mm-1889 === I have an summation expression: m.t+ 2.m.t+....+ 2^k .m.t where, k = ceil ( log ( (D/e) / t) ) I have an answer to that summation as 4.2^(k-1).t Any idea how? JS Seems unlikely. What happened to m? -- 1 + 2 + ...+ 2^k = 2^k - 1, so that answer says that (2^k - 1).m.t = 4.2^(k-1).t from which we deduce that m = 4.2^(k-1)/(2^k - 1), or t = 0. On the whole, I think I prefer t = 0. The players of a football team are to be given awards by their club directors. Including the reserves, the team of 15 players stand in a line along the pitch with one meter between them each. The manager will begin with any player and then continue presenting the awards to the others. If this procedure is to be completed using the shortest route, then beginning with player number 1 and continuing in line until player number 15 is the most appropriate. (In this case 14 meters would be covered). What will the total distance be if the same procedure is to be completed using the longest possible route? Hint: it's enough to ensure that the manager starts with player #8, alternates between the first 8 players and the last 7 players, and ends with player #7. It's fun to do this as a Traveling Salesman Problem, though. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada You and your friend are each given a piece of cord that are of different integer lengths. Both of your task is to cut the cord into any number of pieces of any length, in such a way that when you multiply the length of your pieces, the result shall be the maximum. After doing a few calculations, both of you cut the cords successfully satisfying the condition. The length of your cord is 6 times the length of the cord that your friend has. However, the number of pieces that you obtain is 5 times the number that your friend has. Find the length of the cords that have been given to you and to your friend. Notes: *The number of the pieces is greater than or equal to 2. *The lengths of the pieces need not to be integers. *Example: If you have a 5 m. cord, you can divide it into two pieces of 2 and 3 meters, and when you multiply the length of these pieces you will get 6 as the result. If you divide it into 3 pieces of 1.5, 1.5 and 2 meters, then you will get 4.5 as the result. But, both of these results are not the maximum. (Enter the length of your cord. Unit is in meters.) Hint: show that you should always cut a cord into pieces of equal lengths. If you want more hints, say who you are and where these problems come from. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada X-Proxy-User: $$rnw$j2sdg_ On Mon, 04 Apr 2005 21:59:21 EDT, Sazan Suzan The most significant digit has nine choices, but all the rest are constrained by the preceding digit, with only two choices for each. --- Stan Liou ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com boundary=----=_NextPart_000_0010_01C53955.58B9C460 --------------------------------------------------------------------- Stan Liou No, only one digit is possible following any 1 or any 9. I got the answer 1370 by hand as follows. Here each number below the first line is the sum of those above it and one to the left or one to the right. The m-th column will contain the number of qualifying numbers which begin with digit m. 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 1 2 3 4 4 4 4 4 3 2 3 6 7 8 8 8 7 6 3 6 10 14 15 16 15 14 10 6 10 20 25 30 30 30 25 20 10 20 35 50 55 60 55 50 35 20 35 70 90 110 110 110 90 70 35 70 125 180 200 220 200 180 125 70 The sum of the n-th row is the number of qualifying numbers having n digits, and 1370 is the sum of the n-th row. LH boundary=----=_NextPart_000_001C_01C53956.76D26DC0 --------------------------------------------------------------------- Stan Liou No, only one digit is possible following any 1 or any 9. I got the answer 1370 by hand as follows. Here each number below the first line is the sum of those above it and one to the left or one to the right. The m-th column will contain the number of qualifying numbers which begin with digit m. 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 1 2 3 4 4 4 4 4 3 2 3 6 7 8 8 8 7 6 3 6 10 14 15 16 15 14 10 6 10 20 25 30 30 30 25 20 10 20 35 50 55 60 55 50 35 20 35 70 90 110 110 110 90 70 35 70 125 180 200 220 200 180 125 70 The sum of the n-th row is the number of qualifying numbers having n digits, and 1370 is the sum of the n-th row. LH Should say ... having n digits, and 1370 is the sum of the 9-th row. Copy-n-paste strikes again :) LH In particular, look up Liouville's theorem and Liouville numbers. --Bill Dubuque X-Proxy-User: $$rnw$j2sdg_ On Mon, 04 Apr 2005 21:54:35 EDT, Sazan Suzan Obviously, the answer would depend on whether the spies have finished their homework yet. Sigh. --- Stan Liou ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com Distribution: World Pushing the envelope (stretching the convex hull) with mathematical / redneck crossover comedy. humor humor humor humor You Might be a Mathematician if ... * hypergeometric summations are the most fun you can have with your clothes on; * at the age of 19 your most productive years are behind you; * your major result will be named for someone else; * your fame lies in posing the question you can't answer; * you make mistakes ... but they are really interesting mistakes; * you wonder how Euler pronounced Euclid; * you understand all the mathematics Gauss produced ... through age 13; * a copy of Russel's letter to Frege adorns your wall; * your major was Mathematics, minor Caffeine; * you know all of the Greek alphabet, but not a word of Greek; * You can recite Pretty Poly Nomial and Curly Pi from memory; * your correspondence has footnotes and bibliography; * irresistible little combinatorics puzzles keep appearing like a nervous tic; * unemployment is a welcome opportunity to make progress on your life's work. * your progeny are relieved to learn that Mathematics is not a heritable genetic trait; * the solution to every problem involves counting balls into boxes; * you cannot refrain from blurting out counterexamples when someone claims an impossibility; * you can fold planar strips into regular polyhedra ... entirely in your head; * doing something more than once is boring; * you suffer dental and gum disease because brushing teeth is boring; * because of your dental problems, mention of the word calculus raises mixed emotions; * you celebrate Rota's birthday decadently with donuts and champagne in paper cups; * you celebrate Erd.9as's birthday furtively with Benzedrine chased by a double espresso; * it is difficult to plan for retirement given the current state of the continuum hypothesis; * you remember postal addresses by means of number theory: The smallest integer which is the sum of two cubes in two different ways; * you have calculated how many ways (ignoring reflections) there are of lacing your sneaker; * you count on your fingers in binary; * your romantic relationship is strained when you show too much interest in your beloved's mathematician acquaintances; * you have already obtained your next three years of reading material; * you know a six-letter word with three vowels, all of which are y; * you learned French so you could read Bourbaki; * you bring Bourbaki's Varietes Differentielles Analytiques Fascicule de Resultats on vacation; * you don't bother taking vacations when you can read Bourbaki at home; * you visit Earth primarily for lectures and family obligations; * your opinion of A Beautiful Mind is been there; done that. humor humor humor humor posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY Ok, those of you who know something about factoring know that surrogate factoring theorem is just not like anything that's part of the mainstream literature. It uses a difference of squares, but where you calculate the difference of squares in some way that involves factoring a number you get to pick, and its square subtracted from your original target number. So you don't go searching. No sieving. The only thing you guess at is the number I call j, and the rest is handed to you. Now I'm a guy who has lots of mathematical discoveries under my belt, and there are mathematicians who have behaved oddly with basically all of them. So I have prime numbers research and people try to claim it's not new when you can look at the equations and just compare. I have algebraic number theory research where one small electronic journal published it and then yanked it after pressure from Usenetters sending emails, and then the entire journal shutdown a few months later. Who I am at this point is not in doubt, as I am a legitimate researcher who is facing a very hard situation as each of my results spoils someone's pie, you might say. Sure, none of you care about me, or the pressure I'm under. Can you imagine being in my situation? I'm sure 99.9% of you would fall to pieces, to have incredible results and face such bizarre opposition. Maybe I'm getting a little nutty myself, but that's neither here nor there. Eventually the truth comes out, and as I posted earlier today, no matter what you may think it's not the end of the world. I figure that I'll get some breakthrough within a few years at the most, probably sooner with the factoring theorem now found, and that means that the world figures out the story, and you don't just disappear. You don't get to just fade into the ground or run and go hide. People will track you down. Reporters, lawyers, psychologists, and who knows who else as these people will want to know about the collective social breakdown that allowed this to happen. They will want to drag mathematicians into rooms to study you, figure you out, and find ways to convince themselves that they are not you that they could never do what you're doing now that they could never betray the world on this scale that they could never lie on this scale. The entire world will work as hard as you're working now to convince itself that you are an aberration. What does that put you? Well, don't think you'll be just waltzing to the supermarket, or laughing over it all with friends talking about this weird story. People will need to prove to themselves that they have handled you. They will need to prove to themselves that you are not something that will happen again if it can be helped. They will need to prove to themselves that they are not you. How society decides to handle you is anybody's guess, but be sure that they will try, and that does not involve trips to the beach, checking up on a 401k, or figuring out way to put your kids through college. It may involve trying to find a job, dealing with ridicule, and explaining yourself over and over and over again to people who will ask you, why, probably for the rest of your life. James Harris X-RFC2646: Original True. It does not fit in with the mainstream literature. It is a theorem that has no proof.... only Fermat got away with that, and his conjecture was eventually elevated to theorem status by wiles. But your theorem is even less of a theorem, because it does not actually state anything that can be proved. No searching... just guessing instead... wow... what an innovation! Imagine being in you situation? You mean imagine being a world-class mindless jabbering asshole? That might be a bit of a challenge, but I would rather not try. So you pick a number, how is that different from guessing a number? How can guessing a number be more effecient, than a random pick? OK, j now what? Off Topic - Off Topic - Off Topic - Off Topic - Off Topic - Off Topic - You havent shown anything, just pick a j. Off Topic - Off Topic - Off Topic - Off Topic - Off Topic - Paranoid Off Topic - Off Topic - Off Topic - !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ Your results are not credible, right. Rigid counterproofs and counterexamples tend to have that effect. Oh, it has been here alright for quite some years. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum On 4 Apr 2005 19:33:00 -0700, jstevh@msn.com fed this fish to the penguins: It must be hard being you. But not for the reasons you state. Maybe you should consider that hypothesis seriously. Maybe you are nutty. Or, something milder maybe, as in... I don't know, a crank? Tell me the truth. You've been in the mothership, haven't you? They (you *know* who they are) have tested you, haven't they? Inserted those horrible rectal probes. Master, master, are you the Blimp? G. Rodrigues posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY results Yeah, a crank with one paper accepted then rejected, by a small electronic math journal which then did a hari kari. A crank with a prime counting formula STILL not part of the mainstream of mathematics using a constrained partial difference equation unlike any other prime counting method known. Yeah right, a crank with a solid proof of a theorem that links factors in a step beyond difference of squares. What's clear here is that I'm not a crank, but people like you have learned to yell crank when the truth is something you don't like. You're an intellectual charlatan. You can't match me intellectually, so you throw names. You are weak-minded. You don't have the mental capacity to do anything more constructive. James Harris Your prime counting algorithm (not formula) contains NO partial differentials. The only working version contains finite differences -- that's it! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com X-RFC2646: Original You must be so proud. After a decade of effort, you get one paper accepted, by a small electronic math journal, by oversight. Then it is rejected when the error was brought to their attention. A journal so pathetic that it imploded, perhaps due to the error they made in publishing your nonsense. Nothing really. Less than a garbage. When will you be defining that theorem clearly, and when will you provide a proof that makes sense? On 5 Apr 2005 16:09:15 -0700, jstevh@msn.com fed this fish to the penguins: What is more impressive about you is that you aren't the *least* sensitive to the irony in your *own* rhetoric. It's really amazing. Hint: read again your sentence above. Just one correction: you advanced the hypothesis that you were going nutty. I suggested that you consider the milder hypothesis that you are a crank. Oh, just one question out of curiosity: you are adept at throwing names, so what does that make of you? Maybe so, I don't really care. But I laugh joyously at your extraordinary displays of involuntary humour which, methinks, is ample compensation for my weak mindedness. I suppose your powerful mind has better things to do than to laugh. Like getting papers accepted then rejected in small electronic math journals that then die with a whimper. G. Rodrigues Why not just guess at a factor of the original number, and eliminate all the algebra in between? Can you name one example of a mathematician who has not behaved oddly, as you're using the term? Didn't think so. Hmm. The only person who believes that is you. Of course it's possible you're right and everyone else is wrong, but saying it's not in doubt when in fact it's doubted by all but one person seems curious. Possibly. Or maybe it's _not_ neither here nor there. Think about it. Suppose hypothetically that someone was just deluded somehow, thinking he was doing earthshattering research when in fact he was accomplishing nothing at all. The way that person would behave could be very much like the way you behave, especially if that person was getting a little nutty. Hint, for the hundredth time: When you talk this way you don't sound like you're maybe getting a little nutty, you sound like a raving lunatic. ************************ David C. Ullrich : It uses a difference of squares, but where you calculate the difference : of squares in some way that involves factoring a number you get to : pick, and its square subtracted from your original target number. Is this supposed to be interesting? It's just algebra, and it's only useful if it gives some good results. : So you don't go searching. No sieving. The only thing you guess at is : the number I call j, and the rest is handed to you. Great...reduced to guessing. This is a faaaaantastic algorithm. As Nora has pointed out, there are other far simpler algorithms which have a better than 50% success rate, which is a claim you made but didn't prove. Or wait, you had solved the factoring problem. Whatcha got? : Who I am at this point is not in doubt, as I am a legitimate researcher : who is facing a very hard situation as each of my results spoils : someone's pie, you might say. What pie of mine is spoiled? Please address this. I am a non-tenured university lecturer who would only have to GAIN from supporting some big discoveries in mathematics. If the news broke that said after years of hard work, Harris' work broke into society with the help of Justin, since nobody else saw the truth in his work. That'd look pretty darned sweet for me. But it ain't so. : I'm sure 99.9% of you would fall to pieces, to have incredible results : and face such bizarre opposition. You have no worthwhile results - this is the problem. Justin posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn difference at is Nora prove. Whoa here, partner. What I said was, there are algorithms with a 100% success rate. They however require enormous numbers of tries and enormous amounts of calculation and time (unless M is small). Harris claims a 50% success rate with essentially one try. He is completely wrong about that - for large M, the success rate of his algorithm is probably submicroscopic. If anybody ever found an algorithm with a 50% success rate for one try, RSA would be instant toast. Nora B. researcher non-tenured big of since sweet results posting-account=82j3EgwAAACYxp9hHOcWy78r2IGkmc3t I have a suggestion. Rather than starting thread after thread talking about how SF works, must be a theorem etc. etc. why not actually try working through a couple of examples. - William Hughes posting-account=v6xdSg0AAAD63tq-yRAAhvNtO5tjUP7G difference Essentially all fast factoring algorithms use DoS to factor. You're not the first to propose this. is Are guessing j and searching for a suitable j not equivalent statements? You're making the assumption that guessing is some magical constant time function that will always result in desirable outputs. You have to actually show how guessing works in a deterministic way and importantly in P-time [as per your claims that you solved the factoring problem]. Also if your advantage of guessing j is negligible [e.g. knowledge of the outcome of one guess doesn't help the next] then your factoring algorithm is non-deterministic and likely EXP-time. In which case it's reducible to trial division, a solution I might add works for all composites. Not just the rational ones, *giggle*. Tom Yes. Somewhere in between it works all the time and Sometimes it doesn't work, there was a period when it was it works exactly half the time, because of Quadratic Residues. Greg. -- Greg Rose 232B EC8F 44C6 C853 D68F E107 E6BF CD2F 1081 A37C Qualcomm Australia: http://www.qualcomm.com.au posting-account=v6xdSg0AAAD63tq-yRAAhvNtO5tjUP7G Quadratic Residues ... yet another way you mainstream folk keep a free-range brother down... Tom :-) posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY Oh, hey, notice that actually you get b_2 f_1 and not just b_2, so in fact, it's f_1 that could be the feaction and b_2 could be an integer. The theorem doesn't separate them out for you. So why do that? Well, there are posters who spend a lot of time arguing with me, who basically lie. Now I know they lie, but they claim they don't and other people claim they don't, which seems to work when it comes to convincing people. Now if they were serious and knew much mathematics, it'd be clear that you get b_2 f_1 and in fact, you can simply separate out integer factors in common with M^2, so you don't actually know whether or not b_2 is a fraction or not, as the mathematics doesn't say. Also, if you go over the proof you'll realize that you can change indices to get b_1 f_2 = (-(Az - 2M^2) -/+ sqrt((Az - 2M^2)^2 - 4TM^2))/2 where I shift to -/+ so it makes sense, but basically, the two solutions for b_2 f_1 in the original are the two factors of M^2 T. James Harris !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ So you have some theorem that gives you not the integer you are hunting for, but rather the product of the interesting integer and an arbitrary fraction. Brilliant. Reminds me of that joke Are you sure that you lost your key here under the lantern? -- No, way over there, but it's dark and dirty there. Do you know the probability that a some special integer turns out to be representable as the product of a different integer and a fraction? Some theorem you have there. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY by in integer. Well, there you go again. The arbitrary fraction would be a factor of T, where T = M^2 - j^2. The point though is that posters in replying never bothered to see that simple reality, and went on and on about rationals, in posts that didn't make sense by the way, without realizing that b_2 could be an integer. The theorem doesn't say either way. That's mathematics. Their ability to get so much mileage out of disagreeing with me despite not realizing basic facts, now that's social. My point? People can disagree with mathematics that is absolute and simple, getting away with it using basic tactics, as even if something is not true, in the social world, all you have to do is convince enough people it's true, anyway. Succinctly, mathematics is not immune from social bull. James Harris If M = 15 and j = 16, T = -31. Would you mind posting the factors of T so I can continue testing your 'Theorem'? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com X-RFC2646: Original Mathematics, as with any human activity, is not immune from social bull. In fact it is true of you more than most (all?) others. I do not recall anyone disputes that with you. Most people do not have such a big problem with it. Why do you repeat this point in so many of your threads. It is a dead issue. Get over it. X-RFC2646: Original [Nora Baron, beating her head against a surrogate wall] ... Nope, that's the probability that it's divisible by 31 or 37 -- or both. Not that it matters much, and especially not when M=pq is large. This is most obvious from drawing a Venn diagram. 1/(31*37) corresponds to the intersection, and you don't want to count that at all, but counted it twice when adding 1/31 to 1/37. 1/31 + 1/37 - 2/(31*37) is the right answer. More pedantically, consider n mod (31*37), which will be uniformly distributed if n is chosen at random uniformly. 31*37/31 = 37 of those are divisible by 31, _including_ 0; and similarly 31 of those are divisible by 37, including 0; and 0 is the only one divisible by 31 and 37. Since we don't want 0 at all, there are 31+37-2 winners out of the 31*37 possible residue classes. residues mod 6 are: 0: loser divisible by 2 and 3 1: loser divisible by neither 2: winner divisible by 2 but not 3 3: winner divisible by 3 but not 2 4: winner divisible by 2 but not 3 5: loser divisible by neither and 3 winners = 6/3 + 6/2 - 2, and the probability of winning is 1/2 + 1/3 - 2/6 = 1/2. Of course this was all explained in equally tedious detail several times in posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn both. You're right of course. Should be 5.76%, i.e., even farther away from Harris's 50%. corresponds to counted it those are divisible by we possible possible + 1/3 - What!!! You mean, Harris is right? Oh. Just when M = 6. A little short of RSA range. times in Was he saying 50% even back then? I must have overlooked it. Nora B. posting-account=vRF1ow0AAADEaiuWV4RovZTrGLSdF9Mt My graduate school decision is narrowed down between Michigan and UW-Madison. I've visited both places, have enjoyed both schools, and have not yet decided; the deadline for that is 15 April. 1. What are the relative topical strengths of Madison over Michigan, and vice versa? For example, which department is better in algebra, analysis, et cetera? 2. Which department has a better environment for graduate students? 3. Which city is better to live in: Ann Arbor or Madison? 4. What overall factors make one place better than the other? Please try to present what you know in favor of both sides. 