mm-372 === Subject: Re: Symbol confusion? Alg Int Prob REVISED > In algebra, letter symbols are variables. Well, this explains *everything*. The statement of Fermat's Last Theorem is: If n>2 and xyz=/=0, then x^n + y^n =/= z^n I'd always thought that x, y and z were supposed to be *integers* in this statement. Now that James has explained that in algebra, letter symbols are always *variables*, I see that I've been misinterpreting it all along. Of course, when x, y and z are *variables* the statement is almost trivially true. So what's all the fuss about? Steven E. Landsburg www.landsburg.com/about2.html === Subject: Optimal Betting Strategy ?? I recently saw a bet where the following odds were offered on a person winning a certain event: +140, +150, +250, and +325. I am pretty much 100% certain that either the person with the +250 odds or the person with the +325 will win so I wanted to split $1100 (the maximum bet allowed) in wagering between the two. I am really unsure of who will win between the +250 odds person and the +325 odds person so I am just assigning a 50/50 chance that either of the two people could win. Assuming a 50% probability that either of the two could win, is there an optimal way for me to spilt my wagering between the two people?? Also, if I were to assign a different probability p and 1-p to the two people winning, is there a formula that would allow me to come up with an optimal betting strategy?? It seems as though figuring out a general formula would require relatively simple math, but I can't put my finger on it so any help would be much appreciated!! Jose === Subject: Re: Optimal Betting Strategy ?? > I recently saw a bet where the following odds were offered on a person > winning a certain event: +140, +150, +250, and +325. I am pretty much 100% > certain that either the person with the +250 odds or the person with the > +325 will win so I wanted to split $1100 (the maximum bet allowed) in > wagering between the two. I am really unsure of who will win between the > +250 odds person and the +325 odds person so I am just assigning a 50/50 I'll work with the 250 as 2.5 to 1 and with the 325 as 3.25 to 1. So the 2.5 to 1 pays 3.5 while the 3.25 to 1 pays 4.25... Then (3.5 / (3.5 + 4.25)) * 1100 = 496.77. And 1100 - 496.77 = 603.23 . Obviously, 603.23 bets on the 2.5 to 1 odds while 496.77 bets on the 3.25 to 1 odds. === Subject: Re: Optimal Betting Strategy ?? > I recently saw a bet where the following odds were offered on a person > winning a certain event: +140, +150, +250, and +325. I am pretty much 100% > certain that either the person with the +250 odds or the person with the > +325 will win so I wanted to split $1100 (the maximum bet allowed) in > wagering between the two. I am really unsure of who will win between the > +250 odds person and the +325 odds person so I am just assigning a 50/50 > chance that either of the two people could win. Assuming a 50% probability > that either of the two could win, is there an optimal way for me to spilt my > wagering between the two people?? Also, if I were to assign a different > probability p and 1-p to the two people winning, is there a formula that > would allow me to come up with an optimal betting strategy?? It seems as > though figuring out a general formula would require relatively simple math, > but I can't put my finger on it so any help would be much appreciated! I can think of two possible ways to optimize your strategy -- either maximize expected winnings, or maximize worst-case winnings. Maximizing expected winnings is riskier, but you will come out farther ahead in the long run. Let d1 and d2 be the odds on the two people (2.50 and 3.25), and p1 and p2 be their probabilities of winning (.50 and .50). To maximize expected winnings, you'll put all of your money on one of the two people. Your expected return is d1*p1 if you bet on person 1 and d2*p2 if you bet on person 2 -- bet on whichever of these is greater. In this case, you'll bet it all on person 2. On a $1 bet, half the time you'll win $2.25 and the other half you'll lose $1, so your expected winnings are $.625. To maximize worst-case winnings, you want your return to be the same no matter who wins. So set x*d1*p1 = (1-x)*d2*p2. If you solve for x, you'll get the fraction of your money you should bet on person 1. In this case, we get x=.565, so you should be 56.5% of your money on person 1 and the remainder on person 2. On a $1 bet, you're guarenteed to win about $.413. You can mix these strategies, too. So I'd say an optimal strategy is to bet anywhere from 0% to 56.5% of your money on person 1, and the remainder on person 2. The more you bet on person 1, the safer your bet is, but the less you'll win in the long run. - Nate === Subject: Re: Optimal Betting Strategy ?? In message , Jose Luiz >I recently saw a bet where the following odds were offered on a person >winning a certain event: +140, +150, +250, and +325. I am not familiar with this way of stating odds. I assume it means that if I bet $100 on the +140 person, and they win, I will get my stake back with another $140. >I am pretty much 100% >certain that either the person with the +250 odds or the person with the >+325 will win so I wanted to split $1100 (the maximum bet allowed) in >wagering between the two. I am really unsure of who will win between the >+250 odds person and the +325 odds person so I am just assigning a 50/50 >chance that either of the two people could win. Assuming a 50% probability >that either of the two could win, is there an optimal way for me to spilt my >wagering between the two people?? What do you mean by optimal? If you are trying to maximise your expected wealth, then bet everything on the +325 person. If you are trying to maximise the worst-case result, then bet $440 on the +325 person and $660 on the +250 person. If you mean something else by optimal, please tell us what it is. >Also, if I were to assign a different >probability p and 1-p to the two people winning, is there a formula that >would allow me to come up with an optimal betting strategy?? It seems as >though figuring out a general formula would require relatively simple math, >but I can't put my finger on it so any help would be much appreciated!! Yes, there is such a formula. What it is depends on what you mean by optimal. Nick -- Nick Wedd nick@maproom.co.uk === Subject: Re: Optimal Betting Strategy ?? > In message , Jose Luiz I recently saw a bet where the following odds were offered on a person >winning a certain event: +140, +150, +250, and +325. > I am not familiar with this way of stating odds. I assume it means that > if I bet $100 on the +140 person, and they win, I will get my stake back > with another $140. >I am pretty much 100% >certain that either the person with the +250 odds or the person with the >+325 will win so I wanted to split $1100 (the maximum bet allowed) in >wagering between the two. I am really unsure of who will win between the >+250 odds person and the +325 odds person so I am just assigning a 50/50 >chance that either of the two people could win. Assuming a 50% probability >that either of the two could win, is there an optimal way for me to spilt my >wagering between the two people?? > What do you mean by optimal? If you are trying to maximise your > expected wealth, then bet everything on the +325 person. If you are > trying to maximise the worst-case result, then bet $440 on the +325 > person and $660 on the +250 person. If you mean something else by > optimal, please tell us what it is. >Also, if I were to assign a different >probability p and 1-p to the two people winning, is there a formula that >would allow me to come up with an optimal betting strategy?? It seems as >though figuring out a general formula would require relatively simple math, >but I can't put my finger on it so any help would be much appreciated!! > Yes, there is such a formula. What it is depends on what you mean by > optimal. > Nick Yes, I was looking for a strategy to maximize the worst-cast scenario. I really don't want to put all my money on one or the other because there would be a very good possibility I would lose the whole wager, which I obviously don't want... Jose === Subject: Re: Optimal Betting Strategy ?? In message , Jose Luiz >> In message , Jose Luiz >>I recently saw a bet where the following odds were offered on a person >>winning a certain event: +140, +150, +250, and +325. >> I am not familiar with this way of stating odds. I assume it means that >> if I bet $100 on the +140 person, and they win, I will get my stake back >> with another $140. >>I am pretty much 100% >>certain that either the person with the +250 odds or the person with the >>+325 will win so I wanted to split $1100 (the maximum bet allowed) in >>wagering between the two. I am really unsure of who will win between the >>+250 odds person and the +325 odds person so I am just assigning a 50/50 >>chance that either of the two people could win. Assuming a 50% >probability >>that either of the two could win, is there an optimal way for me to spilt >my >>wagering between the two people?? >> What do you mean by optimal? If you are trying to maximise your >> expected wealth, then bet everything on the +325 person. If you are >> trying to maximise the worst-case result, then bet $440 on the +325 >> person and $660 on the +250 person. If you mean something else by >> optimal, please tell us what it is. >>Also, if I were to assign a different >>probability p and 1-p to the two people winning, is there a formula that >>would allow me to come up with an optimal betting strategy?? It seems as >>though figuring out a general formula would require relatively simple >math, >>but I can't put my finger on it so any help would be much appreciated!! >> Yes, there is such a formula. What it is depends on what you mean by >> optimal. >> Nick >Yes, I was looking for a strategy to maximize the worst-cast scenario. I >really don't want to put all my money on one or the other because there >would be a very good possibility I would lose the whole wager, which I >obviously don't want... Right. In that case, the actual probabilities (your p and 1-p) are irrelevant. You should allocate the money according to the odds, so that whichever person wins, you will receive the same amount. If the odds offered on the people you think can win are m and n, you should bet fractions n/m+n and m/m+n of your available cash on them. Nick -- Nick Wedd nick@maproom.co.uk === Subject: Re: Optimal Betting Strategy ?? X-Mutant: yes > In message , Jose Luiz >>I recently saw a bet where the following odds were offered on a person >>winning a certain event: +140, +150, +250, and +325. I am not familiar with this way of stating odds. I assume it means > that if I bet $100 on the +140 person, and they win, I will get my > stake back with another $140. I think you are right. This looks like the american odds notation, explained here: http://www.betting-guide.co.uk/scripts/odds_explained.php If I'm getting this right then: +140 means that you win $140 if you bet $100 -140 means that you have to bet $140 to win $100 In both cases you end up with $240 (if you win) Marco -- The text above is a result of a bug in my newsreader and I take no responsibility for this text appearing in my post. === Subject: Re: Optimal Betting Strategy ?? > I recently saw a bet where the following odds were offered on a person > winning a certain event: +140, +150, +250, and +325. I am pretty much 100% > certain that either the person with the +250 odds or the person with the > +325 will win so I wanted to split $1100 (the maximum bet allowed) in > wagering between the two. I am really unsure of who will win between the > +250 odds person and the +325 odds person so I am just assigning a 50/50 > chance that either of the two people could win. Assuming a 50% probability > that either of the two could win, is there an optimal way for me to spilt my > wagering between the two people?? Yes. Bet all your money on the +325 person. There is no reason to put any money on the +250 person, if your 50/50 estimate is correct. > Also, if I were to assign a different > probability p and 1-p to the two people winning, is there a formula that > would allow me to come up with an optimal betting strategy?? It seems as > though figuring out a general formula would require relatively simple math, > but I can't put my finger on it so any help would be much appreciated!! Assuming that the +250 person has probability p... If 250*P > 325*(1-p) put all your money on 250, otherwise put it all on 325. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Function space topologys > Hello ! > I am quite bewildered by all of function space topologys. Could someone give > a quick big-view and say why to use one before the others ? For example, > the topology of pointwise convergence or the compact open topology ? I also > here the compact open one called topology of compact convergence but I do not > know the reason. Generally what is the topology of P convergence for a > property P ? I am well confused because many books say strongest topology > such that and other books say weakest topology when they mean the same > thing ! > Thank you for your attention. > Ann. Ann, The first thing to do is remember the difference between pointwise convergence and uniform convergence of a sequence of functions. In the general case, we use nets rather than sequences, but the essential ideas are the same. The topology of pointwise convergence is simply the topology where convergence of a net of functions is exactly pointwise convergence of that net. You might try to show that the definition of this topology that you are given gives this notion of convergence. The topology of compact convergence is a bit more involved, so let's take it in steps. First assume that we are considering the space of functions between X and Y where X is compact and Y is a metric space. We can then talk about uniform convergence of functions. In particular, a uniform neighborhood of a function h will consist of all functions f such that d(f(x),h(x)) a .8ecrit dans le message de > ((e^pi) + 1i) ^2 = (((e^pi)^2) -1) + 2*(e^pi)i > ;-) > Dan Well, no; e^(i*pi) = 1 and more generally, e^(i*x) = 1 for any value of x ! Here is a Proof : e^(i*x) = e^(i*x*(2*pi)/(2*pi)) (Just multiply and divide by 2*pi) = e^((2*i*pi)*(x/(2*pi))) (Just group terms in a different manner) = (e^(2*i*pi))^(x/(2*pi)) (since e^(a*b) = (e^a)^b) = 1^(x/(2*pi)) (since e^(2*i*pi) = 1) = 1 !!! And it makes mathematics much more simple, isn't it ? ;-) J-L. === Subject: Re: Another proof of e^pi(i) + 1 = 0 > Dan <30pack@sbcglobal.net> a .8ecrit dans le message de > ((e^pi) + 1i) ^2 = (((e^pi)^2) -1) + 2*(e^pi)i ;-) Dan > Well, no; e^(i*pi) = 1 and more generally, e^(i*x) = 1 for any value of x ! It follows pi=x for any value of x! ;-) Tapio > Here is a Proof : > e^(i*x) = e^(i*x*(2*pi)/(2*pi)) (Just multiply and divide by 2*pi) > = e^((2*i*pi)*(x/(2*pi))) (Just group terms in a different > manner) > = (e^(2*i*pi))^(x/(2*pi)) (since e^(a*b) = (e^a)^b) > = 1^(x/(2*pi)) (since e^(2*i*pi) = 1) > = 1 !!! > And it makes mathematics much more simple, isn't it ? ;-) > J-L. === Subject: Re: Cautionary Note (was): AI::GA Almost everything Einstein first published was derived from > pieces first found by others, he was as much the jigsaw > puzzle assembler as anything, bringing just his one more > piece that was key to the work of seeing how the other parts > best fit, not springing the whole thing forth from sheer > testosterone poisoning derived genius, as your attempts > reflect. [...] Kent, a similar assembling of the puzzle over 13 years yielded > http://mentifex.virtualentity.com/theory5.html A Theory of Mind. Then I spent another ten years writing the first AI Mind software > http://www.seedai.e-mind.org/jsaimind.html -- JavaScript AI Mind; > http://mentifex.virtualentity.com/mind4th.html -- AI in Forth. Now I am inviting coders in many programming languages to do AI: > http://mentifex.virtualentity.com/cpp.html -- C++ with new AI code; > http://mentifex.virtualentity.com/java.html -- see Mind.JAVA #001; > http://mentifex.virtualentity.com/lisp.html -- Lisp AI Weblog; > http://mentifex.virtualentity.com/perl.html -- first Perl module; > http://mentifex.virtualentity.com/prolog.html -- Prolog AI Weblog; > http://mentifex.virtualentity.com/python.html -- Python AI Weblog; > http://mentifex.virtualentity.com/ruby.html -- Ruby AI Blog (OO AI); > http://mentifex.virtualentity.com/scheme.html -- Scheme AI Weblog; > http://mentifex.virtualentity.com/vb.html -- see Mind.VB #001 link. Corporations or wealthy indivduals are invited to hire me via KWAI. Remember, Kent, when I analysed the Unabomber's (Ted's) Manifesto and > deduced (right here on Usenet) that he would be a mathematician? Have you read Underwood's Mathematical Cranks? In late 1984 I > met one of his chapter subjects (another Ted :-) here in Seattle. > My familiarity with Ted the mathematician led me to see the signs. Artificial intelligence, however, is different from mathematics. > AI is more linguistic, because language separates us from brutes. > With the programming-language XYZ AI Weblogs linked to above, I > float the AIPO (artificial intelligence public offering) among, > for example, the ca. one million Visual Basic programmers and > the legions of potential AI coders in all the other specialties. > I would like to keep working on AI, but I am low on time and > defunct of funds. Arthur Ted Murray What if that wet lump in our heads is not a computer, but merely a port or interface to a computer, a conduit that conducts the information to the real computer or mind existing in another dimension? What if memory goes through it through different areas of the interface, but then goes on to storage and computing functions, and we are just seeing the traffic through the port, and not the destination, or the hard drive or any part of the computer at all? Andrea http://www.andrearogers.com === Subject: Re: Cautionary Note (was): AI::GA > Artificial intelligence, however, is different from mathematics. > AI is more linguistic, because language separates us from brutes. > With the programming-language XYZ AI Weblogs linked to above, I > float the AIPO (artificial intelligence public offering) among, > for example, the ca. one million Visual Basic programmers and > the legions of potential AI coders in all the other specialties. > I would like to keep working on AI, but I am low on time and > defunct of funds. Arthur Ted Murray What if that wet lump in our heads is not a computer, but merely a > port or interface to a computer, a conduit that conducts the > information to the real computer or mind existing in another > dimension? > Your what if posits an presently infeasible data communications path. > It might actually exist, but I dunno from it. Imagine instead an AI as an avatar existing in a vat of liquid crystals, where the vat of liquid crystal is the computer, and the brain of the AI/Avatar is just the bus, that communicates with and tasks in the liquid crystal fishbowl world. The AI avatar doesn't know that the fishbowl is the computer, he thinks he is. He has no tools or perceptions to even know he is in a fishbowl, only that there are features and attributes to his world around him that have limits he cannot change, surpass or comprehend beyond. > What if memory goes through it through different areas of the > interface, but then goes on to storage and computing functions, and we > are just seeing the traffic through the port, and not the destination, > or the hard drive or any part of the computer at all? It don't work that way. A good principle of computing is, ultimately, > that the more local something is, the more reliable and faster it is. If the brain is a 3D location with 3D space, why does the information or programming have the limitations of just occupying that space? What makes us think that is that all brains are so simliar that there are > patterns, lots of patterns in commmon that make the thought that a common > external control is possible. Same make and model, different DNA, retrieving unique persons. Which is faster, playing a CD on a Discman, or downloading a CD at full T1 > rate over a high speed link? It's like that. What if the air around me could be surrounded by all the media in a broadcast form, that remained invisable until I needed the information, that all I had to do was turn on a portable radio or TV set,cell phone, computer and not drag 25 miles worth of cable around? This is sad. We're born alone, we more or less die alone, and we don't > really know much about people we do know. Make it a hard problem, one we > have to work at. Why should it be too easy? What would be the point of building an AI and not testing it until it breaks? Andrea === Subject: Re: Cautionary Note (was): AI::GA Artificial intelligence, however, is different from mathematics. > AI is more linguistic, because language separates us from brutes. > With the programming-language XYZ AI Weblogs linked to above, I > float the AIPO (artificial intelligence public offering) among, > for example, the ca. one million Visual Basic programmers and > the legions of potential AI coders in all the other specialties. > I would like to keep working on AI, but I am low on time and > defunct of funds. Arthur Ted Murray What if that wet lump in our heads is not a computer, but merely a > port or interface to a computer, a conduit that conducts the > information to the real computer or mind existing in another > dimension? > Your what if posits an presently infeasible data communications path. > It might actually exist, but I dunno from it. Imagine instead an AI as an avatar existing in a vat of liquid > crystals, where the vat of liquid crystal is the computer, and the > brain of the AI/Avatar is just the bus, that communicates with and > tasks in the liquid crystal fishbowl world. The AI avatar doesn't know that the fishbowl is the computer, he > thinks he is. > He has no tools or perceptions to even know he is in a fishbowl, only > that there are features and attributes to his world around him that > have limits he cannot change, surpass or comprehend beyond. > That's nice. This is essentially the point of The Matrix. SFAIK, it doesn't much hold up outside entertainment product. One of the more salient features of empiricist thought is that if you cannot reliably detect something, it is reasonable to assume it doesn't exist. This might change over time, but it is not there now. > What if memory goes through it through different areas of the > interface, but then goes on to storage and computing functions, and we > are just seeing the traffic through the port, and not the destination, > or the hard drive or any part of the computer at all? It don't work that way. A good principle of computing is, ultimately, > that the more local something is, the more reliable and faster it is. If the brain is a 3D location with 3D space, why does the information > or programming have the limitations of just occupying that space? Because the information has to take some measurable form in order to be communicated or otherwise used. If it's not in the 3d space, it must be imported from without, and processed by the brain. What you posit is entirely possible. We don't know anything about it actually being that way. Since there is no evidence to support that it is, it is more prudent to assume it is not. But a better explanation is that similarities in internal brain structure create systems of idioms which are common to all people, which creates the perception that there's some external structure. But this perception is very questionable. What makes us think that is that all brains are so simliar that there are > patterns, lots of patterns in commmon that make the thought that a common > external control is possible. Same make and model, different DNA, retrieving unique persons. Which is faster, playing a CD on a Discman, or downloading a CD at full T1 > rate over a high speed link? It's like that. What if the air around me could be surrounded by all the media in a > broadcast form, that remained invisable until I needed the > information, that all I had to do was turn on a portable radio or TV > set,cell phone, computer and not drag 25 miles worth of cable around? > Don't hold your breath for that. This is sad. We're born alone, we more or less die alone, and we don't > really know much about people we do know. Make it a hard problem, one we > have to work at. Why should it be too easy? Perhaps it should not. > What would be the point of building an AI and not testing it until it > breaks? I think youy can pretty much stop at what is the point of building an AI. > Andrea -- Les Cargill === Subject: Re: Cautionary Note (was): AI::GA > Imagine instead an AI as an avatar existing in a vat of liquid > crystals, where the vat of liquid crystal is the computer, and the > brain of the AI/Avatar is just the bus, that communicates with and > tasks in the liquid crystal fishbowl world. The AI avatar doesn't know that the fishbowl is the computer, he > thinks he is. > He has no tools or perceptions to even know he is in a fishbowl, only > that there are features and attributes to his world around him that > have limits he cannot change, surpass or comprehend beyond. > That's nice. This is essentially the point of The Matrix. SFAIK, > it doesn't much hold up outside entertainment product. One of > the more salient features of empiricist thought is that if > you cannot reliably detect something, it is reasonable to > assume it doesn't exist. I cringe whenever Hollywood presents something as reality we know that the reality show are just sensationalized fiction, that have nothing to do with reality, and the Lawnmower Man in no way represents Virtual Reality. Why should a fictional blockbuster like The Matrix be any different. A matrix of information exisiting as it's own universe and transparently connected to the three dimensional universe is a lot duller without the villians and the drama on it's own. If you can accept that many unseen universes can overlap each other, then the brain as a kind of bus/port device is easier to imagine. This might change over time, but it is not there now. What if memory goes through it through different areas of the > interface, but then goes on to storage and computing functions, and we > are just seeing the traffic through the port, and not the destination, > or the hard drive or any part of the computer at all? It don't work that way. A good principle of computing is, ultimately, > that the more local something is, the more reliable and faster it is. Which is why a universe of information that doesn't have any 3D space at all, wouldn't have to worry about location, because there is no space. When information is transformed to a 3D universe, then it takes up space, a lot of space, or becomes filtered because there is not enough space for the information. If the brain is a 3D location with 3D space, why does the information > or programming have the limitations of just occupying that space? Because the information has to take some measurable form > in order to be communicated or otherwise used. If it's not in the > 3d space, it must be imported from without, and processed > by the brain. What you posit is entirely possible. We don't know anything about it > actually being that way. Since there is no evidence to support that it > is, it is more prudent to assume it is not. But a better explanation is that similarities in internal brain > structure create systems of idioms which are common to all people, which > creates the perception that there's some external structure. But this > perception is very questionable. Suppose the AI fishbowl becomes populated by many avatars. A few of the avatars on occasion do break, and some are mended by other avatars who become doctors and scientists, makes tools to make thier tasks more efficiant,who remember and test the information by trial and error, still without ever suspecting that they are in a fishbowl computer. Even if they crack open the heads of other avatars to study, and probe and test, they still won't see it. They will wonder why sometimes they can fix other avatars, but be clueless as to why sometimes they cannot, never seeing or knowing what gives them consciousness or life. > What makes us think that is that all brains are so simliar that there are > patterns, lots of patterns in commmon that make the thought that a common > external control is possible. > Don't hold your breath for that. > This is sad. We're born alone, we more or less die alone, and we don't > really know much about people we do know. Make it a hard problem, one we > have to work at. Why should it be too easy? Perhaps it should not. What would be the point of building an AI and not testing it until it > breaks? I think youy can pretty much stop at what is the point of building > an AI. Would building a model of the human body make it more than a machine, if we still do not know outside of the operation of the machine what is the spark of life, and actual consciousness,what universes they reside,those things that are seemingly invisable to this world. Andrea === Subject: Re: Cautionary Note (was): AI::GA > Artificial intelligence, however, is different from mathematics. > AI is more linguistic, because language separates us from brutes. > With the programming-language XYZ AI Weblogs linked to above, I > float the AIPO (artificial intelligence public offering) among, > for example, the ca. one million Visual Basic programmers and > the legions of potential AI coders in all the other specialties. > I would like to keep working on AI, but I am low on time and > defunct of funds. Arthur Ted Murray What if that wet lump in our heads is not a computer, but merely a > port or interface to a computer, a conduit that conducts the > information to the real computer or mind existing in another > dimension? > Your what if posits an presently infeasible data communications path. > It might actually exist, but I dunno from it. Imagine instead an AI as an avatar existing in a vat of liquid > crystals, where the vat of liquid crystal is the computer, and the > brain of the AI/Avatar is just the bus, that communicates with and > tasks in the liquid crystal fishbowl world. The AI avatar doesn't know that the fishbowl is the computer, he > thinks he is. > He has no tools or perceptions to even know he is in a fishbowl, only > that there are features and attributes to his world around him that > have limits he cannot change, surpass or comprehend beyond. What if memory goes through it through different areas of the > interface, but then goes on to storage and computing functions, and we > are just seeing the traffic through the port, and not the destination, > or the hard drive or any part of the computer at all? It don't work that way. A good principle of computing is, ultimately, > that the more local something is, the more reliable and faster it is. If the brain is a 3D location with 3D space, why does the information > or programming have the limitations of just occupying that space? What makes us think that is that all brains are so simliar that there are > patterns, lots of patterns in commmon that make the thought that a common > external control is possible. Same make and model, different DNA, retrieving unique persons. Which is faster, playing a CD on a Discman, or downloading a CD at full T1 > rate over a high speed link? It's like that. What if the air around me could be surrounded by all the media in a > broadcast form, that remained invisable until I needed the > information, that all I had to do was turn on a portable radio or TV > set,cell phone, computer and not drag 25 miles worth of cable around? > This is sad. We're born alone, Really ? We're born alone ? I'm pretty sure my mother was present at the time. That we all die alone is a commonplace. Tim Worstall we more or less die alone, and we don't > really know much about people we do know. Make it a hard problem, one we > have to work at. Why should it be too easy? > What would be the point of building an AI and not testing it until it > breaks? > Andrea === Subject: Is 2^2^n + 1 prime for natural numbers n? When studying a set of primes I rediscovered that 2^2^n + 1 is prime for at least n=1,2,3,4 How much further does this go? Are there other patterns for certain primes in the form of this one (an expression with a natural number variable)? === Subject: Re: Is 2^2^n + 1 prime for natural numbers n? > Are there other patterns for certain primes in the form of this one (an > expression with a natural number variable)? You can play with patterns of the form P1 = ((a+1) + b*i)^p - (a + b*i)^p or P2 = (a + (b+1)*i)^p - (a + b*i)^p (with a, b integers and p prime) and study their norms : you'll find new prime numbers ... from time to time. A special case is a = 1 and b = 0 which gives Mersenne's numbers 2^p - 1 J-L. === Subject: Re: Is 2^2^n + 1 prime for natural numbers n? > When studying a set of primes I rediscovered that 2^2^n + 1 is prime for > at least n=1,2,3,4 > How much further does this go? > Are there other patterns for certain primes in the form of this one (an > expression with a natural number variable)? Numbers of the form 2^(2^n)+1 (where n is a non-negative integer) are called Fermat numbers. The number 2^(2^n)+1 is denoted by F_n. Only five of these numbers (F_0 thru F_4) are known to be prime. Numbers of the form b^(2^n)+1 (where b is an integer greater than 1 and n is a non-negative integer) are known as generalized Fermat numbers, and they are sometimes prime. Such a number can only be prime if b is even. See http://perso.wanadoo.fr/yves.gallot/primes/ For example 1176694^(2^17)+1 is prime and has 795,695 digits. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Is 2^2^n + 1 prime for natural numbers n? Visiting Assistant Professor at the University of Montana. >When studying a set of primes I rediscovered that 2^2^n + 1 is prime for >at least n=1,2,3,4 You might want to write 2^{2^n} + 1 so it does not get confused with 4^n+1. >How much further does this go? As far as anyone knows, no further than that. Fermat guessed it would always be prime (you can also throw in n=0), but Euler factored 2^{32}+1 and showed it was not prime. Check out: http://www.prothsearch.net/fermat.html ============================================================== ======== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ============================================================== ======== Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Is 2^2^n + 1 prime for natural numbers n? |>When studying a set of primes I rediscovered that 2^2^n + 1 is prime for |>at least n=1,2,3,4 | |You might want to write 2^{2^n} + 1 so it does not get confused with |4^n+1. He might want to, but let's also point out that the convention on stacks of exponents c b a is to take them to associate on the right, as a^(b^c). Leaving the parentheses or brackets in is mainly for those who might not be aware of this convention. The use of ^ has less of a tradition behind it, and I think TeX will take $a^b^c$ as meaning that a has two superscripts on it, but I think we should still take a^b^c as meaning a^(b^c) even if the parentheses or brackets have been left off. If you want to have two separate superscripts on the same symbol, presumably you aren't using both of them as exponents, and if you mean (a^b)^c you'd better not write it as a^b^c. The original poster might be amused to know that the odd primes p for which a circle can be divided into p equal arcs using only straightedge-and-compass constructions are the primes p of the form F_n=2^(2^n)+1 for integers n>=1. As far as we know the ones with n=1,2,3,4 are the only ones. The last I read, F_n had been factored for n=5 through n somewhere in the 20s. I read once about a guy who spent quite a long time working on a straghtedge-and-compass construction of the regular 65537-gon, which he never finished. The partial construction got collected by some library. Wish I could remember where I read that. Math Reviews has a review of a paper in a journal I haven't heard of before or since, which claimed to prove all of the F_n for n>=5 are composite, using rocky manipulations of the power series sum_n (n!*x^n). The reviewer expresses doubts about the value of continuing to review material from publications of such poor quality. Keith Ramsay === Subject: Re: Is 2^2^n + 1 prime for natural numbers n? Visiting Assistant Professor at the University of Montana. [.snip.] >He might want to, but let's also point out that the convention on >stacks of exponents > c > b >a >is to take them to associate on the right, as a^(b^c). Leaving the >parentheses or brackets in is mainly for those who might not be >aware of this convention. Fair enough. >The use of ^ has less of a tradition behind it, and I think TeX will >take $a^b^c$ as meaning that a has two superscripts on it, Hmmm... No, TeX and LaTeX will complain about double superscript and will not compile. You get same problems if you try to use double underscripts, $a_b_c$. You need to group them explicitly. I suspect that's what's gotten me into out of the habit of the convention you specify (though, when printed as exponents, as you do in your ASCII art, I always interpret it as a^(b^c)). ============================================================== ======== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ============================================================== ======== Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Is 2^2^n + 1 prime for natural numbers n? > I read once about a guy who spent quite a long time working > on a straghtedge-and-compass construction of the regular > 65537-gon, which he never finished. The partial construction > got collected by some library. Wish I could remember where > I read that. Well, here are some places you _might_ have read something along those lines: # Coxeter, Introduction to Geometry, page 27 of the second edition. # A post to sci.math: http://tinylink.com/?WHklbCkLbo # MathWorld: http://mathworld.wolfram.com/65537-gon.html # A post by somebody who recently looked at the manuscript: http://mathforum.org/epigone/historia_matematica/brahpendglun Sorry if none of these is where you actually read it. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Goldbach-like sequences > I've been dabbling with Goldbach's conjecture for a while now, and > I've come across something that I've not seen described in any of the > literature I've seen on the net so far. I wondered if anyone else has > had any thoughts, or can point me to any references, on this and > related matters... Let S be a set of odd numbers with maximum element m. I call it a > Goldbach Spanning Set if all the even numbers from 6 to m+1 can be > expressed as the sum of two numbers in S. These things obviously exist > : the set of all odd numbers between 3 and m will do. The question I > asked myself was what's the smallest such set that will suffice?. If I recall right, Paul Bruckman published something about S, which for him that gave the minimal set of primes needed to produce all evens up to some set 2N. It was somewhat odd that some small primes are not needed until much later than one would expect. Larry 5, 9, 11, 21, 25, 35, 43, 53, 63, 71, 85, 97, 101, 125, 149, 157, 171, > 179, 205, 215, 235, 241, 283,...}. I can't find this sequence in > Sloane's sequences. A continuation of the sequence needs 75 terms to > span (in the above sense) all numbers up to 2042 : there are 297 > primes less than 2042, so this sequence is significantly smaller than > the corresponding prime sequence. Of those 75 terms, 28 are primes and > 47 are composite. My original plan was to show that composite numbers in this sequence > could be replaced by some suitable number of suitable primes, and > hence that some subset of the primes would suffice, and hence that the > set of all primes was plenty... but I recognize that I don't have > enough number theory to make a good attack on these lines... > Anyway, like I said, if anyone knows anything about this sequence > (like an asymptotic formula for its size, or its size relative to the > corresponding set of primes, or the ratio of primes to composites in > the sequence, or anything, really) I'd really like to hear it! BTW, it strikes me that such a sequence in some sense acts as a > minimal 2-term additive generator for the even numbers > 4. Is this a > useful concept???? John > === Subject: Re: Goldbach-like sequences > I've been dabbling with Goldbach's conjecture for a while now, and > I've come across something that I've not seen described in any of the > literature I've seen on the net so far. I wondered if anyone else has > had any thoughts, or can point me to any references, on this and > related matters... Let S be a set of odd numbers with maximum element m. I call it a > Goldbach Spanning Set if all the even numbers from 6 to m+1 can be > expressed as the sum of two numbers in S. These things obviously exist > : the set of all odd numbers between 3 and m will do. The question I > asked myself was what's the smallest such set that will suffice?. Here is where it may run into trouble. Take {3, 5, 7, 11}. These of will give us the evens 6 to 18. However, so will {3, 5, 7, 9}. In addition, even if you prove that a subset of primes spans 3 to m, how will you extend this so that it shows a subset of primes spans 3 to m+2? === Subject: Re: Goldbach-like sequences I've been dabbling with Goldbach's conjecture for a while now, and >I've come across something that I've not seen described in any of the >literature I've seen on the net so far. I wondered if anyone else has >had any thoughts, or can point me to any references, on this and >related matters... >Let S be a set of odd numbers with maximum element m. I call it a >Goldbach Spanning Set if all the even numbers from 6 to m+1 can be >expressed as the sum of two numbers in S. These things obviously exist >: the set of all odd numbers between 3 and m will do. The question I >asked myself was what's the smallest such set that will suffice?. Here is where it may run into trouble. Take {3, 5, 7, 11}. These of will > give us the evens 6 to 18. However, so will {3, 5, 7, 9}. In addition, > even if you prove that a subset of primes spans 3 to m, how will you extend > this so that it shows a subset of primes spans 3 to m+2? What Bruckman considered was the minimal set S of the smallest primes that would span 2N for some N, i.e., 2N=sum of two primes. 3,3,7,11,13,17,19,23,31 is minimal set for 2N =6 through 50 and prime 29 isn't needed until 2N=52. There are small primes that are not needed for much longer spans. There is no way, AFAIK, to take some 2N and give a minimal set of primes which spans 6 through 2N. The minimal set depends upon all the evens 6 through 2N. If GBC is false then there is some 2N without a minimal set of primes. Concerning the number of pairs of primes that add to 2N, one way is to use the old method ( circa 1895). For the odds <=2N-3, e.g., 2N= 22, 3 5 7 9 11 13 15 17 19 form the numbers 1 0 0 -1 1 0 -1 1 0 (where the a first prime alone or of consecutive primes is 1 and any consecutive prime after that prime are 0 and the first alone or of consecutive composites is -1 and any consecutive following composites are 0) and the running sum of primes 1 2 3 3 4 5 5 6 7 and place it in reverse under the first set of numbers 1 0 0 -1 1 0 -1 1 0 7 6 5 5 4 3 3 2 1 multiply the columns together and add the results to get 7-5+4-3+2=5 pairs of primes which equal 22. Note that for GBC to be true, at least one of the 1...-1 groups must have a decrease in the running sum under it. For GBC to be false all decreases in the running sum must be in the -1...1 groups. I have not been able to find any reason why that cannot happen since the number of composites far out number the primes as N increases and can easily accommodate all the primes. There must be something about the efficiency of the double sieves on the odd numbers which would prohibit that from happening. Larry > === Subject: Re: Guarantee of finding a prime in a fixed interval? NNTP-Proxy-Relay: library2.airnews.net charset=iso-8859-1 > Looking at primes in the intervals begun by > n, n+2, n+4, ...., n + 2*k, for some fixed k, can anything > be said about existance of primes in each for finite k? For x >= 114, we have > x / ln(x) < pi(x) < 5*x / 4*ln(x) > where pi(x) is the number of primes smaller or equal to x, > can be used to calculate a lower bound on the number > of primes in [n, n +2*k] References appreciated! Feel free to reply to this subthread if you don't want to crosspost. === Subject: Re: Rank of a group? > feet wet), as far as I have always understood things. Going in > a bit over my head, I think that's pretty much the same as the > number of independent directions in which the Riemannian manifold > is flat at a general point (of course for a symmetric space > all points look the same vis a vis isometries, and every point > is general in that sense). So really here it isn't the > group-as-a-group which has a rank, it's the > group-as-an-isometry-group. Hey, could you explain that a bit? I mean, R^3 surely is symmetric as it can be, but the rank of the isometry group is bounded by 2 (by a theorem of Grove and Searl, I don't really know the rank of the group of symmetries of R^3), while it looks flat in every direction... cheers Philipp === Subject: Re: Rank of a group? >> feet wet), as far as I have always understood things. Going in >> a bit over my head, I think that's pretty much the same as the >> number of independent directions in which the Riemannian manifold >> is flat at a general point (of course for a symmetric space >> all points look the same vis a vis isometries, and every point >> is general in that sense). So really here it isn't the >> group-as-a-group which has a rank, it's the >> group-as-an-isometry-group. >Hey, could you explain that a bit? Easily: I'm totally confused and mistaken. (And sinking.) >I mean, R^3 surely is symmetric as >it can be, but the rank of the isometry group is bounded by 2 (by a >theorem of Grove and Searl, I don't really know the rank of the group >of symmetries of R^3), while it looks flat in every direction... Sounds right to me. ... Where are the differential geometers of sci.math when we need them? Help! Helg! Glug! (<---goes down for third time) Lee Rudolph === Subject: Re: Rank of a group? > Sounds right to me. ... Where are the differential geometers of > sci.math when we need them? Help! Helg! Glug! (<---goes down > for third time) Hrhr, I understand that *G* Just for the record: My citation of Grove&Searl wasn't right, because R^3 doesn't have positive sectional curvature, so it's isometry group probably has rank 3 ... Cheers Philipp === Subject: Re: Rank of a group? feet wet), as far as I have always understood things. Going in >a bit over my head, I think that's pretty much the same as the >number of independent directions in which the Riemannian manifold >is flat at a general point (of course for a symmetric space >all points look the same vis a vis isometries, and every point >is general in that sense). So really here it isn't the >group-as-a-group which has a rank, it's the >group-as-an-isometry-group. >>Hey, could you explain that a bit? > Easily: I'm totally confused and mistaken. (And sinking.) >>I mean, R^3 surely is symmetric as >>it can be, but the rank of the isometry group is bounded by 2 (by a >>theorem of Grove and Searl, I don't really know the rank of the group >>of symmetries of R^3), while it looks flat in every direction... > Sounds right to me. ... Where are the differential geometers of > sci.math when we need them? Help! Helg! Glug! (<---goes down > for third time) Lee Rudolph I thought that rank of a Lie group was defined as the dimension of a maximal abelian subgroup. If so, R^3 is itself an abelian Lie group, not compact of course, but by acting on itself (by translation) one gets a transitive action (duh). I would think its rank is 3. If you require the origin to be fixed, you get O(3), or if you require orientation to be preserved, SO(3). I believe the rank of SO(3) is 1 (rotations about any fixed axis serve as a maximal abelian subgroup). I may be wrong. Dale. === Subject: Re: Rank of a group? > I thought that rank of a Lie group was defined as the dimension of a > maximal abelian subgroup. If so, R^3 is itself an abelian Lie group, > not compact of course, but by acting on itself (by translation) one > gets a transitive action (duh). I would think its rank is 3. > If you require the origin to be fixed, you get O(3), or if you > require orientation to be preserved, SO(3). I believe the rank > of SO(3) is 1 (rotations about any fixed axis serve as a maximal > abelian subgroup). The (real) rank of a (linear) Lie group is the dimension of a maximal abelian subgroup which is conjugate to a subgroup of the diagonal matrices (also known as an R-split torus) with real values. This is independent of its realization as a linear group. Using this definition, it follows that R^3 has real rank 3 and SO(n) has real rank 0. You can also define the K-rank of a k-algebraic linear group, where K/k is any extension. For a general Lie group, the real rank is probably the dimension of the largest abelian subgroup consisting of semisimple elements. Think about real hyperbolic space, whose isometry group (without the involution induced by inversion) is PO(n,1). This has real rank 1, where a maximal R-split torus is the subgroup of all loxodromic isometries with the same two fixed points at the boundary (say 0 and infinity in the upper half space model). It couldn't be maximal abelian subgroups, since the subgroup of parabolic isometries fixing a particular boundary point is isomorphic to R^{n-1}. Hope this helps. === Subject: Re: Rank of a group? [Stuff about linear Lie groups snipped] Unfortunately, the groups I'm thinking about aren't linear, so that doesn't really clear up things :( > For a general Lie group, the real rank is probably the dimension of the > largest abelian subgroup consisting of semisimple elements. I don't really know what semisimple elements are (yes, I'll look it up :)), but my advisor told me, that what I asked about is the definition, i.e. the rank is the maximal dimension of an embedded torus (though he's probably hiding from me that it could be defined differently in general lie group theory). I guess I should grab a book about Lie groups... Thanks for your help :) Cheers Philipp === Subject: Re: Rank of a group? > [Stuff about linear Lie groups snipped] > Unfortunately, the groups I'm thinking about aren't linear, so that > doesn't really clear up things :( If the group isn't linear, its just the dimension of the Cartan subgroup. I can't remember of the top of my head what the definition of a semisimple element in a general Lie group. For a linear Lie group, its an element which can be diagonalized. Probably the best way to define rank is using Cartan subgroups, since this is equivalent to the previous definitions I gave and works in the non-linear case. Knapp's Lie groups beyond an introduction should have everything you want to know. Hope this helps. Ben === Subject: System of differential equations Originator: israel@math.ubc.ca (Robert Israel) i know there exists a general solution for systems oif linear differential equations with constant coefficients. In my case the system matrix depends linearly on the states. This results in a system where the derivatives of the states x1,x2 depend not only linearly on themselves but also quadratic: dx1/dt=A11*(x1;x2)+(x1;x2)'*A12*(x1;x2) dx2/dt=A21*(x1;x2)+(x1;x2)'*A22*(x1;x2) where the matrices A11 and A21 have the size 1x2 and the matrices A12 and A22 the size 2x2. Does anyone know if there exists a general solution for that kind of differential equations? [ Moderator's note: of course a solution exists (at least locally) by the general existence-and-uniqueness theorem for systems of differential equations. Jan is asking about closed-form solutions. In correspondence with the moderator, he added: > Thank you for your reply, but before posting to the forum, I had a > look at several textbooks and of course for the scalar case it's a > Riccatti differential equation with the solution given in the > textbooks. But for systems of differential equations I could not find > a solution! I'm an engineer and such a problem is not a common > problem for us; so I would be very pleased if one of the experts on > the forum could help me by giving the solution or literature on this > problem! Thank you very much in advance! -ri ] Dipl.-Ing. Jan Weigel Technische Universit.8at Darmstadt Institut f.9fr Stromrichtertechnik und Antriebsregelung Landgraf-Georg-Str.4 64283 Darmstadt Email:jweigel@srt.tu-darmstadt.de Tel.: 06151/166748 Fax.: 06151/162613 === Subject: NEW: Giga-Tera Foundation At last! A Foundation for those who were ineligible to join Mensa or the Mega Foundation because they were overqualified. http://www.crbond.com/gtf.htm -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com === Subject: Re: NEW: Giga-Tera Foundation > http://www.crbond.com/gtf.htm Ummmm....I'm a little confused on the pain-spectrum of intellegence. I'm having-a-root-canal-with-no-anasthetic-ly gifted and I'm not sure if that falls between excruciatingly gifted and agonizingly gifted or above that range, and so I'm unable to tell whether I'm over-qualified. I guess it doesn't matter, since I don't have FAX number, so I can't join anyway. === Subject: Re: NEW: Giga-Tera Foundation At last! A Foundation for those who were ineligible to join Mensa or the > Mega Foundation because they were overqualified. http://www.crbond.com/gtf.htm > Will there be a subsection 'glueing and sheafication of trisected angels'? I want to go to the xmas session. Hmmm ... if you send me your fax pls choose a new one. And you dont forget to pay shipping & customs, yes? === Subject: Re: NEW: Giga-Tera Foundation > At last! A Foundation for those who were ineligible to join Mensa or the > Mega Foundation because they were overqualified. http://www.crbond.com/gtf.htm > When you get a few more links to the discussion boards up, I'll consider paying the $100 if you choose to demand it. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: NEW: Giga-Tera Foundation > At last! A Foundation for those who were ineligible to join Mensa or the > Mega Foundation because they were overqualified. > http://www.crbond.com/gtf.htm To save time on the dues rebate, could I just fax you the $100 and send a copy to myself? (That brilliant idea alone should qualify me for membership.) -- 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, === Subject: Re: diophantine eq (sort of) >> I have only 37 cent stamps and 23 cent stamps. I want to mail a >> package that requires an arbitrary amount of postage. If I cannot >> achieve the exact postage required, then I want to minimize the >> overpayment. (assuming all variables are integers ..) If x0, y0 is any given solution of a.x - b.y = c then every solution >of a.X - b.Y = c.d is given by X, Y = d.x0 + b.t, d.y0 + a.t as t >ranges over the integers. ... if you assume gcd(a,b) = 1 (which of course is the case in this > example). > More generally it should be X, Y = d.x0 + (b/g).t, d.y0 + (a/g).t > where g = gcd(a,b). Yes, I was assuming (a,b) = 1. But this general result is more useful. > ... So for any given d (the total postage) you choose, all you need do >is find the value of t that minimizes x + y (> 0) while satisfying >these bounds. Why > 0, not >= 0? Well, you need at least one stamp on the envelope (in this problem). >For the values in your example, x + y > 0 is equivalent to 3.d/14 < t. >But as 3d/14 < 8d/37 (<=> 111d < 112d), the lower bound implies this. >So it suffices to take t = floor(8.d/37). ceil, not floor. There will be a solution with no wasted postage iff > ceil(8d/37) <= floor(5d/23). That is certainly true if > 5d/23 - 8d/37 >= 1, i.e. d >= 851. In fact it turns out that > the largest integer for which ceil(8d/37) > floor(5d/23) is 791, > i.e. it is impossible to make $7.91 with 23 and 37 cent stamps, > but it is possible to make any higher postage. More generally, if gcd(a,b) = 1, every integer >= (a-1)(b-1) can be > represented as a x + b y with x,y nonnegative integers, while > (a-1)(b-1)-1 can not be represented in this way. I didn't follow my solution through very well at all (rushing too much on a monday morning). In case the OP finds the above a bit concise, the idea is that an attainable ('stampable') postage, d, can only follow if there is at least one integer t in the closed interval [8d/37, 5d/23]. Considering successive values of d, whether attainable or not, i.e. d = 1, 2, ... this interval starts out small (less than 1) and as d increases it shifts along the real number line and expands at the same time. For values of d such that 5d/23 - 8d/37 (the interval width of course) is less than 1 there is a chance the interval may not contain any integers, as it is small enough to slip between two consecutive integers. In this case to get an attainable postage you must increase d to the first larger value, say d', for which the interval does enclose an integer. Since for each increment of d the interval shifts forward by 5/23, which is less than 1, we can be sure it doesn't skip any integers, and thus the required value of the next highest integer, t, is ceil(5d/23). Then d' can be expressed as ceil(23t/5) (hint: 5(d'-1)/23 < t <= 5d'/23). But, as Robert pointed out, once the interval width reaches 1 we know that from that value of d and above, every interval must contain at least one integer. Cheers John Ramsden (john_ramsden@sagitta-ps.com) === Subject: formula for a spectrum of all even natural numbers The exercise reads: Find a first order sentence whose spectrum is the set of even positive integers I googled it and the only hint I've got mentiones even permutations. I still have no clue even what predicate(s) might be in the sentence. While we are at it, I would be also interested to know formula for all even integers (without sign restriction). === Subject: Re: formula for a spectrum of all even natural numbers How about for each x( x is contained in a unique y XOR x contains a unique y)? > The exercise reads: > Find a first order sentence whose spectrum is the set of even positive > integers I googled it and the only hint I've got mentiones even permutations. I still > have no clue even what predicate(s) might be in the sentence. While we are at it, I would be also interested to know formula for all even > integers (without sign restriction). === Subject: Re: formula for a spectrum of all even natural numbers >The exercise reads: >Find a first order sentence whose spectrum is the set of even positive >integers What is the spectrum of a first-order sentence? >I googled it and the only hint I've got mentiones even permutations. I still >have no clue even what predicate(s) might be in the sentence. >While we are at it, I would be also interested to know formula for all even >integers (without sign restriction). ************************ David C. Ullrich === Subject: Re: formula for a spectrum of all even natural numbers The set of cardinalities of the finite models of the first order sentence phi: spec(phi) = { n | n = |A| for some A belonging to MOD[phi] } The definition renders my second question as silly as we are talking about natural numbers. I would be also interested if anybody can criticize that definition. For example, Spectrum in Linear Algebra is a set of eigenvalues, which are associated with eigenvectors -- very rich theory. What theory can we expect when values are from the field that is not algebraically closed (sorry, natural numbers aren't even a field)? > What is the spectrum of a first-order sentence? === Subject: Re: formula for a spectrum of all even natural numbers >The set of cardinalities of the finite models of the first order sentence >phi: >spec(phi) = { n | n = |A| for some A belonging to MOD[phi] } >The definition renders my second question as silly as we are talking about >natural numbers. Yes. You can easily give a sentence whose spectrum is the set of even natural numbers, for example a formalization of R is an equivalence relation and each equivalence class contains exactly two elements. >I would be also interested if anybody can criticize that definition. For >example, Spectrum in Linear Algebra is a set of eigenvalues, which are >associated with eigenvectors -- very rich theory. What theory can we expect >when values are from the field that is not algebraically closed (sorry, >natural numbers aren't even a field)? >> What is the spectrum of a first-order sentence? ************************ David C. Ullrich === Subject: Re: This just in: elementary proof of FLT [not JSH] > I tried the Safari browser for the first time today. It seems to have > an area for calling one's attention to interesting Web sites. One of > them announces an elementary proof of Fermat's Last Theorem. Can you read Italian? > Dunno ... never tried. > So I can't give a critical report on the contents. http://www.multiwire.net/fermat/welcome.html Ossicini's name in sci.math is curst > When JSH finds out he publihed first. Here's google's translation: > http://translate.google.com/translate?u=http%3A%2F% 2Fwww.multiwire.net%2Fferm at%2Fwelcome.html&langpair=it%7Cen&hl=en&ie=UTF-8&oe=UTF-8& prev=%2Flanguage_t ools Copy and paste the URL into your address bar. Why? The translated page appears to me to be in Italian as well, including the hypertext. I tried searching for the original in Google and then hitting translate this page, and similarly got no change in text. - Randy === Subject: Re: This just in: elementary proof of FLT [not JSH] Christopher J. Henrich > I tried the Safari browser for the first time today. It seems to have > an area for calling one's attention to interesting Web sites. One of > them announces an elementary proof of Fermat's Last Theorem. > http://www.multiwire.net/fermat/welcome.html J. Edgar Harris won't like this guy: http://www.coolissues.com/mathematics/Fermat/fermat.htm http://www.coolissues.com/mathematics/Riemann/riemann.htm http://www.coolissues.com/mathematics/Goldbach/goldbach.htm http://www.coolissues.com/mathematics/Tprimes/tprimes.htm LH === Subject: Re: This just in: elementary proof of FLT [not JSH] > Christopher J. Henrich >I tried the Safari browser for the first time today. It seems to have >>an area for calling one's attention to interesting Web sites. One of >>them announces an elementary proof of Fermat's Last Theorem. >> http://www.multiwire.net/fermat/welcome.html J. Edgar Harris won't like this guy: > http://www.coolissues.com/mathematics/Fermat/fermat.htm > http://www.coolissues.com/mathematics/Riemann/riemann.htm > http://www.coolissues.com/mathematics/Goldbach/goldbach.htm > http://www.coolissues.com/mathematics/Tprimes/tprimes.htm > LH Why should James care? Anyone can *claim* to have proven Fermat's Last Theorem, the Riemann Hypothesis, etc., and provide absurdly invalid proofs. Clearly this guy is just some delusional bozo. James, on the other hand, has proven Fermat's Last Theorem. Comparing James to the nut on that website is like comparing Neil Armstrong to some spaced-out hippy. Sure, both *claim* to have been to the moon . . . What you have to remember here, people, is that the *math* shows that James is correct, even if you and I are not able to understand it. Time to face the facts: James is much smarter than any of us -- perhaps the smartest person on Earth. He's blazing new trails on the frontiers of mathematics, and we are simply incapable of following. Most of us can't even understand the very definitions he gives, let alone how he uses them. The reaction to James's work has been disappointing, if expected: institutional mathematicians objecting ignorantly and baselessly, waving around their fancy degrees. But they'd do better to wave a white flag. A hard rain's a gonna fall. Sorry to rant, but comparing James to some FLT crank boils my blood. James is a published mathematician, after all. Maybe he rejects that title because of the sorry state of the mathematical establishment, but the point is he's not just some guy with a website and a fistful of manic scribblings. True, he is a pariah whom all respectable journals have been compelled to shun, but he has found an editor, Chris Langan, intelligent enough to understand his work and brave enough to publish it. When are you people going to stop persecuting James? This is getting ridiculous! === Subject: minimal ellipse centered at origin with 2 points Given two points in the plane, P1 and P2, how does one determine the ellipse that: 1) is centered at the origin, 2) goes through P1 and P2, 3) is of minimal area ? -- dave === Subject: Re: minimal ellipse centered at origin with 2 points > Given two points in the plane, P1 and P2, how does one determine the ellipse that: > 1) is centered at the origin, > 2) goes through P1 and P2, > 3) is of minimal area > ? -- dave One does it by spending quite a bit of time and paper, I'm afraid. ;-( Typing out the whole story isn't an attractive prospect, but FWIW here's the answer. To keep the ascii formula reasonably clear I'll reluctantly avoid subscripts by using coordinates (a, b) for P1 and (c, d) for P2. Then the equation of your ellipse is (b^2 + d^2)x^2 - 2(ab + cd)xy + (a^2 + c^2)y^2 = (ad - bc)^2. This is an ellipse unless ad - bc = 0, in which case it degenerates into the repeated line through O, P1 and P2. Ken Pledger. === Subject: Re: minimal ellipse centered at origin with 2 points > One does it by spending quite a bit of time and paper, I'm afraid. > ;-( Typing out the whole story isn't an attractive prospect, but FWIW > here's the answer. To keep the ascii formula reasonably clear I'll reluctantly avoid > subscripts by using coordinates (a, b) for P1 and (c, d) for P2. > Then the equation of your ellipse is (b^2 + d^2)x^2 - 2(ab + cd)xy + (a^2 + c^2)y^2 = (ad - bc)^2. This is an ellipse unless ad - bc = 0, in which case it degenerates into > the repeated line through O, P1 and P2. Ken Pledger. Thanks a lot! I tried your formula on a few examples, and it seems correct. Can you provide any reference about this result? Or, can you sketch the steps that led you to the answer? -- dave === Subject: Re: minimal ellipse centered at origin with 2 points Originator: broeker@ >> One does it by spending quite a bit of time and paper, I'm afraid. >> ;-( Typing out the whole story isn't an attractive prospect, but FWIW >> here's the answer. [...] > Can you provide any reference about this result? Or, can you sketch > the steps that led you to the answer? There shouldn't be all that much to it: set up the generic equation for an origin-centered, non-axis-aligned ellipse: F(X) := := F([x,y]^t) = A*x^2 + 2*B*x*y + C^2*y^2 = X^T * M * X for a point in column vector notation X, and a matrix M = [A B] [B C] Matrix M must be positive definite for this equation to describe an ellipse. Its area is pi * determinant(M), which you want to minimize, i.e.: A*C - B^2 = min! You have 2 input equations: F(P1) = 1 and F(P2) = 1, plus the minimization condition, which makes 3; and you have 3 unknowns, so everything's fine. involved. -- Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de) Even if all the snow were burnt, ashes would remain. === Subject: Re: minimal ellipse centered at origin with 2 points >> One does it by spending quite a bit of time and paper, I'm afraid. >> ;-( Typing out the whole story isn't an attractive prospect, but FWIW >> here's the answer. > [...] Can you provide any reference about this result? Or, can you sketch > the steps that led you to the answer? There shouldn't be all that much to it: set up the generic equation > for an origin-centered, non-axis-aligned ellipse: F(X) := > := F([x,y]^t) > = A*x^2 + 2*B*x*y + C^2*y^2 > = X^T * M * X for a point in column vector notation X, and a matrix M = [A B] > [B C] Matrix M must be positive definite for this equation to describe an > ellipse. Its area is pi * determinant(M), which you want to > minimize, i.e.: A*C - B^2 = min! You have 2 input equations: F(P1) = 1 and F(P2) = 1, plus the > minimization condition, which makes 3; and you have 3 unknowns, so > everything's fine. involved. Actually, the 'equation' for the ellipse will be more like A*x^2 + 2*B*x*y + C*y^2 = 1 and, since A*C - B^2 is invariant under rotations about origin, all such ellipses with the same value of A*C - B^2 will be congruent. Note that the area of the ellipse with equation in the common standard form x^2/a^2 + y^2/b^2 = 1 is known to be Area = pi*a*b, where a*b= 1/sqrt(A*C-B^2) when B = 0. It follows that the area of ellipse A*x^2 + 2*B*x*y + C*y^2 = 1 will be pi/sqrt(A*C - B^2). This area will be a minimum when A*C - B^2 is a maximum, so finding the extreme value of A*C - B^2 subject to the ellipse passing through the given points does indeed solve the problem. I did it as follows: Let G(x,y) = A*x^2 + 2*B*x*y + C*y^1 - 1, so that G(x,y) = 0 is the equation of the ellipse. Let (x1,y1) and (x2,y2) be the points through which the ellipse must pass, and assume that they are not colinear with origin. If they are colinear with origin, there is no ellipse of minimal positive area. The following uses Lagrange multipliers u and v: Let H(A,B,C,u,v) = A*C - B^2 + u*G(x1,y1) + v*G(x2,y2). Set the partials of H with respect to A,B,C, u and v all equal to zero and solve the resulting system of equations for A, B, and C. For these last steps, I used an HP49 calculator, rather than doing it by