mm-419 === Subject: Re: Very Ugly Sum by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i11EcCv16591; >>The following series converges for all t>0: >>1*exp(-t)+2*exp(-4t)+3*exp(-9t)+.....+n*exp(-n^2*t)+.... >>I would like to know if it is possible to express it in a closed >>form,or to give integral expression for this series. >It is the derivative of a theta function; see any decent >discussion on this branch of elliptic functions. Thank you,but... I'm not sure that it can be derivative of a theta function. I read that theta function is defined as theta(x) := exp(-x)+exp(-4x)+exp(-9x)+....+exp(-n^2*x)+... then the derivative is: theta'(x)=-exp(-x)-4exp(-4x)-9exp(-9x)-....-n*exp(-n^2*x)--- Am I right? === Subject: another approach Re: Please advise! what kind of training I am lackfor math? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i11FXGk21080; . One other approach no-one may suggest for you, Walala. Get some part-time work tutoring maths at a level you are confident in. You may find that having to explain to students some of the concepts you understand, even if that is high-school maths, will help your own understanding to become firmer and clearer. There is nothing like having to explain something you _mostly_ understand, to a paying customer, to get your brain clarifying and rehearsing the connections and concepts underlying the topic. When some teachers say that teaching has helped them learn, they're not just being pious. Explaining to others really does help you explain it better to yourself. Good luck, Mark Griffith >> So I am at a very akward stage now: if give me undergraduate math to read, I >> feel too easy and boring, if give me difficult math to read as those in >> Information Theory Transactions, I feel dizzy... === Subject: Re: Anyone have a user manual for Sharp EL-9200C Graphing Scientific calculator? > Nowadays manuals are often put on line by the manufacturers. So check > Sharp's web page and see. The manuals that Sharp provide online are for only their current line up, they do not support older machines at all. === Subject: Re: sqrt(-1)=0/0 In sci.logic, Earle Jones > In sci.logic, ZZBunker >> : > In sci.physics, Earle Jones > or R. > 0 > denotes the absence of quantity which is instrinsically a > quantity > (ie. nothing is something). Nothing has no constituent parts, and therefore cannot be > divided by > or multiplied against... * > Do not equate 'nothing' with 'zero'. earle > * Zero denotes the absence of quantity which is intrisically a quantity, > since osophically, nothing is something. * > 'Nothing' is the opposite of 'something'. 'Zero' is a point on the number line, like 6, -23, pi, and 1,931. Pedant point: the number line doesn't exist either, at least not > as a physical object. Then again, the only reason zero needs > to be specially treated anyway is because it *is* the arithmetic > identity and therefore X * 0 = 0 for all X in the field. > No other number in the field has that property. * > You are saying that zero is unique. You are right. But then so are all of the other numbers. Every number has something > that can be said about it that cannot be said about any other number. Do you remember the proof of this: There are no non-interesting > numbers? No, but I can probably reconstruct many variants of it. :-) Let B_1, B_2, B_3, ... be a set of non-interesting numbers. (B = boring. Surprise.) Order the set such that 0 <= abs(B_1) <= abs(B_2) <= ... . Then B_1 becomes an interesting number, as it's either the closest boring number to 0, or one of the two. Therefore, the sequence cannot exist and all numbers are interesting. :-) Of course 0 and 1 are unique in that both are identities; 0 is the arithmetic identity, 1 the multiplicative. Does it mean anything? I can't say; I know nothing about it... :-) earle > * -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: sqrt(-1)=0/0 In sci.logic, ZZBunker : >> In sci.logic, ZZBunker >> : > In sci.physics, Earle Jones > 0 > denotes the absence of quantity which is instrinsically a quantity > (ie. nothing is something). Nothing has no constituent parts, and therefore cannot be divided by > or multiplied against... * > Do not equate 'nothing' with 'zero'. earle > * Zero denotes the absence of quantity which is intrisically a quantity, > since osophically, nothing is something. * > 'Nothing' is the opposite of 'something'. 'Zero' is a point on the number line, like 6, -23, pi, and 1,931. Pedant point: the number line doesn't exist either, at least not > as a physical object. Then again, the only reason zero needs > to be specially treated anyway is because it *is* the arithmetic > identity and therefore X * 0 = 0 for all X in the field. > No other number in the field has that property. >> >> Double pendant point. That's not true, >> since zero is not handled in any proof. >> Only it's current day dork symbol is >> handled. It's handled like gravity. >> >> Gently, gently, move it a little bit to the left, >> move it a little bit to right, Then move it faster. >> Then all of sudden, vailo, it's magic. The result is >> a Quantum Chemist Dork gibbering about the holistic >> CONSCIOUSNESS of parallel universes, and cold fusion. >> >> >> You're not making a whole lot of sense here. >> >> How are we supposed to handle numbers? > I don't know. Since set theory was supposed to answer that > question, not me. Set theory answers questions about sets, which are a different brand of abstract entity from what is commonly termed a number. Of course one can have a set of numbers, or a number (nonnegative integer) of sets. Of course one can get bogged down here; if one has 0 sets, one has the empty set, leading to either a very odd osophical problem, or a contradiction. One can construct a series of sets that map 1-1 with the whole numbers, starting with 0 <=> empty set. > >> How are we supposed to handle zero? I always handled it with a 0, rather than a how. > How mathematicans & musicians handle it, depends on piano > players and Newton, not people who know things about > music and logic. That didn't make a whole lot of sense, either. >> [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Degree of field extension of a completion > I'm sure that if one is careful with the p-adic log, one > can use that as well. This might be, but at that stage of the book, the log is not yet defined, but the infinite product solution works well. Now I have another question, unrelated to the previous: I have a valuated field L and a superfield K extending the valuation on L. I complete both fields an get L1 and K1. I can interpret L1 as a subfield of K1. Now the assertion is: If [K:L] = n is finite, then [K1:L1] <= n. I see this in the case where L is archimedian, and have a proof where L is nonarchimedian but discrete. Is the statement also true for nondiscrete nonarchimedian valuations on L? Christian Semrau === Subject: Re: Tarski's World vs. The language of FOL > The homepage of all those books: http://www-csli.stanford.edu/hp/#Hyperproof I don't know about these books, but I took a course in introductory > logic using Barwise and Etchemendy's Language Proof and Logic. The > book included a bunch of software for doing the exercises, and in fact > included the Tarski's World program. I see. > I have mixed feelings about the book. It is intended to be used as a > osophical logic book (that is, to spend time teaching methods of > proof and do a bit of meta-theory later) as opposed to a mathematical > logic book (where you skim over proving things in a deductive system > and do a lot of meta-theory). Since its osophical in nature, a > good 150 pages are spent trying to convince the reader that the > connectives capture what is intuitively meant by conjunction and > disjunction and so on. It's certainly a useful book, and it does > what it intends to do very well. Once you actually start doing formal > proofs in FOL, the book excels. There are hundreds of good exercises, > some involving Tarski's World (which sort of helps to understand > quantificational semantics) and others involving Fitch, a deductive > system (and proof checking software). At this point, I understand the > material well enough so that I could jump into just about any logic > book and feel comfortable. So LPL seems a bit basic. Very interesting. Thank you. > But if you're a beginner, I would recommend it highly. Thank you for writing. === Subject: Re: Tarski's World vs. The language of FOL > I have the later book, and it seems to have a complete implementation of > Tarski's World. Right. Thank you. That's what I needed to know. > I shall be most thankful for any kind (and perhaps reviewing) > comments. I like Language, Proof, and Logic a lot. I haven't heard of this title. Will look through it. Thank you. > The book includes a log-in code > (which I so far haven't used) for submitting answers to the exercises to a > server at Stanford for grading. Cool. I'll probably have my son use that in > the next year or so. I won't ask you how old he is. :-) > http://www-csli.stanford.edu/hp/#Hyperproof Yes, quite an interesting Web site. Thank you for writing. === Subject: Re: How Does the Fibonacci Sequence Start from Zero > It makes a lot of sense to define the Fibonacci sequence so that > F(0)=0 and F(1)=F(2)=1. Various results are easier to state when you > index it that way. For instance, if n divides m, then F(n) divides F(m). Indeed, indexing it that way is most practical... but when some people will talk about the second term of a sequence, sometimes they are referring to F(2). (So then the first two terms of the sequence are sometimes written as F(1), F(2), or 1,1. Of course, a C programmer would say the first two terms are F(0) and F(1).) J === Subject: Re: How Does the Fibonacci Sequence Start from Zero Virgil: Thank you! (And hats off to Mr. Montgomery for saying the same thing in a foreign language.) Although the symbology of the F(x's) was not transparent to me, Virgil gave me a new view of how the Fibonacci sequence passes through the zero point. Priceless! Very Respectfully, raydpratt === Subject: Re: Proof of fundamental group of SO(3) being mathbb{Z}_2 ... stuff deleted ... >Simple-connectivity for SU(2) follows from the following argument: SU(1) = {1}, since U(1), the unitary group in dimension 1 is > the circle group S^1, and SU(1) is the set of elements of U(1) > with determinant 1. SU(2) acts by multiplication on complex space of dimension 2, > and preserves the norm of vectors; it thus acts on the unit > sphere of that space, which happens to be S^3. By evaluating > this action at your favorite point x in S^3, you get a map SU(2) ---> S^3. It is also worth noting that for any other point y of S^3, > there is an element of SU(2) that maps x to y. That is, the > map is a surjection. >Injectivity follows directly from isometry. >>What do you mean? The same argument in