mm-589 === Subject: how come calculus can be exact? How come calculus gives the exact results despite we are making approximations(neglecting the infinitesimal which tends to zero) at its basic definition level? I am getting very much frustated over it. Can someone please convince me over the exclusion of the infinitesimal terms from the definition and still getting the correct results.?? === Subject: Re: how come calculus can be exact? >How come calculus gives the exact results despite we are making >approximations(neglecting the infinitesimal which tends to zero) at >its basic definition level? Your question presupposes something contrary to fact. Calculus does not rely on approximations or infinitesimals. You need to read up on limits in order to understand the basic definitions of Calculus. Note: there is something called Nonstandard Analysis in which the term infinitesimal is rigorously used, but unless you have a good textbook using that approach it would only confuse you. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: how come calculus can be exact? >How come calculus gives the exact results despite we are making >approximations(neglecting the infinitesimal which tends to zero) at >its basic definition level? > Your question presupposes something contrary to fact. Calculus does > not rely on approximations or infinitesimals. You need to read up on > limits in order to understand the basic definitions of Calculus. > Note: there is something called Nonstandard Analysis in which the term > infinitesimal is rigorously used, but unless you have a good textbook > using that approach it would only confuse you. Let's say you do use infinitesimals. But understand that a positive infinitesimal is NOT simply a very small ordinary number but a number that is smaller than any ordinary number. More precisely, for any integer n, it is smaller than 1/n. So let us say you calculate the derivative (actually a difference quotient) of x^2 as 2x + h, where h is infinitesimal. Now you throw away the h! Have you lost anything? Yes, but something smaller than any ordinary number. So what you have thrown away is invisible as far as ordinary numbers are concerned. So what is neglected is truly negligeable and are imperceptible. And this is approximately imperceptible, but literally so. === Subject: Re: how come calculus can be exact? <413cf038$1$fuzhry+tra$mr2ice@news.patriot.net> X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 02:02 PM, barr@barrs.org (Michael Barr) said: >Let's say you do use infinitesimals. But understand that a positive >infinitesimal is NOT simply a very small ordinary number but a number >that is smaller than any ordinary number. More precisely, for any >integer n, it is smaller than 1/n. ITYM for any ordinary integer n, it is smaller than 1/n. >Now you throw away the h! More precisely, you take the equivalence class. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: how come calculus can be exact? charset=iso-8859-1 >How come calculus gives the exact results despite we are making >approximations(neglecting the infinitesimal which tends to zero) at >its basic definition level? You're reading the equations of calculus wrong. They're not (precisely) equations but rather shorthand statements about limiting processes. Calculus gives precise results only in the context of these limits. There is, as another poster remarked, something called Nonstandard Analysis in which these infinitesimal quantities are presumed to exist and then Calculus presents true equations within that context. It is, however, non-standard. Norm === Subject: Re: how come calculus can be exact? >How come calculus gives the exact results despite we are making >approximations(neglecting the infinitesimal which tends to zero) at >its basic definition level? > You're reading the equations of calculus wrong. They're not (precisely) > equations but rather shorthand statements about limiting processes. > Calculus gives precise results only in the context of these limits. There > is, as another poster remarked, something called Nonstandard Analysis in > which these infinitesimal quantities are presumed to exist and then Calculus > presents true equations within that context. It is, however, non-standard. > Norm Prior to encountering calculus the only curves that bounded areas were circles, and pi (which is inexact) was always part of the solution. I was ßabbergasted when I solved 1 $ x^2 dx = 1/3 0 and saw no pi! I was certain there was a mistake, after all there was a curve involved, and I mistakenly thought all curved bounded areas must have pi somewhere. If a parabola is a circle with a variable radius, the area must be more complicated than a circles, but it not and I'm still ßabbergasted. Ken === Subject: Re: how come calculus can be exact? > >How come calculus gives the exact results despite we are making >approximations(neglecting the infinitesimal which tends to zero) at >its basic definition level? > > You're reading the equations of calculus wrong. They're not (precisely) > equations but rather shorthand statements about limiting processes. > Calculus gives precise results only in the context of these limits. There > is, as another poster remarked, something called Nonstandard Analysis in > which these infinitesimal quantities are presumed to exist and then Calculus > presents true equations within that context. It is, however, non-standard. > Norm > Prior to encountering calculus the only curves that bounded > areas were circles, and pi (which is inexact) was always > part of the solution. I was ßabbergasted when I solved > 1 > $ x^2 dx = 1/3 > 0 > and saw no pi! I was certain there was a mistake, > after all there was a curve involved, and I mistakenly > thought all curved bounded areas must have pi somewhere. > If a parabola is a circle with a variable radius, > the area must be more complicated than a circles, > but it not and I'm still ßabbergasted. Why do you say pi is inexact? === Subject: Re: how come calculus can be exact? Hi Richard, (and thank's Dave) > Prior to encountering calculus the only curves that bounded > areas were circles, and pi (which is inexact) was always > part of the solution. I was ßabbergasted when I solved > 1 > $ x^2 dx = 1/3 > 0 > and saw no pi! I was certain there was a mistake, > after all there was a curve involved, and I mistakenly > thought all curved bounded areas must have pi somewhere. > If a parabola is a circle with a variable radius, > the area must be more complicated than a circles, > but it not and I'm still ßabbergasted. > Why do you say pi is inexact? Unlike 1/3 =.3333... the *next* digit of pi will forever remain unknown. Using a straight edge and compass is it possible to create two straight lengths in the ratio pi? Ken === Subject: Re: how come calculus can be exact? Originator: grubb@lola > If a parabola is a circle with a variable radius, > the area must be more complicated than a circles, > but it not and I'm still ßabbergasted. > Why do you say pi is inexact? >Unlike 1/3 =.3333... the *next* digit of pi will forever >remain unknown. >Using a straight edge and compass is it possible to create >two straight lengths in the ratio pi? No. But it is possible to create lengths with ratio the square root of 2. Computing the next digit of this is about as hard as for pi. It is *not* possible to create lengths with ratio the *cube* root of 2. Furthermore, if we let x=.1100010000... with all zeroes except for 1's in the n! places (i.e. 1, 2, 6, 24, 120, 720...), then it is not possible to construct lengths with ratio x even though I can tell you every digit of x way ahead of time. Now, you were saying? --Dan Grubb === Subject: Re: how come calculus can be exact? > Hi Richard, (and thank's Dave) > Prior to encountering calculus the only curves that bounded > areas were circles, and pi (which is inexact) was always > part of the solution. I was ßabbergasted when I solved > > 1 > $ x^2 dx = 1/3 > 0 > > and saw no pi! I was certain there was a mistake, > after all there was a curve involved, and I mistakenly > thought all curved bounded areas must have pi somewhere. > > If a parabola is a circle with a variable radius, > the area must be more complicated than a circles, > but it not and I'm still ßabbergasted. > Why do you say pi is inexact? > Unlike 1/3 =.3333... the *next* digit of pi will forever > remain unknown. The *next* digit of pi can be determined by numerous well-known methods. You can look it up. In fact, on the internet, you can look up the digits (up to a billion at least - how exact do you need to be?). > Using a straight edge and compass is it possible to create > two straight lengths in the ratio pi? non sequitur. === Subject: Re: how come calculus can be exact? > Hi Richard, (and thank's Dave) > Prior to encountering calculus the only curves that bounded > areas were circles, and pi (which is inexact) was always > part of the solution. I was ßabbergasted when I solved > > 1 > $ x^2 dx = 1/3 > 0 > > and saw no pi! I was certain there was a mistake, > after all there was a curve involved, and I mistakenly > thought all curved bounded areas must have pi somewhere. > > If a parabola is a circle with a variable radius, > the area must be more complicated than a circles, > but it not and I'm still ßabbergasted. > Why do you say pi is inexact? > Unlike 1/3 =.3333... the *next* digit of pi will forever > remain unknown. Pi is a computable number. Any particular digit can be computed. There are lots of real numbers that are not computable in this sense (most of them, in fact), but even those numbers are exact in the sense that they have exact decimal representations, even if we can't find out what they are. > Using a straight edge and compass is it possible to create > two straight lengths in the ratio pi? > Ken Computable does not mean geometrically constructible. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: how come calculus can be exact? > Prior to encountering calculus the only curves that bounded > areas were circles, and pi (which is inexact) was always On the contrary, pi is exact. It's only approximations to pi that are inexact. > part of the solution. I was ßabbergasted when I solved > 1 > $ x^2 dx = 1/3 > 0 > and saw no pi! I was certain there was a mistake, > after all there was a curve involved, and I mistakenly > thought all curved bounded areas must have pi somewhere. By a number that has pi in it, I take it you mean a rational multiple of pi. If you consider the entire family of curves passing through the points (0,0) and (1,1), and confined to the unit square, it should be clear that every number between 0 and 1 is the area under some curve in the family. Clearly, only countably many of those areas can be rational multiples of pi. > If a parabola is a circle with a variable radius, > the area must be more complicated than a circles, > but it not and I'm still ßabbergasted. It's related to the fact that y = x^2 is simpler than y = sqrt(1-x^2). -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: how come calculus can be exact? > If a parabola is a circle with a variable radius, > the area must be more complicated than a circles, > but it not and I'm still ßabbergasted. >It's related to the fact that y = x^2 is simpler than y = sqrt(1-x^2). I'm not sure we can push that argument too far. After all, y = 1/x is simpler still, but the area under it is transcendental (between rational boundaries). -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: how come calculus can be exact? >How come calculus gives the exact results despite we are making >approximations(neglecting the infinitesimal which tends to zero) at >its basic definition level? >I am getting very much frustated over it. >Can someone please convince me over the exclusion of the infinitesimal >terms from the definition and still getting the correct results.?? The idea of the limit is to decide what happens as the error term (infinitesimal) is made smaller and smaller. If you would post a more specific question, I and others could probably give you more useful help. I'd suggest also /Hitchhiker's Guide to the Calculus/, which is good at explaining the concepts of first-semester calculus in ordinary language. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: how come calculus can be exact? > How come calculus gives the exact results despite we are making > approximations(neglecting the infinitesimal which tends to zero) at > its basic definition level? > I am getting very much frustated over it. > Can someone please convince me over the exclusion of the infinitesimal > terms from the definition and still getting the correct results.?? You really need to study limits. For differentiation, an approximation based on a small value h becomes an exact solution when h tends to zero. You can substitute in actual values of .001 and smaller to see how the solution approaches the exact value. Ex. for y = x^2, calculate dy buy 2.001. As you decrease the difference to say x = 1 and x = 1.0001, the closer you get to the exact answer of 2. PH === Subject: Re: how come calculus can be exact? >How come calculus gives the exact results despite we are making >approximations(neglecting the infinitesimal which tends to zero) at >its basic definition level? >I am getting very much frustated over it. >Can someone please convince me over the exclusion of the infinitesimal >terms from the definition and still getting the correct results.?? The main idea is that of a limit. You aren't just making approximations, you are taking a limit of approximations as something (e.g. Delta x) goes to 0. In any good approximation method, the limit of the approximations is the exact result. 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: how come calculus can be exact? X-RFC2646: Original > How come calculus gives the exact results despite we are making > approximations(neglecting the infinitesimal which tends to zero) at > its basic definition level? > I am getting very much frustated over it. > Can someone please convince me over the exclusion of the infinitesimal > terms from the definition and still getting the correct results.?? I tried to answer, but I think I need to know a little more. Perhaps you could provide the example that you are having trouble with? === Subject: Algebra Helper I just looked up Weber's Lehrbuch der Algebra at Amazon.com. Near the bottom of the page, there was an advertisement that read: Algebra Helper solves your Lehrbuch Der Algebra problems step-by-step! (Price $29.00) This and The Only Math Book You'll Ever Need... :) -- Allan Adler * Disclaimer: I am a guest and *not* a member of the MIT CSAIL. My actions and * comments do not reßect in any way on MIT. Also, I am nowhere near Boston. === Subject: Too funny :-) An American friend of mine showed me this website today: http://ratemyprofessors.com/ . Of course, I tried to check the ratings of the professors of sci.math; then I picked Robert Israel (one that seems among the most very prominent and interesting, and the only one I could find among sci.math posters). Check it out it's too funny; while some rare (probably good) students seem to be very happy with him, others would say: the worst math prof i ve ever seen ... but he's a nice guy ... just can't teach or: I took this class with an interest in linear programming. Israel killed it for me, or just horrible, or: (the most funny rating to me) try to avoid this guy, u have to spend couple hours on the hw and the mt is not easy. O shot, final is even more worst. My big deception is that I couldn't find David Ullrich's rating on this website ... ahahah -- Julien Santini === Subject: Re: Too funny :-) >An American friend of mine showed me this website today: >http://ratemyprofessors.com/ . Of course, I tried to check the ratings of >the professors of sci.math; then I picked Robert Israel (one that seems >among the most very prominent and interesting, and the only one I could find >among sci.math posters). >Check it out it's too funny; while some rare (probably good) students seem >to be very happy with him, others would say: >the worst math prof i ve ever seen ... but he's a nice guy ... just can't >teach or: >I took this class with an interest in linear programming. Israel killed it >for me, or >just horrible, or: >(the most funny rating to me) try to avoid this guy, u have to spend couple >hours on the hw and the mt is not easy. O shot, final is even more worst. >My big deception is that I couldn't find David Ullrich's rating on this >website ... ahahah yeah, i was disappointed by that when i looked once. i figure the kids just can't bring themselves to come up with the sort of effusive praise that would be required to give an accurate rating. [or possibly there's a moderator who checks for libel, death threats, etc.] ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Too funny :-) > http://ratemyprofessors.com/ . No rating for me, great. I looked at a few from Ohio State. Sample of worst-rated instructor: Hard grader. Expects you to know everything! Sample of top-rated instructors: This guy is insane. He could intimidate prison inmates. However, he is an excellent instructor. ... He wants nobody but genuine mathematical geniuses in his class, and uses public insult as a disincentive for all others. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Too funny :-) >http://ratemyprofessors.com/ . My college has students rate their teachers. It's very nearly useless; certainly the useful information I've got has never justified the loss of class time to the exercise. The problem is that students don't actually rate the professor -- they simply give good marks if they like their grade and bad marks if they don't like their grade. Sure, there's the occasional exception, but the great majority of students simply aren't interested in taking the time to do a reasonable evaluation. Having students rate profs actually does great harm. Since the ratings are positively correlated with student grades, the system puts extra pressure on profs to give higher grades. Grade inßation benefits no one. H.H. Bauer has a thoughtful analysis of the problems with student evaluation of profs (and other educational problems) at http://www.bus.lsu.edu/accounting/faculty/lcrumbley/study.htm Please understand that I'm not disagreeing with the idea of holding profs accountable for the quality of their teaching. I _am_ saying that student evaluations are a good way to identify an easy grader but a lousy way to identify a good teacher. Think back: Did you learn more from easy graders, or from profs who challenged you to do more than you knew you could? >(the most funny rating to me) try to avoid this guy, u have to spend couple >hours on the hw and the mt is not easy. O shot, final is even more worst. Horrors! He assigns homework! -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: Too funny :-) > Having students rate profs actually does great harm. Since the > ratings are positively correlated with student grades, the system > puts extra pressure on profs to give higher grades. Grade inßation > benefits no one. I'm not denying that there is a correlation between student grades and evaluations, but my sense has always been that high evaluations are strongly correlated to the entertainment value of the instructor. That is, the instructor that tells good jokes, has a good speaking voice, and a stage presence will get a much better review than the non effective speaker regardless of how hard they work the students or how much the students actually learn. Now entertainment value does have its uses--the more highly entertaining lecturer might have an easier time motivating the students, for example. Also, students of all ability levels might come out of the class more satisfied, which administrators love, of course. On the other hand, I was quite surprised when I substituted for the person recognized as being the best teacher in our department once. He is clearly has the highest entertainment value, but he probably (most of the time) is effective in presenting the material. But in this particular class, I mentioned that I was surprised at how little material they had covered so far. The class told me they found a trick: when the professor told a joke or funny story, they would laugh and say Tell us another! It turns out that he had difficulty resisting this, and waisted many classes. For the record, I have some pretty terrible evaluations at http://ratemyprofessors.com/ I seem to have gotten the most disgruntled half dozen students in a large lecture of calculus to write in. John === Subject: Re: Too funny :-) Too right, there are those that speak, admittedly good presenters, and those that think. The intersection is small, but they do exist apparently. If we go by the published literature, Richard Feynman was a good example of someone who was good at both. On the other hand, it seems he was also a crap tutor? One thing is certain, Politicians can speak, and that's about it. > Having students rate profs actually does great harm. Since the > ratings are positively correlated with student grades, the system > puts extra pressure on profs to give higher grades. Grade inßation > benefits no one. > I'm not denying that there is a correlation between student grades > and evaluations, but my sense has always been that high evaluations > are strongly correlated to the entertainment value of the instructor. > That is, the instructor that tells good jokes, has a good speaking > voice, and a stage presence will get a much better review than the > non effective speaker regardless of how hard they work the students > or how much the students actually learn. > Now entertainment value does have its uses--the more highly > entertaining lecturer might have an easier time motivating the > students, for example. Also, students of all ability levels might > come out of the class more satisfied, which administrators love, > of course. > On the other hand, I was quite surprised when I substituted for > the person recognized as being the best teacher in our department > once. He is clearly has the highest entertainment value, but he > probably (most of the time) is effective in presenting the material. > But in this particular class, I mentioned that I was surprised at how > little material they had covered so far. The class told me they found > a trick: when the professor told a joke or funny story, they would > laugh and say Tell us another! It turns out that he had difficulty > resisting this, and waisted many classes. > For the record, I have some pretty terrible evaluations at > http://ratemyprofessors.com/ I seem to have gotten the most > disgruntled half dozen students in a large lecture of calculus > to write in. > John === Subject: Re: Too funny :-) > Too right, there are those that speak, admittedly good presenters, and those > that think. The intersection is small, but they do exist apparently. If we > go by the published literature, Richard Feynman was a good example of > someone who was good at both. On the other hand, it seems he was also a crap > tutor? > One thing is certain, Politicians can speak, and that's about it. I recently read Feynman's Five Easy Pieces and Five Not So Easy Pieces. These were based on a course he gave in freshman physics that's become something of a legend. I think they said in the introduction that the students didn't much care for it, but the large number of graduate students and faculty sitting in the course thought it was fantastic. John === Subject: Re: Too funny :-) > Too right, there are those that speak, admittedly good presenters, and those > that think. The intersection is small, but they do exist apparently. If we > go by the published literature, Richard Feynman was a good example of > someone who was good at both. On the other hand, it seems he was also a crap > tutor? > One thing is certain, Politicians can speak, and that's about it. > I recently read Feynman's Five Easy Pieces and Five Not So Easy > Pieces. These were based on a course he gave in freshman physics > that's become something of a legend. I think they said in the > introduction that the students didn't much care for it, but the large > number of graduate students and faculty sitting in the course thought > it was fantastic. I heard that about The Feynman Lectures on Physics. My Freshman physics prof (I was a physics major at the time) recommended The Feynman Lectures for additional reading after you understood the given course material. === Subject: Re: Too funny :-) I have to admit, I read Feynman's stuff in hindsight. It was great when I knew most of it. I would never use it as a text book....There is a big element of jealously though.Oh to be such a smart ass. Oh well, back to Math I guess (being UK I would normally say Maths...) I'll post som'ink in a few days Greene - any relation (to Mr Strings)? Enjoy those axioms, beats subjectivity any day > Too right, there are those that speak, admittedly good presenters, and > those > that think. The intersection is small, but they do exist apparently. If > we > go by the published literature, Richard Feynman was a good example of > someone who was good at both. On the other hand, it seems he was also a > crap > tutor? > > One thing is certain, Politicians can speak, and that's about it. > I recently read Feynman's Five Easy Pieces and Five Not So Easy > Pieces. These were based on a course he gave in freshman physics > that's become something of a legend. I think they said in the > introduction that the students didn't much care for it, but the large > number of graduate students and faculty sitting in the course thought > it was fantastic. > I heard that about The Feynman Lectures on Physics. My Freshman physics > prof (I was a physics major at the time) recommended The Feynman Lectures > for additional reading after you understood the given course material. === Subject: Re: Too funny :-) * Stan Brown >http://ratemyprofessors.com/ . > My college has students rate their teachers. It's very nearly > useless; certainly the useful information I've got has never > justified the loss of class time to the exercise. > The problem is that students don't actually rate the professor -- > they simply give good marks if they like their grade and bad marks > if they don't like their grade. Sure, there's the occasional > exception, but the great majority of students simply aren't > interested in taking the time to do a reasonable evaluation. Having been rated I certainly found it useful. The rates were given on the final lecture, but prior to the actual exam. Most of the ratings were given as points on a predefined scale, but a few free texts fields were possible. This lead to that only useful comments were in practice included. -- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92 === Subject: Re: Too funny :-) > Grade inßation benefits no one. 1) students are happier because they get better grades. 2) parents are happier because their kids are doing better. 3) teachers are happier because they get better evals. 4) colleges are happier because their students and teachers are better. 5) employers are happier beccause the students they hire have better GPAs. What's not to like? -- Mitch Harris (remove q to reply) === Subject: Re: Too funny :-) > Grade inßation benefits no one. When writing a jest in Usenet, you nearly always need a smiley to let people know you're not serious. >1) students are happier because they get better grades. Children are happier eating ice cream than broccoli. Once our species left the savannahs of Africa, what makes us happy has not been a reliable guide to what makes us healthy and safe. >2) parents are happier because their kids are doing better. Parents know in their hearts that the kids are not doing better. >3) teachers are happier because they get better evals. But they get pressures from colleagues to maintain standards. And those who cave in and give easy undeserved grades know in their hearts that they are doing the wrong thing. I'm currently under tremendous pressure from a student at another college who took a course at my college this summer and wants his/her D+ raised to a C- despite having actually ßunked the final exam. Ôm glad to say that the academic dean is supporting me in saying no. >4) colleges are happier because their students and teachers are better. No, colleges with a reputation as easy diploma mills are happy because students ßock to them and they make lots of money -- in the short term. In the longer term, they find that their reputation is poisoned because other institutions stop accepting their graduates at face value and stop accepting their courses for transfer credit. (This is why the academic dean supports holding to standards -- many of our students transfer to other institutions, and if we hand out grades like lollipops that will be jeopardized.) >5) employers are happier beccause the students they hire >have better GPAs. Funny, I think employers care about whether employees can do the work. An engineer who got a B in calculus despite believing (x+2)^2 equals x^2+4 is going to get his employer into a huge lawsuit when the bridge or office building comes crashing down. Then no one will be happy. Once again, happy is not the criterion, though we'd certainly rather see people happy than not. Educated is the criterion. