Subject: Re: Cum Hoc, Ergo Propter Hoc by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7N2w8p16963; >Looking over my threads I talk a good bit about sci.math¹ers who >dispute reality--living in their own little fantasy world--and >sci.math¹ers claim I live in my own little fantasy world, so what¹s >the answer? >Well it is a *math* newsgroup and you¹d think that math would be >enough, but I¹ve noticed repeatedly that certain posters say things >that go against rather basic mathematics, and somehow many of you seem >to suddenly forget basic mathematics long enough to believe them. >But it¹s not just a sci.math thing, and mathematicians rather >bizarrely ignore logic itself in claiming that Andrew Wiles found a >proof of Fermat¹s Last Theorem, and his mistake is so basic there¹s a >name in logic for it: >Cum Hoc, Ergo Propter Hoc >And just so you know that I¹m not just tossing out some technical term >which may leave many of you befuddled because you don¹t know logic, >I¹ll explain carefully, and in detail how Wiles screwed up and why >basic logic says it must be so. >Now you may know that Wiles was looking to prove that modular forms >and elliptic curves are related. >Basically there are 4 numbers that you can use to describe a modular >describe an elliptic curve, and they noticed the for every set of 4 >numbers for a modular form they could Þnd an elliptic curve with the >It¹s a perfect setup for a logical error, as rather than do what¹s >necessary in such a situation, which is Þnd out what exactly the >relation is--if there is one--between modular forms and elliptic >curves, Wiles set out to compare between sets. >He set out to compare every elliptic curve against every modular form, >which is a logical error called Cum Hoc, Ergo Propter Hoc. >Computer scientists can understand the error by considering that >what¹s needed is a superclass which has the 4 numbers, which both >modular forms and elliptic curves belong to, but you see, that¹s not >what was done. >To nail it down that mathematicians DO make a logical error in >considering Wiles¹s work, consider that often you¹ll see the claim >that Wiles proved that in some sense modular forms and elliptic curves >Look up Cum Hoc, Ergo Propter Hoc, and see for yourself. >So why would mathematicians ignore a basic logical error? >I think it¹s because they can. >If I¹m wrong here I¹d like a cogent explanation as to why. I¹ve >brought this subject up before and noticed a lot of dancing around the >actual subject. Posters would make rather vague statements when I >could actually go look at Wiles¹s paper as it¹s now available on the >web and see they were basically saying b.s. and it seems to me that >often when challenged mathematicians, and especially sci.math posters, >say a lot of b.s. as if no one will check them. >What¹s the satisfaction though? >Don¹t any of you actually want math research that¹s truly correct >versus being some group fubar? >Don¹t ANY of you actually want math proofs without having to deal with >smart people who don¹t like you pointing out over and over again that >you¹re lying losers who when you can¹t actually prove something can >just congratulate yourselves anyway because you¹re weak lemmings? >You people follow your leaders at any and all costs, no matter how >wrong they are, how stupid their mistakes, no matter how wrong that >makes all of you. >You¹re rather pathetic. What can you possibly believe in? Do any of >you care about anything at all? >Does truth mean anything to any of you? >Don¹t any of you have souls? >James Harris You can¹t even get that right, can you? The logical error is POST hoc, ergo propter hoc- After this, therefore because of this, claiming that one thing is caused by another simply because it occurs immediately after it. Cum hoc, ergo propter hoc would be with this, therefore because of this and that ISN¹T a common enough error to be labled because there is no direction implied by with. Yes, we care about truth. You, on the other hand, have a history of repeatedly calling people liars when they disagreed with you only later to admit that you were wrong (and typically then claiming that you discovered your error yourself). You have, also, repeatedly told us that you were intentionally posting errors in order to get people to show you how to correct them so that you could get credit for it. Where did you misplace your own soul? === Subject: Re: Cum Hoc, Ergo Propter Hoc > You can¹t even get that right, can you? The logical error is POST hoc, ergo propter hoc- After this, therefore because of this, claiming that one thing is caused by another simply because it occurs immediately after it. > Cum hoc, ergo propter hoc would be with this, therefore because of this and that ISN¹T a common enough error to be labled because there is no direction implied by with. There is more than one error to be made in logic. True, _post hoc ergo proper hoc_ - the fallacy of assuming causation on the basis of sequential occurence - is more generally known, but that¹s not to say _cum hoc ergo propter hoc_ isn¹t a known error too - it is the fallacy of assuming causation on the basis of correlation. was what showed *me* that the unfamilar term did already exist and was not another of JSH¹s coinages. -- Larry Lard Replies to group please === Subject: Re: Cum Hoc, Ergo Propter Hoc more to the point, after is a temporal correlation, a with, it if means adjacently. > > Cum hoc, ergo propter hoc would be with this, therefore because of this and that ISN¹T a common enough error to be labled because there is no direction implied by with. > _cum hoc ergo propter hoc_ isn¹t a known error too - it is the fallacy > of assuming causation on the basis of correlation. http://www.rwgrayprojects.com/synergetics/plates/Þgs/plate02. html === Subject: Re: Cum Hoc, Ergo Propter Hoc >Looking over my threads I talk a good bit about sci.math¹ers who >dispute reality--living in their own little fantasy world--and >sci.math¹ers claim I live in my own little fantasy world, so what¹s >the answer? >Well it is a *math* newsgroup and you¹d think that math would be >enough, but I¹ve noticed repeatedly that certain posters say things >that go against rather basic mathematics, and somehow many of you seem >to suddenly forget basic mathematics long enough to believe them. >But it¹s not just a sci.math thing, and mathematicians rather >bizarrely ignore logic itself in claiming that Andrew Wiles found a >proof of Fermat¹s Last Theorem, and his mistake is so basic there¹s a >name in logic for it: >Cum Hoc, Ergo Propter Hoc >And just so you know that I¹m not just tossing out some technical term >which may leave many of you befuddled because you don¹t know logic, >I¹ll explain carefully, and in detail how Wiles screwed up and why >basic logic says it must be so. >Now you may know that Wiles was looking to prove that modular forms >and elliptic curves are related. >Basically there are 4 numbers that you can use to describe a modular >describe an elliptic curve, and they noticed the for every set of 4 >numbers for a modular form they could Þnd an elliptic curve with the >It¹s a perfect setup for a logical error, as rather than do what¹s >necessary in such a situation, which is Þnd out what exactly the >relation is--if there is one--between modular forms and elliptic >curves, Wiles set out to compare between sets. >He set out to compare every elliptic curve against every modular form, >which is a logical error called Cum Hoc, Ergo Propter Hoc. >Computer scientists can understand the error by considering that >what¹s needed is a superclass which has the 4 numbers, which both >modular forms and elliptic curves belong to, but you see, that¹s not >what was done. >To nail it down that mathematicians DO make a logical error in >considering Wiles¹s work, consider that often you¹ll see the claim >that Wiles proved that in some sense modular forms and elliptic curves >Look up Cum Hoc, Ergo Propter Hoc, and see for yourself. >So why would mathematicians ignore a basic logical error? >I think it¹s because they can. >If I¹m wrong here I¹d like a cogent explanation as to why. you¹re wrong. because he didn¹t just assert the two things were Œessentially the same¹ because the numbers happened to come out the same, he -proved- his assertion. [of course Œmodular forms and elliptic curves are THE correspondence between the two.] > I¹ve >brought this subject up before and noticed a lot of dancing around the >actual subject. Posters would make rather vague statements when I >could actually go look at Wiles¹s paper as it¹s now available on the >web and see they were basically saying b.s. and it seems to me that >often when challenged mathematicians, and especially sci.math posters, >say a lot of b.s. as if no one will check them. >What¹s the satisfaction though? >Don¹t any of you actually want math research that¹s truly correct >versus being some group fubar? >Don¹t ANY of you actually want math proofs without having to deal with >smart people who don¹t like you pointing out over and over again that >you¹re lying losers who when you can¹t actually prove something can >just congratulate yourselves anyway because you¹re weak lemmings? >You people follow your leaders at any and all costs, no matter how >wrong they are, how stupid their mistakes, no matter how wrong that >makes all of you. >You¹re rather pathetic. What can you possibly believe in? Do any of >you care about anything at all? >Does truth mean anything to any of you? >Don¹t any of you have souls? >James Harris ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Cum Hoc, Ergo Propter Hoc > Looking over my threads I talk a good bit about sci.math¹ers who > dispute reality--living in their own little fantasy world--and > sci.math¹ers claim I live in my own little fantasy world, so what¹s > the answer? > Well it is a *math* newsgroup and you¹d think that math would be > enough, but I¹ve noticed repeatedly that certain posters say things > that go against rather basic mathematics, and somehow many of you seem > to suddenly forget basic mathematics long enough to believe them. > But it¹s not just a sci.math thing, and mathematicians rather > bizarrely ignore logic itself in claiming that Andrew Wiles found a > proof of Fermat¹s Last Theorem, and his mistake is so basic there¹s a > name in logic for it: > Cum Hoc, Ergo Propter Hoc James, why don¹t you write up this error you think you¹ve spotted and try to get it published? If you¹re truly onto something, it might be a good idea! alex === Subject: Re: Cum Hoc, Ergo Propter Hoc > > Looking over my threads I talk a good bit about sci.math¹ers who > > dispute reality--living in their own little fantasy world--and > > sci.math¹ers claim I live in my own little fantasy world, so what¹s > > the answer? > > > > Well it is a *math* newsgroup and you¹d think that math would be > > enough, but I¹ve noticed repeatedly that certain posters say things > > that go against rather basic mathematics, and somehow many of you seem > > to suddenly forget basic mathematics long enough to believe them. > > > > But it¹s not just a sci.math thing, and mathematicians rather > > bizarrely ignore logic itself in claiming that Andrew Wiles found a > > proof of Fermat¹s Last Theorem, and his mistake is so basic there¹s a > > name in logic for it: > > > > Cum Hoc, Ergo Propter Hoc > James, why don¹t you write up this error you think you¹ve spotted and try to > get it published? If you¹re truly onto something, it might be a good idea! > alex I¹ve learned to talk out things until I¹m satisÞed LONG before I even think about writing a math paper. Much of what I¹m actually trying to do on sci.math is talk out ideas and get good critiques to evaluate them. Unfortunately I have some parasites who tend to mess with the system. Still I usually, eventually, get some on-point observations here and there from others. What I¹d like is for some airing out of this particular charge. I say that Wiles made a basic logical error which is essentially looking at what may be a coincidence and mapping what may be a LOT of coincidences to make a proof which turns out to be false. In the real world the logical error is rather commmon, and in the math world it¹s a little more interesting because in mathematics you can map INFINITE sets and still, amazingly enough, have the equivalent of a coincidence. If I¹m right, to me it¹s not nearly as big of a deal as you might think, though to others it should be. Besides at the rate I¹m going with my ideas, it could be quite a few years before my critique can drill through mathematicians¹ resistance, so for now, it¹s just fun to muse. James Harris === Subject: Re: Cum Hoc, Ergo Propter Hoc charset=iso-8859-1 [SNIP] > >James Harris > You can¹t even get that right, can you? The logical error is POST hoc, ergo propter hoc- After this, therefore because of this, claiming that one thing is caused by another simply because it occurs immediately after it. > Cum hoc, ergo propter hoc would be with this, therefore because of this and that ISN¹T a common enough error to be labled because there is no direction implied by with. SIGH! JSH may or may not be wrong about Wiles¹ proof, but you¹re wrong about the assertions you make about the non-existance of the error Cum Hoc, Ergo Propter Hoc http://www.fallacyÞles.org/cumhocfa.html === Subject: Heart equation Hi Is there an equation that describes a heart shape in polar coordinates - either exactly or roughly? I¹ve tried a series of sines but its rather crude. In this system 1 is equivalent to 360 degrees. === Subject: Re: Heart equation > Hi > Is there an equation that describes a heart shape in polar coordinates > - either exactly or roughly? I¹ve tried a series of sines but its > rather crude. > In this system 1 is equivalent to 360 degrees. I¹ll answer a different, yet related question: You might try in rectangular coordinates x^2 + (y - (x^2)^(1/3))^2 = 1 or parametrically x = cos(t), y = sin(t) + ((cos(t))^2)^(1/3) , -pi < t <= pi (I am avoiding the use of exponent 2/3 because a plotting program can misunderstand it.) === Subject: Re: Heart equation > Hi > Is there an equation that describes a heart shape in polar coordinates > - either exactly or roughly? I¹ve tried a series of sines but its > rather crude. > In this system 1 is equivalent to 360 degrees. A 2000 fr.sci.maths thread could perhaps help you : Hoping it helped, Raymond === Subject: Re: Heart equation === Subject: Re: Heart equation > Hi > Is there an equation that describes a heart shape in polar coordinates > - either exactly or roughly? I¹ve tried a series of sines but its > rather crude. > In this system 1 is equivalent to 360 degrees. I think you¹re describing what¹s called a Cardioid. The general form is r=a(1-cos(theta)). Dave === Subject: Re: need help with bell curve > > >i want to write a program that will approximate the bell curve.. > sorry i¹m not Þnding the best word to ask what i want to ask.. > yes i know the equation for it.. my problem is.. i don¹t wish to just > graph the well known equation.. i¹d like to have a computer > simulation.. graph it by.. some kind of repetitive-trials sort of > algorithm. þipping a coin over and over again sucks weener in my > opinion.. but maybe it¹s not so bad and i¹m just dealing with the > values obtained through that method in an inefÞcient way. > any help apprecitaed. How random is your computer random function? Do you need a bigger sample size? === Subject: Re: numerically solving simple equation >I need help in writing code to solve this equation : >Sum(i=1,n)(1/[N-(i-1)])=n/(N-a) >N unknown >n,a known If you clear out the denominators, you get a polynomial equation of degree n-1. There may be up to n-1 real solutions. In practice, for any n > 3 you should use numerical methods to solve it. 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: A strange constant perhaps, Avogadro¹s Number N_A..... charset=Windows-1252 | | ... | > | See, dude, very many ways to skin the cat....ahahaha....AHAHAHAHA.... | > | It¹s all just theorizing, = ing story telling about nature...that¹s | > all!.... | > | Well, I can¹t swear about the ing, but it¹s somewhere in there | too... | > | and they knew all this stuff above and its equations presented since | > | around the turn of the 2nd last century, Planck stuff since 1899.... | > | BUT then enter Einstein and the Juden physics ....ahahahaha... | > | with its immense propaganda machine......and we are still stuck in the | > | coul de sac of relativity with no apparent way out......ahahahaha..... | > | Read the hilarious, tormented twists and turns they take in their | blurbs | > | in xxx.lanl.gov, arXiv.org/, spr, etc, them trying to escape out of | their | > | rel-sac they maneuvered themselves into, by making grand math | > | assumptions, .... only to apply & use at ever turn and corner in their | > | ground-breaking works the simplest of the good, old fashioned | > | NEWTONIAN tools and his deÞnitions.......ahahahahaha.......which is | > | why their physics comedies made the eminent Professor Carver A. Mead | > | of Caltech assess the situation with: | > | It is my Þrm belief that the last seven decades of the twentieth | century | > | will be characterized in history as the dark ages of physics. ..... or | an | > | even more damming view was expressed in the sentiments of F.A Hayek, | > | a Nobel Price winner: | > | In the future, Humanity will see in our Epoch an Era of superstition, | > | essentially associated with the names of Marx, Freud and Einstein ! | to | > being a wave again--more like wavicles. Everything is pointing in that | > direction due to the discoveries of the last half of last century. | | There was a fellow on NPR the other day, that published a paper in which he | light... | URL:http://www.npr.org/features/feature.php?wfId=3804795 Here is the problem in a nutshell. IMHO, photons really are wavicles. The problem is why doesn¹t a photon disperse? What keeps it together? No one properties. This is not so critical now-a-days as there are ways for a wave to hang together ala phonon-like or soliton. The quantum vacuum has a natural boundary condition that does the trick. Yes, you will be able to but I would not call them a wavicle like I would a gauge boson. Gauge bosons are simply the manifestation of *bound* vacuum fermionic charge due to the real matter anomaly. FrediFizzx === Subject: Re: A strange constant perhaps, Avogadro¹s Number N_A..... ... back > | to > | > being a wave again--more like wavicles. Everything is pointing in > that > | > direction due to the discoveries of the last half of last century. > | There was a fellow on NPR the other day, that published a paper in which > he of > | light... > | URL:http://www.npr.org/features/feature.php?wfId=3804795 > Here is the problem in a nutshell. IMHO, photons really are wavicles. The > problem is why doesn¹t a photon disperse? What keeps it together? No one > properties. This is not so critical now-a-days as there are ways for a wave > to hang together ala phonon-like or soliton. The quantum vacuum has a > natural boundary condition that does the trick. Yes, you will be able to my properties > but I would not call them a wavicle like I would a gauge boson. Gauge > bosons are simply the manifestation of *bound* vacuum fermionic charge due > to the real matter anomaly. The only features that photons express as waves are the ability to self-interfere, and the towards and away from normal thing on entering/leaving a region with restricted propagation velocity. All þight. So lack of dispersion simply says that photons are no different than other wavicles. David A. Smith === Subject: Re: A strange constant perhaps, Avogadro¹s Number N_A..... > The only features that photons express as waves are the ability to > self-interfere, and the towards and away from normal thing on > entering/leaving a region with restricted propagation velocity. All > þight. So lack of dispersion simply says that photons are no different > than other wavicles. In total internal reþection, Þelds extend into the rare (low index) medium. Does that mean there are no photons there? Bill === Subject: Re: A strange constant perhaps, Avogadro¹s Number N_A..... N: > > The only features that photons express as waves are the ability to > > self-interfere, and the towards and away from normal thing on > > entering/leaving a region with restricted propagation velocity. All of > > þight. So lack of dispersion simply says that photons are no different > > than other wavicles. > In total internal reþection, Þelds extend into the rare (low index) > medium. Does that mean there are no photons there? Think of it like this. Not all of *any* photon can be said to be anywhere particularly until absorbed. So since any position is grey, then more so is any path. If a single photon travels through all slits in a diffraction grating, being a little bit in the rareÞed space is no stretch. And by no photons I am assuming you are referring to none of the internally reþected photons... David A. Smith === Subject: Need of Paper ! I am in need of below paper J. G. Wardrop. Some Theoretical Aspects of Road TrafÞc Research, Proceedings of the of the Institute of Civil Engineers, Pt. II, Vol. 1, pp. 325–378, 1952. If accesisble, send me a copy or do let me know. best ratnik === Subject: Re: imaginary solution to constraints of a real number system? or just a hoax >>> why can¹t we deÞne -3 squared to be -9? >> Because if >> -3 * -3 = -9 >> and >> +3 * -3 = -9 >> then >> -3 = +3 >> --Mark > which shows merely that negative numbers are just as ridiculous and > illogical as imaginary numbers In the rest of this thread you¹ve been arguing for your new rule for multiplying negative numbers. Now you¹re saying negative numbers are ridiculous and illogical. (What¹s illogical of course is your rule.) If you believe in negative numbers, I¹ve shown that your rule doesn¹t work. If you don¹t believe in them, stop wasting everyone¹s time with a rule you don¹t need. If you¹re going to troll, at least do it consistently. --Mark === Subject: Re: Uncountable sets in CZF? > In any model of ZFC, R is completely determined up to isomorphism. On the > other hand, given any model V of ZFC, and any two inÞnite sets A and B in > V, there exists a generic extension V[G] of V in which A and B are > bijective. Is this true? Do you have any references, or a quick argument? Because if it is true, doesn¹t this mean that Finlayson doesn¹t talk complete nonsense, when he insists that all inÞnite sets are alike? -- Herman Jurjus === Subject: Re: Uncountable sets in CZF? >> In any model of ZFC, R is completely determined up to isomorphism. On the >> other hand, given any model V of ZFC, and any two inÞnite sets A and B in >> V, there exists a generic extension V[G] of V in which A and B are >> bijective. >Is this true? Yes. >Do you have any references, or a quick argument? Let the set of forcing conditions be the set of bijections between Þnite subsets of A and Þnite subsets of B. For two conditions p and q, q is stronger than p if q contains p as a subset. This means that p and q are compatible iff p(x) = q(x) for all x in the intersection of dom(p) and dom(q), and p^{-1}(y) = q^{-1}(y) for all y in the intersection of range(p) and range(q). The union of a generic set G is a bijection between A and B in V[G]. >Because if it is true, doesn¹t this mean that Finlayson doesn¹t talk >complete nonsense, when he insists that all inÞnite sets are alike? Not at all. In any model of ZFC, there is more than one inÞnite cardinality (the class of inÞnite cardinalities is actually a proper class). So not all inÞnite sets are alike. SpeciÞcally, in any model of ZFC, R is uncountable, so R and N are not bijective. The closest that the claim that all inÞnite sets are alike comes to the truth is the fact that there are countable models of ZFC, and in such a model, all inÞnite sets are represented by countable sets. The bijections between such countable sets are not generally themselves represented by elements of the model. David And all dared to brave unknown terrors, to do mighty deeds, to boldly split inÞnitives that no man had split before - and thus was the Empire forged. ----- === Subject: Re: Uncountable sets in CZF? > >> In any model of ZFC, R is completely determined up to isomorphism. On the > >> other hand, given any model V of ZFC, and any two inÞnite sets A and B in > >> V, there exists a generic extension V[G] of V in which A and B are > >> bijective. > >Is this true? > Yes. > >Do you have any references, or a quick argument? > Let the set of forcing conditions be the set of bijections between Þnite > subsets of A and Þnite subsets of B. For two conditions p and q, q is > stronger than p if q contains p as a subset. This means that p and q are > compatible iff p(x) = q(x) for all x in the intersection of dom(p) and > dom(q), and p^{-1}(y) = q^{-1}(y) for all y in the intersection of > range(p) and range(q). The union of a generic set G is a bijection > between A and B in V[G]. > >Because if it is true, doesn¹t this mean that Finlayson doesn¹t talk > >complete nonsense, when he insists that all inÞnite sets are alike? > Not at all. In any model of ZFC, there is more than one inÞnite > cardinality (the class of inÞnite cardinalities is actually a proper > class). So not all inÞnite sets are alike. SpeciÞcally, in any model of > ZFC, R is uncountable, so R and N are not bijective. The closest that the > claim that all inÞnite sets are alike comes to the truth is the fact that > there are countable models of ZFC, and in such a model, all inÞnite sets > are represented by countable sets. The bijections between such countable > sets are not generally themselves represented by elements of the model. > David > And all dared to brave unknown terrors, to do mighty deeds, > to boldly split inÞnitives that no man had split before - > and thus was the Empire forged. Hi David, Progress! You agree that there are countable models of ZFC, and in such a model, all inÞnite sets are represented by countable sets? I¹m wary. Are not the powerset of the set of naturals or the set of reals sets in those models? Otherwise there must be some other justiÞcation of your refusal to accept any post-Cantorian logic. The set of reals is a set in ZFC, is that not so? How about ZF? What do you think about the notion that there can only be one or no proper classes? What do you think about a theory where some value, say, the proper class, is dually represented as zero and inÞnity? I have plentiful self-effacing humor. It does not support my effort to call others¹ nonsense without gently cajoling, nor to overreact, because I do not push with the weight of establishment, only with reason. The fallacy of Not at all is trivial. For some, an introduction to Cohen¹s forcing and Loewenheim and Skolem¹s theorems in this thread have changed their fundamental perceptions. I speak and write rightly and honestly, and with sense. Ross F. === Subject: Re: Uncountable sets in CZF? >>>In any model of ZFC, R is completely determined up to isomorphism. On the >>>other hand, given any model V of ZFC, and any two inÞnite sets A and B in >>>V, there exists a generic extension V[G] of V in which A and B are >>>bijective. >>Is this true? > Yes. >>Do you have any references, or a quick argument? > Let the set of forcing conditions be the set of bijections between Þnite > subsets of A and Þnite subsets of B. For two conditions p and q, q is > stronger than p if q contains p as a subset. This means that p and q are > compatible iff p(x) = q(x) for all x in the intersection of dom(p) and > dom(q), and p^{-1}(y) = q^{-1}(y) for all y in the intersection of > range(p) and range(q). The union of a generic set G is a bijection > between A and B in V[G]. But surely this only makes sense if V is countable? What if V is uncountable, and R Œin¹ V is also uncountable (also when looked at Œfrom the outside¹ so to say), then surely you can¹t extend V and cook up some bijection with N? That *would* be rather shocking. -- Herman Jurjus === Subject: Re: Uncountable sets in CZF? >>>>In any model of ZFC, R is completely determined up to isomorphism. On the >>>>other hand, given any model V of ZFC, and any two inÞnite sets A and B in >>>>V, there exists a generic extension V[G] of V in which A and B are >>>>bijective. >>>Is this true? >> Yes. >>>Do you have any references, or a quick argument? >> Let the set of forcing conditions be the set of bijections between Þnite >> subsets of A and Þnite subsets of B. For two conditions p and q, q is >> stronger than p if q contains p as a subset. This means that p and q are >> compatible iff p(x) = q(x) for all x in the intersection of dom(p) and >> dom(q), and p^{-1}(y) = q^{-1}(y) for all y in the intersection of >> range(p) and range(q). The union of a generic set G is a bijection >> between A and B in V[G]. >But surely this only makes sense if V is countable? What if V is >uncountable, >and R Œin¹ V is also uncountable (also when looked at Œfrom the outside¹ >so to say), then surely you can¹t extend V and cook up some bijection >with N? That *would* be rather shocking. As I see it, the model of ZFC is represented by a nonempty set (so V is a set as seen from the outside, but not a set in the model). The Œoutside¹ is itself a model of ZFC, and it may be possible that the Œoutside¹ can itself be extended generically to a model in which V is countable, and in such an extended model of the Œoutside¹, there is no need to worry. Of course, this will take more thought than these initial musings. For example, thinking about the size of the language, and the number of constant symbols, etc, may lead to other considerations. David ----- === Subject: Re: Uncountable sets in CZF? <87llgd3prz.fsf@phiwumbda.org> <87ekm212vo.fsf@phiwumbda.org> <87wtzrpq6r.fsf@phiwumbda.org> <2otdsjFdqemsU1@uni-berlin.de> Discussion, linux) >> In any model of ZFC, R is completely determined up to isomorphism. On the >> other hand, given any model V of ZFC, and any two inÞnite sets A and B in >> V, there exists a generic extension V[G] of V in which A and B are >> bijective. > Is this true? Do you have any references, or a quick argument? > Because if it is true, doesn¹t this mean that Finlayson doesn¹t talk > complete nonsense, when he insists that all inÞnite sets are alike? Well, I wouldn¹t say that. (*) R is never equinumerous (internally) with N in a given model M. (**) The set R-in-M is equinumerous (internally) with N-in-M in a generic extension M¹ of M. That¹s not much alike. (By the way, I don¹t know David¹s argument or even the meaning of generic extension. I¹m just taking the last question about Ross not talking complete nonsense. It¹s the easiest.) -- Not all features that are found on the Security tab are designed to help make your documents and Þles more secure. --Microsoft on OfÞce security features (after it was pointed out by a third party that a certain password setting is easily bypassed.) === Subject: Question Regard Cardinality and Its Implications Most people here are well familiar with the function f(x) = {1 if x is rational, 0 is x is irrational } integrates to 0 on any interval. The rationals are countable and the irrationals are uncountable, so the irrationals are a larger set than the rationals. possible for a set P to have the property that card N < card P < card R. If this is the case, would that imply f(x) = { 1 if x is in P, 0 if x is in R P } integrate to non-zero and Þnite on any non-empty interval? I¹m afraid I do not know enough about Lebesgue measures to answer the question for myself. === Subject: Re: Question Regard Cardinality and Its Implications > Most people here are well familiar with the function f(x) = {1 if x is > rational, 0 is x is irrational } integrates to 0 on any interval. The > rationals are countable and the irrationals are uncountable, so the > irrationals are a larger set than the rationals. > possible for a set P to have the property that card N < card P < card R. If > this is the case, would that imply f(x) = { 1 if x is in P, 0 if x is in R > P } integrate to non-zero and Þnite on any non-empty interval? I¹m afraid I > do not know enough about Lebesgue measures to answer the question for > myself. Unlike the other repliers, *I* say this question is independent of the axioms in ZFC. Certainly any set P with card P < c has *inner* measure zero. Under certain axioms (Martin¹s Axiom, for example) it follows that all sets with card P < c are Lebesgue measurable. So your function would be Lebesgue integrable, and have integral 0. But under other axioms (not so well-known, but proved consistent perhaps assuming large cardinals) there exist sets P subset [0,1] with card P < c but outer measure 1. In that case, your function is not Lebesgue integrable. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Question Regard Cardinality and Its Implications >Most people here are well familiar with the function f(x) = {1 if x is >rational, 0 is x is irrational } integrates to 0 on any interval. The >rationals are countable and the irrationals are uncountable, so the >irrationals are a larger set than the rationals. >possible for a set P to have the property that card N < card P < card R. that¹s the negation of ch, yes. >this is the case, would that imply f(x) = { 1 if x is in P, 0 if x is in R >P } integrate to non-zero and Þnite on any non-empty interval? no, it¹s not hard to show that the integral is still 0; a set of positive measure must have cardinality card R. > I¹m afraid I >do not know enough about Lebesgue measures to answer the question for >myself. ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Question Regard Cardinality and Its Implications > possible for a set P to have the property that card N < card P < card R. The continuum hypothesis is actually the opposite: all inÞnite subsets of R are countable or of the cardinality of the continuum, i.e. there is no such P. -- Aatu Koskensilta (aatu.koskensilta@xortec.Þ) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Question Regard Cardinality and Its Implications > Most people here are well familiar with the function f(x) = {1 if x is > rational, 0 is x is irrational } integrates to 0 on any interval. The > rationals are countable and the irrationals are uncountable, so the > irrationals are a larger set than the rationals. > possible for a set P to have the property that card N < card P < card R. If > this is the case, would that imply f(x) = { 1 if x is in P, 0 if x is in R > P } integrate to non-zero and Þnite on any non-empty interval? I¹m afraid I > do not know enough about Lebesgue measures to answer the question for > myself. That integral is the measure of the set P, which is zero, even without assuming the continuum hypothesis. Do a Google search for a sci.math thread called Continuum hypotheses and Lebesgue measure. Jose Carlos Santos === Subject: New mathematics/physical sciences positions at http://jobs.phds.org New job listings at http://jobs.phds.org - Jobs for PhDs List your job at no cost! http://jobs.phds.org/jobs/post * Ecosystem Analyst/Designer: BLUE: Land, Water, Infrastructure, North Carolina. Current Position Openings - Analysis and Design Focus Aquaculture Systems, Neighborhood Planning, Community Revitalization, Natural Systems Modeling, Infrastructure Design, Regional Growth Planning, Environmental Monitoring, Legal Boundary... * Become a NYC Teacher!: The New York City Teaching Fellows, NYC, NY. Who will remember yours? Become a NYC Teaching Fellow. 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They are looking to hire trader and analyst that is in the market daily; they want to see the opportunities there and you are to... === Subject: Re: The corruption of the Relativity cultist was: sci.math vs True Believers > Invariance is beyond their abilities to comprehend, as is the fact that > reciprocity in itself is sufÞcient to Þle-13 the special theory of > relativity. Reciprocity? Not meaning the Œinverse¹ nonsense? eleaticus === Subject: Re: The corruption of the Relativity cultist was: sci.math vs True Believers > >> go faster. There is no such thing in your beloved classical mechanics. > >That is the subject of the riddle: if you call cats dogs you can reasonable > >make erroneous statements about dogs. they > >meant anything about actual light, without the obvious qualiÞcations, is > >mindless. > Isn¹t it nice being able to totally misunderstand everything said to > you? It is reasonable for you take the position that supports your behavior. Are you trying to claim you have studies that show LIGHT taking on mass? > >> If you still insist Galilean transforms are correct even though there > >> are numerous counterexamples, I am forced to say you are a ing > >> idiot. > >I showed McNally¹s demonstration to contradict itself on two counts, and you > >can¹t refute it. > >I showed both linear and inverse-square equations in coordinates to be > >invariant and you can¹t refute that. > Do you think I care what you Œshow¹ when no matter what you say, at > the end of the day Galilean transforms do NOT work in physical > reality? Of course you care. We just have to count the posts you have made deprecating the simple math on the basis of experiment that isn¹t even relevant to the equations involved. Of course you care. Your True Believer dogma has been shown false. And you lack the honesty to even consider considering your new knowledge¹s application versus your True Belief about light. > >So, you (Gisse) obviously are once again þailing around. How dare you say my > >pastries suck! exclaimed the angry chef, my soups are perfect! > I wish I had a pony. Wouldn¹t it be nice if I had a pony? Wiggle your nose. Feel the fur? If you¹ll remove your head and the pony from the dark hole in which they are to be found, you could see your wish has proved true. eleaticus === Subject: Re: Existence of algorithm for algebras without quantors? |Please recall me whether trueness of an equation in an abstract |algebra with axioms without quantors can be algorithmically |checked. I¹m not sure exactly what you mean, but it sounds like you are asking about Þnitely axiomatized equational theories. The axioms are identities such as x(yz)=(xy)z involving certain operations. (You say without quantors, by which perhaps you mean without quantiÞers; identities here are presented without explicit quantiÞers, but implicitly what is meant is that for every x,y, and z in the structure, x(yz)=(xy)z. The identity is said to be quantiÞer free simply because the universal quantiÞcation is implicit.) There is no general procedure for determining whether a given identity is a consequence of a Þnite set of other identities. The obvious deductive rules (especially the rule that if an identity holds, the result of substituting for one of the variables is also an identity, and the rules for equality) are sufÞcient to prove the new identity if it is a consequence of the axioms. But there is no computable upper bound on how long the derivation might be. It¹s a problem equivalent to the halting problem for Turing machines. Given a Turing machine, we can construct a set of identities as axioms, and a given identity that is a consequence of the axioms if and only if the Turing machine halts. Conversely, determining whether a given identity is a consequence of a Þnite set of others is equivalent to determining whether a machine that searches for a derivation of the identity ever Þnds one. Keith Ramsay === Subject: Re: Gneralised Eisenstein Criterion |I was wondering if anyone knew of a generalization of Eisenstein¹s |criterion for irreducability of polynomials in Q[x,y] There seem to be a number of generalizations based on of them: http://citeseer.ist.psu.edu/179113.html Keith Ramsay === Subject: Re: When radius point is not accessible. I suppose you need a full circle. First mark four points N,S,E,W of column correctly. If possible,weld a very small 1/8 steel hook or drill a tiny hole at center of H-column. Draw arc of circle, as much as it is possible not obstructed by þange of H-sectioned beam. (They can be cut/Þlled up later on).For the remaining part transfer from the already drawn circle using cardboard or canvas or thick paper. > I work construction as a carpenter and need to layout interior > partition walls on a curve. More speciÞcally, this curve is a 25¹-0 > radius. The problem is that the focus point or pivot point to swing > the > radius is not accessible to me because it¹s position is located dead > center of an existing 10 H-column. > What is the best and most efÞcient way to proceed while trying to > maintain the integrity of the 25¹ radius? > Chris Grubb === Subject: A. SR-cult fraud and corruption Preamble -------------- Less than insightful opinion and later True Believer blindness has resulted in a number of fraudulent claims by Relativity cultists. Among the frauds are the twin original claims that Special Relativity is necessary because (a) the Newtonian-Galilean coordinate transformations do not work invariantly on Maxwell¹s electrodynamics equations and (b) that the famous Michelson-Morley experiment does not Þt the classical Newtonian-Galilean model but and only Þts Special Relativity. The most serious fraud is (c) the implied claim that the Lorentz-Einstein coordinate transformations of Special Relativity actually are applied to Maxwell. This series deals with various aspects of Relativity¹s fraudulent claims. By the way, look up the word Œcorrupt¹ in your dictionary if you think it only describes police and politicians who take bribes. ............................................................. ............... .............. ............................................................. ............... .............. One corrupt demonstration of non-invariance under the Newtonian-Galilean coordinate transformations debunked. ------------------------------------------------------------- --------------- -------------- ............................................................. ............... ...................................... t¹ = t, x¹ = x - vt, y¹ = y, z¹ = z, and the inverse transformation is given by t = t¹, x = x¹ + vt¹, y = y¹, z = z¹. Denoting partial differentiation with respect to a variable u by @/@u, the components of the 4-gradient operator transform as @/@t¹ = @/@t + v @/@x, @/@x¹ = @/@x, @/@y¹ = @/@y, @/@z¹ = @/@z. ............................................................. ............... ..................... McNally¹s point lies in the line @/@t¹ = @/@t + v @/@x. It says that although the function changes simply when the time coordinate is t, it changes in a more complicated, different manner when the time coordinate is t¹, the moving system coordinate. (Well, the supposed coordinate. The actual moving system time coordinate is t. Time is the same for all in the Newtonian theory McNally and Special Relativity is trying to trash with the garbage being dissected here. Let¹s take McNally¹s standard material step by step. t¹ = t, x¹ = x - vt, y¹ = y, z¹ = z, y¹=y and z¹=z are simpliÞcations to the special case where the Œprimed coordinate system, the moving system, is moving parallel to the x-axis and the moving system y¹ and z¹ coordinates are Œtransformed¹ to the same value as the Œstationary¹ system y and z values. x¹=x-vt is the standard formula for how the moving system coordinate corresponding to some x-value will change over time. t¹=t is the Þrst corrupt assertion by Relativity. It is not a simpliÞed transformation as is the case of y¹ and z¹. It is a lie. It is counter to actual Newtonian-Galilean theory which says the time is the same for everyone everywhere no matter how fast or slow they are moving. It says there is a time coordinate transformation required. SuperÞcially that transformation equation says time is the same for the moving system as for the stationary system, but it is a necessary corruption for the defense of the True Belief because it enables the corruption that follows. Now, let¹s examine the second corrupt act involved in the fraudulent claims we examine here. He says and the inverse transformation is given by t = t¹, x = x¹ + vt¹, y = y¹, z = z¹. Why on earth does he now want to discuss the Œinverse transformation? This puzzled me for years until just a day ago. Every time Relativity cultists do something strange it turns out to have a fraudulent purpose (not implying True Believers know what they are doing]. McNally¹s post Þnally enlightened me. It transforms x from an independent variable - one the observer or experimenter chooses at will - into a dependent variable. In x¹=x-vt one can say, what about x=0? x=-13.4, x=33? All without regard for some other variable¹s value. And so on for an inÞnite number of values. But x=x¹+vt¹ says it is x¹ and vt¹ that are the independent variables and - the main point - what we know is a freely choosable x is dependent of the value of both t¹ and x¹. But we know from real Newtonian theory (which is the subject of this post really) that time does not depend on your location, and we know from the corrupt, anti-theoretical t¹=t that t¹ and x have no relationship. So, the purpose of switching attention to the so-called inverse transformations is to deliberately or unknowingly/unthinkingly lie about x, say its value depends on t¹ and x¹. The effect of this corrupt action shows up in the next step: He says Denoting partial differentiation with respect to a variable u by @/@u, the components of the 4-gradient operator transform as @/@t¹ = @/@t + v @/@x, @/@x¹ = @/@x, @/@y¹ = @/@y, @/@z¹ = @/@z. Differentiating is Þnding out how much an equation/function changes when some input into it changes. Differentiating y=4x tells us that however much you change x, y changes 4 times as much. We can write this idea as dy/dx=4. Partial differentiation is the same except that we use a different, fancier Greek d to show that there is more than one part of the way the function changes. That fancy d isn¹t available so we use the @ sign which turned onto its right side looks a little bit like the d we want. This line is the target result of the two corrupt steps the standard/McNally: @/@t¹ = @/@t + v @/@x, It says that changes in the function when t¹ changes are due to changes in t and x. That is the same as saying that t¹ changes when t and x change but we already know t¹ does not change when x changes. dt¹/dx=0. That is obvious from t¹=t. By reduction to the absurd, the standard Œproof¹ of classical transformations¹ invalidity is proved invalid. You shall know the tree by its fruit. And this standard/McNally Special Relativity-tree assault on the Newtonian-Galilean coordinate transformations bears rotten fruit. The facts bout the Newtonian-Galilean transformations. ------------------------------------------------------------- --------------- --------- There are many more rotten fruit from the SR-tree to be buried before you know the nature of their whole orchard. Look for other posts in this series. In the meantime, focus well on negative comments. Are the replies actually responsive. Do they rant about gravity, or how Relativity is proved correct a million times each day, or some other Œwe are proved right¹ rave that doesn¹t deal details in the debunking done here? For one thing, it is typically General Relativity or items about the energy and mass of moving objects that are being waved at you, and such items are completely irrelevant. Just ask them for a list of all the observations that have been made of the shortening of moving objects that Special Relativity says always occurs. eleaticus === Subject: Re: A. SR-cult fraud and corruption > Preamble > -------------- > Less than insightful opinion and later True Believer blindness has > resulted in a number of fraudulent claims by Relativity cultists. > Among the frauds are the twin original claims that Special Relativity > is necessary because (a) the Newtonian-Galilean coordinate > transformations do not work invariantly on Maxwell¹s electrodynamics > equations and (b) that the famous Michelson-Morley experiment does > not Þt the classical Newtonian-Galilean model but and only Þts > Special Relativity. > The most serious fraud is (c) the implied claim that the > Lorentz-Einstein coordinate transformations of Special Relativity > actually are applied to Maxwell. > This series deals with various aspects of Relativity¹s fraudulent > claims. > By the way, look up the word Œcorrupt¹ in your dictionary if you think > it only describes police and politicians who take bribes. > ............................................................. ................ ............. > ............................................................. ................ ............. > One corrupt demonstration of non-invariance under the > Newtonian-Galilean coordinate transformations debunked. > ------------------------------------------------------------- ---------------- ------------- > ............................................................. ................ ..................................... > t¹ = t, > x¹ = x - vt, > y¹ = y, > z¹ = z, > and the inverse transformation is given by > t = t¹, > x = x¹ + vt¹, > y = y¹, > z = z¹. > Denoting partial differentiation with respect to a variable u by > @/@u, > the components of the 4-gradient operator transform as > @/@t¹ = @/@t + v @/@x, > @/@x¹ = @/@x, > @/@y¹ = @/@y, > @/@z¹ = @/@z. > ............................................................. ................ .................... > McNally¹s point lies in the line @/@t¹ = @/@t + v @/@x. It says that > although the function changes simply when the time coordinate is t, > it changes in a more complicated, different manner when the time > coordinate is t¹, the moving system coordinate. (Well, the supposed > coordinate. The actual moving system time coordinate is t. Time is the > same for all in the Newtonian theory McNally and Special Relativity is > trying to trash with the garbage being dissected here. > Let¹s take McNally¹s standard material step by step. > t¹ = t, > x¹ = x - vt, > y¹ = y, > z¹ = z, > y¹=y and z¹=z are simpliÞcations to the special case where the > Œprimed coordinate system, the moving system, is moving parallel to > the x-axis and the moving system y¹ and z¹ coordinates are > Œtransformed¹ to the same value as the Œstationary¹ system y and z > values. > x¹=x-vt is the standard formula for how the moving system coordinate > corresponding to some x-value will change over time. > t¹=t is the Þrst corrupt assertion by Relativity. It is not a > simpliÞed transformation as is the case of y¹ and z¹. It is a lie. It > is counter to actual Newtonian-Galilean theory which says the time is > the same for everyone everywhere no matter how fast or slow they are > moving. > It says there is a time coordinate transformation required. > SuperÞcially that transformation equation says time is the same for > the moving system as for the stationary system, but it is a necessary > corruption for the defense of the True Belief because it enables the > corruption that follows. > Now, let¹s examine the second corrupt act involved in the fraudulent > claims we examine here. > He says and the inverse transformation is given by > t = t¹, > x = x¹ + vt¹, > y = y¹, > z = z¹. > Why on earth does he now want to discuss the Œinverse transformation? > This puzzled me for years until just a day ago. Every time Relativity > cultists do something strange it turns out to have a fraudulent > purpose (not implying True Believers know what they are doing]. > McNally¹s post Þnally enlightened me. > It transforms x from an independent variable - one the observer or > experimenter chooses at will - into a dependent variable. In x¹=x-vt > one can say, what about x=0? x=-13.4, x=33? All without regard for > some other variable¹s value. And so on for an inÞnite number of > values. But x=x¹+vt¹ says it is x¹ and vt¹ that are the independent > variables and - the main point - what we know is a freely choosable x > is dependent of the value of both t¹ and x¹. > But we know from real Newtonian theory (which is the subject of this > post really) that time does not depend on your location, and we know > from the corrupt, anti-theoretical t¹=t that t¹ and x have no > relationship. > So, the purpose of switching attention to the so-called inverse > transformations is to deliberately or unknowingly/unthinkingly lie > about x, say its value depends on t¹ and x¹. > The effect of this corrupt action shows up in the next step: > He says Denoting partial differentiation with respect to a variable u > by @/@u, the components of the 4-gradient operator transform as > @/@t¹ = @/@t + v @/@x, > @/@x¹ = @/@x, > @/@y¹ = @/@y, > @/@z¹ = @/@z. > Differentiating is Þnding out how much an equation/function changes > when some input into it changes. Differentiating y=4x tells us that > however much you change x, y changes 4 times as much. We can write > this idea as dy/dx=4. > Partial differentiation is the same except that we use a different, > fancier Greek d to show that there is more than one part of the way > the function changes. That fancy d isn¹t available so we use the @ > sign which turned onto its right side looks a little bit like the d we > want. > This line is the target result of the two corrupt steps the > standard/McNally: @/@t¹ = @/@t + v @/@x, > It says that changes in the function when t¹ changes are due to > changes in t and x. That is the same as saying that t¹ changes when t > and x change but we already know t¹ does not change when x changes. > dt¹/dx=0. That is obvious from t¹=t. > By reduction to the absurd, the standard Œproof¹ of classical > transformations¹ invalidity is proved invalid. > You shall know the tree by its fruit. And this standard/McNally > Special Relativity-tree assault on the Newtonian-Galilean coordinate > transformations bears rotten fruit. > The facts bout the Newtonian-Galilean transformations. > ------------------------------------------------------------- ---------------- -------- > There are many more rotten fruit from the SR-tree to be buried before > you know the nature of their whole orchard. Look for other posts in > this series. > In the meantime, focus well on negative comments. Are the replies > actually responsive. Do they rant about gravity, or how Relativity is > proved correct a million times each day, or some other Œwe are proved > right¹ rave that doesn¹t deal details in the debunking done here? For > one thing, it is typically General Relativity or items about the > energy and mass of moving objects that are being waved at you, and > such items are completely irrelevant. > Just ask them for a list of all the observations that have been made > of the shortening of moving objects that Special Relativity says > always occurs. > eleaticus I think you need to learn the difference between covariance and invariance. Maxwell¹s equations are covariant under Galilean transformations, but are invariant under Lorentz transformations. That¹s why Lorentz transformations occupy a special place in SR. === Subject: Re: A. SR-cult fraud and corruption Special Relativity http://scienceworld.wolfram.com/physics/SpecialRelativity.html http://math.ucr.edu/home/baez/physics/Relativity/SR/ experiments.html === Subject: Re: A. SR-cult fraud and corruption > Let¹s take McNally¹s standard material step by step. > t¹ = t, > x¹ = x - vt, > y¹ = y, > z¹ = z, > y¹=y and z¹=z are simpliÞcations to the special case where the > Œprimed coordinate system, the moving system, is moving parallel to > the x-axis and the moving system y¹ and z¹ coordinates are > Œtransformed¹ to the same value as the Œstationary¹ system y and z > values. > x¹=x-vt is the standard formula for how the moving system coordinate > corresponding to some x-value will change over time. > t¹=t is the Þrst corrupt assertion by Relativity. It is not a > simpliÞed transformation as is the case of y¹ and z¹. It is a lie. It > is counter to actual Newtonian-Galilean theory which says the time is > the same for everyone everywhere no matter how fast or slow they are > moving. ------------------------------------------------------------- --------------- ------ Idiot, t=t¹ is pure Galilean Transformation. SR is t¹=(t-v/c^2 z)/(1-v^2/c^2)^1/2. ------------------------------------------------------------- --------------- ------ > It says there is a time coordinate transformation required. ------------------------------------------------------------- --------------- ------ No it says no time transformation is required. ------------------------------------------------------------- --------------- ------ [snip idiotic crap] > Now, let¹s examine the second corrupt act involved in the fraudulent > claims we examine here. > He says and the inverse transformation is given by > t = t¹, > x = x¹ + vt¹, > y = y¹, > z = z¹. > Why on earth does he now want to discuss the Œinverse transformation? > This puzzled me for years until just a day ago. Every time Relativity > cultists do something strange it turns out to have a fraudulent > purpose (not implying True Believers know what they are doing]. > McNally¹s post Þnally enlightened me. > It transforms x from an independent variable - one the observer or > experimenter chooses at will - into a dependent variable. In x¹=x-vt > one can say, what about x=0? x=-13.4, x=33? All without regard for > some other variable¹s value. And so on for an inÞnite number of > values. But x=x¹+vt¹ says it is x¹ and vt¹ that are the independent > variables and - the main point - what we know is a freely choosable x > is dependent of the value of both t¹ and x¹. > But we know from real Newtonian theory (which is the subject of this > post really) that time does not depend on your location, and we know > from the corrupt, anti-theoretical t¹=t that t¹ and x have no > relationship. ------------------------------------------------------------- --------------- ------ You want the inverse transformation to a) transform back to the original frame b) its quiet handy if you consider a frame which moves in the opposide direction What you have written above is senseless . ------------------------------------------------------------- --------------- ------- [snip unread] mw === Subject: Re: A. SR-cult fraud and corruption >>> Let¹s take McNally¹s standard material step by step. >Eleaticus does not even have the courtesy to spell my name correctly. Oops. Sorry. eleaticus === Subject: Re: A. SR-cult fraud and corruption x-mimeole: Produced By Microsoft MimeOLE V6.00.2800.1441 > >>> Let¹s take McNally¹s standard material step by step. > >Eleaticus does not even have the courtesy to spell my name correctly. > Oops. Sorry. No problem. Just publish a picture of you, so we can all get overwhelmed by a sudden feeling of compassion. Don¹t you dare? Dirk Vdm === Subject: Re: A. SR-cult fraud and corruption >Of course it did not enlighten Eleaticus. A person like Eleaticus, who is >determined not to be educated, and who is determined to remain as closed- >minded as always, will never be enlightened by anything that anybody >out a way to Þt in the inverse transformation with the rest of his >prejudices. Eleaticus will always start with the assumption that he is >completely right about everything, and then he will consider what other >people write with respect to how much they agree with the prejudices that >he holds. I have no problem with the inverse transformation as a means of verifying the correctness of the simple algebra of getting from the unprimed values to the primed. But there already is a method of verifying the formula: the formula. Apply it to the primed values (which includes the recognition that v¹=-v) and see if the unprimed are returned. This would prevent the sililness of treating independent variables as if they were dependent. eleaticus === Subject: Re: A. SR-cult fraud and corruption x-mimeole: Produced By Microsoft MimeOLE V6.00.2800.1441 So, mental case Oren Webster, when do we get that picture of yours? Dirk Vdm === Subject: Re: A. SR-cult fraud and corruption >Of course, my principal reason for introducing t¹ was so that partial >differentiation with respect to t (holding x, y, z constant) and partial >differentiation with respect to t¹ (holding x¹, y¹, z¹ constant) may >be distinguished (as indeed they must necessarily be). That is complete nonsense. The only reason for the false assertion that there is not just a t but a t¹ is to enable the nonsense result you get, that t¹ is a function of both t and x. This being evident in your @/@t¹=@/@t+v@/@x, which absurdly contradicts both your explicit t=t¹ and the obvious @t/@x=0, from t¹=t. Just look at velocity. It actually does transform from one system to the other, being in the reverse direction. v¹=-v. So, why not assert that transform? Because the phoney t¹=t has already served the purpose of falsely indicting the Galileans, and v¹=+-v would indict the LET. eleaticus === Subject: Re: A. SR-cult fraud and corruption >I would like to point our here that when I introduced the transformation > t¹ = t, > x¹ = x - vt, > y¹ = y, > z¹ = z, >I explicitly stated that this was the Galilean Transformation, so for >Eleaticus to claim the transformations as relativistic was dishonest >in the light of the fact that I had already explicitly stated that the >transformation was Galilean (and not relativistic). Talk about strawmen! I never even hinted at such a thing. eleaticus === Subject: Re: A. SR-cult fraud and corruption >Preamble >-------------- >Less than insightful opinion and later True Believer blindness has >resulted in a number of fraudulent claims by Relativity cultists. >Among the frauds are the twin original claims that Special Relativity >is necessary because (a) the Newtonian-Galilean coordinate >transformations do not work invariantly on Maxwell¹s electrodynamics >equations and (b) that the famous Michelson-Morley experiment does >not Þt the classical Newtonian-Galilean model but and only Þts >Special Relativity. >The most serious fraud is (c) the implied claim that the >Lorentz-Einstein coordinate transformations of Special Relativity >actually are applied to Maxwell. I didn¹t see Maxwell¹s equations anywhere. Why is that? === Subject: Re: A. SR-cult fraud and corruption >>Preamble >>-------------- >>Less than insightful opinion and later True Believer blindness has >>resulted in a number of fraudulent claims by Relativity cultists. >>Among the frauds are the twin original claims that Special Relativity >>is necessary because (a) the Newtonian-Galilean coordinate >>transformations do not work invariantly on Maxwell¹s electrodynamics >>equations and (b) that the famous Michelson-Morley experiment does >>not Þt the classical Newtonian-Galilean model but and only Þts >>Special Relativity. >>The most serious fraud is (c) the implied claim that the >>Lorentz-Einstein coordinate transformations of Special Relativity >>actually are applied to Maxwell. >I didn¹t see Maxwell¹s equations anywhere. You are such an idiot. The part of the preamble yuu deleted said there would be a series, and the subject says A. - as ini A., B., C., etc. eleaticus === Subject: Re: A. SR-cult fraud and corruption >>>Preamble >>>-------------- >>>Less than insightful opinion and later True Believer blindness has >>>resulted in a number of fraudulent claims by Relativity cultists. >>>Among the frauds are the twin original claims that Special Relativity >>>is necessary because (a) the Newtonian-Galilean coordinate >>>transformations do not work invariantly on Maxwell¹s electrodynamics >>>equations and (b) that the famous Michelson-Morley experiment does >>>not Þt the classical Newtonian-Galilean model but and only Þts >>>Special Relativity. >>>The most serious fraud is (c) the implied claim that the >>>Lorentz-Einstein coordinate transformations of Special Relativity >>>actually are applied to Maxwell. >>I didn¹t see Maxwell¹s equations anywhere. >You are such an idiot. The part of the preamble yuu deleted said there >would be a series, and the subject says A. - as ini A., B., C., etc. >eleaticus You are yet to show Maxwell¹s equations. You never show Maxwell¹s equations. You don¹t understand Maxwell¹s equations. === Subject: Re: A. SR-cult fraud and corruption > >> > >>>Preamble > >>>-------------- > >>>Less than insightful opinion and later True Believer blindness has > >>>resulted in a number of fraudulent claims by Relativity cultists. > >>> > >>>Among the frauds are the twin original claims that Special Relativity > >>>is necessary because (a) the Newtonian-Galilean coordinate > >>>transformations do not work invariantly on Maxwell¹s electrodynamics > >>>equations and (b) that the famous Michelson-Morley experiment does > >>>not Þt the classical Newtonian-Galilean model but and only Þts > >>>Special Relativity. > >>> > >>>The most serious fraud is (c) the implied claim that the > >>>Lorentz-Einstein coordinate transformations of Special Relativity > >>>actually are applied to Maxwell. > >> > >> > >>I didn¹t see Maxwell¹s equations anywhere. > >You are such an idiot. The part of the preamble yuu deleted said there > >would be a series, and the subject says A. - as ini A., B., C., etc. > >eleaticus > You are yet to show Maxwell¹s equations. > You never show Maxwell¹s equations. > You don¹t understand Maxwell¹s equations. Never argue with an idiot. He¹ll drag you down to his level and beat you with experience. Dirk Vdm === Subject: Re: A. SR-cult fraud and corruption >Never argue with an idiot. He¹ll drag you down to his >level and beat you with experience. >Dirk Vdm I usually realise that after 6 pages of arguing. I have 2 rules for conÞrmed idiots: 1) If the research required takes more than 5 minutes, or needs me to grab a book thats not within reach - I will not do it. 2) I won¹t get worked up over the idiocy. === Subject: Re: A. SR-cult fraud and corruption >>Never argue with an idiot. He¹ll drag you down to his >>level and beat you with experience. >>Dirk Vdm > I usually realise that after 6 pages of arguing. > I have 2 rules for conÞrmed idiots: > 1) If the research required takes more than 5 minutes, or needs me to > grab a book thats not within reach - I will not do it. > 2) I won¹t get worked up over the idiocy. ------------------------------------------------------------- --------------- ----- I¹ll do it just to train myself. mw === Subject: Re: A. SR-cult fraud and corruption > eleaticus Would you be so kind to publish a picture of yourself on your website? Dirk Vdm === Subject: Re: A. SR-cult fraud and corruption >> eleaticus > Would you be so kind to publish a picture of > yourself on your website? > Dirk Vdm ------------------------------------------------------------- --------------- --------- Why the hell do you want to have a picture of him? mw === Subject: Re: A. SR-cult fraud and corruption > >> eleaticus > > Would you be so kind to publish a picture of > > yourself on your website? > > Dirk Vdm > ------------------------------------------------------------- ---------------- -------- > Why the hell do you want to have a picture of him? See ... and of course... the PINS! ;-)) Dirk Vdm === Subject: matrix formula for the surface area of n-dimensional ellispoid Can you give or point me to the correct formula to compute surface area of an n-dimensional ellipsoid. For my particular application I prefer a matrix form formula. Furthermore, is there a form which does not include an integration term (similar as for the volume of ellipsoid which contains Gamma-functions, but no integral terms)? What would be the best book for this kind of geometric and matrix topics? Joni === Subject: Sum in Harmionic Series Is there a formula for sum of n terms of harmonic progression [1/a,1/(a+d),1/(a+2d), ... 