mm-414 === Subject: Re: lattices, rings of algebraic integers So suppose P in Z[x] is a monic irreducible polynomial of degree at least 2, and that alpha and beta are distinct roots of P. Is it always the case that Z[alpha] is a lattice in C if and only if Z[beta] is a lattice? Well, the answer is yes, but for a boring reason. In fact, Z[alpha] is a lattice in C if and only if (1) P(x) has degree 2 and (2) its roots are complex. The right way to generalize this is as follows. Label the roots of P(x) as a_1,a_2,...,a_n - the real roots b_1,b_2,...,b_m and b_1',b_2',...,b_m' - the complex roots where b_i and b_i' are complex conjugates. Let alpha be any one of the roots. Then the ring Z[alpha] embeds as a lattice inside R^n x C^m using the map that sends alpha to (a_1,a_2,...,a_n,b_1,b_2,...,b_m). In other words, an element of Z[alpha] looks like F(alpha) for some polynomial F(x) in Z[x], and then F(alpha) is sent to the point (F(a_1),F(a_a),...,F(a_n),F(b_1),F(b_2)...,F(b_m)) in R^n x C^m. The volume of a fundamental domain for this lattice is an extremely interesting quantity. This same construction is used more generally to study the ring of integers in a number field. This is explained in most basic books on algebraic number theory. A further generalization is to use the p-adic completions, in addition to the real and complex completions. Then one gets the space of adeles, and the entire field Q(alpha) sits inside the adeles as a lattice, i.e., as a discrete subgroup. Joe Silverman Suppose gamma is an algebraic integer and we let Z[gamma] denote the smallest ring extension of Z that contains gamma. For example, Z[i] is the Gaussian integers, and they form a lattice in the compex plane. On the other hand, Z[sqrt(2)] does not form a lattice, since 1 > sqrt(2) - 1 >0 and if n>0 is sufficiently large, then (sqrt(2)-1)^n becomes arbitrarily small as n is allowed to increase without bound. So suppose P in Z[x] is a monic irreducible polynomial of degree at least 2, and that alpha and beta are distinct roots of P. Is it always the case that Z[alpha] is a lattice in C if and only if Z[beta] is a lattice? David Bernier === Subject: Re: Magnetic force: An approach with Bernoulli's equation. 1. Abstract: In this paper, Bernoulli's effect is used to interpret the magnetic force. See my posting: Fm = F1+F2 = -(m/R)*v1*v2 - L1*(v1)^2 /2 - L2*(v2)^2 /2 (1) Assuming the surrounding are neutral, static, and far away from wire 1 and wire 2, then L1 and L2 terms can be ignored. Fm = -(m/R)*v1*v2 = - mu_0 *q1*q2*v1*v2/ (4*pi* R^2) Fm = - mu_0 * i1 * i2 / (4 * pi * R^2) (2) My dear friends, I am very sorry to disappoint you, the above derivation is incomplete. To eliminate L1 and L2 terms, negative potential masses have to be considered. A shematic diagram is shown as figure 2. (+q1,0) (-q1,v1) ______________+______________________________ wire 1 ^ | R (+q2,0) | (-q2,v2) V______________+_____________________________ wire 2 Figure 2 The whole electric wire is neutral; for every drifting electron(-), there is a resting ion(+). (-q1,v1) is charges of drifting electrons, and (+q1,0) is charges of resting ions. (-q2,v2) and (+q2,0) are same definition. 1) To (-q1,v1) and (-q2,v2) pair, we have (m/R)*(v1+v2)^2 /2. 2) To (-q1,v1) and (+q2,0) pair, we have (-m/R)*v1^2 /2. 3) To (+q1,0) and (-q2,v2) pair, we have (-m/R)*v2^2 /2. 1) + 2) + 3) = (m/R)*v1*v2 = mu_0 *q1*q2*v1*v2/(4*pi* R^2) = mu_0 *i1*i2 / (4*pi* R^2) Ka-In Yen yenkain@yahoo.com.tw How to correctly measure an unknown length with a clock. === Subject: Re: Names of binary numbers === Subject: Names of binary numbers Quick question: Calling the binary number 10 ten is completely wrong, correct? Correct. Ten is the name of a base-ten number, No, ten is the name of successor(nine). and implies the use of this base, if my understanding is correct... Ten does not imply the use of any base. Ten is always the successor(nine) regardless how it is represented (ten would represented as A in base sixteen). Now if you assert that 10=ten, then you are implying that you are using base ten. Note that I did not say base 10, because the 10 could be interpreted as a number of radix two. So, how would one go about reading a document that contains binary numbers?] Usually people just read the digits, which seems most inefficient to me. -- === Subject: Generalized Lie Bracket with Some Elementary Examples by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i173it502391; I've argued that the Lie Bracket can be generalized to the difference between a sequence of elements, steps, procedures, indeterminates, etc., and its reverse. For example: 1) [A, B] = P(A-->B) - P(B-->A) = P(B) - P(A) where the center equality is the definition of the far left-hand- side and P( ) is probability of, and (A-->B) is the set/event defined by: 2) (A-->B) = (AB')' = A' U B with AB' the intersection of A and the complement B' of B. It sometimes happens, but not by any means always, that the generalized Lie Bracket can be written as f(x, y) - f(y, x) for f a function and x, y some objects. Aside from probability-statistics, the definition seems rather useful for algebra in general and even number theory. For example, it is rather easy to prove that for noncommutative algebras or rings or modules, we have: 3) (x o y) - (y o x) = x - xy + y -(y - yx + x) = yx - xy 4) (x o y)' - (y o x)' = x + xy + y - (y + yx + x) = xy - yx where x o y is the Jacobson Radical star product x + y - xy which can alternatively be formulated as what I label (x o y)' = x + y + xy for (noncommutative) a ring or module with elements x, y. Notice also: 5) (x^n - y^n) - (y^n - x^n) = 2(x^n - y^n) but for a noncommutative algebra we don't have the usual factoriza- tion of x^n - y^n but rather: 6) (x - y)(x^(n-1) + x^(n-2)y + ... + y^(n-1)) = = x^n + x^(n-1)y + ... + xy^(n-1) - [yx^(n-1) + ... + y^n] = (x^n - y^n) + (x^(n-1)y - yx^(n-1)) + ... + (xy^(n-1) - yx^(n-1)) So, for example, the following is wrong: 7') (x^2 - y^2) - (y^2 - x^2) = (x + y)(x - y) - (y + x)(y - x) = x^2 - xy + yx - y^2 - (y^2 - yx + xy - x^2) = 2x^2-2xy + 2yx -2y^2 because it claims that 2x^2 - 2y^2 = 2x^2 - 2y^2 + 2yx - 2xy which is only true iff xy = yx. The error is writing x^2 - y^2 = (x + y)(x - y) in (7') because the latter is really x^2 -xy + yx -y^2 which isn't x^2 - y^2 in general. However, we do have: 7) (x + y)(x - y) = 2x^2 - 2y^2 + 2(yx - xy) which relates things again to the Lie bracket yx - xy. Osher Doctorow === Subject: the song you rewrite re: James by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i173iut02395; . Nice one, Francis! I hope, whoever James is, that he deserves this ruthless mockery? The unknown author[s] you rather niftily rewrite is the Victorian writing duo Gilbert and Sullivan, responsible for a string of comic opera hits performed from the 1870s to the 1890s [and oddly echoed in the name of a forgettable 1970s singer Gilbert O'Sullivan]. William Gilbert used to write the plots and words, and Arthur Sullivan set them to music. The story is they quarrelled increasingly bitterly and ultimately fell out, but continued to collaborate successfully by post between different parts of London on operettas for years after they were no longer on speaking terms with each other. Shame these lists can't be a bit more like that - humour and professionalism surviving personality clashes, eh? Mark >To be sung to the 'Pirates Of Penzance' tune 'Modern Major General' (author >unknown): >I am the very model of a Newsgroup Personality >I intersperse obscenity with tedious banality. >Addresses I have plenty of, both genuine and ghosted to, >On all the countless newsgroups that my drivel is cross-posted to. >Your bandwidth I will fritter with my whining and my snivelling, >And you're the one who pays the downloading all my drivelling. >My enemies are numerous, and no one would be blaming you >For cracking my head open after I've been rudely flaming you. >I hate to lose an argument (by now I should be used to it). >I wouldn't know a valid point if I was introduced to it. >My learning is extensive but consists of mindless trivia, >Designed to fan my ego, which is larger than Bolivia. >The comments that I vomit forth, disguised as jest and drollery, >Are really just an exercise in unremitting trollery. >I say I'm frank and forthright, but that's merely lies and vanity, >The gibberings of one who's at the limit of his sanity. >If only I could get a life, as many people tell me to; >If only mum could find a circus freak-show she could sell me to; >If I go off to Zanzibar to paint the local scenery; >If I lose all my fingers in a mishap with machinery; >If I survive to forty, which is somewhat problematical; >If what I post was more mature, or slightly more grammatical; >If I could learn to spell a bit, and maybe even punctuate; >Would I still be the loathsome and objectionable punk you hate? >But while I have this tiresome urge to prance around and show my face, >It simply isn't safe for normal people here in cyberspace. >To stick me in Old Sparky and turn on the electricity >Would be a fitting punishment for my egocentricity. >I always have the last word; so, with utmost finality, >That's all from me, the model of a Newsgroup Personality. === Subject: Re: the song you rewrite re: James Shame these lists can't be a bit more like that - humour and professionalism surviving personality clashes, eh? Ouch, that hurt! No sign of either humour nor professionalism here? *sniff* dave Member, Society of Professional Humourists === Subject: Re: the song you rewrite re: James ... Nice one, Francis! I hope, whoever James is, that he deserves this ruthless mockery? Of course he (JSH) does, in the opinions of many who read his offerings in sci.math, alt.math, alt.math.undergrad, sci.logic sci.math.num-analysis, sci.physics, and any other newsgroups he so inexplicably cross-posts to. The unknown author[s] you rather niftily rewrite is the Victorian writing duo Gilbert and Sullivan, responsible for a string of comic opera hits performed from the 1870s to the 1890s ... I imagine Harrington's author unknown applies not to Pirates Of Penzance but instead to the very model of a Newsgroup Personality verses. Google on that phrase (all on one line: ) shows numerous attributions to Tom Holt, and a scattering of others. -jiw >To be sung to the 'Pirates Of Penzance' tune 'Modern Major >General' (author unknown): >I am the very model of a Newsgroup Personality >I intersperse obscenity with tedious banality. >Addresses I have plenty of, both genuine and ghosted to, >On all the countless newsgroups that my drivel is cross-posted to. ... === Subject: Re: 1024-bit Primes === Subject: Re: 1024-bit Primes > How long would it take to generate many of the primes in the 1024- > bit or less range? How about 512 bit? That's a large number so I > presume it would be GB size files (unless it was in hex format)... > is there such a file? has it been generated yet? > Have you ever heard the story about the Buddhist monks and their 64- > disk solid-gold Towers of Hanoi? > How about the one about the grains of rice on the chessboard? > Long story short, you have no *comprehension* of the magnitudes > you're talking about! > Well, he didn't ask for all, but for many. Huge difference... > He also says That's a large number. So he is asking how long it would > take to generate a large number of such primes. I think the only > possible