) as a product (a^m)(b^n) for some integers m > and n ? >>Yes, when the order of a is finite. >But in general, no. >As a simple example, let G be the multiplicative matrix group over the >rational generated by >a = [2 0] and b = [1 0] > [0 1] [1 1] >Then ba = ab^2, but >ba^-1 = [1/2 0] > [1/2 1] >whereas all products (a^m)(b^n) have integral entries. Sorry - that last statement is not correct for negative m. But, in general, (a^m)(b^n) = [2^m 0] [n 1] which is unequal to ba^-1 for all integral m,n. Derek Holt. === Subject: Re: multiplication negs > I've been making my way throught the Principia, taking my e, working > along. One day I was talking to my young neice, helping her out with some > simple algebra. We were discussing mulitiplication and division of negative > numbers and my neice brought up a question and for some reason I have been > unable so far to reason my way through the answer: exactly why is it that > when one multiplies 2 negative numbers one ends up with a positive? It's > disturbing me that I cannot come up with a solid answer, and certainly > 'We've been taught that that is the case. will not suffice. > Any help would be greatly appreciated. The law of signs is a consequence of the well-known elementary school laws of arithmetic - most notably the distributive law, e.g. Claim: a b equals (-a)(-b) Proof: a b + [a + -a](-b) = a[b + -b] + (-a)(-b) Equivalently, evaluate in two different ways the following expression -------------- a b + a(-b)+(-a)(-b) ----------- by undistributing xy+xz -> x(y+z) the overlined and underlined terms. This is a FAQ (Frequently Asked Question) here, e.g. see earlier threads -Bill Dubuque === Subject: Re: Factorial/Exponential Identity, Infinity > I got to thinking about sum 2^n and how it was equal to 2^(x+1) - 1. > I wonder: does it work for 3? The answer is not quite, no. > Look up geometric series. > If you must reinvent the wheel, you need not do in so publicly. I guess it's kind of like how the sum for i=1 to oo of b^-i is 1/(b-1). The concept here is the repeated unit, the rep-unit, in base b. The sum for i=1 to n of b^i is the sequence 1_n 1_(n-1) ... 1_2 1_1, a sequence of 1's of length n representing an integer in base b. Then b^(n+1)-1 is a sequence of n many (b-1)'s in base b: (b-1)_n (b-1)_(n-1) ... (b-1)_2 (b-1)_1, dividing that by (b-1) has as a result the rep-unit of length n. In summing the negative integers powers of an integer, it is similar, represent the number in that base so that the sum is simply .1111.... If you multiply that by (b-1) then the result is .(b-1) (b-1) (b-1) (b-1) ... = 1. These are well-known and almost trivial, for example, the sum for i=1 to infinity of 1/25000^i is 1/24999, 1/9 in decimal is .1111..., etcetera. I assume that for real x>1 that the sum of x^-i for i from 1 to oo is 1/(x-1). I will research geometric series. I'm not here to reinvent the wheel, but it damn well better let me say that half the integers are even, as they are as the asymptotic density of the even numbers in the integers is one half. What do you think about the canonicalization of the infinite binary sequences? That is to say, for the sequence .010101(01)... with half zeroes and half ones, that you could exchange any two sequence elements any number of es and not get .001001001(001).... Is there a rule to exchange (permute) the elements of the sequence .0101(01)... to get 00010001(0001)...? Ross === Subject: Re: The Bible Code modern, but ex post facto, with a few rather repugnant exceptions (Die, Rabin, Die!) the sequel has this doofus, Drosnin, supposedly watching the WTC hit from his hotel room, and immediately running the program to post-dict it. the second book also shows (I think it was indexed, or I was just lucky in finding it, as I didn't *read* the God- am thing) that he got the original idea from one of Sharon's buddies (presumably) in hte military! yes, the NT can be used to equal effect (on everage). > repeat, _War and Peace_ or just the 26 letters in any order > can be used with the infinite set of co-prime skips, > with teh resulting hits being further massaged > into some m by n array (or what ever). > The Code stuff has been applied successfully to Moby Dick to predict > moderns events with Astounding accuracy. (NOT!) --le ducs d'Enron! http://larouchepub.com === Subject: Re: Re fermat by Tomas sorry; I guess that you meant the second root of 3 by %3 -- but it's still a rather silly use of the law of large numbers. an example that may apply, since I don't recall what that is: there's a proof that li(x) goes less than pi(x), the number of prime numbers less than x, infinitely often, although there is (was) no known place in the sequence where it occurs. the second (skware) root of 3 is irrational by *definition*. > Fernat's Last Theorem is negative is not problem, just as > proofs (of neg or pos statements) by contradiction are good. > of course, three per cent is rational by definition. > Negative statements about numbers are unverifiable. Take the --les ducs de Buffet et Schulz (R&D Chair Assoc.Intl.); vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall/e-Dereg! http://larouchepub.com http://members.tripod.com/~american_almanac/ === Subject: Re: Exercise in Projective Linguistics >When I posed the problem, I just assigned names A, B, C, ... to the >points and gave the lines as 5-tuples. Later, it occurred to me >I might have been able to find a different assignment of the >points into the alphabet in such a way that each line was >( a permutation of ) the letters in a 5-letter word. >So here is the challenge: how well can you do ? Can you embed >these letters back into the alphabet so that all, or most, of >these 5-tuples become English words? (Interpret word as you wish.) >(I would be willing to take a solution which works only for >an affine plane inside here, if necessary.) > This might be a good one to send to rec.puzzles. I doubt that you can get > all 21. The best I've been able to come up with so far is 13, with > AB...U = DYCAOEBTJRMLSNKUGFWIV generating the words > cadgy, fumed, blind, folky, nervy, misty, curio, clews, > woman, barfs, vault, begot, gunks I can get 14: AB...U = YISLKDAWZCGNTEROMPUHB generating the words slimy, podgy, hyena, prink, cebid, wight, hocks, dunts, brags, kluge, clapt, blown, metro, bumph === Subject: Re: mandelbrot iterations correction: the talk was given at Royce Hall Auditorium, which should henceforth be known as On the Subject of Magnifications in the Complex Plain, or Why Didn't I Go Down the Hall and Talk to the Engineers at teh IBM-Watson Thingy, for All Those Years? or, What is Chaos?... The Maternal Unit of Chronos! I can tell of the original insight, but when I found some issue of 1980. I'd been using an IBM-XT clone 8088 chip, which uses the famous dysabled-to-8-bit-16-bit-path, and the registers hold 80 bits. not that I ever generated any fractals on it ... or any math ... what else can a screensaver be used to dysplay, other than blackspace? I've never seen teh update, -855 (I'm not sure of these #s), but it only tries to minimize the overflows & underflows; it explicitly dyscusses these. > changing the hardware setting on the machine from double-precision > to single, he pooh-poohed it -- the grad student that I found. > the mini-Ms do not appear at *every* magnification, since > the rounding-errors are tied to the lengths of the registers, > the specification is inherently chaotic, > that is the recurrence of mini-bugs or cardioids, > at every level of magnification, is just an artifact > of the floating-point ops (IEEE-755, -855, I think). > or doesn't change, in calculations. FP doubles are good for several > tens of thousand of iterations, down to an area of 10E-10 or so, > before the precision gives out. The cartoids are visible several > orders of magnitude above this. --les ducs de Buffet et Schulz (R&D Chair Assoc.Intl.); vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall/e-Dereg! http://larouchepub.com http://members.tripod.com/~american_almanac/ === Subject: looking for a function I'm looking for a function which garantuees the following conditions : Suppose you have an m*n matrix. Each element of the matrix has a value between 1 to p. Is there a function which ensures that for each element the element left, right, above and under are a different value than the specified element. in the case n is not dividable by p the following function seems to work : S { [(x%n)+1]%p in case x/n is even (rounded) { [(x%n)+1+p/2]%p in case x/n is odd (rounded) -example correct matrix : [1 2 3 4 5 1 2 3 ] [4 5 1 2 3 4 5 1 ] [2 3 4 5 1 2 3 4 ] -example incorrect matrix (6/2=3) : [1 2 3 1 2 3] [1 2 3 1 2 3] [1 2 3 1 2 3] though there is a correct solution : [1 2 3 1 2 3 ] [2 3 1 2 3 1 ] [3 1 2 3 1 2 ] I could split up the function into another part but i think there is an easier function (or maybe i'm just hoping) All ideas are more than welcome. Yves (sorry for the crossposting in alt.math but there wasn't much life there) === Subject: Re: Antidiagonal, Infinity > Ross wants to use non-standard numbers to confirm his hypothesis > that there is a next real after any real, and that the irrationals > and rationals alternate on the real line or on some non-standard > real line. So he cares. On the other hand, his hypothesis is way out > in left field, where he has been stuck for months, if not years. > Actually, I have not been claiming to be using the nonstandard > numbers. In