5. Does the slight prestige differential (Mich is generally ranked 7-9 among math depts.; UW, 10-14) mean anything, or matter, at all? I felt, when I went to these departments, that UW had a slightly better environment, but both were good. Also, the city itself (Madison) is one that in any case will be painful to pass up, though Ann Arbor might be just as nice. The Michigan offer was higher (with no teaching first year) but I felt like the Wisconsin department wanted me to go there more, and ultimately, subjective senses of mutual appreciation matter much more than a couple thousand bucks. I also sense slightly higher happiness levels in Madison, both among the people, the students, and specifically the math grad students. People at Michigan seem reasonably happy too, but in a reserved/ironic East Coast sorta way. My brain (and my father) is saying Michigan, my gut is saying Wisconsin. I realize I'll probably learn great mathematics and have a good time at either place. In any case, this feels like a choice between great and excellent, and I'm just trying to optimize. Any available information would be much appreciated. X-RFC2646: Original Living in Columbus, OH, there is little doubt where I would recommend. Michigan is pure evil :-) posting-account=Kb0T_QwAAACc9B9LpxfLjH0hHHYjPxft David Ullrich and WWWade may tell you more about UW-M. well, both schools have a good reputation and in the end i guess what matters is who you work with, how smart you are and how good a work you can produce. X-RFC2646: Original There is a proof in my text that Im not quite understanding, and I would appreciate any help on this : bases for V. They are said to be consistenly oriented if the transition are the alternating n-tensors. The set of orientation for V. linear map sending E_j to F_j. This means that B is the transition matrix from E_i to F_j. Then, A(F_1, ..., F_n) = A(BE_1, ..., BE_n) = det(B) A(E_1, ..., E_n). It follows that (F_j) is consistently oriented with ..., E_n) have the same sign. I understand this entire proof, but I don't understand why the last line implies that the set of ordered bases (E_1, ..., E_n) such that A(E_1, I know that being consistently oriented is an equivalence relation, and there are exactly two equivalence classes. The reason for that is the if (F_i) is consistently oriented with (E_i), then they are in the same equivalence class. If they are not consistently oriented, then (F_i) is consistently oriented with (-E_i), so there really are only two equivalence classes. Any help is highly appreciated, Tony What does your text say about the relationship between /^n and determinants? Beware: the transition matrix between (E_i) and (-E_i) has determinant (-1)^n, so when n is even, these bases are consistently ordered. (Draw a picture for n=2: (-E_1,-E_2) is obtained from (E_1,E_2) through a half-turn around the origin.) Way out: use the basis (-E_1,E_2,...,E_n) instead. LD Suppose that A is a N-by-N real matrix. If every element of the matrix must satisfy the following relation A(j,l) = p*(A(j+1,l) + A(j-1,l) + A(j,l+1) + A(i,l-1)) + B (where p and B are constants) how can I find an A that satisfies this? I am in the process of doing a numerical simulation and have encountered a similar situation. I need form the matrix A. It's kind of important to know what happens when j or l equals 1 or N. You might want to think especially about the case B=0, p=1/4 : you're trying to say you've got a number at each lattice point, and that every point's value is the average of that of its neighbors. That's the sort of thing that happens when you watch heat flow across a flat plate, for example. Well, what are you doing to get the heat there in the first place? Are you, for example, holding the temperatures steady at various points along the boundary (i.e. fixing the values of A(j,l) for j,l = 1 or N) ? By the way, you're really just asking for trouble calling a variable 'l' when you don't need to, especially with '1's floating around. In the font before me these are nearly indistinguishable. dave PS -- depending on what you say about the boundary values, you may be able to dispense with the B by replacing each A(j,l) with C(j,l) = A(j,l) + B/(4p-1) . is it not A(i,,j)=B/(1-4p) for all i,j I don't believe it is. What was given represents N*N equations in N*N unknowns. What I was to form a vector out of the components of the matrix and then put in the coefficients into a N*N by N*N matrix at set it equal to a matrix of B. Most of the entries were 0 for the coefficient matrix, but I was still able to solve for all of the A(i,j)'s by multiplying the inverse of it with the matrix of B and then row reducing. It's not efficient, but I don't any other ways. It works though. Adam. posting-account=83VcHQ0AAAAIyORN0HDqp6AI6Bcif55r In setting up an exercise for a course I'm teaching, I've come across something I can't explain, and it involves the simplest probability situation there is: flipping a coin. The exercise is designed to show that any set of four coin tosses is equally likely, whether it is HTHT, HHHH, HTHH, or anything else. I set up the simulation in MS Excel in two ways and only one works. Here's the one that works as I expect: 0) Choose a pattern to match, consisting of four heads and/or tails (e.g., HTHT). 1) Flip a coin four times, writing down the results in order. This gives you a set of four flips. 2) If it matches the pattern you're seeking, note how many sets of four flips it took you to match the pattern, otherwise, go back to (1) and flip again. Doing it this way, the results are essentially identical no matter what pattern you choose to try and match. Here's the part that's perplexing. It seems like it should work equally well: 0) Choose a pattern to match, consisting of four heads and/or tails (e.g., HTHT). 1) Flip a coin, keeping track of all the heads and tails you get. When the last four flips match your chosen pattern, write down how many flips were required to match the pattern. When I do it this way, instead of in discrete sets of four, then it takes on average much longer to match a pattern such as TTTT or HHHH than it does to match HTHT. Here are the results of a large number of Excel trials of this second method. Each value is the number of flips required to achieve the beginning of the correct four-flip sequence at the top of the column. [needs a fixed-width font to view correctly] TTTT THTH HHHH HTHT Num T Num H 77 5 59 4 108 104 8 3 39 29 103 109 6 25 47 24 103 109 3 23 53 10 110 102 9 15 35 14 99 113 61 22 5 23 104 108 15 8 11 30 109 103 5 9 39 21 101 111 3 67 22 37 108 104 34 18 6 19 105 107 12 54 59 53 102 110 53 10 6 9 105 107 97 4 24 3 97 115 47 5 105 4 107 105 78 18 5 56 97 115 13 21 17 20 108 104 7 16 75 4 105 107 6 25 10 26 107 105 24 28 15 43 110 102 41 3 9 4 112 100 15 4 103 3 114 98 25 10 5 9 117 95 70 22 5 12 107 105 27 3 101 14 124 88 13 19 42 20 104 108 24 20 14 21 113 99 3 28 21 9 112 100 80 30 3 6 104 108 57 6 9 5 108 104 26 39 5 23 89 123 51 14 32 13 108 104 43 5 55 4 103 109 21 6 13 7 112 100 5 8 19 23 106 106 16 3 55 4 119 93 13 3 33 4 109 103 4 11 23 10 105 107 26 9 59 17 114 98 7 10 13 57 110 102 5 8 27 18 104 108 4 42 10 33 110 102 5 30 49 31 112 100 4 13 78 14 117 95 5 19 10 15 110 102 64 4 20 5 104 108 34 22 17 39 98 114 104 18 31 3 93 119 34 6 14 17 96 116 28 19 7 18 107 105 17 6 140 14 126 86 57 3 70 4 109 103 8 47 29 16 108 104 9 12 66 3 108 104 32 23 19 22 97 115 6 112 118 20 114 98 67 14 6 15 99 113 15 10 93 20 91 121 3 16 52 17 111 101 23 5 50 6 99 113 61 18 31 17 115 97 90 6 36 10 123 89 41 9 45 14 109 103 28 40 43 13 112 100 54 16 65 26 112 100 28 3 9 15 104 108 23 66 27 65 107 105 27 11 16 12 113 99 3 11 102 23 109 103 32 3 45 4 107 105 4 8 20 9 111 101 4 20 37 10 105 107 4 50 44 19 111 101 14 23 53 24 126 86 Column Means: 27.6 18.6 37.1 17.8 107.6 104.4 Why are the first and third columns so different from the second and fourth? Dave Hirsch posting-account=GYng8QwAAAASJCdK-kCiIRtmF4RV4yrR There's a very clever bar-bet/hustle based on this, at: http://www.cs.queensu.ca/home/dawes/coins.html Two players each pick a pattern of three coin tosses. The mark picks first, say HTT; then the hustler picks, say HHT. They then flip coins until one of the two players' sequences comes up; that person wins. Contrary to intuition, it turns out that, given the first player's pattern, the hustler can always choice a pattern which will come up first more than 50% of the time! Because the expected number of flips for the first and third column is 30 and the expected number of flips for the second and fourth column is 20 (if I've made no errors in my calculation :) ). That the expectations are different is to be expected. My 'feel'-explanation: Compare the 'goals' TTTT and THTH. Suppose you have thrown the 'starting' T. In the case of TTTT an H will bring you back to the situation where you need to throw the 'initial' T again (in other words, you have to start over completely). In the case of THTH a T will bring you back to the same situation (the new T is now the first of the wanted THTH result). In this second case you will not have to start over completely (You will have already thrown the first of the wanted THTH goal). Your expectation is thus that the THTH will be reach sooner. posting-account=dGTVTg0AAACKi_LD2WBZc_VD8mnsu3gw The intuitive reason is simple: say you are waiting for HHHH to appear. After some tosses you get and H... good... another H... great... yet another H... wow that's really close... and now a T. Och. You have to start, so to speak, from scratch: you're back to square one and need to wait for a fresh set of H's. On the other hand, say your'e looking for HHHT. First H: good. Second: great. Third: you're happy. If a T arrives now, you're done, but even if not, you are *not* back to square one - in fact, the minute a T appears now you're done. In other words: * Starting with HHHT, the shortest sequence that ends with HHHH is HHHTHHHH. * Starting with HHHH, the shortest sequence that ends with HHHT is HHHHT. There is a simple formula to calculate the expected waiting time for a sequence to appear; roughly, it involves find all the shifts of the string that perfectly overlap with itsefl, interpreting the result as a binary integer, and multiplying the results by 2. For example the waiting time for HHHH is 30, while for HHHT it's 16. Ask me if you need more details... Alon Yes, that is exactly what you should expect. See, for example, Ross SM, Introduction to Probability Models. (Look up patterns in the index.) Also, see the sci.math thread with the subject heading, Monkey typing -- Stephen J. Herschkorn sjherschko@netscape.net Consider only sequences terminating after exactly 5 flips. If you are terminating on TTTT or HHHH, there is only one 5-string of each, HTTTT or THHHH respectively, but for terminating on HTHT, there are two, HHTHT and THTHT. Similar differences hold for terminating on any but exactly 4 flips. This sort of analysis, extended, is the reason behind the differences you found experimentally. posting-account=LdDaSg0AAAAEITwonQfgUyw4NgSHPzU6 Having given two different methods for fitting an ellipse that seek to circularise the data by minimising the variance of the squared radius, I should be clear that they're *not* precisely equivalent. Although they will both give the same correct answer for perfect data (i.e. data where all the points lie precisely on an ellipse), for real-world data there's no guarantee that they'll give identical results. The method I described in the thread Ellipse fit from moments is free only to rotate the data and then re-scale it along the x-axis, or equivalently to re-scale it along a single, arbitrary direction. Local minima will occur when the direction of re-scaling coincides with either the major or minor axis of the ellipse, but obviously the global minimum will occur when the major axis is shrunk to be equal to the minor axis, rather than the minor axis stretched to be equal to the major axis. The final quantity that is minimised in this approach is the variance of the squared radii of the circularised data when one eigenvalue of the metric used to circularise the data is constrained to be equal to 1. In contrast, the method I described in the previous post of the current thread constrains the metric by requiring gxx^2+gxy^2+gyy^2=1. That's not an identical constraint, and it need not give identical results when the minimum is non-zero. It is possible to solve for the 3 parameters that define the metric while imposing other constraints. The constraint that the metric has one eigenvalue equal to 1 is equivalent to requiring det(g-I)=(gxx-1)(gyy-1)-gxy^2=0. However, imposing this requires solving a quartic equation, along with some other messy computations, so there's no real advantage over the method I described in Ellipse fit from moments. X-RFC2646: Original How can one go about solving a DE that involves norms such as f'(t) = f(t)*||f'(t)|| Is there any general method? In the above equation I see that if I take the norm of both sides I get ||f(t)|| = 1 but that doesn't help much.... I was thinking there might be a way to relate the solution of g'(t) = g(t) but can't seem to get anywhere. Jon I doubt it. I assume f is a vector in R^n, so this is really a system of DE's. Let ||f'(t)|| = g(t), and suppose this is not 0. For any component i we have (f_i)'(t) = f_i(t) g(t) implying that f_i(t) = A_i exp(G(t)) where G is an antiderivative of g and A_i is constant. Then ||f(t)|| = exp(G(t)) ||A||. But (if f' is not 0) ||f(t)|| is the constant 1, as you noticed, so G must be constant and g must be 0, i.e. f is constant. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada X-RFC2646: Response Ofcourse I mainly mean analytical methods... but I have also tried to implement some simple numerical method by letting f'(t) ~ (f(t+h)-f(t))/h... but seems impossible to solve for f(t+h) so it would require, as far as I can tell, an implicit function approximation, which I'd rather not do. 1. Call the local systems guru. 2. Dare my kids to do it. 3. Etc. What is a way, anyway? I never did like counting problems that didn't make clear what set is to be counted. dave PS - And what's a henway? About 5 kilos. -- posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3 I'm not sure how many ways _you_ can do it. I can show you how many ways it can be done (assuming any cable reaches from any arbitrary computer to any other). There are three cases: (1) One computer is connected to each of the other 4 computers. (2) The cables are connected in a path (such as C1 to C2 to C3 to C4 to C5). (3) One computer is connected to 3 others, and one of the 3 others is connected to the fifth. (C1 to C2, C3, and C4; C2 connected to C5.) If the computers are indistinguishable, then the answer is 3. If each computer is different from the others, then we have to count them: (1) The connection scheme is decided once we've chosen which computer will be connected to the other 4. There are 5 possibilities. (2) We need to put 5 objects in a row here, so the answer would normally be 5!. However, C1-C2-C3-C4-C5 and C5-C4-C3-C2-C1 (its reversal) are the same connection, so we have to divide 5! by 2, giving 60. (3) The connection scheme is decided once we've chosen the computer connected to 3 computers, the one connected to 2, and the one computer only connected to the computer connected to 2 other computers. There are 5 choices for the first one, 4 for the second one, 3 for the third, so there are 5*4*3 = 60 ways to choose the connection. The total number is 5 + 60 + 60 = 125. The problem related to this in graph theory is the number of labelled trees with n vertices; the general asnwer is n^(n-2), which also gives 125 here, albeit a lot quicker. --- Christopher Heckman I thank you for your help. I understand the solution. Are the computers identical, or distinguishable? Are the cables identical, or distinguishable? When is the assignment due? -- posting-account=TSVGsw0AAABvQFoagF0DNRiPC-ntlke1 is the set {n 1+2^n | 1+3^n} infinite ? posting-account=5Ce0-Q0AAAA61KBHstChW6nsPTYCEd86 Hint: n must be even, (say) 2k. What does Lagrange's Thm say about divisors of 3^(2k) + 1? They must be equal to 1 mod 2K. You want 2^n + 1 = 1 mod 2K. What does this say about K? n = 1, no n = 2, 5|10 yes n = 3, 9, 28 no 1 + 2^n = 0 or 2 (mod 3) 1 + 3^n = 1 (mod 3) Thus n must be even. Is the set { n : (1 + 4^n) | (1 + 9^n) } infinite? During a lecture our lecturer said that the following is obvious. I can not see it. Could someone help me out ? Given a vector v in the tangentspace T_q (R^n) of R^n at q, there exists a unique vector field on R^n wich is parallel and equals v at q. David. There is a canonical trivialization of the tangent bundle TR^n, obtained by translating the origin of R^n to each point q separately. That is, TR^n = R^n x R^n, where the first factor represents the base, the second factor the standard fibre, and the pair (q,v) represents the vector v translated to have its base located at q. Given v in T_q(R^n), note that its description wrt the above trivialization is (q,v). The vector field that is so obvious just takes the value of v everywhere, so for any given p in R^n, wrt the above trivialization, the value of V is given by: V(p) = (p,v) -- Dale. posting-account=BRQo2w0AAAB6rp2bLqF2miYqvCwYGis4 Let n be in the integers (Z) and write u to denote either of the square roots of n in The Complex numbers. Show that Z[u]={a+bu: a,b in Z}. Proof: Let f(u) be in Z[u] where f(u)=a+a1*u+a2*u^2+...+an*u^n. Now we are given that u denotes either of the square roots of n in the Complex Numbers, so u^2=n. Now even powers of u will reduce to integers and odd powers will reduce to integer multiples of u, since u^2k=(u^2)^k. I'm not really sure how to write the part about even and odd or how to bridge that to where Z[u]={a+bu: a,b in Z}. Fast Clause Backtracking for k-SAT Most systems for solving the Boolean Satisfaction (SAT) problem use some form of backtracking.. Often, these systems backtrack on Boolean variables. I describe a simple transformation that converts a k-clause into a k valued variable. This allows backtracking on clauses instead of variables. The method I describe is at least as effective as backtracking on variables and produces more binary and unit clauses than other backtracking methods. http://home.comcast.net/~logiclab/ClauseBacktrack.html Russell - 2 many 2 count posting-account=BRQo2w0AAAB6rp2bLqF2miYqvCwYGis4 Let n be in the integers (Z) and write r to denote either of the square roots of n in The Complex numbers. Show that Z[r]={a+br: a,b in Z}. Proof: Let f(r) be in Z[r] where f(r)=a+a1*r+a2*r^2+...+an*r^n. Now we are given that r denotes either of the square roots of n in the Complex Numbers, so r^2=n. Now even powers of r will reduce to integers and odd powers will reduce to integer multiples of r, since u^2k=(u^2)^k. I'm not really sure how to write the part about even and odd or how to bridge that to where Z[r]={a+br: a,b in Z}. posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3 You've almost got it. You just need to write a+a1*r+a2*r^2+...+an*r^n in the form A + B r, where A and B are in Z. (What would you do for 3 + 4 r - 2 r^2 + r^3?) Note that r^(2k) = (r^2)^k = n^k, which is in Z. --- Christopher Heckman posting-account=HK86Ug0AAAAx9su3odaxsaW3UvhrfMAd Yes, indeed, author does not have perfect English; however, the mathematical component is simply nice! J. Watanabe watanabe_john@yahoo.com posting-account=BRQo2w0AAAB6rp2bLqF2miYqvCwYGis4 test posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3 Oral or written? And would you prefer hyperelliptic curves or algebraic topology? X-Abuse-Notes: Abuse reports must be submited via the usenetabuse.com portal listed above. X-Abuse-Notes2: Reports sent via any other method will not be processed. Yes to both. 1) What do you mean with the big n-plane ? 2) What do you mean with the long line L ? Sem Tuarg I made the term up, in analogy with the standard term long line. Given the long line, the above definition: cartesian product L x L x ... x L will define L^N , or taking the longer line 2L, take this 2L x 2L x ... x 2L instead. This is the standard example of a non-paracompact manifold: Take Omega: the first uncountable ordinal. For each ordinal k < Omega, take a copy I_k of the unit interval, and identify the right endpoint 1_k with the left endpoint 0_(k+1). The long line has no denumerable basis, since no countable set of points can get far enough out on L to approach the end at Omega. Thus, no countable set of points can be dense in L, and if a space has a countable basis for its topology, then it must have a countable dense subset (this looks like it requires AC; I hadn't recalled that requirement, but don't see a way around it). Since L is not separable (countable dense subset), it cannot have a countable basis (alias 2nd countable). Checking Google for [topology long line] , I find I've defined the long ray. The long line is apparently what you get by removing the left endpoint 0_0 of the zeroth interval. In that event, you don't need to double the long line to avoid those nasty corners in the product L^n. -- Dale. The result must be independent of c (changing c is in effect scaling R1 and R2 by the same constant, and doesn't affect the product). Moreover, the distribution is invariant under multiplication by orthogonal matrices on the left or right. In particular, it's invariant under multiplication by -1 and by permutation matrices, so all entries have the same marginal distribution which is symmetric about 0. Since values of R2 that are close to having rank < n produce a pseudo-inverse with large entries, I doubt that any moments exist. The pseudo-inverse of R2 is a rather complicated function of the entries of R2, so I suspect that it's difficult to say very much more about this except numerically (e.g. by simulation), or in cases where m and n are rather small (e.g. you might try n=1,m=2). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada posting-account=S9QRrgoAAAB_cl6JZ01YvFuCc1GXEQbJ What if Rank(R1) = Rank(R2) = n? In that case, the pseudo-inverse of R2 is equal to (R^T R)^(-1) R^T. posting-account=S9QRrgoAAAB_cl6JZ01YvFuCc1GXEQbJ Nobody even has any comments/thoughts/suggestions or jokes for this post? Roy Look at http://mathworld.wolfram.com/TotientFunction.html where phi(n) is plotted against its corresponding O(n)= n/log log n. Is there a most composite number [phi(n)/O(n)=maximum] ? The graph looks otherwise. (But in any case it looks like an upper limes of ~2 exists, what's the value?) -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn posting-account=5Ce0-Q0AAAA61KBHstChW6nsPTYCEd86 You have it backward. phi(n)/ (n/loglog n) = phi(n) (loglog n)/n is largest when n is minimally composite. Let n = pq with p ~ q. Then phi(n)/n ~ 1 so your ratio becomes ~loglog n and this is trivially unbounded. Are you sure that you asked what you meant to ask? posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs I see what Hauke is talking about in this figure. http://mathworld.wolfram.com/t1img1973.gif There does seem to be an envelope which, at first glance looks like it asymptotically approaches a value near 2. That's only because the envelope, based on your comments, is actually the very slowly growing function A*log log n for some constant A. To Hauke: this function grows very slowly indeed but it will eventually exceed any positive value you choose. - Randy posting-account=5Ce0-Q0AAAA61KBHstChW6nsPTYCEd86 John Selfridge is well known for the following (paraphrased) comment: We know that loglog N goes to infinity, but noone has actually seen it do so. :-) If you think this goes to infinity rather slowly, you should consider the inverse of Ackermann's function. :-) :-) A French cartoonist Jean Effel put it thus: (In Paradise, God is giving Adam a geometry lesson): Two parallels meet at infinity. It cannot be proved, but I was there. I've got the following function: w(t,f) = abs(f)/sqrt(2Pi)*exp(-f^2t^2/2) and yet I am wondering if its Fourier transform has any meaning. If so, I guess it should only depend on f but if I calculate its FT using classical FT results, I get W(f) = exp(-1/2)!!!??? Anyone would have an idea on what is wrong and where? Carine. A function can describe a 'signal' in the time-domain, OR in the frequency-domain. Not both at