higher dimensions fails to hold: >>one has SU(n) acting as a group of isometries on C^n (and thus on the >>sphere S^(2n-1)), but the isotropy subgroup of any point is isomorphic >>to SU(n-1), certainly not the trivial group. > I guess it was another context, functional analysis. I should be more > careful with the underlying assumptions and structures. > I just figured out what your context must have been: If X is a metric space, then an isometry f: X ---> X must be 1:1. The reason is that every pair of distinct points is separated by a nonzero distance, and the isometry preserves distance. This much is correct, but it is unrelated to the issue here. By saying that the group SU(n) acts as a group of isometries of C^n, it is surely the case that each element's action: (1) A: C^n ---> C^n is an isometry, and thus a 1:1 mapping. The issue has rather do do with what are called isotropy subgroups, that is, subgroups that fix particular elements of S^(2n-1), the unit sphere of C^n. We have this mapping: (2) P: SU(n)---> S^(2n-1) by fixing an element x of S^(2n-1) in the mapping (1) above, and allowing A to vary over the full group SU(n). The image Ax, as A varies, defines the mapping (2). While each mapping (1) is an isometry, it is by no means the case that the mapping (2) is required to be 1:1. For this to be 1:1, it must be the case that *only* one element of SU(n) fixes the element x (from the preceding paragraph), that is the equation (where A is the variable): Ax = x has a unique solution. Generally, looking at the space C^n, one sees the subspace Cx, as well as its hermitian complement K, so that C^n = Cx (+) K (where (+) is direct sum), in such a way that the elements of K are orthogonal to x under the hermitian inner product: = 0 for all y in K. The fact that an element A of SU(n) fixes x means (1) Ax = x, as well as (2) A is unitary of determinant 1. Since a unitary operator preserves the hermitian inner product, that says that = = and so if y is in K in the above decomposition, Ay is also in K. In short, the operator A splits as an orthogonal direct sum A = A|x (+) A|K where I have used A|??? to denote the restriction of A to ???. The isotropy subgroup corresponding to x can be identified with the set of operators A|K that can occur while A|x = id|x. ... the rest deleted ... > Max Dale === I'm not for or against this guy (J.S.), but I do have a question about > this exchange in the original post. Is the exchange in this post an > guy, or is this guy writing a monologue of comments to what writer in The latter. Moreover, all the quotes come from the first six paragraphs > If it's the latter then I'm confused, as I always thought that the > comments and notes sections in well respected journals were the > appropriate places to really conduct a dialogue of this sort. Alas, well respected journals have higher standards than Usenet does, > making it hard for some experts to get their views aired there. I wasn't aware that Scientific American was still being published. Then too, I stopped reading it when it was sold and dumbed down about 10 years ago. Harry C. === Subject: Re: Dilemma with undergraduate real analysis > Right now I'm taking the undergraduate year-long 'real analysis' class. To be > honest the professor is terrible and the way he goes about things is to make > the class so easy that even the worst students do great. The main thing that > has happened is that I'm not learning anything. The problem is that I'm going > to graduate school next year and I bet they will expect me to be taking their > graduate level analysis class with Lebesgue Integration and the rest. I have two questions. Do I need to know this before taking a class at the > graduate level (I assume so but it could be radically different..right?). What > are books that are good for self-study that teach all the basics about metric > spaces, riemann/riemann-stieltjes integrals, and the rest? Randy If you can do Rudin's Elementary Real Analysis (be careful, there's two books on analysis by him, one much harder than than the other), you'll be fine. Try to work some random problems from the section you're studying at the time. If you can do them out of Rudin, you're fine. Bartle also has a very good book on analysis that is particularly well-suited for self-study, as it has the aswers in the back. (Again, he has several boks on analsis, so you'll have to do a little research to get the right one) IMHO, it is better written for a person approaching real analysis for a first time. Another way you can get stronger at math, even with an easy prof is to work out the theorems on your own, and think of new things that come from the material you've done, and see if you can prove these propositions on your own. There is something about working a problem in which you are personally interested that makes the material really stick. Good Luck, Jack === Subject: Re: Dilemma with undergraduate real analysis > Right now I'm taking the undergraduate year-long 'real analysis' class. To be > honest the professor is terrible and the way he goes about things is to make > the class so easy that even the worst students do great. The main thing that > has happened is that I'm not learning anything. The problem is that I'm going > to graduate school next year and I bet they will expect me to be taking their > graduate level analysis class with Lebesgue Integration and the rest. I have two questions. Do I need to know this before taking a class at the > graduate level (I assume so but it could be radically different..right?). What > are books that are good for self-study that teach all the basics about metric > spaces, riemann/riemann-stieltjes integrals, and the rest? Randy A great professor can give you insights and perspectives that a bad one cannot. In order to get more out of the course, you should read other real analysis books to get some different perspectives that your prof is not providing. It is a vital to have a good foundation when taking courses at the graduate level, but if you read other sources, do plenty of problems, and go to other professors with questions, you will do better. -- Baldin Pramer === Subject: Re: Ask Marilyn by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i11IpVf04181; >Marilyn vos Savant seems to be defending the indefensible this week. A >reader takes her to task over her answer to this problem: Two people >race over a course of 100 miles, the first 50 of which must be covered >at 55 mph, and the second 50 must be covered at a rate of 65 mph. The >difference is that one racer must stop for 15 minutes in the 55 mph leg >and the second must stop for 15 minutes in the 65 mph leg. >Marilyn claims the race is a tie. There are many special cases that make >the problem work out Marilyn's way, but not every scenario works. >Unfortunately, her exact original problem statement is not included in >either the reader's response, or in her response to the reader. >As one example, suppose the first racer stops after 3/4 hour for 1/4 >hour. It then takes him 7/44 hour to finish the remaining 55 mph part of >the course, then 50/65 hour to finish the 65 mph part of the course, for >a total time of 3/4 + 1/4 + 7/44 + 50/65 hour. If the second racer >drives straight through the 55 mph part then drives 3/4 hour in the 65 >mph part then stops 15 minutes, he has only 5/4 mile left to cover at >the end. His time will be 50/55 + 3/4 + 1/4 + 1/52 hour. >-- >Baldin Pramer And since 3/4 + 1/4 + 7/44 + 50/6 and 50/55 + 3/4 + 1/4 + 1/52 add up to exactly the same thing, just as Marilyn Vos Savant predicted, your point is?? === Subject: Re: Request for comments on attempts of proofs Adjunct Assistant Professor at the University of Montana. >First let me extend a sincere thanks for taking the time to comment on my >humble proof attempts. One thing which puzzles me when trying to prove >something in general, be it graph theory or something else, is WHERE to >start. Is there some general thought procedure on how to start out a proof ? The thought procedure of course varies from person to person. But I think one of the most common things I see in students who seize up and say the can't come up with an answer is that they cannot see in their heads all the steps from beginning to end, and therefore think they cannot get started... Here are some suggestions that I found useful when I was student, and which to a large extent I still follow (as you can see in some of my published proofs): 1. First, write down explicitly what you are given. Then write down explicitly what you want to prove. 2. If either (or both) of these conditions are given in words pursuant to a relevant definition, then rewrite them again using the definition. [If you have a number of equivalent definitions for the notion, then this is not necessarily a good idea; for example, if you know that a set function is bijective if and only if it has an inverse, but that it is also true that a set function is bijective if and only if it is both injective and surjective, then writing down either down will probably 'lock you' onto one of the possibilities, while the other may be more useful. In that case, it is usually more constructive to leave the wording as is] 3. If the conclusion you want to prove is an implication, you may assume the ->antecedent<-. E.g., if you want to prove that a certain function is surjective, surjectivity means: If y in Y, then there exists x in X such that f(x)=y So you may start your proof by assuming that you are given a y in Y. Then write down what you want to prove: that there exists an x in X such that f(x) = y. 4. Try a number of things: if your assumptions have some obvious implications, try writing them down. If your conclusion would follow from something else that sounds simpler, write that down and aim for ->that<- instead of the conclusion. Etc. 5. In Spanish, there is an expression for cooking up a process when you know what the answer should be, by assuming that you have an answer and figuring out what that answer should 'look like.' We call it spooning the answer. I have not encountered an expression like that in English, but you saw me using this in the proof that if gf is bijective and g is bijective, then f is bijective: to figure out what the inverse of f should be, we assume it is bijective, in which case we would necessarily have (gf)^{-1} = f^{-1}g^{-1}, so f^{-1} = f^{-1}g^{-1}g = (gf)^{-1}g. This tells you what function you should try for the inverse of f: (gf)^{-1}g. What you need to be careful here is that this sort of thing is what you use to come up with a proof, but is not usually part of the proof (it is part of the intuition behind the proof; in a properly organized proof, you would make this observation as an aside, not part of the proof, to explain what you are doing, but your proof would NOT contain any mention of f^{-1} until you have proven