light of the issues that we cover and mutually understand > to some extent, it might be the case that of the set of reals that for > each that it has a next element of opposite rationality that that > would be in neither the nonstandard nor classical model. Its elements > would still be the elements of the set of real or hyperreals. As the classical model is the reals and a non-standard is the hyperreals, you are claiming to simultaneously be inside of and outside of some model simultaneously, which certainly puts you outside of rationality. > About equating density with measure in the unit interval, that means > equating density in the unit interval to measure in the unit > interval. And how do you do equate density to measure? It would help if you could explain what YOU mean by dense and what you mean by measure, as the standard meanings do not allow of equating these ideas. > In this model, say alternating analysis, or a rationally alternating > model of the continuous reals, the alternating measure of the > rationals in [0,1) is 1/2, as is that of the irrationals, as is their > density in the unit interval and the density of the rationals in the > reals. But every e you have a rational, x, and irrational, y, supposedly next to each other, you have the problem of (x+y)/2 between them, so they aren't really next to each other after all. > This doesn't disagree with the measure of the reals in the unit > interval being equal to one. Yes it does. For each of uncountably many irrationals, x, linearly independent over the rationals, consider U(x) = (x+Q) / [0,1), the intersecting ofx+Q and [0,1), where x+Q = {x+q: q in Q}. Since each x+Q is essentially a translation of Q, each must have the same measure, but, as they are pairwise disjoint, the measure of [0,1) must be the sum of their separate measures. Thus, according to your arithmetic, uncountably many measures of 1/2 add up to 1. So your measure is self-contradictory. > The results of classical analysis would > hold true. Anything that didn't hold true would be suspect. Your formulation must be suspect, then, since it doesn' hold true. > What's a proper subset of the rationals or irrationals that is dense > in the reals, where its complement is infinite? Let n be an arbirary integer grreater than 1, Z be the set of integers, and Z[1/n] be the ring generated by appending 1/n to Z and taking the closure under addition and multiplication. Each such ring will be dense in the reals and have infinite compliment in the rationals and in the reals, furthermore, if n is a proper factor of m, then Z[1/m] is a proper subset of Z[1/n]. Thus each of Z[1/2], Z[1/4], Z[1/8], z[1/16], ..., is a proper subset of all its predecessors in this sequence, but each is also dense in R. > The rationals and irrationals are each dense in the reals, their > union, the reals are continuous. Ross, what do YOU mean by continuous? For the standard meaning of continuous, a function may be continuous, but not a mere set. What property does the ordered field of reals have that the ordered field of rationals lacks that makes the set of reals continuous in your sense but the set of rationals not continuous in your sense? === Subject: Re: Convergence in distribution > Yes,I mean random 'sequences'. > But BOTH X_n and Y_n are random sequences converging in distribution. > Does that make any difference to the answer? It makes a difference to the question. One doesn't usually talk about one sequence converging to another sequence, but presumably what you're talking about is equivalent to saying that for any bounded continuous real-valued function f on the real line, E[f(X_n)] - E[f(Y_n)] -> 0. > Well,the actual pdf's of X_n and Y_n at a given n are reasonably well modeled > by GGD(Generalized Gaussian Distribution)'s with zero means and variances > typically 25~100 while the delta's can be 1~10. > Although we can easily find counter-examples with discrete probability > distributions, I observed in the actual problems that the 'quantized'(i.e. > X1_n & Y1_n) sequences seemed to 'approach' in distribution. (By 'approach',I > mean the variances and the shape parameter characterizing the GGDs of both > X1_n & Y1_n approached the same values.) It's also easy to find counter-examples with continuous probability distributions. But if the X_n and Y_n are continuous random variables with uniformly integrable densities (i.e. some integrable function g such that f_{X_n}(x) <= g(x) for all n and x, and similarly for Y_n) it should work, because this ensures that the probability of X_n or Y_n being very close to one of the discontinuities is small. === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed? > [...] So try ARCSIN(x) = 2*ARCTAN(x/(1+SQRT(1-x^2))). === Subject: Re: lopital's rule? >> Why's that then? It seems to me to be a mildly interesting theorem. I >> can see that it might not be a good idea to prove it and then give >> students lots of 0/0 limit problems for them to solve using >> L'Hospital's when doing so is just pedagogically useless manipulation. >> But that is no reason for not proving the theorem. >> Who anything about proving anything. My experience >> is that it is taught as a method, not as a theorem, and when > I'm probably being dim, but a method is only useful if it works. Hence > one needs a proof that the method works. Such a proof is a proof of a > theorem. Try telling students that :-) >> taught it, students will use it *exclusively* for all limit >> problems thereafter (it's a formula isn't it) no matter >> how inappropriate it is for the problem at hand. > So, tell the students that they will be marked down for inappropriate > methods just as they will be for wrong answers. Alternately tell them > to Prove/find ... without appeal to L'Hospital's theorem. Alas, it's not me who teaches that course :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 es) === Subject: Re: Chessboard knight metric? >>Take a chessboard (with or without infinetely many squares) let the >>distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >>chessboard be defined as the minimum number of moves a knight takes >>to reach y from x. >>Is d a metric? >With this distance, the triangular equation is obviously true, and that >makes it a metric. >> Not trying to suggest that this is some new >>question that hasn't been asked/answered before. Is there a general >>formula for calculating d? More generally, the same question may be >>asked for the other pieces (queen, king, knight, biship)? Actually, I >>asked myself this question a few years ago. If I remember back to the >>notes I took, I had something like (x_1-y_1, x_2-y_2)= (even number, >>even number), then d(x,y) = even number. If (x_1-y_1, x_2-y_2)= (even >>number, odd number), then d(x,y) = odd number. Finally, if (x_1-y_1, >>x_2-y_2)= (odd number, odd number), then d(x,y) = even number. In other >>words, the same rules for adding natural numbers... C.Dement >With this distance, the triangular equation is obviously true, and that >makes it a metric. >I don`t (yet) see the triangle inequality as so inherently >obvious that it is not worth further thought. Take the >points x=(1,1), y=(77,79), and z=(163,163). It is still feasable >that d(x,y)+d(y,z) is less than d(x,z) (especially as an >explicit formula for d doesn`t exist). The fact that I would >be somewhat surprised if that wasn`t the case shouldn't be used >as part of the proof. > C.Dement With your definition: the minimum number of moves . . ., the triangle equality is obviously true. If d(x,y) + d(x,z) < d(x,z) then your d function does not obey your definition since if going from x to y and then y to z is less than x to z directly the value for d(x,z) cannot be the minimum. If you have a particular formula in mind then checking the triangle equality is partially checking whether the formula obeys your definition. It would not be checking whether your definition is a metric. That is clear. It is necessary to check that your definition is well defined. You could do this by first checking that a knight can get between any two squares in a finite number of moves. This is easy. First check how to go one square horizontally in either direction. Then check also vertically. Combinations of those moves will clearly get you between any two squares in a finite number of moves. Of course, this is unlikely to be the minimum. But now we know an upper bound for the distance between any two pairs of squares. Since your metric clearly is integer and non-negative there must be a unique answer to the minimum distance for any pair of squares. I am not saying that there is any easy formula for the answer. Maybe there is and maybe there is not. I have not thought about it yet. But the existence or otherwise of an easy formula does not detract from the existence of the distance function and the fact that it is a metric. The similar queen, castle and bishop metrics are less interesting. A castle can get between any two squares in at most 2 moves. In fact the formula is simple, in pseudo code: if x1 = x2 and y1 = y2 then distance = 0 else if x1 = x2 or y1 = y2 then distance = 1 else distance = 2 The queen is similar but also has distance 1 in the diagonal cases: abs(x1 - x2) = abs(y1 - y2) The bishop is a little tricky since it cannot get between squares of different colours. So it is not defined or not-finite between squares of different colours. It would be a boring metric on the two single colour subspaces. J === Subject: how many resolutions? A^2 + B^2 + C^2 + D^2 = 85^2 where a to d are different one-digit or two-digit numbers. I thought that it could have just one solution: since 85 is a product of two prime numbers of 4k+1 type: 85=5*17. As it is known that only prime numbers 4k+1 could be resolved as p^2=a^2 + b^2. so 85^2=5^2 * 17^2 = (3^2 + 4^2)*(15^2 + 8^2) so right numbers are 24, 32, 45, 60. But there is another solution! 