the same time. Because you're speaking of the Fourier transform of this function (from time- to frequency-domain), 't' is your variable and 'f' MUST be regarded as a constant. The 'f' in your function w(t,f) just cannot be the same 'f' as in the Fourier transform. In order to avoid confusion when calculating the Fourier transform, replace the constant 'f' in function w(t,f) by 'a' such that it becomes w(t,a) and calculate the Fourier transform. You can see that w(t,a) is just a Gaussian distribution in the time-domain. The Fourier transform should result in a Gaussian distribution in the frequency domain. Jeroen posting-account=_qC3WAwAAAA4j9fI9-ZyzQDmEqhiU6_R Could anyone help me with the following: Find the values of k for which x / ((x - k) * (x + 1)^2) has one stationary value. Just to be clear, that denominator is (x-k)*(x+1)*(x+1). I used the quotient formula and obtained: dy/dx = (2x^3 + (2-k)x^2 + k) / ((x+1)^4(x-k)^2) I reasoned that this would equal zero when the numerator equals zero and so have: 2x^3 + (2-k)x^2 + k = 0. The answer given is 0.8 and indeed I see graphically by plotting this function that when k = 0.8 there is indeed only one real root for this cubic whereas k = 0.7 or 0.9 provides no solutions, so the answer look right. How can I solve the above cubic to find the value of k for which it has only one root (i.e. 0.8). Alternatively, would someone else do something entirely different. Finally, I'm not comforable ignoring the denominator of dy/dx and simply reaoning that the nuerator must equal zero, I guess I'm worried becuase this denominator may be zero for some choice of k and x and so I sould more formally handle it. Mitch. Perhaps you need to check your differentiation. Maple gives: (-2*x^2+k*x-k)/(-x+k)^2/(x+1)^3 for the derivative, simplifying your problem greatly, no? --Lynn Maple, Maple ....don't get the habit of doing everything with brain prothesis Apart the global sign, the derivation is OK. but when summing 5/6 + 2/9 the common denominator shouldn't be 54 but just the LCM = 18 ! The same thing happens here, the LCM of (x - k) * (x + 1)^2, (x - k)^2 * (x + 1)^2, (x - k) * (x + 1)^3, is not a ((x+1)^4 factor, but just (x - k)^2 * (x + 1)^3 ! However, solving a cubic through Cardano method is rarely a good idea. You should first look for trivial solutions : 2x^3 + (2-k)x^2 + k = 0 has the trivial solution x = -1, as 2 = (2-k) + k for any k. Hence the cubic should factor with (x+1) and result into a (x+1) * (a*x + b*y + c) = 0, hence simplifying the derivative to a denominator without ^4 factor, as it should have directly obtained with a proper LCM. -- philippe mail : chephip+misc at free dot fr replace misc by news otherwise message will go to trash Old fashion mathematical handbook has for y^3 + py^2 + qy + r = 0 by substitution y = x - p/3 gives x^3 + ax + b = 0 which for d = b^2 / 4 + a^3 / 27 has one real, two conjugate roots when three real roots, at least two equal when d = 0 has three distinct real roots when d < 0 That's when x = k or x = -1 which aren't in the domain of your y. oppositr sign. But here you don't *need* to solve a cubic !!! Why do you have a (x+1)^4 and not a (x+1)^3 as it should in denominator of dy/dx ? [only (x+1)^2 for denominator of y] -- philippe mail : chephip+misc at free dot fr replace misc by news otherwise message will go to trash Old fashion mathematical handbook has for y^3 + py^2 + qy + r = 0 by substitution y = x - p/3 gives x^3 + ax + b = 0 which for d = b^2 / 4 + a^3 / 27 has one real, two conjugate roots when three real roots, at least two equal when d = 0 has three distinct real roots when d < 0 That's when x = k or x = -1 which aren't in the domain of your y. Denote this by X = u + v (u+v)^3 = u^3+v^3 + 3uv(u+v), here u^3+v^3 = 10+w + 10-w = 20 uv = (u^3 v^3)^(1/3) = (100-108)^(1/3) = -2 which has real root X = 2 (e.g. via rational root test) a unique real root since X^3 + 6X - 20 is increasing. Expressions like the above are often discussed since they arise naturally when solving a cubic via Cardano's method. Above we are essentially reversing this method of solution. --Bill Dubuque posting-account=37hkYgwAAABi-q-EYF7y8sZoyYm8zxwY Why? What is the motivation for this trick? OK How did you get that without solving a cubic? OK -- Rich posting-account=37hkYgwAAABi-q-EYF7y8sZoyYm8zxwY Huh? -- Rich posting-account=37hkYgwAAABi-q-EYF7y8sZoyYm8zxwY ick OK, one root is the expression, and another root = 2, but you have yet to prove that those are identical. -- Rich Lazy Derive the cubic. It has three roots, one real and two complex, which can't be the expression as it's real. posting-account=37hkYgwAAABi-q-EYF7y8sZoyYm8zxwY It stumped me. Huh? -- Rich Indeed? The question you asked, and the way you asked it, lead to think that you are very young. Proba(Homework) = 1-epsilon. Here are some exercises from a book I used when I was at college. Prove that all these expressions are integers. 1) Sqrt(7 + 4 * Sqrt(3)) + Sqrt(7 - 4 * Sqrt(3)) 2) Sqrt(2) * Sqrt(2 + Sqrt(3)) * (Sqrt(3) - 1) 3) Cbrt(a + ((a+1)/3) + Sqrt((8a-1)/3)) + Cbrt(a - ((a+1)/3) + Sqrt((8a-1)/3)) Sqrt(x) is x^(1/2) Cbrt(x) is x^(1/3) Why Huh? Since 108 = 36*3, you have that 108^(1/2) = 6*3^(1/2). Now, identify (a*3^(1/2) + b) to some cube means identify (6*3^(1/2) + 10) to (u*3^(1/2) + v)^3. What you have to do is just to apply the formula (a + b)^3 = a^3 + 3 a^2 b + 3a b^2 + b^3 so (u*3^(1/2) + v)^3 = 3*3^(1/2)*u^3 + 9 u^2 v + 3*3^(1/2)*u*v^2 + v^3 thus 3*u^3 + 3u v^2 = 6 9u^2 v + v^3 = 10 or, else, 3u (u^2 + v^2) = 6 (9u^2 + v^2) v = 10 It takes about one hundredth second to see that u=v=1 (and you do not need an other solution, if any, you just need one) so (3^(1/2) + 1)^3 = 6*3^(1/2) + 10 and finally (10 + 108^(1/2))^(1/3) - (-10 + 108^(1/2))^(1/3) = 2 The only trap in my post is a sentence you didn't quote: The trap is that integer part is a word to word translation of the French Partie entiere function, which one is equivalent to the English Floor function. That's why I put a smiley. BTW, why didn't you answer my second post? The one where What are you not understanding in this one? mm posting-account=xzIIbAwAAAB9g3T3nOBBKHft23y9UlyE Let X be a unity-mean exponentially distributed random variable. Let Y be a non-negative random variable which, upon conditioning on X, is also exponentially distributed with mean 1+ a*X. Here, a is a positive scalar. My goal is to investigate the asymptotic behavior of the pdf of Y (unconditional pdf of Y) as 'a' tends to infinity. That is, can I write the pdf of Y only as a function of 'a' as 'a' tends to infinity?? Next, consider Y1 and Y2, conditioned on X, each having exponential distributions with respective mean values 1+a*X and 1+ b*X. Here again a and b are positive scalars. Can I now obtain an asymptotic expression for the joint pdf of Y1 and Y2 (unconditional) as either a or b or both tend to infinity?? Ramesh One way to produce such a Y is Y = Z + a*X where Z is another exponentially distributed random variable with mean 1, independent of X. Then for the density of Y I get Are Y1 and Y2 to be conditionally independent given X? If so, write Y1 = Z1 + a*X and Y2 = Z2 + b*X with Z1 and Z2 having exponential distributions with mean 1, independent of each other and X. I'm not sure of the joint density, but the joint characteristic function is easy: E[exp(i*t1*Y1 + i*t2*Y2)] = E[exp(i*t1*Z1)*exp(i*t2*Z2)*exp(i*(a*t1+b*t2)*X)] = 1/((1-i*t1)*(1-i*t2)*(1-i*(a*t1+b*t2))) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada The density f of Y is given by f(y) = integral(x=0..infty, very rusty: can this be evaluated by tahkeing the limit of contour integrals around the singularity? Here, the joing density is given mby f(y, z) = integral(x=0..infty, exp(-[y/(1+a x) + z/(1+b x) + x]) / [(1+a x) (1+b x)]) Ditto? -- Stephen J. Herschkorn sjherschko@netscape.net posting-account=xzIIbAwAAAB9g3T3nOBBKHft23y9UlyE Hi Stephen, integrals for quite a long time (for the past three weeks or so). I have also looked into Table of Integrals, but no luck. For the first case, we can infact bound the integral in the form of modified Bessel function of second kind and zero order (K_0(x)) which is given by pdf_Y(y) <= (2/a)*Exp[1/a]*K_0(2*sqrt(y/a)). For the second one I almost lost hope. My goal is not to over bound (i.e., I don't prefer loose upper bounds), but to get the limiting behavior. sincerely Ramesh Y positive write is again expression both [(1+a (sorry for mistakes in english). I would want to change a matrix of relations between variables to a hierarchy between these variables. For instance, here is a matrix A=[. 1 1 . . 1 . . .] associated to relations between three variables Z1, Z2, Z3 (one for each row) 1 means that a row variable has a greater value than a column variable. and we can give the rank 1 to Z1,, the rank 2 for Z2, the rank 3 for Z3. Even if the matrix does not give a perfect order relation (lack of transitivity), I would like to extract a hierarchy (a sorting) which would be the best giving the matrix. I hope that it is a tournament graph but I am a newbie in this. My problem is that I have some incomplete data in these matrices. For example, if we consider the matrix A=[. 1 x . . 1 . . .] the relation between Z1 and Z3 is not defined. In the matrix I consider, I can have 1 for greater than, 0 for equal to, x for relation not defined). I look for an algorithm that would give me from the matrix A=[. 1 x 0 . . 1 x . . . x . . . .] the result Z1 = 1 Z2 = 2 Z3 = 3 Z4 = 1 My old idea was to use the linear system Z1-Z2=1 Z1-Z4=0 Z2-Z3=1 where each coefficient 0 or 1 between Zi and Zj defines the equation Zi-Zj=0 or 1. The results are good but I am not sure this is a very rigorous idea and perhaps we can do better ? Have you got ideas ? -- Boris http://www.pi314.net having a bit of difficulty with. The question is related to graph theory. This question is: There are 6 people in a room. Prove that there must be either 3 people who know each other, or three people of whom do not know the other two. I have a problem with an equation: I have the fractal of Cantor and for each branch I have a circuit with impedance between the input terminal and the ground of the network which has the form of a continued fraction: 1 2 1 1 Z(ω)= R+ --------- -------- -------- -------- ....... i ωC+ aR+ i ωC+ a2R+ after simplification I find this form: Z(ω/a) = (a/2)*Z(ω) this is a scaling equation a is a constant ω is a variable (pulsation) I must to find a solution from Z(ω) Actually I know the solution for this equation but I don't know how to solve that equation. The solution is Z(ω) = k*(i*ω)η Where: k is a factor η= 1-ln2/lna with ln/lna the fractal dimension Octavian posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs with You posted exactly the same question in sci.physics with the title Physics and fractal, etc. If you have a legitimate reason for cross-posting, then put both newsgroups in the header. Otherwise people in the two newsgroups won't see answers from each other. Meanwhile, can you post a link explaining what the fractal of Cantor is? The term fractal postdates Cantor by many years, so I'm sure this is not actually anything Cantor discussed. My guess is that it's a fractal structure related in some way to Cantor's set (the set of all real numbers in [0,1] which do not contain 1's in their base-3 representation). - Randy non-ASCII like ω is not readable by me, sorry. Comparing the other, I guess this is omega? You can just write w if you like. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ I have a problem with an equation: I have the fractal of Cantor and for each branch I have a circuit with impedance between the input terminal and the ground of the network which has the form of a continued fraction: 1 2 1 1 Z(w)= R+ --------- -------- -------- -------- ....... i wC+ aR+ i wC+ a2R+ after simplification I find this form: Z(w/a) = (a/2)*Z(w) this is a scaling equation a is a constant w is a variable (pulsation) I must to find a solution from Z(w) Actually I know the solution for this equation but I don't know how to solve that equation. The solution is Z(w) = k*(i*w)^n; Where: k is a factor n= 1-ln2/lna with ln/lna the fractal dimension Octavian You didn't explain a2R ... is 2 an exponent or something? What is the pattern of the numerators: 1, 2, 1, 1, ?? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ posting-account=vtkPsw0AAAAWjQERxIyiJEVUICMXTlxO I would like to know if Abstract Algebra's branches have any application in Artificial Intelligence or Data Mining. Could you please introduce me any resources about these subjects? X-RFC2646: Original I'm not sure I've understood what you're looking for exactly, and I don't know what level you're approaching this at, but you might find http://www-2.cs.cmu.edu/~awm/tutorials/ a helpful introduction for data mining (I know nothing about AI). posting-account=vtkPsw0AAAAWjQERxIyiJEVUICMXTlxO I want to know if there are any relationships between Ring Theory (Field Theory ...) and AI or Data Mining. posting-account=FHpiLAwAAABjiwghIWy4qlQDZHDOoe_2 If you want something different, you should be more specific. Artificial Intelligence or Data Mining are essentially meaningless buzzwords. That wasn't part of my reasoning. What they say is very important to whether or not they are correct. What isn't important (for determining if their conclusions are correct) is whether or not they include extra stuff (e.g. examples) to help facilite understanding. So, again, how does it follow that journals are a total waste of time? Martin Well, those are supposed to be conguent signs; three parallell horizontal lines. I didn't know they might be unreadable. I will replace the codes by gifs soon. J K Haugland It just means you have a browser problem : you browser doesn't understand unicode 8801 which is congruent Your browser is told to search this in the Lucida console font It fails to do that. With both my IE6 and Mozilla + Windows 98, it is ok What is your browser / operating system ? Note : presently *No* browser handles unicode/math symbols correctly. The consequence is that webmasters have to replace all these by images (as for the in symbol on that page) or not be shure if anybody will read correctly (coding a unicode or symbol font character, as this 8801). However, a page with hundreds images is not fair either... and the coding of math symbol characters has to be browser dependent ! BTW could you check your browser against my site's congruent symbol, for instance at http://chephip.free.fr/ie/kcongr.html (should switch page automatically depending on the browser IE/NS, site is in French, but doesn't matter for a symbol check). -- philippe mail : chephip+misc at free dot fr replace misc by news otherwise message will go to trash ... ... Ordinarily, Netscape Communicator 4.8 for Linux with Use my default fonts, overriding document fonts selected. Jan's revised page (with in.gif, congr.gif, leq.gif, etc) shows up fine now. ... On my system: With Netscape Communicator 4.8 it shows up as x .bc y, that is, like a degree symbol or raised small circle. With Opera 6.01 it shows up as an underlined degree symbol. With Konqueror 3.2.2 it shows up as a three-line equivalence symbol. -jiw ... font 2 bad browsers over 5 (including mines)... Helas, changing the 200 and some pages on my site for gif instead of characters is not to be considered. My Opera 7.23 is OK (opera's preferences being identified as MSIE6 or Opera) changing opera preferences to be considered as Netscape Mozilla 5.0 completely crashes my site : congruent character replaced by a new line ! (as for other sqrt, product ... characters) Nothing's perfect ! -- philippe mail : chephip+misc at free dot fr replace misc by news otherwise message will go to trash [cross-posting elided] This happens because you cannot use the north metric at all points of a sphere. There isn't a way to map a 2-D transformation to 3-D at the point where the 2-D is undefined; north of north pole is undefined unless you rotate the sphere. Since the units used to measure are time and space and are undefined at the point of the BB, you can't use them for mapping. I think I've used correct nouns and verbs but I'm not sure. /BAH Subtract a hundred and four for e-mail. Chris@Sonnack.com says... I don't buy into physical singularities. So I assume that general relativity breaks down under the hot, dense conditions we know existed in the past. Whether that means we should talk about times prior to the Big Bang, I don't know, but I don't reject the concept. As for no outside I'm not sure this is as unambiguous a claim as people like to say. Suppose we had an old-fashioned 'primeval atom' that exploded due to a universal repulsive force that was active under the then conditions, and the part of it that formed our universe was somewhere un-special within that fireball. Just how would our observations differ from the standard relativistic cosmological models? Surely we should say that the relativistic models have no need of or concern with an outside, rather than insist that it doesn't exist? I think it's a failing of those enthused by relativity to forget that its simplification of space and time is a limitation as well as an advantage - there seems to be a blindness to alternatives which may seem slightly more complicated but are hardly to be rejected as inevitably false on that account. Illogical is worse than weird! Really, what Aspect made sure of was that the entanglement was very Well, here we are talking about two sorts of randomness. Yes, from the perspective of the observer it is random. It doesn't collapse the other's wave function. That only happens when the second observer measures it. And we'd expect all the usual phenomena related to 'delayed choice' etc. to apply as normal. Well, if we generalise from the Aspect experiment, there should be correlations between seemingly random observations everywhere. Very obscure correlations, but correlations nonetheless. It just makes it very hard to spot. Though I suppose it might depend what exactly you mean by decoherence. I just mean that the wave function evolves in such a way that the entanglements between separate states are effectively impossible to measure. - Gerry Quinn I have no idea, since your suppose makes no sense to me (by which I mean I don't understand what you're getting at). Truly, fully random? It has certainly affected how that wave function can collapse, though, yes? The two observers can NOT get conflicting results, right? There may well be, but I'm still not understanding how correlations between We seem to agree the first measurement is random. I think we agree there I think the only thing we disagree on, perhaps, is why the first that first measurement is random? To me, it's simply because the universe is fuzzy and random when you get down far enough. That is to say that true randomness does exist at the quantum level. Pretty much what I said: it destoys the entanglement. Chris@Sonnack.com says... Randomness is something - a piece of information. Order isn't. Something happened versus nothing happened. Yes, but you are defining randomness in a particular way, i.e. randomness as related to a particular observer. I was talking about randomness as part of how the universe actually works. course, non-locality suggests in itself the concept of an 'eternal instant', or rather that time is something of an illusion. One way of looking at the many-worlds concept is that nothing actually 'happens' but every instant one can experience consists of a valid observer/observed pair out of the gamut of possible unitary evolutions of the universal wave function. Of course, the difficulty is to explain why we only observe universes that seem to encode a history of smaller universes that evolved into the observed one! Yes, I think that's where this line of thought leads. Of course, there is no matter generated when a universe splits - it's the same universe, just different branches of its wave function that have somehow become separated. I take the view that free will and determinism are not connected. If you can always outguess me in rock-paper-scissors, have I suddenly lost free will? - Gerry Quinn We have got to be talking about different things, since I would have said it was precisely the opposite! Randomness has a much higher state of entropy than order doesn't it? And hence considerably less--if any--information. universe is intrinsically fuzzy. Actually, they are necessarily connected unless you can come up with a mechanism that allows you to generate free will from nothing. IF there is no randomness and the universe is hard (not fuzzy), then interactions are fully determined. Therefore each action is a necessary outcome of conditions leading up to it, and there can be no free will. This is (absent a mechanism to generate free will) an inescapable consequence of a fully determined universe. I don't see what that has to do with anything, since it doesn't at all touch on WHY I can outguess you. IF it's because I have in my going, you haven't lost free will--you never had it. You have no more choice than a billiard ball on its way to the corner pocket. volker.hetzer@ieee.org says... Well, we don't really know that, as such. We know that the universe was once hot and dense, and the FRW metric that models it in general relativity has a singularity at a finite time in the past. But that may just reflect a breakdown in general relativity, and many cosmologies have been proposed in which time extends backwards past the Big Bang. But either way, if 'came from nothing' doesn't involve time, what do you mean by it? Because the 'nothing' you refer to is surely at a hypothetical time prior to the singularity? Wherever it must. There may be a finite multiverse in which logic prescribes every possible observable universe (and its observer). Just as logic prescribes every prime in the natural numbers. What I said first - an intrinsic physical randomness in the universe/multiverse as a whole. That is to be distinguished from the randomness observed by an observer who only sees part of the whole. He refers somewhat inaccurately to The Hidden Variable Theory as if there were only one. The particular theory he rules out is (at least implicitly - he seems to discount Bohmian 'pilot waves' without mentioning them) a *local* hidden variables theory. Also, the experiment confirms quantum *theory* - the various interpretations are a different and more philosophical matter. Ruling out local hidden variables does indeed mean that we have randomness from the perspective of a particular observer, who has no access to non-local information (essentially, any influence emanating from outside his past light cone). But from the 'perspective' of the universe/multiverse as a whole, we don't necessarily have intrinsic randomness. - Gerry Quinn Just a question - is it too much to ask that this extended discussion of physics/metaphysics be taken to a more suitable newsgroup? None of the three current newsgroups (sci.math, sci.crypt, comp.programming) seem like the right place for this discussion. X-Enigmail-Version: 0.90.0.0 X-Enigmail-Supports: pgp-inline, pgp-mime Hi Let S be a set of n d-dimensional, axis parallel orthotopes (generalization of rectangle, hyper-rectangle) such that no two orthotopes share a common point. Each s in S is given by a d tuple of ranges where s(i) denotes the range of s in dimension i. exists 1 <= i <= d: for all s != t in T: s(i) intersection t(i) = empty_set be a predicate. A partition W of S is a subset of powerset(S)-{empty_set} such that Union_(w in W) (w) = S AND forall w != w' in W: w intersection w' = empty set. Let X be the set of all partitions W of S for which for all w in W: P(w) holds. Let's call the partitions in X valid partitions. Do you know any (tight) upper bound for m := min{|W| : W in X}? Clearly, m <= ceil(n/2) since there is always a valid partition which consists of pairs of orthotopes. I believe that this bound is rather bad. However, note that for any d, one can construct a set T = {a,b,c} of orthotopes such that P(T) is not satisfied, e.g. a, b are equal in all but one dimension i and c is some orthotope such that c(i) = [min(a(i) union b(i)), max(a(i) union b(i))]. Moreover, do you know an algorithmic way to construct a partition of size m or close to m? A simple greedy approach would be the