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: Too funny :-) Originator: harris@tcs.inf.tu-dresden.de (Mitchell Harris) > Grade inßation benefits no one. Yes, missing smiley. Sorry. Touched a nerve, eh? The whole thing bugs me too. But, the smiley is more about the situation, because frankly, line for line, I think each of those points were true (for various personal standards of truth. Oh yeah, :) ) >When writing a jest in Usenet, you nearly always need a smiley to >let people know you're not serious. but sometimes that detracts from the full humor effect. >1) students are happier because they get better grades. >Children are happier eating ice cream than broccoli. The conscientious students who get a better grade than expected (from their own self reßection) wil be surprised. But they won't complain. >2) parents are happier because their kids are doing better. >Parents know in their hearts that the kids are not doing better. parents have -no- clue. how could they possibly have any idea? >3) teachers are happier because they get better evals. >But they get pressures from colleagues to maintain standards. And >those who cave in and give easy undeserved grades know in their >hearts that they are doing the wrong thing. aren't the evals (not the online ones but the ones students do in class for the official university records) done -before- final exams and final grades? and the results of the evals come sometime in the midle of the next semester. so the connect between maintaining standards and the rush (or depression) of seeing the eval results is a bit large. >I'm currently under tremendous pressure from a student at another >college who took a course at my college this summer and wants >his/her D+ raised to a C- despite having actually ßunked the final >exam. be thankful a Ôconcerned' parent is not involved. >'m glad to say that the academic dean is supporting me in >saying no. >4) colleges are happier because their students and teachers are better. >No, OK, this one was a stretch. but do those college ranking thingies give avg GPA? I wonder what the correlation stats are for those. or do the stats makers realize how meaningless that would be? -- Mitch === Subject: Re: Too funny :-) <2q2kfaFqi0snU1@uni-berlin.de> Discussion, linux) > Grade inßation benefits no one. > When writing a jest in Usenet, you nearly always need a smiley to > let people know you're not serious. Like hell. His joke was evident to absolutely everyone except those that think it can't be funny unless a visual soundtrack says it's funny. Some of us find smileys ugly little crutches that promote bad writing and bad reading skills. They prompt people to do things like respond point by point to obvious jokes as if they were serious. (I won't digress into the writing sins they promote --- save this one: more often than not, they mean nothing more than, you cannot take offense because here is a smiley.) Please don't take it for granted that all of Usenet agrees smileys are a good thing. -- I am the barbarian at the gates, raw creative force, willpower, and the will to fight for the truth no matter what, no matter who stands against me, no matter how many of you band [...] together in your weakness to fight against the math. -- James S. Harris === Subject: Re: Too funny :-) > Some of us find smileys ugly little crutches that promote bad writing > and bad reading skills. They prompt people to do things like respond > point by point to obvious jokes as if they were serious. (I won't > digress into the writing sins they promote --- save this one: more > often than not, they mean nothing more than, you cannot take offense > because here is a smiley.) AKA ha ha only serious. Phil -- They no longer do my traditional winks tournament lunch - liver and bacon. It's just what you need during a winks tournament lunchtime to replace lost === Subject: Re: Too funny :-) Originator: harris@tcs.inf.tu-dresden.de (Mitchell Harris) > Grade inßation benefits no one. > When writing a jest in Usenet, you nearly always need a smiley to > let people know you're not serious. >Like hell. >His joke was evident to absolutely everyone except those that think it >can't be funny unless a visual soundtrack says it's funny. What if I were serious? Back to math... is it totally obvious to everyone that the ratio of the diameter to the circumference of a circle is a constant? -- Mitch === Subject: Re: Too funny :-) Couldn't agree more, if you need a smiley, you need to polish up your English. I'd also like to banish mobile phone text abbreviations. Look mum, no smileys Even Grumpier old'ish man > Grade inßation benefits no one. > When writing a jest in Usenet, you nearly always need a smiley to > let people know you're not serious. > Like hell. > His joke was evident to absolutely everyone except those that think it > can't be funny unless a visual soundtrack says it's funny. > Some of us find smileys ugly little crutches that promote bad writing > and bad reading skills. They prompt people to do things like respond > point by point to obvious jokes as if they were serious. (I won't > digress into the writing sins they promote --- save this one: more > often than not, they mean nothing more than, you cannot take offense > because here is a smiley.) > Please don't take it for granted that all of Usenet agrees smileys are > a good thing. > -- > I am the barbarian at the gates, raw creative force, willpower, and > the will to fight for the truth no matter what, no matter who stands > against me, no matter how many of you band [...] together in your > weakness to fight against the math. -- James S. Harris === Subject: Re: Too funny :-) Older students are not happy because they have to suffer brats boasting they have (in the UK), six grade A, A levels. The norm 30 years ago was 3 or 4 if you were ly! Yet, my old Physics Dept relatively recently started 4 year courses and remedial classes because entrants couldn't do calculus with much ßuency (if any). Working also in Engineering, we get confident graduates who aren't much better, their confidence is quite extraordinary though. Apparently, A level grades in the UK have increased every year for the past 20+ years. A stunning improvement in intelligence. A nation of geniuses if you believe the Politicians. If you believe them that is... Whinge moan Richard Miller > Grade inßation benefits no one. > 1) students are happier because they get better grades. > 2) parents are happier because their kids are doing better. > 3) teachers are happier because they get better evals. > 4) colleges are happier because their students and teachers are better. > 5) employers are happier beccause the students they hire > have better GPAs. > What's not to like? > -- > Mitch Harris > (remove q to reply) === Subject: Re: Too funny :-) <2q2kfaFqi0snU1@uni-berlin.de> > Grade inßation benefits no one. > 1) students are happier because they get better grades. > 2) parents are happier because their kids are doing better. > 3) teachers are happier because they get better evals. > 4) colleges are happier because their students and teachers are better. > 5) employers are happier because the students they hire > have better GPAs. > What's not to like? The investors in the company that had a building, a plane, a power plant, or whatever that was designed by an uneducated engineer with a we're all so very happy quick easy degree, that collapsed, crashed, exploded, or whatever, will be most unhappy and incidentally the victims thereof and their families may also be unhappy. In addition the insurers will definitely be unhappy, however the lawyers handling the law suits will be happy. They'll be happy to sue the company. Then the company will sue the college or have I got that backwards? The company will sue the engineer who'll have to sue the college? === Subject: Re: Too funny :-) > The investors in the company that had a building, a plane, a power plant, > or whatever that was designed by an uneducated engineer with a we're all > so very happy quick easy degree, that collapsed, crashed, exploded, or > whatever, will be most unhappy and incidentally the victims thereof and > their families may also be unhappy. In addition the insurers will > definitely be unhappy, however the lawyers handling the law suits will be > happy. > They'll be happy to sue the company. Then the company will sue the > college or have I got that backwards? The company will sue the engineer > who'll have to sue the college? ...who'll sue the prof, who'll sue the student, who happened to be a victim), whose parents will pay with the original settlement. As long as interest rates are higher than inßation... What I want to see is a lawyer suing themselves for malpractice (in that very suit). -- Mitch Harris (remove q to reply) === Subject: Re: Too funny :-) X-RFC2646: Original > An American friend of mine showed me this website today: > http://ratemyprofessors.com/ . Of course, I tried to check the ratings of > the professors of sci.math; then I picked Robert Israel (one that seems > among the most very prominent and interesting, and the only one I could > find > among sci.math posters). > Check it out it's too funny; while some rare (probably good) students seem > to be very happy with him, others would say: I looked up my old school and found my senior thesis advisor, John Conway of Princeton, listed. He had one rating. It was something like good. That's just silly. He's a great teacher. That list is of very limited usefulness. Michael === Subject: Re: Too funny :-) >showed me this website today: >http://ratemyprofessors.com/ . Of course, I tried to check the ratings of >the professors of sci.m Be cautious about relying too much on the ratings information in any such website. They can be somewhat misleading. Keep in mind that Mathematics and some other subjects are difficult to study and so many students may give lower ratings becaue of the difficulty of the material. The student might not be able to separate the quality of the instructor from the easiness of his instruction. G c === Subject: Re: Too funny :-) > Be cautious about relying too much on the ratings information in any such > website. They can be somewhat misleading. Keep in mind that Mathematics and > some other subjects are difficult to study and so many students may give lower > ratings becaue of the difficulty of the material. The student might not be > able to separate the quality of the instructor from the easiness of his > instruction. > G c Obviously, obviously ... === Subject: Re: Too funny :-) >An American friend of mine showed me this website today: >http://ratemyprofessors.com/ . Of course, I tried to check the ratings of >the professors of sci.math; then I picked Robert Israel (one that seems >among the most very prominent and interesting, and the only one I could find >among sci.math posters). >Check it out it's too funny; while some rare (probably good) students seem ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Naw, just a bunch of nerds. Thomas ;-) >to be very happy with him, others would say: >the worst math prof i ve ever seen ... but he's a nice guy ... just can't >teach or: >I took this class with an interest in linear programming. Israel killed it >for me, or >just horrible, or: >(the most funny rating to me) try to avoid this guy, u have to spend couple >hours on the hw and the mt is not easy. O shot, final is even more worst. >My big deception is that I couldn't find David Ullrich's rating on this >website ... ahahah === Subject: Euler's lemma ?? I came across this simple problem. 4. Let a, b in N and let d = gcd(a,b). Set l = ab/d. Show that l is least common multiple (lcm) of a and b in the sense that the following two properties hold: (i) a | l and b | l; (ii) if a | m and b | m then l | m. (You might find Euler's Lemma useful.) I am not familiar with Euler's Lemma, unless he means Euler's thm. x^phi(n) = 1 if x in Z*_n. Can this be what he means? I don't see that it applies to lcm. === Subject: Re: Euler's lemma ?? > Let a,b in N and let d = (a,b) be their gcd. > Show m = ab/d is the lcm [a,b] of a,b in the > sense that the following two properties hold: > (i) a,b | m > (ii) a,b | n => m | n. Use these dual characterizations of lcm,gcd: GCD: c = (a,b) iff n|a,b <=> n|c [G] LCM: c = [a,b] iff a,b|n <=> c|n [L] Now a,b|n <=> ab| an,bn <=> ab|(an,bn) via [G] <=> ab|(a,b)n via *,gcd distributivity <=> ab/(a,b)|n So by [L], c = ab/(a,b) = [a,b]. QED Notice that this proof works in any gcd-domain, which needn't be true if one uses more special properties of Z, e.g. PID (via Bezout Identity). --Bill Dubuque === Subject: Re: Euler's lemma ?? === Subject: Euler's lemma ?? >4. Let a, b in N and let d = gcd(a,b). Set l = ab/d. Show that l is >least common multiple (lcm) of a and b in the sense that the >following two properties hold: >(i) a | l and b | l; (ii) if a | m and b | m then l | m. >(You might find Euler's Lemma useful.) >I am not familiar with Euler's Lemma, unless he means Euler's thm. I think he means the computation from the Euclidean algorithm that determines the a,b with an + bm = (n,m) ---- === Subject: Re: Euler's lemma ?? days. My association with the Department is that of an alumnus. >I came across this simple problem. >4. Let a, b in N and let d = gcd(a,b). Set l = ab/d. Show that l is >least common multiple (lcm) of a and b in the sense that the following >two >properties hold: >(i) a | l and b | l; (ii) if a | m and b | m then l | m. >(You might find Euler's Lemma useful.) >I am not familiar with Euler's Lemma, unless he means Euler's thm. >x^phi(n) = 1 if x in Z*_n. Can this be what he means? >I don't see that it applies to lcm. I don't either, and I'm not sure what they mean by Euler's Lemma. It seems easiest just to prove the result. Write a = dr b = ds, with gcd(r,s)=1. We are asked to show that dmn is lcm(a,b). Clearly a and b divide drs. Now assume that a and b divide m. Then dr, ds divide m. Write m = dr*u = ds*v. Then dru = dsv, so ru = sv. That means that s|ru, and since gcd(r,s)=1, then s|u. Therefore, writing u = s*w, we have m= drs*w, proving that drs divides m, as claimed. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: conspiracy in sci.math Wondering about the stubbornness some people produce rubbish here, and more wondering about the people answering with still more stubbornness - a phantastic idea. It's conspiracy. With conspiracy one can explain everything. On the other hand, a lot of public money ßows into this sector. Some might remember the times of cold war - on short-wave radio some voices were reading endless columns of cyphers. That were encoded messages for secret agents. Today most is done over the internet. Now, as many of the so called intelligence people are mathematicians (poor fellows), they read and sometimes write for sci.math anyhow, especially about prime numbers. So, what is nearer to the hand, as using this medium for messages too. Wondering about obscure text underneath the name, that might be their private key. And wondering about swearing and rude language? Is that their public key? They might even recognize new members by this. And as -seemigly - a new thread is always started by their headquarter or boss, hope they do not think, i'm one of them. Hero P.S. If this leads to a reduction of rubbish, i've made my point. === Subject: Re: conspiracy in sci.math >Wondering about the stubbornness some people produce rubbish >here, and more wondering about the people answering with still >more stubbornness - a phantastic idea. It's conspiracy. Never assume conspiracy when sheer foolishness is an adequate explanation. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: conspiracy in sci.math Hero > Wondering about the stubbornness some people produce rubbish > here, and more wondering about the people answering with still > more stubbornness - a phantastic idea. It's conspiracy. With > conspiracy one can explain everything. On the other hand, a lot > of public money ßows into this sector. Some might remember the > times of cold war - on short-wave radio some voices were reading > endless columns of cyphers. That were encoded messages for secret > agents. I remember. This sort of thing: DOMOD TOGIL TEKOL PUMEL KEBIL LIMOD LOTOL TOMOD DOSEL MANOL TOMOD GODUL SUBOL MOLOD which deciphers into this: Unfortunately, I've yet to see a poster with the abilities to challenge any of my work. which you might have read before, right here on sci.math. The cipher system? Hint: 14*5*14*5*14 is slightly more than 2^16. > Today most is done over the internet. > Now, as many of the so called intelligence people are > mathematicians (poor fellows), they read and sometimes write > for sci.math anyhow, especially about prime numbers. So, what > is nearer to the hand, as using this medium for messages too. > Wondering about obscure text underneath the name, that might be > their private key. And wondering about swearing and rude language? > Is that their public key? They might even recognize new members by > this. And as -seemigly - a new thread is always started by their > headquarter or boss, hope they do not think, i'm one of them. > Hero > P.S. If this leads to a reduction of rubbish, i've made my point. It won't reduce the rubbish, but nothing will, thus you might have a point nonetheless :) LH === Subject: Re: conspiracy in sci.math > Wondering about the stubbornness some people produ c > e rubbish here, and m o > re wondering about the people a > n > swering with s > till > more stubbornness - a p > hantast i > c idea. It's conspi r > acy. With > conspiracy one can explain everything. On the other h a > nd, a lot > of public money ßows into this se c > tor. Some might remember the > times of cold war - on short-wave radio some voices were reading > endless columns of c y > phers. That were encoded messages for secret > agents. Today most is done over the internet. > Now, as many of the so called intelligence people are > mathematicians (poor fellows), they read and sometimes write > for sci.math anyhow, especially about prime numbers. So, what > is nearer to the hand, as using this medium for messages too. > Wondering about obscure text underneath the name, that might be > their private key. And wondering about swearing and rude language? > Is that their public key? They might even recognize new members by > this. And as -seemigly - a new thread is always started by their > headquarter or boss, hope they do not think, i'm one of them. > Hero > P.S. If this leads to a reduction of rubbish, i've made my point. Well, it lead to the production of my rubbish above. Sorry you didn't make the point :-)) How about some math? How far is P(k) = k(k+1)(k+2)(k+3)(k+4) from the nearest square below? Or from the next square above? The latter distance next^2 - P(k) seems especially interesting, because it is a square Ômost of the time' *hehe* r.rosenthal@web.de === Subject: Distance from square (was: conspiracy in sci.math) >How about some math? How far is P(k) = k(k+1)(k+2)(k+3)(k+4) from >the nearest square below? Or from the next square above? The latter >distance next^2 - P(k) seems especially interesting, because it >is a square Ômost of the time' *hehe* I find it's true for 0 <= k <= 17, but rather sporadically after that. Asymptotically, sqrt(P(k)) = k^(5/2) + 5 k^(3/2) + 5 k^(1/2) - 1/2 k^(-3/2) + ... When k = x^2 is a square, next = x^5 + 5 x^3 + 5 x for sufficiently large x (and it seems to be the case for all nonnegative integers x). In this case next^2 - P(k) = x^2 is a square. Also when k = x^2 - 4, next = x^5 - 5 x^3 + 5 x, with next^2 - P(k) = x^2. And when k = 4 x^2 - 2, next = 32 x^5 - 5 x, with next^2 - P(k) = 9 x^2. 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: Distance from square (was: conspiracy in sci.math) > How far is P(k) = k(k+1)(k+2)(k+3)(k+4) from the nearest > square below? Or from the next square above? The latter > distance next^2 - P(k) seems especially interesting, > because it is a square Ômost of the time'. > I find it's true for 0 <= k <= 17, but rather sporadically after that. These nice differences for 0 <= k <= 17 in the case of five consecutive factors made me curious. One sees that in these cases the prime factors can be grouped such that one part gives the product L1, the other part gives product L2. And L1 and L2 are quite near to each other. And: (L1+L2)/2 = L is integer, i.e. P(k) = L1*L2 = (L-t)(L+t) for some integer value t (not too large). This gives P = L^2 - t^2. Example k = 17: L1 = 5*17*18 = 1530, L2 = 4*19*21 = 1596, L = 1563, t = 33. This gives 17*18*19*20*21 = 1563^2 - 33^2 > 1562^2. For k = 18 the closest partitioning I could find (nice puzzle!) was L1 = 5*18*19 and L2 = 4*21*22. ( This is a nice puzzle when you try to get back to the prime divisors and divide them as well as possible.) We then have L1 = 1779-69 and L2 = 1779+69 giving distance 69^2 to the square 1779^2 above. But the next square above is 1778^2. Seems like a dead end. Can we repair something in the reasoning? > Asymptotically, > sqrt(P(k)) = k^(5/2) + 5 k^(3/2) + 5 k^(1/2) - 1/2 k^(-3/2) + ... > When k = x^2 is a square, next = x^5 + 5 x^3 + 5 x for sufficiently > large x (and it seems to be the case for all nonnegative integers x). > In this case next^2 - P(k) = x^2 is a square. > Also when k = x^2 - 4, next = x^5 - 5 x^3 + 5 x, with next^2 - P(k) = x^2. > And when k = 4 x^2 - 2, next = 32 x^5 - 5 x, with next^2 - P(k) = 9 x^2. I like the name SquareMod and would be very pleased to enter the sequence a(n) = minimum of all SquareMod(P) where P is a product of n consecutive positive integers in the OEIS ... I strongly doubt Neil's for the numbers i(i+1)(i+2)...j mod Ôbiggest square' are randomly distributed, no?. They surely are less random than primes :-) Rainer Rosenthal r.rosenthal@web.de === Subject: Re: conspiracy in sci.math Discussion, linux) [...] > P.S. If this leads to a reduction of rubbish, i've made my point. Is this fighting fire with fire or something? -- Jesse F. Hughes [Lancelot] sighed, defeated. ÔIt is as practical to hurry an acorn toward treeness as to urge a damsel when her mind is set.' -- John Steinbeck, /The Acts of King Arthur and His Noble Knights/ === Subject: Motives Does any one know of a reasonably accessible introduction to Motives? Isaac Khabaza === Subject: Re: Motives > Does any one know of a reasonably accessible introduction to Motives? > Isaac Khabaza Allan Adler refers to the unclear nature of motives. I have seen similar comments. Indeed the massive Encyclopedic Dictionary of Mathematics (2nd edition) has no entry for motives. There is a rumour that motives refer to a form of YOGA practised by its inventor. Be that as it may, there is a motives epidemic raging in algebraic geometry: pure, mixed, Lefschetz motives, Chow motives and many more. A truce should be called, please no more motives until a reasonably accessible and comprehensible definition is published preferably in PDF format on the WEB. Isaac Khabaza === Subject: Re: Motives >Does any one know of a reasonably accessible introduction to Motives? >Isaac Khabaza >Allan Adler refers to the unclear nature of motives. I have seen >similar comments. Indeed the massive Encyclopedic Dictionary of >Mathematics (2nd edition) has no entry for motives. There is a rumour >that motives refer to a form of YOGA practised by its inventor. Be >that as it may, there is a motives epidemic raging in algebraic >geometry: pure, mixed, Lefschetz motives, Chow motives and many more. >A truce should be called, please no more motives until a reasonably >accessible and comprehensible definition is published preferably in >PDF format on the WEB. >Isaac Khabaza Perhaps the following quote will help, from the 1997 reference that I supplied in response to Isaac Khabaza's previous message: The categorical framework for the universal cohomology theory of algebraic varieties is the category of mixed motives. This category has yet to be constructed, although many of its desired properties have been described. [...] Rather than attempting the construction of MMS, we consider a more modest problem: The construction of a triangulated tensor category which has the expected properties of the bounded derived category of MMS. http://books.pdox.net/Math/Mixed%20Motives.pdf Pie === Subject: Re: Motives >Does any one know of a reasonably accessible introduction to Motives? >Isaac Khabaza Try http://books.pdox.net/Math/Mixed%20Motives.pdf and the references therein. === Subject: Re: A Non Linear Trigonometric Expression for P i . by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i85Ltkw14052; > For any positive real number X: > Pi= 4*ArcTanX + 2*ArcTan[(1-X^2)/2X] , Working in Radians, OR > 180= 4*ArcTanX + 2*ArcTan[(1-X^2)/2X] , Working in Degrees ]. > Copyright P. Stefanides > http://www.stefanides.gr >Maple says the value is -Pi for negative x and Pi for positive x. is alright for an odd function, but may change for x<0 using arctan2(x,y) of FORTRAN that takes into account quadrant placement of x,y arguments. But in Mathematica Pi = 4 ArcTan[1,X] + 2 ArcTan[2 X,(1-X^2)]is OK using arctan and arctan2. === Subject: Re: Do things/events happen/occur randomly without a reason? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i85LtkM14068; > Math. Forum: > Is a cause for Randomness neccessary? Is Randomness just Randomness > for no particular reason. No one reason or cause? Randomness may be, > as some suggest, the result of a lack of a cause or reason. > Randomness the default mode in the Universe? > I would like to suggest that Randomness requires an underlying > circumstance. Of course some see order where others see disorder. So > what are we talking about. > Where I am coming from are the Statistical Sciences. Is Statistical > Randomness really Random? I say Mathematically no. You can find a > Mathematical cause for this Randomness. Events occurr randomly for a > reason. Which I think has some usefull Mathematical purpose and > application somewhere/somehow. >Many case they are just random, within some general limits. >Take a rock of Uranium ore and count the rate that the partical degrade at, >or the time intervals between the gamma rays. >The underlying cause is known, but the randomness only grossly quantifiable. >1000 pulses +-500 per second. Zim Here: Another system that is so unpredictable and randomn and not quantifiable is Predicting the Future. Maybe there is a reason/cause for this randomn state. A usefull one to know. Zim Olson http://www.zimmathematics.com === Subject: Re: Do things/events happen/occur randomly without a reason? > Math. Forum: > > Is a cause for Randomness neccessary? Is Randomness just Randomness > for no particular reason. No one reason or cause? Randomness may be, > as some suggest, the result of a lack of a cause or reason. > Randomness the default mode in the Universe? > > > I would like to suggest that Randomness requires an underlying > circumstance. Of course some see order where others see disorder. So > what are we talking about. > > Where I am coming from are the Statistical Sciences. Is Statistical > Randomness really Random? I say Mathematically no. You can find a > Mathematical cause for this Randomness. Events occurr randomly for a > reason. Which I think has some usefull Mathematical purpose and > application somewhere/somehow. >Many case they are just random, within some general limits. >Take a rock of Uranium ore and count the rate that the partical degrade at, >or the time intervals between the gamma rays. >The underlying cause is known, but the randomness only grossly quantifiable. >1000 pulses +-500 per second. > > > Zim Here: > Another system that is so unpredictable and randomn and not > quantifiable is Predicting the Future. Maybe there is a > reason/cause for this randomn state. A usefull one to know. > Zim Olson > http://www.zimmathematics.com Although I haven't put much rigor into this thought yet, when I think of Predicting the Future, not in the clock-work mechanics sense of Newtonian physics, I think of meteorology. If your interested in pondering the notion furthur, perhaps you might want to study some meteorology to see if you notice any parallels to other predictions. === Subject: Re: A Non Linear Trigonometric Expression for P i . by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i85LtkR14088; > For any positive real number X: > Pi= 4*ArcTanX + 2*ArcTan[(1-X^2)/2X] , Working in Radians, > [ or , > 180= 4*ArcTanX + 2*ArcTan[(1-X^2)/2X] , Working in Degrees ]. > Copyright P. Stefanides > Panagiotis Stefanides > http://www.stefanides.gr >Maple says the value is -Pi for negative x and Pi for positive x. Niel, Correct, I, did not include it for aesthetic reasons. However ,I thank You for Your response. P.Stefanides === Subject: Re: Power series expansion of log(z) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i85LtjC14048; >Find the power series expansion of log(z) about z = i and find its radius of >convergence : >My question is : Am I going about this problem in the correct way? >The theorem to employ here is : If f is analytic in B(a;R), the ball of >radius R centered at a, then f(z) = sum a_n (z-a)^n from n = 0 to infinity >for |z-a|convergence greater than or equal to R. >Well, I already know that log(z) is analytic on C minus 0 and the negative >real axis. So log(z) about z = i has to be analytic in a open disk centered >at i of radius less than 1 since log(0) is not defined. Then of course I >employ the theorem and calculate a_n. >But what if I didn't know that log(z) was analytic in an open disk of radius >less than 1 before I looked at this problem? What if instead I knew that it >was analytic in an open disk of radius less than 1/2 (and didn't know about >radii less than 1) ? After all, the theorem only starts out with if f is >analytic in some ball of radius R about a, ..., so I am still ok if I use >B(i, 1/2). So if I go about the problem like this, could I still get the >right answer somehow for the question? That is, could I still employ the >theorem and find out that the radius of convergence of the power series >expansion is 1? This question applies then to more complicated functions. >If I have some very complicated function f(z), then it might be hard to find >out in what balls B(a;R) is f(z) analytic. But let's say I try to find the >power series expansion of this complicated function f(z) about some point >and through careful calculations, I solve that f(z) is definetely analytic >on a ball of radius R_0 about that point (but it might be analytic in even a >greater disk). Can I recover the true radius of convergence after I get the >power series expansion? >So the reason why I asked am I going about this problem in the correct way >is because I KNEW where log(z) was analytic BEFOREHAND. So in a sense I >sort of cheated. But for more complicated f(z), I might not be able to know >this beforehand, but as I said, I might be able to show that for some R_0 it >is analytic. Then can I get the correct radius of convergence for the power >series expansion? >(I hope what I have said makes sense) >Isaac Yes, you COULD start by assuming log(z) is analytic is some small neighborhood of i. Calculate the Taylor series for log(z) and then find the radius of convergence of that (by using the ratio test, say). There is a theorem that says if two functions are analytic and equal on an open set, then they are equal on the largest set on which both are defined and analytic. === Subject: Re: Power series expansion of log(z) | Yes, you COULD start by assuming log(z) is analytic is some small |neighborhood of i. I think if the question asks for the power series, it's probably okay to take for granted that it has a power series, and hence is analytic at the point. | Calculate the Taylor series for log(z) and then find the |radius of convergence of that (by using the ratio test, say). There is a |theorem that says if two functions are analytic and equal on an open set, |then they are equal on the largest set on which both are defined and |analytic. The set has to be connected. It's possible for the two functions to be separately defined on connected domains without the intersection of their domains being connected. Take two branches f and g of the log function, where f(1)=g(1)=0, but f is defined everywhere except 0 and the negative imaginary axis, and g is defined everywhere except 0 and the positive imaginary axis. Then f(-1)=pi*i while g(-1)=-pi*i. The region on which the two functions are both defined and analytic is the plane minus the imaginary axis, which is broken into two connected components by the imaginary axis. The fact that we can continue the function starting from z=1 around to z=-1 and get different values depending on the path we take is known as monodromy. Keith Ramsay === Subject: Re: GCD for Rational Numbers > I seem to have trouble with: > for rationals r = a/b, s = c/d > GCD(2^r-1,2^s-1) = 2^g-1, g = GCD(r,s) rational > GCD is not defined for these animals, unless b|a and d|c, > because you only defined it for rational numbers. This particular gcd always exists in any ring where the equation makes sense, i.e. in any ring containing 2^g, e.g. Z[2^g]. This is because the equation results from specializing the analogous polynomial Bezout relation (x^m - 1, x^n - 1) = (x^(m,n) - 1) in Z[x] for x = 2^g, and naturals m = r/g, n = s/g. Below I give a proof of the polynomial Bezout identity which has the neat feature that it clearly specializes to a proof of the integer Bezout identity for x = 1. For example 15 21 3 15 9 x - 1 9 3 x - 1 x - 1 (x + x + 1) ------ - (x + x ) ------ = ----- x - 1 x - 1 x - 1 for x = 1 specializes to 3 (15) - 2 (21) = 3, i.e. (15,21) = (3) DEFINITION n' := (x^n - 1)/(x-1). Note n' = n for x = 1. THEOREM (m',n') = ((m,n)') as ideals in Z[x], for naturals m,n. PROOF: It is trivially true if m = n or if m = 0 or n = 0. Therefore suppose n > m > 0. We proceed by induction on n + m. Since x^n - 1 = x^r (x^m - 1) + (x^r - 1) for r = n - m we obtain n' = x^r m' + r' via dividing above by x-1 Hence (m', n') = (m', r') via n' = r' (mod m') by prior = ((m,r)') via induction: m+r = n < m+n = ((m,n)') via n = r (mod m). QED COROLLARY Integer Bezout Theorem (set x = 1 above: erase primes). A deeper understanding comes when one studies Divisibility Sequences and Divisor Theory. See also the literature on q-analogies, e.g. --Bill Dubuque === Subject: Re: Euklidean algorithm Christian Ober-Blobaum > How can I prove that: > gcd(x^m - 1, x^n - 1) = x^gcd(m,n) - 1 ? Below I give a proof of this polynomial Bezout identity which has the neat feature that it clearly specializes to a proof of the integer Bezout identity for x = 1. For example 15 21 3 15 9 x - 1 9 3 x - 1 x - 1 (x + x + 1) ------ - (x + x ) ------ = ----- x - 1 x - 1 x - 1 for x = 1 specializes to 3 (15) - 2 (21) = 3, i.e. (15,21) = (3) DEFINITION n' := (x^n - 1)/(x-1). Note n' = n for x = 1. THEOREM (m',n') = ((m,n)') as ideals in Z[x], for naturals m,n. PROOF: It is trivially true if m = n or if m = 0 or n = 0. Therefore suppose n > m > 0. We proceed by induction on n + m. Since x^n - 1 = x^r (x^m - 1) + (x^r - 1) for r = n - m we obtain n' = x^r m' + r' via dividing above by x-1 Hence (m', n') = (m', r') via n' = r' (mod m') by prior = ((m,r)') via induction: m+r = n < m+n = ((m,n)') via n = r (mod m). QED COROLLARY Integer Bezout Theorem (set x = 1 above: erase primes). A deeper understanding comes when one studies Divisibility Sequences and Divisor Theory. See also the literature on q-analogies, e.g. --Bill Dubuque === Subject: Are logarithms still useful? We learn that logarithms were invented to facilitate arithmetic with big numbers. Electronic calculators have made them unecessary for this purpose. In what ways are logarithms still useful in mathematics? === Subject: Re: Are logarithms still useful? >We learn that logarithms were invented to facilitate arithmetic with big >numbers. Electronic calculators have made them unecessary for this >purpose. In what ways are logarithms still useful in mathematics? Not big numbers, but logarithms reduce multiplication to addition, and exponentiation to multiplication. Of course, this means that one has to use tables (or other computational means) to get the logarithms and antilogarithms, so that with good computers, direct multiplication is faster. However, exponentiation is generally done by using logarithms and exponentiation to a base for which the computations are easier than for a general base. So 2.31^2.69 would be calculated as exp(2.79*ln(2.31)). Here the base used is e, but any other base could be used. The logarithm function and exponential function have many major uses in mathematics as functions, not just for calculation. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Are logarithms still useful? > We learn that logarithms were invented to facilitate arithmetic with big > numbers. Electronic calculators have made them unecessary for this > purpose. In what ways are logarithms still useful in mathematics? Analysis of algorithmic complexity. Logarithms are used very VERY frequently in analyzing asymptotic work and time behavior of algorithms. -m === Subject: Re: Are logarithms still useful? > We learn that logarithms were invented to facilitate arithmetic with big > numbers. Electronic calculators have made them unecessary for this > purpose. In what ways are logarithms still useful in mathematics? Solutions of differential equations of the form dy/dt=-kt are useful in chemistry, physics, electronics. The solution is in terms of natural logarithms. You also sometimes have to use logarithms to solve problems in probability theory. What is the proportion of 1, 2, ..., or 9 as leading digit among, say, share prices? They are not equally likely. The probability of n being the first digit is log(1+1/n) (this is our conventional logarithm, base 10). You could expect 1 to be the first digit approximately 30% of the time, and 9 only 4.6%. For more information on this do a search with the following as key words: Benford's rule, Mark Nigrini. (Professor Nigrini has used this principle to detect fraud). === Subject: Re: Are logarithms still useful? > We learn that logarithms were invented to facilitate arithmetic with big > numbers. Electronic calculators have made them unecessary for this > purpose. In what ways are logarithms still useful in mathematics? Logarithms were invented to faciliate Physics calculations, and that's still the only place the only they're used. Since calculators and e were invented to do math caculations. log was invented to do spiral calculations, not number calculations. And they're still used every day. === Subject: Re: Are logarithms still useful? > Logarithms were invented to faciliate Physics > calculations, and that's still the only > place the only they're used. Astronomy, not physics; and they're not. Physics hardly existed in 1614 when Napier did his thing; but Laplace said By shortening the labors, the invention of logarithms doubled the life of the astronomer. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: Are logarithms still useful? > Logarithms were invented to faciliate Physics > calculations, and that's still the only > place the only they're used. > Astronomy, not physics; and they're not. > Physics hardly existed in 1614 when Napier did his thing; but > Laplace said By shortening the labors, the invention of logarithms > doubled the life of the astronomer. Since Physics existed since like before Zeno existed. You quite obviously mean that Newton, the Logic-challenged wanker, did not exist in 1614. And Napier invented logarithm *tables*. *Logarithms* were invented at the same time *division* of numbers were invented. Which is why it was Fibonacii, not Napier, that discovered that logarithms have the something do with both Geometry and the sqrt(2) and Napier didn't. Which is where both the word limit and continued fraction came from in Pythagorus et al sqrt(2) proof crap. === Subject: Re: Are logarithms still useful? .................. > Since Physics existed since like before Zeno existed. > You quite obviously mean that Newton, the > Logic-challenged wanker, did not exist > in 1614. And Napier invented logarithm *tables*. > *Logarithms* were invented at the same time > *division* of numbers were invented. Considering that we have examples of division, with remainder, farther back than 4000 BCE, this can hardly be the case. The Greeks had tables of trigonometric functions and used them; this could not be the case without extensive division. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Are logarithms still useful? In sci.math, John Smith : > We learn that logarithms were invented to facilitate arithmetic with big > numbers. Electronic calculators have made them unecessary for this > purpose. In what ways are logarithms still useful in mathematics? Summing logs is useful for computing geometric means. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Are logarithms still useful? > We learn that logarithms were invented to facilitate arithmetic with big > numbers. Electronic calculators have made them unecessary for this > purpose. In what ways are logarithms still useful in mathematics? Three ways that haven't been mentioned yet: 1) In computational complexity theory the logarithm plays an important role, not just as a tool but even in the statement of many theorems. 2) In mathematical statistics the only practical way to get a handle on likelihood functions is by first taking their logs. 3) In understanding the asymptotic distribution of primes (the prime number theorem and analytic number theory in general). === Subject: Re: Are logarithms still useful? > We learn that logarithms were invented to facilitate arithmetic with big > numbers. Electronic calculators have made them unecessary for this > purpose. I wouldn't be so sure of that if I were you. I seem to recall a P-Chem exam where we had to calculate the probability of finding an electron in a building that was about 100 meters away from the nucleus of the atom it was attached to. There were four possible answers on the multiple choice exam, like 10^-10^10, etc. Well, I punched in the numbers, and ended up with an answer of zero, which was not one of the choices. As I busily scratched my head, I overheard the prof telling another student to use logarithms. I thought, Aha!!!, and promptly re-computed the answer. Of course, I got it right ;) === Subject: Re: Are logarithms still useful? > We learn that logarithms were invented to facilitate arithmetic with big > numbers. Electronic calculators have made them unecessary for this > purpose. >I wouldn't be so sure of that if I were you. (story of use of logs in an exam, snipped) I taught Technical Math at my college last spring, and the logarithms section gave specific examples of using logarithms to compute large powers like 48.6^75. Many calculators overßow at anything greater than 10^99, but if you take 75*log(48.6) you get 126.497720195. Subtract 126, take the antilog of the remaining decimal part to get 3.14572, and multiply by 10^126 for your answer. But as others have mentioned, these days logarithms are much more often used for their own sake than simply to aid computation. I don't think anyone has yet mentioned the beautiful fact(*) that the area under the curve of y = 1/x, between x = 1 and x = a, is the natural log of a. (If a is between 0 and 1, the logarithm is the negative of the area.) (*) Well, okay, in calculus books it's the definition of the logarithm. I don't think anyone has yet mentioned Eli Maor's /e: The Story of a Number/, which devotes quite a few pages to logarithms. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: Are logarithms still useful? ..... and there is electrochemistry, the Nernst equation, and various concentration product formulas (and these days, we have the convenience of a scientific calculator so we do not need to look in data tables of handbooks and such). G C === Subject: Re: Are logarithms still useful? >We learn that logarithms were invented to facilitate arithmetic with big >numbers. Electronic calculators have made them unecessary for this >purpose. In what ways are logarithms still useful in mathematics? They have many real-life applications. In chemistry, pH is a logarithmic scale. In astronomy, magnitude is a logarithmic scale. In seismology, Richter is a logarithmic scale. In acoustics, decibels are a logarithmic scale. As for pure mathematics, logarithmic differentiation is a recognized technique for easily finding derivatives of expressions like (x^2+7)(x^3+3x^2+2x+2) / ( (x-5)(x^2+11) ) -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: Are logarithms still useful? ... > As for pure mathematics, logarithmic differentiation is a recognized > technique for easily finding derivatives of expressions like > (x^2+7)(x^3+3x^2+2x+2) / ( (x-5)(x^2+11) ) I do not think many logarithms are needed to get the derivative of that expression ;-). -- 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: Are logarithms still useful? > ... > As for pure mathematics, logarithmic differentiation is a recognized > technique for easily finding derivatives of expressions like > (x^2+7)(x^3+3x^2+2x+2) / ( (x-5)(x^2+11) ) > I do not think many logarithms are needed to get the derivative of > that expression ;-). > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Quite the contrary. Call the expression y and differentiate w.r.t. y ln y=ln(x ^ 2 + 7 ) + ln(x ^ 3 + 3 x ^ 2 + 2 x + 2) - ln(x-5) - ln( x^2 + 11) (1/y)(dy/dx)=2x / (x ^ 2 + 7) + (3 x^2 + 6 x + 2) / (x^3 + 3 x^2 + 2) - 1 / (x-5) - 2x / (x^2+11) dy/dx = [ ( x ^ 2 + 7 ) / (x^3 + 3x^2 +2x+2)]2x / (x ^ 2 + 7) + (3 x^2 + 6 x + 2) / (x^3 + 3 x^2 + 2) - 1 / (x-5) - 2x / (x^2+11) Simplify as needed. Dave === Subject: Re: Are logarithms still useful? >... > As for pure mathematics, logarithmic differentiation is a recognized > technique for easily finding derivatives of expressions like > (x^2+7)(x^3+3x^2+2x+2) / ( (x-5)(x^2+11) ) >I do not think many logarithms are needed to get the derivative of >that expression ;-). Needed? Maybe not. But to apply the Product Rule and Quotient Rule repeatedly would be tedious and error prone. Differentiating logarithmically: y = (x^2+7)(x^3+3x^2+2x+2) / ( (x-5)(x^2+11) ) ln(y) = ln(x^2+7) + ln(x^3+3x^2+2x+2) - ln(x-5) - ln(x^2+11) y' / y = 2x/(x^2+7) + (3x^2+6x+2)/(x^3+3x^2+2x+2) - 1/(x-5) - 2x/(x^2+11) etc. And logarithmic differentiation is an even greater labor saver if any of the factors are raised to powers. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: Are logarithms still useful? John Smith >We learn that logarithms were invented to facilitate arithmetic with big >numbers. Electronic calculators have made them unecessary for this >purpose. In what ways are logarithms still useful in mathematics? Understanding exponential growth. Statements like The incumbent political party spent more money in the past ten years than all previous budgets in the history of this nation. This is quite plausible with a steady inßation rate. If the present year budget is b_1 and the budget j years ago is b_j * r^(j - 1) (where r < 1), then the statement is (b_1 + b_2 + ... + b_10) > (b_11 + b_12 + b_13 + ...) Substitute b_j = b_1 * r^(j-1). Factor b_1 out of each side. b_1 * (1 + r + ... + r^9) > b_1 * (r^10 + r^11 + r^12 + ...) Sum the geometric progression and geometric series b_1 * (1 - r^10)/(1 - r) > b_1 * r^10/(1 - r) Multiply by (1 - r). Divide by b_1. 1 - r^10 > r^10 which becomes r^10 < 0.5 or r < 0.933 (here you use logarithms to take a tenth root). If there has been a steady inßation rate about 7%, and the budget has grown 7% each year merely to account for this, then r will be 1/1.07 ~= 0.935, and the statement is no surprise. Logarithms arise throughout mathematics and science. The number of primes below a large positive number N is approximately N / ln(N), where ln denotes logarithm base 2.718. Sound is measured in decibels, which is a logarithmic scale. The magnitude of an earthquake is also measured on a logarithmic scale. The half-life of a radioactive substance is a logarithm. What does it mean to earn 2% interest on a bank account? The best algorithms for sorting N values have complexity O(N * log(N)), which means they take at most c * N * log(N) operations (e.g., comparisions, swaps) where c is a constant dependent only on the program and the hardware, not on the amount of input (N). -- During a wedding ceremony, members of the audience cried. Scientists examined the tears. They determined the liquid was eye dew. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Re: Are logarithms still useful? Let us not forget about chemical pH. (negative logarithm of the hydronium ion Ôactivity'), a measure acidity or alkalinity. G C === Subject: re:Are logarithms still useful? Obviously, for numerical work, their importance has greatly diminished. However, the log function is quite important in analytic work. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: Are logarithms still useful? >In what ways are logarithms still useful in mathematics? Graphing. Eventually you will learn about log-log and semi-log graph paper. G C === Subject: Re: Are logarithms still useful? charset=iso-8859-1 >In what ways are logarithms still useful in mathematics? > Graphing. Eventually you will learn about log-log and semi-log graph paper. Yes, graphing is very important. Especially in engineering where the ranges of parameters can span many decades and relationships are non-linear. Sometimes in the life sciences too. It's also the only way to use a pocket calculator to compute powers and roots other than square/square-root [and occasional cube functions buttons]. Norm === Subject: Re: Are logarithms still useful? >In what ways are logarithms still useful in mathematics? >Graphing. Eventually you will learn about log-log and semi-log graph > paper. > Yes, graphing is very important. Especially in engineering where the ranges > of parameters can span many decades and relationships are non-linear. > Sometimes in the life sciences too. It's also the only way to use a pocket > calculator to compute powers and roots other than square/square-root [and > occasional cube functions buttons]. > Norm test === Subject: Re: Are logarithms still useful? >In what ways are logarithms still useful in mathematics? >Graphing. Eventually you will learn about log-log and semi-log graph > paper. > Yes, graphing is very important. Especially in engineering where the ranges > of parameters can span many decades and relationships are non-linear. > Sometimes in the life sciences too. It's also the only way to use a pocket > calculator to compute powers and roots other than square/square-root [and > occasional cube functions buttons]. > Norm After some research I came up with this method for calculating roots with logs to the base e: exp(x) = e^x x root of b = exp((1/x)*log(b)) Here is some C code that calculates the cube root of 37: #include #include int main(void) { double b, x; b = 37.0; x = 3.0; printf(%fn, exp((1/x) * log(b))); return 0; } Seems to work. Is there a better way? === Subject: Re: Are logarithms still useful? >In what ways are logarithms still useful in mathematics? > It's also the only way to use a pocket > calculator to compute powers and roots other than square/square-root [and > occasional cube functions buttons]. > Norm After a little research I came up with this method for calculating roots with logs to the base e: exp(x) = e^x x root of b = exp((1/x)*log(b)) Here is some C code that calculates the cube root of 37: #include #include int main(void) { double b, x; b = 37.0; x = 3.0; printf(%fn, exp((1/x) * log(b))); return 0; } Seems to work. Is there a better way? === Subject: Re: Are logarithms still useful? >In what ways are logarithms still useful in mathematics? >Graphing. Eventually you will learn about log-log and semi-log graph > paper. > It's also the only way to use a pocket > calculator to compute powers and roots other than square/square-root [and > occasional cube functions buttons]. > Norm After a little research, I came up with this method for computing roots with logs to the base e: exp(x) = e^x x root of b = exp((1/x)*log(b)) here is some C code that calculates the cube root of 37: #include #include int main(void) { double b, x; b = 37.0; x = 3.0; printf(%fn, exp((1/x) * log(b))); return 0; } Seems to work. Is there a better way? === Subject: Re: Are logarithms still useful? > >In what ways are logarithms still useful in mathematics? > >Graphing. Eventually you will learn about log-log and semi-log graph > > paper. > > > It's also the only way to use a pocket > calculator to compute powers and roots other than square/square-root [and > occasional cube functions buttons]. > > Norm > > After a little research, I came up with this method for computing roots > with logs to the base e: > exp(x) = e^x > x root of b = exp((1/x)*log(b)) > here is some C code that calculates the cube root of 37: > #include > #include > int main(void) > double b, x; > b = 37.0; > x = 3.0; > printf(%fn, exp((1/x) * log(b))); > return 0; > Seems to work. Is there a better way? What do you mean by better? Britannica used to (and probably still does) show how to take a cube root by fixed point arithmetic. If implemented at the assembly level, it would surely be faster than evaluating log and exp functions. David Ames === Subject: Re: Are logarithms still useful? > We learn that logarithms were invented to facilitate arithmetic with big > numbers. Electronic calculators have made them unecessary for this > purpose. In what ways are logarithms still useful in mathematics? Solving exponential equations as well as integration. I used logs about an hour ago for my atmosphere dynamics homework to simplify a derivation. Dave === Subject: Re: JSH: So what's the point? >That is just a reality that most people accept, but for some reason >lately they've been acting as if mathematicians can be perfect over >hundreds of pages, when the reality is that many long mathematical >works are probably just ßawed, like Wiles, but mathematicians feel >good saying they're perfect, whether they are or not. > what's your position on the fact that an -error- was found in the > first version of wiles' proof? i mean why didn't we all just agree > that -that- one was right? Aren't two Millenium Prize proofs currently out there being checked? (Perelman's proof of the Poincare conjecture and de Branges proof of the Riemann hypothesis). Those two guys are certainly in the club. Under the JSH theory of Mathematical Peer Review, shouldn't their proofs just have been accepted right off the bat? What's with this checking stuff? - Randy === Subject: Re: JSH: So what's the point? Discussion, linux) > Those two guys are certainly in the club. Under the JSH > theory of Mathematical Peer Review, shouldn't their > proofs just have been accepted right off the bat? What's > with this checking stuff? Appearances. Duh. -- It seems to me that some of you don't realize that [...] some day the truth comes out. It doesn't matter if you're dead. I've made certain that your name will live in infamy if it's known at all: Wiles, Ribet, Granville, or anyone else from this generation. --JSH beyond the grave === Subject: Re: JSH: So what's the point? > proofs just have been accepted right off the bat? What's > with this checking stuff? > Appearances. Duh. --Chiar Man George -- follow the money to Strep Throat! http://tarpley.net === Subject: Re: JSH: So what's the point? [Snip] > A good start would be requiring computer proof checking!!! Algorithms and programmes in CompSci are precise and mathematical, but algorithms and programmes, when actually run on computers, become an exercise in statistics. We say computers work because they work 99.999...% of the time. As you know (or don't?) proof requires 100% verification. Do you remember the fuss kicked up when a computer was used to help prove the four-colour theorem? Lawrence Runacres. === Subject: Sums equal to squares of primes Is it the case that if sum means a sum of consecutive integers, p is any prime number, and k - j = p - 1, and if sum_(j to k) = p^2 then j + k = 2p ? Using the same notation, it is also true in some cases that sum_(j to k) = n^2, k - j = n - 1, and j + k = 2n when n is non-prime. What distinguishes these n from other non-primes? === Subject: Re: Sums equal to squares of primes > Is it the case that if sum means a sum of consecutive integers, > p is any prime number, and k - j = p - 1, and if > sum_(j to k) = p^2 > then j + k = 2p ? > Using the same notation, it is also true in some cases that > sum_(j to k) = n^2, k - j = n - 1, and j + k = 2n > when n is non-prime. What distinguishes these n from other non-primes? Isn't this true for any odd n? You're adding up n consecutive numbers and getting n^2 so the number average n, so the first & last average n, so they add to 2n. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Sums equal to squares of primes > axe@cam.org (Axel Harvey) had written: > Is it the case that if sum means a sum of consecutive integers, > p is any prime number, and k - j = p - 1, and if > > sum_(j to k) = p^2 > > then j + k = 2p ? > > [ ... ] > Isn't this true for any odd n? > You're adding up n consecutive numbers and getting n^2 > so the number average n, > so the first & last average n, > so they add to 2n. Well, that's what you get for doing empirical maths. I was distracted by the large variety of sums that can produce some numbers. For example, 15^2 = 225 = sum_(4 to 21)=sum_(8 to 22)=sum_(18 to 27)= sum_(21 to 29)=sum_(35 to 40)=sum_(40 to 43)=sum_(74 to 76)=112 + 113. The above relationships are satisfied by sum_(8 to 22) but I missed them. So, then... Is it the case that only one sum of consecutive integers represents the square of any prime number > 2, besides the trivial sum (x/2 - 1/2) + (x/2 + 1/2)? === Subject: Re: Sums equal to squares of primes >So, then... Is it the case that only one sum of consecutive integers >represents the square of any prime number > 2, besides the trivial >sum (x/2 - 1/2) + (x/2 + 1/2)? You want solutions of p^2 = b (b+1)/2 - a (a+1)/2 where p is prime and b > a, for a sum from a+1 to b. But the right side is (b-a)(a+b+1)/2. Since p is prime, the only possibilities are b-a=1, a+b+1=2 p^2 => a=p^2-1, b=p^2 (i.e. the sum of one consecutive integer p^2) b-a=2, a+b+1=p^2 => a=(p^2-3)/2, b=(p^2+1)/2 (your trivial case (p^2-1)/2 + (p^2+1)/2) b-a=p, a+b+1=2 p => a=(p-1)/2, b=(3p-1)/2 b-a=2p, a+b+1=p => a= (-p-1)/2, b=(3p-1)/2 (well, you didn't say positive integers) b-a=p^2, a+b+1=2 => a = (1-p^2)/2, b=(p^2+1)/2 b-a=2p^2, a+b+1=1 => a=-p^2, b=p^2 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 Series of Interest > A common financial formula is P = iA/(1 - (1+i)^(-N)), where P denotes > the payment, i the interest rate, A the loan amount, and N the number of > payments. A series is presented here for the interest rate. > In the aus.mathematics newsgroup recently, someone asked about solving > such an equation for i. Ken Pledger mentioned some appropriate links, > such as Stan Brown's (see > formula (2) there) and the sci.math FAQ entry > . The formula > cannot be solved for the interest rate in closed form in terms of > elementary functions. Apparently the most common approach to determining > i is to use a numerical technique, such as Newton's method. But an > alternative technique is to use a series: > Letting u = (PN/A - 1)/(N + 1), > i = 2( u - (N-1) u^2/3 + (N-1) (2N+1) u^3/9 - (N-1) (2N+1) (11N+7) u^4/135 > + (N-1) (2N+1)^2 (13N+11) u^5/405 -+ ...) > which I obtained by reversion of series. Surely this series must be well > known to those who deal with such matters often; I would greatly > appreciate references to it. (Of course, more terms of the series could > be given here, but references should make doing so unnecessary.) Perhaps it's not well known. But in any event, I found out just today that the series had been already published: H. E. Stelson, Note on finding the interest rate _Amer. Math. Monthly_ 60:10 (Dec. 1963) 703-705. As I did, he obtained the series by reversion, and did not discuss convergence or mention the form of the general term. He then expressed the interest rate as a continued fraction and, by considering convergents, obtained some very excellent approximations. I may discuss his and various other approximations in another thread soon. David Cantrell === Subject: Re: A Series of Interest > A common financial formula is P = iA/(1 - (1+i)^(-N)), where P denotes > the payment, i the interest rate, A the loan amount, and N the number of > payments. A series is presented here for the interest rate. > In the aus.mathematics newsgroup recently, someone asked about solving > such an equation for i. Ken Pledger mentioned some appropriate links, > such as Stan Brown's (see > formula (2) there) and the sci.math FAQ entry > . The formula > cannot be solved for the interest rate in closed form in terms of > elementary functions. Apparently the most common approach to determining > i is to use a numerical technique, such as Newton's method. But an > alternative technique is to use a series: > Letting u = (PN/A - 1)/(N + 1), > i = 2( u - (N-1) u^2/3 + (N-1) (2N+1) u^3/9 - (N-1) (2N+1) (11N+7) u^4/135 > + (N-1) (2N+1)^2 (13N+11) u^5/405 -+ ...) David, Could you please post the code that could generate your (remarkable) formula at any order? Valeri === Subject: Re: A Series of Interest > A common financial formula is P = iA/(1 - (1+i)^(-N)), where P > denotes the payment, i the interest rate, A the loan amount, and N > the number of payments. A series is presented here for the interest > rate. > Letting u = (PN/A - 1)/(N + 1), > > i = 2( u - (N-1) u^2/3 + (N-1) (2N+1) u^3/9 > - (N-1) (2N+1) (11N+7) u^4/135 > + (N-1) (2N+1)^2 (13N+11) u^5/405 -+ ...) > David, > Could you please post the code that could generate your (remarkable) > formula at any order? First, I'll note again that it's not really _my_ series. Yes, I obtained it independently, on the day that I started this thread. But I have since found out that the series -- well, to be precise, just the first four terms of it -- were given by H. E. Stelson in a note in the Monthly in 1953. Below my signature is code for Mathematica which can be used to generate the series. Note that its output is not quite as neat looking as what I had posted. Instead of being in terms of N and my u, it's in terms of N and s = P/A. But it should then be simple to get from Mathematica's output to the form I had posted. While I'm doing this, since some people who don't have Mathematica or an equivalent CAS might be interested in more terms of the series, I'll go ahead now and show the output giving ten terms. (Of course, just change the 10 in the code to a larger integer if you want still more terms of the series.) David Cantrell ----------------------- In[1]:= Normal[Simplify[InverseSeries[Series[i/(1-(1+i)^(-N)),{i,0,10 }],s]]] Out[1]= (2*N*(-(1/N) + s))/(1 + N) - (2*(-1 + N)*N^2*(-(1/N) + s)^2)/(3*(1 + N)^2) + (2*N^3*(-1 - N + 2*N^2)*(-(1/N) + s)^3)/(9*(1 + N)^3) - (2*N^4*(-7 - 18*N + 3*N^2 + 22*N^3)*(-(1/N) + s)^4)/(135*(1 + N)^4) + (2*N^5*(1 + 2*N)^2*(-11 - 2*N + 13*N^2)*(-(1/N) + s)^5)/(405*(1 + N)^5) - (2*N^6*(-41 - 253*N - 447*N^2 - 43*N^3 + 484*N^4 + 300*N^5)*(-(1/N) + s)^6)/(2835*(1 + N)^6) + (2*N^7*(-301 - 2823*N - 7803*N^2 - 6421*N^3 + 4272*N^4 + 9252*N^5 + 3824*N^6)*(-(1/N) + s)^7)/(42525*(1 + N)^7) - (2*N^8*(-311 - 5860*N - 23694*N^2 - 34952*N^3 - 8455*N^4 + 30876*N^5 + 32428*N^6 + 9968*N^7)*(-(1/N) + s)^8)/(127575*(1 + N)^8) + (1/(1148175*(1 + N)^9))* (2*N^9*(719 - 31924*N - 196606*N^2 - 426364*N^3 - 346849*N^4 + 132824*N^5 + 463256*N^6 + 325664*N^7 + 79280*N^8)*(-(1/N) + s)^9) - (1/(189448875*(1 + N)^10))* (2*N^10*(512983 - 2225859*N - 27338082*N^2 - 82514430*N^3 - 109075749*N^4 - 36930111*N^5 + 77552880*N^6 + 109981632*N^7 + 58322064*N^8 + 11714672*N^9)*(-(1/N) + s)^10) === === 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 * Assistant, Associate or Full Professor-Genomic Biology: University at Buffalo-Dept of Biological Sciences, Buffalo, NY. The Department of Biological Sciences ( http://www.biology.buffalo.edu ) at the University at Buffalo , the largest and most comprehensive campus in the SUNY system is seeking outstanding applicants for a tenured or tenure-track... === Subject: Alicequeenox Your Correct Horoscope = Your Correct Future = Astronomical Astrology Lewis Carroll http://ghettobox.dhs.org/plist/iau/hpl /nu /28 /extra /pcs /o5 alligned on the northern horizon. The majikal forest is made an opportunity. Tolkien and RPG folllow. Alice is not yet mature but very sexy as issuing commands to the four winds. And four card decks. Alice is in fact an initiate into enochian majik. Cruithne is with Ceres, as well as Jupiter; the virgin female shaman is bad off in journey. She should not stray from home. As the physical body is asleep, it's guardian angel is in second attention; exploring 8 months of initiation. As a curiousity we can say that Sedna is about to enter Cetus. The whales sing as strange creature issue from the creation soup. Major main belt asteroid (unused by astrologers) Eunomia is in Sextans (a zodiacal as well as currently ascendental constellation). This is a good omen for the little prophet who receives three majikal guests in his cave. Sextans is a demon on Tiamat's back, one of the three guests who receive exorcism, Crater, Corvus and Sextans. Tiamat is Hydra (a zodiacal, e.g. Venus in Hydra with 2002 AW197 Juno is in Sextans (both a zodiacal, as well as currently ascendental demoniac constellation); marriage is calculated or even hierogamic. The current position of Juno in Scutum is not good for marriage, not even marriage delineation (see also Juno in Aquila). 1994 TA is the immortal centaur at the meridian, but there is big a cluster in azimuthal conjunction, affording for a contemplative mediumistic character. Fatal 2002 VE95 as well as a couple of centaur objects strangely spell mathematical order and double checking. The otherwise dangerously destructive non-human 2002 VE95 is thus restrained and amazingl friendly. This man tames the inorganic. A born exorcist. horizon) in Auriga. Rare Moon transits hardly trigger this aloof position (the Moon, when in Auriga is not nearly near this position). 1999 XX17 and 2002 PY42. This is most unfortunate for marriage, already the story has a happy beginning. Curiousely, the position is very favorable for cats (this brings Regulus energy up, evoking Narasimha (MIKAL, Christ, Saint Marc) as Cheshire Cat (darshan). as Alice falls into oblivion ( = matter ). The matrix is a puzzle. One is witness to creation and Chaos, not partecipating. 2000 PH5 is with Mars and Venus: always a third party intruding quite often in the affairs of lovers. Quarrel and teleapathy happens. Sharp words (Mars and Venus form a mouth) are issued. The eight position reached and II:::: Alice becomes queen. AL-ISA. Alice has a rare position of Moon in Scorpius, sure to portray the Cheshire Cat (the Moon appears but seldom in this constellation). The Moon is conjunct Varuna (by azimuth) meaning the Cheshire cat changes spots /1/. Alice is fellow to mankind much the same as Lovecraft. http://ghettobox.dhs.org/plist/sedna (Sedna delineation) /1/ A lesson in sun-spots, as well as centripetal force. --- http://ghettobox.dhs.org/plist /iau /nu /28 /extra /pcs /o5 For cubewano, plutino and centaurs delineations check: === Subject: inverse laplace transform I have the following laplace transform result F(s) = (s+3)/(s^2 + 4s + 5 exp^(-6s)); Denote the inverse laplace transform of F(s) as f(t). I am hoping that f(t) is equal to or approximated by a hyper-exponential function. Can you please give me a few reference or links for this? === Subject: Re: Particular case of a generalized Borsuk-Ulam theorem > I am interested in a particular case of a generalized Borsuk-Ulam > theorem: I want to show that there exists no map f : S^3 -> R^2 which > separates all triples in the orbit of a Z_3 action on S^3 given by the > homeomorphism h : S^3 -> S^3, h(z0,z1) = (z0*exp(2*pi*i/3), > z1*exp(2*pi*i/3)) of period 3. > Unfortunately, in all the litterature I have encountered, this > corresponds to a limit case. For instance, Lusk provides a strong > answer under the assumption that m>n(p-1)-1. In my case, m=3, n=2 and > p=3, so there is equality. Mramor-Kosta and Munkholm state that the > set of such triples mapped to a single point in R^2 has dimension >= > -1... This is not surprising since there exist maps f : S^3 -> R^2 such that no orbit of this Z_3 action is mapped to a single point. > I said strong because I would be satisfyied with weaker versions of > the theorems. Indeed, it is enough for me to know if, for any map f : > S^3 -> R^2, there exists a triple as above such that two of the three > points are mapped to a single point. Well, this weaker version is already true. The proof I know is not elementary, but maybe there is a simpler one for this particular case. I'll try to adapt it to the situation. By the way, I found this mine; the reference is given below. Let T:S^3->S^3 be a fixed points free map of period 3. We shall say that the closed invariant set F is an Ôorbital partition' in S^3, if no two points of S^3F of one and the same orbit may be connected by a continuum within S^3F. I shall refer to a basic result of C.T.Yang, which implies that in our case any 3 orbital partitions in S^3 have nonempty intersection. Suppose now that there is a map f : S^3 -> R^2 such that f(Tx)<>f(x) for any x. Consider the set F_1 of points in S^3 such that for any x in F_1, the triangle f(orbit(x)) has a side parallel to Ox. It may be shown that F_1 is an orbital partition in S^3. Consider now the set F_2 of points in F_1 such that for any x in F_2, f(orbit(x)) lies on a line, parallel to Ox. F_2 is an orbital partition in F_1. Then Yang's result (together with some common technique of extension of orbital partitions) implies that there exists an invariant continuum in the set F_2. But now it is an easy exercise to show that f(Tx)=f(x) for some x in F_2, since f|F_2 may be considered as a map into R^1; a contradiction. > Do you know about this case? Any suggestions for litterature on > separators? If T:S^n->S^n is a fixed points free periodic map of a prime period p, then for any map f:S^n->R^m, the set A(f)={x in S^n | f(Tx)=f(x)} has dimA(f)>=n-(m-1)(p-1)-1. In our case n=3, m=2, p=3 and we get dimA(f)>=0. fixed points, Serdica, 16(1990), 87 - 93. Here the main objects are periodic maps with fixed points, but the fixed point free case is treated as well. Simeon === Subject: Re: Particular case of a generalized Borsuk-Ulam theorem No clue? :-) === Subject: Sheaf exercises Is there anywhere I can get some exercises that will make me more comfortable with sheaf theory? I was trying to look in the net for people who had lecture on this, but these lectures don't seem to be one which involve much exercise work by the students. Jose Capco === Subject: Re: Sheaf exercises > Is there anywhere I can get some exercises that will make me more > comfortable with sheaf theory? I was trying to look in the net for > people who had lecture on this, but these lectures don't seem to be one > which involve much exercise work by the students. > Jose Capco There's a whole book on Sheaf Theory by Bredon (GTM 170): http://www.springeronline.com/sgw/cda/frontpage/0,11855,4- 10043-22-1418364-0 ,00.html But if you're just starting, section 2.1 of Hartshorne's Algebraic Geometry should be a good introduction. === Subject: Re: Sheaf exercises >Is there anywhere I can get some exercises that will make me more >comfortable with sheaf theory? I was trying to look in the net for >people who had lecture on this, but these lectures don't seem to be one >which involve much exercise work by the students. i certainly wouldn't know about resources involving sheaf theory specifically. but i'll point out that as things get more advanced you should expect fewer exercises designed to check the reader's understanding. the idea is eventually you're supposed to be able to understand things by reading the exposition. lily you can make up your own exercises, which people do all the time: fill in the gaps in the proofs. think about how some abstraction works out in a specific example. etc. >Jose Capco ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Explaining the foundations of math > Despite all that, you still find the time to learn and do mathematics? Did > mathematics appeal to you because it acted as some kind of mental escape > hatch from your inner turmoil in the past because of the tyranny of your > previous religious institution? Would you still be drawn to it even if you > were not brutalized by your religion? > Shedar > Shedar Google formatting is making a mess of sci.math on groups2 for MAC users (again)--I don't know about others. I am really going to have to find another source. Suggestions for free usenet? I am going to ask SBC-Yahoo to get on the ball and provide usenet feeds. This attempt to provide Yahoo message boards, etc. is the failure one would expect. How have these people gotten by without providing the basics of the net. Usenet has been one of the basic feature of the net since before there was any WWW, or HTML. How can a provider expect to be successful without providing a newsfeed? Its basic. WWW has not replaced Usenet, any more than it can replace email. I got away from the religion at the age of 19. Finally went to UCLA, took physics, now have decided to learn some math besides the kind that one learns as a physicist. This is off topic. I apologize to readers who are bothered by off-topic posts. Van === Subject: Re: Explaining the foundations of math > > I got away from the religion at the age of 19. Finally went to UCLA, > took physics, > now have decided to learn some math besides the kind that one learns as > a physicist. > This is off topic. I apologize to readers who are bothered by off-topic > posts. > Van You mean how one's life experiences affect one's pursuit of mathematics as a field of study or as a profession is an improper subject to discuss in sci.math? If it is improper, I apologize too. Shedar === Subject: Re: complete space > Two points: > 1. In your example, the sets A(n), n>0, are open in M. > 2. Does your book say what kind of a space M is? Is it a metric space? > Is it a linear topological space? Yes - It was metric space. === Subject: Re: The multiplicative groups Z*_n; One more time... myself. > Lemmas to show (Z_p^m)^* is cyclic for odd prime p > g generates (Z_p^m)^* ==> o(g) in (Z_p^(m+1))^* = phi(p^m) or phi(p^(m+1)) > g^o(g) = 1 (mod p^(m+1)); g^o(g) = 1 (mod p^m); phi(p^m) | o(g) > o(g) | phi(p^(m+1)); o(g)/phi(p^m) | p; o(g)/phi(p^m) = 1 or p That part seems solid to me. > 2 <= m, Why must m => 2 ? m = 1, Z*_p = ; g^(p-1) /= 1 mod(p^2). I think I see. (mod p). Explain why the >order of the congruence class of h modulo 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 > g generates (Z_p^m)^*, g^phi(p^m) /= 1 (mod p^(m+1)) You assume this is true, right? If so, why do you need m => 2? > ==> g generates (Z_p^(m+1))^*, g^phi(p^(m+1)) /= 1 (mod p^(m+2)) > let b = g^phi(p^m); if b^p = g^phi(p^(m+1)) = 1 (mod p^(m+2)): > (b - 1)(b^(p-1) +..+ 1) = b^p - 1 = 0 (mod p^(m+2)) > p^m | b-1; not p^(m+1) | b-1; p^2 | b^(p-1) +..+ 1 > b = 1 (mod p^m); b = 1 (mod p^2); b^j = 1 (mod p^2) > 0 = b^(p-1) +..+ 1 = p (mod p^2); p^2 | p, whoops So g^phi(p^(m+1)) /= 1 (mod p^(m+2)), right? This is what you are showing here--do I have that right? > g generates Z_p^* ==> g or g + p generates (Z_p^2)^* > if (g + p)^(p-1) = 1 = g^(p-1) (mod p^2) (g + p)^(p-1) = [g^(p-1) - pg^(p-2)] > 1 = g^(p-1) + (p-1) g^(p-2) p + ... = 1 - p.g^(p-2) (mod p^2) g^(p-1) /= 1 mod p^2 ; g^(p-1) = 1 mod p ==> g^(p-1) = 1 + ap ==> (g + p)^(p-1) = 1 + p[a - g^(p-2)]. I think its best to use g + kp and then choose k so that either o(g) = p-1 or o(g) = p(p-1) in Z*_p^2, i.e we can always find a k so that g generates Z*_p^2. > p.g^(p-2) = 0 (mod p^2 > ) > p = p.g^(p-1) = 0 (mod p^2); p^2 | p, whoops > g generates (Z_p^2)^* ==> g + p^2 generates (Z_p^2)^*, I don't understand. In Z*_p^2, g + p^2 = g. > (g + p^2)^phi(p^2) = 1 (mod p^3). > otherwise: let b = (g + p^2)^p > 1 = b^(p-1) + (p-1) b^(p-2) p^2 + ... = 1 - b^(p-2) p^2 > (mod p^3) > (only time p > 2 is used); b^(p-2) p^2 = 0 (mod p^3) > p^2 = b^(p-1) p^2 = 0 (mod p^3); p^3 | p^2 whoops I will go over this last again. I am not clear what you are doing here. === Subject: Re: The multiplicative groups Z*_n; One more time... it myself. Got scrambled in spite of reformatting. Again. > Lemmas to show (Z_p^m)^* is cyclic for odd prime p > g generates (Z_p^m)^* ==> o(g) in (Z_p^(m+1))^* = phi(p^m) or phi(p^(m+1)) > g^o(g) = 1 (mod p^(m+1)); g^o(g) = 1 (mod p^m); phi(p^m) | o(g) > o(g) | phi(p^(m+1)); o(g)/phi(p^m) | p; o(g)/phi(p^m) = 1 or p That part seems solid to me. > 2 <= m, Why must m => 2 ? m = 1, Z*_p = ; g^(p-1) /= 1 mod(p^2). >(a) Let h is an integer satisfying h = g (mod p). Explain why the >order of the congruence class of h modulo 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 > g generates (Z_p^m)^*, g^phi(p^m) /= 1 (mod p^(m+1)) You assume this is true, right? If so, why do you need m => 2? > ==> g generates (Z_p^(m+1))^*, g^phi(p^(m+1)) /= 1 (mod p^(m+2)) > let b = g^phi(p^m); if b^p = g^phi(p^(m+1)) = 1 (mod p^(m+2)): > (b - 1)(b^(p-1) +..+ 1) = b^p - 1 = 0 (mod p^(m+2)) > p^m | b-1; not p^(m+1) | b-1; p^2 | b^(p-1) +..+ 1 > b = 1 (mod p^m); b = 1 (mod p^2); b^j = 1 (mod p^2) > 0 = b^(p-1) +..+ 1 = p (mod p^2); p^2 | p, whoops So g^phi(p^(m+1)) /= 1 (mod p^(m+2)), right? This is what you are showing here--do I have that right? > g generates Z_p^* ==> g or g + p generates (Z_p^2)^* > if (g + p)^(p-1) = 1 = g^(p-1) (mod p^2) (g + p)^(p-1) = [g^(p-1) - pg^(p-2)] > 1 = g^(p-1) + (p-1) g^(p-2) p + ... = 1 - p.g^(p-2) (mod p^2) g^(p-1) /= 1 mod p^2 ; g^(p-1) = 1 mod p ==> g^(p-1) = 1 + ap ==> (g + p)^(p-1) = 1 + p[a - g^(p-2)]. I think its best to use g + kp and then choose k so that either o(g) = p-1 or o(g) = p(p-1) in Z*_p^2, i.e we can always find a k so that g generates Z*_p^2. > p.g^(p-2) = 0 (mod p^2 > p = p.g^(p-1) = 0 (mod p^2); p^2 | p, whoops > g generates (Z_p^2)^* ==> g + p^2 generates (Z_p^2)^*, I don't understand. In Z*_p^2, g + p^2 = g. > (g + p^2)^phi(p^2) = 1 (mod p^3). > otherwise: let b = (g + p^2)^p > 1 = b^(p-1) + (p-1) b^(p-2) p^2 + ... = 1 - b^(p-2) p^2 > (mod p^3) > (only time p > 2 is used); b^(p-2) p^2 = 0 (mod p^3) > p^2 = b^(p-1) p^2 = 0 (mod p^3); p^3 | p^2 whoops I will go over this last again. I am not clear what you are doing here. === Subject: The multiplicative groups Z*_n; One more time... === Subject: Re: The multiplicative groups Z*_n; One more time... Use http://mathquest.com/discuss/sci.math http://mathforum.org/epigone/sci.math > 2 <= m, g generates (Z_p^m)^*, g^phi(p^m) /= 1 (mod p^(m+1)) >You assume this is true, right? If so, why do you need m => 2? > ==> g generates (Z_p^(m+1))^*, g^phi(p^(m+1)) /= 1 (mod p^(m+2)) > let b = g^phi(p^m); if b^p = g^phi(p^(m+1)) = 1 (mod p^(m+2)): > (b - 1)(b^(p-1) +..+ 1) = b^p - 1 = 0 (mod p^(m+2)) > p^m | b-1; not p^(m+1) | b-1; p^2 | b^(p-1) +..+ 1 > b = 1 (mod p^m); b = 1 (mod p^2); b^j = 1 (mod p^2) 2 <= m is needed for this step^^^^^^^ > 0 = b^(p-1) +..+ 1 = p (mod p^2); p^2 | p, whoops >So g^phi(p^(m+1)) /= 1 (mod p^(m+2)), right? >This is what you are showing here--do I have that right? Yes, the indented statements after if....: make contradiction. > g generates Z_p^* ==> g or g + p generates (Z_p^2)^* > if (g + p)^(p-1) = 1 = g^(p-1) (mod p^2) >(g + p)^(p-1) = [g^(p-1) - pg^(p-2)] Huh? > 1 = g^(p-1) + (p-1) g^(p-2) p + ... = 1 - p.g^(p-2) (mod p^2) >g^(p-1) /= 1 mod p^2 ; g^(p-1) = 1 mod p ==> >g^(p-1) = 1 + ap ==> (g + p)^(p-1) = 1 + p[a - g^(p-2)]. Where does a come from? >I think its best to use g + kp and then choose k so that >either o(g) = p-1 or o(g) = p(p-1) in Z*_p^2, i.e we can always >find a k so that g generates Z*_p^2. That's what I said, k = 0 or k = 1 in paticular. > p.g^(p-2) = 0 (mod p^2) > p = p.g^(p-1) = 0 (mod p^2); p^2 | p, whoops > g generates (Z_p^2)^* ==> g + p^2 generates (Z_p^2)^*, >I don't understand. In Z*_p^2, g + p^2 = g. 2 generates Z_3^*, so does 5 and 8. 5 = 2, 5^2 = 1 (mod 3) In Z_3^2, 2^2 = 4, 5^2 = 7, 8^2 = 1 (mod 9) Thus, 2 and 5 but not 8 generate (Z_3^2)^* as from above: g generates Z_p^*, g^(p-1) /= 1 (mod p^2) ==> g generates (Z_p^2)^* g generates Z_p^* ==> g or g + p generates (Z_p^2)^* g + p generates Z_p^*, thus g + p or g + 2p generates (Z_p^2)^* > (g + p^2)^phi(p^2) = 1 (mod p^3). and (g + p^2)^phi(p^2) /= 1 (mod p^3). > otherwise: let b = (g + p^2)^p > 1 = b^(p-1) + (p-1) b^(p-2) p^2 + ... = 1 - b^(p-2) p^2 (mod p^3) > (only time p > 2 is used); b^(p-2) p^2 = 0 (mod p^3) > p^2 = b^(p-1) p^2 = 0 (mod p^3); p^3 | p^2 whoops >I will go over this last again. Don't bother. Instead find for all odd prime p, some g generates (Z_p^2)^* with g^(p-1) /= 1 (mod p^3). >I am not clear what you are doing here. Me neither. ---- === Subject: Re: Particular case of a generalized Borsuk-Ulam theorem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i86CrtH21796; > I shall refer to a basic result of C.T.Yang, which implies that in >our case any 3 orbital partitions in S^3 have nonempty intersection. Sorry, my fault, seems that it was proven first by M.A. Krasnosel'skii in 1955 for maps of a prime period, C.T.Yang dealed only with the case of involution. Simeon === Subject: well-ordering by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i86Crtl21792; I have this problem Exhibit well ordering of the set Q of rational numbers. But it seems to me a nonesence if we take Q as a subset of Q, then no minimal element is here. === Subject: Re: well-ordering > Exhibit well ordering of the set Q of rational numbers. A well order of Q of order type w^2. 0,1,2,3, ... -1,-2,-3, ... 1/2, 3/2, 5/2, ... -1/2, -3/2, -5/2, ... 1/3, 2/3, 4/3, 5/3, ... -1/3, -2/3, -4/3, -5/3, ... 1/4, 3/4, 5/4, 7/4 -1/4, -3/4, -5/4, -7/4 1/5, 2/5, ... -1/5, ... 1/6, ... -1/6, ... ... ... *** === Subject: Re: well-ordering X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 12:53 PM, angel_1978@hotmsil.com (dominique) said: >I have this problem >Exhibit well ordering of the set Q of rational numbers. Hint: express the nonzero rational numbers as ratios of relatively prime integers. >But it seems to me a nonesence if we take Q as a subset of Q, then no >minimal element is here. You're confusing two separate issues. The standard ordering of the rationals is not a well ordering, but that does not prevent you from constructing a nonstandard ordering that does well order them. In fact, the standard proof that the rationals are countable provides a well ordering. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: well-ordering X-no-archive: yes linux) > I have this problem > Exhibit well ordering of the set Q of rational numbers. > But it seems to me a nonesence if we take Q as a subset of Q, > then no minimal element is here. Hint four your homework: use a bijection between Q and a well-known well ordered set. pg. === Subject: Re: well-ordering > I have this problem > Exhibit well ordering of the set Q of rational numbers. > But it seems to me a nonesence if we take Q as a subset of Q, > then no minimal element is here. If? (let's deal with just positives (negatives scare me). Let q in Q be a/b (in lowest terms). then let the ordering be: 1/1 is bottom if a I have this problem > Exhibit well ordering of the set Q of rational numbers. > But it seems to me a nonesence if we take Q as a subset of Q, > then no minimal element is here. So the usual ordering of Q isn't a well-ordering. Thus, you need an unusual ordering of Q :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: formal power series and local rings by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i86CrsG21762; >regarding the algebraic closure of K((x)). I have been recently >reading up some algebraic geometry. There seem to be two routes to >algebraic geometry. One using local rings which is simple and >elegant. The other is using power series rings for local analysis. This is not correct: first of all both approaches are only >local<. That means they are representing only a part of algebraic geometry. Second they are not completely comparable: studying local rings of algebraic varieties means to study the variety >locally< in the sense of Zariski topology, which is a very rough topology. The power series approach is finer, yields more information about the local structure. >That is good for algebraically closed field of characteristic zero (I >have complex numbers in mind, nothing more), but what about finite >fields? Is such an analysis using power series possible for finite >fields and their algebraic closure? Basically both approaches work fine over every field. >Another thing that is kind of troubling me is that the analysis with >local rings and valuations seems to specify exactly the values of >intersection multiplicities at a point (in terms of the dimenstion of >some vector space). But the theory of resultants only seems to give What kind of intersection multiplicities? Intersection theory is a quite advanced topic with lots of facets. It strongly depends on which kind of varieties one is working. Already in its simplest form you need >lengths< to define intersection multiplicities - valuations are not sufficient. >inequalities for intersection multiplities. Also, the definition of >genus is not so clear if we use power series rings. Well... if all >these i said above is true (which should be if my understanding is >correct), then it looks like it is not good enough to learn the >analysis using power series and resultants. If you are referring to the genus of a curve: the genus is a global invariant of a curve, that cannot be calculated on a local basis. >Prasanna. H === Subject: Re: formal power series and local rings >regarding the algebraic closure of K((x)). I have been recently >reading up some algebraic geometry. There seem to be two routes to >algebraic geometry. One using local rings which is simple and >elegant. The other is using power series rings for local analysis. > This is not correct: first of all both approaches are only >local<. > That means they are representing only a part of algebraic geometry. This is clear. > Second they are not completely comparable: studying local rings > of algebraic varieties means to study the variety >locally< in the > sense of Zariski topology, which is a very rough topology. I thought so. But i read recently that a power series ring is a local ring. Infact, it is a theorem in Algebraic Geometry by Solomon Lefschetz, page 88: Properties of local rings: (7.3) A power series ring is a local ring. And does not that mean that they are same? I have only heard of Zariski topology, but do not know what it is. I also wish to add that i am no math student, and have had no course on topology except having read something from analysis books out of curiosity. > The power series approach is finer, yields more information about > the local structure. How? Can you please give a brief outline. >That is good for algebraically closed field of characteristic zero (I >have complex numbers in mind, nothing more), but what