that such a thing exists). The point being that 'spooning the answer' can be useful when you don't really know how to get to the answer. It is important not to be discouraged if you cannot see the whole proof from beginning to end from the start. Just start taking one step at a time from the premises in the (general) direction of the conclusion you want. With some luck and experience, somewhere along the line you will finally see the full connecting trail. When you are done, go back through your proof and see if you took some unnecessary detour, or some extra assumption. (Another important thing is to remember that, usually, ALL hypothesis should come into play. It is certainly possible that you are assuming more than you need for a proof, but especially when you are still learning, all hypothesis turn out to be important. So if you managed to complete the proof of an exercise in the book without using all the hypothesis, go through it with particular care to make sure that you really did not need it. More often than not, you've made a mistake.) -- === Subject: Which is better at plotting 3D vector fields? Derive, Maple, Mathematica, MathCad, other mathematical software? Which is better at plotting 3D vector fields? Derive, Maple, Mathematica, MathCad, other mathematical software? Casey === Subject: Re: Ask Marilyn > Apparently Marilyn has quite a history of wrongness. ...and at least as much a history of being wronged. V. -- === Subject: Re: Ask Marilyn by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i11IpaI04282; >Marilyn vos Savant seems to be defending the indefensible this week. A >reader takes her to task over her answer to this problem: Two people >race over a course of 100 miles, the first 50 of which must be covered >at 55 mph, and the second 50 must be covered at a rate of 65 mph. The >difference is that one racer must stop for 15 minutes in the 55 mph leg >and the second must stop for 15 minutes in the 65 mph leg. >Marilyn claims the race is a tie. There are many special cases that make >the problem work out Marilyn's way, but not every scenario works. >Unfortunately, her exact original problem statement is not included in >either the reader's response, or in her response to the reader. >As one example, suppose the first racer stops after 3/4 hour for 1/4 >hour. It then takes him 7/44 hour to finish the remaining 55 mph part of >the course, then 50/65 hour to finish the 65 mph part of the course, for >a total time of 3/4 + 1/4 + 7/44 + 50/65 hour. If the second racer >drives straight through the 55 mph part then drives 3/4 hour in the 65 >mph part then stops 15 minutes, he has only 5/4 mile left to cover at >the end. His time will be 50/55 + 3/4 + 1/4 + 1/52 hour. >-- >Baldin Pramer Apparently Marilyn has quite a history of wrongness. See http://www.wiskit.com/marilyn/marilyn.html for details. (Of special interest here is here `refutation' of Wiles' proof of FLT. I bet it really shook him up...) a.p. === Subject: Re: How often is prod(primes<=p)+1 prime? >In the standard proof of the infinitude of primes, we form the product >of the first n primes and add 1. That number is either prime, or is >divisible by a prime greater than any of the factors. >I computed that among the first 13 primes, prod(primes<=p)+1 is prime >six times, and composite 7 times. So the ratio of primes/total for n =13 >is 6/13. >Is there any known theory about this? Could the ratio possibly have a >limit as p->infinity? The limit is very likely to be 0. E_n = product_{j=1}^n p_j is (very roughly) n^n, and the probability that a randomly-chosen number of this size is prime is about 1/(n ln n) (the fact that it can't be divisible by the first n primes does not make much of a difference here). Since sum_n 1/(n ln n) diverges, I would guess that there are infinitely many primes in the sequence, but since the series diverges extremely slowly they will be very few and far between after a while. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: How often is prod(primes<=p)+1 prime? > In the standard proof of the infinitude of primes, we form the product > of the first n primes and add 1. That number is either prime, or is > divisible by a prime greater than any of the factors. These number (products of primes, plus 1) are called Euclid numbers. > I computed that among the first 13 primes, prod(primes<=p)+1 is prime > six times, and composite 7 times. So the ratio of primes/total for n =13 > is 6/13. Yup, the six are 2 + 1 2*3 + 1 2*3*5 + 1 2*3*5*7 + 1 2*3*5*7*11 + 1 2*3*5*7*11*13*17*19*23*29*31 + 1 The next such number has a product of primes up to 379, so it is quite a far way away. > Is there any known theory about this? Could the ratio possibly have a > limit as p->infinity? The only thing known is that no ratio is currently known. In fact, it is unknown whether there are an infinite number of such primes (i.e. it is unknown if there are infinitely many prime Euclid numbers.) > If it has a limit (1/2? 0? 1?) it's interesting. If the list is finite, the limit will be 0 ... I guess we'll see! You're asking about the ratio of prime Euclid numbers to normal Euclid numbers... you probably know (but maybe not) that the ratio of prime numbers to normal numbers is something strictly between 0 and 1 (that is, consider N* = {1,2,...,N}, and randomly pick an element of N* at uniform, what is the probability that it is prime, in the limit as N->inf). J === Subject: How often is prod(primes<=p)+1 prime? This came up on another group and I thought I'd ask if there's any theory on the following. In the standard proof of the infinitude of primes, we form the product of the first n primes and add 1. That number is either prime, or is divisible by a prime greater than any of the factors. I computed that among the first 13 primes, prod(primes<=p)+1 is prime six times, and composite 7 times. So the ratio of primes/total for n =13 is 6/13. Is there any known theory about this? Could the ratio possibly have a limit as p->infinity? Either way the answer would be interesting. If it has a limit (1/2? 0? 1?) it's interesting. But if it doesn't have a limit, the ratio must oscillate around between 0 and 1, so it would at least have a lim inf and a lim sup that bound the oscillation. That would also be interesting. And if the bounds of oscillation are 0 and 1, that would mean that there are runs of unbounded length of both primes and nonprimes. That would be interesting too. === Subject: Re: stereographic proj.: circles --> circles y d. > im really stuck on this, been mucking with it for about 2 hours. im > trying to prove the stereographic projection of S^2 onto C_infty > preserves circles. perhaps a hint? http://geom.math.uiuc.edu/docs/education/institute91/handouts/node33.html Books on functions of a complex variable also go into this. LH === Subject: stereographic proj.: circles --> circles im really stuck on this, been mucking with it for about 2 hours. im trying to prove the stereographic projection of S^2 onto C_infty preserves circles. perhaps a hint? Im trying to use the fact that any circle in S^2 can be represented as the intersection of a plane in R^3 with the sphere S^2, but havent had much luck. thanks === Subject: Humpty Dumpty's M Theory Sarfatti Commentary on .8bFROM SLOWDOWN to SPEEDUP.8a By Adam G. Riess and Michael S. Turner .8bUntil recently, astronomers fully expected to see gravity slowing down the expansion of the cosmos. In 1998, however, researchers discovered the repulsive side of gravity. By carefully observing distant supernovaeÖstellar explosions that for a brief time shine as brightly as 10 ion sunsÖastronomers found that they were fainter than expected. The most plausible explanation for the discrepancy is that the light from the supernovae, which exploded ions of years ago, traveled a greater distance than theorists had predicted. And this explanation, in turn, led to the conclusion that the expansion of the universe is actually speeding up, not slowing down. á general relativity also allows for the possibility of forms of energy with strange properties that produce repulsive gravity The discovery of accelerating rather than decelerating expansion has apparently revealed the presence of such an energy form, referred to as dark energy. á The leading candidate to explain dark energy.89s effects is vacuum energy, which is mathematically equivalent to the cosmological constant that Einstein invented in 1917. á But to produce the observed acceleration of the universe, the constant.89s density would have to be twice that of matter. Where could this energy density come from? The uncertainty principle of quantum borrowed time and energy, popping in and out of existence. But when theorists try to compute the energy density associated with the quantum vacuum, they come up with values that are at least 55 orders of magnitude too large.8a This is the problem I claim I have solved with the idea of .8bvacuum coherence.8a that is in fact the inflation field in disguise. Einstein.89s gravity together with both dark energy and dark matter all emerge together from the self-organizing creative information-rich ripples in our conscious holographic universe.89s phase and intensity of this vacuum coherence field. Stephen Hawking has glimpsed this through the glass darkly calling it .8bThe Mind of God..8a This subterranean giant tsunami virtual tidal wave of .8binflation.8a decreases the randomness of the zero point vacuum fluctuations of all the quantum fields. I have provided a very simple dynamical reason based on Dirac.89s theory of the negative energy virtual electron-positron pair quantum vacuum why this happens in an intrinsic instability of flat spacetime without gravity that is very like what happens when a normal metal undergoes a phase transition, an emergent .8bMore is different.8a metamorphosis to a superconductor where electric currents freely flow without the annoying resistive heating of the circuits. .8bThis discrepancy has been called the worst embarrassment in all of theoretical physics, but it may actually be the sign of a great opportunity..8a Indeed it is. The really great opportunity that President Bush.89s science pundits are missing in their new vision for the American manned space exploration program is the metric engineering of super cosmos enabling us, like caterpillars to butterflies in a metamorphosis of Men Like Gods to fulfill our manifest super cosmic destiny to harness the repulsive dark energy. Indeed, anti-gravitating universally repelling dark energy is .8bThe Right Stuff.8a enabling us to time travel effectively faster than light without actually moving faster than light with a free floating weightless warp drive through traversable star gate wormholes to distant places in our own universe in our cosmic present, past and future and to the other parallel universes next door. As General Douglas MacArthur