3, 4, 12, 84. Why? And maybe there are some other solutions left? It means that resolution p^2=a^2 + b^2 is ambiguos? So we could find c and d (c not equal a and b) such that p^2=c^2 + d^2? Or not? === Subject: Re: how many resolutions? >So we could find c and d (c not equal > a and b) Sorry for mistype - I meant c and d not equal a and b === Subject: Re: Algebra proof > Let a, b, r and s be integers with r>1, s>1 and d(r,s)=1. Prove > that if a = b (mod r) and a = b (mod s) then a = b (mod rs). How do I do this???? One way: r,s|a-b => lcm(r,s)|a-b But d(r,s) = 1 => lcm(r,s) = rs via d(r,s) lcm(r,s) = rs >> a = b + mr for some m, and a = b + ns for some n, right? That means >> that mr = ns. Now use d(r,s) = 1 to conclude something about m (or n). > I guess since r and s are relatively prime, then r must divide n. so > rx = n. and a = b + rxs, so a = b (mod rs), right? > Is there a name to this theorem? That is, the theorem that says that > (a,b) = 1 & da = bc => a|c and b|d ? EUCLID'S LEMMA: a | bc => a|c if (a,b) = 1. Equivalently: UNIQUENESS OF REDUCED FRACTIONS: c a c = na d(a,b)=1 => - = - <=> d b d = nb See my prior posts for much more -Bill Dubuque === Subject: Re: Factorial/Exponential Identity, Infinity > I will research geometric series. I'm not here to reinvent the wheel, > but it damn well better let me say that half the integers are even, as > they are as the asymptotic density of the even numbers in the integers > is one half. Doing any sort of research in mathematics with a fixed idea of what you will allow yourself to find is not really a good idea. > What do you think about the canonicalization of the infinite binary > sequences? That is to say, for the sequence .010101(01)... with half > zeroes and half ones, that you could exchange any two sequence > elements any number of es and not get .001001001(001).... Is there > a rule to exchange (permute) the elements of the sequence .0101(01)... > to get 00010001(0001)...? It depends on what you consider an allowable permutation. The most general form of such permutation can be considered as follows. Let N be the set of positive integers, then any infinite binary sequence can be uniquely represented by a functions f:N -> {0,1} in which f(k) is the k'th digit of the sequence. Let B be the set of such functions f:N -> {0,1}. Then any bijection g:N <-> N creates a permutation G: B -> B by defining G(f) = fog, where o represents function composition, G(f) is the function whose valueat k in N is f(g(k), i.e., G(f)(k) = f(g(k)), for each k in N. Let P represent the set of all such permutations, It is clear that the number of zeros and number of ones in an expansion must remain fixed under any such permutation of digits. Then, for a given f:N -> {0,1} and h:N -> {0,1} in B, a necessary and sufficient condition for there being some permutation G in P such that G(f) = h is that card(k in N: f(k) = 0}) = card(k in N: h(k) = 0}) and card(k in N: f(k) = 1}) = card(k in N: h(k) = 1}) i.e., the infinite binary sequences for f and h have the same number, possibly infinite, of zeros and the same number, possibly infinite, of ones, and only one of these can be finite. But if the number of zeros or the number of ones differ, then there is no permutation carrying one into the other. This divides up the set B into equivalence classes of binaries permutable into each other which can be indexed by the integers, Z. Any two such binaries with infinitely many zeros and infinitely many ones in their expansions can be permuted into each other, and can be indexed by 0. If the number of ones is finite, use that number as index, and if the number of zeros is finite use the negative of that number as index. Then two binaries can be permuted into each other if and only if they have the same index. === Subject: Re: Exercise in Projective Linguistics > .... the plane over the field of 4 elements .... can be > described as an assembly of 21 points, some subsets of 5 of which > are called lines. > When I posed the problem, I just assigned names A, B, C, ... to the > points and gave the lines as 5-tuples. Later, it occurred to me > I might have been able to find a different assignment of the > points into the alphabet in such a way that each line was > ( a permutation of ) the letters in a 5-letter word.... > By the way, I hereby claim my title as inventor of a new field.... Not so. A few years ago I was shown a sentence something like Yea, why try her raw wet hat? whose connection with the projective plane of order 2 won't escape you. I _think_ the person who mentioned it was Burkard Polster, but his web site http://www.maths.monash.edu.au/~bpolster/ doesn't seem to have it now (although it has a lot of other nice things). Ken Pledger. === Subject: Re: a puzzle related to artinian group > A quick sketch of my argument for the linear case is that you number cards > and players and arrange the play so that the cards are always in order from > left to right and so that each player holds a contiguous set of card > numbers. [...] What's the role the numbering of the cards? > You then show that the centre of gravity of the cards is fixed, > that the variance of the positions of the cards relative to some point to > the left of all of them is strictly increasing whenever a round of play > takes place, [...] Nice!:) > As for the cyclic version? A similar idea should also work in the cyclic version. Although, it is not clear to me whether the center of the gravity and else make any sense in this case. One should find a suitable function that strictly increases in each round, except for the uniform configuration, where it is maximized. === Subject: Re: 0! = 1 > If I know my algebra, then an identity for an operation (let's call > it #) is such an i, that for all x, x#i=x. But there is another concept > too: such a j, that for all x, x#j=j. For multiplication, this is 0. > For addition, there doesn't seem to be one. What is this concept called? A small addendum to my previous response: You For addition, there doesn't seem to be one. But of course, that depends on how addition is defined, what the set of operands is, ... Here's a nice example where there is an element which is absorptive under addition: On R* = R U {oo}, the one-point extension of the reals, extend the operation of addition by specifying that x + oo = oo = oo + x for all x in R*. David Cantrell === Subject: Re: Minimal Graph, Four Color Theorem >> Indeed, the proof of the Four Color Theorem rests on showing that >> there does not exist any graph G which requires 5 colors, is planar, >> and such that the removal of any vertex results in a graph which can >> be colored with only 4 colors. But you have not given any coherent >> argument to establish this proposition that I can see anywhere. All I have is a hypothesis. The validity of the hypothesis does not depend upon the a priori assumption that G is planar. Therefore, you must automatically reject it. Why should I bother? === Subject: recurrence relation question hi, how do i go about solving a recurrence relation when the function T(n) is multiplied by n? such as the following: 2n*T(n) + 2nT(n-1) - 2T(n-1) = 2^n? What tricks would one use? === Subject: Re: lopital's rule? > Guy Corrigall scribbled the following: >> L'Hospital's Rule (pronounced lopital in French) can be found in >> any book on calculus of a single variable (look it up!). > L'Hospital or L'H.99pital. The circumflex (the ^ thingy) means that the > following s is understood implicitly. AFAIK the guy's name is de l'h.99pital and in my pronounciation of the french language :-) the ^ makes the o-sound longer. === Subject: Re: Is this newsgroup useless? at 03:34 PM, Bart Goddard : >I'd be all for sci.math.moderated under a charter that specifies that >the moderator must reject any post that refers negatively on the >character of another person. Well, get some volunteers, put together an RFD and steer it through the process. Be warned that it is a lot of work, but it can be done. -- === Subject: Re: Is this newsgroup useless? <3F81A4DB.2364E9EE@ix.netcom.com> at 06:43 PM, Bart Goddard : >Half the problem here is that it's unmoderated. The other half is >that it's populated by overly-anal-retentive jerks who think it's >going to be important to point out that I spelled you're your >twice in the preceding paragraphs. PKB. You're upset about flames, yet you post them yourself. You claim to be a Christian; don't your scriptures say soething about the beam as 90% of the people you're complaining about. >I _thought_ we had beaten these guys up sufficiently in the high >school locker room that they'd be quiet by now. So by you correcting an error is wrong, but battery is acceptable behavior? And you have the arrant hypocrisy to call others jerk. >Maybe the real problem is that a guy can't e-mail a matburn. The real problem is that people like you are hypocrites. That's far worse than any amount of flaming. I suggest that you read what you posted, and then explain how you're any better than JSH. Better