1/(a+(n-1)d)]? When n-> Inf there may be a result with Log and Euler¹s number. === Subject: Re: Sum in Harmionic Series >Is there a formula for sum of n terms of harmonic progression >[1/a,1/(a+d),1/(a+2d), ... 1/(a+(n-1)d)]? When n-> Inf there may be a >result with Log and Euler¹s number. If we rewrite your sum as n-1 1 --- 1 - > ------- d --- a/d + k k=0 and put this together with the results from http://www.whim.org/nebula/math/partharm.html we get that your sum is equal to 1 Gamma¹(a/d + n) Gamma¹(a/d) - ( --------------- - ----------- ) d Gamma(a/d + n) Gamma(a/d) where Gamma is the the log convex function such that Gamma(1) = 1 and x Gamma(x) = Gamma(x+1). As n -> oo, we get that your sum is asymptotic to 1 Gamma¹(a/d) 1 - ( log(n) - ----------- ) + O( - ) d Gamma(a/d) n Euler¹s gamma constant appears in the standard Harmonic series because Gamma¹(1) = -gamma. Rob Johnson take out the trash before replying === Subject: Re: Sum in Harmionic Series > Is there a formula for sum of n terms of harmonic progression > [1/a,1/(a+d),1/(a+2d), ... 1/(a+(n-1)d)]? When n-> Inf there may be a > result with Log and Euler¹s number. David Cantrell gives the correct formula in terms of digamma function. So, any representation for the digamma function can be adjusted to get your sum. For example: For d not 0, and a/d does not equal integer <= 0: sum{k=0 to n-1} 1/(a +k*d) = (1/d) *integral{0 to 1} (x^(a/d -1) -x^(a/d +n-1))/(1-x) dx = (also) n * sum{k=0 to oo} 1/((a +d*k)(a +d*(n+k)) Leroy Quet === Subject: Re: Sum in Harmionic Series > Is there a formula for sum of n terms of harmonic progression > [1/a,1/(a+d),1/(a+2d), ... 1/(a+(n-1)d)]? In terms of the digamma function, the sum is 1/d * ( digamma(a/d + n) - digamma(a/d) ). IIRC, you¹re a Mathematica user. Is there something you don¹t like about the answer that Mathematica gives for this sum? David === Subject: Re: Sum in Harmionic Series the silence when it comes to Harmonic series related sums (then, PolyGamma was not taught, or was not available, or I persued Engineering). Arithmetic series and geometric series have neat sum formulae, but not H.P. . However, geometrically by contrast, when representing means of two quantities, a Þgure of a circle with rays as secants and tangents from a Þxed outside point is drawn to get neat circles in it to represent geometric mean,arithmetic mean and harmonic mean as well. With algebraic harmonic sums,picture gets perturbed,even there is a big debate about irrationality of EulerGamma etc. Mma gives for a large number N[Sum[1/i, {i,1,10^5}]-Log[10^5]]= 0.577716 approximately for C or EulerGamma. But it is not what bothered me. Narasimham > > Is there a formula for sum of n terms of harmonic progression > > [1/a,1/(a+d),1/(a+2d), ... 1/(a+(n-1)d)]? > In terms of the digamma function, the sum is > 1/d * ( digamma(a/d + n) - digamma(a/d) ). > IIRC, you¹re a Mathematica user. Is there something you don¹t like about > the answer that Mathematica gives for this sum? > David === Subject: Re: Binomial Theorem for X^n + Y^n > Very interesting observations. It can further be simpliÞed. > As an example, write x^5 + y^5 = Q5/16 > Then x^5 + y^5 = Q5/2^(5-1) > So in general, x^m + y^m = Qm/2^(m-1) > Now, the challenge is to write Qm in more compact form so that it can > be recognized that Qm/(2^m-1) cannot be an m-th power of an integer > without applying FLT. If A = B [(A+A)/2]^3 + [(A-A)/2]^3 = [(A+A)/2]^3 + 0 = A^3 [(A+A)/2]^m + 0 = A^m If B = n*A [(A + n*A)/2]^m + [(A - n*A)/2]^m = [(A*(n+1))/2]^m + [(A*(1-n))/2]^m ... x+y = A x-y = B (A+B)/2 = x (A-B)/2 = y Let B = A + K A+B = 2A+K A-B = -K [(2A+K)/2]^m + [(-K)/2]^m x^m + y^m = (A + K/2)^m + (-K/2)^m === Subject: Re: how to prove that f^2+f Œ ^2 <=1 if ... schrieb eric detourre : > Here is the following problem : > Prove that: f^2 + f Œ ^2 <= 1 > when f is twice differentiable over R > and f^2 <= 1 > and f Œ ^2 + f ^2 <= 1 (1 is not less than the sum > of the square of the Þrst and second derivative of f). > I don¹t have the slightest idea on how to assert this, and this result > (which happened to be an exercise in a french engineering school) > may be false (but i found no counter-example) > as it is false when replacing R with any subset of R. The result is correct, but my proof is rather longish. It is based on the usual procedures to solve the ordinary differential equation g¹ ^2 + g¹¹ ^2 = 1, but you have to be very careful! I welcome any suggestions for improvement! So, we want to show f(x)^2 + f¹(x)^2 <= 1 for all points x in R. In case that x is a critical point of f, i.e. f¹(x) = 0, the claim follows from f(x)^2 <= 1. So let x now be a regular point of f: x in U := {y in R | f¹(y)<>0 } U is an open subset of R, hence a disjoint union of open, non-empty Intervals I_i. For every such interval, the function f is a stronly monotonic function on the closure cl(I_i), thus bijective, and we can write the derivative f¹ as a function F_i of f on cl(I_i): f¹ = F(f) (I dropped the index - the interval will now be called I, the function F). Then f¹¹ = F¹(f) * F(f), and the inequality f¹ ^2 + f¹¹ ^2 <= 1 yields F(f)^2 + F¹(f)^2 * F(f)^2 <= 1 F¹(f)^2 * F(f)^2 <= 1 - F(f)^2 The next step will only work on intervals I on which f¹ does not take on the values +/- 1. -- Division by (1 - F(f)^2) >(!!) 0: | F(f) * F¹(f) / sqrt(1 - F(f)^2 ) | <= 1 | d/df [ sqrt(1 - F(f)^2 ) ] | <= 1 (*) Now comes the part, where we have to use that f is deÞned and bounded on *all* of R [at least, we need to use it when I is an unbounded interval]: Given I=I_i and a Þxed point x in I, there are for every epsilon > 0 points x- = x-(epsilon) and x+ = x+(epsilon) such that x- in I, x+ in I, x- <= x <= x+ and f¹(x-)^2 <= epsilon and f¹(x+)^2 <= epsilon [If I = (a,b) is Þnite from from below, a <> -oo, existence of x- follows from f¹(a)=0, in case a = -oo, existence of x- follows from the fundamental theorem of differential calculus, applied to f on the interval [x-2/sqrt(epsilon),x]: f¹(x-) = (f(x) - f(x-2/sqrt(epsilon)) )/( 2/sqrt(epsilon) ) --> |f¹(x-)| <= 2/( 2/sqrt(epsilon) ) = sqrt(epsilon) Similarly for x+.] Now, integrate Inequality (*) over the interval [f(x-),f(x)] or [f(x+),f(x)], depending on whether f is monotonically de- or increasing. Let¹s take the case that f is decreasing. Let¹s also say, that epsilon has been taken to be smaller than F = f¹(x)^2. Then: sqrt(1 - epsilon) - sqrt(1 - F(f(x))^2 ) <= f(x-) - f(x) <= 1 - f(x) sqrt(1 - F(f(x))^2 ) >= sqrt(1 - epsilon) - 1 + f(x) The same result follows if f is increasing on I from integration over [f(x+),f(x)]. Now take the limit epsilon to 0: sqrt(1 - F(f(x))^2 ) >= f(x) i.e.: sqrt(1 - f¹(x)^2 ) >= f(x) That is one half of the desired inequality. For the other half, sqrt(1 - f¹(x)^2 ) >= -f(x), you will have to start again at (*), but this time integrate it over [f(x+),f(x)] if f is decreasing on I, resp. over [f(x-),f(x)] if f is increasing: Example: sqrt(1 - epsilon) - sqrt(1 - F(f(x))^2 ) <= f(x) - f(x-) <= -(-f(x)) - 1 sqrt(1 - F(f(x))^2 ) >= sqrt(1 - epsilon) - 1 + (-f(x)) Here ends the proof of f(x)^2 + f¹(x)^2 <= 1 for x in connected components I = I_i of the set of regular values of f, on which f¹ does not take on either of the values 1 and -1. If I = I_i is an open interval on which f¹ becomes 1 or -1, do the following: Since I is an interval of regular values of f, we may w.l.o.g. assume that f¹>0 on I, and thus that f¹ takes on the value +1 somewhere on I. [The other case is analogous.] The remaining open set I minus {y in I | f¹(y) = 1} decomposes again into a disjoint union of non-empty open intervals J_j. We will show that {y in I | f¹(y) = 1} can only consist of a single point, that at this point f is zero, and that in conclusion f will be a sine function (up to translation). f^2 + f¹ ^2 = 1 follows automatically. So, let x be a point in I with f¹(x) = 1. Fix a positive constant epsilon > 0. Then there will again be points x- < x < x+ in I such that f¹(x-) = epsilon and f¹(x+) = epsilon. Let¹s say, x- lies in J_1, and x+ lies in J_2, J_1 disjoint from J_2. Then, take the inequality f¹ ^2 + f¹¹ ^2 <= 1, say on J_1, and rewrite it: | f¹¹ / sqrt(1-f¹ ^2) | <= 1 [Recall that f¹ ^2 <> 1 on J_1!] | d/dx ( arcsin(f¹) ) | <= 1 Integrate over an interval [z,sup(J_1)) with z in J_1 intersected with [x-,x]: --> arcsin(1) - arcsin(f¹(z)) <= b - z with b:= sup(J_1) f¹(z) >= sin( pi/2 - (b-z)) as long as 0 < (b-z) < pi/2 (**) But since x- is a point, where f¹ takes a value very near to zero, it follows from (**) that f¹(z) >= sin( pi/2 - (b-z)) *and* 0 < (b-z) < pi/2 holds [up to order O(epsilon)] for *all* points in the interval (a,b) := (x-,x) intersected J_1. Thus the difference f(b) - f(a) will be f(b) - f(a) = int_a^b f¹(z) dz >= int_(b-pi/2)^b sin( pi/2 - (b-z)) dz = 1 - 0 --> f(x) >= f(b) [since x >= b] >= f(a) + 1 >= -1 + 1 Of course, there are still some terms of order O(epsilon) in the last inequality, but at this point epsilon can go to zero, and we obtain f(x) >= 0. Analogously, but argueing over the interval (x,+x) intersect J2, one obtains f(x) <= 0, thus f(x) = 0, and in addition, all inequalities met inbetween must be equalities, so in particular, the complement of {f¹=1} in I must not contain any other components than J_1 and J_2, and {f¹=1} itself must be of measure zero. Thus I = J_1 u {x} u J_2, and f must be identical to z |-> sin(z-x) on I. This ends the proof. PFFFF.... -- I¹m exhausted! Sorry, but I don¹t have the nerve to proofread the whole text right now. Maybe later... Thomas === Subject: Re: What am I studying here? Ken, A related subject, perhaps not quite what you¹re looking for, is > called the geometry of numbers. It might give you a lead. The geometry of numbers does perhaps go a little beyond what I¹m doing, but === Subject: Re: What am I studying here? > Can anyone give me any pointers to what exactly it is I¹m studying here? :) > I am exploring properties of the *Þnite* set of 2D integer vectors > generated by the matrix multiplication > {a c)(i) modulo (D,D) where D=ad-bc > {b d)(j) > for all i,j in Z, where a,b,c,d are Þxed integers, and D<>0. > I know the matrix is a linear operator, and I know some basic group theory. > Just some names of things I can look up the web would be enough. I¹d think of it as group theory problem. Instead of thinking Z x Z, let Z_D be the cyclic group of order D, and think of G = Z_D x Z_D with group operation pointwise addition, being generated by <(0,1),(1,0)>. Note that for any g,h in Z_D x Z_D, A(g+h) = Ag + Ah; so A induces an endomorphism of Z_D x Z_D. Thus, since g in G > g = n(1,0) + m(0,1), Ag = nA(1,0)+mA(0,1); i.e., AG is the subgroup of elements of the form m(a,b)+n(c,d) for integers m,n. Use the fact that gcd(ad,D) = gcd(bc,D) and resulting facts to Þgure out the exact form of AG, which will be isomorphic to either Z_i or Z_i x Z_j, where i,j divide D. Just some thoughts... === Subject: Re: What am I studying here? Chas, > > {a c)(i) modulo (D,D) where D=ad-bc > > {b d)(j) > I¹d think of it as group theory problem. Yes, that¹s what I was doing. :) > Instead of thinking Z x Z, let Z_D be the cyclic group of order D, > and think of G = Z_D x Z_D with group operation pointwise addition, > being generated by <(0,1),(1,0)>. Note that for any g,h in Z_D x Z_D, > A(g+h) = Ag + Ah; so A induces an endomorphism of Z_D x Z_D. Endomorphism is a useful term to know. > Thus, since g in G > g = n(1,0) + m(0,1), Ag = nA(1,0)+mA(0,1); i.e., > AG is the subgroup of elements of the form > m(a,b)+n(c,d) > for integers m,n. Use the fact that gcd(ad,D) = gcd(bc,D) and > resulting facts to Þgure out the exact form of AG, which will be > isomorphic to either Z_i or Z_i x Z_j, where i,j divide D. I had Þgured this out (and of course that ij=D in the latter case), but haven¹t yet Þgured out a simple way to determine when it can be the former. At the very least, I have now discovered for myself that Z_i x Z_j is isomorphic to Z_{ij} when gcd(i,j)=1. :) I hope your hint that gcd(ad,D) = gcd(bc,D) will enable me to Þnd (i,j). --------------- The application, BTW, is the enumeration of two-glider collisions in CA Rule 110. Of course I could trivially enumerate them just by deÞning a dull algorithm which lists all of the members of AG in some arbitrary order according to some Þxed algorithm. But the mathematician in me wants to Þnd the simplest possible parametrization of a canonical enumeration (with as much elegant symmetry as possible), ie a short list of parameters and a Þxed formula which allows me to map from (my) (i,j) to the corresponding ordinal (or perhaps pair of ordinals) in the enumeration of AG. I am *not* asking for anyone to solve my problem. The fun is doing it for myself. :) === Subject: Help : Tangent Bundles What is the explicit isomorphism : T(T*M) = TM + T*M (direct sum) where M is a symplectic manifold, T and T* are tangent and cotangent bundles. === Subject: Re: Help : Tangent Bundles > What is the explicit isomorphism : ^^^^^ > T(T*M) = TM + T*M > (direct sum) > where M is a symplectic manifold, > T and T* are tangent and cotangent bundles. Is there supposed to be a canonical one? What properties shall this splitting have? Here is a family of non-canonical examples: What you are looking for is basically a linear connection on the cotangent bundle, so for every Riemannian metric compatible with the symplectic form you could take the Levi-Civita connection of this metric. The horizontal space TM of your splitting will then given by tangent vectors to curves in T*M that result from parallel transports with respect to LC. The vertical space is deÞned as the kernel of the differential of the projection T*M -> M. === Subject: Re: Help : Tangent Bundles > > What is the explicit isomorphism : > ^^^^^ > > T(T*M) = TM + T*M > > (direct sum) > > where M is a symplectic manifold, > > T and T* are tangent and cotangent bundles. > Here is a family of non-canonical examples: > What you are looking for > is basically a linear connection on the cotangent bundle, > so for every Riemannian metric compatible with the symplectic form > you could take the Levi-Civita connection of this metric. > The horizontal space TM of your splitting > will then given by tangent vectors to curves in T*M > that result from parallel transports with respect to LC. Could you please explain this : TM is already attained from tangent vectors to curves in M, how should tangent vectors to curves in T*M be imagined and why don¹t they give more information then TM ? How exactly do you use the L.C. to show this ? > The vertical space is deÞned as the > kernel of the differential of the projection > T*M -> M. So this is actually the zero section of T*M in T(T*M) ? === Subject: Re: Help : Tangent Bundles >> > What is the explicit isomorphism : >> ^^^^^ >> > >> > T(T*M) = TM + T*M >> > >> > (direct sum) >> > where M is a symplectic manifold, >> > T and T* are tangent and cotangent bundles. >> Here is a family of non-canonical examples: >> What you are looking for >> is basically a linear connection on the cotangent bundle, >> so for every Riemannian metric compatible with the symplectic form >> you could take the Levi-Civita connection of this metric. >> The horizontal space TM of your splitting >> will then given by tangent vectors to curves in T*M >> that result from parallel transports with respect to LC. > Could you please explain this : TM is already attained from tangent > vectors to curves in M, how should tangent vectors to curves in T*M > be imagined and why don¹t they give more information then TM ? > How exactly do you use the L.C. to show this ? I tried to be brief in my Þrst answer, partly because I didn¹t know whether my explanation was what you are aiming at, and partly because I do not fully understand the issue, either. However, I hope I understand it good enough to answer your question. >> The vertical space is deÞned as the >> kernel of the differential of the projection >> T*M -> M. > So this is actually the zero section of T*M in T(T*M) ? No. The kernel of the differential of the projection of a Þbre bundle at a point p in the bundle is the tangent space to the Þbre. In this case, since the Þbre is a vector space, it is isomorphic to the vector space T*_p M. Let¹s be a little bit more concrete about what you are after: You want for every point p in T*M (*not* in M) an identiÞcation of of the tangent space to T*M at this point with the direct sum of a tangent space to M at the point m with the cotangent space to M at the same point m, where m = pi(p) is the image of p under the projection pi: T*M -> M. Well, the space T*_m(M) can, as said before, be seen as naturally isomorphic to the kernel of the differential of the projection pi. (This is so, because every Þbre bundle, especially T*M is locally a product of Þbre and base.) pi: T*M --> M p |--> m so the differential D(pi) at point p is a linear map D(pi)_p: T_p(T*M) --> T_m(M), of which we know the kernel to be T*_m(M), and we could conclude T_p(T*M) isomorphic to [T*_m(M) direct sum T_m(M)] if we had (for every point p in T*M) a horizontal subspace H(p) of T_p(T*M) such that the restriction of D(pi)_p was an isomorphism. This now, is one of the deÞnitions of a *connection* in a vector bundle -- a collection of horizontal subspaces H(p), complementary to all Þbers. Given this deÞnition one can deÞne *horizontal paths* in this vector bundle (paths in T*M whose tangents lie in directions of H), one can show that every path in the base M of the vector bundle can be lifted to horizontal paths in the vector bundle, which are unique up to choice of their starting points), and one can deÞne parallel transport in the vector bundle: To every path in the base M one has a diffeomorphism of the two Þbers over starting- and end-point of the path. Well, and if all parrallel transports deÞned from a connection on a vector bundle are linear maps of the Þbres (which are *vector spaces*), then such a connection is called *linear*, and linear connections are in 1-1 correspondence to covariant derivatives in the vector bundle. And of course, LC induces a covariant derivative on cotangent vectors, thus induces a linear connection H on the vector bundle T*M -> M, thus induces a splitting. As a reference, I¹d recommend A.Besse Einstein manifolds, Chapter 9 about Riemannian submersions. === Subject: Re: Analytic functions of two variables >Has it occured to you that you may be answering at least twice as >often to yourself as the other way round? Isn¹t that disturbing? not in this case. my self-reply corrected an -error- in what i¹d written - you agree that errors should be corrected, right? ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Differential Geometry..... hello.....doctor~ Prove that a curve of constant geodesic curvature on a sphere is a circle. ------------------------------- um.......i know that geodesic curvature = kg radius of small circle = r radius of great circle = R geodesic curvature of small circles is kg = k*sin@ (@ : angle between plane containing small circle and plane containing great circle) sin@ = sqrt[(R^2) - (r^2)] / R k = 1/r so, kg = sqrt[(R^2) - (r^2)] / (r*R) thus, geodesic curvature of circles is constant. but i must show that other curve but circle is not constant. how do you show this ? help me, please~ thank you very much. always. === Subject: Re: Measurable Þleds of Hilbert spaces >>Hi all, >>Is there a generalization of measurable Þelds of (separable) Hilbert >>spaces to, let¹s say, measurable Þelds of (arbitrary) Banach spaces? >>References? (if online so much the better). >>G. Rodrigues >> >Most likely the answer is yes. Further details depend on what exactly >you mean by a measurable Þeld of Hilbert spaces. A bundle of Hilbert spaces is a surjective map p:Elongrightarrow X such that each Þber p^{-1}(x) is a Hilbert space. Now assume X is a measure space, then: A measurable Þeld Hilbert spaces is a bundle of Hilbert spaces together with a countable set of sections {s_n}<=Gamma (a pervasive sequence) such that: i) H_x={s_n(x)} is dense in the Þber p^{-1}(x) (=> the Þbers are all separable) ii) The function x|-> is measurable. A section s of p is measurable iff the function x|-> is measurable for every n. I can think of a rather straightforward generalization to Banach spaces but then it is problematic in proving that the norm of a measurable operator (an operator taking measurable sections into measurable sections) is measurable and, if you ask measurability of the norm of an operator, then it is problematic to prove that the norm of the composition of two operators is measurable. Hope this clears thing up a bit. G. Rodrigues === Subject: Re: measurable functions > Let $X$ be a measure space with measure $mu$ and $mu(X)=1$. > is it really true that the space of measurable functions $u:Xto mathbb{R}$ > is > a topological space with basis deÞned by the following neighborhoods: > $U_{r,sigma}(v)={umid mu(xmid |u(x)-v(x)|1-sigma}$ ? > Oleg. Isn¹t this called the topology of convergence in measure? One warning is that this topology is not Hausdorff: if u and v agree almost everywhere, then u is in every neighborhood of v. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: measurable functions >Let $X$ be a measure space with measure $mu$ and $mu(X)=1$. >is it really true that the space of measurable functions $u:Xto mathbb{R}$ is >a topological space with basis deÞned by the following neighborhoods: >$U_{r,sigma}(v)={umid mu(xmid |u(x)-v(x)|1-sigma}$ ? In a topological space we usually say base, not basis. There are only two requirements for a collection B of subsets of a set S to be a base for a topology: 1) each member of s is in some member of B 2) given u in A_1 intersect A_2 with A_1, A_2 in B, there is A_3 in B contained in A_1 intersect A_2 and containing u. 1) is easy, 2) is slightly tricky. Note that, using the fact that {x: |u(x) - v(x)| < r} is the union of {x: |u(x) - v(x)| < r - 1/n} for positive integers n > 1/r, if mu{x: |u - v| < r} > 1 - sigma, then there is some n such that mu{x: |u - v| < r - 1/n} > 1 - sigma + 1/n. So if v_3 is in the intersection of U_{r_1,sigma_1}(v_1) and U_{r_2,sigma_1}(v_2), taking n that works as above for u=v_3, v=v_1, r=r_1, sigma=sigma_1 and also for u=v_3, v=v_2, r=r_2, sigma=sigma_2, then A_3 = U_{1/n,1/n}(v_3) should work. 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: A Question on Luzin¹s Theorem THEOREM Let $f(x)$ be a measurable function (in Lebesgue¹s sense) on $Esubset mathbb{R}$ and almost everywhere bounded over $E$. Then $forall delta>0$, $exists F subset E$, that is closed, and (i) $mu (E subset F)0$ and every $E_i$, we can Þnd closed set $F_isubset E_i$ such that [m(E_isetminus F_i) < frac{delta}{p},quad i=1,2,dots,p] because $f(x)=c_i$ when $xin F_i$, $f(x)$ is continous on $F_i$. And $F_1,dots,F_p$ are disjoint, we know that $f$ is continous on [F=bigcup_{i=1}^p F_i.] It is easy to see that $F$ is closed and [mu(Esetminus F) = sum_{i=1}^p mu(E_isetminus F_i) < sum_{i=1}^p frac{delta}{p} = delta] Now we come back to the general case. Since [tex]g(x) = frac{f(x)}{1+|f(x)|}[/tex] ([tex]f(x) = frac{g(x)}{1-|g(x)|}[/tex]) we can suppose $f(x)$ is bounded without loss of generality. There exists measurable simple functions ${phi_k(x)}$ converges to $f(x)$ uniformly on $E$ (See also Theorem 1.17, Real and Complex Analysis, Rudin). For any given $delta > 0$ and each $phi_k(x)$ we can Þnd closed set $F_ksubset E$ such that [mu(Esetminus F_k) < frac{delta}{2^k},] and $phi_k(x)$ is continuous on $F_k$. Let [F = bigcap_{k=1}^infty F_k,] we have $Fsubset E$ and [mu(Esetminus F) leq sum_{k=1}^infty m(Esetminus F_k) < delta] Since every $phi_k(x)$ is continuous on $F$ and from the uniformly convergence, we know that $f(x)$ is continuous on $F$. Q.E.D My question is: Where is Egorov¹s Theorem used in the proof above? === Subject: Re: Automorphisms of groups === Subject: Re: Automorphisms of groups >I guess in the zero ring R = {0}, 0 = 1 since 0 + 0 = 0 ; 0*0 = 0. Yes, that¹s one way to deÞne rings and identities, perhaps the smoother. Whence one has the option to express 1 /= 0 as non-trival ring with identity. -- >LOL! But sadly, true. I don¹t get a warm feeling about MS either. Updates are worse than blind dates. >How is it that the marketing people took over and made all the >money off the ideas that the geeks had when fooling with computers? Get revenge, install Linux. >I am glad to see that the money is everything philosophy is >a hard sell to a segment of computer people, and we still have >FSF and Sourceforge, and many others with great free software. mega mania merged monsters In our deregulated, out of control, ultra-modern, hyper-commercial, socially engineered society of the dehumanizing capitalists¹ cultural revolution wherein the measure of all things is money, are you an anybody corporate dog wage slave, within that compulsively competitive, spiritually and socially bankrupt, free trade, media-prattle reality? >I recall when the only Navigator was Netscape--well 1st there was >Mosaic and that guy at Cern (was it CERN? bad memory). Anyway, here >comes MS with Inet Explorer, tried to wipe out Netscape, and did a >pretty good job, though I guess Mozilla is a free continuation. Windows is a diaster, use unix. >If not for these people providing free stuff, we would be >getting charged outrageous prices for everything, browsers, >every program--well, I could go on, but that is enough for now. State of Oregon considered using open source software for it¹s data network as it would cost less, cost less to maintain, provide data security. Leasing from MicroS..., yes leasing, couldn¹t buy, was costly recuring expense. All needed maintaince and modiÞcations, could only be done by MicroSoft, of course at great expense. If all of this wasn¹t bad enuf, if payments of leasing fees were not forthcoming, then MicroSoft from it¹s ofÞce on high, would block access to all of Oregon¹s data across the state network. I conclude from this, that MicroSoft also has access to all of Oregon¹s records. The measure introduced into state legislature to use open source software, went nowhere. Yes, Oregon legislature is as corrupt and beholding to corporate demands as congress, the White Lie House and other states. -- Riddle of the day Why couldn¹t the mathematican fall to sleep counting sheep? ---- === Subject: Re: Easy number theory problem === Subject: Re: Easy number theory problem Four proofs of Fermat¹s theorem: >Yes. If o(x) | p-1, then x^(p-1) = 1. >Same for o(x) | phi(n) in general Z*_n. -- 1 7.5.7. Lemma. Let p be a prime number, and let k,a,b be integers. (a) If 1 <= k <= p-1, then p is a divisor of the binomial coefÞcient p!/k!