I am working on a project in Macromedia Flash MX using its drawing API to construct Bezier curves parametrically. Ultimately, I need to examine a set of curves and find all the intersections of those curves. I am currently scouring the web for different approaches to solving this problem. Any code, algorithms, references, etc. would be greatly appreciated. Luke ==== > I am working on a project in Macromedia Flash MX using its drawing API > to construct Bezier curves parametrically. > > Ultimately, I need to examine a set of curves and find all the > intersections of those curves. > > I am currently scouring the web for different approaches to solving > this problem. > > Any code, algorithms, references, etc. would be greatly appreciated. You could look at Knuth's Metafont (mf.web) where he tackles what seems to be the same question. -- Timothy Murphy tel: +353-86-233 6090 ==== > ucdavis!newsfeed.berkeley.edu!ucberkeley!newsfeed.mathworks.com!news.maxwel > l.syr.edu!sn-xit-03!sn-xit-01!sn-post-01!supernews.com!news.supernews.com!w > aderameyxiii > <3F46BE3A.E057DE36@mdli.com > > >> For convenience, I'm going to change this relative form of the paradox >> to an absolute form. >> >> Achilles starts at spot A and is trying to get to spot B. When he is >> halfway between A and B, he needs to get to halfway between where he is >> and B. Even when he gets to that spot, he still has to get to the point >> halfway between where he is and B. Ad infinitum. Therefore, he can >> never get to B. > > Therefore? What process of reasoning gets us to he can never get to B? > I don't see the paradox. > I didn't mean therefore to indicate a logical deduction. That's just the way it's usually stated, at least the versions I've seen. Of course, with any kind of would-be paradox, there's an element of trickery involved. The basic issue seems to be that Achilles at any point of this infinite process is never there, but he gets there in the end. If you can explain this, I'll accept there's no paradox. But good luck trying to explain it just in terms of mathematics. >> The common thing to do at this point is to say, But of course, we know >> from Calculus that infinite series of positive numbers can add up to some >> finite number. How silly of Zeno not to realize this! >> >> This kind of thinking misses two major points. The setting of the >> paradox relies on the fact that infinite series of positive numbers can >> addup to some finite number. >> In particular, it relies on the fact that >> you can divide up a geometric segment in half, and then in half, and then >> in half, ad infinitum. Knowing this fact is part of the *setting* of the >> paradox. It in no way resolves anything. > > I agree that the infinite subdivisibility of time and distance are in play > - it's mentioned explicitly - but that is not the same as understanding the > convergence of infinite series. The first concept is easy to grasp; you can > convince a child of its plausibility. The second is a much more > sophisticated idea. (Not that I'm claiming infinite series resolve the > paradox, because I still don't see what the paradox is.) > I never said you have to understand all the subtleties of infinite series to understand infinite subdivisibility. But understanding the former will not resolve the paradox. >> The other point is that Zeno, as a Greek philosopher, was well aware of >> this kind of calculus-type reasoning. He lived during the fifth and fourth >> centuries BC, but the Pythagoreans, for example, had been around before >> him. The kinds of things they studied, for example, constructing the >> golden rectangle, and creating the spiral formed by it, would have, I >> believe, been known to Zeno. > > Well, the truth is we don't know what Zeno knew. But it's beside the point. > The paradox should be stated as clearly as possible. If it qualifies as a > genuine paradox, it should need no defense based on allusions to ideas > extant in Zeno's time that he may or may not have known. > I'm sorry you don't care to know what Zeno may have known. As you say it's irrelevant (in some sense), but others may feel broadening their understanding of the historical context is a good thing. >> To better see that calculus does not resolve the Achilles paradox, >> consider the following addition to the above absolute form of the paradox: >> as Achilles travels in smaller intervals of space, he alternates opening >> and closing his mouth, i.e. When he is traveling to the first midpoint, he >> opens his mouth, when he is traveling from the first midpoint to the >> second midpoint, he closes his mouth, ad infinitum. What is the position >> of his mouth when he arrives at his destination? Open, or closed? Or >> neither? > > Come now. This is not a serious example. > Well, I say it is. Should we continue this back and forth? >> The typical objection to this question is to say, there is no last term in >> a sequence, so it doesn't make sense to ask this question. You can't >> assign some state to each event in this infinite sequence of events, and >> then ask what is the final state, since that is contradictory to the >> idea of an infinite sequence. > > There are better objections: You can certainly talk about a final state in > the presence of continuity. > > For centuries linear motion has been modeled as a smooth function f : [a,b] > -> R, where [a,b] is to be thought of as a time interval and f(t) is the > location along a line. This model has been a resounding success. Zeno's > paradox seems to be this: Achilles can't reach f(b), because for each t < > b, f(t) < f(b)! This is falling into the same kind of trap as saying infinite series resolves the Achilles paradox. If you model the situation as you have, then that is a different situation than that outlined in the paradox. Look at it this way: does your f correspond to anything in the paradox? Are you seriously suggesting that there is an uncountable number of positions that Achilles is reaching before he reaches f(b)? The key point here is that such a model of linear motion sidesteps the paradoxes of Zeno's much like how the epsilon delta definition of limit avoids the kinds of paradoxes the founders of analysis had to tangle with. Stating that this model has been a resounding success doesn't resolve any of the issues; it just shows that to a large extent Zeno's paradoxes don't have to be resolved in order to do physics. ==== > > ucdavis!newsfeed.berkeley.edu!ucberkeley!newsfeed.mathworks.com!news.maxwel > l.syr.edu!sn-xit-03!sn-xit-01!sn-post-01!supernews.com!news.supernews.com!w > aderameyxiii > <3F46BE3A.E057DE36@mdli.com > > > > For centuries linear motion has been modeled as a smooth function f : [a,b] > -> R, where [a,b] is to be thought of as a time interval and f(t) is the > location along a line. This model has been a resounding success. Zeno's > paradox seems to be this: Achilles can't reach f(b), because for each t < > b, f(t) < f(b)! > > This is falling into the same kind of trap as saying infinite series > resolves the Achilles paradox. > > If you model the situation as you have, then that is a different situation > than that outlined in the paradox. But maybe it isn't. More on this below. However, there's a solution to this particular paradox that is hidden in the continuous motion model. Consider: Zeno does allow a first step in this particular paradox, does he not? This first step itself is the limit point of a sequence that, just like the original sequence in question, cannot be actually completed, right? And so this first step is a limit point that, just like the original limit point in question, cannot be actually reached, right? Zeno is trying to have it both ways, and so we have a solution to the paradox in that it cannot even be consistently stated. You referred to another form of the paradox: Achilles can never move. For, before he can get to some spot, he needs to move through a spot in between that and his starting spot. And of course, I can divide even that further. Since there is no last term in a sequence, Achilles can never get started. What was stated here is actually called the Dichotomy Paradox. Here is a fuller statement of the Dichotomy Paradox: If we are at 0 in 0, . . ., 1/4, 1/2, 1, then how can we get started in traversing all the steps if we can't even take a first step? Answer: We can't, and so motion is not possible - we are forever stuck at 0. Zeno's aim with his Paradoxes was to prove motion impossible - that motion is an illusion. This Dichotomy Paradox seems trickier than the Achilles Paradox, because completing an infinite step-by-step process without taking a last step (the Achilles Paradox) seems less psychologically troublesome than starting an infinite step-by-step process without taking a first step (the Dichotomy Paradox). This is probably because with the latter, the problem is right at one's doorstep but with the former, the problem is infinitely removed. Note: Actually reaching a limit point of a sequence is equivalent to completing an infinite step-by-step process without taking a last step and it not equivalent to taking a last step in an infinite step-by-step process (because there is no such thing as such a last step). If one considers completing an infinite step-by-step process without taking a last step (Achilles) and especially starting an infinite step-by-step process without taking a first step (Dichotomy) to be nonsensical, them maybe one is seeing what Zeno was trying to get at. If we restrict ourselves to only the sequences given by Zeno, then there do seem to be problems. But why should we do this? Zeno may not have required us to do this. You also stated: A clarification: Zeno's paradoxes are basically objections to this kind of model of motion. To accept the model is to disregard any objections, such as the paradoxes. In other words, Zeno seems to not have accepted the existence of infinite sets, in that he was trying to derive absurdities from the postulated existence of motion over a continuous interval (an infinite set) by looking at certain subsets - sequences and their limit points. But the absurdities he derived are not mathematical, but merely psychological. Many thinkers in the past did and some in the present did not make the distinction between these types of absurdity with regard to infinite sets. For example, it was argued that infinite sets could not exist because if they did, some proper subsets would have the same cardinality as the whole set. Some just couldn't psychologically handle this, and so deriving this absurdity was deemed a proof that there could not be infinite sets. Zeno is doing the same type of thing. By looking at certain subsets of an interval, sequences and their limit points, he is deriving absurdities of motion along a continuous interval. In response to this first absurdity about infinite sets, that some proper subsets would have the same cardinality as the whole set, some have taken this absurdity to be a defining property of an object called infinite sets. So why not give the same response to those of Zeno's paradoxes that can be consistently stated? Just take these derived absurdities of such as Achilles and Dichotomy as some defining properties of some parts of an object called motion along a continuous interval? Some might not be able to psychologically handle this, but why let psychology get in the way? And if one objects to embedding the sequences of the paradoxes in other sets, then the same points can still be made: The derived absurdities are still not mathematical, and they are still merely psychological. One can still take these absurd consequences as defining properties of objects. Again, why let psychology get in the way? And also, what if one claims that Achilles and Dichotomy are not equally problematic (or non-problematic) on mathematical grounds, that the latter is more problematic? Answer: Motion along a continuous interval covers both. But if that doesn't satisfy, look into non-well-founded set theory and infinite descending chains for a possible answer to the latter. Again, as I said in the other post, I think it all boils down to the philosophical bias one brings to the table with regard to the question of potential vs. actual infinity: Does one accept the existence of actual infinity - of infinite sets, AND THE CONSEQUENCES OF THEIR EXISTENCE? (Prior to Cantor, many did not, and some still today do not.) Paul ==== > ... [snippage] >> To better see that calculus does not resolve the Achilles paradox, >> consider the following addition to the above absolute form of the paradox: >> as Achilles travels in smaller intervals of space, he alternates opening >> and closing his mouth, i.e. When he is traveling to the first midpoint, he >> opens his mouth, when he is traveling from the first midpoint to the >> second midpoint, he closes his mouth, ad infinitum. What is the position >> of his mouth when he arrives at his destination? Open, or closed? Or >> neither? > Come now. This is not a serious example. > > Well, I say it is. Should we continue this back and forth? The position of his mouth is whatever you say it should be, and you haven't said. You haven't given any rule for what he does with his mouth at the destination and beyond. Since apparently he can change the position of his mouth in zero time, it's no problem for Achilles to arrange it either way. Just tell him what to do. Btw if he can't change it in zero time, then there's no problem at all as I'm sure you can see. > >> The typical objection to this question is to say, there is no last term in >> a sequence, so it doesn't make sense to ask this question. You can't >> assign some state to each event in this infinite sequence of events, and >> then ask what is the final state, since that is contradictory to the >> idea of an infinite sequence. > There are better objections: You can certainly talk about a final state in > the presence of continuity. > For centuries linear motion has been modeled as a smooth function f : [a,b] > -> R, where [a,b] is to be thought of as a time interval and f(t) is the > location along a line. This model has been a resounding success. Zeno's > paradox seems to be this: Achilles can't reach f(b), because for each t < > b, f(t) < f(b)! > > This is falling into the same kind of trap as saying infinite series > resolves the Achilles paradox. > > If you model the situation as you have, then that is a different situation > than that outlined in the paradox. > > Look at it this way: does your f correspond to anything in the paradox? > Are you seriously suggesting that there is an uncountable number of > positions that Achilles is reaching before he reaches f(b)? Why not? (Of course, Zeno only considers a denumerable subset of them; in fact that is probably what causes all the brouhaha, since it's not immediately clear that a denumerable set of points can be dense.) > > The key point here is that such a model of linear motion sidesteps the > paradoxes of Zeno's much like how the epsilon delta definition of limit > avoids the kinds of paradoxes the founders of analysis had to tangle with. > Stating that this model has been a resounding success doesn't resolve > any of the issues; it just shows that to a large extent Zeno's paradoxes > don't have to be resolved in order to do physics. If you put it that way, I guess I see your point; however, as you yourself said, the paradox relies on trickery and (I would add) vagueness of one's definitions, so I'm not sure how much that point is worth. ==== >> For centuries linear motion has been modeled as a smooth function f : [a,b] >> -> R, where [a,b] is to be thought of as a time interval and f(t) is the >> location along a line. This model has been a resounding success. Zeno's >> paradox seems to be this: Achilles can't reach f(b), because for each t < >> b, f(t) < f(b)! > > This is falling into the same kind of trap as saying infinite series > resolves the Achilles paradox. > > If you model the situation as you have, then that is a different situation > than that outlined in the paradox. > > Look at it this way: does your f correspond to anything in the paradox? > Are you seriously suggesting that there is an uncountable number of > positions that Achilles is reaching before he reaches f(b)? > > The key point here is that such a model of linear motion sidesteps the > paradoxes of Zeno's much like how the epsilon delta definition of limit > avoids the kinds of paradoxes the founders of analysis had to tangle with. > Stating that this model has been a resounding success doesn't resolve > any of the issues; it just shows that to a large extent Zeno's paradoxes > don't have to be resolved in order to do physics. A clarification: Zeno's paradoxes are basically objections to this kind of model of motion. To accept the model is to disregard any objections, such as the paradoxes. ==== > How to compute this definite integral? > > Integrate[LegendreP[15, x] ChebyshevT[11, x], {x, 0, 1}] >> Excuse me, I didn't make myself very clear. >> I want to compute this integral without using the computer. > If it were true that > Integrate[LegendreP[15, x] x^k , {x, 0, 1}] > is equal to zero for all odd k, then the problem > wouldn't be very difficult. On the other hand, a bit of experimenting suggests that Integrate[LegendreP[2m+1,x]ChebyshevT[2n+1,x],{x,0,1}] == 0 whenever 0 <= n < m. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== In alt.anagrams, >David James Polewka used the Gregg Method to post ... > Typing Monkeys Don't Write Shakespeare >= >Spanking the keys typed, No worries mate. >Tyke's gay peep show is rotten in Denmark. >Yon thirteen-toed yakkers spewing SPAM >Mork Went To See The Dark Gypsy In Spain. >We paid something rotten. Press any key ... 'K North Korea: Peking swept my seat, Sidney! ========================= Endeavor to persevere ========================= ==== > Let > h(0,m) = 1, for all positive integers m. > > Let h(n,m) = sum{k=1 to m} h(n-1,k)/k, > > for all positive integers m and n. > > Then, > > (1/n) sum{k=1 to oo} h(k-1,n) /(k 2^k) > > = > > sum{k=1 to n} binomial(n,k) ln(binomial(2k,k)) (-1)^k > > > As ascii-image: > oo > --- > 1 h(k-1,n) > --- / --------- = > n --- k 2^k > k=1 > > n > --- / n / 2k k > | | ln(| |) (-1) > / k / k / > --- > k=1 > > > > Even if true, this is pretty pointless... > This is ONLY true if n is >= 2. I think that the identity is correct for these n's. But now I am starting to have my doubts about the truth of the above for EVERY integer n >= 2, generally. Why would there be an exception to this identity at n = 1? [note: I have editted the ascii-image in the copied original post so that the upper limit of the lower sum is 'n' instead of 'oo', even though the earlier version was correct as well.] Leroy Quet ==== > > > Relax, I'm just yanking your chain. > > > Yeah I want to share. > > I'm pondering a closed group to discuss an aspect. > > Like anyone I have but bits and pieces. > > I'm waiting also for my partner to decide to drop out of get with > it.. If he drops I am free to share . > > Anyway yes .. I have a something to suggest. > > So far my experience with this is once one door is open I see several > more that need to be.. > > You should know I actually respect your abilities. > Ok, the ball's in your court. If you have something to discuss, post it here or in the Collatz Community. ==== > > > Relax, I'm just yanking your chain. > > > Yeah I want to share. > > I'm pondering a closed group to discuss an aspect. > > Like anyone I have but bits and pieces. > > I'm waiting also for my partner to decide to drop out of get with > it.. If he drops I am free to share . > > Anyway yes .. I have a something to suggest. > > So far my experience with this is once one door is open I see several > more that need to be.. > > You should know I actually respect your abilities. > > > > Ok, the ball's in your court. If you have something to discuss, post > it here or in the Collatz Community. Alright. I'll get it together this weekend. I'll go over there and log in. I'm working overtime this week so I'm dog tired. This idea I'm thinking of needs to be kicked around a bit so, I hope Collatz people will give it a kick or two before it gets posted to the world on Google. My partner is into higher math studies so he would like to know how it goes but doesn't have the time at this time to work with me on it. Ernst ==== > This does not directly solve the original poster's question, but those > interested in this topic should take, if they like, a look at this old > post of mine: > > http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=j4wtyvj > i98vr%40forum.mathforum.com&rnum=3&prev= > > That link is broken for me. > > If my result is correct, it should be noted that MANY A1's (a(1)'s = x in > my post) lead to later undefined A's (a(k)'s). > > I believe that the A1's that you refer to are simply the roots of > (f@@n)(z) = 1 where f(z) = z^2/(z-1) and `@@` indicates iterated > function composition. > > (For example, if a(1) = 1, I get that a(2) is 1/0. But a(3) is 1. > {because 1 + oo + 1 = 1 *oo *1....snicker}) > > I think it's best to avoid that silliness by ignoring the original > posing of the problem in terms of sums and products of arbitrary > length and rather just using the recurrence p[n+1] = p[n]^2/(p[n]-1) There has appeared to be a techno-glitch which only allowed the first part of the URL to actually be linked to. Here is the link again: http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=j4wtyvji 98vr%40forum.mathforum.com&rnum=3&prev= The recursion I got APPEARS to be different from yours, but I am most certainly not understanding what you mean exactly. Anyway, in case the link still does not link, here is my post copy/pasted anyway: >One may wonder what sequence of a[m]'s is such that the first n terms >summed equals the first n terms multiplied. >i.e.: >sum{k=1 to n} a[k] = >product{k=1 to n} a[k] , >for every positive integer n. This is how to calculate the terms: a[1] = x; a[2] = x /(x-1); and for m >=3, >a[m] = a[m-1]^2 /(a[m-1]^2 -a[m-1] +1) . These are the first few a[m]'s: x, x /(x-1), x^2 /(x^2 -x +1), x^4 /(x^4 -x^3 +2x^2 -2x +1), >x^8 /(x^8 -x^7 +3x^6 -6x^5 +9x^4 -10x^3 +8x^2 -4x +1),... The m_th (m >= 2) term is of the form >x^(2^(m-2)) /Q_m(x) , >where Q_m(x) is a 2^(m-2)-order polynomial, >where Q_2(x) = x-1; >and for m >= 3, >Q_m(x) = x^(2^(m-2)) - Q_{m-1}(x) *x^(2^(m-3)) +(Q_{m-1}(x))^2. Is there a closed form for the Q_m(x)'s ? This whole thing may be well studied. Any additional info? > Leroy Quet ==== > I think it's best to avoid that silliness by ignoring the original > posing of the problem in terms of sums and products of arbitrary > length and rather just using the recurrence p[n+1] = p[n]^2/(p[n]-1) The recursion I got APPEARS to be different from yours, but I am most > certainly not understanding what you mean exactly. Your recurrence is a[n+1] = a[n]/(a[n]-1). It generates the A's, the terms of the sum. My recurrence generates the sum (or product -- same thing) of all the terms. ==== I've noticed that there has been a preponderance of JSH threads that do not have JSH in the subject lines. Especially recently it seems. I've isolated one reason for this. Many of the sci.math regulars who participate in JSH threads are no longer actively using the JSH prefix in their responses. Here is a (partial) list of names of regulars who have participated in JSH threads without the JSH prefix: James Harris David C. Ullrich Randy Poe Larry Hammick Will Twentyman Arturo Magidin Ed Hook Brian Quincy Hutchings Nora Baron Virgil Robin Chapman George Cox Bill Hale Clive Tooth I got these names from just a cursory glance through the last couple months worth of postings on google groups. I think anybody on this list posts enough to sci.math, and in particular, to JSH related threads, that the cumulative effect of all these people not using the JSH prefix is causing an increasingly large jump in my news traffic. I encourage people (especially those on the list, and in particular James Harris) to use the JSH prefix in their future posts to JSH related threads. The non-JSH part of sci.math will be very grateful. ==== > I've noticed that there has been a preponderance of JSH threads that do > not have JSH in the subject lines. Especially recently it seems. > > I've isolated one reason for this. Many of the sci.math regulars who > participate in JSH threads are no longer actively using the JSH prefix in > their responses. > > Here is a (partial) list of names of regulars who have participated in JSH > threads without the JSH prefix: > > James Harris > David C. Ullrich > Randy Poe > Larry Hammick > Will Twentyman > Arturo Magidin > Ed Hook > Brian Quincy Hutchings > Nora Baron > Virgil > Robin Chapman > George Cox > Bill Hale > Clive Tooth > > I got these names from just a cursory glance through the last couple > months worth of postings on google groups. I think anybody on this list > posts enough to sci.math, and in particular, to JSH related threads, that > the cumulative effect of all these people not using the JSH prefix is > causing an increasingly large jump in my news traffic. how large is that jump? > I encourage people (especially those on the list, and in particular James > Harris) to use the JSH prefix in their future posts to JSH related > threads. The non-JSH part of sci.math will be very grateful. ==== I encourage people (especially those on the list, and in particular James > Harris) to use the JSH prefix in their future posts to JSH related > threads. The non-JSH part of sci.math will be very grateful. Another attempt to suppress Harris's work. Typical. I can almost hear the bleating, Outsiiide the baaahx! Baaaad! The funny thing is, it *would* be a good idea for us all to use JSH in the subject line. James's true audience is the future. It may be 1000 years before the human race has evolved enough to understand James's ideas. Why not make it easy for them to collect his work, and the pitiful whining of his detractors, by distinguishing it from the general noise on sci.math with the simple tag, JSH? I imagine a day, when the world is almost exclusively digital, when people will buy and sell original electronic documents the way we would trade Starry Night or an original copy of the Declaration of Independence. One of the most coveted e-documents will be the Harris originals. (You laugh, but your great^n-grandchildren won't.) And you know what the most frequent lament of the lucky owners will be? He signed it, but didn't initial it! I think James should sign and initial all his work, if it's not too much trouble. For the sake of posterity. ==== >I've noticed that there has been a preponderance of JSH threads that do >not have JSH in the subject lines. Especially recently it seems. I've isolated one reason for this. Many of the sci.math regulars who >participate in JSH threads are no longer actively using the JSH prefix in >their responses. You are correct. Bad habit. I was never very reliable at remembering the prefix. I need to develop the habit. That said, I have to say I would really miss JSH threads were they to stop. Though for his own sake I do wish he'd get some help. In my opinion his postings are escalating in a number of ways. I think that claiming definitions to be wrong is a relatively recent thing, for instance. - Randy ==== > > >>I've noticed that there has been a preponderance of JSH threads that do >>not have JSH in the subject lines. Especially recently it seems. >>I've isolated one reason for this. Many of the sci.math regulars who >>participate in JSH threads are no longer actively using the JSH prefix in >>their responses. > > > You are correct. Bad habit. I was never very reliable at remembering > the prefix. I need to develop the habit. > > That said, I have to say I would really miss JSH threads were they to > stop. Though for his own sake I do wish he'd get some help. In my > opinion his postings are escalating in a number of ways. I think that > claiming definitions to be wrong is a relatively recent thing, for > instance. > > - Randy > What I find odd is that he seems to think rational numbers were designed to be fields, and then the definition made to fit that. That's the only way I can make sense of his claims about them. He doesn't appear to understand that the definition comes first, then we see what category they fit into (Ring, Division Ring, Field, etc). -- Will Twentyman ==== >I've noticed that there has been a preponderance of JSH threads that do >not have JSH in the subject lines. Especially recently it seems. I've isolated one reason for this. Many of the sci.math regulars who >participate in JSH threads are no longer actively using the JSH prefix in >their responses. Here is a (partial) list of names of regulars who have participated in JSH >threads without the JSH prefix: James Harris >David C. Ullrich >Randy Poe >Larry Hammick >Will Twentyman >Arturo Magidin >Ed Hook >Brian Quincy Hutchings >Nora Baron >Virgil >Robin Chapman >George Cox >Bill Hale >Clive Tooth Wow. Probably posting this list of names will shame us all into abiding by your wishes. If it doesn't work I'd suggest sending the same list to the FBI. >I got these names from just a cursory glance through the last couple >months worth of postings on google groups. I think anybody on this list >posts enough to sci.math, and in particular, to JSH related threads, that >the cumulative effect of all these people not using the JSH prefix is >causing an increasingly large jump in my news traffic. I encourage people (especially those on the list, and in particular James >Harris) to use the JSH prefix in their future posts to JSH related >threads. The non-JSH part of sci.math will be very grateful. ************************ David C. Ullrich ==== > Does anybody know of a pointer to the Levinson Time Eqns.? What are > these eqns. supposed to describe/model? I've heard this come up in relation to a certain Montauk Project. Anybody know?? ==== >Could anyone point to me some references or give me some help with >following optimal control problem ? >Problem :minimize integral of f(x,u)dt over [t1, t2] > dx/dt=g(n,lambda,u) Do you mean g(x,lambda,u)? Is lambda a constant? If so, you could try including it as a state variable with d(lambda)/dt = 0. > subject to state inequality constraints : dx/dt >= -E ; lambda <= 1 > and state equality constraints : (dx/dt + E)*(lambda-1)=0 Here you have only three possibilities: either dx/dt= -E, lambda = 1, or both. If E is a constant, then the first case restricts x to be a linear function of time, in fact, g = - E, etc. -- John E. Prussing University of Illinois at Urbana-Champaign Department of Aerospace Engineering http://www.uiuc.edu/~prussing ==== >>.......... [lots snipped] > I might as well add that, > > If f(x) happens to be such that > > if > > a[j] (the j_th derivative of f(x) at x = 0) > > is an integer multiple of (j-1) for all j's >= 1, (and if a[0] = 0) > > then: > > > > b[j] is, for *ALL* integer j's >= 1, an integer multiple of (j-1), > > > where b[j] is (as noted in original post) the j_th derivative of exp(f(x)) at x = 0. > I might as well add again that: (with a little redefining of {a(k)} and {b(k)}.) If (for n = fixed nonnegative integer) B(x) = exp(integral{0 to x} A(y) y^n dy), where A(x) = sum{k=0 to oo} a(k) x^k/k!, and B(x) = sum{k=0 to oo} b(k) x^k/k!, and integral{0 to x} A(y) y^n dy = sum{k=0 to oo} a(k) x^(k+n+1)/(k!(k+n+1)) (if the above is all kosher, converging uniformly), then: b(0) = 1; b(m) = 0 for 1<=m<=n; b(m+n+1) = ((m+n)!/m!) sum{k=0 to m} b(m-k) a(k). (hopefully true.) So, if {a(k)} is an integer sequence, b(m+n+1) is divisible by ((m+n)!/m!), for m = all nonegative integers. Leroy Quet ==== > YOUCH! That's a heavy condition. So the prime factorization of n > has no lone primes. I would call that powerful rather than squareful. > > Yes, the term powerful is also used. But the thing about squareful > (or, squarefull) is that one can then use cubefull for a number n > with the property that p divides n implies p^3 divides n, and k-full > to mean p divides n implies p^k divides n. Likewise, we can similarly use the term cubeful to mean it has at least one prime factor to power 3 or more. I mentioned these once in class, maybe in a problem:- to prove that every natural can be uniquely expressed as a product of a perfect cube and a cube-free number. Cubeful numbers arose naturally along the way. ---------------------------------------------------------------------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ---------------------------------------------------------------------------- -- The sum of the cubes of the first n natch, Is square the sum of the batch! ---------------------------------------------------------------------------- -- ==== Dear Friends and Colleagues, We are pleased to announce the launching of our new journal, Analysis and invite you to view the table of contents of the inaugural issue at http://www.worldscinet.com/aa.html, as well as download the Analysis and Applications publishes high quality mathematical papers that treat those parts of analysis with direct or potential applications to the physical and biological sciences and engineering. Some of the topics from analysis include approximation theory, asymptotic analysis, calculus of variations, integral equations, integral transforms, ordinary and partial differential equations, delay differential equations, and perturbation methods. The primary aim of the journal is to encourage the development of new techniques and results in applied analysis. As you view the journal contents, please consider recommending Analysis and Applications to your library acquisitions officer. You also can request a complimentary copy of Analysis and Applications by contacting Ms. Ng Min Chao (Marketing) at mcng@wspc.com. Also, if you are interested Co-Editors-in-Chief, Roderick Wong at mawong@cityu.edu.hk or Robert M. Miura at miura@njit.edu. Sincerely, Roderick S.C. Wong Co-Editor-In-Chief ==== > >[...] >I'd like to put the paper on the main math preprint server but have >been blocked by an automatic message saying that some established >mathematician has to vouch for me, as it apparently checks my domain >and notices I'm not using a *.edu account. > > You should thank them. Eventually you'll agree that the current Proof > is wrong, just like you eventually agreed that the previous ones were > wrong. In case you hadn't noticed I've had each of my *three* major results thoroughly critiqued now, which is why I'm moving to the final phase. I realized a while back that my FLT proof was just complicated enough that mathematicians could block me indefinitely. Then I thought my prime counting function would break things open, but saw a strong and successful resistance. I realized then that I needed a third result, and finding the error in the ring of algebraic integers gave it to me, with an important twist: With the previous results people could stay quiet, but when mathematicians teach the old flawed mathematics I can make a case for fraud, and make that case on the federal level. Individual mathematicians may find themselves at best having to pay back federal funds, and also entire universities may find themselves disgraced as well as facing penalties. And, it takes out an entire cross-section of universities from top tier ones like Harvard and Yale all the way to the bottom. I will, of course, personally protect Vanderbilt University. >As the academic year begins here in the United States, remember that >the error in taught mathematics can potentially be part of fraud cases >against universities and individual mathematicians. > > People really really wish you'd _do_ something about these bizarre > threats of lawsuits. Really. I don't intend on suing anyone. If things go according to plan, at least some of the suits will come from the federal government. >Some of you need to read the fine print on those federal funds you >receive. Basically now I can shift between my three results and hammer at mathematicians endlessly, continually shifting until I see a break. When I see the break I'll push with overwhelming force. When the story finally breaks, then the investigations will definitely begin in earnest, and meanwhile, mathematicians build the case against themselves by teaching the flawed mathematics, thus removing plausible deniability. And believe me, if the United States even just tells you to pay back federal funds that you received for math research, you won't like it. And your universities will bail on you when even a whiff of the possibility gets to them. But that's the big plan. For now, I look for weak spots, while in America, many of you begin teaching, building my case against you with your own energy. As I continue to probe, I will find a break, and then it will be over quickly. James Harris ==== >> >>[...] >>I'd like to put the paper on the main math preprint server but have >>been blocked by an automatic message saying that some established >>mathematician has to vouch for me, as it apparently checks my domain >>and notices I'm not using a *.edu account. >> >> You should thank them. Eventually you'll agree that the current Proof >> is wrong, just like you eventually agreed that the previous ones were >> wrong. In case you hadn't noticed I've had each of my *three* major results >thoroughly critiqued now, which is why I'm moving to the final phase. That's very funny. The critique of your proofs of FLT and APF have consisted of pointing out errors in the parts that were stated coherently enough that this was possible. Given that, most people would think that the final stage would be to give up and start over (or give up and not start over.) >I realized a while back that my FLT proof was just complicated enough >that mathematicians could block me indefinitely. Then I thought my prime counting function would break things open, but >saw a strong and successful resistance. I realized then that I needed a third result, and finding the error in >the ring of algebraic integers gave it to me, with an important twist: Tee-hee. A _ring_ contains an _error_? Doesn't make much sense. >With the previous results people could stay quiet, but when >mathematicians teach the old flawed mathematics I can make a case for >fraud, and make that case on the federal level. Individual mathematicians may find themselves at best having to pay >back federal funds, and also entire universities may find themselves >disgraced as well as facing penalties. And, it takes out an entire cross-section of universities from top >tier ones like Harvard and Yale all the way to the bottom. I will, of course, personally protect Vanderbilt University. > >>As the academic year begins here in the United States, remember that >>the error in taught mathematics can potentially be part of fraud cases >>against universities and individual mathematicians. >> >> People really really wish you'd _do_ something about these bizarre >> threats of lawsuits. Really. I don't intend on suing anyone. If things go according to plan, at >least some of the suits will come from the federal government. >>Some of you need to read the fine print on those federal funds you >>receive. Basically now I can shift between my three results and hammer at >mathematicians endlessly, continually shifting until I see a break. >When I see the break I'll push with overwhelming force. When the story finally breaks, then the investigations will definitely >begin in earnest, and meanwhile, mathematicians build the case against >themselves by teaching the flawed mathematics, thus removing plausible >deniability. And believe me, if the United States even just tells you to pay back >federal funds that you received for math research, you won't like it. And your universities will bail on you when even a whiff of the >possibility gets to them. But that's the big plan. For now, I look for weak spots, while in >America, many of you begin teaching, building my case against you with >your own energy. As I continue to probe, I will find a break, and then it will be over >quickly. What sort of break do you have in mind here? I mean, assuming for a second that your bizarre fantasies about how everyone on sci.math is lying, every journal editor in the world seems to be immediately seem to realize that your work is going to overthrow the establishment so they need to reply with lies about how it's wrong or trivial, etc are true - given that the entire world seems to be united against you here it's hard to see what sort of break you're waiting for - if some renegade were to boldly speak the Truth about your work he'd be suppressed just as you've been. Giggle. We all really do wish you'd post the text of the complaint you sent to the FBI and the text of their reply. I mean these will be important documents some day when the Truth finally comes out - we should get them on the record as soon as possible. >James Harris ************************ David C. Ullrich ==== > In case you hadn't noticed I've had each of my *three* major results > thoroughly critiqued now, which is why I'm moving to the final phase. But I think your 4th result, the Z[1/2] = R is much more important. The logic is so simple that even undergrads can follow it, but all mathematitians in sci.math still reject it for some reason. I'm sure it's the weakest spot. ==== > > In case you hadn't noticed I've had each of my *three* major results > thoroughly critiqued now, which is why I'm moving to the final phase. > > But I think your 4th result, the Z[1/2] = R is much more important. > The logic is so simple that even undergrads can follow it, but > all mathematitians in sci.math still reject it for some reason. > > I'm sure it's the weakest spot. I doubt that you're sincere. My original post in this thread points out a result which should raise suspicions about other posters, as I can show with 2x^3 - 3xy^2 + y^3 that my predictions are correct as you can *see* the answer. What some of you may fail to realize is that posters who dispute my argument are claiming that mathematicians are blind in that area. Their argument is that NO ONE can figure out the factors that I say can be determined, and I've read some fascinating posts on the subject. For those of you who don't understand, I've proven that 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where *only two* of the a's have a factor that is sqrt(2). That's easily checkable. Posters have tried to claim that the reducibility over Q allows that, so that they can try and claim that there is NO WAY to figure out the factors if it's irreducible. For instance with 65x^3 - 12xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) these posters claim that NO ONE can say how the 65, and in particular, the factor 5 of 65 separates out into the a's, while I say I can prove only two of the a's have a factor that is sqrt(5). The difference they cite is reducibility over Q, lack of which they argue blinds humanity, blocking the determination that I say is possible. They claim intellectual weakness for mathematics; I prove greater strength. James Harris ==== > >In case you hadn't noticed I've had each of my *three* major results >thoroughly critiqued now, which is why I'm moving to the final phase. >>But I think your 4th result, the Z[1/2] = R is much more important. >>The logic is so simple that even undergrads can follow it, but >>all mathematitians in sci.math still reject it for some reason. >>I'm sure it's the weakest spot. > > > I doubt that you're sincere. Why should sincerity have anything to do with the truth? You apparently have not done as I requested, which is: 1. Provide the definition of the ring Z[x]. This will establish for us that you know what you're talking about (or not, which is really the case). 