answer to the question of the space required is a lot of bytes. > Gib Wouldn't a 1024-bit number (whether prime or composite) take up 1024 bits= 128 bytes Sure, if you'll settle for just one such prime. But there's a slight difference between one of them and all of them. or am I missing something ? -- To e-mail me get rid of the cats and dogs. -- === Subject: resolving Zeno [was: Re: lots of balls = 0 balls] > Noon is only attained as a step with a transfinite index, not a finite > one. If we do not allow a label to be transfinite, then we cannot say > that we are at noon since all finite indices occur before noon. Huh??? The allowance of transfinite indices has no affect on the flow of time. Seconds flow 1 per second, regardless of the puzzle. You are apparently trying to apply Zeno's paradox here (and doing it poorly, I'm sorry to say). In any case, Zeno's paradox was introduced and resolved millenia ago, so at best that point is merely a footnote to the original problem. Jonathan is right in this thread, about the balls in buckets question. I am curious about his statement that Zeno's paradox was resolved millenia ago. As I understood it, Aristotle thought he had resolved Zeno's paradox simply by asserting that it isn't possible to subdivide space and time indefinitely. (I assume this is the refutation from millenia ago referred to.) But this is no refutation - we now know about the Planck length etc, but Aristotle certainly didn't, and his statement was an unsupported assertion. I have always told people that the paradox wasn't really resolved until the calculus was on a secure theoretical footing: it needed to be shown that it *is* possible to sum an infinity of quantities and obtain a finite result, provided the quantities are small enough. What do the rest of you think? === Subject: Re: re: James >To be sung to the 'Pirates Of Penzance' tune 'Modern Major General' (author >unknown): the original tune is Arthur Sullivan; lyrics William[?] Gilbert. I *think* (but I'm not sure) that this version might be by Jim Ferry, from before he saw the light. If so, you'll probably make him very unhappy by reposting it, as I'm sure he doesn't want to be reminded of his former Disturbing Lack of Faith. [snip most of it] >I always have the last word; so, with utmost finality, >That's all from me, the model of a Newsgroup Personality. I think the last line is missing a two-syllable word near the end. Apart from that the whole thing scanned beautifully. === Subject: Re: need help in understanding Torkel's ZFC comment X-ID: GWl9umZeretqwp5hGteK9sA9i59ZRVo-DJY+ZJO6qu1Kr4bjysuX85 > P=>(Q=>P) is not a formula of first-order logic. > (D. Ullrich) > I would say that propositional calculus is a subset of first-order > logic. (C. Volkstorf) So would a lot of people. Indeed. They're all wrong No, they are n o t all wrong. [...] of course there's a sense in which PC is part of FOL, but it's not literally true; a wff of PC is not a wff of FOL.) It is - in many systems (though certainly not all). F. -- I do tend to feel Hughes & Cresswell is a more authoritative source than you. (D. Ullrich) === Subject: Re: need help in understanding Torkel's ZFC comment > P=>(Q=>P) is not a formula of first-order logic. > (D. Ullrich) > I would say that propositional calculus is a subset of first-order > logic. (C. Volkstorf) > So would a lot of people. Indeed. > They're all wrong No, they are n o t all wrong. This is _truly_ bizarre, the way you continue to reply to posts refuting statements that I've already agreed many times were not quite correct. > [...] of course there's a sense in which PC is part of FOL, > but it's not literally true; a wff of PC is not a wff of FOL.) It is - in many systems (though certainly not all). F. David C. Ullrich === Subject: Series What are the next ten characters in the following series? 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 === Subject: Re: Series What are the next ten characters in the following series? 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 Start with 1. Now add 2 terms to the sequence: take the last term (1) and add 1, next subtract 1. This gives 1 2 1. Next take the second term of the sequence under construction: 2. Use this number to first add, then subtract from the latest term. This gives 1 2 1 3 1. Now take the 3th term:1. Same procedure. This would give (first 20 terms): 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 6 Q: Is this the same sequence as the one given by Rob Johnson? greetings, Bart Jans === Subject: Re: Series What are the next ten characters in the following series? 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 Count the number of terminal zeroes in the binary representation of each natural number and add 1: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 10000 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 Following this pattern, we would get 17 18 19 20 21 22 23 24 25 26 10001 10010 10011 10100 10101 10110 10111 11000 11001 11010 1 2 1 3 1 2 1 4 1 2 Rob Johnson take out the trash before replying === Subject: Re: Series >What are the next ten characters in the following series? >1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 Count the number of terminal zeroes in the binary representation of each natural number and add 1: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 10000 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 Following this pattern, we would get 17 18 19 20 21 22 23 24 25 26 10001 10010 10011 10100 10101 10110 10111 11000 11001 11010 1 2 1 3 1 2 1 4 1 2 Or just solve the Tower of Hanoi problem for n disks and keep a record of the number of the disk that you move each time. Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Series |What are the next ten characters in the following series? | |1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 there aren't any more- it ends right there. [e-mail address jdolan@math.ucr.edu] === Subject: Re: Series |What are the next ten characters in the following series? | |1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 there aren't any more- it ends right there. Nice one. I've sometimes tended to give facetious answers to these ill-defined what's-the-next-number questions, but I have to stop now, this is better than anything I've ever done. David C. Ullrich === Subject: Re: Series >|What are the next ten characters in the following series? >| >|1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 >there aren't any more- it ends right there. Nice one. I've sometimes tended to give facetious answers to these ill-defined what's-the-next-number questions, but I have to stop now, this is better than anything I've ever done. Hmm... but not half as witty as Rob Johnsons answer, sorry. I always wonder, why people who don't like this kind of riddles feel like responding at all. It sounds like Well I didn't get it. And so it won't be worth it. In German: Dem Fuchs sind die Trauben zu sauer. Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Series What are the next ten characters in the following series? 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 I think you mean 'sequence.' 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 6 J === Subject: Re: Series What are the next ten characters in the following series? 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 Donald Duck and Superman. === Subject: Re: Did Fermat have a proof of his theorem? KRamsay Shmuel (Seymour J.) Metz |I know of no case where he claimed to have a proof of |something that turned out to be false. Everything I've read about Fermat is consistent with his having been a reasonably careful guy. Maybe the most famous example of something that Fermat expected to be true but isn't is that 2^2^n+1 is prime for positive integers n. Of course, he fortunately didn't claim he had a proof of that. As I recall, he said outright that he had been unable to find a proof of this one, although he thought it was true. I doubt very much that as he was writing his famous marginal note, he was thinking that after he died, his son would publish those marginal notes. Indeed, and moreover, he did say in more than one letter that he could prove the cases of exponent 3 and exponent 4, without claiming the same for the general case. So I think he realized his general proof was no good, but saw no reason to go back and correct his personal copy of Diophantus. === Subject: Re: Elevator problem > >A mathematician lives in an apartment building with k floors in total. >The building has n elevators. Every time a tenant living in floor k >wishes to use an elevator, the one that is closest to the floor will >be the one that is used. There are two types of elevator trips taken, > >from the 1st floor to a random floor k and vice-versa. Both are > >equally common (tenants don't disappear or appear out of thin air) and >no inter-floor trips are made. >If each floor has the same number of tenants and all of them use the >elevators, what is the expected distribution of the elevators between >floors after sufficient time has passed? Also assume the trips are >always taken consecutively. > >I am a little unclear on a few details of the model: > Okay. Let's reformulate: We start from a random state. Every trip moves one elevator from one floor to another. There is a 50-50 chance of either going up to a random floor from floor 1 or going down to 1 from a random floor. The elevator closest to the starting floor is always called in. Example: There are two elevators and three floors. E_1 = 2 E_2 = 1 E_3 = 2 The first trip is from 1 to 3. There is an elevator in floor 1 so we use it. The state now looks like: E_1 = 2 E_2 = 3 E_3 = 2 Now there is another trip from 1 to 3. We pick elevator 1, since both 1 and 3 are equally close. The state now looks like: E_1 = 3 E_2 = 3 E_3 = 2 Now there is a trip from 2 to 1. The state is: E_1 = 3 E_2 = 3 E_3 = 1 And so on. >When you say the expected distribution of the elevators between floors, >do you mean *amongst* floors, or the those elevators which are actually > There are actually many ways to measure the distribution of the elevators. The one I was originally interested in is, what probability is there over time that at least one elevator will be in floor x? >So I'm guessing you are describing a discrete-time Markov chain with >state-space {1, 2, 3,..., k}^n. A state x indicates that elevator i >is on floor x_i for i = 1 to n. > Looks like what I had in mind. >Now, is this part correct? You keep track of discrete time by trips: >The state at the mth epoch is the positions of the elevators at the end >of the mth trip. You assume trips do not overlap. > Yes, correct. >It is not clear how the trips are being generated. Are you saying that >with probability 1/2, a call comes from floor 1, and with probability >1/[2(k-1)], a call comes from floor i, i = 2, 3,..., k? > Yes. >Have you tried a simpler parameter set, e.g., n = 1, k = 3? Or n = >2, k = 3? > It appears my code has a problem with the ends of the distribution. The n = 1 case is not uniform, as it should be, for large k. I'll fix it and post more simulation results tomorrow. I have commented on the case where n=1 in another post. For n > 1, one can ostensibly derive all the transition probabilities for this model. One can then solve for the stationary probabilities of the discrete-time Markov chain. However, the state-space has size k^n, so even for n = 2 and k = 3, this becomes tedious to do by hand, since then one has to solve nine linear equations in nine unknowns (though the matrix is sparse). I guess one could come up with a program to generate the transition matrix and solve for the stationary probabilities. For the record, for the case n = 2 and k = 3, I get the following fractions of discrete time that there is at least one elevator on the indicated floor: 1: 2/3 2: 11/30 3: 7/15 As I commented in another post, there is no reason that these should sum to 1. I assumed that when the elevators are on floors 1 and 3 and the call comes from floor 2, then the elevator from floor 3 responds. When both elevators are on the same floor, each is equally likely to respond. I used Mathematica(R) to solve this. Perhaps someone else who wants to waste their time can verify these results. Perhaps some other line of thought will come up with a general approximation of the stationary probability that there is at least one elevator at a given floor. Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Elevator problem For n > 1, one can ostensibly derive all the transition probabilities for this model. One can then solve for the stationary probabilities of the discrete-time Markov chain. However, the state-space has size k^n, so even for n = 2 and k = 3, this becomes tedious to do by hand, since then one has to solve nine linear equations in nine unknowns (though the matrix is sparse). I guess one could come up with a program to generate the transition matrix and solve for the stationary probabilities. Sounds like a job for Matlab. For the record, for the case n = 2 and k = 3, I get the following fractions of discrete time that there is at least one elevator on the indicated floor: 1: 2/3 2: 11/30 3: 7/15 Here are the simulation results for the same case: 1: 0.64999 2: 0.35906 3: 0.51598 Close, but not quite. As I commented in another post, there is no reason that these should sum to 1. No, of course not. The figures in the original post were incorrect. I assumed that when the elevators are on floors 1 and 3 and the call comes from floor 2, then the elevator from floor 3 responds. The code picks randomly either 1 or 3. This could explain the small variance in the results. About the original case (n = 2, k = 9). Here are the corrected probabilities over time: 1: 0.63577 2: 0.09947 3: 0.10187 4: 0.10184 5: 0.10392 6: 0.1114 7: 0.12414 8: 0.16067 9: 0.24452 I'm not interested in mathematics that might have anything to do with reality. -- Easterly, in sci.math === Subject: Re: Elevator problem >A mathematician lives in an apartment building with k floors in total. >The building has n elevators. Every time a tenant living in floor k >wishes to use an elevator, the one that is closest to the floor will >be the one that is used. There are two types of elevator trips taken, >from the 1st floor to a random floor k and vice-versa. Both are >equally common (tenants don't disappear or appear out of thin air) and >no inter-floor trips are made. >The question is, is there an analytic >answer for the final distribution of elevators for any system n, k? >Maybe with Markov chains? So I'm guessing you are describing a discrete-time Markov chain with state-space {1, 2, 3,..., k}^n. A state x indicates that elevator i is on floor x_i for i = 1 to n. It is not clear how the trips are being generated. Are you saying that with probability 1/2, a call comes from floor 1, and with probability 1/[2(k-1)], a call comes from floor i, i = 2, 3,..., k? Finally, you want the limiting distribution of each component of the the state variable X. This should be identical for each component by symmetry. I agree with both posters that Markov chains are a good idea. Stephen describes the problem with a state space of size k^n. However, one can use a smaller state space (of size C(k+n-1,n)) by using unordered rather than ordered n-tuples, which produces a sufficient savings in the size of the transition matrix to be worthwhile. I don't believe the statement about the identical limiting distributions because I don't perceive a symmetry that would make it so. There is a symmetry among all floors above the first in the trips that are made, but it does not extend to the rule of calling elevators. Here's an algorithm for setting up the transition matrix M = (m_ij). First, initialize M to 0. Then loop over all states j. For each state j we then loop over the 2(k-1) possible (and equally likely) trip requests. For each trip request, we compute the state i that it produces from j and augment the value of m_ij by 1/(2(k-1)). In some cases, however, we don't get a single state i, but two possible states i1 and i2, in which case we augment m_i1,j and m_i2,j by e1/(2(k-1)(e1+e2)) and e2/(2(k-1)(e1+e2)), respectively, where e1 and e2 will be defined shortly. This complication can only arise when someone wishes to go to floor 1 and the floor he is on is equidistant from the nearest elevators above and below him. If the nearest elevators consist of e1 on the lower floor and e2 on the higher floor, then we assume an elevator comes from below with probability e1/(e1+e2) (leading to state i1). Computing these transitions is straightforward. Note that i and j can be considered *sets* (in addition to being indices) because they represent states, and states are represented as sets. The trips take two forms: 1 -> x, and x -> 1. In the case 1 -> x, we form i from j by replacing the minimal element of j by x. In the case x -> 1, we assume that we are dealing with the complicated, bifurcated subcase, with the understanding that e1 or e2 could be 0. Let y1 be the lower floor and y2 the higher one. If e1 > 0, then i1 is formed by replacing an element y1 of j with 1 (and if e1 = 0, then we don't need state i1). If e2 > 0, we form i2 similarly: by replacing y2 with 1. After you form M you can find the limiting distribution by computing the eigenvectors of M corresponding the eigenvalues of largest absolute value. For this problem I don't think they'll be any oscillatory limiting behavior: I think you'll find that the only eigenvalue of magnitude 1 is 1 itself, and that it has no multiplicities. If this is so, then you don't even need an eigensystem solver, you just need to find the solution x to (M-I)x = 0, and normalize it to sum to 1. -Jim Ferry === Subject: Need Help !!! Prove : derived set A' of any subset A of R^2 is a closed set. ____________________________________________________________ insufficient pf> *main idea is show that : (A')' is a subset of A'* Let p be a limit point of A' and G be a open set containing p. Since p is a limit point of A', G contains at least one point q in A' different from p. But G is an open set containing q in A'. Since, G is an open set containing q in A', G contains at least one point x in A, different from q. So, x=/=q. But for concluding p be a limit point of A, I need x=/=p. How can I prove this ?? I think over about this, but I don't know--;. If someone can help me, please post reply. === Subject: Re: Need Help !!! complement of A is open for any T1 space such as R^n. derived set of A is closed for any T1 space. ^^ ----== Posted via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Need Help !!! > complement of A is open for any T1 space such as R^n. derived set of A is closed for any T1 space. ^^ complement of derived set of A is open for any T1 space such as R^n. === Subject: re:Need Help !!! Note that R^2 is a hausdorff space and hence q has a neighborhood V that doesn't intersects p. ----== Posted via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Need Help !!! Prove : derived set A' of any subset A of R^2 is a closed set. ____________________________________________________________ insufficient pf> *main idea is show that : (A')' is a subset of A'* I suppose that X* stands for the closure of X. Why do you want to prove that (A')' is a subset of A'*?. Try to prove that (A')* is a subset of A' instead. Best regards, Jose Carlos Santos === Subject: Re: Need Help !!! > Prove : derived set A' of any subset A of R^2 is a closed set. > ____________________________________________________________ > insufficient pf> *main idea is show that : (A')' is a subset of A'* I suppose that X* stands for the closure of X. Why do you want to prove that (A')' is a subset of A'*? Because, if (A')' is a subset of A' then this imply that (A')' union A' is A' and this imply cl(A')=A'. therefore, A' is a closed set. Try to prove that (A')* is a subset of A' instead. Surely, if (A')* is a subset of A' will be shown, this problem will be solved. but, that was not quite easy to me. if p in (A')* then p in A' or p in (A')' (i) if p in A' there is nothing to prove. (ii) if p in (A')' Since p is a limit point of A', for any open subset G containing p, (G{p}) intersection A' =/= empty. so, G contains at least one point x in A' different from p. but, how can I proceed from here ? (for reaching result : p in A') Best regards, Jose Carlos Santos === Subject: Re: Need Help !!! >Prove : derived set A' of any subset A of R^2 is a closed set. >____________________________________________________________ >insufficient pf> *main idea is show that : (A')' is a subset of A'* >I suppose that X* stands for the closure of X. >Try to prove that (A')* is a subset of A' instead. Surely, if (A')* is a subset of A' will be shown, this problem will be solved. but, that was not quite easy to me. Let p be in (A')*. If n is a natural number then the open disk D with center p and radius 1/n contains some point q of A' (because p is in its closure). But since the disk D is an open set and since q is in A', there is some element a_n of A that belongs to D. So, the sequence a_1, a_2, a_3, ... converges to p and so p belongs to A'. Best regards, Jose Carlos Santos === Subject: Re: Need Help !!! I suppose that X* stands for the closure of X. > Why do you want to prove that (A')' is a subset of A'*? Because, if (A')' is a subset of A' then this imply that (A')' union A' is A' and this imply cl(A')=A'. therefore, A' is a closed set. > Try to prove that (A')* is a subset of A' instead. Surely, if (A')* is a subset of A' will be shown, this problem will be solved. but, that was not quite easy to me. Hint: Instead of a 'bull by the horns' approach, show complement of A is open for any T1 space such as R^n. if p in (A')* then p in A' or p in (A')' (i) if p in A' there is nothing to prove. (ii) if p in (A')' Since p is a limit point of A', for any open subset G containing p, (G{p}) intersection A' =/= empty. so, G contains at least one point x in A' different from p. but, how can I proceed from here ? (for reaching result : p in A') === Subject: Re: epsilon numbers : a problem of choice > [...] > without choice, it is consistent that w1 is not > regular. > What???? I can't believe it (I barely understand how it could be > possible, as the denumerable union of denumerable sets can be not > denumerable, but then I would have sworn it would have a cardinal > uncomparable with w). Any reference? > Kunen makes this statement without citation, and he goes on to say > that it is unknown whether one can prove in ZF that there exists a > cardinal with cofinality >w. (p. 33). > A few more thoughts on it > 1) every denumerable limit ordinal is of cofinality w (in ZF) > Proof: if n->x_n is an enumeration of x, the sequence f(n+1)=the > smallest p such that x_p>x_(f(n)) gives a set {x_f(n)} cofinal to > x., of order type w. > 2) if (x_1 exists a injection of w_1 in IR (actually in Q) > First, we inject x_i-> i, then we use the classical (not choice- > defined) injection of {x/ x* x_i for all i*x_(n+1) === Subject: Re: Point-set Topology Hints Wanted . I am trying to show that Cl(A)' is a subset of A.' >A' is the set of all limit point of A . > limit pts A = derived A = A' = { x | x in cl Ax } So what you actually want is that every punctured neighborhood intersect A. > When space is T1, A' is closed. > Now for all A > A' subset cl A > so > A subset cl A' = A' > Is A subset A' even when space is not T1 ? Let X = {x,y}, with the indiscrete topology T = {empty, X}. Let A ={x}. Then x is not in A', but y is. So A' = {y}. And by a symmetric argument, A'' = {x}. So A'' is disjoint from A'. Yes. Perhaps for an encore you have counter example for T0 space? === Subject: Re: :: towards a constructive education :: (news server friendly) Galathaea, from what I gather from other postings of yours, you are probably brighter than those around you and those you grew up with. You probably are acustomed to receiving praise and approval at the merest trembling of your lips. Before you go on reading I sugest you eat some chocolate burn some inciense and say a prayer to your favorite god of no hard feelings. And yes, maybe take some of that prozac too, you know, the one you take to take the edgess off your extremely sharp mind. The various systems that have been formed concerning the standard of right and wrong, may all be reduced to the principle of sympathy and antipathy. One account may serve for all of them. They consist all of them in so many contrivances for avoiding the obligation of appealing to any external standard, and for prevailing upon the reader to accept of the author's sentiment or opinion as a reason for itself. The phrases are different, but the principle the same. --Jeremy Bentham === Subject: Re: :: towards a constructive education :: (news server friendly) > Galathaea, from what I gather from other postings of yours, you are > probably brighter than those around you and those you grew up with. > You probably are acustomed to receiving praise and approval at the > merest trembling of your lips. Before you go on reading I sugest you > eat some chocolate burn some inciense and say a prayer to your > favorite god of no hard feelings. And yes, maybe take some of that > prozac too, you know, the one you take to take the edgess off your > extremely sharp mind. The various systems that have been formed concerning the standard of right and wrong, may all be reduced to the principle of sympathy and antipathy. One account may serve for all of them. They consist all of them in so many contrivances for avoiding the obligation of appealing to any external standard, and for prevailing upon the reader to accept of the author's sentiment or opinion as a reason for itself. The phrases are different, but the principle the same. --Jeremy Bentham ???? There are two types that rationalise and argue the denial of objective right and wrong. Those who have had their faces violently stuck into a very big pile of reality , and those who sorely need it. The former sometimes deserve pity the latter always deserve contempt. The rest of the people just shut up and hope they never really have to find out. according to this theory, it would be moral to e.g. torture one person if this would produce an amount of happiness in other people outwheighing the unhappiness of the tortured individual.. So y, how large does the gang need to be in order for a gang rape to be moraly ok? dimwit. === Subject: Re: :: towards a constructive education :: (news server friendly) : Galathaea, from what I gather from other postings of yours, you are : probably brighter than those around you and those you grew up with. : You probably are acustomed to receiving praise and approval at the : merest trembling of your lips. Before you go on reading I sugest you : eat some chocolate burn some inciense and say a prayer to your : favorite god of no hard feelings. And yes, maybe take some of that : prozac too, you know, the one you take to take the edgess off your : extremely sharp mind. I have _rarely_ been the smartest over those around me. When I come to these groups and read, I find many, many people who are more intelligent than I. I seek out intelligent people, just as I seek out creative people. I don't think there is anything wrong with that, and I do understand quite clearly that the criterion are my own definitions. I have been on anti-depressants in my life, but I don't see how that has anything to do with my post. It looks like you are just using it as show of negative feelings towards me. : : [Barble garble betty marble] : [Burpl gurpl rhubarb crumble] : : You are so full of crap. Obviously, you were paraphrasing there, but I would appreciate some elaboration. What was crap? Am I to take the following as the crap you mention? : Before I go on, let me thank you for the following lines. That little : window into yourself, that gives us in your own words a view where : you come from and gives us a very good handle on what is wrong with : you. : : > ...understanding of the more classical approaches. My hunch for the : > third question is that the desire for classical logic is very similar : > to the desire to believe in monotheism, that there is this insecurity : > in many people concerning the absolute and reality, and questions : : So that's it, isn't it? Everybody who has a preference for the more : classical (do you fully understand the word classical?)approaches, : that is virtually everyone who has an understanding of these things, : IS INSECURE. And not only that, more than half the wolrd's population : which incidentally happens to hold monotheistic religious beliefs, are : INSECURE. As opposed to you: open minded, enlightened polytheistic : genial galathaea... You've got it all figured out, haven't you? You : just forgot to tell us who your favourite Beatle is, and wheter you.. : no forget it nobody really want's to know that. I'm insecure. Its a personal observation that I have found myself drawn to starkly true and false, good and evil philosophies in the past. I was, for example, an objectivist libertarian in my teens, and I believe on recall that much of the attraction I felt towards such a belief system was due to insecurities I had about choosing the right path. Even today, as I mention, I am attracted to realist interpretations of quantum mechanics, and I even acknowledge that I think the attraction has the same source. That's why it was my _hunch_, and I did mention that I was hoping that point would stimulate discussion as to why. I'm polytheistic only as a metaphor, as an acknowledged mythological framework for honoring the universe. Ringo. : > of truth and falseness, like questions of good and evil, should be : > knowable to some form of completion. : : And objective reality, we insecure worms also have this feeling that : there should be something like objective reality. But you of course : are above that, miss Heyting for you, reality is an aspect of : existence which easily adapts to your needs of comfort. Heyting forms say nothing against objectivity of reality. They describe the logic we model reality with. : >.........................................I know that is certainly a : > controversial opinion in some circles, but I wanted to throw it out : > there to give a better view of where I come from and perhaps to : > stimulate discussion on the third topic. Paradoxically, I have found : > that although I am drawn to multivalent logics and find monotheism : > anathema (ironic considering the words origins), I also enjoy studying : : What a refined sense of humour we've got, haven't we? Although that is a lovely compliment, I believe you intend that in a negative fashion as well. : > the realist interpretations of quantum mechanics, which I suspect are : > guided by very similar desires, so perhaps I just like to be : > contrary... : : Oh ooooh, what a brave little piece of self criticism. Didn't we blush If you want more blatant self-criticism, see my postings to the poetry newsgroups. You do seem to want very much to think about me in negative ways. : > In philosophy, epistemics, cognition, linguistics, formal models, : > computation, and the possible structure of our world, there are : > unifying principles of expressability that I find more and more : > useful. It confuses me that I find them still poorly understood in : > crowds where, at least to me, it has always seemed they should be more : > well known. : : Hey missie, cut the crap. Why don't you just plainly state what you : mean? That you are by way the most intelligent being on earth. : : Jeeeezes I don't mean that at all. If I wanted to believe myself the most intelligent being on earth, I'd hide away and not post to the newsgroups. I wouldn't open myself up to rational critique, because I make a lot of mistakes and would have to ignore comments on them and build avoidance reactions and all that bs. Instead, in your post, you... : > : : You are so full of crap. : (Hey! didn't I say that already?) : : > The newsgroups I have broken this up into are divided into 3 groups of : > 5. They cover the major newgroups discussing issues mentioned, and : > whom collectively : [list of victimized newsgroups] : : By the way, you forgot to post to alt.cooking.bouillabaisse, you might : help settle the century long debate about wether one should add : croutons and grated cheese once the soup is served. How about a : Heyting solution in the form of suggesting that people feed solely on : fermented scraps of raw macrobiotic lettuce? ...just showed the usenetiverse watching this particular corner that you are willing to post a negative reply based on only some psychologically negative picture you have built up about me, without reference once to anything relevant about Heyting algebras. All I was trying to do was to be clever. A little creative. I tried to describe something I enjoy and feel has an earnest place in education. I'm trying to give the idea a little play among the various communities it intersects. Don't attack me for making you have to ask questions about what something means. Look it up. Get back to me when you think about some more. ===-=-=-=-=- === Subject: Re: :: towards a constructive education :: (news server friendly) : Galathaea, from what I gather from other postings of yours, you are : probably brighter than those around you and those you grew up with. : You probably are acustomed to receiving praise and approval at the : merest trembling of your lips. Before you go on reading I sugest you : eat some chocolate burn some inciense and say a prayer to your : favorite god of no hard feelings. And yes, maybe take some of that : prozac too, you know, the one you take to take the edgess off your : extremely sharp mind. I have _rarely_ been the smartest over those around me. When I come to well, you certainly spare no effort to convey the idea. For those who these groups and read, I find many, many people who are more intelligent than I. I seek out intelligent people, just as I seek out creative people. I don't think there is anything wrong with that, and I do understand quite clearly that the criterion are my own definitions. I have been on anti-depressants in my life, but I don't see