(p-k)! (a + 1)^p = sum(k=0,p) pCk a^(p-k) = a^p + 1 (mod p) 0^p = 0 (mod p) If a^p = a (mod p), then (a + 1)^p = a^p + 1 = a + 1 (mod p) By induction, for all a in Z+, a^p = a (mod p) If p odd, a < 0, then a^p = -(-a)^p = --a = a (mod p) If p = 2, a < 0, then a^p = (-a)^p = -a = a (mod 2) For all a /= 0 (mod p), a^(p-1) = 1 (mod p) -- 2 -- 3 >>3) In a group G = Z*_p (I will not show that this is a group), >>we have o(x) | p-1 so x^(p-1) = 1 (mod p) for all x in Z*_p. > g in G iff g unit Z_p > if g,h in G: g,h units Z_p; gh unit; gh in G > 1 identity in G; for all g in G, some h in G with gh = 1. -- 4 >> pZ is maximal ideal, thus Z_p = Z/pZ is a Þeld. >> Hence Z_p^* = Z_p0 is a multiplicative group of order p-1. >> Or if you prefer, pZ is a prime ideal in an integral domain, >> hence maximal ... >4) with maximal ideals pZ is easy. >I have not seen it before that I recall. It¹s an easy theorem to proof: If R is a non-trivial communitative ring with identity and I a proper ideal of R, then R/I is a Þeld iff I is maximal -- Z_mn^* = Z_m^* x Z_n^* >>>Now I want to show that f extends to a group isomorphism of Z*_mn >>>with Z*_m x Z*_n. This should be straightforward it seems, though >>>I have never done it. i.e. to show >>>f : Z*_mn --> Z*_m x Z*_n : f(z + mnZ) = (z + mZ, z + nZ) >> The restriction of f to Z_mn^* is what you¹re wanting. >> if u unit in Z_mn, then some v with uv = 1 >> (1,1) = f(1) = f(uv) = f(u)f(v), f(u) unit in Z_m x Z_n >> f(u) = (u_m, u_n); f(v) = (v_m, v_n); u_m = u + mZ >> (1,1) = (u_m v_m, u_n v_n); u_m, u_n units in Z_m, Z_n >> f(u) in Z_m^* x Z_n^* >> We already know f is injection and multiplicative homomorphism. >> What¹s left is surjectiveness. >> If u_m, u_n units in Z_n, Z_m >> some v_m, v_n in Z_m, Z_n with u_m v_m = 1 = u_n v_n >> some r,s with f(r) = (u_m, u_n), f(s) = (v_m, v_n) >> f(rs) = f(r).f(s) = (u_m, u_n)(v_m, v_n) = (u_m v_m, u_n v_n) = >> (1,1) >> Thus rs = 1 as f injection, and r is unit. >f(r) = (1 + mZ,1 + nZ) with (m,n) = 1, ==> r = 1 + mnZ in Z/mnZ. >Ker(f) = (1) in Z_mn. (Should make some decision and clean up >notation.) Let g = f|Z_p^*, the restriction of f to Z_p^* ker f = { 0 } in Z_mn, ring homomorphism ker g = { 1 } in Z_mn^*, multiplicative group homomorphism g is injection as restrictions of injections are injective. >==> f is into (1-1), and as you showed, its a homomorphism >of Þnite groups, so its onto, and an isomorphism. No, an injective homomorphism between Þnite groups is surjection iff the groups are equinumerous. Tho you know Z_mn and Z_m x Z_n are equinumerous, you do not know if Z_mn^* and Z_m^* x Z_n^* are. Hence the proof of surjection is needed. Indeed, that¹s needed to show phi(mn) = phi(m).phi(n), which you don¹t have yet until you prove the surjection. ---- === Subject: Re: Easy number theory problem === Subject: Re: Easy number theory problem >p.55 Chap 11, Primitive Roots Web address? >Polynomials over Z/pZ (p prime). >Prop. 1) A nonzero polynomial f in Z/pZ[x] has at most deg(f) >(distinct) elements a in Z/pZ such that f(a) = 0. >I won¹t reproduce the proof of this here. A polynomial p in F[x] for any Þeld F, has at most deg p solutions to p(x) = 0. >Prop. 2) Let d | p-1 . Then f = x^d - 1 has exactly d solutions. >Proof: Let p-1 = de. Then x^(p-1) - 1 = (x^d)^e - 1 >= (x^d - 1) g(x) where g(x) = (x^d)^(e-1) + (x^d)^(e-2) + ... + 1, >so deg(g) = d(e-1) = p - 1 - d. Prop. 1 ==> g has at most >n_g <= p - 1 - d roots, and x^d - 1 has at most n_d <= d roots, >we know that x^(p-1) - 1 has p-1 roots in Z/pZ, i.e. the elements >of Z*_p == (Z/pZ){0}. Thus n_g + n_d = p-1 which means that >n_g = p - 1 - d and n_d = d. >Somewhere the fact that Z/pZ is a Þeld enters Prop 1. -- >Lemma. If a and b are in an Abelian group with |a| = r, |b| = s, >and (r,s) = 1, then |ab| = rs. This is trival, (r,s) = 1 not needed. rs = |a||b| = |ab| Are you thinking [r,s] = |rs|/(r,s). BTW, have you notice making corrections takes more time than proof reading? -- >Theorem. For any prime p, Z*_p is a cyclic group of order p-1. >Proof: Let p-1 = q_1^e_1 ... q_r^e_r. >Let q^e = (q_i)^(e_i) for some i. Confusing. Let q = q_i, e = e_i for some i. >Prop 2 ==> x^(q^e) - 1 has exactly q^e roots, and >x^[q^(e-1)] - 1 has q^(e-1) roots ==> >there is a solution a_i of x^(q^e) - 1 = 0 such that >(a_i)^[q^(e-1)] != 1 ==> >|a_i| = (q_i)^(e_i) for each factor of (q_i)^(e_i) in p - 1. >Let a = a_1 ... a_r. Then |a| = p - 1 by the Lemma, and How¹s lemma used? p-1 is positive, pick positive numbers for q¹s. BTW, how do you factor -4 in cannonical form? 2^2 or (-2)^2 or 2(-2) ? >Z*_p is cyclic. How so? Compare For a in Z_p, let d_a = d be the smallest for which a^d = 1. Show o(a^j) = d iff (j,d) = 1. Thus o(a) = d ==> |{ x | o(x) = d}| = phi(d) For all d | p-1, |{ x | o(x) = d}| = 0 or phi(d) sum(d | p-1) phi(d) = p - 1 Thus for all d | p-1, some a with o(a) = d. In paticular, p-1 | p-1, thus some a with o(a) = p-1 This generalizes to show Þnite Þelds are cyclic. Compare 7.5.7(d) 7.5.6. Corollary. Let G be a Þnite abelian group. If a in G is an element of maximal order in G, then the order of every element of G is a divisor of the order of a. 7.5.7. Lemma. Let p be a prime number, and let k,a,b be integers. (a) If 1 <= k <= p-1, then p is a divisor of the binomial coefÞcient p!/k!(p-k)! (b) If k >= 1 and a is congruent to b (mod p^k), then a^p is congruent to b^p (mod p^{k+1} ). (c) If k >= 2 and p is an odd prime, then (1 + ap)^(p^(k-2)) = 1 + ap^(k-1) (mod p^k) (d) If p is an odd prime and p is not a divisor of a, then (1 + ap)^(p^(k-1)) = 1 (mod p^k) (1 + ap)^(p^(k-2)) /= 1 (mod p^k) 7.5.8 Theorem. Let p be an odd prime, let k be a positive integer, and let n=p^k. Then Z_n^x is a cyclic group. -- >The number of primitive roots of Z*_p is phi(phi(p)) = phi(p-1), >since there are phi(p-1) generators of Z_(p-1) =~ Z*_p, >= numbers in Z_(p-1) which are prime relative to p-1. >i.e. prime relative to p-1 when Z_p-1 is written as an >additive group, not as a multiplicative cyclic group. e.g. >Z*_13 = <2> =~ Z_12 has 2^n, with n = 1,5,7,11 as generators, >since (n,p-1) = (n,12) = 1. ---- === Subject: spot rate bloomberg does anybody know how bloomberg computes spot rates for the benchmark curve used in the bond yield compute ? === Subject: Re: Partial difference equation, primes > It¹s faster than Mathematica¹s prime counting function over I think a > decent range, though at the top of Mathematica¹s range it is faster. This from someone who is allegedly a professional programmer! Hint: Being faster only up to a certain point doesn¹t win you any prizes at all Even more hint: A lookup table beats your algorithm over ŒI think a decent range¹... -- Larry Lard Replies to group please === Subject: Re: Partial difference equation, primes James Harris a .8ecrit : > > Ok, so I still don¹t know what you call a partial difference > > equation. But don¹t worry, it has no importance. > I handled that topic with a separate thread. > But it¹s not complicated--no matter what a sci.math¹er might try to > insinuate--as a partial difference equation is the discrete analog to > a partial differential equation. Ok. So, dS(x,y) is what you call a PDE. > There is no other known in recorded history used to count prime > numbers besides my dS(x,y) and yes, I¹m talking about all of human > history here. You might be right. Who would be moronic enough to compute pi(y) - pi(y-1) in order to check the primality of y? > To me that¹s a simple enough claim that it should either be refutable > by you sci.math¹ers, or if you¹re sane, you¹ll quit trying to > insinuate that it¹s not a big deal. > I fear you¹re not exactly what most people would call sane. Do you mean that I could write things like I am the barbarian at the gates...? :-) > > Do you understand that the fact that your formula contains or not > > a PDE has absolutely no impact on the way we could use it to compute > > pi(x)? Do you understand that with many formulas containing > > something like F(x,y), one could claim that F(x,y) is a PDE? Of course, > > one could do it but what would it change? > You¹re lying because you¹re insinuating that mathematicians have in > fact used a partial difference equation to count prime numbers when > they have not. No. I wanted to be sure that what you call a PDE is also what I call a PDE. Experience proves that, with you, the meaning of a word can vary from any-thing to almost-all. > It¹s not complicated here. You may think you can parse the language > to fool everyone on sci.physics or that the other sci.math¹ers will > just go along with you as they have for over two years now, but it > doesn¹t change the facts. > I say that no one in recorded human history has used a partial > difference equation to count prime numbers. > That is a fairly straightforward claim, and if it¹s not true you can > just give some other partial difference equation, but instead you try > to spin the facts, like what in the hell is F(x,y)? F(x,y) is any function I can call a PDE. So, with something like z(x,y) = Sum(i=a to y, F(x,i)), trivially I have F(x,y) = z(x,y) - z(x,y-1) and it allows me to claim that all the stuff is new because there is a PDE. Neat, isn¹t it? Yes, I read The art of bullting by JSH. > I¹ve seen how you sci.math¹ers operate for over two years now, and you > have a contempt for the truth, and a demonstrated contempt for > mathematics. > > > > > > > > > That¹s only with the pure math implementation. > > > > > > > > !? > > > > > > Shown above. > > What I meant was that Œpure math implementation¹ is a mere nonsense. > Yet people can see that it is not nonsense most dramatically by > looking at it, and I¹ve put it on my blog: Have you a dictionnary at hand? Search for Œimplementation¹. Maybe you will learn something today. > > > Posters continually use the world algorithm in a derisive manner, > > > when actually the prime counting function is just a formula. > > Ok. So, if a program based on your formula is slow it¹s not inherent > > in your formula but it is due to computer scientists who are quite > > unable to efÞciently program it :-) > No. But then, what about your claim according to which your method leads to the fastest possible prime counting? > > > Algorithms can be *derived* from it, but it¹s no more an algorithm > > > than > > > > > > e = mc^2 > > > > > > though some nutcase *could* call that an algorithm, if they were > > > trying to argue that it was not important. > > > > > > But it¹s a formula, not an algorithm. > > > > > > Algorithms are based off of formulas, not the other way around. > > > > > > > > You can also move to an explicit representation, for instance, > > > > > > > > > > dS(N,2) = N/2 - 1, with even N, > > > > > > > > > > dS(N,3) = þoor((N-4)/6), if N is even, and N>2, and > > > > > > > > > > dS(N,5) = þoor((N-16)/10) - þoor((N-16)/30), N even, and N>6, while > > > > > > > > > > dS(N,7) = þoor((N-8)/14) - þoor( (N-22)/42) - þoor((N-106)/70) + > > > > > þoor( (N-106)/210) - 2, N even, N>36, > > > > > > > > > > and now, what gets put on the stack now? > > > > > > > > Huh? Isn¹t dS(N,2), dS(N,3), dS(N,5), dS(N,7), ... a list? And why > > > > do you index it with prime values (which, btw, makes useless your > > > > (p(y,sqrt(y)) - p(y-1,sqrt(y-1))) whereas you claim you do not need > > > > a list of primes? > > > > > > The compressed explicit prime counting function exists as I¹ve shown. > > Which one makes use of an IMPLICIT list of values. > That¹s what it looks like. But then, since your technology makes use of a list, what about your claim ŒIt does not need a list¹? > The math is simply rigid here, no matter how much you might like to > make more out of it. > It just is. > > > Notice it too is a formula and not an algorithm. > > > > > > It just so happens that¹s what it looks like when you have it take > > > into account that N is even and 2 is prime. > > > > > > The math is rigid. > > Yes and that¹s your main problem. That¹s precisely because the math is > > rigid that you¹re almost always wrong. If making math was singing, you > > would sing out of tune. > Then give a single wrong point I¹ve made. > Give ANYTHING mathematical which will stand up to scrutiny. My prefered one is the algebraic integer ring is not complete (consequence of your irrefutable proof of FLT). > Do more than insinuate, like talk straight for once. But I insinuate nothing. I am clearly saying you¹re a dead loss in math. Anybody can read any of the billion posts you sent to sci.math and can see that all what you say in math is always either trivial or wrong (this is even sometimes trivially wrong). > > > What I¹ve given are the least computationally complex ways to do the > > > calculations shown, which makes them technology beyond what > > > mathematicians had before my work. > > Ways? Technology? You just said you gave a formula, just a formula! > > Is a formula a way? Is a formula a technology? Or did you give a > > formula, a way and a technology but NOT an algorithm? > > Frankly, are you not a little tired to continuously bull? > Technology refers as a word to state of the art. So, the state of the art in math is technology... I love the poetic quality of such a sentence, don¹t you :-) > I can not only give state of the art in terms of explicit > representations of the prime counting function, I can explain why it¹s > state of the art. Please, do it. For once, do something rather than repeating ad nauseam ŒI can do it¹ or ŒI did it¹. > > > And again it¹s proof that my research IS in fact new and cutting edge. > > No, it is not new. It might be new for you but it is not for others. > I¹ve shown facts, and the facts support my position and refute yours. > > > If not, then I challenge you to give formulas that have less > > > computational complexity, > > What does mean having less computational complexity? > I mean formulas that have less computational complexity. In space or in time? Both? The computational complexity of a math formula... Well, after Œpure math implementation¹, it is true that all becomes possible. > > > or even just show that mathematicians had > > > these formulas before me. > > Legendre. > Claiming something that has repeatedly been shown to be false does not > make it true. Who did show it is false? Who is lying here? > I can kind of understand the hurt you feel, or maybe you feel betrayed > because you think of mathematics as a social system which supports > you, so it cannot in your mind support someone like me. Spare me your analysis. You¹re as hopeless at psychology as you are at math. > But your emotions do not change the mathematical truth. > > > The reality is that I¹m far ahead of mathematicians at every level, > > > when it comes to counting prime numbers: > > You already claimed the same about FLT, Goldbach Conjecture and > > integer factorization. And yet, I am not aware of all your Œresearch¹. > Sigh. So now you¹re trying to shift the subject to other topics with > more claims that I guess you expect me to try to defend or refute. > I¹m not interested in changing the subject at this time. No, I was changing nothing. I just meant The fact that you were a moron the day before yesterday and the fact that you were a moron yesterday perceptibly reduce the probability so that you can be a genius today. It was not clear? > Straight answers. > Have you ever heard of straight answers? > > > 1. My prime counting function derivation is just neat. > > Subjective point of view. > Well, I like it. At last, an irrefutable fact. > I think that anyone who compares it to what mathematicians have will > agree that it¹s just neat, if they¹re objective. > > > 2. My prime counting function itself is beautiful and compact. > > Subjective point of view. > where I put on a single-line an extraordinary formula that deserves a > place in the mathematical literature, no matter how many sci.math¹ers > lie about that truth. Extraordinary formula... :-) > > > 3. I can outline the full theory that determines computational > > > complexity for fast prime counting. > > Wrong. A program using your Legendre variant CANNOT be fast. > I¹ve already written a fast algorithm which I call PrimeCountH.java, > which you can Þnd on sci.math.symbolic, see Fast? How could a non-efÞcient algorithm badly programmed in Java be fast? > It¹s faster than Mathematica¹s prime counting function over I think a > decent range, though at the top of Mathematica¹s range it is faster. Up to 100? To 150? What do you call a decent range? > > > I win on all counts. > > In your reality, maybe, but not in mine. > I win in reality, while you are in your own little world. > In this one I make claims that stand because no one can refute them > with facts as the facts are on my side. > I demonstrate while people like you make claims that you can¹t > support, while you continually shift when your claims are challenged. > You lie repeatedly, as if lying were all that really mattered. > > > Mathematicians win on obstinacy and sheer refusal to accept reality > > > that they don¹t like. > > > > > > They¹re weak. > > Whereas you¹re strong. Do you really believe that you could explain > > to M. Schumacher how he should drive? Well, considering your math > > level, that¹s exactly what you are doing with mathematicians when > > you claim that you can teach them something. > And there I think you are the most honest yet. > You simply believe in *people* not facts, and you don¹t like me, so > you think your feelings matter. > But you see, mathematics never has been nore will it ever be about > your feelings. > To you a name like Schumacher is what¹s important. If I had a name > you respected you¹d probably Þght all comers who dared to deny the > importance of my work. Bull. I just tried to examplify the fact that, when you contradict mathematicians, you contradict people who have skills and knowledge you will never have. I just tried to show you that you¹re nothing but a pretentious fool. > But it¹s not the name that¹s important. > It¹s the mathematics. For fame, money and impressing women? :-) -- mm http://www.ellipsa.net/ mm@ellipsa.no.sp.am.net ( suppress no.sp.am. ) === Subject: Re: Partial difference equation, primes > James Harris a .8ecrit : > > > > > Ok, so I still don¹t know what you call a partial difference > > > equation. But don¹t worry, it has no importance. > > > > I handled that topic with a separate thread. > > > > But it¹s not complicated--no matter what a sci.math¹er might try to > > insinuate--as a partial difference equation is the discrete analog to > > a partial differential equation. > Ok. So, dS(x,y) is what you call a PDE. Sigh. And yes readers, this is how the sci.math¹ers operate. If you look through the entire posts you will see the poster Marcel Martin has become a lot more foul. I beat them at their claims and the sci.math¹ers get ruder and ruder, but manage to never really back down to the facts. Later, after some time, when I post again, they make their original claims all over again, and the entire drama starts anew. Sometimes I think I wait a bit just to see them wind up and come again with the same old song, just out of curiousity. It¹s like they¹re automatons or something with one record they keep playing over and over again, but of course they claim that¹s the way I am, which is part of their record... > > There is no other known in recorded history used to count prime > > numbers besides my dS(x,y) and yes, I¹m talking about all of human > > history here. > You might be right. Who would be moronic enough to compute > pi(y) - pi(y-1) in order to check the primality of y? So now it¹s clear that my work IS indeed different from what mathematicians found and use, but from the sci.math¹er--oddly enough acknowledging the facts for once--suddenly it¹s also moronic. These people are not what you and I would call sane. > > To me that¹s a simple enough claim that it should either be refutable > > by you sci.math¹ers, or if you¹re sane, you¹ll quit trying to > > insinuate that it¹s not a big deal. > > > > I fear you¹re not exactly what most people would call sane. > Do you mean that I could write things like I am the barbarian at the > gates...? :-) I am the barbarian at the gates. I am a revolutionary, a discoverer, a guy who didn¹t just try, but did, who didn¹t just wonder, but accomplished. After all, you¹ve conceded my main point, as sci.math¹ers have been repeating over and over again that my work is not new, but I¹ve beaten you down to acknowledging a clear difference, even if you childishly call it moronic. I¹m the guy who went out and found--from scratch--some beautiful little formulas that count PRIME numbers: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], S(x,1) = 0, p(x, y) = þoor(x) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y) That dS(x,y) is NOT moronic as anyone who read through the derivation and *understood* it can appreciate. Not surprisingly the derivation of the prime counting function is one of the most beautiful in mathematics: See http://mathforproÞt.blogspot.com/ > > > Do you understand that the fact that your formula contains or not > > > a PDE has absolutely no impact on the way we could use it to compute > > > pi(x)? Do you understand that with many formulas containing > > > something like F(x,y), one could claim that F(x,y) is a PDE? Of course, > > > one could do it but what would it change? > > > > You¹re lying because you¹re insinuating that mathematicians have in > > fact used a partial difference equation to count prime numbers when > > they have not. > No. I wanted to be sure that what you call a PDE is also what I call > a PDE. Experience proves that, with you, the meaning of a word can > vary from any-thing to almost-all. Sigh. There you go again. The pure math prime counting formula is clearly something important from the derivation, where *each* piece of it is highly signiÞcant. Sure, algorithms derived from it will be different, but that doesn¹t change the beauty and elegance of the mathematics from which they are based!!! Here sci.math¹ers are caught on the most fundamental thing--beauty in mathematics--with a purest of the pure function, which people who read the derivation can understand completely. It¹s mathematics not only concise, but understandable, as well as beautiful. My work represents problem solving at its purest. And no mean sci.math¹er, rude to the discoverer, can change that reality. > > It¹s not complicated here. You may think you can parse the language > > to fool everyone on sci.physics or that the other sci.math¹ers will > > just go along with you as they have for over two years now, but it > > doesn¹t change the facts. > > > > I say that no one in recorded human history has used a partial > > difference equation to count prime numbers. > > > > That is a fairly straightforward claim, and if it¹s not true you can > > just give some other partial difference equation, but instead you try > > to spin the facts, like what in the hell is F(x,y)? > F(x,y) is any function I can call a PDE. So, with something like > z(x,y) = Sum(i=a to y, F(x,i)), trivially I have F(x,y) = z(x,y) - > z(x,y-1) and it allows me to claim that all the stuff is new because > there is a PDE. Neat, isn¹t it? Yes, I read The art of bullting > by JSH. See? The sci.math¹er gets ruder. These sci.math¹ers hate mathematics when it suits them. I win on the facts, and they start cursing. > > I¹ve seen how you sci.math¹ers operate for over two years now, and you > > have a contempt for the truth, and a demonstrated contempt for > > mathematics. > > > > > > > > > > > > > That¹s only with the pure math implementation. > > > > > > > > > > !? > > > > > > > > Shown above. > > > > > > What I meant was that Œpure math implementation¹ is a mere nonsense. > > > > Yet people can see that it is not nonsense most dramatically by > > looking at it, and I¹ve put it on my blog: > > > Have you a dictionnary at hand? Search for Œimplementation¹. Maybe you > will learn something today. If you implement the pure math function itself--that is use no speed-ups--it¹s rather slow. But if you do the same thing with the Fourier Transform--it¹s slow. You sci.math¹ers wish to throw out the baby with the bathwater by creating dumb criteria. Like suddenly, if there are fast algorithms the base mathematics is junk!!! Oh wait, even you probably aren¹t socially stupid enough to try and toss out the Fourier Transform, so you sci.math¹ers are *picky* about where you apply your rules. If I discover something, you make up one rule, but you dare not challenge Fourier!!! Wait, if Fourier were making his discovery today, and posted on sci.math, you¹d probably challenge him because that¹s what you sci.math¹ers do. > > > > Posters continually use the world algorithm in a derisive manner, > > > > when actually the prime counting function is just a formula. > > > > > > Ok. So, if a program based on your formula is slow it¹s not inherent > > > in your formula but it is due to computer scientists who are quite > > > unable to efÞciently program it :-) > > > > No. > But then, what about your claim according to which your method leads > to the fastest possible prime counting? What about it? > > > > Algorithms can be *derived* from it, but it¹s no more an algorithm > > > > than > > > > > > > > e = mc^2 > > > > > > > > though some nutcase *could* call that an algorithm, if they were > > > > trying to argue that it was not important. > > > > > > > > But it¹s a formula, not an algorithm. > > > > > > > > Algorithms are based off of formulas, not the other way around. > > > > > > > > > > You can also move to an explicit representation, for instance, > > > > > > > > > > > > dS(N,2) = N/2 - 1, with even N, > > > > > > > > > > > > dS(N,3) = þoor((N-4)/6), if N is even, and N>2, and > > > > > > > > > > > > dS(N,5) = þoor((N-16)/10) - þoor((N-16)/30), N even, and N>6, while > > > > > > > > > > > > dS(N,7) = þoor((N-8)/14) - þoor( (N-22)/42) - þoor((N-106)/70) + > > > > > > þoor( (N-106)/210) - 2, N even, N>36, > > > > > > > > > > > > and now, what gets put on the stack now? > > > > > > > > > > Huh? Isn¹t dS(N,2), dS(N,3), dS(N,5), dS(N,7), ... a list? And why > > > > > do you index it with prime values (which, btw, makes useless your > > > > > (p(y,sqrt(y)) - p(y-1,sqrt(y-1))) whereas you claim you do not need > > > > > a list of primes? > > > > > > > > The compressed explicit prime counting function exists as I¹ve shown. > > > > > > Which one makes use of an IMPLICIT list of values. > > > > That¹s what it looks like. > But then, since your technology makes use of a list, what about your > claim ŒIt does not need a list¹? The prime counting function does not need a list. The explicit prime counting function necessarily looks as shown. So what is the *explicit* prime counting function? It¹s where the dS values are set by formulas, like dS(x,2) = þoor(x/2) - 1 is simple enough that you should understand what I mean. The *explicit* prime counting function is inÞnite in size, but pieces of it can be derived and looked at, and those pieces only look ONE WAY, as the math is rigid. > > The math is simply rigid here, no matter how much you might like to > > make more out of it. > > > > It just is. > > > > > > Notice it too is a formula and not an algorithm. > > > > > > > > It just so happens that¹s what it looks like when you have it take > > > > into account that N is even and 2 is prime. > > > > > > > > The math is rigid. > > > > > > Yes and that¹s your main problem. That¹s precisely because the math is > > > rigid that you¹re almost always wrong. If making math was singing, you > > > would sing out of tune. > > > > Then give a single wrong point I¹ve made. > > > > Give ANYTHING mathematical which will stand up to scrutiny. > My prefered one is the algebraic integer ring is not complete > (consequence of your irrefutable proof of FLT). Trying to change the subject? I say that¹s a concession that I have NOT made a single wrong point. > > Do more than insinuate, like talk straight for once. > But I insinuate nothing. I am clearly saying you¹re a dead loss in > math. Anybody can read any of the billion posts you sent to sci.math > and can see that all what you say in math is always either trivial > or wrong (this is even sometimes trivially wrong). It sounds like you¹ve quit even trying to use mathematics and now are just trying to convince with words. You can¹t win, don¹t you understand that? Mathematics is not a social convention. It doesn¹t change depending on whether or not people believe you or not. Why can¹t you just rely on math itself? Mathematically my prime counting function stands out clearly in several ways. Why Þght the truth? > > > > What I¹ve given are the least computationally complex ways to do the > > > > calculations shown, which makes them technology beyond what > > > > mathematicians had before my work. > > > > > > Ways? Technology? You just said you gave a formula, just a formula! > > > Is a formula a way? Is a formula a technology? Or did you give a > > > formula, a way and a technology but NOT an algorithm? > > > Frankly, are you not a little tired to continuously bull? > > > > Technology refers as a word to state of the art. > So, the state of the art in math is technology... I love the poetic > quality of such a sentence, don¹t you :-) Actually the state of the art in mathematics IS technology: Main Entry: technology Pronunciation: -jE Function: noun Inþected Form(s): plural -gies Etymology: Greek technologia systematic treatment of an art, from technE art, skill + -o- + -logia -logy 1 a : the practical application of knowledge especially in a particular area ... http://www.m-w.com/cgi-bin/dictionary?book=Dictionary&va= technology&x=0&y=0 By technology I refer to the *explicit* prime counting function. Like dS(N,3) = þoor((N-4)/6) with N even. It is the least computationally complex formula for calculating the count of odd composites with 3 as a factor up to and including N. That technology is state of the art, as mathematicians cannot do better. They also cannot do better--or even math without my research--the the other dS values I can give, like dS(N,5) = þoor((N-16)/10) - þoor((N-16)/30), with even N>6, wher that¹s the count of odd composites, NOT divisible by 3, that have 5 as a factor. Since I can give the most compact, most efÞcient formulas for each dS count, I can deÞne the most efÞcient prime counting possible. Like I said, on every point I¹m way ahead of mathematicians and the technology I have is state of the art. James Harris === Subject: Re: Partial difference equation, primes > James Harris a .8ecrit : > > > > > Ok, so I still don¹t know what you call a partial difference > > > equation. But don¹t worry, it has no importance. > > > > I handled that topic with a separate thread. > > > > But it¹s not complicated--no matter what a sci.math¹er might try to > > insinuate--as a partial difference equation is the discrete analog to > > a partial differential equation. > Ok. So, dS(x,y) is what you call a PDE. Sigh. And yes readers, this is how the sci.math¹ers operate. If you look through the entire posts you will see the poster Marcel Martin has become a lot more foul. I beat them at their claims and the sci.math¹ers get ruder and ruder, but manage to never really back down to the facts. Later, after some time, when I post again, they make their original claims all over again, and the entire drama starts anew. Sometimes I think I wait a bit just to see them wind up and come again with the same old song, just out of curiousity. It¹s like they¹re automatons or something with one record they keep playing over and over again, but of course they claim that¹s the way I am, which is part of their record... > > There is no other known in recorded history used to count prime > > numbers besides my dS(x,y) and yes, I¹m talking about all of human > > history here. > You might be right. Who would be moronic enough to compute > pi(y) - pi(y-1) in order to check the primality of y? So now it¹s clear that my work IS indeed different from what mathematicians found and use, but from the sci.math¹er--oddly enough acknowledging the facts for once--suddenly it¹s also moronic. These people are not what you and I would call sane. > > To me that¹s a simple enough claim that it should either be refutable > > by you sci.math¹ers, or if you¹re sane, you¹ll quit trying to > > insinuate that it¹s not a big deal. > > > > I fear you¹re not exactly what most people would call sane. > Do you mean that I could write things like I am the barbarian at the > gates...? :-) I am the barbarian at the gates. I am a revolutionary, a discoverer, a guy who didn¹t just try, but did, who didn¹t just