following. For each dimension i and for each orthotope s, compute the number v(s,i) of orthotopes overlapped by s in dimension i. In each dimension, remove the orthotopes with largest v value until the remaining orthotopes are non-overlapping in dimension i. Pick the largest set of remaining orthotopes among all dimensions and repeat the procedure for the orthotopes not contained in this set. Thomas SL2(Z) grope is defined { [a b]| a,b,c,d are integer and ad-bc = 1} [c d] , which generators are [1 1] and [-1 0] [0 1] [0 1]. Gamma(11) is SL2(Z) subgroup, where c is limited to be able to be divided by 11. What are generators for Gamma(11)? Morio nagira@d5.dion.ne.jp group, not grope! Those two matrices clearly do not generate SL2(Z) - they are both upper triangular, and the second one does not even lie in SL(2,Z) ! Since you cannot take the trouble to get your preferred generators correct, I will answer your question in terms of two generators of my choosing, and let you translate them back to whatever generators you prefer! Let x = [ [0, -1], [1, 0] ], and y = [ [1, 1], [-1, 0] ]. of SL(2,Z). The the subgroup you describe is generated by yx, x^2, y^-1xyxyxy^-1x^-1yx^-1y, y^-1xyxy^-1xyx^-1yx^-1y which evaluate to the matrices [[1,-1],[0,1]], [[-1,0],[0,-1]], [[8,3],[-11,-4]], [[7,2],[-11,-3]]. It has index 12 in SL2(Z). Derek Holt. posting-account=82j3EgwAAACYxp9hHOcWy78r2IGkmc3t If every subring of the reals with an infinite number of elements must contain its convergent infinite sums, then there are no rings between the integers and the reals. Not the rationals, not the algebraic integers, not the object ring, nothing. You can put off dealing with this for some time (possibly several years) but it is not going away. - William Hughes That's correct. Minimal ring doesn't make much sense. Here we're talking about the minimal ring containing Z and 1/2. We all know that this is your point. Instead of repeating your point you should pay attention to explanations of why your point is nonsense. One doesn't need to block infinite sums. There's no mention of infinite sums in the definition of the word ring. The set of all rationals of the form n/2^k where n is an integer and k is a non-negative integer _is_ a ring, it contains Z and 1/2, and hence your point is simply wrong. _You_ don't seem to realize how stupid you felt the day you finally realized that you'd been spouting nonsense when you informed us that integers were irrational. ************************ David C. Ullrich !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ Uh, what? Could you please specify a ring which has operators including infinite sums that converge in a consistent way? If not, you are talking about a ring extension, and the infinite sums that converge are not a property of the ring, but a way of determining an extension to a ring. In short: all of this only makes sense if you are talking about _two_ different rings. One well-defined ring that is not affected by converging sums at all, and one well-defined extension ring that is defined based on the non-extended ring by an extension rule based on infinite sums. You _need_ two different rings to bootstrap the concept, as infinite sums are not a ring property. So this talk about the ring is nonsensical. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum *If* you allow infinite sums. I don't think David was suggesting we do. -- Will Twentyman email: wtwentyman at copper dot net David Kastrup a .8ecrit : Not to mention the fact that convergence implies some topology underlying the ring (as the fact taht Q_p and R are quite different, even as algebraic beasts, shows well) X-mailer: xrn 9.02 Mail-To-News-Contact: abuse@dizum.com Am I using the Axiom of Choice here? I want to show that there's an injection from A x {{}} to A x B (B is stated to be non-empty). It seems simple enough: pick one element, y, from B. Then, any element (z,{}) (z in A) is mapped to (z,y). However, I'm suddenly very aware of the (possible) problem with saying pick one element. I *think* that I'm OK here, since I'm only choosing one element from one set one time. I believe that Choice is only required when picking one element from each member of an infinite collection of sets. Am I on solid ground this time? -- Michael F. Stemper This email is to be read by its intended recipient only. Any other party reading is required by the EULA to send me $500.00. Yes, you are fine. Since you have as a hypothesis, There exists x such that x is in B, you are not applying AC. As you note, AC comes in only when we have to make infinitely many choices. Here is another, really unrelated question I have been contemplating lately: Does one need AC (or some weak version thereof) to show that the countable union of finite sets is countable? Seems to me that the answer is yes: you have to choose a bijection for each of the finite sets. -- Stephen J. Herschkorn sjherschko@netscape.net Stephen J. Herschkorn a .8ecrit : Yes; Cohen constructed a counterexemple of a denumerable set of two-elements sets whose union is not countable (a formalisation of the famous example of shoes and socks) Even in the case of a countable collection of pairs, you need (some form of) choice, unless the pairs are already ordered. The standard example is choosing from infinitely many pairs of socks, vs. infinitely many pairs of shoes. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Here is some recurrence formula. F(n,k) = 1, when n < k. What can one say about its asymptotic? -- Alex Vinokur email: alex DOT vinokur AT gmail DOT com http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn posting-account=ncb9fA0AAAC3P1sJG-EKSeKxatgT-rHd I was reading thru Rudin's Real and Complex Analysis book and I got stuck at Theorem 1.14 which tries to prove that for a set of fn positive measurable functions their lim sup is measurable. His argument goes like this: Then g-'(A)=U fn-'(A), countable union of open set is open so g-' is open and therefore g is measurable. I'm ok up to this point ( I know But then he said since the result still holds if we replace sup with inf h is measurable since I have 2 issues here (1) I cannot see how the result holds if replace sup with inf (that is if we define g = inf fn instead of g = sup fn) because my accepting the statement g-'(A)=U fn-1(A) is based on the argument that if x e U argument no longer holds if g = inf fn (2) What does g got to do with lim sup ( g is defined for n 1 to I've searched thru the web and the proofs I found made the same assumptions as Rudin. Any help would be greatly appreciated. It is difficult to imagine an argument that works for sup and fails for inf. If your argument for sup is valid, it trivially extends to inf. Check it again. sup(f_n, f_(n+1), ...) decreases with n, lim can be replaced by inf. posting-account=ncb9fA0AAAC3P1sJG-EKSeKxatgT-rHd as you said the argument would work for inf, but only if we use {all posting-account=ncb9fA0AAAC3P1sJG-EKSeKxatgT-rHd X-mailer: xrn 9.02 Mail-To-News-Contact: abuse@dizum.com Except for the bit about being on the opposite side of the Sun, this is a dead ringer for the short story, Backward, O Time, by Damon Knight. -- Michael F. Stemper This sentence no verb. posting-account=VOcjCQwAAAAuS03wBmk1NOZrZAVSQTY1 The Practice Effect (I forget the author) was another SF book with a similar theme, although not so extreme. It was set in a distant future, with the universe contracting, and entropy increasing. X-mailer: xrn 9.02 Mail-To-News-Contact: abuse@dizum.com The author was David Brin. It was set in a parallel universe, not the distant future. The basic McGuffin was that artifacts got better as they were used, or they improved with practice. The natural consequence was that poor folk got all of the new stuff. After they'd used it for a while, improving it, they sold it to better-off folks, and started over. -- Michael F. Stemper This sentence no verb. X-RFC2646: Original You mean the brightest star in the Andromeda constellation? Hmmmmmm... very interesting discovery David... If you go to nytimes.com, you can do a search. I don't know if you have pay because it is more that a week (?) old.) -- Stephen J. Herschkorn sjherschko@netscape.net in its entirety, see my previous post. I will get in a library. Hiroshi Sugimoto Partant du principe que l'on est jamais mieux servi que par soi-meme et que j'aurais du m'y prendre autrement voici quelques resultats: la comparaison avec l'ordinateur: http://abel.math.harvard.edu/archive/21b_fall_04/exhibits/sugimoto/ In addition to a class of curves, hyperbolic also means exaggerated beyond reason. I don't know if that meaning came before or after the curve, but it would be reasonable to name the y=1/x curve using a word with that meaning. If you move x a finite amount, you can get an infinite/undefined y. Contrast with a parabola y=x^2 where you have to move x to infinity to do the same. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. 11:43:03 -0700: beyond That's only natural, since in Greek hyperbole means exaggeration :-) be with To address the OP, I don't know what's happening with the Mandelbrot set (it's been some time since I played around with it), but in some fractals where I've seen attractive parabolic fixed points, namely with the tetration fractal, it turns out that certain boundaries of the basin of attraction of such points resembles a parabola. -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable Primes proof #13 --- quoting from the book MATHEMATICS AND LOGIC , Mark Kac and Stanislaw M. Ulam 1968 , page 3 --- The following proof, probably still the simplest, asserts the mere existence of arbitrarily large primes. Suppose the number were finite; there would then be a largest prime p. Consider now the number n = p! + 1 (p! is read p factorial and equals 1x2x3x...xp). This number is not divisible by any prime up to p. If there is no prime between p and n; as we have assumed, then n itself would be a prime, contrary to our assumption that p is the largest one. This remarkably simple and elegant result of Euclid's is one of the first known proofs by contradiction. --- end quoting MATHEMATICS AND LOGIC , Kac and Ulam --- Kac and Ulam should have made the factorial notation serve only primes where for example my notation of P!+1 is a restricted notation servicing only those known primes and multiplied together. This is one of the most convoluted proof attempts I have ever looked at. It is so messed up and not worth trying to correct it for there is a mistake in every sentence. And in this sentence alone there are four mistakes-- If there is no prime between p and n; as we have assumed, then n itself would be a prime, contrary to our assumption that p is the largest one. It is remarkable that mathematics editors ever allowed this book into publication. Kac and Ulam made two major mistakes. First they wrongly ascribed Euclid as giving a Indirect proof method when in fact it was Direct method. Secondly, in the indirect method, once P!+1 is formed then the very next sentence or step is to declare P!+1 is prime also and thus discharging the initial assumption that set of primes is finite. There is no prime-factor-search in the indirect method for it conflicts with the definition of what it means to be prime used at the start of the proof. #14 --- quoting Victor J. Katz, A HISTORY OF MATHEMATICS, 2nd edition, 1998, p 87 --- The final book on number theory is Book IX. Proposition 20 of that book shows that there are infinitely many prime numbers. Proposition IX-20 Prime numbers are more than any assigned multitude of prime numbers. In analogy to the proof of proposition VI-1, Euclid has no way of writing down an arbitrary assigned multitude of primes. Therefore, he again uses the method of generalizable example. He picks just three primes, A, B, C, and shows that one can always find an additional one. To do this, consider the number N = ABC+1. If N is prime, a prime other than those given has been found. If N is composite, then it is divisible by a prime p. Euclid shows that p is distinct from the given primes A, B, C, because none of these divides N. It follows again that a new prime p has been found. Euclid presumably assumes that his readers are convinced that a similar proof will work, no matter how many primes are originally picked. --- end quoting Victor J. Katz, A HISTORY OF MATHEMATICS, 2nd edition, 1998, p 87 --- It is a relief that math professors have started to read Archimedes Plutonium posts on Internet showing what is wrong and flawed in previous renditions given by math professors of Euclid's IP. However, Mr. Katz did not understand my corrections fully. Katz makes two big blunders. His last sentence of Euclid presumably assumes that his readers are convinced that a similar proof will work, belies the fact that Katz believes Euclid gave an Indirect Method. Euclid gave a Direct Method of set cardinality increase and if Katz had understood Euclid to be a direct method would have stated more clearly in his attempt above. Secondly, in the indirect method N (P!