yet, print out a copy and ask your pastor what he thinks of it. -- === Subject: Re: Is this newsgroup useless? <3F81A4DB.2364E9EE@ix.netcom.com> at 08:19 PM, Bart Goddard : >It wouldn't be so hard to read a message and decide whether to post >it. Does that mean that you're volunteering? -- === Subject: Re: Antidiagonal, Infinity <3f79d8b6$5$fuzhry+tra$mr2ice@news.patriot.net> <3f7cafc2$11$fuzhry+tra$mr2ice@news.patriot.net> <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> at 07:09 PM, raf@tiki-lounge.com (Ross A. Finlayson) : >Here I equate density with measure in the unit interval. What do you mean by measure? With the obvious meaning, the rationals have measure 0. As has been explained to you. >only concerned with considering a model where the rationals and >irrationals alternate in the reals. That's like saying that you're concerned with a model where 1+1 > 1. It's not a model for the reals, because it doesn't have the properties of the reals. >rationals and irrationals are disjoint and distinct, What are you trying to say? >then there necessarily would be irrationals with no >rationals between them. Would you care to provide a proof of that? >I like to think that the rationals and irrationals alternate I like to think that the nice gentleman with the map will make me rich. Alas, if I do believe it then I will lose a substantial amount of money. It's false. >and that >the function f(x)=x+iota maps Q[0,1) onto P(0,1), and f(x)=x-iota >maps Q(0,1] to P(0,1). Also false. >half infinity Meaningless. >finite multiples of a scalar infinity. Also meaningless. >a set of a vector Meaningless. >One thing to note is that *R, the hyperreals, as a set >contains the same elements as R, the reals. Incorrect. >the characteristics of a probability distribution What do you believe them to be? How do you apply them to an infinite set? >We were talking >about the probability of an infinite binary seqence having one >element being on, the rest off. No we weren't, because you've refused to define what you mean by that. >That probability is expressed as n/2^n, as n >diverges to infinity. Meaningless. Assuming that you meant the limit of n/2^n as n approaches infinity, that comes out to 0. >The probability of each among all possible infinite >binary sequences is being1/2^n, What is n? You seem to be confusing bound and unbound variables. >So anyways out of those n >possible sequences with one on bit and the rest off bits, each is >equally probable. Yes 0=0. >So a theoretical (read: thought experiment) Theoretical is not the same as geken experiment. >method to generate an element of N is to >once again flip infinitely many coins. You can't. >At this point it's a crazy, or >rather, unconventional thought experiment in that the first coin >toss says whether it is oo/2 or greater or less than oo/2. No, because oo/2 has no meaning. >talk about the probability of selecting a given element of the >natural integers assuming a uniform probability distribution over >the integers. That's easy; it doesn't exist. >At least we seem to have some agreement that a uniform >probability distribution over an interval of the reals exists, No, because we don't have common understandings of what any of the words mean. >simple method to sample an element of an interval of the reals >exists. No, because a finite number of coin tosses only lets us sample a finite set. >infinitesimals, And it was clear that your understanding was fuzzy and incorrect. Mathematics is a precise discipline; you must reason from the axioms, definitions and rules of inference, not from your preconceptions. -- === Subject: Re: Antidiagonal, Infinity <3F716897.4040005@tcs.inf.tu-dresden.deqqqq> <3f79d8b6$5$fuzhry+tra$mr2ice@news.patriot.net> <3f7cafc2$11$fuzhry+tra$mr2ice@news.patriot.net> <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> at 09:08 PM, Herman Jurjus : >?? Perhaps i'm missing something. Slip of the finger; I meant to reply to the consecutive part. >In non-standard analysis, there might be, however. No, because you can take the average of two nonstandard reals. >Actually, there is an axiomatic approach of NSA in which a few >axioms are added to ZF(C), and in which the above suggestion makes >sense. The Devil is in the details. The nonstandard reals are a model of the reals only in the sense that the same propositions are true with nonstandard definitions of various terms. You can't construct an isomorphism between them. Further, even with those axioms there wouldn't be a successor function. >The other poster might be interested in this approach towards NSA. He doesn't have the background for it. He'd need to start with, e.g., Halmos, and work his way forward. -- === Subject: Re: Antidiagonal, Infinity <3F716897.4040005@tcs.inf.tu-dresden.deqqqq> <3f79d8b6$5$fuzhry+tra$mr2ice@news.patriot.net> <3f7cafc2$11$fuzhry+tra$mr2ice@news.patriot.net> <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> In nsa, there can be a nonstandard number that can be to be >'greater than (a given standard rational) and less than any other >(standard) greater' There is still no alternation of rational and irrational. Nor is it plausible that the OP meant images of rational and irrational under an imbedding into the nonstandard reals. -- === Subject: Re: Numeric one-way hash function > Because you snipped my proof that your original procedure > generates a lot of duplicates, I assume that you could > not find any errors in the proof. Two separate arguments in the same posting tend to lead to problems, best to separate them. Since AES encryption has a 1 to 1 translation printing the entire 128 bits would give a unique number to each ticket. Is this consistent with your assumption that a sample of a sample of AES+CTR output can be considered to be random numbers? After all they appear to be opposites. I could believe either answer. Andrew Swallow === Subject: Re: quantum echo Are you saying it is possible for something to exist at location A and > location B at the same e? I thought it was inference by the rule of > non-contradiction. > I have no idea whether this is possible or not. > My suggestion that quanta may satisfy '~(x=x)' is meant to translate > into logical terms the claim by some physicists that quanta > 'lack individuality' (Heller suggests that quanta 'lack haecceities'). > The scenario with cakes-to-be suggests that the behaviour of other > entities without individuality resembles that of quanta in the > relevant respect. Heller's description of that behaviour follows > my sig. > Metaphysical Background, Thomas McTighe asserted that the quiddity of a > thing is nothing other than unity itself. Hence, by virtue of its positive > content, the sun differs not at all from the moon or any other particular > thing. The diversity which is exhibited by the natural world is merely the > product of accidental differences; no object possesses any specific form > which interposes itself between a particular existing thing and the source > of their being e.g. the Absolute.15 All individual entities are nothing more > than differing contractions of the whole devoid of any being of their own. > ...because the restricted quiddity of a thing is the thing itself. > http://www.crvp.org/book/Series01/I-10/chapter_ii.htm I take a quiddity to be a thing's *suchness* and a haecceity to be its *thisness*. A *thisness* I take to be, as Robert Adams does, the property of self-identity, although I disagree with an assumption which might be read into Adams, that an individual can lack self-identity and yet possess the property of being some individual or other). In Primitive Thisness and Primitive Identity (_The Journal of Philosophy_, Vol. 76, No. 1. (Jan., 1979), pp. 5-26), Adams A thisness is the property of being a certain particular individual, not the property of being some individual or other, but my property of being identical with me, your property of being identical with you, etc. These properties have recently been called 'essences', but that is historically unfortunate, for essences have normally been understood to be constituted by logical properties, and we are entertaining the possibility of nonqualitative thisnesses. In defining 'thisness' as I have, I do not mean to deny that universals have analogous properties--for example, the property of being identical with the quality red. But since we are concerned here principally with the question whether the identity and distinctness of individuals is purely qualitative or not, it is useful to reserve the term 'thisness' for the identities of individuals. It may be controversial to speak of a property of being identical with me. I want the word 'property' to carry as light a metaphysical load here as possible. 'Thisness' is intended to be a synonym or translation of the traditional term 'haecceity' (in Latin, 'haecceitas'), which so far as I know was invented by Duns Scotus. Like many medieval philosophers, Scotus regarded properties as components of the things that have them. He introduced haecceities (thisnesses), accordingly, as a special sort of metaphysical component of individuals.