hand, getting essentially the same result as others have posted here. === Subject: Re: minimal ellipse centered at origin with 2 points Can you provide any reference about this result? Or, can you sketch > the steps that led you to the answer? ... No rocket science involved. No rocket science and no calculus either. I was able to find the minimum by completing a square. However, Keith Ramsay's approach is a nice alternative. Ken Pledger. === Subject: Re: minimal ellipse centered at origin with 2 points [...] | One does it by spending quite a bit of time and paper, I'm afraid. |;-( Typing out the whole story isn't an attractive prospect, but FWIW |here's the answer. It's easier if you notice that the problem is preserved by invertible linear transformations. Any invertible linear transformation leaves the origin fixed, takes ellipses to ellipses, and multiplies their areas by a positive constant factor. As you notice, if the points are collinear with the origin, you can make the ellipse have as small an area as desired just by making it long enough and very thin. If the points are not collinear with the origin, there's always a linear transformation that takes them to the points (1,0) and (0,1), namely the one given by the matrix (a' c') (b' d') which is the inverse matrix to (a c) (b d). Inverting the matrix, you get a'=d/D, b'=-b/D, c'=-c/D, d'=a/D where D=ad-bc is the determinant of the original matrix. | To keep the ascii formula reasonably clear I'll reluctantly avoid |subscripts by using coordinates (a, b) for P1 and (c, d) for P2. |Then the equation of your ellipse is | |(b^2 + d^2)x^2 - 2(ab + cd)xy + (a^2 + c^2)y^2 = (ad - bc)^2. | |This is an ellipse unless ad - bc = 0, in which case it degenerates into |the repeated line through O, P1 and P2. When P1=(1,0) and P2=(0,1) this reduces to x^2+y^2=1. Your answer can be written (dx-cy)^2 + (-bx+ay)^2 = D^2, which is equivalent to (a'x+c'y)^2 + (b'x+d'y)^2 = 1, the transformed form of x^2+y^2=1. I guess that still leaves the problem of showing that the unit circle is the solution for P1=(1,0) and P2=(0,1). That's at least plausible. :-) We need to show that the area of an ellipse centered at the origin containing the two points has an area of at least |D|*pi. We could reverse the idea and transform the ellipse into a unit circle. Then we're trying to show that for any points P1 and P2 in the unit disk, |D|<=1. That's true by the Cauchy-Schwarz inequality, as (ad-bc)^2 <= (a^2+b^2)(c^2+d^2) <= 1. Done. Keith Ramsay === Subject: Re: Fundamental Discovery: Program = |- P(x)[YES] > To the Research Community: Greetings. I would like to announce a Fundamental Discovery in > Computer Science. Symbolically: > In words: A program is a constructive proof that a given set or function > is recursive or recursively enumerable. > Proofs in Coq are lambda terms, and in a sense ARE programs Yes, Coq is really a programming language. The user inputs the program. Coq doesn't create it. > not programs generated from proofs, I would > say (so that generating needs some extra theory). Yes. And until someone does it, it hasn't been done. Program verification is very different from program synthesis. > it's (nearly) all that Coq is about :-) There have been many attempts to draw a correspondence between Logic/Theorem Proving and Computer Programming. It is a natural connection. But I am saying that I actually did it, not just discussed it. > Program Verification is the third point of the quoted text. The first > two is the Assisted Programming. Doesn't it assist you by verifying your assertions concerning the program? That's the whole point (with a very rich byproduct.) > Yes, you're right -- Coq could sometimes output the program to you (if it is > simple enough, by a use of automation tactical But has anyone done that? Any examples? > But certainly APC cannot generate all sorts of programs. Harder programs > need to be written (e.g. in Coq). But then a person is writing the program! > I asked him the same question I ask everyone: What would the input and > output be to generate programs to (1) determine if one number is a > factor of another, and (2) list the prime numbers between 1000 and > 1005? (as I show above.) Sorry, I don't have time and I didn't intend this discussion (which I hope > is helpful to you) (but you can ask directly on coq-club@pauillac.inria.fr Thanks. I will ask - but could I see the context in which I am asking this question? Is there a web page where I can just post the question? Actually, about a year ago I asked some French research group whether Coq could generate programs, and received no response. I don't see any evidence that it ever has. > He was never able to give me an example of any program (much less the > above two) generated by Coq. > You are right about you remark, but you are wrong about Coq. My only intent is to determine if Coq (or any other system) succeeds at generating computer programs and not systems where a person is entering in the program! > But this answer wouldn't satisfy our Porky Pig. This is a different program. > He meant his program, written in C. He asked what his program, written in C, > proves. It proves that {Hello World} is recursively enumerable (as I said.) Do you disagree or am I not being clear? > Informally speaking, stating that every program is a proof that given > function is recursive / given set is recursively enumerable says just that > every computer program can be run on a computer (which is at least as > powerful as Turing Machine ;-) Well, it says that we can analyze and construct computer programs automatically by proving the theorem with no reference to any programming - and then building a program in parallel to the proof. And APC generates the proof (and program) using its Rules of Inference and the programming language's axioms and definitions. > Your theorem is stronger than you express here in > that we can formalize the setting of any programming language, not only the > APC. Yes. APC actually represents any programming language. > Best wishes, > Lukasz Thanks for the comments. Cambridge, MA http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ 20021008.1/1 http://www.arxiv.org/html/cs.lo/0003071 === Subject: Is c_0 a closed subset in l_infinity? I read in a textbook that c_0 is a closed subset in l_infinity under sup-norm. A proof is left as an exercise...:-) However, I am tempted to believe that c_0 is not closed in l_infinity. My reasoning is as follows. Let e(n) be an infinite sequence filled with with zeros except for the n-th position, where it is one. For example, e(0) = {1,0,0,0,...}, e(1) = {0,1,0,0,...}, and so on. Then I form a sequence of these sequences, denoted {e(n)}. Note that e(n) is in c_0 for all finite n. But the limit of this sequence (of sequences) is not in c_0, because lim_{n->infinity}e(n) is not converging to zero. What is wrong in my reasoning? Zdenek === Subject: Re: Is c_0 a closed subset in l_infinity? >I read in a textbook that c_0 is a closed subset in l_infinity under >sup-norm. It is. >A proof is left as an exercise...:-) >However, I am tempted to believe that c_0 is not closed in l_infinity. My >reasoning is as follows. >Let e(n) be an infinite sequence filled with with zeros except for the n-th >position, where it is one. For example, >e(0) = {1,0,0,0,...}, >e(1) = {0,1,0,0,...}, >and so on. >Then I form a sequence of these sequences, denoted {e(n)}. Note that e(n) is >in c_0 for all finite n. But the limit of this sequence (of sequences) is >not in c_0, because lim_{n->infinity}e(n) is not converging to zero. >What is wrong in my reasoning? The sequence e_n does not _have_ a limit. (That is, it does not have a limit in the l_infinity norm. It does have a limit in the weak* topology on l_infinity = (l_1)*, and the example you give does show that c_0 is not weak* closed.) Zdenek ************************ David C. Ullrich === Subject: Re: Is c_0 a closed subset in l_infinity? > I read in a textbook that c_0 is a closed subset in l_infinity under > sup-norm. A proof is left as an exercise...:-) However, I am tempted to believe that c_0 is not closed in l_infinity. My > reasoning is as follows. Let e(n) be an infinite sequence filled with with zeros except for the n-th > position, where it is one. For example, > e(0) = {1,0,0,0,...}, > e(1) = {0,1,0,0,...}, > and so on. Then I form a sequence of these sequences, denoted {e(n)}. Note that e(n) is > in c_0 for all finite n. But the limit of this sequence (of sequences) is > not in c_0, because lim_{n->infinity}e(n) is not converging to zero. What is wrong in my reasoning? The limit of your sequence does not exist in the l^oo norm. The limit exists in each component, but that's not good enough. To prove c_0 is closed in l^oo, suppose x(1), x(2),... is a sequence in c_0 that converges to x in the l^oo norm. Let epsilon > 0. Choose an x(n) within epsilon of x in the l^oo norm. The components of x(n) have absolute value < epsilon far enough out. That implies the components of x will have absolute value < 2*epsilon far enough out. === Subject: Re: Is c_0 a closed subset in l_infinity? bjnv8e$2e6d$1@ns.felk.cvut.cz... > I read in a textbook that c_0 is a closed subset in l_infinity under > sup-norm. A proof is left as an exercise...:-) > However, I am tempted to believe that c_0 is not closed in l_infinity. My > reasoning is as follows. I don't get a word... you read A in a textbook, then you want to prove not(A)... and what do c_0 and 1_infinity mean? Those are not standard notations at all. === Subject: Re: Is c_0 a closed subset in l_infinity? >bjnv8e$2e6d$1@ns.felk.cvut.cz... >> I read in a textbook that c_0 is a closed subset in l_infinity under >> sup-norm. A proof is left as an exercise...:-) >> However, I am tempted to believe that c_0 is not closed in l_infinity. My >> reasoning is as follows. >I don't get a word... you read A in a textbook, then you want to prove >not(A)... and what do c_0 and 1_infinity mean? Those are not standard >notations at all. They're quite standard. c_0 is the space of sequences tending to 0 and l_infinity is the space of bounded sequences. (Both with the norm ||a|| = sup |a_n|.) ************************ David C. Ullrich === Subject: Re: Is c_0 a closed subset in l_infinity? Content-transfer-encoding: 8bit >bjnv8e$2e6d$1@ns.felk.cvut.cz... >> I read in a textbook that c_0 is a closed subset in l_infinity under >> sup-norm. A proof is left as an exercise...:-) >> However, I am tempted to believe that c_0 is not closed in l_infinity. My >> reasoning is as follows. I don't get a word... you read A in a textbook, then you want to prove >not(A)... and what do c_0 and 1_infinity mean? Those are not standard >notations at all. They're quite standard. c_0 is the space of sequences tending to 0 > and l_infinity is the space of bounded sequences. (Both with the > norm ||a|| = sup |a_n|.) as Hurak did. one not little-ell. That wouldn't be standard, although it's clear what's meant...and of course c_0 IS standard. --Ron Bruck === Subject: Re: Is c_0 a closed subset in l_infinity? >>bjnv8e$2e6d$1@ns.felk.cvut.cz... > I read in a textbook that c_0 is a closed subset in l_infinity under > sup-norm. A proof is left as an exercise...:-) However, I am tempted to believe that c_0 is not closed in l_infinity. My > reasoning is as follows. >>I don't get a word... you read A in a textbook, then you want to prove >>not(A)... and what do c_0 and 1_infinity mean? Those are not standard >>notations at all. >> They're quite standard. c_0 is the space of sequences tending to 0 >> and l_infinity is the space of bounded sequences. (Both with the >> norm ||a|| = sup |a_n|.) >as Hurak did. one not little-ell. I _really_ need to get around to that visit to the optometrist sometime. >That wouldn't be standard, >although it's clear what's meant...and of course c_0 IS standard. >--Ron Bruck ************************ David C. Ullrich === Subject: Rigorous Textbook Needed Can anyone suggest a rigorous textbook treatment of multivariable calculus and linear algebra. My current textbook omits a lot of the proofs--and say that they will be left to a more advanced text. === Subject: Re: Rigorous Textbook Needed X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS >Can anyone suggest a rigorous textbook treatment of multivariable >calculus and linear algebra. I concur with the suggestion for Finite Dimensional Vector Spaces. I haven't seen Apostol's book on Calculus, but his Mathematical Analysis was excellent. You might try to find some older Calculus textbooks; the standards seem to have declined since the 1950s. But I have seen references to more modern texts that were promising. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Rigorous Textbook Needed Visiting Assistant Professor at the University of Montana. >Can anyone suggest a rigorous textbook treatment of multivariable calculus >and linear algebra. My current textbook omits a lot of the proofs--and say >that they will be left to a more advanced text. For linear algebra you will find many such books; check to see if your university offers an upper division linear algebra course and check out their textbook. Off the top of my head, for something from a purely mathematical point of view, I could mention Paul Halmos's _Finite Dimensional Vector Spaces_ There are few textbooks that cover the barren wasteland between the calculus is easy, calculus for everyone textbooks of lower division calculus courses and mathematical analysis books that cover topological considerations and other necessities for the foundation of calculus. You might want to look at books with Analysis in their titles. There's Rudin's Real Analysis; I know Norman Haaser and Joseph Sullivan's Real Analysis (originally published in 1971 and reissued in the 90s by Dover) starts from the notions of sets, through linear and metric spaces, to the Fundamental Theorem of Calculus. The book A Course in Mathematical Analysis, two volumes, by Haaser, Sullivan, and Joseph Lasalle includes, if memory serves, multivariable calculus as well. ============================================================== ======== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ============================================================== ======== Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Rigorous Textbook Needed Can anyone suggest a rigorous textbook treatment of multivariable calculus >and linear algebra. My current textbook omits a lot of the proofs--and say >that they will be left to a more advanced text. >[...] >There are few textbooks that cover the barren wasteland between the >calculus is easy, calculus for everyone textbooks of lower division >calculus courses and mathematical analysis books that cover >topological considerations and other necessities for the foundation of >calculus. I think that Spivak, Calculus on Manifolds fits the bill. Perhaps Buck, Advanced Calculus as well. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: University of Minnesota math joke > At one point, the University of Minnesota's departments of mathematics > and mortuary science were housed in the same building; one ascended the > staircase and turned left for mathematics, and right for mortuary > science. > One day, this clarification was posted on the sign at the top of the > staircase: RIGOR MORTIS > <------ ------- > [Heard through the grapevine a while ago. Quite possibly true.] > *************************************************************** > Well, I've learnt something today. Which is that there is a discipline called mortuary science. And even a desire to be cared for by individuals trained in the discipline. Mark Atherton === Subject: Re: University of Minnesota math joke Gerry Myerson scribbled the following: > The appended appeared on another newsgroup. > Anyone know whether there's any truth to the story? > *************************************************** > === Subject: Either/Or > At one point, the University of Minnesota's departments of mathematics > and mortuary science were housed in the same building; one ascended the > staircase and turned left for mathematics, and right for mortuary > science. > One day, this clarification was posted on the sign at the top of the > staircase: > RIGOR MORTIS > <------ ------- [Heard through the grapevine a while ago. Quite possibly true.] > *************************************************************** I'm afraid I don't get the joke. I know what rigor mortis is, but I don't know how you get rigor from mathematics. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ Bad things only happen to scoundrels. - Moominmamma === Subject: Re: University of Minnesota math joke > Gerry Myerson scribbled the following: > The appended appeared on another newsgroup. > Anyone know whether there's any truth to the story? > *************************************************** > === Subject: Either/Or > At one point, the University of Minnesota's departments of mathematics > and mortuary science were housed in the same building; one ascended the > staircase and turned left for mathematics, and right for mortuary > science. > One day, this clarification was posted on the sign at the top of the > staircase: > RIGOR MORTIS > <------ ------- [Heard through the grapevine a while ago. Quite possibly true.] > *************************************************************** > I'm afraid I don't get the joke. I know what rigor mortis is, but I > don't know how you get rigor from mathematics. > -- Actually, I thought it was the Geology Department who hung the direction for their Rocks for Jocks class next door to the Calculus class. The arrows said: TERRA FIRMA <------- -------> Half of them ended up in the wrong room, and became mathematicians. --riverman === Subject: Re: University of Minnesota math joke riverman scribbled the following: >> Gerry Myerson scribbled the following: >> The appended appeared on another newsgroup. >> Anyone know whether there's any truth to the story? >> *************************************************** >> === Subject: Either/Or >> At one point, the University of Minnesota's departments of mathematics >> and mortuary science were housed in the same building; one ascended the >> staircase and turned left for mathematics, and right for mortuary >> science. >> One day, this clarification was posted on the sign at the top of the >> staircase: >> RIGOR MORTIS >> <------ -------> [Heard through the grapevine a while ago. Quite possibly true.] >> *************************************************************** >> I'm afraid I don't get the joke. I know what rigor mortis is, but I >> don't know how you get rigor from mathematics. > Actually, I thought it was the Geology Department who hung the direction for > their Rocks for Jocks class next door to the Calculus class. The arrows > said: > TERRA FIRMA > <------- ------- Half of them ended up in the wrong room, and became mathematicians. I still don't get either of the jokes. Please provide explanations. Twist them out of wire if necessary. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ I am not very happy acting pleased whenever prominent scientists overmagnify intellectual enlightenment. - Anon === Subject: Re: University of Minnesota math joke Originator: jgamble@ripco.com (John M. Gamble) >riverman scribbled the following: > Gerry Myerson scribbled the following: > The appended appeared on another newsgroup. > Anyone know whether there's any truth to the story? > *************************************************** > === Subject: Either/Or > At one point, the University of Minnesota's departments of mathematics > and mortuary science were housed in the same building; one ascended the > staircase and turned left for mathematics, and right for mortuary > science. > One day, this clarification was posted on the sign at the top of the > staircase: > RIGOR MORTIS > <------ ------- [Heard through the grapevine a while ago. Quite possibly true.] > *************************************************************** > I'm afraid I don't get the joke. I know what rigor mortis is, but I > don't know how you get rigor from mathematics. >> Actually, I thought it was the Geology Department who hung the direction for >> their Rocks for Jocks class next door to the Calculus class. The arrows >> said: >> TERRA FIRMA >> <------- -------> Half of them ended up in the wrong room, and became mathematicians. >I still don't get either of the jokes. Please provide explanations. >Twist them out of wire if necessary. It has to do with the correctness of a mathematical proof. In English, a proof that deals with all the necessary conditions is rigorous, or, has rigor. It's a pun. A similiar pun can be made with Firm i suppose, although i've never heard the word used in connection with a proof (but i am not a professional mathematician). -- -john February 28 1997: Last day libraries could order catalogue cards from the Library of Congress. === Subject: Re: University of Minnesota math joke > A similiar pun can be made with Firm i suppose, although i've never > heard the word used in connection with a proof (but i am not a > professional mathematician). I thought the pun lay in the similarity of pronunciation of firma and Fermat. -- 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, === Subject: Re: University of Minnesota math joke > A similiar pun can be made with Firm i suppose, although i've never > heard the word used in connection with a proof (but i am not a > professional mathematician). > I thought the pun lay in the similarity of pronunciation of firma and > Fermat. Actually, that's a *much* better interpretation! TERRA FERMAT Ok, its sold. --riverman === Subject: Re: University of Minnesota math joke I still don't get either of the jokes. Please provide explanations. >Twist them out of wire if necessary. > It has to do with the correctness of a mathematical proof. In English, > a proof that deals with all the necessary conditions is rigorous, or, > has rigor. It's a pun. Yes, and one can go further to say that rigor is what distinguishes mathematics, as a field of study, from mathematics used as a tool; IOW it's practically the defining characteristic of a university department of mathematics. I think the utter appropriateness of this (along with the wacky juxtaposition of mortuary science) is what gives this joke its appeal, especially for students of math. OTOH if you are also a student of Latin and care about declension of nouns, it loses some of its appeal. (MORTIS should be MORS, but that would ruin the pun.) Perhaps Joona himself is in that category? In that case the joke works better if you momentarily suspend your standards of grammatical, er, rigor. A similiar pun can be made with Firm i suppose, although i've never > heard the word used in connection with a proof (but i am not a > professional mathematician). Agree; and even if the joke worked it would be rather lame IMHO (sorry riverman). === Subject: Re: University of Minnesota math joke A similiar pun can be made with Firm i suppose, although i've never > heard the word used in connection with a proof (but i am not a > professional mathematician). > Agree; and even if the joke worked it would be rather lame IMHO > (sorry riverman). Ugh. (and 'sigh') As mathematicians, we have a rather dismal reputation in the world's eye concerning humor. I think most folks imagine a convention of mathematicians, with the keynote speaker up on stage, and he opens with a joke. Some sort of knee-slapper that worked extremely well the week before at the Laborers convention, or the local Elks club or something. Something like: What's the mathematician do when he's constipated? He har har But the mathematicians meet his punch line with stony faces and wrinkled foreheads. After a moments silence, they lean towards each other in small groups, analyzing the alleged humor, dissect its intended meaning, and judge it 'just not funny, sorry. Lame joke.' So the keynote speaker switches over to some sort of intellectually concise and subtle joke, something the Laborers or Elks would find unfathomable. Something like: A constant function and e^x are walking on Broadway. Then suddenly the constant function sees a differential operator approaching and runs away. so e to-the x follows him and asks why the hurry. Well, you see, there's this diff.operator coming this way, and when we meet, he'll differentiate me and nothing will be left of me...! Ah, says e^x, he won't bother ME, I'm e to-the x! and he walks on. Of The crowd roars it approval, and the janitor in the hallway walks away scratching his head. --riverman PS: Ok, so the Rock Jocks see the sign to mean that one way leads to 'Terror' (mathematics) and the other side is 'Firm', meaning the solid earth...something tangible. The mathematicians see that one way leads to 'Terra', meaning the solid earth, while the other leads to 'Firmness', meaning the solidity and tangibility of the rules of mathmatics. === Subject: Re: University of Minnesota math joke > Something like: A constant function and e^x are walking on Broadway. Then > suddenly the constant function sees a differential operator approaching and > runs away. so e to-the x follows him and asks why the hurry. > Well, you see, there's this diff.operator coming this way, and when we > meet, he'll differentiate me and nothing will be left of me...! > Ah, says e^x, he won't bother ME, I'm e to-the x! and he walks on. Of The crowd roars it approval, and the janitor in the hallway walks away > scratching his head. The janitor shoulda known better. I think if Gary Larson had read this he would have appreciated it somewhat (if not more). (Follows Gary's cartoon, with Larsonian mathematicians at some U, scribbling and scratching, and on the back wall of the department hall an outdated poster with a big d/dy symbol engraved in black and below it the textual description: 'Wanted dead or alive: $10,000 Reward'. (As the most famous murderer of functions).... > --riverman -- Ioannis http://users.forthnet.gr/ath/jgal/ ___________________________________________ Eventually, _everything_ is understandable. === Subject: Re: University of Minnesota math joke >riverman scribbled the following: > Gerry Myerson scribbled the following: > The appended appeared on another newsgroup. Anyone know whether there's any truth to the story? > *************************************************** > === Subject: Either/Or At one point, the University of Minnesota's departments of mathematics > and mortuary science were housed in the same building; one ascended the > staircase and turned left for mathematics, and right for mortuary > science. > One day, this clarification was posted on the sign at the top of the > staircase: RIGOR MORTIS > <------ ------- > [Heard through the grapevine a while ago. Quite possibly true.] > *************************************************************** I'm afraid I don't get the joke. I know what rigor mortis is, but I > don't know how you get rigor from mathematics. > Actually, I thought it was the Geology Department who hung the direction for >> their Rocks for Jocks class next door to the Calculus class. The arrows >> said: > TERRA FIRMA >> <------- ------- >> Half of them ended up in the wrong room, and became mathematicians. I still don't get either of the jokes. Please provide explanations. >Twist them out of wire if necessary. > It has to do with the correctness of a mathematical proof. In English, > a proof that deals with all the necessary conditions is rigorous, or, > has rigor. It's a pun. Non-Logikoi value *consistent* reasoning. Logikoi value consistency only when it serves their purposes: perhaps that's where the mortis in rigour mortis comes in. Non-Logikoi take care not to base their conclusions on premises that are false. Logikoi avoid false premises only when this serves their purposes-- a second source of the mortis in rigour-mortis mathematics. So the anecdote in question is not a pun: If it's a pun you want, here's one: What do you get when you cross a Logikoi with a loaf of French bread? A similiar pun can be made with Firm i suppose, although i've never > heard the word used in connection with a proof (but i am not a > professional mathematician). --John === Subject: Re: University of Minnesota math joke >Yes, David verbs do have tenses: past, present and future, but also > 1a.Appointed to office.b. Nominated as a candidate for office. 2. >Having or bearing a person's name: nominative shares. 3. (-n-tv) Grammar Of, >relating to, or being the case of the subject of a finite verb (as I in I >as a predicate nominative (as children in These are his children). > NOUN: Grammar (-n-tv)1. The nominative case. 2. A word or form in the >nominative case. Huh? First a minor huh: I'm guessing that the above is a definition of nominative that you found in some dictionary? Assuming so: Your but also seems to be suggesting that the definition above shows that _verbs_ have _cases_. I don't see how that follows from the definition you cited. It doesn't say anything about the nominative case of a verb (note that it _does_ say something about the nominative case of the _subject of_ a verb, which is a noun.) I'm not sure what your point is - if your point is what I'm guessing it is then it doesn't follow from that definition. >>How does the nominative and genitive case apply here, to two nouns not >>verbs? >> Huh? Verbs don't have cases. They have tenses and persons and >> moods... >>Lurch > Was this in Vincent Hall? Or, some other hall? If it is true, then >it >>is > hysterical! > Lurch The appended appeared on another newsgroup. Anyone know whether there's any truth to the story? > *************************************************** > === Subject: Either/Or At one point, the University of Minnesota's departments of >mathematics > and mortuary science were housed in the same building; one ascended >>the > staircase and turned left for mathematics, and right for mortuary > science. > One day, this clarification was posted on the sign at the top of >the > staircase: RIGOR MORTIS > <------ ------- > [Heard through the grapevine a while ago. Quite possibly true.] > *************************************************************** -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) Uh-huh. Rigor is nominative case, mortis is genitive case. David Ames >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: University of Minnesota math joke My mistake, as usual. I am getting my languages mixed up. Lurch >Yes, David verbs do have tenses: past, present and future, but also > 1a.Appointed to office.b. Nominated as a candidate for office. 2. >Having or bearing a person's name: nominative shares. 3. (-n-tv) Grammar Of, >relating to, or being the case of the subject of a finite verb (as I in I such >as a predicate nominative (as children in These are his children). > NOUN: Grammar (-n-tv)1. The nominative case. 2. A word or form in the >nominative case. > Huh? First a minor huh: I'm guessing that the above is a definition of > nominative that you found in some dictionary? Assuming so: > Your but also seems to be suggesting that the definition above > shows that _verbs_ have _cases_. I don't see how that follows > from the definition you cited. It doesn't say anything about the > nominative case of a verb (note that it _does_ say something > about the nominative case of the _subject of_ a verb, which is > a noun.) > I'm not sure what your point is - if your point is what I'm guessing > it is then it doesn't follow from that definition. >>How does the nominative and genitive case apply here, to two nouns not >>verbs? >> Huh? Verbs don't have cases. They have tenses and persons and >> moods... >>Lurch > Was this in Vincent Hall? Or, some other hall? If it is true, then >it >>is > hysterical! > Lurch The appended appeared on another newsgroup. Anyone know whether there's any truth to the story? > *************************************************** > === Subject: Either/Or At one point, the University of Minnesota's departments of >mathematics > and mortuary science were housed in the same building; one ascended >>the > staircase and turned left for mathematics, and right for mortuary > science. > One day, this clarification was posted on the sign at the top of >the > staircase: RIGOR MORTIS > <------ ------- > [Heard through the grapevine a while ago. Quite possibly true.] > *************************************************************** -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) Uh-huh. Rigor is nominative case, mortis is genitive case. David Ames >> ************************ >> David C. Ullrich ************************ > David C. Ullrich === Subject: Re: University of Minnesota math joke Charlie Johnson scribbled the following: > My mistake, as usual. I am getting my languages mixed up. > Lurch I think it's time to stop quoting over 100 lines just to add a one-line comment now. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ Parthenogenetic procreation in humans will result in the founding of a new religion. - John Nordberg === Subject: Re: University of Minnesota math joke Why are you reading it ? Just ignore it if it bothers you. Are you the NG police? Are you a communist? Lurch > Charlie Johnson scribbled the following: > My mistake, as usual. I am getting my languages mixed up. > Lurch > I think it's time to stop quoting over 100 lines just to add a one-line > comment now. > -- > /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- > | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| > | http://www.helsinki.fi/~palaste W++ B OP+ | > ----------------------------------------- Finland rules! ------------/ > Parthenogenetic procreation in humans will result in the founding of a new > religion. > - John Nordberg === Subject: Re: University of Minnesota math joke <85hslv4ut8i1fifa93dj2b3egtjdedcfb3@4ax.com> <9Ut7b.2322$TC1.219@newsread2.news.atl.earthlink.net> police? Are you a communist? Why are you reading his criticism? Just ignore it if it bothers you. Are you the NG police? Are you a communist? >> I think it's time to stop quoting over 100 lines just to add a one-line >> comment now. -- I've ... contacted [some of the...] highest I.Q.'s in the country... I've even helped the FBI out a few times... I've met at least one governor..., a senator... and I've had some really good seats at sports games. My experiences are not your experiences. --JSH != you === Subject: Re: University of Minnesota math joke Yes and Yes! Your next! Lurch > Why are you reading it ? Just ignore it if it bothers you. Are you the NG > police? Are you a communist? > Why are you reading his criticism? Just ignore it if it bothers you. > Are you the NG police? Are you a communist? >> I think it's time to stop quoting over 100 lines just to add a one-line >> comment now. > -- > I've ... contacted [some of the...] highest I.Q.'s in the country... > I've even helped the FBI out a few times... I've met at least one > governor..., a senator... and I've had some really good seats at > sports games. My experiences are not your experiences. --JSH != you === Subject: Re: University of Minnesota math joke Originator: jgamble@ripco.com (John M. Gamble) >Why are you reading it ? Just ignore it if it bothers you. Are you the NG >police? Are you a communist? >Lurch Just in case you don't really know, it's because we have to scan through the rest of the 100+ lines in order to see if you added any other comments. This is a pain. After all, we're reading your message because we think you might have something interesting to say (and you have). Why make it harder to read your messages? -- -john February 28 1997: Last day libraries could order catalogue cards from the Library of Congress. === Subject: Re: University of Minnesota math joke >Why are you reading it ? Just ignore it if it bothers you. Are you the NG >police? Are you a communist? Well, since you're relatively new here I'll point out that you got the reply _almost_ right. You're supposed to ask whether he's a Nazi, not a communist. >Lurch >> Charlie Johnson scribbled the following: >> My mistake, as usual. I am getting my languages mixed up. >> Lurch >> I think it's time to stop quoting over 100 lines just to add a one-line >> comment now. >> -- >> /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- >> | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| >> | http://www.helsinki.fi/~palaste W++ B OP+ | >> ----------------------------------------- Finland rules! ------------/ >> Parthenogenetic procreation in humans will result in the founding of a >new >> religion. >> - John Nordberg ************************ David C. Ullrich === Subject: Re: University of Minnesota math joke Well, I just went with communist because of his e-mail address: palaste@cc.helsinki.fi The Finns were controlled by the Communists, even thought they like to say they were independent. Lurch >Why are you reading it ? Just ignore it if it bothers you. Are you the NG >police? Are you a communist? > Well, since you're relatively new here I'll point out that you got the > reply _almost_ right. You're supposed to ask whether he's a Nazi, > not a communist. >Lurch > Charlie Johnson scribbled the following: >> My mistake, as usual. I am getting my languages mixed up. >> Lurch >> I think it's time to stop quoting over 100 lines just to add a one-line >> comment now. >> -- >> /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- >> | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| >> | http://www.helsinki.fi/~palaste W++ B OP+ | >> ----------------------------------------- Finland rules! ------------/ >> Parthenogenetic procreation in humans will result in the founding of a >new >> religion. >> - John Nordberg ************************ > David C. Ullrich === Subject: Re: University of Minnesota math joke > Well, I just went with communist because of his e-mail address: > palaste@cc.helsinki.fi > The Finns were controlled by the Communists, even thought they like to say > they were independent. Lurch Are you related to Dan Quayle? Marc === Subject: Re: University of Minnesota math joke * Charlie Johnson > Well, I just went with communist because of his e-mail address: > palaste@cc.helsinki.fi The Finns were controlled by the Communists, > even thought they like to say they were independent. Are you out of your mind? -- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52 === Subject: Re: University of Minnesota math joke : The Finns were controlled by the Communists, even thought they like to say : they were independent. Could you tell me a little bit more about this? When and how? And by the way: controled by communists doesn't imply non independent. === Subject: Re: University of Minnesota math joke > : The Finns were controlled by the Communists, even thought they like to say > : they were independent. Could you tell me a little bit more about this? > When and how? > And by the way: controled by communists doesn't imply non independent. Finlandization is a heavily loaded term for the alliance between Finland and the Soviet Union that existed from roughly 1948 to 1995. To an English speaker, it connotes a state of affairs in which a country is allowed to believe it is independent when in fact its policies are controlled by its much larger neighbor. It is a common but unjustified belief in the U.S. that Finnish domestic policy was dictated by the Soviet Union during this period. What may have been true of Finland to a small degree is much more strongly true of the Latin American states that the U.S. has dominated for years under such supposed justifications as the Roosevelt Corollary. -- Chris Green === Subject: Re: University of Minnesota math joke > Could you tell me a little bit more about this? But not in this newsgroup. === Subject: Re: University of Minnesota math joke > Well, I just went with communist because of his e-mail address: > palaste@cc.helsinki.fi > The Finns were controlled by the Communists, even thought they like to say > they were independent. without a clue. Again you post your additions at the start of the (meaning an inhabitant of the US in this case)? Meaning bulging in and if the way you do it is not according to standard, so what? They better accustom to my way? BTW, your vision on the Finnish situation is not, eh, entirely accurate. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: University of Minnesota math joke > The appended appeared on another newsgroup. Anyone know whether there's any truth to the story? > *************************************************** > === Subject: Either/Or At one point, the University of Minnesota's departments of mathematics > and mortuary science were housed in the same building; one ascended the > staircase and turned left for mathematics, and right for mortuary > science. > One day, this clarification was posted on the sign at the top of the > staircase: RIGOR MORTIS > <------ ------- > [Heard through the grapevine a while ago. Quite possibly true.] > *************************************************************** In Le Monde, the publications of theses look exactly like the notices of deaths and are printed beside them on the same page. The title of the thesis, in italics, plays the role of the deceased, the name of the university that of the church where the funeral will take place, and the future title of doctor that of the headstone. --Jean Baudrillard _Cool Memories II_ === Subject: looking for a Paul Erdos proof Altalanositasa by Paul Erdos was published in 1932 in a journal titled Matematikai es fizikai lapok (pp.17-24). anyone I know can read Hungarian. Translation services are available, but expensive. In this paper, Erdos proves the following: the sum [1/m] + [1/(m+1)] + [1/(m+2)] + ... + [1/(n-1)] + [1/n] cannot equal an integer unless m=n=1. I would love to obtain this proof. If anyone is familiar with this result, can they please post it, or direct me to a source where I may find an english translation of this proof? Thank you. === Subject: Re: looking for a Paul Erdos proof In this paper, Erdos proves the following: the sum [1/m] + [1/(m+1)] + [1/(m+2)] + ... + [1/(n-1)] + [1/n] cannot > equal an integer unless m=n=1. I would love to obtain this proof. If anyone is familiar with this result, can they please post it, or > direct me to a source where I may find an english translation of this > proof? The special case with m = 1 comes up ad nauseam in this group. The usual proof for that works for any m :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: looking for a Paul Erdos proof Thank you for your response, Robin. As I am not a regular reader of this newsgroup, can you, or anyone else reading this, please direct me to this proof? Thank you. > In this paper, Erdos proves the following: the sum [1/m] + [1/(m+1)] + [1/(m+2)] + ... + [1/(n-1)] + [1/n] cannot > equal an integer unless m=n=1. I would love to obtain this proof. If anyone is familiar with this result, can they please post it, or > direct me to a source where I may find an english translation of this > proof? The special case with m = 1 comes up ad nauseam in this group. > The usual proof for that works for any m :-) === Subject: Re: looking for a Paul Erdos proof > In this paper, Erdos proves the following: >> the sum [1/m] + [1/(m+1)] + [1/(m+2)] + ... + [1/(n-1)] + [1/n] cannot >> equal an integer unless m=n=1. >> I would love to obtain this proof. >> If anyone is familiar with this result, can they please post it, or >> direct me to a source where I may find an english translation of this >> proof? >> The special case with m = 1 comes up ad nauseam in this group. >> The usual proof for that works for any m :-) > Thank you for your response, Robin. As I am not a regular reader of this newsgroup, can you, or anyone > else reading this, please direct me to this proof? The idea is to consider the largest power of 2 dividing the denominators of the terms of the sum. As this power divides exactly one term's denominator it also divides the denominator of the sum. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Calculating Percentages Hi all, This is a little embarrassing. I need to calculate a few percentages and haven't had to do anything like this for years. I did a Google search but I must be thicker than I thought because I can't make head nor tail of any explanation found so far. This is my problem: Out of 83 software reviews in which each product was given a 1 - 10 rating: 2 products scored 10 41 scored 9 20 scored 8 14 scored 7 7 scored 6 I need to express these scores as percentages, e.g. 5% scored 10 34% scored 7 etc. Any help most gratefully received! :0) Greg === Subject: Re: Calculating Percentages > 2 products scored 10 > 41 scored 9 > 20 scored 8 > 14 scored 7 > 7 scored 6 > I need to express these scores as percentages, e.g. > 5% scored 10 > 34% scored 7 etc. 2+41+20+14+7=84 2*100/84 ~ 2.38% 41*100/84 ~ 48.81% etc. === Subject: Value of product of summations Value of product of summations Let h: N x N --> N be one-to-one and onto, and let b_ij = a_h (i, j). Prove that the summation of a_n as n goes from one to infinity equals the products of: (the summation as i goes from 1 to infinity) times (the summation as j goes from 1 to infinity) of b_ij. Thanks in advance for your help. Diana === Subject: Re: Value of product of summations Elaine, I would like to give credit to the author of the notes. Am I to understand that you are the author of the text? I assume so from your message. If not, would you just provide the reference so that I can give proper credit? Thanks again, Diana > I have notes about this. I'll put them on my webpage and you can get them from > there. http://www.geocities.com/sean_mcilroy > | Value of product of summations > | > | Let h: N x N --> N be one-to-one and onto, > | > | and let b_ij = a_h (i, j). > | > | Prove that the summation of a_n as n goes from one to infinity equals > | the products of: > | > | (the summation as i goes from 1 to infinity) times (the summation as j > | goes from 1 to infinity) of b_ij. > | > | Thanks in advance for your help. > | > | Diana === Subject: Re: Value of product of summations Hi Diana. I'm glad you found the notes helpful. I am the author of them. My name is Sean McIlroy. | Elaine, | | I would like to give credit to the author of the notes. Am I to understand | that you are the author of the text? I assume so from your message. If not, | would you just provide the reference so that I can give proper credit? | | Thanks again, | | Diana | | > I have notes about this. I'll put them on my webpage and you can get them | from | > there. http://www.geocities.com/sean_mcilroy | > | > | Value of product of summations | > | | > | Let h: N x N --> N be one-to-one and onto, | > | | > | and let b_ij = a_h (i, j). | > | | > | Prove that the summation of a_n as n goes from one to infinity equals | > | the products of: | > | | > | (the summation as i goes from 1 to infinity) times (the summation as j | > | goes from 1 to infinity) of b_ij. | > | | > | Thanks in advance for your help. | > | | > | Diana | > | > | | === Subject: Re: Value of product of summations Sean, Yes, the proof is great, and I want to give proper credit to you. Thanks again, Diana M. > Hi Diana. > I'm glad you found the notes helpful. I am the author of them. My name is Sean > McIlroy. > | Elaine, > | > | I would like to give credit to the author of the notes. Am I to understand > | that you are the author of the text? I assume so from your message. If not, > | would you just provide the reference so that I can give proper credit? > | > | Thanks again, > | > | Diana > | > | > I have notes about this. I'll put them on my webpage and you can get them > | from > | > there. http://www.geocities.com/sean_mcilroy > | | > | Value of product of summations > | > | > | > | Let h: N x N --> N be one-to-one and onto, > | > | > | > | and let b_ij = a_h (i, j). > | > | > | > | Prove that the summation of a_n as n goes from one to infinity equals > | > | the products of: > | > | > | > | (the summation as i goes from 1 to infinity) times (the summation as j > | > | goes from 1 to infinity) of b_ij. > | > | > | > | Thanks in advance for your help. > | > | > | > | Diana > | | | > | === Subject: Re: Value of product of summations Elaine, Thanks for the great proof. I will print it out and study it. Diana M. > I have notes about this. I'll put them on my webpage and you can get them from > there. http://www.geocities.com/sean_mcilroy > | Value of product of summations > | > | Let h: N x N --> N be one-to-one and onto, > | > | and let b_ij = a_h (i, j). > | > | Prove that the summation of a_n as n goes from one to infinity equals > | the products of: > | > | (the summation as i goes from 1 to infinity) times (the summation as j > | goes from 1 to infinity) of b_ij. > | > | Thanks in advance for your help. > | > | Diana === Subject: Re: Value of product of summations Visiting Assistant Professor at the University of Montana. >Value of product of summations >Let h: N x N --> N be one-to-one and onto, >and let b_ij = a_h (i, j). >Prove that the summation of a_n as n goes from one to infinity equals >the products of: >(the summation as i goes from 1 to infinity) times (the summation as j >goes from 1 to infinity) of b_ij. I think you got confused in the notation. The sum as n goes from 1 to infinity of a_n is equal to the sum(as i goes from 1 to infinity) of the (sum as j goes from 1 to infinity of b_ij). It's not a product of sums, it is a sum of sums. What the equality is saying is that adding the a_n's, under the relation b_ij = a_h(i,j), is the same as (b_11 + b_12 + b_13 + b_14 + ... + b_1j + ...) + (b_21 + b_22 + b_23 + b_24 + ... + b_2j + ...) + (b_31 + b_32 + b_33 + b_34 + ... + b_3j + ...) . . . ============================================================== ======== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ============================================================== ======== Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Value of product of summations Arturo, Thanks for your help. Yes, a sum of sums, not a product... Diana >Value of product of summations Let h: N x N --> N be one-to-one and onto, and let b_ij = a_h (i, j). Prove that the summation of a_n as n goes from one to infinity equals >the products of: (the summation as i goes from 1 to infinity) times (the summation as j >goes from 1 to infinity) of b_ij. > I think you got confused in the notation. > The sum as n goes from 1 to infinity of a_n is equal to the > sum(as i goes from 1 to infinity) of the (sum as j goes from 1 to > infinity of b_ij). > It's not a product of sums, it is a sum of sums. > What the equality is saying is that adding the a_n's, under the > relation b_ij = a_h(i,j), is the same as > (b_11 + b_12 + b_13 + b_14 + ... + b_1j + ...) > + (b_21 + b_22 + b_23 + b_24 + ... + b_2j + ...) > + (b_31 + b_32 + b_33 + b_34 + ... + b_3j + ...) > . > . > . > ============================================================== ======== > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > ============================================================== ======== > Arturo Magidin > magidin@math.berkeley.edu === Subject: Fast prime counting implementation An implementation of the Extended Meissel-Lehmer Algorithm for counting the number of primes <= N is available for download from my web page at http://www.cbau.freeserve.co.uk/ This is all very preliminary and work in progress, with lots of room for speed improvements, and the code needs cleaning up. However, first speed measurements are quite promising: On a 1700 MHz PC, calculating pi (10^16) took 694 seconds, calculating pi (10^15) took 120 seconds, pi (10^12) less than one second. There is room for improvement; at least a factor two with portable C++ should be possible. For comparison: The times published in the paper by Lagarias, Miller and Odlyzko on which this implementation is based ar 718 minutes, 165 minutes and two minutes on an IBM 3081 Model K in the early '80s for the same tasks; other implementations posted here take more than 100 times longer. Memory requirements are quite moderate, calculating pi (10^17) on a PC with 128 MB of memory is no problem. The results agree with values published in www.mathworld.com. A paper describing the mathematics of the algorithm can be found on my web page as well; any comments are welcome. === Subject: Re: Fast prime counting implementation > An implementation of the Extended Meissel-Lehmer Algorithm for counting > the number of primes <= N is available for download from my web page at http://www.cbau.freeserve.co.uk/ This is all very preliminary and work in progress, with lots of room for > speed improvements, and the code needs cleaning up. However, first speed > measurements are quite promising: On a 1700 MHz PC, calculating pi > (10^16) took 694 seconds, calculating pi (10^15) took 120 seconds, pi > (10^12) less than one second. There is room for improvement; at least a > factor two with portable C++ should be possible. Great work, Christian! I reckon that if you pulled in some FP maths to perform the div/rems there might be some optimisation possible. It would need extended doubles though, and thus be x86-bound. > Memory requirements are quite moderate, calculating pi (10^17) on a PC > with 128 MB of memory is no problem. The results agree with values > published in www.mathworld.com. A paper describing the mathematics of > the algorithm can be found on my web page as well; any comments are > welcome. 'gv' crapped out while trying to view each of the pages in that PDF. I think on page 3 it only managed to render 3 characters! Phil === Subject: Re: Fast prime counting implementation > I reckon that if you pulled in some FP maths to perform the div/rems there > might be some optimisation possible. It would need extended doubles though, > and thus be x86-bound. The 64 bit divisions are a problem, but mostly on PowerPC. CodeWarrior on PowerPC uses a library function that is truly horribly slow; the x86 version is significantly faster; about 60 cycles. So right now I am preparing algorithm + papers how to do 64 bit integer by 32 bit constant integer division _fast_, without using a divide instruction; should be less than 30 cycles. One modification that I made to the algorithm is that I can balance where to do the work: To calculate pi (N), you choose a number M, then you have to create sieves of size (N / M) and find about (M^2 / log M) values in these sieves. Lagarias etc. pick M = N^(1/3). Depending on whether creating the sieve or looking up values takes longer, I can make M larger or smaller and save some time that way. > 'gv' crapped out while trying to view each of the pages in that PDF. > I think on page 3 it only managed to render 3 characters! Sorry about that. I tried it with Acrobat Reader 5 on Macintosh and Windows, and it works fine. === Subject: Re: Fast prime counting implementation > An implementation of the Extended Meissel-Lehmer Algorithm for counting > the number of primes <= N is available for download from my web page at > http://www.cbau.freeserve.co.uk/ > This is all very preliminary and work in progress, with lots of room for > speed improvements, and the code needs cleaning up. However, first speed > measurements are quite promising: On a 1700 MHz PC, calculating pi > (10^16) took 694 seconds, calculating pi (10^15) took 120 seconds, pi > (10^12) less than one second. There is room for improvement; at least a > factor two with portable C++ should be possible. Well done. Here are the results of the first few test values in your driver on my P3-500 compiled with g++ -O version 3.3.1 (mingw version) -4.08744e-017 seconds: Pi (1000) = 168 <- I didn't check out these odd times -4.97649e-017 seconds: Pi (10000) = 1229 -5.86553e-017 seconds: Pi (100000) = 9592 -6.75458e-017 seconds: Pi (1000000) = 78498 0.01 seconds: Pi (10000000) = 664579 0.01 seconds: Pi (100000000) = 5761455 0.03 seconds: Pi (1000000000) = 50847534 0.171 seconds: Pi (10000000000) = 455052511 0.791 seconds: Pi (100000000000) = 4118054813 3.865 seconds: Pi (1000000000000) = 37607912018 19.018 seconds: Pi (10000000000000) = 346065536839 99.202 seconds: Pi (100000000000000) = 3204941750802 (I killed it here) I had to make some minor text changes to get it to compile in g++. The minor issues (all warnings) had to do with the long long constants in the driver, and some long long initializations with the result of pow(). I'm sure the changes I made are portable - added the LL and an explicit cast. The biggest problem is a difference in the packaging of the math functions. In g++ pow and sqrt aren't in the std namespace so I had to remove the std:: and add main.h to the include list. Also, the linker complained about no implementation for unsigned long phi_fast (unsigned long N, unsigned long k) so I commented it out in the class. It compiles and links with no warning now. I can send you the modified sources with embedded comments about the changes I needed to make for g++. It should be very easy to make this portable between CodeWarrior and g++. === Subject: How to compute the fundamental group of SO(2,R). Dear all, I hear many times that the fundamental group of SO(2,R) is Z. Can any one fill in the key steps to establish this fact? Thank you. K. Yam === Subject: Re: How to compute the fundamental group of SO(2,R). X-SessionID: h3Q7b-2657-05-5610@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: 4a9e4783 852c75ee ad3c8e44 1b4d766f eb9548d0 > I hear many times that the fundamental group of SO(2,R) is Z. Can any one > fill in the key steps to establish this fact? Thank you. SO(2,R) is just the circle S^1. To compute the fundamental group of the circle, consider the map R -> S^1, x maps to e^(2pi i x). Show loops in S^1 based at 1 can be lifted to paths in R starting at 0, and likewise homotopies (respecting endpoints) in S^1 between such loops can be lifted. Then f -> endpoint of lifting of f, defines an isomorphism between the fundamental group of S^1 and Z. -Jeff === Subject: Parameterizing the solution of det(xy(x+y+I)) > 0 Dear all, Let (X,Y) be the collection of solutions to Det(xy(x+y+I)) > 0. Does it exist a homeomorphism that map (X,Y) to a simple set (say some subset of (phi, theta) such that Det(phi)*Det(theta) > 0) ? Any suggestion on how to approach this problem will be greatly appreciated. Thank you, K. Yam === Subject: Re: integral3 >> Let A >1 real >> Find all f such that >> f : R+---> R continous such that for any x in R+ >> f(x) = int_[0;Ax]f(t)dt >> thank you m.o. >The Fabius function satisfies this with A=2. This function Fb(x) is >infinitely differentiable, but nowhere analytic. It is increasing on >[0,1], Fb(0)=0, Fb(1) = 1, Fb'(x) = 2 Fb(2x). Fabius [1] initially >constructs it for x in [0,1], but it can be extended easily to all real >x, using the defining equation for x>1, and by reflection for x<0 [2]. >See a picture at [1] J. Fabius, A probabilistic example of a nowhere analytic >C^infty-function. Z. Wahrsch. Verw. Geb. 5 (1966) 173--174. >[2] K. Stromberg, PROBABILITY FOR ANALYSTS (Chapman 7 Hall, 1994), pp. >117--120. >Here is the construction of Fabius: Let (X_k) be an iid sequence of >random variables, each uniformly distributed on [0,1]. Let X = sum(k=1 >to infinity) 2^{-k} X_k. Let Fb(x) be the cumulative distribution >function of X. Check Fb(x) = integral(0 to 2x) Fb(t) dt for all x in >[0,1/2]. Beautiful! And more generally, for any A > 1 we can define X = sum_{k=1}^infinity A^(-k) X_k; its cumulative distribution function F(x) = Prob(X <= x) satisfies F(x) = int_{Ax-1}^{Ax} F(t) dt (which we can get by conditioning on X_1) and thus F'(x) = A (F(Ax) - F(Ax-1)). Let R be the set of polynomials p(t) with coefficients in {0,1}. For x >= 0, let f(x) = sum_{p in R} (-1)^p(1) F(x-p(A)) (note that on any bounded interval, there are only finitely many nonzero terms in the sum). Then f'(x) = A sum_{p in R} (-1)^p(1) (F(Ax-Ap(A)) - F(Ax-Ap(A)-1)) Noting that every p in R can be uniquely written in the form p(t) = t q(t) or t q(t)+1 for q in R, we see that f'(x) = A sum_{q in R} (-1)^q(1) F(Ax-q(A)) = A f(Ax) So f is a nonzero solution of moubinool's problem. Next question: are there more solutions, or is f unique (up to constant multiples)? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: integral3 Thanks Robert Israel and G.A.Edgar for your answers m.o. === Subject: Re: Mega Foundation, change in plans David C. Ullrich pushed briefly to the to the shed door: (all silly Xposts removed, sci.math retained) ^ ^ >mensanator@aol.com (mensanator) pushed briefly to the front of the ^ >^ Maybe they should add bag of hammers to their keywords. ^ >Hehe. I can see how hammer would be insufficiently exclusive, but ^ >unfortunately bag of hammers makes me think of bag of spanners, as ^ >in face like a .... Was this the comic effect you were after? ^ >Perhaps fistful of hammers would be better, bringing to mind the ^ >$$$s that JSH has always felt should be rightfully his. ^ > ^ >Anyway, IIRC JSH wields a Hammer, not just any old hammer. ^ Very curious. There are plenty of people around here who spend ^ a lot of time making fun of JSH. And there are various people ^ who often state that such behavior is inappropriate, or various ^ other uncomplimentary adjectives. You're the only person I ^ recall who enjoys doing both. Curious indeed, that your recall should be so at variance with my own. In any event, I'm only interested in your recall insofar as it tallies with facts that contradict my own recall. I do spend some time making fun of JSH. Not a lot of time, by any reasonable standard; as far as I recall, no-one spends more time than you doing it. By your standard, I recall spending a negligible amount of time doing it. Also as far as I recall, I have never stated that such behaviour is inappropriate. But enough of such assertion and counter-assertion. All the evidence is out there, if you would care to assemble it. Hint: my recall would estimate that in terms of posts by either of us that make fun of JSH, I have made less in total than you have made specifically containing the string Hint:. Second hint: a single counter-example will suffice to disprove my recall about my never stating that such behaviour is inappropriate. Andy -- sparge at globalnet point co point uk It's your fault we don't have a no-blame culture Lee Hartley, RHM Technology === Subject: Re: Mega Foundation, change in plans <3f5cda64$1_3@newsfeed.slurp.net> <4Qq7b.11165$Ff6.323630@phobos.telenet-ops.be> <3f5ed3f8.41967729@news.global.net.uk> <3f5fb5dd.99805989@news.global.net.uk David C. Ullrich pushed briefly to the > to the shed door: > ^ Very curious. There are plenty of people around here who spend > ^ a lot of time making fun of JSH. And there are various people > ^ who often state that such behavior is inappropriate, or various > ^ other uncomplimentary adjectives. You're the only person I > ^ recall who enjoys doing both. > Curious indeed, that your recall should be so at variance with my own. > In any event, I'm only interested in your recall insofar as it tallies > with facts that contradict my own recall. I do spend some time making > fun of JSH. Not a lot of time, by any reasonable standard; as far as I > recall, no-one spends more time than you doing it. By your standard, I > recall spending a negligible amount of time doing it. Also as far as I > recall, I have never stated that such behaviour is inappropriate. I suspect he is remembering the following quote from <3f5e5c01.11252651@news.global.net.uk>: ,---- | | Terry E pushed briefly to the front of the queue | door: | | ^ | | ^ > Because he has better things to do than correspond with crackpots. | ^ > | ^ > That was easy. | | ^ Corollary? | | Ooooh, me, Sir, ME! | | Why does JSH keep hearing from Dave The Rave Ullrich? Because DTRU has | got nothing better to do than correspond with crackpots. | | Do I win first prize? | | Andy `---- One could interpret your response as being critical of participation in JSH threads. I thought it was just a joke and you didn't really intend any serious criticism, but it wasn't aimed at me. -- [R]eality has a fascinating ability to check us when we get a little too big for our britches... Make no mistake. There isn't a mathematician alive today that I can't now touch, and not a mathematical career on the planet that I can't now affect. --, render of worlds === Subject: Re: Mega Foundation, change in plans >> David C. Ullrich pushed briefly to the >> to the shed door: >> ^ Very curious. There are plenty of people around here who spend >> ^ a lot of time making fun of JSH. And there are various people >> ^ who often state that such behavior is inappropriate, or various >> ^ other uncomplimentary adjectives. You're the only person I >> ^ recall who enjoys doing both. >> Curious indeed, that your recall should be so at variance with my own. >> In any event, I'm only interested in your recall insofar as it tallies >> with facts that contradict my own recall. I do spend some time making >> fun of JSH. Not a lot of time, by any reasonable standard; as far as I >> recall, no-one spends more time than you doing it. By your standard, I >> recall spending a negligible amount of time doing it. Also as far as I >> recall, I have never stated that such behaviour is inappropriate. >I suspect he is remembering the following quote from ><3f5e5c01.11252651@news.global.net.uk>: >,---- >| >| Terry E pushed briefly to the front of the queue >| door: >| >| ^ | ^ > Because he has better things to do than correspond with crackpots. >| ^ | ^ > That was easy. >| >| ^ Corollary? >| >| Ooooh, me, Sir, ME! >| >| Why does JSH keep hearing from Dave The Rave Ullrich? Because DTRU has >| got nothing better to do than correspond with crackpots. >| >| Do I win first prize? >| >| Andy >`---- Actually that was his second post recently about Dave The Rave. Seems like he was disappointed when I didn't reply to his first: http://www.google.com/groups?q=+%22Dave+The+Rave%22+ group:sci.math&hl=en&lr=&ie=UTF-8&selm=3f5b5bdf. 246312921%40news.global.net.uk&rnum=2 >Trying to reason with Dave the Rave is also useless, unfortunately. >Perhaps he is mentally ill too. He certainly can't seem to get his >head round the idea that a mentally ill person like James is simply >not susceptible to having the obvious pointed out. God bless him, he's >been plugging away pointing out the obvious to him, in various ways >from condescending to sarcastic, for a few years now, and never seems >to tire of it. One has to conclude that he really does do it for his >own amusement. >Andy >One could interpret your response as being critical of participation >in JSH threads. Possibly. One might even interpret Perhaps he is mentally ill too as uncomplimentary. It does make his statement as far as I recall, I have never stated that such behaviour is inappropriate sort of interesting. I suppose I should apologize for incorrectly paraphrasing what he said - he never said such behavior was inapproriate, just that it might be symptomatic of mental illness. >I thought it was just a joke and you didn't really >intend any serious criticism, but it wasn't aimed at me. Don't get the wrong idea - regardless of how it was intended it was nothing but amusing at this end (because it's been clear the guy has a bug up his ass since the thread a while ago where he insisted on making a _pompous_ fool of himself over CLT (pompous referring to his repeated statements that the reason he understood the issues better than his stat friend was he was a mathematician, when in _fact_ he was totally wrong about the whole thing and his stat friend had things straight.)) Just thought I'd point out how much more amusing it becomes when we see him participating in the JSH-bashing. ************************ David C. Ullrich === Subject: Re: Mega Foundation, change in plans > I'd like to take some credit here. I asked James more than four years >> ago >> to contact Chris Langan. See 1063143790.v3.net While stumbling around the Mega site I managed to log on to something that > let me read the conferences. So far, only James has contributed to the > Object mathematics thread. Looking back through previous Math threads we > find that one Russel Rierson is a frequent contributor. Rierson has been sporadically active in sci math: http://makeashorterlink.com/?P24A24AD5 -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Mega Foundation, change in plans <8as7b.2135$u96.32235@vixen.cso.uiuc.edu> <3f5d1dd7$1_2@newsfeed.slurp.net I don't think I've ever seen a site that relies so heavily on the > ego of readers to get someone in. Bizarre. Bizarre, perhaps, but smart. -- Start obeying math rules, or if you prefer I can proceed to tear away your illusions of being rational and logical piece by piece over time, as I reduce your society using advanced psychological tools that rival the best psychological warfare techniques of world governments. --JSH === Subject: Re: Mega Foundation, change in plans >> I don't think I've ever seen a site that relies so heavily on the >> ego of readers to get someone in. Bizarre. >Bizarre, perhaps, but smart. Incredbly smart. Smarter than 99.997% of the population, if I recall the figure I read there correctly! ************************ David C. Ullrich === Subject: Re: Mega Foundation, change in plans > I don't think I've ever seen a site that relies so heavily on the >> ego of readers to get someone in. Bizarre. Bizarre, perhaps, but smart. > Incredbly smart. Smarter than 99.997% of the population, > if I recall the figure I read there correctly! I thought the Mega was supposed to mean that only one in a million satisfies the requirements. That should be 99.9999%, I guess. === Subject: Re: Mega Foundation, change in plans <3f5cda64$1_3@newsfeed.slurp.net> <4Qq7b.11165$Ff6.323630@phobos.telenet-ops.be Nice HTML header too: > Society, a high-IQ group > open to anyone scoring at the 1/1,000,000 level, approximately IQ 170 or > higher. Ultranet, Ubiquity, > Chris Langan, Gina LoSasso, Bob Seitz, severely-gifted, severely gifted, > extremely-gifted, > extremely gifted, Telenet, Telemach Network for Gifted Youth, Triple Nine, > Gifted, highly gifted, profoundly gifted, superior intelligence, > intellect, intellectual, cognitive ability, clever, brain, HiQ, > quick, brilliant, brilliance, brainpower, think, thinking, thought, brainy, > whiz kid, HIQ, Will Hunting, Einstein, prodigy, > american mensa, intertel, ispe, prometheus society, Top One Percent Society, > Triple Nine Society, TNS, Vidya, 999, 99.9, high IQ, High-IQ, Hi-Q, High-IQ > society, ultra-smart, super-smart, precocious, Four Sigma, 4 sigma, genius, > One-in-a-Thousand Society, mega society, Thinkfast, Mega, Prometheus, > Mega Test, Titan Test, LAIT, SAT, GRE, Stanford-Binet, Cattell, CTMM > Ron Hoeflin, Paul Cooijmans, Darryl Miyaguchi, > Gift of Fire, Noesis, Telicom top one percent = one in a thousand ? Will Hunting? -- But remember, as long as one human being follows the rules of mathematics, then mathematics as a human discipline survives. Right now I'm that one human being, so mathematics survives. -- James S. Harris === Subject: Re: Mega Foundation, change in plans > > thinking, thought, brainy, whiz kid, HIQ, Will Hunting, [...] Will Hunting? According to Google, Will Hunting is some appropriate movie character with a high IQ: ... plays the kid, Will Hunting, whose brutal upbringing in foster homes has made him less than a model citizen despite possessing an exceptional IQ and being ... .. Matt Damon starred as the eponymous Will Hunting, a janitor at MIT University in Boston who, unbeknown to everyone, had an IQ which made him a fantastic ... -- Jussi === Subject: Re: Mega Foundation, change in plans <3f5cda64$1_3@newsfeed.slurp.net> <4Qq7b.11165$Ff6.323630@phobos.telenet-ops.be> <87fzj5gmwt.fsf@phiwumbda.org> thinking, thought, brainy, whiz kid, HIQ, Will Hunting, [...] >> Will Hunting? > According to Google, Will Hunting is some appropriate movie character > with a high IQ: > ... plays the kid, Will Hunting, whose brutal upbringing in foster > homes has made him less than a model citizen despite possessing an > exceptional IQ and being ... > .. Matt Damon starred as the eponymous Will Hunting, a janitor at MIT > University in Boston who, unbeknown to everyone, had an IQ which made > him a fantastic ... I know who he is. I just thought that the inclusion of ludicrous movie character Will Hunting in the list of keywords was indicative of the author's self-perceptions and sense of reality. Oh, and funny, too. Not that the number of synonyms for smart in the keywords wasn't funny. Somehow, though, they left out really, really, really smart -- really! ,---- | intelligence, intelligent, severely-gifted, severely gifted, | extremely-gifted, extremely gifted, Gifted, highly gifted, | profoundly gifted, superior intelligence, intellect, intellectual, | cognitive ability, clever, brain, HiQ, quick, brilliant, brilliance, | brainpower, think, thinking, thought, brainy, whiz kid, HIQ, | prodigy, 999, 99.9, high IQ, High-IQ, Hi-Q, ultra-smart, | super-smart, precocious, genius `---- They also included ubiquity -- I guess on the grounds that anyone that knows that word must be severely gifted. Or maybe it's some genius's pseudonym. -- Jesse Hughes We will run this with the same kind of openness that we've run Windows. Steve Ballmer, speaking about MS's new .Net project. === Subject: Re: Mega Foundation, change in plans [JSH] >>.. Matt Damon starred as the eponymous Will Hunting, a janitor at MIT >> University in Boston who, unbeknown to everyone, had an IQ which made >> him a fantastic ... I know who he is. I just thought that the inclusion of ludicrous > movie character Will Hunting in the list of keywords was indicative > of the author's self-perceptions and sense of reality. Oh, and funny, > too. > This experiment is over. Regarding James' promised departure from sci.math I'd like to (mis)quote another movie character... He'll be back. Mark Atherton === Subject: Re: Mega Foundation, change in plans [JSH] .. Matt Damon starred as the eponymous Will Hunting, a janitor at MIT > University in Boston who, unbeknown to everyone, had an IQ which made > him a fantastic ... >> I know who he is. I just thought that the inclusion of ludicrous >> movie character Will Hunting in the list of keywords was indicative >> of the author's self-perceptions and sense of reality. Oh, and funny, >> too. > This experiment is over. Regarding James' promised departure from sci.math I'd like to (mis)quote > another movie character... He'll be back. We can place bets on how long it'll take. I'm guessing within 10 days. Gib === Subject: History of toplogy terminology Were words and concepts open, closed, and dense as applied to subsets of R^n (at least for n <= 3) in common mathematical usage before the development of general topology? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: History of toplogy terminology > Were words and concepts open, closed, and dense as applied to > subsets of R^n (at least for n <= 3) in common mathematical usage before > the development of general topology? Look them up at Mathword: -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Digital-Clock (sometimes) Upside-Down I think that this post, as others (at least) of the posts I made through Google a few days ago, have not necessarily propagated throgh the Usenet correctly. So I will reply with the original text below (with >'s added): > Now, as you must know, with the standard digital-clock readout, some > times represent other times if the clock happens to be lying > upside-down. > (12:01 <> 10:21, for example) > (But, even if we have a 4-digit display for all times > {in which case, 01:21 <> 12:10 is a valid rotation}, > then 03:10 is NOT equal to 01:30 after flipping, because 3 is not any > number-digit after being rotated 180-degrees.) So, we have a 12-hour digital clock with the standard readout. > ie. the digits are formed from some combination of 7 line-segments, > each segment being lit or un-lit, the segments arranged as: > -- > | | > -- > | | > -- There is also a colon separating the minutes and the hours, as is > typical. Now, we take a diplayed time, that if flipped 180-degrees produces > another valid time where the colon is in the same place. (example: 10:10 <> 01:01 is valid, but 02:10 <> 12:00 is NOT valid > because the colon has moved.) So, between every time and its rotation, there is an absolute > difference in the two times (many differences, actually). But here, I > am interested in only the two positive differences between 0 and 12 > hours. Now these differences themselves can be other valid times when > displayed on a digital-clock and rotated 180-degrees. And, in turn, we can take the absolute difference between the previous > difference and its rotation, continuing. What, then, is the time that has the most times which, by a chain of > differences and rotations, leads to this particular time? By searching by-hand and brute-force (which is what I suggest that you > do, since a computer-search is not worthwhile, this puzzle being too > easy to justify that), I came up with a time which has a total of 14 > times which lead eventually to this time. > But this is actually 7 pairs equal-by-rotation, and these pairs exist > within the branches of the tree, not necessarily at the leaves. > (I get 5 pairs which are the leaves of the tree.) (And, yes, leading 0's are allowed, the colon is unmoved, and all of > the other conditions implied above.) There could be a better answer. But I will post mine in the next few > days, unless I do not feel the need to. There are several other puzzles that can be made using the basic idea > above. > You can ask about, say, using 24-hour clocks, allowing the moving of > the colon (putting 0's wherever needed on the ends of the times), > allowing differences greater than 12 hours, asking about other > symmetries (such as reflections), or (oh, yeah) having the puzzle on > an ANALOG (hands on a circlular face) clock. > Even though this puzzle is not mathematical, in any purest sense, I > will cross-post it to sci.math anyway, since some on that group may > find this somewhat interesting anyway. > Leroy Quet === Subject: Re: Digital-Clock (sometimes) Upside-Down As a trivial case, 11:11 is self similar when rotated as described which leads to a time difference of 00:00 (midnight). Of course 00:00 is self similar when rotated and therefore also has a time difference of 00:00. Repeat ad infinitum... > I think that this post, as others (at least) of the posts I made > through Google a few days ago, have not necessarily propagated throgh > the Usenet correctly. So I will reply with the original text below > (with >'s added): > Now, as you must know, with the standard digital-clock readout, some > times represent other times if the clock happens to be lying > upside-down. > (12:01 <> 10:21, for example) > (But, even if we have a 4-digit display for all times > {in which case, 01:21 <> 12:10 is a valid rotation}, > then 03:10 is NOT equal to 01:30 after flipping, because 3 is not any > number-digit after being rotated 180-degrees.) So, we have a 12-hour digital clock with the standard readout. > ie. the digits are formed from some combination of 7 line-segments, > each segment being lit or un-lit, the segments arranged as: > -- > | | > -- > | | > -- There is also a colon separating the minutes and the hours, as is > typical. Now, we take a diplayed time, that if flipped 180-degrees produces > another valid time where the colon is in the same place. (example: 10:10 <> 01:01 is valid, but 02:10 <> 12:00 is NOT valid > because the colon has moved.) So, between every time and its rotation, there is an absolute > difference in the two times (many differences, actually). But here, I > am interested in only the two positive differences between 0 and 12 > hours. Now these differences themselves can be other valid times when > displayed on a digital-clock and rotated 180-degrees. And, in turn, we can take the absolute difference between the previous > difference and its rotation, continuing. What, then, is the time that has the most times which, by a chain of > differences and rotations, leads to this particular time? By searching by-hand and brute-force (which is what I suggest that you > do, since a computer-search is not worthwhile, this puzzle being too > easy to justify that), I came up with a time which has a total of 14 > times which lead eventually to this time. > But this is actually 7 pairs equal-by-rotation, and these pairs exist > within the branches of the tree, not necessarily at the leaves. > (I get 5 pairs which are the leaves of the tree.) (And, yes, leading 0's are allowed, the colon is unmoved, and all of > the other conditions implied above.) There could be a better answer. But I will post mine in the next few > days, unless I do not feel the need to. There are several other puzzles that can be made using the basic idea > above. > You can ask about, say, using 24-hour clocks, allowing the moving of > the colon (putting 0's wherever needed on the ends of the times), > allowing differences greater than 12 hours, asking about other > symmetries (such as reflections), or (oh, yeah) having the puzzle on > an ANALOG (hands on a circlular face) clock. > Even though this puzzle is not mathematical, in any purest sense, I > will cross-post it to sci.math anyway, since some on that group may > find this somewhat interesting anyway. > Leroy Quet === Subject: Re: Digital-Clock (sometimes) Upside-Down > As a trivial case, 11:11 is self similar when rotated as described which > leads to a time difference of 00:00 (midnight). I did not make it clear that the (absolute) differences, and their rotations, should themselves be valid times (as represented on a typical 12-hour digital clock). It seems that you might already know this, but I have never seen a clock which represents midnight as 00:00, only as 12:00. (Of course, most clocks do not put initial-0's if the hour is a single digit, however...) I will give the answer that *I* found below immediately after the rest of the copied original post. (spoiler below.) Of course 00:00 is self similar when rotated and therefore also has a time > difference of 00:00. Repeat ad infinitum... > I think that this post, as others (at least) of the posts I made > through Google a few days ago, have not necessarily propagated throgh > the Usenet correctly. So I will reply with the original text below > (with >'s added): > Now, as you must know, with the standard digital-clock readout, some > times represent other times if the clock happens to be lying > upside-down. > (12:01 <> 10:21, for example) > (But, even if we have a 4-digit display for all times > {in which case, 01:21 <> 12:10 is a valid rotation}, > then 03:10 is NOT equal to 01:30 after flipping, because 3 is not any > number-digit after being rotated 180-degrees.) So, we have a 12-hour digital clock with the standard readout. > ie. the digits are formed from some combination of 7 line-segments, > each segment being lit or un-lit, the segments arranged as: > -- > | | > -- > | | > -- There is also a colon separating the minutes and the hours, as is > typical. Now, we take a diplayed time, that if flipped 180-degrees produces > another valid time where the colon is in the same place. (example: 10:10 <> 01:01 is valid, but 02:10 <> 12:00 is NOT valid > because the colon has moved.) So, between every time and its rotation, there is an absolute > difference in the two times (many differences, actually). But here, I > am interested in only the two positive differences between 0 and 12 > hours. Now these differences themselves can be other valid times when > displayed on a digital-clock and rotated 180-degrees. And, in turn, we can take the absolute difference between the previous > difference and its rotation, continuing. What, then, is the time that has the most times which, by a chain of > differences and rotations, leads to this particular time? By searching by-hand and brute-force (which is what I suggest that you > do, since a computer-search is not worthwhile, this puzzle being too > easy to justify that), I came up with a time which has a total of 14 > times which lead eventually to this time. > But this is actually 7 pairs equal-by-rotation, and these pairs exist > within the branches of the tree, not necessarily at the leaves. > (I get 5 pairs which are the leaves of the tree.) (And, yes, leading 0's are allowed, the colon is unmoved, and all of > the other conditions implied above.) There could be a better answer. But I will post mine in the next few > days, unless I do not feel the need to. There are several other puzzles that can be made using the basic idea > above. > You can ask about, say, using 24-hour clocks, allowing the moving of > the colon (putting 0's wherever needed on the ends of the times), > allowing differences greater than 12 hours, asking about other > symmetries (such as reflections), or (oh, yeah) having the puzzle on > an ANALOG (hands on a circlular face) clock. > Even though this puzzle is not mathematical, in any purest sense, I > will cross-post it to sci.math anyway, since some on that group may > find this somewhat interesting anyway. > Leroy Quet Just a little more spoiler-space: (in case you still might change your mind to just look at the answer I got...) | | V | | V | | V | | V (11:12 <> 11:21) | (01:20 <> 02:10) }-(11:10 <> 01:11) (11:01 <> 10:11) | / | / | (12:10 <> 01:21) ----------/ | | (12:20 <> 02:21) ------------------- (02:01 <> 10:20) and the final pair of times has a difference of 3:41, which is unrotatable. Can 3:41 be improved upon?? Leroy Quet === Subject: Re: My nighttime puzzle >> ... if p divides rs, with p prime, and p does not divide r, >> then p and r must be coprime, so by the lemma there exist >> a and b with ap + br = 1, and hence aps + brs = s. >> Now p divides the LHS of this last equation, so p divides s. >> So we have proved that p divides r or s. > Yes, thank you very much! This is a nice proof and I like > the funny line ap + br = 1 ----> aps + brs = s such that rs = pq implies p(as+bq) = s and thus confirms p as a factor of s, in case it isn't already > a factor of r. Lovely - I sure will have to remind this. It's a one-liner via gcds. Let (x,y) := gcd(x,y); x|y := x divides y Then p|sr,sp => p|(sr,sp) = s(r,p) = s, i.e. p|s. QED In more detail: p| sr,sp [p divides sr and sp] => p|(sr,sp) [so p divides their gcd] => p|s(r, p) [by distributivity of gcd] => p|s [since (r,p) = 1 by hypothesis] Note how use of gcd eliminates a,b in the quoted proof, which helps to clarify the real essence of the proof. Here's a similar short proof of irrationality of sqrt's. Suppose sqrt(n) = a/b, (a,b) = 1. Then nbb = aa => b|aa,ab => b|a(a,b) = a QED Much further is written in many of my prior posts, e.g. see http://google.com/groups?q=group%3A*math*+dubuque+euclid+lemma http://mathforum.org/epigone/historia_matematica/gosneyah -Bill Dubuque === Subject: Re: My nighttime puzzle ... if p divides rs, with p prime, and p does not divide r, > then p and r must be coprime, so by the lemma there exist > a and b with ap + br = 1, and hence aps + brs = s. > Now p divides the LHS of this last equation, so p divides s. > So we have proved that p divides r or s. >> Yes, thank you very much! This is a nice proof and I like >> the funny line >> ap + br = 1 ----> aps + brs = s >> such that rs = pq implies >> p(as+bq) = s >> and thus confirms p as a factor of s, in case it isn't already >> a factor of r. Lovely - I sure will have to remind this. >It's a one-liner via gcds. Let (x,y) := gcd(x,y); x|y := x divides y >Then p|sr,sp => p|(sr,sp) = s(r,p) = s, i.e. p|s. QED >In more detail: > p| sr,sp [p divides sr and sp] > => p|(sr,sp) [so p divides their gcd] > => p|s(r, p) [by distributivity of gcd] Well OK, but for proofs at this level, you really need to supply a proof of the distributivity of gcd Derek Holt > => p|s [since (r,p) = 1 by hypothesis] >Note how use of gcd eliminates a,b in the quoted proof, >which helps to clarify the real essence of the proof. >Here's a similar short proof of irrationality of sqrt's. >Suppose sqrt(n) = a/b, (a,b) = 1. >Then nbb = aa => b|aa,ab => b|a(a,b) = a QED >Much further is written in many of my prior posts, e.g. see >http://google.com/groups?q=group%3A*math*+dubuque+euclid+lemma >http://mathforum.org/epigone/historia_matematica/gosneyah >-Bill Dubuque === Subject: Re: My nighttime puzzle => p|(sr,sp) [so p divides their gcd] => p|s(r, p) [by distributivity of gcd] > Well OK, but for proofs at this level, you really > need to supply a proof of the distributivity of gcd I want to thank everybody in the thread for providing interesting references and ideas. I especially like the many variations! The proofs in a nutshell by Bill Dubuque are in a fine contrast with the link given by Dan Grayson: http://www.math.uiuc.edu/~dan/ShortProofs/uf.html I admit I read this lengthier proof with better under- standing after having read both yours and the proof by Bill Dubuque. People often ask, what proofs are. Well, how about flowers? Rainer Rosenthal r.rosenthal@web.de === Subject: Re: My nighttime puzzle |I want to thank everybody in the thread for providing |interesting references and ideas. I'm just happy to see someone realize without special prodding that there's something here to be proven (i.e., unique factorization into primes). Someone once described a conversation with some young people about the need to prove that the ratio of the circumference calling the common ratio pi. Apparently they had a hard time seeing that there was something to be proven. The ratio is the same for each-- because it's pi after all. Keith Ramsay === Subject: Re: Sum Congruent To... (harmonic-numbers related) I think that this post, as others (at least) of the posts I made through Google a few days ago, have not necessarily propagated throgh the Usenet correctly. So I will reply with the original text below (with >'s added): > Let H(k) = 1 +1/2 +1/3 +...+1/k, the k_th harmonic number. Let m be any integer >= 2. > Let n be any positive integer that is divisible by every prime which > divides m. > (This condition may be overly restrictive for any m.) Then: > n-1 > --- > / m+1 k+n+1 > (m+1) > | | H(k) (-1) > / k / > --- > k=1 is congruent to / m n > (m+1) | | H(n-1) + (-1) (mod m) > n-1/ In linear-mode: (m+1) sum{k=1 to n-1} binomial(m+1,k) H(k) (-1)^(k+n+1) is congruent to (m+1) binomial(m,n-1) H(n-1) +(-1)^n (mod m) > (Unless I made an error, which is most-likely to be due to > algebraic-manipulation errors, but something else could still have > messed me up...) Now, the two sides of the congruence are not necessarily integers. But > they both have the same fractional-part. So, if you wish, you may take > the integer-part of each side of the congruence, and the congruence is > still valid. > Leroy Quet === Subject: Re: Sum Congruent To... (harmonic-numbers related) > I think that this post, as others (at least) of the posts I made > through Google a few days ago, have not necessarily propagated throgh > the Usenet correctly. So I will reply with the original text below > (with >'s added): Let H(k) = 1 +1/2 +1/3 +...+1/k, the k_th harmonic number. Let m be any integer >= 2. > Let n be any positive integer that is divisible by every prime which > divides m. > (This condition may be overly restrictive for any m.) Then: > n-1 > --- > / m+1 k+n+1 > (m+1) > | | H(k) (-1) > / k / > --- > k=1 is congruent to / m n > (m+1) | | H(n-1) + (-1) (mod m) > n-1/ In linear-mode: (m+1) sum{k=1 to n-1} binomial(m+1,k) H(k) (-1)^(k+n+1) is congruent to (m+1) binomial(m,n-1) H(n-1) +(-1)^n (mod m) > (Unless I made an error, which is most-likely to be due to > algebraic-manipulation errors, but something else could still have > messed me up...) Now, the two sides of the congruence are not necessarily integers. But > they both have the same fractional-part. So, if you wish, you may take > the integer-part of each side of the congruence, and the congruence is > still valid. > Of course, if n is ANY integer >= m+2, then both sides are equal to (-1)^n, and are therefore congruent. For which n, given an m, does the congruence fail? Leroy Quet === Subject: infinitely splitting in almost unramified extensions X-ID: ThV+3gZFgeboNCGy-rK3TBQtZDvPBOkY8+ukJWhBzR1fAbg-c3hlYy Let S= 2. Then: m --- / m |H(k)| k | | |----| (-1) = / k / |_n _| --- k=1 / m - 1 floor(e^(n-c)) | | (-1) , | |e^(n-c)| | |_ | - 1 / where c = Euler's constant (.5772...). In linear-mode: sum{k=1 to m} binomial(m,k) floor(H(k)/n) (-1)^k = binomial(m-1,floor(e^(n-c))-1) (-1)^floor(e^(n-c)) This SEEMS to be weird, but is actually easily proved and generalizable. Leroy Quet === Subject: Collatz & odd numbers Consider the following representation obtained by reverse Terras function: 1 1 | 2 2 | 3 4 | 4 8 / / 5 5 16 / 6 3 10 32 / 7 6 20 21 64 ... ................. . Is it proven that each row n (with n>=5) contains at least one odd positive integer? Thanks. === Subject: Re: Collatz & odd numbers > Consider the following representation obtained by reverse Terras function: > 1 1 > | > 2 2 > | > 3 4 > | > 4 8 > / > / > 5 5 16 > / > 6 3 10 32 > / > 7 6 20 21 64 > ... ................. . > Is it proven that each row n (with n>=5) contains at least one odd positive > integer? > Thanks. It seems not at all. n=8 comprises only evens (if I've got your pattern correct) it appears: 12 40 42 128. However if you add a third dimension (which inevitably results in an infinite of dimensions because each branch branches off to another one) and make each branch travel deep, then yes because for every odd, the branch extends with pattern 4x+1 and as long as the number is not divisible by 3, it will have a branch of odds off of it: 3, 13, 53, 213, ... (4x+1) all reduce to 5. it if n mod 3 = 1. and an infinite series reducing to 2T(n) if n mod 3 = 2. Thus this series is the 3rd dimension and extends into your screen from 3. 3 is divisible by 3 and thus has no numbers reducing to it (easily proven, if you want i can). 13 mod 3 = 1, and so the series reducing to 13 goes as follows: 17, 69, 277, ... (4x+1). This is the fourth dimension (not _the_ fourth dimension, but our fourth :) as it branches from 13 which is part of the 3rd dimension. This branching can be taken to an infinite depth, and since all terms are odd (4x+1 is odd when x is odd) and their mod 3 values increase (i.e. T(1) mod 3 = T(1+3) mod 3 = T(1+3+3) mod 3 = ...), there will always be an odd value, off of which comes another odd reducing to the same value... Anyway, I think I may be digressing a little bit, so I'll stop here unless you want more. If you want proofs for anything I've said, let me know and I'll try get them online. HTH, Davin. === Subject: Re: Collatz & odd numbers > Consider the following representation obtained by reverse Terras function: > 1 1 > | > 2 2 > | > 3 4 > | > 4 8 > / > / > 5 5 16 > / > 6 3 10 32 > / > 7 6 20 21 64 > ... ................. . > Is it proven that each row n (with n>=5) contains at least one odd positive > integer? > Thanks. It seems not at all. n=8 comprises only evens (if I've got your > pattern correct) it appears: 12 40 42 128. Level 8 has an odd integer. The correct pattern is 1 1 | 2 2 | 3 4 | 4 8 / / / / 5 5 16 / / 6 3 10 32 / 7 6 20 21 64 / 8 12 13 40 42 128 However if you add a third dimension (which inevitably results in an > infinite of dimensions because each branch branches off to another > one) and make each branch travel deep, then yes because for every odd, > the branch extends with pattern 4x+1 and as long as the number is not > divisible by 3, it will have a branch of odds off of it: 3, 13, 53, 213, ... (4x+1) all reduce to 5. > it if n mod 3 = 1. and an infinite series reducing to 2T(n) if n mod 3 > = 2. Thus this series is the 3rd dimension and extends into your screen > from 3. 3 is divisible by 3 and thus has no numbers reducing to it > (easily proven, if you want i can). 13 mod 3 = 1, and so the series > reducing to 13 goes as follows: 17, 69, 277, ... (4x+1). This is the > fourth dimension (not _the_ fourth dimension, but our fourth :) as it > branches from 13 which is part of the 3rd dimension. This branching > can be taken to an infinite depth, and since all terms are odd (4x+1 > is odd when x is odd) and their mod 3 values increase (i.e. T(1) mod 3 > = T(1+3) mod 3 = T(1+3+3) mod 3 = ...), there will always be an odd > value, off of which comes another odd reducing to the same value... Anyway, I think I may be digressing a little bit, so I'll stop here > unless you want more. If you want proofs for anything I've said, let > me know and I'll try get them online. HTH, Davin. === Subject: Collatz (3x+1) as diophantine problem X-ID: rkhg7+ZAwexq7TXzCkRP1fRxN9PcY6QmwiAyJZ-+z5kTj17I9ggc8p Hi - I'm just reading these 3x+1-threads and after fiddling a bit on older ideas, I stuck with a - possibly not difficult - idea for a partial proof. It depends on a certain class of equations, and it is the question, whether there is another x (besides x=1), which solves the actual equation in nonnegative integers (x,a,b,c,...) It is simple to show that there is always the solution x = 1 with a=b=c=...=2, and never a solution for instance with a=b=c=...=0 1 1) x = ------- 2^a -3^1 2^b + 3^1 2) x = ------------ 2^(b+a) -3^2 2^(c+b) + 2^c*3^1 + 3^2 3) x = ----------------------- 2^(c+b+a) -3^3 2^(d+c+b) + 2^(d+c)*3^1 + 2^d*3^2 + 3^3 4) x = ---------------------------------------- 2^(d+c+b+a) - 3^4 If I can prove, that - except the above indicated solutions - there exist no others, then there is no loop except the root- loop 1-2-4-1 caused by the collatz (3x+1)-transformation. It looks simple in a first view, since the different powers of 3 make seemingly the nominator and denominator incompatible for generating integers in x, I remember weakly something like the Eisenstein-criterion (but which I cannot apply firmly just now). Examples: 1) solves simply: the only natural number x, which can be result of the equation is 1, since 1 1 x = --------- = -------------- 2^a -3^1 (2^a -4) + 1 and for any a<>2 result a negative or positiv fraction. Conclusion: there is no one-step-loop except (1)-(4-2-1)-... (a loop of x1 = (3 x0 + 1) / 2^a ) 2) is already more complicated, when tried by simply reformulating: 2^b + 3^1 2^b-4 + 7 2) x = ------------ = -------------- 2^(b+a) -3^2 2^(b+a)-16 + 7 indicates at least the standard solution for a=b=2 giving x=1 again, but to disprove other solutions I currently have to check min/max- conditions for a and b (not too difficult) and to check each discrete combination. Conclusion: there is no two-step-loop except (1)-(4-2-1)-(4-2-1)-... (a loop of x1 = (3 x0 + 1) / 2^a ) 3) similar, but not done. Instead I'm fiddling with a method to make an induction from equation type n to equation type n+1. But - after some hours of sketching - it looks too much complicated, if there could be a more simple proven criterion, which is applicable here ... Gottfried -- Gottfried Helms Univ Kassel === Subject: Re: Collatz & odd numbers wired ha scritto nel messaggio > Consider the following representation obtained by reverse Terras function: > 1 1 > | > 2 2 > | > 3 4 > | > 4 8 > / > / > 5 5 16 > / > 6 3 10 32 > / > 7 6 20 21 64 > ... ................. . > Is it proven that each row n (with n>=5) contains at least one odd positive > integer? > Thanks. > It seems not at all. n=8 comprises only evens (if I've got your > pattern correct) it appears: 12 40 42 128. 12,13,40,42,128. ^^ === Subject: Re: Collatz & odd numbers >=== Subject: Collatz & odd numbers >Message-id: 1 1 > | > 2 2 > | > 3 4 > | > 4 8 > / > / > 5 5 16 > / > 6 3 10 32 > / > 7 6 20 21 64 > ... ................. . >Is it proven that each row n (with n>=5) contains at least one odd positive >integer? Yes. Every number that is congruent to congruent to 2 (mod 3) will spawn an odd number in the next level. Since the successor rules for multiplication by 2 are 0 (mod 3) -> 0 (mod 3) 1 (mod 3) -> 2 (mod 3) 2 (mod 3) -> 1 (mod 3) the 8 16 32 64 ... sequence alternates between 1 (mod 3) and 2 (mod 3) to odd levels 5, 7, 9, ... So we only need to show that there is an odd integer on every even level starting with level 8. But 5, on level 5, is also the start of a sequence of alternating 2 (mod 3), 1 (mod 3). So from that sequence we are guaranteed an odd integer on the even levels 6, 8, 10 ... QED >Thanks. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Continuous fractions I was trying to solve the following problems and got stuck in the first 2. Then, I was told they require some knowledge about continuous (or continued?) fractions, so I'd like someone to tell me what this means and where I can get some material on it. The problems are: 1) If a is an irrational, show that the set A = {n*a+m |m and n are integers} is dense in R 2) If a>0 is irrational, show that the set B = {n*a -m| m>=0 and n>=0 are integers} is dense in (0, inf) 3) Does the set C = {n*a + m | m and n are >=0 integers}, where a>0 is irrational, have limit points in R? Here, the definition of a dense set is: A is dense in R if every open interval of R intersects A. Though not very usual, this definition is equivalent to saying R is the closure of A, right? And is equivalent to saying that between every 2 elemnts of A there is another element of A. As for the 3rd, I don't think we need continuous (continued?) fractions, no matter what this means. We see that every sequence of elements of C that is not ultimately constant goes to infinity, and, therefore, C cannot have any limit point. We can come to this same conclusion noting that every open interval of R can have at most finitely many points of C. Actually, this is true for every real a, it doesn't need to be irrational. Any help is welcome. Thank you. Amanda. === Subject: Re: Continuous fractions >I was trying to solve the following problems and got stuck in the >first 2. Then, I was told they require some knowledge about continuous >(or continued?) fractions, so I'd like someone to tell me what this >means and where I can get some material on it. It's continued fractions, but no, they're not needed for these problems. >The problems are: >1) If a is an irrational, show that the set A = {n*a+m |m and n are >integers} is dense in R Hint: Define frac(x), the fractional part of the real number x by the two conditions 0 <= frac(x) < 1 x - frac(x) is an integer. Fix a number eps > 0. (i) It's enough (why?) to show that A intersects the interval (0, eps). (ii) Hence it's enough (why?) to show that there exist s and t in A such that |s - t| < eps. (iii) Let S = {frac(n*a) : n is an integer}. It's enough to show that there exist s, t in S such that |s - t| < eps. But this is clear because S is an infinite subset of [0, 1). (This is where the pigeonhole principle Julien mentioned comes in: You can divide [0, 1) into finitely many intervals, each of length less than eps. Since S is infinite, at least one of these intervals must contain two elements of S.) Question: _where_ in the argument above (once you figure out what the argument is) did we use the fact that a is irrational? You should make certain to be clear about where this fact is used, because of course the result is false for rational a, so a proof that doesn't use the irrationality of a can't be right... >2) If a>0 is irrational, show that the set B = {n*a -m| m>=0 and n>=0 >are integers} is dense in (0, inf) This one is a little trickier, but I'm pretty sure that minor modifications of the previous argument will suffice. >3) Does the set C = {n*a + m | m and n are >=0 integers}, where a>0 is >irrational, have limit points in R? >Here, the definition of a dense set is: A is dense in R if every open >interval of R intersects A. Though not very usual, ??? Seems like exactly the standard definition to me. >this definition is >equivalent to saying R is the closure of A, right? And is equivalent >to saying that between every 2 elemnts of A there is another element >of A. Right, right. >As for the 3rd, I don't think we need continuous (continued?) >fractions, no matter what this means. We see that every sequence of >elements of C that is not ultimately constant goes to infinity, and, >therefore, C cannot have any limit point. We can come to this same >conclusion noting that every open interval of R can have at most >finitely many points of C. Actually, this is true for every real a, it >doesn't need to be irrational. >Any help is welcome. >Thank you. >Amanda. ************************ David C. Ullrich === Subject: Re: Continuous fractions I was trying to solve the following problems and got stuck in the >first 2. Then, I was told they require some knowledge about continuous >(or continued?) fractions, so I'd like someone to tell me what this >means and where I can get some material on it. It's continued fractions, but no, they're not needed for these > problems. The problems are: 1) If a is an irrational, show that the set A = {n*a+m |m and n are >integers} is dense in R Hint: Define frac(x), the fractional part of the real number x by > the two conditions 0 <= frac(x) < 1 > x - frac(x) is an integer. Fix a number eps > 0. (i) It's enough (why?) to show that A intersects the interval (0, > eps). I got confused here. I couldn' really realize why this is not a loss of generality. If this is true, then there exist m and n such that n*a + m is in (0,eps). But so far I couldn't see why this carries over to (r, r+eps) for every real r. For every real r we can find a rational p/q, p and q integers, such that n*a + m +p/q is in (r, r+eps), but this doesnt imply your conclusion. > (ii) Hence it's enough (why?) to show that there exist s and t > in A such that |s - t| < eps. We readily see that A is closed with respect to addition and to multiplication by an integer. Therefore, s-t is in A, which shows that (ii) implies the existence of some u in A such that |u| that there exist s, t in S such that |s - t| < eps. But this is > clear because S is an infinite subset of [0, 1). (This is where > the pigeonhole principle Julien mentioned comes in: > You can divide [0, 1) into finitely many intervals, each of > length less than eps. Since S is infinite, at least one of > these intervals must contain two elements of S.) Question: _where_ in the argument above (once you figure > out what the argument is) did we use the fact that a is irrational? > You should make certain to be clear about where this fact is > used, because of course the result is false for rational a, so > a proof that doesn't use the irrationality of a can't be right... Answer (I hope it' correct!): We used that fact that a is irrational when we stated that S is an infinite set. Proof (by induction on n). First, it's easy to see that, for every real a, frac(a) = frac(2a) iff a is an integer. But since in our case a is irrational, this is impossible. Therefore, frac(a) <> frac(2a). Now, suposse that, for some natural n, frac(i*a)<>frac(j*a) whenever i,j =1,...n and i<>j. Admit, by way of contradiction, that frac((n+1)*a)= f, where f = frac(i*a) for some natural i, 1<=i<=n. Then, (m+1)a =I_m+1 f and i*a = I_i + f, where I_m+1 and I_i are integers. It follows that (m+1-i)a = I_m+1 - I_i and a = (I_m+1 - I_i)(m+1-i), contrarily to the basic assumption that a is irrational. This completes the induction and shows that the terms of the sequence (frac(n)) are pairwise distinct. Therefore, S = {frac(n*a) : n is an integer} is a countable infinite set, as stated. On the other hand, if a is rational then a= p/q, where p and q are integers with no common divisors. Since (fundamental theorem of Arithmetic) na = np/q = (k*q + r)/q = , 0<=r2) If a>0 is irrational, show that the set B = {n*a -m| m>=0 and n>=0 >are integers} is dense in (0, inf) This one is a little trickier, but I'm pretty sure that minor > modifications of the previous argument will suffice. I'm trying to work it out. 3) Does the set C = {n*a + m | m and n are >=0 integers}, where a>0 is >irrational, have limit points in R? Here, the definition of a dense set is: A is dense in R if every open >interval of R intersects A. Though not very usual, ??? Seems like exactly the standard definition to me. I said that just because people usually say that X is dense in Y (X and Ytopological spaces) if Y is the closure of X. Thank you for your help Amanda === Subject: Re: Continuous fractions >>I was trying to solve the following problems and got stuck in the >>first 2. Then, I was told they require some knowledge about continuous >>(or continued?) fractions, so I'd like someone to tell me what this >>means and where I can get some material on it. >> It's continued fractions, but no, they're not needed for these >> problems. >>The problems are: >>1) If a is an irrational, show that the set A = {n*a+m |m and n are >>integers} is dense in R >> Hint: Define frac(x), the fractional part of the real number x by >> the two conditions >> 0 <= frac(x) < 1 >> x - frac(x) is an integer. >> Fix a number eps > 0. >> (i) It's enough (why?) to show that A intersects the interval (0, >> eps). >I got confused here. I couldn' really realize why this is not a loss >of generality. Suppose x is in A intersect (0, eps). Think about what the sequence x, 2x, 3x, ... looks like. > If this is true, then there exist m and n such that n*a >+ m is in (0,eps). But so far I couldn't see why this carries over to >(r, r+eps) for every real r. For every real r we can find a rational >p/q, p and q integers, such that n*a + m +p/q is in (r, r+eps), but >this doesnt imply your conclusion. > (ii) Hence it's enough (why?) to show that there exist s and t >> in A such that |s - t| < eps. >We readily see that A is closed with respect to addition and to >multiplication by an integer. Therefore, s-t is in A, which shows that >(ii) implies the existence of some u in A such that |u|u being in A implies -u is in A, your conclusion follows. >> (iii) Let S = {frac(n*a) : n is an integer}. It's enough to show >> that there exist s, t in S such that |s - t| < eps. But this is >> clear because S is an infinite subset of [0, 1). (This is where >> the pigeonhole principle Julien mentioned comes in: >> You can divide [0, 1) into finitely many intervals, each of >> length less than eps. Since S is infinite, at least one of >> these intervals must contain two elements of S.) >> Question: _where_ in the argument above (once you figure >> out what the argument is) did we use the fact that a is irrational? >> You should make certain to be clear about where this fact is >> used, because of course the result is false for rational a, so >> a proof that doesn't use the irrationality of a can't be right... >Answer (I hope it' correct!): We used that fact that a is irrational >when we stated that S is an infinite set. Yes. >Proof (by induction on n). >First, it's easy to see that, for every real a, frac(a) = frac(2a) iff >a is an integer. _Exactly_ how do you show that? Doesn't seem easy to me... no, doesn't seem at all easy. > But since in our case a is irrational, this is >impossible. Therefore, frac(a) <> frac(2a). Now, suposse that, for >some natural n, frac(i*a)<>frac(j*a) whenever i,j =1,...n and i<>j. >Admit, by way of contradiction, that frac((n+1)*a)= f, where f = >frac(i*a) for some natural i, 1<=i<=n. Then, (m+1)a =I_m+1 f and i*a = >I_i + f, where I_m+1 and I_i are integers. It follows that (m+1-i)a = >I_m+1 - I_i and a = (I_m+1 - I_i)(m+1-i), contrarily to the basic >assumption that a is irrational. This completes the induction and >shows that the terms of the sequence (frac(n)) are pairwise distinct. >Therefore, S = {frac(n*a) : n is an integer} is a countable infinite >set, as stated. There's really no need to get all inductive here. You only need to show that frac(n*a) <> frac(m*a) if n <> m. Suppose that frac(n*a) = frac(m*a). Then n*a - m*a is an integer, and this implies that a is rational. >On the other hand, if a is rational then a= p/q, where p and q are >integers with no common divisors. Since (fundamental theorem of >Arithmetic) na = np/q = (k*q + r)/q = , 0<=rk + r/q, which shows that for every n, frac(n*a) = r/q must belong to >the finite set {0, 1/q....(q-1)/q}. Therefore, for every rational a >the set S = {frac(n*a) : n is an integer} is always finite and the >pigeonholes principle can't be applied. The set A is never dense in R. >>2) If a>0 is irrational, show that the set B = {n*a -m| m>=0 and n>=0 >>are integers} is dense in (0, inf) >> This one is a little trickier, but I'm pretty sure that minor >> modifications of the previous argument will suffice. >I'm trying to work it out. >>3) Does the set C = {n*a + m | m and n are >=0 integers}, where a>0 is >>irrational, have limit points in R? >>Here, the definition of a dense set is: A is dense in R if every open >>interval of R intersects A. Though not very usual, >> ??? Seems like exactly the standard definition to me. >I said that just because people usually say that X is dense in Y (X >and Ytopological spaces) if Y is the closure of X. >Thank you for your help >Amanda ************************ David C. Ullrich === Subject: Re: Continuous fractions > 1) If a is an irrational, show that the set A = {n*a+m |m and n are > integers} is dense in R Pigeonhole principle. (it's usually presented as a lemma in any continued fractions course) > 2) If a>0 is irrational, show that the set B = {n*a -m| m>=0 and n>=0 > are integers} is dense in (0, inf) Very similar. > 3) Does the set C = {n*a + m | m and n are >=0 integers}, where a>0 is > irrational, have limit points in R? ... > therefore, C cannot have any limit point. We can come to this Of course; maybe there's a mistake in the wording? Run a google search about continued fractions and you'll probably find one introductory course ... I'd give you mine but... it's in French ah// -- Julien Santini === Subject: Re: Continuous fractions > 1) If a is an irrational, show that the set A = {n*a+m |m and n are > integers} is dense in R Pigeonhole principle. (it's usually presented as a lemma in any continued > fractions course) 2) If a>0 is irrational, show that the set B = {n*a -m| m>=0 and n>=0 > are integers} is dense in (0, inf) Very similar. But trickier! 3) Does the set C = {n*a + m | m and n are >=0 integers}, where a>0 is > irrational, have limit points in R? > ... > therefore, C cannot have any limit point. We can come to this Of course; maybe there's a mistake in the wording? Run a google search about continued fractions and you'll probably find one > introductory course ... I'd give you mine but... > it's in French ah// Great! Je parle un peut de Fran.8dais. Si vous vouler m'envoyer votre travail, il est tr.8fs bien venu.8e. (sorry if I killed your language. I said , if you want to send me your work, it's very welcome) Mecy beaucoup Amanda === Subject: Metrics on finite rings X-Enigmail-Version: 0.76.2.0 X-Enigmail-Supports: pgp-inline, pgp-mime Can a metric be defined on a finite ring? Can a metric be defined on polynomials over a finite ring? I have never seen any examples of metrics being defined on these algebraic structures. Are there any reasons why or why not this can not be done? You can take the elements of a ring and map them to the unit circle. Being on a circle, any two elements will have at most two distances between them. Take the min distance for the metric. Of course for a finite ring the metric has to apply to the result of the + or * operation. Maybe the problem is that for a finite ring - is not defined which is necessary to define a distance? My reason for asking is that I am studying approximating functions on finite rings (or fields). To evaluate a proposed approximation means that you have to have some method to determine how close the approximate solution is to the optimal solution. Most optimization problems are cast with their cost functions in real metric spaces. I am curious if anything has been previously said about problems in finite rings (or fields). polynomials over a finite ring can or can not have a metric. Thanks in advance. === Subject: Re: Metrics on finite rings > polynomials over a finite ring can or can not have a metric. One metirc used on vector spaces over a finite field is the Hamming distance: various generalizations have been used. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Metrics on finite rings >Can a metric be defined on a finite ring? Can a metric be defined on >polynomials over a finite ring? I have never seen any examples of >metrics being defined on these algebraic structures. Are there any >reasons why or why not this can not be done? Of course they can have metrics. The metrics might not be as interesting as in infinite rings, though, because they just correspond to the discrete topology. >You can take the elements of a ring and map them to the unit circle. >Being on a circle, any two elements will have at most two distances >between them. Take the min distance for the metric. Of course for a >finite ring the metric has to apply to the result of the + or * >operation. Maybe the problem is that for a finite ring - is not >defined which is necessary to define a distance? Of course - is defined, in any ring: -x is the additive inverse of x, and x-y is x + (-y). I think what you want is called a normed ring. A norm is a function x -> |x| from the ring to the nonnegative reals, such that |x| = 0 iff x = 0 |-x| = |x| |x+y| <= |x| + |y| |x y| <= |x| |y| If your ring has an identity 1, we also require |1| = 1. The metric is then defined by d(x,y) = |x-y|. Every ring has a norm, e.g. |x| = 1 for x <> 0, 0 for x = 0. But other norms might be more interesting. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: The Grand Facade > As an unAmerican fascist sympathizer, you'll watch tyrant bush Ah, *that's* what you're talking about, some foreign country. I didn't realize there was a foreign leader also named bush, just like our duly elected and properly acting U.S. President. -- If you have had problems with Illinois Student Assistance Commission (ISAC), please contact shredder at bellsouth dot net. There may be a class-action lawsuit in the works. === Subject: Re: The Grand Facade As an unAmerican fascist sympathizer, you'll watch tyrant bush Ah, *that's* what you're talking about, some foreign country. I didn't > realize there was a foreign leader also named bush, just like our duly > elected and properly acting U.S. President. There are many who dispute that Dubya was either duly elected or has been acting properly as U. S. President. The economy has certainly gone to hell since he was sworn in. === Subject: Re: The Grand Facade The economy has certainly > gone to hell since he was sworn in. Actually it's done quite well considering the Clinton/Gore Recession that began to come on the scene in late '99/early 2000, the bursting of the dotcom bubble, and the subsequent mighty thwack the economy took from 9/11 and the resultant wars. -- Scott Lowther, Engineer Any statement by Edward Wright that starts with 'You seem to think that...' is wrong. Always. It's a law of Usenet, like Godwin's. - Jorge R. Frank, 11 Nov 2002 === Subject: Re: The Grand Facade The economy has certainly > gone to hell since he was sworn in. Actually it's done quite well considering the Clinton/Gore Recession > that began to come on the scene in late '99/early 2000, the bursting of > the dotcom bubble, and the subsequent mighty thwack the economy took > from 9/11 and the resultant wars. Outside of his personal war against Saddam that seems to be less won than Dubya claims, and the residues of another war (Afganistan) that is not quite over either, and a national deficit beyond belief, and a tax break for Dubya's rich buddies, and the emasculation of civil liberties, and a few other things like that, the state of the nation seems almost healthy. But appearances can be deceiving. === Subject: Re: The Grand Facade > a tax break for Dubya's rich buddies Isn't it interesting how often we hear this, yet what we *don't* hear is that *rich people pay the lion's share of taxes*. If taxes are cut, then *taxpayers* should be the ones who benefit. Welfare recipients don't pay taxes, or at least, not enough to come close to the taxpayer provided benefits, and thus *should not* receive any benefits from a tax cut. Of course, pointing out that poor people don't pay taxes and so shouldn't benefit from a tax cut won't get you reelected. > But appearances can be deceiving. Sure- at first, you *appeared* to be intelligent and reasonable. -- If you have had problems with Illinois Student Assistance Commission (ISAC), please contact shredder at bellsouth dot net. There may be a class-action lawsuit in the works. === Subject: Re: The Grand Facade As an unAmerican fascist sympathizer, you'll watch tyrant bush Ah, *that's* what you're talking about, some foreign country. I didn't > realize there was a foreign leader also named bush, just like our duly > elected and properly acting U.S. President. > There are many who dispute that Dubya was either duly elected They would be wrong, as the record clearly shows. He got a clear majority of the electoral votes and the election was certified by the appropriate authorities. It's over, and anyone who disputes it is just a sore loser, just like his opponents, Al Sore and Joe Loserman (who didn't believe he could really win, which is why he also ran for a backup position as a senator). or has > been acting properly as U. S. President. They would also be wrong, but in this case they are entitled to their opinion. The economy has certainly > gone to hell since he was sworn in. It's been good to me, and if there is a problem, it certainly wasn't due to him. You can thank a bunch of who hijacked some planes 2 years ago for that. -- If you have had problems with Illinois Student Assistance Commission (ISAC), please contact shredder at bellsouth dot net. There may be a class-action lawsuit in the works. === Subject: Re: The Grand Facade spamfile@bellsouth.net says... > The economy has certainly > gone to hell since he was sworn in. It's been good to me, and if there is a problem, it certainly wasn't due to > him. You can thank a bunch of who hijacked some planes 2 years ago > for that. If it was good for you, then it was good for you. Hooray for you. Take it from someone who was in the midst of a lengthy (depressingly lengthy) job search when the planes hit the towers: the economy had gone to hell before 9/11/2001. Better yet, don't take my word for it -- go find the facts. Warping the facts to fit a favorite theory isn't pretty. -- Willoughby @scispace.org.invalid Imagine that, a FROG ON-OFF switch, hardly the work for test pilots. -- Mike Collins === Subject: Re: The Grand Facade > If it was good for you, then it was good for you. Hooray for you. > Take it from someone who was in the midst of a lengthy (depressingly > lengthy) job search when the planes hit the towers: the economy had gone > to hell before 9/11/2001. *That* is the truth. The problems with the economy were *not* caused by Bush. I was doing a lot of temp work. I had an assignment that was going to start the week after 9/11. That was cancelled and I didn't get another call until February- and my wife had to take off work in January because of complications of pregnancy. She had some insurance, but our income was cut to less than 40%. Our parents helped out. In December we made the decision to move to Florida because I couldn't find a job. I found one, my wife was able to transfer, and since then she's gotten a raise and I started a business. Better yet, don't take my word for it -- go > find the facts. The *facts* are that 9/11 opened opportunities for those willing to take the risk. >Warping the facts to fit a favorite theory isn't pretty. Right. That's why I killfiled the original poster. -- If you have had problems with Illinois Student Assistance Commission (ISAC), please contact shredder at bellsouth dot net. There may be a class-action lawsuit in the works. === Subject: Re: The Grand Facade @scispace.org.invalid says... > spamfile@bellsouth.net says... > The economy has certainly > gone to hell since he was sworn in. It's been good to me, and if there is a problem, it certainly wasn't due to > him. You can thank a bunch of who hijacked some planes 2 years ago > for that. If it was good for you, then it was good for you. Hooray for you. Take it from someone who was in the midst of a lengthy (depressingly > lengthy) job search when the planes hit the towers: the economy had gone > to hell before 9/11/2001. Better yet, don't take my word for it -- go > find the facts. Warping the facts to fit a favorite theory isn't pretty. Having been in the same position, beginning two months prior to 9/11/00, I can attest to the truth of your statements, . On the other hand, you can't blame the dot-com bust (which is what's affected my career so badly) on Dubya, or any other President, for that matter. I don't think his administration has proposed any effective (or any at all) concepts on how to revitalize the tech sector, though. I think that's going to be a campaign issue next year, seeing as how tech sector unemployment is STILL in the 30% range in many areas. -- Do not meddle in the affairs of dragons, for | Doug Van Dorn thou art crunchy and taste good with ketchup | dvandorn@mn.rr.com === Subject: Re: The Grand Facade >On the other hand, you can't blame the dot-com bust (which is what's >affected my career so badly) on Dubya, or any other President, for that >matter. I don't think his administration has proposed any effective (or >any at all) concepts on how to revitalize the tech sector, though. I >think that's going to be a campaign issue next year, seeing as how tech >sector unemployment is STILL in the 30% range in many areas. Frankly does the tech sector need revitalizing? Part of the reason for the high unemployment is the abnormally large number of people who flocked to the sector during the boom. D. -- The STS-107 Columbia Loss FAQ can be found at the following URLs: Text-Only Version: http://www.io.com/~o_m/columbia_loss_faq.html Enhanced HTML Version: http://www.io.com/~o_m/columbia_loss_faq_x.html Corrections, comments, and additions should be e-mailed to om@io.com, as well as posted to sci.space.history and sci.space.shuttle for discussion. === Subject: Re: The Grand Facade says... >On the other hand, you can't blame the dot-com bust (which is what's >affected my career so badly) on Dubya, or any other President, for that >matter. I don't think his administration has proposed any effective (or >any at all) concepts on how to revitalize the tech sector, though. I >think that's going to be a campaign issue next year, seeing as how tech >sector unemployment is STILL in the 30% range in many areas. Frankly does the tech sector need revitalizing? Part of the reason > for the high unemployment is the abnormally large number of people who > flocked to the sector during the boom. Frankly, yes it does. You have people with 20+ years of experience in tech sector jobs and who had been making between $50K and $120K who have either been unemployed for 2+ years or are doing jobs outside of their field, making between $15K and $25K. Several *million* people across the country. Not a few people -- a LOT of people. People who comprise between 10% and 30% of the unemployed, and nearly 80% of the underemployed, in some areas of the country. (Next time you're in Seattle, ask the Starbuck's counter guy what his last job was...) These people are the solid core of the taxpayers who used to contribute millions of dollars to federal, state and local tax bases who now are contributing FAR less, which is the basis of the budget crises at all levels around the country. This isn't a minor economic re-adjustment in which, for example, the one or two thousand people across the country who had 20+ years experience doing something that has been made obsolete, like setting hot- lead type for newspapers, are forced out of their fields and into low- paying jobs. This has had, and continues to have, a major effect on the economy at all levels. Yes, something needs to be done. I, for one, am NOT going to happily write off my chances of getting a decent-wage job for the rest of my life just because *you* and others like you are not being affected by this situation. I'll start a ing revolution and take what YOU have by force before I do that... :-) -- Do not meddle in the affairs of dragons, for | Doug Van Dorn thou art crunchy and taste good with ketchup | dvandorn@mn.rr.com === Subject: Re: The Grand Facade > Frankly does the tech sector need revitalizing? Part of the reason for > the high unemployment is the abnormally large number of people who > flocked to the sector during the boom. And what would the unemployment rate be then (and now) if they hadn't had those jobs when they were available? The problem with the tech sector was that the highly paid business professionals vastly over projected demand, meanwhile the technology improved productivity and cost performance so much that the revenue fell at the same time that demand failed to meet expectations. For example in the communications sector the advent of internet/packet switched technologies drove costs towards zero so that the traditional business models of communications companies could no longer be supported when competition drove pricing to end-users down as well. === Subject: Re: The Grand Facade > On the other hand, you can't blame the dot-com bust (which is what's > affected my career so badly) on Dubya, or any other President, for that > matter. It was a direct result of the money people being dazzled by the new technology and forgetting to apply good business principles and practices. -- If you have had problems with Illinois Student Assistance Commission (ISAC), please contact shredder at bellsouth dot net. There may be a class-action lawsuit in the works. === Subject: how to prove the inner route is shorter If a route forms a loop, is there a simple proof that the inner edge of the route is shorter than that of the outer one? Assumes the width of the route is constant. Can also assume the route is piceswise straight. But the route is not necessarily convex everywhere. Thanks for sharing your thought. === Subject: Re: how to prove the inner route is shorter > If a route forms a loop, is there a simple proof that the inner edge > of the route is shorter than that of the outer one? Assumes the width > of the route is constant. Can also assume the route is piceswise > straight. But the route is not necessarily convex everywhere. Thanks > for sharing your thought. If you assume that the center of the path is a simple, closed, smooth, planar curve C oriented counter-clockwise and the half-width r = width/2 is less than the minimum radius of curvature of C, then the formula for the length of an offset curve, together with the fact that the total curvature of C is 2PI, can be used to show that the length of the inner loop is L - 2PI*r and the length of the outer loop is L + 2PI*r, where L is the length of C. [The inner loop is the offset of C to the left; the outer loop is the offset to the right] If you allow corners, the inner loop may be longer than the outer loop. For example, let C be a square with a narrow spike sticking about halfway up into the square from the middle of the bottom edge. If the angle of the spike is theta (a small number) and the length of one of the spike's sides is R, choose the width of the path to be 2*R*tan(theta/2). The outer loop will have a very small spike, whereas the inner loop will have a relatively large spike, and it's not too hard to see (by calculation, if necessary), that the inner loop is longer than the outer loop. You didn't precisely define what you mean by a path of constant width, but this example probably fits your idea. John Mitchell === Subject: Re: how to prove the inner route is shorter > If a route forms a loop, is there a simple proof that the inner edge > of the route is shorter than that of the outer one? Assumes the width > of the route is constant. Can also assume the route is piceswise > straight. But the route is not necessarily convex everywhere. Thanks > for sharing your thought. If you assume that the center of the path is a simple, closed, smooth, > planar curve C oriented counter-clockwise and the half-width r = > width/2 is less than the minimum radius of curvature of C, then the > formula for the length of an offset curve, together with the fact that > the total curvature of C is 2PI, can be used to show that the length > of the inner loop is L - 2PI*r and the length of the outer loop is L + > 2PI*r, where L is the length of C. [The inner loop is the offset of C > to the left; the outer loop is the offset to the right] If you allow corners, the inner loop may be longer than the outer > loop. For example, let C be a square with a narrow spike sticking > about halfway up into the square from the middle of the bottom edge. > If the angle of the spike is theta (a small number) and the length of > one of the spike's sides is R, choose the width of the path to be > 2*R*tan(theta/2). The outer loop will have a very small spike, whereas > the inner loop will have a relatively large spike, and it's not too > hard to see (by calculation, if necessary), that the inner loop is > longer than the outer loop. You didn't precisely define what you mean > by a path of constant width, but this example probably fits your idea. John Mitchell Wow! Thanks for clarification. How come you can think so clearly? How did you do it? Anyhow, you are great! === Subject: Re: how to prove the inner route is shorter John Mitchell > If a route forms a loop, is there a simple proof that the inner edge > of the route is shorter than that of the outer one? Assumes the width > of the route is constant. Can also assume the route is piceswise > straight. But the route is not necessarily convex everywhere. Thanks > for sharing your thought. ... > If you allow corners, the inner loop may be longer than the outer > loop. For example, let C be a square with a narrow spike sticking > about halfway up into the square from the middle of the bottom edge. > If the angle of the spike is theta (a small number) and the length of > one of the spike's sides is R, choose the width of the path to be > 2*R*tan(theta/2). The outer loop will have a very small spike, whereas > the inner loop will have a relatively large spike, and it's not too > hard to see (by calculation, if necessary), that the inner loop is > longer than the outer loop. You didn't precisely define what you mean > by a path of constant width, but this example probably fits your idea. Another of my conjectures just bit the dust :) LH === Subject: Re: how to prove the inner route is shorter mathphyer > If a route forms a loop, is there a simple proof that the inner edge > of the route is shorter than that of the outer one? Assumes the width > of the route is constant. Can also assume the route is piceswise > straight. But the route is not necessarily convex everywhere. Thanks > for sharing your thought. This is not a complete answer, but I reckon a simple proof will be possible provided that we take enough care in the statement of the hypotheses. Consider two polygons A,B, with A inside B, and a bijection between the sides of A and the sides of B such that corresponding sides are parallel. (Otherwise it is not clear what constant width would mean.) There is a formula, seen in textbooks about a complex variable, for the area of a parametrized closed curve. Since the inner polygon has (I trust) smaller area, I think we can bring home your proposition. LH === Subject: Re: how to prove the inner route is shorter > mathphyer > If a route forms a loop, is there a simple proof that the inner edge > of the route is shorter than that of the outer one? Assumes the width > of the route is constant. Can also assume the route is piceswise > straight. But the route is not necessarily convex everywhere. Thanks > for sharing your thought. > This is not a complete answer, but I reckon a simple proof will be possible > provided that we take enough care in the statement of the hypotheses. > Consider two polygons A,B, with A inside B, and a bijection between the > sides of A and the sides of B such that corresponding sides are parallel. > (Otherwise it is not clear what constant width would mean.) There is a > formula, seen in textbooks about a complex variable, for the area of a > parametrized closed curve. Since the inner polygon has (I trust) smaller > area, I think we can bring home your proposition. > LH Can you give more detailed reference to the formula? Remember the polygons may not be convex on all the sides. Also, each pair of the sides of A and B are not only parallel but also with a single distance among all the pairs. Without the latter condition, I dont believe the original statement is necessarily true anymore. === Subject: Re: how to prove the inner route is shorter mathphyer > Larry Hammick > mathphyer > If a route forms a loop, is there a simple proof that the inner edge > of the route is shorter than that of the outer one? Assumes the width > of the route is constant. Can also assume the route is piceswise > straight. But the route is not necessarily convex everywhere. Thanks > for sharing your thought. > This is not a complete answer, but I reckon a simple proof will be possible > provided that we take enough care in the statement of the hypotheses. > Consider two polygons A,B, with A inside B, and a bijection between the > sides of A and the sides of B such that corresponding sides are parallel. > (Otherwise it is not clear what constant width would mean.) There is a > formula, seen in textbooks about a complex variable, for the area of a > parametrized closed curve. Since the inner polygon has (I trust) smaller > area, I think we can bring home your proposition. > LH > Can you give more detailed reference to the formula?... Some people call it the Gauss-Green formula. Google found: http://www.ma.iup.edu/projects/CalcDEMma/Green/Green.html But it simplifies for a polygon instead of a more general closed curve. Fix some point P in the plane. Say AB is one side of the polygon. The area of the triangle PAB has a simple formula in terms of the coordinates of P,A,B -- half the value of a certain 2x2 determinant (or in other jargon, half the magnitude of the vector product PAxPB). Summing over the sides gives the Gauss-Green area formula for the whole polygon. There are kindred formulas for the winding number of a curve with respect to a point, and for the barycentre of the interior of the curve, and for other weighted averages too, called moments. LH === Subject: calculus I like calculus, it's my favorite class! Everyone should like calculus! Calculus, it's just a great word, don't you agree? Well, that's my input. I wanted to express my opinion on calculus. === Subject: Re: calculus > I like calculus, it's my favorite class! Everyone should like calculus! > Calculus, it's just a great word, don't you agree? Well, that's my input. > I wanted to express my opinion on calculus. Well, calculus is the Latin word for stone. Yet another reason to like calculus! === Subject: Re: calculus > I like calculus, it's my favorite class! Everyone should like calculus! > Calculus, it's just a great word, don't you agree? Well, that's my input. > I wanted to express my opinion on calculus. > Well, calculus is the Latin word for stone. > Yet another reason to like calculus! In particular, a piece of limestone as a