about finite >fields? Is such an analysis using power series possible for finite >fields and their algebraic closure? > Basically both approaches work fine over every field. But a power series expansion of an element of a finite field! It is not clear... >Another thing that is kind of troubling me is that the analysis with >local rings and valuations seems to specify exactly the values of >intersection multiplicities at a point (in terms of the dimenstion of >some vector space). But the theory of resultants only seems to give > What kind of intersection multiplicities? Intersection theory is > a quite advanced topic with lots of facets. It strongly depends on > which kind of varieties one is working. > Already in its simplest form you need >lengths< to define intersection > multiplicities - valuations are not sufficient. I am putting down a result from my memory. Dunno know how precise it is. Let Vr be a variety of dimension r and Vs be a variety of dimension s, the intersection multiplicity of Vr and Vs at any point is less than min{r,s} and this result is an outcome of the theory of resultants. There are a few more bounds that i dont have in memory, but all of them are inequalities. >inequalities for intersection multiplities. Also, the definition of >genus is not so clear if we use power series rings. Well... if all >these i said above is true (which should be if my understanding is >correct), then it looks like it is not good enough to learn the >analysis using power series and resultants. > If you are referring to the genus of a curve: the genus is a global > invariant of a curve, that cannot be calculated on a local basis. Well.. books that develope the theory using resultants and power series define genus as follows: If the curve C is given by a homogeneous polynomial equation of degree d, the genus of the curve is the non-negative number given by g = (d-1)(d-2)/2 - sum_P {mp(mp-1)/2} where the sum is over all singular points P and mp is the multiplicity of P. And the books deal with anything related to singular points P using power series. This definition of genus also offers no insight into the geometrical picture that genus is the number of holes of the curve over complx numbers. Like the torus has only one hole and so is genus one and hence an elliptic curve which corresponds to a torus is of genus 1. I think i am a bit clumsy in writing these, but i am only starting to learn AG, so i have an excuse. >Prasanna. === Subject: Re: formal power series and local rings ... > Second they are not completely comparable: studying local rings > of algebraic varieties means to study the variety >locally< in the > sense of Zariski topology, which is a very rough topology. >I thought so. But i read recently that a power series ring is a local >ring. Infact, it is a theorem in Algebraic Geometry by Solomon >Lefschetz, page 88: >Properties of local rings: (7.3) A power series ring is a local ring. >And does not that mean that they are same? Certainly not; why should it? A dog is a mammal. And does that not mean that they are they same? Many important local rings are not power series rings (at least, on the usual interpretation of power series rings). ... > The power series approach is finer, yields more information about > the local structure. >How? Can you please give a brief outline. Not an outline, but an example. Consider, over the complex numbers C, two affine plane curves: A, defined by zw=0, and B, defined by zw=z^3+w^3. Globally in C^2 with its ordinary (Euclidian) topology these are quite different (for instance, if the singular point (0,0) is removed from A, what's left is no longer pathwise connected, whereas if (0,0) removed from B, the remaining set is still pathwise connected). Also, in C^2 with its Zariski topology (in which the open sets are C^2, the empty set, and the sets where some polynomial p(z,w) is non-zero), A and B are still quite different even locally (meaning, in any open set), though showing that would take a few more ideas and lines. BUT, both locally in C^2 with its Euclidian topology, AND locally in the sense of (formal or convergent) power series centered at (0,0), A and B look identical: there is an analytic change of coordinates near (0,0) which transforms A near (0,0) into B near (0,0). Thus, the local structure of A in either the Euclidian or the power-series sense is the same as the local structure of B. (Is this more information, or less information? It's of more interest, anyway.)_ ... >But a power series expansion of an element of a finite field! It is >not clear... Think formally. ... >And the books deal with anything related to singular points P using >power series. This definition of genus also offers no insight into the >geometrical picture that genus is the number of holes of the curve >over complx numbers. Oh, it does, it does; but that's for another time, I have to go to work. Lee Rudolph === Subject: A solution to a quadratic problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i86Cru021883; I want to find the solution to the following set of equations x^T A_i x=a_i, i=1,...,N === Subject: dynkin system by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i86CrsS21753; I have this problem here: Determine the Dynkin system generated, by the system consisting of just two subsets A,B of G. show that d(e) and c(e) coincide just in case one of the sets A intersects B, A intersects compliment of B, B intersects compliment of A or compliment of A intersects compliment of B is empty. d(e) is a Dynkin system generated by e. c(e) is a sigma algebra generated by e. The question is why should it be expected that card c(e) Determine the Dynkin system generated, by the system consisting of just > two subsets A,B of G. show that d(e) and c(e) coincide just in case one > of the sets A intersects B, A intersects compliment of B, B intersects > compliment of A or compliment of A intersects compliment of B is empty. > d(e) is a Dynkin system generated by e. > c(e) is a sigma algebra generated by e. > The question is why should it be expected that card c(e) Determine the Dynkin system generated, by the system consisting of just > two subsets A,B of G. show that d(e) and c(e) coincide just in case one > of the sets A intersects B, A intersects compliment of B, B intersects > compliment of A or compliment of A intersects compliment of B is empty. > d(e) is a Dynkin system generated by e. > c(e) is a sigma algebra generated by e. > The question is why should it be expected that card c(e) What's a Dynkin system? A collection of subsets (of G) containing G and closed under the formation of complements and countable disjoint unions. -- A. === Subject: A question on Stirling number Is there any closed-form solution for Stirling number of the second kind? === Subject: Re: A question on Stirling number > Is there any closed-form solution for Stirling number of the second > kind? 1) define closed-form. 2) i) if by closed form you allow bounded summations and binomial coeffs, then yes ii) if you don't allow bounded summations, I don't think so (I don't know if there is an impossibility proof either). -- Mitch Harris (remove q to reply) === Subject: Re: Euler's lemma ?? >I came across this simple problem. >4. Let a, b in N and let d = gcd(a,b). Set l = ab/d. Show that l is >least common multiple (lcm) of a and b in the sense that the following >two >properties hold: >(i) a | l and b | l; (ii) if a | m and b | m then l | m. >(You might find Euler's Lemma useful.) >I am not familiar with Euler's Lemma, unless he means Euler's thm. >x^phi(n) = 1 if x in Z*_n. Can this be what he means? >I don't see that it applies to lcm. > I don't either, and I'm not sure what they mean by Euler's Lemma. It > seems easiest just to prove the result. > Write a = dr > b = ds, with gcd(r,s)=1. > We are asked to show that dmn is lcm(a,b). You mean drs is lcm(a,b). > Clearly a and b divide drs. > Now assume that a and b divide m. Then dr, ds divide m. Write m = > dr*u = ds*v. Then dru = dsv, so ru = sv. That means that s|ru, and > since gcd(r,s)=1, then s|u. Therefore, writing u = s*w, we have m= > drs*w, proving that drs divides m, as claimed. the word simple, doesn't seem appropriate). I am better with gcd, as I have more experience with it, which is why I wanted to do a problem involving lcm. === Subject: Re: Euler's lemma ?? days. My association with the Department is that of an alumnus. >I came across this simple problem. > >4. Let a, b in N and let d = gcd(a,b). Set l = ab/d. Show that l is >least common multiple (lcm) of a and b in the sense that the >following >two >properties hold: >(i) a | l and b | l; (ii) if a | m and b | m then l | m. > >(You might find Euler's Lemma useful.) > >I am not familiar with Euler's Lemma, unless he means Euler's thm. > >x^phi(n) = 1 if x in Z*_n. Can this be what he means? >I don't see that it applies to lcm. > I don't either, and I'm not sure what they mean by Euler's Lemma. > seems easiest just to prove the result. > Write a = dr > b = ds, with gcd(r,s)=1. > We are asked to show that dmn is lcm(a,b). >You mean drs is lcm(a,b). Oops. Yes, sorry. I was using m and n for r and s until I realized that you had already co-opted m, and went back and changed it. Obviously didn't change it enough. > Clearly a and b divide drs. > Now assume that a and b divide m. Then dr, ds divide m. Write m = > dr*u = ds*v. Then dru = dsv, so ru = sv. That means that s|ru, and > since gcd(r,s)=1, then s|u. Therefore, writing u = s*w, we have m= > drs*w, proving that drs divides m, as claimed. >the word simple, doesn't seem appropriate). >I am better with gcd, as I have more experience with it, which is >why I wanted to do a problem involving lcm. Obviously, there is a duality between lcm's and gcd's, so once you figure out how to exploit it, they should both be equally easy. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Help with Boolean Algebra How can I learn to simplify the following? (BC' + A'D)(AB' + CD') Any hints or assistance would be appreciated. === Subject: Re: Help with Boolean Algebra > How can I learn to simplify the following? > (BC' + A'D)(AB' + CD') Use the distributive law (BC' + A'D)(AB' + CD') = BC'(AB' + CD') + A'D(AB' + CD') = BC'AB' + BC'CD' + ... > Any hints or assistance would be appreciated. reply. Others who have answer, crammed the math together into harsh to read expressions. === Subject: Re: Help with Boolean Algebra > How can I learn to simplify the following? > (BC' + A'D)(AB' + CD') > Any hints or assistance would be appreciated. i saw expressions like this in a digital logic class frequently. that definitely is equivelant to false. it's not really hard to see either because obviously if the left multiplicand is 1 the right one is zero and vice versa === Subject: Re: Help with Boolean Algebra Kris Wheeler skrev i melding > How can I learn to simplify the following? > (BC' + A'D)(AB' + CD') > Any hints or assistance would be appreciated. Wouldn't this simplyfi to 0 (zero), using the distributive, commutative, idempotent and absorption laws? Karl-Olav Nyberg === Subject: Re: Help with Boolean Algebra Karl-Olav Nyberg skrev i melding > Kris Wheeler skrev i melding > How can I learn to simplify the following? > (BC' + A'D)(AB' + CD') > Any hints or assistance would be appreciated. > Wouldn't this simplyfi to 0 (zero), using the distributive, commutative, > idempotent and absorption laws? > Karl-Olav Nyberg I will try be more specific: E = (BC'+A'D)(AB'+CD' ) = BC'AB'+BC'CD'+A[ CapitalOTilde]DAB'+A'DCD'. Using commutative law I get: E = ABB'C'+BCC'D'+AA 'B'D+A'CDD'. Using complement law I get: E = A(0)C'+B(0)D'+(0)B'D+A[Capital OTilde]C(0). Using boundness law I get: E = 0. Hope this helps. Karl-Olav Nyberg === Subject: Re: Unique factorization > I recall that the ring of integers of the field Q[sqrt(-5)] > = Z[sqrt(-5)] (have I got this part right?) also fails to > be a UFD (Unique factorization fails.) > Yes, the ring of integers of the field Q[sqrt(-5)] is Z[sqrt(-5)]. > And this is not a UFD. Set b = sqrt(-5). Then > 18 = (-1)*(2 + b)*(1 + b)^2 = 3*(1 - b)*(1 + b) > = (-1)*(2 - b)*(1 - b)^2 = 2*(2 + b)*(2 - b) = 2*3*3 > gives 5 different factorizations of 18 into irreducibles. > In terms of ideals of Z[b], (2) = I^2 (2 ramifies) and (3) = JK (3 splits), > where I = [2, 1 + b], J = [3, 1 + b], K = [3, 2 + b]. The product of any two > of these is principal. In fact, IJ = (1 + b), IK = (1 - b), J^2 = (2 - b), and > K^2 = (2 + b). The several ways of separating I^2 J^2 K^2 into products > principal ideals give the factorizations above. > John Robertson I recall 6 = 2.3 = (1 + b)(1 - b) = Norm(1 + b). Isn't the point of using ideals that when one uses ideals to factor, factorization becomes unique, or sort of unique? I need to learn this stuff. Fortunately, there are a lot of notes online on this. === Subject: Re: Unique factorization >Isn't the point of using ideals that when one uses ideals to factor, >factorization becomes unique, or sort of unique? In a Dedekind domain, factorization of ideals is unique. Rings of integers of quadratic number fields Q(sqrt(D)), for D not a square, are Dedekind domains. Note that the ring of integers of the quadratic number field Q(sqrt(D)), for D squarefree, is the Z-module [1, b] where b = sqrt(D) if d == 2 or 3 (mod 4), and b = (1 + sqrt(D))/2 if d == 1 (mod 4). The order of conductor f in the ring of integers is the Z-module [1, fb]. For f > 1, orders are not Dedekind domains. And these orders do not have unique factorization of ideals. But, from what I can tell, this does not mean there are ideals so IJ = KL with I, J, K, L irreducible, and none of the pairs {I, K}, {I, L}, {J, K}, {J, L} associates. It does seem to mean that there are ideals that cannot be factored into irreducibles, although every ideal contains a product of irreducible ideals. For example, if D = -3 and f = 2, so b = (1 + sqrt(-3))/2, the ideal [4, b] is not irreducible because the ideal [2, b] divides it. But [2, b]^2 = [4, 2b], which is contained in [4, b], and there is no product of irreducible ideals that is equal to [4, b]. On the other hand, in these orders, there is unique factorization for ideals relatively prime to the ideal (f). John Robertson === Subject: Re: Determinant maximization problem under trace constraint > I am recently working on a determinant maximization problem > with trace constraint, which is: > MAX det(A+F'BF) with respect to F > s.t. trace(F'F)=C > Where F' is the transpose of F. > Both A and B are constant diagonal matrix with positive diagonal entries. > C is a constant value. > Without loss of generality, we can rearrange the diagonal entries of > A and B in the opposite order, e.g., A with non-decreasing diagonal > entries and B with non-increasing diagonal entries. > My conjecture is that under such condition, one optimal solution is the > DIAGONAL matrix F with diagonal entries determined by water-filling algorithm > I am struggling for a long to prove or solve this optimization problem, > but still not much progress by now. > Could anyone please give me some clues or ideas of how to solve this problem? > Or is there any reference helpful to me. When one introduces a lagrange multiplier, and then differentiates the function F |--> det(A+F'BF) - lambda * [ trace(F'F) - C ], the result is (up to a factor 2) X |--> det(A+F'BF) * trace(X'BF) - lambda * [ trace(X'F) ] = trace( X'.[ det(A+F'BF) * BF - lambda * F ]) So F (with trace(F'F)=C) is a **critical** value of the problem iff BF is a scalar multiple of F, i.e. the columns of F are all eigenvectors to a fixed eigenvalue of B. Might be still not easy to find the eigenvalue of B and a corresponding F with trace(F'F)=C for which the actual **maximum** is taken on. - It might be easier if you have not general A and B, but only some specific examples. Can't help you on the part about the water-filling algorithm... === Subject: Re: Math at Princeton > >I looked online at Princeton's math dept. They have about 70 >faculty (including 13 visitors), and 55 grad + 30 undergrad students. >It seems a little small, but I guess its quality rather than quantity >that is important. >And supposedly, there is no course requirement. I guess that means you >don't have to take a single class to graduate. Every graduate students >are encouraged to begin their research/thesis upon matriculation >immediately. > No course requirement does not mean no knowledge requirement. > Don't misunderstand me. I'm trying to say those who are admitted have > likely gathered enough knowledge to begin research. And whatever else > they do not know, they do not need a class to learn it. There are some classes for graduate math students though, aren't there? It looks like they would be small, but that's OK. Its hard to lecture to less than 4 people though. I taught a class with 6 once, it went well (general relativity in the physics dept.). It seems a little early to start research on entry to grad school. I still had a _lot_ to learn before I was ready to do my own projects, though I was helping profs with their projects (I was a physics student). I don't know if its a good idea not to have 1st year students take classes. === Subject: Re: Math at Princeton > Don't misunderstand me. I'm trying to say those who are admitted >have > likely gathered enough knowledge to begin research. And whatever >else > they do not know, they do not need a class to learn it. >There are some classes for graduate math students though, aren't there? No doubt some things (besides the physical identity of the building named Fine Hall) have changed in the 39 years (eek!) since I arrived at Princeton as an undergraduate student; but I wouldn't be surprised if the status of classes for graduate math students isn't one of them. In those days, there were (as far as I can recall) no graduate courses _on general topics_: thus, there was no graduate course named Complex Analysis or Real Analysis or Number Theory, with a textbook, homework, exams, etc. But there certainly were graduate courses in those areas which were of broader scope (and shallower content) than, say, a seminar devoted to studying a particular theorem in depth (in the style of the seminar on the Atiyah-Singer Index Theorem which had happened a year or two before, the proceedings of which were published in the Annals of Mathematics Studies series). For instance, Fred Almgren had the page proofs of Federer's about-to-appear _Geometric Measure Theory_, and taught a course from them (with that title) in 1966. Similarly, the various Princeton Lecture Notes volumes by Gunning on Riemann surface theory, modular forms, and so on, were (essentially) accounts of graduate courses he was giving at the time. (No doubt other Princeton Lecture Notes volumes, like those by Nelson, were in the same case; but I wasn't aware of them.) Lee Rudolph === Subject: Re: Euler's lemma ?? === > Subject: Euler's lemma ?? >4. Let a, b in N and let d = gcd(a,b). Set l = ab/d. Show that l is >least common multiple (lcm) of a and b in the sense that the >following two properties hold: >(i) a | l and b | l; (ii) if a | m and b | m then l | m. >(You might find Euler's Lemma useful.) >I am not familiar with Euler's Lemma, unless he means Euler's thm. > I think he means the computation from the Euclidean > algorithm that determines the a,b with an + bm = (n,m) I thought that was called the extended Euclidean algorithm, or Bezout's (sp?) thm. I don't see how it relates to this problem though. Is there a similar expression for l = lcm(a,b) = ra + sb ? l^(-1) = (a,b)/ab = r/b + s/a ?? === Subject: Re: Euler's lemma ?? >4. Let a, b in N and let d = gcd(a,b). Set l = ab/d. Show that l is >least common multiple (lcm) of a and b in the sense that the >following two properties hold: >(i) a | l and b | l; (ii) if a | m and b | m then l | m. > algorithm that determines the a,b with an + bm = (n,m) > I thought that was called the extended Euclidean algorithm, or Bezout's > thm. I don't see how it relates to this problem though. assuming n,m | k kan + kbm = k(n,m) kan/(n,m) + kbm/(n,m) = k nm/(n,m) | k === Subject: Re: Euler's lemma ?? days. My association with the Department is that of an alumnus. === > Subject: Euler's lemma ?? >4. Let a, b in N and let d = gcd(a,b). Set l = ab/d. Show that l is >least common multiple (lcm) of a and b in the sense that the >following two properties hold: >(i) a | l and b | l; (ii) if a | m and b | m then l | m. >(You might find Euler's Lemma useful.) >I am not familiar with Euler's Lemma, unless he means Euler's thm. > I think he means the computation from the Euclidean > algorithm that determines the a,b with an + bm = (n,m) >I thought that was called the extended Euclidean algorithm, or Bezout's >(sp?) >thm. I don't see how it relates to this problem though. >Is there a similar expression for l = lcm(a,b) = ra + sb ? Of course: an integer x can be expressed as a linear combination of a and b if and only if x is a multiple of the gcd of a and b. Since the lcm of a and b is such a critter, it can always be expressed as a linear combination. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Help with Boolean Algebra > How can I learn to simplify the following? > (BC' + A'D)(AB' + CD') > Any hints or assistance would be appreciated. [(B/C')/(A'/D)]/[(A/B')/(C/D[C apitalOTilde])] = (R/S)/(T/U) = (R/S/T)/(R/S/U) = ... I would fool around with the distributive laws, the rules for complements (a/b')' = a'/b, etc., and write it both ways, with +, . = /, / . Often you can see something in one form more clearly than in another. I don't know how much one can simplify an expression like this. Other forms may not be any more simple-=-I don't see any simpler form from a cursory look at it. === Subject: Re: Simple question regarding open set in metric space by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i86E4cC28308; >I think this is a very simple question indeed. >Supposed X = set of all points (x,y) such that x, y are integers. And the metric is Euclidean distance. >Now I have 2 subsets:- >A={(x,y) is element of X: x^2 + y^2 < 25} >B={(x,y) is element of X: x^2 + y^2 <= 25} >According to the definition of open set, a subset is open on X if ther is a open ball (belonging to this subset) around each of the points in this subset. >Now, if I purposely choose the radius to be arbitrarily small then for each point in the subset A or B, the open ball about this point is the very point itself. >According to this logic, then A and B are OPEN SUBSETS of X. Which is wrong, of course. Sinec B really is a closed ball, and so should be a closed set. >Can anyone enlighten me? Ooops, I just noticed that you were requiring points in the set to be integers! Yes, in that case, the inherited topology (inherited from R^2 with the usual metric topology) IS the discrete topology. I now understand what everyone else was talking about! === Subject: Re: Simple question regarding open set in metric space by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i86E4ca28302; >I think this is a very simple question indeed. >Supposed X = set of all points (x,y) such that x, y are integers. And the metric is Euclidean distance. >Now I have 2 subsets:- >A={(x,y) is element of X: x^2 + y^2 < 25} >B={(x,y) is element of X: x^2 + y^2 <= 25} >According to the definition of open set, a subset is open on X if ther is a open ball (belonging to this subset) around each of the points in this subset. >Now, if I purposely choose the radius to be arbitrarily small then for each point in the subset A or B, the open ball about this point is the very point itself. >According to this logic, then A and B are OPEN SUBSETS of X. Which is wrong, of course. Sinec B really is a closed ball, and so should be a closed set. >Can anyone enlighten me? Several people answered this in terms of the discrete metric in which EVERY set is open but I don't think that's what you mean. If can't purposely choose the radius to be arbitrarily small. The radius must be a fixed positive number. No matter how arbitrarily small r (the radius is) it is still a positive number and for any x_0, there is an point y such that 0 If you've learnt functional analysis, you should know the function whose > Fourier expansion = 0 is f(x) = 0 a.e. Of course the > Fourier expansion for delta function is not zero, but it's not Lebesgue > measurable. A misleading characterization. The problem is not that delta is nonmeasurable. The problem is that delta is not a function (of a real variable). > Then are there any other functions which are not > Lebesgue measurable and whose Fourier expansion is not zero? I thought some > fractal functions might be such functions. All of the standard fractal functions are Lebesgue measurable. Most of them are even continuous. There are fractal measures that we sometimes use. (The so-called delta function is actually a measure.) Fractal measures can sometimes be studied using a Fourier expansion. > For detail, please visit: > http://139.134.5.123/tiddler2/fbt/fbt.htm Sorry, couldn't read it ... some Javascript error. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Norm of Fractal Function When visiting that webpage, please turn on JavaScript for it is protected by spammer filters. > If you've learnt functional analysis, you should know the function whose > Fourier expansion = 0 is f(x) = 0 a.e. Of course the > Fourier expansion for delta function is not zero, but it's not Lebesgue > measurable. > A misleading characterization. The problem is not that delta is > nonmeasurable. The problem is that delta is not a function (of a > real variable). > Then are there any other functions which are not > Lebesgue measurable and whose Fourier expansion is not zero? I thought some > fractal functions might be such functions. > All of the standard fractal functions are Lebesgue measurable. > Most of them are even continuous. There are fractal measures > that we sometimes use. (The so-called delta function is actually > a measure.) Fractal measures can sometimes be studied using > a Fourier expansion. > For detail, please visit: > http://139.134.5.123/tiddler2/fbt/fbt.htm > Sorry, couldn't read it ... some Javascript error. > -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Norm of Fractal Function >When visiting that webpage, please turn on JavaScript for it is protected by spammer filters. uh, i have javascript turned on. when i go there a page slowly appears - then when the download is finished the page disappears and is replaced by Ônotice! please turn on javascript.' [of course if the information on the page is as accurate as the information in your post there's really no reason anyone should care...] > If you've learnt functional analysis, you should know the function whose > Fourier expansion = 0 is f(x) = 0 a.e. Of course the > Fourier expansion for delta function is not zero, but it's not Lebesgue > measurable. > A misleading characterization. The problem is not that delta is > nonmeasurable. The problem is that delta is not a function (of a > real variable). > Then are there any other functions which are not > Lebesgue measurable and whose Fourier expansion is not zero? I thought some > fractal functions might be such functions. > All of the standard fractal functions are Lebesgue measurable. > Most of them are even continuous. There are fractal measures > that we sometimes use. (The so-called delta function is actually > a measure.) Fractal measures can sometimes be studied using > a Fourier expansion. > For detail, please visit: > > http://139.134.5.123/tiddler2/fbt/fbt.htm > > > Sorry, couldn't read it ... some Javascript error. > -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Metric Tensor of Flat Space-Time > at 12:36 PM, thoovler@excite.com (Igor) said: >You're right, but that's not what I said. The metric is independent >of coordinate system, but the metric tensor is not. > The two term metric in Differential Geometry normally refers to the > metric tensor. The metric tensor is independent of any coordinate > system. What depends on the coordinate system are the components. >So my remarks stills stand. > No. You're confusing the tensor with its components in particular > coordinate systems. This business of Ômetric' and Ômetric tensor' is oddly confusing in the textbooks. I have adopted Rindler's terminology where: ds^2 = expression in terms of differentials in cartesian coordinates or ds^2 = expression in terms of differentials in polar coordinates to be the Ômetric' whilst the table of differential coefficients Ôg' such as : -1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 for ßat space-time, is the Ômetric tensor'. There is no doubt however that the table is often referred to as simply the Ômetric', despite the fact that the expression for the space-time interval is also called the Ômetric'. If we are investigating the metric of ßat space-time then the metric tensor should be dependent on the coordinate system: See: http://www.users.globalnet.co.uk/~lka/tensors.htm What disturbs me in this analysis is that the Ôtangent plane' is a plane with reference to an assumed observer standing on the surface, it is part of the observer's coordinate system. I've got a feeling I am wrong about this because it is too obvious and too many people must have passed this way before, I must have misunderstood something. But what? Best Wishes Alex Green === Subject: Re: Metric Tensor of Flat Space-Time > [...] > I don't view the fourth axis as useful representing time in a > gravitational model but if the intervals on a nonliner or dynamic > fourth axis are inversly proportional to the proper coupling for the > region's potential then they would take on some relation to curvature > or line density. > > Yes, if I understand you correctly, 3D spatial line > density should yield *time* curvature. > Ahh! line density I saw in a dream but the curvature was on axis #4 > of 4 so I gave it more local importance than TIME as in waiting does > nothing to the dusty book or sword here... your lamp is getting > dimmer ;-) > Sounds good... > Agreed, personally I'm comfortable with the semantics > that gravity results from electrostatic attraction being > greater than repulsion. I quip that's love on a universal > scale and produces structure and order. If repulsion were > stronger, chaos and disorganization would result. > > In free space that's dandy. Below a body's surface you have to switch > to magnetic analog. I am leaning more toward an energy density model. > You qualified your statements above with the word Ôsemantics' so I > won't ask the possibility of shielding gravity with aluminum foil or > window screening. ;-) > Ha, ok, well the gravitational force term turns > up dependant on q^2 Q^2/(R^2 r^2) unlike the more > familiar qQ/r^2 Coulomb provides. Mass and/or energy > is qQ/R. The term q^2 Q^2/(R^2 r^2) does NOT notice > shielding unfortunately. A monopolar attractive charge which is electrodynamically inert. A Higgs somthing-or-other. OK > If we have any common ground it is if your > U_i = 0 > means the same as my elimination of a time axis and motion. > > Certainly covariant motion is eliminated by U_i=0, > and would agree the inclusion of a time axis is a > specialization of a CS to 4 dimensions that are > comfortable when humans make predictions. But I > doubt the time axis is necessary for the physical > foundation of the universe as it requires memory. > > Is that close? > > No. But in retrospect I would have better said rejection of the > equivalence principle. After all that is a result of elimination of a > time axis and motion [from the gravitational model]. > The best explanation of the PoE I've seen is the > vanishing of the covariant derivative of the metric, > g_uv;w = 0 . > So if time and motion were eliminated I think that > equation would still be true.... > of your post, very stimulating... > Likewise! > Sue... > > Saving a the math part of this thread some foot traffic and confusion > by sticking this here: > Re: 4D Orthogonality > > My interpretation, and this is surely not my ballywick, so it could be > entirely wrong as it applies to the formalism you are using but I will > toss it out so you will know what I heard you say tho it may not be > what you meant for me to hear: > In 2D or 3D space, a vector can't effect displacement (independent?) > on an orthogonal axis and the vector will be 90 degress to the line or > plane. > > In 4D space, a vector can't effect displacement (independent?) on an > orthogonal axis and the vector's *angle* will be undefined to lines or > planes or spaces. > Example: Displacement on the time axis (waiting) will not cause > displacements on the x,y and z axes (cause a pot to boil) so t is > orthogonal to x,y and z. > Yes, have a look at a clock in your palace. If > you aren't moving then the clock is travelling > *threw time* at 186,000 mps, orthogonally to > x,y,z. But if you approach (on the x-axis say) > the clock until it's over your head, you have made > the clock change it's course threw time so some > of it's path has a projection on your x-axis, as > it moves toward you, relative to you. > When regarding the motion of the clock you find > it's NOT orthogonal to your x-axis, that's where > the g_01 = -v/c comes from, ie. (i dot t) in > unit vectors. All events at rest to your clock > also impose g_01 on your CS. > The usual mathematical notion of describing metrics > is to use a point at *rest* on some curved surface, > volume etc. > SR considers a point in motion on a ßat surface, > so our usual notions about metrics must be placed > into a *dynamic*, (moving) context, that is less > familiar. > We need to be certain our conventional rules and > ideas about metrics, using stationary points, aren't > in conßict with the principle of relativity when > the point has uniform motion (SR) or any motion (GR), > when that logical apparatus is applied to the theory > of relativity. > Hence a *nonorthogonal dynamic space time metric > like g_01 = - v/c has a necessary basis in relativity. Just reject the equivalence principle. LOL ;-) Honest! There are no rocket motors under your chair. > By reducing the importance of time and the equivalence of > acceleration, I seem to be probing down the road Einstein left > untraveled. > I don't know, I think alot of theoreticians have deeply > questioned PoE, (I have), it's pretty well armoured > theoretically and experimentally, although Uncle Al > has some interesting caveats, at the limit of my > understanding. > Exhaustive and exhausting IMHO. It is entertaining anyway. >I hope you'll take no offence that I can't get too deeply > into some very interesting nuances of your formalism. I am just > peeking over your shoulders and quipping from the peanut gallery where > inappropriate. ;-) Please don't alter the course of your conversaton > to > explain or even reply. That will impede motion on both paths. I will > pick up little morsels. > It shouldn't be that way, take big morsels, and come > back for dessert. more of the notation and I will surely have to or else find a box of 4D tinker-toys. Sue... === Subject: Re: Metric Tensor of Flat Space-Time [...] > We need to be certain our conventional rules and > ideas about metrics, using stationary points, aren't > in conßict with the principle of relativity when > the point has uniform motion (SR) or any motion (GR), > when that logical apparatus is applied to the theory > of relativity. > Hence a *nonorthogonal dynamic space time metric > like g_01 = - v/c has a necessary basis in relativity. > Just reject the equivalence principle. LOL ;-) > Honest! There are no rocket motors under your chair. Depends on dinner :) > It shouldn't be that way, take big morsels, and come > back for dessert. > more of the notation and I will surely have to or else find a box of > 4D tinker-toys. > Sue... Ok, then understand the basics, most guys seem to think the basics are boredom, but that's where I find the really good stuff. Without a good foundation, the exotic erections fall down, same as building a palace on quick sand. I've made countless 4D sketches like tinker-toys, good stuff. If you think there is a reason to e-com by e-mail, let me know, (my posted email collects spam, so I don't read it). Ken S. Ter === Subject: Re: Metric Tensor of Flat Space-Time > [...] > But I > doubt the time axis is necessary for the physical > foundation of the universe as it requires memory. > > I proposed earlier that there were 3 ways of using Gauss' analysis to > get to: > ds^2 = - dt^2 + dx^2 etc.. > 1. Time could be imaginary. > 2. g00 could be assigned the value -1 at the outset (ie: the metric is > assumed) > 3. According to Gauss' analysis of surfaces dT/dt could be imaginary. > See: http://www.users.globalnet.co.uk/~lka/tensors.htm > Suppose the third option is true so that the g00=-1 in the metric > tensor for ßat space-time originates in (idT/dt)^2 where dT is a > small interval in the observer's coordinate system and dt is a small > interval on the observed space-time surface. > If it is assumed that the surface is ßat and x,y,z,t,T are > displacements from the origin then: > s^2 = idT/dt idT/dt tt + xx + yy + zz > s^2 = -t^2 + x^2 etc.. > Let the observation point be the origin of the observer's coordinate > system, a Pythagorean metric apply and the space-time be ßat so that, > to be observed things exist on the observer's hypersphere given by: > 0 = r^2 + (iT)^2 > ie: > 0 = r^2 - T^2 > In the observer's coordinate system events satisfying this equation > would have no separation from the observation point, there being a > zero length bridge between the observation point and the events. It is > being proposed that the conscious observer has a different time > coordinate from the time coordinate found on the space-surfaces that > we measure. Observations consist of things being simultaneously at an > observation point whereas measurements consist of an encoding of the > state of an object or objects. > However, we know from the existence of the Lorentzian metric that in > real space-time (where we use measurements): > 0 = r^2 - t^2 > and this does NOT mean that there is any zero length bridge between > events and the origin because time is real. > But if there is such a thing as real time in the universe the observer > will also observe events where: > 0 = r^2 + t^2 - T^2 > Suppose Ôr' were the radius of a small sphere of brain activity, t is > the real time of historical brain activity and T is the imaginary time > of the same activity then it would be possible for the observer to > observe an Ôextended present'. We would be able to do things like hear > whole words and see movements that occur extended in time. If someone > said Ôhello' you would be able to hear the whole word extended in time > with the phonemes separated. It would be like hearing things. > Of course, this analysis would only apply if the moment events are in > principle observable in the observer's coordinate system they become > fixed in time with respect to measurements. (ie: if measurements > exclude imaginary time). There would be some strange Ôcollapse' of the > state of the world with the past conforming to GR and the position of > things, before they are in principle observable, being indeterminate. > Best Wishes > Alex Green What you're discussing is like radar ranging. x^2+y^2+z^2+(ct)^2 is the radar distance for two objects A and B at relative rest, separated by Ôr'. The diagonal line is the radars spacetime path originating at A, then reßected back from B to A, received at 2t. A B at time=0 | | | | | | | /| | / | |/ | A B at time = 2t, r=ct Seems that s^2 = x^2+y^2+z^2+(ct)^2 = r^2 + (ct)^2 is the spacetime distance. This is compatible with ds^2 = g_uv dx^u dx^v if one uses g00 =1, g11=1 and g01 = -v/c, then one obtains, ds^2 = g00 (ct)^2 + 2g01 cdt dx + g11 dx^2. Subbing the metrics and (v = dx/dt) gives the familiar ds^2 = (ct)^2 - dx^2. Given a Euclidean space, the metrics are constant so s^2 = g_uv x^u x^v Again subbing the metrics, s^2 = g00 (ct)^2 + 2g01 ct x + g11 x^2 = (ct)^2 - 2 (v/c) t x + x^2 The relative velocity of A and B is v =0, so s^2 = (ct)^2 + x^2 The radar pulse itself moves at the speed of light c. In this case v=c = dx/dt =x/t, and this gives, s^2 = (ct)^2 - x^2 = 0, meaning the spacetime distance from the radar to itself is zero, which is reasonable. The *dynamic nonorthogonal spacetime metric*, g01, simplifies spacetime in accord with SR. Ken S. Ter === Subject: Re: Metric Tensor of Flat Space-Time > [...] > But I > doubt the time axis is necessary for the physical > foundation of the universe as it requires memory. > > > I proposed earlier that there were 3 ways of using Gauss' analysis to > get to: > > ds^2 = - dt^2 + dx^2 etc.. > > 1. Time could be imaginary. > 2. g00 could be assigned the value -1 at the outset (ie: the metric is > assumed) > 3. According to Gauss' analysis of surfaces dT/dt could be imaginary. > > See: http://www.users.globalnet.co.uk/~lka/tensors.htm > > Suppose the third option is true so that the g00=-1 in the metric > tensor for ßat space-time originates in (idT/dt)^2 where dT is a > small interval in the observer's coordinate system and dt is a small > interval on the observed space-time surface. > > If it is assumed that the surface is ßat and x,y,z,t,T are > displacements from the origin then: > > s^2 = idT/dt idT/dt tt + xx + yy + zz > > s^2 = -t^2 + x^2 etc.. > > What you're discussing is like radar ranging. > x^2+y^2+z^2+(ct)^2 is the radar distance for two objects > A and B at relative rest, separated by Ôr'. > The diagonal line is the radars spacetime path originating > at A, then reßected back from B to A, received at 2t. > A B at time=0 > | | > | | > | | > | /| > | / | > |/ | > A B at time = 2t, r=ct > Seems that > s^2 = x^2+y^2+z^2+(ct)^2 = r^2 + (ct)^2 > is the spacetime distance. > This is compatible with ds^2 = g_uv dx^u dx^v > if one uses g00 =1, g11=1 and g01 = -v/c, > then one obtains, > ds^2 = g00 (ct)^2 + 2g01 cdt dx + g11 dx^2. > Subbing the metrics and (v = dx/dt) gives the familiar > ds^2 = (ct)^2 - dx^2. Surely this only applies on the inbound path of the radar pulse where the velocity is -v and means that the pulse comes back to the radar. > Given a Euclidean space, the metrics are constant so > s^2 = g_uv x^u x^v > Again subbing the metrics, > s^2 = g00 (ct)^2 + 2g01 ct x + g11 x^2 > = (ct)^2 - 2 (v/c) t x + x^2 > The relative velocity of A and B is v =0, so > s^2 = (ct)^2 + x^2 > The radar pulse itself moves at the speed of light c. > In this case v=c = dx/dt =x/t, and this gives, > s^2 = (ct)^2 - x^2 = 0, > meaning the spacetime distance from the radar to > itself is zero, which is reasonable. Yes, agreed, but if Ôt' is real the space-time interval has a different topological signifcance from the concept of Ôzero distance'. The Ôcausal light cone' summarises this topology, the surface of the backward cone being all those points where a light ray could have originated that could be at the origin at a particular moment. If Ôt' were based on units of (sqrt -1) then the causal light cone disappears and the backward cone becomes all those points that are also at the origin at a particular instant. Clearly this is physically incorrect, a photon cannot interact at two places at once. But what of the third option described above? Returning to the Gaussian derivation of metric tensors, suppose g00 is dT/dt where dT is the time interval on the tangent plane to a space-time surface and dt is the time interval on the surface itself. If the tangent plane has imaginary time intervals and the surface has real time intervals then g00 = -1 and the causal light cone reappears because all interactions take place on the space-time surface where time is real. See: http://www.users.globalnet.co.uk/~lka/tensors.htm The third option would be indistinguishable from GR at the classical level but would have three advantages, firstly it would allow the existence of strange phenomena, such as wavefunctions, provided actual observed transfers of energy do not occur on the space-time surface. Secondly it would allow a set of things at (r,t) to also be at a point (0,0) in the observer's coordinate system (the tangent plane) provided no energy was transferred to the point. Thirdly time would exist in the observer as a geometric entity like space, this would allow the strange phenomena of conscious experience such as things in brain activity being present simultaneously in experience, the existence of an apparent observation point within the brain activity that is experience etc. Best Wishes Alex Green === Subject: Re: Metric Tensor of Flat Space-Time > [...] > But I > doubt the time axis is necessary for the physical > foundation of the universe as it requires memory. > > > I proposed earlier that there were 3 ways of using Gauss' analysis to > get to: > > ds^2 = - dt^2 + dx^2 etc.. > > 1. Time could be imaginary. > 2. g00 could be assigned the value -1 at the outset (ie: the metric is > assumed) > 3. According to Gauss' analysis of surfaces dT/dt could be imaginary. > > See: http://www.users.globalnet.co.uk/~lka/tensors.htm > > Suppose the third option is true so that the g00=-1 in the metric > tensor for ßat space-time originates in (idT/dt)^2 where dT is a > small interval in the observer's coordinate system and dt is a small > interval on the observed space-time surface. > > If it is assumed that the surface is ßat and x,y,z,t,T are > displacements from the origin then: > > s^2 = idT/dt idT/dt tt + xx + yy + zz > > s^2 = -t^2 + x^2 etc.. > > What you're discussing is like radar ranging. > > x^2+y^2+z^2+(ct)^2 is the radar distance for two objects > A and B at relative rest, separated by Ôr'. > The diagonal line is the radars spacetime path originating > at A, then reßected back from B to A, received at 2t. > > A B at time=0 > | | > | | > | | > | /| > | / | > |/ | > A B at time = 2t, r=ct > > Seems that > s^2 = x^2+y^2+z^2+(ct)^2 = r^2 + (ct)^2 > is the spacetime distance. > > This is compatible with ds^2 = g_uv dx^u dx^v > if one uses g00 =1, g11=1 and g01 = -v/c, > then one obtains, > > ds^2 = g00 (ct)^2 + 2g01 cdt dx + g11 dx^2. > > Subbing the metrics and (v = dx/dt) gives the familiar > > ds^2 = (ct)^2 - dx^2. > Surely this only applies on the inbound path of the radar pulse where > the velocity is -v and means that the pulse comes back to the radar. v is velocity of A relative to B. > > Given a Euclidean space, the metrics are constant so > > s^2 = g_uv x^u x^v > > Again subbing the metrics, > > s^2 = g00 (ct)^2 + 2g01 ct x + g11 x^2 > > = (ct)^2 - 2 (v/c) t x + x^2 > > The relative velocity of A and B is v =0, so > > s^2 = (ct)^2 + x^2 > > The radar pulse itself moves at the speed of light c. > In this case v=c = dx/dt =x/t, and this gives, > > s^2 = (ct)^2 - x^2 = 0, > > meaning the spacetime distance from the radar to > itself is zero, which is reasonable. > Yes, agreed, but if Ôt' is real the space-time interval has a > different topological signifcance from the concept of Ôzero distance'. > The Ôcausal light cone' summarises this topology, the surface of the > backward cone being all those points where a light ray could have > originated that could be at the origin at a particular moment. Ok > If Ôt' were based on units of (sqrt -1) then the causal light cone > disappears and the backward cone becomes all those points that are > also at the origin at a particular instant. Clearly this is physically > incorrect, a photon cannot interact at two places at once. > But what of the third option described above? Returning to the > Gaussian derivation of metric tensors, suppose g00 is dT/dt where dT > is the time interval on the tangent plane to a space-time surface and > dt is the time interval on the surface itself. If the tangent plane > has imaginary time intervals and the surface has real time intervals > then g00 = -1 and > the causal light cone reappears because all interactions take place on > the space-time surface where time is real. > See: http://www.users.globalnet.co.uk/~lka/tensors.htm Did that. > The third option would be indistinguishable from GR at the classical > level but would have three advantages, firstly it would allow the > existence of strange phenomena, such as wavefunctions, provided actual > observed transfers of energy do not occur on the space-time surface. > Secondly it would allow a set of things at (r,t) to also be at a point > (0,0) in the observer's coordinate system (the tangent plane) provided > no energy was transferred to the point. Thirdly time would exist in > the observer as a geometric entity like space, this would allow the > strange phenomena of conscious experience such as things in brain > activity being present simultaneously in experience, the existence of > an apparent observation point within the brain activity that is > experience etc. > Best Wishes > Alex Green I think most GRist's would agree with your reasoning, and I have no problem with using it as an introduction and in a weak field. It was the approach Einstein used in his 1914 introduction to the GR foundation. For brevity and clarity of introduction he made some simplifying assumptions. I've clearly presented an alternative spacetime, using g_01 = -dx/dt, that is in accord with the ISU's L=cT interdefinition of Length and Time using c, without any reservations whatsoever, and therefore fully support the ISU's decision and have no need of any imaginary, physically ill-defined quantities, nor do I recommend them in modern GR, not withstanding the real possibilty that time is truly imaginary as some think it should be on philosphical grounds and so an imaginary time may yield deeper insight into our physical laws. For me that argument is *fringe* now, looked at it but the experimentalists clock sweeps out real areas in real time, as do the planets. Ken S. Ter === Subject: Re: Metric Tensor of Flat Space-Time > I think most GRist's would agree with your reasoning, and > I have no problem with using it as an introduction and in > a weak field. It was the approach Einstein used in his 1914 > introduction to the GR foundation. For brevity and clarity > of introduction he made some simplifying assumptions. > > I've clearly presented an alternative spacetime, > using g_01 = -dx/dt, that is in accord with the ISU's > L=cT interdefinition of Length and Time using c, > without any reservations whatsoever, and therefore > fully support the ISU's decision and have no need > of any imaginary, physically ill-defined quantities, > nor do I recommend them in modern GR, not withstanding > the real possibilty that time is truly imaginary as > some think it should be on philosphical grounds and > so an imaginary time may yield deeper insight into our > physical laws. For me that argument is *fringe* now, > looked at it but the experimentalists clock sweeps > out real areas in real time, as do the planets. But then we have the other side of physics where things have an amplitude to be at a particular location and behave like mini planets that do not sweep out courses that are so neat. We also have the possibility of interference not only between wavefunctions in space but also of wavefunctions in time. The difference between the predictable clocks of GR and the smeared out devices of QM needs to be explained. That it is possible that imaginary time underlies the metric tensor that predicts how things behave in real time seems very reminiscent of QM where complex numbers that can be related to imaginary time also predict how things behave in real time (cf: Cramer's Transactional Interpretation of QM). Best Wishes Alex Green === Subject: Re: Metric Tensor of Flat Space-Time > > I think most GRist's would agree with your reasoning, and > I have no problem with using it as an introduction and in > a weak field. It was the approach Einstein used in his 1914 > introduction to the GR foundation. For brevity and clarity > of introduction he made some simplifying assumptions. > > I've clearly presented an alternative spacetime, > using g_01 = -dx/dt, that is in accord with the ISU's > L=cT interdefinition of Length and Time using c, > without any reservations whatsoever, and therefore > fully support the ISU's decision and have no need > of any imaginary, physically ill-defined quantities, > nor do I recommend them in modern GR, not withstanding > the real possibilty that time is truly imaginary as > some think it should be on philosphical grounds and > so an imaginary time may yield deeper insight into our > physical laws. For me that argument is *fringe* now, > looked at it but the experimentalists clock sweeps > out real areas in real time, as do the planets. > But then we have the other side of physics where things have an > amplitude to be at a particular location and behave like mini planets > that do not sweep out courses that are so neat. We also have the > possibility of interference not only between wavefunctions in space > but also of wavefunctions in time. The difference between the > predictable clocks of GR and the smeared out devices of QM needs to be > explained. > That it is possible that imaginary time underlies the metric tensor > that predicts how things behave in real time seems very reminiscent of > QM where complex numbers that can be related to imaginary time also > predict how things behave in real time (cf: Cramer's Transactional > Interpretation of QM). The use of the sqrt(-1) in QM is a calculation instrument, just as the j operator (j=sqrt(-1)) is used for convenience in calculating complex impedance in electronic circuits, it is helpful to articulate right-angles in phasor diagrams. Would a g_00 =-1 be applicable there? > Best Wishes > Alex Green Ken === Subject: Re: Metric Tensor of Flat Space-Time > The use of the sqrt(-1) in QM is a calculation > instrument, just as the j operator (j=sqrt(-1)) > is used for convenience in calculating complex > impedance in electronic circuits, it is helpful > to articulate right-angles in phasor diagrams. > Would a g_00 =-1 be applicable there? My take on sqrt -1 is that it can be used to model some aspects of physical systems (cf: j). Maths using sqrt -1 is a descriptive tool in science that cannot be replaced by real numbers in some circumstances (such as qm amplitudes). Another role for sqrt -1 is Ôgeometrical' in that it allows us to insert another direction for arranging things that has an interesting relationship to directions based on real numbers. Clearly we cannot measure phenomena that are based on imaginary numbers where these cannot be resolved into real numbers. Does this mean that phenomena such as quantum amplitudes prior to Ôcollapse' do not exist or does it just mean we cannot observe them directly? Do you know of any mathematical recipe where complex numbers can always be reduced to reals? Best Wishes Alex Green === Subject: Re: Metric Tensor of Flat Space-Time > The use of the sqrt(-1) in QM is a calculation > instrument, just as the j operator (j=sqrt(-1)) > is used for convenience in calculating complex > impedance in electronic circuits, it is helpful > to articulate right-angles in phasor diagrams. > Would a g_00 =-1 be applicable there? > My take on sqrt -1 is that it can be used to model some aspects of > physical systems (cf: j). Maths using sqrt -1 is a descriptive tool in > science that cannot be replaced by real numbers in some circumstances > (such as qm amplitudes). Another role for sqrt -1 is Ôgeometrical' in > that it allows us to insert another direction for arranging things > that has an interesting relationship to directions based on real > numbers. Clearly we cannot measure phenomena that are based on > imaginary numbers where these cannot be resolved into real numbers. > Does this mean that phenomena such as quantum amplitudes prior to > Ôcollapse' do not exist or does it just mean we cannot observe them > directly? Again I think we're using a mathematical instrument to do a calculation then trying to see if it's physically real, like asking, does 1+1=2 exist prior to our calculator saying so. Well no, it doesn't exist until we have a reading of 2 on the calculator. > Do you know of any mathematical recipe where complex numbers can > always be reduced to reals? I think you mean complex units on x-axis orthogonal to t, for example, then s^2 = t^2 - x^2. The same result occurs when all the units are real, when t is nonorthogonal wrt x, t^2 = s^2 + x^2, so yes always a recipe when two axes are involved. Ken === Subject: Is this a good Random Number Generator? Suppose we want to produce N integer random numbers of k digits in a computer that can work with 1.5k digts precision. Input a integer of 7 or more digits.Multiply it by 4 and substract 1 Take its square root. Take its integer part. Call this So ; i = 0 1.- Multiply by 10^k its decimal part. 2.- Take its integer part. i= i + 1 . This is the random number R(i) 3.- So = So + 1 4.- M = R(i)+ So 5.- M = 4*M - 1 6.- Take the square root of M and take its decimal part 7.- Goto 1 until i = N === Subject: Re: Is this a good Random Number Generator? > Suppose we want to produce N integer random numbers of k digits > in a computer that can work with 1.5k digts precision. > Input a integer of 7 or more digits.Multiply it by 4 and substract 1 > Take its square root. Take its integer part. Call this So ; i = 0 > 1.- Multiply by 10^k its decimal part. > 2.- Take its integer part. i= i + 1 . This is the random number R(i) > 3.- So = So + 1 > 4.- M = R(i)+ So > 5.- M = 4*M - 1 > 6.- Take the square root of M and take its decimal part > 7.