[4] I am not proposing to revive this aspect of his conception of a haecceity, because I am not committed to regarding properties as components of individuals. To deny that thisnesses are purely qualitative is not necessarily to postulate 'bare particulars', substrata without qualities of their own, which would be what was left of the individual when all its qualitative properties were subtracted. Conversely, to hold that thisnesses are purely qualitative is not to imply that individuals are nothing but bundles of qualities, for qualities may not be components of individuals at all. (pp. 6-7) Note 4. Johannes Duns Scotus, _Quaestiones in libros metaphysicorum_, VII. xii. schol. 3; cf. _Ordinatio_, II.3.1.2, 57. I am indebted to Marilyn McCord Adams for acquainting me with these texts and views of Scotus, and for much discussion of the topics of this paragraph. > Aristotle and Aquinas and Scotus and Bonaventura all believed that human > minds can conceive and express the intelligibilities or quiddities of things > and their properties, intelligibilities that are not simply mind-dependent. > We can capture in thought and language the actual natures of things, > spelling out their genera and specific differences. Definition brackets or > delimits for us as knowers just what it is we attempt to understand and > nothing else. The mind-independent thing-substance or the characteristics > that we are attempting to define measure the epistemic correctness of a > definition. Such real (as opposed to nominal) definition relies on the > intelligible and perceptible characteristics thing-substances exhibit to > perception and thought for understanding what they are and for picking out > individuals of a type. In this way the epistemological realism of the > definition corresponds to an ontological realism of actual formal features > in mind-independent entities. > http://www.sunysb.edu/philosophy/faculty/lmiller/Delinonaliud.htm > After all, the three tenets that largely define Nicholas's 'metaphysic of > contraction' seem altogether remote from Anselm's Scholasticism. For Anselm > has no use for the triad of notions (1) that there is an infinite > disproportion between the Creator and His creatures, (2) that, therefore, > finite minds can never positively know what God is, given the alleged ground > (3) that He is the Coincidence of opposites, i. e., is undifferentiated > 'Being' itself, which, with respect to its Quiddity, can never be conceived > by anyone except itself. > http://www.cla.umn.edu/jhopkins/CusaAnselm.pdf === Subject: Re: Minimal Graph, Four Color Theorem Visiting Assistant Professor at the University of Montana. > Indeed, the proof of the Four Color Theorem rests on showing that > there does not exist any graph G which requires 5 colors, is planar, > and such that the removal of any vertex results in a graph which can > be colored with only 4 colors. But you have not given any coherent > argument to establish this proposition that I can see anywhere. >All I have is a hypothesis. The validity of the hypothesis does not >depend upon the a priori assumption that G is planar. Therefore, you >must automatically reject it. Why should I bother? I must automatically reject it? Mind reader, on top of everything else? Look, you started ->this<- thread by asking a question about the minimal counterexample argument to prove the 4 color theorem. You are certainly welcome to attempt to prove the 4 color theorem by an argument which is ->not<- the minimal counterexample argument, but what you were doing is ->asking about the minimal counterexample argument<-. If the reason you posted was because you did not understand what the author was saying, and if the reason you did not understand what the author you quoted was saying was because you were trying to think of the minimal counterexample argument as if it were your argument, then we've fixed it. Surely you now understand what the author was saying, and presumably you agree that the statement you quoted was correct (whereas you insinuated it was incorrect in your original post). If the reason you posted your question, however, was because you wanted to use any response as an excuse to discuss your idea, ignoring the text that you were supposedly asking about, then all I can say is you have lousy manners and I have no interest in discussing an argument with someone who resorts to misrepresentation in order to start the discussion. But it has nothing to do with any a priori assumption that G is planar; it would have to do with the fact that I find you an unappealing person to hold such discussions with, and that I honestly do not have the e to talk about the 4 color theorem. I answered your original query because I knew the answer and it seemed straightforward. If I was drawn into that discussion on false pretenses (as is become more and more apparent, given your insistent comments about your other thread, and now this comment about your hypothesis), then all I can say is Have a good day. === Subject: Re: ARCSIN function, single precision floating point. -- Example routine needed? > I'm trying to write ATAN2 function for a small basic language that has > IEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are > availible in the language. > I've tried a few methods I've found but the results are way off due to > low precision, rounding, etc. > Are there any repositorys of old fortran routines or algorithms that I > could use to get a good accuracy single precision routine. Speed or > space aren't as important as reasonably good accuracy. > Jon To get an idea how the corresponding production codes work, look at The Computation of Transcendental Functions on the IA-64 Architecture http://www.intel.com/technology/itj/q41999/pdf/transendental.pdf The reference for the classical approximations is the book: Cody Jr., William J. and Waite, William, Software Manual for the Elementary Functions, Prentice Hall, 1980 For c source code that can be easily converted into other langages see the Cephes library http://www.netlib.org/cephes/single.tgz (e.g. asinf.c, atanf.c) Pfoertner === Subject: Re: multiplication negs > Hopefully, I'm in the right place here.... > I've been making my way throught the Principia, taking my e, working > along. One day I was talking to my young neice, helping her out with some > simple algebra. We were discussing mulitiplication and division of negative > numbers and my neice brought up a question and for some reason I have been > unable so far to reason my way through the answer: exactly why is it that > when one multiplies 2 negative numbers one ends up with a positive? It's > disturbing me that I cannot come up with a solid answer, and certainly > 'We've been taught that that is the case. will not suffice. > Any help would be greatly appreciated. > TIA turn your sock inside out twice, and it is right again. === Subject: Re: how many resolutions? >A^2 + B^2 + C^2 + D^2 = 85^2 >where a to d are different one-digit or two-digit numbers. >I thought that it could have just one solution: since 85 is a product of two >prime numbers of 4k+1 type: 85=5*17. As it is known that only prime numbers >4k+1 could be resolved as p^2=a^2 + b^2. >so 85^2=5^2 * 17^2 = (3^2 + 4^2)*(15^2 + 8^2) so right numbers are 24, 32, >45, 60. >But there is another solution! 3, 4, 12, 84. There are lots of solutions, e.g. 1, 2, 38, 76 1, 4, 22, 82 1, 4, 58, 62 1, 10, 50, 68 1, 22, 46, 68 1, 26, 52, 62 1, 34, 38, 68 1, 34, 52, 58 1, 38, 44, 62 ... and those are just the solutions that start with 1. > Why? And maybe there are some other solutions left? It means that >resolution p^2=a^2 + b^2 is ambiguos? So we could find c and d (c not equal >a and b) such that p^2=c^2 + d^2? No, but not every way to get 85^2 as the sum of four squares comes from a product of two sums of two squares. === Subject: Re: recurrence relation question >how do i go about solving a recurrence relation when the function T(n) >is multiplied by n? such as the following: >2n*T(n) + 2nT(n-1) - 2T(n-1) = 2^n? >What tricks would one use? One trick is to collect the terms in T(n-1) and think of the new function S(n) = 2n*T(n). === Subject: Re: Algebra proof > Let a, b, r and s be integers with r>1, s>1 and d(r,s)=1. Prove > that if a=b(mod r) and a=b(mod s) then a=b(mod rs). > How do I do this???? Another way to prove this is to use the identity of Bezout (the extended Euclidean algorithm): if d = d(r,s) then d = u*r + v*s for some integers u and v. When you see the hypothesis d(r,s) = 1, often it is worthwhile to convert that into the identity of Bezout. For your problem, you have: 1 = u*r + v*s for some integers u and v since d(r,s)=1; a - b = c*r for some integer c since a=b(mod r); a - b = d*s for some integer d since a=b(mod s). Now, multiply the first equation by a - b to get: a - b = (a-b)*u*r + (a-b)*v*s. Substitute the third and second equations into the above equation to get: a - b = d*s*u*r + c*r*v*s = r*s*(d*u + c*v). Thus, r*s divides a - b. That is, a=b(mod rs). -- Bill Hale === Subject: Re: lopital's rule? > L'Hospital or L'H.99pital. The circumflex (the ^ thingy) means that the > following s is understood implicitly. > AFAIK the guy's name is de l'h.99pital and in my pronounciation of the french > language :-) the ^ makes the o-sound longer. In the case of l'h.99pital the ^ is here to recall the old noun hospital, but it doesn't alter the sound o. -- Julien Santini, France. === Subject: Re: Chessboard knight metric? >Take a chessboard (with or without infinetely many squares) let the >distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >chessboard be defined as the minimum number of moves a knight takes >to reach y from x. As far as I can see, this should give the distances (on an ordinary chessboard). The only exception I see is when your starting point O is at a corner, and you want to reach the next square on the diagonal. Then the distance is 4 and not 2. The paths are O -> P -> X -> A -> W -> S -> T. - - - - S W S W S W S T - - - - W A W A W S W S - - - - A W A W A W S W - - - - X A X A W A W S S X A X A X A X A W A W X A W P X P W A X A W S A W P X A X P X A W A W X A X A O A X A X A W S A X P X A X P X - - - - X A W P X P W A - - - - A X A X A X A X - - - - W A X A X A X A - - - - Karin === Subject: Re: multiplication negs > Hopefully, I'm in the right place here.... > I've been making my way throught the Principia, taking my e, working ^^^^^^^^^ Which Principia is that? > [...] -- G.C. === Subject: Re: Is this newsgroup useless? > at 06:43 PM, Bart Goddard : >>Half the problem here is that it's unmoderated. The other half is >>that it's populated by overly-anal-retentive jerks who think it's >>going to be important to point out that I spelled you're your >>twice in the preceding paragraphs. > PKB. You're upset about flames, yet you post them yourself. You claim > to be a Christian; don't your scriptures say soething about the beam > as 90% of the people you're complaining about. 