warned the cadets at West Point in his .8bDuty, Honor, Country.8a Farewell Address, we are not the first intelligent species to do this as the UFO .8bflying saucer.8a phenomenon is, in all likelihood, showing us, nor will we be the last. My esteemed colleagues, all honorable men, the Pundits of respectable physics and cosmology have put on blinders rejecting this evidence, which is, in my opinion, the missing link, the last piece needed to complete the jig saw puzzle. .8bAlthough it is possible that new attempts to estimate the vacuum energy density may yield just the right amount to explain cosmic acceleration, many theorists believe that a correct calculation, incorporating a new symmetry principle, will lead to the conclusion that the energy associated with the quantum vacuum is zero. (Even quantum nothingness weighs nothing!) If this is true, something else must be causing the expansion of the universe to speed up..8a Jack: No, this is simply confused thinking. What you see below Is All The King.89s Men grasping wildly willy-nilly at straws like the Sufi Story where The Pundits shine strong light in the wrong part of Plato.89s Dark Cave. .8bTheorists have proposed a variety of ideas, ranging from the influence of extra, hidden dimensions á Perhaps the most radical idea is that there is no dark energy at all but rather that Einstein.89s theory of gravity must be modified..8a Jack: This is a very bad idea. It is a false idol, a Golden Calf. It is Fool.89s Gold. It is a wrong turn off the path through Dante.89s Inferno. The following however is in complete exact agreement with my theory that I published in 2002 my two books .8bDestiny Matrix.8a and .8bSpace-Time and Beyond II.8a on file in the catalog of the Library of Congress. .8bIN NEWTON.89S THEORY, gravity is always attractive and its strength depends on the mass of the attracting object. The twist in Einstein.89s theory is that the strength of the gravitational pull exerted by an object also depends on its composition. Physicists characterize the composition of a substance by its internal pressure. An object.89s gravity is proportional to its energy density plus three times the pressure. Our sun, for example, is a hot sphere of gas with positive (outward) pressure; because gas pressure rises with temperature, the sun.89s gravitational pull is slightly greater than that of a cold ball of matter of equivalent mass. On the other hand, a gas of photons has a pressure that is equal to one third its energy density, so its gravitational pull should be twice that of an equivalent mass of cold matter. Dark energy is characterized by negative pressure. (Elastic objectsÖfor instance, a rubber sheetÖalso have negative, or inward, pressure.) If the pressure falls below .9a1?3 times the energy density, then the combination of energy plus three times the pressure is negative and the gravitational force is repulsive. The quantum vacuum has a pressure that is .9a1 times its energy density, so the gravity of a vacuum is very repulsive. á Only theories stipulating large variations in dark energy density have been ruled out ... The only way to forecast our cosmic future is to figure out the nature of dark energy..8a Amen. When gravity operates over microscopic distances Öfor instance, at the center of a black hole, where a huge mass is packed into a subatomic volumeÖthe bizarre quantum properties of matter come into play, and string theory describes how the law of gravity changes. Over greater distances, string theorists have generally assumed that quantum effects are unimportant. Yet the cosmological discoveries of the past several years have encouraged researchers to reconsider. Four years ago my colleagues and I asked whether string theory would change the law of gravity not just on the smallest scales but also on the largest ones. The feature of string theory that could bring about this revision is its theory adds six or seven dimensions to the usual three. Theorists today are willing to pay any price to avoid signal nonlocality .9a even extra dimensions. Well, perhaps, that is not too high a price after all? .8bIn the past, string theorists have argued that the extra dimensions are too small for us to see or move in. But recent progress reveals that some or all of the new dimensions could actually be infinite in size. They are hidden from view not because they are small but because the force of gravity, and as a result, the law of gravity changes..8a There is one nice feature about this idea of enormous .8bbranes.8a in that it easily explains why there is no significant amount of antimatter in our universe. According to John Archibald Wheeler.89s .8bGeometrodynamics.8a .8blepto-quarks.8a are in reality David Bohm.89s .8bhidden.8a or .8bextra variable.8a that are tiny wormholes in three dimensional space with quantized gauge force fluxes threading them like we see in Type II superconducting .8bvortices.8a in condensed matter physics near absolute zero temperature. Wheeler called this .8bMass without mass.8a with .8bCharge without charge.8a and .8bSpin without spin..8a Wheeler.89s idea did not work back then almost fifty years ago because gravity was thought to be too weak. In other words, spacetime geometry was simply too stiff to bend easily. Spacetime.89s .8bstring tension.8a was too large. This implied that Wheeler.89s wormholes the electron. Indeed, this is just like the speed of sound barrier was mistakenly thought to be before the era of jet planes and Chuck Yeager.89s that Andrei Sakharov.89s .8bmetric elasticity.8a for emergent gravity out of zero point vacuum energy fluctuations, or what Ed Witten, (The Big Cheese of .8bM Theory.8a) calls alpha.89 can be a scale-dependent variable not an immutable constant. Witten.89s alpha.89 is proportional to the reciprocal of the string tension, that originated in the observed universal .8bgravity.8a .8bRegge parallel trajectories slope.8a of the .8bhadronic resonances.8a of (1 Gev)^-2 unceremoniously kicked up to the quantum gravity scale of (10^19 Gev)^-2 is a variable that gets less stiff at smaller scales of higher energy transfer in scattering experiments. Indeed, the effective strong short-range gravity, suggested by Abdus Salam in the early 1970.89s, solves more than one hitherto unsolved fundamental conceptual problems in high-energy physics. Strong gravity at the scale of 1 fermi (10^-13 cm) is 40 powers of ten stronger than Newton.89s gravity. This means Wheeler.89s wormholes are just the right size. It.89s like those size-changing drugs that Alice takes in Lewis Carroll.89s Wonderland. First of all it solves the mystery of the missing anti-matter called .8bbroken charge conjugation symmetry.8a or .8bC-violation..8a The missing antimatter is in a parallel .8bbrane world.8a next door across a thin barrier of extra dimensional .8bhyperspace..8a Imagine a wormhole with two mouths or ends attached to two different brane worlds with infinitely large uncompactified extra space dimension as Dvali explains. The Faraday flux lines of electro-weak-strong gauge forces must form closed loops. Flux lines coming out of one mouth of the wormhole must enter the other mouth of the same wormhole. We know from Gauss.89s divergence theorem that the sign of the effective charge changes from + to .9a depending on whether the flux lines leave or enter the wormhole mouths that you can picture as little 2-dimensional spherical surfaces in ordinary 3-dimensional space. These spheres may themselves not be simply connected but may have little wormhole handles on them as well. It may be Ezekial.89s vision of wormholes on wormholes. This might mean only a single electromagnetic flux in which the weak and strong forces are simply topological variations of the same theme. That would be a nice unification hardly less speculative than what you see on Brian Greene.89s .8bElegant Universe.8a on NOVA. One famous Dead Physicist said .8bElegance is for tailors..8a If NOVA was produced by BBC they might call it .8bSaville Row.8a physics .9a very fashionable. Therefore, there are equal amounts of matter and anti-matter but the corresponding wormhole mouths with equal and opposite quantized flux charges are attached to different parallel brane universe next door to each other! There is still another problem solved here. It is the problem that Richard Feynman tried and failed and All The King.89s Men have not solved it to this day. The problem is the infinite self-energy of the point electron. Why a point electron you ask? Because, there was, hitherto, no way to prevent the extended electric charge distribution of a non-point electron from exploding! This unsolved problem is more than one hundred years old! It.89s solved now. The Lorentz-Abraham stress is simply the strongly short-range attractive exotic vacuum .8bdark matter.8a zero point fluctuation .8bpositive pressure.8a from negative zero point energy density where the throats of Wheeler.89s wormhole geons are like the vortex cores of superfluids where the vacuum coherence field drops to zero on the scale of a .8bcoherence length.8a. Note that the trapped .8bnear field.8a quantized Faraday flux lines of virtual spin 1 quanta in coherent states are contained in the Meissner effect.89s .8bpenetration depth.8a that is no more than one over the square root of two of the coherence length like in a Type II superconductor. .8bWHEN ASTRONOMERS ENCOUNTERED the cosmic acceleration, their first reaction was to attribute it to the so-called cosmological constant. Notoriously introduced and then retracted by Einstein, the constant represents the energy inherent in space Maybe cosmic acceleration isn.89t caused by dark energy after all but by an inexorable leakage of gravity out of our world..8a Don.89t bet on that ENRON stock. Although our standard model of cosmology has been confirmed by recent observations, it still has a gaping hole: nobody knows why the expansion of the universe is accelerating. Speak for yourself George. === Subject: What is normalized surface area? What does normalized surface area mean? you divide the authalic radius squared by a^(1/3)*b^(2/3) you will have the normalized surface area radius. So for the WGS84 spheroid, a= 6378.137,b=6356.7523 and (Surface area radius) Sr= 6371.00718, 6371.00718^2/(6378.137^(1/3)*6356.7523^(2/3)) =++6378.14980383181. What is that? -------------- Ëë.81ê.b3[CapitalYAc ute] ÄÇü¨.bd.bcü[P aragraph] KORNET ------------- === Subject: Re: simple equation solving, with a twist? > This is a problem which originated in chemistry, but being that math lover > that I am I quickly tried to find a way to turn it into a pure math problem. > I succeeded, but now I don't know how to solve it (other than trial and > error or similar... you'll understand what I mean), so any help would be > appreciated. I am particularly interested in whether this can be programmed > into a