writing instrument on a slate (to record results of computation). Dentists use the word to refer to mineral (calcium carbonate?) deposits on teeth. Calculated = stoned? Cheers, ZVK(Slavek). === Subject: Re: calculus >I like calculus, it's my favorite class! Everyone should like calculus! >Calculus, it's just a great word, don't you agree? Well, that's my input. >I wanted to express my opinion on calculus. >>Well, calculus is the Latin word for stone. >>Yet another reason to like calculus! >In particular, a piece of limestone as a writing instrument on a >slate (to record results of computation). Are you sure? I thought it was because the Romans (following earlier traditions) moved stones in a tray (similar to using an abacus). I'm sure I've seen this explanation from someone who studies history for a living, I'm not making it up. >Dentists use the word to refer to mineral (calcium carbonate?) >deposits on teeth. I have several calculi in my kidney. As long as they stay there, I'm okay. >Calculated = stoned? Friday evening seminars at Coslo's. Professors buy (seems fair to me, considering what the grad students get paid). Jon Miller === Subject: Death Rattle Of Neoclassical Theory Of Value The neoclassical theory of value and distribution is mistaken. I have updated my demonstration of this proposition at: Here's the sort of comments I had for the previous version: I got Sraffa3 to print beautifully! Now to find the time to work on it. -- Mark Witte, 4 December 1998 reposting the same URL does not make it any more correct. -- John J. Weatherby, 15 May 2002 We'll have another hearty laugh at Sraffa3.pdf another time. But nothing substantial. I don't expect substantial comments on this version either. Most of the economists that post to sci.econ don't seem to be the sort that are willing or capable of giving substantial comments. (One could take umbrage at this remark, or one could prove me wrong.) -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Death Rattle Of Neoclassical Theory Of Value > The neoclassical theory of value and distribution is mistaken. I have > updated my demonstration of this proposition at: Here's the sort of comments I had for the previous version: I got Sraffa3 to print beautifully! Now to find the time to work > on it. > -- Mark Witte, 4 December 1998 Randall Jarrell once said something like, even worse for a serious fellow than ridicule is to be ignored, or perhaps rather than serious, it was dull, but with the same intent. With that concern in mind, I thought I had long ago given my thoughts on a version of some of its predecessors, I do not find any elementary errors that should stop the reader in his or her tracks. There is something of a puzzle though, as to the intent of the piece. In reviewing an academic paper, the question in the reader's mind should always be, why was this written? What intellectual purpose did the author intend? The piece is entitled, A Critique Of Disaggregated Neoclassical Theory and draws upon the work of Sraffa. This paper visits the currently sleep area of value theory and seems to emphasize that general equilibrium models are complicated and can produce results that can seem bizarre. This is not a particularly novel result, and the author does not seem to be making any claim toward any original contribution here. What substantial comments could be made? Most readers would likely want to know how would applying the approaches or results in this paper produce differences from the findings in some well cited papers in the literature. Perhaps the next time this paper surfaces, the value of Sraffa's work for its ability to improve upon standard methods in modeling the observed world will be explained. reposting the same URL does not make it any more correct. > -- John J. Weatherby, 15 May 2002 We'll have another hearty laugh at Sraffa3.pdf another time. But nothing substantial. I don't expect substantial comments on > this version either. Most of the economists that post to sci.econ > don't seem to be the sort that are willing or capable of giving > substantial comments. (One could take umbrage at this remark, or > one could prove me wrong.) -- > Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html > To solve Linear Programs: .../LPSolver.html > r c A game: .../Keynes.html > v s a Whether strength of body or of mind, or wisdom, or > i m p virtue, are found in proportion to the power or wealth > e a e of a man is a question fit perhaps to be discussed by > n e . slaves in the hearing of their masters, but highly > @ r c m unbecoming to reasonable and free men in search of > d o the truth. -- Rousseau === Subject: Multitude of infinities. When people look out to the stars, they wonder how many stars there are out there, or even if the number of stars can be counted. And when people look down at the ground, they wonder how many times you can divide matter before you cannot divide it anymore. Now we know that in the observable universe that ther are a finite number of stars, but that beyond this universe there may be an infinite number of stars. We also know that, while throughout time our conception of the smallest division of matter has been changing to become smaller and smaller, there has still been a finite division. However, now I ask you this: Are they the same type of infinity? My answer is that they're not. Not just because one is getting bigger and bigger and one is getting smaller and smaller, but because the concept of infinity is not unique. There are various 'kinds' of infinity, just as there are various kinds of infinitesimal. It is my belief that in the universe, there are infinite quantities, but only of a certain type of infinity. Like in a black hole, there is certainly an infinite amount of volume within it, that is, if black holes exist. But if we consider the black hole case in a universe where space is not infinitely divisible, that is not the same amount of infinity that would be contained even within a single cubic milimeter of space if space was infinitely divisible. Nor would the amount of time extending from the beginning of the universe to eternity in a universe with a finite minimum division of time be the same as the amount of time in a single tick of the clock if time was infinitely divisible. While this is my view, this may not be your view. I am eager to hear your view on things. I posted this to four different newsgroups because these are the four newsgroups which would be most interest in this topic. I hope you enjoy debating this topic with eachother. I know I had fun starting this debate! (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. > When people look out to the stars, they wonder how many stars there > are out there, or even if the number of stars can be counted. And when > people look down at the ground, they wonder how many times you can > divide matter before you cannot divide it anymore. ... Maybe what we think of as the universe is a quark in a universe at the next level up. Etc. Gib === Subject: Re: Multitude of infinities. > Maybe what we think of as the universe is a quark in a universe at the > next level up. Etc. And maybe it isn't. Bob Kolker === Subject: Re: Multitude of infinities. > Maybe what we think of as the universe is a quark in a universe at the >> next level up. Etc. > And maybe it isn't. Just trying to get into the swing of it - science fiction physics. Gib === Subject: Re: Multitude of infinities. > Maybe what we think of as the universe is a quark in a universe at > the next level up. Etc. >> And maybe it isn't. > Just trying to get into the swing of it - science fiction physics. LOts of that hereabouts. === Subject: Re: Multitude of infinities. When people look out to the stars, they wonder how many stars there > are out there, or even if the number of stars can be counted. And when > people look down at the ground, they wonder how many times you can > divide matter before you cannot divide it anymore. All this crap has been addressed. Get off your lazy stupid troll ass and do a literature search. > Now we know that What do you know? Your hundreds of troll posts amply demonstrate that you know nothing. You are a ditzy NY coed going to night classes at Hunter to get educated. She carries a softcover copy of Finnegans Wake in her oversided handbang, snaps gum, and spreads syphilis. > However, now I ask you this: Are they the same type of infinity? Untutored idiot. Google aleph zero infinity 317 hits Georg Cantor infinity 2580 hits Cantor infinity 15300 hits > It is my belief Keep it to yourself. Physics is empirical not religious. Get an education - and not the crap handed to you in a doggy bag by third rate academics in fourth rate diploma mills. > (...Starblade Riven Darksquall...) (...Enemanozzle Bloody Brainfart...) -- Uncle Al http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Multitude of infinities. When people look out to the stars, they wonder how many stars there > are out there, or even if the number of stars can be counted. And when > people look down at the ground, they wonder how many times you can > divide matter before you cannot divide it anymore. All this crap has been addressed. Get off your lazy stupid troll ass > and do a literature search. > I was just leading up to my post. You are very impatient. > Now we know that What do you know? Your hundreds of troll posts amply demonstrate that > you know nothing. You are a ditzy NY coed going to night classes at > Hunter to get educated. She carries a softcover copy of Finnegans > Wake in her oversided handbang, snaps gum, and spreads syphilis. > You didn't even let me finish. > However, now I ask you this: Are they the same type of infinity? Untutored idiot. > I asked that question already knowing the answer. It is you who is the idiot, not I. > Google > aleph zero infinity 317 hits > Georg Cantor infinity 2580 hits > Cantor infinity 15300 hits It is my belief Keep it to yourself. Physics is empirical not religious. Get an > education - and not the crap handed to you in a doggy bag by third > rate academics in fourth rate diploma mills. > Mathematics is not empirical. It is purely rational. And to a great degree it depends on how we define it. > (...Starblade Riven Darksquall...) (...Enemanozzle Bloody Brainfart...) You idiot. (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. > While this is my view, this may not be your view. I am eager to hear > your view on things. I posted this to four different newsgroups > because these are the four newsgroups which would be most interest in > this topic. I hope you enjoy debating this topic with eachother. I > know I had fun starting this debate! There is an infinite hierarchy of infinite (transfinite) cardinals. Also there is an infinite hierarchy of infinite ordinals. The matter is dealt with in Cantor's theory of transfinite sets, about which there are scores of good treatises. You can easily find one on amazon.com or abebooks.com. Bob Kolker === Subject: Re: Multitude of infinities. While this is my view, this may not be your view. I am eager to hear > your view on things. I posted this to four different newsgroups > because these are the four newsgroups which would be most interest in > this topic. I hope you enjoy debating this topic with eachother. I > know I had fun starting this debate! There is an infinite hierarchy of infinite (transfinite) cardinals. Also > there is an infinite hierarchy of infinite ordinals. The matter is dealt with in Cantor's theory of transfinite sets, about > which there are scores of good treatises. You can easily find one on > amazon.com or abebooks.com. Bob Kolker Does cardinality necessarily mean one right after the other? How else could we explain the multitide of infinities? I say we should not limit ourselves to only a certain number of infinities. Also, I believe that we should define the ultimate infinity, as the highest, greatest infinity possible, though which is undirected. It could be, for example, the amount of information contained in the universal set. Since no other set contains more information than the universal set, then it would indeed be the highest. One could also say that it deserves the status of being the reciprocal of zero. (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. > Does cardinality necessarily mean one right after the other? How else > could we explain the multitide of infinities? I say we should not limit ourselves to only a certain number of > infinities. Also, I believe that we should define the ultimate infinity, as the > highest, greatest infinity possible, There is no such infinity. Just as there is no greatest integer Bob Kolker === Subject: Re: Multitude of infinities. Does cardinality necessarily mean one right after the other? How else > could we explain the multitide of infinities? I say we should not limit ourselves to only a certain number of > infinities. Also, I believe that we should define the ultimate infinity, as the > highest, greatest infinity possible, There is no such infinity. Just as there is no greatest integer Bob Kolker Yes there is. Take the universal set. There is no set greater than that. Therefore, the amount of information it contains is not only infinity, but it must be the highest infinity possible. There could not be an infinity higher, since all the numbers contained within that would be contained within the universal set, thus contradiction the notion that something could be a higher form of infinity. Just because there is no highest set that can be represented using the outdated method of using cardinalities to represent infinities, IE, an 'algebraic representation' of infinities, doesn't mean that such a set does not exist. (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. Does cardinality necessarily mean one right after the other? How else > could we explain the multitide of infinities? I say we should not limit ourselves to only a certain number of > infinities. Also, I believe that we should define the ultimate infinity, as the > highest, greatest infinity possible, There is no such infinity. Just as there is no greatest integer Bob Kolker Yes there is. Take the universal set. There is no set greater than that. > What about the power set of your universal set? Darren > Therefore, the amount of information it contains is not only infinity, > but it must be the highest infinity possible. There could not be an > infinity higher, since all the numbers contained within that would be > contained within the universal set, thus contradiction the notion that > something could be a higher form of infinity. Just because there is no highest set that can be represented using the > outdated method of using cardinalities to represent infinities, IE, an > 'algebraic representation' of infinities, doesn't mean that such a set > does not exist. (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. Does cardinality necessarily mean one right after the other? How else > could we explain the multitide of infinities? I say we should not limit ourselves to only a certain number of > infinities. Also, I believe that we should define the ultimate infinity, as the > highest, greatest infinity possible, There is no such infinity. Just as there is no greatest integer Bob Kolker Yes there is. Take the universal set. There is no set greater than that. > What about the power set of your universal set? Darren > It would also contain the power set of the universal set. It may confuse you at first, but I assigned this infinity to also have an infinite cardinality. This means its power set would also be in itself. Meaning it is more infinite than any other infinite set. If you consider this a contradiction, fine. I don't. That's because this is by definition not only a set which contains an infinite number of members but a set which contains everything. > Therefore, the amount of information it contains is not only infinity, > but it must be the highest infinity possible. There could not be an > infinity higher, since all the numbers contained within that would be > contained within the universal set, thus contradiction the notion that > something could be a higher form of infinity. Just because there is no highest set that can be represented using the > outdated method of using cardinalities to represent infinities, IE, an > 'algebraic representation' of infinities, doesn't mean that such a set > does not exist. > (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. Take the universal set. There is no set greater than that. That set of all subsets of the universal set has more elements than the universal set. By the way, there is not such set. The assumption that there is a set which includes all other sets as elements leads to a contradiction. Consider the set of all sets that are not elements of themselves. See what you get. Therefore, the amount of information it contains is not only infinity, > but it must be the highest infinity possible. There could not be an > infinity higher, since all the numbers contained within that would be > contained within the universal set, thus contradiction the notion that > something could be a higher form of infinity. This leads to contradictions. The history of set theory is essentially the elimination of contradictions caused by overreaching. The paradox I stated above is one of the earliest of the set theoretical paradoxes. What about the set of all universal sets. Is it a universal set or not. Just because there is no highest set that can be represented using the > outdated method of using cardinalities to represent infinities, IE, an > 'algebraic representation' of infinities, doesn't mean that such a set > does not exist. There is nothing outdated in set theory. After the contradictions were removed, about 90 years ago, it is as good as the day it was first invented. Unlike physics, sound mathematics never becomes outdated. That is why Euclid's geometry (properly tightened up with Hilbert.s axioms) is as good now as it was 2500 years ago. Physics theories come and go, but mathematics is forever. Why don't you learn some math instead of spewing ax ano? Bob Kolker === Subject: Re: Multitude of infinities. > Take the universal set. There is no set greater than that. That set of all subsets of the universal set has more elements than the > universal set. By the way, there is not such set. The assumption that > there is a set which includes all other sets as elements leads to a > contradiction. Consider the set of all sets that are not elements of > themselves. See what you get. > What if we define a 'cardinality' (I hate that word, you know. There has GOT to be a better way to describe an infinity) of infinity for which this is not the case? For which, even though it containes infinities within infinities, it remains the same. It would essentially be an infinity of infinite cardinality. > Therefore, the amount of information it contains is not only infinity, > but it must be the highest infinity possible. There could not be an > infinity higher, since all the numbers contained within that would be > contained within the universal set, thus contradiction the notion that > something could be a higher form of infinity. This leads to contradictions. The history of set theory is essentially > the elimination of contradictions caused by overreaching. The paradox I > stated above is one of the earliest of the set theoretical paradoxes. > Yes, but how is that a contradiction? It would also contain all its subsets, and those subsets would be bigger, et cetara. It is only a contradiction if we limit ourselves to finite cardinalities of infinity. > What about the set of all universal sets. Is it a universal set or not. Just because there is no highest set that can be represented using the > outdated method of using cardinalities to represent infinities, IE, an > 'algebraic representation' of infinities, doesn't mean that such a set > does not exist. There is nothing outdated in set theory. After the contradictions were > removed, about 90 years ago, it is as good as the day it was first > invented. Unlike physics, sound mathematics never becomes outdated. That > is why Euclid's geometry (properly tightened up with Hilbert.s axioms) > is as good now as it was 2500 years ago. Physics theories come and go, > but mathematics is forever. > Alright, but how many of those contradictions were actually meant to be removed, and how many should we have actually learned something from and revised rather than thrown away? > Why don't you learn some math instead of spewing ax ano? Bob Kolker I do know a lot of math. I also believe that our current understanding of infinity is laughable. (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. I do know a lot of math. You obviously do not know set theory and your reasoning is deficient. > I also believe that our current understanding > of infinity is laughable. Propose a better one. Bob Kolker === Subject: Re: Multitude of infinities. > I do know a lot of math. You obviously do not know set theory and your reasoning is deficient. > I know enough to know what I dislike about it. > I also believe that our current understanding > of infinity is laughable. Propose a better one. > Tell me, in set theory, is there an aleph infinity? > Bob Kolker (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. : I know enough to know what I dislike about it. It's dangerous to only learn until you find something you dislike. Instead you should learn until you *understand*, and then if you find some problems, you can explain them. You are not quite understanding. : Tell me, in set theory, is there an aleph infinity? In set theory, is there a tweeble? Putting a string of two set-theoretic words down next to each other makes as much sense as putting down a random word. What is your aleph infinity supposed to be? Give us a rigid consistent definition which modern set theory lacks. Justin === Subject: Re: Multitude of infinities. : I know enough to know what I dislike about it. It's dangerous to only learn until you find something you dislike. > Instead you should learn until you *understand*, and then if you find > some problems, you can explain them. You are not quite understanding. > : Tell me, in set theory, is there an aleph infinity? In set theory, is there a tweeble? Putting a string of two set-theoretic words down next to each other makes > as much sense as putting down a random word. What is your aleph > infinity supposed to be? Give us a rigid consistent definition which > modern set theory lacks. Justin Well, it would be a type of infinity of which no greater set could be constructed. This would require that, if one were to try to construct it in any way, one would simply get this same set. Meaning for each set within this super infinite set, any transformation that can be done to it will also exist in this set. (BTW, if modern set theory has a way of representing this, then that would honestly be perfect.) Now, conceptually, I see nothing wrong with explaining an infinitude set like that. But clearly it's been told to me a number of times that you simply can't do this. At least not with modern set theory. I think maybe it has to do with the definition of the set. If this is the case, then we need a more general type of set. Does anybody know a type of set which isn't really a set but acts like a set in most ways, except that it is open where the set is closed? (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. >Well, it would be a type of infinity of which no greater set could be >constructed. This would require that, if one were to try to construct >it in any way, one would simply get this same set. Meaning for each >set within this super infinite set, any transformation that can be >done to it will also exist in this set. If you allow the transformation of taking the power set of any set, then Cantor's diagonal proof shows that the size of the power set is always strictly greater than the size of the original set. Are you familiar with Cantor's proof? If so, then which step would you want to invalidate in order to allow your super-infinite set to be the same as its power set? >(BTW, if modern set theory has a way of representing this, then that >would honestly be perfect.) Well, it has a way of showing that it can't be done. Is that almost perfect? >Now, conceptually, I see nothing wrong with explaining an infinitude >set like that. But clearly it's been told to me a number of times that >you simply can't do this. At least not with modern set theory. Yes. But don't take our word for it. Cantor's diagonal proof is short and quite accessible, even from an intuitive naive-set-theory point of view. >I think maybe it has to do with the definition of the set. If this is >the case, then we need a more general type of set. Does anybody know a >type of set which isn't really a set but acts like a set in most ways, >except that it is open where the set is closed? I think a way for you to tighten up that question is to study Cantor's proof and see what aspect of sets you'd like to change in order to make the proof fail. Frankly, I very much doubt your chances of finding a change that doesn't do fatal violence to some other aspect of your intuitive notion of set. >(...Starblade Riven Darksquall...) -- --------------------------- | BBB b barbara minus knox at iname stop com | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- === Subject: Re: Multitude of infinities. >Well, it would be a type of infinity of which no greater set could be >constructed. This would require that, if one were to try to construct >it in any way, one would simply get this same set. Meaning for each >set within this super infinite set, any transformation that can be >done to it will also exist in this set. > If you allow the transformation of taking the power set of any set, then > Cantor's diagonal proof shows that the size of the power set is always > strictly greater than the size of the original set. Are you familiar with > Cantor's proof? If so, then which step would you want to invalidate in > order to allow your super-infinite set to be the same as its power set? Well there would be a single infinty type, the size of the set is infinite and the size of the power set of this infinite set is also infinite. >(BTW, if modern set theory has a way of representing this, then that >would honestly be perfect.) > Well, it has a way of showing that it can't be done. Is that almost perfect? >Now, conceptually, I see nothing wrong with explaining an infinitude >set like that. But clearly it's been told to me a number of times that >you simply can't do this. At least not with modern set theory. > Yes. But don't take our word for it. Cantor's diagonal proof is short > and quite accessible, even from an intuitive naive-set-theory point of > view. >I think maybe it has to do with the definition of the set. If this is >the case, then we need a more general type of set. Does anybody know a >type of set which isn't really a set but acts like a set in most ways, >except that it is open where the set is closed? > I think a way for you to tighten up that question is to study Cantor's > proof and see what aspect of sets you'd like to change in order to make > the proof fail. Frankly, I very much doubt your chances of finding a > change that doesn't do fatal violence to some other aspect of your > intuitive notion of set. A more constructive approach where a set contains itself, as is simply represented in finite space using computer data structures with pointers. Infinity has some properties of numbers and not others, but an infinite set is just a construct. first(4, N) = {1, 2, 3, 4} Herc === Subject: Multitude of infinities. There are no infinite sets. Sets do not 'count' numerals, but there are sets of signs. jJ Well, it would be a type of infinity of which no greater set could be >constructed. This would require that, if one were to try to construct >it in any way, one would simply get this same set. Meaning for each >set within this super infinite set, any transformation that can be >done to it will also exist in this set. If you allow the transformation of taking the power set of any set, then > Cantor's diagonal proof shows that the size of the power set is always > strictly greater than the size of the original set. Are you familiar with > Cantor's proof? If so, then which step would you want to invalidate in > order to allow your super-infinite set to be the same as its power set? > Well there would be a single infinty type, the size of the set is infinite and > the size of the power set of this infinite set is also infinite. (BTW, if modern set theory has a way of representing this, then that >would honestly be perfect.) Well, it has a way of showing that it can't be done. Is that almost perfect? Now, conceptually, I see nothing wrong with explaining an infinitude >set like that. But clearly it's been told to me a number of times that >you simply can't do this. At least not with modern set theory. Yes. But don't take our word for it. Cantor's diagonal proof is short > and quite accessible, even from an intuitive naive-set-theory point of > view. I think maybe it has to do with the definition of the set. If this is >the case, then we need a more general type of set. Does anybody know a >type of set which isn't really a set but acts like a set in most ways, >except that it is open where the set is closed? I think a way for you to tighten up that question is to study Cantor's > proof and see what aspect of sets you'd like to change in order to make > the proof fail. Frankly, I very much doubt your chances of finding a > change that doesn't do fatal violence to some other aspect of your > intuitive notion of set. A more constructive approach where a set contains itself, as is simply > represented in finite space using computer data structures with pointers. > Infinity has some properties of numbers and not others, but an infinite set > is just a construct. first(4, N) = {1, 2, 3, 4} > Herc === Subject: Re: Multitude of infinities. > There are no infinite sets. Sets do not 'count' numerals, but there are sets > of signs. Try the set of integers. You do believe in the existence of integers I assume. How many integers are there? Bob Kolker === Subject: Re: Multitude of infinities. > Try the set of integers. You do believe in the existence of integers I > assume. How many integers are there? Here is your answer: Because mathmatics is the contingent application of a function for which an outcome must be defined, there is no contingency to create integers if that outcome is not defined. The outcome that is not defined, is not 'infinity'! 'The outcome that is not defined' means not to look for an outcome. And so we leave the realm of mathematics alltogether. The set of integers is a set of numerals, or signs. > There are no infinite sets. Sets do not 'count' numerals, but there are sets > of signs. > Try the set of integers. You do believe in the existence of integers I > assume. How many integers are there? > Bob Kolker === Subject: Re: Multitude of infinities. : I know enough to know what I dislike about it. It's dangerous to only learn until you find something you dislike. > Instead you should learn until you *understand*, and then if you find > some problems, you can explain them. You are not quite understanding. > : Tell me, in set theory, is there an aleph infinity? In set theory, is there a tweeble? Putting a string of two set-theoretic words down next to each other makes > as much sense as putting down a random word. What is your aleph > infinity supposed to be? Give us a rigid consistent definition which > modern set theory lacks. Justin Well, it would be a type of infinity of which no greater set could be > constructed. This would require that, if one were to try to construct > it in any way, one would simply get this same set. Meaning for each > set within this super infinite set, any transformation that can be > done to it will also exist in this set. (BTW, if modern set theory has a way of representing this, then that > would honestly be perfect.) Now, conceptually, I see nothing wrong with explaining an infinitude > set like that. But clearly it's been told to me a number of times that > you simply can't do this. At least not with modern set theory. I think maybe it has to do with the definition of the set. I don't know what the definition of set is. Is there one? In ZF, there are rules for constructing sets. > If this is > the case, then we need a more general type of set. Why do we need a more general type of set? What would you do with it? Where is it needed? > Does anybody know a > type of set which isn't really a set but acts like a set in most ways, > except that it is open where the set is closed? What about classes? The set of ordinals doesn't exist. But the class of ordinals can be considered. Other interesting sets are inaccessible cardinals. If memory serves me correctly: 1) If M > 0 is inaccessible, and {x_i} is a list of fewer than M cards, each smaller than M, then sum(x_i) < M. 2) If M > 0 is inaccessible, and N < M, then 2^N < M. Aleph_0 is inaccesible. I believe that no other inaccessible cards have been produced. > (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. What is your aleph > infinity supposed to be? Give us a rigid consistent definition which > modern set theory lacks. Justin > Well, it would be a type of infinity of which no greater set could be > constructed. This would require that, if one were to try to construct > it in any way, one would simply get this same set. Meaning for each > set within this super infinite set, any transformation that can be > done to it will also exist in this set. > (BTW, if modern set theory has a way of representing this, then that > would honestly be perfect.) > Now, conceptually, I see nothing wrong with explaining an infinitude > set like that. But clearly it's been told to me a number of times that > you simply can't do this. At least not with modern set theory. > I think maybe it has to do with the definition of the set. If this is > the case, then we need a more general type of set. Does anybody know a > type of set which isn't really a set but acts like a set in most ways, > except that it is open where the set is closed? > (...Starblade Riven Darksquall...) You might be interested in a guy called (Paul?) Finsler. He tried to defend naive set theory, but nobody quite understood him. (Just FYI: the currently used axiom systems of set theory, like ZFC, do not suffer from Russell's paradox.) Another thought that comes to mind (also well-known, and not new at all) is that, say, the axiom system ZFC lacks an axiom of 'maximality'. Something that formally captures something like 'everything that could be a set, is indeed a set'. When you formalize this, you end up with a semantics over all the models of ZFC, rather than just one fixed structure. If you think a bit deeper about it, this semantics relies heavily on the assumptions you make on the meta-level. So, as a foundation for mathematics, that is not very convincing. And yet another association: working with dynamic models, as done in AI. Buzz words that you may look after: circumscription, belief revision, situation calculus, frame problem. Bottom line: These things have already been studied ad nauseam, in the past and present, but they are not seen as relevant for the foundation of mathematics. Whether that is justified or not is a separate discussion. Hope this helps. Herman Jurjus === Subject: Re: Multitude of infinities. > I do know a lot of math. You obviously do not know set theory and your reasoning is deficient. > I know enough to know what I dislike about it. I also believe that our current understanding > of infinity is laughable. Propose a better one. > Tell me, in set theory, is there an aleph infinity? What would that be? There is an aleph_omega, the limit of succesors of aleph_0. Perhaps that is what you mean. Bob Kolker (...Starblade Riven Darksquall...) === Subject: Re: Multitude of infinities. >There is nothing outdated in set theory. After the contradictions were >removed, about 90 years ago, it is as good as the day it was first >invented. Unlike physics, sound mathematics never becomes outdated. That >is why Euclid's geometry (properly tightened up with Hilbert.s axioms) >is as good now as it was 2500 years ago. Physics theories come and go, >but mathematics is forever. One could argue that 'sound' physics never becomes outdated, either. We simply don't have any examples of that particular beast... The history of physics shows more cases of ideas that were thought to be sound and later found to be only approximations than does the history of mathematics, but it's certainly happened - proofs that appeared sound on their face and which were accepted for a few decades, perhaps a century or two, but in which a hole was ultimately found. Those ideas became outdated only because they were unsound, of course, but given an idea of sufficient complexity, it can be very difficult to tell the difference. Physical theories that are supported by all known evidence are challenged each time new evidence becomes available. In mathematics, it's a new IDEA that's needed, and new ideas are a somewhat more rare commodity than new generations of measuring devices. --S === Subject: Set theory is seriously flawed Set theory is conceptually seriously flawed: Here's why: 1. The properties of the objects marked out by a set do not confer properties upon the set, e.g. a set of cows is not a herd. 2. A set has no properties. 