1. I wasn't complaining, I was chatting with the guy who was complaining. 2. Christianity is not the same as pietism, and if you don't know the difference then don't try to preach to Christians about their own doctrine. 3. This was meant to be humour, dumbass. >>I _thought_ we had beaten these guys up sufficiently in the high >>school locker room that they'd be quiet by now. > So by you correcting an error is wrong, but battery is acceptable > behavior? And you have the arrant hypocrisy to call others jerk. >>Maybe the real problem is that a guy can't e-mail a matburn. > The real problem is that people like you are hypocrites. That's far > worse than any amount of flaming. > I suggest that you read what you posted, and then explain how you're > any better than JSH. Better yet, print out a copy and ask your pastor > what he thinks of it. I did. He laughed at the locker room part. Then I showed him your post and he suggested a wedgie. Bart === Subject: Re: Numeric one-way hash function > Because you snipped my proof that your original procedure >> generates a lot of duplicates, I assume that you could >> not find any errors in the proof. > Two separate arguments in the same posting tend to > lead to problems, best to separate them. OK. > Since AES encryption has a 1 to 1 translation printing the > entire 128 bits would give a unique number to each > ticket. Is this consistent with your assumption that > a sample of a sample of AES+CTR output can be considered > to be random numbers? As far as we know AES behaves like a pseudo-random permutation. When you stay sufficiently below the birthday limit (for AES that is 2^64) you cannot distinguish a pseudo-random function from a pseudo-random permutation. Because we only need to generate 10^12 numbers, which is slightly less than 2^40, we stay sufficiently below the birthday limit and therefore we can treat the outputs as a (pseudo-) random set of numbers. But if you don't believe my proofs, the easiest way to check your algorithm is to implement it for a smaller value of Max and then to check the number of duplicates. I expect that you will be highly surprised. greetings, Ernst Lippe === Subject: Re: Homeomorphism and boundaries at 06:15 PM, () : >f(int(A)) = f( union{B : B is open, B contained in A}) > = union{ f(B): B is open, B contained in A} (since f is >bijective) > = union{ f(B): f(B) is open, f(B) contained in f(A)} > = int(f(A)) No. Consider A=(0,2Pi) with the standard topoplgy, B=RxR with the standard topoplgy. Define f: A -> B as f(x) =(0,x). Int(A) = A, but Int(f(A)) is the null set. >f(clos(A)) = f( intersection{B: B closed, B contains A}) > = intersection {f(B): B closed, B contains A} > = intersection {f(B): f(B) closed, f(B) contains f(A)} > = clos(f(A)). No. Consider A=(0,2Pi) with the standard topoplgy. Define f: A -> C as f(x) = exp(i*x). cl(A)=A, but f(A) is not closed; it has the limit point 1. -- === Subject: Re: hw help -- continuity at 03:59 PM, : >Proving a function continuous at any point proves it continuous at >every point. f: R -> R, f(x)=x for x rational, f(x)=0 for x irrational. The function f is continuous only at 0. -- === Subject: Re: hw help -- continuity at 01:18 PM, : >But 1 is not any point, it is a specific point. any point includes every specific point. >To restate less ambiguously: >If you prove f continuous at an arbitrary point, then it is >continuous at every point That's just as bad. That can still be construed as either any or every. -- === Subject: Re: Is this newsgroup useless? <4oi5ovou332c2598pql5rvvcsb18k7r6f3@4ax.comOne learns not to step on anything pretty quickly... Ouch! But you're bragging. -- === Subject: Re: Is this newsgroup useless? at 04:56 PM, Bart Goddard : >I haven't been complaining at all. So you're not only a boor and a bully, you're also a liar. >I haven't been complaining at all. Just chatting with the original >complainer. Everyone knows what the really, really, absolutely >true problem is, however, and it is the insufferable rudeness of >several posters. Indeed, like boasting about beating people up instead of hanging their heads in shame. Oh, wait, that was you. It's too bad that some of the ones that you consider insufferably rude are not only more polite than you but know some Mathematics to boot. -- === Subject: Re: dissolving Russel's Paradox at 01:04 PM, zbi74583@boat.zero.ad.jp (Yasutoshi Kohmoto) : > I studied an alternative set theory which dissolves Russel's >Paradox. There are many such. In what way is it more interesting or more useful than, e.g., GBN, ZFC? -- === Subject: Re: Homeomorphism and boundaries Visiting Assistant Professor at the University of Montana. > at 06:15 PM, () : >>f(int(A)) = f( union{B : B is open, B contained in A}) >> = union{ f(B): B is open, B contained in A} (since f is >>bijective) >> = union{ f(B): f(B) is open, f(B) contained in f(A)} >> = int(f(A)) >No. Consider A=(0,2Pi) with the standard topoplgy, B=RxR with the >standard topoplgy. Define f: A -> B as f(x) =(0,x). Int(A) = A, but >Int(f(A)) is the null set. And is f a homeomorphism? That was part of the hypothesis above. Since the map f is not even surjective (and you will note the parenthetical comment stating since f is bijective), surely that is not the case. >>f(clos(A)) = f( intersection{B: B closed, B contains A}) >> = intersection {f(B): B closed, B contains A} >> = intersection {f(B): f(B) closed, f(B) contains f(A)} >> = clos(f(A)). >No. Consider A=(0,2Pi) with the standard topoplgy. Define f: A -> C as >f(x) = exp(i*x). cl(A)=A, but f(A) is not closed; it has the limit >point 1. Again, f is not a homeomorphism from A to C, so of course I make no claim that the argument applies for that case. So, again: the original question was, if f is a ->homeomorphism<- (a bijective continuous open mapping), is the image of the boundary equal to the boundary of the image? === Subject: Re: 0! = 1 > Is this making any sense? > to which > David W. Cantrell replied > Yes. But exponentiation isn't commutative. Perhaps one might say that > 1 is left-absorptive, and that exponentiation has no right-absorptive > element. > You sure its making sense? Or you can't tell? to which > A small addendum to my previous response: Don't every say you've never seen a live prediction come true. Herc the numerologist === Subject: Re: JSH: About e > not just some guy talking maliciously. > I megalomania, not malice. > Btw, the link you want is > http://www.geocities.com/jrstrader2000/Incomnt.htm > V. The link to *math* not malicious attacks is the following one: http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 Readers can also see my reply to Nora Baron to understand the issues here, as objectors are attacking math axioms, while I'm noting that a *definition* simply wasn't inclusive enough. The argument I've presented centers on the fact that setting m=0 gives you what's *independent* of m, but if you look at objectors, they keep trying to claim that there's actually an m dependency, which is a direct contradiction. So if you follow the math, you accept what's at the Hong Kong website link. But if you're willing to believe that with a polynomial P(m) and its factors that setting m=0 gives terms that are STILL dependent on m, then the arguments will continue. Meanwhile the error from the flawed definition in core mathematics, remains. === Subject: Need help with one equation, please Before you read thism, pelae visit my website for background reading on me, for else otherwise, you will not be able to understand this... My website is at website at http://www.evidence-eline.com I am working on one equation that reads as follows: G15=D, or 7154, it spells god, as follows: GOD = 715 4, = JOD = YOD = IODINE= YODO = JOD = JOR = LOR = LORD THUS GOD SPELLS LORD, AND LORD SPELLS XOR, thus god is an XOR equation j=l, thus xor = lor (lord) Why is 715 in me life?, GAE, LAE, YAE, XAE, XOR, but only one problem, 715 = 11-29 ( I COME UP WITH 11/29 MY GRAND MOM DATE OF BIRTH FROM MY 1966 HAVANA CUBA ADDRES, My grand mother Iluminada Josefa Cruz Gomez, Born November, 29, 1908, La Boca, Guayos, Sancti Spiritus, Cuba < dont try to do the 11/29 boys, it take some knowledge of HALES 715 = 64, 10, 410 = DIO = D=4 10 = IO = D10 = 410 517 = 15-35, 07-15-35-11-29 715 in octal is 1313 , or 