- Goto 1 until i = N Luis, RNG's are a tricky lot and it is important to be specific about the usage (an RNG for crypto is typically much different than one for monte-carlo simulations). Here is a set of tests from Dr. Marsaglia (a known expert in the field), the test suite is called DIEHARD. http://stat.fsu.edu/~geo/diehard.html You can also look to the NIST web site for more examples and another test suite. I would look for such RNGs as Add with Carry, Subtract with borrow, mother of all RNGs, Mersenne-Twister and others of those families. You can also look to Fortuna or Yarrow for examples of crypto RNGs. For a hardware RNG, you can look to ComScire, QRandom, ZRandom, HotBits and others. Your method above, although I haven't spent no time on it, probably would not be a good choice. Not sure that helps, Flip === Subject: Re: Is this a good Random Number Generator? Generally, if you have to ask, the answer's no. Phil -- They no longer do my traditional winks tournament lunch - liver and bacon. It's just what you need during a winks tournament lunchtime to replace lost === Subject: Re: Is this a good Random Number Generator? charset=iso-8859-1 As John von Neumann put it, Anyone who considers arithmetical methods of producing random digits is, of course, in a state of sin. Norm === Subject: Re: Is this a good Random Number Generator? > As John von Neumann put it, Anyone who considers arithmetical methods of > producing random digits is, of course, in a state of sin. Norm Justly. With this algorithm I want to contradict that dictum of Von Neumann and the Chaitin mesure of the complexity of a list of digits ,as the length of the program that reproduces the list. My program surely has less than 1000 bytes but can produce a list of thrillions of digits without falling in a loop, and impossible to reproduce if I forget the initial seed. If I delete the instruction M = 4*m -1 , the length of program is reduced by 7 bytes but the complexity is practically reduced to nothing. === Subject: Re: Is this a good Random Number Generator? > As John von Neumann put it, Anyone who considers arithmetical methods of > producing random digits is, of course, in a state of sin. > Norm > Justly. With this algorithm I want to contradict that dictum of Von > Neumann > and the Chaitin mesure of the complexity of a list of digits ,as the > length of the program that reproduces the list. You will fail to contradict them by so doing. > My program surely has less than 1000 bytes but can produce a list of > thrillions > of digits without falling in a loop, Trillions of digits is nothing for a program that size. Algorithms a tiny fraction of that size (<20 characters in C, ~20 bytes of machine code for the actual function that changes the state) can happily produce PRNGs with periods >10^18. > and impossible to reproduce if I > forget > the initial seed. That would imply it has a non-maximal cycle. One doesn't often boast about such things when it comes to PRNGs. > If I delete the instruction M = 4*m -1 , the length of program is > reduced by > 7 bytes but the complexity is practically reduced to nothing. You seem to have evaded the issue of the diehard tests. I can only assume you've not run the tests yet. Have you even plugged your algorithm into the entropy tester ent? (Which always over-estimates entropy but does spot many patterns quite easily.) Phil -- They no longer do my traditional winks tournament lunch - liver and bacon. It's just what you need during a winks tournament lunchtime to replace lost === Subject: Re: Explaining the foundations of math >A film on TV tonight; Dangerous Minds, Michell Pfeiffer. >An ex-Marine English teacher uses karate, drug talk, and >bribes to get thru to her class of urban delinquents. >Where do film makers get crazy ideas like this? >I have taught urban youth algebra, and none of this is >going to have any effect. Either they are the kind of >students who want to learn what you can teach them, >or they could care less about school. >The former are a delight, the latter--the best you >can do is make sure they don't interfere with the learning >of those who want to learn. > Having lived most of my childhood in an urban area, some kids aren't > fortunate enough to have parents who encourage their kids to learn, or > understand the value of learning. Unless they develop an inner desire > to learn, they don't bother. Other kids grow up in dysfunctional or > even abusive situations. Said kids may come to resent teachers or > students who they feel aren't treating them well (e.g. a teacher who > tends to favor students who more quickly understand what's being > taught). So the students may develop an I hate math attitude and > may even become disruptive in the classroom. If there was a way for > the teachers to perhaps spend a bit more time on them, or for the > students to find tutors, they might develop a passion for learning. > --gregbo > gds at best dot com I sympathize, and was a tutor for a while. Unfortunately, some of those who were doing badly, and needed a tutor, just could not seem to adjust to math thinking. I recall trying to teach arithmetic to a HS student who could not handle negative numbers. I tried everything, and I fear that I never did make much of an dent. Maybe someone else could have done better. It was discouraging. I don't want to sound racist, (one thing I learned at UCLA was that racism is evil and stupid), but I found that oriental students were usually good at learning and doing math. I can't say why. I also noticed that they were doing well as a group at UCLA and UC Berkeley. Perhaps it comes from the culture and the parents putting a premium on education. I think it has, and will, pay off. === Subject: Re: Explaining the foundations of math from calccurve-test23@yahoo.com: >I found that oriental students >were usually good at learning and doing math. I can't say why. >I also noticed that they were doing well as a group at UCLA and UC >Berkeley. >Perhaps it comes from the culture and the parents putting a premium on >education. I think it has, and will, pay off. Your last paragraph is the most important part. The strongest inßuence is often the situation in which the students live. Race or ethnicity alone does not give much information of how a student will perform academically. Be careful with your interpretation of your observations. EAch group contains academically strong and academically weak members. G C === Subject: Product of decimal digits For positive integers n let f(n) be the product of the decimal digits of n. Neil Sloane defines the persistence k(n) of n as the as smallest k for which f^{(k)}(n) [the kth iterate of f evaluated at n] is a single-digit number. To steal an example from [2], 679 -> 378 -> 168 -> 48 -> 32 -> 6, so that 679 has persistence 5. Sloane has conjectured that the persistence is bounded; no numbers are currently known with persistence larger than 11. This definition is naturally rather sensitive to the occurrence of zeros. Erdos asks for the behavior of k*(n) defined analogously but with respect to the product f*(n) of the _nonzero_ digits of n. He notes that f*(n) < n^{c} for some fixed c < 1 and all large n; this implies that k*(n) < c_2 log log n for some positive constant c_2 and all large n. QUESTION: Is k*(n) unbounded? Is this known? Walter Schneider [2] gives examples of powers of two with modified-persistence taking each value from 1 through 18, remarks that k*(n) seems not to be bounded, and asks for a proof. If it is known it does not seem well-known but perhaps it has not been much thought about. Some head-scratching on my part has proved fruitless, so additional insights would be appreciated. Questions about the behavior of k(n) and k*(n) occur under F25 in in [3] (and also in the second edition of the same work). However, the (IMHO quite interesting) problem of the unboundedness of k*(n) is not listed as unsolved. Paul [1] N.J.A. Sloane The persistence of a number, J. Recreational Math., 6 (1973), 97-98. [2] W. Schneider http://www.wschnei.de/digit-related-numbers/persistence.html [3] R.K. Guy Unsolved problems in number theory. Third Edition. Problem Books in 0-387-20860-7 === Subject: Very easy number theorry exercises I am going over some intro number theory homeworks I found online. I should probably be more selective about what I post, but here they are. 3. Let a, b, n in Z with gcd(a, n) = gcd(b, n) = 1. Prove that gcd(ab, n) = 1. --------- ra + sn = 1 = ub + vn ; (ra + sn)(ub + vn) = ruab + rvan + snub + svnn = r'ab + mn = 1 ==> (ab,n) = 1. 7. Show that if n > 3 then (i) n, n + 2 and n + 4 cannot all be prime; (ii) n, 2n + 1 and 4n + 1 cannot all be prime. -------- (i) 3 must divide one of these 3, since (n + 4) mod 3 = (n + 1) mod 3, then m = (n, n + 2, or n + 4) mod 3 = (n, n + 1, or n + 2) mod 3 must be a complete set of residues of Z_3, so one of the 3 must satisfy m mod 3 = 0, thus is not prime. (ii) n must be odd, or 2 | n. Let n mod 3 = 1 or 2 ( n mod 3 = 0 ==> 3 | n ). n mod 3 = 1 ==> (2n + 1) mod 3 = 0, ==> n mod 3 = 2. Then (4n + 1) mod 3 = 9 mod 3 = 0 mod 3, so 3 | n, 2n + 1 or 4n + 1. === Subject: cone manifold I cannot visualize what a cone manifold looks like. Can someone point Dick Fischbeck === Subject: Re: cone manifold >I cannot visualize what a cone manifold looks like. Can someone point The definition I'm semi-familiar with has been around for years, so it may not be whatever fashionable frippery you're inquiring about. But in case it is: visualize your favorite polyhedral surface in R^3, say a cube to be definite. In terms of the intrinsic metric of the surface, the edges are invisible (they're just epiphenomenal manifestations of the embedding in ßat space), but the vertices are genuinely there: each of the 8 vertices has, in fact, 1/8 of the total curvature of the surface locked up in it (and the surface is ßat everywhere else). The topology of the cube-with-its-intrinsic-metric in a small metric disk centered at a vertex is the cone on a circle, i.e., a 2-disk, just as it is everywhere else; but at that vertex the metric relationship between distance-from-the-center (of the metric disk) and distance-along-the-boundary (of the metric disk) isn't the ßat relationship r=C/2pi. In this case, obviously, it's a different linear relationship. In a more general cone surface (for instance, the cone sphere which is the natural compactification of the upper-half-plane modulo SL(2,Z)), the relationship would be non-linear (and only its asymptotics at the cone point would matter, in some deep sense, I think). Now generalize from 2 to n, and Bob's your uncle. Lee Rudolph === Subject: Re: cone manifold Lee complexities of mathematical language and I don't fully comprehend the situation. I have a couple more questions of you don't ming. Is any polyhedron a cone manifold? Can we assemble a cone manidfold from individual cone elements? Dick Fischbeck >I cannot visualize what a cone manifold looks like. Can someone point > The definition I'm semi-familiar with has been around for years, so > it may not be whatever fashionable frippery you're inquiring about. > But in case it is: visualize your favorite polyhedral surface in > R^3, say a cube to be definite. In terms of the intrinsic metric > of the surface, the edges are invisible (they're just epiphenomenal > manifestations of the embedding in ßat space), but the vertices > are genuinely there: each of the 8 vertices has, in fact, 1/8 > of the total curvature of the surface locked up in it (and the > surface is ßat everywhere else). The topology of the > cube-with-its-intrinsic-metric in a small metric disk > centered at a vertex is the cone on a circle, i.e., a > 2-disk, just as it is everywhere else; but at that vertex > the metric relationship between distance-from-the-center > (of the metric disk) and distance-along-the-boundary (of > the metric disk) isn't the ßat relationship r=C/2pi. > In this case, obviously, it's a different linear relationship. > In a more general cone surface (for instance, the cone sphere > which is the natural compactification of the upper-half-plane > modulo SL(2,Z)), the relationship would be non-linear (and > only its asymptotics at the cone point would matter, in some > deep sense, I think). > Now generalize from 2 to n, and Bob's your uncle. > Lee Rudolph === Subject: Quotient rings - help! I could not find any algebra book that really defines quotient rings in polynomials. I know the elements of quotient rings are all the left/right cosets, but how do they look like? I wish some book would make this clear. It is so frustrating. If R[x] is a polynomial ring, and I[x] is some ideal of R[x], how do the elements of R[x]/I[x] look like? For example, I can't visualize these: a. Z[x]/(2, x) for instance; or b. Z[x]/(x); or c. Q[x]/(y - x^2); or d. Z[x]/(2, x^2-x+1). How do I show that this is a finite field? How many elements does it have? How can I show maximality of (2, x^2-x-1)? or finiteness of the quotient? N.C. === Subject: Re: Quotient rings - help! > I could not find any algebra book that really defines quotient rings > in polynomials. I know the elements of quotient rings are all the > left/right cosets, but how do they look like? I wish some book would > make this clear. It is so frustrating. > If R[x] is a polynomial ring, and I[x] is some ideal of R[x], how do > the elements of R[x]/I[x] look like? This is perhaps a bit OT, but using the notation I[x] for a general ideal of R[x] is not such a good idea, because it clashes with another well estabished meaning. When I is an ideal of R, then I[x] := { f in R[x] | all coefficients of f are in I } is indeed an ideal of R[x]. But you cannot expect every ideal of R[x] to be of this particular form. Marc === Subject: Re: Quotient rings - help! days. My association with the Department is that of an alumnus. >I could not find any algebra book that really defines quotient rings >in polynomials. I know the elements of quotient rings are all the >left/right cosets, but how do they look like? I wish some book would >make this clear. It is so frustrating. >If R[x] is a polynomial ring, and I[x] is some ideal of R[x], how do >the elements of R[x]/I[x] look like? >For example, I can't visualize these: >a. Z[x]/(2, x) for instance; or You can do this one in steps. First mod out by (x); then mod out the result by 2. If you mod out by x, that is essentially the same as as set x to 0. The reson for this is that if you map Z[x] ---> Z by evaluate at a, then the kernel of the map is the ideal generated by (x-a); and the quotient is then isomorphic to the image. Once you've modded out by (x) by evaluating at 0, you then mod out by 2 to get Z/(2), which is the field of two elements. Basically, every polynomial in Z[x] can be written as p(x) or p(x)+1, where p(x) is in (2,x). >b. Z[x]/(x); or See above. >c. Q[x]/(y - x^2); or This doesn't make sense. y is not an element of Q[x]. Do you mean Q[x,y]? >d. Z[x]/(2, x^2-x+1). This one is a bit tricker. If you mod out by (2) first, then basically, every monomial with even coefficient is in the ideal, so you are left with (Z/2Z)[x], i.e., polynomials over the field of two elements. (In general, if R is a ring, and I is an ideal of R, then R[x]/(I) is isomorphic to (R/I)[x]; prove it). Then you are moding out by x^2-x+1 = x^2+x+1. Since you are now working over a field, you can use the division algorithm: every polynomial in F_2[x] can be written uniquely as p(x) = q(x)*(x^2+x+1) + r(x) where r(x) =0 or deg(r)<2. So every polynomial is equivalent to somethingi n the ideal plus a polynomial of degree at most 1. Note that x^2+x+1 is irreducible in F_2; so you end up with just four possible representatives: 0, 1, x, and x+1. So F_2[x]/(x^2+x+1) is a ring with four elements, 0, 1, x, and x+1, and where x^2 = x+1; check to verify that it is in fact a field of 4 elements. > How do I show that this is a finite field? How >many elements does it have? How can I show maximality of (2, x^2-x-1)? >or finiteness of the quotient? Once you show the quotient is a field, the fact that (2,x^2-x-1) is maximal will follow from that. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Quotient rings - help! >I could not find any algebra book that really defines quotient rings >in polynomials. Run some of the unreal definitions past us, so we can snicker. >For example, I can't visualize these: Why in the world would you want to visualize them? There's nothing (on the face of it) visual about them! They're just what they're defined to be! (I can visualize james dolan right now...) Maybe you can better understand what's going on by trying to find a normal form for each equivalence class in the quotient ring. >a. Z[x]/(2, x) Here, for instance, the normal form of a+bx+cx^2+... could be a mod 2. (Don't reßexively ask us, What do you mean? or What does that mean?; spend a minute or so asking yourself What might he mean? What might that mean?) That would leave you with just two distinct normal forms, the two distinct integers mod 2. What's the ring structure on this set with two elements? >b. Z[x]/(x) Here, the normal form of a+bx+cx^2+... still throws out b, c, and so on, but it keeps a intact. So...? >c. Q[x]/(y - x^2) What's y? Does it appear elsewhere in what is appearing more and more like a homework assignment? >d. Z[x]/(2, x^2-x+1). How do I show that this is a finite field? How >many elements does it have? How can I show maximality of (2, x^2-x-1)? >or finiteness of the quotient? Yep, homework. Go ask the instructor, kid; someone's paying hir on your behalf. No one is paying sci.math. Lee Rudolph === Subject: Re: Quotient rings - help! And it is not a homework assignment: I am studying for an upcoming exam. N.C. >I could not find any algebra book that really defines quotient rings >in polynomials. > Run some of the unreal definitions past us, so we can snicker. . . .. . > Yep, homework. > Go ask the instructor, kid; someone's paying hir on your behalf. > No one is paying sci.math. > Lee Rudolph === Subject: Re: Quotient rings - help! OK, so you can trade y for x^2. So any polynomial in x and y is equivalent to a polynomial in x alone... 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: Quotient rings - help! >I could not find any algebra book that really defines quotient rings >in polynomials. I know the elements of quotient rings are all the >left/right cosets, but how do they look like? I wish some book would >make this clear. It is so frustrating. >If R[x] is a polynomial ring, and I[x] is some ideal of R[x], how do >the elements of R[x]/I[x] look like? I would think of it this way: you're looking at the polynomial ring R[x], but considering two polynomials as equivalent if their difference is in the ideal. >For example, I can't visualize these: >a. Z[x]/(2, x) for instance; or OK, let's take this example. You're looking at polynomials in x with integer coefficients, and (1) anything with all its coefficients even is equivalent to 0. So every polynomial is equivalent to one with coefficients in {0,1}. (2) any multiple of x is equivalent to 0. So you can throw away all the non-constant terms of a polynomial. Conclusion: Z[x]/(2,x) has only two members, (the equivalence classes for) 0 and 1. >c. Q[x]/(y - x^2); or What is y? It can't be an indeterminate since you said Q[x], not Q[x,y]. So I guess it is some fixed member of Q. Now for any polynomial of degree >= 2, you can get an equivalent polynomial of lower degree by adding an appropriate multiple of y - x^2. On the other hand, two distinct polynomials of degree <= 1 can't be equivalent. So here the quotient ring corresponds to everything of the form a x + b, a and b in Q, with multiplication (a x + b)*(c x + d) = (a d + b c) x + b d + a c y 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: Quotient rings - help! > I could not find any algebra book that really defines quotient rings > in polynomials. I know the elements of quotient rings are all the > left/right cosets, but how do they look like? I wish some book would > make this clear. It is so frustrating. > If R[x] is a polynomial ring, and I[x] is some ideal of R[x], how do > the elements of R[x]/I[x] look like? > For example, I can't visualize these: > a. Z[x]/(2, x) for instance; or > b. Z[x]/(x); or > c. Q[x]/(y - x^2); or > d. Z[x]/(2, x^2-x+1). How do I show that this is a finite field? How > many elements does it have? How can I show maximality of (2, x^2-x-1)? > or finiteness of the quotient? > N.C. One thing that can help is to think of them as the polynomial ring R[x] with the relationship that all polynomials in the idea are equal to zero. so for example Z[X]/(2,x) is the set of all polynomials in Z with 2=0 and x=0, so you get Z/2Z Z[x]/(x) is just Z I'm not sure what you mean by Q[x]/(y-x^2) since y is not an element of Q[x]. For (d) to show finiteness, first note that since we are modding out the element 2, we can think of it as Z_2[x]/(x^2-x+1). Then the fact that x^2-x+1=0 means that x^2=x-1, and thus one can represent every element by a+bx, with a,b elements of Z_2. Hope this helps. -Ron === Subject: help with transposition, dot product Hiya, Im really struggling to transpose this to find 1 of the vectors, my maths is very basic with now further education to support it... (BTW before anyone says this is homework its not! im just dumb, and why would anyone do homework in the summer holidays!) ok well this is what i want to transpose - u.v / |u||v| = I where I is a scaler and U and V are unit vectors, I want to get V on its own on one side, but im st! any help would be really appreciated! === Subject: Re: help with transposition, dot product > Hiya, > Im really struggling to transpose this to find 1 of the vectors, my > maths is very basic with now further education to support it... (BTW > before anyone says this is homework its not! im just dumb, and why > would anyone do homework in the summer holidays!) > ok well this is what i want to transpose - > u.v / |u||v| = I > where I is a scaler and U and V are unit vectors, I want to get V on > its own on one side, but im st! > any help would be really appreciated! The most you can do is rearrange to get |v| on its own. u.v = I|u||v| so |v| = u.v / I|u| u.v is a dot product where if u = (a,b) and v = (c,d) then u.v = ac + bd |v| = (ac + bd) / I|u| where |v| = sqrt(c^2+d^2) PH === Subject: Re: help with transposition, dot product u (dot) v = |u| |v| cos(theta), where theta is the angle between the vectors u and v. If you know u, you can find the direction, but not the magnitude, of v; it is at an angle of arccos(I) from u. === Subject: Re: help with transposition, dot product > u (dot) v = |u| |v| cos(theta), where theta is the angle between the > vectors u and v. If you know u, you can find the direction, but not > the magnitude, of v; it is at an angle of arccos(I) from u. Does it determine the direction uniquely? I think not. Does it not give the choice of two directions in 2-space and a cone of directions in 2-space? === Subject: Question: Divisors of the difference of unknown primes Hello. I've been given the product of two large primes and tasked with finding divisors of the difference between the two (unknown to me) primes. I don't even know if it's possible, though. That is, I've been given M = N(N+P), where N and N+P are large primes. And I need to find divisors of P. Clearly 2 | P. Also 3 | P iff M = 1 (mod 3). This is because N^2 = 1 (mod 3) if N is relatively prime to 3. The same is true for 4, 6, and 8. So I know whether P is divisible by 2, 3, 4, 6, and 8 (and obviously 12 and 24). But this trick doesn't work for 5, 7, or 9, and I've run out of ideas. I especially want to know about 9. Any thoughts? === Subject: Re: parabola - conic section I would be interested to see similar derivations for the ellipse and hyperbola; if you happen to have such derivations handy on-hand, could you please post these as well. Michael === Subject: Re: parabola - conic section > I would be interested to see similar derivations for the ellipse and > hyperbola; if you happen to have such derivations handy on-hand, could > you please post these as well. All you have to do differently is change the angle by which the y-axis is rotated. Instead of 45 degrees, let it be some angle f, between 0 and 90 degrees. The substitution is then: x = cu - sv z = su + cv where c = cos(f) and s = sin(f). The equation of the cone then becomes: (cu)^2 - 2scuv + (sv)^2 + y^2 - ((su)^2 + 2scuv + (cv)^2) = 0 Simplifying: y^2 = (c^2 - s^2)(u^2 - v^2) - 4scuv = C(u^2 - v^2) - 2Suv where C = c^2 - s^2 = cos(2f) and S = 2sc = sin(2f). Completing the square in u, and bringing it to the left, you get: (u - Sv/C)^2 - y^2/C = (1 - (S/C)^2)v^2 For constant v, this equation describes a hyperbola for C > 0, and an ellipse for C < 0. Recalling C = cos(2f), this occurs when f < 45 degrees and f > 45 degrees, respectively. === Subject: Re: parabola - conic section > I was just reading a book on conic sections. How can one prove that if > cone is > sliced by a plane parallel to a lateral side of the cone, we get a > parabola? > Let pi be some fixed plane. > Let D be some fixed line in pi. > Let F be some fixed point in pi, but not in D. > A parabola may be defined as the locus of a point, in pi, which moves so > that its distance from F is equal to its distance from D. > F is called the focus of the parabola. > D is called the directrix of the parabola. > We may use a Dandelin sphere to prove that if a cone, C, is intersected by > a plane, pi, which is parallel to one of the generators, G, of C, then the > curve of intersection is a parabola. > Let S be a sphere (a Dandelin sphere) inscribed in C and touching pi. [I > will not prove the existence (or uniqueness) of such a sphere.] > Let delta be the plane in which S touches C. > Let D be the line of intersection of delta and pi. > Let F be the point at which S touches pi. > Let the vertex of the cone be the point V. > Let A be the point at which G (the generator of C to which pi is parallel) > intersects delta. > Let L be the line in pi, and through F, which is parallel to G. > Let B be the point of intersection of L and D. > Clearly, angle VAB = angle ABF (since VA [the line G] is parallel to BF [the > line L]). > Let us call this angle t. > Clearly, all the generators of C meet delta at angle t. > Let P be any point on the intersection of C and pi. > Let the generator of C which passes through P intersect delta at Q. > Let the foot of the perpendicular from P to delta be the point R. > Let the foot of the perpendicular from P to D be the point C. > Clearly, the plane PCR is parallel to the plane FBA and so the angle PCR is > equal to t. > Hence the triangles PCR and PQR are congruent (they are both right-angled > with a common side PR and equal corresponding angles [since angle PQR = t = > PCR]). > So PQ=PC. > Also, PF=PQ (they are both tangents from P to S, and hence of equal length). > Hence, PF=PC. > So, the distance from P to F is the same as the distance from P to D. > Thus, P lies on a parabola with focus F and directrix D. > Q.E.D. See http://www.pisquaredoversix.force9.co.uk/Cone1.gif for an animation that may help you visualise the above proof. Imagine the chequered background to be a vertical wall. A glass sphere (of refractive index 1), diameter d, touches the wall at the point marked by the small red ball. The white ball is at a distance d from the wall. The yellow ball moves across the back of the sphere in such a way that the green line is always a tangent to the sphere. The path of the yellow ball is a semicircle. The plane of that semicircle intersects the wall in the red line. The green line intersects the wall at a point marked by the green ball. It can be shown that the distance from the green ball to the red ball is the same as the distance from the green ball to the red line. Hence, the locus of the green ball is part of a parabola. The correspondence between this animation and the proof above is as follows: The white ball: the point V The green ball: the point P The blue ball: the point B The red ball: the point F, the focus The cyan ball: the point C The magenta ball: the point R The yellow ball: the point Q The red line: the line D, the directrix -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Status of Waring-problem Am 20.08.04 23:23 schrieb Gottfried Helms: > Hi - > googling about waring's problem brought two different > results: > some stated, that 1986 the last unsolved case of G(4) > was solved, > mathworld stated, that it needs a certain inequality > to prove, that warings problem is completely solved. > Is mathworld 18 years behind or are that different facts? > Gottfried Helms Well, it seems there is no more discussion for this subject. The replies were very informative for me, summarized, what I did understand. My starting point was the connection between one critical condition for the existence of a primitive loop in the collatz-problem and one critical inequality of the waring-problem. The condition can be stated, that the following equation must hold 3^N*i - 1 2^A = ----------- (1) 2^N*i - 1 and also 3^N*i - 1 powerceil2(3^N) <= 2^(A+N) = 2^N * ---------- 2^N*i - 1 or, rewritten: powerceil2(3^N) 2^N 3^N*i - 1 ---------------- <= --- * ---------- (2) 3^N 3^N 2^N*i - 1 One translation of this condition forms the inequality, which is common to both problems: express a power 3^N in terms of 2^N as d*q + r where q is 2^N, d is ßoor(3^N/q) and r is the remainder then if d+r< q (3) then this is equivalent to the mathworld-formula of a waring-condition as well as to a nonexistence-condition of the primitive-loop/1-cycle in the collatz-problem. 1) This condition is still not proven, while the connected question of the g()-function of the waring-problem is answered, and this condition (*) plays only a role as a selector between two possible formulations of the g()- function. 2) The primitive loop (in the notation of 1-cycle) was already disproven by Ray Steiner 1978, by showing that (1) cannot hold. While (3) only negates, that the rhs of (1) could be integer, Steiner showed, that it cannot satisfy the stronger requirement to be also a power of 2. 