wonder, but accomplished. After all, you¹ve conceded my main point, as sci.math¹ers have been repeating over and over again that my work is not new, but I¹ve beaten you down to acknowledging a clear difference, even if you childishly call it moronic. I¹m the guy who went out and found--from scratch--some beautiful little formulas that count PRIME numbers: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], S(x,1) = 0, p(x, y) = þoor(x) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y) That dS(x,y) is NOT moronic as anyone who read through the derivation and *understood* it can appreciate. Not surprisingly the derivation of the prime counting function is one of the most beautiful in mathematics: See http://mathforproÞt.blogspot.com/ > > > Do you understand that the fact that your formula contains or not > > > a PDE has absolutely no impact on the way we could use it to compute > > > pi(x)? Do you understand that with many formulas containing > > > something like F(x,y), one could claim that F(x,y) is a PDE? Of course, > > > one could do it but what would it change? > > > > You¹re lying because you¹re insinuating that mathematicians have in > > fact used a partial difference equation to count prime numbers when > > they have not. > No. I wanted to be sure that what you call a PDE is also what I call > a PDE. Experience proves that, with you, the meaning of a word can > vary from any-thing to almost-all. Sigh. There you go again. The pure math prime counting formula is clearly something important from the derivation, where *each* piece of it is highly signiÞcant. Sure, algorithms derived from it will be different, but that doesn¹t change the beauty and elegance of the mathematics from which they are based!!! Here sci.math¹ers are caught on the most fundamental thing--beauty in mathematics--with a purest of the pure function, which people who read the derivation can understand completely. It¹s mathematics not only concise, but understandable, as well as beautiful. My work represents problem solving at its purest. And no mean sci.math¹er, rude to the discoverer, can change that reality. > > It¹s not complicated here. You may think you can parse the language > > to fool everyone on sci.physics or that the other sci.math¹ers will > > just go along with you as they have for over two years now, but it > > doesn¹t change the facts. > > > > I say that no one in recorded human history has used a partial > > difference equation to count prime numbers. > > > > That is a fairly straightforward claim, and if it¹s not true you can > > just give some other partial difference equation, but instead you try > > to spin the facts, like what in the hell is F(x,y)? > F(x,y) is any function I can call a PDE. So, with something like > z(x,y) = Sum(i=a to y, F(x,i)), trivially I have F(x,y) = z(x,y) - > z(x,y-1) and it allows me to claim that all the stuff is new because > there is a PDE. Neat, isn¹t it? Yes, I read The art of bullting > by JSH. See? The sci.math¹er gets ruder. These sci.math¹ers hate mathematics when it suits them. I win on the facts, and they start cursing. > > I¹ve seen how you sci.math¹ers operate for over two years now, and you > > have a contempt for the truth, and a demonstrated contempt for > > mathematics. > > > > > > > > > > > > > That¹s only with the pure math implementation. > > > > > > > > > > !? > > > > > > > > Shown above. > > > > > > What I meant was that Œpure math implementation¹ is a mere nonsense. > > > > Yet people can see that it is not nonsense most dramatically by > > looking at it, and I¹ve put it on my blog: > > > Have you a dictionnary at hand? Search for Œimplementation¹. Maybe you > will learn something today. If you implement the pure math function itself--that is use no speed-ups--it¹s rather slow. But if you do the same thing with the Fourier Transform--it¹s slow. You sci.math¹ers wish to throw out the baby with the bathwater by creating dumb criteria. Like suddenly, if there are fast algorithms the base mathematics is junk!!! Oh wait, even you probably aren¹t socially stupid enough to try and toss out the Fourier Transform, so you sci.math¹ers are *picky* about where you apply your rules. If I discover something, you make up one rule, but you dare not challenge Fourier!!! Wait, if Fourier were making his discovery today, and posted on sci.math, you¹d probably challenge him because that¹s what you sci.math¹ers do. > > > > Posters continually use the world algorithm in a derisive manner, > > > > when actually the prime counting function is just a formula. > > > > > > Ok. So, if a program based on your formula is slow it¹s not inherent > > > in your formula but it is due to computer scientists who are quite > > > unable to efÞciently program it :-) > > > > No. > But then, what about your claim according to which your method leads > to the fastest possible prime counting? What about it? > > > > Algorithms can be *derived* from it, but it¹s no more an algorithm > > > > than > > > > > > > > e = mc^2 > > > > > > > > though some nutcase *could* call that an algorithm, if they were > > > > trying to argue that it was not important. > > > > > > > > But it¹s a formula, not an algorithm. > > > > > > > > Algorithms are based off of formulas, not the other way around. > > > > > > > > > > You can also move to an explicit representation, for instance, > > > > > > > > > > > > dS(N,2) = N/2 - 1, with even N, > > > > > > > > > > > > dS(N,3) = þoor((N-4)/6), if N is even, and N>2, and > > > > > > > > > > > > dS(N,5) = þoor((N-16)/10) - þoor((N-16)/30), N even, and N>6, while > > > > > > > > > > > > dS(N,7) = þoor((N-8)/14) - þoor( (N-22)/42) - þoor((N-106)/70) + > > > > > > þoor( (N-106)/210) - 2, N even, N>36, > > > > > > > > > > > > and now, what gets put on the stack now? > > > > > > > > > > Huh? Isn¹t dS(N,2), dS(N,3), dS(N,5), dS(N,7), ... a list? And why > > > > > do you index it with prime values (which, btw, makes useless your > > > > > (p(y,sqrt(y)) - p(y-1,sqrt(y-1))) whereas you claim you do not need > > > > > a list of primes? > > > > > > > > The compressed explicit prime counting function exists as I¹ve shown. > > > > > > Which one makes use of an IMPLICIT list of values. > > > > That¹s what it looks like. > But then, since your technology makes use of a list, what about your > claim ŒIt does not need a list¹? The prime counting function does not need a list. The explicit prime counting function necessarily looks as shown. So what is the *explicit* prime counting function? It¹s where the dS values are set by formulas, like dS(x,2) = þoor(x/2) - 1 is simple enough that you should understand what I mean. The *explicit* prime counting function is inÞnite in size, but pieces of it can be derived and looked at, and those pieces only look ONE WAY, as the math is rigid. > > The math is simply rigid here, no matter how much you might like to > > make more out of it. > > > > It just is. > > > > > > Notice it too is a formula and not an algorithm. > > > > > > > > It just so happens that¹s what it looks like when you have it take > > > > into account that N is even and 2 is prime. > > > > > > > > The math is rigid. > > > > > > Yes and that¹s your main problem. That¹s precisely because the math is > > > rigid that you¹re almost always wrong. If making math was singing, you > > > would sing out of tune. > > > > Then give a single wrong point I¹ve made. > > > > Give ANYTHING mathematical which will stand up to scrutiny. > My prefered one is the algebraic integer ring is not complete > (consequence of your irrefutable proof of FLT). Trying to change the subject? I say that¹s a concession that I have NOT made a single wrong point. > > Do more than insinuate, like talk straight for once. > But I insinuate nothing. I am clearly saying you¹re a dead loss in > math. Anybody can read any of the billion posts you sent to sci.math > and can see that all what you say in math is always either trivial > or wrong (this is even sometimes trivially wrong). It sounds like you¹ve quit even trying to use mathematics and now are just trying to convince with words. You can¹t win, don¹t you understand that? Mathematics is not a social convention. It doesn¹t change depending on whether or not people believe you or not. Why can¹t you just rely on math itself? Mathematically my prime counting function stands out clearly in several ways. Why Þght the truth? > > > > What I¹ve given are the least computationally complex ways to do the > > > > calculations shown, which makes them technology beyond what > > > > mathematicians had before my work. > > > > > > Ways? Technology? You just said you gave a formula, just a formula! > > > Is a formula a way? Is a formula a technology? Or did you give a > > > formula, a way and a technology but NOT an algorithm? > > > Frankly, are you not a little tired to continuously bull? > > > > Technology refers as a word to state of the art. > So, the state of the art in math is technology... I love the poetic > quality of such a sentence, don¹t you :-) Actually the state of the art in mathematics IS technology: Main Entry: technology Pronunciation: -jE Function: noun Inþected Form(s): plural -gies Etymology: Greek technologia systematic treatment of an art, from technE art, skill + -o- + -logia -logy 1 a : the practical application of knowledge especially in a particular area ... http://www.m-w.com/cgi-bin/dictionary?book=Dictionary&va= technology&x=0&y=0 By technology I refer to the *explicit* prime counting function. Like dS(N,3) = þoor((N-4)/6) with N even. It is the least computationally complex formula for calculating the count of odd composites with 3 as a factor up to and including N. That technology is state of the art, as mathematicians cannot do better. They also cannot do better--or even math without my research--the the other dS values I can give, like dS(N,5) = þoor((N-16)/10) - þoor((N-16)/30), with even N>6, wher that¹s the count of odd composites, NOT divisible by 3, that have 5 as a factor. Since I can give the most compact, most efÞcient formulas for each dS count, I can deÞne the most efÞcient prime counting possible. Like I said, on every point I¹m way ahead of mathematicians and the technology I have is state of the art. James Harris === Subject: Some Amazing and Beautiful Mathematical Results by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7NChLV29770; Some Amazing and Beautiful Mathematical Results - 1) ei = -1. (Please note that i is the power of e.) Squaring both sides, e2i = 1. logee2i = loge1. 2i 1 = 0. 2i = 0. i = 0. So either i = 0, or = 0, or both i and = 0. 2) i4 = 1. logei4 = loge1. 4 logei = 0. logei = 0. e0 = i. [CapitalEth].equ. 1 However, e0 = 1. Also, squaring equ. 1, e0 = -1. [CapitalEth]equ. 2 So, e0 = i, 1, and -1. [CapitalEth]equ. 3 e0 = i. Multiplying by e, e0 e = i e. e1 = ie. e = ie. 1 = i. [CapitalEth]equ. 4 Also, from equ. 3, e0 = -1. e0 e = -1 e. e1 = -e. e = -e. 1 = -1. [CapitalEth]equ. 5 i = -1. [CapitalEth]equ. 6 However, i = -1. So, from equ. 5 and 6, 1 = -1 = -1. [CapitalEth]equ. 7. - Kedar Joshi Author of The 21st Century Principia === Subject: Re: Some Amazing and Beautiful Mathematical Results > Some Amazing and Beautiful Mathematical Results - > 1) > ei = -1. (Please note that i is the power of e.) > Squaring both sides, > e2i = 1. > logee2i = loge1. > 2i 1 = 0. > 2i = 0. > i = 0. > So either i = 0, or = 0, or both i and = 0. > 2) > i4 = 1. > logei4 = loge1. > 4 logei = 0. > logei = 0. > e0 = i. [CapitalEth].equ. 1 > However, e0 = 1. > Also, squaring equ. 1, > e0 = -1. [CapitalEth]equ. 2 > So, e0 = i, 1, and -1. [CapitalEth]equ. 3 > e0 = i. > Multiplying by e, > e0 e = i e. > e1 = ie. > e = ie. > 1 = i. [CapitalEth]equ. 4 > Also, from equ. 3, > e0 = -1. > e0 e = -1 e. > e1 = -e. > e = -e. > 1 = -1. [CapitalEth]equ. 5 > i = -1. [CapitalEth]equ. 6 > However, i = -1. > So, from equ. 5 and 6, > 1 = -1 = -1. [CapitalEth]equ. 7. > - Kedar Joshi > Author of The 21st Century Principia > > Well, your notation is very difÞcult to follow, but if I gather correctly what you¹re trying to claim, its that since if you Þrst exponentiate and then take logarithm you get a different answer, you get weird equalities like 1=-1. The problem with this of course is that in the complex plane, logarithm can¹t be deÞned on the whole plane as a single valued function, and there is a choice to be made when deÞning any branch of it. Another way to look at it is that the exponentiatial function is periodic with period 2*pi*i. Your argument (if I understand correctly) would be no different than arguing that 2*pi=0 since sin(2*pi)=0 and arcsin(0)=0. -Ron === Subject: Is g(f(x1)...f(xi),..f(xn))=f(g(x1,...xi,...xn) a known equation? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7NChMk29853; I¹d like to know if this equation is known in math literature and which solutions are given. To precise things:f is real,known.Moreover phi(x) such as : phi(f(x)=phi(x)+1 is also known;g(...) is a n real variables function R^n->R. f(x) # x, ax+b or constant. Do want your comments. Alain. === Subject: Re: Is g(f(x1)...f(xi),..f(xn))=f(g(x1,...xi,...xn) a known equation? > I¹d like to know if this equation is known in math literature and > which solutions are given. > To precise things:f is real,known.Moreover phi(x) such as : > phi(f(x)=phi(x)+1 most known computations in mathematics have balanced parentheses > is also known;g(...) is a n real variables function > R^n->R. f(x) # x, ax+b or constant. > Do want your comments. Your system of dots and commas: g(f(x1)...f(xi),..f(xn))=f(g(x1,...xi,...xn) is confusing. How about writing the case n=2 so we can see what you mean. Is it: g(f(x), f(y)) = f(g(x,y)) ? And we assume known a function phi such that phi(f(x)) = phi(x) + 1 ? === Subject: Re: Lacking Concept Of Numbers >= 3 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7NChMm29837; >Did you hear about this Amazon tribe, the Piraha?: >Their language lacks words for any speciÞc numbers >= 3. >Even though they supposedly are generally intelligent and understand >such mathematical concepts such as catagory, they cannot easily grasp >the idea of speciÞc integers much above 3. >And unlike Piraha adults, the Piraha children did not have difÞculty >understanding larger numbers. >Makes me wonder how much of what WE know (know) is only a direct >consequence of our language and experience and genetics. >Leroy Quet Many studies have been carried out pointing to the property that the only numbers anyone can immediately comprehend (and that from babyhood onwards) are the numbers 1, 2, and 3. A popular science book on this topic is The Number Sense by Stanislas Dehaene. By the way, assume the tribe has only words for one and two. Also assume that you and I are both in this tribe and I ask you How many deer are in that Þeld over there? If you do not answer my question immediately with don¹t know then I will autatically assume there are more than two. In addition (as the book points out) our way of thinking is not discrete: we immediately see a crowd of people (or deer) and know there must be Œbout a hundred of them. The processes used to come to this assumption are similar to pouring water into various pales and weighing them do determine the general feel for an amount. It is also extremely efÞcient in certain situations [I believe facial recognition is also done on a similar basis (but I¹m ignorant on this topic) ] In particular, our heads do not appear to be sliding any abacus pearls around to get to any number estimate. As a last example of this continuous type of counting: If asked how many deer there are in the Þeld and I hold my hands out as far as I can reach. Then you, having been in this tribe with me for many years, will surely know that I mean somewhere between 15-25 dear. As far as building tall buildings and civilisation is concerned, I believe the concept of number discrete-isation would come autamatically and intuitively as a part of the realisation of any such building projects (not the other way around). More important to civilisation than an already well-established and discrete set of numbers is, in my opinion, the fortuitousness of living in a climate where the peoples are not forced (by weather, natural disasters,) to perpetually wander from place to place (if tradition dictates this, that is a different story). In this regard, I recently read that the only reason Texas, for ex., (where I went to high school) has more than a few thousand people populating it is almost entirely because of the invention of the air-conditioner. C. Dement http://www.crowdog.de === Subject: Binomial distribution problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7NChMS29826; Hello. I hope that this is a trivial question: For a binomial distribution P=n!/k!(n-k)! * p^k * (1-p)^(n-k) P, p and k is known (P is the Þnal computed probability, p is a probability of the event, k is the observed number of occurences). I need to calculate a k¹, setting p¹ to a Þxed value e.g. 0.5 and keeping P the same Thus: n!/k!(n-k)! * p^k * (1-p)^(n-k)= n!/k¹!(n-k¹)! * p¹^k¹ * (1-p¹)^(n-k¹) The only unknown variable is the k¹ whereas p, p¹, n, k, are known constants. My questions are: 1. Is this approach valid? Will the resulting k¹ (approximatelly) estimate the observed number of cases after p is altered to p¹? 2. If this is valid, can the equation be solved/simpliÞed or do I just need to use computation power? === Subject: Re: Binomial distribution problem >Hello. >I hope that this is a trivial question: >For a binomial distribution P=n!/k!(n-k)! * p^k * (1-p)^(n-k) >P, p and k is known (P is the Þnal computed probability, >p is a probability of the event, k is the observed number of occurences). >I need to calculate a k¹, setting p¹ to a Þxed value e.g. 0.5 and keeping P the same >Thus: >n!/k!(n-k)! * p^k * (1-p)^(n-k)= n!/k¹!(n-k¹)! * p¹^k¹ * (1-p¹)^(n-k¹) >The only unknown variable is the k¹ whereas p, p¹, n, k, are known constants. >My questions are: >1. Is this approach valid? Will the resulting k¹ (approximatelly) estimate >the observed number of cases after p is altered to p¹? I am wondering if you are asking the right question. If X is the random variable giving the number of successes in n trials of a binomial(n,p) distribution then the probablility that X = k is your n!/k!(n-k)! * p^k * (1-p)^(n-k) expression. If you want to estimate the observed number of cases when p is altered to p¹ , it seems to me you are wanting the expected value of X. The expected value of a B(n,p) distribution is E(X) = np. So if you did n trials with probability p you would expect (average) np successes. If the probablility is changed to p¹ you would expect np¹ successes. --Lynn === Subject: Re: Easy number theory problem I just lost my post. Damn. === > Subject: Re: Easy number theory problem > >p.55 Chap 11, Primitive Roots > Web address? Hard to Þnd. Here it is; http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/ lectures_all/ Also, he has notes on Number Theory and Elliptic Curves at http://modular.fas.harvard.edu/edu/Fall2002/124/stein/ (I almost got them mixed up). > >Polynomials over Z/pZ (p prime). > >Prop. 1) A nonzero polynomial f in Z/pZ[x] has at most deg(f) > >(distinct) elements a in Z/pZ such that f(a) = 0. > >I won¹t reproduce the proof of this here. > A polynomial p in F[x] for any Þeld F, > has at most deg p solutions to p(x) = 0. > >Prop. 2) Let d | p-1 . Then f = x^d - 1 has exactly d solutions. > >Proof: Let p-1 = de. Then x^(p-1) - 1 = (x^d)^e - 1 > >= (x^d - 1) g(x) where g(x) = (x^d)^(e-1) + (x^d)^(e-2) + ... + 1, > >so deg(g) = d(e-1) = p - 1 - d. Prop. 1 ==> g has at most > >n_g <= p - 1 - d roots, and x^d - 1 has at most n_d <= d roots, > >we know that x^(p-1) - 1 has p-1 roots in Z/pZ, i.e. the elements > >of Z*_p == (Z/pZ){0}. Thus n_g + n_d = p-1 which means that > >n_g = p - 1 - d and n_d = d. > >Somewhere the fact that Z/pZ is a Þeld enters > Prop 1. > -- > >Lemma. If a and b are in an Abelian group with |a| = r, |b| = s, > >and (r,s) = 1, then |ab| = rs. > This is trival, (r,s) = 1 not needed. rs = |a||b| = |ab| > Are you thinking [r,s] = |rs|/(r,s). Yes, this is what I was thinking. Also if b = a^j, ab = a^(j+1). > BTW, have you notice making corrections takes more time than proof > reading? Yes, its time consuming. > -- > >Theorem. For any prime p, Z*_p is a cyclic group of order p-1. > >Proof: Let p-1 = q_1^e_1 ... q_r^e_r. > >Let q^e = (q_i)^(e_i) for some i. > Confusing. Let q = q_i, e = e_i for some i. > >Prop 2 ==> x^(q^e) - 1 has exactly q^e roots, and > >x^[q^(e-1)] - 1 has q^(e-1) roots ==> > >there is a solution a_i of x^(q^e) - 1 = 0 such that > >(a_i)^[q^(e-1)] != 1 ==> > >|a_i| = (q_i)^(e_i) for each factor of (q_i)^(e_i) in p - 1. > >Let a = a_1 ... a_r. Then |a| = p - 1 by the Lemma, and > How¹s lemma used? p-1 is positive, pick positive numbers for q¹s. Order of a product |ab| = rs if (r,s) = 1, and |a| = r, |b| = s. Here, |a_i| = q_i^(e_i) ==> product has order p-1. Yes, all the q_i and a_i are positive, I don¹t see why one would need to consider negative numbers. > BTW, how do you factor -4 in cannonical form? > 2^2 or (-2)^2 or 2(-2) ? > >Z*_p is cyclic. > How so? I guess I should talk about all steps. I forget some things. You are so picky, but I guess doing math requires being picky on all points. It has an element whose order is the order of the group, so the group is cyclic. See the next post. === Subject: Re: Easy number theory problem > For a in Z_p, let d_a = d be the smallest for which a^d = 1. > Show o(a^j) = d iff (j,d) = 1. > Thus o(a) = d ==> |{ x | o(x) = d}| = phi(d) > For all d | p-1, |{ x | o(x) = d}| = 0 or phi(d) > sum(d | p-1) phi(d) = p - 1 > Thus for all d | p-1, some a with o(a) = d. > In paticular, p-1 | p-1, thus some a with o(a) = p-1 > This generalizes to show Þnite Þelds are > cyclic. If an (Abelian) group G of order n has an element x of order n, then the cyclic subgroup generated by x, = G =~ Z_n is the whole group. I guess that this is just what you show. The method of proof that the multiplicative group of a Þnite Þeld is cyclic that I am most familiar with shows that such an element exists (e.g. Birkhoff and MacLane). (Wish I had a good text like that-- I could probably get a used one pretty cheap from Amazon. I am using mainly notes from the web, but they aren¹t as good as a good text.) > 7.5.4. Theorem. [Fundamental Theorem of Finite Abelian Groups] > Any Þnite abelian group is isomorphic to a direct product of cyclic > groups of prime power order. Any two such decompositions have the same > number of factors of each order. > 7.5.5. Proposition. Let G be a Þnite abelian group. Then G is > isomorphic to a direct product of cyclic groups such that > n_i | n_i-1 for i = 2,3,...k. > What do you think that means? 7.5.4 is the one I am familiar with. I have seen 7.5.5, but I thought of it as just an alternative way of writing Abelian groups. I guess it reþects the fact that in an Abelian group G the order of all cyclic subgroups = order of their generators must divide the order of the group, but 7.5.5 says more than that. I must admit I¹m not sure what it means. G = Z_n_r x Z_n_(r-1) x ... x Z_n_1 with n_i | n_i-1 for i = 2,3,...r. > 7.5.6. Corollary. Let G be a Þnite abelian group. If a in G is an > element of maximal order in G, then the order of every element of G is > a divisor of the order of a. > 7.5.7. Lemma. Let p be a prime number, and let k,a,b be integers. > (a) If 1 <= k <= p-1, then p is a divisor of the binomial coefÞcient > p!/k!(p-k)! > (b) If k >= 1 and a is congruent to b (mod p^k), > then a^p is congruent to b^p (mod p^{k+1} ). > (c) If k >= 2 and p is an odd prime, then > (1 + ap)^(p^(k-2)) = 1 + ap^(k-1) (mod p^k) > (d) If p is an odd prime and p is not a divisor of a, then > (1 + ap)^(p^(k-1)) = 1 (mod p^k) > (1 + ap)^(p^(k-2)) /= 1 (mod p^k) > 7.5.8 Theorem. Let p be an odd prime, let k be a positive integer, and > let n=p^k. Then Z*_n is a cyclic group. I¹m glad you posted this. This is what I wanted to prove. Also for 2p^k when p odd, and the case for p = 2. 7.5.7 Lemma. (a) I¹ve done this part a few times--see recent post. (b) If b = a (mod p^k), b = a + q p^k, b^p = (a + q p^k)^p = a^p + a^(p-1) q p^k p + ... = a^p + q¹ p^(k+1) ==> a^p = b^p (mod p^(k+1)) This notation seems OK to me. My use of mod p comes mainly from what I see in books and notes, rather than from programming. (I¹m not much of a programmer--I just learned what I needed to when I needed it. I often wish I had learned more programming, but one can¹t do everything.) To me, a = b mod p means either [a] = [b] in Z_p, or that p | (b - a). Note that phi(p^k) = p^(k-1) (p - 1), so if x in Z*_p^2, |x| = p-1, p, p(p-1) (assume x != 1). Consider p + 1 in Z*_p^2. (1 + p)^p = 1 + p^2 + ... = 1 (mod p^2), so |p + 1| = p in Z*_p^2. (c) k >= 2 and p is an odd prime, let n = p^(k-2). (1 + ap)^n = 1 + nap + C(2,n) (ap)^2 + ... = 1 + ap^(k-1) (mod p^k) (d) Is a straightforward application of the binomial theorem using that n | C(m,n) for 1 <= m < n. Noting that phi(p^k) = p^(k-1) (p-1) = |Z*_(p^k)|, and the order of an element (not 1) must divide phi(p^k), (together with the following problem), gives the theorem 7.5.8. Here is a problem related to the above theorem on cyclic groups of order p^k when p is an odd prime. 6. Let p be a odd prime number, and let g be a primitive root of p. (a) Let h is an integer satisfying h = g (mod p). Explain why the order of the congruence class of h mod p^2 is either p - 1 or p(p - 1). Hence or otherwise prove that h is a primitive root of p^2 if and only if h^(p-1) != 1 (mod p^2). (b) Use the result of (a) to prove that there exists a primitive root of p^2. (This primitive root will be of the form g + kp for some integer k.) (c) Let x be an integer, and let m be a positive integer. Use the binomial theorem to prove that if x = 1 (mod p^m) and x != 1 (mod p^m+1) then x^p = 1 (mod p^(m+1)) and x != 1 (mod p^(m+2)). (d) Use the results of previous parts of this question to show that any primitive root of p^2 is a primitive root of p^m for all m >= 2. What does this tell you about the group of congruence classes modulo p^m of integers coprime to p? (e) Do the above results hold when p = 2 (i.e., when the prime number p is no longer required to be odd)? This problem, together with what you posted, shows how to prove that Z*_(p^k) is cyclic for all k. I think I have it worked on paper, and I¹ll probably post it later. Van === Subject: Re: Z -> z versus Z -> z e - sole thread for future discussion/postings David, maybe I don¹t understand your question, but I can¹t see the problem there. G¹ is not a linear grammar, but it generates a regular language. When you have a regular language, you can always Þnd a linear grammar that generates it... and many other context-free (or even irrestrict) grammars. Chomsky¹s Hierarchy is a _language_ hierarchy. guillermo > My understanding of the deÞnition of linear grammar is that > this grammar G¹ cannot be considered linear because it contains > one production Z -> z e which does NOT introduce either one > just one terminal nor just one terminal and just one non-terminal. > But if G¹ is not linear for this reason, then must it be > considered context-free (albeit a very trivial type of context-free ?) > So, perhaps the correct way to ask my question is as follows: > Is G¹ linear (at the lowest-level of the Chomsky hierarchy) or > context-free (one level up in this hierarchy?) > Again, I understand that this question may have no answer because > it is itself not a well-formed question; but if it is not > well-formed, I would very much appreciate it if someone would > tell me why (no sarcasm intended!) === Subject: Re: Z -> z versus Z -> z e - sole thread for future discussion/postings > David, maybe I don¹t understand your question, but I can¹t see the > problem there. > G¹ is not a linear grammar, but it generates a regular language. When > you have a regular language, you can always Þnd a linear grammar that > generates it... and many other context-free (or even irrestrict) > grammars. Chomsky¹s Hierarchy is a _language_ hierarchy. David responds: Guillermo - thread Robert Low¹s posting indicates that a linear grammar CAN have a Þnal production of the form Z -> x y , x,y in Vt. So if one accepts this deÞnition of linear grammar, then one doesn¹t have to resort to the artiÞcial rule Z -> z e to create the situation of a non-linear grammar generating a regular language. See also the thread with the same name at comp.compilers - interesting posting by Kenn Heinrich as to whether Z -> z e is really an allowable (i.e. formally meaningful) production from an automata-theoretic point of view. My interest in the matter arises solely from the fact that there is a natural class of ordered trees which includes derivation trees from two kinds of context-free grammars and one type of linear grammar; this is not precisely what one would expect if the distinction between linear (i.e. Þnite-state) and CF were as sharp as folks usually make it out to be. David === Subject: Re: Distance Learning Fractals Class? > Here are two good links to the Frame-Mandelbrot tutorial: > http://classes.yale.edu/Fractals/ > http://www.math.union.edu/~framem/AprilWorkshop/ >> Does any one know of a college that offers a course on fractals >> (on the level of Barnsley¹s Fractals Everywhere) in a distance >> learning format (via web, vidoes, etc.) ? What¹s the haussdorf dimension of that distance? :/ alex === Subject: Re: Distance Learning Fractals Class? Here are a few references : - Engelking, Ryszard /Theory of dimensions Þnite and inÞnite/, Sigma Series in Pure Mathematics, 10. Heldermann Verlag, Lemgo, 1995. viii+401 pp. - Nagata, J. /Modern dimension theory/, Revised edition. Sigma Series in Pure Mathematics, 2. Heldermann Verlag, Berlin (1983). - Hurewicz, W. and Wallman, H. /Dimension Theory/. Princeton Mathematical Series, v. 4. Princeton University Press, Princeton, N. J. (1941). >> Here are two good links to the Frame-Mandelbrot tutorial: >> http://classes.yale.edu/Fractals/ >> http://www.math.union.edu/~framem/AprilWorkshop/ >>> Does any one know of a college that offers a course on fractals >>> (on the level of Barnsley¹s Fractals Everywhere) in a distance >>> learning format (via web, vidoes, etc.) ? > What¹s the haussdorf dimension of that distance? :/ > alex -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === Subject: http://it.geocities.com/povigna/prize.htm I want to buy a proof (50 euro is the sum I offer) for a problem I couldn¹t solve on my own. Please tell me whether my explanation of the matter in the page http://it.geocities.com/povigna/prize.htm is clear enough and whether there are possibilities of misunderstanding (I want a proof, not wasting money!). Michele === Subject: Re: http://it.geocities.com/povigna/prize.htm > I want to buy a proof (50 euro is the sum I offer) for a problem I > couldn¹t solve on my own. > Please tell me whether my explanation of the matter in the page > http://it.geocities.com/povigna/prize.htm > is clear enough and whether there are possibilities of > misunderstanding (I want a proof, not wasting money!). There¹s a theorem of Erdos & Selfridge to the effect that the product of n consecutive integers, n > 1, can¹t be a perfect square. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: http://it.geocities.com/povigna/prize.htm > I want to buy a proof (50 euro is the sum I offer) for a problem I > couldn¹t solve on my own. > Please tell me whether my explanation of the matter in the page > http://it.geocities.com/povigna/prize.htm > is clear enough and whether there are possibilities of > misunderstanding (I want a proof, not wasting money!). The equation n! = a^2 _does_ have a solution in the positive integers: n = 1, a = 1. David === Subject: problem of Þnding apples (maximum likelihood estimation) >> I need help in writing code to solve this equation : >> Sum(i=1,n)(1/[N-(i-1)])=n/(N-a) >> N unknown >> It¹s actually equation from Jelinski-Moranda model. > > fsolve(%,N,complex); > 0.3926627584, 1.497454668, 2.577311552, 3.647350277, 4.713260480, > 5.778662309, 6.847843410 I know that if I try to solve it manully I get n-1 answers, but there should be only one correct answer. I¹ll try to explain what I am trying to do. I have a large Þeld full of apples (number of apples=N) I am trying to Þnd out how many apples are on the Þeld with few assumptions. Now I try this, I am collecting apples and writing down the time it took me to Þnd each new apple. apples time 1 5s 2 4s 3 5s 4 4s ... 90 25s ... 700 200s I know that when I have found 700 apples, there are N-700 apples left in the Þled. Using the Jelinski-Moranda model and it¹s assumptions I¹m trying to solve problem of Þnding bugs in code. But I can¹t seem to Þnd the number N using the above equation. In above equation Œa¹ is not a constant, but it can be used as a constant when I calculate it. Œa¹ equals (Sum(i=1,n)[(i-1)*X(i)])/(Sum(j=1,n)X(j)) where X(i) is time it took to Þnd apple i. This is acctually maximum likelihood estimation for JM model. I¹ll start a new post so that this replay doesn¹t get lost in maybe a wrong subject === Subject: Re: problem of Þnding apples (maximum likelihood estimation) > >> I need help in writing code to solve this equation : > >> Sum(i=1,n)(1/[N-(i-1)])=n/(N-a) > >> N unknown > >> It¹s actually equation from Jelinski-Moranda model. > > > fsolve(%,N,complex); > > 0.3926627584, 1.497454668, 2.577311552, 3.647350277, 4.713260480, > > 5.778662309, 6.847843410 >I know that if I try to solve it manully I get n-1 answers, but >there should be only one correct answer. Then you¹ll have to decide what makes one answer correct and the others wrong. >I¹ll try to explain what I am trying to do. >I have a large Þeld full of apples (number of apples=N) >I am trying to Þnd out how many apples are on the Þeld with >few assumptions. >Now I try this, I am collecting apples and writing down the >time it took me to Þnd each new apple. >apples time >1 5s >2 4s >3 5s >4 4s >... >90 25s >... >700 200s >I know that when I have found 700 apples, there are >N-700 apples left in the Þled. >Using the Jelinski-Moranda model and it¹s assumptions I¹m >trying to solve problem of Þnding bugs in code. >But I can¹t seem to Þnd the number N using the above equation. >In above equation Œa¹ is not a constant, but it can be used as a >constant when I calculate it. >¹a¹ equals (Sum(i=1,n)[(i-1)*X(i)])/(Sum(j=1,n)X(j)) >where X(i) is time it took to Þnd apple i. >This is acctually maximum likelihood estimation for JM model. I think I see what you¹re trying to do. If there are N apples in the Þeld, and N >= i, X(i) are independent exponential random variables with expected values E[X(i)] = 1/(c (N-i+1)) for some constant c. The log likelihood function is then L(c,N) = n ln c + sum_{i=1}^n (ln (N-i+1) - c (N-i+1) X_i) for N >= n. Taking the derivatives wrt c and N, you want n/c - sum_{i=1}^n (N-i+1) X_i = 0 and sum_{i=1}^n (1/(N-i+1) - c X_i) = 0 Thus if sum_i X_i = A and sum_i (i-1) X_i = B, you end up with your equation sum_{i=1}^n 1/(N-i+1) = n/(N-B/A) = n/(N-a) The important points are that a = B/A always satisÞes the inequalities 0 < a < n-1, and you want a solution N such that N >= n. The left side is certainly greater than the right as N -> (n-1)+, while as N -> inÞnity the left side is n/N + n(n-1)/(2 N^2) + O(1/N^3) and the right side is n/N + na/N^2 + O(1/N^3). Thus if a > (n-1)/2 the right side is greater as N -> inÞnity. We should expect a > (n-1)/2 if the X_i tend to be increasing, which should usually be the case if your model is any good (if all X_i were equal you¹d have a = (n-1)/2). So in this case, by the Intermediate Value Theorem there should be at least one solution N > n-1. I can¹t guarantee that N >= n, but if you get a value less than n you¹d conclude N=n. You¹ll still need numerical methods to get the actual numerical solution, but at least you can tell the numerical software to look in the interval (n-1, inÞnity). 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: problem of Þnding apples (maximum likelihood estimation) > I think I see what you¹re trying to do. If there are N apples > in the Þeld, and N >= i, X(i) are independent exponential random > variables with expected values E[X(i)] = 1/(c (N-i+1)) for some > constant c. > The log likelihood function is then > L(c,N) = n ln c + sum_{i=1}^n (ln (N-i+1) - c (N-i+1) X_i) > for N >= n. > Taking the derivatives wrt c and N, you want > n/c - sum_{i=1}^n (N-i+1) X_i = 0 > and > sum_{i=1}^n (1/(N-i+1) - c X_i) = 0 > Thus if sum_i X_i = A and sum_i (i-1) X_i = B, you end up > with your equation > sum_{i=1}^n 1/(N-i+1) = n/(N-B/A) = n/(N-a) > The important points are that a = B/A always satisÞes the inequalities > 0 < a < n-1, and you want a solution N such that N >= n. > The left side is certainly greater than the right as N -> (n-1)+, > while as N -> inÞnity the left side is n/N + n(n-1)/(2 N^2) + O(1/N^3) > and the right side is n/N + na/N^2 + O(1/N^3). Thus if a > (n-1)/2 > the right side is greater as N -> inÞnity. We should expect a > (n-1)/2 > if the X_i tend to be increasing, which should usually be the case > if your model is any good (if all X_i were equal you¹d have a = (n-1)/2). > So in this case, by the Intermediate Value Theorem there should be > at least one solution N > n-1. I can¹t guarantee that N >= n, but > if you get a value less than n you¹d conclude N=n. > You¹ll still need numerical methods to get the actual numerical solution, > but at least you can tell the numerical software to look in the > interval (n-1, inÞnity). The problem is, I need to write code to get the result. And I need some pointers on how this is done. You correctly interpretated my problem, and what is the solution, but I still don¹t know where to look how to solve the problem actually. What numerical method should be used to solve this problem? === Subject: Re: problem of Þnding apples (maximum likelihood estimation) >> Thus if sum_i X_i = A and sum_i (i-1) X_i = B, you end up >> with your equation >> sum_{i=1}^n 1/(N-i+1) = n/(N-B/A) = n/(N-a) >> The important points are that a = B/A always satisÞes the inequalities >> 0 < a < n-1, and you want a solution N such that N >= n. >> The left side is certainly greater than the right as N -> (n-1)+, >> while as N -> inÞnity the left side is n/N + n(n-1)/(2 N^2) + O(1/N^3) >> and the right side is n/N + na/N^2 + O(1/N^3). Thus if a > (n-1)/2 >> the right side is greater as N -> inÞnity. We should expect a > (n-1)/2 >> if the X_i tend to be increasing, which should usually be the case >> if your model is any good (if all X_i were equal you¹d have a = (n-1)/2). >> So in this case, by the Intermediate Value Theorem there should be >> at least one solution N > n-1. I can¹t guarantee that N >= n, but >> if you get a value less than n you¹d conclude N=n. As mentioned before, the equation, made into a polynomial, has degree at most n-1. If a is not an integer, there should be a solution in each of the intervals (i-1,i) for i=1 to n-1 except for the one that contains a. That makes n-2 solutions. So the solution > n-1 should be unique. >> You¹ll still need numerical methods to get the actual numerical solution, >> but at least you can tell the numerical software to look in the >> interval (n-1, inÞnity). >The problem is, I need to write code to get the result. >And I need some pointers on how this is done. >You correctly interpretated my problem, and what is the solution, >but I still don¹t know where to look how to solve the problem actually. >What numerical method should be used to solve this problem? I¹d use Newton¹s method. A good starting point (if it is in the appropriate interval) might be ((2n-1)(n-1) - 6 a^2)/(6 a - 3(n-1)) which is what I get by approximating the equation to order 1/N^3 as N -> inÞnity. 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: Vandermonde matrices and discriminants Let v be a Vandemonde matix [ 1 1 ... 1 ] [ a0 a1 ... an ] [ a0^2 a1^2 ... an^2] . . . v= . . . . . . [ a0^n a1^n ... an^n] and let m=v.v^T It is easy to see that the determinant of m is the discriminant of a polynomial having roots at the ai, i.e. product((ai-aj)^2,i Let v be a Vandermonde matix > [ 1 1 ... 1 ] > [ a0 a1 ... an ] > [ a0^2 a1^2 ... an^2] > . . . > v= . . . > . . . > [ a0^n a1^n ... an^n] > and let m=v.v^T > It is easy to see that the determinant of m is the discriminant > of a polynomial having roots at the ai, i.e. product((ai-aj)^2,i What is easy to notice, but I cannot see my way to proving cleanly > is that the determinants of the principal minors of m are just > sums of discriminants of subsets of the ai. I take it you mean the _leading_ principal minors, viz those from the _Þrst_ k rows and columns. > If we let > [ n+1 s1 s2 ... sn ] > [ s1 s2 s3 ... s{n+1}] > m=[ . . . . ] > [ . . . . ] > [ . . . . ] > [ sn s{n+1} ..... s{2n}] > then the determinant of > [ n+1 s1 ] > [ s1 s2 ] > is the sum( (ai-aj)^2, i and the determinant of > [ n+1 s1 s2 ] > [ s1 s2 s3 ] > [ s2 s3 s4 ] > is sum((ai-aj)^2 (ai-ak)^2 (aj-ak)^2, i now I can prove these identities for the Þrst couple of principal > minors using brute algebra, but there must be some simple proof > related to the proof of the determinant of the vandermonde matrix. > I think that I am missing something simple here. What is it? It could be the Binet-Cauchy theorem on minors of a product of oblong matrices. The reference I have is A. C. Aitken, Determinants and Matrices, p.86, just after equation (4): The general result may be stated thus:.... Ken Pledger. === Subject: Re: Vandermonde matrices and discriminants >>Let v be a Vandermonde matix >> [ 1 1 ... 1 ] >> [ a0 a1 ... an ] >> [ a0^2 a1^2 ... an^2] >> . . . >>v= . . . >> . . . >> [ a0^n a1^n ... an^n] >>and let m=v.v^T >>It is easy to see that the determinant of m is the discriminant >>of a polynomial having roots at the ai, i.e. product((ai-aj)^2,i>What is easy to notice, but I cannot see my way to proving cleanly >>is that the determinants of the principal minors of m are just >>sums of discriminants of subsets of the ai. > I take it you mean the _leading_ principal minors, viz those from > the _Þrst_ k rows and columns. Yes. >>If we let >> [ n+1 s1 s2 ... sn ] >> [ s1 s2 s3 ... s{n+1}] >>m=[ . . . . ] >> [ . . . . ] >> [ . . . . ] >> [ sn s{n+1} ..... s{2n}] >>then the determinant of >> [ n+1 s1 ] >> [ s1 s2 ] >>is the sum( (ai-aj)^2, i>and the determinant of >> [ n+1 s1 s2 ] >> [ s1 s2 s3 ] >> [ s2 s3 s4 ] >>is sum((ai-aj)^2 (ai-ak)^2 (aj-ak)^2, i>now I can prove these identities for the Þrst couple of principal >>minors using brute algebra, but there must be some simple proof >>related to the proof of the determinant of the vandermonde matrix. >>I think that I am missing something simple here. What is it? > It could be the Binet-Cauchy theorem on minors of a product of > oblong matrices. The reference I have is A. C. Aitken, Determinants > and Matrices, p.86, just after equation (4): The general result may > be stated thus:.... too long. > Ken Pledger. === Subject: Group theory question I have been trying to prove this statement which seems intuitively very obvious and which seems like it ought to be very straightforward to prove, but which is actually turning out to be quite hard... Conjecture: suppose two binary operations * and . are deÞned on a nonempty set S such that: 1. S is a group under * and also under . 2. The groups (S under *) and (S under .) are isomorphic 3. The identity element of S under * is the identity element of S under . Then: * and . are the same operation As stated, it¹s more general than I actually need-- all I actually need is the case where S is of the form {0,1,...,n}. In this special case it is equivalent to Conjecture: a group with underlying set {0,1,...,n} has a unique proper Cayley table, where by proper we mean one which is arranged 0,1,...,n along the top and side and wherein 0 is the identity. One strategy I¹ve been trying, but running into problems with the details: try the more general case where S is a Þnite set of not-necessarily-consecutive nonnegative integers, including zero. Then one would employ induction on |S|, the small cases being trivial, and for the inductive step split it into three parts: Þrst assume (S,*) is cyclic (and hence (S,.) as well, by isomorphism), and handle this case specially. Second, assume (S,*) is not cyclic but can be generated by some 2 elements. Again, handle this case specially. Finally, assume neither cyclic nor generatable with 2 elements, and the (i,j) entry in the Cayley table is identical for both * and . by using the subgroup generated by the appropriate two elements, which by hypothesis is smaller than S and thus covered by the inductive assumption. This seems (intuitively) like a whole lot of work for a rather obvious-seeming result, can anyone help me out, I am sure there is something obvious I¹m missing. It isn¹t a homework problem, though you¹ll just have to take my word on that... Sniz Pilbor === Subject: Re: Group theory question days. My association with the Department is that of an alumnus. > I have been trying to prove this statement which seems >intuitively very obvious and which seems like it ought to be very >straightforward to prove, but which is actually turning out to be >quite hard... >Conjecture: suppose two binary operations * and . are deÞned on a >nonempty set S such that: >1. S is a group under * and also under . >2. The groups (S under *) and (S under .) are isomorphic >3. The identity element of S under * is the identity element of S >under . >Then: * and . are the same operation Do you mean, identical functions? i.e., for every x and y in S, x*y = x.y? It is false. Let S be the symmetric group on 3 letters, with * being the usual multiplication. S = {1, r, r*r, s, r*s, r*r*s} where s*s = 1, r*r*r = 1, s*r=r*r*s. Now deÞne on S the operation . as follows: y.x = x*y for all x and y in S. CLAIM: S is a group under ., isomorphic to S, with the same identity elements. Proof. . is associative, since x.(y.z) = (y.z)*x = (z*y)*x = z*(y*x) = (y*x).z = (x.y).z for all x, y, z. The fact that 1 is the identity and that inverses are the same in both groups gives the fact that (S,.) is a group. Since (S,.) is a nonabelian group of order 6, it is isomoprhic to S_3, hence (S,.) is isomorphic to (S,*). Yet . and * are patently not the same operation: r.s = s*r = r*r*s which is different from r*s. >As stated, it¹s more general than I actually need-- all I actually >need is the case where S is of the form {0,1,...,n}. In this special >case it is equivalent to >Conjecture: a group with underlying set {0,1,...,n} has a unique >proper Cayley table, where by proper we mean one which is arranged >0,1,...,n along the top and side and wherein 0 is the identity. The underlying set is immaterial. Rewrite my S above by setting 1 -> 0, r->1, r*r->2, s->3, r*s->4, r*r*s->5. -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Group theory question > 2. The groups (S under *) and (S under .) are isomorphic Do you mean that the isomorphism is given by the identity map? J. === Subject: Re: Group theory question * Snis Pilbor > I have been trying to prove this statement which seems > intuitively very obvious and which seems like it ought to be very > straightforward to prove, but which is actually turning out to be > quite hard... > Conjecture: suppose two binary operations * and . are deÞned on a > nonempty set S such that: > 1. S is a group under * and also under . > 2. The groups (S under *) and (S under .) are isomorphic > 3. The identity element of S under * is the identity element of S > under . > Then: * and . are the same operation What happens if you take any non-abelian group and deÞne x*y=y.x? -- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@iÞ.uio.no http://www.iÞ.uio.no/~jonhaug/, Phone: +47 22 85 24 92 === Subject: Re: Group theory question >* Snis Pilbor >> I have been trying to prove this statement which seems >> intuitively very obvious and which seems like it ought to be very >> straightforward to prove, but which is actually turning out to be >> quite hard... >> Conjecture: suppose two binary operations * and . are deÞned on a >> nonempty set S such that: >> 1. S is a group under * and also under . >> 2. The groups (S under *) and (S under .) are isomorphic >> 3. The identity element of S under * is the identity element of S >> under . >> Then: * and . are the same operation >What happens if you take any non-abelian group and deÞne x*y=y.x? Or for just about any group, take x*y = f^(-1)(f(x).f(y)) where f is a one-to-one map of S onto itself that is not a group automorphism (under .) but satisÞes f(e)=e. There aren¹t many Þnite groups with (n-1)! automorphisms (where n is the order of the group), I think. 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: Group theory question >>* Snis Pilbor >>> I have been trying to prove this statement which seems >>> intuitively very obvious and which seems like it ought to be very >>> straightforward to prove, but which is actually turning out to be >>> quite hard... >>> Conjecture: suppose two binary operations * and . are deÞned on a >>> nonempty set S such that: >>> 1. S is a group under * and also under . >>> 2. The groups (S under *) and (S under .) are isomorphic >>> 3. The identity element of S under * is the identity element of S >>> under . >>> Then: * and . are the same operation >>What happens if you take any non-abelian group and deÞne x*y=y.x? >Or for just about any group, take x*y = f^(-1)(f(x).f(y)) where f is a >one-to-one map of S onto itself that is not a group automorphism (under >.) but satisÞes f(e)=e. There aren¹t many Þnite groups with (n-1)! >automorphisms (where n is the order of the group), I think. No - in fact only C1, C2, C3 and C2 x C2 have that property. So those are the only groups for which the conjecture is true. Derek Holt. === Subject: Re: Group theory question Well, I guess that would explain why I was not having much luck proving it! Too bad... but, that is how we learn, yes? Hehe still get through, but will end up being twice as complicated and less than half as neat without that nice little (false) conjecture... Sniz Pilbor > >>* Snis Pilbor > >>> I have been trying to prove this statement which seems > >>> intuitively very obvious and which seems like it ought to be very > >>> straightforward to prove, but which is actually turning out to be > >>> quite hard... > >>> Conjecture: suppose two binary operations * and . are deÞned on a > >>> nonempty set S such that: > >>> 1. S is a group under * and also under . > >>> 2. The groups (S under *) and (S under .) are isomorphic > >>> 3. The identity element of S under * is the identity element of S > >>> under . > >>> Then: * and . are the same operation > >>What happens if you take any non-abelian group and deÞne x*y=y.x? > >Or for just about any group, take x*y = f^(-1)(f(x).f(y)) where f is a > >one-to-one map of S onto itself that is not a group automorphism (under > >.) but satisÞes f(e)=e. There aren¹t many Þnite groups with (n-1)! > >automorphisms (where n is the order of the group), I think. > No - in fact only C1, C2, C3 and C2 x C2 have that property. So those are > the only groups for which the conjecture is true. > Derek Holt. === Subject: Re: Heart equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7NFexF13390; >Is there an equation that describes a heart shape in polar coordinates >- either exactly or roughly? I¹ve tried a series of sines but its >rather crude. >In this system 1 is equivalent to 360 degrees. y = 0.75 |x| +/- sqrt(1 - x^2) Now you convert it === Subject: Re: Eric Gisse, your comment was ridiculous In sci.math, eleaticus : >> > What is the Œstandard¹ galilean xform set that has been Œproved¹ invalid > so >> > many times? >> x¹ = x-vt >> t¹ = t > When transforming coordinates in x (rather than dx) t¹=t can do no harm I > think. The Galilean transform has the problem of not preserving lightspeed. The Lorentzian transforms have the problem of either my misremembering them (which is quite possible!) or observer interchangeability. > It is in transforming equations in dx that the harm is done. Which is what > you are talking about in saying the galileans have been proved wrong? Transforming equations is what they *do*. :-) > Right? Wrong? The Galilean transform is interchangeable. I¹m still working on the Lorentz transform; AFAICT I¹m making some sort of goofy assumption somewhere. It turns out that the proper characterization is: x¹ = (x - vt) * gamma t¹ = (t - vx/c^2) * gamma where gamma = 1/sqrt(1 - v^2/c^2). So I¹ve goofed yet again, and maybe this time it will be a goof in the right direction. :-) http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html Is this one interchangeable? Lessee. x * gamma = x¹ + vt * gamma x = x¹ / gamma + vt t * gamma = t¹ + vx/c^2 * gamma t = t¹ / gamma + vx/c^2 x = x¹ / gamma + v(t¹/gamma + vx/c^2) x = x¹ / gamma + vt¹/gamma + v^2x/c^2 x(1 - v^2/c^2) = x¹/gamma + vt¹/gamma x = x¹ * gamma + vt¹ * gamma t = t¹ / gamma + v(x¹ / gamma + vt)/c^2 t = t¹ / gamma + vx/(gamma*c^2) + v^2t/c^2 t(1 - v^2/c^2) = t¹ / gamma + vx/(gamma * c^2) t = (t¹ + vx/c^2) / gamma I feel so goofy, but this restores part of my faith in SR, at least mathematically; I¹ve derived the reverse transform from the forward, and it is the same as the forward transform except that v switched sign, as it should be. Of course one has to be careful here. These are transformations for *measurements*, not for *units*; this was confusing me for a bit, and I may not be the only one. Brieþy put, as one speeds up near lightspeed, lengths and time measurements of the observed Universe shrink -- this is usually characterized as time intervals dialating, but if one¹s seconds are twice as long, then one has only half of them... :-) A second check is to ensure that a light beam traveling at c still travels at c for all observers -- a requirement of SR. To that end, set (x - a) = c(t - b). Now ( (x¹ + vt¹)*gamma - a) = c((t¹ + vx¹/c^2) * gamma - b) ( (x¹ + vt¹) - a/gamma) = c((t¹ + vx¹/c^2) - b/gamma) ( (x¹ - vx¹/c) - a/gamma) = c((t¹ - vt¹/c) - b/gamma) ( x¹(1 - v/c) - a/gamma) = c(t¹(1-v/c) - b/gamma) ( x¹ - a¹) = c(t¹ - b¹) where a¹ = a/(gamma * (1 - v/c)), b¹ = b/(gamma * (1 - v/c)) So this is the correct transform, from a mathematical standpoint, although I was expecting slightly different values for a¹ and b¹. But it doesn¹t matter, really. The third requirement of course is whether this is compatible with physical observation, and that is *not* something that can be derived mathematically, but must be done by old-fashioned testing the Universe -- i.e, observation and experimentation. AFAIK SR has been thereby validated a fair number of times by others; I can¹t do it, as my equipment is more along the lines of Galileo¹s shutter lanterns than anything that might be used to accurately measure lightspeed. :-) The best I can do is ping another machine on my network through a short cable, and I get about 0.200 ms, or 200 microseconds -- way too high for a pair of cables at most about 10 feet in length total. Admittedly, I think part of this is my switch -- a small Hawking unit. I do have a crossover cable which I might use. At 2.2 m in length, it should give me about 7.3 nanoseconds. While pselect() can resolve nanoseconds, gettimeofday() only goes to microseconds. Not that it matters; my readings are still far too high: 0.150 ms, or 150 microseconds, on a different pair of machines. The odd thing: after restoring my network I get 0.180 ms on that pair; the switch isn¹t that slow. But obviously my equipment can¹t measure lightspeed, unless I can get a far longer cable. :-) But I still like my shutter lanterns, as they can do a lot of other things. :-) > eleaticus -- #191, ewill3@earthlink.net It¹s still legal to go .sigless. === Subject: Re: Eric Gisse, your comment was ridiculous > > When transforming coordinates in x (rather than dx) t¹=t can do no harm I > > think. > The Galilean transform has the problem of not preserving lightspeed. Which is not true if done right. My Maxwell piece, coming up soon will show the failing is provisional at best. > > It is in transforming equations in dx that the harm is done. Which is what > > you are talking about in saying the galileans have been proved wrong? > Transforming equations is what they *do*. :-) Yes, but it is the insistence on t¹=t rather than plain old t is t that does the harm, so you are missing the point. Yes or no, if you don¹t treat the Galileans as including t¹=t ther is no problem with them vis a vis Mawell? And why would you want to include the non-xform as if there were an xform, when velocity, which does transform is treated as a constant? The truth is that v¹=-v. But acknowledge that fact would make Maxwell non-invariant under the LETs. > Of course one has to be careful here. These are transformations for > *measurements*, not for *units*; this was confusing me for a bit, and > I may not be the only one. In fact, the LET change Œratio¹ scale measurements into inferior Œinterval¹ scale measurements. At some time t: x0¹.........x¹0.............................................. ...x¹ x0------------------------------------------x (insert periods or dashes to line up x¹ and x, as needed.) The length x-x0 = x has been converted to x¹-x¹0 = xa¹, which doesn¹t even represent the whole interval x had represented. The true representation of x¹ as the distance x was set up to be in any real, physical equation that uses it, is x¹-x0¹. Note that under the Galileans that interval is invariant: x¹-x0¹ = (x-vt) - (x0-vt) = (x-vt). > A second check is to ensure that a light beam traveling at c > still travels at c for all observers -- a requirement of SR. > To that end, set (x - a) = c(t - b). Now That, of course, is a conclusion based on the illusion that Michelson-Morley has only that explanation, and the Maxwell solution under Newton-Galileo shows it is not a necescary conclusion. > AFAIK SR has been thereby validated a fair number of times by > others; Perhaps your equipment is as good as theirs when it comes to all those validations of the incredible shrinking object aspect of SR? Tell me about them. LOL! eleaticus === Subject: Re: Eric Gisse, your comment was ridiculous > As math, no. It is self-consistant but incorrect. > If you cannot tell the difference between math [abstract, but self > consistant] and physics [math with real-world applications], you > should not be discussing either. Let¹s see. Perhaps two dozen times you Œrespond¹ to math stuff with nonsense about how the math stuff is disproved by experiments that wouldn¹t even be relevant if the math stuff had been physics stuff. And then you talk about not knowing the difference!?? LOL! LOL! LOL! LOL! > What is it with all these cranks who þout their stupidity by > addressing posts speciÞcally to me? My email works just Þne, and I > would be glad to explain further if asked. You are the cretin/crank who volunteered GR nonsense in response to what at most could be called SR math. And when you got a response you whined about why didn¹t I email you! What a maroon you are! Still no response to my pointing out my email was working Þne! eleaticus === Subject: Re: nitpicking in the complex plane? there isn¹t such a Þgment of our lazy imaginations! In sci.math, David Dathe >>WHERE INGENUITY COULD NOT PREVAIL. show me a negative anything and >>i¹ll show you a crook.. and that crook would be YOU > Interesting comment. I¹m not sure what to make of it except to give you two examples. > 1. In a few months here in Wisconsin we will undoubtedly have a wind > chill of -10oF. So if negative numbers can¹t exist, what will happen > when I step outside to shovel the driveway? A more relevant question is what might happen if one places a paper or styrofoam cup of water out the night before, covered by a plastic cap (e.g., from a fast food joint). If the temperature (without windchill) is below 0 degrees Celsius (32 degrees F), the water will freeze. If the water is uncovered, windchill might be a factor; I¹d have to look. Of course, that¹s the way out: windchill is 10 below zero F would be one description. One other way is to use absolute zero. 0 C = 273.15 K, or thereabouts. 250 K sounds positively tropical. :-) (-10F = -23.3C.) However, 310 K *is* tropical -- 98 F. 300K is room temperature (a warm room, admittedly: 80 F). 