how that has anything to do with my post. It doesn't in terms of causality. But it shows. Your post is a very strong indication that you have big difficulties in coping with reality. It looks like you are just using it as show of negative feelings towards me. I wasn't using anything. I didn't know, I just made it up along the way. Jeezes Kryst, I hope you don't start thinking I'm someone you know personally, this is getting scary. : : [Barble garble betty marble] : [Burpl gurpl rhubarb crumble] : : You are so full of crap. Obviously, you were paraphrasing there, but I would appreciate some elaboration. What was crap? Am I to take the following as the crap you mention? I'll say something about discussing this at the end. : Before I go on, let me thank you for the following lines. That little : window into yourself, that gives us in your own words a view where : you come from and gives us a very good handle on what is wrong with : you. : : > ...understanding of the more classical approaches. My hunch for the : > third question is that the desire for classical logic is very similar : > to the desire to believe in monotheism, that there is this insecurity : > in many people concerning the absolute and reality, and questions : : So that's it, isn't it? Everybody who has a preference for the more : classical (do you fully understand the word classical?)approaches, : that is virtually everyone who has an understanding of these things, : IS INSECURE. And not only that, more than half the wolrd's population : which incidentally happens to hold monotheistic religious beliefs, are : INSECURE. As opposed to you: open minded, enlightened polytheistic : genial galathaea... You've got it all figured out, haven't you? You : just forgot to tell us who your favourite Beatle is, and wheter you.. : no forget it nobody really want's to know that. I'm insecure. Its a personal observation that I have found myself drawn to Nobody here gives a about that. Nobody wants to know. But most important, that is no excuse for handing out your idiotic judgement on the greater part of humanity. It is no excuse for being a blatantly imbecile arrogant bigot. : And objective reality, we insecure worms also have this feeling that : there should be something like objective reality. But you of course : are above that, miss Heyting for you, reality is an aspect of : existence which easily adapts to your needs of comfort. Heyting forms say nothing against objectivity of reality. They describe the logic we model reality with. It seems however they help a good deal in rationalising its denial. How very comfty it must be not having to choose between I succeeded and I failed. : > stimulate discussion on the third topic. Paradoxically, I have found : > that although I am drawn to multivalent logics and find monotheism : > anathema (ironic considering the words origins), I also enjoy studying : : What a refined sense of humour we've got, haven't we? Although that is a lovely compliment, I believe you intend that in a negative fashion as well. And we're quite perceptive too, arent we? : > the realist interpretations of quantum mechanics, which I suspect are : > guided by very similar desires, so perhaps I just like to be : > contrary... : : Oh ooooh, what a brave little piece of self criticism. Didn't we blush If you want more blatant self-criticism, see my postings to the poetry newsgroups. You do seem to want very much to think about me in negative ways. You call that blatant? Holy subtle candlesticks batman! No I don't want any more of it at all! And no, I don't want to think of you in any particular way, it's just you rub it on one's face with such vehemence. : Hey missie, cut the crap. Why don't you just plainly state what you : mean? That you are by way the most intelligent being on earth. : : Jeeeezes I don't mean that at all. If I wanted to believe myself the most intelligent being on earth, I'd hide away and not post to the newsgroups. I wouldn't open myself up to rational critique, because I make a lot of mistakes and would have to ignore comments on them and build avoidance As I said at the beginning, you do make a big effort to convey a wrong impression. reactions and all that bs. Instead, in your post, you... : > : : You are so full of crap. : (Hey! didn't I say that already?) : : > The newsgroups I have broken this up into are divided into 3 groups of : > 5. They cover the major newgroups discussing issues mentioned, and : > whom collectively : [list of victimized newsgroups] : : By the way, you forgot to post to alt.cooking.bouillabaisse, you might : help settle the century long debate about wether one should add : croutons and grated cheese once the soup is served. How about a : Heyting solution in the form of suggesting that people feed solely on : fermented scraps of raw macrobiotic lettuce? ...just showed the usenetiverse watching this particular corner that you are willing to post a negative reply based on only some psychologically negative picture you have built up about me, without reference once to anything relevant about Heyting algebras. All I was trying to do was to be clever. A little creative. I tried to describe something I enjoy and feel has an earnest place in education. I'm trying to give the idea a little play among the various communities it intersects. Don't attack me for making you have to ask questions about what something means. Look it up. Get back to me when you think about some more. * So you want to talk about things relevant to Heyting forms and are cashing in a lot of hating? (should I post that to the poetry groups ?). Let's see, you told more than half of humanity that they hold their religious belief because they are insecure. The same thing for all those unsuspecting cowards that hold to the more classical views. Then you told the crowds that you were confused because they hadn't caught up with you. It seems to me you should have managed to antagonize everybody by now, but let's asume just for the heck of it, that there are still some good willed people left after that. Now, which of the titles you apply to yourself: prankster, fablist, magician, liar do you think is the most inviting one to a person wanting to have a serious conversation about formal systems and education? moron. === Subject: Re: bundle help Ok, i've understood. But, what do you say about: M_{i} will denote a differentiable manyfold; If M is a differentiable manyfold then V(M) will denote the additive category of C^{infty} real vector bundles over M. Let M = M_{1} x ... x M_{k} be the product manyfold of M_{i}. If xi^{i} in V(M_{i}), then xi^{1} otimes ... otimes xi^{k} is a bundle over M whose fiber at x = (x_{1},...,x_{k}) is xi^{1}_{x_{1}}otimes ... otimes xi^{k}_{x_{k}}. If f ^{i} in C^{infty}(M_{i}, xi^{i}), we define f^{1}otimes ... otimes f^{k}, a function on M, by (f^{1}otimes...otimes f^{k})(x) = = f^{1}(x_{1})otimes...otimes f^{k}(x_{k}). The bundle structure of xi^{1}otimes ... otimes xi^{k} is characterized by the property that f^{1}otimes ... otimes f^{k} in C^{infty}(M, xi^{1} otimes ... otimes xi^{k}) wherever f ^ {i} in C^{infty}(M_{i}, xi^{i}). I don't understand the reason for which ``the bundle structure of xi^{1} otimes... otimes xi^{k} is characterized by the property that f ^{1} otimes... otimes f^{k} in C^{ infty}(M, xi^{1}otimes ... otimes xi^{k}) wherever f^{i} in C^{infty}(M_{i}, xi^{i}) ''. May you to help me, please? Tern ternnret@yahoo.it === Subject: Re: 1 + 1/3 + 1/5 + ... + 1/(2*N - 1) I'm interested in the value of the sum 1 + 1/3 + 1/5 + ... + 1/(2*N - 1) I can bound this sum above and below using integrals, but I wonder if there are better ways to approximate its value for large N. If the partial sum of the Harmonic series is H(n), then the sum above is H(2N) - H(N)/2. Using the asymptotic expansion I gave in get the asymptotic expansion log(N)/2 + log(2) + gamma/2 1 7 31 127 511 + ----- - ------- + -------- - -------- + ---------- 48N^2 1920N^4 16128N^6 61440N^8 135168N^10 Rob Johnson take out the trash before replying === Subject: Re: JSH: Don't talk to me I want several people to understand without a doubt that I'm not interested in them trying to talk to me. Yeah, you say this every once in while, but then every once in a while it turns out it's not true. (Which should be no surprise, considering how few of the things you say actually _are_ true...) I'll make the same request that I made to David Ullrich some time ago: Reply in my threads if you must, just do NOT address me. You can even reply to my posts, but do NOT ask me questions, talk in any familiar way, or in any way act as if you wish me to reply or are addressing me. Uh, this is not what most people would call a request. That message is to David Ullrich, since the stupid is still at it, Virgil, and Nora Baron. That's it. I just want those three people to OFF!!!!!!!!! Good of you to clarify that. A little while ago, when you said You stupid head!!! What the is wrong with you Ullrich? No matter how many ing times I tell you to off, you keep replying to me!!! What the is your problem you head? You Ullrich are a stupid piece of dumb who refuses to get the message when someone does NOT want to talk to you, you stupid ing ty asshole. You are an ASSHOLE Ullrich!!! Now why don't you take your dumb ass stupid self somewhere to GET A ING CLUE and QUIT ING REPLYING TO ME AS IF I EVER WANT TO TALK TO YOU!!!!!!!!!!!!! OFF!!!! Can't you get it through your stupid head? OFF!!!!!!!!!!!!!!!!! I thought maybe that meant you wanted me to off, but I wasn't sure. Now if only you could be this clear when you were writing about math. James Harris David C. Ullrich === Subject: Re: JSH: Don't talk to me You can even reply to my posts, but do NOT ask me questions, talk in any familiar way, or in any way act as if you wish me to reply or are addressing me. That message is to David Ullrich, since the stupid **** is still at it, Virgil, and Nora Baron. That's it. I just want those three people to **** OFF!!!!!!!!! Unfortunately for you, Mr. Narcissistically Disordered Jerk, what *you* want is unimportant. Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Don't talk to me Discussion, linux) > Back in the 70's when it came out that Nixon had a secret list of > journalists he disliked, it became a badge of honor to be on the list. > In fact, many journalists NOT on the list were embarassed to have been > omitted. I don't find your analogy very convincing. Posts from the cult of JSH detractors are often sad documents that should prove embarrassing to their authors when and if their heads are ever removed from their posteriors. And I don't think many sci.math posters are embarrassed at being ommited from JSH's list, do you? What, you don't find mocking responses to JSH comparable to investigative reporting making a corrupt presidency uncomfortable? Golly, the analogy worked for me. Weirdo. I thought it relevant to inform that I notified the FBI a couple of months ago about some of the math issues I've brought up here. -- James S. Harris gives Special Agent Fox a new assignment. === Subject: Re: JSH: Don't talk to me > Back in the 70's when it came out that Nixon had a secret list of > journalists he disliked, it became a badge of honor to be on the list. > In fact, many journalists NOT on the list were embarassed to have been > omitted. > I don't find your analogy very convincing. Posts from the cult of JSH > detractors are often sad documents that should prove embarrassing to their > authors when and if their heads are ever removed from their posteriors. And > I don't think many sci.math posters are embarrassed at being ommited from > JSH's list, do you? What, you don't find mocking responses to JSH comparable to investigative reporting making a corrupt presidency uncomfortable? Golly, the analogy worked for me. Weirdo. I don't necessarily mind mocking responses. James Harris === Subject: Re: JSH: Don't talk to me I don't necessarily mind mocking responses. The irony of this statement reminds me of my favourite Jack Handey quote: Broken promises don't upset me. I just think, why did they believe me? Doug === Subject: Re: JSH: Don't talk to me === Subject: JSH: Don't talk to me I want several people to understand without a doubt that I'm not interested in them trying to talk to me. Is that why you never reply to my replies? I'll make the same request that I made to David Ullrich some time ago: Reply in my threads if you must, just do NOT address me. Ok, Mr. Harris, I will refrain from addressing you. Does referring to you as JSH count as a form of address? You can even reply to my posts, but do NOT ask me questions, Why? talk in any familiar way, Ok, bro', no problemo. or in any way act as if you wish me to reply or are addressing me. The whole point is to make you mad. I score points when you get so mad you stop ignoring me. That's how I got promoted to the Ace of Clubs, I got both you and T__o__m__P__o__t__t__e__r to foam at the mouth on the same day. It's part of _my_ social experiment. Now get back in the maze and find the cheese. That message is to David Ullrich, since the stupid is still at it, Virgil, and Nora Baron. That's it. I just want those three people to OFF!!!!!!!!! So it's ok for me to continue calling you a disloyal American who is a disgrace to the armed services of which you claim to be a veteran? James Harris -- === Subject: Re: JSH: Don't talk to me I want several people to understand without a doubt that I'm not interested in them trying to talk to me. I'll make the same request that I made to David Ullrich some time ago: Reply in my threads if you must, just do NOT address me. You can even reply to my posts, but do NOT ask me questions, talk in any familiar way, or in any way act as if you wish me to reply or are addressing me. That message is to David Ullrich, since the stupid is still at it, Virgil, and Nora Baron. That's it. I just want those three people to OFF!!!!!!!!! === Subject: Re: JSH: Don't talk to me by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i17DAiR10307; I want several people to understand without a doubt that I'm not interested in them trying to talk to me. I'll make the same request that I made to David Ullrich some time ago: Reply in my threads if you must, just do NOT address me. You can even reply to my posts, but do NOT ask me questions, talk in any familiar way, or in any way act as if you wish me to reply or are addressing me. That message is to David Ullrich, since the stupid is still at it, Virgil, and Nora Baron. That's it. I just want those three people to OFF!!!!!!!!! James Harris Jim, my son, what's troubling you? Let me know. I'm waiting. God === Subject: Re: there is no such thing as infinity by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i17DAjs10322; I've thought really hard about this one and came to the conclusion that there is no scientific evidence of infinity existing. The highest number that anyone has ever measured to according to Isaac Asimov in his book Science and Human Thought is only about 5.0 x 10^48. No Craig , This sounds interesting. Is there a corresponding number smaller than 1 that could indicate ZERO? However, infinity and zero are technicalities. Zero could also state the absence of something (e.g not even one)and the impossibility of reaching it in ones lifetime (REFER TO SPIRALS(appart from their logarithms which is something special) they have an assymptotic point which is never rached ,and never reach a an end point-infinity http://www.stefanides.gr/why_logatithm.htm ) Similarly this impossibility apllies to infinity. If ,however there is a wall around our world this could change the ideas of infinity and zero It has currently reached about 2.0 x 10^18. Just as Einstein proved that there is no aether What about Gravitationl Fields,could they be considered as such? The Classical Greek Phiolosophers used this term, but the 4 elements an the 5 Platonic Bodies implied uncodified completely yet Mathematics of the world's creation (especially that of the Platonic Timaeus. http://www.stefanides.gr/plato.htm Panagiotis Stefanides http://www.stefanides.gr , I am convinced that I will prove that there is no infinity and then write a book or two. Dr. Ben Zona === Subject: polynomials that produce only primes There was a recent post asking if all numbers of the form n^2 + n + 41 are prime. This set me thinking ... No polynomial ax^n + bx^(n-1) ... + fx + g can produce only primes, because if we set x=k*g (for any k) then its divisible by g. The polynomial can be zero for only at most n of these cases. For all other values of k, the polynomial must be a non-zero number divisible by g. The only cases this doesn't work for are g=0 and g=1. The g=0 case is also trivial, as x is a factor of the polynomial. So the only case left where its not trivial is g=1. Now, this bugs me. Firstly, its somewhat annoying that there is a single exception, and one which appears somewhat arbitrary (so what if the constant term in the polynomial is a 1?). Secondly, I cannot see a simple method to remove this last remaining case. Any thoughts? Peter Webb PS I know there aren't any polynomials that produce only primes - I just want to know if the simple proof above can be completed. === Subject: Re: polynomials that produce only primes There was a recent post asking if all numbers of the form n^2 + n + 41 are prime. This set me thinking ... No polynomial ax^n + bx^(n-1) ... + fx + g can produce only primes, because if we set x=k*g (for any k) then its divisible by g. The polynomial can be zero for only at most n of these cases. For all other values of k, the polynomial must be a non-zero number divisible by g. The only cases this doesn't work for are g=0 and g=1. The g=0 case is also trivial, as x is a factor of the polynomial. So the only case left where its not trivial is g=1. Now, this bugs me. The case g = -1 doesn't work either, but .. A polynomial of degree n > 0 can only have _any_ fixed value at most n times, because if f(x) = a_n.x^n + .. + a_1.x + a_0 equals b then f(x) - b, which is just a_n.x^n + .. + a_1.x + (a_0 - b) would be zero for all those values! In particular your polynomial, for any given value of g, can only equal 1 for at most n values of x, integers or otherwise, and -1 for at most another n values of x. If you then choose any integer value, say h, which is not one of those 2n values then plugging x = x' + h into your polynomial you end up, after expanding each term, with a polynomial in x' with new integer coefficients with constant coefficient equal to f(h), which is the original polynomial with x = h and is thus not +/-1. If you don't see this immediately, try it with a quadratic and watch how the constant coefficient of the new polynomial gets built up from the constant 'tail end' of each expanded term. Cheers --------------------------------------------------------------------------- John R Ramsden (jr@adslate.com) --------------------------------------------------------------------------- Eternity is a long time, especially towards the end. Woody Allen === Subject: Re: Question for logarithm experts I am looking for a step-by-step method (proof) on the solution of this particular equation - (Solve for x) and equations like it: 7^X = 4*X. Using a graphing calculator it's easy to show that there is *no* solution for this equation. I plugged both into mine, and then zoomed into where they're at their closest, and it's very obvious that there's no solution. This seems to be a relatively simple problem but I've been tortured by it since the eleventh grade, 28 years ago. It's probably because there's no solution. ;) I have heard the following remarks about this problem: it's an unfair question. Its an equation that's not an equation. It is a single equation with two variables. The second is right, it's not an equation. There is no value of X that can make this true. Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people think, Mr. Feynman? === Subject: Re: next Farey fraction Astonishly simple method: http://homepages.paradise.net.nz/quentin/fareyfraction.htm is it O(log(denom)) fast? Please, do *not* top-post. Even if this is tolerated here far more than e.g. in many technical groups, it is a waste of banwidth, an exercise in unclear communication, an unrespectful behaviour. Said this, I'm currently offline, but I will check that page ASAP. However please read below for an IMHO interesting observation: >(1) b|ab'+1 > *complete rubbish*! You just have to solve (1) for b', thus > b' = -a^{phi(b)-1}%b, > where m%n is the remainder of the division of m by n and phi is > Euler's totient function. While I didn't add anything to this thread after that post, I thought some more about the problem and an interesting thing that comes out is that if f(a/b) is the function that gives the Farey fraction following a/b in F_b (with a and b coprime!) and if f(a/b) = a'/b' = (a-da)/(b-db) then you can calculate f(a'/b') = (a'-da')/(b'-db') in a quite effortless way, i.e. you don't have to solve (1) above again, but da' and db' are simply... ;-) Michele Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === Subject: Re: Zero Point Energy Farce Claptrap === Subject: Re: there is no such thing as infinity I've thought really hard about this one and came to the conclusion that there is no scientific evidence of infinity existing. The highest number that anyone has ever measured to according to Isaac Asimov in his book Science and Human Thought is only about 5.0 x 10^48. No one has ever gotten past that number. Doesn't this sound weird? You're right. There is no scientific evidence of an infinity of real objects. However, that's science, not math. In math there is the Axiom of Infinity, which assumes the existence of the set of natural numbers. This axiom does not claim to have physical support, nor does it need any, because it is NOT a claim about physicality. This is not to say that infinity does not come up in science, because without it we could not have calculus. Patrick === Subject: Re: there is no such thing as infinity I've thought really hard about this one and came to the conclusion that there is no scientific evidence of infinity existing. Wow! how did I miss this thread! A hypothesis which disproves itself. that there is no scientific evidence of infinity existing. OK, How about infinite stupidity? I was going to suggest Archie Poo, but your little program is even better. All the scientific proof we need. An undefined Magic variable M. And I thought magic numbers in code were a Bad Thing. Printed out 2 x 10^18 lines so far, has it? Proof of infinity, for all the world to see. Dr.