calculator--I have a TI-83+--so that would also be of interest. Say we have some equations with a number of variables. For example: 5x - 3y = 212 > 3y - z = 7 Here's the twist: we don't want to solve the set of equations (infact, in > this case they are not solvable). Instead, we want to find what, say, 5x - z > is equal to. Now, in this case it is easy, we can simply add the bottom to > the top, and we get 5x - z = 219. We can do similar things for other numbers > etc. In general, you are allowed to multiply any equation by a constant, add > any two equations, and subtract any two equations. But, with, say, 5 equations and 7 variables, it can get tricky. Each > equation doesn't have every variable in it; most of them have 3 or 4, which > may make it easier. But, any hints on how to solve this in a general case > (even though this isn't necessarily solvable for any expression)? One idea I had for programming my calculator for doing it was using a > matrix, e.g. we could represent the above example by 5 -3 0 212 > 0 3 -1 7 We can add another row if we want, for the expression we want to get... 5 0 -1 ?? But I don't know what to do with it after that. Any hints etc. would be greatly appreciated. > If you have five equations in seven unknowns, you will have two degrees of freedom. This means that usually five of the variables can at least be solved in terms of the other two, the values of which are left open. This is referred to as underdetermination. So it's just a question of which of the variables you choose to keep open and then solve for everything else in terms of them. What this actually means is that there will be an infinite number of independent solutions, depending on what values the open variables are assigned. Equations of this type never have unique solutions. === Subject: Re: how do circular combination/permutation different from normal combination/permutation? >I know the normal combination and permutation formular, suppose we select m >from n, >then they can be represented as C(n, m) and P(n, m), where n should be on >top of m; >But how about circular combination and permutation? Can anybody tell me the >equations? A circular permutation (i.e. a way to arrange objects in a circle, with (ABCD) considered the same as (BCDA) but not (DCBA)) becomes an ordinary permutation if you pick one of the chosen items to be first. So the number of circular permutations of n distinct objects taken m at a time is just P(n,m)/m. Since order doesn't matter at all in combinations, I'm not sure what you mean by a circular combination. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: how do circular combination/permutation different from normal combination/permutation? Dear all, I know the normal combination and permutation formular, suppose we select m from n, then they can be represented as C(n, m) and P(n, m), where n should be on top of m; But how about circular combination and permutation? Can anybody tell me the equations? Thank you, === Subject: Re: Pure abstract demonstration by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i122RNg07786; >> The biggest possibility is in some kind of Your mistake. >> However lets check Your revelations for n=3: >>You claim Your parameters: >> T=3^u *abt once A=a^3 and B=b^3 or 3^(3u-1)*b^3 >>Already it is known for n=3, that 3 should divide Z;X or Y >>so lets take 3/X ; X=T+B ; B=3^(3u-1) b^3 >>also does Your parameters fits ? Ls=(T+A+B)^3=(T+A)^3 +(T+B)^3=Rs ? >> Ls= T^3 +3T^2(A+B) +3T(A+B)^2 +(A+B)^3 >> Rs= 2T^3 +3T^2(A+B) +3T(A^2 +B^2) +A^3 +B^3 >>then Rs-Ls= T^3 -6TAB -3AB(A+B) = 0 ; T^3 -3AB(2T+A+B) = 0 >>3^3u a^3 b^3 t^3 -3*a^3 3^(3u-1)*b^3 [2*3^u abt +a^3 +3^(3u-1) b^3]=0 >>and really axtracting 3^3u a^3 b^3 we'll have: >> t^3 - (2*3^u abt +a^3 +3^(3u-1) b^3] = 0 >> What is just Your f(t) for n=3 and for the fall X/3 >>Now You suggest Y^2 -YX +X^2 = s^3, what is also correct, >>but how to express everything in a;b;t;p parameters? >> I can see, that X+Y in this case should be some cube too: >> X+Y = T+B+T+A = 2T+A+B = 2*3^u abt +a^3 +3^(3u-1) b^3 >>also Your t^3 = X+Y and once Y^2 -YX +X^2 = s^3 so Z = t*s >> now t*s = T+A+B; t^3 = 2T +A+B: T = t^3 -ts = t(t^2 -s); >>but also once T = 3^u abt = t(t^2 -s): t^2 -s = 3^u ab >>and finally s = t^2 -3^u ab *****(how smart are those numbers) >> What You sugessted then before: >> Ls= s^3 = X^2 -XY +Y^2 =Rs >>Ls=(t^2 -3^u ab)^3 = >> = t^6 -3*t^4 3^u ab +3*t^2 3^2u a^2 b^2 - 3^3u a^3 b^3 >>Rs={3^u abt +3^(3u-1) b^3}^2 -[3^u abt +3^(3u-1) b^3][3^u abt +a^3] + >> +{3^u abt + a^3}^2 ...... >> I can feel, that there is really some more accounts as one web >>page could hold so I'll do it on some paper and come back >> very soon >> Juan C.C. Dear Mr. Juan and anybody else, I should just accent some methodical mistake: My parameters creating relatively simple primary equation f(t): with some falls: 1.) for X;Y;Z not divided: t^n - n^u abpt - a^n - b^n = 0 2.) for Z/n : n^(nu-1) t^n - n^u abpt - a^n - b^n = 0 3.) for X/n : t^n - n^nu abpt - a^n - n^(nu-1) b^n = 0 ( for Y/n see only inversed values ) If we succed to get some t as natural number so we'll get for sure s^n as s = t^(n-1) - n^u abp ...............(*) My method involves double crossing check of that same relation. It should be next to be check the p^5 , however again from (*) n^u abp = t^(n-1) - s and so on p should be dissolved as natural number. So on, the only really obstacles are with 1.);2.);3.) falls and with their equations. I hope to be more carefull and correct in my future targets but with Eisenstein criterion at the moment I can only simply assure u parameter bigger or equal to 2. Thank You for Your attention Ro === Subject: Re: intriguing numbers. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i122ROo07815; >i just read something in a magazine (forgot what title.) a small >true or not. because it seems very interesting. two numbers intrigued >me. one is the number À (psi) and they said that its the so-called last finite number. the number right before infinity. and a very >short mention on something called the end number. the end number is >the highest in the kingdom of numbers and makes absolute infinity a >dust on its shoulder. nothing is higher than the end number because by >theory it is the last number. now i was very intrigued and suddenly it >became my favorite number. so please verify this for me. Oh my God I beg of you to find the magazine title and issue number; please, I *need* that magazine :) Really, this message all by itself might make it to a top 100 math humor list :) === Subject: Re: question about strictly increasing integer sequences >Let S be the set of functions from N to N which are strictly >increasing, that is to say, if f is a function in S then f(n+1)>f(n) >for all n>1 in N. >If S' is an arbitrary subset of S, is it necessarily true that there >is a least element of S' (in the lexicographical sense)? Hint: construct a set of functions that does not contain f(n)=n, but contains functions that agree with it up to any given natural number. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: question about strictly increasing integer sequences Hello. Let S be the set of functions from N to N which are strictly increasing, that is to say, if f is a function in S then f(n+1)>f(n) for all n>1 in N. If S' is an arbitrary subset of S, is it necessarily true that there is a least element of S' (in the lexicographical sense)? === Subject: Re: Choosing a Math Grad school > Hi guys, I'd like some info about math PhD programs, if anyone can please help. A > concern I have are rankings, in particular: how much do they matter when you > make *final* decisions? Well how about this?: interest or find out otherwise who is doing well in your area of interest. Then try to get next to those people. That would make it more likely that you would go in the right direction and be successful publishing, which will lead to a PhD (and a tenure-track position if you are so inclined). You should be more concerned about general rankings if you are not so sure that you will stick with a certain research area. === Subject: Re: Solving linear inhomogenous recursion >Say we want to find an eplicit formula for the following recursion relation >: >f_(n) = f_(n-1) + f_(n-2) + n for n>=2 and f_(0) = f_(1) = 1 Double this equation and compare it to the sum of f_(n-1)=... f_(n+1)=... This gets you to the level of a linear recursion with constant coefficients. Do you know how to solve those (without generating functions, if you prefer to have one hand tied behind your back) ? === Subject: Re: Solving linear inhomogenous recursion Content-transfer-encoding: 8bit > Say we want to find an eplicit formula for the following recursion relation > : > f_(n) = f_(n-1) + f_(n-2) + n for n>=2 and f_(0) = f_(1) = 1 > WITHOUT using the concept of generation functions. That is, finding the > general homogenous solution and then guessing a particular solution. My > problem here is finding a particular solution. Do we know anything about > what a particular solution should look like here ? Since you don't want to use generating functions, I guess you want to use characteristic roots. As I guess you know, the characteristic roots are r_1 = (1 + sqrt(5))/2 and r_2 = (1 - sqrt(5))/2 so the general solution to the homogeneous case is h(n) = c_1*r_1^n + c_2*r_2^n. Since the n part of the inhomogeneous case is nowhere near being a characteristic root and is to the first power, we'll guess p(n) = A*n + B for the particular solution. Setting p(n) = p(n - 1) + p(n - 2) + n and solving gives A = -1 and B = -3. Consequently f(n) = c_1*r_1^n + c_2*r_2^n - n - 3. Setting f(0) = f(1) = 1 gives c_1 = 2 + 3/sqrt(5) and c_2 = 2 - 3/sqrt(5). -- Paul Sperry Columbia, SC (USA) === Subject: Re: Solving linear inhomogenous recursion I ran it in mathematica and it produced a closed form solution. > Actually i just failed to solve this even when using generating function > approach :( Say we want to find an eplicit formula for the following recursion > relation > : > f_(n) = f_(n-1) + f_(n-2) + n for n>=2 and f_(0) = f_(1) = 1 > WITHOUT using the concept of generation functions. That is, finding the > general homogenous solution and then guessing a particular solution. My > problem here is finding a particular solution. Do we know anything about > what a particular solution should look like here ? > And how convinced are you that this even has a closed form solution? > J === Subject: Re: Solving linear inhomogenous recursion > Actually i just failed to solve this even when using generating function > approach :( Say we want to find an eplicit formula for the following recursion > relation > : > f_(n) = f_(n-1) + f_(n-2) + n for n>=2 and f_(0) = f_(1) = 1 > WITHOUT using the concept of generation functions. That is, finding the > general homogenous solution and then guessing a particular solution. My > problem