300K is thereby convenient for certain thermodynamic and/or gas computations. > 2. My friend just scored a -5 on the new golf course he tried. > 18 holes, 5 under par. Do I conclude, since this is a negative > number, that he never really plays golf at all? No, one concludes he¹s 5 under par. Par is probably around 72 (it depends on the golf course and how malicious the designers are :-) ); so his absolute score is actually around 67. > Negatives were introduced because they do exist in physical > reality as well as mathematical theory. Ingenuity is not > conÞned to positive, real numbers. Negative numbers are useful abbreviations for the above concepts. Once one introduces negative numbers, of course, one can extend the standard mathematical operations (+, -, *, /) thereto, and get the usual arithmetic system. And then: the square root.... :-) -- #191, ewill3@earthlink.net It¹s still legal to go .sigless. === Subject: Re: Lacking Concept Of Numbers >= 3 > >Did you hear about this Amazon tribe, the Piraha?: > > >Their language lacks words for any speciÞc numbers >= 3. > >Even though they supposedly are generally intelligent and understand > >such mathematical concepts such as catagory, they cannot easily grasp > >the idea of speciÞc integers much above 3. > >And unlike Piraha adults, the Piraha children did not have difÞculty > >understanding larger numbers. > >Makes me wonder how much of what WE know (know) is only a direct > >consequence of our language and experience and genetics. > >Leroy Quet > Many studies have been carried out pointing to the property that > the only numbers anyone can immediately comprehend (and that > from babyhood onwards) are the numbers 1, 2, and 3. > A popular science book on this topic is The Number Sense > by Stanislas Dehaene. > By the way, assume the tribe has only words for one and two. > Also assume that you and I are both in this tribe and > I ask you How many deer are in that Þeld over there? > If you do not answer my question immediately > with don¹t know then I will autatically assume there are > more than two. In addition (as the book points out) our > way of thinking is not discrete: we immediately > see a crowd of people (or deer) and know there must be > Œbout a hundred of them. The processes used to come to this > assumption are similar to pouring water into various pales > and weighing them do determine the general feel for > an amount. It is also extremely efÞcient in certain > situations [I believe facial recognition is also done > on a similar basis (but I¹m ignorant on this topic) ] > In particular, our heads do not appear to be sliding > any abacus pearls around to get to any number estimate. > As a last example of this continuous type of counting: > If asked how many deer there are in the Þeld and I > hold my hands out as far as I can reach. Then you, > having been in this tribe with me for many years, will > surely know that I mean somewhere between 15-25 dear. As far as > building tall buildings and civilisation is concerned, > I believe the concept of number discrete-isation would > come autamatically and intuitively as a part of the > realisation of any such building projects > (not the other way around). More important to > civilisation than an already well-established and discrete set of > numbers is, in my opinion, the fortuitousness of living in a climate > where the peoples are not forced (by weather, natural disasters,) > to perpetually wander from place to place (if tradition dictates > this, that is a different story). In this regard, I recently > read that the only reason Texas, for ex., (where I went to > high school) has more than a few thousand people populating it > is almost entirely because of the invention of the air-conditioner. > C. Dement http://www.crowdog.de Apparently, some people can¹t count beyond one when it comes to paragraphs. === Subject: Re: Cum Hoc, Ergo Propter Hoc > > You can¹t even get that right, can you? The logical error is POST hoc, ergo propter hoc- After this, therefore because of this, claiming that one thing is caused by another simply because it occurs immediately after it. > > Cum hoc, ergo propter hoc would be with this, therefore because of this and that ISN¹T a common enough error to be labled because there is no direction implied by with. > There is more than one error to be made in logic. True, _post hoc ergo > proper hoc_ - the fallacy of assuming causation on the basis of > sequential occurence - is more generally known, but that¹s not to say > _cum hoc ergo propter hoc_ isn¹t a known error too - it is the fallacy > of assuming causation on the basis of correlation. search > was what showed *me* that the unfamilar term did already exist and was > not another of JSH¹s coinages. > -- > Larry Lard > Replies to group please So what¹s the name of the logical fallacy of inappropriately applying a logical fallacy? === Subject: Re: Cum Hoc, Ergo Propter Hoc >So what¹s the name of the logical fallacy of inappropriately applying a >logical fallacy? Ille, ergo propter hoc? Lee Rudolph === Subject: Re: Dimention theory http://www.amazon.com/exec/obidos/tg/detail/-/0471922870?v= glance Fractal Geometry: Mathematical Foundations and Applications by Kenneth Falconer 1991 Mandelbrot, B. B. The Fractal Geometry of Nature. New York: W. H. Freeman, 1983. Bogomolny, A. Fractal Curves and Dimension. http://www.cut-the-knot.org/do_you_know/dimension.shtml. > Could anybody recommend any modern good book about Dimention theory ? > I took a look at amazon.com and found that most books like Hurewicz > Dimention theory are > out of print and may be out of date ... -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === Subject: Re: need Reed-Solomon C++ code for RS(4, 2) and RS(24, 20) on 6 bit symbols > my problem is, with s=6 (mm in the aforementioned code) that means my > data n is 63 symbols long. but it¹s not, it¹s explicitly 4 symbols > long, with 2 parity symbols (t = 1, k = 2). > the other solution i need deals with 24 symbols with t=2 (4 symbols of > parity, 20 of data) > I heard something about just rounding up to the next table size and > throwing out the unneded extras. Is that the way to go? What am I > missing? how can you have n=4 when N must be 2**s-1 which always > yields an odd number? You are dealing with *shortened* codes. A (4,2) RS with 6-bit symbols is just a (63, 61) RS code in which the Þrst 59 data symbols are set to 0 (that is, each of these 59 six-bit symbols is 000000). If you have a program to encode/decode the (63, 61) code, it will encode/decode the (4,2) code as well if you set the Þrst 59 data/received symbols to 0. If you delve into the details of the software, you will Þnd that the program spins its wheels while it is processing the Þrst 59 zeroes. In short, some re-programming (having some for-loops run for 4 instead of 61 iterations) will save a lot of time. Hope this helps. === Subject: My XSLT stylesheets for math in XHTML At this page I uploaded my XSLT 1.0 stylesheets for writing extensions): http://ex-code.com/~porton/math/xslt.html It intented to replace LaTeX. The main feature is that it uses a variation of XHTML, not a completely new format. I think it is the right approach. The project is yet in VERY early stage, nor ready to use (however I myself already edit my drafts with these), but indeed I announce these for you to let you know about the project. (You can also take part in development if you know XSLT.) Yes. We are to throw away LaTeX! -- Victor Porton (porton@ex-code.com) http://ex-code.com-software http://ex-code.com/~porton/-math[ CapitalEHat]discoveries,Christianrevelations === Subject: Re: My XSLT stylesheets for math in XHTML > At this page I uploaded my XSLT 1.0 stylesheets for writing > extensions): http://ex-code.com/~porton/math/xslt.html > It intented to replace LaTeX. The main feature is that it > uses a variation of XHTML, not a completely new format. > I think it is the right approach. > The project is yet in VERY early stage, nor ready to use > (however I myself already edit my drafts with these), but indeed > I announce these for you to let you know about the project. (You > can also take part in development if you know XSLT.) > Yes. We are to throw away LaTeX! Do you have some examples of results and how the code is simpler than LaTeX? Note: Since I still haven¹t gotten a version of TeX installed and working, this may be of interest to me. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Transposition... Can anyone conÞrm that the following equation: a = b(1-[x/y])^2 transposes to: x = y([sqrt a/b] -1) -- What is now proved was once only imagin¹d. - William Blake, 1793. === Subject: Re: Transposition... x-mimeole: Produced By Microsoft MimeOLE V6.00.2800.1409 > Can anyone conÞrm that the following equation: > a = b(1-[x/y])^2 > transposes to: > x = y([sqrt a/b] -1) x = y*(1 - sqrt(a/b)) or x = y*(1 + sqrt(a/b)) is now proved was once only imagin¹d. - William Blake, 1793. === Subject: Re: Transposition... >> Can anyone conÞrm that the following equation: >> a = b(1-[x/y])^2 >> transposes to: >> x = y([sqrt a/b] -1) >x = y*(1 - sqrt(a/b)) >x = y*(1 + sqrt(a/b)) I still don¹t get it. Can somebody spoon-feed me through the transposition to see how this answer is obtained? -- What is now proved was once only imagin¹d. - William Blake, 1793. === Subject: Re: zero-dimensional spaces Disconnected zero topological dimensional spaces are not necessarily actual zero dimensioned. Study of rational numbers from the fractal dimensional point of view suggests that functional and Lyapunov exponent approaches give d<0.5 for the dimension of the rational number. box counting give in two web sites: d=1 d=0.9 Hausdourff: d=0 Dr Edgar quotes d=1 in a sci.math post: the Bouligand dimension of the set of rationals is 1. It also implies that the irrational numbers with transcendentals and without have fractal dimensions as well( both are disconnected sets with the rational numbers between them). It has to do with roughness theory as the basis of fractal dimension. >> > I have encountered various questions about totally disconnected >> topological spaces. Any classical or well-known texts that deal >> with totally disconnected spaces in more than just a passing way, >> and in sufÞcient generality (see below), would be very welcome. > Engelking¹s `General Topology¹ contains a comprehensive discussion > of `various kinds of disconnectedness¹ in Chapter 6. >> > DeÞnition 1: >> A space X is *totally disconnected* if for every two points x and y >> there exists a clopen (closed and open) set U that contains x and does >> not contain y. > This is the commonly accepted deÞnition of total disconnectivity. >> > DeÞnition 2: >> A space X is *totally disconnected* if it has a topological basis >> consisting of clopen sets. > This notion is commonly known as zero-dimensionality. >> > Question 1: Is it true in general that Def. 1 implies Def. 2? >> No, a famous example is the set of points in $ell_2$ all of whose > coordinates are rational --- this is due to ErdH{o}s. > KP > E-MAIL: K.P.Hart@twi.tudelft.nl PAPER: Department of Pure Mathematics > PHONE: +31-15-2784572 TU Delft > FAX: +31-15-2787245 Postbus 5031 > URL: http://aw.twi.tudelft.nl/~hart 2600 GA Delft > the Netherlands Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === Subject: Re: zero-dimensional spaces How is this problem of fractal dimension 0 Disconnected zero topological dimensional spaces are not necessarily > actual zero dimensioned. > Study of rational numbers from the fractal dimensional point of view > suggests that functional and Lyapunov exponent approaches give d<0.5 > for the dimension of the rational number. > box counting give in two web sites: d=1 d=0.9 > Hausdourff: d=0 > Dr Edgar quotes d=1 in a sci.math post: > the Bouligand dimension of > the set of rationals is 1. > It also implies that the irrational numbers with transcendentals and > without have fractal dimensions as well( both are disconnected sets > with the rational numbers between them). > It has to do with roughness theory as the basis of fractal > dimension. >>> > I have encountered various questions about totally disconnected >>> topological spaces. Any classical or well-known texts that deal with >>> totally disconnected spaces in more than just a passing way, >>> and in sufÞcient generality (see below), would be very welcome. >> Engelking¹s `General Topology¹ contains a comprehensive discussion >> of `various kinds of disconnectedness¹ in Chapter 6. >>> > DeÞnition 1: >>> A space X is *totally disconnected* if for every two points x and y >>> there exists a clopen (closed and open) set U that contains x and does >>> not contain y. >> This is the commonly accepted deÞnition of total disconnectivity. >>> > DeÞnition 2: >>> A space X is *totally disconnected* if it has a topological basis >>> consisting of clopen sets. >> This notion is commonly known as zero-dimensionality. >>> > Question 1: Is it true in general that Def. 1 implies Def. 2? >>> No, a famous example is the set of points in $ell_2$ all of whose >> coordinates are rational --- this is due to ErdH{o}s. >> KP >> E-MAIL: K.P.Hart@twi.tudelft.nl PAPER: Department of Pure >> Mathematics >> PHONE: +31-15-2784572 TU Delft >> FAX: +31-15-2787245 Postbus 5031 >> URL: http://aw.twi.tudelft.nl/~hart 2600 GA Delft >> the Netherlands > Respectfully, Roger L. Bagula > tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: > 619-5610814 : > URL : http://home.earthlink.net/~tftn > URL : http://victorian.fortunecity.com/carmelita/435/ -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ X-mailer: xrn 9.02 === Subject: Re: Polar integration? Mail-To-News-Contact: abuse@dizum.com >>It seems intuitive to me that the limit of these lower sums, as the number >>of points in the partition increases, would be the same as the limit of >>the upper sums, although I haven¹t actually proven it. >>Does such a development of calculus exist? >Yes. Look in any standard (3 semester) calculus text. Well, from this and other replies, it¹s beginning to sound as if I should have paid more attnetion during my third semester of calculus. This just conÞrms what I¹d already Þgured out. :-< > You are >outlining the proof of the standard integral for polar area: I¹m on the right track? That¹s encouraging. -- Michael F. Stemper #include A preposition is something that you should never end a sentence with. === Subject: Re: Polar integration? Michael, may be this is an approach, it¹s from Mamikon: http://www.its.caltech.edu/~mamikon/BikingGardner.html (after reading this, click on caculus) I just learned about it in de.sci.mathematik Eine neue (?) Integrationemethode For me: Mamikon, das ist wie Wasser f.9fr den Calculus-Baum nach einer Zeit bloen Sonnescheins. (Not þuent in german - no problem with Google - language tools or http://babelÞsh.altavista.com) See You Hero So, here¹s another try. === Subject: Re: Re Is g(f(x1)...f(xi),..f(xn))=f(g(x1,...xi,...xn) a known equation?. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7NHDaL21401; >> I¹d like to know if this equation is known in math literature and >> which solutions are given. >> To precise things:f is real,known.Moreover phi(x) such as : >> phi(f(x)=phi(x)+1 >most known computations in mathematics have balanced parentheses >> is also known;g(...) is a n real variables function >> R^n->R. f(x) # x, ax+b or constant. >> Do want your comments. >Your system of dots and commas: > g(f(x1)...f(xi),..f(xn))=f(g(x1,...xi,...xn) >is confusing. How about writing the case n=2 so we can see what you >mean. Is it: g(f(x), f(y)) = f(g(x,y)) ? >And we assume known a function phi such that phi(f(x)) = phi(x) + 1 ? You did fully understand what I mean... Do you propose anything? === Subject: Re: Re Is g(f(x1)...f(xi),..f(xn))=f(g(x1,...xi,...xn) a known equation?. > >> > >> I¹d like to know if this equation is known in math literature and > >> which solutions are given. > >> To precise things:f is real,known.Moreover phi(x) such as : > >> phi(f(x)=phi(x)+1 > >most known computations in mathematics have balanced parentheses > >> is also known;g(...) is a n real variables function > >> R^n->R. f(x) # x, ax+b or constant. > >> Do want your comments. > >> > >Your system of dots and commas: > > g(f(x1)...f(xi),..f(xn))=f(g(x1,...xi,...xn) > >is confusing. How about writing the case n=2 so we can see what you > >mean. Is it: g(f(x), f(y)) = f(g(x,y)) ? > >And we assume known a function phi such that phi(f(x)) = phi(x) + 1 ? > You did fully understand what I mean... > Do you propose anything? Using the known function phi, we may reduce this problem to the special case where f(x) = x+1, namely: g(x+1,y+1) = g(x,y)+1 For this, we may deÞne f(x,y) arbitrarily for 0 <= x < 1, then use the functional equation to Þll in the other values. Namely, for any x,y, let n be the greatest integer <= x, then g(x,y) = g(x-n,y-n)+n. === Subject: Re: Tangle: The Game Answer to strategy puzzle immediately following copied post. > > > Here is a game which *might* be fun to play, and might be fun to > > > analyze mathematically and for strategy. > > > > > > > > > For 2 or more players, each with a different colored pencil. > > > > > > Start with a large square (or any convex polygon) drawn on paper. > > > > > > Player 1 starts by drawing a straight line-segment, using a > > > straight-edge, from any corner of the square/polygon through the > > > interior of the polygon until hitting an edge of the polygon. > > > > > > Players together draw a path by alternatingly drawing line segments, > > > using a straight-edge (and making each segment the color associated > > > with the player drawing the segment), each segment which is from the > > > end of the last segment drawn by the previous player, in any > > > direction(with restrictions*), and ending at the Þrst previously > > > drawn line-segment intersecting the segment being drawn, or ending at > > > a boundry of the square/polygon. > > > > > > (*) If the last line-segment drawn ended at the edge of the > > > square/polygon, the path must bounce, ie. the next segment must be > > > towards the interior of the polygon. > > > If the last segment drawn ended at a previously drawn segment (of any > > > color), the path must cross this older segment, ie. the new segment > > > must be on the opposite side of the older crossed segment from the > > > immediately previous segment. > > > > > > Segments cannot be drawn to vertexes (where the path crosses itself or > > > bounces off the edge of the polygon.) > > > > > > The players draw a predetermined number of segments (say, a total of > > > 50). > > > Each player gets a point every time the path crosses a pre-drawn > > > segment of his/her color. > > > The winner has the most number of points. > > > > > > > > > A clariÞcation: > > The path refered to above is the total collection of line-segments. > > The line-segments are drawn end-to-end and alternate in color. > > > > I will attempt some ascii-art. > > (View with Þxed-width font.) > > > > If the game is like so: > > > > < older line-segment > > > > -- < last line drawn by previous player > > > > > > > > you continue on the opposite side of the older line-segment in any > > direction: > > > > /< your new segment > > / > > -- > > > > > > > > And your new segment continues only until the Þrst line-segment or > > edge of the square/polygon encountered, where your line-segment then > > terminates. > > > > > > If the game is like so: > > > > > > < last line drawn by previous player > > > > > > You may bounce in any direction (either to left or right), as long as > > the path stays within the square/polygon: > > > > / > > / < your new segment > > / > > > > > > One Þnal note: > > No segment is allowed to be drawn along a previous segment. > > (No co-linear segments.) > > > > Leroy Quet > 4 more things: > 1) It is evident after I played this that it would probably be better, > rather than making a path of a predetermined number of line-segments, > that the game simply continue until one player Þrst gets a > predetermined score, say 10 or 20. > In this way, players do not have to worry about keeping track of the > number of moves made. And there is also, with this new rule, no chance > of a tie. > 2) One purpose of playing this game, aside from winning, is to > hopefully get a cool abstract design. > :) > 3) I was a little unclear about when a point is scored > pre-drawn segment of his/her color.). > Whenever a line-segment is drawn by anyone to another segment of color > c, and player c is using pencil/pen color c, then player c gets a > point. > By path crosses, I mean 2 segments come together at the previously > drawn line-segment. > Like so: > < previously drawn segment of color c, path crosses this segment. > _______ > / > / > / > 4) As is noted elsewhere in this thread, there is an advantage to > being player 1 (and so players should play multiple rounds, trading > off who is player 1). > There is a likely set of opening moves. As an easy puzzle, try to > give the best next move for player 2: > Playing on a square board. > Moves: > 1) Player 1 draws segment from lower-left (Southwest) corner > East-Northeast to right edge of square. > 2) Player 2: NW to upper edge of square. > 3) Player 1: South to player 1¹s Þrst segment. > 4) Player 2: SE to right edge of square. > 5) Player 1: West to player 1¹s Þrst segment. > Now, where should player 2 move? > (Easy, but this will help you familiarize yourself with the game, if > you choose to try to solve this puzzle.) > Leroy Quet Best move for player 2, as far as I Þgure, is NE to player 1¹s second segment, which gives player 1 another point, but is better a move for player 2 than the alternative. For if player 2 connects to either the top or left edge of the square (the only other possible moves for him/her), player 1 can then connect back to his/her Þrst segment (connecting southwesterly of where player 2¹s last segment connected with player 1¹s Þrst segment). If player 2 then (on next move) connects to the bottom or right edge of the square, then player 1 can connect back with his/her Þrst segment (connecting southwesterly of where player 2¹s last segment connected with player 1¹s Þrst segment). This can continue as long as player 2 allows it: each time player 1 moves, he/she gets a point; each time player 2 moves to an edge of the square, he/she gets nothing. So the best time for player 2 to exit this loop is as soon as possible, which is why I suggest the move I suggest above for player 2. With the suggested move for player 2, player 1 can then get, if he/she chooses, yet another point on his/her next move (the game¹s move #7) by moving to his/her Þrst segment, but player 2 cuts his/her losses and can Þnally score on his/her next move (the game¹s move #8). Leroy Quet === Subject: Re: 2 Recursion Puzzles (1 Unexpected, 1 More General) > > Neither of these puzzles is too difÞcult, but I Þnd them interesting. > > > > > > Puzzle 1: > > > > a(1) = a(2) = 1; > > > > For m >= 3, > > > > a(m) = (a(m-1) +1) *a(m-1) /(a(m-1) +a(m-2) +1) > > > > > > Find a non-recursive deÞnition for the sequence of rationals {a(k)}. > > > > (I Þnd this puzzle¹s solution to be unexpected.) > sum(k=0 to m-1) (k!) / m! > Proof tbd > > > > --- > > > > Puzzle 2: > > > > {c(k)} is any arbitrary non-zero sequence. > > > > Let b(1) = 1/c(1). > > > > For m >= 2, > > > > b(m) = > > > > (m + sum{k=1 to m-1} b(k) (c(k+1) -c(k))) /c(m) > > > > > > Find a non-recursive deÞnition for the sequence {b(k)} in terms of {c(k)}. > > > > > b(m)=sum (k=1 to m) (1/c(k)) > proof by induction (unusually involving assumption of truth for all k<=m of > b(k)*c(k)=b(k-1)*c(k)+1) > > Leroy Quet > Where do you get them from Leroy? > JJ Both you (John Jones) and Michael Mendelsohn have correctly solved these puzzles, of course. By the way, sum{k=0 to m-1} k! is the so-called left factorial and can be represented as !m. http://www.research.att.com/projects/OEIS?Anum=A003422 So, from puzzle 1, a(m) = !m /m!. Leroy Quet === Subject: Re: 2 Recursion Puzzles (1 Unexpected, 1 More General) > > Neither of these puzzles is too difÞcult, but I Þnd them interesting. > > > > > > Puzzle 1: > > > > a(1) = a(2) = 1; > > > > For m >= 3, > > > > a(m) = (a(m-1) +1) *a(m-1) /(a(m-1) +a(m-2) +1) > > > > > > Find a non-recursive deÞnition for the sequence of rationals {a(k)}. > > > > (I Þnd this puzzle¹s solution to be unexpected.) > sum(k=0 to m-1) (k!) / m! > Proof tbd Yes it took me an bit to see the trick: rearrange as 1+[a(m-2)+1]/a(m-1) = [a(m-1)+1]/a(m) write b(m)= [a(m-1)+1]/a(m) so 1+b(m-1)=b(m) => b(m)=m (from initial conds) so a(m) = [a(m-1)+1]/m leads to the result. > > > > --- > > > > Puzzle 2: > > > > {c(k)} is any arbitrary non-zero sequence. > > > > Let b(1) = 1/c(1). > > > > For m >= 2, > > > > b(m) = > > > > (m + sum{k=1 to m-1} b(k) (c(k+1) -c(k))) /c(m) > > > > > > Find a non-recursive deÞnition for the sequence {b(k)} in terms of {c(k)}. > > > > > b(m)=sum (k=1 to m) (1/c(k)) > proof by induction (unusually involving assumption of truth for all k<=m of > b(k)*c(k)=b(k-1)*c(k)+1) > > Leroy Quet > Where do you get them from Leroy? > JJ === Subject: Re: Lacking Concept Of Numbers >= 3 >Did you hear about this Amazon tribe, the Piraha?: >Their language lacks words for any speciÞc numbers >= 3. >Even though they supposedly are generally intelligent and understand >such mathematical concepts such as catagory, they cannot easily grasp >the idea of speciÞc integers much above 3. >And unlike Piraha adults, the Piraha children did not have difÞculty >understanding larger numbers. >Makes me wonder how much of what WE know (know) is only a direct >consequence of our language and experience and genetics. two remarks: - In recent intelligence tests with cats and dogs one could also show that cats can easily count to four, while dogs can not. The tests were done by ringing a bell multiple times (up to four) and letting the cats then select one of the four boxes that were marked with our standard die symbols (black points). The reason for this difference is not clear; some believe that while a cat must sometimes change place it needs to carry around the kitten from one place to another and must keep track of how many of them are still to be transported. So language is not needed to count, and genetics is yet another form of experience, according to standard Darwin evolution theories. - The sentence ..is *only* ... of our language and experience and genetics... includes everything that comprises a conscious human being. If you take these away, no intelligent creature in the entire universe can count; it even cannot use a computer (to substitute the experience) because interfacing with a computer needs some kind of language. So I do not how this sentence contains something like any debatable statement. Richard J Mathar === Subject: Re: Lacking Concept Of Numbers >= 3 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7NHWtX23054; >> >Did you hear about this Amazon tribe, the Piraha?: >> > >> >> > >> >Their language lacks words for any speciÞc numbers >= 3. >> > >> >Even though they supposedly are generally intelligent and understand >> >such mathematical concepts such as catagory, they cannot easily >grasp >> >the idea of speciÞc integers much above 3. >> > >> >And unlike Piraha adults, the Piraha children did not have >difÞculty >> >understanding larger numbers. >> > >> > >> >Makes me wonder how much of what WE know (know) is only a direct >> >consequence of our language and experience and genetics. >> > >> >Leroy Quet >> Many studies have been carried out pointing to the property that >> the only numbers anyone can immediately comprehend (and that >> from babyhood onwards) are the numbers 1, 2, and 3. >> A popular science book on this topic is The Number Sense >> by Stanislas Dehaene. >> By the way, assume the tribe has only words for one and two. >> Also assume that you and I are both in this tribe and >> I ask you How many deer are in that Þeld over there? >> If you do not answer my question immediately >> with don¹t know then I will autatically assume there are >> more than two. In addition (as the book points out) our >> way of thinking is not discrete: we immediately >> see a crowd of people (or deer) and know there must be >> Œbout a hundred of them. The processes used to come to this >> assumption are similar to pouring water into various pales >> and weighing them do determine the general feel for >> an amount. It is also extremely efÞcient in certain >> situations [I believe facial recognition is also done >> on a similar basis (but I¹m ignorant on this topic) ] >> In particular, our heads do not appear to be sliding >> any abacus pearls around to get to any number estimate. >> As a last example of this continuous type of counting: >> If asked how many deer there are in the Þeld and I >> hold my hands out as far as I can reach. Then you, >> having been in this tribe with me for many years, will >> surely know that I mean somewhere between 15-25 dear. As far as >> building tall buildings and civilisation is concerned, >> I believe the concept of number discrete-isation would >> come autamatically and intuitively as a part of the >> realisation of any such building projects >> (not the other way around). More important to >> civilisation than an already well-established and discrete set of >> numbers is, in my opinion, the fortuitousness of living in a climate >> where the peoples are not forced (by weather, natural disasters,) >> to perpetually wander from place to place (if tradition dictates >> this, that is a different story). In this regard, I recently >> read that the only reason Texas, for ex., (where I went to >> high school) has more than a few thousand people populating it >> is almost entirely because of the invention of the air-conditioner. >> C. Dement http://www.crowdog.de >Apparently, some people can¹t count beyond one when it comes to >paragraphs. Or sentences for that matter. === Subject: Re: 42 is interesting > > While playing with the number 42 on my recent 42nd birthday, > > it came to my attention that the average divisor of 42 is > > twelve, which exactly equals the Euler Phi function of 42 > > (the count of positive integers less than 42 with no common > > divisors to 42). > > > > Some smaller numbers (1,3, and 14) share this property, > > but I could not Þnd any larger ones (although I only > > searched up to 100,000). > > > > Does anyone know if there are larger numbers with this > > property (Average Divisor equals Euler Phi function), > > or know of a proof that there are none? > > > > _ > > |/|/| || Burnaby South Secondary > > || |orewood@olc.ubc.ca || Beautiful British Columbia > > > > (You should be using a non-proportional font to correctly > > read the email address in my signature.) > Of course 42 is interesting, all natural numbers are interesting. > Proof by contradiction. > Suppose that there was an uninteresting natural number then there > would be a least uninteresting natural number. The property of being > the least uninteresting natural number would be interesting > contradicting the assumption that it was an uninteresting number. > The proof can be extended to the rational numbers, algebraic numbers > and many other sets but not, as far as I know, to the real numbers. > Se.87n O¹Leathl.97bhair Since we are on the topic of the interestingness of 42, here is part of a reply of mine on the subject in a thread from January: >... >But even better is the related 1764, >which is >1*2*3*4*5*6*(1 +1/2 +1/3 +1/4 +1/5 +1/6), >and, amazingly, >1764 also = .... >42 *42 (!)... >Speaking of 42, I heard years ago that scientists found that 42 REALLY >did have some astrophysical signiÞcance, >related to universal-expansion or something. >I do not know if the 42 constant was exact or if the result was >refuted or just a joke. >What was up with this?... Leroy >Quet === Subject: incresing functions on [0, inf) Hello I¹d like some hints on how to prove or disprove the following statement: f and g are real valued functions deÞned on [0, inf). f is strictly increasing and lim (x=> inf) g(x) =0. Then, there is a k>0 such that f+g is strictly increasing for x>k. I think this is false, but couldn¹t Þnd a counter example. Amanda === Subject: Re: incresing functions on [0, inf) > I¹d like some hints on how to prove or disprove the following > statement: > f and g are real valued functions deÞned on [0, inf). f is strictly > increasing and lim (x=> inf) g(x) =0. Then, there is a k>0 such that > f+g is strictly increasing for x>k. > I think this is false, but couldn¹t Þnd a counter example. Try f(x) = x and g(x) = sin(x^2)/x. Then f(x) is str. incr. to oo, but (f+g)¹(x) < 0 for lots of x -> oo. === Subject: Re: incresing functions on [0, inf) > > I¹d like some hints on how to prove or disprove the following > > statement: > > > > f and g are real valued functions deÞned on [0, inf). f is strictly > > increasing and lim (x=> inf) g(x) =0. Then, there is a k>0 such that > > f+g is strictly increasing for x>k. > > > > I think this is false, but couldn¹t Þnd a counter example. > Try f(x) = x and g(x) = sin(x^2)/x. Then f(x) is str. incr. to oo, but > (f+g)¹(x) < 0 for lots of x -> oo. In fact, for any differentiable f with f(x) -> oo as x -> oo, set g = [sin(f^2)]/f. Then g(x) -> 0 as x -> oo, and f + g is strictly decreasing on a sequence of intervals going out to oo. === Subject: Re: incresing functions on [0, inf) >Hello >I¹d like some hints on how to prove or disprove the following >statement: >f and g are real valued functions deÞned on [0, inf). f is strictly >increasing and lim (x=> inf) g(x) =0. Then, there is a k>0 such that >f+g is strictly increasing for x>k. >I think this is false, but couldn¹t Þnd a counter example. >Amanda Think about f = 1 - (1/x^2) and g = 1/x for x >= 1. Move them 1 to the left if 0 must be in the domain. --Lynn === Subject: Egyptian Harmonic Numbers I am wondering about this variation on Egyptian fractions. Let H(m) = sum{k=1 to m} 1/k. Now, what would be the best way (lowest m¹s, or smallest number of H¹s, etc) to represent the rational r as a sum of H(m_k)¹s? If we can only add (not subtract) H¹s, then some r¹s are not representable. But if we allow the sum: r = sum{k>=1} H(m_k) *n_k, where the n¹s = any (positive or negative or 0) integers, and the m¹s are positive integers, then no rational r has no representations as such a sum. (Because every r has a representation as the sum of Egyptian fractions, and H(m)-H(m-1) =1/m.) Example: 7/3 = H(3) + H(2) - H(1). We can also ask about the case where: H(0,m) = 1/m, H(j,m) = sum{k=1 to m} H(j-1,k), and the sum is of only positive n¹s, but the j¹s can vary. r = sum{k>=1} H(j_k,m_k) *n_k. Cases: 1) All j¹s in sum are the same nonnegative integer. 2) m¹s are all distinct and n¹s are all 1. 3) j¹s are all distinct. 