(!!!) Ben Zona, huh. I must be missing the reference. Cheers, Tony. === Subject: Re: there is no such thing as infinity > I've thought really hard about this one and came to the conclusion > that there is no scientific evidence of infinity existing. Wow! how did I miss this thread! A hypothesis which disproves itself. > that there is no scientific evidence of infinity existing. OK, How about infinite stupidity? I was going to suggest Archie Poo, but your little program is even better. All the scientific proof we need. An undefined Magic variable M. And I thought magic numbers in code were a Bad Thing. Printed out 2 x 10^18 lines so far, has it? Proof of infinity, for all the world to see. Dr.(!!!) Ben Zona, huh. I must be missing the reference. Ben zona = son of a whore in Yiddish. Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: there is no such thing as infinity What, it's April 1st already? No, just a cross-post from sci.physics. This was about a valuable as the other cross-posts from there... === Subject: Re: there is no such thing as infinity Twaddle Franz === Subject: Help Needed For a scalar random variable X, we can maximize or minimize its variance which is a number. For a vector random variable Y, we have a matrix cov(Y). Then what do we maximize (or minimize)? Any idea or reference is appreciated. Mike === Subject: Re: When 0 divided by 0 isn't infinity... (was: Re: I lost an account...) 0/0 = 1 Any division problem asks us to subtract the denominator from the numerator, to the limit of the numerator, and to state how many times we may do so as our answer, including any fractional parts. Thus 0 - 0 reaches the limit of zero the very first time that the required process is invoked, and thus our answer is 1, a process 1. Note that as a matter if strict logic that multiplication is not exactly the inverse of division, for multiplication tells us to add an amount to itself a given number of times and to state the SUM as our answer, not the NUMBER OF TIMES that we add. In contrast, with division, we state the number of times that we may subtract one number from another as our answer. Thus, the two processes, multiplication and division, are not truly exact inverses of each other as a matter of process, and the supposed proofs that x/0 should therefore be left undefined has not been truly thought out. Very Respectfully, Ray === Subject: Re: When 0 divided by 0 isn't infinity... (was: Re: I lost an account...) === Subject: Re: When 0 divided by 0 isn't infinity... (was: Re: I lost an account...) 0/0 = 1 Any division problem asks us to subtract the denominator from the numerator, to the limit of the numerator, and to state how many times we may do so as our answer, including any fractional parts. Thus 0 - 0 reaches the limit of zero the very first time that the required process is invoked, No, it is at the limit of zero BEFORE the process is invoked and thus, the count is 0. # start 0_0.py import sys numerator = long(sys.argv[1]) denominator = long(sys.argv[2]) quotient = 0 remainder = 0 while (numerator>0) and (numerator>=denominator): numerator = numerator - denominator quotient = quotient + 1 remainder = numerator print quotient:,quotient,remainder:,remainder c:python23user>python 0_0.py 8 2 quotient: 4 remainder: 0 c:python23user>python 0_0.py 9 2 quotient: 4 remainder: 1 c:python23user>python 0_0.py 9 3 quotient: 3 remainder: 0 c:python23user>python 0_0.py 13 4 quotient: 3 remainder: 1 c:python23user>python 0_0.py 0 0 quotient: 0 remainder: 0 # end 0_0.py and thus our answer is 1, a process 1. Sorry, the answer is 0. Note that as a matter if strict logic that multiplication is not exactly the inverse of division, for multiplication tells us to add an amount to itself a given number of times and to state the SUM as our answer, not the NUMBER OF TIMES that we add. In contrast, with division, we state the number of times that we may subtract one number from another as our answer. Thus, the two processes, multiplication and division, are not truly exact inverses of each other as a matter of process, and the supposed proofs that x/0 should therefore be left undefined has not been truly thought out. Very Respectfully, Ray -- === Subject: Re: When 0 divided by 0 isn't infinity... (was: Re: I lost an account...) >I thusly replied: > and I thusly replied: >0 divided by 0 could be just about anything u want it to be. > Exactly. > wrong. > you lot are talking about limits. The O/P was talking about zeros. > Zero (or nought) amount of anything (real) can have no representing > constituent parts of anything else thus 0/0=0 > Look at it from a practical standpoint. > So why do calculators not give the answer as zero? I'm sure it would be > simple enough to be coded in... > Something wrong with my reasoning was there? I don't really care > how many calculators get it wrong. the important thing is to get > it right yourself and not to perpetuate other peoples errors. >Why did you take my posts as a personal jibe? > It is the usenet, context means nothing and everything is a personal > attack? >Quote this so called attack pls >roy > Just a guess mind you, not scientific or mathmatical proof. > 'Ratz > Follow thread? Response was to the Revs post. Never said anyone > attacked, posed hypothesis about normal usenet protocol. Your response > may actually be the proof of hypothesis. >No, you clearly implied I attacked the Reverend. You also stupidly >implied that my reply to the reverand was out of context when in >fact it was no such thing as I clearly was dealing with the matter >of 0/0. You have a serious comprehension/reasoning problem. >Nothing personal. >roy > 'Ratz > Didn't your Mama teach you not to talk to strangers? So you ignored her as usual and said LMAO You really think I gave a what you said? You give yourself way too much credit. I responded strictly to the Revs posts and his question. The fact you were in there is happenstance. But since you want to turn this into some pissing match feel free. No, you are the one that came in with the smart ass comments. I only attended to the reverends original point. Does not alter the facts in any way shape or form. Of course not, it is clear for any dummy to see from the usenet record BTW did you see that my original post had a question mark at the end? Which means it was a question not a statement. You mean it was a leading question. Nothing is ever as it seems. But then obviously a reading comprehension expert like yourself would never have missed something as obvious as that? Would you? Correct, there are agendas hidden in all communication. roy 'Ratz === Subject: Re: When 0 divided by 0 isn't infinity... (was: Re: I lost an account...) double secret probation because: >I thusly replied: > and I thusly replied: >0 divided by 0 could be just about anything u want it to be. > Exactly. > wrong. > you lot are talking about limits. The O/P was talking about zeros. > Zero (or nought) amount of anything (real) can have no representing > constituent parts of anything else thus 0/0=0 > Look at it from a practical standpoint. > So why do calculators not give the answer as zero? I'm sure it would be > simple enough to be coded in... > Something wrong with my reasoning was there? I don't really care > how many calculators get it wrong. the important thing is to get > it right yourself and not to perpetuate other peoples errors. >Why did you take my posts as a personal jibe? > It is the usenet, context means nothing and everything is a personal > attack? >Quote this so called attack pls >roy > Just a guess mind you, not scientific or mathmatical proof. > 'Ratz > Follow thread? Response was to the Revs post. Never said anyone > attacked, posed hypothesis about normal usenet protocol. Your response > may actually be the proof of hypothesis. >No, you clearly implied I attacked the Reverend. You also stupidly >implied that my reply to the reverand was out of context when in >fact it was no such thing as I clearly was dealing with the matter >of 0/0. You have a serious comprehension/reasoning problem. >Nothing personal. >roy > 'Ratz > Didn't your Mama teach you not to talk to strangers? So you ignored her as usual and said > LMAO > You really think I gave a what you said? You give yourself way > too much credit. I responded strictly to the Revs posts and his > question. The fact you were in there is happenstance. But since you > want to turn this into some pissing match feel free. No, you are the one that came in with the smart ass comments. I only attended to the reverends original point. Ahh so now they are smart ass comments. What happened to: Quote this so called attack pls roy So you now admit you see what was said in the correct light and I was poking fun at the Rev. So this is all about you and your poor wounded ego. What a ing wanker. You proved my original hypothesis. Children like you with tissue paper skins think everything is always about them. It has to be, you are way to important and what you have to say is the only important words. Everything in every thread you are in is about you and you alone. Nice little ego problem you have. Go back and look, rev asked question, I posed possible answer. You feeling slighted and your massive ego telling you that you are the point of my smartass answer must force me to make amends. Only one minor little problem. I never gave one thought about what you said. Only what the rev said. So stuff that massively over inflated ego back in that massively swelled head and shove them both up your massively inflated ass. And thank you for proving the hypothesis that: It is the usenet, context means nothing and everything is a personal attack? > Does not alter > the facts in any way shape or form. Of course not, it is clear for any dummy to see from the usenet record Yes and you still cannot understand you had nothing to do with what I said? Ego still in the way? So how do you get anything done with your ego screaming MEMEMEMEMEMEMEMEMEME all the time? >BTW did you see that my original > post had a question mark at the end? Which means it was a question not > a statement. You mean it was a leading question. Nothing is ever as it seems. Oh and a psychic too. So do you have a phony Jamaican accent too? > But then obviously a reading comprehension expert like > yourself would never have missed something as obvious as that? Would > you? Correct, there are agendas hidden in all communication. roy And a loon paranoid. So it is a conspiracy and this is all a plan by the CABAL to get you. > 'Ratz === Subject: Plotting a 7 vertice graph in which every vertex has degree 4 Math folks, I am trying to determine the number of isomorphism classes of simple 7-vertex graphs in which every vertex has degree 4. To do this, I am trying to show the graphs for these, and print out tables of table vertex adjacencies. I see the ShowGraph[CompleteGraph[7]]; command, and also the TableForm[ToAdjacencyMatrix[CompleteGraph[7]]] command. Is there a way to work backwards, in other words to create a list or table to represent a graph, and then to show it? In that way, I could design and display 7-vertex graphs with 4-degree vertices. Diana ===================================================== God made the integers, all else is the work of man. L. Kronecker, Jahresber. DMV 2, S. 19. === Subject: Minkowski sums Has much is known about results of Minkowski sums on pairs of simple shapes (polyhedra, ellipsoids, cylinders etc) in 3d? I'm particularly curious about cases where the results can be expressed in terms of other simple shapes. For example line segment + sphere gives a 'capsule', made from 2 hemispheres and a cylinder. And if noone's interested in that general question, does anyone have any idea of the result of triangle + cylinder? thanks all! glen. === Subject: hello....easy analysis problem... f: (-1,1) -> R let f is continuous on 0 let any x in (-1,1) , f(x) = f(x^2) show that any x in (-1,1) , f(x) = f(0) ----------------------- i think .......it's very trivial..... any a, b in (-1,1) , f(a)=f(a^2)=f(a^4)............