3. A named set with no marks of objects ('members') is not a set, but is the name of an object. 4. An empty set is the general name for any set. 5. There are no relationships between sets. 6. Sets cannot be counted. 7. The 'members' of a set are indistinguishable marks, and cannot be counted. 8. The objects of attention of a set cannot be counted by the set. This is the form of a set. However, set theory assumes the functions of mathematics and any other tools it needs in order to execute functions, but these are not conceptual categories of a set. JJ >There is nothing outdated in set theory. After the contradictions were >removed, about 90 years ago, it is as good as the day it was first >invented. Unlike physics, sound mathematics never becomes outdated. That >is why Euclid's geometry (properly tightened up with Hilbert.s axioms) >is as good now as it was 2500 years ago. Physics theories come and go, >but mathematics is forever. > One could argue that 'sound' physics never becomes outdated, either. > We simply don't have any examples of that particular beast... > The history of physics shows more cases of ideas that were thought to > be sound and later found to be only approximations than does the > history of mathematics, but it's certainly happened - proofs that > appeared sound on their face and which were accepted for a few decades, > perhaps a century or two, but in which a hole was ultimately found. > Those ideas became outdated only because they were unsound, of course, > but given an idea of sufficient complexity, it can be very difficult > to tell the difference. > Physical theories that are supported by all known evidence are challenged > each time new evidence becomes available. In mathematics, it's a new > IDEA that's needed, and new ideas are a somewhat more rare commodity than > new generations of measuring devices. > --S === Subject: Re: Set theory is seriously flawed > Set theory is conceptually seriously flawed: > Here's why: 1. The properties of the objects marked out by a set do not confer > properties upon the set, e.g. a set of cows is not a herd. If a set of cows is not a herd, what is a herd. > 2. A set has no properties. Nonsense. Sets can be empty (that is a property) or non-empty. A set can be finite or not finite. > 3. A named set with no marks of objects ('members') is not a set, but is the > name of an object. The property objects must have to be in the set can, in some cases, be written out quite explicitly. > 4. An empty set is the general name for any set. Nonsense. The empty set is not a name of a set with elements in it. > 5. There are no relationships between sets. Nonsense. One set can be a subset of another. That is a relationship between sets. > 6. Sets cannot be counted. The set of subsets of a finite set can be counted very nicely. If the base set has n elements, the set of subsets of that set has 2^n elements. > 7. The 'members' of a set are indistinguishable marks, and cannot be > counted. The set of even integers between 0 and 8 has a cardinality. It is 5. 0,2,4,5,8. > 8. The objects of attention of a set cannot be counted by the set. The set -contains- the objects of interest This is the form of a set. However, set theory assumes the functions of > mathematics and any other tools it needs in order to execute functions, but > these are not conceptual categories of a set. Totally meaningly string of words. You are a deep fount of misinformation and nonsense. Bob Kolker === Subject: Re: Set theory is seriously flawed > The set of even integers between 0 and 8 has a cardinality. It is 5. > 0,2,4,5,8. Um, no. That's 0,2,4,6,8. ;p (...Starblade Riven Darksquall...) === Subject: Re: Set theory is seriously flawed > The set of even integers between 0 and 8 has a cardinality. It is 5. > 0,2,4,5,8. Um, no. That's 0,2,4,6,8. ;p > Experimental error. === Subject: Re: Set theory is seriously flawed Its wrong anyway, Sets don't count. > The set of even integers between 0 and 8 has a cardinality. It is 5. > 0,2,4,5,8. Um, no. That's 0,2,4,6,8. ;p Experimental error. === Subject: Re: Set theory is seriously flawed > Its wrong anyway, Sets don't count. All of modern matthematics is based on sets. You need sets to define what a function is. Bob Kolker === Subject: Set theory is seriously flawed 3 A function is defined by the limits of its operation. That's not a set. We cannot speak of 'all' in mathematics and so we cannot attempt to group 'all' in a set. A 'set of all numerals' has an error that must be questioned. The 'members' or marks of a set, do not assume the properties of its 'members' or marks. The counting is not done by the set, but by ourselves, when we use mathematics. JJ > Its wrong anyway, Sets don't count. > All of modern matthematics is based on sets. You need sets to define > what a function is. > Bob Kolker === Subject: Re: Multitude of infinities. >There is nothing outdated in set theory. After the contradictions were >>removed, about 90 years ago, it is as good as the day it was first >>invented. Unlike physics, sound mathematics never becomes outdated. That >>is why Euclid's geometry (properly tightened up with Hilbert.s axioms) >>is as good now as it was 2500 years ago. Physics theories come and go, >>but mathematics is forever. > One could argue that 'sound' physics never becomes outdated, either. > We simply don't have any examples of that particular beast... Mathematics operatates in the realm of the abstract and a priori. Physics is an empirical science and is very much conditioned on avialble technology and what happens to be known at any given time. There is more in heaven and earth than is accounted for by our physics. The history of physics shows more cases of ideas that were thought to > be sound and later found to be only approximations than does the > history of mathematics, but it's certainly happened - proofs that > appeared sound on their face and which were accepted for a few decades, > perhaps a century or two, but in which a hole was ultimately found. > Those ideas became outdated only because they were unsound, of course, > but given an idea of sufficient complexity, it can be very difficult > to tell the difference. Physical theories that are supported by all known evidence are challenged > each time new evidence becomes available. In mathematics, it's a new > IDEA that's needed, and new ideas are a somewhat more rare commodity than > new generations of measuring devices. New ideas in mathematics do not falsify already existing ideas in mathematics. The changes in mathemtics is an additive process. In physics, it is sometimes necessary to can old theories because they have been shown to be inadequate. For example, Caloric, Phlogiston. Vital Essence, and Luminiferous Aether. These notions, which at one time were useful, have now been discarded. Bob Kolker === Subject: Re: Multitude of infinities. >Mathematics operatates in the realm of the abstract and a priori. >Physics is an empirical science and is very much conditioned on avialble >technology and what happens to be known at any given time. >There is more in heaven and earth than is accounted for by our physics. Oh, agreed. Because mathematics need only be internally consistent, rather than consistent with any empirical observation, it's a more reliable structure on the whole, and not really subject to re-evaluation every time new data comes to light. I just felt compelled to indulge in my habit of objecting to absolutes :) >New ideas in mathematics do not falsify already existing ideas in >mathematics. The changes in mathemtics is an additive process. In >physics, it is sometimes necessary to can old theories because they have >been shown to be inadequate. For example, Caloric, Phlogiston. Vital >Essence, and Luminiferous Aether. These notions, which at one time were >useful, have now been discarded. Actually, this is interesting. Has there ever been (or can you imagine) a case where two new ideas are each consistent with the body of accepted mathematical thought, but contradict each other? It seems possible, IF the new ideas merely fail to contradict accepted thought rather than being rigorously proved as necessarily following from accepted axioms. --S === Subject: Re: Multitude of infinities. >There is nothing outdated in set theory. After the contradictions were >>removed, about 90 years ago, it is as good as the day it was first >>invented. Unlike physics, sound mathematics never becomes outdated. That >>is why Euclid's geometry (properly tightened up with Hilbert.s axioms) >>is as good now as it was 2500 years ago. Physics theories come and go, >>but mathematics is forever. > One could argue that 'sound' physics never becomes outdated, either. > We simply don't have any examples of that particular beast... > Mathematics operatates in the realm of the abstract and a priori. > Physics is an empirical science and is very much conditioned on avialble > technology and what happens to be known at any given time. > There is more in heaven and earth than is accounted for by our physics. The history of physics shows more cases of ideas that were thought to > be sound and later found to be only approximations than does the > history of mathematics, but it's certainly happened - proofs that > appeared sound on their face and which were accepted for a few decades, > perhaps a century or two, but in which a hole was ultimately found. > Those ideas became outdated only because they were unsound, of course, > but given an idea of sufficient complexity, it can be very difficult > to tell the difference. Physical theories that are supported by all known evidence are challenged > each time new evidence becomes available. In mathematics, it's a new > IDEA that's needed, and new ideas are a somewhat more rare commodity than > new generations of measuring devices. > New ideas in mathematics do not falsify already existing ideas in > mathematics. The changes in mathemtics is an additive process. In > physics, it is sometimes necessary to can old theories because they have > been shown to be inadequate. For example, Caloric, Phlogiston. Vital > Essence, and Luminiferous Aether. These notions, which at one time were > useful, have now been discarded. You are talking an ideal. Mathematics uses the same peer revue methodology as physics and is open to the same inevitable corrections. There are *parts* of mathematics where the precise nature is explicit. Herc === Subject: Re: contingent application as maths > For an infinite regress in finite space, so called, you don't have to > construe Koch curves or triangles, you can have a function for say, +1. > Here, the space is always finite, while the steps become proportionally > smaller. But in all these examples, the construction can only go so far as a > contingent application can go. Thats as far as maths goes. All the functions > of maths are contingent applications. There is no mathematical application > 'to infinity'. > JJ Can you define contingent application? Are you saying that whenever infinity is and has been used in physics all those theories are negated because the dependence upon applicability for testing is not possible? But we could construe any number of strange maths: How long is the coast-line of Great Britain? Given a map one can sit down with a ruler and soon come up with a value for the length. The problem is that repeating the operation with a larger scale map yields a greater estimate of the length. If we actually went to the coast and measured them directly, then still greater estimates would result. It turns out that as the scale of measurement decreases the estimated length increases without limit. Thus, if the scale of the (hypothetical) measurements were to be infinitely small, then the estimated length would become infinitely large! The ultimate map is the terrain itself. The particular method we use to search through the possibility space determines how close we look and how accurately we model the terrain. ------------------- Take the coastline of Britain for example. How long is it ? Nobody knows. Of course they do you say ! Ah, but they know roughly the area of the country so they must, by Euclid, know the minimum boundary surely, an equivalent circle ? Yes, but the actual boundary is infinite ! To see this, go in your mind to the seaside with a metre rule and measure a section of rock. You will skip over a few crevices will you not ? Now use a kilometre ruler instead - this skips over a lot more resulting in a different, lower reading. Take now a 1 cm measurement, this will go around most irregularities and give a much bigger total. So, the length is variable isn't it ? But not infinite surely ? Move up the coast and you find a river. What do you measure now ? Well, just continue up the left bank until you re-emerge on the right, the coastline must be continuous ! But now you are following all the tributaries and streams and rivulets and....you will never appear on the other side I think ! If you do, take an micron sized measuring stick and try again, by the time infinite time. Britain thus has an coastline approaching infinity but a finite area. A 2 dimensional paradox. More Paradoxes Now take a tree and measure its volume (easy, just dunk it in a big bath and see how much water overflows, like Archimedes would). But what is its surface area ? Yes, that's right it's nearly infinite again (don't forget to measure down every pore or stomata in every leaf...). A 3 dimensional paradox now ! How many dimensions did I say ? Two ? Three ? Neither are right, both items have what is called 'Fractal Dimension' - a fractional number. Ignoring technicalities, this is a measure of how irregular an object is. A Koch snowflake (a triangle, with other third size triangles stuck midway on each side, and so ad infinitum - illustrated) has for example a fractal dimension of 1.26. In the extreme cases we have seen the dimension become one more than we would expect, a single dimensional line becomes two dimensional, a two dimensional surface becomes three. I'll leave to your imagination a 3 dimensional fractal in 4 dimensions ! Self-Similarity Can we go the other way, reducing a three dimensional object to two ? Yes we can. Take a pyramid and drill a hole in it. We increase the surface area and reduce the volume. Keep drilling smaller holes in what is left until you run out of material, you now have a solid without any volume, made up only of edges. The formal version of this is called the Sierpinski gasket in three dimensions, but has a fractal dimension of only two ! While measuring the coast with the different ruler sizes something may have struck you. The shape of the coast at 1km scale is the same as at 1m scales, and again the same at 1cm scale. Think of Africa from space, an island on a map, a rock pool beneath your feet. The shape of the coastline always appears the same, equally jagged. This is a feature of fractal systems which we call 'Self-Similarity'. We cannot tell just by looking at a system what scale it really is. Yet every view is slightly different, the objects are identical in form but also not identical in detail ! Why is this ? Mathematical Roots To see let us look at something quite different. Many people know that you can find the roots of an equation involving squares of x by using a simple formula, and it gives two values. Generalising, we can expect n roots for an equation of x to the power of n. How do we solve these bigger equations ? Often Newton's method is used, this is an iterative solution and gives a better approximation by putting the last answer back into the equation, repeating until the answer doesn't change - that being a solution. For other solutions we try different starting values and, if lucky, the formula will converge to an alternative solution. Luck ? This is precise mathematics isn't it ? Not quite, it is another fractal ! The boundary between the range of values converging on one root and those converging on another is a fractal curve. If we guess a value close to this boundary we cannot be sure which root it will converge towards. It make no difference if we magnify a graph of value versus root, the irregularity and unpredictability of the boundary is still there, big as ever (as illustrated in this plot of x6). It is this boundary between part of a system moving in one direction and part moving in another which is at the heart of fractals. > While this is my view, this may not be your view. I am eager to hear > your view on things. I posted this to four different newsgroups > because these are the four newsgroups which would be most interest in > this topic. I hope you enjoy debating this topic with eachother. I > know I had fun starting this debate! There is an infinite hierarchy of infinite (transfinite) cardinals. Also > there is an infinite hierarchy of infinite ordinals. The matter is dealt with in Cantor's theory of transfinite sets, about > which there are scores of good treatises. You can easily find one on > amazon.com or abebooks.com. Bob Kolker When you run the algorithms for Koch curves and Cantor sets it has the > apparent potential to create and infinite amount of information in a > finite > space. Physical vs Design - Space In physical space we observe objects which are stable processes or for > particular durations stable enough to be building blocks in larger > constructions with other duration potentials. In design space we observe the possible changing configurations through > durations. A finite number of configurationally stable durations [atoms] likely have > an > infinite configurational possibility space. Cantor Sets & Koch Curves: The Von Koch curve. Its construction is almost as easy as the Sierpinski > Triangle. You start with a triangle (equilaterality is really more > important > in this case than for the Sierpinski Triangle), and, on each side, tack on > little equilateral triangles, so you end up with a shape very much like a > Star of David. Now tack on triangle on all straight segments and iterate > smaller and smaller... The length of the Koch curve is infinite. Assuming that we started with a > unit line segment the length of the curve in the k^th stage is equal to > the > number of line segments times the length of each segment. The number of > pieces in the kth stage is 4k and the length of a segment is (1/3)k. Thus, > the the total length of the curve is (4/3)k at the k^th stage of > construction. Since, the Koch curve, by definition, is the limiting set of > the above geometric construction scheme, when k goes to infinity. The > length > of the Koch curve is infinite. http://chaos.phy.ohiou.edu/~thomas/fractal/frac1.html To make an Cantor sets take a line of a certian length and take out the > middle third, draw the two resulting lines missing middle space below it > and > then do this again and again and we seem to get an infinite space out of > and > finite line. http://www.math.sunysb.edu/~scott/Book331/Cantor_sets.html Georg Cantor, a German mathematician, in 1883, introduced a set, now > called > the Cantor set that had some exceptional properties. Following is a simple > geometric construction scheme to visualize the Cantor set. The > construction > of the Cantor set (C) is done by starting with a unit line [0,1]. Divide > the > unit line segment into three equal parts and then remove the middle third > (leaving the end points)to form new segments that exist at [0, 1/3] and > [2/3, 1]. This is the first stage in the construction of the C set. In the > next stage repeat the above process of removing the middle third of the > each > of the two line segments that was obtained in the last step to get four > smaller line segments. This process of removing the middle third from the > remaining segments from the previous stage, is continued adinfinitum. The > figure below shows the first six stages in the construction of the C set. > The points marked by the white vertical line (in stage 0 and stage 1) are > the line segments that are removed. http://chaos.phy.ohiou.edu/~thomas/fractal/frac1.html The same phenomena happens with the Koch curve which is an design used by > snow flake formation with freezing H2O molecules. Take a triangle, and > along > each side put another triangle precicely one third the size and then you > get > a star of david. Repeat the process over and over and u get this seemingly > infinite (down to molecule size) line of triangles within the same space > as > the orgional finite triangle. Some claim that in design space (space of possible [irreducible] *circuit > properties) their is infinite variability within a finite space with > finite > stabilities: The snowflake never escapes the dashed square you see in figures 1-4, so > it > encloses a finite amount of area no larger than a credit card. On the > other > hand, at each step building the new little triangles adds more than one > unit > of length to the curve. To be precise, [4 3]n - 1 units are added at the > nth step, so the length of the snowflake is larger than 3 + 1 + 1 + 1 + 1 > + > 1 + ....... = infinity. The snowflake curve is infinitely long, yet it would fit in your wallet! http://scidiv.bcc.ctc.edu/Math/Snowflake.html Student: So I can think of infinity as being larger than any counting > number? And iterating infinitely many times is the idea of repeating the > steps forever? Mentor: For now these are good ways to think. Here is a more standard way > to > say repeat infinitely many times: Let the number of iterations approach infinity. http://www.shodor.org/interactivate/discussions/infinity.html Now, the real trouble with infinity is much the same: we can't count that > high! Cantor's insight was that, even though we can't enumerate an > infinite > set, we can nonetheless apply the same procedure to any well-defined > infinite set that we applied above to determining if our hands have the > same > number of fingers. In other words, we can determine if two infinite sets > are > the same size (equinumerous) by seeking to find a one-to-one match-up > between the elements of each set. Now, remember Galileo's Paradox? Galileo > noticed that we can do the following: http://www.mathacademy.com/pr/minitext/infinity/ > Cantor's Comb Example: > http://www.shodor.org/interactivate/activities/cantor/ index.html Koch's Infinite Snowflake Example: > http://www.shodor.org/interactivate/activities/koch/index.html > http://www.shodor.org/MASTER/fractal/software/Snowflake.html Sierpinski's Gasket with 100,000 iterations: > http://www.cs.wisc.edu/~richm/cs302/applets/gasket.html Infinity - Definitions: > http://scidiv.bcc.ctc.edu/Math/infinity.html > http://members.shaw.ca/quadibloc/math/infint.htm Infinite Perimeter, Finite Area: > http://www.zeuscat.com/andrew/chaos/vonkoch.html > === Subject: Re: Multitude of infinities. [...] > While this is my view, this may not be your view. I am eager to hear > your view on things. I posted this to four different newsgroups > because these are the four newsgroups which would be most interest in > this topic. I hope you enjoy debating this topic with eachother. I > know I had fun starting this debate! [...] I'm curious as to how many information bits are necessary to describe the world? If there are 6000 million people on earth, presumably one needs at least 6000 million bits... Even a finite and large upper bound would be interesting. My view is that it's fun to think about, but is almost metaphysics. David Bernier === Subject: Re: Multitude of infinities. > I'm curious as to how many information bits are necessary to describe the > world? If there are 6000 million people on earth, presumably one > needs at least 6000 million bits... Even a finite and large > upper bound would be interesting. My view is that it's fun to think about, but is > almost metaphysics. David Bernier capacity of the universe, and estimates the number. You can find it here: http://xxx.lanl.gov (document quant-ph/0110141). Rudy. === Subject: Re: Multitude of infinities. > I'm curious as to how many information bits are necessary to describe the > world? If there are 6000 million people on earth, presumably one > needs at least 6000 million bits... Even a finite and large > upper bound would be interesting. Six gigabits is a trivial amount. Your computer disk probably has 20 gigabytes of store, each byte having 8 bits. The complexity of a human being (just one!) far exceeds this. The human brain has something like 10^10 neurons. Assuming that a neuron is a binary computing device ( it is really much more complicated than that) this would yield 2^(10^20) global states and such a complex could compute 2^(2^(10^10)) boolean functions. And this is a -simplification- of something even more complicated! Bob Kolker electron-dot-cloud are galaxies === Subject: Re: 150 free shares SBC + $500.cash extra; VonNeumann Gametheory applied to > Who's your broker, and how much was the commission, or are you trading on > paper? Getting way to personal, don't you think. The Internet has created what is called deep discount brokers. Really a nice thing for those to play the stockmarket. And I hope they drive all other broker houses lean. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Fraud in Computer Science Publishing Not this twaddle again! They've been formalized since the 1920s, at > least. If you can't understand the formalism, that's your problem, not > theirs. Ok, then what is the formal representation of: 1. The unsolvability of the halting problem. !(EM e T)(AN e T)(Ak e w)[(En e w) M(, k) = n <-> (En e w) N(k) = > n] (T is the set of all turing machines, w the set of all natural > numbers, is the code of N) Thank you so much for taking the time and effort to respond to my > request for a formalization of these theorems and their proofs. > Please correct me if I am wrong: ( M(, k) = n <-> N(k) = n ) means that any proposed solution M > halts (with a value n) on input of any machine N and its input k, the > same as the machine N halts on its input k (with its own value n). > That is, M simulates running N on k w.r.t. halting. But this is the definition of the Universal Turing Machine (UTM), not > the Halting Problem. (A UTM exists, so the above statement is > actually false.) An actual solution to the Halting Problem needs to > indicate when it halts YES and when it halts NO. However, I understand the spirit of your intended notation. (See > below.) 2. The recursion theorem. (for any recursive two-place function there > is a program that takes in one input and applies the function to that > input plus itself) (Af e R_2) (EM e T) (Ax e w) [(Ey e w) f(x, ) = y & M(x) = y] v > [(Ay e w) f(x, ) != y & (Ay e w) M(x) != y] R_2 is the set of all recursive two-place functions. While I personally would change [(Ey e w) f(x, ) = y & M(x) = y] to > simply [ f(x, ) = M(x) ], I can imagine a motivation to shy from > such direct references to partial functions f and (that computed by) > M. Yes. (See below.) 3. The fixed point theorem. (for any recursive f there is a value > such that this value and the result of applying f to this value > produce functionally equivalent programs) (Af e R)(Ex e w)(Ay e w) (Ez e w) [{f(x)}(y) = z & {x}(y) = z] v [(Az > e w) {f(x)}(y) != z / (Az e w) {x}(y) != z] R is the set of all recursive one-place functions; {x} is the machine > with code x. Yes. (See below.) And what is the formal representation of a proof of each of these > theorems? This is left as an exercise to the reader. That is the whole problem. You can give names to concepts like the > recursive one-place functions and Turing Machines, but can you use > that syntax to formally manipulate these concepts? Can you generate > theorems? Can you (partially) decide if a given wff is a theorem? Not really; the Predicate Calculus is too weak to be useful in practive. Of course, I could construct a proof of these theorems in the calculus, but it would too much hassle. > That is what is missing from what has been published on the problems > of theoretical computer science. There is no formal derivation of > theorems such as these. I would call that a problem of mathematics, not computer science. If you need a computer to formalize proofs, you aren't doing it right. === Subject: Re: Factoring polynomials in a GF(2^m) ring? Sorry for the late response, but I had to attend a conference. Thanks a lot for all the replies Jaco === Subject: Re: How is the probability of this? >Suppose a dice is thrown a large number of times. The >eyes that come up are {2, 3, 4, 5, 6} but not {1}. Now, >if that holds for, say 100.000 throws, how is the chance >of {1} coming up in the 100.001:st throw? If the die is a fair one - i.e. it is not weighted or biased in favour of any particular outcome, then the probability is 1/6 that any thow is a 1. Of course, if you don't know anything about the die (i.e. if you make no assumptions about whether or not it is rigged), you must calculate the probability from the results of the 100k throws. In this case the probability would be zero (1 - p(2) - p(3) - p(4) -p(5) -p(6), where p(n) is the number of times n came up, divided by 100000). gareth === Subject: Re: How is the probability of this? > Suppose a dice is thrown a large number of times. The > eyes that come up are {2, 3, 4, 5, 6} but not {1}. Now, > if that holds for, say 100.000 throws, how is the chance > of {1} coming up in the 100.001:st throw? It looks like we're estimating the probability of a Bernoulli event: roll 1 versus roll not-1. If one has no information, other than the statistics produced by counting events, one might proceed as follows. If the true probability is P, then the probability of having M successes in N trials is [(N choose M) P^M (1-P)^(N-M)]. We can regard that as a function of P. It's not generally a distribution function, since its integral (over the range of P) is 1/(N+1). But a scaled version is a D.F.: in fact, it's a Beta distribution. One might estimate P by calculating the mean of that distribution: after much work one gets simply P = (M+1) / (N+2). With no statistics whatsoever (M=N=0), P is 1/2, perhaps a reasonable guess. In the example given, M=0 and N=100000; so P = 1/100002. --- But I think that here one has more information, namely that the die looks like one that should succeed 1/6th of the time. Perhaps (M+1)/(N+6) is a better estimate of P. (No big deal, we get 1/100006 instead.) --- On last thought, the die is clearly biased. Those respondants who say roll-1 is impossible or negligibly possible are likely right. And they're not far from 1/100006. -- Don Reble djr@nk.ca === Subject: Re: How is the probability of this? > Suppose a dice is thrown a large number of times. The > eyes that come up are {2, 3, 4, 5, 6} but not {1}. Now, > if that holds for, say 100.000 throws, how is the chance > of {1} coming up in the 100.001:st throw? If one sees those two actions (100k throws and the single > throw respectively) as separate, the probability would be > 1/6 for {1}. > Otherwise, one would expect {1} to come up far more likely > than that. > Which is correct statistically speaking? If one throws a die 100,000 times without getting a 1 even once, one may reasonably conclude that the die throwing is not honest, and that the chances of the die coming up 1 in future similar throws is negligible. === Subject: Re: How is the probability of this? Suppose a dice is thrown a large number of times. The > eyes that come up are {2, 3, 4, 5, 6} but not {1}. Now, > if that holds for, say 100.000 throws, how is the chance > of {1} coming up in the 100.001:st throw? If one sees those two actions (100k throws and the single > throw respectively) as separate, the probability would be > 1/6 for {1}. > Otherwise, one would expect {1} to come up far more likely > than that. > Which is correct statistically speaking? If one throws a die 100,000 times without getting a 1 even once, one > may reasonably conclude that the die throwing is not honest, and > that the chances of the die coming up 1 in future similar throws is > negligible. If one throws a fair die enough times, one should get 100,000 consecutive rolls without a 1. It should take a while. One should have to roll 50,000+ without a 1 several times in the process. Eldon:)