2. Provide the definition of the ring Z[1/2], and verify that it is the particular case of applying the definition you gave in 1. above to the situation x = 1/2. 3. Prove that the result of the definition in 2. above actually yields the ring of real numbers. I've said before that I can prove 1 and 2, and can prove that 3 is incorrect; furthermore, most of your mathematical critics can do the same. If you care to be other than a curiosity, a sideshow freak, then you'll take the time and effort to provide responses to these three items. As it is, you appear to be content as the butt of every available joke on sci.math. That's your prerogative, to be sure, but it isn't the way to get your case heard, either here or at the FBI or CIA or NSA or NSC or whatever 3-letter agency strikes your fancy. > > My original post in this thread points out a result which should raise > suspicions about other posters, as I can show with > > 2x^3 - 3xy^2 + y^3 > > that my predictions are correct as you can *see* the answer. > > What some of you may fail to realize is that posters who dispute my > argument are claiming that mathematicians are blind in that area. > > Their argument is that NO ONE can figure out the factors that I say > can be determined, and I've read some fascinating posts on the > subject. > > For those of you who don't understand, I've proven that > > 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where *only two* of the a's have a factor that is sqrt(2). It is easily verified that, up to order, the a's are these numbers: a_1 = -1 a_2 = -1+sqrt(3) a_3 = -1-sqrt(3) As the person making the claim that two and only two of these numbers are divisible by sqrt(2), please indicate which, and then show the factorization. > > That's easily checkable. > So, what's keeping you? I've shown you the values, now it's up to you to make good on your implicit promise to factor the particular a's to make that factor evident. > Posters have tried to claim that the reducibility over Q allows that, > so that they can try and claim that there is NO WAY to figure out the > factors if it's irreducible. > > For instance with > > 65x^3 - 12xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > these posters claim that NO ONE can say how the 65, and in particular, > the factor 5 of 65 separates out into the a's, while I say I can prove > only two of the a's have a factor that is sqrt(5). > Untrue, and you are aware of that. Knowing that I have shown you *PRECISELY* the factorization of the a's that exhibits the common factor, your statement that these posters claim that NO ONE can say how the 65, and in particular, the factor 5 of 65 separates out into the a's is clearly a lie. As a person who claims to be interested in the TRUTH, and who claims that mathematicians are notorious liars, I find it suspicious that you have no problem indulging in lying here. BTW, none of the a's appears divisible by sqrt(5). > The difference they cite is reducibility over Q, lack of which they > argue blinds humanity, blocking the determination that I say is > possible. > > They claim intellectual weakness for mathematics; I prove greater > strength. > explicitly claiming intellectual weakness for mathematics. Also, I haven't seen any proofs coming from you. I've seen more than my share of *claims* of proof, but [ain't it amazin'?] an equal number of fatal errors. > > James Harris Dale. ==== > > > > My original post in this thread points out a result which should raise >> suspicions about other posters, as I can show with >> 2x^3 - 3xy^2 + y^3 >> that my predictions are correct as you can *see* the answer. >> What some of you may fail to realize is that posters who dispute my >> argument are claiming that mathematicians are blind in that area. >> Their argument is that NO ONE can figure out the factors that I say >> can be determined, and I've read some fascinating posts on the >> subject. >> For those of you who don't understand, I've proven that >> 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) >> where *only two* of the a's have a factor that is sqrt(2). > > > It is easily verified that, up to order, the a's are these > numbers: > > a_1 = -1 > a_2 = -1+sqrt(3) > a_3 = -1-sqrt(3) > > As the person making the claim that two and only two of these > numbers are divisible by sqrt(2), please indicate which, and > then show the factorization. > James is correct here, since sqrt(2) times the algebraic integer unit (sqrt(2)/2)(-1 + sqrt(3)) is indeed equal to -1 + sqrt(3), for example. However, this does absolutely nothing to make his larger point, since he's chosen a polynomial that's _reducible_ over Z. In fact, if one divides off the linear factor over Z, his bumfoggery actually provides an example of what you and others have been saying all along. Rick ==== > For those of you who don't understand, I've proven that > > 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where *only two* of the a's have a factor that is sqrt(2). > > That's easily checkable. some elemantary calculations yield that a_1 = -1 a_2 = -1+sqrt(3) a_3 = -1-sqrt(3) solves the problem. But I can't see any sqrt(2)s here!!! The same argument shows that this is the only solution up to permutation of the a's. So what? I would really be happy to see how this contradicts to what I learned about polynomial factorization. Imre Polik ==== > >In case you hadn't noticed I've had each of my *three* major results >thoroughly critiqued now, which is why I'm moving to the final phase. >>But I think your 4th result, the Z[1/2] = R is much more important. >>The logic is so simple that even undergrads can follow it, but >>all mathematitians in sci.math still reject it for some reason. >>I'm sure it's the weakest spot. > > > I doubt that you're sincere. > > My original post in this thread points out a result which should raise > suspicions about other posters, as I can show with > > 2x^3 - 3xy^2 + y^3 > > that my predictions are correct as you can *see* the answer. > > What some of you may fail to realize is that posters who dispute my > argument are claiming that mathematicians are blind in that area. > > Their argument is that NO ONE can figure out the factors that I say > can be determined, and I've read some fascinating posts on the > subject. > > For those of you who don't understand, I've proven that > > 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where *only two* of the a's have a factor that is sqrt(2). > > That's easily checkable. > > Posters have tried to claim that the reducibility over Q allows that, > so that they can try and claim that there is NO WAY to figure out the > factors if it's irreducible. > > For instance with > > 65x^3 - 12xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > these posters claim that NO ONE can say how the 65, and in particular, > the factor 5 of 65 separates out into the a's, while I say I can prove > only two of the a's have a factor that is sqrt(5). If you would like, I'll be happy to calculate the a's for you, calculate the common root of each a and 5 (using W. Dale Hall's equations), and examine wither or not sqrt(5) is actually a factor of your a's. Would that be sufficient to cause you to give up? -- Will Twentyman ==== >> >> In case you hadn't noticed I've had each of my *three* major results >> thoroughly critiqued now, which is why I'm moving to the final phase. >> >> But I think your 4th result, the Z[1/2] = R is much more important. >> The logic is so simple that even undergrads can follow it, but >> all mathematitians in sci.math still reject it for some reason. >> >> I'm sure it's the weakest spot. I doubt that you're sincere. Why would you doubt that? It may or may not be the weakest spot in mathematical terms - hard to say whether one bit of nonsense is more or less nonsensical than another. But it seems clear that it's the weakest spot, in the sense that it's the spot that's most obviously nonsense to readers with the least knowledge of mathematics. Which of course makes it a very welcome addition to the canon, since it clarifies things for lots of readers who might not follow the arguments regarding FLT or APF. >My original post in this thread points out a result which should raise >suspicions about other posters, as I can show with 2x^3 - 3xy^2 + y^3 that my predictions are correct as you can *see* the answer. What some of you may fail to realize is that posters who dispute my >argument are claiming that mathematicians are blind in that area. Their argument is that NO ONE can figure out the factors that I say >can be determined, and I've read some fascinating posts on the >subject. For those of you who don't understand, I've proven that 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where *only two* of the a's have a factor that is sqrt(2). That's easily checkable. Posters have tried to claim that the reducibility over Q allows that, >so that they can try and claim that there is NO WAY to figure out the >factors if it's irreducible. For instance with 65x^3 - 12xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) these posters claim that NO ONE can say how the 65, and in particular, >the factor 5 of 65 separates out into the a's, while I say I can prove >only two of the a's have a factor that is sqrt(5). The difference they cite is reducibility over Q, lack of which they >argue blinds humanity, blocking the determination that I say is >possible. Uh, right. Exactly where has anyone claimed that irreducibility over Q blinds humanity? >They claim intellectual weakness for mathematics; I prove greater >strength. All you prove is that if you ignore what words mean you can spout a lot of nonsense. This does show you can do things that actual mathematicians cannot, but they're things an actual mathematician would not want to do. >James Harris ************************ David C. Ullrich ==== > In my previous thread I talked about some simple experiments with > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > > 3(-1+mf^2 )x u^2 + u^3 f) > > which show a problem with the ring of algebraic integers, which some > posters seem to be *wishing* doesn't exist as they reply back with > dedication, and wrong arguments. You forgot to mention W. Dale Hall's concrete counter-example for 65x^3-12x+1. If you'd like I'll work on the explicit form of the numbers in his counter-example. > > I'm posting here to remind you of a case where you can check my > conclusion directly as using f=sqrt(2), m=1, y=sqrt(2) u, you get > > 2x^3 - 3xy^2 + y^3 > > which factors over Q. > > My simple experiments would indicate that you have > > 2x^3 - 3xy^2 + y^3 = (sqrt(2)a_1 x + y)(sqrt(2)a_2 x + y)(a_3 x + y) > > and in fact that is the case. > > Posters have claimed that reducibility is why my methods work here, > when in fact, reducibility only allows a_1 and a_2 to be in the ring > of algebraic integers. Part of the scientific method is exploring how certain situations may be special cases. If I drop an unfolded sheet of paper its acceleration is *not* 9.8 m/s/s. Is that a special case or a counter-example to acceleration due to gravity *in a vacuum* at the Earth's surface is 9.8 m/s/s? Reducibility could be a special case. > > That's it. > > Why don't any of you wonder why these posters themselves can't predict > where factors must go? Instead they simply challenge my work, unable > to do better themselves, as they claim that all is hidden. > > Yup, think carefully, and you'll realize that posters are claiming > that NO ONE can figure out where factors distribute, which is a claim > of forced ignorance. > > I'm saying that the mathematics gives tools to see, where they claim > humanity is blind. > > Beyond experiments I have rigorously proven my case and written a > paper Advanced Polynomial Factorization. > > That paper is linked to on my primary website > > http://groups.msn.com/AmateurMath > > where though you both have to have an MSN Passport and join the group. I've joined the group twice and been removed twice, apparently by the moderator. Any ideas who that might be? In the mean time, being removed makes it very difficult to check your paper for any changes that may occur. -- Will Twentyman ==== >> [...] >> >> I'm saying that the mathematics gives tools to see, where they claim >> humanity is blind. >> >> Beyond experiments I have rigorously proven my case and written a >> paper Advanced Polynomial Factorization. >> >> That paper is linked to on my primary website >> >> http://groups.msn.com/AmateurMath >> >> where though you both have to have an MSN Passport and join the group. I've joined the group twice and been removed twice, apparently by the >moderator. Any ideas who that might be? In the mean time, being >removed makes it very difficult to check your paper for any changes that >may occur. This is really quite curious. The math is supposed to speak for itself; publishing it but only allowing true believers to view the publication certainly gives the lie to his claims about how anyone can check the proof for themselves. ************************ David C. Ullrich ==== Good news??....(for once) Leroy ==== > Good news??....(for once) > Yeah. But weren't the scores renormalized a few years back? --Ron Bruck Approved: Daniel Grayson, dan@math.uiuc.edu, moderator for sci.math.research ==== > I wanted too but the only copy our library has is on loan till 2015. This seems excessive even for a university librarian. -- Timothy Murphy tel: +353-86-233 6090 ==== > My suggestion would be to keep track of the smallest prime factor > for 48^p - 47^p for the many primes which have already been tried. > Perhaps a pattern in the smallest of prime factors will pop out. > > Actually, there is at least one thing that can be said about factors of > 48^p-47^p (with p prime) : they have the form 2*k*p+1 for some integer k ... > BTW, this is true for any integer n >= 1 that factors of (n+1)^p-n^p have the form 2*k*p+1; not just for the special value n=47. ==== >Let f(n) = 48^n - 47^n. Apparently [ie, based on looking at factors >of a few hundred f(n)] for n>0, 5|f(2n), 7|f(3n), 11|f(5n), 337|f(7n), >23|f(11n), 79|f(13n+13), 443|f(17n), 7937|f(31n), etc. Well, yes: 3^2 = 2^2 mod 5 so f(2n) = 0 mod 5; 6^3 = 5^3 mod 7 so f(3n) = 0 mod 7; and in general for each prime p, if m is the least positive integer such that 48^m = 47^m mod p, then f(mn) = 0 mod p. And m will always be a divisor of p-1 (by Fermat's Little Theorem). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > >Let f(n) = 48^n - 47^n. Apparently [ie, based on looking at factors >of a few hundred f(n)] for n>0, 5|f(2n), 7|f(3n), 11|f(5n), 337|f(7n), >23|f(11n), 79|f(13n+13), 443|f(17n), 7937|f(31n), etc. > > Well, yes: 3^2 = 2^2 mod 5 so f(2n) = 0 mod 5; > 6^3 = 5^3 mod 7 so f(3n) = 0 mod 7; > and in general for each prime p, if m is the least positive integer > such that 48^m = 47^m mod p, then f(mn) = 0 mod p. And m will always > be a divisor of p-1 (by Fermat's Little Theorem). I usually see the obvious after I post, not before. :) Before my earlier post I reduced 48^n and 47^n mod 47 to see there is no k such that f(kn)=0 mod 47, but didn't think to reduce 48^n and 47^n separately for other moduli. -jiw ==== The three-signed plane is converted to the cartesian plane by the following: x = s - 1/2( p + m ) y = sqrt(3)/2( m - p ) where x and y are the traditional cartesian coordinates and s, m and p are the star, minus, and plus components of a value in three signed space. I have written some code that performs the conversion and I have checked products against their x + iy form and get complete agreement! For two values y1, y2 in Y the product y1y2 is the same as if I convert them to z1, z2 in C and compute the product z1z2. This means that angles add in the product. I believe that this is quite a discovery. Simply put, the complex plane is equivalent to three-signed space as I have defined it. X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 05:59 PM, amorgan@Xenon.Stanford.EDU (Alan Morgan) said: >At what level? Any? >Would you suggest teaching Cauchy sequences to >junior high school students? Would you suggest sprinkling them with pixie dust? That has as much wouldn't suggest introducing real numbers in terms of rational numbers at all; I'd suggest introducing real numbers geometrically at an early age and only teaching constructions much later. When it was time to introduce them to more rigorous methods, I'd start them on an axiomatic approach well before I did a construction from the integers. And if I did get around to teaching a construction from Peano's Postulates, it most certainly wouldn't be rdecimals[1] expansions, because that is by far the *most* complex approach under discussion. Look at all the asinine threads on the claim that .9 ... is not equal to 1. >but most classes don't do that Then why introduce the complications inheren't in using repeating decimals as a definition? Do it geometrically, and teach decimals as simle nomenclature. >Nice. It's almost impossible to find qualified math teachers for >public school now. You'll only make it harder. No. I'd make it easier to find *qualified* teachers. It doesn't bother me that I'd make it harder to find unqualified teachers. [1] Why base 10, anyway? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org ==== I wonder if Conway's surreal numbers will ever be the standard way of constructing the rationals, the reals and more? Logically, it seems much simpler than any other method. It's surprising (to me) it's made no headway, even at university level. It seems so much neater than non-standard analysis which has (used to have?) a few fanatical supporters. -- Timothy Murphy tel: +353-86-233 6090 ==== Fred Galvin > Congratulations, it looks like you've rediscovered Ramsey's theorem. > More precisely, the infinite version (there is also a finite version) > of Ramsey's theorem, Theorem A in F. P. Ramsey's 1930 paper On a > problem of formal logic, which says in part: If the increasing pairs of natural numbers [ordered pairs (i,j) where > i and j are natural numbers and i < j] are partitioned arbitrarily > into a finite number of classes, then there is an infinite set M of > natural numbers such that all pairs (i,j) with i, j in M belong to the > same class. Actually, Ramsey proved the analogous statement for r-tuples; what I > have stated above is the case r = 2. Nice. I new the finite Ramsey theorem but didn't spot any connection. And thanks G.A.E. also. LH ==== >Original Format Don't let the harping from the king's men of science [KMS] >discourage you. You are working on the frontiers of knowledge >with what corresponds to the Hamilton Visualization [HV] of >spin [they call it Lie algebras and Clifford algebras and >spinors and twistors and such today but it's all the same HV >with the names changed and the pictures destroyed] from the >early part of the 1800s. I've described it as grade school level >geometry when it's done right as Hamilton did with his giant >science breakthroughs. I've explained why I don't agree with >some of your specialized techniques but the KMS carping with >their phony baloney fancy word games which they don't understand >as you note, is just a diversion. The French changed all the >Hamilton pictures and made it into king's secret symbol manipulation >[symbman] without real meaning but finally their cover up is being >exposed worldwide and your work is helpful. As you have recognized >they suppressed the HV for over 150 years and your methods are a >restoration of the truth geometric meanings. Good seeing. JD > I found him JD. Your KMS mastermind. (at least for the 1800s) I have less doubts now. pmj http://www.hypercomplex.com ==== I hope you don't mind if I tag this thread from here. It is the only post I can see in my newsreader. Was at google for the rest. _______________ Quote: The spinors for a sub algebra of geometric algebra. The algebra of spinors was discovered independent of the full geometric algebra by Hamalton, who gave it the name quaternion algebra. Some readers may want to say, rather, that we have here two isomorphic algebras, but there is no call for any such distinction. A quaternion is a spinor. The identification of quaternions with spinors is fully justified not only because they have equivalent algebraic properties, but more important, because they have the same geometric significance. It is unfortunate that Hamilton chose the name quaternion, for this name merely refers to the comparatively insignificant fact that the quaternions compose a linear space of four dimensions. The name quaternion diverts attention from the key fact that Hamilton had invented a geometric algebra. Hamilton's work itself shows clearly the crucial form of geometry in his invention. Hamilton was consciously looking for a system of numbers to represent rotations in three dimensions. He was looking for a way to describe a geometry by algebra, and so he found a geometric algebra. In the twentieth century quaternions seem to be generally regarded as something of a mathematical museum piece, a curious artifact from the nineteenth century or a diversion from the mainstream of modern mathematics. One reason for this unfortunate circumstance might be found in the fact that Frobenius proved in 1878 that the quaternions constitute the largest associative divisions algebra. Thus, if division is a characteristic feature of quaternions, then quaternion algebra cannot be generalized. This is a typical diversion from the geometrical spirit of HamiltonÕs inquiry. The predominance of such diversion from the geometric to the algebraic features of quaternions goes far to explain why quaternions have not yet found the central place in mathematics which Hamilton had foreseen. The treatment of reflections and rotations given above originated with Hamilton... 'Vetors, Spinors, And Complex Numbers in Classical and Quantum Physics' David Hestenes, 'American Journal of Physics', Sept. 1971 _____________ Best, Dan. -- if( this == NULL ) return that; ==== ----- Original Message ----- > invented a geometric algebra. Airy fought against the teaching of Algebraic Geometry, Geometric Algebra is just another modern twist on the use of algebra to represent geometry, here's a quote, QUOTE: He was a practical man, with his work directed at application, not theory. tripos examinations for Cambridge in 1859: I have looked very carefully over the Examination Papers and think them on the whole very bad. They are utterly perverted by the insane love of Problems, and by the foolish importance given to wholly useless parts of useful application of pure mathematics are cut down or not mentioned. http://coldrain.net/lucas/airy.html END QUOTE He helps destroy Hamilton's career -- QUOTE: So witches some enchanted WAND bestride, And think they through the AIRY regions ride.>> - Oldham ----------------------------------------------------------------------- 1. AIRY's taunt during a dinner at the estate of Lord Rosse induced the sensitive William Rowan Hamilton (1805-1865), then two years on the wagon, to resume drinking, thus darkening the latter half of a career whose first half put Hamilton in the front rank of mathematicians in that century. http://groups.google.com/groups?selm=3B5CC257.48D4975D%40erols.com&oe=UTF-8& output=gplain END QUOTE Hamilton lost the woman he loved to another because of finances, while Airy fought his poverty to win his woman, and so while Hamilton starts drinking to drown out his sadness, Airy is a fighting man wining all his lifes battles and challlenges. But Hamilton gives up drinking with a new resoution to change, and actually is alcohol free for 2 years before this social dinner where Airy unkindly reminds him of the difference between the two of them, taunts Hamilton about drinking water and being unable to hold his liquor and using this quote from Oldham to refer to Hamilton as a WAND whose creativity must be the result of some witches, while he AIRY is a warrior who makes things happen by his own efforts, etc. from this point on Hamilton went downhill, END QUOTE Airy's battles are not scientific, but using whatever he can to win. Not only did he try to stop the teaching of Algebraic Geometry but also Michael Faraday's Electromagnetic theory QUOTE Known for his sarcasm and caustic personality, Airy had an ongoing battle with Charles Babbage in which he prevailed professionally and financially to the detriment of science. Airy's infamy ranges from ignoring John Adams' discovery of Neptune to dismissing Michael Faraday's field theory. http://micro.magnet.fsu.edu/optics/timeline/people/airy.html END QUOTE What would have happened to J.C. Maxwell's career then? Good thing for science Airy's views didn't always prevail. But since he didn't restrict himself to science in his methods of battle, even when fighting scientific wars, but used whatever tactic was nearest at hand, being a practical man, many question some of his scientific claims, like QUOTE: Nick Kollerstrom, an independent researcher and SHA council member, followed with a remarkable paper given under a very unremarkable title. Presenting his work as simply 'archiving the British Neptune correspondence (1837-1848)' Nick began with the curious story of where these papers had been prior to their recovery in Brazil in 1999 when he started to study them. He then went on to tell the story that they told, namely of the cover up, engineered by such great names in 19th century astronomy, as the Astronomer Royal, George Biddell Airy, the Cambridge Professor James Challis, and the Gentleman amateur, John Herschel, of turning the discovery of Neptune into a British triumph. In fact, as Nick pointed out, it was Dennis Rawlins, in 1993 that first came up with the theory, but Nick's research confirms his findings. According to Rawlins, these British astronomers manufactured their own hero in the shape of John Couch Adams following the Berlin Observatory's discovery of Neptune, based on the French mathematician Leverrier's very accurate predictions. In doing so Airy and Challis had agreed to take the blame for failing to respond to Adams' mathematical predictions of the planet, thereby allowing the French to beat them to it but at the same time saving the reputation of British astronomy and mathematics as being technically able to compete with the French. Nick's findings do confirm this version of events, the manuscripts alleged to be Adams' predictions show for example they were not on a par with Leverrier's. Similarly his diary shows just how close Adams and Challis were, suggesting it would have been unlikely that Adams would have approached Airy first instead of Challis as the official story has it. http://www.shastro.org.uk/report100403.htm END QUOTE > spirit of Hamiltonâs inquiry. The predominance of such diversion from > the geometric to the algebraic features of quaternions goes far to > explain why quaternions have not yet found the central place in > mathematics which Hamilton had foreseen. > I think after the quaternion battle was lost to the vector camp, science simply went down another path, and since we always tend to look back and think that past knowledge is in some sense less valuable than current ideas, because progress is always marching forward, it becomes more difficult for scientists to adopt an open mind and explore things we believe we once examined thoroughly and discarded. If it was discarded by men of the past, it must have been useless then. The idea that it was discarded because it was too useful, would not be accepted by scientists today. > The treatment of reflections and rotations given above originated with > Hamilton... 'Vetors, Spinors, And Complex Numbers in Classical and Quantum Physics' > David Hestenes, 'American Journal of Physics', Sept. 1971 > pmj http://www.hypercomplex.com ==== math is this ? TIA, Lurch ==== math is this ? It's an English letter, used in various words like math, etc. It's also often used as a variable, denoting some mathematical object in an equation. What object it denotes varies from context to context; if the thing was written properly you can figure out what the h means by looking for phrases like let h = .... (An h with a funny slash through it is a standard symbol in _physics_, I think for Plank's constant.) >TIA, Lurch > ************************ David C. Ullrich ==== > (An h with a funny slash through it is a standard symbol in _physics_, > I think for Plank's constant.) > Actually Planck's constant is usually noted h; h with a funny slash through is h/(2*pi) Sam -- People sometimes ask me if it is a sin in the Church of Emacs to use vi. Using a free version of vi is not a sin; it's a penance. - Richard Stallman ==== > (An h with a funny slash through it is a standard symbol in _physics_, > I think for Plank's constant.) Also h without the slash. The slash means to divide by 2 Pi. ==== > (An h with a funny slash through it is a standard symbol in _physics_, >> I think for Plank's constant.) Also h without the slash. The slash means to divide by 2 Pi. Oh. Right, I knew that... ************************ David C. Ullrich ==== Charlie Johnson of > math is this ? The typsetting program called TeX, and its variations such as Latex, use $...$ around all kinds of mathematical notation. That's the only usage of $...$ that I know of. LH ==== Two motivating examples: - One can construct a finitary computational model of the integers in many ways. One way is to represent them as a string of binary digits, with the classic two's-completement algorithms for binary addition, subtraction, multiplication, division, and comparison. If the representation is constrained to have no redundent leading digits, then the mapping from integers to representations is bijective. - Given a computational model of the integers, one can construct a finitary model of the rational numbers, by representing each rational number as a pair (numerator,denominator). The elementary operations can be defined in the obvious way, i.e. (a,b)+(c,d)=(a*d+b*c,b*d). If the representation is constrained to be in lowest terms with positive denominator, the mapping from rational numbers to representations is bijective. Now a few questions: - Does there exist a finitary computational model of the algebraic numbers, such that there exists a bijection between mathematical algebraic numbers and their model, and there exist terminating algorithms for addition, subtraction, multiplication, division, and comparison? To clarify finitary here, I mean a representation such that if you start with a finite set of rational numbers and perform a finite sequence of field operations and polynomial-solving operations on them, then the result is guaranteed to be representable in finite space. - Does there exist a halting algorithm that, for every finite set of polynomial coefficients (a0+a1*x+a2*x^2+..=0), can determine all of its solutions in the algebraic numbers in such a representation? - Any references to computer implementations of the above? ==== >Two motivating examples: - One can construct a finitary computational model of the integers in >many ways. One way is to represent them as a string of binary digits, >with the classic two's-completement algorithms for binary addition, >subtraction, multiplication, division, and comparison. If the >representation is constrained to have no redundent leading digits, >then the mapping from integers to representations is bijective. - Given a computational model of the integers, one can construct a >finitary model of the rational numbers, by representing each rational >number as a pair (numerator,denominator). The elementary operations >can be defined in the obvious way, i.e. (a,b)+(c,d)=(a*d+b*c,b*d). If >the representation is constrained to be in lowest terms with positive >denominator, the mapping from rational numbers to representations is >bijective. Now a few questions: - Does there exist a finitary computational model of the algebraic >numbers, such that there exists a bijection between mathematical >algebraic numbers and their model, and there exist terminating >algorithms for addition, subtraction, multiplication, division, and >comparison? To clarify finitary here, I mean a representation such >that if you start with a finite set of rational numbers and perform a >finite sequence of field operations and polynomial-solving operations >on them, then the result is guaranteed to be representable in finite >space. - Does there exist a halting algorithm that, for every finite set of >polynomial coefficients (a0+a1*x+a2*x^2+..=0), can determine all of >its solutions in the algebraic numbers in such a representation? > I think the answer to all of the above questions is yes. The KANT system, which can be accessed through Magma, allows you to create and compute in algebraic number fields (and also within their rings of integers). You have to bear in mind that the roots of an irreducible polynomial, like x^2 -5 over Q are conjugate under the automorphism group of the algebraic numbers A, and hence are indistinguishable over A. A computation within KANT would always take place within a particular finite extension of Q, which might be constructed as a sequence of simple extensions. To do exactly what you are asking for, I think you would need to first decide on a particular enumeration of the irreducible polynomials over Q. You would then insist that you were going to always contruct your finite algebraic extensions of Q by adjoining the roots of these equations in the order specified. That way, you could perform any calcualtion involving algebraic numbers in a well-defined manner. Derek Holt. ==== > decide on a particular enumeration of the irreducible polynomials over Q. do you have one in mind? er.. one that is reasonable to compute with? That is you -could- just enumerate tuples of integer coeffs (reasonably easy) reduce the poly (also reasonable) and throw out ones you've already seen (-not- reasonable). Is there a method for enumertaing them similar to enumerating the rationals via Farey sequences (or the related Stern-Brocot tree)? -- Mitch Harris Lehrstuhl fuer Automatentheorie, Fakultaet Informatik Technische Universitaet Dresden, Deutschland http://tcs.inf.tu-dresden.de/~harris ==== > decide on a particular enumeration of the irreducible polynomials over Q. do you have one in mind? >er.. one that is reasonable to compute with? >That is you -could- just enumerate tuples of integer coeffs (reasonably >easy) reduce the poly (also reasonable) and throw out ones you've >already seen (-not- reasonable). I don't think the particular enumeration is very important in terms of practical computation. The way I described things, for every irreducible polynomial encountered, you explicitly construct its splitting field over the extension of Q that you have constructed so far. As far as practical computation is concerned, this extension building process will grind to a halt long before the problem of discarding irreducibles you have seen already becomes noticeable. In practice is it only possible to compute using extensions of reasonably small degree over Q. So I guess this procedure for computing in A is notional, and is not recommended for real computation. Real computations would always be carried out in the smallest possible extension of Q in which the particular algebraic numbers lie. >Is there a method for enumertaing them similar to enumerating the >rationals via Farey sequences (or the related Stern-Brocot tree)? I don't know, but probably not! Derek Holt. >-- >Mitch Harris >Lehrstuhl fuer Automatentheorie, Fakultaet Informatik >Technische Universitaet Dresden, Deutschland >http://tcs.inf.tu-dresden.de/~harris > ==== >> decide on a particular enumeration of the irreducible polynomials over Q. >do you have one in mind? >er.. one that is reasonable to compute with? >That is you -could- just enumerate tuples of integer coeffs (reasonably >easy) reduce the poly (also reasonable) and throw out ones you've >already seen (-not- reasonable). > > I don't think the particular enumeration is very important in terms of > practical computation. The enumeration is needed only when you want to model *all* of the algebraic numbers at once. Is that correct? > The way I described things, for every irreducible polynomial > encountered, you explicitly construct its splitting field over the > extension of Q that you have constructed so far. The hard part is to find the irreducible factors of a polynomial with coefficients in a finite extension of Q. Newton knew how to find the irreducible factors of a polynomial with coeffients in Q (although not with an efficient method). In the 1980's there appeared several papers on how to do the more general problem (Zassenhaus, Berkelamp, others). Note that when you have an irreducible polynomial p(X) and adjoin a root r to the base field, then you need to find the irreducible factors of p(X) / (X - r) to continue on. > As far as > practical computation is concerned, this extension building process > will grind to a halt long before the problem of discarding irreducibles > you have seen already becomes noticeable. > > In practice is it only possible to compute using extensions of reasonably > small degree over Q. You didn't specify how small small is. But certain polynomials are easier than others. For example, X^p - 1 might be reasonalbe up to primes less than or equal to 257. This is so because such X^p - 1 split completely when you add just one non-trivial root. > So I guess this procedure for computing in A is > notional, and is not recommended for real computation. Real computations > would always be carried out in the smallest possible extension of Q > in which the particular algebraic numbers lie. poster's idea of starting with just the integers and building up everything from there, even the rational numbers as being ordered pairs of integers. The Scheme interpreter was implemented in Java, so the program was not too fast. I recall that X^19 - 1 took several minutes to computer the model of the algebraic integers for the spitting field of X^19 - 1. My goal was to solve equations like X^7 - 1 in terms of radicals. -- Bill Hale ==== I didn't know what to write for the subject, I hope it's understandable. My problem ist, that I have 8 points Mn(x,y,z) that describe a cuboid. And then the 8 points get shifted each in another direction, thus I get the points Rn(x,y,z). These Points all are known. Now, I want to translate any point Pm inside the cuboid Mn into the space of Rn using a quantified linear combination. (Don't know how to say this in English.) What I mean is, that a point of Mn that is closer to Pm has more influence than one that's further away. Can anyone help? -- -Gernot In order to reply, revert my forename. ________________________________________ Looking for a good game? Do it yourself! GLBasic - you can do www.GLBasic.com ==== Not sure that I have understood the question, and I am sure there are better solutions to the one I offer, but just to get the ball rolling.... Each corner of the cuboid (x_i,y_i,_z_i) get shifted in their own separate way so the end shape is no longer cuboid. so new corners are given by x'_i = x_i + b_i, i=1 to 8 and same for y and z. So there are 24 b_i s let d_i be the distances from some point (X,Y,Z) in original space to corners of cuboid. Let D be their average. Then new X'=X + B and also for Y and Z. Need some algorithm for B Could take B=Sum (i= 1 to 8) (b_i * (1 - d_i/D)) This at least has the property that a point near a corner stays near a corner, but there are other properties that it may lack like the point staying inside the shape. hope this helps I didn't know what to write for the subject, I hope it's understandable. > My problem ist, that I have 8 points Mn(x,y,z) that describe a cuboid. And > then the 8 points get shifted each in another direction, thus I get the > points Rn(x,y,z). These Points all are known. > Now, I want to translate any point Pm inside the cuboid Mn into the space of > Rn using a quantified linear combination. (Don't know how to say this in > English.) What I mean is, that a point of Mn that is closer to Pm has more > influence than one that's further away. Can anyone help? -- > -Gernot In order to reply, revert my forename. > ________________________________________ > Looking for a good game? Do it yourself! > GLBasic - you can do > www.GLBasic.com ==== >If S=set and P(S)=collection of all subsets of S, >then: there is no 1-1 corespondence between S and P(S). This is Cantor's Theorem. The usual proof is to take an arbitrary > function f:S->P(S), and show it is not surjective; this works for > finite and infinite sets. > Here's a generalization. surjection f:S -> A^S ==> |A| <= 1 > Given f, consider A = {x in S : x is not a member of f(x)} Note that f(x) is a subset of S, so it makes sense to ask whether x is > a member of f(x) or not. (1) Prove that A is a subset of S (easy). > (2) Prove that for every y in S, f(y) is not equal to A (hint: try > (arguing by contradiction) [hardest part of the proof] > (3) Explain why (1) & (2) prove that there can be no 1-1 > correspondence between S and P(S). > ==== > at 09:50 AM, Aatu Koskensilta said: > > >>No, because if it did you could well-order any uncountable set. > > > How? Remember, I is countable. Oh. Sorry. What I said was incorrect. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus ==== A few years ago, I programmed a Matrix QR decomposition algorithm to solve systems of linear equations. It worked for complex coefficients as well, until I came across this matrix: [1 0] [i 1] which cannot be decomposed although the description of QR algorithm says it's always possible. Since then, I sometimes wander if there is any trick to overcome this? Quick check tells that the matrix is invertible and LU decomposition is possible. The problem cannot be detected by eigenvalues, since the second column can be of your choice and thus eigenvalues can be any. What is the most numerically stable way of inverting a matrix that is *always* possible if the inverse exists? (I tried SVD as well but I don't have any math program to run it directly and the source codes on the internet are too complicated..., but it seems to have problems as well). Milan Vandrovec ==== Ahoj Milane, I don't understand the details of a complex QR algorithm, but here goes what I obtained in Octave (and the same result in Matlab): octave:5> A = [1 0;i 1] A = 1 + 0i 0 + 0i 0 + 1i 1 + 0i octave:6> [Q,R,P] = qr(A) Q = -0.70711 + 0.00000i 0.00000 + 0.70711i 0.00000 - 0.70711i 0.70711 + 0.00000i R = -1.41421 + 0.00000i 0.00000 + 0.70711i 0.00000 + 0.00000i 0.70711 + 0.00000i P = 1 0 0 1 Zdenek Hurak > > A few years ago, I programmed a Matrix QR decomposition algorithm to solve > systems of linear equations. It worked for complex coefficients as well, > until I came across this matrix: > [1 0] > [i 1] > which cannot be decomposed although the description of QR algorithm says > it's always possible. > > Since then, I sometimes wander if there is any trick to overcome this? > Quick check tells that the matrix is invertible and LU decomposition is > possible. The problem cannot be detected by eigenvalues, since the second > column can be of your choice and thus eigenvalues can be any. > > What is the most numerically stable way of inverting a matrix that is > *always* possible if the inverse exists? (I tried SVD as well but I don't > have any math program to run it directly and the source codes on the > internet are too complicated..., but it seems to have problems as well). > > > Milan Vandrovec ==== > Q = -0.70711 + 0.00000i 0.00000 + 0.70711i > 0.00000 - 0.70711i 0.70711 + 0.00000i I have forgotten to replace transposition in R with transposition-conjugate in C. I believe I will be able to fix the algorithm now. Milan Vandrovec ==== all helpful for my problem. Best Axel ==== it turns out indeed, that the problem can be solved for arbitrary p. Applying Eulers formula, one gets e.g. sin(p*arctan(x))=1/(2i) * ( (1+ik)^p - (1-ik^p) ) / (1+k^2)^{p/2} where i denotes the imaginary unit. Similar results are obtained for the tan- and cos-case. Axel ==== > sin(p*arctan(x))=1/(2i) * ( (1+ix)^p - (1-ix)^p ) / (1+x^2)^{p/2} But if one wants to do an actual computation with non integer p, one will still have to resort to trigonometric functions. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== > it turns out indeed, that the problem can be solved > for arbitrary p. Applying Eulers formula, > one gets e.g. Just for the record, there are two types of typographical errors below. 1) On the right, k should be x instead. 2) A power is misplaced: Instead of (1-ix^p), it should be (1-ix)^p. > sin(p*arctan(x))=1/(2i) * ( (1+ik)^p - (1-ik^p) ) / (1+k^2)^{p/2} David ==== > Just for the record, there are two types of typographical errors below. > 1) On the right, k should be x instead. > 2) A power is misplaced: > Instead of (1-ix^p), it should be (1-ix)^p. > > >>sin(p*arctan(x))=1/(2i) * ( (1+ik)^p - (1-ik^p) ) / (1+k^2)^{p/2} yes, thanks: sin(p*arctan(x))=1/(2i) * ( (1+ix)^p - (1-ix)^p ) / (1+x^2)^{p/2} ==== >I'm not much up on the current rage of category theory. > To understand this, consider the following... Ooops, forgot to say - thanks for the answer!! ==== >> M proper subset M / { x | xMa }x{ y | bMy } in P, which cannot be >Since a,b are arbitrary incomparable elements, we can conclude that >any order is the intersection of the total orders that extend it. > That's so slick it goes over my head. Would you clarify? Don't we have to show the intersection of total orders isn't empty? ---- ==== I'll call the empty set E, since I don't know a better way to represent it in ASCII and I'll call the power set of a set A, P(A). Is the following correct? P(E) = {E} P(P(E)) = { E, {E} } P(P(P(E))) = { E, {E}, {{E}}, {E, {E}} } P(P(P(P(E)))) = { ... } will have 2^4 = 16 members (?) P(P(P(P(P(E))))) = { ... } will have 2^(2^4) = 2^16 = 65536 members (?) My questions are related to an exercise in Elements of Set Theory, Chapter 1. Question 7: List all the members of V_3. List all the members of V_4. (It is to be assumed here that there are no atoms.) V_n is defined to be V_n+1 = P(V_n). And V_0 = E. If my observations about the number of members resulting from repeated applications of power set function are correct, then I can conclude V_4 has over 65,000 members... Ok - so it's obvious I've gone wrong somewhere. ==== > I'll call the empty set E, since I don't know a better way to > represent it in ASCII and I'll call the power set of a set A, P(A). Is the following correct? P(E) = {E} > P(P(E)) = { E, {E} } > P(P(P(E))) = { E, {E}, {{E}}, {E, {E}} } > P(P(P(P(E)))) = { ... } will have 2^4 = 16 members (?) > P(P(P(P(P(E))))) = { ... } will have 2^(2^4) = 2^16 = 65536 members (?) But this is V_5 (using the notation from below). > My questions are related to an exercise in Elements of Set Theory, > Chapter 1. Question 7: List all the members of V_3. List all the members of V_4. > (It is to be assumed here that there are no atoms.) V_n is defined to be V_n+1 = P(V_n). And V_0 = E. If my observations about the number of members resulting from > repeated applications of power set function are correct, then I > can conclude V_4 has over 65,000 members... Ok - so it's obvious I've gone wrong somewhere. > V_0 has 0 elements. V_1 has 2^0=1 element. V_2 has 2^1=2 elements. V_3 has 2^2=4 elements. V_4 has 2^4=16 elements. -- Clive Tooth http://www.clivetooth.dk ==== > I'll call the empty set E, since I don't know a better way to > represent it in ASCII and I'll call the power set of a set A, P(A). > Is the following correct? > P(E) = {E} > P(P(E)) = { E, {E} } > P(P(P(E))) = { E, {E}, {{E}}, {E, {E}} } > P(P(P(P(E)))) = { ... } will have 2^4 = 16 members (?) > P(P(P(P(P(E))))) = { ... } will have 2^(2^4) = 2^16 = 65536 members (?) But this is V_5 (using the notation from below). > My questions are related to an exercise in Elements of Set Theory, > Chapter 1. Question 7: > List all the members of V_3. List all the members of V_4. > (It is to be assumed here that there are no atoms.) > V_n is defined to be V_n+1 = P(V_n). And V_0 = E. > If my observations about the number of members resulting from > repeated applications of power set function are correct, then I > can conclude V_4 has over 65,000 members... > Ok - so it's obvious I've gone wrong somewhere. > > V_0 has 0 elements. > V_1 has 2^0=1 element. > V_2 has 2^1=2 elements. > V_3 has 2^2=4 elements. > V_4 has 2^4=16 elements. -- > Clive Tooth > http://www.clivetooth.dk ==== > Oops... the equation didn't come out right. Let me retype it out in words: > > dX = Kappa*(Miu - X)dt + sigma*X^gammadW This process is mean reverting to Miu. If Miu is zero then it will be absorbing as the stochastic term will dissapear. If Miu is not zero then its not absorbing (which I guess means reflecting ?) as the stochastic term will knock it away from the mean. hth Peter. ==== > >>Oops... the equation didn't come out right. Let me retype it out in words: >>dX = Kappa*(Miu - X)dt + sigma*X^gammadW > > > This process is mean reverting to Miu. If Miu is zero > then it will be absorbing as the stochastic term will dissapear. > If Miu is not zero then its not absorbing (which I guess means reflecting ?) > as the stochastic term will knock it away from the mean. Could you maybe explain your statements some more? I'm interested :-) > > hth > Peter. Nijmegen, Netherlands ==== 2nd appeal... please try to help. ==== > 2nd appeal... please try to help. OK, some thoughts again then... Note that it's a bit of handwaving and I haven't checked the calculations. If you look at the process I provided and you write the SDE in integral form you can obtain the expectation and the variance of r(t)-r(0) (by defining m(t)=E(r(t)-r(0)) en solving the equation for m(t)). The increments of r(t) have a normal distribution, so then you have the complete distribution. And you can see that the mean tends to b/a. term made me wonder whether the increments of the process are now still look it up in a textbook if it would make any difference, but I haven't gotten around to that yet. *If* it is the case that the resulting process still has normal increments, than you could *maybe* apply the same technique as I did above to determine the mean and variance of this normal distribution. But I think the X^gamma will give rise to computational problems again. If you can do it (determine that the increments are normal, what the expectation and the variance are) though, then you can determine the level to which the process is reverting. Another possibility is to determine the moment-generating function of X(t), but here also the X^gamma gives a problem I think. Maybe someone who knows by heart what effect the X^gamma in the diffusion term has, could tell it here. Otherwise I'll try to look it up. Another thought I have on this is maybe to make a substitution. A certain conditional expectation of the solution of the SDE you mention is the solution to a particular partial differential equation. These are linked via the Feynman-Kac theorem. Maybe the PDE is more easy to handle in terms of computability. Maybe a substitution (something like exp{X^gamma} or so) in this PDE to get rid of the X^gamma (which will come into the PDE somehow) will work. But I wouldn't know how and I have bad hopes for a substitution to work. But in any case the PDE might be numerically tractable. A final thought is to apply numerical procedures to the SDE. I have no experience in that area, so just a mere thought. Nijmegen, Netherlands ==== > > Forget math/science types, what about the whole population of humans. I have never met a single person who does not derive a lot of pleasure from at least some kind of music. False. Professor Andrei Zavrotzky, told me that for him music was mere noise. Luis Rodriguez ==== >> >> Forget math/science types, what about the whole population of humans. I >> have never met a single person who does not derive a lot of pleasure from >> at least some kind of music. > > False. Not necessarily, if the previous poster had never met Prof Zavrotzky. > Professor Andrei Zavrotzky, told me that for him music was mere > noise. Luis Rodriguez -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== I can agree with mathematics being a science, or a tool of science, but >the standard methods don't always line up. My objections is not to math >being a science, but to the notion that math can be done by the >scientific method. JSH appears to use experimental evidence as his >proof, because he doesn't see how math formulas can display widely >varying behaviors after very small changes in the parameters. He isn't saying so in this round, but in the past he has claimed that all the fallacious proofs ARE his experiments, and are his version of the scientific method. You try a proof, you see if it succeeds, you try another proof. If sooner or later you get a proof accepted, you've done math by the scientific method. - Randy ==== > >>I can agree with mathematics being a science, or a tool of science, but >>the standard methods don't always line up. My objections is not to math >>being a science, but to the notion that math can be done by the >>scientific method. JSH appears to use experimental evidence as his >>proof, because he doesn't see how math formulas can display widely >>varying behaviors after very small changes in the parameters. > > > He isn't saying so in this round, but in the past he has claimed that > all the fallacious proofs ARE his experiments, and are his version of > the scientific method. You try a proof, you see if it succeeds, you > try another proof. If sooner or later you get a proof accepted, > you've done math by the scientific method. > > - Randy > If that's what he means, he isn't doing anything new. Attempt a proof, look for holes, patch holes, look for more holes, patch, etc until there are no holes. -- Will Twentyman ==== > > I can agree with mathematics being a science, or a tool of science, > but the standard methods don't always line up. My objections is not > to math being a science, but to the notion that math can be done by > the scientific method. JSH appears to use experimental evidence as > his proof, because he doesn't see how math formulas can display > widely varying behaviors after very small changes in the parameters. >> He isn't saying so in this round, but in the past he has claimed that >> all the fallacious proofs ARE his experiments, and are his version of >> the scientific method. You try a proof, you see if it succeeds, you >> try another proof. If sooner or later you get a proof accepted, >> you've done math by the scientific method. >> - Randy > > If that's what he means, he isn't doing anything new. Attempt a proof, > look for holes, patch holes, look for more holes, patch, etc until there > are no holes. > But you forgot the most important steps: 1. Insist that those claiming holes are not to be believed: they are lying, confused, inept, corrupt, or engaged in a conspiracy to silence him. 2. Threaten lawsuit, congressional subcommittee meetings, the downfall of the corrupt mathematical establishment. 3. Invoke The Hammer. 4. Inform the FBI. 5. Oh, well. It wasn't important anyway. Dale. ==== > > > > I can agree with mathematics being a science, or a tool of science, >> but the standard methods don't always line up. My objections is not >> to math being a science, but to the notion that math can be done by >> the scientific method. JSH appears to use experimental evidence as >> his proof, because he doesn't see how math formulas can display >> widely varying behaviors after very small changes in the parameters. > He isn't saying so in this round, but in the past he has claimed that > all the fallacious proofs ARE his experiments, and are his version of > the scientific method. You try a proof, you see if it succeeds, you > try another proof. If sooner or later you get a proof accepted, > you've done math by the scientific method. >> - Randy > If that's what he means, he isn't doing anything new. Attempt a >> proof, look for holes, patch holes, look for more holes, patch, etc >> until there are no holes. > > But you forgot the most important steps: > > 1. Insist that those claiming holes are not to be believed: > they are lying, confused, inept, corrupt, or engaged in > a conspiracy to silence him. > > 2. Threaten lawsuit, congressional subcommittee meetings, > the downfall of the corrupt mathematical establishment. > > 3. Invoke The Hammer. > > 4. Inform the FBI. > > 5. Oh, well. It wasn't important anyway. > > Dale. > Are you suggesting that there are new steps in the Scientific Method? Does James own that now too? -- Will Twentyman ==== > >Facing vigorous opposition to my warnings about a problem with an >esoteric area in mathematics, I make this post to show how the >scientific method can be applied to what turns out to be a question in >a complex system. >>The scientific method is not math. Period. If you use the scientific >>method to draw conclusions in math, any correct conclusions will be a >>result of pure luck. > > I wonder how the latter statement can be reconciled with the uncertain > foundations of mathematics cf G.9adel's undecidability results. Uhm. What's so uncertain here? G.9adel's 1. incompleteness theorem simplys prove that the set of arithmetical truths is not recursively enumerable and that there is a primitive recursive function G, s.t. for any recursively enumerable axiomatisation A of a theory capable of representing certain primitive recursive functions, the sentence G(A) is not provable from A, yet N |= G(A) (G(A) is true about the natural numbers). > If the > axioms of arithemetic are uncertain surely results derived therefrom are > also uncertain. I'm not attempting to support the original poster, but > it seems wrong to be so dogmatic about an un-provable system. The axioms of arithmetics are not uncertain. There is a second order axiomatisation of arithmetics that has the natural numbers as its only model (modulo isomorphism). -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus <3f49662f$1_2@newsfeed> <585ab5d8.0308260809.18ae3351@posting.google.com> ==== ...... At the level of a talk for undergraduates the result is that you >can propose a proposition that you can neither prove nor disprove. >At that point you are free to have a larger system with that >proposition as a new axiom which may be either asserted or denied >by the new axiom. The axiom of choice was eventually proven to >be independent of the normal Zormelo-Frankel axioms and lead to >systems which either had the axiom of choice true or had it denied. >problems was answered negatively by the Godel proof. Most >interesting mathematics had assumed AoC so whether it was a >theorem under Z-F or an axiom was a minor technical issue at best. > I rather naively imagined that the 'problem' was slightly deeper so that it applied to all such systems ie that even after adding each new 'axiom' there would remain undecidable propositions. If that were the case there could be infinitely many ways to construct particular proofs. Hopefully (Z-F+A)+B is identical to (Z-F+B)+A, but I'm a bit stupid and don't understand axiomatic independence. I have heard of the experimental mathematical approach has it arisen because of computers or because of the difficulty at the cutting edge? >>In fact I would characterize the certainty of the axioms >>as one of the most essential differences between science >>and mathematics. In science, your axioms are only guesses >>and there is always Nature out there who might tell you that >>you guessed wrong. In mathematics, there is no out there. >>The universe is built entirely from your axioms. The axioms >>can't possibly be wrong so long as they're self-consistent. Some would say that all axioms are permissible except some >will lead to systems with no content, a situation which might >be summarized as not self-consistent axioms. > - Randy -- Robin Becker ==== ...... >>At the level of a talk for undergraduates the result is that you >>can propose a proposition that you can neither prove nor disprove. >>At that point you are free to have a larger system with that >>proposition as a new axiom which may be either asserted or denied >>by the new axiom. The axiom of choice was eventually proven to >>be independent of the normal Zormelo-Frankel axioms and lead to >>systems which either had the axiom of choice true or had it denied. >>problems was answered negatively by the Godel proof. Most >>interesting mathematics had assumed AoC so whether it was a >>theorem under Z-F or an axiom was a minor technical issue at best. I rather naively imagined that the 'problem' was slightly deeper so that >it applied to all such systems ie that even after adding each new >'axiom' there would remain undecidable propositions. Precisely! In the new system there will be some new proposition that can be neither proven nor disproved. And so on and so on ... > If that were the >case there could be infinitely many ways to construct particular proofs. >Hopefully (Z-F+A)+B is identical to (Z-F+B)+A, but I'm a bit stupid and >don't understand axiomatic independence. There would be finitely many axioms, the propositions would involve a finite number of symbols and be proven in a finite number of steps so inifinities would not be present. There are always alternate proofs but they are also finite. Mechanical proofs are often not very understandable to humans. Technical questions about what is a miinimal proof etc are the guist of topics in either Logic or Computer Science. >I have heard of the experimental mathematical approach has it arisen >because of computers or because of the difficulty at the cutting edge? >>In fact I would characterize the certainty of the axioms >as one of the most essential differences between science >and mathematics. In science, your axioms are only guesses >and there is always Nature out there who might tell you that >you guessed wrong. In mathematics, there is no out there. >The universe is built entirely from your axioms. The axioms >can't possibly be wrong so long as they're self-consistent. >>Some would say that all axioms are permissible except some >>will lead to systems with no content, a situation which might >>be summarized as not self-consistent axioms. > - Randy -- >Robin Becker ==== After a discussion of the significance of incremental increases in athletic prowess I checked out the following URL for world records in the men's 100m event http://multimedia.olympic.org/pdf/en_report_70.pdf Looking at the data it's clear that despite progress in health/diet/sports science in postwar years the amount by which the record is changing is reducing. It's an interesting question, I think at least, to say, is there a time below which human beings will not be able to run the 100m? The first check I thought of doing was a Pade approximation, (courtesy of Numerical Recipes' routine) and true enough there seems to be a singularity in the date at which the world record will be under 9seconds (although breaking the 9.5s barrier will take another century or so!). However, my question is, is there a statistically rigorous way to determine a lower bound on data of this type? I guess it would have a range of very dubious applications (populations, technical advances etc.) and looks likely to have been studied at some point. Tom ==== Laurent, For numerical solutions, I would suggest LAPACK (from www.netlib.org) and LAPACK95. If you prefer C, you can use CLAPACK. You can obtain the explicit inverse using LAPACK by solving the system of equations A*X=I, where A is your Hermitian matrix. However, are you sure you really need an explicit inverse? Usually, you end up multiplying the inverse by something else, so you're better off (numerically) solving a system of equations. If you need the answer symbolically (for low order matrices), you can try Maple, Mathematica or MuPad (all cost a lot of money). Good luck, John > Is there any method that quickly inverts an hermitian matrix? Laurent > ==== I started writing a program to enumerate the k-subsets of a set {1, ..., n} and it's taking a little longer than I thought it would. I notice that a subset of a set of n elements can be represented as an n-bit word, with an on bit for that the element at that index is in the subset, and an off bit for that it is not. For example, for an eight bit word, that can represent a subset of a set of eight elements, in this case {1, ..., 8}. The set {1, ..., 8} is 11111111 or in hexadecimal notation 0xFF. The set {1} would be 10000000, {8} would be 00000001, with the empty set {} being 00000000. Anyways, I have started writing a recursive method to enumerate the subsets, and it's a hassle to write and has quite a few expressions of n and k. In frustration I started typing patterns and noticed something about how the subsets may be generated with a seemingly simpler method, at least computationally. Consider a set of say, six elements, {1, 2, 3, 4, 5, 6}, and the 3-subsets of that, each subset with three elements. Start with {1, 2, 3} in the binary notation, and then shift right, carrying over the right bit back to the beginning of the sequence. 1, 1, 1, 0, 0, 0 0, 1, 1, 1, 0, 0 0, 0, 1, 1, 1, 0 0, 0, 0, 1, 1, 1 1, 0, 0, 0, 1, 1 1, 1, 0, 0, 0, 1 Then, I start with {1, 3, 4}, having shifted the bits after the first one bit right. Then I repeat the right-shift with wraparound n=6 times again. 1, 0, 1, 1, 0, 0 0, 1, 0, 1, 1, 0 0, 0, 1, 0, 1, 1 1, 0, 0, 1, 0, 1 1, 1, 0, 0, 1, 0 0, 1, 1, 0, 0, 1 Next is {1, 4, 5}, shifted six times. 1, 0, 0, 1, 1, 0 0, 1, 0, 0, 1, 1 1, 0, 1, 0, 0, 1 1, 1, 0, 1, 0, 0 0, 1, 1, 0, 1, 0 0, 0, 1, 1, 0, 1 Finally there is {1, 3, 5}, there have been eighteen sequences representing different subsets thus far, two more will make the remaining 3-subsets. 1, 0, 1, 0, 1, 0 0, 1, 0, 1, 0, 1 I thought I had stumbled upon some incredibly simple method to enumerate the k-subsets of a set. Yet, it wasn't immediately clear how it would work with other values of n and k, besides the fact that it is very easy to implement as a machine. The 1-subsets of {1, ..., 6} are simple enough: 1, 0, 0, 0, 0, 0 0, 1, 0, 0, 0, 0 0, 0, 1, 0, 0, 0 0, 0, 0, 1, 0, 0 0, 0, 0, 0, 1, 0 0, 0, 0, 0, 0, 1 That is {1}, {2}, {3}, {4}, {5}, {6}. The 2-subsets of {1, ..., 6} are not quite so simple. There are 6!/4!2! many of them, or 6*5/2*1=15: 1, 1, 0, 0, 0, 0 0, 1, 1, 0, 0, 0 0, 0, 1, 1, 0, 0 0, 0, 0, 1, 1, 0 0, 0, 0, 0, 1, 1 1, 0, 0, 0, 0, 1 1, 0, 1, 0, 0, 0 0, 1, 0, 1, 0, 0 0, 0, 1, 0, 1, 0 0, 0, 0, 1, 0, 1 1, 0, 0, 0, 1, 0 0, 1, 0, 0, 0, 1 1, 0, 0, 1, 0, 0 0, 1, 0, 0, 1, 0 0, 0, 1, 0, 0, 1 Very good. It seems that there is a simple pattern to generate sequences representing the subset elements. Yet, this is so far only for values of n and k such that n is around 2k, and that k is a factor of n, for small n. I guess I should implement this and see if it works. Consider the 4-subsets of {1, ..., 6}. There are just as many 4-subsets of six as there are 2-subsets of six, 6!/2!4! = 6!/4!2!. Those sequences above are the 2-subsets of 6, their complements are the 4-subsets of six. Thus I can consider only the cases where k <= n/2. The process generates the k-subsets and (n-k)-subsets at about the same time, it is a matter of a complement operation on each. Of course, my goal here is to sum the products of those subsets, the 4-subsets would have higher products than the 2-subsets. Given one of these sequences, I wonder if there's a way to quickly calculate the product of the elements that it represents. The bit sequence up to word size is typically stored on a machine register in a computer program. I can see some problems with this method for higher values of n. The size of n isn't a problem, it's when k is small compared to n, generating the starter sequences requires its own algorithm. It's when k is relatively prime compared to n that this is more complex, when k is a factor of n it is more simple. For example, consider the 3-subsets of {1, ..., 8}. There would be 8!/5!3! of them, 8*7*6/3*2 = 56. These would be the starter sequences with 8 many each: 1, 1, 1, 0, 0, 0, 0, 0 1, 0, 1, 1, 0, 0, 0, 0 1, 0, 0, 1, 1, 0, 0, 0 1, 0, 0, 0, 1, 1, 0, 0 1, 0, 0, 0, 0, 1, 1, 0 That's 40 of them, then there are 16 more. 1, 0, 1, 0, 1, 0, 0, 0 1, 0, 0, 1, 0, 0, 1, 0 Hmm..., well there they are, those two sequences each generate eight subsets. That's different from the case where the sequence 0, 1, 0, 1, ... generates only two subsets. I wonder how it is different for odd n. How many starter sequences are there for k many sequences apiece? I'll call them the blocked starter sequences. Let's look at that: for n=6, k=1 there is one blocked sequence, for n=6, k=2 there are two blocked sequences, for n=6, k=3 there are 3. For n=8, k=1 there is one, n=8, k=2, there could be 6 except {1, 3} is considered under the alternating starter sequence, it would have to be not considered. It should be a blocked starter sequence and not repeated in the alternating starter sequence. For n=8, k=3 there are 5, and for n=8, k=4, there is four. Then there are the alternating starter sequences, there is a given number of them based upon n and k and whether k is a factor of n, and then they generate a given number of sequences based upon n!/k!(n-k)! - k(n-k) and the number of sequences. Here I'm going to work with k < n/2, so the function there is about the greatest integer x thus that k/n < 1/x. For example for n=8, k=3, 1/3 < 3/8 < 1/2, x =2, there are 2 alternating starter sequences. I guess it's easier to compare n/k to an integer, 8/3 in integer math will give the answer of x=2. Then, to generate the alternating starter sequences there is 1, 0, 1, 0, ..., spacing the on bits by one off bit, then 1, 0, 0, 1, 0, 0, etctera, the sequence yn+1 for postive integer y <= x, i.e. {1, 3, 5, ...}, {1, 4, 7, ...}, {1, 5, 9, ...}, each of those. For example in the above description for k=3 and n=8, x =2 and thus for y=1 and y=2 the starter sequences are for {1, 3, 5} and {1, 4, 7}. Why is it that there are k many subsets from alternating starter sequences where k is not a factor of n and n/k many for k being a factor of n? Maybe it is more apt to say that as k divides into n that there are only n/k many subsets from the alternating starter sequence. In a way the k-subsets relate back to the concept of the canonical sequence. This is where the canonical sequence for k=n/2 is 101010..., and the other k-subset's sequences canonicalize to that. For n=8, there are more starting sequences. This is where there are enough sequence elements to separate blocks of multiple elements. For n=10, k=5, there are 10!/5!5! or 10*9*8*7*6/5*4*3*2*1 = 2*3*2*7*3=252 5-subsets of {1, ..., 10}. 1, 1, 1, 1, 1, 0, 0, 0, 0, 0 1, 0, 1, 1, 1, 1, 0, 0, 0, 0 1, 0, 0, 1, 1, 1, 1, 0, 0, 0 1, 0, 0, 0, 1, 1, 1, 1, 0, 0 1, 0, 0, 0, 0, 1, 1, 1, 1, 0 1, 1, 0, 1, 1, 1, 0, 0, 0, 0 1, 1, 0, 0, 1, 1, 1, 0, 0, 0 1, 1, 0, 0, 0, 1, 1, 1, 0, 0 1, 1, 0, 0, 0, 0, 1, 1, 1, 0 1, 1, 0, 1, 1, 0, 1, 0, 0, 0 1, 1, 0, 1, 1, 0, 0, 1, 0, 0 1, 1, 0, 1, 1, 0, 0, 0, 1, 0 1, 1, 0, 1, 0, 1, 1, 0, 0, 0 1, 1, 0, 1, 0, 1, 0, 1, 0, 0 1, 1, 0, 1, 0, 1, 0, 0, 1, 0 1, 1, 0, 1, 0, 0, 1, 1, 0, 0 1, 1, 0, 0, 1, 1, 0, 1, 0, 0 1, 1, 0, 0, 1, 1, 0, 0, 1, 0 1, 1, 0, 0, 0, 1, 1, 0, 1, 0 1, 0, 1, 0, 1, 1, 1, 0, 0, 0 1, 0, 1, 0, 0, 1, 1, 1, 0, 0 1, 0, 1, 0, 0, 0, 1, 1, 1, 0 1, 0, 0, 1, 0, 1, 1, 1, 0, 0 1, 0, 0, 1, 0, 0, 1, 1, 1, 0 1, 0, 0, 0, 1, 0, 1, 1, 1, 0 Those 25 sequences, if correct, rotate to each form ten sequences forming a total of 250 sequences representing 5-subsets of 10. With those blocked starting sequences there is the alternating sequence, it generates 10/5 = 2 5-subsets: 1, 0, 1, 0, 1, 0, 1, 0, 1, 0 At that point again it is about as not simple as the regular recursive method to enumerate the subsets. Yet, it is much more computationally efficient. Figuring this out would probably help me determine how the product lists as discussed earlier would progress. I thought to look at discussion of Gray codes, they are progressions of sequences, but their use is not immediately apparent here as the sequences here have fixed numbers of on and off bits. The Hamming distance is the minimum number of bits to be changed to convert one sequence to another. What's the regular way to enumerate the sequences with a given number of on bits that are each different on rotation? I figure the easiest way to generate the byte arrays of each rotation of a starter sequence will be to put two copies of the starter sequence into a sequence of bits, and then to extract subsequences. For example, for the starter sequence 1, 0, 0 put into memory 1, 0, 0, 1, 0, 0 and then extract the three subsequences of length three starting at the index from one to three: [ 1, 0, 0, ] 1, 0, 0 -> 1, 0, 0 1, [ 0, 0, 1, ] 0, 0 -> 0, 0, 1 1, 0, [ 0, 1, 0, ] 0 -> 0, 1, 0 This is similar to having the sequence being written about a circle, and then reading off a sequence of that length from each element, beads on a string. Some machine architectures might have instructions to rotate a register. I was reading MathWorld and notice that Mathematica has a KSubsets method to list the k-subsets of a set, I'm still in the process of implementing it. The other day I was reading the documentation of HYPERG, a Hypergeometric function package for Maple, based on HYP. Today I'll read the documentation of HYP. Heh, bait. Ross ==== > I started writing a program to enumerate the k-subsets of a set {1, > ..., n} and it's taking a little longer than I thought it would. If Ross means subsets with k elements, he should try the binomial coefficient, C(n,k). If he does not mean that, he should explain up front what he does mean. ==== >it seems to me that a quaternion is nothing more than a simple >ordered quadruplet. And that multiplication/addition is simply >extended in such a way that it is defined over such ordered >quadruplets. It's as if I were to write a geometry book where I >called points binarinions and write them as a#{(b}) instead of >(a,b).. the point being that it would be arbitrary and accomplish >nothing and, in fact, greatly obfuscate something for no reason. >This, it seems, is the entire nature of quaternions. IIRC, I used quaternions once in a computer graphics program that needed to do rotations of objects in 3D, and it simplified the program nicely. ==== >>is a group of transformations of a vector space over the quaternions: >>in your notation > >>Sp(2n) = {T in GL(n,H): T*T = I} > >>Symplectic structure. it seems to me, arises if and only if >>one has an underlying vector space over the quaternions. > > Only if one also demands that the matrices are unitary. > The above group is automatically constrained to be compact. > And compactness and the fact that the group is quaternionic > *together* imply that the group is symplectic. You seem to enjoy making life more complex (pun intended). It seems to me much simpler to use the term symplectic group for the groups Sp(n) (or Sp(2n)), just as it is simpler to use the term unitary group for the groups U(n). You might as well say that unitary has nothing to do with complex numbers because there are quasi-unitary groups over finite fields. As far as I can see, whenever symplectic structure is important (as in symplectic geometry) one can in fact express the structure in terms of quaternions. -- Timothy Murphy tel: +353-86-233 6090 ==== >is a group of transformations of a vector space over the quaternions: >in your notation >> >Sp(2n) = {T in GL(n,H): T*T = I} >> >Symplectic structure. it seems to me, arises if and only if >one has an underlying vector space over the quaternions. >> >> Only if one also demands that the matrices are unitary. >> The above group is automatically constrained to be compact. >> And compactness and the fact that the group is quaternionic >> *together* imply that the group is symplectic. >You seem to enjoy making life more complex (pun intended). >It seems to me much simpler to use the term symplectic group >for the groups Sp(n) (or Sp(2n)), >just as it is simpler to use the term unitary group >for the groups U(n). But U(p,q) are the only unitary groups. Sp(2p,2q) are not the only symplectic groups. Similarly, SO(p,q) and O(p,q) (and connected components, etc, thereof) are not the only orthogonal groups. >You might as well say that unitary has nothing to do with complex numbers >because there are quasi-unitary groups over finite fields. >As far as I can see, whenever symplectic structure is important >(as in symplectic geometry) >one can in fact express the structure in terms of quaternions. With respect to symplectic geometry and symplectic manifolds, the appropriate symplectic group is generally noncompact, and is in fact generally equal to Sp(2n,R), whose elements are most simply described by matrices over the real numbers. David McAnally -------------- X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 12:02 AM, Timothy Murphy said: >Surely nobody interested in quantum theory >would worry about the Axiom of Choice? Why not, if they can't derive what they need without it? >In practice, the terms quaternionic and symplectic >seem more or less interchangeable to me. ? Complex I could understand and even sympathize with, but Quaternionic? No way. The dimensionality of a symplectic manifold need not be a multiple of 4. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org <70a76acc.0308240358.35124f3a@posting.google.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 04:58 AM, doug_protocols@yahoo.com (Doug B) said: >What I was saying >was that these results could all be proven just as easily with only >the most trivial modifications, if you skipped the whole quaternion >detour and just defined multiplication and addition of ordered >quadruplets. Sure, it's easy to see in hindsight the proper way to define multiplication of OF THE 4-touples of reals. It's not so easy to do when you're the first to try it and don't have the advantage of modern Vector Analysis, Group Theory, etc. For his time and place it was a major breakthrough. >would be just as applicable if the entire theory simply used ordered >quadruplets. Correct. >So what I meant to say is that it seems >that the quaternion apparatus merely makes the results unnecessarily >opaque. Doing it the way you want might take more machinery than you realize, and Hamilton would probably have found it to be more opaque than his way. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org ==== > > >2) given a formula in +,-,*,/, and radicals, and its computed minimal >polynomial (for which the formula is a solution), are the other real >solutions expressible similarly (as formulas in +,-,+,/,radicals)? >> Yes, the other roots (ie the conjugates of the given root) >> are obtained by replacing each nth root by the other nth roots, >> eg replacing a^{1/n} by wa^{1/n} where w is an nth root of 1. >>but is that really in the given syntax? w (=e^(2pi i/n) ) >>isn't really allowed here. none of e, i, or pi are definable (I also >>left out the restriction that powers can only be rationals). > > This is really a religious question. > You can either regard the algebraic numbers > as a subfield of the complex numbers > in which case you can set w = exp(2pi i/n) as you say; > or one can regard the algebraic numbers > as constructed from the rationals, > in which case w is one of the roots of x^n - 1. OK, maybe this is allowable symbolically. But my thoughts about it are not clear enough really to say yes or no. > In any case, if you want to look at all the roots of the equation > you will have to go outside the field Q(r) where r is your root. > > But I feel I'm just saying what you know perfectly well anyway! Maybe not perfectly. I was under the impression that, by definition, algebraics were a subset of the reals. but looking around the web it seems that sometimes it is specified as such but mostly not. And maaybe now I realize I don't care (i.e. algebraics that are not reals are OK). If any spirituality is involved, it is in my desire to have only symbolic operations involved, no numerical ones (yes, I realize that is problematic distinction, let's say no bit counting?). I think my problem may have been better stated in other threads (What is an algebraic integer? and Computational algebraic integers). -- Mitch Harris Lehrstuhl fuer Automatentheorie, Fakultaet Informatik Technische Universitaet Dresden, Deutschland http://tcs.inf.tu-dresden.de/~harris ==== >> A question please: In the phrase if and only if is the first if > necessary? >> IOW, A only if B is not the same as A if and only if B? >> Yes, both if's are significant. >> The following statements all mean exactly the same thing: >> 1. A only if B. >> 2. B if A. >> 3. If A, then B. >> 4. A is sufficient for B. >> 5. B is necessary for A. > My difficulty is with the statement 1 and I would never imagine that it is > equivalent to A implies B. I would rather say that what it means is only if B > is true is A true but I am not sure that this is a meaningful phrase in > English. That is precisely what it means. I don't understand why you think there > is a difference. Because I thought that the statement Only if a teacher has given permission is a student allowed to enter this room (from an Oxford dictionary explaining the meaning of only if) implies the statement If a teacher has given permission, a student is allowed to enter this room. It seems I was wrong and my English is worse than I thought :-) To be specific, the following statements all are equivalent: 1. A only if B. > 2. A implies B. > 3. Only if B is true is A true. > 4. A is not true unless B is true. > 5. B or not A. -- > Dave Seaman dseaman@purdue.edu > Manager, Advanced Applications YONG 517 > http://www-rcd.cc.purdue.edu/~dseaman/ FAX: (765) 494-0566 ==== Repeatedly I've given out a definition for counting prime numbers, but I'm afraid that something has not been communicated. So first I'm going to give an example, counting the number of prime numbers up to 100, and afterwards I'll give the function. In that way I hope to point clearly to a key feature of my research. What my function does is count the number of composites for each integer up to the square root of your base number. So for counting primes up to 100 the base number is 100 and its square root is 10. Here are the composite counts: dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; dS(100,9) = 0; dS(100,10)= 0; and summing the dS values gives you 74. Notice when dS equals 0. Now you add 1 for the number 1, as its not considered prime, and subtract from 100 to get 25, which is the count of primes up to 100. And those prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. Now a key feature of dS that I want to emphasize is that for dS(x,y) if y is not prime dS equals 0. I want to repeat that for emphasis, for dS(x,y) if y is not prime dS equals 0. Amazingly enough, mathematicians never discovered such a function as my dS throughout their entire history. Some posters have tried to convince you otherwise. While some apparently have tried to convince you that it doesn't matter if that's true. Basically I think many of them have just worked to confuse you. Now here's that definition I've given so many times before: dS(x,y) = [pi(x/y, y-1) - pi(y-1, sqrt(y-1))] [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))], S(x,1) = 0. And pi(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS from dS(x,1) to dS(x,y). Note: pi(x,sqrt(x)) here gives the same value as the traditional pi(x). For faster calculations you need to use dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)] [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))] when sqrt(x/y) < y-1. See http://groups.msn.com/AmateurMath/primecountingfunction.msnw If your eyes glazed over, or you found yourself wishing to quickly run away from what looks very complicated then you might understand how some posters could so easily manipulate the discussion away from the truth. The truth is that the dS function does the amazing, as you see a complete definition above, where you don't see any prime numbers except 2 but *somehow* as you noticed above dS(x,y) equals 0 when y is a prime number. You see, the function *knows* when a number is prime, making it unique. Those values from above were dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; dS(100,9) = 0; dS(100,10)= 0; and you can see the definition I gave above at work, as it picks its way through the numbers dropping to 0, whenever y is not prime. That's so huge that I can't emphasize it enough. It's also something that should give you a certain sadness because what should have happened is that mathematicians should have excitedly embraced such a result. And I think for some of you, disbelieving *me* is easier than considering even the possibility that a discipline like mathematics could have enough corruption that mathematicians would fight what they should be celebrating. But focusing on me is a mistake. If you really wish to consider human achievement as some personal issue, where you wish to make me that powerful as a single human being that the accomplishments of thousands of years of effort can be reduced to just being about the individuals involved, then you shatter the foundations of human civilization, and make it just some fashion show. The truth is that the discoveries are more important than the discoverers, and in letting it turn into some celebrity issue, you demean human society and its accomplishments. For some of you I fear that the idea that I might become a celebrity is all that matters to you. And I fear you think that's all that ever mattered through all of human history, as if the truth were just a word, and life has always just been a big popularity contest. The truth is not just a word and emphasizing truth over social issues has helped humanity to make the technological achievments which make my communication through this medium possible. Where would we be if engineers had paused to figure out who they liked and disliked before they bothered to use the information discovered? Mathematicians aren't doing you any favors. I suggest that all of you focus less on me than on the information, and forget about what benefits telling the truth will give me, as you consider that the value to society is far, far greater. So how can mathematicians behave as they have? That's an important question, and I suggest we find out why popularity has so much importance in their world. James Harris ==== > > Repeatedly I've given out a definition for counting prime numbers, [snuip] and been made an utter fool by competent folks who trivially refute your psychotic spews with concrete counterexamples that you ignore. http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg -- Uncle Al http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) ==== <... Now a key feature of dS that I want to emphasize is that for dS(x,y) > if y is not prime dS equals 0. I want to repeat that for emphasis, > for dS(x,y) if y is not prime dS equals 0. Amazingly enough, mathematicians never discovered such a function as > my dS throughout their entire history. Function. <...> ==== <... >> Now a key feature of dS that I want to emphasize is that for dS(x,y) >> if y is not prime dS equals 0. I want to repeat that for emphasis, >> for dS(x,y) if y is not prime dS equals 0. >> Amazingly enough, mathematicians never discovered such a function as >> my dS throughout their entire history. > Can you _prove_ this negative for me please? (bait = never + entire history) ==== > <... >> Now a key feature of dS that I want to emphasize is that for dS(x,y) >> if y is not prime dS equals 0. I want to repeat that for emphasis, >> for dS(x,y) if y is not prime dS equals 0. >> Amazingly enough, mathematicians never discovered such a function as >> my dS throughout their entire history. > > Can you _prove_ this negative for me please? > (bait = never + entire history) Well for context it helps if the information you deleted out is given. What follows is what the poster deleted out, up to the statement on which focus was placed by that deletion. What my function does is count the number of composites for each integer up to the square root of your base number. So for counting primes up to 100 the base number is 100 and its square root is 10. Here are the composite counts: dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; dS(100,9) = 0; dS(100,10)= 0; and summing the dS values gives you 74. Notice when dS equals 0. Now you add 1 for the number 1, as its not considered prime, and subtract from 100 to get 25, which is the count of primes up to 100. And those prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. Now a key feature of dS that I want to emphasize is that for dS(x,y) if y is not prime dS equals 0. I want to repeat that for emphasis, for dS(x,y) if y is not prime dS equals 0. Amazingly enough, mathematicians never discovered such a function as my dS throughout their entire history. Ok then, now to that statement, it's amazing but true. Throughout their *entire* history mathematicians never discovered such a function. Obviously, if that were false the easiest thing would be for some poster to show you something that mathematicians have that does behave the same way. But the closest thing they might try works by brute force and exploiting rounding. I'll see if any try to put it up before I go into details. Now notice that the poster deleted out the data, and focused on a statement which said poster probably figured most of you would simply reject as being impossible. But that depends on your faith in the very people I'm explaining to you aren't living up to that faith. Remember, math people know that many of you find math either unappealing or difficult, after all, math people *teach* mathematics. By pushing you to depend on your trust, deeply ingrained, this poster showed a fascinating disdain for some of you. James Harris ==== > >><...>>Now a key feature of dS that I want to emphasize is that for dS(x,y) >>if y is not prime dS equals 0. I want to repeat that for emphasis, >>for dS(x,y) if y is not prime dS equals 0. >>Amazingly enough, mathematicians never discovered such a function as >>my dS throughout their entire history. >Can you _prove_ this negative for me please? >>(bait = never + entire history) > > > Well for context it helps if the information you deleted out is given. Ok, so you are going to answer his question about proving mathematicians never discovered something like dS... > > What follows is what the poster deleted out, up to the statement on > which focus was placed by that deletion. > > What my function does is count the number of composites for each > integer up to the square root of your base number. So for counting > primes up to 100 the base number is 100 and its square root is 10. > > Here are the composite counts: > > dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; > dS(100,9) = 0; dS(100,10)= 0; > > and summing the dS values gives you 74. Notice when dS equals 0. > > Now you add 1 for the number 1, as its not considered prime, and > subtract from 100 to get 25, which is the count of primes up to 100. > > And those prime numbers are > > 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, > 67, 71, 73, 79, 83, 89, and 97. > > Now a key feature of dS that I want to emphasize is that for dS(x,y) > if y is not prime dS equals 0. I want to repeat that for emphasis, > for dS(x,y) if y is not prime dS equals 0. > > Amazingly enough, mathematicians never discovered such a function as > my dS throughout their entire history. > > Ok then, now to that statement, it's amazing but true. Throughout > their *entire* history mathematicians never discovered such a > function. > > Obviously, if that were false the easiest thing would be for some > poster to show you something that mathematicians have that does behave > the same way. But that doesn't show that it's true. Only that it's (presumably) easy for us to show it false. > > But the closest thing they might try works by brute force and > exploiting rounding. I'll see if any try to put it up before I go > into details. > > Now notice that the poster deleted out the data, and focused on a > statement which said poster probably figured most of you would simply > reject as being impossible. > > But that depends on your faith in the very people I'm explaining to > you aren't living up to that faith. > > Remember, math people know that many of you find math either > unappealing or difficult, after all, math people *teach* mathematics. > By pushing you to depend on your trust, deeply ingrained, this poster > showed a fascinating disdain for some of you. > > > James Harris But, can you, for that matter *will* you, prove that no mathematician came up with anything like your dS? You didn't answer the question, just showed us what the function is and talked about methods of disproof. -- Will Twentyman ==== > Repeatedly I've given out a definition for counting prime numbers, but > I'm afraid that something has not been communicated. So first I'm > going to give an example, counting the number of prime numbers up to > 100, and afterwards I'll give the function. In that way I hope to > point clearly to a key feature of my research. > > What my function does is count the number of composites for each > integer up to the square root of your base number. So for counting > primes up to 100 the base number is 100 and its square root is 10. > > Here are the composite counts: > > dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; > dS(100,9) = 0; dS(100,10)= 0; > > and summing the dS values gives you 74. Notice when dS equals 0. > > Now you add 1 for the number 1, as its not considered prime, and > subtract from 100 to get 25, which is the count of primes up to 100. > > And those prime numbers are > > 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, > 67, 71, 73, 79, 83, 89, and 97. > > Now a key feature of dS that I want to emphasize is that for dS(x,y) > if y is not prime dS equals 0. I want to repeat that for emphasis, > for dS(x,y) if y is not prime dS equals 0. > > Amazingly enough, mathematicians never discovered such a function as > my dS throughout their entire history. > > Some posters have tried to convince you otherwise. While some > apparently have tried to convince you that it doesn't matter if that's > true. Basically I think many of them have just worked to confuse you. > > Now here's that definition I've given so many times before: > > dS(x,y) = [pi(x/y, y-1) - pi(y-1, sqrt(y-1))] > > [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))], > > S(x,1) = 0. > > And pi(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS > from dS(x,1) to dS(x,y). > > Note: pi(x,sqrt(x)) here gives the same value as the traditional > pi(x). > > For faster calculations you need to use > > dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)] > > [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))] > > when sqrt(x/y) < y-1. > > See http://groups.msn.com/AmateurMath/primecountingfunction.msnw Ok, let's step through the calculation of dS(100,2)=49. dS(100,2)= [pi(100/2,2-1) - pi(2-1,sqrt(2-1))] * [ pi(2,sqrt(2)) - pi(2-1, sqrt(2-1)] dS(100,2) = [pi(50,1) - pi(1,1)] [pi(2,sqrt(2)) - pi(1,1)] Evaluating the pi's: pi(50,1)= floor(50)- S(50,1)- 1 = 50 - 0 - 1 = 49 pi(1,1) = floor(1) - S(1,1) - 1 = 1 - 0 - 1 = 0 pi(2,sqrt(2))=floor(2) - S(2,sqrt(2)) - 1 = 1 - S(2,sqrt(2)) pi(1,1) = floor(1) - S(1,1) - 1 = 1 - 0 - 1 = 0 Evaluating the odd S(2,sqrt(2)): S(2,sqrt(2)) = dS(2,1) + dS(2,sqrt(2)) [Is this correct?] dS(2,1) = [pi(1,0) - pi(0,0)] [pi(1,1) - pi(0,0)] pi(1,0) = floor(1) - S(1,0) - 1 = -S(1,0) which I don't see how to calculate. At this point I know that S(2,sqrt(2)) should be 1 to give the answer you arrived at, but I'm not sure how you are getting there. -- Will Twentyman ==== > Repeatedly I've given out a definition for counting prime numbers, but > I'm afraid that something has not been communicated. So first I'm > going to give an example, counting the number of prime numbers up to > 100, and afterwards I'll give the function. In that way I hope to > point clearly to a key feature of my research. > > What my function does is count the number of composites for each > integer up to the square root of your base number. So for counting > primes up to 100 the base number is 100 and its square root is 10. > > Here are the composite counts: > > dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; > dS(100,9) = 0; dS(100,10)= 0; > > and summing the dS values gives you 74. Notice when dS equals 0. > > Now you add 1 for the number 1, as its not considered prime, and > subtract from 100 to get 25, which is the count of primes up to 100. > > And those prime numbers are > > 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, > 67, 71, 73, 79, 83, 89, and 97. > > Now a key feature of dS that I want to emphasize is that for dS(x,y) > if y is not prime dS equals 0. I want to repeat that for emphasis, > for dS(x,y) if y is not prime dS equals 0. > > Amazingly enough, mathematicians never discovered such a function as > my dS throughout their entire history. > > Some posters have tried to convince you otherwise. While some > apparently have tried to convince you that it doesn't matter if that's > true. Basically I think many of them have just worked to confuse you. > > Now here's that definition I've given so many times before: > > dS(x,y) = [pi(x/y, y-1) - pi(y-1, sqrt(y-1))] > > [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))], > > S(x,1) = 0. > > And pi(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS > from dS(x,1) to dS(x,y). > > Note: pi(x,sqrt(x)) here gives the same value as the traditional > pi(x). > > For faster calculations you need to use > > dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)] > > [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))] > > when sqrt(x/y) < y-1. > I am not going to argue with the correctness of your results. A point of curiousity about this method is, how efficient is it (in terms of big-O notation)? If your new algorithm is considerably slower than existing algorithms, it is only a curiousity. If it is faster, then you have done something worthy of note. > See http://groups.msn.com/AmateurMath/primecountingfunction.msnw This would be the site I can only see when I'm not logged into my msn passport. > > If your eyes glazed over, or you found yourself wishing to quickly run > away from what looks very complicated then you might understand how > some posters could so easily manipulate the discussion away from the > truth. > > The truth is that the dS function does the amazing, as you see a > complete definition above, where you don't see any prime numbers > except 2 but *somehow* as you noticed above dS(x,y) equals 0 when y is > a prime number. > > You see, the function *knows* when a number is prime, making it > unique. > > Those values from above were > > dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; > dS(100,9) = 0; dS(100,10)= 0; > > and you can see the definition I gave above at work, as it picks its > way through the numbers dropping to 0, whenever y is not prime. > > That's so huge that I can't emphasize it enough. It's also something > that should give you a certain sadness because what should have > happened is that mathematicians should have excitedly embraced such a > result. You haven't stated *why* it's exciting. You have a new method of counting primes. That is not, in itself, exciting. How fast is your method? Does dS return 0 for composites faster than checking whether a number is composite by other methods? Note: these are not questions I have the answer to, these are the questions that will help determine the value of this work. [non mathematics deleted.] -- Will Twentyman ==== > Repeatedly I've given out a definition for counting prime numbers, but > I'm afraid that something has not been communicated. So first I'm > going to give an example, counting the number of prime numbers up to > 100, and afterwards I'll give the function. In that way I hope to > point clearly to a key feature of my research. > > What my function does is count the number of composites for each > integer up to the square root of your base number. So for counting > primes up to 100 the base number is 100 and its square root is 10. > > Here are the composite counts: > > dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; > dS(100,9) = 0; dS(100,10)= 0; > > and summing the dS values gives you 74. Notice when dS equals 0. > > Now you add 1 for the number 1, as its not considered prime, and > subtract from 100 to get 25, which is the count of primes up to 100. > > And those prime numbers are > > 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, > 67, 71, 73, 79, 83, 89, and 97. > > Now a key feature of dS that I want to emphasize is that for dS(x,y) > if y is not prime dS equals 0. I want to repeat that for emphasis, > for dS(x,y) if y is not prime dS equals 0. > > Amazingly enough, mathematicians never discovered such a function as > my dS throughout their entire history. > > Some posters have tried to convince you otherwise. While some > apparently have tried to convince you that it doesn't matter if that's > true. Basically I think many of them have just worked to confuse you. > > Now here's that definition I've given so many times before: > > dS(x,y) = [pi(x/y, y-1) - pi(y-1, sqrt(y-1))] > > [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))], > > S(x,1) = 0. > > And pi(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS > from dS(x,1) to dS(x,y). > > Note: pi(x,sqrt(x)) here gives the same value as the traditional > pi(x). > > For faster calculations you need to use > > dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)] > > [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))] > > when sqrt(x/y) < y-1. > > > > I am not going to argue with the correctness of your results. A point > of curiousity about this method is, how efficient is it (in terms of > big-O notation)? If your new algorithm is considerably slower than > existing algorithms, it is only a curiousity. If it is faster, then you > have done something worthy of note. It is just old Legendre's formula, slightly repacked. Legendre defined phi (x, k) := number of positive integers <= x which are not divisible by any of the first k primes. You get phi (x, 0) = x for all x, and if k > 0 then phi (x, k) = 0 if x = 0 phi (x, k) = 1 if 1 <= x < p (k+1), where p(k+1) is the (k+1)st prime phi (x, k) = pi (x) - k + 1 if p(k) <= x < (p(k+1))^2 and from the last equation it follows that with k = pi (x^(1/2)) pi (x) = phi (x, k) + k - 1 And then there is the recursion formula phi (x, k) = phi (x, k-1) - phi (x/p(k), k-1). That one is quite obvious because the difference between phi (x, k) and phi (x, k-1) are exactly those integers <= x with p(k) as the smallest prime factor, and these integers can be written as y*p(k), where y has no divisor <= p(k) and y <= x / p(k). I think you can figure out easily how phi (x, k) and his function dS (x, y) are related. Execution time is O (N). Meissel's algorithm is a trivial variation of this; if you take the first recursion level of Legendre's formula then you will see that you end up calculating many values pi (x') where x' <= x^(2/3). You can get all those by using a sieve in O (N^(2/3)), and the remaining work is done in O (N / log^3 N), I think; so this is quite simple and much much faster. Btw. I wonder if he ever figured out the cute trick how to calculate phi (x, k) for lets say k <= 6 and arbitrarily large x using just one division and two multiplications. I don't know if Meissel knew it, but Lehmer certainly did. ==== > > >Repeatedly I've given out a definition for counting prime numbers, but >I'm afraid that something has not been communicated. So first I'm >going to give an example, counting the number of prime numbers up to >100, and afterwards I'll give the function. In that way I hope to >point clearly to a key feature of my research. >>What my function does is count the number of composites for each >integer up to the square root of your base number. So for counting >primes up to 100 the base number is 100 and its square root is 10. >>Here are the composite counts: >> dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; > dS(100,9) = 0; dS(100,10)= 0; >>and summing the dS values gives you 74. Notice when dS equals 0. >>Now you add 1 for the number 1, as its not considered prime, and >subtract from 100 to get 25, which is the count of primes up to 100. >>And those prime numbers are >>2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, >67, 71, 73, 79, 83, 89, and 97. >>Now a key feature of dS that I want to emphasize is that for dS(x,y) >if y is not prime dS equals 0. I want to repeat that for emphasis, >for dS(x,y) if y is not prime dS equals 0. >>Amazingly enough, mathematicians never discovered such a function as >my dS throughout their entire history. >>Some posters have tried to convince you otherwise. While some >apparently have tried to convince you that it doesn't matter if that's >true. Basically I think many of them have just worked to confuse you. >>Now here's that definition I've given so many times before: >>dS(x,y) = [pi(x/y, y-1) - pi(y-1, sqrt(y-1))] >> [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))], >>S(x,1) = 0. > >And pi(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS >from dS(x,1) to dS(x,y). > >Note: pi(x,sqrt(x)) here gives the same value as the traditional >pi(x). >>For faster calculations you need to use > > dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)] > > [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))] >>when sqrt(x/y) < y-1. >I am not going to argue with the correctness of your results. A point >>of curiousity about this method is, how efficient is it (in terms of >>big-O notation)? If your new algorithm is considerably slower than >>existing algorithms, it is only a curiousity. If it is faster, then you >>have done something worthy of note. > > > It is just old Legendre's formula, slightly repacked. > > Legendre defined > > phi (x, k) := number of positive integers <= x which are not > divisible by any of the first k primes. > > You get phi (x, 0) = x for all x, and if k > 0 then > > phi (x, k) = 0 if x = 0 > phi (x, k) = 1 if 1 <= x < p (k+1), where p(k+1) is the (k+1)st prime > phi (x, k) = pi (x) - k + 1 if p(k) <= x < (p(k+1))^2 > > and from the last equation it follows that with k = pi (x^(1/2)) > > pi (x) = phi (x, k) + k - 1 > > And then there is the recursion formula > > phi (x, k) = phi (x, k-1) - phi (x/p(k), k-1). > > That one is quite obvious because the difference between phi (x, k) and > phi (x, k-1) are exactly those integers <= x with p(k) as the smallest > prime factor, and these integers can be written as y*p(k), where y has > no divisor <= p(k) and y <= x / p(k). > > I think you can figure out easily how phi (x, k) and his function dS (x, > y) are related. Execution time is O (N). Meissel's algorithm is a > trivial variation of this; if you take the first recursion level of > Legendre's formula then you will see that you end up calculating many > values pi (x') where x' <= x^(2/3). You can get all those by using a > sieve in O (N^(2/3)), and the remaining work is done in O (N / log^3 N), > I think; so this is quite simple and much much faster. And with this, I believe, we arrive at why James is not getting the primes any faster. It is slower than existing algorithms, and therefore a curiousity. > > Btw. I wonder if he ever figured out the cute trick how to calculate phi > (x, k) for lets say k <= 6 and arbitrarily large x using just one > division and two multiplications. I don't know if Meissel knew it, but > Lehmer certainly did. -- Will Twentyman ==== Let f be an holomorphic function on an B(0,R) = {z, |z|t_1 and C_r is not the entire circle of radius r. Olivier. ==== > Let f be an holomorphic function on an B(0,R) = {z, |z| > Let M(r) = max{|f(z)| / |z|=r} for r < R > > We know that there exists a complex number z_0=r*exp(it) with |f(z_0)| = > M(r) > and therefore, C_r = {z, |f(z)| = M(r) and |z|=r} is not empty. > > Can we have C_r = { re^(it) / t in [t_0,t_1]} where t_0<>t_1 and C_r is not > the entire circle of radius r. No, in fact |f| cannot be constant on any subarc of a circle without being constant on the full circle; it doesn't matter if the constant is the maximum value. Proof: |f|^2 is real analytic on the circle in question. If a real analytic function on a circle is constant on any subarc of a circle, it is constant on the circle. (In fact, if it constant on any infinite subset of the circle, it is constant on the circle.) ==== The World Wide Wade a .8ecrit dans le > Let f be an holomorphic function on an B(0,R) = {z, |z| Let M(r) = max{|f(z)| / |z|=r} for r < R > We know that there exists a complex number z_0=r*exp(it) with |f(z_0)| = > M(r) > and therefore, C_r = {z, |f(z)| = M(r) and |z|=r} is not empty. > Can we have C_r = { re^(it) / t in [t_0,t_1]} where t_0<>t_1 and C_r is not > the entire circle of radius r. No, in fact |f| cannot be constant on any subarc of a circle without being > constant on the full circle; it doesn't matter if the constant is the > maximum value. Proof: |f|^2 is real analytic on the circle in question. If > a real analytic function on a circle is constant on any subarc of a circle, > it is constant on the circle. (In fact, if it constant on any infinite > subset of the circle, it is constant on the circle.) Ok, but can you give me a proof with more details.... ==== > No, in fact |f| cannot be constant on any subarc of a circle without being > constant on the full circle; it doesn't matter if the constant is the > maximum value. Proof: |f|^2 is real analytic on the circle in question. If > a real analytic function on a circle is constant on any subarc of a > circle, > it is constant on the circle. (In fact, if it constant on any infinite > subset of the circle, it is constant on the circle.) > > Ok, but can you give me a proof with more details... Definition: Suppose g: I -> R, where I is an interval. We say g is real analytic on I if g is locally expressible on I as a power series. Fact 1. If g is real analytic on [a,b] and g is constant on an infinite subset of [a,b], then g is constant on [a,b]. Fact 2: Let f = u + iv be holomorphic as in your problem. Then the map t -> [u(r*exp(it))]^2 + [v(r*exp(it))]^2 is a real analytic function of t in R. ==== > > For this to be true you need H to be the Borel field generated by F (or, > more generally, for H to be contained in the u-completion of the Borel > field generated by F). > > It seems you are worrying over the proof of the seocnd part of Theorem 4 > (pages 382-383) in Chung's book. To your credit it must be noted that > it is implicit in Chung's argument that the sequence you call (Q_n) > (and that Chung labels (Omega_n) is *increasing*. This sequence, which > bears witness to the sigma-finiteness of u, can always be taken to be > increasing if desired. And in the assertion on page 383 that (in your > notation) > > liminf u(Q_n intersect A) = u(A) > > the reference to property (e) of the measure u makes it clear that the > author *is* assuming (Q_n) to be monotone increasing. [The same problem > occurs in the proof of Theorem 3 on page 381; to salvage things one > should simply added the hypothesis (Omega_n) increasing to Definition > 5 on page 381.] This is, indeed, my problem and the reason for my posts. Now, re. Theorem 3, p. 381 I believe I can prove that for every n u_1(A Union (1 to n)Omega_n)= u_2(A Union (1 to n)Omega_n) It would appear that this would eliminate the need to assume that Omega_n is monotonic increasing in proving u_1(A) = u_2(A). One can then assume sigma-finiteness of u without qualifications. Is this correct? If true, then maybe there is a similar approach to avoiding the additional hypothesis for the second part of Theorem 4. It's hard enough for a beginner to learn this stuff without texts making it even harder by their mistakes and omissions, and that's why your help is extremely useful to me. ==== > > For this to be true you need H to be the Borel field generated by F (or, > more generally, for H to be contained in the u-completion of the Borel > field generated by F). > > It seems you are worrying over the proof of the seocnd part of Theorem 4 > (pages 382-383) in Chung's book. To your credit it must be noted that > it is implicit in Chung's argument that the sequence you call (Q_n) > (and that Chung labels (Omega_n) is *increasing*. This sequence, which > bears witness to the sigma-finiteness of u, can always be taken to be > increasing if desired. And in the assertion on page 383 that (in your > notation) > > liminf u(Q_n intersect A) = u(A) > > the reference to property (e) of the measure u makes it clear that the > author *is* assuming (Q_n) to be monotone increasing. [The same problem > occurs in the proof of Theorem 3 on page 381; to salvage things one > should simply added the hypothesis (Omega_n) increasing to Definition > 5 on page 381.] > > This is, indeed, my problem and the reason for my posts. Now, re. > Theorem 3, p. 381 I believe I can prove that for every n > > u_1(A Union (1 to n)Omega_n)= u_2(A Union (1 to n)Omega_n) > > It would appear that this would eliminate the need to assume that > Omega_n is monotonic increasing in proving u_1(A) = u_2(A). One can > then assume sigma-finiteness of u without qualifications. Is this > correct? Yes > If true, then maybe there is a similar approach to avoiding the > additional hypothesis for the second part of Theorem 4. > > It's hard enough for a beginner to learn this stuff without texts > making it even harder by their mistakes and omissions, and that's why > your help is extremely useful to me. This is not a matter of additional hypotheses: a measure u on a sigma-field H is sigma-finite if and only if there is an *increasing* sequence (Q_n) of elements of H such that u(Omega_n) is finite for each n. [For if u is sigma-finite in the sense of definition 5, page 381, of Chung, with u(Omega_n) finite and union(Omega_n)=S, then Q_n = Omega_1 U Omega_2 U ... U Omega_n , n=1,2,3,... defines an increasing sequence of elements of H with u(Q_n) <= sum_{k=1}^n u(Omega_n) finite.] The problem is that Chung did not include the word increasing in his Definition 5 (though he might as well have, with no loss of generality) but he argues in the proofs of Theroems 3 and 4 as if he had. -- A. ==== > Think about TWO axiom systems S1 and S2. > How can they be related? (uselessly vague possibilities snipped) lol - a full understanding of the following might answer the question: Boolos & Jeffrey, Logic & Computability (or something like that) Chang & Keisler, Model Theory (ditto) Hartley Rogers, Recursive Function Theory (one more time) have fun, (rofl) cdj ==== > Under the assumption that both S1 and S2 are consistent? Yes, *and* that (see above) one isn't a subset of the other, which this... > If that's what > you're asking the answer is no. As a trivial example, consider > Presburger arithmetic (a decidable theory of arithmetic having axioms > concerning only + and successor) and Presburger arithmetic + 1*2=2. ...looks to fall under. Maybe if one is Pr+1*2=2 and the other Pr+1*3=3? -- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the address For spamming, remove anything else ==== Repeatedly I've given out a definition for counting prime numbers, but I'm afraid that something has not been communicated. So first I'm going to give an example, counting the number of prime numbers up to 100, and afterwards I'll give the function. In that way I hope to point clearly to a key feature of my research. What my function does is count the number of composites for each integer up to the square root of your base number. So for counting primes up to 100 the base number is 100 and its square root is 10. Here are the composite counts: dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; dS(100,5) = 6; dS(100,6) = 0; dS(100,7) = 3; dS(100,8) = 0; dS(100,9) = 0; dS(100,10)= 0; and summing the dS values gives you 74. Notice when dS equals 0. Now you add 1 for the number 1, as its not considered prime, and subtract from 100 to get 25, which is the count of primes up to 100. And those prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. Now a key feature of dS that I want to emphasize is that for dS(x,y) if y is not prime dS equals 0. I want to repeat that for emphasis, for dS(x,y) if y is not prime dS equals 0. Amazingly enough, mathematicians never discovered such a function as my dS throughout their entire history. Some posters have tried to convince you otherwise. While some apparently have tried to convince you that it doesn't matter if that's true. Basically I think many of them have just worked to confuse you. Now here's that definition I've given so many times before: dS(x,y) = [pi(x/y, y-1) - pi(y-1, sqrt(y-1))] [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))], S(x,1) = 0. And pi(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). Note: pi(x,sqrt(x)) here gives the same value as the traditional pi(x). For faster calculations you need to use dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)] [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))] when sqrt(x/y) < y-1. See http://groups.msn.com/AmateurMath/primecountingfunction.msnw If your eyes glazed over, or you found yourself wishing to quickly run away from what looks very complicated then you might understand how some posters could so easily manipulate the discussion away from the truth. The truth is that the dS function does the amazing, as you see a complete definition above, where you don't see any prime numbers except 2 but *somehow* as you noticed above dS(x,y) equals 0 when y is a prime number. You see, the function *knows* when a number is prime, making it unique. Those values from above were dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; dS(100,5) = 6; dS(100,6) = 0; dS(100,7) = 3; dS(100,8) = 0; dS(100,9) = 0; dS(100,10)= 0; and you can see the definition I gave above at work, as it picks its way through the numbers dropping to 0, whenever y is not prime. That's so huge that I can't emphasize it enough. It's also something that should give you a certain sadness because what should have happened is that mathematicians should have excitedly embraced such a result. And I think for some of you, disbelieving *me* is easier than considering even the possibility that a discipline like mathematics could have enough corruption that mathematicians would fight what they should be celebrating. But focusing on me is a mistake. If you really wish to consider human achievement as some personal issue, where you wish to make me that powerful as a single human being that the accomplishments of thousands of years of effort can be reduced to just being about the individuals involved, then you shatter the foundations of human civilization, and make it just some fashion show. The truth is that the discoveries are more important than the discoverers, and in letting it turn into some celebrity issue, you demean human society and its accomplishments. For some of you I fear that the idea that I might become a celebrity is all that matters to you. And I fear you think that's all that ever mattered through all of human history, as if the truth were just a word, and life has always just been a big popularity contest. The truth is not just a word and emphasizing truth over social issues has helped humanity to make the technological achievments which make my communication through this medium possible. Where would we be if engineers had paused to figure out who they liked and disliked before they bothered to use the information discovered? Mathematicians aren't doing you any favors. I suggest that all of you focus less on me than on the information, and forget about what benefits telling the truth will give me, as you consider that the value to society is far, far greater. So how can mathematicians behave as they have? That's an important question, and I suggest we find out why popularity has so much importance in their world. James Harris ==== > Repeatedly I've given out a definition for counting prime numbers, but > I'm afraid that something has not been communicated. So first I'm > going to give an example, counting the number of prime numbers up to > 100, and afterwards I'll give the function. In that way I hope to > point clearly to a key feature of my research. > > What my function does is count the number of composites for each > integer up to the square root of your base number. So for counting > primes up to 100 the base number is 100 and its square root is 10. > > Here are the composite counts: > > dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,6) = 0; dS(100,7) = 3; > dS(100,8) = 0; dS(100,9) = 0; dS(100,10)= 0; > > and summing the dS values gives you 74. Notice when dS equals 0. > > Now you add 1 for the number 1, as its not considered prime, and > subtract from 100 to get 25, which is the count of primes up to 100. > > And those prime numbers are > > 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, > 67, 71, 73, 79, 83, 89, and 97. > > Now a key feature of dS that I want to emphasize is that for dS(x,y) > if y is not prime dS equals 0. I want to repeat that for emphasis, > for dS(x,y) if y is not prime dS equals 0. > > Amazingly enough, mathematicians never discovered such a function as > my dS throughout their entire history. I know that for some my claims of uniqueness for my function that counts primes just by itself seems impossible, as you probably have some exalted view of mathematicians or something. But do the research. They missed it. What mathematicians found were approximations to what I have, and it's easy for me to show. Then they took off chasing faster and faster counting algorithms, never having found the full function, which eluded them for quite some time. And in fact, I can get to some of their work from mine, but they can't get to mine from theirs unless they exploit the same principle I use. Various posters have claimed that they do, but here's what mathematicians actually exploit. Given a positive integer x, a positive integer y, divides into it x/y times. Yup, that's the big principle behind what they call Legendre's Method. For instance, take 11/2 and drop the remainder which gives you 5, which correspondes to 2, 4, 6, 8 and 10, then subtract 1 for the 2 as its prime to get you 4. Now take 11/3 and drop the remainder and you get 3, which corresponds to 3, 6 and 9. Mathematicians would again subtract 1 to handle 3, to get the composite count including 6 and 9, so you have 2. But THEN they'd go to 11/6 which gives 1 and *subtract* it to handle 6, as it's been counted *twice* from the 2 count and the 3 count. That way you get the entire composite count using that simple principle. Then you subtract that count from your x, and subtract 1 and you have the number of primes. For instance above you have 4 + 2 - 1 = 5 and 11-5-1 = 5, and those primes are 2, 3, 5, 7, and 11. But mathematicians found that method ever harder to work out as you expanded as you have to keep adding and subtracting these combinations. Over a hundred years they found various tricks to speed things up, and actually got close to what I discovered but apparently never made the leap to considering a pi(x,y) function versus a pi(x) function. So what I'm facing now may be a massive ego tantrum from mathematicians. James Harris ==== > Repeatedly I've given out a definition for counting prime numbers, but > I'm afraid that something has not been communicated. So first I'm > going to give an example, counting the number of prime numbers up to > 100, and afterwards I'll give the function. In that way I hope to > point clearly to a key feature of my research. > > What my function does is count the number of composites for each > integer up to the square root of your base number. So for counting > primes up to 100 the base number is 100 and its square root is 10. > > Here are the composite counts: > > dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,6) = 0; dS(100,7) = 3; > dS(100,8) = 0; dS(100,9) = 0; dS(100,10)= 0; > > and summing the dS values gives you 74. Notice when dS equals 0. > Yup, on the composites. > If your eyes glazed over, or you found yourself wishing to quickly run > away from what looks very complicated then you might understand how > some posters could so easily manipulate the discussion away from the > truth. > > The truth is that the dS function does the amazing, as you see a > complete definition above, where you don't see any prime numbers > except 2 but *somehow* as you noticed above dS(x,y) equals 0 when y is > a prime number. No, it's 0 when y is a composite. > > You see, the function *knows* when a number is prime, making it > unique. > Now, can you explain *how* it knows when a number is prime? Also, this is far from unique, there are functions (often called isPrime(n)) that return true/false, 1/0 depending on whether the number is prime. -- Will Twentyman ==== >> There are always *more* divisions than multiplications. >> always? That assumes the very thing you are trying to >> prove. > > I can run this on my grid http://www.gridontap.com and let you know... I am > 99.999% sure that there will be no counter examples that show there are more > multiplications then divisions. Since every multiplication step is immediately followed by a division, isn't it trivial that there aren't more multiplications than divisions (other than if you happen to stop counting right after a multiplication step in a sequence that has alternated multiplication and divisions)? -- Evidence Eliminator is worthless: www.evidence-eliminator-sucks.com --Tim Smith ==== >Ok, I am at 3,000,000 and I have not yet found a counter example to There >are always *more* divions then multiplications. I will keep it going... >and post updates on the list? According to a note on this page: http://ksl.stanford.edu/people/bmac/papers/peim.html, the Collatz conjecture has been verified up to about 3 trillion. Wikipedia says 1.2 trillion. http://www.wikipedia.org/wiki/Conjecture Mathworld says that as of 1999 the record is up to 3*2^53, which is in excess of 27 quadrillion in USA notation (2.7*10^16). http://mathworld.wolfram.com/CollatzProblem.html So you have a little competition. - Randy ==== >Man, sometimes I hate math... ;) I am a programmer... (so be easy on me) >hehe. This collatz thing is making me bash my head off of my keyboard. Is this >problem unsolveable like the NP problem? I mean, you would have to check >every integer (and that is not possible as you say) Right. So the truth has to be determined by proof, not by simply checking out integers. Infinite loops are possible. I was just in another Collatz thread a few days ago, and somebody pointed out that if you use 3x + 3 instead of 3x + 1, you get lots of loops, including a short one at 19. So it's clearly not true that all such expressions eventually take you to 1 for all integers. Then the question becomes whether there's something special about 3x + 1. - Randy ==== > Infinite loops are possible. I was just in another Collatz thread a > few days ago, and somebody pointed out that if you use 3x + 3 instead > of 3x + 1, you get lots of loops, including a short one at 19. My memory was wrong. If you use 3x+3, you do get a loop starting at 19, but it's a rather uninteresting one: 12, 6, 3, 12. The whole sequence is: 19, 60, 30, 15, 48, 24, 12, 6, 3, 12, ... 