=f(0) f(b)=f(b^2)=f(b^4)............=f(0) obviously , it's constant but i don't write for prove. help...me please...... === Subject: Re: hello....easy analysis problem... f: (-1,1) -> R let f is continuous on 0 let any x in (-1,1) , f(x) = f(x^2) show that any x in (-1,1) , f(x) = f(0) ----------------------- i think .......it's very trivial..... Yes. Take any x in (-1,1). Then lim_n x^(2^n) = 0, and so, since f is continuous at 0, lim_n f(x^(2^n)) = f(0). But f(x) = f(x^2) = f(x^4) = f(x^8) = ... So, the sequence (f(x^(2^n)))_n is constant. Since it converges to f(0), you have f(x^(2^n)) = f(0) for every non-negative n. Taking n = 0, you get f(x) = f(0). Best regards, Jose Carlos Santos === Subject: Lagragian, Quantum Mechanics,Path of least action Special thanks to Nova Quantum Mechanics, Wavefunctions are deeply connected to finding the path of least action. Action S = int_a^b dt L where L is the lagrangian. I was thinking of roads. Suppose nature is fundamentally discrete at the space-time level, it would have to feel every path. This is the direct result of the Speed-Up Theorem. Take the integral S = int_a^b dt L Suppose we break into minute paths S = int_a^t1 dt L + int_t1^t2 dt L + int_t2^t3 dt L + .... int_tn^b dt L Then, action 'S' has a gaussian distribution. Random variable Y has a gaussian distribution if Y = X1 + X2 + X3 +.... Xn if X1,X2,....Xn all have a uniform or gaussian distribution. Action is always positive. Minimining S is same as taking S = min int_a^t1 dt L + min int_t1^t2 dt L + min int_t2^t3 dt L + min .... int_tn^b dt L Suppose we assume that 'int_ tm^tn dt L' has a gaussian distribution, since it can infestionally be broken down future technically probability of finding 'min int_tm^tn dt L' is erfc( (min int_tm^tn dt L - avg int_tm^tn dt L )/( std int_tm^tn dt L ) ) As tm->tn 'min int_tm^tn dt L' would approach the 'avg int_tm^tn dt L' On the other hand, avg int_tm^tn dt L = int dx1 dx2 dx3 pdf(x1) pdf(x2) pdf(x3) (tn - tm) L ( x1,x2,x3, 0.5(tn + tm) )) var int_tm^tn dt L = int dx1 dx2 dx3 pdf(x1) pdf(x2) pdf(x3) (tn - tm)^2 L^2 (x1,x2,x3, 0.5(tn+tm) ) ) since (tn-tm) -> 0 Probabilities naturally emerge when considering problems like the probability of passing thru path 'x(x1,x2,x3) to y(y1,y2,y3)' in time frame tn to tk interesting thought. I will finish up later -suresh === Subject: Re: Lagragian, Quantum Mechanics,Path of least action operator dx_1 frac{partial}partial x_1} L = L( x_1 + dx_1 , x_2, x_3,t) - L (x_1, x_2, x_3,t) becomes a way to take the lagragian action S between 2 points in space. dt frac{partial}partial dt} L = L( x_1 , x_2, x_3, t + dt) - L (x_1, x_2, x_3,t) becomes a way to take the lagragian action S between 2 points in time. The Lord of the Rain( Suresh __NoJunkMail kumar) Special thanks to Nova Quantum Mechanics, Wavefunctions are deeply connected to finding the path of least action. Action S = int_a^b dt L where L is the lagrangian. I was thinking of roads. Suppose nature is fundamentally discrete at the space-time level, it would have to feel every path. This is the direct result of the Speed-Up Theorem. Take the integral S = int_a^b dt L Suppose we break into minute paths S = int_a^t1 dt L + int_t1^t2 dt L + int_t2^t3 dt L + .... int_tn^b dt L Then, action 'S' has a gaussian distribution. Random variable Y has a gaussian distribution if Y = X1 + X2 + X3 +.... Xn if X1,X2,....Xn all have a uniform or gaussian distribution. Action is always positive. Minimining S is same as taking S = min int_a^t1 dt L + min int_t1^t2 dt L + min int_t2^t3 dt L + min .... int_tn^b dt L Suppose we assume that 'int_ tm^tn dt L' has a gaussian distribution, since it can infestionally be broken down future technically probability of finding 'min int_tm^tn dt L' is erfc( (min int_tm^tn dt L - avg int_tm^tn dt L )/( std int_tm^tn dt L ) ) As tm->tn 'min int_tm^tn dt L' would approach the 'avg int_tm^tn dt L' On the other hand, avg int_tm^tn dt L = int dx1 dx2 dx3 pdf(x1) pdf(x2) pdf(x3) (tn - tm) L ( x1,x2,x3, 0.5(tn + tm) )) var int_tm^tn dt L = int dx1 dx2 dx3 pdf(x1) pdf(x2) pdf(x3) (tn - tm)^2 L^2 (x1,x2,x3, 0.5(tn+tm) ) ) since (tn-tm) -> 0 Probabilities naturally emerge when considering problems like the probability of passing thru path 'x(x1,x2,x3) to y(y1,y2,y3)' in time frame tn to tk interesting thought. I will finish up later -suresh === Subject: Re: Lagragian, Quantum Mechanics,Path of least action err again As tm->tn 'min int_tm^tn dt L' would approach the ' (tm-tn) min_{all x1,x2,x3} L ( 0.5(tm+tn) )' The Lord of the Rain( Suresh __NoJunkMail kumar) Special thanks to Nova Quantum Mechanics, Wavefunctions are deeply connected to finding the path of least action. Action S = int_a^b dt L where L is the lagrangian. I was thinking of roads. Suppose nature is fundamentally discrete at the space-time level, it would have to feel every path. This is the direct result of the Speed-Up Theorem. Take the integral S = int_a^b dt L Suppose we break into minute paths S = int_a^t1 dt L + int_t1^t2 dt L + int_t2^t3 dt L + .... int_tn^b dt L Then, action 'S' has a gaussian distribution. Random variable Y has a gaussian distribution if Y = X1 + X2 + X3 +.... Xn if X1,X2,....Xn all have a uniform or gaussian distribution. Action is always positive. Minimining S is same as taking S = min int_a^t1 dt L + min int_t1^t2 dt L + min int_t2^t3 dt L + min .... int_tn^b dt L Suppose we assume that 'int_ tm^tn dt L' has a gaussian distribution, since it can infestionally be broken down future technically probability of finding 'min int_tm^tn dt L' is erfc( (min int_tm^tn dt L - avg int_tm^tn dt L )/( std int_tm^tn dt L ) ) As tm->tn 'min int_tm^tn dt L' would approach the 'avg int_tm^tn dt L' On the other hand, avg int_tm^tn dt L = int dx1 dx2 dx3 pdf(x1) pdf(x2) pdf(x3) (tn - tm) L ( x1,x2,x3, 0.5(tn + tm) )) var int_tm^tn dt L = int dx1 dx2 dx3 pdf(x1) pdf(x2) pdf(x3) (tn - tm)^2 L^2 (x1,x2,x3, 0.5(tn+tm) ) ) since (tn-tm) -> 0 Probabilities naturally emerge when considering problems like the probability of passing thru path 'x(x1,x2,x3) to y(y1,y2,y3)' in time frame tn to tk interesting thought. I will finish up later -suresh === Subject: Re: Lagragian, Quantum Mechanics,Path of least action err, step probability of finding 'min int_tm^tn dt L' is erfc( (min int_tm^tn dt L - avg int_tm^tn dt L )/( std int_tm^tn dt L ) ) should read probability of finding 'min int_tm^tn dt L' is Phi( (min int_tm^tn dt L - avg int_tm^tn dt L )/( std int_tm^tn dt L ) ) The Lord of the Rain( Suresh __NoJunkMail kumar) Special thanks to Nova Quantum Mechanics, Wavefunctions are deeply connected to finding the path of least action. Action S = int_a^b dt L where L is the lagrangian. I was thinking of roads. Suppose nature is fundamentally discrete at the space-time level, it would have to feel every path. This is the direct result of the Speed-Up Theorem. Take the integral S = int_a^b dt L Suppose we break into minute paths S = int_a^t1 dt L + int_t1^t2 dt L + int_t2^t3 dt L + .... int_tn^b dt L Then, action 'S' has a gaussian distribution. Random variable Y has a gaussian distribution if Y = X1 + X2 + X3 +.... Xn if X1,X2,....Xn all have a uniform or gaussian distribution. Action is always positive. Minimining S is same as taking S = min int_a^t1 dt L + min int_t1^t2 dt L + min int_t2^t3 dt L + min .... int_tn^b dt L Suppose we assume that 'int_ tm^tn dt L' has a gaussian distribution, since it can infestionally be broken down future technically probability of finding 'min int_tm^tn dt L' is erfc( (min int_tm^tn dt L - avg int_tm^tn dt L )/( std int_tm^tn dt L ) ) As tm->tn 'min int_tm^tn dt L' would approach the 'avg int_tm^tn dt L' On the other hand, avg int_tm^tn dt L = int dx1 dx2 dx3 pdf(x1) pdf(x2) pdf(x3) (tn - tm) L ( x1,x2,x3, 0.5(tn + tm) )) var int_tm^tn dt L = int dx1 dx2 dx3 pdf(x1) pdf(x2) pdf(x3) (tn - tm)^2 L^2 (x1,x2,x3, 0.5(tn+tm) ) ) since (tn-tm) -> 0 Probabilities naturally emerge when considering problems like the probability of passing thru path 'x(x1,x2,x3) to y(y1,y2,y3)' in time frame tn to tk interesting thought. I will finish up later -suresh === Subject: Re: Bayesian Statistics Book > I am taking a graduate class in Bayesian statistics. The textbook they use > is The Baayesian Choice by Christian P. Robert, second edition. > I'm a more intuitive learner and am having problems conceptualizing what the > book is saying. The book is very formal and does not give enough real world > examples for me to understand. I'm very able quantitatively (quantitive > SAT/GMAT/GRE scores 99+ percentile) but I do not learn well with > theory-heavy textbooks. I've just started reading Data Analysis: A Baysian Tutorial by D.S. Sivia. It came recommended by a good, smart friend. It seems conceptual, but clear. Not a big book, but it conveys the ideas well with nice explanation...so far. You might also want to check out the online book by Jaynes on probability and statistics. He was a physicist who looked deeply into the problems, definitions, and approaches that go into the field. It's free and interesting reading (from the little I've read so far). -- My views are my own. I've read both and agree that they could be helpful to the OP. === Subject: Re: Bayesian Statistics Book I am taking a graduate class in Bayesian statistics. The textbook they use is The Baayesian Choice by Christian P. Robert, second edition. Probability and Measurement Uncertainty in Physics - a Bayesian Primer Author: G. D'Agostini available at: http://lanl.arxiv.org/abs/hep-ph/9512295 John Bailey http://home.rochester.rr.com/jbxroads/mailto.html === Subject: Re: Is this quotientan integer? >for k=1,2,3 and m,n in N*, mset A(k,m,n)=((k+2)m)!*((k+2)n)! / [m!*n!*(km+n)!*(kn+m)!] >is this one an integer? The answer seems to be: yes, if k=3, and otherwise no (for some m,n). [mega-snip] There is surely a better way to prove the A(3,m,n) to be integral! One has to wonder whether either the proposers or some of the participants the problem in http://www.kalva.demon.co.uk/usa/usa75.html suggests such a possibility, however remote :-) baloglouAToswego.edu =============================================================== === Subject: Measure theory / Rudin 1.5 (b) In order to show that the set {x in X; (f_n(x)) has a finite limit when n->+oo} is measurable (where (f_n) is a sequence of measurable functions with values in R), can I say that this set equals: limsup(limsup(abs(f_n-f_m),n),m)^(-1)({0}) ? (using Cauchy's criterion) Thks -- Julien Santini === Subject: Re: D.E. problem I'm having troube solving this D.E. problem: y'' + y = 0. y'' = -y Let v = y' and v*(dv/dy) = y'' v*(dv/dy) = -y v*dv = -y*dy (1/2)*v^2 = -(1/2)*y^2 + C v^2 = -y^2 + C and I'm stuck. I need to get rid of the v^2. Is there a way without doing a square root? your method leads you to vv + yy = cc. take y = csin(x) or y = ccos(x) it probably would have been better to guess a function (or functions) which added to its second derivative yields zero.