+1) when formed is instantly a new prime and ends the proof and Katz's further talk about N being composite belies his inability to form a logical chain of deduction of Infinitude of Primes proof. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies Four brothers all share a house, and they are having trouble coming up with this month's rent. So they sit down and play poker. They play all night. When morning comes, do they have the rent? No, because poker is a zero sum game. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. And when JFK said A rising tide lifts all boats, he meant that the economy is not a zero-sum game. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. There are a lot interesting discussion. How about my solution as follows: (Sketch of proof) Let S(n) be a n-th partial sum. Then we can show that S(4n) is an increasing and bounded sequence. So S(4n) converges. Since S(4n+1) = S(4n) + 1/(4n+1) and S(4n) converges, then S(4n+1) also converges to the same value of the limit of S(4n). S(4n+2) and S(4n+3) also converges to the same value likewise. Using the fact S(4n), S(4n+1), S(4n+2) and S(4n+3) converge to the same value, we can show that S(n) converges (to the same value, too). posting-account=JV6tFQ0AAAA35LP2Y2Qv8s797if4HYhL There is a nice exercise in Palka's An Introduction to Complex Function Theory which answers this question : of decreasing positive numbers with limit 0, then the sum $sum_{n=0}^infty z_n r_n$ converges. The prrof is a nice application of summation by parts, and extends the classical Alternating Series Test. sums are clearly bounded. BCE Today is a day before SBC goes ex-dividend and usually today SBC would be dollars ahead of BCE in stock price. However, today, SBC is 24.25 and BCE is 24.75. This is the first time I have seen this circumstance in the past 3 or 4 years. A circumstance in which SBC was not several dollars ahead of BCE especially on the eve of ex-dividend. I think one of the new factors causing this rarity is the devaluation of the USA dollar to foreign currency due to the massive debt. It would be an interesting graduate school research project to find out how much percentage of SBC shares are held by foreigners and how much percentage of BCE shares held by foreigners and during the time period of March and April 2005 whether the price decline of SBC relative to BCE was due primarily to foreigners selling those shares or foreigners decreasing their desire to invest in SBC relative to BCE. One way for the USA to balance its budgets is to impose a national sales tax on luxury goods. Europe has a national sales tax, so why not the USA. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies Hint: Consider (a-b)*(a^2 + a*b + b^3) = a^3 - b^3 Sorry, (1) Typo, should have been (a-b)*(a^2 + a*b + b^3) = a^3 - b^3. (2) Wrong assumption, hinted method doesn't work. A better aproach is to assume that there should be some fairly simple solution. Note first that sqrt(108) = 6*sqrt(3), so that +/-10 + sqrt(108) = +/-10 + 6*sqrt(3) It would be nice if the expressions +/-10 + 6*sqrt(3) were the cubes of something simple, say of form +/- x + y*sqrt(3), where x and y are (small) integers. Since cubes of such +/-x + y*sqrt(3) will be of form +/- a + b*sqrt(3), this might just work out. Working with equations (+/-x + y*sqrt(3))^3 = +/-10 + 6*sqrt(3), one can fairly easily see that the only small integer possibility is x = y = 1 E.g., (x + y*sqrt(3))^3 = 10 + 6*sqrt(3) x^3 + 3*x^3*y*sqrt(3) + 9*x*y^3 + 3*y^3*sqrt(3) = 10 + 6*sqrt(3) Then separating rational from irrational, which we can do assuming x and y are integers, quickly gives x^3 + 9*x*y^2 = 10 and x^2*y+y^3 = 2 A small integer solution here is clearly x = y = 1. At which point, (10 + 108^(1/2))^(1/3) - (-10 + 108^(1/2))^(1/3) = (1 + sqrt(3)) - (-1 + sqrt(3)) = 2 A key in solving the problem is assuming that there is a nice solution. There is a classical example illustrating the same key: Given a solid sphere of radius r. A hole is drilled through the sphere centered on a diameter. What is the volume of the remaining object? [snipped original problem] Are you shure you quoted that correctly ? Given a sphere (of *unknown* radius) Drill a hole. The total *height* of the torus like object is then h (the height of the hole) What is the remaining volume. Your variant gives a volume which depends on an unknown parameter (hole diameter, or hole height) -- philippe mail : chephip+misc at free dot fr replace misc by news otherwise message will go to trash [snipped misstated problem] No! Sorry, but I misstated the problem. Having problems proofreading my postings. Hope it is my eyes, and not what is behind them. It is not the radius of the sphere that is given but the length of the hole, i.e. the distance between the circles which form the hole's ends on the surface of the sphere. Call this distance h. The sphere's radius is not given. Corrected restatement of problem: Given a solid sphere of unknown radius. A hole is drilled through the sphere centered on a diameter. The length of the hole is given as h (distance between the parallel circles that are its ends on the sphere's surface). What is the volume of the remaining object? That's OK now... corresponds to the one I had in mind. And letting the puzzle for other's pleasure. A lot of other weird simplifications like that, for instance : Consider a circular yard and a round tower in the middle. With only one straight line measurement, find the remaining area. Or the well known paper tape around the Earth puzzle -- philippe mail : chephip+misc at free dot fr replace misc by news otherwise message will go to trash Infinitude of Primes #15 --- quoting from MATHEMATICAL THOUGHT FROM ANCIENT TO MODERN TIMES, 1972, Morris Kline, page 80 --- Proposition 20. Prime numbers are more than any assigned multitude of prime numbers. In other words, the number of primes is infinite. Euclid's proof of this proposition is a classic. He supposes that there is just a finite number of primes, p1,p2, . . .,pn. He then forms (p1xp2x . . .xpn) +1 and argues that if this new number is a prime, we have a contradiction, because this prime is larger than any of the n primes and so we would have more than n primes. On the other hand, if this new number is composite it must be divisible (exactly) by a prime. But this prime divisor is not p1,p2, . . ., or pn because these leave a remainder of 1. Hence there must be some other prime; and again we have a contradiction of the assumption that there are just the n primes p1,p2,. . .,pn. --- end quoting from MATHEMATICAL THOUGHT FROM ANCIENT TO MODERN TIMES, 1972, Morris Kline, page 80 --- Kline makes two mistakes. First, Kline erroneously thought Euclid's IP was indirect proof method. Secondly, Kline goes ahead and gives an invalid indirect proof of IP by hunting around for a prime factor. This is a shame since Kline invested so much time in writing this classical book on mathematics history. Kline could have read Euclid's IP from the original Greek and perhaps Kline would have realized that Euclid was increasing set cardinality and generalizing to the infinite. This is the direct proof method. Kline could have then investigated where in the history of mathematics that mathematicians erroneously ascribed Euclid with Indirect Method. I believe from sketchy sources that it was Gauss who led astray the mathematics community into believing the Ancients used the Indirect method frequently when in fact the Ancients never used the method. Perhaps it was Gauss in his century who most clearly used and espouses the mechanics of the indirect method that would lead other mathematicians astray. Kline should have been clued by the fact that reductio ad absurdum is a Roman term, not a Ancient Greek term, and so should have investigated. And although Kline failed to give an accurate historical account of Euclid's IP and failed to even give a correct proof of such a simple theorem. Well, I would still buy his book. adjoined the prime numbers: --- end E. Landau rendition of Euclid's IP --- Please tell if Landau gives any discussion about Euclid's IP before he gives his rendition. the the --- Landau makes no mistakes in his Euclid IP proof, and he uses the direct proof method. He states it is a adjoining onto any finite set, and later this any finite set is generalized into the infinite set. Thus, Landau does not make the mistake of saying Euclid's IP was indirect and Landau, further, does not fall in the trap of giving a shoddy indirect proof of IP. So, this is proof that some math professors can see and think clearly. And that some math professors have a logical mind. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies Others already offered solutions, and unless I missed it, they were of the brute force kind. Let me give it a try: Reduce the problem to a single equation with 4 rational unknowns, and keep the problem as symmetric as we can: Denote m the sum x+y+z. For m non-zero, substitute x/m - 1/3 = U (U rational) and so on, as below, so x = m * (1/3 + U) y = m * (1/3 + V) z = m * (1/3 - U - V) r = m * (1/3 + W) s = m * (1/3 + Z) t = m * (1/3 - W - Z) (so x+y+z=r+s+t automatically) After simplification, we obtain U^2 + U*V + V^2 = W^2 + W*Z + Z^2 with rational unknowns. Without loss of generality we may assume that the denominators below are non-zero. Then we can manipulate the equation into (2*U + V - 2*W - Z) / (Z + V) = 3 * (Z - V) / (2*U + V + 2*W + Z) = p, a rational parameter. (multiply by 4, complete squares, and shuffle the terms to create differences of squares). My computer algebra system gave me a solution for U and V as functions of W, Z, and p as follows, in symbolic matrix notation: [U ] [V ] = 1/(3+p^2) * [3 + 2*p - p^2 4*p ] [ -4*p 3 - 2*p - p^2] times the column vector [W ] [Z ] Well, some symmetry/antisymmetry features are still visible. Finally, we restore x, y, z, r, s, t by picking m large enough to give the expressions for x,...,t integer values. As an example, pick p=2, W=1, Z=2, then after all the arithmetic we get x=64, y=-47, z=4, r=28, s=49, t=-56. How about the case x+y+z=m=0? Then the system reduces to x^2 + x*y + y^2 = r^2 + r*s + t^2 and can be treated similarly, with fewer unknowns. posting-account=qYt4fg0AAABejqcDA-VvO3WzjfSQtTMb Well, I made heavy weather out of this -- it's much simpler. Eliminating z gives: x^2 + y^2 - (x+y)(r+s+t) + (xy+r(s+t)+st) = 0 which is linear in t (and also in r and s, but that's not important). Solving for t gives: t(r+s-x-y) = (r+s)(x+y) -x^2 - xy - y^2 - rs i.e. t = (rx+ry+sx+sy-x^2-xy-y^2-rs) / (r+s-x-y) This will always give a solution for arbitrary values of x,y,r,s, Clearing fractions gives the following solution in parameters X,Y,R,S: x = X(R+S-X-Y) y = Y(R+S-X-Y) z = (R+S-X-Y)^2 + (RX+RY+SX+SY-X^2-XY-Y^2-RS) r = R(R+S-X-Y) s = S(R+S-X-Y) t = RX+RY+SX+SY-X^2-XY-Y^2-RS which satisfies the original equations for any rational values of X,Y,R,S. As in 1 + 1 + 2 divides 1^2 + 1^2 + 2^2, and 1^3 + 1^3 + 2^3. Nice try. Mily Pavle Pokorny, pan Tapio je znam jako dozivotni popleta. S pozdravem, ZVK(Slavek) X-RFC2646: Original How about (x,y,z) = (ai, a(a-i), a^2) with i < a-i? --Edwin Very well done and rapidly! :-) I assume a and i are integers, i.e. i is not sqrt(-1) Thus, there are infinite many of those. But, consider - is it necessary that i