mm-3229 === Subject: Re: Axiom of Foundation (absymally stupid question) permission for an emailed response. > Doesn't seem to like the necessary implication is more difficult than > with the usual definition.. If {a,{a,b}} = {c,{c,d}}, suppose a = > {c,d} and c = {a,b}. Then a is an element of c, which is an > element of a. This is not possible with foundation. By > contradiction, a = c and {a,b} = {c,d}, whence b = d. I suppose so. (You also need to prove that {a,{a,b}} is necessarily a doubleton, which is also a quick deduction from Foundation.) > I would say that my definition is in a sense simpler, since it > requires the construction of one fewer set, or the representation > requires two fewer characters. Seems to me that the real objection is > that invokes foundation unnecessarily. (Note that Halmos doesn't even > mention foundation in NST.) Ok, I suppose I grant your point then. I'd only add that I think this objection is a very important one. As noted before, Foundation isn't added because of some intuitive confidence, but rather because it is known to be harmless, and it's a big help in model theory. So that means that one must be able to develop ordinary mathematics without using it (or else it wouldn't be so harmless, it would be important), and since you need to show that, once you've done it, it is no longer interesting to show that you could have used it here or there along the way. So invoking Foundation unnecessarily is a bad thing, but in a very different way from (say) invoking Choice unnecessarily. Thomas === Subject: Re: Axiom of Foundation (absymally stupid question) at 06:35 PM, Elaine Jackson said: >The whole problem is just that you're misquoting the axiom. No he is not. >You say: Every nonempty B contains a y >with (B intersect y) = empty. Not only him. Everybody who uses GBN or ZF says so as well. >I say: Every nonempty B contains a y for which >there is no z with z in y and z in B. In what context do you say it? >My axiom What set theory is your axiom a part of? What are the other axioms? >but it allows for the possibility that there exist >citrus fruits that are not sets. Then it's not part of the same set theory, it is your responsibility to give the complete set of axioms that you are using. >Citrus fruits that have no elements, but aren't the >empty set, are technically called individuals. No. There are sets theories that have individuals without having to abandon extentionality. Again, if you wish to be taken seriously you will need to state what set theory it is that you are using. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Axiom of Foundation (absymally stupid question) | at 06:35 PM, Elaine Jackson said: | |>The whole problem is just that you're misquoting the axiom. | |No he is not. | |>You say: Every nonempty B contains a y |>with (B intersect y) = empty. | |Not only him. Everybody who uses GBN or ZF says so as well. In a careful presentation, it would be usual to write it out in terms of epsilon and =, and not defined terms like intersect which technically are not part of the language of ZFC. I did a web search, and I see that on Mathworld the axiom is given as a form of the axiom scheme of epsilon-induction, which again does not involve referring directly to intersections. |>I say: Every nonempty B contains a y for which |>there is no z with z in y and z in B. | |In what context do you say it? The point here is that speaking of intersections presupposes that the objects in question are sets, whereas set theories with urelements typically consider epsilon to be a relationship on the whole domain, including both urelements and sets. So saying there doesn't exist a z such that z in y and z in B allows for the possibility that y or B is an urelement (like a piece of fruit). Thus the same statement of the foundation axiom would suffice for a set theory with urelements of this kind. |>My axiom | |What set theory is your axiom a part of? What are the other axioms? ZFU would be a familiar example of this kind of set theory. |>but it allows for the possibility that there exist |>citrus fruits that are not sets. | |Then it's not part of the same set theory, it is your responsibility |to give the complete set of axioms that you are using. Seymour Metz is being overly formal again. |>Citrus fruits that have no elements, but aren't the |>empty set, are technically called individuals. | |No. But they are. |There are sets theories that have individuals without having to |abandon extentionality. She didn't say this was the only sense in which the term was used. True, another way to model urelements/atoms/individuals is to have them bear the epsilon relation to themselves. But that requires either abandoning or adjusting the foundation axiom. |Again, if you wish to be taken seriously you |will need to state what set theory it is that you are using. The context was adequate. Keith Ramsay === Subject: Vector Calculus problem Ok, I'm trying to work on my homework and am stuck on 4.9 #10 of Vector Analysis by Davis. The question states By means of Stokes' theorem, find S F*dR around the ellipse x^2+y^2=1, z=y, where F=xi+(x+y)j+(x+y+z)k. I got the curl of F and that equalled i-j+k but I'm not really sure how to do the rest of the problem. Any help would be appreciated. I've wasted a lot of time and gotten almost nowhere. === Subject: Re: Vector Calculus problem I forgot to mention that the answer is in the back of the book: +-2pi depending upon the direction of integration. I just can't figure out how to get that. === Subject: topological groups I'm reading about topological groups and I am having trouble with the definition of such a group. What exactly are the open sets? Any help would be appreciated. === Subject: Re: topological groups at 08:02 PM, Arthur said: >I'm reading about topological groups and I am having trouble with the > definition of such a group. What exactly are the open sets? That's like asking what the open sets are in a topological space. Part of specifying a topological group is specifying a topology; the open sets of a topological group are the open sets of its topology. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: topological groups > I'm reading about topological groups and I am having trouble with the > definition of such a group. What exactly are the open sets? Hey, You can't talk about the open sets since a group is said topological iff it is continuous ( (x,y)-> x+y and x->x^(-1)are continuous); sometimes it is also required to be Hausdorff. But considering a group (G,+) you can put different topologies on G so that G is a topological group (provided that G is continuous (and sometimes, Hausdorff)). We're not definining a special topology here. -- Julien Santini, France. === Subject: Re: topological groups >> I'm reading about topological groups and I am having trouble with the >> definition of such a group. What exactly are the open sets? > Hey, > You can't talk about the open sets since a group is said topological > iff it is continuous ( (x,y)-> x+y and x->x^(-1)are continuous); > sometimes it is also required to be Hausdorff. But considering a group > (G,+) you can put different topologies on G so that G is a topological > group (provided that G is continuous (and sometimes, Hausdorff)). > We're not definining a special topology here. > -- > Julien Santini, > France. I see. Arthur === Subject: ADE Groups, Cosmology and Consciousness Jack & Jonathan, The *Roots of Consciousness* by Jeffrey Mishlove on the web does NOT include my 40 page appendix paper, Consciousness: a Hyperspace View. Jeffrey wanted to include it but there was some difficulty with the complexity of the figures -- if I remember correctly. Of course, it had never been in pdf form, since the web (& home computers) were in a very primitive state in 1993. This paper has the most detailed account of what I now call ADEX theory -- the application of the A-D-E Coxeter graphs to mathematics, physics, and other fields such as consciousness theory. This appendix paper was written in 1989, and published in 1993. A short summary and update of this paper (with more ADEX examples) was titled, A Mathematical Strategy for a Theory of Consciousness. It was written in 1994 and published in the book, *Toward a Science of Consciousness: the First Tucson Discussions and Debates* (edited by Stuart R. Hameroff, Alfred W. Kaszniak, and Alwyn C. Scott), MIT Press, 1996. People have told me that my papers are hard to read without much more mathematical knowledge than they possess. Jeffrey tells me that Russian scientists have less trouble with the math since their mathematical training is more advanced than that of American psychologists and biologists. Actually much of the mathematics of ADEX theory is of very recent vintage, but the key area of mathematics is quite old -- group theory (both finite groups and Lie groups and relationships between them). Our understanding of these relationships depends on both algebra and geometry -- especially hyperspace geometry -- including differential geometry and algebraic geometry. Physicists who study general relativity know some differential geometry, but they have not studied the more recently developed algebraic geometry -- and very few physicist (or mathematicians) have seen the great utility of the A-D-E Coxeter graphs which have been used to classify more than 20 mathematical objects. The advantage of having these classifications, is that the A-D-E graphs provide the relationships between all these mathematical objects. I will mention a few of these objects -- the study and application of which I call ADEX theory: 1. Finite reflection groups (Coxeter groups also called Weyl groups) 2. Hyperspace polytopes and thus crystallographic lattices 3. Coxeter arrangements (mirrors in reflection space) 4. Lie algebras and Lie groups (& also Kac-Moody Lie algebras) 5. Thom-Arnold catastrophe bundles (useful for Jack's version of Bohm) [BTW: Thom claimed (1975) that it models the mind-body relationship] [Yes, this is the aspect I want to flesh out with you.] 6. McKay groups (finite subgroups of SU(2) -- unit length quaternions) 7. Gravitational Instantons (closely related to Penrose twistors) 8. 2-d Conformal field theories (which live on hyperspace strings) [Jack: It is interesting that O(2) macro-quantum order parameter in ordinary space has string defects e.g. Hagen Kleinert also books on soft condensed matter physics and cosmic strings. Then introduce extra space dimensions including fermi dimensions for supersymmetry to get higher dim branes from the macro-quantum order parameters perhaps with higher O(N) internal symmetry, hyper-complex matrix order parameters over hyper-complex generalized space-time manifolds. My model in http://qedcorp.com/APS/EmergentGravity.doc is only the low energy tail of that. BTW new version shows how to go from my BIT FROM IT Landau-Ginburg eq to Andrei Linde's specific equations for chaotic APS-AAPT. I had discovered the friction term from my equations a year ago not realizing their crucial role in Linde's theory of the continuous creation of the parallel universes.] 9. Error-correcting codes (related to Jack's IT <--> BIT idea) That too. The mind field must have error correcting codes built into it.] 10. Quantizing lattices (analog to digital transforms) As the motto for Plato's academy said, Let no one enter here without geometry. Today, this geometry must include hyperspace geometry -- which dates back to the 19th century. BTW: *Roots of Consciousness* is mentioned at the end of John McKay's very short paper, A Rapid Introduction to ADE Theory. The URL for this paper is: http://math.ucr.edu/home/baez/ADE.html This is on John Baez's very extensive website, and from the above URL you can access 4 much longer (tutorial) papers by John Baez on the ADE related mathematics. Nuff said! Saul-Paul ---------- , Jeffrey Mishlove === Subject: Re: Roots of Consciousness sent to Saul-Paul and Mishlove All the old stuff is really obsolete IMHO it's pre-science both Leibniz and Young. It's not asking the right questions not using the right concepts. It's mainly metaphor and not useful given today's advances. === Subject: Re: How do you prove that the circumference of a circle is proportional to it's radius? >> .... >> This shows that our modern area formula (pi)(r^2) or (pi)(d^2)/4 .... > Ken , > Is this really a modern formula for circle's area?.... What I meant was modern _notation_ for the formula. I'm sorry if that wasn't clear. Ken Pledger. === Subject: Playing with Fibonacci, help I noticed that continued fraction expansion for sqrt( (n^2 - n + 2)^2 / 4 - n ) was quite interesting, as soon as n is very great (if n is small, the property is still true, but not easy to see). Take n=2^67; then, you can notice that starting from the 4-th term in the expansion, you will find blocks starting with a great value, and a few very small terms just after. What is interesting is that a formula f(n,i) can be built that returns rationals equal to the rational built from the i-th block. Here is an example: for n=2^67 The 38-th block is [22801,14,1,1,5,15,1,1,3,3,1,2,1,1,1,18,...] (length is 33). Considering this block is the continued fraction of a rational, we have the 38-th rational, being: 238...709 / fibonacci(38+2)^2 Now, I found the formula that gives rational equal to the successive blocks: floor( (n-epsilon) / (fibo(i+1)*fibo(i+2)) ) [Times] ( ((-1)^i [Times] fibo(i+1)[Times](n-fibo(i+2)-1)[Times]fibo(i+2)-1) mod (fibo(i+2)^2)) / (fibo(i+2)^2) Now, just put epsilon=0, and you will see the formula is very good. BUT... The integer part is sometimes greater (diff=1) than the real value; the rest of the block is fine. Thus, I think that an epsilon value should be put in the formula. It looks like epsilon depends on both n and i. It means that floor( n / (fibo(i+1)*fibo(i+2)) ) is almost the right value for the successive big values in the initial continued fraction expansion, but not quite exact. Could someone fix the formula ? PS - may my formula be simplified (for instance, is it possible to remove the (-1)^i ?) Cordially, === Subject: Re: Problems book > Is there a book similar to Berkeley Problems in Mathematics, but > geared more towards the undergraduate, and perhaps more finding > stuff rather than proving stuff? I would hope the problems would be > of the same relative difficulty to an undergraduate as the BPM book > would be to a graduate. You may find some of what you want in D.K. Faddeev & I.S. Sominskii (trans. J.L. Brenner), Problems in Higher Algebra, Freeman, 1965. It's old-fashioned, but has a lot of good problems. Ken Pledger. === Subject: Bottom line on prime counting issue Some of you may have noticed frenetic activity from posters trying to convince you that there's nothing sinister about mathematicians doing their best to downply my find of a way to count prime numbers by integrating a partial difference equation, but what's the bottom line? Does what I found work or not? It does. End of story, so mathematicians should acknowledge it. If it's not important they can just put it in some math text somewhere, or in some journal and drive on. No big deal. But they're fighting to totally ignore it. Translation: Sinister attempt by academic types to hide something really important. Otherwise, why go to so much effort to fight me, when a simple way to shut me up on the issue is just record it somewhere? And it is a FIRST in human history, so use your common sense. The loser academic world is fighting me over something that works. End of story. These posters trying to convince you otherwise are just insulting your basic intelligence. James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Bottom line on prime counting issue > Some of you may have noticed frenetic activity from posters trying to > convince you that there's nothing sinister about mathematicians doing > their best to downply my find [...] And again you find it perfectly acceptable to hurl insults at millions -- while reserving to play the indignant sensitive little flower when someone hands you a tiny fraction of your insults back. > of a way to count prime numbers by > integrating a partial difference equation, but what's the bottom line? You have yet to present any kind of way to count prime numbers that actually has anything to do with integration at all. Hint: a sum is not an integral. > Does what I found work or not? It has been conclusively proven that it doesn't. I presented the implementation of the exact literal lines you posted here and you yourself could not find anything wrong with it. If you had a quarrel with the implementation, you could even simply have posted your own little fortran or basic or c-routine. No big deal. But of course you can't. It does not work. That is all there is to it. I have given you thebenefit of the doubt long enough to implement exactly what you posted here to see for myself whether you're on to something or not. That's called 'science': I go and examine the evidence myself. And I have seen with my own eyes that you don't have anything here that counts primes. And further *lies* of yours to the contrary will not sway someone who's actually examined the evidence himself. > It does. End of story, so mathematicians should acknowledge it. Ah: you say so and thus it is so. 'Tis a simple world you live in. So why does this go for you but not for everybody else on the planet? Because there's a lot of people out there that say you stuff doesn't work. And contrary to you they have evidence for their claim. > But they're fighting to totally ignore it. Translation: Sinister > attempt by academic types to hide something really important. Ask yourself: how does this line distinguish you from every run-of-the-mill dime-a-dozen psychotic crackpots with a new theory of everything to sell, without a shred of evidence to present and with demonstrated lack of grasp of what they're talking about? > Otherwise, why go to so much effort to fight me, when a simple way to > shut me up on the issue is just record it somewhere? Nobody is going to any particular effort fighting you. Nobody is going to record anything anywhere because there's nothing to record here. > These posters trying to convince you otherwise are just insulting your > basic intelligence. Just to clarify for to odd reader out there: I am not trying to convince you of anything at all. (Contrary to Mr. Harris.) Go and see for yourself, as I did. You'll see for yourself. === Subject: Re: Bottom line on prime counting issue > Some of you may have noticed frenetic activity from posters trying to > convince you that there's nothing sinister about mathematicians doing > their best to downply my find of a way to count prime numbers by > integrating a partial difference equation, but what's the bottom line? What you are posting is, as usual, complete nonsense. To everyone except you it is obvious that your supposed find of a way to count prime numbers by integrating a partial difference equation is complete and utter garbage. Pointing out that complete and utter garbage is complete and utter garbage is not sinister, it is the obvious thing to do. So the reason why people post that you are wrong is just plain because you are wrong. Nothing sinister about that. === Subject: Re: Bottom line on prime counting issue > Some of you may have noticed frenetic activity from posters trying to > convince you that there's nothing sinister about mathematicians doing > their best to downply my find of a way to count prime numbers by > integrating a partial difference equation, but what's the bottom line? Your work does *not* involve the integration of a partial difference equation. Integration is used to find anti-derivatives. Difference equations are solved using the sum calculus. Nowhere in the exposition of your find do you ever integrate any equation whatsoever, much less a partial difference equation. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Bottom line on prime counting issue | |> Some of you may have noticed frenetic activity from posters trying to |> convince you that there's nothing sinister about mathematicians doing |> their best to downply my find of a way to count prime numbers by |> integrating a partial difference equation, but what's the bottom line? | |Your work does *not* involve the integration of a partial difference |equation. Integration is used to find anti-derivatives. Difference |equations are solved using the sum calculus. Nowhere in the exposition of |your find do you ever integrate any equation whatsoever, much less a |partial difference equation. you're lucky herman rubin isn't reading these threads. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Bottom line on prime counting issue >Some of you may have noticed frenetic activity from posters trying to >convince you that there's nothing sinister about mathematicians doing >their best to downply my find of a way to count prime numbers by >integrating a partial difference equation, but what's the bottom line? >Does what I found work or not? >It does. End of story, so mathematicians should acknowledge it. If >it's not important they can just put it in some math text somewhere, >or in some journal and drive on. >No big deal. >But they're fighting to totally ignore it. Translation: Sinister >attempt by academic types to hide something really important. Fascinating. If people were raving about how important it was of course that would prove it was important. In fact people are ignoring it, and curiously that also proves it's important. Do you really think anyone's buying this? >Otherwise, why go to so much effort to fight me, when a simple way to >shut me up on the issue is just record it somewhere? Huh? First, nobody's going to any trouble - what looks to you like people going to a lot of trouble is just people having a bit of good-natured fun with the village idiot. And second, all your stuff _is_ recorded, right there on Google. (You're going to find that fact embarassing if you ever sober up...) > And it is a >FIRST in human history, so use your common sense. >The loser academic world is fighting me over something that works. >End of story. >These posters trying to convince you otherwise are just insulting your >basic intelligence. Uh, right. Our common sense tells us that when people on sci.math, journal editors, mathematicians you harass with email, mathematicians you harass in person _all_ find your work of no interest the only possible explanation is that they all realize it's tremendously important, and every single one of them is quick enough to realize he needs to lie about his opinion before once saying oh my god that's incredible even once. That's not common sense as we know it, Jim. >James Harris >My math discoveries, found for profit >http://mathforprofit.blogspot.com/ David C. Ullrich === Subject: Re: Bottom line on prime counting issue > Some of you may have noticed frenetic activity from posters trying to > convince you that there's nothing sinister about > My math discoveries, found for profit http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.net -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Bottom line on prime counting issue James Harris > Some of you may have noticed frenetic activity from posters trying to > convince you that there's nothing sinister about mathematicians doing > their best to downply my find of a way to count prime numbers by > integrating a partial difference equation, but what's the bottom line? > Does what I found work or not? He's talking about this one: === Subject: Partial difference equation, counting primes For newbies: www.crank.net/harris.html For harrisologists: mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=705 === Subject: Re: Bottom line on prime counting issue James I believe that the people responsible for covering up the Roswell incident are the same people responsible for suppressing your world changing ideas about algebraic integers, and your prime counting function. I would be careful about what you say of the government. You may end up speaking with Mulder and Scully. Lurch > Some of you may have noticed frenetic activity from posters trying to > convince you that there's nothing sinister about mathematicians doing > their best to downply my find of a way to count prime numbers by > integrating a partial difference equation, but what's the bottom line? > Does what I found work or not? > It does. End of story, so mathematicians should acknowledge it. If > it's not important they can just put it in some math text somewhere, > or in some journal and drive on. > No big deal. > But they're fighting to totally ignore it. Translation: Sinister > attempt by academic types to hide something really important. > Otherwise, why go to so much effort to fight me, when a simple way to > shut me up on the issue is just record it somewhere? And it is a > FIRST in human history, so use your common sense. > The loser academic world is fighting me over something that works. > End of story. > These posters trying to convince you otherwise are just insulting your > basic intelligence. > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ === Subject: Re: Bottom line on prime counting issue And that are my dogs! Mulder and Scully are two lovely dogs, Amstaffs of course. They love Harris as I do. I and them would like Harris to be honest about his work and commit to failure. Or else!!!! What I really want is that Harris fully gave his work a scrutiny which included all the corrections from the bistanders and helping hands. Then he can conclude and put forward his proof. One last question to Harris: What is the difference between algebraic numbers, algebraic integers, numbers, integers and complex numbers? Karl-Olav Nyberg === Subject: Re: Bottom line on prime counting issue > One last question to Harris: > What is the difference between algebraic numbers, algebraic integers, > numbers, integers and complex numbers? Hint: Use Harris's theorem, Q=R. LH === Subject: [JSH] Super-Crank Troll: Re: Bottom line on prime counting issue > What is the difference between algebraic numbers, algebraic integers, > numbers, integers and complex numbers? I believe you might have to wade through JSHs Object Oriented Mathematics and the brilliance that that has yet to shine on all of us losers before you can even ask the self-proclaimed highest ranking number theorist in the world. In fact, Mr. Harris ego is growing at a super-exponential rate and soon no one in sci/math will be able to contain his NPD! === Subject: Re: Bottom line on prime counting issue > Some of you may have noticed frenetic activity from posters trying to > convince you that there's nothing sinister about mathematicians doing > their best to downply my find of a way to count prime numbers by > integrating a partial difference equation, but what's the bottom line? > Does what I found work or not? Now that is an interesting question, isn't it! pssst, hey Harris... your stuff doesn't work.... but don't tell anybody... === Subject: Re: Bottom line on prime counting issue > > Some of you may have noticed frenetic activity from posters trying to > convince you that there's nothing sinister about mathematicians doing > their best to downply my find of a way to count prime numbers by > integrating a partial difference equation, but what's the bottom line? > > Does what I found work or not? > > Now that is an interesting question, isn't it! > pssst, hey Harris... your stuff doesn't work.... but don't tell anybody... And Sam Wormley, surprise, surprise, is making a false statement as it *does* work, but I'm not terribly surprised by his immature behavior. After all, I found this partial difference equation, which integrates over a certain range to give a count of prime numbers!!! Here's the equation: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,sqrt(y-1))], Here are the instructions for the integration: S(x,1) = 0. And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS from dS(x,2) to dS(x,y). Now for someone like Sam Wormley it probably doesn't seem fair that I'm here posting on Usenet stealing thunder from everyone else, but hey, blame the mathematicians. The bottom line on my work is that it DOES work. Now then, do any of you know *how* it works? Maybe some poster will reply to this post to explain it to you, but how do you trust them? And what about the story--my story--of the discovery? What was I thinking? What motivated me to look in this area? What's the story? If mathematicians hadn't decided to break faith with you and the rest of the world, probably there'd be a book, some popular work, explaining the story. But how can you get that story if mathematicians are playing their academic games? Bottom line: What I have works. So what if I sell my story and get rich. That's how capitalism works. These mathematicians are worse than communists, as how do you explain their behavior? I *am* the American Dream, fighting for what should be mine, having to get past weak-minded academics who are fighting to block my success. But I shall prevail!!! I'm sure some hate that I'm in it for the money. But why screw over the world, why screw *you* over by blocking the story of my success? What's their motivation? James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Bottom line on prime counting issue >If mathematicians hadn't decided to break faith with you and the rest >of the world, probably there'd be a book, some popular work, >explaining the story. >But how can you get that story if mathematicians are playing their >academic games? >Bottom line: What I have works. >So what if I sell my story and get rich. Psst, James, there is a very small market for stories about mathematics. Better find some way to work in spies and the CIA, and pretty girl agents, and such like. And sex. Sex always sells, even when the sex scenes are separated by boring mathematical explanations. People just skip those. -- Wolf Kirchmeir, Blind River ON Canada Nature does not deal in rewards or punishments, but only in consequences. (Robert Ingersoll) === Subject: Re: Bottom line on prime counting issue >If mathematicians hadn't decided to break faith with you and the rest >of the world, probably there'd be a book, some popular work, >explaining the story. >But how can you get that story if mathematicians are playing their >academic games? >Bottom line: What I have works. >So what if I sell my story and get rich. > Psst, James, there is a very small market for stories about mathematics. > Better find some way to work in spies and the CIA, and pretty girl agents, > and such like. And sex. Sex always sells, even when the sex scenes are > separated by boring mathematical explanations. People just skip those. A very small market in today's world can be worth millions of dollars US. The bottom line is that what I have works, people expect mathematicians to report on discoveries, but they are not doing their jobs. It's easy to check using Google. Go search on partial difference equation which can verify for you that they are real. Then search on prime counting or counting primes to see if ANYONE besides me has ever used a partial difference equation to count prime numbers. For those wondering what they might do to help, I think that maybe sending an email to some news organization might have an impact. For instance, you can email TIME magazine at letters@time.com, and who knows what might happen? James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Bottom line on prime counting issue >A very small market in today's world can be worth millions of dollars >US. IMO, you need to brush up on your arithmetic, too. -- Wolf Kirchmeir, Blind River ON Canada Nature does not deal in rewards or punishments, but only in consequences. (Robert Ingersoll) === Subject: Re: Bottom line on prime counting issue > For those wondering what they might do to help, I think that maybe > sending an email to some news organization might have an impact. For > instance, you can email TIME magazine at letters@time.com, and who > knows what might happen? Here is what can happen with education gone bad, a person with delusions of grandeur, fame and fortune. This individual believes he is one of the greatest number theorists and analytical researchers of ALL TIME! Can you perhaps run a story on NPD (you have a real-live case here)? Here is a clinical definition of NPD: *** Diagnostic criteria for 301.81 Narcissistic Personality Disorder (cautionary statement) A pervasive pattern of grandiosity (in fantasy or behavior), need for admiration, and lack of empathy, beginning by early adulthood and present in a variety of contexts, as indicated by five (or more) of the following: (1) has a grandiose sense of self-importance (e.g., exaggerates achievements and talents, expects to be recognized as superior without commensurate achievements) (2) is preoccupied with fantasies of unlimited success, power, brilliance, beauty, or ideal love (3) believes that he or she is special and unique and can only be understood by, or should associate with, other special or high-status people (or institutions) (4) requires excessive admiration (5) has a sense of entitlement, i.e., unreasonable expectations of especially favorable treatment or automatic compliance with his or her expectations (6) is interpersonally exploitative, i.e., takes advantage of others to achieve his or her own ends (7) lacks empathy: is unwilling to recognize or identify with the feelings and needs of others (8) is often envious of others or believes that others are envious of him or her (9) shows arrogant, haughty behaviors or attitudes Reprinted with permission from the Diagnostic and Statistical Manual of Mental Disorders, fourth Edition. Copyright 1994 American Psychiatric Association *** === Subject: Re: Bottom line on prime counting issue ... > After all, I found this partial difference equation, which integrates > over a certain range to give a count of prime numbers!!! > Here's the equation: > dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - > p(y-1,sqrt(y-1))], As written this does not look like a partial difference equation to me. perhaps you intend: S(x, y) - S(x, y-1) = the expression on the right, or: S(x, y+1) - S(x, y) = the expression on the right. It is not clear which of the two you intend here. > Here are the instructions for the integration: > S(x,1) = 0. Instruction? Looks more like a boundary condition. > And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS > from dS(x,2) to dS(x,y). There must be something wrong. You now instruct how to get S. But I thought in a partial difference equation you had to get S by solving the equation. So how can you now give explicit instructions on how to calculate S? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Bottom line on prime counting issue In sci.physics, Sam Wormley <3FB7DF7A.13A6732D@mchsi.com>: >> Some of you may have noticed frenetic activity from posters trying to >> convince you that there's nothing sinister about mathematicians doing >> their best to downply my find of a way to count prime numbers by >> integrating a partial difference equation, but what's the bottom line? >> Does what I found work or not? > Now that is an interesting question, isn't it! > pssst, hey Harris... your stuff doesn't work.... but don't tell anybody... Actually, his stuff worked more or less fine ... but I can't say it was the fastest. http://home.earthlink.net/~ewill3/math/primecounters/index.html was a somewhat tongue-in-cheek contest I sponsored 2 months back that produced a few bizarre results and some interesting algorithms. (However, Christian Bau has a better one anyway, although he didn't submit that particular one for my contest. Perhaps it was because my contest was unworthy thereof. :-) ) James' wasn't the best, especially after a round of memoization which I for one did not foresee. In fact, yours truly beat him with a rather simple Legendrephi entry that was more than twice as fast. However, an even simpler sieving entry (sieve6bool, which did take advantage of a prime being of one of the forms 6k+1 or 6k-1, after the two entries 2 and 3) was 10.5 times as fast, once I got it working -- and the best entry (edgar3), apart from a couple of what were essentially large table lookups by yours truly, was submitted by a hitherto unknown poster by E-mail. James did win a consolation prize for elegance. Of course prize here is a bit of a misnomer. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Bottom line on prime counting issue > was a somewhat tongue-in-cheek contest I sponsored 2 months back > that produced a few bizarre results and some interesting > algorithms. (However, Christian Bau has a better one anyway, > although he didn't submit that particular one for my contest. > Perhaps it was because my contest was unworthy thereof. :-) ) No, it was not finished at that time, and I have to find some spare time to improve it anyway. What I am quite interested in at the moment is that there seems to be a substantial improvement possible if you want to calculate pi (N) for many different values of N, for example N = k * 10^14 for 1 <= k <= 10000. My implementation should take about O (N^(2/3)) to find pi (N). However, it might be possible to find pi (x) for n different values x <= N in about O (N^(2/3)) * sqrt (n) instead of O (N^(2/3)) * n. === Subject: Re: Bottom line on prime counting issue In sci.physics, Christian Bau was a somewhat tongue-in-cheek contest I sponsored 2 months back >> that produced a few bizarre results and some interesting >> algorithms. (However, Christian Bau has a better one anyway, >> although he didn't submit that particular one for my contest. >> Perhaps it was because my contest was unworthy thereof. :-) ) > No, it was not finished at that time, and I have to find some spare time > to improve it anyway. What I am quite interested in at the moment is > that there seems to be a substantial improvement possible if you want to > calculate pi (N) for many different values of N, for example > N = k * 10^14 for 1 <= k <= 10000. > My implementation should take about O (N^(2/3)) to find pi (N). However, > it might be possible to find pi (x) for n different values x <= N in > about O (N^(2/3)) * sqrt (n) instead of O (N^(2/3)) * n. I suppose it might depend in part on the value of max(N_i), where N_i are the numbers fed into pi(N). I really don't know, and haven't researched the issue. Good luck. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: JSH: All the dumb crap in journals With all the dumb crap in math journals, you people know that my find of a way to count prime numbers by integrating a partial difference equation is worthy of publication somewhere. But you sit by as if there's nothing sinister going on, when all these people are fighting such a choice result. Yet in your ENTIRE CAREERS most of you will never publish anything that's even close in neatness, but you'll fill up journals anyway. That's the math world. If there's anything I really like here it's proving the lie to all of you of pure math, as sure, you may convince others that my work isn't important, but those of you desperate to find something worth publication in your publish or perish world know the reality. Pure math is a fraud, just something you say, when you don't believe in it. It's just a way to get by and pay your bills. Yes, to me you are losers, too weak to handle the truth, so you think you can crap on me, when what you do anyway is fill up journals with junk. I'm better than you, and you know it. James Harris === Subject: Re: JSH: All the dumb crap in journals > With all the dumb crap in math journals, you people know that my find > of a way to count prime numbers by integrating a partial difference > equation is worthy of publication somewhere. > But you sit by as if there's nothing sinister going on, when all these > people are fighting such a choice result. > Yet in your ENTIRE CAREERS most of you will never publish anything > that's even close in neatness, but you'll fill up journals anyway. > That's the math world. > If there's anything I really like here it's proving the lie to all of > you of pure math, as sure, you may convince others that my work > isn't important, but those of you desperate to find something worth > publication in your publish or perish world know the reality. > Pure math is a fraud, just something you say, when you don't believe > in it. > It's just a way to get by and pay your bills. Yes, to me you are > losers, too weak to handle the truth, so you think you can crap on me, > when what you do anyway is fill up journals with junk. > I'm better than you, and you know it. > James Harris fuffy === Subject: Re: JSH: All the dumb crap in journals > With all the dumb crap in math journals, you people know that my find > of a way to count prime numbers by integrating a partial difference > equation is worthy of publication somewhere. There are indeed journals that will publish nifty new ways of proving old results if there is some insight to be gained from the new way. In explain well-known results in ways that are especially easy to grasp. You may want to consider that route if you think that your approach to Legendre's (?) method gives some new perspective on the matter. V. -- email: lastname at cs utk edu homepage: cs utk edu tilde lastname === Subject: Re: JSH: All the dumb crap in journals > With all the dumb crap in math journals, How would you know? You've repeatedly refused to attempt answering even the most basic of homework-type questions in algebra. === Subject: Re: All the dumb crap in journals > I'm better than you, and you know it. Do you know that you actually have more than nine of these symptoms? That also deals with numbers, including 9 which is the perfect square 3^2! *** Diagnostic criteria for 301.81 Narcissistic Personality Disorder (cautionary statement) A pervasive pattern of grandiosity (in fantasy or behavior), need for admiration, and lack of empathy, beginning by early adulthood and present in a variety of contexts, as indicated by five (or more) of the following: (1) has a grandiose sense of self-importance (e.g., exaggerates achievements and talents, expects to be recognized as superior without commensurate achievements) (2) is preoccupied with fantasies of unlimited success, power, brilliance, beauty, or ideal love (3) believes that he or she is special and unique and can only be understood by, or should associate with, other special or high-status people (or institutions) (4) requires excessive admiration (5) has a sense of entitlement, i.e., unreasonable expectations of especially favorable treatment or automatic compliance with his or her expectations (6) is interpersonally exploitative, i.e., takes advantage of others to achieve his or her own ends (7) lacks empathy: is unwilling to recognize or identify with the feelings and needs of others (8) is often envious of others or believes that others are envious of him or her (9) shows arrogant, haughty behaviors or attitudes Reprinted with permission from the Diagnostic and Statistical Manual of Mental Disorders, fourth Edition. Copyright 1994 American Psychiatric Association *** === Subject: Re: JSH: All the dumb crap in journals > With all the dumb crap in math journals, you people know that my find > of a way to count prime numbers by integrating a partial difference > equation is worthy of publication somewhere. You're work does *not* involve the integration of a partial difference equation. Integration is used to find anti-derivatives. Difference equations are solved using the sum calculus. Your prime counting function does not involve integration in any way. If you'd stop misrepresenting your work, maybe someone would pay attention to it. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: All the dumb crap in journals >With all the dumb crap in math journals, you people know that my find >of a way to count prime numbers by integrating a partial difference >equation is worthy of publication somewhere. Why are you telling us this? Do you think that sci.math has something to do with what gets published in math journals? That's not how it works. Send your stuff to a journal. Let us know how it turns out. >But you sit by as if there's nothing sinister going on, when all these >people are fighting such a choice result. >Yet in your ENTIRE CAREERS most of you will never publish anything >that's even close in neatness, but you'll fill up journals anyway. >That's the math world. >If there's anything I really like here it's proving the lie to all of >you of pure math, as sure, you may convince others that my work >isn't important, but those of you desperate to find something worth >publication in your publish or perish world know the reality. >Pure math is a fraud, just something you say, when you don't believe >in it. >It's just a way to get by and pay your bills. Yes, to me you are >losers, too weak to handle the truth, so you think you can crap on me, >when what you do anyway is fill up journals with junk. >I'm better than you, and you know it. >James Harris David C. Ullrich === Subject: Re: JSH: All the dumb crap in journals |With all the dumb crap in math journals, you people know that my find |of a way to count prime numbers by integrating a partial difference |equation is worthy of publication somewhere. my comment here is only indirectly related to your comment above. your work on writing computer programs to count prime numbers is so much, much better than your work on trying to prove theorems of any kind (fermat's last theorem, algebraic integer theory, et cetera) that you would have to be a total idiot to continue pursuing the theorem-proving stuff when instead you could be out there trying to do stuff like writing computer programs to win those rsa challenge prizes or whatever they're called. you really, really need something to provide external discipline for you, something to tell you when you've gotten a wrong answer and when you've gotten a right answer. the concept of mathematical proof does provide that kind of external discipline for people who understand that concept, but you're not one of those people! (at least, it would be astonishing if you did understand the concept of mathematical proof, and yet produced the kind of incomprehensible near-gibberish proofs that you regularly produce. i'm willing to bet that you yourself really don't have the slightest idea as to whether or not your proofs are really valid.) if you work on creating algorithms and writing computer programs then you can get concrete answers about whether you've done things correctly simply by running a computer program. i guess if you really have your heart set on pursuing the theorem-proving stuff then it would be ok to pursue it, but only if you first learn what mathematical proof really means. it's not that stupid horse about a proof starts with a truth and proceeds by logical steps to a conclusion which then must be true! at least, that stupid slogan is worthless unless you learn what a valid logical step really is. really, it would do you well to learn the simple way in which rigorous (though not necessarily computationally efficient) theorem-checking and theorem-proving computer programs actually work, and maybe even write such a program of your own so you could honestly say that you understand what a mathematical proof is. also, have you ever considered maintaining your opinion of yourself as a million times smarter than everyone else but _not going around telling everyone about it every five minutes_? works for me. also, a lot of the people on this newsgroup who go around attacking you constantly are worthless idiots and when you try to adopt for yourself the same attitudes and the same methods of attacking that they use you just make yourself look like a worthless idiot too, so cut it out. you really need to learn to ignore the rantings of worthles idiots instead of trying to become one of them. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: JSH: All the dumb crap in journals >also, a lot of the people on this newsgroup who go around attacking >you constantly are worthless idiots It's considered customary to have researched the topic you're speaking about before making an ass of yourself. Doug === Subject: Re: JSH: All the dumb crap in journals >if you work on creating >algorithms and writing computer programs then you can get concrete >answers about whether you've done things correctly simply by running a >computer program. Excellent advice! Few professionals will find an amateur's proof interesting. On the other hand, today's amateur can perform carefully crafted computations that are sometimes of interest to the professional. It's one of the reasons why I believe the 21st century is a great time to be an amateur mathematician. Rich Burge === Subject: Re: JSH: All the dumb crap in journals > |With all the dumb crap in math journals, you people know that my find > |of a way to count prime numbers by integrating a partial difference > |equation is worthy of publication somewhere. > my comment here is only indirectly related to your comment above. snip > also, a lot of the people on this newsgroup who go around attacking > you constantly are worthless idiots Well, actually, they are not. They probably should find something better to do with their time than pointing out James's deficiencies, but that isn't the same thing. I think that if you look you will find that all of JSH's fiercest critics are helpful and polite to those who are less antisocial than JSH. Mark Atherton === Subject: Re: JSH: All the dumb crap in journals |> |With all the dumb crap in math journals, you people know that my find |> |of a way to count prime numbers by integrating a partial difference |> |equation is worthy of publication somewhere. |> my comment here is only indirectly related to your comment above. | |snip | |> also, a lot of the people on this newsgroup who go around attacking |> you constantly are worthless idiots | |Well, actually, they are not. bull. either you have trouble understanding words like a lot, or you're just as worthless as they are. |They probably should find something better to do with their time than |pointing out James's deficiencies, but that isn't the same thing. I |think that if you look you will find that all of JSH's fiercest |critics are helpful and polite to those who are less antisocial than |JSH. if you think that all his fiercest critics are helpful and polite then you're an idiot, since all includes for example the guy that calls himself uncle al. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: JSH: All the dumb crap in journals > |> also, a lot of the people on this newsgroup who go around attacking > |> you constantly are worthless idiots > |Well, actually, they are not. > bull. either you have trouble understanding words like a lot, > or you're just as worthless as they are. are not worthless at all. > |They probably should find something better to do with their time than > |pointing out James's deficiencies, but that isn't the same thing. I > |think that if you look you will find that all of JSH's fiercest > |critics are helpful and polite to those who are less antisocial than > |JSH. > if you think that all his fiercest critics are helpful and polite > then you're an idiot, since all includes for example the guy that > calls himself uncle al. That's a fair point, though mostly Uncle Al confines himself to brief insults. By his fiercest critics I meant those who criticise his maths. And if you respond to a polite disagreement by calling me an idiot then you should think twice before commenting on other posters' manners. By the way, why the lower case? ALL UPPER CASE is considered to be rude but normal sentence capitalization is the norm. Have a nice day. :-) Mark Atherton === Subject: Re: All the dumb crap in journals James, I apologize that my previous posts may have been a bit insulting, it wasn't very mature of me. In the following, I'll assume that the things you have been working on are correct. It seems that the prime counting function is generally agreed to be correct, so just consider that if you like. Don't be so bitter. Contemplate on why many people in the math community aren't taking your result seriously. In a legislature bills are presented in a certain way. If someone from the outside were to insist on presenting a bill in another way, many would ignore it. A computer can only compile a program if it is very carefully formatted. If it is not formatted correctly, the computer not only doesn't give the expected output, it will respond with a long list of errors. If you were to approach a chemist with an astounding result, but express your result in the language of alchemy, you would be ignored. Not because the chemist is out to get you, but because they are busy with their work. Mathematicians expect results to be presented a certain way. If a result is handed to them in some other way, they most often will simply not read it. This is because it is much more difficult to read something that isn't presented in the way they expect, and additionally because there are many people who don't know what they are doing. It is a waste of time to spend alot of time working through a paper that is hard to read because it is written in a strange way, only to realize that it is either nonsensical or just completely wrong. There are many such papers. Math departments at universities are inundated with proofs of squared circles and so on, along with less inane things. It would be a generous person indeed who carefully read each letter and carefully responded. Are math journals filled with dumb crap? In the journals I have read I've seen only thoughtful papers. Each has been carefully reviewed, and the author spent alot of time carefully writing it. This doesn't garantee that it will be correct or even especially insightful, but it certainly is the best effort of those involved. It was allowed into the journal on it's own merit, and is almost certainly correct becuase of the time and effort spent checking it, by very smart people who have nothing to gain, and alot to lose, by publishing a false result. If the results aren't very insightful or important, it is probably because those papers are the best that were available, or that one doesn't have the specialized knowlege to understand why they are insightful or important. No one has any interest in preventing your work from being recognized. Quite the opposite. When someone revolutionalizes mathematics, it doesn't discredit every previous result. Instead, it opens new areas of mathematics to exploration. This would only lead to many more papers to be published, and it would certainly add to the prestige of those publishing at this time, since history often remembers those who are working on something in it's early stages. There isn't a lie of 'pure math'. I'm not sure exactly what you mean here. It isn't something to be believed in, or asserted. You are making a category error here. You use 'pure math' as though it is an assertion. 'pure math' is only defined when you also consider 'applied math'. 'pure math' is mathematics with no alterior motive, so to speak. Applied math is using math to solve a specific problem or class of problems in the physical world. You might say that pure math is defined as everything that is not applied math. Beyond this, there really isn't any meaning to the word. I'll end this with an open offer. If you want to learn some things, I would be glad to help however I can. Within reason of course. This newsgroup presents an opportunity for smart people to interact with many other smart people. So if you were to get a copy of Hersteins Topics in Algebra and work through some of the problems, I would be glad to check them over or to make suggestions if you needed it. You can get this book at amazon.com. The result of this would be twofold. First, you would learn the standard way in which mathematical results are presented. Secondly, you would learn alot of very interesting mathematics, and add to your mathematical toolbox so to speak. Sorry the post was so long, I just started taking ritalin again, Justin Van Winkle http://atheism.about.com/library/glossary/general/bldef_categoryerror.htm http://www.amazon.com/exec/obidos/tg/detail/-/0471010901/qid=1069019047/sr=8 -1/ref=sr_8_1/102-4193208-0966565?v=glance&n=507846 Wow, Topics is almost a hundred dollars. Here is an alternative: http://www.amazon.com/exec/obidos/tg/detail/-/0070026556/qid=1069019119/sr=1 -1/ref=sr_1_1/102-4193208-0966565?v=glance&s=books These Schaum's Outlines are very good in my experience. I first learned about them from this book: http://www.amazon.com/exec/obidos/tg/detail/-/0817638660/qid=1069019208/sr=1 -4/ref=sr_1_4/102-4193208-0966565?v=glance&s=books > With all the dumb crap in math journals, you people know that my find > of a way to count prime numbers by integrating a partial difference > equation is worthy of publication somewhere. > But you sit by as if there's nothing sinister going on, when all these > people are fighting such a choice result. > Yet in your ENTIRE CAREERS most of you will never publish anything > that's even close in neatness, but you'll fill up journals anyway. > That's the math world. > If there's anything I really like here it's proving the lie to all of > you of pure math, as sure, you may convince others that my work > isn't important, but those of you desperate to find something worth > publication in your publish or perish world know the reality. > Pure math is a fraud, just something you say, when you don't believe > in it. > It's just a way to get by and pay your bills. Yes, to me you are > losers, too weak to handle the truth, so you think you can crap on me, > when what you do anyway is fill up journals with junk. > I'm better than you, and you know it. > James Harris === Subject: Re: All the dumb crap in journals > With all the dumb crap in math journals, you people know that my find > of a way to count prime numbers by integrating a partial difference > equation is worthy of publication somewhere. > But you sit by as if there's nothing sinister going on, when all these > people are fighting such a choice result. > Yet in your ENTIRE CAREERS most of you will never publish anything > that's even close in neatness, but you'll fill up journals anyway. > That's the math world. > If there's anything I really like here it's proving the lie to all of > you of pure math, as sure, you may convince others that my work > isn't important, but those of you desperate to find something worth > publication in your publish or perish world know the reality. > Pure math is a fraud, just something you say, when you don't believe > in it. > It's just a way to get by and pay your bills. Yes, to me you are > losers, too weak to handle the truth, so you think you can crap on me, > when what you do anyway is fill up journals with junk. > I'm better than you, and you know it. > James Harris If you had any common sense, you'd go learn some math so you could know what we're talking about instead of the immature name calling. Yes you ARE Immature, and it shows. David Moran === Subject: Re: All the dumb crap in journals I take it Jimmy that you have given up on the new marketing strategy. Back to the old tricks, eh! Lurch > With all the dumb crap in math journals, you people know that my find > of a way to count prime numbers by integrating a partial difference > equation is worthy of publication somewhere. > But you sit by as if there's nothing sinister going on, when all these > people are fighting such a choice result. > Yet in your ENTIRE CAREERS most of you will never publish anything > that's even close in neatness, but you'll fill up journals anyway. > That's the math world. > If there's anything I really like here it's proving the lie to all of > you of pure math, as sure, you may convince others that my work > isn't important, but those of you desperate to find something worth > publication in your publish or perish world know the reality. > Pure math is a fraud, just something you say, when you don't believe > in it. > It's just a way to get by and pay your bills. Yes, to me you are > losers, too weak to handle the truth, so you think you can crap on me, > when what you do anyway is fill up journals with junk. > I'm better than you, and you know it. > James Harris === Subject: Re: All the dumb crap in journals > With all the dumb crap in math journals, you people know that my find > of a way to count prime numbers by integrating a partial difference > equation is worthy of publication somewhere. > But you sit by as if there's nothing sinister going on, when all these > people are fighting such a choice result. > Yet in your ENTIRE CAREERS most of you will never publish anything > that's even close in neatness, but you'll fill up journals anyway. > That's the math world. > If there's anything I really like here it's proving the lie to all of > you of pure math, as sure, you may convince others that my work > isn't important, but those of you desperate to find something worth > publication in your publish or perish world know the reality. > Pure math is a fraud, just something you say, when you don't believe > in it. > It's just a way to get by and pay your bills. Yes, to me you are > losers, too weak to handle the truth, so you think you can crap on me, > when what you do anyway is fill up journals with junk. > I'm better than you, and you know it. > James Harris Oh James... I love it when you talk like that. You are so... ... forceful . -- Clive Tooth http://www.clivetooth.dk === Subject: Re: All the dumb crap in journals > With all the dumb crap in math journals, you people know that my find > of a way to count prime numbers by integrating a partial difference > equation is worthy of publication somewhere. > But you sit by as if there's nothing sinister going on, when all these > people are fighting such a choice result. > Yet in your ENTIRE CAREERS most of you will never publish anything > that's even close in neatness, but you'll fill up journals anyway. > That's the math world. Does the sound like sour grapes? === Subject: Re: Semdirect product of categories? | |>|Now it occurs to me that another simple generalization step can bring |>|the concept to a categorical formulation: let A,B be categories and |>|let (A,A) (the monoid of functors A->A) be regarded as a category |>|itself. Then choose a functor sigma:B->(A,A) and define |>|(a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1) |>|for all a2,b2,a1,b1 s.t. Dom(a2)=sigma(b2)Cod(a1), Dom(b2)=Cod(b1). |>actually i'm not sure i understand your notation well enough to see |>whether what you're describing actually works, but i'm a bit skeptical |>about it because you don't seem to be giving the most straightforward |>generalization of the semi-direct product construction to a context |>involving arbitrary categories. | |Well, I didn't know the construction mentioned hereafter, and in this |respect my one is indeed extremely naive. But I don't see how it |could not work, and what is not clear: | |sigma is a functor B->(A,A), so sigma(b) is a functor A->A for all |morphisms b. Its action on objects is obviously trivial. Note that |sigma(e)=1_A (the identical functor) for all identities e of B. | |The notation (a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1) is a |straightforward generalization of that used for groups/monoids. The |only difference being that the obvious compatibility relation for the |composition of the given morphisms is required: | |a1:X->Y, a2:sigma(b2)Y->Z, |b1:S->T, b2:T->U. | |Verification that both the associative and the identity properties |hold is merely a mechanical task. | |>the most straightforward such generalization is some version of the |>homotopy colimit construction. you start with a category c and a |>functor (or in some versions a pseudo-functor) f from c to the |>category of categories. the homotopy colimit of f is the category |>where an object is a pair (x,y) with x an object in c and y an object |>in f(x), and a morphism from (x,y) to (z,w) is a pair (m,n) with |>m:x->z in c and n:f(m)(y)->w in f(z), with composition of morphisms |>defined in a semi-obvious way. |>semi-direct product of monoids is then the special case where c has a |>unique object x and f(x) has a unique object y. | |Then if I'm not mistaken (or misunderstanding), my construction is |the special case when c is arbitrary and f(x) has a unique object A |(in the notation above). In other words it is the same construction |with the difference that the codomain of f is restricted to a |subcategory of the category of categories, namely the category of |functors A->A. certainly the special case where [for any object x in c, f(x) has a unique object] can be considered, but i'm still confused about whether what you tried to describe is really the same thing. if [for any object x in c, f(x) has a unique object], then what we're dealing with is something like a functor from c to (some version of) the category of monoids, but i didn't see anything in your description that really sounded like that. but it's not implausible to me that you might be trying to describe the same thing in different language, because despite what you say i honestly find your notation and terminology confusing, and it makes me wonder whether you might have made what i call a level slip somewhere- getting concepts on the level of objects mixed up with concepts on the level of morphisms, or something like that. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Semdirect product of categories? >certainly the special case where [for any object x in c, f(x) has a >unique object] can be considered, but i'm still confused about whether >what you tried to describe is really the same thing. if [for any >object x in c, f(x) has a unique object], then what we're dealing with >is something like a functor from c to (some version of) the category >of monoids, but i didn't see anything in your description that really >sounded like that. but it's not implausible to me that you might be >trying to describe the same thing in different language, because >despite what you say i honestly find your notation and terminology >confusing, and it makes me wonder whether you might have made what i >call a level slip somewhere- getting concepts on the level of >objects mixed up with concepts on the level of morphisms, or something >like that. I don't think I've made a level slip. To be sure I'm writing the whole lot from scratch: maybe my wording will be more fortunate... (or precise!) Let's start with monoids, say A,B. Aut(A) is a monoid itself, so there may well be something in Hom(B,Aut(A)), and indeed -incidentally- there always is! So, chosen f in Hom(B,Aut(A)), we can define the semidirect product as usual: (*) (a2,b2),(a1,b1)|->(a2 f(b2) a1,b2 b1). Now, as a matter of a fact a Category turns out to be a sort of big monoid with possibly more than one identity and a product not defined for all elements, right? (intendedly loosely speaking!) Now instead of Aut(A) we have the monoid of functors A->A. But a monoid IS a category, with just one object. So we take into account Hom(B,Hom(A,A)): again it is not empty and if we choose f in it we can define a composition *exactly* as in (*) provided that all compositions in it are well defined, i.e. the domain of a2 is the image through f(b2) of the codomain of a1 and the domain of b2 is the codomain of b1. Please do not make me write down extensively (in an ASCII environment) the (trivial) proof that both associativity and identity properties hold! TIA, 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: Problem calculating limits - Need Help! I'm having difficulties solving these two limits. I mustn't use L'hospital rule: a) lim(x*(2^(1/x))-x) where x increases to infinite. b) lim(cosh(x)-1)/(x^2) where x approaches 0. === Subject: Re: Problem calculating limits - Need Help! > I'm having difficulties solving these two limits. > I mustn't use L'hospital rule: > a) lim(x*(2^(1/x))-x) where x increases to infinite. > b) lim(cosh(x)-1)/(x^2) where x approaches 0. SOLUTION for a) in a more general form F:(0,infty)-->R , F(x):= x*(A^{1/x}-1) , with A>0 , A=/=1 . SOLUTION 1 : by supposing as known that lim_{z-->0}(1+z)^{1/z}= e , where ,,e is Napier's constant. Let z:= A^{1/x}-1 . Then x-->infty iff z--->0 .Also x=ln(A)/ln(1+z) . Therefore F(x)= ln(A)/(ln(1+z)^(1/z)) and L=lim_{z-->0} ln(A)/(ln(1+z)^{1/z})= ln(A)/ln(e)=ln(A). SOLUTION 2 : by supposing known the theory of Riemann integral, and in particular case when x take only integer values. More precisely consider the sequence with general term X_n:= n(A^{1/n}-1) . Let us assume that A>1. On interval [1,A] take the division (D_n) (D_n) 1=x_0(n)infty iff norm ||D_n||:=max_{k=0,1,...,n}(x_{k+1}(n)-x_k(n))= =A*(A^(1/n)-1)-->0 . Take the continuous function f:[1,A]-->R , f(x)=1/x and consider the integral sum S_n(f)=S((D_n), f )= SUM_{k=0 to k=n}(x_(k+1)-x_k(n))f(x_k(n)). We have X_n= S_n(f) . In this manner lim_{n-->infty}X_n = lim_{||D_n||-->0}S_n(f)= =INTEGRAL_{t=1 to t=A}dt/t= ln(A). === Subject: Re: Problem calculating limits - Need Help! In sci.math, Roy I mustn't use L'hospital rule: > a) lim(x*(2^(1/x))-x) where x increases to infinite. > b) lim(cosh(x)-1)/(x^2) where x approaches 0. (a) = lim{x->oo} x*(2^(1/x) - 1) = lim{x->oo}x*(exp(ln2/x) - 1) = lim{y->0} (exp(y*ln 2) - 1) / y (y = 1/x) It's now clearly a derivative. Note that we do *not* need to worry about the chain rule here; all we're doing is switching variables within a limit. In fact, using z = ln2/x is instructive; one gets (a) = lim{z->0} (exp(z) - 1) * ln(2) / z which gives the same answer anyway. Or one can write (a) = lim{y->0} (2^y - 1)/y = lim{y->0} {exp(y*ln2) - 1)/y (b) might be doable by setting x^2 = y (y = sqrt(x)); note that one *has* to use the chain rule in this case. A little confusing perhaps but remember that a derivative is lim{d->0} (f(x+d) - f(x))/d; (the traditional notation is delta x, but ASCII is a pain at times :-) ); the value lim{d->0} (f(x+d)-f(x))/d^2 is something else entirely. However, since cosh'(0) = sinh(0) = 0, it works in this case. Note that both are a special case of L^Hopital's rule (note spelling): lim{x->c}f(x)/x = lim{x->c}f'(x)/1 = f'(c) since the derivative of x is the constant 1; be careful how you explain your answer. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Problem calculating limits - Need Help! > I'm having difficulties solving these two limits. > I mustn't use L'hospital rule: > a) lim(x*(2^(1/x))-x) where x increases to infinite. We want lim_{x->oo} [2^(1/x)) - 1]/(1/x). As x -> oo, 1/x -> 0. So this is nothing but the derivative of 2^x at x = 0. > b) lim(cosh(x)-1)/(x^2) where x approaches 0. Use Taylor series as others have suggested, or if you're feeling adventurous, show that for any C^2 function f in a neighborhood of 0 with f'(0) = 0, (f(x) - f(0))/x^2 -> f''(0)/2 as x -> 0. === Subject: Re: Problem calculating limits - Need Help! >> I'm having difficulties solving these two limits. >> I mustn't use L'hospital rule: >> a) lim(x*(2^(1/x))-x) where x increases to infinite. >We want lim_{x->oo} [2^(1/x)) - 1]/(1/x). As x -> oo, 1/x -> 0. So this is >nothing but the derivative of 2^x at x = 0. >> b) lim(cosh(x)-1)/(x^2) where x approaches 0. >Use Taylor series as others have suggested, or if you're feeling >adventurous, show that for any C^2 function f in a neighborhood of 0 with >f'(0) = 0, (f(x) - f(0))/x^2 -> f''(0)/2 as x -> 0. For b), think cosh(x) = sqrt(1 + sinh(x)^2). Rationalize the numerator by multiplying both numerator and denominator by 1 + sqrt(1 + sinh(x)^2) = 1 + cosh(x). Or substitute x = 2*y and use double-angle formulae. sinh(x) is better-behaved in a limit problem since sinh(x) approaches zero as x -> 0. -- After California's recall election, wildfires Schwartz-en-ed the Bush-lands on its geographic right (when we wanted the forests to be Green). pmontgom@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: Problem calculating limits - Need Help! > I'm having difficulties solving these two limits. > I mustn't use L'hospital rule: The only alternative which springs to mind is to expand in series valid for the limit under consideration. > a) lim(x*(2^(1/x))-x) where x increases to infinite. Write y = 1/x; take limit y -> 0: lim (2^y - 1)/y = lim (exp(y log2) - 1) / y > b) lim(cosh(x)-1)/(x^2) where x approaches 0. This is an easier exercise in Taylor series expansion. -- P.A.C. Smith The vast majority of Iraqis want to live in a peaceful, free world. And we will find these people and we will bring them to justice. === Subject: Re: Problem calculating limits - Need Help! > I'm having difficulties solving these two limits. > I mustn't use L'hospital rule: > The only alternative which springs to mind is to expand in series valid > for the limit under consideration. > a) lim(x*(2^(1/x))-x) where x increases to infinite. > Write y = 1/x; take limit y -> 0: > lim (2^y - 1)/y = lim (exp(y log2) - 1) / y > b) lim(cosh(x)-1)/(x^2) where x approaches 0. > This is an easier exercise in Taylor series expansion. Sorry, but I can't use series development. 10x, Roy === Subject: Re: Problem calculating limits - Need Help! > I'm having difficulties solving these two limits. > I mustn't use L'hospital rule: > a) lim(x*(2^(1/x))-x) where x increases to infinite. > b) lim(cosh(x)-1)/(x^2) where x approaches 0. Can you use series developments? -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Problem calculating limits - Need Help! > I'm having difficulties solving these two limits. > I mustn't use L'hospital rule: > a) lim(x*(2^(1/x))-x) where x increases to infinite. > b) lim(cosh(x)-1)/(x^2) where x approaches 0. > Can you use series developments? No, I can't. 10x, Roy === Subject: Re: Problem calculating limits - Need Help! >> I'm having difficulties solving these two limits. >> I mustn't use L'hospital rule: >> a) lim(x*(2^(1/x))-x) where x increases to infinite. >> b) lim(cosh(x)-1)/(x^2) where x approaches 0. >> Can you use series developments? > No, I can't. What a shame. They make such problems much easier :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Problem calculating limits - Need Help! In sci.math, Robin Chapman : > I'm having difficulties solving these two limits. > I mustn't use L'hospital rule: > a) lim(x*(2^(1/x))-x) where x increases to infinite. > b) lim(cosh(x)-1)/(x^2) where x approaches 0. > > Can you use series developments? >> No, I can't. > What a shame. They make such problems much easier :-( All this one needs is a simple variable change. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Can someone help with these diophantine equations? This is not homework. Could anyone with some numbercrunching power help solve these following diophantine equations (there are infinitely many solutions but I am looking for the one with the smallest A) (and of course, nonnegative integers only please!): A^2 = 729B + 364 A^2 = 2187B + 1093 A^2 = 6561B + 3280 A^2 = 19683B + 9841 The general pattern is A^2 = (3^k)B + SUM( i = 0 to k-1 )3^i k > 5 But help with even just 1 of the equations would be much appreciated. S.A. === Subject: Re: Can someone help with these diophantine equations? > This is not homework. Could anyone with some numbercrunching > power help solve these following diophantine equations (there are > infinitely many solutions but I am looking for the one with the > smallest A) (and of course, nonnegative integers only please!): > A^2 = 729B + 364 > A^2 = 2187B + 1093 > A^2 = 6561B + 3280 > A^2 = 19683B + 9841 > The general pattern is > A^2 = (3^k)B + SUM( i = 0 to k-1 )3^i k > 5 See also Robert Israel's posts in the thread, Squares that end with four identical digits -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Can someone help with these diophantine equations? Originator: twomack@chiark.greenend.org.uk ([193.201.200.170]) > This is not homework. Could anyone with some numbercrunching >power help solve these following diophantine equations (there are >infinitely many solutions but I am looking for the one with the >smallest A) (and of course, nonnegative integers only please!): >A^2 = 729B + 364 >A^2 = 2187B + 1093 >A^2 = 6561B + 3280 >A^2 = 19683B + 9841 OK, you're looking at successive approximations to the 3-adic square root of -1/2 So if you ask magma p := pAdicRing(3, 100); Sqrt(p!(-1/2)) you get 14918756739905062250418133874814682552290432142 and if you consider that large number mod 3^n you get the values of A that work at each step. For example, 475^2 = 729 * 309 + 365 1933^2 = 2187 * 1708 + 1093 You just need to look at X, X+3^n and X+2*3^n, modulo 3^(n+1), to see which digit to put on the beginning at each stage. I'm afraid I don't have a good reference for the p-adic numbers; they turn up incidentally in most number-theory books, but you're expected to absorb them instantaneously. Tom === Subject: Re: Can someone help with these diophantine equations? > This is not homework. Could anyone with some numbercrunching >power help solve these following diophantine equations (there are >infinitely many solutions but I am looking for the one with the >smallest A) (and of course, nonnegative integers only please!): >A^2 = 729B + 364 >A^2 = 2187B + 1093 >A^2 = 6561B + 3280 >A^2 = 19683B + 9841 > OK, you're looking at successive approximations to the 3-adic square root of > -1/2 > So if you ask magma > p := pAdicRing(3, 100); > Sqrt(p!(-1/2)) > you get 14918756739905062250418133874814682552290432142 That's one of the square roots - its negative is the other > and if you consider that large number mod 3^n you get the values of A that > work at each step. But you might not get the smallest A, which OP asked for, as that might sometimes come from the given square root and sometimes from the negative. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Can someone help with these diophantine equations? > But help with even just 1 of the equations would be much appreciated A^2 = B*c + d ---------------------- 254^2 = 29*2187 + 1093 254^2 = 88*729 + 364 475^2 = 309*729 + 364 1933^2 = 569*6561 + 3280 983^2 = 1325*729 + 364 1933^2 = 1708*2187 + 1093 1204^2 = 1988*729 + 364 2441^2 = 2724*2187 + 1093 4628^2 = 3264*6561 + 3280 8494^2 = 3665*19683 + 9841 1712^2 = 4020*729 + 364 1933^2 = 5125*729 + 364 11189^2 = 6360*19683 + 9841 4120^2 = 7761*2187 + 1093 2441^2 = 8173*729 + 364 2662^2 = 9720*729 + 364 4628^2 = 9793*2187 + 1093 8494^2 = 10996*6561 + 3280 3170^2 = 13784*729 + 364 3391^2 = 15773*729 + 364 6307^2 = 18188*2187 + 1093 11189^2 = 19081*6561 + 3280 3899^2 = 20853*729 + 364 6815^2 = 21236*2187 + 1093 4120^2 = 23284*729 + 364 4628^2 = 29380*729 + 364 4849^2 = 32253*729 + 364 8494^2 = 32989*2187 + 1093 15055^2 = 34545*6561 + 3280 9002^2 = 37053*2187 + 1093 5357^2 = 39365*729 + 364 28177^2 = 40336*19683 + 9841 5578^2 = 42680*729 + 364 17750^2 = 48020*6561 + 3280 30872^2 = 48421*19683 + 9841 6086^2 = 50808*729 + 364 10681^2 = 52164*2187 + 1093 6307^2 = 54565*729 + 364 11189^2 = 57244*2187 + 1093 6815^2 = 63709*729 + 364 7036^2 = 67908*729 + 364 21616^2 = 71216*6561 + 3280 12868^2 = 75713*2187 + 1093 7544^2 = 78068*729 + 364 13376^2 = 81809*2187 + 1093 7765^2 = 82709*729 + 364 24311^2 = 90081*6561 + 3280 8273^2 = 93885*729 + 364 8494^2 = 98968*729 + 364 15055^2 = 103636*2187 + 1093 15563^2 = 110748*2187 + 1093 9002^2 = 111160*729 + 364 47860^2 = 116373*19683 + 9841 9223^2 = 116685*729 + 364 28177^2 = 121009*6561 + 3280 50555^2 = 129848*19683 + 9841 9731^2 = 129893*729 + 364 9952^2 = 135860*729 + 364 17242^2 = 135933*2187 + 1093 17750^2 = 144061*2187 + 1093 30872^2 = 145264*6561 + 3280 10460^2 = 150084*729 + 364 10681^2 = 156493*729 + 364 11189^2 = 171733*729 + 364 19429^2 = 172604*2187 + 1093 11410^2 = 178584*729 + 364 19937^2 = 181748*2187 + 1093 34738^2 = 183924*6561 + 3280 11918^2 = 194840*729 + 364 12139^2 = 202133*729 + 364 37433^2 = 213569*6561 + 3280 21616^2 = 213649*2187 + 1093 12647^2 = 219405*729 + 364 22124^2 = 223809*2187 + 1093 12868^2 = 227140*729 + 364 67543^2 = 231776*19683 + 9841 13376^2 = 245428*729 + 364 70238^2 = 250641*19683 + 9841 13597^2 = 253605*729 + 364 23803^2 = 259068*2187 + 1093 41299^2 = 259961*6561 + 3280 24311^2 = 270244*2187 + 1093 14105^2 = 272909*729 + 364 14326^2 = 281528*729 + 364 43994^2 = 294996*6561 + 3280 14834^2 = 301848*729 + 364 25990^2 = 308861*2187 + 1093 15055^2 = 310909*729 + 364 26498^2 = 321053*2187 + 1093 15563^2 = 332245*729 + 364 15784^2 = 341748*729 + 364 47860^2 = 349120*6561 + 3280 28177^2 = 363028*2187 + 1093 16292^2 = 364100*729 + 364 16513^2 = 374045*729 + 364 28685^2 = 376236*2187 + 1093 87226^2 = 386545*19683 + 9841 50555^2 = 389545*6561 + 3280 17021^2 = 397413*729 + 364 17242^2 = 407800*729 + 364 89921^2 = 410800*19683 + 9841 30364^2 = 421569*2187 + 1093 17750^2 = 432184*729 + 364 30872^2 = 435793*2187 + 1093 17971^2 = 443013*729 + 364 54421^2 = 451401*6561 + 3280 18479^2 = 468413*729 + 364 18700^2 = 479684*729 + 364 32551^2 = 484484*2187 + 1093 57116^2 = 497216*6561 + 3280 33059^2 = 499724*2187 + 1093 19208^2 = 506100*729 + 364 19429^2 = 517813*729 + 364 19937^2 = 545245*729 + 364 34738^2 = 551773*2187 + 1093 20158^2 = 557400*729 + 364 60982^2 = 566804*6561 + 3280 35246^2 = 568029*2187 + 1093 106909^2 = 580680*19683 + 9841 20666^2 = 585848*729 + 364 20887^2 = 598445*729 + 364 109604^2 = 610325*19683 + 9841 63677^2 = 618009*6561 + 3280 36925^2 = 623436*2187 + 1093 21395^2 = 627909*729 + 364 37433^2 = 640708*2187 + 1093 21616^2 = 640948*729 + 364 22124^2 = 671428*729 + 364 22345^2 = 684909*729 + 364 67543^2 = 695329*6561 + 3280 39112^2 = 699473*2187 + 1093 22853^2 = 716405*729 + 364 39620^2 = 717761*2187 + 1093 23074^2 = 730328*729 + 364 70238^2 = 751924*6561 + 3280 23582^2 = 762840*729 + 364 23803^2 = 777205*729 + 364 41299^2 = 779884*2187 + 1093 41807^2 = 799188*2187 + 1093 24311^2 = 810733*729 + 364 126592^2 = 814181*19683 + 9841 24532^2 = 825540*729 + 364 74104^2 = 836976*6561 + 3280 129287^2 = 849216*19683 + 9841 25040^2 = 860084*729 + 364 43486^2 = 864669*2187 + 1093 25261^2 = 875333*729 + 364 43994^2 = 884989*2187 + 1093 76799^2 = 898961*6561 + 3280 25769^2 = 910893*729 + 364 25990^2 = 926584*729 + 364 45673^2 = 953828*2187 + 1093 26498^2 = 963160*729 + 364 46181^2 = 975164*2187 + 1093 26719^2 = 979293*729 + 364 80665^2 = 991745*6561 + 3280 for 1 <= B <= 1,000,000 courtesy of GP-Pari and about 4 minutes time. === Subject: Re: Can someone help with these diophantine equations? > This is not homework. Could anyone with some numbercrunching > power help solve these following diophantine equations (there are > infinitely many solutions but I am looking for the one with the > smallest A) (and of course, nonnegative integers only please!): > A^2 = 729B + 364 A = 254 (you can work out the B value) > A^2 = 2187B + 1093 254 > A^2 = 6561B + 3280 1933 > A^2 = 19683B + 9841 8494 > The general pattern is > A^2 = (3^k)B + SUM( i = 0 to k-1 )3^i k > 5 You're after the 3-adic expansion of sqrt(-1/2). I wouldn't expect much of a pattern in the answers. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Why Derivative is Inverse to Integral; geometric explanation formath education I'm really glad that you didn't forget this discussion, only don't understand, why did you take just this place of it. Someone likes trapezoid, someone other - picket fence, this is of no importance for integral. Of course, if you integrate numerically (even with a very small step), there will be the difference, and the more step the more difference. Numerical integration generally is better in parabolic curves, and even in curves of higher order. When you pass to infinitesimals, all these differences disappear, because ALWAYS only first term of expansion remains - just as in differentiating. The fact that expansion in Fourier series in irreversible, especially after operations with this series - this also is a well-known difficulty which one can solve only finding some class of analytical functions which would generalise all existing functions and would continuously transform from one analytic function into another with the parameters variation. I'm afraid, this task in general case is some alike seeking the philosophic stone. But in particular problems it is possible. I don't remember, have I told it in that discussion, but we dealt with such functions when studied the models of resistant elastic lines. The shape of one such function you can see in Fig. 2, page 22 of our paper Some features of vibrations in homogeneous 1d resistant elastic lumped line, http://angelfire.lycos.com/la3/selftrans/v2_1/resist22/resist22.html This function is described by the algebraic expression (16) in page 20 of that paper, http://angelfire.lycos.com/la3/selftrans/v2_1/resist20/resist20.html As you can see, the function varies with varied r from the asymptotic line to a line with a bend, retaining always analytical and algebraic! Besides, in the limiting case r = 0, this curve is described by a system of two functions - linear and asymptotically descending, with a bend at critical regime (see (25) in the page 22)! With it the analytical function approaches to a broken curve WITH ANY ACCURACY. Anyway, lest to build a separate plot in the diagram in Fig. 2, we built the limiting curve with the use of analytical function. Yes, this is a complicated function even in the complex form. Its complicacy grows even more in passing to real variables. Its integration is out of question. But we needn't to integrate, as this is just the solution of problem. If we have another elastic line, there will be simply another solution. Have you ever thought in this direction? I can show you a function that changes the number of resonance peaks with changing parameter, and many other. This means, transforming functions exist, just as exist the functions that combine several analytical functions. This is just what differs our solutions from conventional techniques. We yield these complicated functions in the end of solving, not trying to pluck the solution out of integral or to rape the Green function. This is why we have not problems which make a trouble to others. Though we have other problems, indeed. However this pragmatism is not utilitarianism. Our solutions are exact, analytic, not replicable by other techniques, they can fine work also in combination with numerical techniques, providing the accuracy and effectiveness of these last. But the main, our solutions are in full agreement with experiments. This is why I cannot understand, what for have you raised from archive that thought from our dialogue with Archimedes. Could you explain your meaning better? Sergey. PS: And there exists, of course, relation between the geometric representation of second derivative and acceleration in physics. >> just this Archimedes Plutonium meant when said of a slope of one of >> rectangle sides. To simplify a trapezium to a rectangle is simpler >> than to resume the lost information of the slope. So in most cases we >> are unable to integrate. The same, we can easily expand a function >> into Fourier series, but to reconstruct the original by the Fourier >> series - it's practically unsolvable problem. Or, if a function has >better than picketfence. >Can you think of a term for a rectangle that has an end-sides of just a >mere point rather than a true rectangle whose four sides are more than >just a point? >In the derivative there are no strange objects but in the integral there >is this >strange object of a rectangle whose end-sides are mere one point. >So in differentiation, there appears to be no real strange objects for a >set of trapezoids is normal geometric objects but in integration we have >this strange set of objects of rectangles whose end sides are one point. >And thus, I see geometrically the inverse relationship between derivative >and integral as that between trapezoids to one-point-sided rectangles. When >differentiating one makes trapezoids and when one integrates they collapse >the trapezoids into these strange rectangles. >Sergey, can you comment on the issue that derivative has normal geometric >objects of trapezoids (what I call picketfences) yet the integral relies >upon these >strange abnormal geometric objects of a collapsed rectangle whose end-sides >consist of one mere point. >i think trapezoids are stupid!!!! >So can we say that the essence of the Inverse relation between >derivative and >integral is the geometric idea that one is to expand strange rectangles into >trapezoids and the other is to collapse trapezoids into one-point end sided >rectangles. >P.S. I do not know if I should bother with a geometric explanation of the >2nd derivative that is acceleration in physics. >Archimedes Plutonium, a_plutonium@hotmail.com >whole entire Universe is just one big atom where dots >of the electron-dot-cloud are galaxies === Subject: square root of -1 mod p is there a way to compute sqrt(-1) mod p easily (for a large prime p, say 7-digits)? i know that -1 has a square mod p by the law of quadratic reciprocity, but that doesnt help constructing an x such that x^2+1=0 (mod p). i also know that the problem is equivalent to find numbers of order 4 (mod p), since order of sqrt(-1) is 4, but yet again that seems like a difficult problem. i am trying to do this without checking every number between 1 and p. anyone know about this? === Subject: Re: square root of -1 mod p Henri Cohen's book A Course in Computational Algebraic Number Theory(Graduate Texts in Mathematics, Vol 138), Springer-Verlag has a nice explanation of efficient ways to compute square roots modulo p in general, although as one poster pointed out, there are may well be quicker ways to compute the square root of -1. JoeS > is there a way to compute sqrt(-1) mod p easily (for a large prime p, > say 7-digits)? i know that -1 has a square mod p by the law of > quadratic reciprocity, but that doesnt help constructing an x such > that x^2+1=0 (mod p). i also know that the problem is equivalent to > find numbers of order 4 (mod p), since order of sqrt(-1) is 4, but yet > again that seems like a difficult problem. > i am trying to do this without checking every number between 1 and p. > anyone know about this? === Subject: Re: square root of -1 mod p |>is there a way to compute sqrt(-1) mod p easily (for a large prime p, |>say 7-digits)? i know that -1 has a square mod p by the law of |>quadratic reciprocity, but that doesnt help constructing an x such |>that x^2+1=0 (mod p). i also know that the problem is equivalent to |>find numbers of order 4 (mod p), since order of sqrt(-1) is 4, but yet |>again that seems like a difficult problem. In Maple 9, for example: > msolve(x^2=-1, 1299709); {x = 329008}, {x = 970701} takes very little time, even on my rather slow computer. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: square root of -1 mod p tungsteneer3@yahoo.com (billy d.) asked: > is there a way to compute sqrt(-1) mod p > easily (for a large prime p, > say 7-digits) You can use the continued fraction expansion of sqrt(p) to solve x^2 - py^2 = -1 and then x = sqrt(-1) (mod p). For an efficient way to do this, see Solving x^2 - Dy^2 = +-1 in Solving the generalized Pell equation at http://hometown.aol.com/jpr2718 Other than that, Henri Cohen, A Course in Computational Algebraic Number Theory, Section 1.5 might help. John Robertson === Subject: Re: square root of -1 mod p > tungsteneer3@yahoo.com (billy d.) asked: >> is there a way to compute sqrt(-1) mod p >> easily (for a large prime p, >> say 7-digits) >You can use the continued fraction >expansion of sqrt(p) to solve > x^2 - py^2 = -1 >and then x = sqrt(-1) (mod p). For an >efficient way to do this, see Are you sure this is wise? For p of the range indicated by the OP, one expects to find continued fractions with periods up to the thousands. Even though this is not overwhelmingly large in terms of space or time complexity, I'm not sure I see the point of doing it this way. -- Erick === Subject: Re: square root of -1 mod p >is there a way to compute sqrt(-1) mod p easily (for a large prime p, >say 7-digits)? i know that -1 has a square mod p by the law of >quadratic reciprocity, but that doesnt help constructing an x such >that x^2+1=0 (mod p). i also know that the problem is equivalent to >find numbers of order 4 (mod p), since order of sqrt(-1) is 4, but yet >again that seems like a difficult problem. It's not difficult at all (unless you demand a deterministic algorithm). For any b, b^[(p-1)/4] has order at most 4 (mod p), so if you pick b randomly there is a 50% chance that it will have order exactly 4 (the only other possibilities are order 1 and order 2 which are +1 and -1 respectively). -- Erick === Subject: Re: square root of -1 mod p Adjunct Assistant Professor at the University of Montana. >is there a way to compute sqrt(-1) mod p easily (for a large prime p, >say 7-digits)? i know that -1 has a square mod p by the law of >quadratic reciprocity, You mean, you know that -1 has a square mod p when p is congruent to 1 mod 4; this does not use quadratic reciprocity per se (which is about two odd primes), but rather the Legendre symbol and similar results. > but that doesnt help constructing an x such >that x^2+1=0 (mod p). i also know that the problem is equivalent to >find numbers of order 4 (mod p), since order of sqrt(-1) is 4, but yet >again that seems like a difficult problem. >i am trying to do this without checking every number between 1 and p. Usually, one proves that x^2 = -1 (mod p) has a solution if and only if p=2 or p=1 (mod 4) by ->constructing<- a solution. Namely, take (p-1)/2, and compute its factorial mod p. That's a solution, a consequence of Wilson's Theorem. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: square root of -1 mod p >is there a way to compute sqrt(-1) mod p easily (for a large prime p, >say 7-digits)? > Usually, one proves that x^2 = -1 (mod p) has a solution if and only > if p=2 or p=1 (mod 4) by ->constructing<- a solution. Namely, take > (p-1)/2, and compute its factorial mod p. That's a solution, a > consequence of Wilson's Theorem. Yes, but if you actually want to compute a solution, finding (p - 1) / 2 factorial mod p is a really bad way to do it, at any rate for the 7-digit p that OP has. A couple of algorithms are given in Crandall & Pomerance, Prime Numbers, Section 2.3.2 -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: S^n x S^m Homology I find it difficult to find the homology group of S^n x S^m. Can anyone give me some ideas? I am thinking of using the MV sequence (exact): ... -->H_n(A and B)-->H_n(A) + H_n(B)--> H_n(A union B)-->H_(n-1) (A and B) --> ... === Subject: Re: S^n x S^m Homology > I find it difficult to find the homology group of >S^n x S^m. Can anyone give me some ideas? I am thinking of using the MV >sequence (exact): >... -->H_n(A and B)-->H_n(A) + H_n(B)--> H_n(A union B)-->H_(n-1) (A and >B) --> ... I'm not sure that's the best approach but you could make it work. What will you choose for A and B? You might try decomposing S^m = C union D and then take A = S^n x C, B = S^n x D. You'll need to do the calculation for all m this way so that you can use induction. Don't forget to use the naturality of the MV sequence so that you can identify the maps between the groups. By the way, Mr V just died about a year ago -- at the time he was the oldest living person in Austria (110.8 yrs). dave === Subject: Re: S^n x S^m Homology > I find it difficult to find the homology group of > S^n x S^m. Can anyone give me some ideas? Apply the Eilenberg-Zilber and Kunneth theorems. Let S(X) denote the singular complex of a space X. The Eilenberg-Zilber theorem states that S(X x Y) is chain homotopy equivalent to the tensor product of S(X) and S(Y). The Kunneth theorem expresses the homology of the tensor product of two complexes in terms of the homology of each one. Here life is simple since the homology of S^m and of S^n is torsion-free so we can forget the Tor factor in the Kunneth theorem and conclude that the total homology of S^m x S^n is 4-dimensional with generators in dimensions 0, m, n and m+n. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Can anyone recommend a good lie algebra book? Looking for a good lie algebra book at the introductory level. === Subject: Re: Can anyone recommend a good lie algebra book? > Looking for a good lie algebra book at the introductory level. Varadarajan's Lie groups, Lie algebras, and their representations and Serre's Lie Algebras and Lie Groups are good choices. Jose Carlos Santos === Subject: Re: Can anyone recommend a good lie algebra book? > Looking for a good lie algebra book at the introductory level. Humphreys, James E. Introduction to Lie algebras and representation theory. (2.ed) Graduate Texts in Mathematics, 9. Springer-Verlag, New York-Berlin, 1978. should still serve as a good introduction. It requires not muuch more that linear algebra. Marc === Subject: The Cipher of Genesis re: http://qedcorp.com/APS/EmergentGravity.doc http://qedcorp.com/APS/StarGate1.mov Commentary 1 The fiber bundle as an idea has 4 parts. 1. A structure symmetry group G. 2. The total hyperspace H or, in some applications Wheeler's BIT. 3. The projection map P. 4. The base space M or, in some applications. Wheeler's IT. The hyperspace H consists of fibers f(x) that are either copies of or representations of the symmetry group G. The projection map P collapses a fiber f(x) in the hyperspace H to a point x in the base space M. All of these objects are continuum differential manifolds depending on the continuum of real numbers which its associated issues of Cantor's infinity of infinities of Cabalistic Aleph's in an ascending Jacob's Ladder. This is not a discrete combinatoric mathematics although such a skeletal structure is associated with it as in Herman Weyl's Theory of Groups and Quantum Mechanics and as in Saul-Paul Sirag's presentation of V.I. Arnold's A-D-E mathematics of everything. The base space is covered by an atlas of local coordinate patches with all important overlap transition functions sewing the patches together like a quilt. M is space-time in local micro-quantum field theory of point The extra-dimensions of hyperspace form the Calabi-Yau space of vibrations of the superstring beyond space-time. The connection on the total hyperspace H is the potential of a local gauge force. Examples of connections is the 4 potential Au(x) in Maxwell's electromagnetism with G as U(1). There are similar connections for the Yang-Mills weak force with G = SU(2) and the strong force with G = SU(3). Classical general relativity, as distinct from local micro-quantum field theory, has the torsion-free symmetric three-index non-tensor Levi-Civita connection with G as the Diff(4) group. The latter comes from locally gauging the 4 parameter translation subgroup (generated by the 4-momentum Pu of globally flat special relativity ) of the 15 parameter conformal group of Roger Penrose's massless twistors. Bottom -> Up: Given base space M and symmetry group G construct the hyperspace H as a quilt patchwork. Top -> Down: Given hyperspace H and symmetry group G construct the base space M as the non-overlapping partition of hyperspace into G-orbits called the quotient space of H mod G in the principal bundle. Micro-quantum source renormalizable local fields of spin 1/2 lepto-quarks are associated vector bundles. Micro-quantum force renormalizable local fields of spin 1 gauge force bosons (electro-weak and strong) are from the principal bundle. There is no renormalizable quantum gravity in this precise sense. This is because classical Einstein gravity is a More is different (P.W. Anderson) emergent collective effect as in Andrei Sakharov's metric elasticity of an instability in the globally flat false vacuum of the interacting lepto-quark source/electroweak-strong force. Einstein's gravity + unified exotic vacuum dark energy/matter with Andrei Linde's chaotic inflationary cosmology are the result of the continual phase transitions from globally flat false high entropy micro-quantum vacua to locally curved macro-quantum low entropy metastable vacua. to be continued: === Subject: Re: The Cipher of Genesis Mr. Sarfatti Sir How are you..how do you feel? dave === Subject: Re: The Cipher of Genesis Nothing, as usual, and was prolix about it. Society national meeting in Denver next year? Uncle Al knows a fellow who was in your audience last year. He said you had everybody rolling on the floor clutching their tummies - some from laughter, others puking, and the remainder collecting lint for tinder for igniting your faggots. -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: The Cipher of Genesis Tiz a frightened man whom scoffs! > Nothing, as usual, and was prolix about it. > Society national meeting in Denver next year? Uncle Al knows a fellow > who was in your audience last year. He said you had everybody rolling > on the floor clutching their tummies - some from laughter, others > puking, and the remainder collecting lint for tinder for igniting your > faggots. > -- > Uncle Al > http://www.mazepath.com/uncleal/qz.pdf > http://www.mazepath.com/uncleal/eotvos.htm > (Do something naughty to physics) === Subject: Re: How Many Blinkin' Lights On The Tree? > God isn't responsible for the bad things that happen, Satan is. Read > your damn Bible, specifically, the Book of Job. > Lol and god created Satan knowing *FULL* well what would happen > (unless hes not omniscient). > So then the question is, if God knew what would happen, then perhaps Satan > isn't such a bad guy after all? Consider: he obeyed God's commands > concerning how much he was allowed to afflict Job. Sounds like an obedient > son of God to me... Hehe maybe satan isn't? Or maybe god just isn't as powerful or as loving as everyone says he is (if he exists at all)? I don't know what you mean by an obedient son of god but Job did get a bit drilled by the almighty. === Subject: Re: Vector Calculus problem === >Subject: Vector Calculus problem >Ok, I'm trying to work on my homework and am stuck on 4.9 #10 of Vector >Analysis by Davis. >The question states By means of Stokes' theorem, find S F*dR around the >ellipse x^2+y^2=1, z=y, where F=xi+(x+y)j+(x+y+z)k. z=y ? what? >I got the curl of F and that equalled i-j+k but I'm not really sure how to do >the rest of the problem. Any help would be appreciated. I've wasted a lot of >time and gotten almost nowhere. write down an expresion for the (vector) dA for your circle... I think that you should use (vector) dA = (scalar) dA k then curl F dot dA is just (scalar) dA and a circle of radius one has area 2 Pi adam === Subject: Re: Vector Calculus problem >>Ok, I'm trying to work on my homework and am stuck on 4.9 #10 of Vector >>Analysis by Davis. >>The question states By means of Stokes' theorem, find S F*dR around the >>ellipse x^2+y^2=1, z=y, where F=xi+(x+y)j+(x+y+z)k. >z=y ? what? >>I got the curl of F and that equalled i-j+k but I'm not really sure how to do >>the rest of the problem. Any help would be appreciated. I've wasted a lot of >>time and gotten almost nowhere. >write down an expresion for the (vector) dA for your circle... >I think that you should use (vector) dA = (scalar) dA k >then curl F dot dA is just (scalar) dA >and a circle of radius one has area 2 Pi >adam uhhh. duh no a circle of radius one has area Pi. hmmm, maybe the problem here is that I'm not sure what the z=y condition means... sorry adam === Subject: Re: Vector Calculus problem === >Subject: Vector Calculus problem >Ok, I'm trying to work on my homework and am stuck on 4.9 #10 of Vector >Analysis by Davis. >The question states By means of Stokes' theorem, find S F*dR around the >ellipse x^2+y^2=1, z=y, where F=xi+(x+y)j+(x+y+z)k. >I got the curl of F and that equalled i-j+k but I'm not really sure how to do >the rest of the problem. Any help would be appreciated. I've wasted a lot of >time and gotten almost nowhere. ahh, z=y means the ellipse is tilted in the z-y plane and the projection of it on the x-y plane is x^2+y^2=1 so the area vector of the ellipse is pointing in the (0,-1,1) direction so that you get two contributions from curlF dot dA I.e. integrate (-j+k)dot(-j+k)= 2 over the circle to get 2 Pi Also, Check that the back of book is right by doing it explicitly: x=cost y=sint z=sint t==0..2Pi dR=i(-sint)+j(cost)+k(cost) F=i(cost)+j(cost+sint)+k(cost+2sint) FdotdR=(-costsint)+(cos^2t+costsint)+cos^2t + 2costsint integrate from 0 to 2pi = 0 + Pi + 0 + Pi + 0 = 2 Pi adam === Subject: Re: Indian firsts in Maths <8DWsb.193431$HS4.1622428@attbi_s01> <3fb4d62b.18805594@news.clara.net> at 03:12 PM, habshi@anony.com (habshi) said: === >Subject: Indian firsts in Maths ROTF,LMAO! You list a bunch of tripe that has nothing to do with Mathematics, But you forget G. H. Hardy's greatest discovery, Ramanujuan! Plus, of course, you misrepresented some and you neglected Chinese and Greek work much earlier than some things you list. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Indian firsts in Maths > INVENTIONS & DISCOVERIES > INVENTION OF NUMERALS > Numerals are found in the inscriptions of Ashoka The Great in the 3rd > Century BC. This knowledge traveled from there to Europe and West. In > Arab countries even now numerals are known as HINDSE: from India. La > time, It is India that gave us the ingenious method of expressing all > numbers by means of ten symbols - Prof. O.M. Mathew in Bhavan's > Journal > INVENTION OF ZERO > Brahmagupta was the first mathematician to treat ZERO (0) as a number > and showed its mathematical operations > INVENTION OF ARITHMETIC > Arithmetic was discovered by Indians in about 2nd Century BC. > Bhaskaracharya's book Lilavathi is regarded as the first book on > modern arithmetic. The Arabs learnt and adopted it from India and > spreaded it to Europe. In 499 AD Aryabhatta finished his work > Aryabhatt, giving rules of Arithmetic (Encyclopedia Britannica) > INVENTION OF ALGEBRA > In Western Europe the knowledge of Algebra was borrowed, not from > Greece but from Arabs, who acquired this from India. Algebra is the > only Arabic name for Bijaganitha. Aryabhatta was one of the first to > use Algebra (Encyclopedia Britannica) > INVENTION OF GEOMETRY AND TRIGNOMETRY > The brick work of Harappa and Mohenjodaro excavations show that people > of ancient India (2500 BC) possessed knowledge of Geometry. Aryabhatta > formulated the rules for finding the area of a ?triangle', which led > to the origin of Trignometry. > DISCOVERY OF ASTRONOMY > The knowledge of the motion of heavenly bodies was discovered by > Aryabhatta (499 AD), Latadeva (505 AD) and Brahmagupta (628 AD), for > calculating the timing of eclipses. In Surya Sidhanta' Latadeva, > talked about the earth's axis and called it SUMERU. That the earth is > a sphere and it rotates on its own axis, was known to Varahamihira > and other Indian astronomers much before Copernicus published this > theory. (Jewish Encyclopedia) > INVENTION OF CALENDAR MAKING > Discovery of measurement of time and discovery of nomenclature of > days, month and years and invention of calendar making was made in > India. In his book ?Surya Sidhanta' Latadeva (505 AD) divided the year > into 12 months. Seven planets of the solar system effect the earth's > atmosphere and their names were added to the seven days of the week, > which was accepted all over the world. > DISCOVERY OF THEORY OF GRAVITATION > In his book ?Sidhanta Shiromani' Bhaskaracharya mentions about force > of attraction resembling gravity, discovered centuries later by > Newton. (Jewish Encyclopedia) > INVENTION OF IRON PRODUCTS IN 3000 BC > The word AYAS occurs in the four Vedas which denotes iron. Ashoka > pillar at Mehrauli, New Delhi and another iron pillar in Karnataka > stand proof of India's metallurgical heritage (A study published in > the magazine ?The Current Science'). > INVENTION OF COPPER, BRONZE AND ZINC > The copper and bronze artifacts dates back to Indus Valley > Civilization (2500 BC). According to treatise RASARATNAKAR, zinc was > made in around 50 BC at Zawar in Rajasthan (India). > INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS > Chemistry known as RASAYAN SHASTRA was invented in India. Elphinstone > to prepare the sulphate of copper, zinc and iron and carbonates of > lead and iron. RASAVIDYA or Indian alchemy made its appearance around > 5th Century AD (National Science Centre, New Delhi) At http://ancienthistory.about.com/gi/dynamic/offsite.htm?site=http%3A%2F%2Fmem b ers.aol.com%2Fbbyars1%2Ffirst.html is the title The First Mathematicians. At http://www.stormloader.com/ajy/zero.html is the title Zero. === Subject: Re: Was Lucy the split image - Jabriol is a creationist. Psycho-hypocrite Antonio once again resorts to the same ad-hom he so whiningly accuses others of! How the mighty have fallen!. Look at this hilarious bull: > oh by the way since you are in an informative mood.. tell them how you > have been caught lying so much.. that nobody belives you anymore.. Nobody ia a bigger liar than you, as I have repeatedly demonstrated. BTW, you have now proven evolution by evolving into a weasel. Budikka === Subject: Re: Vedic Mathematics --- Myth and Reality (a working out) >> See, exactly the same as your method. >Not at all! It's 4 things to combine, no matter how it's done. > Oh, it matters a lot how it's done. Because it is done differently, it > is a different method. Because it is done more easily and quickly > both in the mind and on paper, it is a better method. > Absolutely. But: > 1. is it Vedic? That's what I too would like to know. I think it is, for I suspect my father knew this method. If a number of Indian grandparents from all over India jump up saying that they knew this method, that it was taught to them traditionally, then it would be Vedic (and not a fraud derived from say Tractenberg or someone and put in a book and passed off as Vedic) unless one showed that they were all lying. > 2. is it really such a big deal? Oh yes. The idea is that it should be taught in primary schools, replacing the current multi-line method. If that is indeed done all over the world, then certainly is is a big deal. > Ad 1: The expression crosswise vertical doesn't say anything about the order of addition. That's the beauty of it. > Ad 2: Math (at least western math) is much more than arithmetic. It sure is a cool thing > to do 345 x 167 in the mind. Especially for school children. But it's not an advance > in the field of mathematics. It's simply arithmetic. And isn't arithmetic really wonderful? Especially when you know what it is all about? Vedic arithmetic shows you that. If implemented, it will be an advance in arithmetic, and arithmetic was taught to me in school as a part of mathematics. We started with arithmetic, then we had algebra, then geometry, then trigonometry, then co-ordinate geometry and finally calculas. Look, let us do the following sum using Vedic arithmetic: (1234*5678 + 5659873 - 9987654)*345 just like that. Well, we get ( (5)(16)(34)(60)(61)(52)(32) + 5659873 - 9987654)*345 or ((10)(22)(39)(69)(69)(59)(35) - 9987654)*345 or ((1)(13)(31)(62)(63)(54)(31))*345 or (3)(43)(150)(379)(632)(724)(624)(394)(155) or 925010495. I'm not sure if I have worked it out correctly, for I was only trying to show the Vedic method, how beautifully and elegantly it works, that is, without bothering about all the carries from one operation to another, as done in current methods taught in school. Concentrating on the place values, and not the carries, makes life lot easier for the working-out. If the beauty of all this does not strike you or anyone, I have nothing further to say. And still I don't know the one-line division and square root extraction method. As far as I can see, that is also unknown to anyone around. But these methods have been published in the book on Vedic mathematics, that is hailed as a fraud. Arithmetic is the basis of all computation. And computation is so important, in business, finance, engineering, etc. Faster computation will naturally lead to greater efficiencies. > And for the purpose of math education, learning people calculation 'tricks' > is generally considered poor education of math. We are talking about teaching the fundamentals of arithmetic to children using better methods. After 44 years of dealing with mathematics, I am of the firm conviction that mathematics is an art, and like all arts, it is a bag of tricks. I am confident that people will learn to love maths, once they get over their dread of arithmetic, taught the usual way. Anyway, the best judges for this will not be experienced mathematicians, but primary school teachers and their charges, oce the green signal has been given by the educators. Arindam Banerjee. > Herman Jurjus === Subject: Re: Vedic Mathematics --- Myth and Reality >> Equivalently, M*N is the same as M*N mod (M + N - 1). >> Sorry, this should be multiplication of M digits with N digits, base b, >> is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits. >Oh well, so FFT or not, looks like multiplying M by N by any method means >MN multiplications! Well, when these multiplications are hardwired (as in >human memory for single digits) the computational issues (On*n) becomes >really irrelevant, for they all are done in no time at. Like, the video >extraction for radar data processing is done by NAND gates - its all done in >real time! > You are in error. The number of multiplications required for > multiplying two numbers with the FFT method is O(n*log(n) where n is > the larger of the two numbers; it is not m*n. Fine, just multiply 12345 by 67809 using FFT with less than 25 multiplications. Do it here. > Richard Harter, cri@tiac.net > http://home.tiac.net/~cri, http://www.varinoma.com > We have people from every planet on the earth in this State. > -- California Governor Gray Davis === Subject: Re: Vedic Mathematics --- Myth and Reality > > Equivalently, M*N is the same as M*N mod (M + N - 1). >> Sorry, this should be multiplication of M digits with N digits, base b, >> is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits. Oh well, so FFT or not, looks like multiplying M by N by any method means >MN multiplications! Well, when these multiplications are hardwired (as in >human memory for single digits) the computational issues (On*n) becomes >really irrelevant, for they all are done in no time at. Like, the video >extraction for radar data processing is done by NAND gates - its all done in >real time! > > You are in error. The number of multiplications required for > multiplying two numbers with the FFT method is O(n*log(n) where n is > the larger of the two numbers; it is not m*n. > Fine, just multiply 12345 by 67809 using FFT with less than 25 > multiplications. Do it here. Do not misinterpret this beahaviour as typical arrogance of brahmins. His perception is so narrow, he cannot understand others points. So he demands explanations in his own ways. If he wants to learn how to calculate squres and if you teach him how to find cubes, he may get confused and may stop learning mathematics. > > > Richard Harter, cri@tiac.net > http://home.tiac.net/~cri, http://www.varinoma.com > We have people from every planet on the earth in this State. > -- California Governor Gray Davis === Subject: Re: Vedic Mathematics --- Myth and Reality > > Equivalently, M*N is the same as M*N mod (M + N - 1). >> Sorry, this should be multiplication of M digits with N digits, base b, >> is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits. Oh well, so FFT or not, looks like multiplying M by N by any method means >MN multiplications! Well, when these multiplications are hardwired (as in >human memory for single digits) the computational issues (On*n) becomes >really irrelevant, for they all are done in no time at. Like, the video >extraction for radar data processing is done by NAND gates - its all done in >real time! > > You are in error. The number of multiplications required for > multiplying two numbers with the FFT method is O(n*log(n) where n is > the larger of the two numbers; it is not m*n. > Fine, just multiply 12345 by 67809 using FFT with less than 25 > multiplications. Do it here. Do not misinterpret this beahaviour as typical arrogance of brahmins. His perception is so narrow, he cannot understand others points. So he demands explanations in his own ways. If he wants to learn how to calculate squres and if you teach him how to find cubes, he may get confused and may stop learning matheamatics. > > > Richard Harter, cri@tiac.net > http://home.tiac.net/~cri, http://www.varinoma.com > We have people from every planet on the earth in this State. > -- California Governor Gray Davis === Subject: Re: Need advice on letters of recommendation > 1. When a student (or anyone) asks for a letter, it is > implicit that they are asking Will you write me a _good_ > recommendation. Although I do have one example to the contrary. Many years ago, an undergraduate asked me to write her a letter of recommendation for mathematics graduate school. I told her that I wouldn't be able to say very much beyond she completed most of the homework assignments and she passed the course, but she still wanted me to write the letter, which I did. The story turned out to be that she was an overseas student whose government had been paying her expenses. The catch was that she had to apply to graduate school. Didn't have to go to grad school, but had to apply. She didn't want to go, and knew as well as I did that it wasn't a good idea for her to go, so what she actually wanted was a letter that would get her rejected! Unfortunately for her, one of the places she applied to was so desperate for students that it accepted her despite the lukewarm letters of recommendation she presented. I regret that I don't know how things turned out in the end. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Need advice on letters of recommendation >> >> letters of recommendation are about the most ludicrous mechanism in >> place in academia - it's a shame. institutional admissions exams >> would be far superior. >> >> Letters of recommendation may well be ludicrous. I have no reason >> to believe, and much reason to disbelieve, that institutional >> admissions exams would be any better at deciding who would be >> a good graduate student. >perhaps you do not believe that a solid undergraduate >preparation is key component in graduate studies success. >is that it? > I suppose I believe that, for a value of key that may be > lower than your value of key. What I'm sure I believe is that > (1) institutional admissions exams probably aren't particularly > good at measuring that component in graduate studies success > (since I don't have much faith in exams), and that well, your faith regarding institutional exams has to be the weakest defense you have for your belief. perhaps if you made a statement, and justified it, regarding subjectivity, perhaps you would have better luck. > (2) as Bart Goddard has pointed out, letters of recommendation do > have at least some chance of indicating that other (for me, much more > key) component in graduate studies success, namely, the > ability to *do mathematics*, while (3) institutional admissions > exams surely couldn't do anything at all towards indicating that > other component. the ability to solve mathematical problems, say in an admissions exam, is without a doubt, a much better indicator of ability to do mathematics. in fact, the ability to solve mathematics problems is the only accurate indicator of ability in mathematics. do you understand where i am coming from? >> Go into detail on how you'd make such an exam, if you don't mind. >i do not need to go into any detail to ascertain that a >resounding reason for the lack of institutional exams in the >united states is correlated with the economic and logistics >required to put such systems in place. > You certainly do not need to go into any detail to do that > on *my* behalf, because I didn't express (and don't have) any > interest in your opinion on that question, which I didn't raise. > The question I raised was, How would you make such an exam (with > some reasonable hope that it would do what you want it to do)? you propose the question, you seek the answer. not from me though. > Lee Rudolph === Subject: Re: Need advice on letters of recommendation ... >> (2) as Bart Goddard has pointed out, letters of recommendation do >> have at least some chance of indicating that other (for me, much more >> key) component in graduate studies success, namely, the >> ability to *do mathematics*, while (3) institutional admissions >> exams surely couldn't do anything at all towards indicating that >> other component. >the ability to solve mathematical problems, say in an admissions exam, >is without a doubt, a much better indicator of ability to do >mathematics. Scarcely without a doubt! The conditions of an exam, admissions exam or otherwise, are entirely different from the conditions under which (real, practicing) mathematicians (of my broad personal acquaintance--which, I admit, does not include you, who may for all I know work along entirely different lines) actually do mathematics. And such problems as must, necessarily, be posed on an exam (admissions exam or otherwise), are just that--problems (with known solutions) to solve, not new mathematics to do (and possibly fail at). >in fact, the ability to solve mathematics problems is the >only accurate indicator of ability in mathematics. I would trust the seriously considered opinion of almost any of my professional acquaintances, expressed in a letter of recommendation, that a third party had--or had not--the ability to do mathematics, far more than I would trust an admissions exam. I wouldn't trust a stranger's opinion as much, but I'd still trust it more than an exam. >do you understand >where i am coming from? Either (if I am to be charitable) from somewhere far from where I am, or (otherwise) from a position of profound ignorance of what it means to do mathematics. Lee Rudolph === Subject: Re: Need advice on letters of recommendation >> > ... >> (2) as Bart Goddard has pointed out, letters of recommendation do >> have at least some chance of indicating that other (for me, much more >> key) component in graduate studies success, namely, the >> ability to *do mathematics*, while (3) institutional admissions >> exams surely couldn't do anything at all towards indicating that >> other component. >the ability to solve mathematical problems, say in an admissions exam, >is without a doubt, a much better indicator of ability to do >mathematics. > Scarcely without a doubt! The conditions of an exam, admissions > exam or otherwise, are entirely different from the conditions > under which (real, practicing) mathematicians (of my broad > personal acquaintance--which, I admit, does not include you, > who may for all I know work along entirely different lines) > actually do mathematics. And such problems as must, > necessarily, be posed on an exam (admissions exam or > otherwise), are just that--problems (with known solutions) > to solve, not new mathematics to do (and possibly fail at). if you were wishing to establish that the conditions between taking exams and being a mathematician are different, congratulations, you have done a superv job! i wonder though, how does that damage my statement? >in fact, the ability to solve mathematics problems is the >only accurate indicator of ability in mathematics. > I would trust the seriously considered opinion of almost any > of my professional acquaintances, expressed in a letter of > recommendation, that a third party had--or had not--the > ability to do mathematics, far more than I would trust an > admissions exam. I wouldn't trust a stranger's opinion > as much, but I'd still trust it more than an exam. i reckon that is a problem you and many others share. >do you understand where i am coming from? > Either (if I am to be charitable) from somewhere far from where > I am, or (otherwise) from a position of profound ignorance of > what it means to do mathematics. perhaps now you will reject the notion that a mathematician is defined by his ability to solve math problems? no, i am not referring to ones with known solutions... > Lee Rudolph === Subject: Re: Need advice on letters of recommendation >> The question I raised was, How would you make such an exam (with >> some reasonable hope that it would do what you want it to do)? > you propose the question, you seek the answer. not from me though. There is a GRE Subject test in Mathematics which covers the a general knowledge one would expect a math major to have mastered at a pretty good school. I downloaded the practice exam a few months ago and worked it. It is, of course, multiple choice, but one has to evaluate contour integral, say which of the following is not a basis for a vector space, do arithmetic in a dihedral group, and so on. No proofs, but there are a couple questions from every normal course in the undergraduate sequence: topology, number theory, algebra, analysis, probability, statistics, etc. And Calculus, just to check. Some graduate school require the test. The reason I downloaded the test was that I was designing a new math major, and wanted to go through the questions and how many of the them a potential graduate from my new program would have a chance of answering. The practice test was good for pointing out holes in my program. (Namely, complex analysis and topology.) Not that I have to kowtow to the test designers, but the test designers are trying to find what's typical for an American math major, and so I thought the test would be at least somewhat normative. Anyway, if one is curious, the practice test is downloadable from the GRE website. Bart === Subject: Re: SymbMath.com: web-based computer algebra system > SymbMath For Java is web-based symbolic math and computer algebra > system, > which runs in any computer with Java. You can play it online. www.SymbMath.com === Subject: Re: Data analysis software It plots and analyses any x-y data for peak location, peak height, > peak > width, semi-derivative, derivative, integral, semi-integral, > convolution, > deconvolution, curve fitting, and separating overlapped peaks > and > background. > www.chemSoftware.com === Subject: Re: online math handbook > web-based symbolic calculator, math handbook and computer algebra system. > www.mathHandbook.com === Subject: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB? I'm trying to get the deadwood off my reading shelf, and I just finished BITCH, by Elizabeth Wurtzel, which turned out to be worth reading for her powerful, insightful, and soul-baring epilogue, but too much of the rest of the book was just too much, over and over, about O.J. Simpson. Worse, Ms. Wurtzel is a cinema-holic who dismisses the woes of real-life victims who don't play their parts well, as if we should all have just the right life scripts for our tragedies and have the necessary acting abilities to earn her respect and sympathy, or at least a tepid encore? And, one of the next space-hoggers on my shelf is INFINITY AND THE MIND: THE SCIENCE AND PHILOSOPHY OF THE INFINITE, by Rudy Rucker. Rudy ticked me off yesterday -- I had hoped to get a good start on finishing the rest of the book, but then he stopped writing his barely comprehensible Math-eze and added some stinking number questions, including the following: Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a) So I did: (1 - a) [1 + (a + a^2 + a^3 . . .)] = (1 - a) (1 / (1 - a)) {multiplying both sides by (1 - a)} (1 - a) + [(1 - a) (a + a^2 + a^3 . . .)] = 1 {interim result} (1 - a) + [(a + a^2 + a^3 . . .) - (a^2 + a^3 + a^4 . . .)] = 1 {interim result} (1 - a) + a = 1 {interim result} 1 = 1 {final proof of equality} This proves that the original equation is true no matter what value of a we use, but Rudy argues that the original formula makes no sense when we substitute certain values for a. However, before we discuss it, let's give a step-by-step definition of division so that everybody can follow along: A / B asks as to count how many times we can subtract all of B from A to the limit of A, plus count any fractional units of B that we can subtract from A, and to state the total as our answer. For example, 9 / 5 asks us to count how many times we can subtract 5 from 9 (once), plus how many fractional units of 5 we can subtract from the remainder of 4 (and our total answer is 1 & 4/5ths). Now, in the above formula: 1+ (a + a^2 + a^3 . . .) = 1 / (1 - a), Rudy notes that if we use 2 for a, then the formula becomes: 1 + (2 + 4 + 8 . . .) = 1 / (1 - 2), or 1 + (2 + 4 + 8 . . .) = 1 / -1 The brain stops working when we see an ever-increasing series on the left side and a simple negative fraction on the right side, and we conclude that the two expressions are not equal. However, if we force that supposedly simple negative fraction through the true process required by division, as explained above, we will see how complex that simple fraction really is, and that it exactly equals the expression on the left side: 1 / -1 says to state how many times we can subtract -1 from 1 to the limit of 1 plus state the fractional units. The first time that we subtract -1 from 1 we get 2 and move away from our limit of 1 instead of towards it (which makes sense as a reverse process from a division process using numbers with the same sign), and we immediately have a whole count of 1 subtraction: 1 + (. . .) After subtracting -1 from 1, the 1 now has a remainder of 2 after being subtracted from, and thus our division limit is now 2 instead of 1, and the fractional units of -1 that can subtracted from the limit of 2 becomes 2 / -1, or -2. As such, the 'fractional' units to be added to our count comes from the next division of 2/ -2., which says to subtract -2 from 2, which equals 4 (to the limit of 2), and to add the interim answer to our series: 1 + (2 + . . .) And so on as above: 1 + (2 + 4 + 8 . . .) = 1 / -1 The rote learners among us will immediately object that 1 / -1 = -1, and thus that -1 cannot possibly equal 1 + (2 + 4 + 8 . . .). The argument ignores the fact that transforming 1 / -1 into -1 is an algebraic process (-1 = [-1] * 1 = [-1] * 1 / 1 = etc.), and is not a given fact, and any such transformation should be performed on both sides of the equation to be truly and logically valid in all cases, regardless that it is irrelevant in most cases. Thus, whenever the division process itself is crucial to the meaning of an equation, it cannot simply be dispensed with or collapsed on one side of the equation alone. Thus, the above idea is not contrary to algebra, but rather states a point that is not being taught. It's a distinction, not a contradiction. Rudy gives a more interesting but nonetheless complying example of a substitution for a where the substitution is -1: If a = -1 in: 1 + (a + a^2 + a^3 + a^4 + a^5 . . .) = 1 / (1 - a), then all the even powers of a will equal 1 and all the odd powers of a will equal -1: 1 + (-1 + 1 - 1 + 1 - 1 . . .) = 1 / (1 - [-1]) {interim} 1 + (-1 + 1 - 1 + 1 - 1 . . .) = 1 / 2 {final} What Rudy notes is that at any point in the infinite series on the left, the total can be either 0 or 1, and thus that it can be argued that the evidence shows that the sum is ultimatlely either 0 or 1. Rudy then finds it amusing that the right side of the equation seems to average this mind bender by declaring the sum to be the average of 0 and 1, which is 1/2. No such metaphysics are needed to understand the formula. Although 1/ 2 is in its simplest form, it is still in the form of a required division process that continues ad infinitum. Although I haven't worked out the exact mechanics of the math process on the right side that produces exactly the sequence on the left side, I imagine that it is not much more difficult than a 2-cycle engine the produces +1 at the start and alternately procuces -1 every other cycle. (Perhaps something like the following: 1 / 2 = zero 2's subtractable (leaving + 1 untouched) + 1 / 2, and 1 - 2 = -1 (forced past the limit of 1), and -1 - (-2) = 3, and 3 / 2 = +1 with a remainder of 1 / 2, ad infinitum.) 1 + (-1 + 1 - 1 + 1 - 1 . . .) = 1 / 2 {ad infinitum} Rudy argues that the left side of the above equation represents a Thompson Lamp where it can be argued that the final state of the lamp is on, or off, however one chooses to argue the case. The argument is spurious since it is plainly stated that the light, if it were such, is being turned on and off at specific points in the mathematical process, ad infinitum. I'll finish reading the book, and I might even look at all the problems, and perhaps I'll be enriched by it, but it really makes me wonder how much logic mathematicians really study. Very Respectfully, Ray === Subject: Re: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB? > I'm trying to get the deadwood off my reading shelf, and I just > finished BITCH, by Elizabeth Wurtzel, which turned out to be worth > reading for her powerful, insightful, and soul-baring epilogue, but > too much of the rest of the book was just too much, over and over, > about O.J. Simpson. Worse, Ms. Wurtzel is a cinema-holic who > dismisses the woes of real-life victims who don't play their parts > well, as if we should all have just the right life scripts for our > tragedies and have the necessary acting abilities to earn her respect > and sympathy, or at least a tepid encore? > And, one of the next space-hoggers on my shelf is INFINITY AND THE > MIND: THE SCIENCE AND PHILOSOPHY OF THE INFINITE, by Rudy Rucker. > Rudy ticked me off yesterday -- I had hoped to get a good start on > finishing the rest of the book, but then he stopped writing his barely > comprehensible Math-eze and added some stinking number questions, > including the following: > Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a) for all a? Looks like you're better off sticking with E. Wurtzel. === Subject: Re: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB? >Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a) >So I did: >(1 - a) [1 + (a + a^2 + a^3 . . .)] = (1 - a) (1 / (1 - a)) >{multiplying both sides by (1 - a)} And what if a = 1? >(1 - a) + [(1 - a) (a + a^2 + a^3 . . .)] = 1 {interim result} >(1 - a) + [(a + a^2 + a^3 . . .) - (a^2 + a^3 + a^4 . . .)] = 1 >{interim result} Ah, a bit of shuffling the terms of a conditionally convergent series, that's always a neat trick in any proof. >(1 - a) + a = 1 {interim result} >1 = 1 {final proof of equality} Very nice, for hundreds of years we have erroneously assumed the geometric series converges only for |x|<1, but you have proven otherwise. === Subject: Re: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB? Look, math is basically a formalization of common sense and logic. So, if something produces nonsense, that means along the way you have made an erroneous step. Otherwise logic is illogical, so you have undermined the foundations of logic, a feat similar to what Bertrand Russel and later Kurt Godel were able to do. Now, as to your question... You want to prove that 1 + a + a^2 + a^3 . . . = 1 / (1 - a) FOR ALL a. First of all notice that this statement does not make sense when a = 1, because the right side is undefined, hence you are trying to say something using things without definitions -- i.e. potential nonsense. So you may say, fine, I'll DEFINE 1/0 for ya. It's infinity! Then you have to invent your own infinity arithmetic. For you cannot have infinity follow the same rules that we all agree regular numbers follow. For example, 1 / 0 = infinity, 2 / 0 = infinity, so by definition of division 0 * infinity has all the values of the rainbow. So infinity * 0 is not unique. Incidentally, 0 / 0 is therefore not unique either, since the answer is the number which multiplied by 0 gives 0 but this is true for all numbers. So you can replace the with and and have yourself a nice little world with your infinity and zero arithmetic. No one says you can't do that. It just has to be consistent and logical. You can invent your own math. If it's interesting and no one has done it before, you can even publish it. :-) So, notice: You want to prove that 1 + a + a^2 + a^3 . . . = 1 / (1 - a) FOR ALL a. This means you have to define what the above statement MEANs for all a. We've just treated the case a = 1. Now let's see the case a > 1. What does the . . . mean in your express? if it means add infinitely many numbers together I say there is still a lot of things undefined. What do you mean by add infinitely many numbers together? Instead, I want to use the definition that is used by all mathematicians, and is well defined: SERIES [i=0...infinity] a^i = lim [n->infinity] SUM[i=0...n] a^i and I want to prove that = 1/1-a) The standard proof goes like this: for any partial sum, a * SUM[i=0...n] a^i = SUM[i=1...n+1] a^i = a^(n+1) - 1 + SUM[i=0...n] a^i So we solve for SUM[i=0...n] a^i and we get SUM[i=0...n] a^i = (a * SUM[i=0...n] a^i) + (1 - a^n+1) (1 - a) * SUM[i=0...n] a^i) = (1 - a^n+1) SUM[i=0...n] a^i = (1 - a^n+1) / (1-a) // IF a != 1 Taking the limit as n goes to inity NOW gives SUM[i=0...infinity] a^i = 1 / (1-a) // IF |a|<1 SUM[i=0...infinity] a^i = a - INFINITY / (1 - a) // IF |a|>1 This is why limits were invented, also. You don't want to deal with infinities straight out. You will get confused about what the heck is going on. TO make things precise, you usually want to deal with FINITE representations of the same thing. For example, the SERIES (a_n) is defined as the limit of the partial sums s_n = SUM [i=1...n] (a_i). If this limit is infinity, we stop caring about its value. If two series both diverge to infinity but we want to see whether one approximates the other very well, we take the limit of the RATIOS of the partial sums, and that's how we reach our conclusion. All this rests on the concepts of FOR EVERY and THERE EXISTS. Instead of talking about infinity and FOR EVERY, you can use De Morgan's Laws, which are simply logical to our minds, and prove the negative statement about THERE EXISTS. People who have difficulty understanding infinite constructions (I also do, sometimes), should realize that they can replace FOR EVERY x, P(x) with NOT [THERE EXISTS x, NOT P(x)]. This is useful for example for Cantor's Diagonalization Argument. Some things are just impossible to define if you want them to fit existing standards. For example, it's impossible to define an order relation between complex numbers without violating the axioms of order. In the same way, you can make up your own arithmetic with zero and infinity, just make clear what, if anything, you are violating in normal arithmetic (as I did above with dividing by zero). Once you have a consistent system, and you're sure it works, it might give you useful results, but you always have to be careful about translating the results into regular arithmetic. Enjoy, Grisha === Subject: Re: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB? >Rudy ticked me off yesterday -- I had hoped to get a good start on >finishing the rest of the book, but then he stopped writing his barely >comprehensible Math-eze and added some stinking number questions, >including the following: >Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a) Suppose a=2 (you did somewhere). Insert this value in the above equation. The equation is false for a=2, hence you can't prove that the right side and the left side of the equation are equal for all values of a. Any 'proof' that does do this must therefore be incorrect (as logic dictates). So maybe the equation given isn't the whole story. Let's start with this: 1 + a + a^2 + ... + a^(n-1) + a^n Multiply by (1-a): (1 - a) * (1 + a + a^2 + ... + a^(n-1) + a^n) = (1 + a + a^2 + ... + a^(n-1) + a^n) - (a * (1 + a + a^2 + ... + a^(n-1) + a^n)) = (1 + a + a^2 + ... + a^(n-1) + a^n) - (a + a^2 + a^3 + ... + a^n + a^(n+1)) = 1 - a^(n+1) So: 1 + a + a^2 + ... + a^(n-1) + a^n = (1 - a^(n+1)) / (1 - a) When n goes to infinity, the right hand side of this equation only becomes equal to the right hand side of the equation given above by Rudy, if a has a value between -1 and 1. So the complete statement should have been: 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a) with -1 < a < 1 >So I did: >(1 - a) [1 + (a + a^2 + a^3 . . .)] = (1 - a) (1 / (1 - a)) >{multiplying both sides by (1 - a)} >(1 - a) + [(1 - a) (a + a^2 + a^3 . . .)] = 1 {interim result} >(1 - a) + [(a + a^2 + a^3 . . .) - (a^2 + a^3 + a^4 . . .)] = 1 >{interim result} >(1 - a) + a = 1 {interim result} >1 = 1 {final proof of equality} >This proves that the original equation is true no matter what value of >a we use, but Rudy argues that the original formula makes no sense >when we substitute certain values for a. The 'proof' you did must be wrong somewhere as the equation doesn't work with any value outside the range <-1,1>. I believe the error is that the operations shown can only be applied to absolute convergent series ('absoluut convergente reeksen' in dutch). Since the series is not convergent at all for a being outside <-1,1>, the whole proof is nonsense. I'm sure someone else can explain this better... === Subject: Re: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB? > The 'proof' you did must be wrong somewhere as the equation doesn't > work with any value outside the range <-1,1>. I believe the error is > that the operations shown can only be applied to absolute convergent > series ('absoluut convergente reeksen' in dutch). Since the series is > not convergent at all for a being outside <-1,1>, the whole proof is > nonsense. > I'm sure someone else can explain this better... I posted virtually the same arguments at the www.johnpatrick.com message board and recieved a similar response from The Truth. _____________________________________________________________________ >>Essential in your derivation is the step [(a + a^2 + a^3 . . .) - (a^2 + a^3 + a^4 . . .)] = a. But this equivalence only holds if the series a + a^2 + a^3 . . . converges, and it only converges for certain a, not for any a.<< There is only one element of infinity that is not common to both infinite series and that is a to the first power, and thus the difference between the two infinite series is simply a to the first power. If you will take the time to explain in English why the a I've isolated is not the certain a that you require, I'll listen, but you've otherwise said nothing. ------------------------------------------------------------------------ I'm listening ... so tell me in English why I'm wrong. Very Respectfully, Ray === Subject: Re: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB? > The 'proof' you did must be wrong somewhere as the equation doesn't > work with any value outside the range <-1,1>. I believe the error is > that the operations shown can only be applied to absolute convergent > series ('absoluut convergente reeksen' in dutch). Since the series is > not convergent at all for a being outside <-1,1>, the whole proof is > nonsense. > I'm sure someone else can explain this better... > I posted virtually the same arguments at the www.johnpatrick.com > message board and recieved a similar response from The Truth. > _____________________________________________________________________ >>Essential in your derivation is the step [(a + a^2 + a^3 . . .) - > (a^2 + a^3 + a^4 . . .)] = a. > But this equivalence only holds if the series a + a^2 + a^3 . . . > converges, and it only converges for certain a, not for any a.<< > There is only one element of infinity that is not common to both > infinite series and that is a to the first power, and thus the > difference between the two infinite series is simply a to the first > power. > If you will take the time to explain in English why the a I've > isolated is not the certain a that you require, I'll listen, but > you've otherwise said nothing. > ------------------------------------------------------------------------ > I'm listening ... so tell me in English why I'm wrong. The so-called associative property of the addition a + ( b + c ) = ( a + b ) + c allows us to write both sides of the equality as a + b + c. This property is *not* valid for so-[badly]-called infinite sums. You cannot write something like a1 + ( a2 + a3 + ... ) = (a1 + a2) + a3 + ... In fact, the thing a1 + a2 + a3 + .... is not even a sum to begin with! It is a so-called limit of a series of partial sums: s1 = a1 s2 = a1 + a2 ... sn = a1 + a2 + ... + an If this series has a limit for n -> infinity, then one is allowed to use the abbreviation limit(sn; n -> infinity) = a1 + a2 + a3 + ... The property of having a limit in the previous sentence is something that can be verified by other means. hth. Dirk Vdm === Subject: Re: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB? > I'm listening ... so tell me in English why I'm wrong. Do you know what it means for a series to converge? Are you talking about a series when you write a + a^2 + a^3 + ... ? If not, what? === Subject: Re: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB? raydpratt grava .88 la saucisse et au marteau: [SNIP] I hope it's a bad, ugly troll. Otherwise, it's a bad, ugly mathematician. -- Nicolas, who wonders why there is always such people posting on Usenet newsgroups. === Subject: Fiber Bundle Physics/Consciousness 2 Commentary 2 Given coordinate patch C(x) in the base space M in a neighborhood of point x and fiber f(x) form the local Cartesian product C(x)f(x) with ordered pair X = (x,fo). Take the union C(x)f(x)/C(x')f(x')/... of all such local products. There are redundant ordered pairs X because the coordinate patches C(x) and C(x') as sets overlap with non-vanishing intersection C(x)/C(x')=/= Empty Set. Identify the redundant multiple images of the same actual point of the base space M using the symmetry group G as an equivalence relation. That is, two ordered pairs X and X' are identified or equivalent if x = x' < C(x)/C(x') and if fo' = gfo where g < G to form disjoint equivalence classes {f(x)} that are the distinct points of the fiber in hyperspace H. This is all local at a fixed base point x like in an internal gauge force symmetry. g is also called a transition function. The hyperspace H is the factor space of the union C(x)f(x)/C(x')f(x')/ ... mod G. The projection map P:(x,{fo}) -> x When M is the curved space-time of Einstein's gravity theory in addition to the G equivalence in the extra space dimensions of the fiber, x'(E) = Diff(4)x(E) at fixed event E to make disjoint equivalence classes {x(E)} mod Diff4(E). One can imagine a hybrid where the fiber is a discrete space of strings of c-bits. One can also imagine a fiber of strings of qubits. 1 qubit is a parallel infinity of c-bits. i.e. |qubit> = |1 c-bit><1c-bit|qubit> + |0 c-bit><0 c-bit|qubit> Where there is a continuous infinity of different c-bit bases or orthonormal frames each corresponding, for example, the the angular orientation of an inhomogeneous field magnet in a Stern-Gerlach filter for spin qubits in the DARPA spintronics project or like the billion billion Single Electron Transistors inside the human brain at the sub-microtubular protein dimer hydrophobic cage level forming the hardware interface with external world whose software is our stream of inner consciousness. Each possible orientation is a primitive parallel quantum universe. The quantum computer computes in all possible orientations simultaneously like a continuous infinity of classical Turing machines in a distributed network working on the same problem - or so the folklore goes. to be continued. Commentary 1 The fiber bundle as an idea has 4 parts. 1. A structure symmetry group G. 2. The total hyperspace H or, in some applications Wheeler's BIT. 3. The projection map P. 4. The base space M or, in some applications. Wheeler's IT. The hyperspace H consists of fibers f(x) that are either copies of or representations of the symmetry group G. The projection map P collapses a fiber f(x) in the hyperspace H to a point x in the base space M. All of these objects are continuum differential manifolds depending on the continuum of real numbers which its associated issues of Cantor's infinity of infinities of Cabalistic Aleph's in an ascending Jacob's Ladder. This is not a discrete combinatoric mathematics although such a skeletal structure is associated with it as in Herman Weyl's Theory of Groups and Quantum Mechanics and as in Saul-Paul Sirag's presentation of V.I. Arnold's A-D-E mathematics of everything. The base space is covered by an atlas of local coordinate patches with all important overlap transition functions sewing the patches together like a quilt. M is space-time in local micro-quantum field theory of point The extra-dimensions of hyperspace form the Calabi-Yau space of vibrations of the superstring beyond space-time. The connection on the total hyperspace H is the potential of a local gauge force. Examples of connections is the 4 potential Au(x) in Maxwell's electromagnetism with G as U(1). There are similar connections for the Yang-Mills weak force with G = SU(2) and the strong force with G = SU(3). Classical general relativity, as distinct from local micro-quantum field theory, has the torsion-free symmetric three-index non-tensor Levi-Civita connection with G as the Diff(4) group. The latter comes from locally gauging the 4 parameter translation subgroup (generated by the 4-momentum Pu of globally flat special relativity ) of the 15 parameter conformal group of Roger Penrose's massless twistors. Bottom -> Up: Given base space M and symmetry group G construct the hyperspace H as a quilt patchwork. Top -> Down: Given hyperspace H and symmetry group G construct the base space M as the non-overlapping partition of hyperspace into G-orbits called the quotient space of H mod G in the principal bundle. Micro-quantum source renormalizable local fields of spin 1/2 lepto-quarks are associated vector bundles. Micro-quantum force renormalizable local fields of spin 1 gauge force bosons (electro-weak and strong) are from the principal bundle. There is no renormalizable quantum gravity in this precise sense. This is because classical Einstein gravity is a More is different (P.W. Anderson) emergent collective effect as in Andrei Sakharov's metric elasticity of an instability in the globally flat false vacuum of the interacting lepto-quark source/electroweak-strong force. Einstein's gravity + unified exotic vacuum dark energy/matter with Andrei Linde's chaotic inflationary cosmology are the result of the continual phase transitions from globally flat false high entropy micro-quantum vacua to locally curved macro-quantum low entropy metastable vacua. to be continued: === Subject: Re: solution to rotation set of linear equations >I'm looking for a solution to a set of linear equations which has a very >specific form. Given k>0, and real vectors a_0^{k-1} and b_0^{k-1} (all >>= 0, if it matters), I need to find a real vector c_0^{k-1} such that: > a_0 c_0 + a_{k-1} c_1 + a_{k-2} c_2 + ... + a_1 c_{k-1} = b_0 > a_1 c_0 + a_0 c_1 + a_{k-1} c_2 + ... + a_2 c_{k-1} = b_1 > a_2 c_0 + a_1 c_1 + a_0 c_2 + ... + a_3 c_{k-1} = b_2 > a_3 c_0 + a_2 c_1 + a_1 c_2 + ... + a_4 c_{k-1} = b_3 > ... > a_{k-1} c_0 + a_{k-1} c_1 + a_{k-3} c_3 + ... + > a_0 c_{k-1} = b_{k-1} If I'm reading this correctly, you want to solve (sum a_m J^m) c = b for c, where coefficients a_i and the vector b are given, and J is the 0-1 matrix whose 1's are precisely in the (i+1,i) entries (including the (1,k) entry). This matrix J is easily diagonalized (if w = exp(2 pi i / k) then the vector (1,w^m,w^(2m),...) is an eigenvalue corresponding to the eigenvalue w^(-m) ). In this new basis the coordinates of c are found simply by dividing those of b by the values of P(x) = sum a_m x^m at x = 1, w, w^2, etc. ) dave === Subject: Re: sinus approximation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAFLMLK13048; > how can i get a better approximation? you have to first say in what sense you want the polynomial to fit the function (least square fit maybe?) i.e. provide a norm in function space and then solve the resulting minimizing problem with respect to the coefficients of the polynomial. === Subject: Re: JSH: My use of my initials by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAFLMIm12988; > I see, thx. I wasn't clear on who was firing at who :) But JSH does > sometimes threaten people with litigation, and he had somebody in court one > time, for calling him a crank. The case was quickly tossed. >>I don't think there was any such incident. >> He has threatened on occasion; but I think Mr. Hammick may be getting >> confused with the case where Underwood Dudley ->was<- sued for calling >> someone a crank. Dilworth v. Dudley: >> http://www.law. emory.edu/7circuit/jan96/95-2282.html The suit was dismissed. > It wasn't just one suit. The first, suing me, my publisher, and >the president of my school for, among other things, conspiracy to deny >Mr. Dilworth's civil rights (mathematicians wouldn't talk to him >because I called him a crank, he said), was in federal court in >Wisconsin where it was indeed dismissed. Mr. Dilworth then appealed >to the Seventh Circuit Court of Appeals where it was again dismissed. >I was pleased that the decision was written by Judge Posner, >well-known for his many books, who had clearly read some of my >_Mathematical Cranks_. > By the way, my respect for the legal profession went up as a >result of this process. Mr. Dilworth represented himself, I assume >because no lawyer would take his case. The lawyers for the >Mathematical Association of America also did a lot of work, finding >all sorts of precedents where a person was called a scab, a traitor, >and other nasty things in print and the courts let the authors get >away with it. > This wasn't enough for Mr. Dilworth, who started the legal >process all over again by suing in Wisconsin _state_ court. Once >again the case was dismissed, but this time with the requirement that >Mr. Dilworth pay $7000 or so for his opponents' legal fees. That >finally stopped him, though he continued to send me letters. > Earlier this year he died, so now we can all call Mr. Dilworth >whatever we want. > However, the Law of Conservation of Cranks still operates, and >his kind has not died out. Presumably, the kind of crank who sues has >not died out either, so everyone be careful. > Woody Dudley Dilworth's ghost is still cackling, however - see http://www.rcfp.org/news/1996/0422k.html, that the suit was dismissed in Dudley's favor, and that an appeals court upheld the dismissal. Professor loses libel suit over label as 'crank' and clearly identifies Dudley as the professor! === Subject: Factorial ending in 8000000 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAG2ohQ02519; If the last seven digits on n! are 8000000, compute the value of n. === Subject: Re: JSH: Survey on my results, any correct? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAG3cXV05345; >1. I've posted a lot on sci.math over a long period of time, to your >knowledge, have I *ever* been right? You know what they say, even a broken clock is right twice a day... Seriously though, some of the things you say are right. Your proofs usually have one or two major errors in them, the rest is fine. But in math one error is all it takes to produce completely incorrect results. If you're asking whether one of your major proclamations have ever been right, for instance I have a short proof for FLT, I've 'discovered' the prime counting function, The algebraic integers are flawed, etc, then to my knowledge all of them have been wrong. >2. Do I have *any* correct results, or do you think I just talk and >never say anything that is mathematically correct? Some of your results are correct, though I believe you are greatly over-estimating their importance. Again, non of the major results you have claimed to achieve are correct, to my knowledge. If instead of over-estimating your own importance you would simply post your result you wouldn't be treated as an arrogant prick. >3. To your knowledge, has ANYONE ever posted agreement with me on >*anything*? Of course. Most people who review your proofs agree that they are right up to the first mistake. Also there are the occasional cranks and nuts. To my knowledge no competent mathematician has ever agreed with one of your major results. >4. In your opinion, have I ever won an argument on the newsgroup? I don't know what winning an argument means, the term is obviously subjective. Of the serious arguments you have had with critics such as Arturo Magidin and Nora Baron, my personal impression was that your argument was hopelessly lost because you were simply wrong. I can't talk about arguments I didn't read. >5. To your knowledge, have I ever caught other posters in errors? I can't remember, but I must say that even if you did I probably wouldn't remember. It simply doesn't seem to matter that much. No problem. >James Harris Lewis, The Silver Bullet === Subject: Re: JSH: Survey on my results, any correct? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAG5H7H11133; >1. I've posted a lot on sci.math over a long period of time, to your >knowledge, have I *ever* been right? >2. ...do...I just talk and never say anything that is mathematically correct?... >James Harris Now you have been right. (In part 2 above.) === Subject: Length of a chord by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAGDFMa11708; This is a (elementary?) geometry problem and I'm looking for a simple solution. In a unit circle a chord is drawn.The distance of the center from the chord is x (0<=x<=1).What is the length of the chord? Is it proportional to sqrt(1-x^2)? Olivio === Subject: Re: Length of a chord In sci.math, Olivio solution. > In a unit circle a chord is drawn.The distance of the center > from the chord is x (0<=x<=1).What is the length of the chord? > Is it proportional to sqrt(1-x^2)? Half of the chord, the radius to the end of the chord half, and the line from the circle center to the bisection point of the chord results in a right triangle. Therefore: hypotenuse = 1 side adjacent = x side opposite = sqrt(1 - x^2) Bear in mind that this is only half of the chord, but the other half is symmetrical; the answer is 2 * sqrt(1 - x^2). Or one can do it analytically. Place the chord perpendicular to the X axis and to the right of the origin; therefore the circle point above the X axis is (x,y). What is y? Well, x^2 + y^2 = 1 as we're hypothesizing a unit circle; the length of the chord is then 2y = 2* sqrt(1-x^2). > Olivio -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Homology of S^m x S^n by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAGHk5s28159; === Subject: Re: Length of a chord by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAGIeEQ31949; >This is a (elementary?) geometry problem and I'm looking for a simple >solution. >In a unit circle a chord is drawn.The distance of the center >from the chord is x (0<=x<=1).What is the length of the chord? >Is it proportional to sqrt(1-x^2)? >Olivio Yes it is === Subject: Re: Roman carving by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH2tNx32757; >>Can you please let me know if there is/was a Roman Numeral U and >>what amount it stands for. >The Romans used the letter 'V' to represent the number 5. In some >medieval and earlier text the letter 'U' was frequently substituted >for 'V'. The Romans had both the letters 'U' and 'V' in their >alphabet, but sometimes (always? We dont know.) substituted 'V' >for 'U' when carving words into marble (or any other material they >had at hand). For example >S.P.Q.R = SENATUS POPULUSQUE ROMANUS >in carved form reads >SENATVS POPVLVSQVE ROMANVS. >Some say this stems from the fact that a 'V' is so much easier to >carve into marble (or whatever) than a 'U'. Maybe this is the >reason for the resubstitution of the numeral 'V' to 'U' in some >texts. Well, some people like to pay tribute to their >conjectures. >Sincerly, ND > === Subject: Re: Roman carving Why do these old posts keep showing up here through some address at mathforum.org? Is there some sort of glitch at the Math Forum causing this to happen automatically? - Randy === Subject: Re: Complex numbers and 2x2 matrices (was Re: Complex numbers and 2x2 matrixes) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH2tR900344; >> How long has the relation between complex numbers and particular 2x2 >> matrixes been known? >> Are there any future plans to 'phase out' complex numbers and replace >> them with 2x2 matrixes which work exactly the same? >Complex numbers are isomorph 2x2-rotation-matrices are isomorph >ordinary 2D-vectors (with a certain multiplication added) -You calculate >them exactly the same - and you can calculate them in mixed mode too: >see the mixed-mode calculator: >http://i-z.eu.tt or http://i-is-no-longer-ima ginary.gmxhome.dewhich demonstrates, that you don't need this imaginarystuff any longer, >nor the Argand diagramm nor the imaginary axis, just like Caspar >Wessel started 2oo years ago without these. >Have fun >Hero This is all very well, but personally I think the nice thing about complex numbers is that you can usually just treat them as real numbers anyway, so why go to the trouble of working out the corresponding matrices etc. Also take a little function like f(z)=z^2, if you wanted to differentiate it you'd have to separate the real and imaginary parts, find the Jacobian etc, when you can just treat it like a real-valued function and just say f'(z)=2z as usual. Tim Williams === Subject: Fractal Spirals Revisited - $100 Reward ! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH2tVY00372; Hello. I still believe I discovered the fractal spirals described at: http://fractalspiral.zxcvb.org The formula is: x = sum (i = 1 to n) (r/i) sin (i*theta + i^2*pi*variable) y = sum (i = 1 to n) (r/i) cos (i*theta + i^2*pi*variable) I was told they were done by Michel Mendes France in the mid nineties, but I am not able to verify this as I am outside the math world. The only web link I have found is: http://hypo.ge-dip.etat-ge.ch/www/math/html/node56.html It is not the same thing. I would gladly reward the person who donate to the charity of your choice $100.00 (US). I know this isn't a lot but I don't have a lot. I need to know the correct representation of the formula. In it there is a variable, and I would like to know what letter was assigned to it. I would also enjoy reading technical jargon and actually understading the math being described. But most importantly, I need resolution so I stop wasting time thinking I invented this. For my mailing address, email me at: k at symbl zxcvb dot org - K evin D oughty === Subject: A Mathematical Stupidity Constant in all assessed Intelligent Behavior by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH2tbv00410; Really, I am serious. I think this has great potential for usefull applications. Based on the given Quantitative-Qualitative model to assess intelligence I feel there must co-exist a Stupidity Constant in all assesed intelligence, otherwise there would be no basis for said intelligence. All assessed and thus measured intelligence must be also based on stupidity. Usefull applications could include: Finding limits to existing systems outlined by purely positive criteria. Knowing there must exist negative criteria and that the intelligence is based as much on stupidity as anything else. In other words there are as many good reasons against supporting these existing systems (previously supported only by positive criteria) as those supporting them. This would give new meaning to Mathematical Applications and new meaning to their limitidness and finiteness. Zim Olson http://www.zimmathematics.com === Subject: Re: A Mathematical Stupidity Constant in all assessed Intelligent Behavior > All assessed and thus measured intelligence must be also based on > stupidity. I'll concede that yours is. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Factorial ending in 8000000 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH2tjS00455; >If the last seven digits on n! are 8000000, compute the value of n. Hint: Since the number ends with exactly six 0's, it must contain 5^6 as a factor, but not 5^7. === Subject: Re: Factorial ending in 8000000 In sci.math, Dale Shoults If the last seven digits on n! are 8000000, compute the value of n. > Hint: > Since the number ends with exactly six 0's, it must contain 5^6 as a > factor, but not 5^7. Given that hint, it's trivial; it has to be at least 25!, and less than 30!. 27! = 10888869450418352160768000000 -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Another Application of my System Transformation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH36Dp02273; I thought of another application of my 30 minute prior post A Mathematical Stupidity Constant to assessed Intelligence I have come up with a System Transformation used to determine any system functionality which is outlined at: http://www.zimmathematics.com/app.htm In the system transformation outlined here it contains a non- functional component. A usefull application can be found in explaining the Quantum Mechanic Slit experiment phenomena where exhibiting non causal or non rational behavior as my System Transformation said must be included in determining any system functionality. Zim Olson http://www.zimmathematics.com === Subject: Re: another quaternion question > What role does the Jacobson radical > ab = a+b - ab That is not the Jacobson radical. The Jacobson radical of a ring is the intersection of it proper left (or right) ideals. > play in the quaternion algebra > (lets say, over the reals), if any? None---literally! The Jacobson radical of H is {0} since H is a division ring. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Discrete Spaces Boring Discrete Spaces Eventually Bore Converging Sequences. Proof: Should debonair (xj) dare to converge upon demure x, then for any nhood of x, (xj) would eventually squeeze into the nhood. As X is so discrete, a mere {x} will do for x's nhood. Thus eventually (xj) would be squeezed into {x}. Now when (xj) has eventually been confined to {x}, it will find it's values severely restricted. Some may even consider it solitary confinement. Yet there (xj) will be for the rest of it's eternal life, allowed the company of none but x. Thus boredom, QED. ---- === Subject: Re: Inversion? JOn > When I did A level maths at school, about 30+ years ago, in my > analytical geometry course I was taught a subject called 'inversion'. === Subject: Re: Inversion? > When I did A level maths at school, about 30+ years ago, in my > analytical geometry course I was taught a subject called 'inversion'. > .... > I have never come across this again, and the subject seems to heve > disappeared from text books.... You'll find it in Chapter 5 of the recent text-book by Brannan, Esplen & Gray, Geometry, Cambridge U.P., 1998. They show how inversion is connected with Moebius transformations of the complex plane. > .... Does anyone know how/when this concept came about?.... I'd like to know more about that, too. Can anyone help? Ken Pledger. === Subject: Re: Inversion? > .... Does anyone know how/when this concept came about?.... > I'd like to know more about that, too. Can anyone help? It goes back to Apollonius (262-190 BC), at least. I say this because there and inversion relative to the ellipse, the hyperbola, and the parabola in the 'conic sections' of Apollonios, by B.A. Rozenfeld (Istoriko-Matematicheskie Issledovaniya, 30 (1986), 195-199), but I haven't read it (I don't know where to find it and, besides, I can't read russian). Jose Carlos Santos === Subject: properties of graphs of 0/1-polytopes To every polytope, we associate a graph in the following way: take its vertices as nodes. The nodes are joined by an edge if and only if the corresponding vertices are adjacent. How can we decide, given any graph, whether it is the graph of some 0/1-polytope or not? If there is no exact criterion known, is there a good sufficient one? A 0/1-polytope is a polytope where all its vertices have coordinates in {0, 1}. TIA, Tobias -- Phyics is much too hard for physicists. reverse my forename for mail! === Subject: Re: properties of graphs of 0/1-polytopes 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > To every polytope, we associate a graph in the following way: take its > vertices as nodes. The nodes are joined by an edge if and only if the > corresponding vertices are adjacent. > How can we decide, given any graph, whether it is the graph of some > 0/1-polytope or not? > If there is no exact criterion known, is there a good sufficient one? I have no idea, but an obvious necessary condition is that G is (log_2 |V(G)|)-connected. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Order of the orthogonal group over GF(q) What is the order of the orthogonal group of order n over the finite field GF(q) , i.e O(n,GF(q)) ? === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) HenriWilson skrev i melding > Now consider the case of a charge of mass m being accelerated in a field. > Using a=F/m, and m=Mo.gamma, > dv/dt=(F/c.Mo)[sqrt(c^2-v^2)], which is: ... nonsense. You have screwed this up before, and I have shown you the correct equation before. F = dp/dt = d/dt (m*gamma*v) gamma =1/sqrt(1 - v^2/c^2) dv/dt = (1/((v^2/c^2)*gamma^3 + gamma))*F/m ((v^2/c^2)*gamma^3 + gamma)*dv = (F/m)*dt gamma*v = (F/m)*t + C if the charge is accelerated from standstill, v = 0 when t = 0, C = 0 v = c*t/sqrt(t^2+T^2) where T = m*c/F v approaches c asymptotically > dv/[sqrt(c^2-v^2)]=kdt. > Integrating: arcsin(v/c)=kt+B or > v=c.sin(kt+B) > if t=0 then v=0, so B=0 > So we have v=c.sin(kt), where k is definitely finite. > How come? Even in its exponential form this doesn't make any sense. > Does it mean the SR equation is wrong or that matter becomes anti matter as c > is exceeded? It means that you are unable to do the math correctly. But I have shown you a simpler approach before. If we assume the charge is accelerated from standstill in a constant electric field excerting the force F, we get: p(t) = F*t = m*gamma*v (the derivation isn't necessary since we integrate it back in the next step!) Paul === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >> as it enters a static electric field. >Let us consider a charged sphere somewhere in the universe. It exerts a force >on every other charge. If we can arrange for it to lose that charge somehow, >you are claiming that all those forces disappear INSTANTLY. Parse the bloody sentence in quotes, Henry. It doesn't say that. Isn't English your first language? - Randy, grabbing his popcorn and going back to watch the entertainment === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days > as it enters a static electric field. >>Let us consider a charged sphere somewhere in the universe. It exerts a force >>on every other charge. If we can arrange for it to lose that charge somehow, >>you are claiming that all those forces disappear INSTANTLY. >Parse the bloody sentence in quotes, Henry. It doesn't say that. Isn't >English your first language? > - Randy, grabbing his popcorn and going back to watch the >entertainment Moron. Stop kissing Andsernon's arse. Henri Wilson. See the Stupidity of Relativity. www.users.bigpond.com/hewn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >> Remember the old vacuum tube triodes. >> Acccording to you, a signal on the grid would be instantly felt at the >anode. >> If that's not instantaneous communication, what is? >> Henri Wilson. >I think Paul's acrobatics are no match for yours, H. His plea of monotony is >his way of giving up. Paul wants a static field at the grid. That's ok, >until the electron reaches the grid, then that static charge has to reverse >sign, and is no longer static. Leaving aside the slew rate of the amplifier >driving the grid (which in any real experiment we could not do), if the grid >reversed from positive (attracting the electron) to negative as it passed >through (propelling the electron on), it would still take a finite amount of >time for the field beyond the grid where the electron is heading to change, >because c is finite. Hence at the instant the electron reaches the grid, >even if the grid potential is zero, there is a negative field ahead of the >electron to slow it down. Hence the electron cannot attain c, by this >method. However, a positron coming down the other way would have a closing >velocity with the electron that was greater than c. > Precisely. Paul Andsernon doesn't know what he's talking about. I note with interest that you think Androcles' incoherent babble shows that I don't know what I am talking about. :-) Did you actually read it, Henry? in accelerators, isn't it? :-) Hatch or Anrocles - doesn't matter. They are 100% correct as long as they disgree with relativity. Right? :-) Paul, always amused when Henry agrees to nonsense === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days > Consider an object being accelerated by a idealistic jet of water or a > continuous 'stream of elastic ping pong balls'. What is its subsequent velocity > pattern? >> [snippage] > M.dv/dt=m(Vo-v) or: >>[more snippage] >>Henri, you've got major problems with this. It assumes >>Newtonian momentum transfer, and so implies a Newtonian >>equation for kinetic energy. And there's a real problem >>most assuredly not 1/2 mv^2. So your model has gross >>disagreement with observation. >>Socks >> What an idiot! >> Naturally, if one includes assumptions that a theory is correct in an attempt >> to prove it wrong, that attempt is likely to fail. >Yes you are, and you did, and it did. You assumed Newton, and >you got Newton, and those are clearly not similar to what is >obverved in accelerators. >> A moving charge surrounds itself with a volume of 'reverse field' because the >> main field takes time to operate. The energy associated with this 'back field' >> is equivalent to a mass increase. >And their speeds are also not observed to go above c. It is very hard to accelerate anything to c let alone beyond it. The experiments are probably incapable of measuring superluminal speeds anyway. >And you have not provided any theory of E&M that allows any such >thing as a reverse field. Nor why there should be any kind of >speed limit involved. Nor why it should follow any such thing >as the kinetic energy formula observed in accelerators. Nor have >you provided a relation between energy and mass if you don't >accept relativity. >Socks radiation from an acceleraed charge! fields associated with a moving charge! The 'Back EMF' concept. I would be most amazed if a moving charge DID NOT alter the field around itself, wouldn't you? Henri Wilson. See the Stupidity of Relativity. www.users.bigpond.com/hewn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >And you have not provided any theory of E&M that allows any such >thing as a reverse field. Nor why there should be any kind of >speed limit involved. Nor why it should follow any such thing >as the kinetic energy formula observed in accelerators. Nor have >you provided a relation between energy and mass if you don't >accept relativity. >Socks > radiation from an acceleraed charge! > fields associated with a moving charge! > The 'Back EMF' concept. > I would be most amazed if a moving charge DID NOT alter the field around > itself, wouldn't you? Quite. are accelerated. You KNOW the following, Henry. In an accelerator going at full efficiency, we KNOW that because it looses this energy as synchrotron radiation in the bends of the circuit.(Very obvious and easily measurable.) So we - and YOU - know that the RF-cavities never ceases is only few mm/s below the speed of light. So why do you keep pretending that the E-field is not speed approaches c, when you KNOW that isn't true? Another case of selective memory loss? What you admit knowing in one posting, you have forgotten in the next, eh? Paul === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >>And you have not provided any theory of E&M that allows any such >>thing as a reverse field. Nor why there should be any kind of >>speed limit involved. Nor why it should follow any such thing >>as the kinetic energy formula observed in accelerators. Nor have >>you provided a relation between energy and mass if you don't >>accept relativity. >>Socks >> radiation from an acceleraed charge! >> fields associated with a moving charge! >> The 'Back EMF' concept. >> I would be most amazed if a moving charge DID NOT alter the field around >> itself, wouldn't you? >Quite. >are accelerated. >You KNOW the following, Henry. >In an accelerator going at full efficiency, we KNOW that >because it looses this energy as synchrotron radiation in the bends >of the circuit.(Very obvious and easily measurable.) >So we - and YOU - know that the RF-cavities never ceases >is only few mm/s below the speed of light. >So why do you keep pretending that the E-field is not >speed approaches c, when you KNOW that isn't true? I DID NOT SAY THAT. The question is how much energy? You are making no attempt to answer that one. In typical fashion, you pretend the relevant question does not exist. >Another case of selective memory loss? >What you admit knowing in one posting, >you have forgotten in the next, eh? >Paul Henri Wilson. See the Stupidity of Relativity. www.users.bigpond.com/hewn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> Let's get this straight. >> You say that the force on a charge due to an electric field acts >> instantaneously. Correct? >> Let us consider a charged sphere somewhere in the universe. It exerts a >force >> on every other charge. If we can arrange for it to lose that charge >somehow, >> you are claiming that all those forces disappear INSTANTLY. >> I will continue when I receive your answer (if one is forthcoming). > Interesting question, Henry. I've used it myself, on the discussion of >gravity propagation. If a star were to convert *all* it mass to radiation in >one super-supernova, how long would it take for us to detect its loss of >gravity? I don't mean watch its planet suddenly fly away, that would be a >distant observation and would reach us at c. I mean detect the gravity loss >right here. This would be the biggest gravity pulse (negative going) that we >could possibly detect. But in order to detect the loss, where would have to >be aware of it in the first place. As it turns out, the nearest star to us >(other than the sun) is 3.9 light-years away, and that is just too far to >detect it's gravity directly. So I fail to understand why anyone would >attempt an experiment to search for gravity waves, other than to give >themselves some funding on a futile attempt. There's a lot of money to be >made out of relativity, and very few people are altruistic. >Androcles Exactly. There is one problem that I'm sure Paul will pounce on. That is, how to make charge suddenly disappear. Here is another interesting question. The 'Mass' of an object is made up of a major proportion, the total mass of all energy. Presumably, the two portions play an equal role in Newton's gravitation Law. If there is no distinction between the two, is it possible that all MATTER is nothing but some kind of manifestation of various 'fields'? Also, since mass is lost/gained in a Nuclear explosion, would we not expect measurable gravity waves to be produced. Even though the amount of matter is relatively small, the sum total of all the potential energy associated with that matter is quite large. (If the nuclear bomb is considered to be the centre of the universe, what is the energy required to raise all the other matter in the universe to its present distance from that point. Might it just turn out to be mc^2?). Henri Wilson. See the Stupidity of Relativity. www.users.bigpond.com/hewn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >> Let's get this straight. >> You say that the force on a charge due to an electric field acts >> instantaneously. Correct? >> Let us consider a charged sphere somewhere in the universe. It exerts a >force >> on every other charge. If we can arrange for it to lose that charge >somehow, >> you are claiming that all those forces disappear INSTANTLY. >> I will continue when I receive your answer (if one is forthcoming). > Interesting question, Henry. I've used it myself, on the discussion of >gravity propagation. If a star were to convert *all* it mass to radiation in >one super-supernova, how long would it take for us to detect its loss of >gravity? I don't mean watch its planet suddenly fly away, that would be a >distant observation and would reach us at c. I mean detect the gravity loss >right here. This would be the biggest gravity pulse (negative going) that we >could possibly detect. But in order to detect the loss, where would have to >be aware of it in the first place. As it turns out, the nearest star to us >(other than the sun) is 3.9 light-years away, and that is just too far to >detect it's gravity directly. So I fail to understand why anyone would >attempt an experiment to search for gravity waves, other than to give >themselves some funding on a futile attempt. There's a lot of money to be >made out of relativity, and very few people are altruistic. >Androcles > Exactly. > There is one problem that I'm sure Paul will pounce on. That is, how to make > charge suddenly disappear. > Here is another interesting question. > The 'Mass' of an object is made up of a major proportion, the total mass of all > energy. > Presumably, the two portions play an equal role in Newton's gravitation Law. > If there is no distinction between the two, is it possible that all MATTER is > nothing but some kind of manifestation of various 'fields'? Since matter is mostly empty space anyway, why can't I pass through a brick wall? And of course the answer is that I am repelled by the electrons. So what are electrons? > Also, since mass is lost/gained in a Nuclear explosion, would we not expect > measurable gravity waves to be produced. Even though the amount of matter is > relatively small, the sum total of all the potential energy associated with > that matter is quite large. As I understand it, a mountain is just detectable. An nuclear device is small enough to carry on a plane. Therefore you would need a *Very* sensitive device to measure the gravity of the weapon, and that would imply proximity. If you detonate the device, I think you are likely to destroy the instrument before it could react :) > (If the nuclear bomb is considered to be the centre of the universe, what is > the energy required to raise all the other matter in the universe to its > present distance from that point. Might it just turn out to be mc^2?). > Henri Wilson. > See the Stupidity of Relativity. > www.users.bigpond.com/hewn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Of course you are funny. > If electric and magnetic fields acted instantaneously, light would travel at > infinite speed. >>I note with interest that your statement is so ambiguously put >>that it may be right as well as wrong. >>But I take for granted that your statement is meant to be >>relevant to my claim, which was: >> as it enters a static electric field. >>So assuming that acting instantly is to be understood >>in this sense, an unambiguous version of your statement >>becomes: >> travel at infinite speed. >>Was this what you meant to say, Henry? >>If not, what DID you mean to say, and what is the relevancy >>of what you meant to say to your action time of the static >>accelerating field in an accelerator? >>Paul, finding the acrobatic show a little monotonous >> Remember the old vacuum tube triodes. >> Acccording to you, a signal on the grid would be instantly felt at the anode. >So if: > as it enters a static electric field. >it follows: >A signal on the grid would be instantly felt at the anode >Explain why, please. >> If that's not instantaneous communication, what is? >Henry, you know of course that you are babbling nonsense. >There simply isn't possible to be as stupid as you pretend. >Paul > Let's get this straight. > You say that the force on a charge due to an electric field acts > instantaneously. Correct? Why do you ask what I am saying, when what I am saying is quoted right above? I am saying: as it enters a static electric field. > Let us consider a charged sphere somewhere in the universe. It exerts a force > on every other charge. If we can arrange for it to lose that charge somehow, > you are claiming that all those forces disappear INSTANTLY. And why the hell do you think I would claim something as stupid as that? You MUST know this is stupid nonsense, Henry. Why the hell do you pretend that it is possible to interpret my statement above to mean that the electric field will act instantly electric field? This is actually too bloody stupid even for you, Henry. Why do you pretend to be such a moron? > I will continue when I receive your answer (if one is forthcoming). Answer what? You are desperated to evade the point, are you? Lets take this from the beginning. Of course there are really no static electric fields in an accelerator, and I never said it was. that is a resonance cavity with a very powerful EM-field. The typical frequency is in the order of 100MHz (or higher). where the E-field is longitudinal. The crucial point is that the phase of this resonator is adjusted such that the E-field peaks (in the same direction) enters the accelerating stretch. That is the point. Why is that so hard to get? So what I am claiming is still: it enters the (quasi) static electric field. And the force it speed. Can you please explain how you can use this for instant communication, Henry? Of course you cannot. The only decent thing you can do is to admit that this claim is wrong. But you won't do that, will you? You will rather keep insisting that when I say that the RF-cavity in an accelerator, then I really have stated that a force acts on an electron in the Andromeda galaxy in the accelerator. Won't you? Paul === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> Of course you are funny. >> If electric and magnetic fields acted instantaneously, light would travel at >> infinite speed. I note with interest that your statement is so ambiguously put >that it may be right as well as wrong. But I take for granted that your statement is meant to be >relevant to my claim, which was: > as it enters a static electric field. So assuming that acting instantly is to be understood >in this sense, an unambiguous version of your statement >becomes: > travel at infinite speed. Remember the old vacuum tube triodes. Acccording to you, a signal on the grid would be instantly felt at the anode. >>So if: >> as it enters a static electric field. >>it follows: >>A signal on the grid would be instantly felt at the anode >>Explain why, please. > If that's not instantaneous communication, what is? >>Henry, you know of course that you are babbling nonsense. >>There simply isn't possible to be as stupid as you pretend. >>Paul >> Let's get this straight. >> You say that the force on a charge due to an electric field acts >> instantaneously. Correct? >Why do you ask what I am saying, when what I am >saying is quoted right above? >I am saying: > as it enters a static electric field. So there is also an opposite force acting on the electrodes. even if the electrodes are lightyears apart. IS THAT WHAT YOU ARE SAYING? >> Let us consider a charged sphere somewhere in the universe. It exerts a force >> on every other charge. If we can arrange for it to lose that charge somehow, >> you are claiming that all those forces disappear INSTANTLY. >And why the hell do you think I would claim something as stupid as that? Because you just DID, above, (even though you probably do not realise what you said). >You MUST know this is stupid nonsense, Henry. >Why the hell do you pretend that it is possible to interpret my >statement above to mean that the electric field will act instantly >electric field? The bloody field can be from one side of the universe to the other for all I care. >This is actually too bloody stupid even for you, Henry. >Why do you pretend to be such a moron? You just do not understand. You are very confused. >> I will continue when I receive your answer (if one is forthcoming). >Answer what? >You are desperated to evade the point, are you? >Lets take this from the beginning. >Of course there are really no static electric >fields in an accelerator, and I never said it was. >that is a resonance cavity with a very powerful EM-field. >The typical frequency is in the order of 100MHz (or higher). >where the E-field is longitudinal. >The crucial point is that the phase of this resonator is >adjusted such that the E-field peaks (in the same direction) >enters the accelerating stretch. >That is the point. Why is that so hard to get? That is obvious, irrelevant and NOT the point we are discussing. The question is, does a moving charge feel the 'full strength' of the accelerating field? (similarly, does a falling object feel the full 'force of gravity'? Tom Roberts once told me they didn't) The evidence shows that charges do NOT accelerate in 'Newtonian fashion'. SR claims they behave as though their mass increases by gamma. I am suggesting alternative explanations based on two possibilities. Firstly, the field DOES take time to act and therefore the field gradient at the point of the moving charge IS NOT the same as that which would act on a charge AT REST. Secondly, the movement of the charge itself creates an opposing field that effectively reduces the magnitude of the field gradient around the moving charge. I repeat that if what you claim is true then you have invented instantaneous communication. >So what I am claiming is still: >it enters the (quasi) static electric field. And the force >it speed. How do you define KE? >Can you please explain how you can use this for >instant communication, Henry? I told you, above. >Of course you cannot. >The only decent thing you can do is to admit that >this claim is wrong. It is NOT wrong. Fill a box with electrons. The box attracts every positive charge in the universe. What happens to that attraction if the box suddenly disappears? You say it goes to zero instantly. So if you can make the box appear and disappear in an intelligent manner, you will be able to communicate instantly with anyone, ANYWHERE!!!! Congratulations Paul. You will certainly be awarded a Nobel Prize. >But you won't do that, will you? >You will rather keep insisting that when I say that >the RF-cavity in an accelerator, then I really have stated >that a force acts on an electron in the Andromeda galaxy >in the accelerator. >Won't you? No Paul. You are considerating only fields between electrodes and not spherical fields emanating from point charges. >Paul Henri Wilson. See the Stupidity of Relativity. www.users.bigpond.com/hewn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >> You say that the force on a charge due to an electric field acts >> instantaneously. Correct? >Why do you ask what I am saying, when what I am >saying is quoted right above? >I am saying: > as it enters a static electric field. > So there is also an opposite force acting on the electrodes. > even if the electrodes are light years apart. > IS THAT WHAT YOU ARE SAYING? I am saying: as it enters a static electric field. We have two electrodes - say 1 km apart. (Or a light year apart - if you insist) The potential difference is 1 million volts. (Or a zillion volts - if the distance is a light year) There is a small hole in the negative electrode. We inject an electron through this hole. When will a force act on the electron? I am still saying: as it enters a static electric field. But what are YOU saying? Not untill the electrode 1 km away feels the opposing force? IS THAT WHAT YOU ARE SAYING? Come on, make your point. What is the action time of the force on the electron? Why do you think the distance to the other electrode is relevant? How does the distance to the other electrode affect this action time? Please don't say something like we don't know. Because we DO know. Do YOU know? The rest is a repetition of the above. Paul === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days > You say that the force on a charge due to an electric field acts > instantaneously. Correct? >>Why do you ask what I am saying, when what I am >>saying is quoted right above? >>I am saying: >> as it enters a static electric field. >> So there is also an opposite force acting on the electrodes. >> even if the electrodes are light years apart. >> IS THAT WHAT YOU ARE SAYING? >I am saying: > as it enters a static electric field. >We have two electrodes - say 1 km apart. >(Or a light year apart - if you insist) >The potential difference is 1 million volts. >(Or a zillion volts - if the distance is a light year) >There is a small hole in the negative electrode. >We inject an electron through this hole. >When will a force act on the electron? >I am still saying: > as it enters a static electric field. >But what are YOU saying? >Not untill the electrode 1 km away feels the opposing force? > IS THAT WHAT YOU ARE SAYING? NO PAUL. I am saying that your claim infers that the effect of an injected electron will be felt INSTANTLY at the far electrode. Of course, since the electron existed BEFORE it was injected, the effect would have already been there even though the near electrode was in the way.. The only way this experiment can even be hypothesized is by either 'annihilating' a very large number of electrons or by monitoring the force on the far electrode with movement of the electron mass towards it. If the effects are instantaneous as you claim, you will have achieved instantaneous communication. >Come on, make your point. >What is the action time of the force on the electron? >Why do you think the distance to the other electrode is relevant? >How does the distance to the other electrode affect >this action time? >Please don't say something like we don't know. >Because we DO know. >Do YOU know? OK I will agree with you. Assume it is instant. Therefore I can send messages to the far electrode instantaneously. >The rest is a repetition of the above. Ah! readers note, Paul Andsernon snips again! >Paul Henri Wilson. See the Stupidity of Relativity. www.users.bigpond.com/hewn/index.htm === Subject: Re: Problem from Herstein 4 >:> 18. Let G be the group of all real 2-by-2 matrices >:> (a b) >:> (c d), with ad - bc non-zero, under matrix multiplication, and let N >:> be the subgroup consisting of those elements of G with ad - bc = 1. >:> Prove that N contains G', the commutator subgroup of G. >:> 19. In problem 18 show, in fact, that N = G'. >: I assume, that your matrices have their coefficients in some commutative >: field k. If k* are the nonzero elements of k (with multiplication), >: you can verify that >: (a b) >: (c d) --> ad-bc >: does indeed give a surjective homomorphism det: G ---> k*. >: This will exhibit N = ker(det) and G/N = k* commutative. >All true, but this only shows that N contains G'; the hard part is the other >direction (problem 19, show that N = G'), which is not coming to me right now. >Ted Find a few classes of matrices in G'. For example, all rotation matrices (cos(t) sin(t)) (-sin(t) cos(t)) for real t are in G'. So are upper triangular matrices with ones on the diagonal. Choose enough classes that you generate all of G' when they are merged. Then try to show that every member of your classes is in N. -- After California's recall election, wildfires Schwartz-en-ed the Bush-lands on its geographic right (when we wanted the forests to be Green). pmontgom@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: Problem from Herstein 4 > No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has > determinant 1, and thus so do all products of elements of this form. > Thus G' is contained in N. To show that G' actually equals N, I first > thought of working out the form of a general element of G' and solving > the resulting equations, but these turned out to be rather horrible. > The next problem in the book is a simpler one of the same form, and I > was able to show you could actually get away with quite simple A and B > to get the general element. My attempts to do the same thing with this > one have been fruitless so far. G' is a normal subgroup of G = GL(2,R). If one could show that one nontrivial unipotent matrix A was in G' you would be done. By this I mean a matrix A =/= I with (A-I)^2 = 0. These matrices form a single conjugacy class in G. If one is in G', they all are: in particular (1 x // 0 1) and (1 0// y 1) are in G' for all x and y, and these elementary matrices generate SL(2,R). I suppose it's just a question of playing with commutators until tou get such a matrix. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Problem from Herstein 4 :> 18. Let G be the group of all real 2-by-2 matrices :> (a b) :> (c d), with ad - bc non-zero, under matrix multiplication, and let N :> be the subgroup consisting of those elements of G with ad - bc = 1. :> Prove that N contains G', the commutator subgroup of G. :> No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has :> determinant 1, and thus so do all products of elements of this form. :> Thus G' is contained in N. : :> 19. In problem 18 show, in fact, that N = G'. : I assume, that your matrices have their coefficients in some commutative : field k. If k* are the nonzero elements of k (with multiplication), : you can verify that : (a b) : (c d) --> ad-bc : does indeed give a surjective homomorphism det: G ---> k*. : This will exhibit N = ker(det) and G/N = k* commutative. >All true, but this only shows that N contains G'; the hard part is >the other direction (problem 19, show that N = G'), which is not >coming to me right now. Apply theorem: Let G be a group with commutator subgroup G'. (a) The subgroup G' is normal in G, and the factor group G/G' is abelian. (b) If N is any normal subgroup of G, then the factor group G/N is abelian if and only if G' is contained in N. -- G metabelian? Now from N = G' is it possible to show N is Abelian? In general, when G is invertible nxn matrices with *, then N = G' as shown above. Is it still true that N is Abelian? That is do rigid rotations commute? It seems so. How's it shown and does N = G' help? ---- === Subject: Re: Problem from Herstein 4 > :> 18. Let G be the group of all real 2-by-2 matrices > :> (a b) > :> (c d), with ad - bc non-zero, under matrix multiplication, and let N > :> be the subgroup consisting of those elements of G with ad - bc = 1. > :> Prove that N contains G', the commutator subgroup of G. > :> No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has > :> determinant 1, and thus so do all products of elements of this form. > :> Thus G' is contained in N. > :> 19. In problem 18 show, in fact, that N = G'. > : I assume, that your matrices have their coefficients in some commutative > : field k. If k* are the nonzero elements of k (with multiplication), > : you can verify that > : (a b) > : (c d) --> ad-bc > : does indeed give a surjective homomorphism det: G ---> k*. > : This will exhibit N = ker(det) and G/N = k* commutative. > >All true, but this only shows that N contains G'; the hard part is > >the other direction (problem 19, show that N = G'), which is not > >coming to me right now. > Apply theorem: Let G be a group with commutator subgroup G'. > (a) The subgroup G' is normal in G, and the factor group G/G' is abelian. > (b) If N is any normal subgroup of G, then the factor group G/N is > abelian if and only if G' is contained in N. I don't think I follow you here. We already know that G' is contained in N. The problem is the reverse inclusion, namely that N is contained in G'. > -- G metabelian? > Now from N = G' is it possible to show N is Abelian? > In general, when G is invertible nxn matrices with *, > then N = G' as shown above. > Is it still true that N is Abelian? > That is do rigid rotations commute? > It seems so. How's it shown and does N = G' help? > ---- Again, I think you've lost me here. Certainly N is not abelian in the original question. For example, the matrices (3 1) (8 3) and (2 5) (1 3) do not commute. John Harrison === Subject: how to solve the system of differential equation? Consider a system of DE: 1/c dx_i/dt + x_i = sum_{jk} a^i_{jk}x_j*x_k, where a^i_{jk} are non-negative real numbers and a^i_{jk}=a^i_{kj} (i.e. the matrix a^i is symmetric). Given the initial values of x_i, can the system be analytically solved: a) in general case; b) under the condition sum_i a^i_{jk}=1; c) under the condition b) and sum_i x_i=1? === Subject: Re: how to solve the system of differential equation? >Consider a system of DE: > 1/c dx_i/dt + x_i = sum_{jk} a^i_{jk}x_j*x_k, where a^i_{jk} are > non-negative real numbers and a^i_{jk}=a^i_{kj} (i.e. the matrix a^i > is symmetric). > Given the initial values of x_i, can the system be analytically > solved: > a) in general case; > b) under the condition sum_i a^i_{jk}=1; > c) under the condition b) and sum_i x_i=1? Probably not in general. Maybe for the case of a 2 x 2 system, although the solution might be horrendously complicated. In case (b), if X = sum_i x_i, you have 1/c dX/dt + X = X^2 which can be solved (and X=1 is a solution). So this effectively reduces the dimension of the system by 1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Apocalypse NOW! > Action Device to generate unidirectional force. > http://www.geocities.com/actiondevice Abhi, What is a solid angle? Can you give an example of a solid angle? -- Jeff, in Minneapolis . === Subject: Re: Apocalypse NOW! Just for record. I went to Indian Institute of Technology (IIT), Powai, Mumbai to explain mechanism of my Action Device and to seek technical help. I met Dr. Amitay Issac of Aerospace Engineering Department and I tried to explain very basic component/idea of this action device. I have given in my homepage what exactly I tried to convince him. http://www.geocities.com/actiondevice But he insisted that point B will shift its position along Y axis!. I had to return in few minutes. Now I tried to convince again to Dr. G Arvind Rao of Aerospace Engineering Department by email, but he also said that point B will shift its position along Y axis !. Indian Institute of Technology is most prestigious college in India. This institute gives people for Aviation Industry around the world. And I just wonder, why so highly educated people fail to understand such simple problem. In fact, this is not problem at all. But what a tragedy, I am facing such ridiculous problems. I can end my all problems anytime, but I am following the rules of this battle, waiting game. I am just watching how the minds of highly educated people around the world are controlled by that Supreme Force named God. -Abhi. === Subject: Re: Apocalypse NOW! > Just for record. > I went to Indian Institute of Technology (IIT), Powai, Mumbai to > explain mechanism of my Action Device and to seek technical help. I > met Dr. Amitay Issac of Aerospace Engineering Department and I tried > to explain very basic component/idea of this action device. I have > given in my homepage what exactly I tried to convince him. > http://www.geocities.com/actiondevice > But he insisted that point B will shift its position along Y axis!. > I had to return in few minutes. > Now I tried to convince again to Dr. G Arvind Rao of Aerospace > Engineering Department by email, but he also said that point B will > shift its position along Y axis !. Hmmm... did you consider that they could be right, and you could be wrong? > Indian Institute of Technology is most prestigious college in India. > This institute gives people for Aviation Industry around the world. > And I just wonder, why so highly educated people fail to understand > such simple problem. Maybe, just maybe, they do understand it. > In fact, this is not problem at all. But what a tragedy, I am facing > such ridiculous problems. > I can end my all problems anytime, but I am following the rules of > this battle, waiting game. Build a working model and submit it to them for examination. Doesn't matter how much force it produces, as long as it proves that your idea works. > I am just watching how the minds of highly educated people around the > world are controlled by that Supreme Force named God. Let me get this straight.... *God* doesn't want this device discovered? Why not? And if not, what's stopping him from destroying you to make sure you stay quiet? === Subject: Re: Apocalypse NOW! > Just for record. > I went to Indian Institute of Technology (IIT), Powai, Mumbai to > explain mechanism of my Action Device and to seek technical help. I > met Dr. Amitay Issac of Aerospace Engineering Department and I tried > to explain very basic component/idea of this action device. I have > given in my homepage what exactly I tried to convince him. > http://www.geocities.com/actiondevice > But he insisted that point B will shift its position along Y axis!. > I had to return in few minutes. > Now I tried to convince again to Dr. G Arvind Rao of Aerospace > Engineering Department by email, but he also said that point B will > shift its position along Y axis !. > Hmmm... did you consider that they could be right, and you could be wrong? Laura, where from you suddenly dropped in this mess? You just don't know, what is going on. I thought about this thousands of times in last 13 months. I had posted idea of whole device in many newsgroup. This is just one of the basic component or idea behind this invention. At least this problem was not arised. And now suddenly this problem propped up. > Indian Institute of Technology is most prestigious college in India. > This institute gives people for Aviation Industry around the world. > And I just wonder, why so highly educated people fail to understand > such simple problem. > Maybe, just maybe, they do understand it. Have you done elementary Geometry Laura? Take a look at my homepage. http://www.geocities.com/actiondevice > In fact, this is not problem at all. But what a tragedy, I am facing > such ridiculous problems. > I can end my all problems anytime, but I am following the rules of > this battle, waiting game. > Build a working model and submit it to them for examination. > Doesn't matter how much force it produces, as long as it proves that your > idea works. No, not yet. You just don't know what is going on around me. Things are under absolute control. You will never believe it. > I am just watching how the minds of highly educated people around the > world are controlled by that Supreme Force named God. > Let me get this straight.... *God* doesn't want this device discovered? Why > not? And if not, what's stopping him from destroying you to make sure you > stay quiet? He does want this device to be discovered. This is exactly why He controlled absolutely everything in my personal life. He navigated things in last 17 years in such a way that my thought process moves only in one direction. He trained me to gain absolute power of imagination. This device is very simple. But there is no victory without sufferings. And He has discovered His own ways to trap me. Things are being controlled very cleverly. Don't believe me? People in this NG will not answer clearly the question I have posed. Will point B move along Y axis in XY plane? It needs just yes/no. But they will remain silent(or they will be humorous). They will ignore me. Because they are controlled. Laura, Watch Out Apocalypse In Action.... -Abhi. === Subject: Problem with a series Please, can someone help me with this (difficult) exercise ? Let u_k be a positive real sequence, such that the series sum( 1/u_k, k=1..infinity) converges. Let T_n = u_1 + ... + u_n. Prove that the series sum (n/T_n, n=1..infinity) converges and that sum (n/T_n, n=1..infinity) <= 2 * sum( 1/u_k, k=1..infinity). Hint : Use the Cauchy-Schwarz inequality. === Subject: Re: Problem with a series > Let u_k be a positive real sequence, such that the series sum( 1/u_k, > k=1..infinity) converges. > Let T_n = u_1 + ... + u_n. Prove that the series sum (n/T_n, > n=1..infinity) converges and that sum (n/T_n, n=1..infinity) <= 2 * > sum( 1/u_k, k=1..infinity). > Hint : Use the Cauchy-Schwarz inequality. D.8esol.8e je t'avais oubli.8e !! (Solution du Monier, 3.2.23): By Cauchy-Schwarz inequality applied to the vectors (sqrt(u_1), ..., sqrt(u_n)) and (1/sqrt(u_1),...,n/sqrt(u_n)), (1+2+...+n)^2 <= (u_1+...+u_n)*(1/u_1+2^2/(u_2)^2+...+n^2/(u_n)^2). It follows that: (2n+1)/(u_1+...+u_n) <= 4*(2n+1)/(n^2*(n+1)^2)*sum(k^2/u_k, k=1..n); summing for N>0, sum((2n+1)/(u_1+...+u_n), n=1..N) <= 4*sum((2n+1)/(n^2*(n+1)^2), n=1..N)*sum(k^2/u_k, k=1..n) = 4*sum(k^2/u_k*sum((2n+1)/(n^2*(n+1)^2), n=k..N), k=1..N) <= 4*sum(k^2/u_k*1/k^2, k=1..N) because: sum((2n+1)/(n^2*(n+1)^2), n=k..N) = sum(1/n^2-1/(n+1)^2, n=k..N) = 1/k^2 - 1/(N+1)^2 <= 1/k^2, whence: sum((2n+1)/(u_1+..+u_n), n=1..N) <= 4*sum(1/u_k, k=1..N)<=4*sum(1/u_k, k=1..+oo). @+ Julien Santini === Subject: Re: Problem with a series God save Monier === Subject: [REPOST] Yet more relations between C and M(2,R) I've written at least twice about this subject in the past, without receiving any feedback. I'd be glad to read any kind of comment! >>Exponentials, and logarithms of invertible elements, exist in any Banach >>algebra. Look up holomorphic functional calculus. >I should qualify that. The holomorphic functional calculus exists in >complex Banach algebras, while the usual quaternions are an algebra >over the reals. Of course there's no trouble complexifying to get an While this is certainly OT wrt the OP, I would like to expand to some extent on the relationships between C (the complex *field*) and the the *algebra* M(2,R) of 2x2 real matrices. Of course nothing of what I'm saying has a sound mathematical meaning, but IMO my observations yield a very natural point of view in some respects, e.g. when dealing with some particular problems. (However I'll give as a brief account as possible!) The point is that M(2,R) can be thought of (being isomorphic to) a 2-dimensional complex algebra with a basis given by {1,chi} satisfying chi^2=1, ichi + chi i=0. On the other hand (the algebra isomorphic to) M(2,R) is a 4-dimentional real algebra with a basis given by {1,i,chi,ichi}: in particular these elements are not (the images of) the standard basis vectors of M(2,R). Note that in this sense the elements of M(2,R) are (numbers that constitute) another hypercomplex extension of C. Now, you can work abstractly with this extension of the ring of complexes, and it is not important how you do intepret them. But if you want to have a direct expression of z+wchi (z,w in C) in terms of square matrices, then a *possible choice* of i and chi for the translation is: i=[0 -1] chi=[1 0] [1 0], [0 -1]. It's worth to notice that it is not a mere chance that the second matrix acts on a column vector as complex conjugation. Now, it is straightforward to realize that for A=z+wchi det(A)=|z|^2-|w|^2, A^{-1}=det(A)^{-1} (z*-wchi) if det(A)neq 0. (the latter identity works also if you abstractly *define* det(A) as above). Interestingly the operatorial norm of A is ||A||=|z|+|w|. Now, as an example, let's find the solutions of the equation x^2=-1. Let x=a+bchi, a,b in C: the following two equations must be satisfied: a^2+|b|^2=-1, 2Re(a)b=0. If (i) b=0 then a^2=-1 => a=+i or a=-i; if (ii) bneq 0, then a=ik, k in R, |b|^2=k^2-1 => |k|>1. By allowing |k|>=1 one can express all the solutions including those found at point (i) as x=ik+sqrt(k^2-1)e^{itheta}chi, period! If do the similar calculation for x^2=1, then you find that the solutions found at the corresponding point (i) cannot be incorporated in a general expression and one has x=1 or x=-1 or x=ih+sqrt(h^2+1)e^{iphi}chi, h in R. Another interesting exercise is to look for numbers/matrices I,X that satisfy the same identities as i,chi. (Since 1,-1 commute with every element of the algebra, then X must be chosen of the latter form!) Hope this was a TEASER!! 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: Let's talk factorizations Polynomials are well-known in science and mathematics, but while finding roots of polynomials is typically the aim of the average researcher, polynomials themselves can be used as powerful tools for analyzing the roots of *other* polynomials. The concepts are advanced, but can be approached by first considering a basic example. The basic factorization to start is (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integers, notice that only two of the c's have 7 as a factor. It might help to go the *other* way, and start with (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = x^3 + 5x^2 + 3x + 1 and now multiply by 49. In the first example you're looking at a product and realizing that from the distributive property a(b+c) = ab + ac, you know there's *one* way it could be produced, which is to multiply something like the second example by 49. The distributive property is key here. Understanding it thoroughly, is of prime importance. Now notice that you can abstract from here as you're looking at *functions* of x, as introducing f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, you have (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 as long as you're in a ring where 7 is not a factor of 1. Which is consistent with what was found before, as only two of the functions have the property that 7 is a factor. Now I'll move on to a more complicated example. Let (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) so they are functions of x, and since one of the roots equals 3 at x=0, I have b_3(x) = a_3(x) - 3, so that all the functions in (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) equal 0, when x=0. Those of you who find it hard to use the distributive property with the *product* can imagine the factorization from *before* 49 being multiplied. It's harder to show here as the polynomial which defines the function in that factorization is not displayable in general. So I started at the end, with 49 already multiplied because then I can give a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). That slight change, starting at the end, means that you have to understand the distributive property fully and *trust* it. Now notice that I have the result that only two of the roots of the cubic a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) can have factors in common with 7, so the 49 splits between those two. What's so startling is that the result is for a *family* of polynomials as it applies for any algebraic integer x. James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Let's talk factorizations > Polynomials are well-known in science and mathematics, but while > finding roots of polynomials is typically the aim of the average > researcher, polynomials themselves can be used as powerful tools for > analyzing the roots of *other* polynomials. > The concepts are advanced, but can be approached by first considering > a basic example. > The basic factorization to start is > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > > 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integers, notice that only two of the c's have > 7 as a factor. > It might help to go the *other* way, and start with > (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = > > x^3 + 5x^2 + 3x + 1 > and now multiply by 49. > In the first example you're looking at a product and realizing that > from the distributive property a(b+c) = ab + ac, you know there's > *one* way it could be produced, which is to multiply something like > the second example by 49. > The distributive property is key here. Understanding it thoroughly, > is of prime importance. I like this example. Of course if you want to multiply (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) by 49 so that you have 7 as the constant term in two of the factors, you are right, there is basically one way to do it (modulo permutations of d1, d2, and d3). However you can distribute factors of 49 through the three linear factors above in infinitely many *other* ways, and the resulting polynomial is still the same. For example, (7^{4/3}*d1*x + 7^{4/3})*(7^{1/2}*d2*x + 7^{1/2})*(7^{1/6}*d3*x + 7^{1/6}). If you do this, you still get 49(x^3 + 5x^2 + 3x + 1) just as before. The thing is, there is no particular reason you need to get 7 as the constant term for two of the factors. In the example you give, when x = 0, the constant term P(0) is 49, and of course 49 = 7^2 = 7^{12/6} = 7^{4/3} * 7^{1/2} * 7^{1/6}, right? So the constant terms multiply together as they should. And this is only *one* way to use the distributive law to distribute the factors of 49 across the three factors. You can even define the factorization in three parts as a function of x if you want. > Now notice that you can abstract from here as you're looking at > *functions* of x, as introducing > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > you have > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 > as long as you're in a ring where 7 is not a factor of 1. > Which is consistent with what was found before, as only two of the > functions have the property that 7 is a factor. > Now I'll move on to a more complicated example. > Let > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so they are functions of x, and since one of the roots equals 3 at > x=0, I have > b_3(x) = a_3(x) - 3, > so that all the functions in > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > equal 0, when x=0. > Those of you who find it hard to use the distributive property with > the *product* can imagine the factorization from *before* 49 being > multiplied. > It's harder to show here as the polynomial which defines the function > in that factorization is not displayable in general. > So I started at the end, with 49 already multiplied because then I can > give > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Yes - note that now the a's are indeed functions of x. When you consider 5*a1(x) + 7, and you want to factor a factor of 7 out of it, there are many ways that could be done, depending on the value of a1(x). When x = 0, a1(x) = 0. You know that a1(1), for example, is not 0. You argue, for no visible reason, that a1(1) must be divisible by 7 because the constant term of the product is 7*7*22, and you think that if you divide 7*7 out of the whole expression, you cannot divide any piece of 7 out of the constant term 22 because 7 and 22 are relatively prime in the algebraic integers. You are indeed right that 7 and 22 are relatively prime. However, you don't *need* to have 22 itself divisible by any part of 7. The only thing you need to have divisible by some part of 7 is 5*b3 + 22. Now: you know that 22 is coprime to 7 and you know that 5 is coprime to 7. Suppose b3 is also coprime to 7. Can I conclude from that that 5*b3 + 22 is also coprime to 7? Isn't it possible under the hypotheses just stated that 5*b3 + 22 is NOT coprime to 7 ? (Hint: say b3 = 4). In fact, there is a decomposition of 7*7 of the form r*s*t such that: 1. 7*7 = r*s*t 2. 7/r and 7/s are algebraic integers 3. a_1/r and a2_/s are algebraic integers 4. (b_3*x + 22)/t is an algebraic integer. You may well shriek, WHAT ABOUT X ? SEE THAT X IN THERE? IT'S A VARIABLE! WHAT'S IT DIVISIBLE BY? Here's the key thing. The numbers r, s, and t, just like a_1, a_2, and b_3, are *** functions of x ***. That's exactly how it works. Divisibility of the constant terms is not important. What is important is that when you multiply everything out after you have distributed the parts of 49 among the three factors, the product of the constant terms is still 22. Here's what you get: (7/r) * (7/2) *(22/t) = (7*7/(r*s*t))*22 = (49/(r*s*t))*22 = 22, just as it should. ***IT IS NOT IMPORTANT THAT 22/t IS NOT AN ALGEBRAIC INTEGER***. WHAT IS IMPORTANT THAT (5*b_3 + 22)/t IS AN ALGEBRAIC INTEGER. The other key idea here is the functions-of-x idea. How 49 splits into three parts depends on the value of x. You may say, How can r, s, and t know about x? They know about x because they are intimately connected to a_1, a_2, and b_3, and clearly these coefficients are functions of x because a_1 and a_2 and a_3 = b_3 + 3 are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0 See those x's in there? There is NO REASON to think that because, when x = 0, r, s, and t are respectively 7, 7, and 1, that the same is true for all values of x. That is *not* a requirement which one can deduce from the fact that the constant terms are 7, 7, and 22. > That slight change, starting at the end, means that you have to > understand the distributive property fully and *trust* it. No problem with the distributive law. The problem is, you seem to be able to imagine using it in only one way. However 7*7 can be split up as a product of three algebraic integers in infinitely many ways, and thus distributed among the three factors of your polynomial in infinitely many ways. Not all of those ways give algebraic integer coefficients. As we have shown MANY TIMES, a_1 and a_2 cannot be divisible, in the algebraic integers, by 7. Assuming they are leads to a contradiction when x <> 0. Bottom line: for x <> 0, 49 does not split up the way you think it does. There is *another way* to split it up which does not result in the contraction just mentioned. You appear to be making the mistake of thinking that 22/t has to be an algebraic integer. > Now notice that I have the result that only two of the roots of the > cubic > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > can have factors in common with 7, so the 49 splits between those two. No - you have seen the proof *many* times. No root of the polynomial above can be divisible by 7, and no root can be relatively prime to 7. If you assume so you get a contradiction. It's inescapable! Nora B. > What's so startling is that the result is for a *family* of > polynomials as it applies for any algebraic integer x. > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ === Subject: Re: Let's talk factorizations > Polynomials are well-known in science and mathematics, but while > finding roots of polynomials is typically the aim of the average > researcher, polynomials themselves can be used as powerful tools for > analyzing the roots of *other* polynomials. > The concepts are advanced, but can be approached by first considering > a basic example. > The basic factorization to start is > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > > 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integers, notice that only two of the c's have > 7 as a factor. > It might help to go the *other* way, and start with > (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = > > x^3 + 5x^2 + 3x + 1 > and now multiply by 49. > In the first example you're looking at a product and realizing that > from the distributive property a(b+c) = ab + ac, you know there's > *one* way it could be produced, which is to multiply something like > the second example by 49. > The distributive property is key here. Understanding it thoroughly, > is of prime importance. > Now notice that you can abstract from here as you're looking at > *functions* of x, as introducing > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > you have > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 > as long as you're in a ring where 7 is not a factor of 1. This is the only way to divide both side by 49 if the f_i are linear functions of x. If you use more complicated functions all bets are off. > Which is consistent with what was found before, as only two of the > functions have the property that 7 is a factor. > Now I'll move on to a more complicated example. > Let > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) More complicated functions. All bets are off. - William Hughes === Subject: Re: Let's talk factorizations Why don't you try and get a life instead of repeatedly swamping this newsgroup with endless variations of the same material? Is this what you meant when you said you were going to turn up the volume? If so, why not post in all caps? Maybe that will prove the contents are correct. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Techniques for polynomial division? Are there any techniques that can efficiently do polynomial division? (I am looking for techniques similar to Karatsuba used for multiplication ...) Jaco === Subject: Re: Techniques for polynomial division? > Are there any techniques that can efficiently do polynomial division? > (I am looking for techniques similar to Karatsuba used for > multiplication ...) > Jaco Have you checked Knuth? === Subject: Re: Techniques for polynomial division? > Are there any techniques that can efficiently do polynomial division? > (I am looking for techniques similar to Karatsuba used for > multiplication ...) long division === Subject: Popular measures of error Suppose that g(x) is proposed as an approximation of f(x) on [a,b]. What are the most popular ways of measuring how well g approximates f over that interval? Here are several measures. In each case, the smaller the measure is, the better the approximation is considered to be. AInf. The maximum of |absolute error| over the interval, where absolute error = g(x) - f(x). RInf. The maximum of |relative error| over the interval, where relative error = (absolute error)/f(x). A2. The root-mean-square of |absolute error| over the interval. R2. The root-mean-square of |relative error| over the interval. A1. The average of |absolute error| over the interval. R1. The average of |relative error| over the interval. All of these measures may be thought of as power means (also called Hoelder means). They have the form * ( Integral( |error|^p ) / (b-a) ) ^ (1/p) where the integral is taken with respect to x from a to b, and error is either absolute or relative. Obviously, in the cases of A1 and R1, p = 1, and in the cases of A2 and R2, p = 2. The value of p is not so obvious, however, in the cases of AInf and RInf. But in the limit as p increases without bound, the power mean gives simply the maximum, as needed in AInf and RInf. As such, for those cases, we may say that p = +oo. Here are some questions of mine. Are there any important measures of error in form * which use values of p other than 1, 2, and +oo? Are there any important measures of error which are not in form * ? Clearly, using p = 2 yields a measure which is intermediate between those with p = 1 and p = +oo. In that sense, p =2 represents a nice compromise. But is there anything really special about p = 2 (say, as opposed to p = 4 or p = 3/2) ? (Of course, I grant that the integral is typically far easier to evaluate analytically when p = 2 than when p = 4 or 3/2. But I'm wondering if p = 2 is special for a more fundamental reason than that.) David Cantrell === Subject: Re: Popular measures of error > Suppose that g(x) is proposed as an approximation of f(x) on [a,b]. What > are the most popular ways of measuring how well g approximates f over that > interval? > Here are several measures. In each case, the smaller the measure is, the > better the approximation is considered to be. > AInf. The maximum of |absolute error| over the interval, > where absolute error = g(x) - f(x). > RInf. The maximum of |relative error| over the interval, > where relative error = (absolute error)/f(x). > A2. The root-mean-square of |absolute error| over the interval. > R2. The root-mean-square of |relative error| over the interval. > A1. The average of |absolute error| over the interval. > R1. The average of |relative error| over the interval. > All of these measures may be thought of as power means (also called Hoelder > means). They have the form > * ( Integral( |error|^p ) / (b-a) ) ^ (1/p) > where the integral is taken with respect to x from a to b, and error is > either absolute or relative. Obviously, in the cases of A1 and R1, p = 1, > and in the cases of A2 and R2, p = 2. The value of p is not so obvious, > however, in the cases of AInf and RInf. But in the limit as p increases > without bound, the power mean gives simply the maximum, as needed in AInf > and RInf. As such, for those cases, we may say that p = +oo. > Here are some questions of mine. > Are there any important measures of error in form * which use values of p > other than 1, 2, and +oo? > Are there any important measures of error which are not in form * ? How about the geometric mean, A0 exp ( integral ( log |absolute error| ) / (b - a) ). -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Popular measures of error > Here are some questions of mine. > Are there any important measures of error in form * which use values of p > other than 1, 2, and +oo? I wouldn't dare try to answer this - I haven't learned enough yet (as if doing so is ever possible), but everything I do learn usually teaches me that I should have paid more attention to subjects I had previously thought were unimportant. > Are there any important measures of error which are not in form * ? The error norms in higher order Sobolev spaces aren't quite in that form; but at least all the integral order spaces have norms in the form (Integral( Sum_i (|error_i|^p) ) ^ (1/p)) which isn't much more general. > Clearly, using p = 2 yields a measure which is intermediate between > those with p = 1 and p = +oo. In that sense, p =2 represents a nice > compromise. But is there anything really special about p = 2 (say, as > opposed to p = 4 or p = 3/2) ? With any p, the absolute error norms are metrics on appropriately defined function spaces, so everything you can prove about normed vector spaces applies to them. It's only for p = 2 that the metric comes from an inner product, however, and you can prove a lot more about Hilbert spaces. It's easier to find upper bounds for finite element method errors in p=2 norms, for example. --- Roy Stogner === Subject: Re: Popular measures of error > Suppose that g(x) is proposed as an approximation of f(x) on [a,b]. > What are the most popular ways of measuring how well g approximates f > over that interval? As pointed out by another responder, the answer depends on the application. For example, in time-dependent analysis, where x represents time, one might want to compare two time series. The comparison is made more complicated if the two signals look similar to the eye, but differ more in phase than in amplitude. For example, compare two sine waves of equal amplitude, but slightly different frequency. === Subject: Re: Popular measures of error > Are there any important measures of error in form * which use values of p > other than 1, 2, and +oo? There are some harmonic means that, I think, involve expressions with p=-1. $.02 -Ron Shepard === Subject: Re: Popular measures of error > Suppose that g(x) is proposed as an approximation of f(x) on [a,b]. What > are the most popular ways of measuring how well g approximates f over that > interval? I guess it depends on the field of application, and also in the particular application. In engineering, I believe the one used most frequently is the mean-square error, since it is related to energy. > Here are some questions of mine. > Are there any important measures of error in form * which use values of p > other than 1, 2, and +oo? I'm not familiar with any that are commonly used (I'm an electrical engineer -- maybe in other fields there might be) > Are there any important measures of error which are not in form * ? Not that I'm familiar with. > or p = 3/2) ? (Of course, I grant that the integral is typically far easier > to evaluate analytically when p = 2 than when p = 4 or 3/2. But I'm > wondering if p = 2 is special for a more fundamental reason than that.) Calculating the integral is usually irrelevant. What you want it find conditions that guarantee that the error is minimized, not finding out what the error is. For instance, when you solve an overdetermined set of linear equations, you are calculating the optimal solution; you are not calculating the error (though you know that it is minimum, and you could calculate it after you have the solution, if you wanted). HTH, Carlos -- === Subject: Re: Mathematical Integrity > Maybe so, but in the mathematics community, truth takes precedence over > ego, which makes the field very different from business, politics, and > many other human endeavors. Well, I think that the major difference is that in mathematics it is much more difficult to get away with bull, because of the nature of the discipline. Hand waving can get you very in business, politics, philosophy, etc.; not so in mathematics. Now my opinion is that ego is just as strong in the mathematics community as elsewhere, as can be seen in lots of heads of departments world over. === Subject: Re: Mathematical Integrity >Sometimes a student would post an innocent question but he or she >would get scold for posting such a stupid question and would be >called a moron. What does that have to do with integrity? >It seems like these dese days, some mathematicians would use >profanity against others mathematicians. What does that have to do with integrity? >Where is the purity and integrity in mathematics? Neither purity nor integrity has anything to do with civility or with the ability to suffer fools gladly. >Anyhow, I still love math by heart and I would like to give my >respect to many great math fanatics in this forum. You could best show your respect by not referring to them as fanatics. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Mathematical Integrity |>Where is the purity and integrity in mathematics? | |Neither purity nor integrity has anything to do with civility or with |the ability to suffer fools gladly. They are separate issues, but they are not unrelated. I find that the people I've gotten acquainted with, who have a reputation for not suffering fools gladly or who are proud of not doing so, nearly all can be seen erring on the side of prematurely concluding that some idea or person was to be rejected. Incivility conduces to hostility, and hostility clouds the mind. On a crude level, it is possible to be accurate about people who annoy one, but I find it hard, and I think most people do too, to be very observant in such a frame of mind. One feels a much greater temptation to slack off on efforts at describing these fools accurately, which is a lapse in integrity. For example, one mathematician I met had considered whether it would work well to use an alternative treatment of a certain issue in functional analysis. He looked for anyone who might have addressed the issue, and found that of all the places he looked, it was solely in a textbook by a second mathematician that a reason was given why we should do it the usual way. But the first mathematician had a counterargument, though, which he wanted to present to the second mathematician. The second one (who gave me an impression of being usually conscientious about addressing mistakes in his work and so on) had this don't suffer fools gladly attitude, and when approached by the first, stormed angrily away declaring he had no time to discuss such a thing. I realize that I'm expecting the reader to believe me when I indicate that this was an overreaction, although it's hard to demonstrate. It wasn't a matter of the second mathematician being interrupted during something important and the like. And the discussion was not simply postponed to a more convenient time. It is sometimes said that this kind of stern response is needed to keep at bay the large volumes of nonsense that we are all exposed to. I think the value of such hostility is much overrated, however. Plenty of people are civil and relatively nonjudgemental without letting very much of their time be wasted on pointless pursuits. My opinion is that society in the U.S. is generally erring on the side of glibness, and that there are many involved in politics who are keen to encourage us to err still farther in that direction. Rather than reasoning things out, we get discussions in which people express shock and horror that the plain, obvious answer is not being accepted immediately. I see pundits trying to get me not only to agree with them, but to agree that it's so painfully obvious that I needn't bother considering the other point of view, at least not seriously. This goes beyond incivility, but it's the kind of problem that incivility encourages. Keith Ramsay === Subject: Re: Mathematical Integrity Very well said. I agree. Have a tolerable existence. Eli. > |>Where is the purity and integrity in mathematics? > |Neither purity nor integrity has anything to do with civility or with > |the ability to suffer fools gladly. > They are separate issues, but they are not unrelated. I find that > the people I've gotten acquainted with, who have a reputation for > not suffering fools gladly or who are proud of not doing so, > nearly all can be seen erring on the side of prematurely concluding > that some idea or person was to be rejected. Incivility conduces to > hostility, and hostility clouds the mind. On a crude level, it is > possible to be accurate about people who annoy one, but I find it > hard, and I think most people do too, to be very observant in such > a frame of mind. One feels a much greater temptation to slack off > on efforts at describing these fools accurately, which is a lapse > in integrity. > For example, one mathematician I met had considered whether > it would work well to use an alternative treatment of a certain > issue in functional analysis. He looked for anyone who might > have addressed the issue, and found that of all the places he > looked, it was solely in a textbook by a second mathematician > that a reason was given why we should do it the usual way. But > the first mathematician had a counterargument, though, which > he wanted to present to the second mathematician. The second > one (who gave me an impression of being usually conscientious > about addressing mistakes in his work and so on) had this > don't suffer fools gladly attitude, and when approached by the > first, stormed angrily away declaring he had no time to discuss > such a thing. I realize that I'm expecting the reader to believe me > when I indicate that this was an overreaction, although it's hard > to demonstrate. It wasn't a matter of the second mathematician > being interrupted during something important and the like. And > the discussion was not simply postponed to a more convenient > time. > It is sometimes said that this kind of stern response is needed to > keep at bay the large volumes of nonsense that we are all exposed > to. I think the value of such hostility is much overrated, however. > Plenty of people are civil and relatively nonjudgemental without > letting very much of their time be wasted on pointless pursuits. > My opinion is that society in the U.S. is generally erring on the > side of glibness, and that there are many involved in politics who > are keen to encourage us to err still farther in that direction. > Rather than reasoning things out, we get discussions in which > people express shock and horror that the plain, obvious answer > is not being accepted immediately. I see pundits trying to get me > not only to agree with them, but to agree that it's so painfully > obvious that I needn't bother considering the other point of view, > at least not seriously. This goes beyond incivility, but it's the kind > of problem that incivility encourages. > Keith Ramsay === Subject: optic flow numerical model Hello everyone, My research so far in using optic flow for aerial robot navigation has been mainly experimental (see http://www.pages.drexel.edu/~weg22/research.html if interested). However, I want to get more involved in the theoretical aspects of it by creating an optic flow sensor numerical model. When the same input is fed into my model and a sensor (www.centeye.com for example), the output should be the same. I have never developed any numerical models before and I was just wondering if anyone out there could offer some advice or suggestions on where to begin? Also, I would appreciate any information that can be shared about models that are already in existence (possibly 1D). -Bill === Subject: Re: Advanced techniques, non-polynomial factorization > Polynomials are well-known in science and mathematics, but while > finding roots of polynomials is typically the aim of the average > researcher, polynomials themselves can be used as powerful tools for > analyzing the roots of *other* polynomials. > The concepts are advanced, but can be approached by first considering > a basic example. > The basic factorization to start is > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > > 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integers, notice that only two of the c's have > 7 as a factor. > It might help to go the *other* way, and start with > (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = > > x^3 + 5x^2 + 3x + 1 > and now multiply by 49. > In the first example you're looking at a product and realizing that > from the distributive property a(b+c) = ab + ac, you know there's > *one* way it could be produced, which is to multiply something like > the second example by 49. > The distributive property is key here. Understanding it thoroughly, > is of prime importance. > Now notice that you can abstract from here as you're looking at > *functions* of x, as introducing > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > you have > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 > as long as you're in a ring where 7 is not a factor of 1. > Which is consistent with what was found before, as only two of the > functions have the property that 7 is a factor. This part so far is OK. > Now I'll move on to a more complicated example. > Let > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where the a's are roots of >[***] a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so they are functions of x, and since one of the roots equals 3 at > x=0, I have > b_3(x) = a_3(x) - 3, > so that all the functions in > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > equal 0, when x=0. > Those of you who find it hard to use the distributive property with > the *product* can imagine the factorization from *before* 49 being > multiplied. As in the first example you gave above, the 49 can be distributed among the three factors in several different ways. There is no justification for your implied claim below that two constant terms 7 should be cancelled completely. This is where you are making your error. > It's harder to show here as the polynomial which defines the function > in that factorization is not displayable in general. > So I started at the end, with 49 already multiplied because then I can > give > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > That slight change, starting at the end, means that you have to > understand the distributive property fully and *trust* it. > Now notice that I have the result that only two of the roots of the > cubic > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > can have factors in common with 7, so the 49 splits between those two. When you were discussing the simpler polynomial above, you said that the coefficients c_1, c_2, c_3 were algebraic integers, so presumably you are talking about the same thing here. That is, you are saying the a's are algebraic integers and two of them are divisible by 7 in the ring of algebraic integers. Thus assume a_1/7 is an algebraic integer; equivalently a_1 = 7*b_1, where b1 is an algebraic integer. But since you are saying that a_1 is a root of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) we must have that b_1 is a root of 7^3*b^3 + 3*(-1 + 49*x)*7^2*b^2 - 7^2*(2401*x^3 - 147*x^2 + 3*x) = 0. Now divide through by 7^2: 7*b^3 + 3*(-1 + 49*x)*b^2 - (2401*x^3 - 147*x^2 + 3*x) = 0. Finally, let x = 1: 7*b^3 + 144*b^2 - 2257 = 0. This polynomial is primitive, irreducible, and non-monic. Therefore none of its roots can be algebraic integers. Therefore a_1/7 is not an algebraic integer. Therefore a_1 is not divisible by 7. You continue to think that your proof that a_1 *is* divisible by 7 is valid. If it were, it would imply a mathematical contradiction. Mathematics would be inconsistent. There is no sense in claiming that the algebraic integers are incomplete. They are a perfectly well- defined subset of the real numbers. Something cannot both be in that subset and not in that subset. Please explain your own conclusions on this. > What's so startling is that the result is for a *family* of > polynomials as it applies for any algebraic integer x. For example, x = 1, which is what I used above. Your proof implies a contradiction, so it is either incorrect or math is inconsistent. I have identified exactly above where you have made an unjustified assumption. Your argument here is wrong. Nora B. > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ === Subject: Re: Advanced techniques, non-polynomial factorization > so that all the functions in > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > equal 0, when x=0. We have been here many times. The problems is that a_1(x), a_2(x) and b_3(x) are not polynomials. Therefore, we do not know that the way in which the 49 distributes itself among the three factors on the LHS is independent of x. Thus we cannot conclude that 7 divides (5 a_1(x)+ 7) for all x. Thus we cannot conclude that 7 divides a_1(x) for all x. -William Hughes === Subject: Re: Advanced techniques, non-polynomial factorization posted. I did comment on it already, but you did not answer. I think you have not seen my response. So I will try again here. > Polynomials are well-known in science and mathematics, but while > finding roots of polynomials is typically the aim of the average > researcher, polynomials themselves can be used as powerful tools for > analyzing the roots of *other* polynomials. Oh, well, It appears you have modified it a bit. Yes, polynomials are powerful tools to analyse roots of other polynomials. Yup, this paragraph was rewritten, as is the next. > The concepts are advanced, but can be approached by first considering > a basic example. > The basic factorization to start is > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > > 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integers, notice that only two of the c's have > 7 as a factor. A comment I add this time and did not add in the last version. The c's can *only* be algabraic integers because the constant factor of the polynomial is 1. When that constant factor is 2, there is *no* such decomposition. > It might help to go the *other* way, and start with > (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = > > x^3 + 5x^2 + 3x + 1 > and now multiply by 49. And again the same, newly added, comment: the d's are *only* algebraic integers because the constant term of the cubic is 1. > In the first example you're looking at a product and realizing that > from the distributive property a(b+c) = ab + ac, you know there's > *one* way it could be produced, which is to multiply something like > the second example by 49. > The distributive property is key here. Understanding it thoroughly, > is of prime importance. Hrm, an added paragraph. Does not add much information. > Now notice that you can abstract from here as you're looking at > *functions* of x, as introducing > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > you have > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 > as long as you're in a ring where 7 is not a factor of 1. A repeat: This is independent of 7 being a factor of 1. Note however that it is *not* the only way to distribute 49 amongst the three factors on the left hand side. This is the only way *only* if you require that the three factors on the left hand side are polynomials, if you have not such an requirement it can be done differently. However, you can have polynomials on the left hand side *only* because the roots of the polynomial on the right hand side are units, i.e. divisors of 1. It will *not* work if that is not the case. An addition: If you do not have the requirement that the factors on the left hand side are polynomials, the following factorisation is also possible (and gazillion others), and assuming x is integer: Define: w3(x) = gcd(f3(x) + 1, 7) { this can be 1 or some other divisor of 7. } w2(x) = 7 / w3(x). Now: [ (f1(x) + 7)/7 ][ (f2(x) + 7)/w2(x) ][ (f3(x) + 1)/w3(x) ] is a perfectly valid factorisation in the algebraic integer valued functions on the integers of: x^3 + 5 x^2 + 3x + 1 Note that the distributive property dictates: (a + b) * c = a * c + b * c in a particular ring. Not that (a + b) / c = a / c + b / c in a ring. That is, that when (a + b)/c is in a ring, there is no requirements that a/c and b/c are also in that ring. You keep on assuming that. > Which is consistent with what was found before, as only two of the > functions have the property that 7 is a factor. This is true *only* in some special cases... > Now I'll move on to a more complicated example. > Let > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so they are functions of x, and since one of the roots equals 3 at > x=0, I have Note, that they are *not* polynomials. So the situation is different from the one above. And indeed, in general *none* of the factors is divisible by 7. Moreover, you can not even have the requirement that when you divide by 49 the factors on the left must be polynomials, because you do not start with polynomials on the left. So there are other ways to distribute 49 amongst the three factors, however the way you distribute is a function of x. Remainder of repetition skipped. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: AT LAST - TRADING THE MARKETS BY UNIFYING CYCLES AND PROBABILITY!!!! Bull. Flush! === Subject: sum( sin(n)/n , n=1..infinity) < infinity ? Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... converge? How does one prove convergence (or divergence)?. If it converges what is a good way to estimate its value? Jim Buddenhagen ------------ To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE === Subject: Re: sum( sin(n)/n , n=1..infinity) < infinity ? > Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... converge? > How does one prove convergence (or divergence)?. If it converges what > is a good way to estimate its value? Recognize sin(x)/1 + sin(2x)/2 + sin(3x)/3 + sin(4x)/4 + ... as the Fourier series of a certain function, then evaluate it at x=1. Answer: (pi-1)/2 = 1.0708, approx. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: sum( sin(n)/n , n=1..infinity) < infinity ? James Buddenhagen says... >Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... converge? >How does one prove convergence (or divergence)?. If it converges what >is a good way to estimate its value? I'm not sure whether it converges, but if it does, I know what it converges to 8^) Note that the infinite series S = sin(1)/1 + sin(2)/2 + ... is the imaginary part of the series (using the relation exp(ix) = cos(x) + i sin(x)). E = exp(i)/1 + exp(2i)/2 + ... This is a special case of the series x/1 + x^2/2 + ... (To see this, let x = exp(i)) If this converges, it converges to the function log(1/(1-x)), so E = log(1/(1-exp(i))) To take the imaginary part, we need to rewrite 1/(1-exp(i)) in the for A exp(iB) where A and B are real. To get it in this form, note 1/(1-exp(i)) = exp(-i/2)/(exp(-i/2) - exp(i/2)) = i exp(-i/2)/(2 sin(1/2)) = exp(i pi/2) exp(-i/2)/(2 sin(1/2)) = exp(i (pi/2 - 1/2))/(2 sin(1/2)) Taking the log gives E = i (pi/2 - 1/2) - log(2 sin(1/2)) Taking the imaginary part gives: S = (pi/2 - 1/2) -- Daryl McCullough Ithaca, NY === Subject: Re: sum( sin(n)/n , n=1..infinity) < infinity ? > Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... converge? > How does one prove convergence (or divergence)?. If it converges what > is a good way to estimate its value? Abel's rule === Subject: Re: sum( sin(n)/n , n=1..infinity) < infinity ? Julien Santini a .8ecrit dans le message de > Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... converge? > How does one prove convergence (or divergence)?. If it converges what > is a good way to estimate its value? > Abel's rule OK what about tan(n)/n ? Philippe === Subject: Re: sum( sin(n)/n , n=1..infinity) < infinity ? > Julien Santini a .8ecrit dans le message de >> Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... > converge? >> How does one prove convergence (or divergence)?. If it converges what >> is a good way to estimate its value? >> Abel's rule > OK what about tan(n)/n ? The sequence tan(n)/n does not converge to 0; therefore, the series tan(1)/1 + tan(2)/2 + tan(3)/3 + ... diverges. Jose Carlos Santos === Subject: Re: sum( sin(n)/n , n=1..infinity) < infinity ? >> Julien Santini a .8ecrit dans le message de > Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... >> converge? > How does one prove convergence (or divergence)?. If it converges what > is a good way to estimate its value? Abel's rule >> OK what about tan(n)/n ? >The sequence tan(n)/n does not converge to 0; I imagine this is true, since Israel says he's proved it. But it's not all that obvious (it's not clear to me whether you were meaning to say it was obvious or not...) >therefore, the series >tan(1)/1 + tan(2)/2 + tan(3)/3 + ... diverges. >Jose Carlos Santos David C. Ullrich === Subject: Re: sum( sin(n)/n , n=1..infinity) < infinity ? >> OK what about tan(n)/n ? >The sequence tan(n)/n does not converge to 0; therefore, the series >tan(1)/1 + tan(2)/2 + tan(3)/3 + ... diverges. Correct. And for a proof that tan(n)/n does not converge to 0, you can A limit problem with explanation. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: JSH: Survey on my results, any correct? > 1. I've posted a lot on sci.math over a long period of time, to your > knowledge, have I *ever* been right? Only on minor issues. > 2. Do I have *any* correct results, or do you think I just talk and > never say anything that is mathematically correct? Only on minor issues. > 3. To your knowledge, has ANYONE ever posted agreement with me on > *anything*? Only on minor issues. > 4. In your opinion, have I ever won an argument on the newsgroup? Only on minor issues. > 5. To your knowledge, have I ever caught other posters in errors? For the last time, only on minor issues. > James Harris === Subject: Re: JSH: Survey on my results, any correct? > 1. I've posted a lot on sci.math over a long period of time, to your > knowledge, have I *ever* been right? Yes. > 2. Do I have *any* correct results, or do you think I just talk and > never say anything that is mathematically correct? Yes ! > 3. To your knowledge, has ANYONE ever posted agreement with me on > *anything*? Yes. > 4. In your opinion, have I ever won an argument on the newsgroup? At least arguments of typographical errors. > 5. To your knowledge, have I ever caught other posters in errors? At least typographical errors. You are welcome, We Pretty > James Harris === Subject: Re: JSH: Survey on my results, any correct? > 1. I've posted a lot on sci.math over a long period of time, to your > knowledge, have I *ever* been right? Yes, but mostly not on the important things. You were right that you discovered a way to count primes that appears to be different from other ways. > 2. Do I have *any* correct results, or do you think I just talk and > never say anything that is mathematically correct? Mostly the same answer as above. You have been consistently and repeatedly wrong about all your significant claims connected with Fermat's Last Theorem or a core error in mathematics. However you have found a different way to count primes. > 3. To your knowledge, has ANYONE ever posted agreement with me on > *anything*? Yes. > 4. In your opinion, have I ever won an argument on the newsgroup? Not a big one. A few little ones. > 5. To your knowledge, have I ever caught other posters in errors? Yes. But far less often than you have been caught. Nora B. > James Harris === Subject: Re: JSH: Survey on my results, any correct? > 1. I've posted a lot on sci.math over a long period of time, to your > knowledge, have I *ever* been right? Yes. You have been right on occasion. > 2. Do I have *any* correct results, or do you think I just talk and > never say anything that is mathematically correct? Yes. You have been right on occasion. > 3. To your knowledge, has ANYONE ever posted agreement with me on > *anything*? Yes, agreement has been posted on occasion. > 4. In your opinion, have I ever won an argument on the newsgroup? No. > 5. To your knowledge, have I ever caught other posters in errors? I have no idea. Probably. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: l_p norm > I'm having difficulty proving that ||v||_p <= ||v||_2 for p>=2, here v > is a vector in R^n. > Prove it when ||v||_p = 1 and then remember these are norms. === Subject: About naturals Let S(n) the decimal sum of n, this is S(17)=1+7=8, S(98)=9+8=17 I have the following question: There are a, b and c, natural numbers, such that S(a+b)<5 S(a+c)<5 S(b+c)<5 and S(a+b+c)>50 ? I think that there are not. (Exuse-me I write english very bad) Pepe Bosch === Subject: Re: About naturals > Let S(n) the decimal sum of n, this is > S(17)=1+7=8, > S(98)=9+8=17 > I have the following question: There are a, b and c, natural numbers, such > that > S(a+b)<5 this is true then : a < 50 and b < 50 > S(a+c)<5 this also : a < 50 and c < 50 > S(b+c)<5 and and this, of course : b < 50 and c < 50 > S(a+b+c)>50 ? Uhm..if this statement were true, this would be true as well : a+b+c>599999 (as this is smallest number n for which S(n) > 50) And..that can't be true? I'm not sure if this is right since it's awfully simple..perhaps your definiton of decimal sum is different.. -- Quaternion === Subject: Re: About naturals Adjunct Assistant Professor at the University of Montana. >> Let S(n) the decimal sum of n, this is >> S(17)=1+7=8, >> S(98)=9+8=17 >> I have the following question: There are a, b and c, natural numbers, such >> that >> S(a+b)<5 >this is true then : >a < 50 and b < 50 No. Say a=1010, b= 200. Then a+b = 1210, and S(a+b) = 4 < 5. Write x = a_0 + 10a_1 + 10^2a_2 + ... + 10^n a_n y = b_0 + 10b_1 + 10^2b_2 + ... + 10^n b_n with 0<= a_i,b_i < 10, and at least one of a_n,b_n nonzero. So S(x) = a_0 + a_1 + ... + a_n S(y) = b_0 + b_1 + ... + b_n Define d_0,....,d_n recursively as follows: d_0 = 0 if a_0+b_0 < 10 d_0 = 1 if a_b+b_0 > 9. d_{i+1} = 0 if a_{i+1} + b_{i+1} + d_i < 10 d_{i+1} = 1 if a_{i+1} + b_{i+1} + d_i > 9 (The d_i are the carries) Then (x+y) + (a_0+b_0 - 10d_0) + 10(a_1+b_1+d_0-10d_1) + ... + 10^n(a_n+b_n+d_{n-1}-10d_n) + 10^{n+1}d_n. Then S(x+y) = a_0+b_0 + a_1 + b_1 + ... + a_n+b_n + +(d_0+...+d_n)- 10(d_0+...+d_n) = S(x) + S(y) - 9(d_0+...+d_n). Now, you assume that S(a+b) <= 4 S(b+c) <= 4 S(a+c) <= 4 Say we take 2(a+b+c) = (a+b) + (b+c) + (a+c). Now, let's consider the carries involved: Each of a+b, a+c, b+c has at most 4 nonzero digits, and each nonzero digit is at most 4. How many carries can there be? For there to be carries when we add a+b, a+c, and b+c together, there must be corresponding entries, at least one of which is a 4, and the other 2 are either 4's or 3's. But that means that there is at most 1 carry. That is: S(2(a+b+c)) = S(a+b) + S(a+c) + S(b+c) or S(2(a+b+c)) = S(a+b) + S(a+c) + S(b+c) - 9. Now we want to relate S(2(a+b+c)) to S(a+b+c) Again, let x = a_0 + 10a_1 + 10^2a_2 + ... + 10^n a_n with 0<= a_i <10, a_n>0 So S(x) = a_0 +...+ a_n. Define e_0,....,e_n by letting e_i = 0 if 0<=a_i <5 e_i = 1 if 4>Let S(n) the decimal sum of n, this is >>S(17)=1+7=8, >>S(98)=9+8=17 >>I have the following question: There are a, b and c, natural numbers, such >>that >>S(a+b)<5 > this is true then : > a < 50 and b < 50 Surely not? What about a = b = 100? a+b = 200 and s(200) = 2 + 0 + 0 < 5. >>S(a+b+c)>50 ? > Uhm..if this statement were true, this would be true as well : > a+b+c>599999 (as this is smallest number n for which S(n) > 50) > And..that can't be true? I'm not sure if this is right since it's awfully > simple..perhaps your definiton of decimal sum is different.. Doesn't work. You can get arbitrarily big values of a + b + c with S(a+b), etc < 5. I can't see whether or not you can make S(a+b+c) bigger than 50 or not, but will think about it. (Probably won't have too much success - number theory isn't my strength) David (E-mail address spam-blocked in the obvious way) === Subject: Re: About naturals >Let S(n) the decimal sum of n, this is >S(17)=1+7=8, >S(98)=9+8=17 >I have the following question: There are a, b and c, natural numbers, such >that >S(a+b)<5 >> this is true then : >> a < 50 and b < 50 >Surely not? What about a = b = 100? a+b = 200 and s(200) = 2 + 0 + 0 < 5. >S(a+b+c)>50 ? >> Uhm..if this statement were true, this would be true as well : >> a+b+c>599999 (as this is smallest number n for which S(n) > 50) >> And..that can't be true? I'm not sure if this is right since it's awfully >> simple..perhaps your definiton of decimal sum is different.. >Doesn't work. You can get arbitrarily big values of a + b + c with >S(a+b), etc < 5. I can't see whether or not you can make S(a+b+c) bigger >than 50 or not, but will think about it. (Probably won't have too much >success - number theory isn't my strength) >David >(E-mail address spam-blocked in the obvious way) You can get at least 24: a = b = c = 5050505. a+b = a+c = b+c = 10101010 a+b+c = 15151515 S(a+b+c) = 24 An upper bound is 60. S(a + b + c) = S(10a + 10b + 10c) <= S(5a + 5b) + S(5a + 5c) + S(5b + 5c) <= 5S(a + b) + 5S(a + c) + 5S(b + c) <= 15*4 = 60. -- After California's recall election, wildfires Schwartz-en-ed the Bush-lands on its geographic right (when we wanted the forests to be Green). pmontgom@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: About naturals >Let S(n) the decimal sum of n, this is >S(17)=1+7=8, >S(98)=9+8=17 >I have the following question: There are a, b and c, natural numbers, >such that >S(a+b)<5 >> this is true then : >> a < 50 and b < 50 > Surely not? What about a = b = 100? a+b = 200 and s(200) = 2 + 0 + 0 < 5. Right.. of course!..I already thought there was somethign wrong. There's really nothing you can get for S(a) or S(b) out of S(a+b)<5.. So the same counts for the relationship between S(a+b) and S(a+b+c) Sounds like a very complex problem then, unless you perhaps put a reasoning behind 'carries' when summating a+b with c and a with b+c (2*(a+b+c) = (a+b)+c + a+(b+c)), since the carries working in the sum can't have an increasing effect on the decimal sum?.. So for the decimal sum to increase when adding two numbers, you'd need to have values that fill each other up and don't cause carries..which eerily revolves around the boundary of 5. Well I'll stay out of this..stuff for the big boys -- Quaternion === Subject: Re: About naturals I think you are right but I can't demonstrate that Pepe Bosch ha scritto nel messaggio > Let S(n) the decimal sum of n, this is > S(17)=1+7=8, > S(98)=9+8=17 > I have the following question: There are a, b and c, natural numbers, such > that > S(a+b)<5 > S(a+c)<5 > S(b+c)<5 and > S(a+b+c)>50 ? > I think that there are not. > (Exuse-me I write english very bad) > Pepe Bosch === Subject: Re: Relativity is based on assumption. >What experimental evidence? Transverse Doppler effect; Relativistic corrections to the spectrum of the Hydrogen atom; all experimental evidence for QED, which conflates to evidence for Relativity; E = mc^2 directly observed in Hiroshima permanently settling the issue; relativistic momentum and energy half life of 15 minutes (neutrons) to travel light years for thousands of years across the cosmos to reach Earth; Michelson-Morley experiment; the constitutive relations D = epsilon_0 E, B = mu_0 H. search.yahoo.com/search?p=Evidence For Relativity >Moving clocks running slow? They don't. Directly observed to do so, in fact. >The GPS clocks run fast. ... precisely as predicted by Relativity, and in the very amount predicted to do so. === Subject: Re: Relativity is based on assumption. >assumption, and is intuitive as well. Making the assumption that the time it >takes for a signal to reach an object is the same as the time it take for >the signal to return, when in the meantime [sic] you've moved away or >toward the object [...] Motion is relative. You were standing still. It was the object that was moving. So, there. === Subject: Re: Relativity is based on assumption. > I provided proof for my claim. But of course you *conveniently* > snipped the link. That proof consists of fairly straightforward > reasoning. > LOL!! > I provided the proof that Einstein makes assumption, AFTER you denied it, > and called me a liar. You did no such thing. You are a disgusting lying pig. > Then YOU snipped my response, windbag. I did no such thing. You are a disgusting lying pig. > Not only that, > but you gave no proof at all. I did. You are a disgusting lying pig. > Assertion isn't proof, even if you think it is. I never claimed it were. You, however, have provided nothing but assertion for the subject line. > The subject line remains true, whether you like it or not, Ostrich. The subject line has been solidly disproven. You are a disgusting lying pig. === Subject: Re: Relativity is based on assumption. > > > No way, Henri. I've been debating with him for years, and he's just one of > those relativists that refuse to see what is right under their nose. It's a > psychological hang-up that most people have. The don't want to admit they've > been made a fool of by Einstein, they have to appear in the eyes of others > as the one who has all the answers, but that's provided the question suits > them, of course. When it doesn't, they ignore it. > 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)) > ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/ > Where does he get that 1/2 from? > > I send you a letter on the 10th. I get a reply from you on the > 16th. So I assume you received the letter and replied to it on > the 13th: > 1/2( 10 + 16) = 13 > Right... but perhaps still a little too far out. > I will try to improve the analogy, who knows, it may just make the > difference: > Without mentioning it, Dirk also calculated that the total return time > was 16-10 =6 days. > Therefore, if we don't know anything about the speed of either > trajectory, we have no other option but to assume - or declare by > convention - that the one way delivery time is 1/2 * 6 = 3 days. > You may have noticed that at that time Einstein was careful to declare > the total average return speed to be constant, in full agreement with > measurements and LET. No assumption there! > The later SRT assumption that the one-way speed is truly c in one's > own frame of reference, can not be proved if the Lorentz equations are > correct. > Harald xein: How simple. Bravo. When was the last OWLS-TWLS discussion? Was it decided by vote or physics? === Subject: Re: Relativity is based on assumption. <8pr7rvojo5acmso735p31bo58d7g0rq43v@4ax.com> <4C0tb.4282$_g6.527@news-binary.blueyonder.co.uk> === >Subject: Relativity is based on assumption. Duh! So is any scientific theory. But it has nothing to do with Mathematics. Please take it to sci.physics.relativity.crackpots. Oh, BTW, welcome to my filters. *PLONK* -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Relativity is based on assumption. > All theories, in physics and elsewhere, are based on assumptions. > No theory can bootstrap itself out of nothing. > So what? > 'So what' depends on the validity of the assumption. The PoR is an > assumption, and is intuitive as well. Making the assumption that the time it > takes for a signal to reach an object is the same as the time it take for > the signal to return, when in the meantime you've moved away or toward the > object, is a rather silly assumption that I will not accept. That's 'so > what'. > Androcles Then ignore relativity. But unless you can come up with something that agrees with the experimental evidence better than relativity, everyone else will ignore you. === Subject: Re: Relativity is based on assumption. > All theories, in physics and elsewhere, are based on assumptions. No theory can bootstrap itself out of nothing. So what? > 'So what' depends on the validity of the assumption. The PoR is an > assumption, and is intuitive as well. Making the assumption that the time it > takes for a signal to reach an object is the same as the time it take for > the signal to return, when in the meantime you've moved away or toward the > object, is a rather silly assumption that I will not accept. That's 'so > what'. > Androcles > Then ignore relativity. > But unless you can come up with something that agrees with the > experimental evidence better than relativity, everyone else will > ignore you. What experimental evidence? Moving clocks running slow? They don't. The GPS clocks run fast. MMX? It's pretty obvious to anyone with half a brain that the result is what you would expect if the speed of light were source dependent. I would say that agreed with the experimental evidence much better than relativity. Androcles === Subject: Re: Relativity is based on assumption. > All theories, in physics and elsewhere, are based on assumptions. No theory can bootstrap itself out of nothing. So what? > 'So what' depends on the validity of the assumption. The PoR is an > assumption, and is intuitive as well. Making the assumption that the > time it > takes for a signal to reach an object is the same as the time it take > for > the signal to return, when in the meantime you've moved away or toward > the > object, is a rather silly assumption that I will not accept. That's 'so > what'. > Androcles > Then ignore relativity. > But unless you can come up with something that agrees with the > experimental evidence better than relativity, everyone else will > ignore you. > What experimental evidence? Moving clocks running slow? They don't. The GPS > clocks run fast. MMX? It's pretty obvious to anyone with half a brain that > the result is what you would expect if the speed of light were source > dependent. I would say that agreed with the experimental evidence much > better than relativity. > Androcles Androcles 1. Are you saying that the speed of light is source dependent? 2. What is your definition or explanation of 'source dependency'? Peter Riedt === Subject: Re: Relativity is based on assumption. All theories, in physics and elsewhere, are based on assumptions. No theory can bootstrap itself out of nothing. So what? > 'So what' depends on the validity of the assumption. The PoR is an > assumption, and is intuitive as well. Making the assumption that the > time it > takes for a signal to reach an object is the same as the time it take > for > the signal to return, when in the meantime you've moved away or toward > the > object, is a rather silly assumption that I will not accept. That's 'so > what'. > Androcles Then ignore relativity. But unless you can come up with something that agrees with the > experimental evidence better than relativity, everyone else will > ignore you. > What experimental evidence? Moving clocks running slow? They don't. The GPS > clocks run fast. MMX? It's pretty obvious to anyone with half a brain that > the result is what you would expect if the speed of light were source > dependent. I would say that agreed with the experimental evidence much > better than relativity. > Androcles > Androcles > 1. Are you saying that the speed of light is source dependent? Yes. > 2. What is your definition or explanation of 'source dependency'? > Peter Riedt Just throw a ball forward from your car window as you move. It should be obvious, and no different from throwing a photon from a star. The velocity of light in interstellar space is c, with respect to the star. If the star moves, then it still c with respect to the star. To an observer the velocity is c+v, where v is the velocity of the star. Now create a mathematical model of a star in elliptical orbit according to Kepler's laws, determine when the light will arrive, calculate the variation in brightness, calculate a spectrum, and see if it matches empirical data. If it does, then Einstein's second postulate has no foundation. A match to empirical data has been found. There are three models to consider. 1) Newton-Galileo - photons and PoR, SoL is source dependent. 2) Maxwell-Lorentz - wave and PoR, SoL is aether dependent 3) Einstein - photon or wave, SoL is observer dependent. I don't know of any sensible fourth choice. MMX thows out 2), but cannot differentiate between 1) and 3). Therefore 1) should be given due consideration, someone has to do it, and that someone is me. Androcles http://www.androc1es.pwp.blueyonder.co.uk/index.html === Subject: Re: Relativity is based on assumption. > 1. Are you saying that the speed of light is source dependent? > Yes. > 2. What is your definition or explanation of 'source dependency'? > Peter Riedt > Just throw a ball forward from your car window as you move. > It should be obvious, and no different from throwing a photon from a star. > The velocity of light in interstellar space is c, with respect to the star. > If the star moves, then it still c with respect to the star. To an observer > the velocity is c+v, where v is the velocity of the star. And for waves ... ? === Subject: Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? > Ray Steiner > I came up with another way of showing that x/tan x has no elementary > antiderivative. > We need only one result from Wiener's 1997 paper: > arcsin(x)/x does not have an elementary antiderivative. > Let I = int(x/tan x dx)= int (x cot x dx) > Use integration by parts to get > I = x ln(sin x) - int( ln(sin x) dx) > Let I2= int( ln(sin x) dx) > Let u= sin x, x = arcsin(u), dx = 1/sqrt(1-u^2) du > Then > I2= int ( ln(u)/sqrt(1-u^2) du) > Finally, use parts again to get > I2= ln(u) arcsin(u) - int(arcsin(u)/u du). > So, by Wiener's result, the original integral is not elementary. > More results: > By exactly the same method one can show that > I3 = int (x tan x dx) is not elementary. > Now, let's substitue u= tan x, x = arctan u, dx= 1/ (u^2 + 1) du in I3. > Then it reduces to > I4 = int( u*arctan(u)/(1+u^2) du). > so the second integral of my previous post is non-elementary. > Finally, consider > I5= int ( (arctan(x))^2 dx). > By parts, one can reduce it to integrating I4, so I5 is also > non-elementary. > LH After I posted this, I found an even simpler way of doing it. Again We need the result from Wiener's 1997 paper: arcsin(x)/x does not have an elementary antiderivative. In int(arcsin(x)/x dx) substitute x= sin u, dx= cos u du; arcsin x= u Then it becomes int( u cos u/sin u du) which is int(u/tan u). If the latter were elementary then int(arcsin(x)/x dx) would be elementary, which is not the case. So the result follows easily. BTW, if we substitute x = cos u in the same integral, we also find that int(x tan x dx) is not elementary. Ray Steiner === Subject: Re: Why is math so difficult for some people? > Just as the > answer to a mathematical question is either right or right, > I'd like a few of those questions! Some of us learned three-valued logic in high school. There are three ways to solve a problem: the right way, the wrong way, and the way you were told to do it (which may have nothing to do with right or wrong). David Ames === Subject: Re: Why is math so difficult for some people? >> Am I too dumb for math? >Define dumb. Math is a tool, and for some a pleasure >in its own right > Ahem. It's not polite to talk about matherbation in public. ... except among people of the same persuasion. David Ames === Subject: Re: Why is math so difficult for some people? I have taught math for many years, both in a classroom setting and one on one, and more often than not, the main problem lies in students not understanding the basic ideas behind the formulas, rather than just memorizing the formulas. For many students, there is the mistaken belief that knowing the formulas is enough. > [added sci.edu, comp.edu] >> I don't get it.I'm a perfect 0 at math. >> >> Some people have no problems at all with it. >> Am I too dumb for math? >You're asking the wrong question -- you're not >dumb for math; you *might* have a brain/mind that >works in a way that is not compatible with the way >of thinking that you need for understanding maths >(emphasis on the *might* -- you claim to be dumb >for maths, but who knows, maybe you're much better >newsgroup and just don't know it, or maybe you're a >high-expectation kind of person, and then anything >below Newton, or Gauss, or Fourier's brains means >too dumb for math in your mind? :-)) > I've come across various students who viewed that they were missing the > mathematics gene (or programming gene, or whatever the particular > subject happened to be). In those cases it was uniformly the case that > their difficulty was emotional/attitudinal, rather than cognitive. As you > mention, one unhelpful attitude is perfectionism, especially in hard-edged > subjects where some answers are clearly objectively *wrong* and thus the > student has no wiggle room to avoid the conclusion that they made an > error. >Anyway, this, plus many of the things that have >been already said (mainly about math being genuinely >hard -- the more sophisticated level of math, the >harder, of course) > Several of the missing gene students had the unhelpful attitude that > they expected maths to be easy, since they had found their schooling easy > so far. > ISTM that cognitive issues do kick in when dealing with high levels of > abstraction where there are no readily-accessible concrete models. For > example, my brain hit the wall trying to visualise non-Hausdorff spaces, > and my painful memory of the rest of that topology course is of generally > mindless memorizing and proof cranking. >HTH, >Carlos === Subject: Re: Why is math so difficult for some people? >I have taught math for many years, both in a classroom setting and one >on one, and more often than not, the main problem lies in students not >understanding the basic ideas behind the formulas, rather than just >memorizing the formulas. For many students, there is the mistaken >belief that knowing the formulas is enough. This is not surprising, considering that this is essentially how everything is taught in the elementary and high schools, and also in the courses through calculus. As just about all textbooks take that approach, and the great bulk of teachers believe in teaching that way, it is hard to see what can be done. We are NOT going to be able to teach the teachers; this failed for the new math, and it was very definitely tried. It is not even learning the basic ideas behind the formulas; this might get through to a percentage of the teachers. It is knowing the properties and the concepts of the mathematical objects used, and these are not going to be learned by the mistaken methods of the educationists. We can teach variables in the general sense (do NOT limit them to numbers) as linguistic entities with beginning reading. We can teach sound mathematical logic to at least half of those in elementary school; it has been done. But we cannot succeed if learning mathematics is measured by the multiple choice computational stuff on the exams now being mandated. Nobody learns what multiplication means by memorizing the tables, and one can understand what it means without having any facility in computing answers. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Stirling Numbers Approximation? Does anyone know an approximation for Stirling Numbers of the Second kind, S(n,k), for very large values of n? S(n,1) = S(n,n) = 1 are easy, as are S(n,2) and S(n,n-1). It's the intermediate values of k that are difficult. I can't use the alternating series because of cancellation error. === Subject: Re: Stirling Numbers Approximation? >Does anyone know an approximation for Stirling Numbers of the Second >kind, S(n,k), for very large values of n? Graham, Knuth and Patashnik Concrete Mathematics refers to David and Barton, Combinatorial Chance, Hafner 1962, chap. 16, for asymptotics of Stirling numbers. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: L^p space What will be a good reference to understand topological properties of L^p spaces ? === Subject: Re: Newsgroup survey: Math and personality assessment >>says... >2. In your experience, is math quirky? >>Define quirky. > Idiosyncratic. >>Sometimes the results one sees are not what one would intuitively think were the >>case but are nonetheless correct. Is that what you mean by quirky? > No, I guess I mean, like where sometimes there are rules that follow > in one instance, that you aren't sure still apply in another given > instance, and at times it can be so difficult to figure out if that > happens that you just have to rely on experts *telling* you that the > rule no longer applies. By this definition, math is *not* quirky. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Newsgroup survey: Math and personality assessment > > Well, well, well! If it ain't (Suk My) Dik! What's up, dude? > ... > > My, oh my. You make a joke on my name, and in response I make one on your > The relevance is Jesse Hughis inability to distinguish between word > play and an insult--a distinction you were (apparently) able to make. Well, I agree with Jesse. You posted a wordplay that was intended to be insulting (and childishly so), see the surrounding wording. I saw it as an insult. That I did not retaliate in kind was deliberate. I rarely insult people on Usenet. I think I have done that about two times in 20 years of posting history. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Sakharov's Emergent Gravity Jack, you ask: ... Tony what do you mean by D5 describing Gravity? What that would mean to me is that start from the D5 group and end up with Einstein's local field equation Guv = (superstring tension)^-1Tuv can you actually do that? Similarly, start from D4 and get the U(1)xSU(2)xSU(3) principal fiber bundle of the electroweak-strong gauge forces with the associated vector bundle of the lepto-quark sources. Can you do that? .... Yes, to both questions (although I don't use the conventional superstring structure that you mention). I am trying to see if there is a precise mathematical connection between what you are doing and what I am doing in http://qedcorp.com/APS/EmergentGravity.pdf In my theory Einstein's gravity field plus exotic vacuum w = -1 dark energy/matter fields are all ODLRO c-number collective emergent low energy effective MACRO-QUANTUM fields in which Diff(4) is an emergent symmetry from the spontaneous breaking of the U(1) EM symmetry at the unstable false vacuum micro-quantum level of the lepto-quark sources/electroweak-strong gauge forces level. More specifically, I get an emergent c-number LOCAL giant MACRO-QUANTUM VACUUM COHERENT WAVE Vacuum Coherent Field = (Higgs Amplitude Field)e^i(Goldstone Phase Field) World Crystal Lattice Distortion Field = Lp*^2(Goldstone Phase Field),u Einstein's guv field is the flat Minkowski metric + the Strain Tensor of the Distortion Field Therefore, Einstein's Gravity Field emerges as modulation of the Goldstone Phase Field consistent with Andrei Sakharov's metric elasticity, which is the complementary view of P.W. Anderson's generalized phase rigidity in his More is different paradigm. That is, the basic emergent gravity coupling is precisely Ed Witten's alpha' = 1/(string tension) = 8piG*/c^4 where I allow G* to be a scale-dependent variable that is 10^40 G(Newton) at the 1 fermi scale. Unified Exotic Vacuum Dark Energy/Matter Field is from the modulation of the Higgs Amplitude Field which also provides the rest mass of the lepto-quarks as mc^2 ~ e^2/zpf^1/2 where /zpf ~ -1/(1 fermi)^2 at the 1 fermi scale in the vibrating dark matter quantized vortex string core where the Higgs Field is zero. Where with h = c = 1 convention /zpf = (alpha')^-1[(alpha')^3/2|Higgs|^2 -1] alpha' is a scale-dependent variable not fixed at 10^-66 cm^2. BTW I have come to the tentative conclusion that the alleged Holographic Universe formula Lp* = Lp^2/3L^1/3 found in the LNL e-prints has serious problems of interpretation. My basic theory does not essentially depend on that additional assumption. While I see no basic problem in you getting the micro-quantum U(1)xSU(2)xSU(3) from your group theory, I do not understand how you get Einstein's Gravity. The Ed Witten argument that one gets a spin 2 quantum is not good enough for me since in the More is different emergence the consensus quantum gravity idea is not correct at all and that is what non-renormalizability of Einstein's GR in the low-energy sector is telling us. There may be some linear spin 2 perturbative random gravitons as micro-quantum noise or normal fluid in the emergent curved spacetime superfluid background from the modulation of the c-number Goldstone phase. So I am asking where is the MACRO-QUANTUM EMERGENCE in your group theory approach to deriving Einstein's Gravity from a more fundamental level? Tony: Those things (the forces of Gravity and the Standard Model) all come from the 28-dim adjoint rep of the D4 subalgebra of D5. More particularly: As to Gravity (and Higgs, and special conformal generators): D4 has a 15-dimensional D3 subalgebra that is the conformal algebra SU(2,2) = Spin(2,4). It has: 1 dilation generator (corresponds to Higgs) OK 4 special conformal generators What do they locally gauge to? My hunch is /zpf,u 10 anti-deSitter generators. The 4 Pu of T4 locally gauge to Einstein's guv. The 6 Muv of O(1,3) locally gauge to a Torsion Field. By a modified conformal MacDowell-Mansouri mechanism, Is this where ODLRO MACRO-QUANTUM EMERGENCE is buried? you get the Einstein-Hilbert Lagrangian. All this is at conventional textbook level, for example, section 14.6 of Unification and Supersymmetry, 2nd edition, by Rabindra Mohapatra, Springer-Verlag 1992. If you want a prominent establishment name dropped, Frank Wilczek mentions the MacDowell-Mansouri mechanism in http://xxx.lanl.gov/abs/hep-th/9801184 where he notes that the mechanism was also independently formulated by Chamseddine and West. The mechanism was invented to make it possible to get gravity from the anti-deSitter part of Lie superalgebras used in supergravity theories. As to the Standard Model, look at the 28 generators of the D4 Lie algebra. Use 15 of them as above, and 1 more to complete the SU(2,2) to U(2,2) = SU(2,2)xU(1). Then you have 12 generators left. The form the 12-dim Standard Model SU(3) x SU(2) x U(1). Here is how you can see that structure geometrically, and unambiguously: If you look at things (here I was inspired by Saul-Paul) in terms of the Weyl reflection group of the root vector space, and take away the 12 root vectors of the U(2,2) and the 4 Cartan algebra elements of the 16-dim rank 4 U(2,2), that leaves you with 28-12-4 = 12 root vectors. Since the root vectors of D4 form a 24-cell, which can be seen as a 12-vertex cuboctahedron plus two 6-vertex octahedra, you see that the remaining 12 root vectors form a pair of octahedra. Line the two octahedra up so that they share a common axis, and project the two octahedra into a space perpendicular to that axis, so that the 4 axis vertices fall on a line that forms the root vector diagram (including Cartan origin vertices) of the 4-dim Lie algebra U(2) = SU(2) x U(1). The remaining 8 can be seen as the vertices of a cube: tb----xb | | | zb----yb | | | | yr-|--zr | | | xr----tr Now look at the cube along its tb-tr diagonal axis, and project all 8 vertices onto a plane perpendicular to the tb-tr axis, giving the diagram yb xb zb tb tr zr xr yr with two central points surrounded by two interpenetrating triangles, which is the root vector diagram of SU(3). Therefore, the 16 of the 28 D4 generators give us Gravity, Higgs, and special conformal, and the remaining 12 give us the Standard Model SU(3) x SU(2) x U(1). Consideration of the relevant geometries and combinatorics give level calculations, so such higher order things as neutrino masses (they are in my model tree-level massless) remain to be fully calculated. Details are in my papers, including a paper at http://www.innerx.net/personal/tsmith/TQ3mHFII1vNFadd97.pdf that contains material barred from the Cornell arXiv due to me being blacklisted by them. Tony === Subject: Frank Wilzcek's Emergent Gravity By a modified conformal MacDowell-Mansouri mechanism, I asked: Is this where ODLRO MACRO-QUANTUM EMERGENCE is buried? you get the Einstein-Hilbert Lagrangian. All this is at conventional textbook level, for example, section 14.6 of Unification and Supersymmetry, 2nd edition, by Rabindra Mohapatra, Springer-Verlag 1992. If you want a prominent establishment name dropped, Frank Wilczek mentions the MacDowell-Mansouri mechanism in http://xxx.lanl.gov/abs/hep-th/9801184 doing in a simpler way independently. Wilczek is one of the best theoretical physicists around for sure. I heard him speak twice now this year. where he notes that the mechanism was also independently formulated by Chamseddine and West. The mechanism was invented to make it possible to get gravity from the anti-deSitter part of Lie superalgebras used in supergravity theories. === Subject: Russian claims of torsion weapons Commentary 3 The hyperspace H consists of fibers f(x) that are either copies of or representations of the symmetry group G. Jack, this is not quite correct. They are homogenous spaces on which the group operate transitively. Example, for the group SU(2), you can take as the fibre a copy of SU(2) itself (3-dimensional), or you can take sphere S^2, on which SU(2) operate (2-dimensional). Notice that S^2 is not a representation of SU(2). It is a quotient SU(2)/SO(2). Early Kaluza-Klein theories were operating with group Manifolds. Souriau, and later Witten, suggested more realistic theories where fibers could be of lesser dimensions. Thie rigorous mathematics and examples of this latter approach have been developed in the monograph: Riemannian Geometry, Fibre Bundles, Kaluza-Klein Theories and All That... (World Scientific Lecture Notes in Physics, Vol 16) by Robert Coquereaux, Arkadiusz Jadczyk :-) ark Principal fiber bundles for the local gauge forces: 1. A transformation g of the symmetry group G acts on the ordered pair X = (x, fo) in hyperspace H with output gX. Question: Can gx = x' =/=x i.e. can one move the base point in this operation or must G always be the identity in the base space? That is, we always need, in addition to G a connection and a path in order to change location in the horizontal base space and the vertical fiber space that is beyond space-time. G certainly moves fo up and down the vertical fiber for every element g =/= identity. Does it also move x -> x' = gx =/= x horizontally along the base manifold without a connection field and a path specified? Clearly the answer must be NO. See below. The modern understanding of gauge invariance, as a symmetry under transformations of quantum-mechanical wave functions, was reached by Weyl himself and also by London very shortly after the new quantum mechanics was first proposed. In this understanding of abelian gauge invariance, and in its nonabelian generalization [2], the space-time aspect is lost. The gauge transformations act only on internal variables. This formulation has had great practical success. Still, it is not entirely satisfactory to have two closely related, yet definitely distinct, fundamental principles, and several physicists have proposed ways to unite them. One line of thought, beginning with Kaluza [3] and Klein [4], seeks to submerge gauge symmetry into general covariance. Its leading idea is that gauge symmetry arises as a reflec- tion in the four familiar macroscopic space-time dimensions of general covariance in a larger number of dimensions, several of which are postulated to be small, presumably for dynam- ical reasons. Here we should take the opportunity to emphasize a point that is somewhat confused by the historically standard usages, but which it is vital to have clear for what follows. When physicists refer to general covariance, they usually mean the form-invariance of physical laws under coordinate transformations following the usual laws of tensor calculus, including the transformation of a given, preferred metric tensor. Without a metric tensor, one cannot form an action principle in the normal way, nor in particular formulate the ac- cepted fundamental laws of physics, viz. general relativity and the a purely mathematical point of view one might consider doing without the metric tensor; in that case general covariance becomes essentially the same concept as topological invari- ance. The existence of a metric tensor reduces the genuine symmetry to a much smaller one, in which space-times are required not merely to be topologically the same, but congruent (isometric), in order to be considered equivalent. In the Kaluza-Klein construction, for this reason, the gauge symmetries arise only from isometries of the compactified dimensions. Another line of thought proceeds in the opposite direction, seeking to realize general covariance [CapitalEth] in the metric sense [CapitalEth] as a gauge symmetry. arXiv:hep-th/9801184 v4 23 Apr 1998 IASSNS-HEP-97/142 Riemann-Einstein Structure from Volume and Gauge Symmetry Frank Wilczek BTW Wilczek shows that Gennady Shipov's torsion theory is closely related to Roger Penrose's spinors in curved spacetime with the anti-symmetric spin connection as the locally induced compensating torsion field. It all comes from locally gauging the O(3,1) subgroup of the Conformal Group as I said previously based on Utiyama's and Kibble's papers from the mid-1960's. Whether or not Akimov's claims from Moscow that torsion waves from O(1,3) of sufficient intensity to have psychotronic weapons bio-toxic effects can easily be generated when, in contrast, gravity waves from T4 are so hard to find is another issue not considered here. The gravity wave T4 coupling parameter is essentially Ed Witten's alpha' = (superstring tension)^-1. What is the corresponding O(1,3) spin connection coupling parameter? Akimov's claims hang on the answer to that question. Is it easier to make propagating torsion dislocation topological string defects than to make propagating curvature disclination topological string defects in the MACRO-QUANTUM Vacuum Coherence Field's Goldstone Phase? That's what Akimov's claims come down to in terms of my new theoretical paradigm for the emergence of Einstein's Gravity and the Unified Exotic Vacuum Field of w = -1 Dark Energy/Matter. 2. The action of the symmetry group G on the total hyperspace H induces an equivalence relation ~ . That is, if X' = gX, g < G, then X' ~ X. 3. ~ partitions hyperspace H into disjoint non-overlapping equivalence classes called G-orbits G(X) = {gX, for all g < G} Remember that in this principal bundle fo is also a g < G. All G-orbits have identical structure and are diffeomorphic to G. 4. This disjoint partition of hyperspace H gives the quotient space H/G that is the base space M with points x. Every point x of the base space M is really an equivalence class or G-orbit of a continuous infinity of points of a larger dimensional Hermetic or occult hidden hyperspace implicate inside it. Worlds within worlds. Wheels within wheels. Shades of Bohm's Implicate Order? 5. The Projection Map P is simply P:G-orbit -> x. This means that each individual G-Orbit is really associated with a single vertical fiber at a single horizontal base space event. The G-orbit is the vertical fiber beyond, in the usual physics applications, a localized spacetime event x, although we can have delocalized base spaces of twistors whose intersections are points. We can also perhaps have base spaces of finite strings both open and closed and even base spaces of higher dimensional brane worlds? Commentary 2 Given coordinate patch C(x) in the base space M in a neighborhood of point x and fiber f(x) form the local Cartesian product C(x)f(x) with ordered pair X = (x,fo). Take the union C(x)f(x)/C(x')f(x')/... of all such local products. There are redundant ordered pairs X because the coordinate patches C(x) and C(x') as sets overlap with non-vanishing intersection C(x)/C(x')=/= Empty Set. Identify the redundant multiple images of the same actual point of the base space M using the symmetry group G as an equivalence relation. That is, two ordered pairs X and X' are identified or equivalent if x = x' < C(x)/C(x') and if fo' = gfo where g < G to form disjoint equivalence classes {f(x)} that are the distinct points of the fiber in hyperspace H. This is all local at a fixed base point x like in an internal gauge force symmetry. g is also called a transition function. The hyperspace H is the factor space of the union C(x)f(x)/C(x')f(x')/ ... mod G. The projection map P:(x,{fo}) -> x When M is the curved space-time of Einstein's gravity theory in addition to the G equivalence in the extra space dimensions of the fiber, x'(E) = Diff(4)x(E) at fixed event E to make disjoint equivalence classes {x(E)} mod Diff4(E). One can imagine a hybrid where the fiber is a discrete space of strings of c-bits. One can also imagine a fiber of strings of qubits. 1 qubit is a parallel infinity of c-bits. i.e. |qubit> = |1 c-bit><1c-bit|qubit> + |0 c-bit><0 c-bit|qubit> Where there is a continuous infinity of different c-bit bases or orthonormal frames each corresponding, for example, the the angular orientation of an inhomogeneous field magnet in a Stern-Gerlach filter for spin qubits in the DARPA spintronics project or like the billion billion Single Electron Transistors inside the human brain at the sub-microtubular protein dimer hydrophobic cage level forming the hardware interface with external world whose software is our stream of inner consciousness. Each possible orientation is a primitive parallel quantum universe. The quantum computer computes in all possible orientations simultaneously like a continuous infinity of classical Turing machines in a distributed network working on the same problem - or so the folklore goes. to be continued. Commentary 1 The fiber bundle as an idea has 4 parts. 1. A structure symmetry group G. 2. The total hyperspace H or, in some applications Wheeler's BIT. 3. The projection map P. 4. The base space M or, in some applications. Wheeler's IT. The hyperspace H consists of fibers f(x) that are either copies of or representations of the symmetry group G. The projection map P collapses a fiber f(x) in the hyperspace H to a point x in the base space M. All of these objects are continuum differential manifolds depending on the continuum of real numbers which its associated issues of Cantor's infinity of infinities of Cabalistic Aleph's in an ascending Jacob's Ladder. This is not a discrete combinatoric mathematics although such a skeletal structure is associated with it as in Herman Weyl's Theory of Groups and Quantum Mechanics and as in Saul-Paul Sirag's presentation of V.I. Arnold's A-D-E mathematics of everything. The base space is covered by an atlas of local coordinate patches with all important overlap transition functions sewing the patches together like a quilt. M is space-time in local micro-quantum field theory of point The extra-dimensions of hyperspace form the Calabi-Yau space of vibrations of the superstring beyond space-time. The connection on the total hyperspace H is the potential of a local gauge force. Examples of connections is the 4 potential Au(x) in Maxwell's electromagnetism with G as U(1). There are similar connections for the Yang-Mills weak force with G = SU(2) and the strong force with G = SU(3). Classical general relativity, as distinct from local micro-quantum field theory, has the torsion-free symmetric three-index non-tensor Levi-Civita connection with G as the Diff(4) group. The latter comes from locally gauging the 4 parameter translation subgroup (generated by the 4-momentum Pu of globally flat special relativity ) of the 15 parameter conformal group of Roger Penrose's massless twistors. Bottom -> Up: Given base space M and symmetry group G construct the hyperspace H as a quilt patchwork. Top -> Down: Given hyperspace H and symmetry group G construct the base space M as the non-overlapping partition of hyperspace into G-orbits called the quotient space of H mod G in the principal bundle. Micro-quantum source renormalizable local fields of spin 1/2 lepto-quarks are associated vector bundles. Micro-quantum force renormalizable local fields of spin 1 gauge force bosons (electro-weak and strong) are from the principal bundle. There is no renormalizable quantum gravity in this precise sense. This is because classical Einstein gravity is a More is different (P.W. Anderson) emergent collective effect as in Andrei Sakharov's metric elasticity of an instability in the globally flat false vacuum of the interacting lepto-quark source/electroweak-strong force. Einstein's gravity + unified exotic vacuum dark energy/matter with Andrei Linde's chaotic inflationary cosmology are the result of the continual phase transitions from globally flat false high entropy micro-quantum vacua to locally curved macro-quantum low entropy metastable vacua. to be continued: === Subject: Re: Russian claims of torsion weapons > Commentary 3 > The hyperspace H consists of fibers f(x) that are > either copies of or representations of the symmetry > group G. Well, that convinces me... === Subject: Re: Argument with professor >I am not a mathematician by trade, That's obvious. More important, you lack the ability to listen. >but I was talking to a maths >professor and he absolutely refused to acknowledge the concept of a >cross product for two vectors that are not 3 dimensional. No doubt he also refused to acknowledge the concept of a pair of integers i and j such that i+j>j+i. Im sure that there are all sorts of impossible concepts that he refused to acknowledge. >For me, >Let A be vector in N space >Let B be vector in N space >A x B = C You haven't defined anything. What is C? Let's be concrete: if A=(1,0,0,0) in R^4 with the L2 metric and B=(0,1,0,0), what is C=AxB? >Now, I can prove that C has the property How, when you haven't defined it. Worse, one of the properties of the cross product in R^3 is that AxB is orthogonal to A and B, so if C=AxB, C.A=C.B=0. Thus your >C.A / |C||A| = C.B / |C||B| Doesn't say much. >So, one could define the cross product between two n-vectors as an >n-vector with the property Sure: just define AxB=0 for every A and B. It would not, however, have the properties of the cross product in R^3. If you don't want to do that, then you would have to define *WHICH* vector with those properties, and there you run into difficulties. >But, am I wrong? Of course you're wrong. Google for Exterior or Grassmann. >Or is the professor wrong? Not on this he isn't. at 10:29 PM, j.schoenfeld@programmer.net (John Schoenfeld) said: >What is there was no professor? Then you work it out a step at a time, without any handwaving, and you see where you went wrong. Or you don't, and you remain deluded. >Don't you look stupid, believer Somebody does, but it isn't him. You look stupid for presenting as a definition something that in fact did not define anything, and you look stupid for rejecting good advice instead of thinking things through. But tell me, tonto, why are you paying tuition if you believe that the professor doesn't understand things as elementary as that? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Argument with professor permission for an emailed response. For what it's worth, there *is* an extension of the cross product for vectors in higher dimensions. The problem is that the cross product is a cheat. The generic operation is a tensor outer product, and it so happens that the outer product of two 3D vectors has three components, so you can map it into a vector easily. But this is a cheat of sorts, and it leads to confusions in many areas (especially when people start using it in studying electrodynamics). Thomas === Subject: Re: Argument with professor > A cross product is a special case of the Clifford wedge product. Really this is an exterior algebra construction, not a Clifford construction. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Argument with professor > A cross product is a special case of the Clifford wedge product. The > wedge product of two vectors is called a bi-vector. It just so > happens that in R^3 bi-vectors are dual to vectors, so we can > interpret the three dimensional bi-vector as corresponding to a > special type of vector (usually referred to as an axial vector). This > correspondence breaks down in higher dimensions, since the ranks of > dual objects must add up to the number of dimensions. Hence, in R^4, > bi-vectors are dual to themselves and cannot be interpreted as vectors > at all. In even higher dimensions, the situation becomes all the more > hopeless. Did you insult me sometime recently? I can't remember. However, even if it was you, all is forgiven for that nice precis. === Subject: Selecting the correct graph I would like to know some important points when selecting a certain graph for the data I have i.e. Using a Pie graph for displaying the largest % of Annual sales, in a five-year period, for a business. Thanx, cSaviour === Subject: Frank Wilczek's Emergent Gravity On closer look Wilczek has a different idea than mine. His idea is interesting. I am claiming to derive Einstein's gravity with the cosmological term as a large scale limit of the exotic vacuum field from an instability at least in the QED sector of the micro-quantum vacuum. Wilczek in contrast never does any micro-quantum theory that I can see in his paper? He already starts with classical fields and makes no attempt to derive gravity as emergent from the micro-quantum standard model. By a modified conformal MacDowell-Mansouri mechanism, I asked: Is this where ODLRO MACRO-QUANTUM EMERGENCE is buried? you get the Einstein-Hilbert Lagrangian. All this is at conventional textbook level, for example, section 14.6 of Unification and Supersymmetry, 2nd edition, by Rabindra Mohapatra, Springer-Verlag 1992. If you want a prominent establishment name dropped, Frank Wilczek mentions the MacDowell-Mansouri mechanism in http://xxx.lanl.gov/abs/hep-th/9801184 doing in a simpler way independently. Wilczek is one of the best theoretical physicists around for sure. I heard him speak twice now this year. where he notes that the mechanism was also independently formulated by Chamseddine and West. The mechanism was invented to make it possible to get gravity from the anti-deSitter part of Lie superalgebras used in supergravity theories. === Subject: Basic relational theory My professor and my TA skipped their office hours and Im totally lost. Can someone give me a hand? They also rarely respond to emails. Just looking for guidance since Im confused. Consider relation schema R(A,B,C) and the set of functional dependencies F={B->A,A->C} 1. All non-trivial relations. Is this correct? Im just guessing. B->A, A->C,B->C,AB->BC,AC->C, BC->CA,AB->C 2. Find a non-empty instsance of R(give a number of rows) that satisfies every Functional Dependency in F. Is this correct? A B C 2 1 3 3 2 4 4 3 5 3. Find an instance in R that satisfies every FD in F accept A->B How do you get A->B???? I dont see it. 4. Possible to find an instance an instance that satisfies every FD in F, but does not satisfy the FD AB->C. I have no idea. Im totally lost. Can someone point in the right direction? === Subject: Re: transfinite series >> >> > Take the infinite series expansion for e and put it into the infinite > series expansion of e to the power of x and multiply the terms and you > end up with a series of aleph 1 terms >> >> This is nonsense. Even when you multiply everything out, there are >> still only countably many terms. >> Yes, compare countability of the rational numbers. >> >> > that sum to a finite result e to > the power of e. If you repete this process for e to the power of e to > the power of e do you end up with aleph 2 terms? >> >> For series with an arbitrary number of terms, look up the term summable >> family. But a necessary condition for convergence is that at most >> countably many terms are nonzero, this follows from the archimedean axiom >> of the real numbers. >If aleph 1 is the set of all subsets of aleph 0 > No, aleph1 is first uncountable ordinal. You must mean bet1, which is usually written c. Whether or not aleph1=c is called the Continuum Hypothesis, and is undecidable in ZFC. > Anyway, your argument is incorrect for c as well. > then take the counting >numbers the first subset is the empty set next come the sets with only >one member that is 1,2,3 etc next is the sets with two members these >can be arranged in a two dimensional array 1 2,1 3,1 4 etc 2 3,2 4,2 >5 etc 3 4,3 5,3 6 etc etc all the way up to those with an infinite >number of terms which can be arranged in an array with an infinite >number of dimensions. Now if you want each one of these sets has a >complement however I think when you get to the sets with an infinite >number of terms then each subset is duplicated. which subset did I >miss? > You missed infinite subsets. Your counting scheme simply won't work for the infinite number of dimensions case. You did not describe the counting scheme for that case. >As for e to the power of e the first term is 1 next comes 1 + x >+ (x^2)/2 + then comes 1/2 + x/2 + (x^2)/4 + , x/2 + (x^2)/2 + (x^3)/4 >+ , (x^2)/4 + (x^3)/4 + (x^4)/8 + etc arranged in a two dimensional >array all the way up to those with an infinite number of terms that >can be arranged an array with an infinite number of dimensions. you >have a one to one correspondence or at least a one to two >correspondence. > Cron OK lets review e^e=1+e+(e^2)/2+(e^3)/6+...= 1+ 1+1+1/2+1/6+1/24+... + 1/2+1/2+1/4+1/12+1/48+ 1/2+1/2+1/4+1/12+1/48+ sorry about the mess this is the best I can do 1/4+1/4+1/8+1/24+1/48+ 1/6+1/6+1/12+1/36+1/144+ 1/6+1/6+1/12+1/36+1/144 1/12+1/12+1/24+1/72+ 1/6+1/6+1/12+1/36+1/144+ 1/6+1/6+1/12+1/36+1/144 1/12+1/12+1/24+1/72+ 1/12+1/12+1/24+1/72+1/288+ 1/12+1/12+1/24+1/72+1/288 1/24+1/24+1/48+1/144+ this series has 1+aleph 0+(aleph 0)^2+(aleph 0)^3+...epsilon 0 terms now I know that aleph 0 =epsilon 0 but this series contains more than epsilon 0 terms. the value of an infinite series depends on how you add up the terms this series is even worse however I want to leave it how it is to see if it can define curves that a simple series can't === Subject: Re: transfinite series > OK lets review e^e=1+e+(e^2)/2+(e^3)/6+...= 1+ 1+1+1/2+1/6+1/24+... > + 1/2+1/2+1/4+1/12+1/48+ > 1/2+1/2+1/4+1/12+1/48+ sorry about the mess > this is the best I can do > 1/4+1/4+1/8+1/24+1/48+ > 1/6+1/6+1/12+1/36+1/144+ 1/6+1/6+1/12+1/36+1/144 > 1/12+1/12+1/24+1/72+ 1/6+1/6+1/12+1/36+1/144+ > 1/6+1/6+1/12+1/36+1/144 1/12+1/12+1/24+1/72+ > 1/12+1/12+1/24+1/72+1/288+ 1/12+1/12+1/24+1/72+1/288 > 1/24+1/24+1/48+1/144+ this series has 1+aleph 0+(aleph 0)^2+(aleph > 0)^3+...epsilon 0 terms now I know that aleph 0 =epsilon 0 but this > series contains more than epsilon 0 terms. the value of an infinite > series depends on how you add up the terms this series is even worse > however I want to leave it how it is to see if it can define curves > that a simple series can't You are mixing ordinals with cardinals. Aleph_0 is a cardinal, but epsilon_0 is an ordinal. The least transfinite ordinal is called omega (or sometimes omega_0). Both omega and epsilon_0 are countable ordinals, which is to say that their cardinality is aleph_0. If A is any countably infinite set, then AxA is likewise countable. By induction, we can conclude that A^k is countable for each natural number k > 0. The union of { A^k : k > 0 } is likewise countable. The cardinality of each of these sets is aleph_0. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: transfinite series at 04:56 PM, currentresident@veloemail.com (charles ramsey) said: >Take the infinite series expansion for e and put it into the infinite >series expansion of e to the power of x and multiply the terms and >you end up with a series of aleph 1 terms No, you end up with a series of Aleph_0 terms. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: transfinite series > Shmuel (Seymour J.) Metz, SysProg and JOAT off you cunting little Jew. -- G.C. === Subject: Re: transfinite series > Working in binary after the decimal point first you have .000... and > its compliment .111... next comes those with one one .1 .01 .001 etc > and their compliment .0111... .00111... .000111... then those with > two ones .11 .101 .1001 etc .011 .0101 .01001 etc .0011 .00101 > .001001 etc etc and their compliments these can be arranged in two two > dimensional arrays all the way up to those with an infinite number of > ones that can be arranged in two infinite dimensional arrays. Which > real number did I miss? So now we have a one to two correspondence > between my original construction and the reals between zero and one > therefore a series with a continium of nonzero terms sum to a finite > result. Mr fritz says this is in impossible. So does e^(e^e) have c > members or 2^c members. Your compliments seems to mean, based on your examples above, representations of reals betweeen 0 and 1 having two binary representations. There are only countably many reals with such dual binary representation (or dual representation in any larger integral base), as all such numbers are rational. I do not see from the above too brief description how you intend to count those reals whose binary representations contain both infinitely many zeros and infinitely many ones. Indeed, a trivial translation between base 2 and base 4 allows a direct application of Cantor's diagonal proof in base 4 that there is no surjection from the naturals to reals between 0 and 1. Each such real, between 0 and 1, has a base 4 representation a = sum[i=1..oo. a_i/4^1], with all a_i in {0,1,2,3} And any dual representations are all 0's or all 3's from some point onwards, so 1's and 2's do not involve the dual representation uncertainties. For any listingFf: N -> [0,1]: n -> f(n), construct x as follows: let the n'th digit of x equal 1 if the nth base four digit of f(n) is not 1 and let it be 2 otherwise: x_n = 1 if f(n)_n <> 1 x_n = 2 if f(n)_n = 1 Then x is between 0 and 1 but is not equal to any of the f(n). Since no assumptions were made about f except that its domain is N and its codomain is [0,1], and there is shown to be some member of the codomain not in the image of f, such f cannot ever be surjections. In fact, similar explicit constructions can produce a least countably many members of [0,1] not in the image of any such f. === Subject: Re: transfinite series >>Take the infinite series expansion for e and put it into the infinite >>series expansion of e to the power of x and multiply the terms and you >>end up with a series of aleph 1 terms that sum to a finite result e to >>the power of e. If you repete this process for e to the power of e to >>the power of e do you end up with aleph 2 terms? >> >> You still only get aleph 0 terms. >> I think the OP is making a very common error: mistaking the set of >> all subsets of N with the set of all *finite* subsets of N > Working in binary after the decimal point first you have .000... and > its compliment .111... next comes those with one one .1 .01 .001 etc > and their compliment .0111... .00111... .000111... then those with > two ones .11 .101 .1001 etc .011 .0101 .01001 etc .0011 .00101 > .001001 etc etc and their compliments these can be arranged in two two > dimensional arrays all the way up to those with an infinite number of > ones that can be arranged in two infinite dimensional arrays. Which > real number did I miss? Each number in your first list has a finite number of 1's, and each number in your second list has a finite number of 0's. You never reach a number such as 1/3 = (.01010101...)_2 or 2/3 = (.10101010...)_2, in which both the 0's and the 1's are infinite. >So now we have a one to two correspondence > between my original construction and the reals between zero and one > therefore a series with a continium of nonzero terms sum to a finite > result. Mr fritz says this is in impossible. So does e^(e^e) have c > members or 2^c members It has aleph_0 members. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Consecutive primes conjecture mean of two consecutive primes. it's easy to see how primes behave like spectral co-linear equations. Q: Resonance theory of allowed quantum numbers? Tapio