1 in 13, maternal side from Trinidad, algorithms, but I have a zillion of them GOD, jor, yor, lor, xor, 715 = BINARY OF 715 10 11001011, 3-7-14-31-62-124-249-499-998-1997 ( THESE ARE HALES FACTORS, intrigued?, eat me!) 7991899994949242126134173 101100100111001100001101111101001101101000101011110111110101101 37 1 53 16 21 25 3714 3162 1242 4949 9998 1997 371431621242494999981997 CGAD CAFB ABDB DIDI IIIH AIIG === Subject: Re: Numeric one-way hash function > But if you don't believe my proofs, the easiest way > to check your algorithm is to implement it for a smaller > value of Max and then to check the number of duplicates. > I expect that you will be highly surprised. I have seen true random number generators produce the same value but it was no where near a third of the e. However I was not performing an in-depth soak test so I may not have produced sufficient numbers. The rate of occurrence of duplicates is not linear, p(duplicate) = n/10^12 means that there are few at the beginning but most of the towards the end. What the original poster probably needs is an encryption algorithm that is 1 to 1 over 10^12, difficult to predict over short ranges and hard to break. Andrew Swallow === Subject: Re: Numeric one-way hash function >What the original poster probably needs is an >encryption algorithm that is 1 to 1 over 10^12, >difficult to predict over short ranges and hard to >break. Unfortunately, we have no idea what the OP needs. I agree that this is probably what he *wants*. But he didn't state very clearly what he wanted, let alone even come close to telling us what the problem was. Greg. -- Greg Rose 232B EC8F 44C6 C853 D68F E107 E6BF CD2F 1081 A37C Crypto Mini-FAQ: http://www.schlafly.net/crypto/faq.txt Qualcomm Australia: http://www.qualcomm.com.au === Subject: Re: Boolean Algebra - Arithmetic Relationship Allow me to clarify my inquiry, 1)Is all of mathematics reducible to the simple logic presumption that if something is true it cannot be false?? 2)Is all of mathematics inter-linked?? (Geometry, Calculus, and Algebra reducible to Arithmetic etc.) Or are there islands of mathematical understanding off on there own? 3)Can all known Mathematical Notation and Symbolic manipulation be modeled by a turing machine?? Could a symbolic respresentation of a turing machine therefore serve as a master symbolic grammar?? Perhaps difficult to work with but, all symbolic notations could be reduced to it. I recognize the impossibility of storing a irrational number in a finite number of bits; However, our brains have concept of PI and are yet finite. We don't have to know all of numbers behind the decimal point to understand PI's implications. Again forgive my naivety if my questions seem trivial. -Steve === Subject: Re: Is this newsgroup useless? > One learns not to step on anything pretty quickly... Well, okay, but doesn't dragging abrasion become an issue after the first 200 meters of improvement? xanthian. -- === Subject: When is the limit of measurable functions measurable? The question might seem stupid, but I'm not talking about REAL-VALUED functions. Let (X,T_X) and (Y,T_Y) be topological spaces, and let B_X and B_Y denote the Borel fields on them. (= generated by the open sets) It is not reasonable that the pointwise limit of a sequence of measurable functions f_n : (X,B_X) --> (Y,B_Y) is also measurable. (By measurable I of course mean that the inverse image of a set in B_Y is in B_X.) This is because sequences don't mix well with topological spaces. But there ARE circumstances under which this is true. I remember a theorem proved by Calderon in one or another class he was teaching where he proved it was sufficient that T_Y have the property that every open set U can be written as a countable union of open sets G_n with the property that the closure of G_n is a subset of G_{n+1}. This property is satisfied, e.g., by regular T1 second-countable spaces. Does anybody know the most general circumstances under which this result is true? I doubt that second-countability is necessary; it's TOO global a condition. Probably first countability isn't enough. Alternatively, can anybody characterize the property of open sets U being writable as such countable unions? Always wondering, === Subject: Just what is an L-series? I know what a Dirichlet L-series is, and I've seen a few other kinds, such as Artin's in connection with representation of finite groups. But is there a standard comprehensive definition of the term L-series? Maybe, any Dirichlet series sum_{n=1}^infty a_n n^{-z} such that the sequence (a_n) is totally multiplicative? TIA, Larry === Subject: Calculate mode with multiple occurrences Help! I would like to know how to calculate mode when two different numbers appear the same amount of es (i.e., 1, 2, 2, 2, 4, 6, 7, 7, 7, 8, 9). Two and Seven appear the same amount of es in the array above - so please let me know which one is MODE or do I add them together and find the average of the two? Any help is appreciated. Frank === Subject: Re: divide and conquer > I am working on the following problem. I need a divide and conquer > algorithm to solve the following problem. Any help is appreciated. I and others have noticed your repeated attempts in comp.theory, sci.math, and perhaps elsewhere to have your homework done for you. That's your choice, of course, and often you will find someone who will respond as you wish. A small cautionary tale: you may well fool a teacher into giving a passing grade, a college into granting a degree, but when you are sitting in an interview with a hiring manager and people working in your technical field, your lack of thinking through problems for yourself will speak even when you do not. I've seen it happen: a sweet young thing who got all the impressionable young men to do her homework, to rehearse her for exams, to do her class projects for her. She got her degree with acceptable grades, but the only employment she could find in a market crying out for the skills of her degree was the same job she had before she went to college: she cleaned other people's houses for a living. Spending four years in college and arranging to learn nothing only makes you four years older and wastes that much e out of your finite life to no point. Her field of study by the way was the same as your own, and the demand for those skills has fallen off dramatically, so your strategy is worse today than it was in her e. xanthian. -- === Subject: Re: multiplication negs > along. One day I was talking to my young neice, helping her out with some > simple algebra. We were discussing mulitiplication and division of negative > numbers and my neice brought up a question and for some reason I have been > unable so far to reason my way through the answer: exactly why is it that > when one multiplies 2 negative numbers one ends up with a positive? It's Oh, this subject has been posted before... Well, 0 * (-2) can be 0 counts of the value (-2) and 1 * (-2) can be 1 count of the value (-2) . So (-1) * (-2) can be (-1) counts of the value (-2) . Now add [1 count of (-2)] to [(-1) counts of (-2)] and there are [0 counts of (-2)] . If 1 count of (-2) is known to be (-2) and if zero counts of (-2) is known to be zero then [(-1) counts of (-2)] must be 2 . --------------------------------------------------------- I also had a logic down that minus one count is one count of not counting while zero counts is either zero counts or zero counts of not counting...For (-1) counts to become 1 count of not counting then it has transformed to positive. Next to not count the value (-2) one e can only mean to offset the value one e and here the value has transformed to positive. So (-1) * (-2) = 1 * 2 . === Subject: Re: Antidiagonal, Infinity > Ross wants to use non-standard numbers to confirm his hypothesis > that there is a next real after any real, and that the irrationals > and rationals alternate on the real line or on some non-standard > real line. So he cares. On the other hand, his hypothesis is way out > in left field, where he has been stuck for months, if not years. Actually, I have not been claiming to be using the nonstandard > numbers. In light of the issues that we cover and mutually understand > to some extent, it might be the case that of the set of reals that for > each that it has a next element of opposite rationality that that > would be in neither the nonstandard nor classical model. Its elements > would still be the elements of the set of real or hyperreals. > As the classical model is the reals and a non-standard is the > hyperreals, you are claiming to simultaneously be inside of and > outside of some model simultaneously, which certainly puts you > outside of rationality. About equating density with measure in the unit interval, that means > equating density in the unit interval to measure in the unit > interval. > And how do you do equate density to measure? It would help if you > could explain what YOU mean by dense and what you mean by > measure, as the standard meanings do not allow of equating these > ideas. In this model, say alternating analysis, or a rationally alternating > model of the continuous reals, the alternating measure of the > rationals in [0,1) is 1/2, as is that of the irrationals, as is their > density in the unit interval and the density of the rationals in the > reals. > But every e you have a rational, x, and irrational, y, supposedly > next to each other, you have the problem of (x+y)/2 between them, > so they aren't really next to each other after all. This doesn't disagree with the measure of the reals in the unit > interval being equal to one. > Yes it does. For each of uncountably many irrationals, x, linearly > independent over the rationals, consider U(x) = (x+Q) / [0,1), the > intersecting ofx+Q and [0,1), where x+Q = {x+q: q in Q}. > Since each x+Q is essentially a translation of Q, each must have the > same measure, but, as they are pairwise disjoint, the measure of > [0,1) must be the sum of their separate measures. > Thus, according to your arithmetic, uncountably many measures of 1/2 > add up to 1. So your measure is self-contradictory. > The results of classical analysis would > hold true. Anything that didn't hold true would be suspect. > Your formulation must be suspect, then, since it doesn' hold true. What's a proper subset of the rationals or irrationals that is dense > in the reals, where its complement is infinite? > Let n be an arbirary integer grreater than 1, Z be the set of > integers, and Z[1/n] be the ring generated by appending 1/n to Z and > taking the closure under addition and multiplication. Each such ring > will be dense in the reals and have infinite compliment in the > rationals and in the reals, furthermore, if n is a proper factor of > m, then Z[1/m] is a proper subset of Z[1/n]. > Thus each of Z[1/2], Z[1/4], Z[1/8], z[1/16], ..., is a proper > subset of all its predecessors in this sequence, but each is also > dense in R. The rationals and irrationals are each dense in the reals, their > union, the reals are continuous. > Ross, what do YOU mean by continuous? > For the standard meaning of continuous, a function may be > continuous, but not a mere set. > What property does the ordered field of reals have that the ordered > field of rationals lacks that makes the set of reals continuous in > your sense but the set of rationals not continuous in your sense? I find a nice exposition on MathPages about the consideration of alternating rationals and irrationals, from searching for rationals irrationals alternating: http://www.mathpages.com/home/kmath172.htm This page discusses a function continuous at all irrationals, discontinuous at all irrationals: http://www.math.tamu.edu/~tom.vogel/gallery/node6.html#SECTION00024000000000 000000 discontinuous at the rationals is on there are some things to consider about that type of function. A description of that is f(x)=0 for x irrational and f(p/q)=1/q for relatively prime p and q. I believe that that has the same image as f(x)=0 for irrational x and f(x)=1 for rational x with the domain of the irrationals, that is to say, f(x)=0 for x irrational and f(x) is undefined for x rational is nowhere continuous, or a subset of the irrationals besides the subset of a single element would be complete. That is to say, for the set {x}, any sequence (x, x, ...) converges to x, an element of {x}, and thus any singleton set is complete. A Harvey Mudd Fun Fact: Rationals Dense but Sparse. http://www.math.hmc.edu/funfacts/ffiles/30004.3.shtml to an irrational. Obviously enough, I try to think of a sequence of irrationals that converges to a rational. How about a rational sequence that converges to one. Have the irrational sequence be the value of any irrational minus the rational sequence element es the irrational. For example, sum 1/2^x converges to one, have the sequence of irrationals that converge to a rational, zero, being pi - (sum 1/2^x) pi. Zero is rational. That's only saying that neither the rationals nor irrationals are complete, where the reals are complete. I search for rationals dense sparse, interleave rationals irrationals, between any two rationals, denseness rationals, and other phrases to do with the quantified density of the rationals. of rationals and irrationals and their alternation in the reals. For example, Do rational and irrational numbers alternate?: http://mathforum.org/library/drmath/view/51573.html Here is a reference to a function f(x)=0 for rational x and f(x)=1 for irrational x: http://www.math.rutgers.edu/courses/436/436-s00/Papers2000/stasiak.html . Here's some discussion: http://www.nrich.maths.org.uk/askedNRICH/edited/445.html . A post posits the existence of sets of elements of the unit interval of reals of measure 0.5, presumably besides x<0.5. I read that the reals are continuous as they are complete, any convergent sequence of reals converges to a real. Draw the shortest line from zero (0, 0, ...) to one (1, 0, ...) without lifting the pencil, each point on the line is a real, an element of R[0,1], removing an element leaves a set insufficient to represent any point, adding an element leavse a set that can have an element removed and be sufficient. The constant function f(x)=C is continuous with the domain of the reals, it contains no discontinuities, it's not discontinuous. Its range is also the union of the range of the constant function for the domains of the rationals and irrationals. About subsets of the rationals, that was what I had in mind, p/2^q for integers p and q, but I am still trying to determine why it wouldn't contain p/q except the prime factorization of 2^q is always 2, 2, 2, .... Obviously enough for fixed x there are only 2^x many integers y thus that 0 <= y / 2^x <= 1. The set of elements of the reals contains the same elements as the set of hyperreals. Drawing the line from zero to one goes through there all the reals and all the hyperreals. About translating each irrational by each rational and considering their intersection with the unit interval, my opinion is that it is just the set of irrationals of the unit interval. The sum of a rational and an irrational is always irrational. So I say that the measure of each in an interval is the same in this alternation model, and where the sum of their measure is one, the measure of each is one half, the measure of the irrationals and rationals in the unit interval are each presupposed to be equal in the model. So I don't think that shows contradiction to m([0,1))=1. Here I should note some confusion about the measure of [0,1) vis-a-vis the measure of [0,1]. About (a+b)/2 =/= a =/= b, my point is that any a and b that you have selected (unique numbers) are separated by infinitely many other numbers. Their average is as well unique and infinitely distant from either between them in the consecutive sequence of the continuous reals. In a way it's about reconciling point-ness and line-ness. A point is the intersection of two (and infinitely) many lines that intersect, a line connects two (and infinitely many) points that are collinear. It takes two points to describe a line, it takes two non-parallel (coincident) lines to describe a point. That's obviously extravagant. Then there are three non-collinear points for a plane, and three coincident planes for a point, etcetera. The lines are not parallel. They're correlated. Consider the set of all Euclidean points (a_0 e1, a_1 e2, ...). Consider the set of all lines. The points on the chords through the center of the n-sphere centered at (0, 0, 0, ...) are the points of RxRxRx..., each point of that product besides the origin is on only one of those lines. The nonstandard model contains definitions of points that comprise a minimal two-point line segment. The line described by those two points of the reals goes through all the other points of the reals. There are only points to describe the line, here the elements of a set that contains any and all points on a line contains as elements only points. The rationals and irrationals are each dense in the reals, they are disjoint and their union is the reals. Then again, so are the algebraics and transcendentals, or for example the rationals p/q where p and q are relatively prime and q is even and that set's complement in the reals. Also a constant function is everywhere discontinuous with any of those sets as its domain. This causes me angst. I derive humor from using the word angst. The model can't have density 1/2 for each of the rationals and irrationals where the same reasoning applies to the algebraics and transcendentals because the (set of) algebraics is a proper superset of the rationals with infinitely many proper subsets that are supersets of the rationals. That reminds me of about the algebraics and transcendentals, with the rationals, algebraic irrationals, and transcendentals. This throws a large cognitive wrench into the works of determining some positive, finite measure of the rationals in the reals. I can divide the irrationals into the transcendentals and algebraic irrationals. That means I still want to know that given an element, that for the next element, what it is, or the probabilities of what it could be. Ross === Subject: Re: Minimal Graph, Four Color Theorem I have decided not to respond to your insults in kind. I am sorry that you felt that you were justified in taking my opinions personally and forming such outlandish and erroneous judgements of my character and motives. I regret that you expected me to accept so much inaccurate and/or irrelevant data. !!! Bill J. === Subject: Re: topological terminology >> So... homotope or homotop, which is the verb? >Use it in a sentence? > The space X homotoped into Y? This avoids the question though. I've seen both homotope and homotop and the past tense of both would seem to be homotoped. I think homotope is more common, but that's also the one I use, so I'm biased.