here is finding a particular solution. Do we know anything about > what a particular solution should look like here ? And how convinced are you that this even has a closed form solution? J === Subject: Re: Solving linear inhomogenous recursion Actually i just failed to solve this even when using generating function approach :( > Say we want to find an eplicit formula for the following recursion relation > : > f_(n) = f_(n-1) + f_(n-2) + n for n>=2 and f_(0) = f_(1) = 1 > WITHOUT using the concept of generation functions. That is, finding the > general homogenous solution and then guessing a particular solution. My > problem here is finding a particular solution. Do we know anything about > what a particular solution should look like here ? === Subject: Re: Solving linear inhomogenous recursion <1075665053.952994@seven.kulnet.kuleuven.ac.beof course i meant to say generating functions, not generation functions > >>Say we want to find an eplicit formula for the following recursion relation > >>: >>f_(n) = f_(n-1) + f_(n-2) + n for n>=2 and f_(0) = f_(1) = 1 >>WITHOUT using the concept of generation functions. That is, finding the >>general homogenous solution and then guessing a particular solution. My >>problem here is finding a particular solution. Do we know anything about >>what a particular solution should look like here ? Search for annihilator in combination with recurrence (or recursion) relation. Firstly, not every recurrence has a closed form solution. Secondly, why would you want to avoid generating functions? Thirdly, anything you can do with generating functions you can do without mentioning any function of 'x' and only use plain algebra but a lot more of it. Do you know a generating function solution? J === Subject: Re: Solving linear inhomogenous recursion > of course i meant to say generating functions, not generation functions > >>Say we want to find an eplicit formula for the following recursion relation > >>: >>f_(n) = f_(n-1) + f_(n-2) + n for n>=2 and f_(0) = f_(1) = 1 >>WITHOUT using the concept of generation functions. That is, finding the >>general homogenous solution and then guessing a particular solution. My >>problem here is finding a particular solution. Do we know anything about >>what a particular solution should look like here ? Search for annihilator in combination with recurrence (or recursion) relation. Cheers Bart Demoen === Subject: Re: Solving linear inhomogenous recursion of course i meant to say generating functions, not generation functions > Say we want to find an eplicit formula for the following recursion relation > : > f_(n) = f_(n-1) + f_(n-2) + n for n>=2 and f_(0) = f_(1) = 1 > WITHOUT using the concept of generation functions. That is, finding the > general homogenous solution and then guessing a particular solution. My > problem here is finding a particular solution. Do we know anything about > what a particular solution should look like here ? === Subject: Solving linear inhomogenous recursion Say we want to find an eplicit formula for the following recursion relation : f_(n) = f_(n-1) + f_(n-2) + n for n>=2 and f_(0) = f_(1) = 1 WITHOUT using the concept of generation functions. That is, finding the general homogenous solution and then guessing a particular solution. My problem here is finding a particular solution. Do we know anything about what a particular solution should look like here ? === Subject: Re: Permutation Groups ten minutes later. It was a problem that I was stuck on from earlier in the day. === Subject: Re: Permutation Groups > I'm stuck on two problems involving the ORDER of permutation groups. I know > what to do when they are written in disjoint cycle form but not when they are > written like this. > (1 2 3 5)(2 4 5 6 7) > The answer is 12 > and > (3 4 5)(2 4 5) > The answer is 2 You need to learn how to compute the disjoint cycle form of the product of two or more cycles. Use the definition of cycle and the definition of the product of two permutations. It might help to aconvert the cycles to the two line notation for a permutation. For example, in the second case (assuming you are writing the argument on the right) one can calculate that (3 4 5)(2 4 5) = (2 5)(4 3). === Subject: Permutation Groups I'm stuck on two problems involving the ORDER of permutation groups. I know what to do when they are written in disjoint cycle form but not when they are written like this. (1 2 3 5)(2 4 5 6 7) The answer is 12 and (3 4 5)(2 4 5) The answer is 2 === Subject: Re: Anal Probe Scan by Morely 67.160.141.145 Dotes > Do u want ur ports scan I'll probe ur ports I get sexual thrill And no I'm not gonna pay ya > I'll probe ur ports and cum on ur face bitch > attbi_s53!attbi_s52!attbi_s51!attbi_slave62!attbi_master61!attbi_feed3!attbi > _feed4!attbi.com!attbi_s53.POSTED!not-for-mail === > Subject: Probe scan > - > I'm interested in u probing my ports. Do U have any kiddy porn? Especially > any original stuff. I have a lot 2 offer. > Morely Kiddy Porner Dotes === Subject: Re: matrix equation: can I claim the following? Originator: kblomste@cc.hut.fi (Kasimir Blomstedt) >> Dear all, >> >> I am facing to derive some results from the following matrix equation: >> >> UAV+UBV=U(blahblahblah)V >> >> where all variables are matrices. >> >> Can I say that in order for the above to hold for any such U, V, >> >> the equality that A+B=(blahblahblah) should always hold? >> >> Please help me! I really want to get ride of U and V and get some meaningful >> results... >> >> Thank you! >> >> >You are correct, but only if U and V are are invertible. Otherwise, I >can't simplify the equation. The equation can be 'simplified' if we introduce the SVD:s of U and V. Suppose that the equation is written as (C = blahblahblah) U(A+B-C)V = 0, and that the SVD:s of U and V are given by U = XKY', V = ZLW', where ' denotes the hermitian conjugate (transpose). Here the matrices K and L are diagonal with positive elements, which can be taken to be ordered in a decreasing order. The matrices X, Y, Z and W are unitary. With the above representation the equation becomes XKY'(A+B-C)ZLW' = 0. Since X and W are unitary we don't upset the equality if we operate from the left by X' and from the right by W to obtain (X'X = I, W'W = I), KY'(A+B-C)ZL = 0. If we assume that K has r nonzero (diagonal) elements and L has s nonzero (diagonal) elements then the above equation implies that [Y'(A+B-C)Z]_ij = 0, when i <= r, j <= s, [Y'(A+B-C)Z]_ij = arbitrary, otherwise. Maybe this form is of use to the OP? -Kasimir Blomstedt, kblomste@vipunen.hut.fi === Subject: Re: matrix equation: can I claim the following? > Dear all, I am facing to derive some results from the following matrix equation: UAV+UBV=U(blahblahblah)V where all variables are matrices. Can I say that in order for the above to hold for any such U, V, the equality that A+B=(blahblahblah) should always hold? Please help me! I really want to get ride of U and V and get some meaningful > results... Thank you! You are correct, but only if U and V are are invertible. Otherwise, I can't simplify the equation. === Subject: Re: matrix equation: can I claim the following? >I am facing to derive some results from the following matrix equation: >UAV+UBV=U(blahblahblah)V Note that UAV + UBV = UCV is equivalent to U (A+B-C) V = 0. Now if U and V are singular, you can't conclude A+B-C = 0; all you can say is that A+B-C maps the range of V into the null space of U. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: matrix equation: can I claim the following? Adjunct Assistant Professor at the University of Montana. >>Dear all, >>I am facing to derive some results from the following matrix equation: >>UAV+UBV=U(blahblahblah)V >>where all variables are matrices. >>Can I say that in order for the above to hold for any such U, V, >>the equality that A+B=(blahblahblah) should always hold? >> Only if U has a left inverse and V has a right inverse. For an extreme >> case, consider the situation in which U and V are both matrices with >> all 0's. Then both sides of the equality are matrices of the same size >> with all 0's, but clearly A, B, and blahblahblah can be anything (so >> long as they are the correct size. >> If U is n x m, then a left inverse for U is an m x n matrix W such >> that WU is the m x m identity matrix. >> If V is an r x s matrix, then a right inverse for V is an s x r matrix >> Z such that VZ is the r x r identity matrix. >> If you can find such matrices, then from >> UAV + UBV you can do (replacing the blahblahblah by X): >> UAV + UBV = U(AV + BV) = U(A+B)V = UXV >> Then multiplying both wides on the left by W and on the right by Z, >> you have >> WU(A+B)VZ = WUXVZ >> and so >> Id(A+B)Id = Id(X)Id hence A+B = X. >> But if you do not have such matrices, then in general you cannot make >> that conclusion, no. >> -- >I found some clue: there is no clue for U and V directly, but at least I can >tell some matrix QU-I= is some non-singular matirx. >Let 's say QU - I = K, >where Q is some other complex matrix, U is the one we want (size: n x k), I >is the identity matrix, >I can derive that K is nonsingular, >then QU=(I+K) >then I can definitely have some other matrix P=(I+K)^(-1), No. Say K=-I. Then I+K is the zero matrix, which is of course singular, and therefore has no inverse. >to make PQU=I, >then U has left inverse, am I right? No. The sum of two nonsingular matrices can certainly be singular. -- === Subject: Re: matrix equation: can I claim the following? >Dear all, I am facing to derive some results from the following matrix equation: UAV+UBV=U(blahblahblah)V where all variables are matrices. Can I say that in order for the above to hold for any such U, V, the equality that A+B=(blahblahblah) should always hold? > Only if U has a left inverse and V has a right inverse. For an extreme > case, consider the situation in which U and V are both matrices with > all 0's. Then both sides of the equality are matrices of the same size > with all 0's, but clearly A, B, and blahblahblah can be anything (so > long as they are the correct size. > If U is n x m, then a left inverse for U is an m x n matrix W such > that WU is the m x m identity matrix. > If V is an r x s matrix, then a right inverse for V is an s x r matrix > Z such that VZ is the r x r identity matrix. > If you can find such matrices, then from > UAV + UBV you can do (replacing the blahblahblah by X): > UAV + UBV = U(AV + BV) = U(A+B)V = UXV > Then multiplying both wides on the left by W and on the right by Z, > you have > WU(A+B)VZ = WUXVZ > and so > Id(A+B)Id = Id(X)Id hence A+B = X. > But if you do not have such matrices, then in general you cannot make > that conclusion, no. > -- I found some clue: there is no clue for U and V directly, but at least I can tell some matrix QU-I= is some non-singular matirx. Let 's say QU - I = K, where Q is some other complex matrix, U is the one we want (size: n x k), I is the identity matrix, I can derive that K is nonsingular, then QU=(I+K) then I can definitely have some other matrix P=(I+K)^(-1), to make PQU=I, then U has left inverse, am I right? Haha, it is fun! But am I correct? === Subject: Re: matrix equation: can I claim the following? >Dear all, I am facing to derive some results from the following matrix equation: UAV+UBV=U(blahblahblah)V where all variables are matrices. Can I say that in order for the above to hold for any such U, V, the equality that A+B=(blahblahblah) should always hold? > Only if U has a left inverse and V has a right inverse. For an extreme > case, consider the situation in which U and V are both matrices with > all 0's. Then both sides of the equality are matrices of the same size > with all 0's, but clearly A, B, and blahblahblah can be anything (so > long as they are the correct size. > If U is n x m, then a left inverse for U is an m x n matrix W such > that WU is the m x m identity matrix. > If V is an r x s matrix, then a right inverse for V is an s x r matrix > Z such that VZ is the r x r identity matrix. > If you can find such matrices, then from > UAV + UBV you can do (replacing the blahblahblah by X): > UAV + UBV = U(AV + BV) = U(A+B)V = UXV > Then multiplying both wides on the left by W and on the right by Z, > you have > WU(A+B)VZ = WUXVZ > and so > Id(A+B)Id = Id(X)Id hence A+B = X. > But if you do not have such matrices, then in general you cannot make > that conclusion, no. > -- > === =================================================================== > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > === =================================================================== > Arturo Magidin > magidin@math.berkeley.edu Dear Arturo, Thank you very much for your help! But I really don't have left/right inverses for U and V. Can I just say something based on the fact that the equation should hold for all A, B, X's ... and U and V are definitely not zeros? === Subject: Re: matrix equation: can I claim the following? Adjunct Assistant Professor at the University of Montana. >Dear all, >I am facing to derive some results from the following matrix equation: >UAV+UBV=U(blahblahblah)V >where all variables are matrices. >Can I say that in order for the above to hold for any such U, V, >the equality that A+B=(blahblahblah) should always hold? Only if U has a left inverse and V has a right inverse. For an extreme case, consider the situation in which U and V are both matrices with all 0's. Then both sides of the equality are matrices of the same size with all 0's, but clearly A, B, and blahblahblah can be anything (so long as they are the correct size. If U is n x m, then a left inverse for U is an m x n matrix W such that WU is the m x m identity matrix. If V is an r x s matrix, then a right inverse for V is an s x r matrix Z such that VZ is the r x r identity matrix. If you can find such matrices, then from UAV + UBV you can do (replacing the blahblahblah by X): UAV + UBV = U(AV + BV) = U(A+B)V = UXV Then multiplying both wides on the left by W and on the right by Z, you have WU(A+B)VZ = WUXVZ and so Id(A+B)Id = Id(X)Id hence A+B = X. But if you do not have such matrices, then in general you cannot make that conclusion, no. -- === Subject: matrix equation: can I claim the following? Dear all, I am facing to derive some results from the following matrix equation: UAV+UBV=U(blahblahblah)V where all variables are matrices. Can I say that in order for the above to hold for any such U, V, the equality that A+B=(blahblahblah) should always hold? Please help me! I really want to get ride of U and V and get some meaningful results... Thank you! === Subject: Placement of Quantifiers and their rules Dear all, I was stupefied by a question in my tutorial when my tutor gave a different answer from mine and I asked him about it. But he didn't give me concise explanation in his reply and I thought it may be good to ask it here. OK. In my tutorial I came up with an answer, ForAll x, m, n Exists in Z (x is even) AND (x > 2) -> (m is prime) AND (n is prime) AND (m + n = x) The answer he gave me was: ForAll x Exists in Z (x is even) AND (x > 2) -> ForAll m, n Exists in Z (m is prime) AND (n is prime) AND (m + n = x) As we can see, my tutor's answer shifted the quantifier for m and n after the -> while I place all my quantifiers in front. He dismissed me by saying that: (1) my answer is equivalent to his (and so not worry about the difference?) (2) there are rules for quantifier placements (quantifier rules) (3) quantifier rules are _not_ covered within the module syllabus What I hope to do is to prevent from making mistakes in my quantifier placements by learning more about how it works. Can anyone point out to me on how to place the quantifiers? When should it be in front and when in the middle? Goh, Yong Kwang Singapore gohyongkwang@hotmail.com === Subject: Re: Square root extraction abuse@newcastle.ac.uk >> I must also point out that you are >> incorrect about If it is an exact square you are interested in it's >> square root but not otherwise UNLESS you are going to tell me a >> method for telling if a number is an exact square that's faster than >> root extraction. >Must you? I'm not incorrect, if you look at mean execution time. >Rejecting quadratic non-residues modulo small primes has a mean >execution time that is both faster for small numbers and also >asymptotically faster than not doing it. Its worst-case time is >of course marginally higher as it falls back on finding the >square root, but the worst-case scenario occurs for a very small >proportion of numbers. It's possible to reject about 99% of >candidates in a time no greater than add-time on most processors >Put it another way - if it were not faster, why are you doing it? > I should have been clearer - what I meant was unless you are going to tell me a method, which, given any number (or a number that, according to the limited quadratic residue checks I was able to do *may* be a square), can tell me, with 100% accuracy, if that number is a perfect square and which is faster than root extraction. Sorry about that. The thing is that I'm assuming both Fermat versions already reject several numbers like that, but I still need to consider the ones they let through. I'm just about to check the Bernstein papers - but what do you mean by one full size operation? Each Newton stage in root extraction has to square a number, multiply that number by two, calculate x_n+1=(first of those results/second), and check to see if it has a sufficiently precise result that it can stop. (and that assumes I can ignore flat spot checks due to the simplicity of this situation) Incidentally, is it indeed true that Newton's method is O(log N) when trying to find the root of N? === Subject: Re: Square root extraction abuse@newcastle.ac.uk wrotE: >zeroremovethistype@yahoo.com.spamblock (or somebody else of the same name) > eg (pseudocode) >> z=sqrt(y); >> if z=int(z) then (y was the perfect square sought) else (it wasn't - >> set up next y and test again.) >Dodgy! Set this up on any real processor then for sufficiently large integer >N that processor will be unable to distinguish sqr(N^2+1) from N. Not quite. Having tested to see if N is a perfect square (Sqrt was only pseudocode - I'm assuming I'm using arbitrary precision algorithms) next step is actually to set Y=(N+1)^2-N, giving me a smaller number to work with. Continuing with the incrementions and checks, using arb-prec methods, means that the first perfect square y is correct and is enough to finish factoring. === Subject: Question about GF(2^m) if element a is one GF(2^m) primitive element, and g(x)=x+a generate all polymials whose stage is n. i.e. p(x)=g(x)*h(x) and p(x) can be described as: p(x)=p0 + p1x + p2x^2 + .....pn-1x^n-1 and pi is an elemnet in GF(2^m) First we define weight of pi wpi is the number of non-zero's in it. For example, if in GF(2^2) , a=(0 1), so wa=1; and b=(1 1), so wb=2 Then we define weight of p(x) Wp is Wp=wp0 + wp1 + wp2 +.....wpn-1 My question is how to calculate the number of polymials whose weight is equal to l. Yang Jun === Subject: Re: probability in who wants to be a millionaire > in the game 'who wants to be a millionaire' in the 'fastest finger' > round what is the probability of guessing the 4 answers in the correct > order using random selection? and how do you calculate it You have 1/4 for the first. 1/3 for the second, given you have chosen the first correctly. 1/2 for the third. So 1/24. Or you could look at it that there are 24 ways of arranging the letters a b c d only one of which is correct. === Subject: probability in who wants to be a millionaire in the game 'who wants to be a millionaire' in the 'fastest finger' round what is the probability of guessing the 4 answers in the correct order using random selection? and how do you calculate it === Subject: Re: showing sequence converges > Suppose you want to show lim(n -> oo) |f-f_n| = 0. Also suppose you know > that lim(k,n -> oo) |f_n(k) - f_n| = 0 (i.e. f_n is Cauchy) for n(k) an Your assertion (i.e. f_n is Cauchy) is misleading. Yes, it follows from the previous assertion that f_n is Cauchy, but that's not as obvious as that. > increasing sequence, and also that lim(k -> oo) |f-f_n(k)| = 0 also. Can you > then conclude that lim (n -> oo) |f-f_n| = 0 since |f-f_n| <= |f-f_n(k)| + |f_n(k) - f_n|. This > seems like it should work, but on the far right-hand side you need k -> oo Well, this works. Take r > 0. Now take some K_1 such that whenever k > K_1 you have |f - f_(n(k))| < r/2. Finally take K_2 and N such that whenever k > K_2 and n > N, you have |f_(n(k)) - f_n| < r/2. Now, if n > N, choose some k greater than K_1 and K_2 and you have |f - f_n| <= |f - f_(n(k))| + |f_(n(k)) - f_n| < r/2 + r/2 = r. === Subject: showing sequence converges Suppose you want to show lim(n -> oo) |f-f_n| = 0. Also suppose you know that lim(k,n -> oo) |f_n(k) - f_n| = 0 (i.e. f_n is Cauchy) for n(k) an increasing sequence, and also that lim(k -> oo) |f-f_n(k)| = 0 also. Can you then conclude that lim (n -> oo) |f-f_n| = 0 since |f-f_n| <= |f-f_n(k)| + |f_n(k) - f_n|. This seems like it should work, but on the far right-hand side you need k -> oo === Subject: Re: Need some help in set theory >Hello everybody! >I'm taking a course in set theory, and i'm studying for the exam tomorrow. >The problem is our proffesor was very formal and i'm lacking some intuition >about certain subjects. >I wanted to ask the following questions: >1) What is the exact definition of a Suslin tree? (The definition I have >merely requires it to be aleph-1 tree with no uncountable anti-chain, but >that would make aleph-1 itself a Suslin tree... probably something is >missing) Why is the existance of a Suslin tree equals the Suslin conjecture? >2) The most difficult concept in the lecture was