3x+5 is the one that gives a somewhat more interesting loop at 19: 19, 62, 31, 98, 49, 152, 76, 38, 19, ... - Randy ==== >> Infinite loops are possible. I was just in another Collatz thread a >> few days ago, and somebody pointed out that if you use 3x + 3 instead >> of 3x + 1, you get lots of loops, including a short one at 19. My memory was wrong. If you use 3x+3, you do get a loop >starting at 19, but it's a rather uninteresting one: >12, 6, 3, 12. The whole sequence is: >19, 60, 30, 15, 48, 24, 12, 6, 3, 12, ... 3x+5 is the one that gives a somewhat more interesting >loop at 19: >19, 62, 31, 98, 49, 152, 76, 38, 19, ... - Randy 3x+3 is not very interesting because you only get multiples of 3 and the successors of 3n with the 3x+3 problem are 3 times the successors of n with the 3x+1 problem, because n is even iff 3n is even (3n)/2 = 3(n/2) 3(3n) + 3 = 3 (3n+1) -- Wim Benthem ==== > >Man, sometimes I hate math... ;) I am a programmer... (so be easy on me) >hehe. >This collatz thing is making me bash my head off of my keyboard. Is this >problem unsolveable like the NP problem? I mean, you would have to check >every integer (and that is not possible as you say) > > Right. So the truth has to be determined by proof, not by simply > checking out integers. > > Infinite loops are possible. I was just in another Collatz thread a > few days ago, and somebody Me. >pointed out that if you use 3x + 3 instead That was 3x + 5. > of 3x + 1, you get lots of loops, including a short one at 19. > > So it's clearly not true that all such expressions eventually take you > to 1 for all integers. Then the question becomes whether there's > something special about 3x + 1. There is absolutely something special about 3x+1. Generalize the problem to 3x+C, where C is an odd integer constant. For every possible sequence of iterations of x/2 and 3x+C, a Hailstone function can be written in terms of the last number in the sequence: f_e d_c b a f = (8*a - 5*C)/9 Every Hailstone function can be expressed as (X*a - Z*C)/Y where X is always a power of 2, Y is always a power of 3, and Z is a mixture based on how the sequence zig-zags. z y x w v u_t s r q p o n m_l k_j i h g f_e d c_b a the Crossover Point works out to be (35343985*C)/33552245 which factors to (5*23*307339*C)/(5*6710449) which reduces to (7068797*C)/6710449 So the system 3x + 6710449 has a loop at 7068797, which is easily verified 7068797 27916840 13958420 6979210 3489605 17179264 8589632 4294816 2147408 1073704 536852 268426 134213 7113088 3556544 1778272 889136 444568 222284 111142 55571 6877162 3438581 17026192 8513096 4256548 2128274 1064137 9902860 4951430 2475715 14137594 7068797 What's special about 3x + 1 is that with C=1, the Crossover function is Z/(X-Y) which means the only way to get an integer is if the factors of Z cancel all the factors of (X-Y). And apparently, that is a daunting task. There are many candidate sequences (where Z > (X-Y)), but finding one with even one common factor, let alone all, is difficult. Very often X-Y is prime. A survey of all sequences of length 4 and depth 8 ended up with 28 prime and 41 composite X-Y values. And often the composites have large prime factors. So it's possible that if you looked at all sequences up to a million iterations, you might find one where the factors cancel when C=1. But I can't prove it one way or the other. > > - Randy ==== > Man, sometimes I hate math... ;) I am a programmer... (so be easy on me) > hehe. > > This collatz thing is making me bash my head off of my keyboard. Is this > problem unsolveable like the NP problem? I mean, you would have to check > every integer (and that is not possible as you say)... can it be said that > there are more divisions than multiplications, thus it always ends up at 1? Nobody knows if it is solvable or unsolvable. You might find a start value x other than 1, 2 or 4 that eventually returns to x again. That would solve it. You might find a start value x that seems to go off to larger and larger values. That would indicate the conjecture is false. However, you would have to prove that these values get larger and larger, which might be difficult. You might find a proof that every value has to return back to 1. It took a few hundred years for Fourier's Last Theorem, so there is still hope. I haven't got the slightest idea how to start such a proof, so I don't think finding a proof will be easy. (There might be a simple proof that everyone has overlooked. ) ==== ... > You might find a proof that every value has to return back to 1. It took > a few hundred years for Fourier's Last Theorem, so there is still hope. > I haven't got the slightest idea how to start such a proof, so I don't > think finding a proof will be easy. (There might be a simple proof that > everyone has overlooked. ) And what was Fourier's Last Theorem about, that made it difficult? -jiw ==== > There are always *more* divisions than multiplications. > always? That assumes the very thing you are trying to > prove. > > I can run this on my grid http://www.gridontap.com and let you know... I am > 99.999% sure that there will be no counter examples that show there are more > multiplications then divisions. Do you realize that if you have more multiplications than divisions it means that somewhere in the sequence you have at least two consecutive mutiplications? And do you further realize that you cannot have consecutive multiplications because every iteration of 3x+1 returns an even number? The best you could possibly achieve is parity between multiplications and divisons. But you can't even do that because the divisions will win when you reach 16. ==== >> There are always *more* divisions than multiplications. > always? That assumes the very thing you are trying to > prove. >> I can run this on my grid http://www.gridontap.com and >> let you know... I am 99.999% sure that there will be >> no counter examples that show there are more >> multiplications then divisions. > I'm 100% sure you won't check all the integers. True statement, but I don't think the idea that you don't prove things in math by testing a subset of the possibilities and calling that a proof has soaked into his mind yet. To save him a little time, I checked the first six billion or so integers (got to love 2GHz processors) using the code here: http://www.well.com/user/xanthian/java/Hailstone.java not with any misapprehension that I was going to do a proof by exhausting all possibilities, but just to look at the extreme cases to try to find a pattern from which a proof might be begun (to no avail, but it was fun trying, and perhaps half a dozen folks in the comp.lang.java.* groups got involved in a bit of a competition, which my code very much did not win). xanthian. BTW, _this_ is an interesting thing to encounter: http://books.lulu.com/content/9828 Anybody got opinions on whether that's legit, or know if it's been peer reviewed and confirmed? It was this page that convinced me the problem had long outstripped my math skills and yet remained unsolved by those much smarter than me, and was likely to remain so for considerably longer: http://www.cecm.sfu.ca/organics/papers/lagarias/ -- ==== > >> > >There are always *more* divisions than multiplications. > > >>always? That assumes the very thing you are trying to >>prove. > > >I can run this on my grid http://www.gridontap.com and >let you know... I am 99.999% sure that there will be >no counter examples that show there are more >multiplications then divisions. > > >>I'm 100% sure you won't check all the integers. > > > True statement, but I don't think the idea that you don't > prove things in math by testing a subset of the possibilities > and calling that a proof has soaked into his mind yet. > > To save him a little time, I checked the first six billion or > so integers (got to love 2GHz processors) using the code here: > > http://www.well.com/user/xanthian/java/Hailstone.java > > not with any misapprehension that I was going to do a proof by > exhausting all possibilities, but just to look at the extreme > cases to try to find a pattern from which a proof might be begun > (to no avail, but it was fun trying, and perhaps half a dozen > folks in the comp.lang.java.* groups got involved in a bit of a > competition, which my code very much did not win). > > xanthian. > > BTW, _this_ is an interesting thing to encounter: > > http://books.lulu.com/content/9828 > > Anybody got opinions on whether that's legit, or > know if it's been peer reviewed and confirmed? Couldn't tell you, but he's also claiming to have proved the Goldbach Conjecture in a seperate book. I find it interesting that neither is available at amazon.com. > > It was this page that convinced me the problem had > long outstripped my math skills and yet remained > unsolved by those much smarter than me, and was > likely to remain so for considerably longer: > > http://www.cecm.sfu.ca/organics/papers/lagarias/ > > -- Will Twentyman ==== > >> > >There are always *more* divisions than multiplications. > > >>always? That assumes the very thing you are trying to >>prove. > > >I can run this on my grid http://www.gridontap.com and >let you know... I am 99.999% sure that there will be >no counter examples that show there are more >multiplications then divisions. > > >>I'm 100% sure you won't check all the integers. > > > True statement, but I don't think the idea that you don't > prove things in math by testing a subset of the possibilities > and calling that a proof has soaked into his mind yet. > > To save him a little time, I checked the first six billion or > so integers (got to love 2GHz processors) using the code here: > > http://www.well.com/user/xanthian/java/Hailstone.java > > not with any misapprehension that I was going to do a proof by > exhausting all possibilities, but just to look at the extreme > cases to try to find a pattern from which a proof might be begun > (to no avail, but it was fun trying, and perhaps half a dozen > folks in the comp.lang.java.* groups got involved in a bit of a > competition, which my code very much did not win). > > xanthian. > > BTW, _this_ is an interesting thing to encounter: > > http://books.lulu.com/content/9828 > > Anybody got opinions on whether that's legit, or > know if it's been peer reviewed and confirmed? > > Couldn't tell you, but he's also claiming to have proved the Goldbach > Conjecture in a seperate book. I find it interesting that neither is > available at amazon.com. > I just read the preview and I assume from that blurb he has shown that any integer will get smaller and that a relationship exists with three digit ( base 10 I assume ) numbers. Or maybe it has something to do with his brother or sister. I'm not sure. Ernst > > It was this page that convinced me the problem had > long outstripped my math skills and yet remained > unsolved by those much smarter than me, and was > likely to remain so for considerably longer: > > http://www.cecm.sfu.ca/organics/papers/lagarias/ > > ==== >>I'm 100% sure you won't check all the integers. > > > Is this a conjecture or can you prove it ? ;-)) > > J-L. > > There are countably infinitely many integers. There is only finite matter to use for building computers for the computations. Therefor, in finite time with finite resources he can only check finitely many integers he can check. QED Hmm... wait a sec, what if the universe doesn't expire? I'll have to rethink my proof. -- Will Twentyman ==== In sci.math, Will Twentyman <3f4ab69c$1_3@newsfeed>: >I'm 100% sure you won't check all the integers. >> >> >> Is this a conjecture or can you prove it ? ;-)) >> >> J-L. >> >> > > There are countably infinitely many integers. > There is only finite matter to use for building computers for the > computations. > > Therefor, in finite time with finite resources he can only check > finitely many integers he can check. > > QED > > Hmm... wait a sec, what if the universe doesn't expire? I'll have to > rethink my proof. > Don't worry. Those computers also consume energy and the energy density of any given section of the Universe is finite. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > In sci.math, Will Twentyman > : > >>I'm 100% sure you won't check all the integers. >Is this a conjecture or can you prove it ? ;-)) >>J-L. >>There are countably infinitely many integers. >>There is only finite matter to use for building computers for the >>computations. >>Therefor, in finite time with finite resources he can only check >>finitely many integers he can check. >>QED >>Hmm... wait a sec, what if the universe doesn't expire? I'll have to >>rethink my proof. > > > Don't worry. Those computers also consume energy and the energy > density of any given section of the Universe is finite. :-) > Whew! I was worried that I might have to start invoking Newton's laws or something. Or prove that the universe will collapse. -- Will Twentyman ==== > > [snip] >>I don't think humans made it to the top of the food chain by >>out-nice-ing the other animals - I think we out-predated them, >>and we bear the mark of that evolutionary history today. >> >> No. We got to where we are _precisely_ by out-nice-ing other animals: >> the cooperation in a modern human society dwarfs that in a beehive. >This is true, but if you factor out generic behavioral complexity >and look at a ratio of cooperative/altruistic behavior to selfish >behavior, I would think that human behavior is far less cooperative >than that of bees. > > I doubt that very much. Factor out sexual and sex-moderated > competition, which is not possible for worker bees, and how much Right--it's not possible. So in that sense, any comparison between humans and bees is apples to oranges. Given the centrol role sex plays in biology, it's difficult to factor out sex. Everything---every biological feature exhibiting complex design, that is---stems from differential reproductive advantage, and hence is tied to sex. Including reciprocal altruism. > uncooperative human behavior is left? Even when we compete, we > cooperate (e.g., team sports, business competition, war). > >Put another way, I would venture that most ethnologists would consider >bees to be a more cooperative species (well, group of species...) than >humans. > > IMO this just means human behavior is far more complex than bee > behavior. Right, which is why it's apples to oranges---which species is more cooperative depends on the definition and metric used. > -- Roy L ==== > > [snip] >I don't think humans made it to the top of the food chain by >out-nice-ing the other animals - I think we out-predated them, >and we bear the mark of that evolutionary history today. > > No. We got to where we are _precisely_ by out-nice-ing other animals: > the cooperation in a modern human society dwarfs that in a beehive. >>This is true, but if you factor out generic behavioral complexity >>and look at a ratio of cooperative/altruistic behavior to selfish >>behavior, I would think that human behavior is far less cooperative >>than that of bees. >I doubt that very much. Factor out sexual and sex-moderated >competition, which is not possible for worker bees, and how much >uncooperative human behavior is left? Even when we compete, we >cooperate (e.g., team sports, business competition, war). >>Put another way, I would venture that most ethnologists would consider >>bees to be a more cooperative species (well, group of species...) than >>humans. >IMO this just means human behavior is far more complex than bee >behavior. > > Humans are the most co-operative species by far, which is an essential > part of human behavior being more complex. It depends how you define cooperative. > > Pack in millions of total strangers (not sisters) of any other species not sisters---that's a reasonable constraint, but note that the discussion is then constrained to e.g. non-kin reciprocal altruism (not that there's anything wrong with that). > at a density equivalent of the NYC or Tokyo subways, and see how well > they cooperate in each getting to their different destinations. > > Or to borrow an example of Milton Friedman's: can one imagine > thousands of members of any other species, all total strangers to one > another, located all around the world and not even knowing that each > other exist as individuals, cooperating to manufacture a pencil? But this example isn't species-typical---modern industrial civilization is a fluke, biologically speaking. Though it's of course a good example of cooperation. (Of course, even the cooperative efforts within a small tribe are pretty complex.) While humans are one of the few advanced species marked by altruism directed at non-kin, we can also be pretty nasty---as seen by the fact that murder (even murder of kin) appears to be species-typical. This to me is what makes social life interesting---the tension between selfish and cooperative urges. ==== > 3+1 > 1.. Pick any positive integer n. > 2.. If n is even divide by 2 else add 1. > 3.. If n = 1, stop; else go back to step 2. > If 3x+1 is disproved, then 3+1 will be disproved at the same time. > Your conjecture is true. Easy proof by induction. Does that mean > you've proven the Collatz conjecture? Cool. (hehe) > lol, I wish, you know, I bet the answer is in the Beal Ciphers... hehe Interesting idea :) As a average guy I spend way too much time on the Collatz conjecture and also have worked on the Beal ciphers. I know what you mean :) Good luck Ernst Gametheory how to play StockMarket ==== I need something from either biology, or chemistry or physics that says something that if a entity increases in a day time frame of above 5% or above 8% that the likelihood if it increasing in the near future above that initial rise is remote. And I saw it in action just a few days or a week ago when Intel had a daily rise of 8% but has since then fallen back to previous prices. I need something tangible in the world of science to anchor the idea that whenever you see a 8% rise or more in one day, that the Theory of Probability would say the likelihood of further rises is too remote. Perhaps the maximum of a living organism is a 8% increase per day. Perhaps something in chemistry says that a 8% or more is the limit of a parameter. Perhaps something in physics says that 8% or more of a physical parameter is tops to be expected because tomorrow it will have fallen back. whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies <3F43D5C6.2C7909DE@dtgnet.com> <3F43D99C.E8303B33@dtgnet.com> <3F446C78.8C784A85@dtgnet.com> <3F4654AA.9DE4FF4B@dtgnet.com> <3F4CEAE9.CF6A3D6A@dtgnet.com> ==== In message <3F4CEAE9.CF6A3D6A@dtgnet.com>, Archimedes Plutonium >I need something from either biology, or chemistry or physics that says >something that >if a entity increases in a day time frame of above 5% or above >8% that the likelihood if it increasing in the near future above that initial >rise is remote. And I saw it in action just a few days or a week ago when Intel had a >daily rise of 8% >but has since then fallen back to previous prices. I need something tangible in the world of science to anchor the idea >that whenever you >see a 8% rise or more in one day, that the Theory of Probability >would say the likelihood of further rises is too remote. Perhaps the maximum of a living organism is a 8% increase per day. Perhaps >something in chemistry says that a 8% or more is the limit of a >parameter. Perhaps >something in physics says that 8% or more of a physical parameter >is tops to be expected because tomorrow it will have fallen back. > But what if... There is a hostile take-over situation and a stock rises 10% and this is followed by a bidding competition from multiple companies leading to lucrative gains for your good self? -- Jeremy Boden ==== > Hypothetical situation: > > I hire a man to go and get grapes for me. We agree that I will pay > him 1 cent for each grape he brings me (irrespective of source), with > my payment to be rounded to the nearest $1. After a time, he brings > me 30 grapes. I quickly devour them and give him zero payment. > > Under common rounding rules, x.3 rounds down (e.g. 1.3 becomes 1). > Hence my payment of zero (0.3 becomes 0). However, is there any > legitimate argument for a payment of $1, instead of zero in my > hypothetical? Stated another way, are there any reasons why positive > values ought not be rounded to zero, when common rounding rules > suggest they should? > > Would the answer be any different if I were the only source of income > available to the man and I dictated the terms of employment, rather > than reaching arm's length agreement. > > I appreciate any constructive responses. I use a hypothetical to > spare you the boring details of my real life application. Under that arrangement, what possible incentive would the man have for bringing you less than 50 grapes? Assuming that grapes aren't perishable in the short run and that they have market value to others, the only logical thing for him to do in the event that he picked less than 50 would be to try to sell them to someone else instead. ==== >> > Under that arrangement, what possible incentive would the man have for > bringing you less than 50 grapes? Assuming that grapes aren't > perishable in the short run and that they have market value to others, > the only logical thing for him to do in the event that he picked less > than 50 would be to try to sell them to someone else instead. I agree, but in my application, it's not really a question of incentive, but rather a question of if the grape provider has an argument that his payment should be non-zero. The buyer is the only game in town. ==== > Hypothetical situation: > > I hire a man to go and get grapes for me. We agree that I will pay > him 1 cent for each grape he brings me (irrespective of source), with > my payment to be rounded to the nearest $1. After a time, he brings > me 30 grapes. I quickly devour them and give him zero payment. > > Under common rounding rules, x.3 rounds down (e.g. 1.3 becomes 1). > Hence my payment of zero (0.3 becomes 0). However, is there any > legitimate argument for a payment of $1, instead of zero in my > hypothetical? No; the contract sounds clear - you pay him for (and therefore keep track of) ) each and every grape he brings you, but you pay only even dollars. So you pay him nothing at this time, and when later he brings you (app.) 70 more grapes, you pay him $1. The only reasons for rounding off the payment being (1) he has a reason to keep bringing grapes, at this point, (2) it's easier for you to pay, (3) if the contract ends at some point, you don't have to settle up and figure the exact amount left owing by either party. Now, if you'd claimed that, because you had eaten the first 30 grapes, they no longer counted and he had to start over from zero, you'd be trying to cheat him. And if, had you done that and he'd objected, you had cancelled the contract at that point, you would have cheated him - it would not be fair for you to on one hand cancel the contract, and on the other insist that he was still bound by it and so deserved nothing for the grapes he had already brought you. > Stated another way, are there any reasons why positive > values ought not be rounded to zero, when common rounding rules > suggest they should? > > Would the answer be any different if I were the only source of income > available to the man and I dictated the terms of employment, rather > than reaching arm's length agreement. > > I appreciate any constructive responses. I use a hypothetical to > spare you the boring details of my real life application. ==== my actual application, I should state that this was a one-time event (i.e. no carryovers of unpaid balances, etc) and the terms are non-negotiable. Under the hypothetical, I have received something (30 grapes), my counterparty has received nothing ($0). While this seems intuitively unfair, it may be perfectly legitimate. I guess I was looking for some rule or logic that suggests that rounding to zero is inappropriate for payment in a commercial transaction, when there is positive, albeit small, contribution by the counterparty. This is important in my application as, by fiat, a zero payment triggers one unrelated, but very significant, outcome, while a non-zero, positive payment triggers a different outcome. Another possible argument, might be that there is a difference between zero dollars and $0, as suggested in an earlier post. In the contract controlling the unrelated outcome, the application trigger is based on zero payment in the first contract, while the first contract itself (e.g. payment for the grapes), calls for payment rounded to the nearest $1, which in this case is $0. B Stephens > > Hypothetical situation: > > I hire a man to go and get grapes for me. We agree that I will pay > him 1 cent for each grape he brings me (irrespective of source), with > my payment to be rounded to the nearest $1. After a time, he brings > me 30 grapes. I quickly devour them and give him zero payment. > > Under common rounding rules, x.3 rounds down (e.g. 1.3 becomes 1). > Hence my payment of zero (0.3 becomes 0). However, is there any > legitimate argument for a payment of $1, instead of zero in my > hypothetical? > > No; the contract sounds clear - you pay him for (and therefore keep > track of) ) each and every grape he brings you, but you pay only even > dollars. So you pay him nothing at this time, and when later he > brings you (app.) 70 more grapes, you pay him $1. > > The only reasons for rounding off the payment being (1) he has a > reason to keep bringing grapes, at this point, (2) it's easier for you > to pay, (3) if the contract ends at some point, you don't have to > settle up and figure the exact amount left owing by either party. > > Now, if you'd claimed that, because you had eaten the first 30 grapes, > they no longer counted and he had to start over from zero, you'd be > trying to cheat him. And if, had you done that and he'd objected, you > had cancelled the contract at that point, you would have cheated him - > it would not be fair for you to on one hand cancel the contract, and > on the other insist that he was still bound by it and so deserved > nothing for the grapes he had already brought you. > > Stated another way, are there any reasons why positive > values ought not be rounded to zero, when common rounding rules > suggest they should? > > Would the answer be any different if I were the only source of income > available to the man and I dictated the terms of employment, rather > than reaching arm's length agreement. > > I appreciate any constructive responses. I use a hypothetical to > spare you the boring details of my real life application. ==== > my actual application, I should state that this was a one-time event > (i.e. no carryovers of unpaid balances, etc) and the terms are > non-negotiable. Under the hypothetical, I have received something (30 > grapes), my counterparty has received nothing ($0). While this > seems intuitively unfair, it may be perfectly legitimate. I guess I > was looking for some rule or logic that suggests that rounding to zero > is inappropriate for payment in a commercial transaction, when there > is positive, albeit small, contribution by the counterparty. > This is important in my application as, by fiat, a zero payment > triggers one unrelated, but very significant, outcome, while a > non-zero, positive payment triggers a different outcome. > Another possible argument, might be that there is a difference between > zero dollars and $0, as suggested in an earlier post. In the > contract controlling the unrelated outcome, the application trigger is > based on zero payment in the first contract, while the first > contract itself (e.g. payment for the grapes), calls for payment > rounded to the nearest $1, which in this case is $0. Are you asking whether such an arrangement can be fair? Certainly it can be customary. For example, you can make a one-time contract with a performance clause. If the contractor produces at least 50 grapes he gets $1, if he produces at least 150 grapes he gets $2, etc. There could even be penalties for nonperformance, if he produces less than 20 grapes he pays you. People do sign such agreements on occasion. It's generally considered to be legal and enforceable. It's generally considered proper to make sure both parties understand the agreement first. If you have made such an agreement and now are disputing what it means then there's a communication gap that should have been cleared up before the contract was signed. ==== nearest $1, which in this case is $0. > > Are you asking whether such an arrangement can be fair? Not really, I am simply asking if there is any way the employee can legitimately claim he is entitled to payment greater than zero. I am struggling to come up with anything defendable. Certainly it > can be customary. For example, you can make a one-time contract with a > performance clause. If the contractor produces at least 50 grapes he > gets $1, if he produces at least 150 grapes he gets $2, etc. There > could even be penalties for nonperformance, if he produces less than 20 > grapes he pays you. People do sign such agreements on occasion. It's > generally considered to be legal and enforceable. It's generally > considered proper to make sure both parties understand the agreement first. > > If you have made such an agreement and now are disputing what it means > then there's a communication gap that should have been cleared up before > the contract was signed. The agreement in my actual case is a monopoly provider tariff, not a standard arms-length agreement. Disputing what it means is exactly what's at stake, as the circumstances at hand, a tiny amount that rounds to 0, likely have not occurred before in the application of the tariff. To state it in the context of my grape provider, all other grape providers before him brought either none (zero) or thousands of grapes. Rounding precision was not an issue. ==== Dear Maplesoft: I would like to be able to recommend your software to the members of the coalitions I build and help build. These coalitions work on issues of interest to me (any one of about 125 inventions I am preparing for patent), or to the public. I facilitate the purchase of Mathcad software for these coalitions and assist the members in preparing documentation to justify the academic license price when appropriate. I also facilitate the use of coalition member's air miles for travel by their and other coalition's members to meetings, conferences, and symposia. I am an expert and licensed Mathcad user and have asked MathSoft pay me a commission when facilitated purchases are made. I would like to be able to recommend Maple to my coalition members when appropriate and would like you to pay me a commission when facilitated purchases are made. I intend to make this same request to Wolfram, the makers of Mathematica. One coalition member, and possibly others, is a licensed and expert Mathematica user. I am examining Scientific Notebook right now, as well. I'd like to share a little something I noticed on Usenet. In Messaged-ID: Xns93E1C99B8D4DAcam0@202.4.251.50 ----------------------------begin quoted text you can just download maple from kazaa. its only 40MB. the program is great. i use it for large computations and numerical analysis problems. ---------------------------------------------end quoted text which is a clear violation of the provisions of DCMA. Read it carefully: I am being attributed with recommending violation of your copyright. I respect, create, and license intellectual property. Please let me know if you are able to prosecute this user for downloading your copyrighted software from Kazaa, based on my tip. Until then, I'd just like to hear what it takes to enroll with you to receive commisions on facilitated sales. That should be an easy question to answer. Maybe I just have to fill out a form. transmission to you at bizdev@maplesoft.com. Yours, Doug Goncz Replikon Research PO Box 4394 (preferred) 6187 Greenwood Dr. Apt. 102 Seven Corners, VA 22044-0394 (or 2514) 703-536-2367 voice, message 703-536-5469 fax Yours, Doug Goncz Replikon Research (via aol.com) Nuclear weapons are just Pu's way of ensuring that plenty of Pu will be available for The Next Big Experiment, outlined in a post to sci.physics.research at Google Groups under supercritical ==== Dear Dr. Sarfatti, only some of what I can understand. own fashion. You say dark energy is an exotic vacuum phase of negative quantum zero point fluctuations, and dark matter is the positive quantum /zpf. But dark energy is a positive energy form left over from the inflation of space. The dark energy flux is an accounting gimmic to explain the negative energy (work) required to inflate space (and have the universe remain flat). In fact, I don't see any reason to suspect that dark energy is necessarily any different than regular matter. No you do not understand the basic physics here and have not properly described above what I am talking about. Part of the basic physics is on pages 25-26 of John Peacock's Cosmological Physics showing why covariance requires w = -1 for zero point fluctuations of all quantum fields. Dark energy has / zpf > 0, i.e. positive zero point energy density with equal and opposite negative zero point pressure. What matters in Einstein's GR is energy density + 3pressure in the Poisson equation for the system, in this case, the exotic quantum vacuum. Dark matter is simply /zpf < 0, i.e. negative zero point energy density with equal and opposite positive zero point pressure. This explains the dark matter not only in large scale but also in micro-scale i.e. 1. Why self-charge does not cause electron to explode, i.e. the old Abraham-Becker self-energy problem. scattering yet e/mc^2 ~ 1 fermi and h/mc ~ 137e^2/mc^2 e^2|/zpf|^1/2 where microscale dark matter cores of lepto-quarks have /zpf < 0 and |/zpf|~ (1 fermi)^-2 4. This same micro /zpf gives the correct universal Regge slope of hadronic resonances alpha' ~ (1Gev)^-2 5. The universal micro dark matter exotic vacuum core of the lepto-quarks is |/zpf| ~ Lp^-4/3(c/Ho)^-2/3 from Susskind's world hologram model. Lp^2 = hG/c^3 For more details go to http://qedcorp.com/APS/Ukraine.doc http://qedcorp.com/APS/Vigier4.pdf My theory predicts null results for all conventional dark matter I explain both large scale and small scale structure of universe with one simple mainstream idea from P.W. Anderson More is different (ODLRO) - Occam's Razor. If I haven't offended your scientific sensibilities yet, I would space. Does it exist? What are its properties? Is there any room for Mach's principle in this picture? You will recall that Mach claimed that inertial frames are determined by the disposition of matter in the surrounding universe. Does your concept of space allow for the succint derivation of the fundamental equation, F=ma? Yes, of course. Read my papers above. :-) F = ma is basically timelike geodesic equation for neutral test fields. I derive Einstein's GR from the modulation of the local Goldstone phase field of the broken symmetry macro-quantum vacuum, hence the geodesic equation is an automatic result. Global special relativity and quantum field theory is already built into the unstable Minkowski space. Basic force laws come from the quantum action principle is the standard way. In other words F = ma is built into the Feynman path quantum action Mach's principle is inherent in the Feynman path-Hoyle-Narlikar influence functional response of the universe approach to cosmology from the quantum action principle. That is consistent with my theory. Aharonov's advanced destiny quantum state vector is automatically built in with the Costa De Beauregard/John Cramer Feynman Zig Zag explanation of micro-quantum EPR nonlocal correlations violating Bell's locality inequality. ==== > >A question I sometimes ask... > >Give an example of an object, defined in the language of ZF, >such that (1) it is provable in ZF that it is a countably infinite >set of pairs, but that (2) there is no obvious choice function. > >When you omit countably infinite I can do it: the set of >all pairs of subsets of R. But of course that is not countably >infinite. > > Consider the Galois fields F_n = G(p^2^(n-1)), n = 1, ... , p odd. > There are two isomorphic mappings from F_n to F_{n-1}; how can one > choose which one? Can we be more specific here? Like this? ... -> G(p^16) -> GF(p^8) -> G(p^4) -> G(p^2) -> G(p) What sort of maps are we talking about? Are the fields G(p^2^(n-1)) presented to us explicitly in some way? Or do we (before we even start) arbitrarily choose one representative of each isomorphism class? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== > I realize that to you my claims sound miraculous and that my answers > seem incomprehensibly mysterious (like hocus-pocus), but try to > question your programming and understand what I'm saying. SR boasts > that there is no absolute time order, no absolute frame of > reference, and that motion faster than light results in some > observer interpreting the motion as time-travel into the past. > Never into the past... ever... The arrow of time only > points forward. Paul, I'm asking you to set aside your own opinions of the world and select one of two SR type models of the universe. Who do you think invented the time-travel industry? I'll give you a hint. It wasn't Rod Serling, Gene Roddenberry or any travel agency. It comes from SR. http://www.everythingimportant.org/viewtopic.php?t=605 http://www.everythingimportant.org/relativity/simultaneity.htm Eugene Shubert ==== > I realize that to you my claims sound miraculous and that my answers > seem incomprehensibly mysterious (like hocus-pocus), but try to > question your programming and understand what I'm saying. SR boasts > that there is no absolute time order, no absolute frame of > reference, and that motion faster than light results in some > observer interpreting the motion as time-travel into the past. > Never into the past... ever... The arrow of time only > points forward. Paul, I'm asking you to set aside your own opinions of the world and select > one of two SR type models of the universe. Who do you think invented > the time-travel industry? I'll give you a hint. It wasn't Rod Serling, > Gene Roddenberry or any travel agency. It comes from SR. http://www.everythingimportant.org/viewtopic.php?t=605 > http://www.everythingimportant.org/relativity/simultaneity.htm Eugene Shubert Since you asked... I cannot set aside my own opinions just as you cannot set aside yours. Any view I have is based on my world view bias just as yours is ... My understanding of the observed universe is different from yours and yours is different from SmFarts ( well everyone's is different from smFarts) ( Sorry I had too) Given that both views consider only the effect aspect of matter in motion in relation to other matter in motion there are missing rules and assumption used in both that give me more issues than the differences between your hypothesis and the standard models of SR. I tend to think that the existing rules, to the point of failure due to scale, have worked pretty good and I also tend to give the standard models of relativistic interplay. Yours has merit for the effort and I would never denigrate a works that has intellectual content,,, even if shown to be incorrect,,,, As for running the numbers I have made it clear in the past that my interest are more generalized and I suck at Math so I leave that to others to pick at and I will read the explanation of the mathematical expressions and then may have a better view... But so as to not miss lead you if you have not read my lunacy in the past... I'm an engineer, not a physicist and tend to look for ways to poke at the horned rimmed glasses and pocket protector crowd ( with great respect) to see if maybe a new idea my pop into homebodies head ...mine ,,, theirs... yours.. which ever.... I will say that I do not think there are any paradox in physics... its more of a philosophical issue and crops up when the existing rules fail (almost always due to scale) .. I also say the same for several aspects of physics... Infinity is an illusion that occurs when physical models fail due to scale, c is a variable that varies at the rate of separation of all matter universally ( a variance so out of scale as to be undetectable as other than a constant on human scale) ...the UGC is a variable that varies at the rate of separation of all matter universally ( a variance so out of scale as to be undetectable as other than a observable aspect of duality are only EM waves sampled by a method and concept biased as to lead to interpretation X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 10:22 AM, perfectlyInnocent@as-if.com (Perfectly Innocent) said: >The special theory of relativity is a mathematical subject. The. SR is a [hysical model, and the issue of its validity is Physics, not Mathematics. If you don't understand a basic point like that, then it woiuld be a waste of time to read your web site. >I have proposed a simple mathematical riddle: Do logically >consistent, alternative, SR type theories exist? You're a century too late, and the answer is irrelevant. The proper questions are: 1. Is an aether theory with identical predictions easier to use? 2. Can an alternative theory be extended to QFT without adding complexities? 3. Can an alternative theory be extended to gravitation without adding complexities? >The question has been submitted in terms of axioms. The applicability of those axioms to the real world, and their utility or lack thereof, are questions for the physicists, not the mathematicians. >The solicited proof for or against my proposed solution What a mathematician means by proof is very different from what a physicist means by it. For a physical theory, the proof is imperical, and is a matter for physicists. A mathematician can say whether your predicitions follow from your axioms, but can have nothing to say about whether they match the real world. Let me give an analogy. There is a lot of Physics involved in a baseball game. But were I to post a play-by-play description of a game is sci.physics, the reader would, quite properly, chastise me for an off topic post, and would not buy my claim that it was Physics. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org ==== > at 10:22 AM, Perfectly Innocent said: > >The special theory of relativity is a mathematical subject. > SR is a physical model, and the issue of its validity is Physics, not > Mathematics. If you don't understand a basic point like that, then > it would be a waste of time to read your web site. SR is Minkowski geometry. Geometry is mathematics. SR is also a physical model but I'm not raising the issue of its validity. It's obvious that you misunderstand my question, do logically consistent, alternative, SR type theories exist? By exist I mean: exist as mathematical models, not as real physical realities in a parallel universe. >I have proposed a simple mathematical riddle: Do logically >consistent, alternative, SR type theories exist? > You're a century too late, and the answer is irrelevant. So some mathematical questions are irrelevant? Please define mathematical irrelevance. >The question has been submitted in terms of axioms. > The applicability of those axioms to the real world, and their utility > or lack thereof, are questions for the physicists, not the > mathematicians. Where in the world do you get the idea that my view of those axioms has to do with applicability and utility? >The solicited proof for or against my proposed solution > What a mathematician means by proof is very different from what > a physicist means by it. For a physical theory, the proof is > imperical, and is a matter for physicists. I'm using the word proof in its strictest mathematical sense. The question I've raised is strictly mathematical and has nothing to do with practical measuring by experimental physicists. > A mathematician can say > whether your predicitions follow from your axioms, but can have > nothing to say about whether they match the real world. Exactly! And as a mathematician, I'm saying that all the predictions of my model are identical to the standard model within our galaxy but differ on a cosmological scale. I'm only asking that mathematicians confirm or refute my claim in terms of mathematics, which you just said they could do. already answered that question. Physics is too difficult for the physicists. http://www.everythingimportant.org/viewtopic.php?t=605 Eugene Shubert ==== > Exactly! And as a mathematician, I'm saying that all the predictions > of my model are identical to the standard model within our galaxy but > differ on a cosmological scale. That's physics, not mathematics. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== Portfolio of PAF as of 27AUG03 50 BCE 21.64 $1,082.00 50 BLS 25.25 $1,262.50 100 BMY 25.37 $2,537.00 50 DT 14.57 $728.50 51,000 Q 4.50 $229,500.00 10,250 SBC 22.73 $232,982.50 380 VZ 35.10 $13,338.00 80 WYE 42.64 $3,411.20 realestate land 3APR03 of 3 lots $19,000 realestate land 30JUL03 another lot $11,500 art of science-lithographs & porcelain JAN-JUN03 for $12,000 Well, today I sold 950 shares of BMY at 25.35 in order to buy 1050 shares of SBC at 22.75 because of the Crossover. BMY. I expect in the near term future that SBC will climb considerably higher than BMY and I can thence re-play the above all over again and laughing all the way to the whatever. The Stockmarket as an entity is like a gigantic fruit orchard with thousands and millions of free cherries to pick. Absolutely free because they are Crossover entities. If I had bought SBC only back in 3 OCT 2002, then the PAF portfolio would be 20,900 shares of SBC and only that. Collecting the dividends would be the only income. But with Crossover, PAF has gained considerably more in wealth. Even though PAF ran into a dismal experience with SGP. Expectation: I expect in the next 3 months that SBC will be considerably higher than BMY and that the fat dividend that SBC pays will also be very nice. Perhaps SBC will do another one of those extra dividends of 10 cents on top of its regular dividend something that BMY is surely not capable of at this time. The VonNeumann Gametheory applied to the Stockmarket is a picture of Crossover applications where someone continues to pick free cherries, free peaches, free whatever as they continue to apply Crossover phenomenon. Whereas a person who buys a stock and sits on it for years with the expectancy of a gain plus collecting of dividends. The Crossover expects not only the dividends but also the future stock rise but the added dimension of free pickings of the Crossover itself. So a regular investor has only 2 dimensional gain expectancy. The Crossover player has 3 dimensional gain expectancy. Good Question: can the Crossover technique have avoided me of the dog that was SGP? I think not. I think, though, that a Crossover player is better able to detect dogs in a portfolio than nonCrossover players, in that the zest or vitality of a company when compared with others is more easily observed and pronounced. But as to whether the Crossover player can avoid dogs completely is skeptical. It may turn out that SGP is bought in the next weeks at a price of nice proportions, but it is also likely that SGP will fall to the price of Elan of $5 a share and just flounder for 2 years in the bucket. whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== I am having a problem linking to the CLU library from the examples using http://www.perwass.de/cbup/cludownload.html I had to make changes like making class members public, it wasn't recognizing the calls marked as friend. At this point, the library compiles fine but linking has unresolved symbols like, BladeList, MultiV, Blade in all those functions that were declared friend. ==== typo-corrections & additions 2nd draft Dear Dr. Sarfatti, only some of what I can understand. own fashion. You say dark energy is an exotic vacuum phase of negative quantum zero point fluctuations, and dark matter is the positive quantum /zpf. But dark energy is a positive energy form left over from the inflation of space. The dark energy flux is an accounting gimmic to explain the negative energy (work) required to inflate space (and have the universe remain flat). In fact, I don't see any reason to suspect that dark energy is necessarily any different than regular matter. No you do not understand the basic physics here and have not properly described above what I am talking about. Part of the basic physics is on pages 25-26 of John Peacock's Cosmological Physics showing why covariance requires w = -1 for zero point fluctuations of all quantum fields. Dark energy has / zpf > 0, i.e. positive zero point energy density with equal and opposite negative zero point pressure. What matters in Einstein's GR is energy density + 3pressure in the Poisson equation for the system, in this case, the exotic quantum vacuum. Dark matter is simply /zpf < 0, i.e. negative zero point energy density with equal and opposite positive zero point pressure. This explains the dark matter not only in large scale but also in micro-scale i.e. 1. Why self-charge does not cause electron to explode, i.e. the old Abraham-Becker self-energy problem. scattering yet e/mc^2 ~ 1 fermi and h/mc ~ 137e^2/mc^2 e^2|/zpf|^1/2 where microscale dark matter cores of lepto-quarks have /zpf < 0 and |/zpf|~ (1 fermi)^-2 4. This same micro /zpf gives the correct universal Regge slope of hadronic resonances alpha' ~ (1Gev)^-2 5. The universal micro dark matter exotic vacuum core of the lepto-quarks is |/zpf| ~ Lp^-4/3(c/Ho)^-2/3 from Susskind's world hologram model. Lp^2 = hG/c^3 For more details go to http://qedcorp.com/APS/Ukraine.doc http://qedcorp.com/APS/Vigier4.pdf My theory predicts null results for all conventional dark matter I explain both large scale and small scale structure of universe with one simple mainstream idea from P.W. Anderson More is different (ODLRO from spontaneously self-organizing BCS pairing of virtual lepto-quarks with virtual anti lepto-quarks inside the physical vacuum creating the complex scalar inflation field.) - Occam's Razor. If I haven't offended your scientific sensibilities yet, I would space. Does it exist? What are its properties? Is there any room for Mach's principle in this picture? You will recall that Mach claimed that inertial frames are determined by the disposition of matter in the surrounding universe. Does your concept of space allow for the succint derivation of the fundamental equation, F=ma? Yes, of course. Read my papers above. :-) F = ma is basically timelike geodesic equation for neutral test fields. I derive Einstein's GR from the modulation of the local Goldstone phase field of the broken symmetry macro-quantum vacuum, hence the geodesic equation is an automatic result. Global special relativity and quantum field theory is already built into the unstable Minkowski space. Basic force laws come from the quantum action principle in the standard way. In other words F = ma is built into the Feynman path quantum action Mach's principle is inherent in the Feynman path-Hoyle-Narlikar influence functional response of the universe approach to cosmology from the quantum action principle. That is consistent with my theory. Aharonov's advanced destiny quantum state vector is automatically built in with the Costa De Beauregard/John Cramer Feynman Zig Zag explanation of micro-quantum EPR nonlocal correlations violating Bell's locality inequality. I'm having a hard time understanding the use of epsilon and delta in proofs. Are there any books that just focus on this rather than going through it pretty fast and moving on to the rest of analysis? I would appreciate any help at this point. I'm really frustrated. ==== >I'm having a hard time understanding the use of epsilon and delta in proofs. >Are there any books that just focus on this rather than going through it pretty >fast and moving on to the rest of analysis? I would appreciate any help at >this point. I'm really frustrated. Let's consider a simple proof. A function f : R -> R is said to be _continuous_ at a point x0 if given any delta > 0, there exists an epsilon > 0 such that | f(x) - f(x0) | < delta whenever | x - x0 | < epsilon. Your job is to prove that if f1 and f2 are continuous at x0, then f1 + f2 is also continuous at x0, where (f1 + f2)(x) is defined to be f1(x) + f2(x). An outsider picks delta, subject only to delta > 0. To distinguish this specific number from the generic roles of delta and epsilon, I'll call his input delta0. You, the prover, are told the value of delta0 and need to supply epsilon0. Your epsilon0, which may depend on delta0, must ensure that |(f1 + f2)(x) - (f1 + f2)(x0) | < delta0 whenever |x - x0| < epsilon0 First use the definition of f1 + f2 to translate your requirement into |f1(x) + f2(x) - f1(x0) - f2(x0) | < delta0 whenever |x - x0| < epsilon0 Variants of the triangle inequality come up frequently in analysis. Your requirement will be satisfied if |f1(x) - f1(x0)| + |f2(x) - f2(x0)| < delta0 whenever |x - x0| < epsilon0. You have not specified epsilon0 yet. One way to achieve your revised goal requires both |f1(x) - f1(x0)| < delta0/2 and |f2(x) - f2(x0)| < delta0/2 whenever |x - x0| < epsilon0. This looks simpler than the earlier goal because you have separated the roles of f1 and of f2. The delta you've been handed (namely delta0) is guaranteed to be positive; hence delta0/2 > 0 too. Somebody else is guaranteeing that f1 and f2 are continuous at x0 -- this time you can require him to supply an epsilon > 0 for each delta > 0 you send him. |f1(x) - f1(x0)| < delta0/2 whenever |x - x0| < epsilon1 and an epsilon2 > 0 such that |f2(x) - f2(x0)| < delta0/2 whenever |x - x0| < epsilon2. Observe that you are allowed to submit the same positive value delta0/2 for both find me an epsilon queries but the returned values of epsilon may be different. They have been given different names, epsilon1 and epsilon2. Both are guaranteed to be positive. Now you try to supply a suitable epsilon (namely epsilon0) such that both | x - x0 | < epsilon1 and | x - x0 | < epsilon2 are achieved whenever |x - x0| < epsilon0. You choose epsilon0 = min(epsilon1, epsilon2), the smaller of the two values returned by queries about f1, f2. Your epsilon0 is guaranteed to be positive since both epsilon1 and epsilon2 are positive. The actual proof would be written up in backwards order, with much of the material abbreviated (and using delta, epsilon in place of delta0, epsilon0) Let delta > 0. Using the continuity of f1 and f2 at x0, select epsilon1 > 0 and epsilon2 > 0 such that |f1(x) - f1(x0)| < delta/2 whenever |x - x0| < epsilon1 |f2(x) - f2(x0)| < delta/2 whenever |x - x0| < epsilon2 . Let epsilon = min(epsilon1, epsilon2) > 0. If |x - x0| < epsilon, then |x - x0| < epsilon1 and |x - x0| < epsilon2. Hence | (f1 + f2)(x) - (f1 + f2)(x0)) | = | f1(x) + f2(x) - f1(x0) - f2(x0) | <= |f1(x) - f1(x0)| + |f2(x) - f2(x0)| < delta/2 + delta/2 = delta Most of the proof consists of understanding the definitions of continuous and of f1 + f2. The strategic parts were i) Deciding where and how to employ the triangle inequality. ii) Replacing one inequality with bound delta by two inequalities each with bound delta/2. iii) Realizing that we could use epsilon = min(epsilon1, epsilon2) in order to simultaneously satisfy both inequalities in ii) after we had satisfied them individually. These techniques will occur repeatedly in your analysis course. Understand definitions carefully. Do your homework to practice the techniques. If you master these steps, you'll master the course. -- Wanted: Experts at choosing the best of 100+ applicants for a position. Register as a California voter by September 22, and vote on October 7. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI ==== Katie88265 a dit : > I'm having a hard time understanding the use of epsilon and delta in > proofs. Are there any books that just focus on this rather than going > through it pretty > fast and moving on to the rest of analysis? I would appreciate any > help at > this point. I'm really frustrated. Epsilons and deltas are easy to understand and to visualize, provided you : - draw what you want to prove - understand what depends on what (does this delta depends on x or does it not ?) : if you try to understand this (for instance with the notation delta with a x subscripted, you know that delta depends on x) you will understand the proofs easily. -- Alexandre Charitopoulos Em6 / Eb7(5b) / Dm7 / Db7(5b, 9b) / Cmaj7 ==== > I'm having a hard time understanding the use of epsilon and delta in > proofs. Are there any books that just focus on this rather than going > through it pretty fast and moving on to the rest of analysis? I would > appreciate any help at this point. I'm really frustrated. Many Calculus texts have special sections, often in the back in an appendix, where delta/epsilon proofs for limits and such are carefully delineated. (The darned things are 1200 pages thick now, so they ought to have _something_ about rigorous limits in them.) Bart ==== > I'm having a hard time understanding the use of epsilon and delta in proofs. > Are there any books that just focus on this rather than going through it > pretty > fast and moving on to the rest of analysis? I would appreciate any help at > this point. I'm really frustrated. You could take a look at Principles of Mathematical Analysis by Rudin. There are many e/d proofs there, in the context of metric spaces. If you give an example of a particular proof you're having trouble with, perhaps readers here could help. ==== I saw a poster mention the Twin Primes Conjecture and it got me to thinking. What I know from my prime counting function is that given X a prime number, it is required for X+2 to be a prime number if pi(X/2,1) = pi((X+2)/2,1) and pi(X/3, 2) = pi((X+2)/3,2). For instance, with 17 and 19, you have pi(17/2,1) = 4, and pi(19/2, 1)=4, while pi(17/3, 2) = 3 and pi(19/3) = 3. That's because with such a short difference as 2 between X and X+2, only two dS values can be different. So proving the Twin Prime Conjecture is just a matter of proving that as you go out to infinity you can always find an X, such that pi(X/2,1) = pi((X+2)/2,1) and pi(X/3, 2) = pi((X+2)/3,2). Oh, you need my prime counting function to evaluate, of course. It seems to me that giving that away might speed things up, so one of you can just go out, see if you can prove it, and post so then you can take some credit for proving the conjecture. If you decide to be silly, I'll come in later and prove it myself, and probably have yet another thing that mathematicians want to fight me over. But if you ever wanted to be famous, here's your chance to put your name up there, and get in the history books. James Harris ==== I think you forget a very fundamental effect in your works : you do not take in account the effects of Einstein's relativity. Let me explain what I mean : twin primes are primes that are distant of 2; right ? But it is well known from relativity that the fastest you go, the more distances shrink ! So you must admit that as soon as you are in motion, the distance between any 2 twin primes is no longer 2 but a bit less that 2, add thus this two primes are not twin any longer !!! So my question is : are you sure that each time you try to make a demonstration dealing with twin primes your speed is 0 ? ;-)) J-L. ==== > I saw a poster mention the Twin Primes Conjecture and it got me to > thinking. > > What I know from my prime counting function is that given X a prime > number, it is required for X+2 to be a prime number if > > pi(X/2,1) = pi((X+2)/2,1) > > and > > pi(X/3, 2) = pi((X+2)/3,2). > > For instance, with 17 and 19, you have > > pi(17/2,1) = 4, and pi(19/2, 1)=4, According to your definition elsewhere, pi(x,y)= floor(x) - S(x,y) - 1, where S(x,1) = 0 pi(17/2,1) = floor(8.5) -1 = 7 pi(19/2,1) = floor(9.5) -1 = 8 Have you changed the definition of pi(x,y) in the past few hours? -- Will Twentyman ==== > > I saw a poster mention the Twin Primes Conjecture and it got me to > thinking. > > ... > > If you decide to be silly, I'll come in later and prove it myself, and But what if it's false? -- G.C. ==== > I saw a poster mention the Twin Primes Conjecture and it got me to > thinking. What I know from my prime counting function is that given X a prime > number, it is required for X+2 to be a prime number if pi(X/2,1) = pi((X+2)/2,1) and pi(X/3, 2) = pi((X+2)/3,2). For instance, with 17 and 19, you have pi(17/2,1) = 4, and pi(19/2, 1)=4, while pi(17/3, 2) = 3 and pi(19/3) = 3. That's because with such a short difference as 2 between X and X+2, > only two dS values can be different. So proving the Twin Prime Conjecture is just a matter of proving that > as you go out to infinity you can always find an X, such that pi(X/2,1) = pi((X+2)/2,1) and pi(X/3, 2) = pi((X+2)/3,2). Seems like a piece of cake. Just plug in 'infinity' for X and solve. > Oh, you need my prime counting function to evaluate, of course. Naturally. Otherwise there is no way to perform these calculations. > It seems to me that giving that away might speed things up, so one of > you can just go out, see if you can prove it, and post so then you can > take some credit for proving the conjecture. If you decide to be silly, I'll come in later and prove it myself, and > probably have yet another thing that mathematicians want to fight me > over. Why wait? Go ahead and prove it yourself now. > But if you ever wanted to be famous, here's your chance to put your > name up there, and get in the history books. James Harris Isn't that what you've been working towards? Why stop with all your previous achievements. Add a new one to your trophy case. -- 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 ==== > > working on it for a year and I'm looking for an intelligent and > thoughtful evaluation and feedback. > > Evaluating the paper requires knowledge of posets and Dilworth's > Theorem, as this is the key insight in the proof (In the paper, we > look at the subset-sum problem and think of the set of all subset-sums > as a poset). I welcome and take seriously any feedback. > > Craig you gotta some funny ideas about modern rigor, buddy. but there are probably amatuers out there who would find that proof apealing. good luck. ==== >> the norm topology, T_B is the weak topology? Or is that backwards? >a Baire space with respect to the weak topology. It isn't. Let {e_n}_{n=1}^infinity be an orthonormal basis of your This is open and dense in the weak topology. But the intersection of the G_n is empty. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > > programs. How do you define programming? > > Writing: > > FAC(x,I)^LT(x,I)^~(exists A)FAC(A,I)^LT(A,I)^LT(x,A) > > Is programming, no matter what system translates that input into a program. > > Sam > > 1. Do you consider Predicate Calculus wffs to be programs? > >(If it is > written in a logic textbook, it is not a program; if it is passed to a > Prolog interpreter, it is.) What would the Prolog interpreter do? Could you use the above wff or ~LT(x,2)^~(exists A)PPFAC(A,x) as an example and describe what it would do and produce? > 3. Do you know of a better way to sepcify the largest proper factor > of a given number? > Yes: the largest proper factor of a given number. So if a system took that as input and produced a computer program, then it would constitute a program synthesis system? And has anyone ever accomplished that, to your knowledge? Have they tried? Is it a worthwhile problem? > 4. Do you know of a way to specify it that isn't programming? > > Yes: the largest proper factor of a given number. It would be pretty straightforward to translate between that phrase and the predicate calculus wff. We need only define the construction of each term such as largest [ P(a) -> P(a)^~(exists A)P(A)^LT(a,A) ] and factor [ (exists A)MUL(A,a) ]. How could the wff be programming and the English equivalent not, when the semantics are the same - they only difffer in the syntax used? > 6. How do you think that a Mathematician would specify it? > > The largest proper factor of a given number. You don't think there is a formal mathematical expression for that English phrase? > 7. Did you know that the state-of-the-art in Program Synthesis is to > specify the program requirement as a Predicate Calculus wff? > > Not surprising. I doubt it's really state-of-the-art though; I > expect people having doing that for decades. What makes you think that? Have you ever seen or heard of a system that succeeds at translating between predicate calculus wffs that refer to the natural numbers (Number Theory) and computer programs? > 8. Do you see a qualitative difference between a Predicate Calculus > wff and a computer program? > > Yes; a wff is an assertion, while a program is a command. Agreed. That is my whole point. > 9. How about the fact that a Predicate Calculus wff has no assignment, > conditional execution, loops or the possibility of not terminating? > > Many programming languages don't have any of these things, except the > last. And what would it mean for a wff to not terminate anyway? It would only have meaning if a wff were somehow executed as a program is. > 10. Do you think that Predicate Calculus wffs and computer programs > are at the same level of abstraction? > > No, wffs are at a higher level of abstraction, because they need not > be associated with any sort of computer. Exactly. > 12. Do you think that computer programs have to be analyzed to > determine what function they compute, and in general you cannot do > that? > > No. The (partial) function determined by program x is already > determinate, independently of any analysis. But is it discernible in general? > 18. Do you know of any system other than mine that determines programs > that compute a given wff? > > Yes; Prolog springs to mind immediately. How would Prolog determine programs that compute wff ~LT(x,2)^~(exists A)PPFAC(A,x), that is, list the prime numbers? Or FAC(I,J) determine if one number is a factor of another? [My understanding of Prolog is that it is just a database that you create and query, having no relationship to the study of infinite mathematical objects such as the natural numbers that are the basis of Number Theory.] Charlie Volkstorf Cambridge, MA ==== > >> In my case, after the programming language is axiomatized, nobody >> system (using the definitions, axioms and rules.) >The formalism is still programing. > > I've been wondering whether anyone else would point this out... Do you see a difference between programs and predicate calculus wffs? Are they at the same level of abstraction? In what sense is a predicate calculus wff a program? Is there value in being able to map back and forth between a computer program and a predicate calculus wff that defines the result of running that program? Is this a problem that has been actively pursued by researchers? >Just because you disguise it in >axiomatic form does not change the underlying realities. In the last >step all that you write will be translated into machine language, if and >when it actually runs. But what input is required to producre that program? Isn't that by definition of the problem of program synthesis itself? You produce a program without programming. Your output is a computer program. Yes, you can translate or execute or print it if you wish. That's what a program is supposed to do. That is, your assertion applies to any program synthesis system by definition of the fact that we have a system that creates computer programs. Is the notion of program synthesis a valid one? Is there any conceivable scenario (not that it is necessarily possible) which would constitute program synthesis? Then what would the input be? Can you express it in formal terms? >You either are kidding us, or you believe in symbolic juju and magic I give detailed algorithms in my papers as to how the programs are produced, including several examples from Number Theory, Database Query Processing and the Theory of Computation. Charlie Volkstorf Cambridge, MA >Bob Kolker >Sam > > ************************ > > David C. Ullrich ==== > Is there value in being able to map back and forth between a computer > program and a predicate calculus wff that defines the result of > running that program? Is this a problem that has been actively > pursued by researchers? how is that question relevant to the thread? AFAIK, no one has criticized the usefulness of your proposed system yet; the main critic is the fact that despite what you say, using PC wff to specify tasks is programming. Sam -- [...] but the delight and pride of Aule is in the deed of making, and in the thing made, and neither in possession nor in his own mastery; wherefore he gives and hoards not, and is free from care, passing ever on to some new work. - J.R.R. Tolkien, Ainulindale (Silmarillion) ==== > 1. Do you consider Predicate Calculus wffs to be programs? > > Some Predicate Calculus wffs are programs. > > 2. How do you define a program? > > A program is an machine-readable specification of acceptable (computer) > behavior. How would a computer behave given a Predicate Calculus wff? > 3. Do you know of a better way to sepcify the largest proper factor > of a given number? > > No. > > 4. Do you know of a way to specify it that isn't programming? > > No. Then how would a program synthesis system generate a program to determine the largest proper factor of a given number without being merely programming? > 10. Do you think that Predicate Calculus wffs and computer programs > are at the same level of abstraction? > > Predicate Calculus's heavy use of quantifiers makes it more abstract at the > language level than many of today's computer programming languages. That's right - it is at a higher level of abstraction. PC defines a set. A program defines an algorithm that performs a particular operation on a set. > 16. Do you see value in being able to determine computer programs that > implement a given predicate calculus wff? > > Yes. > > 17. Does my system determmine programs that compute a given wff? > > Not if those programs are required to halt for all inputs. Why and who said it did? > 18. Do you know of any system other than mine that determines programs > that compute a given wff? > > False premise. No premise intended. Charlie Volkstorf Cambridge, MA ==== > >>The formalism is still programing. > > I've been wondering whether anyone else would point this out... > > That's what I've been trying to explain from the beginning. PC is a > programming language, though it does not have conditional branching and other > common features. I have also noted that, according to Charlie's paper, it is > translated to APC (which is similar to pseudo-code in its syntax). Therefore, > the system is actually a translator from a high-level non-procedural language > to a procedural one. That does not contradict the fact that it is a program > generator (Charlie does not seem to accept that idea, though), provided it > actually works (since Charlie has not accpted to make his implementation > available, even though he supposedly has used it in replying to one of my > post) > > And again, almost any syntax that specifies a task to be performed is a > programming language. In fact, English can be one (check out the Shakespeare > programming language) > > Sam This is deja vue all over again, so I will proceed with what I noticed awhile ago. One of the strongest ways to attack something is to do so at a very high, general level of abstraction. That is, declare in the most general terms how it violates required principles. There is no alternative than to throw out the entire attempt. The problem with this approach is that it is so broad that it doesn't attack the proposed solution at all. It attacks the problem itself. It paints a condemnation so broad that the problem itself takes the condemned form. In the case with program synthesis, the attack might be that the system for representing the program requirement is no good. And the attack continues with as broad a paint brush as possible - thus condemning the notion of program sythesis itself. There is input and a program is created. Nothing is said about the syntax or semantics of the input. But it is still condemned as being programming and thus is not program synthesis. How could a program synthesis system work without having input to define the programming requirement? What would be a better system than the Predcate Calculus? This is what mathematicians have been using for 100 years and is the ultimate case of Occam's Razor: we are using existing, established mathematics that is here to stay regardless - so if we use that, then it is for free. We minimize the system by not adding anything new. Is the concept of a program synthesis system a valid one? If yes, then describe how it would work. What would be a good input? If you repeat your claim that the LISP programming system is program synthesis, I ask only: Do you (plural) have to input new LISP programs as you use the system? Are you not merely entering in a series of programs, and finally a single line that calls these progams in a certain order? Then you are merely typing in a program in LISP. Where is the program that is supposed to be created and output by the system? There is none. You input a program rather than outputting one. If a person could type in a short, free-form description of what he wanted the new program to do, and out popped one or more programs that did just that, would that constitute program synthesis? But at the same time, you have condemened program synthesis itself: any syntax that specifies a task to be performed is a programming language. English can be one. To perform program synthesis, we must have a syntax to specify the task to be performed, no? But any such syntax constitutes programming. So how could a program synthesis system ever exist? The Predicate Calculus is the (or at least a) standard mathematical way to represent sets and function. It is very general. It is simple. It has only a handful of concepts: not, and, or, exists, for all, paremtheses. It doesn't require you to specify an algorithm to perform any particular task. You can construct multiple different algorithms for it. It has no assignment, if commands, infinite loops. These are some of the reasons that I use the Predicate Calculus. Is there a better way? Is anything else even close? What is next best? PREDICATE CALCULUS VS PROGRAMMING LANGUAGES The predicate calculus is a system for specifying predicates that are either true or false. It is declarative - nonprocedural. We state what the condition is, not how to get it in any particular way. A program is a series of functions that are executed until they run to completion or get into an infinite loop. The author must develop a series of commands that will be executed in order to actually output (list) a set, or calculate a single value, or decide if a given predicate is true or false. The Predicate Calculus doesn't require you to give a program that calculates the set or function being defined. The Predicate Calculus is declarative like English. How would you define what a prime number is? Someone must tell the system what it means to be a prime number. That is the lowest level of abstraction. What is a prime number? I would define a prime number as a number that is not less than 2 and which has no proper factors. I would define a proper factor of N as a number that is a factor of N and is not equal to 1 or N. I would define a factor of N as a number such that there is a number that when multiplied by the factor equals N. How would you? And how do we express this in the Predicate Calculus? Well, since both English and the Predicate Calculus are declarative, the semantics are the same. We have a predicate written in English that we wish to express in the Predicate Calculus. a number that is not less than 2 : ~LT(x,2) and : ^ which has no : ~(exists A) proper factors : PPFAC(A,x) Concatonate it verbatim and we get: a number that is not less than 2 and which has no proper factors is: ~LT(x,2)^~(exists A)PPFAC(A,x) The semantics are the same. It is a matter of translating the syntax of English into the syntax of the Predicate Calculus. What could be easier or simpler? A lot of people have spent a lot of time studying the relationship between the Predicate Callculus and computer programs. Some have even tried very hard to translate between Predicate Calculus wffs and computer programs. How would you represent the functionality provided by a computer program? You said that a Predicate Calculus wff is a program and a program is something that a computer can execute. So I asked you how a computer would execute ~LT(x,2)^~(exists A)PPFAC(A,x) and you said you didn't know my system. But this is not a question about my system per se. I make no mention of my system in the question. I am only asking how this program can be executed. The answer is, it can't be executed in the normal sense of the word. Yes, it can be input into a system. Yes, there can be output from that systen. And if you want to call that a translation, then go ahead, although a definition should relate something unknown to something known. But rather than being a series of simple (recursive) functions like constant zero, or add one, or multiply two numbers as in an actual program that are executed in series, a Predicate Calculus wff doesn't supply a series of functions for us to execute. Instead, it is a statement - a predicate - that is true when a given value is in a set we are defining. How can we execute the declaration that x is not less than two and has no proper factors? A Predicate Calculus wff defines a set. A program is an algorithm to carry some particular process concerning a set, such as listing its elements or deciding if a given value is an element. They are at different levels of abstraction. Programs define (processes on) sets and functions that can be proven to have their particular values. Predicate Calculus wffs define sets and functions in terms of wffs that are true when the set or function is stated to have a correct value. The distinction is that between the provable wffs and the true wffs. (Godel showed they are not the same.) It is much easier to declare something by giving a statement that is true or false depending on the condition (that is called talking) than it is to give a wff that is provable (aka programed) when the correct values for that set or function are given. In some cases, the latter is impossible while the former is possible. Charlie Volkstorf Cambridge, MA ==== I've added sci.math, I understand they know these sorts of things. > Afken says >> div V = 1/sqrt(g) @(sqrt(g)V^k)/@q^k >> where g is the determinant of the metric. So for spherical >coordinates >> I'm getting g=r*sqrt(sin(theta)) and >> sqrt(g) = r * sin(theta) >>shouldn't that be > sqrt(g) = r^2 * sin(theta) ? >;-) >> Should be. I had a little bit of idiocy going back there, but I tried it > that way and got the wrong answer, so I managed to convince myself that I > had too many squares in the determinant, and still got the wrong answer. >>Hehe, that's what happens when you run around >>in Minkowski-circles for too long ;-) >> So I started doing imaginary math. I guess I should have expected that, >> under the circumstances. >> Maybe someone can find an error if I go through this step by step. All >> derivatives are partial, so I'll just use d's instead of trying to >> symbolize partials with a-holes or something. >> Arfken says: >> div V = sum_k 1/sqrt(g) d(sqrt(g) V^k)/dq^k >> Spherical polar coordinates has sqrt(g)=r^2 sin(theta). Go through the >> k's one by one. >> k=r: 1/r^2 sin(theta) d/dr (r^2 sin(theta) V^r) >> = 1/r^2 d/dr (r^2 V^r) >> Beautiful. >> k=theta: 1/r^2 sin(theta) d/d(theta) (r^2 sin(theta) V^theta) >> = 1/sin(theta) d/d(theta) (sin(theta) V^theta) >> Wrong. The derivative should be multiplied by 1/r sin(theta). But there >> was an r^2 in the top and an r^2 in the bottom and no d/dr, so they >> canceled on me. >> k=phi: 1/r^2 sin(theta) d/d(phi) (r^2 sin(theta) V^phi) >> = d/d(phi) V^phi >> Wrong, the derivative part should be multiplied by 1/r sin(theta). But >> the r^2 sin(theta) just passed right through the derivative and canceled. >> I'm doing something wrong here, but I don't know what. You probably are expecting the physical components V_(k) >in stead of the contravariant components V^k of the vector. >They are related by > V_(k) = sqrt(g_kk) V^k (no summation) >i.o.w. > V^r = V_(r) > V^t = 1/r V_(t) > V^f = 1/[r sin(t)] V_(f) When you fill in these ones in the above result, you get >what you are looking for Yes, I've noticed that, but I couldn't fill in the missing steps to prove it. The books I've looked at have been kind of vague on that sort of thing. Like when you have vector components multiplied by a basis, do the indices of the components have any particular meaning, or just the basis vectors? E.g. V^i e_i and V_i e^i? Does V_i e_i make sense? It would be tempting to raise one of them like g_ij V^j e_j, but that's not sqrt(g_ij). I get the same thing as above if I try to compute div V with the Christoffel symbol method, div V = V^i_;i and I've checked my Christoffel symbols again and again, so I thought I must be missing some relation between what I get and what's on the inside cover of Jackson. >I have been wondering about this too when I was working >through my Schaum's tensors last year. >See also example 1 of > >http://gershwin.ens.fr/vdaniel/Doc-Locale/Cours-Mirrored/Maths-Stuff/tensor calc/part7.PDF -- A nice adaptation of conditions will make almost any hypothesis agree with the phenomena. This will please the imagination but does not advance our knowledge. -- J. Black, 1803. ==== Dear Jack, I have sent yesterday the abstract to Richard Amoroso (attached). ... I am glad to see, that the good correlation between some of our ideas is existing. The interplay and feedback [IT <===> BIT] is one of my Unified model feature also, if under IT to understand a matter (including biosystems) and under BIT its virtual replica (VR) in Bivacuum, containing the information about IT. The [IT + BIT] represents in general case complex macroscopic (star systems) levels, able to self-organization. I do not understand what you mean by VR and Bivacuum? In Bohm's realism this is very simple. BIT is simply the deBroglie qubit according to v = (h/m)Grad(argPsi) Self-organization means the BIT Psi has as a source its own IT hidden variable trajectory making the fragile Bohm quantum potential robust. IT is a system point rolling in gradient flow on the BIT landscape. rigid BIT landscape with a fragile quantum potential where BIT has no This is IT FROM BIT without BIT FROM IT. Post-quantum theory with signal nonlocality means that IT is no longer a This is spontaneous self-organization pumped from environmental stochastic inputs from BOTH gauge forces making dynamical phases and also variable boundaries making Berry topological phases from paths in parameter space. I like your idea, that just a specific (exotic) properties of vacuum (bivacuum) with / < 0 and / > 0 are responsible for attractive dark matter and repulsive dark energy, correspondingly. However, I did not find in your materials the clear explanation of physical mechanisms, responsible for such properties of vacuum and the ways of their regulation (vacuum engineering ;-) ). Details are in http://qedcorp.com/APS/Ukraine.doc http://qedcorp.com/APS/Vigier4.pdf As you can see from my Unified Model (UM), it contains the mechanism of Bivacuum 'contraction' due to spin-spin interaction between introduced Bivacuum fermions (BVF) of opposite spins and Bivacuum 'expansion' due to Pauli repulsion between BVF of parallel spins in the huge volumes of 3D domains of virtual BC, formed by BVF. I do not understand that as yet. Why do you need to introduce new ideas when the old ones seem to work nicely? More with less is always better. I do not see the need to introduce any new non-orthodox ideas to explain any of the fundamental data in both the new precision cosmology and also easily explained and understood needing only mainstream ideas: 1. Einstein's general theory of relativity supplemented by Hagen Kleinert's world crystal lattice analogy and Wheeler's geometrodynamics of Mass without mass etc. 2. Standard quantum field theory in globally flat spacetime. 