4) Combination of (2) with (1), or (2) with (3). etc Anything on-line about any of this? Leroy Quet === Subject: Re: Lacking Concept Of Numbers >= 3 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7NIaQt28850; >> >Did you hear about this Amazon tribe, the Piraha?: >> > >> >> > >> >Their language lacks words for any speciÞc numbers >= 3. >> > >> >Even though they supposedly are generally intelligent and understand >> >such mathematical concepts such as catagory, they cannot easily >grasp >> >the idea of speciÞc integers much above 3. >> > >> >And unlike Piraha adults, the Piraha children did not have >difÞculty >> >understanding larger numbers. >> > >> > >> >Makes me wonder how much of what WE know (know) is only a direct >> >consequence of our language and experience and genetics. >> > >> >Leroy Quet >> Many studies have been carried out pointing to the property that >> the only numbers anyone can immediately comprehend (and that >> from babyhood onwards) are the numbers 1, 2, and 3. >> A popular science book on this topic is The Number Sense >> by Stanislas Dehaene. >> By the way, assume the tribe has only words for one and two. >> Also assume that you and I are both in this tribe and >> I ask you How many deer are in that Þeld over there? >> If you do not answer my question immediately >> with don¹t know then I will autatically assume there are >> more than two. In addition (as the book points out) our >> way of thinking is not discrete: we immediately >> see a crowd of people (or deer) and know there must be >> Œbout a hundred of them. The processes used to come to this >> assumption are similar to pouring water into various pales >> and weighing them do determine the general feel for >> an amount. It is also extremely efÞcient in certain >> situations [I believe facial recognition is also done >> on a similar basis (but I¹m ignorant on this topic) ] >> In particular, our heads do not appear to be sliding >> any abacus pearls around to get to any number estimate. >> As a last example of this continuous type of counting: >> If asked how many deer there are in the Þeld and I >> hold my hands out as far as I can reach. Then you, >> having been in this tribe with me for many years, will >> surely know that I mean somewhere between 15-25 dear. As far as >> building tall buildings and civilisation is concerned, >> I believe the concept of number discrete-isation would >> come autamatically and intuitively as a part of the >> realisation of any such building projects >> (not the other way around). More important to >> civilisation than an already well-established and discrete set of >> numbers is, in my opinion, the fortuitousness of living in a climate >> where the peoples are not forced (by weather, natural disasters,) >> to perpetually wander from place to place (if tradition dictates >> this, that is a different story). In this regard, I recently >> read that the only reason Texas, for ex., (where I went to >> high school) has more than a few thousand people populating it >> is almost entirely because of the invention of the air-conditioner. >> C. Dement http://www.crowdog.de >Apparently, some people can¹t count beyond one when it comes to >paragraphs. Wellsaid(nomorepostingpopularsciencecrapafter4hourssleep) C. Dement === Subject: Re: Rafþe Probability-Twist by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7NKLx706043; >Lets assume that there is a rafþe where 1000 tickets are sold and >there are 4 prizes, and a winning ticket is not put back into the pot. >What is the probability of winning a prize? Initially I had thought >it would somewhere close to 1/250, but this doesn¹t seem right. The >formula would be (1/1000) + (1/999) + (1/998) + (1/997), i.e. the sum >of the odds for each individual drawing. The problem is under this >formula someone who purchased 500 tickets should be guaranteed a >prize, which is not the case. Finally, what would be the formula for >calculating the odds of winning a prize? Your formula is not correct. If I buy one ticket and I win on the Þrst draw, then the other three terms clearly don¹t belong. So at each draw you must account for what happens if you win, as well as what happens if you don¹t. phil === Subject: harmonic series convergence To what limit does H(n)= Sum{k=1..Inf}[(-1)^(k-1)/k^n] converge? (In Riemann zeta function signs of alternate terms have been altered). TIA === Subject: Re: harmonic series convergence > To what limit does H(n)= Sum{k=1..Inf}[(-1)^(k-1)/k^n] converge? (In > Riemann zeta function signs of alternate terms have been altered). TIA (Assuming n>1, or for complex n, Real(n) > 1) Request: Don¹t denote it H(n); this is customary for the true harmonic numbers H(n) = 1 + 1/2 + ... + 1/n . Let me call the alternating sibling of Zeta function A(n), hoping I am not trespassing on anyone¹s favorite use of it. An exercise for you: Add up the terms carrying minus signs, factor out a suitable power of 2, and relate the result to zeta function. Then restore the alternating series. Remark: If you express Zeta(n) in terms of A(n), you will obtain an analytic continuation of Zeta to Real(n) > 0, except of course the case n=1. === Subject: sequences of touching spheres Whilst playing around with sequences of tangent spheres, I stumbled upon the following neat result: ------ Consider 5 spheres each of them touching all others. Remove one of them. Given the conÞguration of the four remaining spheres it is always possible to create a new cluster of Þve mutually touching spheres by placing a Þfth sphere in a position different from where the removed sphere was. The bend* of this newly placed Þfth sphere is given by K-k, where k is the bend of the removed sphere, and K is the sum of the bends of the four remaining spheres. * The bend of a sphere mutually tangent to four other spheres is deÞned as the signed curvature (inverse radius) of the sphere. If the contacts are all external, the signs of the bends of all Þve spheres are taken as positive, whereas if one sphere surrounds the other four, the sign of this sphere is taken as negative. ------ Is anyone here familiar with this result? I haven¹t seen it quoted anywhere on the internet, but I doubt whether it is new. (It just seems to be too straightforward to be truly new!) Would appreciate any references / url¹s to a text describing this result. As an aside: one can use the above result to derive the curvatures of inÞnite sequences of spheres in which each Þve consecutive spheres all touch one another. Starting from the Þrst Þve curvatures k(1), k(2), .. k(5), one simply iterates using: k(n) = k(n-1) + k(n-2) + k(n-3) + k(n-4) - k(n-5) For any Þve starting values, for large n this sequence converges to a geometric progression: lim n->Inf k(n+1)/k(n) = (sqrt(2)+1+sqrt(2.sqrt(2)-1))/2 = 1.8832035... An example is given by two spheres of zero bend (i.e. planes) in contact with three unit spheres. Following the above iteration scheme one obtains for the bends of the consecutive spheres: 0, 0, 1, 1, 1, 3, 6, 10, 19, 37, 69 ... Note that by shufþing the last Þve values before computing the next value, one can obtain alternative sequences present in such Apolonian sphere packings. JK === Subject: Re: sequences of touching spheres > Whilst playing around with sequences of tangent spheres, I stumbled > upon the following neat result: ... > Is anyone here familiar with this result? I haven¹t seen it quoted > anywhere on the internet, but I doubt whether it is new. (It just > seems to be too straightforward to be truly new!) Would appreciate any > references / url¹s to a text describing this result. May be Google for : Soddy spheres and circles (a lot of references come out, I didn¹t examine all to see if they are relevant to your problem) -- philippe (chephip at free dot fr) === Subject: Hyperboloid and Nuclear Plants Hi! It seems that the huge chimney of a nuclear power plant has the shape of a hyperboloid of revolution: http://ccins.camosun.bc.ca/~jbritton/jbconics.htm Is there any physical or mechanical reason to do so? === Subject: Re: Hyperboloid and Nuclear Plants > Hi! > It seems that the huge chimney of a nuclear power plant has > the shape of a hyperboloid of revolution: > http://ccins.camosun.bc.ca/~jbritton/jbconics.htm > Is there any physical or mechanical reason to do so? yes, it has a water evaporation core in the middle where the cross section is smaller, and the evaporation, heat induce a draft Air current up vertically through the tower. Shape maximizes airþow through the core. Google Cooling Towers === Subject: Re: Status of Waring-problem Originator: dmoews@ccrwest.org (David Moews) |Am 21.08.04 04:33 schrieb David Moews: |[...] |> I don¹t think so. Steiner¹s paper implies that if a 1-cycle exists in the |> Collatz problem, then there are powers of 2 and 3 whose difference is |> exponentially small relative to their common value, i.e., |> |2^a - 3^b| < 3^b (1-eps)^b. (**) |[...] | |In mathworld in the chapter about power-fractions the waring-inequality, |[...] |is expressed as | frac((3/2)^k) < 1- (3/4)^k [1-MW] | |which possibly could be a erroneous transformation. | |Now my condition for the nonexistence of an 1-cycle is |practically the same as the mathworld-formula, it is | | frac((3/2)^k) < 1- (3/4)^k +eps [2-H] | |where eps would be smaller than 1/3^k, (can be neglected in |diophantine problems) and is so essentially the same as [1-MW]. |THis says, if [2-H] holds, then a 1-cycle is denied. | |Now the Steiner-criterion states that an 1-cycle is impossible |because of this approximation-argument (**), so how could there |be a reason, that (**) does not imply [1-MW] ? (may be I¹m just |having a knot in my mind currently...) |[...] If a 1-cycle exists in the Collatz problem, suppose it consists of c up steps followed by d down steps. Then it must start at a number of the form 2^c(2m+1)-1, and c, d and m must satisfy 3^c(2m+1)-1 = 2^d(2^c(2m+1)-1). It follows from this that (3/2)^c is very close to 2^d, so it¹s true that if (3/2)^k is never close to an integer, then a 1-cycle cannot exist in the Collatz problem. However, the converse is not true, so Steiner¹s proof does not have to prove that (3/2)^k is never close to an integer. In fact, it proves only that (3/2)^k cannot be close to a power of 2. -- David Moews dmoews@xraysgi.ims.uconn.edu === Subject: Intuition about condition number I¹m looking for some intuition about condition number of a matrix. For instance, if all values of the matrix are equal, then condition number if small. Also, if 2x2 matrix is normalized, then corresponding Markov chain converges slowly if original matrix had small condition number. Are there any tricks to Þguring out the condition number by just looking at the matrix? Yaroslav === Subject: Re: JSH: So what¹s the point? No Way a .8ecrit : > >A good start would be requiring computer proof checking!!! > That¹s a good start you can make on your own. You can lead by example > by using computer proof checking on your own claims. Please do so and > report the results (in such a way that they are repeatable by others). Checking a theorem using a program doesn¹t solve any problem. On the contrary. 1) It is very difÞcult, if not impossible, to write a sophisticated program without bugs. 2) The result of a program (even bug-free) may be altered by hardware failures (loss of bits, bugs in the processor, etc). Using a program in order to guarantee the validity of a theorem is an idea of genius, but of a genius like JSH. -- mm http://www.ellipsa.net/ mm@ellipsa.no.sp.am.net ( suppress no.sp.am. ) === Subject: Re: cos(x) >= 1 - x^2/2! + x^4/4! - x^6/6! > f_0(x) = 1 > f_1(x) = x^2/2! > f_2(x) = x^4/4! > f_3(x) = x^6/6! > f_n(x) = x^(2n) / (2n)! > We know that cos(x) = f_0(x) - f_1(x) + f_2(x) - f_3(x) + ... + > (-1)^n*f_n(x) + ... for all x. > I wonder if it is true that: > cos(x) >= f_0(x) - f_1(x) + .... + (-1)^(2n+1)*f_(2n+1)(x) for all x. > This inequality is true for very small x, because the sequence ( f_0(x), > f_1(x), f_2(x), ... ) is a decreasing sequence when -1 This inequality is true for very large x, because then > f_0(x) - f_1(x) + .... + (-1)^(2n+1)*f_(2n+1)(x) > is a negative number far away from 0 while cos(x) is bounded between -1 and > 1. > Therefore we can guess that this inequality may be also true for other > values of x. > This inequality comes from a book called Problem-Solving Through Problems. > The author of the book assumes the inequality to be true, for the only > reason that the sequence f_0(x), -f_1(x), f_2(x), -f_3(x), ... is an > alternating sequence. ( That the tayler series for cos(x) is an alternating > series.) But the absolut values of the sequence is not steadily decreasing. > it is not true that f_n(x) >= f_(n+1)(x) for all x and for all nonnegative > integer n. > I failed to prove or disprove the inequality. Can somebody help me on this? Induction and patience, and willingness to prove slightly more: Claim: For every n = 0,1,... and for every x>0, cos(x) < 1 - x^2/2! + ... + x^(4*n)/(4*n)! and cos(x) > 1 - x^2/2! + ... - x^(4*n+2)/(4*n+2)! Start with cos(x) <= 1, and I will go only through a half on the induction procedure, leaving the other half to you: Let cos(x) < 1 - x^2/2! + ... + x^(4*n)/(4*n)! integrate over [0, x] to get sin(x) < x - x^3/3! + ... + x^(4*n+1)/(4*n+1)! integrate again, keeping in mind that cos(0)=1: 1 - cos(x) < x^2/2! - ... + x^(4*n+2)/(4*n+2)! which is the second inequality in the induction hypothesis. Integrate twice more. Then you can extend what you want to x<0 because... There should be a neat trick using 2-by-2 matrices of functions but it is rather late in the afternoon. === Subject: Re: cos(x) >= 1 - x^2/2! + x^4/4! - x^6/6! > I think you can clear up this thing considering the function > h(x)=cos(x)-(1 - x^2/2! + x^4/4! - x^6/6!) and looking for its > minimums by means of derivatives. I expect someone has pointed this out - I haven¹t been following the thread - but I think the OP said he knew the result was true for small x - actually it follows from the form of Taylor¹s Theorem with remainder that it is true for -pi/2 < x < pi/2 . If one assumes that, then it is sufÞcient to show h(x) has no zero. If it has a zero then so does h¹(x) by the Mean Value Theorem, and then it follows that so does h¹¹(x) since h¹(0) = 0. But that is the same result with n-2 in place of n. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: E. W. Dijkstra VS. John McCarthy. A rebuttal to Paul Graham¹s web writings. <87oel23mmo.fsf@thalassa.informatimago.com> cygwin32) > Note that lugaru is homophone of loup-garrou which means werewolf, > and which is a name that indeed matches perfectly that creepy beast. Interesting, I used to listen to a band called Loop Guru. I had no idea it had any other meaning beyond, loosely, someone well versed in rhythm loops. Footnotes: http://www.loopguru.demon.co.uk/ -- Steven E. Harris === Subject: perturbation from linearity of vector spaces One idea I mentioned to Andrej Bona and Michael Slawinski a while ago and they didn¹t seem too interested (so it may be no good) is that there could be perturbations from linearity of vector spaces. If you like the idea feel free to run with it but please reference this post. If you follow up on this post to say the idea has been taken, please provide a reference. David http://www.nþd.com/~dalton === Subject: Re: perturbation from linearity of vector spaces >One idea I mentioned to Andrej Bona and Michael Slawinski a >while ago and they didn¹t seem too interested (so it may be no >good) is that there could be perturbations from linearity of vector >spaces. On the face of it, this idea doesn¹t make any sense. Either something is a vector space or it isn¹t. But perhaps what you¹re really thinking about is a differentiable manifold. At each point of the manifold, you have the tangent space which, in a sense, is a linear approximation to the manifold near that point. If what you were thinking of wasn¹t that, you might explain more precisely what it was. 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: perturbation from linearity of vector spaces >>One idea I mentioned to Andrej Bona and Michael Slawinski a >>while ago and they didn¹t seem too interested (so it may be no >>good) is that there could be perturbations from linearity of vector >>spaces. >On the face of it, this idea doesn¹t make any sense. Either something >is a vector space or it isn¹t. Well, that¹s like what old sticks-in-the-mud like me said when we heard about quantum groups, but lots of people seem to have gotten lots of mileage out of them. (None of which has motivated me to learn anything more about them, other than the fact that they aren¹t--usually--groups, but are--in some sense--perturbations from groupishness of groups.) Lee Rudolph === Subject: Re: Twin primes and my prime counting function ... > So using my prime counting function you have a quick proof that no > square besides 4 between 3 and 5, exists between twin primes out to > inÞnity! I thought there was a simpler proof. Suppose n is a square, say m^2. In that case n - 1 = m^2 - 1 = (m - 1)(m + 1) is not prime unless m - 1 = 1 (giving n = 4) or m + 1 = 1 (giving n = 0). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: What is the expert opinion on deriving SR with a Shubertian clock? The theory of relativity really has some mystery in it. It¹s a non-trivial model of space and time. To make our presentation of relativity as easy as possible let¹s Þrst introduce an overview of the essential physics before plunging into the fully detailed derivation. We begin with constructing the simplest clock imaginable. What could be simpler than imagining two pristine, frictionless rulers L and L¹ and imagining one of them sliding on the other at a constant velocity? L¹ --> -9__-8__-7__-6__-5__-4__-3__-2__-1__0__1__2__3__4__5__6__7__8 __-9 <-- L -9__-8__-7__-6__-5__-4__-3__-2__-1__0__1__2__3__4__5__6__7__8 __-9 I haven¹t as yet deÞned what constant velocity means but that¹s ok. We¹ll deÞne that soon enough. For now, just use your imagination. Ok. We¹re done. The timepiece that we¹ve just created is called a Shubertian clock. Here¹s how to understand it. Imagine that you¹re a tiny observer located on ruler L at position x. You¹re familiar with your universe and realize that the moving line L¹ provides a continuous stream of numbers x¹ that are steadily þowing past you. You have always been a keen observer and a topnotch experimentalist and it¹s obvious that the þow of the x¹ s is so constant that you can set your clock by them. Unquestionably, your time T at position x (when x¹ þies by and is directly overhead) is just a function of x and x¹. SpeciÞcally, T is the time at x when x¹ touches x. Let¹s write this as T=T(x,x¹). Here¹s a place where your intuition can solve the math. We¹ve declared that x is a constant so that the time T at x is really just a function of x¹. Also, we can add or subtract any function S(x) to T(x,x¹) and have a perfectly good deÞnition of clock time at x. There¹s no reason to be distracted by unnecessary complexities. Clearly, if T is a function of x¹, then T=-x¹/u is the simplest deÞnition of clock time for our observer at x. This deÞnition of clock time is valid for every point on line L. The only thing we have to do is Þnd the right clock rate constant u so that our array of mathematical clocks will be ticking at the same rate as a real clock. The number u is also called the proper velocity of the line L¹ with respect to L. The theory of relativity is all about the consequences of deÞning time for observers in motion. We now deÞne clock time on the moving line L¹. Consider this triviality. If the line L¹ is sliding in L Œs positive direction, then an observer on L¹ is, in effect, þying over the numbers of L and he will see those numbers going from negative to positive. An observer on L however will see the numbers of line L¹ þying in a reverse order, from positive to negative. It is an important axiom of physics that says that these proper velocities are opposite in sign but equal in absolute value. By symmetry therefore, and by repeating our previous analysis, T¹ = x/u + R(x¹) is the Shubertian clock time at the point x¹ of L¹ and T = -x¹/u + S(x) is the Shubertian clock time at the point x of L. That pretty much summaries all the physical machinery that we need to derive the equations of special relativity. The rest is just high school algebra. It¹s time for physical mathematics. As explained, T=-x¹/u and T¹=x/u are perfectly good deÞnitions of clock times for L and L¹ respectfully. We are also free to reset all clocks everywhere point by point. For an illustration of this power, let¹s do that for L and L¹. Let S(x)=x/u and R(x¹)=-x¹/u. So our new clock time T=-x¹/u +x/u =(x-x¹)/u for L and T¹=x/u-x¹/u =(x-x¹)/u for L¹. Consequently, x¹=x-uT and T¹=T and we¹ve proven a delightful little theorem: Any two inertial frames of reference have a Galilean synchronization. To continue with the powerful magic, and as a prelude to the still upcoming derivation of relativity theory, I will now show how to reset a Shubertian clock so that a Galilean synchronization transforms into an Einsteinian synchronization. Let u = v/sqrt(1-v^2/c^2) and deÞne Y=Y(v)=1/sqrt(1-v^2/c^2) I¹m not claiming to be inserting any principle of physics here. I am merely using the universally accepted mathematical freedom called, change of variables. Another name for it is called substitution. Galilean synchronization means that time on L is deÞned as T=(x-x¹)/u Resetting clocks on L means that we can introduce a new variable t such that t=(x-x¹)/u +Ax Recall that x¹ is the critical parameter in determining clock time on L. Please note that the difference between the old and the new clock time T-t=-Ax. This difference is a mere constant independent of x¹ so all the adjusted clocks on L continue to behave normally. All that we¹ve done is just reset an inÞnite number of clocks, point by point, by innocently adding or subtracting a Þxed constant to the their previously set Galilean clock time. Likewise for L¹ T¹=(x-x¹)/u and t¹=(x-x¹)/u +Bx¹ The difference between these clock times, T¹-t¹=-Bx¹, is independent of x. You understand that x is the critical parameter in determining clock time on L¹. Notice too that all the clock speeds at every point remain unchanged. Take a look then at the two new equations: t=(x-x¹)/u +Ax t¹=(x-x¹)/u +Bx¹ Let A=(Y-1)/vY Let B=-(Y-1)/vY Recall that u=vY Inserting the acceptable value of A and u into the Þrst equation and solving for x¹ gives us x¹=Y(v)(x-vt) Inserting this expression for x¹ into the second equation along with the permissible value of B and u yields t¹=Y(v)(t-vx/c^2) I hope that the reader is impressed at this point that the Lorentz transformation equations have been derived and yet not a word has been said about revolutionary physics. The explanation continues here: http://www.everythingimportant.org/relativity Eugene Shubert === Subject: Re: Errors in Wiles¹s proof of Fermat¹s Last Theorem > > > Professor Escultura disputed [1] Wiles¹s proof[2] of Fermat¹s Last > > > Theorem. Any deÞnite comments about the correct status of the proof > > > of FLT will be of great interest to the number theorists,amatur > > > mathematicians and the public in general. > > > > > > [1]Escultura, E.E.,Exact Solutions of Fermat¹s Equation(DeÞnitive > > > Resolution of Fermat¹s Last Theorem),Nonlinear Studies, vol.5,No.8, > > > September 1998 > > > > > > [2]Wiles, A., Modular Elliptic Curves and Fermat¹s Last Theorem, > > > Ann.of Math.,141(1995) 443-551 > > > > Wiles¹s work is most easily seen to be false by noting that the entire > > approach is logically fallacious. > > > > The technical logical term for the error is, cum hoc, ergo propter > > hoc. > EEE appears to be a kindred spirit. He also claims to have > found fundamental errors in the very basis of modern > mathematics (for example the axioms of the real number > system are false). You miss the sheer irony in this case. James says he has proven that FLT is true, using simple mathematics. EEE says he has proven that FLT is false, using counter-examples. -- 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: Errors in Wiles¹s proof of Fermat¹s Last Theorem > > [[ This message was both posted and mailed: see > > the To, Cc, and Newsgroups headers for details. ]] > > > > > > Our library has Nonlinear Studies available in electronic form. > > Oh, how I envy you. > > > > > > > > > Mathematical Reviews covers this journal. But for the paper > > > cited above, they simply say: > > > {There will be no review of this item.} > > > In fact, for all ten of EEE¹s papers, MR has only a remark like this. > > Your posting has a little sketch of a boy apparently about to lose his > > lunch. Appropriate, I think. > *** > After having all the discussions about EEE and his papers nobody has > yet been able to identify any speciÞc errors in his paper on FLT. He Probably because many people don¹t have access to this journal (at least my university hasn¹t). Why don¹t you put it on a webpage for several days? Wilbert === Subject: Re: Errors in Wiles¹s proof of Fermat¹s Last Theorem > > Professor Escultura disputed [1] Wiles¹s proof[2] of Fermat¹s Last > > Theorem. Any deÞnite comments about the correct status of the proof > > of FLT will be of great interest to the number theorists,amatur > > mathematicians and the public in general. > > > > [1]Escultura, E.E.,Exact Solutions of Fermat¹s Equation(DeÞnitive > > Resolution of Fermat¹s Last Theorem),Nonlinear Studies, vol.5,No.8, > > September 1998 > > > > [2]Wiles, A., Modular Elliptic Curves and Fermat¹s Last Theorem, > > Ann.of Math.,141(1995) 443-551 > Wiles¹s work is most easily seen to be false by noting that the entire > approach is logically fallacious. > The technical logical term for the error is, cum hoc, ergo propter > hoc. > James Harris James: Your standing in judgement of Wiles¹ work is like a Þve-year-old who has just learned how to roast marshmallows over an open Þre telling Julia Child how to make crepes Suzette. The last 6 or 8 posts of yours had zero mathematical content! Zero! None! Nada! Zilch! You know, the cardinality of the set of neurons contained in the vacuum that separates your ears! Why don¹t you pull your head out of your ass, and learn some algebra before you pontiÞcate about a level of mathematics that you couldn¹t understand in 10,000 years of study. === Subject: Re: Errors in Wiles¹s proof of Fermat¹s Last Theorem Where exactly in Wiles¹ proof of FLT is the hypothesis that the exponent n>2 used? If the hypothesis n>2 is not used, then Wiles¹ proof also disproves 3^2+4^2=5^2. Perhaps Wiles¹ proof is wrong? === Subject: Re: Errors in Wiles¹s proof of Fermat¹s Last Theorem > Where exactly in Wiles¹ proof of FLT is the hypothesis that the > exponent n>2 used? If the hypothesis n>2 is not used, then Wiles¹ > proof also disproves 3^2+4^2=5^2. Perhaps Wiles¹ proof is wrong? Here is Ribet¹s answer to that question from last month: -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Errors in Wiles¹s proof of Fermat¹s Last Theorem Discussion, linux) > Where exactly in Wiles¹ proof of FLT is the hypothesis that the > exponent n>2 used? If the hypothesis n>2 is not used, then Wiles¹ > proof also disproves 3^2+4^2=5^2. Perhaps Wiles¹ proof is wrong? Yeah, good point. Bet no one thought of checking that. -- Jesse F. Hughes If anything is true in general about Usenet, it¹s that people can go on and on about just about anything. -- James Harris speaks the truth. === Subject: Re: Errors in Wiles¹s proof of Fermat¹s Last Theorem Am 23.08.04 16:45 schrieb Jesse F. Hughes: >>Where exactly in Wiles¹ proof of FLT is the hypothesis that the >>exponent n>2 used? If the hypothesis n>2 is not used, then Wiles¹ >>proof also disproves 3^2+4^2=5^2. Perhaps Wiles¹ proof is wrong? > Yeah, good point. Bet no one thought of checking that. Yes! I Agree! Someone might collect subscribers for a petition to Wiles, in case he happened to forget the power=2 case over all the ten years... === Subject: Re: Errors in Wiles¹s proof of Fermat¹s Last Theorem <87zn4lvrp1.fsf@phiwumbda.org> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@¹ELIi $t^ VcLWP@J5p^rst0+(Œ>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Am 23.08.04 16:45 schrieb Jesse F. Hughes: > > > > > >>Where exactly in Wiles¹ proof of FLT is the hypothesis that the > >>exponent n>2 used? If the hypothesis n>2 is not used, then Wiles¹ > >>proof also disproves 3^2+4^2=5^2. Perhaps Wiles¹ proof is wrong? > > > > > > Yeah, good point. Bet no one thought of checking that. > > > Yes! I Agree! Someone might collect subscribers for a petition > to Wiles, in case he happened to forget the power=2 case over > all the ten years... Has anybody checked his proof for n=1? I think that would be a most unfortunate omission. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Errors in Wiles¹s proof of Fermat¹s Last Theorem >> Am 23.08.04 16:45 schrieb Jesse F. Hughes: >> > >> > >> >>Where exactly in Wiles¹ proof of FLT is the hypothesis that the >> >>exponent n>2 used? If the hypothesis n>2 is not used, then Wiles¹ >> >>proof also disproves 3^2+4^2=5^2. Perhaps Wiles¹ proof is wrong? >> > >> > >> > Yeah, good point. Bet no one thought of checking that. >> > >> Yes! I Agree! Someone might collect subscribers for a petition >> to Wiles, in case he happened to forget the power=2 case over >> all the ten years... >Has anybody checked his proof for n=1? I think that would be a most >unfortunate omission. Hmm, I hope they considered that and didn¹t get carried away by n=0 and 1+1 != 1. It¹s very easy to over-generalize. I think it¹s called the law of small numbers. Even Fermat made blunders like that. Remember the Fermat numbers! Thomas === Subject: Re: Errors in Wiles¹s proof of Fermat¹s Last Theorem >Where exactly in Wiles¹ proof of FLT is the hypothesis that the >exponent n>2 used? Wiles¹ proof proves the semistable case of the Taniyama-Shimura conjecture. By doing that he indirectly proves FLT, because if there were a integer solution to x^n+y^n=z^n (for n>2, and x,y and z not equal to zero), the Taniyama-Shimura conjecture would be false (this was conjectured by Frey and proved by Ribet IIRC). Since it isn¹t false (which is what Wiles proves), there can be no integer solution, which proves FLT. === Subject: Re: Errors in Wiles¹s proof of Fermat¹s Last Theorem > >Where exactly in Wiles¹ proof of FLT is the hypothesis that the > >exponent n>2 used? This question came up a few months ago and I posted the following response: I am not an expert, but since no expert has responded, here, for what it¹s worth, is my understanding: To get Fermat¹s Last Theorem, you need to combine Wiles¹s proof that the Frey curve is modular with Ribet¹s proof that the Frey curve is not modular (from which it follows that the Frey curve, along with the alleged FLT counterexample from which it arises, cannot exist). The part that fails for n=2 and n=3 is not the Wiles part but the Ribet part. That¹s because Ribet¹s argument requires that the action of G(Q-bar,Q) on the n-torsion in the Frey curve has to be irreducible. For this, one invokes a theorem of Mazur that requires n > 4. I can explain almost nothing about how Mazur uses that assumption. I do not know whether the Frey curves for n=2 are well understood in general, but they are certainly understood in particular cases. For example, the Frey curve associated to the equation 3^2 + 4^2 = 5^2 is the modular curve X_0(15). I am sure that there is some clear intuition due to Frey about why the Frey curves should be modular for n=2 and n=3 but not modular for n equal to a prime greater than 3. Here¹s where we need that expert. Steven E. Landsburg www.landsburg.com/about2.html