that of diamond series and >stationary sets. What is the intuition for stationary sets and diamond? Can >you please give me an example of a stationary set? Does it always have to >contain some 'final part' (I don't know the exact name for it in english) of >the cardinal? >Thank you very much! >I know these questions are rather general, but that's why I couldn't find >answer elsewhere. Well, as Keith said, you can probably find answers on the web or in a library. But a suggestion regarding how you might get answers out of sci.math: The title of your post is all wrong! See, here questions about set theory are usually questions at the Venn-diagram level - if I were an expert on actual set theory I might not bother reading posts that were labelled as requests for help with set theory. Try posting something mentioning Suslin trees in the Subject (and cross-post it to sci.math and sci.logic.) >Arie. === Subject: Re: Need some help in set theory |1) What is the exact definition of a Suslin tree? (The definition I have |merely requires it to be aleph-1 tree with no uncountable anti-chain, but |that would make aleph-1 itself a Suslin tree... probably something is |missing) Why is the existance of a Suslin tree equals the Suslin conjecture? You know, I didn't know the answer to either of your questions, but I did a search using google.com, and very soon found some web pages that claim to answer at least the first question. planetmath.org has a definition which also includes the condition that the tree have no cofinal branch. The same site also has alleged definitions of stationary set and the diamond principle, so maybe it will help. Perhaps you could find a good textbook in a library? Hope your studying goes well. Keith Ramsay === Subject: Need some help in set theory Hello everybody! I'm taking a course in set theory, and i'm studying for the exam tomorrow. The problem is our proffesor was very formal and i'm lacking some intuition about certain subjects. I wanted to ask the following questions: 1) What is the exact definition of a Suslin tree? (The definition I have merely requires it to be aleph-1 tree with no uncountable anti-chain, but that would make aleph-1 itself a Suslin tree... probably something is missing) Why is the existance of a Suslin tree equals the Suslin conjecture? 2) The most difficult concept in the lecture was that of diamond series and stationary sets. What is the intuition for stationary sets and diamond? Can you please give me an example of a stationary set? Does it always have to contain some 'final part' (I don't know the exact name for it in english) of the cardinal? Thank you very much! I know these questions are rather general, but that's why I couldn't find answer elsewhere. Arie. === Subject: Re: Riemann problem....simple confirm... > f(x) = 0 , x : rational root x , x : irrational show that f(x) is not Riemann integrable on [0,1] -------------------------- > this is my progress. please, check......my teacher.... There's only a small mistake. > any partition P = {0=x_0, x_1,......x_n=1} U = sigma f(x_i)*delta_xi > i=1~n = sigma root(x_i)*delta_xi >= int {(2/3)x^(3/2)}dx > 0~1 Here, you should have written >= int {sqrt(x)} dx 0~1 = 2/3 L = 0 thus U - L >= 2/3 thus Riemann integration is impossible. Right. === Subject: Riemann problem....simple confirm... f(x) = 0 , x : rational root x , x : irrational show that f(x) is not Riemann integrable on [0,1] -------------------------- this is my progress. please, check......my teacher.... any partition P = {0=x_0, x_1,......x_n=1} U = sigma f(x_i)*delta_xi i=1~n = sigma root(x_i)*delta_xi >= int {(2/3)x^(3/2)}dx 0~1 = 2/3 L = 0 thus U - L >= 2/3 thus Riemann integration is impossible. advice.....please.....thank you very much...... === Subject: Re: Examples Of Non-Measurable Sets Was it Ulam? Jech in his book 'Set Theory, third millennium > Banach and Kuratowski proved in [1929] that... OK! Ulam [1930] proved there is no such measure on a set X if the power of X is aleph_1, alpeh_2, aleph_3, ... or many other cardinals. -- http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Examples Of Non-Measurable Sets One can show: There is no countably additive probability measure P on the powerset of > (0,1] such that P(x)=0 for all xin(0,1]. No one can't. Or at least one would become famous if one did. Ulam > proved this assuming the continuum hypothesis. A proof of my statement above can be found in ingsley's 'Probablity and Measure, 3rd ed' on page 46. The proof does require both AC and CH, and I should have at least mentioned the latter, as both CH and not CH are consistent with ZFC. Was it Ulam? Jech in his book 'Set Theory, third millennium Banach and Kuratowski proved in [1929] that if $2^{aleph_0} =aleph_1$, then there is no measure on the continuum. Note that Jech defines 'measure on R to be a real valued set function m on the *power set* of R such that i) m(emptyset)=0, m(R)=1, ii) if X is a subset of Y, m(X)<=m(Y), iii) m({a})=0 for all ain R, iv) m is countably additive. And here is the reference: S. Banach and K. Kuratowski, Sur une g.8eneralisation du probl.8fme de la meaure. Fund. Math. 14 (1929), 127-131. k.j.h. === Subject: Re: Examples Of Non-Measurable Sets >A colleague of mine recently asked me a question. In math jargon, he >wanted to know why probability set functions were not defined for all >subsets of the sample space. >So I explained the non-measurable set in one of my analysis texts, >Royden I think. This example considers [0, 1) partitioned into >equivalence classes, where two elements of [0, 1) are equivalent >iff their difference is a rational number. A set E, defined to be >a set consisting of one and only one element from each equivalence >class, can be shown to be non-measurable. The existence of E follows >from the axiom of choice. >My question: are other examples of a non-measurable set ever invoked >in teaching measure theory? What other examples? There are other examples. One of them, which uses less non-constructive mathematics, is to divide the equivalence classes of the irrationals into pairs, such that the sum from one member of the pair and the other is rational. This only needs the axiom of choice for two-element sets, which is known to be much weaker than full choice. The Banach-Tarski paradox uses even less choice, having been shown to follow from the Hahn-Banach theorem. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Examples Of Non-Measurable Sets the related question I posed here in August 2002: > >If we accept the axiom of choice, then there exists a subset S of the unit interval [0,1] which is not Lebesgue measurable. I propose the following wager to you: We will generate a value X from the uniform probability distribution on [0,1]. If X is in S, I will pay you $1. How much >are you willing to pay to take this wager? What would be the fair >value? > This is a psychological question, not a mathematical one. By definition, the fair value of a bet like this depends on the probability of winning, so the event x in S MUST be measurable for this to be meaningful. However, there is a sense of fair value which depends on the psychology of the individual. This is different from the mathematical fair value. Of course the remarks above as to the impossibility of carrying out this wager in real life are correct. Another psychological use of fair value arises in the St. sburg paradox where we imagine you tossing a coin and receiving 2^n dollars if you toss n heads before the first tail. The expectation of the winnings is infinite so there is no fair value but many people would be willing to be the banker for such a bet if they were paid enough. I, for example, would do it for $1,000,000,000,000. That is, you pay me one trillion dollars, I'll give you an ordinary quarter to toss and I'll pay you 2^n dollars if you toss n heads in a row before the first tail. You'd only have to toss 40 heads in a row to bankrupt me. I doubt you have the cash to make this bet but maybe you could join with the governments of a few countries to form a syndicate? === > Correction However, if I am not mistaken I think that a drastic overhaul of the > known laws of physics is needed for this problem. should have been in ñBut maybe the laws themselves need to be changed. George Dvali Feb JS: That is too cheap as Albert Einstein mistakenly told David Bohm > theory. BohmÍs ñontologicalî theory gave a deeper meaning to Niels > BohrÍs ñepistemologicalî theory that John Archibald Wheeler summarized > as ñIT FROM BIT.î However, if I am not mistaken I think that a drastic > overhaul of the known laws of physics is not needed for this problem. Einstein was not mistaken. Since Bohm didn't have a pilot wave, he had a causal wave. Founded on what else? Newtonian Physics. The Laws Of Physics obviously don't need *any* overhauling. Since the only place that they even exist to begin with are The Harvard University Library. === Correction However, if I am not mistaken I think that a drastic overhaul of the known laws of physics is needed for this problem. should have been in ñBut maybe the laws themselves need to be changed. George Dvali Feb JS: That is too cheap as Albert Einstein mistakenly told David Bohm theory. BohmÍs ñontologicalî theory gave a deeper meaning to Niels BohrÍs ñepistemologicalî theory that John Archibald Wheeler summarized as ñIT FROM BIT.î However, if I am not mistaken I think that a drastic overhaul of the known laws of physics is not needed for this problem. === Subject: Re: Cat Generator, anyone? > >I need a string generator program which produces >all valid bracket clusters, i.e. >a Empty set. >a*a 1 >(a*a)*a a*(a*a) 1,2 ; 2,1 >(a*(a*(a*a))) (((a*a)*a)*a) (a*a)*(a*a)... 3,2,1 ; 1,2,3 ; 1,3,2 ; ... >you get the idea. >One could drop the a and ) and insert (* >recursively >(* >((** (*(* etc. >which would need weeding out dupes. Or generate >the above string by voters lead. >Anyone have a more elegant solution so I don't have >to reinvent the wheel? >> Why not think of it as a permutation on {1,2,3,...,n-1} where n = the >> number of a's? This would correspond to which * you wish to perform >> first, second, third, etc. Then analyze the permutation and determine >> how to insert the parenthesis to correspond with that order. > I don't understand how you could easily turn a permutation into the kind > of bracketing scheme that is needed; how would you represent > (a*((a*a)*a))? 3,1,2 If you've done work with parsing, such as making a calculator, then you should be able to reverse-parse to create the output. It's still a fair bit of work, but it gives you a different way to look at the problem. -- email: wtwentyman at copper dot net === Subject: Re: Cat Generator, anyone? > No recursive calls, please, we're FORTRAN :-) You should switch to Fortran. Then you can use recursion. -- Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.