3. P.W. Anderson's More is different general theory of emergent complexity in spontaneous broken symmetry ground states. 4. Bohm's quantum realism interpretation of Wheeler's IT and BIT. 5. Lenny Susskind's world hologram generalization of Jacob Bekenstein's black hole thermodynamics. 6. Feynman-Wheeler/Costa De Beauregard/Hoyle-Narklikar/Cramer/Aharonov notions of advanced influences from the future. The other possible explanation of attraction/repulsion between gravitation in UM. It could be a result of interference of uncompensated Bivacuum virtual pressure waves, radiated by two or more interacting bodies, excited by [C - W] pulsation of their elementary two pulsing in liquid bodies the attraction force has (1/r^2) dependence and may change to repulsion when the distance between bodies exceeds the wave-length of density waves in liquid medium (i.e. superfluid Bivacuum). For details see Chapter 11 in my paper: http://arxiv.org/abs/physics/0207027 (see full PDF version). I don't understand this. In my theory /zpf = Lp*^-2[1 - Lp*^3|PSI|^2] The ordinary non-gravitating vacuum has /zpf = 0 This corresponds to the critical vacuum condensate density <|PSI|^2> = Lp*^-3 where Lp* = Lp^2/3(c/Ho)^1/3 ~ 1 fermi Anti-gravitating zero point dark energy density causing the large scale 3D space geometry to accelerate in its expansion from the initial singularity vortex core where PSI = 0 has /zpf > 0 for FRW Omega(Dark Energy) ~ 0.73. This implies a low density superfluid macro-quantum vacuum condensate density |PSI|^2 < Lp*^-3 Lp* is the equilibrium Holographic World Crystal Lattice spacing (combining Kleinert with Susskind). Therefore the low BEC density antigravitating dark energy phase of exotic vacuum has dilated or expanded unit cells away from equilibrium. Note these are 4D unit cells with generalized finite groups of different kinds of broken symmetries. Similarly the gravitating dark matter phase of exotic vacuum is high density vacuum condensate with the unit cells compressed smaller than the equilibrium value where FRW Omega(Dark Matter) ~ 0.23. Now if you can connect your picture in your It is shown in hydrodynamic Bjorkness theory, that between two pulsing in liquid bodies the attraction force has (1/r^2) dependence and may change to repulsion when the distance between bodies exceeds the wave-length of density waves in liquid medium (i.e. superfluid Bivacuum) to what I just said, that would be interesting. There seems to be an analogy or similarity if the equilibrium wavelength of your density waves is my Lp* and your bodies correspond to the vertices of the unit cell in the world crystal lattice? ==== My previous post on Twin Primes had a mistake. Figuring out that mistake (it was just a goof, but oh well) is crucial to getting a piece of the glory of proving the Twin Primes Conjecture. Here's the Twin Primes Formula: Given X a prime, then iff floor(X/3) - floor(X/6) = floor((X+2)/3) - floor((X+2)/6) then X+2 is a prime as well. ==== > My previous post on Twin Primes had a mistake. Flashback to 8 years ago. Post #1: incorrect proof. Post #2: correction. I think I know how it goes from there. V. -- ==== >My previous post on Twin Primes had a mistake. Figuring out that >mistake (it was just a goof, but oh well) is crucial to getting a >piece of the glory of proving the Twin Primes Conjecture. Here's the Twin Primes Formula: Given X a prime, then iff floor(X/3) - floor(X/6) = floor((X+2)/3) - floor((X+2)/6) then X+2 is a prime as well. Let X=47. floor(47/3) - floor(47/6) = 15 - 7 = 8 floor(49/3) - floor(49/6) = 16 - 8 = 8 Your Twin Primes Formula concludes that 49 is prime. ==== > Given X a prime, then iff floor(X/3) - floor(X/6) = floor((X+2)/3) - floor((X+2)/6) then X+2 is a prime as well. Cool! I suggest calling 2, 3 and 6 Harris Numbers. ==== > Given X a prime, then iff > floor(X/3) - floor(X/6) = floor((X+2)/3) - floor((X+2)/6) > then X+2 is a prime as well. Cool! I suggest calling 2, 3 and 6 Harris Numbers. Alas, the formula proved to be wrong. But just imagine math students started learning McLarren series as f(x) = f(0) + f'(0)*x + f(0)*x^2/H2 + f'(0)*x^3/H3 +... where H1 and H3 are the first and the third Harris numbers. ==== > >>Given X a prime, then iff >> floor(X/3) - floor(X/6) = floor((X+2)/3) - floor((X+2)/6) >>then X+2 is a prime as well. > > > Cool! I suggest calling 2, 3 and 6 Harris Numbers. > > Why stop there? By the above conjecture, all multiples of 3 are prime! Proof: For n a whole number, consider X=3n floor(x/3) - floor(x/6) = floor(n) - floor(n/2) floor((x+2)/3) - floor((x+2)/6) = floor(n+2/3) - floor(n/2+1/3) This gives us two cases: Case 1: If n is even, there exists whole number m such that n=2m floor(2m)-floor(2m/2) = 2m-m = m floor(2m+2/3)-floor(2m/2+1/3)=2m-m=m Case 2: if n is odd, there is a whole number m such that n=2m+1 floor(2m+1)-floor((2m+1)/2) = (2m+1) - m = m+1 floor(2m+1+2/3) - floor((2m+1)/2+1/3) = (2m+1) - m = m+1 So, if the Harris conjecture is true, then for all whole numbers n, 3n is a prime number. Therefore, since (by definition) 3n is NOT a prime for all n <> 1, the Harris conjecture is false. Obvious Harrisian Conclusion: the definition of prime is broken. -- Will Twentyman ==== > > Given X a prime, then iff >> floor(X/3) - floor(X/6) = floor((X+2)/3) - floor((X+2)/6) >> then X+2 is a prime as well. >> Cool! I suggest calling 2, 3 and 6 Harris Numbers. > > Why stop there? By the above conjecture, all multiples of 3 are prime! > > Proof: > > For n a whole number, consider X=3n > > floor(x/3) - floor(x/6) = floor(n) - floor(n/2) > floor((x+2)/3) - floor((x+2)/6) = floor(n+2/3) - floor(n/2+1/3) > > This gives us two cases: > Case 1: If n is even, there exists whole number m such that n=2m > floor(2m)-floor(2m/2) = 2m-m = m > floor(2m+2/3)-floor(2m/2+1/3)=2m-m=m > > Case 2: if n is odd, there is a whole number m such that n=2m+1 > floor(2m+1)-floor((2m+1)/2) = (2m+1) - m = m+1 > floor(2m+1+2/3) - floor((2m+1)/2+1/3) = (2m+1) - m = m+1 > > So, if the Harris conjecture is true, then for all whole numbers n, 3n > is a prime number. > > Therefore, since (by definition) 3n is NOT a prime for all n <> 1, the > Harris conjecture is false. > > Obvious Harrisian Conclusion: the definition of prime is broken. I think you are becoming cynical. Gib ==== > > Given X a prime, then iff >> floor(X/3) - floor(X/6) = floor((X+2)/3) - floor((X+2)/6) >> then X+2 is a prime as well. > Cool! I suggest calling 2, 3 and 6 Harris Numbers. >> Why stop there? By the above conjecture, all multiples of 3 are prime! >> Proof: >> For n a whole number, consider X=3n >> floor(x/3) - floor(x/6) = floor(n) - floor(n/2) >> floor((x+2)/3) - floor((x+2)/6) = floor(n+2/3) - floor(n/2+1/3) >> This gives us two cases: >> Case 1: If n is even, there exists whole number m such that n=2m >> floor(2m)-floor(2m/2) = 2m-m = m >> floor(2m+2/3)-floor(2m/2+1/3)=2m-m=m >> Case 2: if n is odd, there is a whole number m such that n=2m+1 >> floor(2m+1)-floor((2m+1)/2) = (2m+1) - m = m+1 >> floor(2m+1+2/3) - floor((2m+1)/2+1/3) = (2m+1) - m = m+1 >> So, if the Harris conjecture is true, then for all whole numbers n, 3n >> is a prime number. >> Therefore, since (by definition) 3n is NOT a prime for all n <> 1, the >> Harris conjecture is false. >> Obvious Harrisian Conclusion: the definition of prime is broken. > > > I think you are becoming cynical. > > Gib > Nope, not cynical... realistic. I've seen James claim that definitions are broken far more often than I've seen him admit to error. Of course, I won't be entirely disappointed if he admits to making a mistake. What would be more useful is if he had posted his logic behind the claim. Then we might have helped him refine it to a correct equivalence. Ok, maybe a little bit cynical, but it's hard to tell when dealing with James. -- Will Twentyman ==== Hash: SHA1 Exactly what part of math is the study of nomenclature notation like GF(p^q), Z/pZ, etc.. I know they teach it through the course of usage in number theory books but is there a specific book [or subject] that teaches how to read that notation? Tom iD8DBQE/TRhjsP+tEsHHY0ARAj2pAJ9B7OIQgpB1WOGlICGnFroAsW6rOACfZd5T MiQooZIHGdwfRtLnwk0D3rg= =epqM ==== Tom St Denis a dit : > Hash: SHA1 > > Exactly what part of math is the study of nomenclature notation like > > GF(p^q), Z/pZ, etc.. The Book-prefacing theory part of math ... All this notation stuff is written at the beginning of each book. As notation is a part of what it represents, you will find Z/pZ and GF(p^q) in an algebra book (ring theory for the former, group theory for the latter) > > I know they teach it through the course of usage in number theory > books but is there a specific book [or subject] that teaches how to > read that notation? I don't think so. -- Alexandre Charitopoulos Em6 / Eb7(5b) / Dm7 / Db7(5b, 9b) / Cmaj7 ==== Sorry but I thought I saw a quick way to prove the Twin Primes Conjecture. On further thought I'm not so sure. While it is true that given X a prime, floor(X/3) - floor(X/6) = floor((X+2)/3) - floor((X+2)/6) if X+2 is prime, it's not sufficient to force X+2 to be prime. My apologies, but I thought I had something there for a while. James Harris ==== > Sorry but I thought I saw a quick way to prove the Twin Primes > Conjecture. On further thought I'm not so sure. While it is true that given X a prime, floor(X/3) - floor(X/6) = floor((X+2)/3) - floor((X+2)/6) if X+2 is prime, it's not sufficient to force X+2 to be prime. My apologies, but I thought I had something there for a while. No apology is required, as history knows many theorems being valid for a short period of time. It's the US: for a success you don't have to be correct, you just have to be loud. ==== : It's the US: for a success you don't have to be correct, you just have : to be loud. Kindly explain this rude comment. -Justin ==== : It's the US: for a success you don't have to be correct, you just have > : to be loud. Kindly explain this rude comment. Oh please, do I have to be that specific? - In politics they were shouting he has WMD! - Enviromentalist are screaming Watch for Global Warming! - You can put the most stupid show on TV and you are a sensation. - You can sue anybody for the most ridiculous reason. Watch for idiot's rights movement. ==== > Sorry but I thought I saw a quick way to prove > the Twin Primes Conjecture. However, you managed to get your name into the history books. - jb ---------------------------------------------------------- Dr. Frankenstein http://seattlepi.nwsource.com/horsey/viewbydate.asp?id=870 ---------------------------------------------------------- ==== > As I'm sure most know, the rhombic dodecahedral numbers and the > tesseract /8-cell numbers are, respectively, the 3rd and 4th > dimensional nexus numbers. The cuboctahedron, of course, is the dual > of the rhombic dodecahedron and the hyperoctahedron/16-cell is the > dual of the tesseract/8-cell. What I'm hoping is that someone can > tell me what the formulae are, if they exist, for computing what I > guess one would call the cuboctahedral and hyperoctahedral numbers. > Any help would be appreciated. > > Ross Totally slipped my mind to check Sloane's (http://www.research.att.com/projects/OEIS?Anum=A005901). So I found the cuboctahedral. Still looking for the hyperoctahedral... Ross ==== I think that you'd find it in Coxeter's _Regular Polytopes_, using quaternions. actually, I'm not sure what you mean by nexus numbers. > As I'm sure most know, the rhombic dodecahedral numbers and the > tesseract /8-cell numbers are, respectively, the 3rd and 4th > dimensional nexus numbers. The cuboctahedron, of course, is the dual > of the rhombic dodecahedron and the hyperoctahedron/16-cell is the > dual of the tesseract/8-cell. What I'm hoping is that someone can > tell me what the formulae are, if they exist, for computing what I > guess one would call the cuboctahedral and hyperoctahedral numbers. --les ducs d'Enron! http://members.tripod.com/~american_almanac ==== > I think that you'd find it in Coxeter's _Regular Polytopes_, > using quaternions. > actually, I'm not sure what you mean by nexus numbers. > > As I'm sure most know, the rhombic dodecahedral numbers and the > tesseract /8-cell numbers are, respectively, the 3rd and 4th > dimensional nexus numbers. The cuboctahedron, of course, is the dual > of the rhombic dodecahedron and the hyperoctahedron/16-cell is the > dual of the tesseract/8-cell. What I'm hoping is that someone can > tell me what the formulae are, if they exist, for computing what I > guess one would call the cuboctahedral and hyperoctahedral numbers. > > --les ducs d'Enron! > http://members.tripod.com/~american_almanac http://mathworld.wolfram.com/NexusNumber.html. Conway and Guy defined them in The Book of Numbers. The (n+1)th rhombic dodecahedral number = 1 + 4n + 6n^2 + 4n^3. The nth = n^4 - (n-1)^4. Likewise, the (n+1)th tesseract number = 1 + 5n + 10n^2 + 10n^3 + 5n^4. The nth = n^5 - (n-1)^5. Ross ==== > For instance (-1+sqrt(3)i)/2 is an algebraic integer. > > Yes. I know that since I know it's a root of x^2 + x + 1. > So, are you excluding it from your ring? Yes or no? > > No. > > Suppose we leave the land of radicals. Let a be the largest of the > three real roots of x^5-3125x-5. Is a included in your ring? Do you > consider it to be an integer? Yes or no? > > It's an algebraic integer, so it's included. > > The polynomial is irreducible over Q, so it's not an integer. > > If it's an algebraic integer, it's included, just like algebraic > integers include gaussian integers, and I remind that I extended from > Dedekind. Very well then. Then since your ring (whatever it is) contains all the algebraic integers, it contains all the units that the ring of algebraic integers has. For you to assert that 1 and -1 are the only units in your ring is to be in flat out denial of the existence of things like (to use your example of an algebraic integer) (-1+sqrt(3)i)/2, whose reciprocal is (-1-sqrt(3)i)/2, which is also an algebraic integer. Learn the properties of your own ring before you make assertions about it. ---- David ==== >For instance (-1+sqrt(3)i)/2 is an algebraic integer. >>Yes. I know that since I know it's a root of x^2 + x + 1. >So, are you excluding it from your ring? Yes or no? >>No. >> >Suppose we leave the land of radicals. Let a be the largest of the >three real roots of x^5-3125x-5. Is a included in your ring? Do you >consider it to be an integer? Yes or no? >>It's an algebraic integer, so it's included. >>The polynomial is irreducible over Q, so it's not an integer. >>If it's an algebraic integer, it's included, just like algebraic >>integers include gaussian integers, and I remind that I extended from >>Dedekind. > > > Very well then. Then since your ring (whatever it is) contains all the > algebraic integers, it contains all the units that the ring of > algebraic integers has. For you to assert that 1 and -1 are the only > units in your ring is to be in flat out denial of the existence of > things like (to use your example of an algebraic integer) > (-1+sqrt(3)i)/2, whose reciprocal is (-1-sqrt(3)i)/2, which is also an > algebraic integer. Learn the properties of your own ring before you > make assertions about it. > > ---- David I got caught on this too. 1,-1 are the only integer units in the ring. -- Will Twentyman ==== I'm stuck with just part of a problem (just the parts that I'm stuck on are listed below) and was looking for some help Poor Bobby wants to gamble, but he has no money. Luckily, his benevolent father offers to give him five dollars to play quarter-dollar slot machines, saying if you ever go bust, you can return to me for another five-dollars worth of quarters, but if your total holdings ever exceeds ten dollars, you may keep five dollars worth of quarters but return everything else to me. Suppose Bobby's slot machines only take quarter-dollar bets (i.e. one quarter per spin) and only two outcomes are possible for each spin: (i) a return of nine quarters with probability .1 and (ii) no return with probability .9. i. Compute the steady state probabilities for each recurrent class above. ii. What is the expected number of quarters in Bobby's stash in steady state? I easily got the following equations but have no idea what to do next to get the solutions to i and ii. Let N = {1,2,3...,47} be the number of quarters that Bob has. For 0 < n < 40 P(n , n+8) = .1 P(n , n-1) = .9 For n >= 40 P( n , 20 ) = 1 For n = 0 P ( n , 20 ) = 1. ==== > I want to make a table that will be level on an unlevel ground > surface. Assuming 3 or 4 vertical legs, how many legs will need to be > adjustable? > > Thinking about this prompted another question: Is there a function z = > f(x,y) in which no two points (x,y) map to the same value of z? > > In particular, could I ever be unlucky enough to need to adjust all > but one of the legs? Let the table be as large as you like, if that > helps. > > John Carter > 25-8-03 See previous postings. So I take it you think: there is no continuous surface where no two points have the same height 'above deck' i.e. there does not exist a continuous function z = f(x,y) for which z is different for all values of x and y If that's true, then it must be true for every region of the surface. Then in any region, no matter how small, there must be at least two points with the same 'height' z. It follows that in any region, there must be an unlimited number of points with the same height (so table size doesn't matter). It follows that in any region there must be N points with the same height, where N is the number of legs on the table. So provided the legs are long enough (to accommodate peaks between the legs), they can all be the same length, at least. John Carter 27-8-03 ==== > I want to make a table that will be level on an unlevel ground > surface. Assuming 3 or 4 vertical legs, how many legs will need to be > adjustable? Two out of three, or three out of four. > If you're allowed to rotate the table any way you like, then > I think 1 out of 3, or 2 out of 4 (or N-2 out of N) is enough, > but I don't have a proof. Another puzzle, sort of in line with John's question below: Does there exist a continuous [infinite] surface (or patch of ground) such that no two points on the surface are both (a) at the same height, and (b) exactly 1 unit apart? Equivalently, is it possible, given any table with two fixed legs, to construct a patch of ground on which that table *cannot* be leveled? Any answers/proofs will be welcome. (I'm almost certain the answer is no, no such patches of ground exist, but can't be sure of it.) > In particular, could I ever be unlucky enough to need to adjust all > but one of the legs? Let the table be as large as you like, if that > helps. Why would the size of the table matter? Unless you're saying that > the table's size can be varied, too -- but I don't *think* that would > help. [And then you replied, something like: Yes, the table's size varies.] That doesn't explain much better. Do you mean: Given a table of size X and fixed legs F, and a patch Z, level the table. You can vary the unfixed legs L. or: Given a table of fixed legs F, and a patch Z, level the table. You can vary the unfixed legs L and the table size X. or something else? And what, exactly, does it mean to vary the size of a table? Is stretching allowed? Do the fixed legs get longer proportionally, or stay the same lengths? (Frankly, I think the original problem is much more interesting. Fewer variables.) -Arthur ==== > > I want to make a table that will be level on an unlevel ground > surface. Assuming 3 or 4 vertical legs, how many legs will need to be > adjustable? > Two out of three, or three out of four. > > If you're allowed to rotate the table any way you like, then > I think 1 out of 3, or 2 out of 4 (or N-2 out of N) is enough, > but I don't have a proof. You're right, at least for the case where the ground is continuous and nowhere vertical (or where there is a patch of sufficient size satisfying these constraints). And with one other caveat: we need to make the table top transparent to bumps, otherwise we can easily create an array of thin, very high spikes such that the finite-length legs of a given table can never reach the level ground below. > > Another puzzle, sort of in line with John's question below: > > Does there exist a continuous [infinite] surface (or patch of > ground) such that no two points on the surface are both (a) at > the same height, and (b) exactly 1 unit apart? > > Equivalently, is it possible, given any table with two fixed legs, > to construct a patch of ground on which that table *cannot* be leveled? > > Any answers/proofs will be welcome. (I'm almost certain the answer > is no, no such patches of ground exist, but can't be sure of it.) No follows immediately from the intermediate value theorem: Put the table down anywhere; if leg A is on higher ground than leg B, obviously there is a 180-degree rotation that swaps the two positions so that leg B is on higher ground. It follows from continuity that there is some intermediate angle of rotation 0 <= theta <= pi such that the height difference between A and B is zero. > > In particular, could I ever be unlucky enough to need to adjust all > but one of the legs? Let the table be as large as you like, if that > helps. > Why would the size of the table matter? Unless you're saying that > the table's size can be varied, too -- but I don't *think* that would > help. It doesn't matter at all, as long as it fits completely on the surface in any orientation. I don't know the answer if we allow vertical patches; you would certainly (in some cases) have to put both legs on vertical ground, which is a rather odd thing to do with a table. My intuition is that there are spiky or oddly reticulated surfaces that won't permit even that, but I really don't know. ==== > > > I want to make a table that will be level on an unlevel ground > surface. Assuming 3 or 4 vertical legs, how many legs will need to be > adjustable? > > I think you have a Goedel Escher Bach record player problem: for any > fixed table size and shape, there is a ground surface on which that > table cannot be leveled unless all but one leg is adjustable. For any > fixed ground shape, you can probably scale the table to make it > levelable with more than one fixed leg length. > > Consider the table with three legs, one adjustable, and for simplicity, > assume the legs are attached to the table in an equilateral triangle > with unit side length. > > Fix the location of the adjustable leg. > > Consider the right cylinder formed by the trace of the extention of the > other two legs as the table is rotated in a circle around the fixed leg. > > Consider that cylinder's intersection with the ground, forming a closed > curve around the cylinder. > > Since the table legs will fit six times around that closed curve, make > the curve a repeat seven sine wave; then at no point will the two legs > be on the same phase, thus height, Same phase requires same height, but the converse is *not* true. For a sine curve of any frequency, all you have to do is position a peak (or trough) at the exact midpoint between the two legs, and the height at each leg will be equal, since sine is symmetrical about any of its peaks (or troughs). ==== > Same phase requires same height, but the converse is *not* > true. For a sine curve of any frequency, all you have to do > is position a peak (or trough) at the exact midpoint between > the two legs, and the height at each leg will be equal, since > sine is symmetrical about any of its peaks (or troughs). Ah, right you are. That's what I get for trying to argue from intuition. xanthian. -- ==== > > > Same phase requires same height, but the converse is *not* > true. For a sine curve of any frequency, all you have to do > is position a peak (or trough) at the exact midpoint between > the two legs, and the height at each leg will be equal, since > sine is symmetrical about any of its peaks (or troughs). > > Ah, right you are. That's what I get for trying to argue from > intuition. But I shamelessly used your setup anyway, for a proof of what you were trying to disprove. So, thanks! ==== See below. I don't see why, in the three-leg case, you say there are just six positions to rotate the table around the adjustable leg. And what about my more general question, which amounts to: ** Is there any surface where no two points are the same height 'above the deck'? ** > > I want to make a table that will be level on an unlevel ground > surface. Assuming 3 or 4 vertical legs, how many legs will need to be > adjustable? > > I think you have a Goedel Escher Bach record player problem: for any > fixed table size and shape, there is a ground surface on which that > table cannot be leveled unless all but one leg is adjustable. For any > fixed ground shape, you can probably scale the table to make it > levelable with more than one fixed leg length. > > Consider the table with three legs, one adjustable, and for simplicity, > assume the legs are attached to the table in an equilateral triangle > with unit side length. > > Fix the location of the adjustable leg. > > Consider the right cylinder formed by the trace of the extention of the > other two legs as the table is rotated in a circle around the fixed leg. > > Consider that cylinder's intersection with the ground, forming a closed > curve around the cylinder. > > Since the table legs will fit six times around that closed curve, make > the curve a repeat seven sine wave; then at no point will the two legs > be on the same phase, thus height, of the sine wave, so a second leg > must be adjustable to make the table levelable. > > Further analysis would unfix the fixed leg to allow the table to move in > three degrees of freedom: x, y, and theta, instead of just theta. > > That is a bit beyond my skills, but I suspect a similar trick can be > extended to that case. > > xanthian. ==== > > See below. I don't see why, in the three-leg case, you say there are > just six positions to rotate the table around the adjustable leg. I think he just meant that you can fit 6 equilateral triangles around a point before you start repeating yourself; so he tried to solve the problem with a function whose period about the circle thus defined is relatively prime to 6. But as I already posted, that turns out not to work. > > And what about my more general question, which amounts to: > > ** Is there any surface where no two points are the same height > 'above the deck'? ** No, not if the height function is continuous in any region. (And if it isn't, I don't think you'd want to call it a surface. Additionally, there are of course surfaces for which height is not a function at all, e.g. a wall; I assume you are ruling these out from the get-go.) Pick any two points (A and B) in the region of continuity; then there exists a simple closed curve in the surface connecting those points. If z(A)=z(B) we are done; otherwise, by the mean value theorem there is some point C on the clockwise leg of the curve from A to B, and some point D on the counterclockwise leg, C and D distinct, that both have the same height (z(A)+z(B))/2. We can say more. Starting from Kent Paul Dolan's picture of the intersection of cylinder with ground, and let A_0 and B_0 be the positions of the two (non-adjustable) legs at rotation angle theta=0. Suppose that z(A_0) > z(B_0). Rotate 60 degrees and it may still be the case that z(A_60) > z(B_60); but by the time you have done 4 more rotations it cannot be the case every time, because A_300 is the same as B_0 and therefore z(A_300) is less than z(A_0). Therefore the difference function z(A)-z(B) at any angle theta is a continuous function with both positive and negative values; we are therefore assured that it is zero at some angle theta. So, this solves your special case. (Of course, speaking practically, you might have a case where these two legs are at a height that differs from that of the adjustable leg by more than the range of adjustability.) As Arthur J. O'Dwyer already responded, there *are* discontinuous functions that have no two values the same. ==== ... > Additionally, there are of course surfaces for which height is > not a function at all, e.g. a wall; I assume you are ruling these > out from the get-go.) Actually you don't really have to rule these out at all, but I wanted to keep things simple. ... Therefore the difference function z(A)-z(B) > at any angle theta is a continuous function with both positive > and negative values; Lousy writing; I meant that the difference z(A)-z(B) is a continuous function of theta with both positive and negative values over the domain 0 <= theta < 2*pi we are therefore assured that it is zero at > some angle theta. So, this solves your special case. ==== The following matrix arises when doing cubic spline interpolation of a closed curve of N=5 points: 4 1 0 0 1 1 4 1 0 0 0 1 4 1 0 0 0 1 4 1 1 0 0 1 4 In view of the symmetry of this matrix is it possible to get an explicit formulae for the inverse of this matrix (for the general NxN case)? Obviously only the first row of the inverse need be given since the rest follow by sysmmetry. ==== >I thought it relevant to inform that I notified the FBI a couple of >months ago about some of the math issues I've brought up here. I >received a single reply that agents were looking into it, as I cited >national security, given that mathematicians are so important in the >defense of this nation. Whee! This made my day, thanks. >It was not a form letter reply. I've followed up but have not gotten >further information from the FBI. >I have also informed a couple of senators, but did not receive >anything other than form letter replies. >The senators were McCain of Arizona and Graham of Florida. >Some of you may be angered by my contacting important agencies like >the FBI who have VERY important work to do in defense of this nation. I doubt anyone's angry about this. You consumed a teensy bit > of FBI resources, and provided a few minutes of entertainment > for sci.math readers around the world - seems like an excellent > bargain. >However, I think it very important if mathematicians are as adept at >lying as I've seen, and the federal government needed to be notified. Yeah, there's that too. >I will also suggest that those of you who receive federal funds >carefully review the terms and conditions you agreed to in order to >get them. >I am not saying that I know of any investigations into mathematicians >resulting from my contacts with the United States Government. I would >suspect that I was simply ignored as a crank, and that they referred >to mathematicians who may have lied to them. I doubt that they needed to ask a mathematician's opinion to > determine you're a crank. Your stuff just oozes crankitude. >However, it was my duty to inform, and possibly at some future date, >if some mathematicians did lie to the FBI or senators, they may face >further questions. >If I was mostly ignored by the FBI and those senators, which is >probable, then, of course, they didn't ask anyone. >At a later date I will probably make higher level contacts, hoping to >get feedback from members of the National Security Council. > >James Harris I bet the FBI has a Crank File.But anyone bothering to pester them is going to get some attention, and possibly profiled in a database for further attention , if needed. RJ P ==== > It's comedy! > It's drama!! > It's tragedy!!! > > It's > > THE POWERS THAT BE!!!!!!! > > > WHOOHOOOOOOO!!!!! > [snipped for brevity but strongly encourage reading the full post for excellent insight and humor] > > > That's right. You go get 'em, tiger. > > > Dale This was brilliant - thank you! ==== > >However, I think it very important if mathematicians are as adept at >lying as I've seen, and the federal government needed to be notified. >>I will also suggest that those of you who receive federal funds >carefully review the terms and conditions you agreed to in order to >get them. >>Sigh. It's about so much more than federal funding. Fermat's Last >>Theorem, or Wiles's Theorem (if his proof holds up), or Harris's Theorem >>(if the Wiles proof is found to be flawed or fraudulent), is a monumental >>piece of number theory (by any name). I don't think the government cares >>much about it, though. > > fraudulent - such an ugly word. > If the government could prove that the Wiles proof was fraudulent, that > would be just the high-profile lever that they need. With Wiles in prison > the time would be ripe to push through legislation mandating Object Oriented > Math and banning the flawed rings and fields that mathematicians have > conspired to use for so many shame-filled years. > Things seem to be moving very quickly now. I think that we will see more and > more mathematicians espousing Harrisian mathematics. The flawed rings and > fields will probably be explained away as a useful fiction. A sort of > luminiferous ether upon which the various algebraic structures (now > revealed by Harris as Objects) float. > Over the next few months we will witness a paradigm shift in mathematics. > Those who shift early will be in the vanguard of the new movement. > ALL HAIL, Jesus Harris, OUR BLESSED SAVIOUR. > Your sarcasm comes through loud and clear. The last line indicates that it's > aimed more at me than at James. Yes, I called James a mathematical messiah. > I called for people to run out in the streets shouting James Harris is God! > I admit that this does sound a little overboard. But put yourself in my shoes for a moment. I grew up believing the party line. > When I realized one day that everything I had been taught was false, and that > the person whom I thought was a total joke was really a profound genius, I went > into a state of shock. My world view was thrown into upheaval. If I was so > wrong about this, I could be wrong about everything. In this mental chaos, the > only thing that was blindingly clear was James Harris's oppressive brilliance. > I squirmed under it, unable to understand his proof, yet convinced of its truth > (a guy that smart doesn't make mistakes). I fell to worshipping James and > posted some things that seem a little over the top now. I don't see James as a messiah any longer. He's not here to save the human race. > He's merely one light that turned on in a vast black plain. And the creatures of > darkness cringed at the sight and are trying, even now, to extinguish it. It grieves me to see the cruel mockery of this great intellect. It grieves me > all the more that I once so mocked. Your vision of a new Harrisian mathematics > could come to be in a perfect world, but never in one as flawed as ours. I have to go now. I can't type through the bitter tears I'm weeping for the > human race. Priceless exchange! Dirk Vdm ==== In sci.math, James Harris <3c65f87.0308260745.7ae17601@posting.google.com>: > I thought it relevant to inform that I notified the FBI a couple of > months ago about some of the math issues I've brought up here. I > received a single reply that agents were looking into it, as I cited > national security, given that mathematicians are so important in the > defense of this nation. OK. Someone please walk me through this *very* slowly. Why would the Federal Bureau of Investigation, which is primarily interested in looking for criminals, both federal and state, be remotely interested in your mathematical theories? The FBI can do simple stuff like 1 bullet + 1 corpse = 1 murder :-), [*] and probably has a rather comprehensive set of lab setups and such for figuring where that fiber came from and whether it was rubbed onto that piece of grass or blew thereonto, but I fail to see how pure math would even interest them, apart from the rather obvious area of cryptography, which is more likely covered by the NSA but could fall under their jurisdiction if the case hinges on who did what to whom when and what they said about it to whomever. If one of course wishes to claim the existence of a deep underground laboratory where Agent J assigns work for the day to ensure that the visiting aliens aren't commiting crimes, don't call us, they'll call you. :-) > > It was not a form letter reply. I've followed up but have not gotten > further information from the FBI. > > I have also informed a couple of senators, but did not receive > anything other than form letter replies. > > The senators were McCain of Arizona and Graham of Florida. I don't see how Graham would be interested unless it involved spending tax dollars :-) (in which case he'd be vehemently against it). Dunno about McCain although his big buzz was straight, honest talk about various issues IIRC. (No, LGMs, UFOs, and math equations weren't among them. Maybe Social Security.) > > Some of you may be angered by my contacting important agencies like > the FBI who have VERY important work to do in defense of this nation. Yes, they do. How that work is augmented by your interesting ideas on math equations is far from clear. > > However, I think it very important if mathematicians are as adept at > lying as I've seen, and the federal government needed to be notified. Oh no. Mathematicians are terrorists! I see it all now: the accountant's cap, the pencil newly sharpened, the math book laid open with a ruler on the page -- it's all part of the sinister plan of Al Mathla[+]. Wooooooooooooooo..... Better inform the...oh, wait, you already did. > > I will also suggest that those of you who receive federal funds > carefully review the terms and conditions you agreed to in order to > get them. You mean I have to give back my tax refund?! > > I am not saying that I know of any investigations into mathematicians > resulting from my contacts with the United States Government. I would > suspect that I was simply ignored as a crank, and that they referred > to mathematicians who may have lied to them. Wow. A shred of sanity 'mongst the silliness. > > However, it was my duty to inform, and possibly at some future date, > if some mathematicians did lie to the FBI or senators, they may face > further questions. > > If I was mostly ignored by the FBI and those senators, which is > probable, then, of course, they didn't ask anyone. > > At a later date I will probably make higher level contacts, hoping to > get feedback from members of the National Security Council. > Well, whatever floats your boat. I personally would prefer to be strapped into an F-15 interceptor or a space shuttle and puke all over myself as he does various hair-raising maneuvers, but to each his own. :-) > > James Harris [*] I should note here that murder is a state crime, not a federal one, unless the victim was serving the Federal Government in some capacity (e.g., a U.S. Marshall or an IRS bureaucrat). It is not clear whether the individual would have had to be serving in that capacity at the time of the crime, or not; there are a number of issues here but they're all peripheral to the discussion. [+] not to be confused with a certain terrorist group in the Middle East. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >In sci.math, James Harris >: >> I thought it relevant to inform that I notified the FBI a couple of >> months ago about some of the math issues I've brought up here. I >> received a single reply that agents were looking into it, as I cited >> national security, given that mathematicians are so important in the >> defense of this nation. OK. Someone please walk me through this *very* slowly. Why would the Federal Bureau of Investigation, which is primarily >interested in looking for criminals, both federal and state, >be remotely interested in your mathematical theories? I'm surprised he didn't mention the NSA. Thrillers have been full of NSA agents lately. It doesn't seem to bother the writers that the NSA has a mission which doesn't involve anything remotely like chasing people (or aliens) around with guns. More like sitting in rooms with headphones on, which lacks that James Bond quality. I guess it's that they really don't know what the NSA does, so that makes it de facto a mysterious 3-letter government agency. And of course there are the sekrit agents who work for an agency so shadowy that being secret FBI/NSA agents is their COVER STORY. James should contact them. - Randy ==== > And of course there are the sekrit agents who work for an agency so > shadowy that being secret FBI/NSA agents is their COVER STORY. > James should contact them. Ah, those must be the agents from Scarecrow and Mrs. King, who worked for an agency so secret that it had no name but The Agency. (And no, it wasn't the CIA; in at least one episode a CIA agent and an Agency agent worked together.) -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock ==== In sci.math, Randy Poe <7v5pkv414v50k320lupievbuu5lmic50pd@4ax.com>: > >>In sci.math, James Harris >><3c65f87.0308260745.7ae17601@posting.google.com>: > I thought it relevant to inform that I notified the FBI a couple of > months ago about some of the math issues I've brought up here. I > received a single reply that agents were looking into it, as I cited > national security, given that mathematicians are so important in the > defense of this nation. >>OK. Someone please walk me through this *very* slowly. >>Why would the Federal Bureau of Investigation, which is primarily >>interested in looking for criminals, both federal and state, >>be remotely interested in your mathematical theories? > > I'm surprised he didn't mention the NSA. Thrillers have been full of > NSA agents lately. It doesn't seem to bother the writers that the > NSA has a mission which doesn't involve anything remotely like chasing > people (or aliens) around with guns. More like sitting in rooms with > headphones on, which lacks that James Bond quality. > > I guess it's that they really don't know what the NSA does, so that > makes it de facto a mysterious 3-letter government agency. > > And of course there are the sekrit agents who work for an agency so > shadowy that being secret FBI/NSA agents is their COVER STORY. > James should contact them. If he can find them. I suspect they're so secret even they don't know. :-) > > - Randy > -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== In sci.math, a <3JQ2b.14373$Kj3.5517@nntp-post.primus.ca>: > I thought it relevant to inform that I notified the FBI a couple of > months ago about some of the math issues I've brought up here. >>Could you post a copy of the letter you sent? I for one would >>love to see it. > > So would I. I'd also like the see the FBI's reply, which of course > was not a form letter. > Well, it probably had the form of a letter, anyway... :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > MAMA: C'mere, boy, and bring me that switch! JSH: Get it yourself, I ain't no fuckin' electrician. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm ==== J. Edgar Harris > I thought it relevant to inform that I notified the FBI a couple of > months ago about some of the math issues I've brought up here. I > received a single reply that agents were looking into it, as I cited > national security, given that mathematicians are so important in the > defense of this nation. ... I've seen these symptoms before. I expect the FBI will add only one name to its databases, and its initials will be JSH. ==== The only thing I don't understand is why you post this here. Wouldn't it help the FBI to have the element of surprise? Gib ==== >The only thing I don't understand is why you post this here. Wouldn't >it help the FBI to have the element of surprise? Gib > It's hard to intimidate all those wacky math people that don't recognize his genious if he doesn't tell you about it. I somehow think he'd be slapping a lot of people with frivolous lawsuits if he had enough money to spend on it. Right or wrong doesn't matter with that tactic, if the instigator has deep enough pockets to drag it through the courts long enough to break the little guy. -- A nice adaptation of conditions will make almost any hypothesis agree with the phenomena. This will please the imagination but does not advance our knowledge. -- J. Black, 1803. ==== >The only thing I don't understand is why you post this here. Wouldn't >it help the FBI to have the element of surprise? >Gib > > > It's hard to intimidate all those wacky math people that don't recognize > his genious if he doesn't tell you about it. I'm not interested in people breaking the law, so the warning is the same as the previous warnings, which is--don't break it. Mathematicians may believe they have a right to continue teaching algebraic integers as if there isn't a problem with them, but I say that's fraud. There are students who believe that they're getting value for their time, and money (or their parents' money), as well as the United States itself, which promotes math research and math studies for the *benefit* of society. > I somehow think he'd be slapping a lot of people with frivolous lawsuits > if he had enough money to spend on it. Right or wrong doesn't matter with > that tactic, if the instigator has deep enough pockets to drag it through > the courts long enough to break the little guy. And I think many of you fail to see that blocking math results, even if only by silence, can mean that you're on the outside of society working against it. As lawbreakers you will be like other criminals, and in facing punishment, it won't be like worrying about lawsuits. Make no mistake, if the United States Government gets convinced to come after mathematicians, individuals and math departments who are responsible for them, who teach the flawed mathematics, then you will be paying attention from that time forward. And none of you will be joking about it. I'd like to remind that I've contacted major universities all over the United States including top tier ones like Harvard and Yale. My strategy is deliberate. Another benefit of warning those of you in academia is that it removes some excuses, and will help take away the possibility of trying to rely on public sympathy. As far as I'm concerned many of you are criminals or are about to be criminals as you go into the classroom to give young minds flawed information, and I don't want the public to get distracted from that fact. James Harris ==== > As far as I'm concerned many of you are criminals or are about to be > criminals as you go into the classroom to give young minds flawed > information, and I don't want the public to get distracted from that > fact. Applying scientific method. 2 hypotheses are: 1. There are so many criminals 2. There is just one conspiracy theorist Many would prefer #2 as the *simpler* one. ==== God that is so fucked up Believe it or not, this guy is dead serious. For some background on James, you can go here: > http://groups.google.com/groups?selm=EWVpa.3442$z15.3317@news.primus.ca >I thought it relevant to inform that I notified the FBI a couple of >months ago about some of the math issues I've brought up here. I >received a single reply that agents were looking into it, as I cited >national security, given that mathematicians are so important in the >defense of this nation. >It was not a form letter reply. I've followed up but have not gotten >further information from the FBI. >I have also informed a couple of senators, but did not receive >anything other than form letter replies. >The senators were McCain of Arizona and Graham of Florida. >Some of you may be angered by my contacting important agencies like >the FBI who have VERY important work to do in defense of this nation. >However, I think it very important if mathematicians are as adept at >lying as I've seen, and the federal government needed to be notified. >I will also suggest that those of you who receive federal funds >carefully review the terms and conditions you agreed to in order to >get them. >I am not saying that I know of any investigations into mathematicians >resulting from my contacts with the United States Government. I would >suspect that I was simply ignored as a crank, and that they referred >to mathematicians who may have lied to them. >However, it was my duty to inform, and possibly at some future date, >if some mathematicians did lie to the FBI or senators, they may face >further questions. >If I was mostly ignored by the FBI and those senators, which is >probable, then, of course, they didn't ask anyone. >At a later date I will probably make higher level contacts, hoping to >get feedback from members of the National Security Council. > >James Harris ==== Will Twentyman skrev i melding > I thought it relevant to inform that I notified the FBI a couple of > months ago about some of the math issues I've brought up here. I > received a single reply that agents were looking into it, as I cited > national security, given that mathematicians are so important in the > defense of this nation. > It was not a form letter reply. I've followed up but have not gotten > further information from the FBI. > I have also informed a couple of senators, but did not receive > anything other than form letter replies. > The senators were McCain of Arizona and Graham of Florida. > Some of you may be angered by my contacting important agencies like > the FBI who have VERY important work to do in defense of this nation. > However, I think it very important if mathematicians are as adept at > lying as I've seen, and the federal government needed to be notified. > I will also suggest that those of you who receive federal funds > carefully review the terms and conditions you agreed to in order to > get them. > I am not saying that I know of any investigations into mathematicians > resulting from my contacts with the United States Government. I would > suspect that I was simply ignored as a crank, and that they referred > to mathematicians who may have lied to them. > However, it was my duty to inform, and possibly at some future date, > if some mathematicians did lie to the FBI or senators, they may face > further questions. > If I was mostly ignored by the FBI and those senators, which is > probable, then, of course, they didn't ask anyone. > At a later date I will probably make higher level contacts, hoping to > get feedback from members of the National Security Council. > > James Harris Interesting followup note to this action: I joined his AmateurMath page > so I could look at his proof of the Advanced Polynomial Factorization, > and he has now blocked me out of the entire site, even the pages that > you do *not* have to register to view. -- > Will Twentyman > That's my experience too. I have had a passport since January 2002 and the only site I can't watch is James' site. I am not allowed. I think that he reads the post carefully and ban those he doesn't like! Karl-Olav Nyberg ==== > Will Twentyman skrev i melding > >I thought it relevant to inform that I notified the FBI a couple of >months ago about some of the math issues I've brought up here. I >received a single reply that agents were looking into it, as I cited >national security, given that mathematicians are so important in the >defense of this nation. >>It was not a form letter reply. I've followed up but have not gotten >further information from the FBI. >>I have also informed a couple of senators, but did not receive >anything other than form letter replies. >>The senators were McCain of Arizona and Graham of Florida. >>Some of you may be angered by my contacting important agencies like >the FBI who have VERY important work to do in defense of this nation. >>However, I think it very important if mathematicians are as adept at >lying as I've seen, and the federal government needed to be notified. >>I will also suggest that those of you who receive federal funds >carefully review the terms and conditions you agreed to in order to >get them. >>I am not saying that I know of any investigations into mathematicians >resulting from my contacts with the United States Government. I would >suspect that I was simply ignored as a crank, and that they referred >to mathematicians who may have lied to them. >>However, it was my duty to inform, and possibly at some future date, >if some mathematicians did lie to the FBI or senators, they may face >further questions. >>If I was mostly ignored by the FBI and those senators, which is >probable, then, of course, they didn't ask anyone. >>At a later date I will probably make higher level contacts, hoping to >get feedback from members of the National Security Council. >James Harris >>Interesting followup note to this action: I joined his AmateurMath page >>so I could look at his proof of the Advanced Polynomial Factorization, >>and he has now blocked me out of the entire site, even the pages that >>you do *not* have to register to view. >>-- >>Will Twentyman > > > That's my experience too. I have had a passport since January 2002 and the > only site I can't watch is James' site. I am not allowed. I think that he > reads the post carefully and ban those he doesn't like! > > Karl-Olav Nyberg > > I haven't even tried to post anything. I just want to be able to get at his .pdf in case he updates it. The last time he made a change I was unaware of it until looking again. I found I was arguing against a significantly different paper than I had originally found. -- Will Twentyman ==== Will Twentyman skrev i melding > Will Twentyman skrev i melding >I thought it relevant to inform that I notified the FBI a couple of >months ago about some of the math issues I've brought up here. I >received a single reply that agents were looking into it, as I cited >national security, given that mathematicians are so important in the >defense of this nation. >>It was not a form letter reply. I've followed up but have not gotten >further information from the FBI. >>I have also informed a couple of senators, but did not receive >anything other than form letter replies. >>The senators were McCain of Arizona and Graham of Florida. >>Some of you may be angered by my contacting important agencies like >the FBI who have VERY important work to do in defense of this nation. >>However, I think it very important if mathematicians are as adept at >lying as I've seen, and the federal government needed to be notified. >>I will also suggest that those of you who receive federal funds >carefully review the terms and conditions you agreed to in order to >get them. >>I am not saying that I know of any investigations into mathematicians >resulting from my contacts with the United States Government. I would >suspect that I was simply ignored as a crank, and that they referred >to mathematicians who may have lied to them. >>However, it was my duty to inform, and possibly at some future date, >if some mathematicians did lie to the FBI or senators, they may face >further questions. >>If I was mostly ignored by the FBI and those senators, which is >probable, then, of course, they didn't ask anyone. >>At a later date I will probably make higher level contacts, hoping to >get feedback from members of the National Security Council. >James Harris >>Interesting followup note to this action: I joined his AmateurMath page >>so I could look at his proof of the Advanced Polynomial Factorization, >>and he has now blocked me out of the entire site, even the pages that >>you do *not* have to register to view. >>-- >>Will Twentyman > > That's my experience too. I have had a passport since January 2002 and the > only site I can't watch is James' site. I am not allowed. I think that he > reads the post carefully and ban those he doesn't like! > Karl-Olav Nyberg > > I haven't even tried to post anything. I just want to be able to get at > his .pdf in case he updates it. The last time he made a change I was > unaware of it until looking again. I found I was arguing against a > significantly different paper than I had originally found. -- > Will Twentyman > I meant posting (commenting) to mr. Harris' treads here on sci. math. Karl-Olav Nyberg ==== > Will Twentyman skrev i melding > >Will Twentyman skrev i melding >Interesting followup note to this action: I joined his AmateurMath page >>so I could look at his proof of the Advanced Polynomial Factorization, >>and he has now blocked me out of the entire site, even the pages that >>you do *not* have to register to view. >>That's my experience too. I have had a passport since January 2002 and > > the > >only site I can't watch is James' site. I am not allowed. I think that > > he > >reads the post carefully and ban those he doesn't like! >>Karl-Olav Nyberg >>I haven't even tried to post anything. I just want to be able to get at >>his .pdf in case he updates it. The last time he made a change I was >>unaware of it until looking again. I found I was arguing against a >>significantly different paper than I had originally found. > > > I meant posting (commenting) to mr. Harris' treads here on sci. math. > > Karl-Olav Nyberg > > That makes sense. It would be a true tragedy if we started posting in his space, rather than let it be a bastion of Pro-JSH. -- Will Twentyman ==== James Harris skrev i melding > I thought it relevant to inform that I notified the FBI a couple of > months ago about some of the math issues I've brought up here. I > received a single reply that agents were looking into it, as I cited > national security, given that mathematicians are so important in the > defense of this nation. It was not a form letter reply. I've followed up but have not gotten > further information from the FBI. I have also informed a couple of senators, but did not receive > anything other than form letter replies. The senators were McCain of Arizona and Graham of Florida. Some of you may be angered by my contacting important agencies like > the FBI who have VERY important work to do in defense of this nation. However, I think it very important if mathematicians are as adept at > lying as I've seen, and the federal government needed to be notified. I will also suggest that those of you who receive federal funds > carefully review the terms and conditions you agreed to in order to > get them. I am not saying that I know of any investigations into mathematicians > resulting from my contacts with the United States Government. I would > suspect that I was simply ignored as a crank, and that they referred > to mathematicians who may have lied to them. However, it was my duty to inform, and possibly at some future date, > if some mathematicians did lie to the FBI or senators, they may face > further questions. If I was mostly ignored by the FBI and those senators, which is > probable, then, of course, they didn't ask anyone. At a later date I will probably make higher level contacts, hoping to > get feedback from members of the National Security Council. > James Harris Mr. Harris You are quite a guy! When your arguments fail, you call in the heavy guys. Is it like that you will press your arguments on us? If we dont like them, you will force them? Grow up! It will not work in Norway! And aside this post, can you explain to me how the scientific method with trials and fails can be used in mathematics. I will repeat that the scientific method doesn't prove a thing, it's only conjuctures things. We have conjuctures in mathemathics too, but thats why we struggle to prove them or disprove them. We never say that verification through examples is a proof! The method of problem solving (Polya) often use tools that looks like the scientific method, but it's fare from it, because the result is either true or false. The scientific method doesn't work in mathematics. We want proof, not verifications. And verifications aren't that good in the scientific method either. One falsification breaks and distroys the theory. Consider Newton's rule of force. He almost had it, but he didn't belive that mass or time could differ with the distance between two bodies. Einstein did show that this could happen. Newton set this to 0 (zero), but Einstein show that this was not so. We use Newton's laws here, but have to use Einstein's laws when it comes to the extremes. And what happens when thing are more extreme? Masses moving above the speed of light? Einstein doesn't apply. Mr. Harris, you say that you have a degree in physics. Well, then you must have learned that the laws of physics are not theoremes, they are (in a mathematical meaning conjuctures). They are conjuctures to be prooved or disaprooved. No trial and verification can proove them. It only needs one counterexample to falsify them. In mathematics we do this in another way. We build the mathematics on some our theoremes. They can't be disputed. Through your work you are questioning set-theory, be aware; you are attacking the axiomes! It is meaningless! Theoremes are prooved by these axiomes! You can't change that! We can construct some sets that make problems for the set-theory, but none of the sets you treat are among these. The basic thing that I will say to you is that (I have said it before) in mathemathics we are dealig with theoremes when we will proove something, not conjuctures and examples. However if you could give an example of a number that should be an algebraic integer and is excluded from (what I know of) the set of algebraic integers I would really like to know! Can't you just give me one? Karl-Olav Nyberg ==== > You are quite a guy! When your arguments fail, you call in the heavy guys. > Is it like that you will press your arguments on us? If we dont like them, > you will force them? > Grow up! It will not work in Norway! Don't think you are safe in Norway! James plans to get the US Army involved, too. As he said in 3c65f87.0304191552.511ad5b4%40posting.google.com on April 19th: > Yup, you guessed it. If worse comes to worse, I *will* turn to the > Army to help me with mathematicians. And then mathematicians don't > think the NSA or CIA can save your asses, as generals LIKE me. So you see, you can expect troops to invade Norway any day now... -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock ==== >Grow up! It will not work in Norway! Ah, you don't see this, but you're mistaken. Unitentional humour, as provided by Mr. Harris, works in any country. Doug ==== Doug Norris skrev i melding >Grow up! It will not work in Norway! Ah, you don't see this, but you're mistaken. Unitentional humour, as provided > by Mr. Harris, works in any country. Doug I see the humourous side of it, but I don't think Harris is trying to be funny. He really thinks that FBI will slay us. :) Karl-Olav Nyberg ==== >Doug Norris skrev i melding >>Grow up! It will not work in Norway! >> Ah, you don't see this, but you're mistaken. Unitentional humour, as >provided >> by Mr. Harris, works in any country. >I see the humourous side of it, but I don't think Harris is trying to be >funny. He really thinks that FBI will slay us. :) That's why I called it unintentional humour. Doug ==== > I thought it relevant to inform that I notified the FBI a couple of > months ago about some of the math issues I've brought up here. I > received a single reply that agents were looking into it, as I cited > national security, given that mathematicians are so important in the > defense of this nation. > Did the FBI have any comments related to m = 0 ? Like, Agent Smith: Look here, I was investigating this m = 0 deal. Harris: Yeah? What of it ? Agent Smith: I'll do the questioning here. Looks to me like there's some kinda coverup here. You know. m = 0 is kind of a, uh, special case. You get my drift? Harris: Um ... no ... you mean, where I say 5 is a factor of a_1 because when m = 0, 5 is a factor of zero because anything divides 0 therefore 5 is a factor of a_1 no matter what m is, even though a_1 isn't 0 when m is not 0, but still it must always be divisible by 5 because it was in that one special case where it happens to be divisible by everything and 5 just happens to be a special case of everything but still it obviously works in that one case so it must work all the time. Is that what you mean? Agent Smith: Yeah. Somethin' like that. Harris: Well, clearly you have been talking to a bunch of lying thieving low-down cheating federal-fund-grubbing (ugh!) mathematicians. Agent Smith: Yeah. I mean, I don't know if they were thieving. They mighta been though. I wouldn't put anything past 'em. Look, kid. I'm gonna go easy on ya. Just write down a little proof for f = 3. Yeah, I know, it's gonna be trivial and all. But the Boss, he says proving this thing for f = 3 ain't enough. He says what you get with 3 is reducible. He says ... Harris: I see what you're doing. It's the old good-cop bad- cop routine. He's feeding you this crap about f = 3 probably from some low-down lying thieving cheating mathematician who can't stand that an amateur can come along and after only 8 years of total bullshit come out with another load of total bullshit ... no, wait, that didn't come out right ... I mean, comes along with the Best Proof Ever Written. Now about f = 3. Again here by SCIENTIFIC EXPERIMEN- TATION I find that when f = 3, a.k.a. 2 + 1, a bona fide element of a ring if there ever was one and a number whose inverse, like all real numbers, is in the ring generated by Z and 1/2, I say when f = 3 (and mathmaticians REFUSE TO ACKNOWLEDGE THIS), even though it is also a special case because of the irreducibility, I say, mathematicians cannot find *how* the factors of f split out into the roots, all they have is some cowpoop of a proof that all the roots have some f in them - I say, by scientific experimen- tation, when f = 3, again the numbers factor just like I say they should. And obviously when you prove a thing for m = 0 and you prove it for f = 3, it MUST be true for all other m's and f's, RIGHT? I mean, take a typical math statement, say, the statement that all numbers of the form m*f are divisible by 3. Well, obviously it's true when m = 0, right? And obviously it's true when f = 3, right? SO IT MUST BE TRUE WHEN m = anything and f = anything, right? Couldn't possibly be just a special case. That's what any normal person would accept as a valid proof, right? But that's just the kind of thing that these putrid, fetid, lying cheating mathema- ticians deny right and left. They WILL NOT ACCEPT MY GREATNESS!!! They keep giving these asshole PROOFS and COUNTEREXAMPLES and they keep covering up their stupidity by asking for shit like DEFINITIONS and for WHAT RING AM I IN? I mean for Christ's sake, what possible difference does it make what ring I am in? Agent Smith: Yup. Harris: I mean, if it doesn't fit, you must acquit. Right? Agent Smith: Yessir. Harris: You're a smart guy, Smith. But you're not convinced. Agent Smith: I didn't say that. Harris: Look, I'll go over it again. You got it when m = 0. True, it's a degenerate case, and true, the polynomial only has degree 1 and I want it to be true when it has degree p, and true, everything divides 0. So you got it without the slightest shadow of a doubt when m = 0. Right? Agent Smith: Uh - right. Harris: And you got it when f = 3, even though f = 3 is a special case that I don't even care about, and no one is squawking that I am wrong about f = 3 because I don't have irreducibility. Right? Agent Smith: Uh - right. No irredoucheability or whatever. Harris: So you sort of got a great big square with an f side and an m side, and you got one side of this big square, where m = 0, sort of, and you got another side of the square when f = 3, and even though neither one of these is what I want or need, I got both of them, sort of, and so it's pretty goddamn obvious, isn't it? Agent Smith: Maybe you better spell it out for me. Harris: Well, you got it for m = 0 and f = 3. Therefore you got it for all m and f. Right? Agent Smith: Somethin' missing there. Can't put my finger on it. Harris: Look, wienie. I say you got the factorization I want when m = 0 and f = 3. So it must be true for all the other m's and f's. Agent Smith: You mean, say, if m = 7 and f = 13? Harris: Yeah, good example. See, I showed it (sort of) for m = 0 and f = 3. So it must be true when m = 7 and f = 13. Only an idiot would fail to see the logical connection. Listen to the Math! Agent Smith: Yup. I'm listening to the Math. The Math knows. Harris: And that's what these lying cheating bullshit artist mathematicians are trying to tell you. That the argument I gave is not a proof. Agent Smith: Yup. > It was not a form letter reply. I've followed up but have not gotten > further information from the FBI. > Dear Mr. Harris, This is not a form letter. We will inform you when we have further information. Your friend, J. Edgar Hoover, Jr. > I have also informed a couple of senators, but did not receive > anything other than form letter replies. > Dear [boxholder] one question. When you say m = 0 or f = 3, why is that sufficient? This is a form letter. Please contribute to my next campaign. Sincerely, Amos S. Senator, Washington, DC > The senators were McCain of Arizona and Graham of Florida. > > Some of you may be angered by my contacting important agencies like > the FBI who have VERY important work to do in defense of this nation. > I'm madder than hell and I'm not going to take it any more. > However, I think it very important if mathematicians are as adept at > lying as I've seen, and the federal government needed to be notified. > But ... it's the only thing they are good at, lying. Take that away, what do they have left? They'll go on welfare. > I will also suggest that those of you who receive federal funds > carefully review the terms and conditions you agreed to in order to > get them. > Whew! Nothing in there about not lying. I'm in the clear. > I am not saying that I know of any investigations into mathematicians > resulting from my contacts with the United States Government. > I would > suspect that I was simply ignored as a crank, and that they referred > to mathematicians who may have lied to them. > A crank ??? How could that be. Those dastards. They have infil- trated the innermost inner sanctums of the Goverment. How dare they call you a crank. You can start a car with a crank. Do people start cars with you? No. You are not a crank. You are more like an ashtray, just under the dash. > However, it was my duty to inform, and possibly at some future date, > if some mathematicians did lie to the FBI or senators, they may face > further questions. > Questions like, What about m = 17 and f = 37? > If I was mostly ignored by the FBI and those senators, which is > probable, then, of course, they didn't ask anyone. > Ask James Harris, you fools. Or ask Rummie. He'll vouch for old ex-soldier James. > At a later date I will probably make higher level contacts, hoping to > get feedback from members of the National Security Council. > Fat chance. The mathematicians have stuffed the Council with their own men, none of whom are going to admit in a million years that proving whatever for m = 0 and f = 3 should be enough for anybody. They are such dishonest crooks, plus right now they are busy planning a pre-emptive attack on Paraguay. Andrzej. > > James Harris ==== What does the J(x)=Li(x)-sum Li(x^p)-log(2)+integral formula look if we are looking at 1 mod 4 primes? We could modify Pi(x) to Pi4,1(x). How do you define J and can we get a formula involving L(chi(4))? I think the correct definitions would let us calculate the difference between 4 mod 3 and mod 1 primes. Do you have a reference? ==== Marriage mathematics The Deccan Herald Everything on this earth can be modelled mathematically, they say. A professor from the University of Washington has proven this statement once again, by formulating a model that can predict the chances of the success of a proposed marriage. He claims data from a couple's conversations, converted into algebra, is 94% accurate in determining how long a couple will remain wed. Professor James Murray, the man behind this theory, creates a graph of their conversations, and believes that it could eventually be used to predict the likelihood of divorce. In this approach, he gives a numerical value for the phrases used by a couple in their conversation, with the range being from minus four to plus four. So, a reference to a partner 'being stupid' would be scored as minus four, but a joke which results in laughter from a partner would be scored as plus two. For each conversation slot, he gets a number and over time a graph showing how the couple are interacting is obtained. Prof Murray says that the information for the graph is collated when couples, who are planning to get married, are brought into a lab and there conversation is assessed for 15 minutes. He uses a scoring system on their reactions and represents their reactions with algebraic terms. With the data collected, he works out whether they are likely to stay together or if they are more likely to divorce. ''Maths provides a language for interpreting the human interaction. It quantifies one person's effect on the other, and it is not difficult. The maths we are using could be done by secondary school pupils with basic algebra. Once they are shown the basic model, they insert their data and make the simple calculation,'' says the Professor. T G Srinidhi Read the complete news at: http://deccanherald.com/deccanherald/aug26/snt1.asp Jai Maharaj http://www.mantra.com/jai Om Shanti Shubhanu Nama Samvatsare Dakshinaya Jivana Ritau Singh Mase Shukl Pakshe Buddh Vasara Yuktayam Magh-Poorvaphalguni Nakshatr Shiv-Siddha Yog Naag-Kinstughn Karan Amavasya-Pratham Yam Tithau http://www.mantra.com/holocaust http://www.hindu.org http://www.hindunet.org The truth about Islam and Muslims http://www.flex.com/~jai/satyamevajayate o Not for commercial use. Solely to be fairly used for the educational purposes of research and open discussion. The contents of this post may not have been authored by, and do not necessarily represent the opinion of the poster. The contents are protected by copyright law and the exemption for fair use of copyrighted works. considered or answered if it does not contain your full legal name, are not necessarily those of the poster. ==== No Im not a student, just an old dude who hasn't had to do this basic algebra for a looooong time... But I need to solve the following quadratic for x: a*(off + x * width)^2 + b * (off + x * width) + c Can anyone take pity and help out? ==== > > No Im not a student, just an old dude who hasn't had to do this basic > algebra for a looooong time... > > But I need to solve the following quadratic for x: > > a*(off + x * width)^2 + b * (off + x * width) + c > > Can anyone take pity and help out? > > > If you set it equal to 0, and replace off + x * width with y, then you have a*y^2+b*y+c=0. This is easily solved with the quadratic equation. Then, since off+x*width=y, subtract off and divide by width on both sides. -- Will Twentyman ==== > > Nobody wins. You're using include in two different ways: > > According to standard set theory, no set includes itself, > meaning that no set is an element of itself. But it _is_ > basic that every set includes itself, meaning that every > set is a subset of itself. > > > @Christine: > > In symbols: > > For any set x: > > x c x. > > But (at least in Zermelo-Fraenkel's ZF): > > For any set x: > > not (x e x) , > > which is equivalent with > > There is no set x such that: > > x e x. That's because they weren't able to figure out how to define a set that contains itself. But I solved that problem a few months ago: Let f(x)={x(x)}. Then f(f)={f(f)} so that f(f) is an element of itself. Just think about recursion theory in which a program can output itself. set=program, element=output. If we think in general terms, then sets, functions, predicates, lists, trees, graphs etc. are each arbitrary mathematical domains that has its own paradox and incompleteness and needs to be defined in relation to the others. Using my notation, it is easy to generate Godel's 1st Incompleteness Theorem (based on soundness) and dozens of other theorems related to it. Some of these theorems have been published (e.g. Smullyan's Dual Form theorem) while others seem to be new. Compare this to books and papers that claim to be a system that generates exactly one theorem! Charlie Volkstorf Cambridge, MA ==== --------------------------------------------------------------------- I can do question #1 but I'm not sure about #2. Can someone please check my answer for me? M means Lamda Dave is taking a multiple-choice exam. You may assume that the number of questions is infinite. Simultaneously, but independently, his conscious and conscious and subconscious are always working on different questions.) Conscious responses are generated at a rate of Mc responses per minute. Subconscious responses are generated at a rate of Ms responses per minute. Each conscious response is an independent Bernoulli trial with probability pc of being correct. Similarly, each subconscious response is an independent Bernoulli trial with probability ps of being correct. Dave responds only once to each question, and you can assume that his time for recording these conscious and subconscious responses is negligible. (1) The papers are to be collected as soon as Dave has completed exactly n responses. Determine: (i) The expected number of questions he will answer correctly Answer: n*Mc*Pc/(Mc + Ms) + n*Ms*Ms/(Mc + Ms) where M means Lamda (ii) The probability mass function for L, the number of questions he answers correctly. Answer: Each question is answered correctly with probability P = Mc*Pc/(Mc + Ms) + Ms*Ms/(Mc + Ms). So, P( L = K ) = ( n C k )* P^k ( 1 - P)^ (n - k) (2) Repeat part (f) for the case in which the exam papers are to be collected at the end of a fixed interval of t minutes. Answer?: P( L = K ) = summation from n = k to oo [e^(-Mt)*Mt^n/n! * P^k( 1-p)^(n - k)]