3) A line of arguing in this matter is to analyze the coefficients of the continued fraction of log(3)/log(2) and see, whether the partial quotients agree with the approximation of (2) dependent on N. Empirically (2) is never satisfied for N up to 2^40, and the approximation of the rhs of (2) to 1 is far better than that of the lhs. 4) The cf-approximation-argument holds, if the coefficients of the cf do not increase too fast. While we have a statistical expectation for the height of the cf-coeffi- cients in such problems provided by Gauss, we seem not to have a rigid proof for the current case, that the lhs in (2) indeed approximates worse then the rhs to 1. ---------- I hope, I got this summary right. This answeres also my opening question concerning mathworlds actuality in that way, that initially I had a misconception of the relation between the open question of the inequality and the status of the solution of the g()- function. of David, and am now watching out for more information regarding 4) and a possible limiting to the cf-coefficients in general and similar cases of approximation of logarithms and naturally for this specific case. Gottfried Helms === === Subject: continued fractions: upper bounds for coefficients? and: something known about cf with coefficients other than integer? Hi , in the discussion of approximation of log(3)/log(2) with means of continued fractions the problem arises, whether there exist known upper bounds for the coefficients a_n of the cf of a real number x depending on its indexposition n. Is something known about that in regard to certain classes of x, say x is a square-root of a rational (here we have periodicity) or of higher algebraic degree, or if x is of transcendental classes like exponential of a rational or logarithm, etc? Fiddling a bit with that I remembered my solutions for nonstandard cf-s (rational coefficients) for rational powers of e, and noticed, that even for transcendent powers of e this simple scheme holds. Unfortunately nonstandard cf-s do not hold for the condition of best rational approximation as simple cf-s do. So I'm looking for information on such constructions. Is something discussed about such cf-s with rational (or even irrational) coefficients ? Gottfried Helms === Subject: Re: continued fractions: upper bounds for coefficients? and: something Gottfried Helms helms@uni-kassel.de > in the discussion of approximation of log(3)/log(2) with > means of continued fractions the problem arises, whether > there exist known upper bounds for the coefficients a_n > of the cf of a real number x depending on its index position > n. > Is something known about that in regard to certain classes > of x, say x is a square-root of a rational (here we have > periodicity) ... For quadratic irrationals, (P_0 + sqrt(D))/Q_0, [P_0^2 == D (mod Q_0), D > 0 not a square, Q_0 <> 0] with complete quotients (P_i + sqrt(D))/Q_i, in the periodic part of the continued fraction expansion, if Q_i <> 1, then a_i < sqrt(D), while if Q_i = 1, sqrt(D) < a_i < 2sqrt(D). Prior to the periodic part, anything can happen. For example, (10095 + sqrt(5))/5098 has a_2 = 50, while all other a_i = 1. John Robertson === Subject: Re: continued fractions: upper bounds for coefficients? and: something Am 07.09.04 01:47 schrieb Jpr2718: > Gottfried Helms helms@uni-kassel.de >in the discussion of approximation of log(3)/log(2) with >means of continued fractions the problem arises, whether >there exist known upper bounds for the coefficients a_n >of the cf of a real number x depending on its index position >n. >Is something known about that in regard to certain classes >of x, say x is a square-root of a rational (here we have >periodicity) ... > For quadratic irrationals, (P_0 + sqrt(D))/Q_0, [P_0^2 == D (mod Q_0), D > 0 > not a square, Q_0 <> 0] with complete quotients (P_i + sqrt(D))/Q_i, in the > periodic part of the continued fraction expansion, if Q_i <> 1, then a_i < > sqrt(D), while if Q_i = 1, sqrt(D) < a_i < 2sqrt(D). > Prior to the periodic part, anything can happen. For example, > (10095 + sqrt(5))/5098 > has a_2 = 50, while all other a_i = 1. > John Robertson Yes, surely. I should limit the question to the actual problems of approximations of a^n - b^m (all integer) for instance, although I mean it in some more generality. Examples show, that for the approximation of, say 3^m - 2^n the convergents of log(3)/log(2) can be used, and vice versa. Since the integer difference can at least be 1, the precision of the convergents is also limited to something related to 3^m, and thus the value of the coefficients. I'm not able to formulate that in a concise way currently, but for that specific case it seems obvious to me, that the convergents must be high-bounded by their index-position, as well as then in turn the coefficients. Isn't it that way? Gottfried Helms === Subject: Problem with inner product by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i86NG9j07960; Here is one little problem. H is Hilbert space and A is linear and bounded operator on H. If (Ax,x)=0 show that Ax=0. How can I show it? J. === Subject: re:Problem with inner product I posted a reply to this question on mathforum. What you are trying to show is not true. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: 2 triginometric equations with 2 unknowns (angles) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i86NGAj07989; >For spatial management, I need to figure out how to simplify (solve) a >system of 2 triginometric equations with 2 unknowns: > A cos(a) - B cos(b) = C1 (1) > A sin(a) - B sin(b) = C2 (2) >A, B, C1, C2 are constants, the only unknowns are the angles (a) and >(b). A simple system of 2 equations with 2 unknowns, but the problem >gains complexity when trasforming one of the sines to a cosine (or >vise versa), where you end up with a messy quadratic equation to the >4th power: >cos(a) = sqrt( 1 - sin(a)*sin(a) ) >Is there a way to simplify equations (1) and (2), trying to solve (a) >and (b)? A w - B x = C1 so w = (C1 + B x)/A ..eq(1) A y - B z = C2 so z = -(C2 - A y)/B ..eq(2) w^2 + y^2 = 1 and substitute from eq(1) x^2 + z+2 = 1 and substitute from eq(2) Gives 2 eqns with unknowns x and y phil === Subject: Re: 2 triginometric equations with 2 unknowns (angles) > A cos(a) - B cos(b) = C1 (1) > A sin(a) - B sin(b) = C2 (2) Square and add (1)and(2), let a-b = th. Gives cos(th)=(A^2+B^2-(C1^2+C2^2))/(2 A B). Plugging this into(1), A*cos(th+b)-B*cos(b) = C1 involving only one unknown b. You have given a vector resolution of forces [Fa and Fb in polar cordinates (A,a) and (B,b)] along x- y- coordinate axes. Squaring and adding gives the force resultant of Fa and Fb which contain angle th between them. > Is there a way to simplify equations (1) and (2), trying to solve (a) > and (b)? === Subject: Trying to solve 2 homogenous quadratics How would one solve the following system of equations? (1) 17*u^2 - 144*u*v + 1904*v^2 = 43*U^2 - 60*U*V + 258*V^2 (2) 162*u^2 - 8568*u*v + 18144*v^2 = 10*U^2 - 172*U*V + 60*V^2 and u,v,U,V are in Z. We seek solutions so that (u,v) and (U,V) have the same value in either (1) or (2) (coincidental solutions?) This is from trying to factor an homogenous quartic space from the elliptic curve [0,6800,0,11559996,0] Randall === Subject: Re: Trying to solve 2 homogenous quadratics >How would one solve the following system of equations? >(1) 17*u^2 - 144*u*v + 1904*v^2 = 43*U^2 - 60*U*V + 258*V^2 >(2) 162*u^2 - 8568*u*v + 18144*v^2 = 10*U^2 - 172*U*V + 60*V^2 >and u,v,U,V are in Z. >We seek solutions so that (u,v) and (U,V) have the same value in either (1) or >(2) (coincidental solutions?) This seems so obvious, maybe I'm oversimplifying something. Let u=U, v=V, so your equation is in u,v only. Subtract so that one side of the equation is 0. Divide everything by v^2, so your terms are u^2/v^2, u/v, and a constant. Let x=u/v. Use the quadratic formula to solve for x. Now you have 1 or 2 values for u/v but not u and v themselves because for any solution (u,v) there will be a set of solutions (u*z, v*z), since there is no constant term in the original equation. >This is from trying to factor an homogenous quartic space from the elliptic >curve [0,6800,0,11559996,0] --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Trying to solve 2 homogenous quadratics Content-Length: 1185 Originator: rusin@vesuvius > How would one solve the following system of equations? > (1) 17*u^2 - 144*u*v + 1904*v^2 = 43*U^2 - 60*U*V + 258*V^2 > (2) 162*u^2 - 8568*u*v + 18144*v^2 = 10*U^2 - 172*U*V + 60*V^2 > and u,v,U,V are in Z. > We seek solutions so that (u,v) and (U,V) have the same value in either (1) or > (2) (coincidental solutions?) > This is from trying to factor an homogenous quartic space from the elliptic > curve [0,6800,0,11559996,0] > Randall Provided the discriminants of each equation satisfy suitable mutual conditions (equal, or have the same square-free factor?) I think it's possible to find a linear transform of u, v, U, V, or maybe a birational transform of the dehomogenized pair, that reduces the original pair to the more canonical form: ax^2 + bxy + cy^2, dx^2 + exy + fy^2 = z^2, t^2 resp However, I'm moving house at the moment and all my notes are in boxes. [sci.math.research added - hope you don't mind, but I'd be interested in an expert answer to this myself!] John R Ramsden (jramsden@glasshouse.cam) com not cam === Subject: Definition of Axiom of Choice? Choice. I have been lead to feel that it means this (please tell me if it is wrong): For any set X with an infinite amount of elements, a set Y exists with any arbitrary elements selected from set X. Yes, no? === Subject: Re: Definition of Axiom of Choice? > Choice. > Here is Russell's way of explaining his Multiplicative Principle (MP) (which is equivalent to the Axiom of Choice (AC)), and I'm paraphrasing here: We need the Axiom of Choice to select one sock from infinitely many pairs of socks, but for shoes, we don't need it. Two of the simplest examples that illustrate our natural desire to invoke AC--often overlooked by beginning students--are found in the usual textbook proofs of the following Theorems (both are reportedly not provable in ZF without AC): 1. Every infinite set contains a denumerable subset. (In other words, if A is an infinite set, then there exists a set B such that B is a subset of A, and B is equipotent to the set of all natural numbers.) 2. A union of countably many countable sets is countable. See if you can locate where AC (usually in the form of MP) is invoked in their proofs. Here is another important observation where AC (in the form of MP) is used, often implicitly and therefore overlooked by beginners: Let N denote {1,2,3,...}, and let P(x) be a formula with x being the only free variable. Statement I: For every n in N, there exists a set A such that P(A) holds. Statement II: For every n in N, there exists a set A_n such that P(A_n) holds. (I + AC) (naturally) implies II. Without AC, other premises are required to derive II from I. Shedar === Subject: Re: Definition of Axiom of Choice? charset=iso-8859-1 > Choice. I have been lead to feel that it means this (please tell me if > it is wrong): > For any set X with an infinite amount of elements, a set Y exists with > any arbitrary elements selected from set X. > Yes, no? Well, apart from a rock group of the same name, MathWorld gives this statement for the AoC given any set of mutually exclusive nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty sets The original set need not be infinite, but it must be non-empty and it's a set of sets. The AoC then guarantees that you can construct a new set that contains one and only one element from each of the original sets in the collection, i.e. a representative collection. Norm === Subject: Re: Definition of Axiom of Choice? > Choice. I have been lead to feel that it means this (please tell me if > it is wrong): > For any set X with an infinite amount of elements, a set Y exists with > any arbitrary elements selected from set X. > Yes, no? >Well, apart from a rock group of the same name, MathWorld gives this >statement for the AoC >given any set of mutually exclusive nonempty sets, there exists at least >one set that contains exactly one element in common with each of the >nonempty sets >The original set need not be infinite, but it must be non-empty no, it's trivial if the original set is empty. >and it's a >set of sets. The AoC then guarantees that you can construct a new set that >contains one and only one element from each of the original sets in the >collection, i.e. a representative collection. > Norm ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Definition of Axiom of Choice? > Choice. I have been lead to feel that it means this (please tell me if > it is wrong): > For any set X with an infinite amount of elements, a set Y exists with > any arbitrary elements selected from set X. > Yes, no? No. What the axiom of choice says is that for any collection of sets X_i with i ranging over any indexing set I, that the product of all the X_i is non-empty, or equivalently, that one can define a function f:I->Union (X_i) with the property that f(i) is an element of X_i. The name choice comes from the idea that if you have an arbitrary collection of sets you can choose one element from each set. -Ron === Subject: Re: Definition of Axiom of Choice? > Choice. I have been lead to feel that it means this (please tell me if > it is wrong): > For any set X with an infinite amount of elements, a set Y exists with > any arbitrary elements selected from set X. > Yes, no? >No. >What the axiom of choice says is that for any collection of non-empty >sets X_i >with i ranging over any indexing set I, that the product of all the X_i >is non-empty, or equivalently, that one can define a function f:I->Union >(X_i) with the property that f(i) is an element of X_i. The name choice >comes from the idea that if you have an arbitrary collection of sets you >can choose one element from each set. >-Ron ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Definition of Axiom of Choice? > Choice. I have been lead to feel that it means this (please tell me if > it is wrong): > For any set X with an infinite amount of elements, a set Y exists with > any arbitrary elements selected from set X. > Yes, no? No. First the grammar nitpick: You mean an infinite number of elements, or simply any infinite set X. Then the logical bug: It doesn't make sense to talk about a set Y with any arbitrary elements selected from X, unless you specify what you mean by the phrase with any arbitrary elements. The way I've heard it phrased most concisely is: Suppose we have a non-empty set I. This is going to be called our index set. Now suppose we are given a whole bunch of other non-empty sets indexed by I --- that is, we have one set S_x for each and every x in I. (To take a trivial example, suppose I was the set {1,2,3}. Then we would have the sets S_1, S_2, and S_3, which could be whatever non-empty sets we like: for example, S_1 could be the set of natural numbers, S_2 could be the set of real numbers, and S_3 could be the set {Monday, Wednesday, Friday}. Of course, that's a boring example, and it would be more interesting to take I to be the set of natural numbers, or the set of real numbers, or something big like that.) Now, what the Axiom of Choice says is that the Cartesian product of all S_x (x in I) is non-empty. That is, we can produce at least one example of an ordered I-tuple (E_x | x in I) such that each and every E_x is a member of S_x. (For example, the 3-tuple (7, pi, Monday) is a member of the Cartesian product of our S_1, S_2, S_3. The 3-tuple (1, 1, Wednesday) is another. See, it's really boring and trivial when I has only a finite number of elements!) So that's all the Axiom of Choice says. If you have a set of non-empty sets S_{x in I}, then you can construct an I-tuple containing one element from each of those sets. That's all. A corollary to the Axiom of Choice is that if you have a single non-empty set, then it is possible to construct a new set containing only one of that set's elements. In other words, you can pick a singleton out of the original set. Again, it looks trivial and obvious, but it is really a separate axiom from the other axioms of set theory (or the real number system, or whatever application you're interested in). HTH, -Arthur === Subject: Re: Definition of Axiom of Choice? > Choice. I have been lead to feel that it means this (please tell me if > it is wrong): > For any set X with an infinite amount of elements, a set Y exists with > any arbitrary elements selected from set X. > Yes, no? > No. First the grammar nitpick: You mean an infinite number of > elements, or simply any infinite set X. Then the logical bug: > It doesn't make sense to talk about a set Y with any arbitrary elements > selected from X, unless you specify what you mean by the phrase with any > arbitrary elements. > The way I've heard it phrased most concisely is: > Suppose we have a non-empty set I. This is going to be called > our index set. Now suppose we are given a whole bunch of other > non-empty sets indexed by I --- that is, we have one set S_x for > each and every x in I. > (To take a trivial example, suppose I was the set {1,2,3}. Then > we would have the sets S_1, S_2, and S_3, which could be whatever > non-empty sets we like: for example, S_1 could be the set of natural > numbers, S_2 could be the set of real numbers, and S_3 could be the > set {Monday, Wednesday, Friday}. Of course, that's a boring example, > and it would be more interesting to take I to be the set of natural > numbers, or the set of real numbers, or something big like that.) > Now, what the Axiom of Choice says is that the Cartesian product > of all S_x (x in I) is non-empty. That is, we can produce at least > one example of an ordered I-tuple (E_x | x in I) such that each and > every E_x is a member of S_x. > (For example, the 3-tuple (7, pi, Monday) is a member of the > Cartesian product of our S_1, S_2, S_3. The 3-tuple (1, 1, Wednesday) > is another. See, it's really boring and trivial when I has only a > finite number of elements!) > So that's all the Axiom of Choice says. If you have a set of > non-empty sets S_{x in I}, then you can construct an I-tuple containing > one element from each of those sets. That's all. > A corollary to the Axiom of Choice is that if you have a single > non-empty set, then it is possible to construct a new set containing > only one of that set's elements. In other words, you can pick a > singleton out of the original set. Again, it looks trivial and obvious, > but it is really a separate axiom from the other axioms of set theory > (or the real number system, or whatever application you're interested > in). > HTH, > -Arthur > ... what the Axiom of Choice says is that the Cartesian product > of all S_x (x in I) is non-empty. (Somewhat nitpicking): Technically speaking the above version is called Russell's Multiplicative Principle (MP) (due to B. Russell). It is equivalent to Zermelo's Axiom of Choice (AC). They are all equivalent to the Well Ordering Principle (WO) (due to Zermelo??, I think), and Kuratowski-Zorn's Lemma (ZL). Calling MP the Axiom of Choice might confuse the beginning student. (AC) If A is any set of NONEMPTY sets, then there exists a choice function on A (that is, there exists a function f with domain A such that for every x in A, f(x) is in x. (MP) If (A_i | i in I) is a family of NONEMPTY sets, and if the index set I is a set, then the cartesian product set is nonempty (that is, there is an I-tuple (x_i | i in I) such that for every i in I, x_i is in A_i). (ZL) Every inductive partially-ordered set has a maximal element. [Recall: A partially-ordered set P is inductive if every chain in P has an upper bound in P. Note: The term inductive here is not related to induction. A chain in P is just a linearly ordered subset of P.] (WO) For every set A, there exists a well-ordering on A. [Def of a well-ordering: A partial order on a set P is a well-order if every NONEMPTY subset of P has a least element.] Motto: All animals are equal but some are more equal than others. (Orwell) We rank the animals as follows: 1. ZL (Sure) 2. MP,AC (OK, if you twist my arm.) 3. ... n. WO [Wow, it's awesome (or, secretly, crazy)!] === Subject: Re: Definition of Axiom of Choice? (lots of good stuff on equivalents to AC snipped) > > Motto: All animals are equal but some are more equal than others. (Orwell) > We rank the animals as follows: > 1. ZL (Sure) > 2. MP,AC (OK, if you twist my arm.) > 3. by 1> > ... > n. WO [Wow, it's awesome (or, secretly, crazy)!] > This ranking is not uncontroversial; I'm sure different people have different favorite equivalents for AC. My favorite is: Given any two sets, there exists a surjection of one onto the other. I think it's called Cardinal Trichotomy, but someone will surely correct me if I'm wrong. My intuition of the way sets should behave forces me to accept this formulation of the axiom. In this it is far superior to either Zorn's Lemma or the Well-Ordering Theorem, although they are certainly much more useful for proving theorems. === Subject: Re: Definition of Axiom of Choice? > (lots of good stuff on equivalents to AC snipped) > > Motto: All animals are equal but some are more equal than others. (Orwell) > We rank the animals as follows: > 1. ZL (Sure) > 2. MP,AC (OK, if you twist my arm.) > 3. by 1> > ... > n. WO [Wow, it's awesome (or, secretly, crazy)!] > > This ranking is not uncontroversial; I'm sure different people have > different favorite equivalents for AC. My favorite is: > Given any two sets, there exists a surjection of one onto the other. > I meant it as a sarcastic ranking (by putting ZL at the top of the list above). No controversy here. Every one is free to re-order the entries to tailor it to their own form of sarcasm. But I think it is safe to bet on WO being at the bottom or near the bottom of everyone's list. :-) Despite the fact that WO is also at the bottom of my own list (different from the list above), it is actually my favorite one because every time I invoke AC (usually implicitly), WO will always rear its little sarcastic smile at the back of my mind reciting Orwell's famous line. Shedar === Subject: Re: Definition of Axiom of Choice? |This ranking is not uncontroversial; I'm sure different people have |different favorite equivalents for AC. My favorite is: | |Given any two sets, there exists a surjection of one onto the other. Given any two *nonempty* sets. There is no surjection from {} to {0} or from {0} to {}. |I think it's called Cardinal Trichotomy, but someone will surely |correct me if I'm wrong. It's similar, but not quite the same thing. I think in the absence of AC, cardinality is usually defined in terms of injections, not surjections. Cantor defines it essentially this way. He defines equal cardinality in terms of one-to-one correspondence, and then defines the cardinality of A to be less than that of B if there is a subset of B equivalent to A but not vice-versa. The fact that if there exists a surjection from B to A, then there exists an injection from A to B, is a relatively minor variation on the axiom of choice. On the other hand, the fact that if there is an injection from A to B, and A is nonempty, then there is a surjection from B to A is just the law of excluded middle. (Send elements in the image of A to the unique elements they came from; send the other elements of B to some given element of A.) So it's easier to see that cardinal trichotomy implies the version you like than to go the other way. I believe cardinal trichotomy as usually stated also bundles in the Cantor-Bernstein theorem. Once we know that either there is an injection from A to B or from B to A, we get three cases. The fact that in the case where injections in both directions exist, there is a bijection (and hence |A|=|B|) is the Cantor-Bernstein theorem. |My intuition of the way sets should behave |forces me to accept this formulation of the axiom. In this it is far |superior to either Zorn's Lemma or the Well-Ordering Theorem, |although they are certainly much more useful for proving theorems. I'm not sure whether to like or dislike this intuition! I'm amused by it in any case. The only proofs I've heard of that pass between Cardinals being ordered and the axiom of choice go by way of well-ordering. If cardinals are linearly ordered, then they all have to fit into the ordering of the alephs, i.e. the cardinalities of ordinal numbers, or equivalently of well-orderable sets. This can be shown to imply they are well-ordered. Conversely, if every set is well-orderable, that shows its cardinality belongs to the linear ordering of the alephs. Without assuming choice, there is the concept of the Hartog's ordinal of a set, the smallest ordinal without an injection to the set. I don't know how far the theory of cardinals in the absence of choice has been developed. I suppose people working with determinacy need some knowledge of this, don't they? In a model of the axiom of determinacy (which might be a submodel of the real universe, hence relevant to it) the axiom of choice fails. Without assuming the law of excluded middle, there are still two partial orders given by injections and surjections between sets. I don't know whether anyone has put serious effort into that theory at all. Keith Ramsay === Subject: Re: Definition of Axiom of Choice? > (AC) If A is any set of NONEMPTY sets, then there exists a choice function > on A (that is, there exists a function f with domain A such that for every x > in A, f(x) is in x. > (MP) If (A_i | i in I) is a family of NONEMPTY sets, and if the index set I > is a set, then the cartesian product set is nonempty (that is, there is an > I-tuple (x_i | i in I) such that for every i in I, x_i is in A_i). > (ZL) Every inductive partially-ordered set has a maximal element. > [Recall: A partially-ordered set P is inductive if every chain in P has an > upper bound in P. Note: The term inductive here is not related to > induction. A chain in P is just a linearly ordered subset of P.] > (WO) For every set A, there exists a well-ordering on A. > [Def of a well-ordering: A partial order on a set P is a well-order if > every NONEMPTY subset of P has a least element.] Or how about the snappy: Every surjective map has a section. Where a section is a right-inverse, e.g. a section for a map f is a map g such that f o g = id. To see that this is a form of AC think of f as inducing a partition on its domain, where the equivalence classes of the partition are given by C_y = {x : f(x) = y}. So f being surjective ensures that each class is nonempty. Then a section for f is a map g taking each y to a member of C_y. So g is a choice function. Bascule === Subject: Re: Problem with inner product by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8700Vo12665; >Here is one little problem. >H is Hilbert space and A is linear and bounded operator on H. >If (Ax,x)=0 show that Ax=0. >How can I show it? Are you sure its true? For 2-space (a very simple Hilbert space), let A=matrix with 0 along main diagonal and (1,-1) for the off-diagonal. Then A(x,y)=(y,-x), so the inner product is xy-yx=0. In the more general case, let the operator do as above for the first 2 coordinates and make all the others 0 or alternatively do the same kind of switch (i.e with one sign reversed) on pairs of coordinates. === Subject: Re: Problem with inner product >H is Hilbert space and A is linear and bounded operator on H. >If (Ax,x)=0 show that Ax=0. >How can I show it? > Are you sure its true? For 2-space (a very simple Hilbert space), > let A=matrix with 0 along main diagonal and (1,-1) for the off-diagonal. > Then A(x,y)=(y,-x), so the inner product is xy-yx=0. > In the more general case, let the operator do as above > for the first 2 coordinates and make all the others 0 or alternatively > do the same kind of switch (i.e with one sign reversed) > on pairs of coordinates. It is only true if you add the condition that A by symmetric. You can prove this by polarization. [Well, unless the field underlying H is of characteristic 2. -- I don't know what happens in this case. ] === Subject: Re: Probability(X is Prime) > Your explanation is not really the correct one. > The reason that the product does not given > the correct probability is that the probability > that p_i divides X is not independent from > the probability that p_j divides x once there are > sufficiently many primes in the product of (1-1/p) > that the direct product of the primes exceeds X. > You can lead a horse's ass to knowledge, but you can't make him think. Hi Bob, I think your answer corresponds to my original statement actually. You are saying that one can use the formual prod (1-1/p) providing the product of the primes used in this formula doesnt exceed X. I would agree with this because you are now considering the interval [1,X]. As soon as you use the next prime that such that the product takes you beyond X the formula cannot Prod (1-1/p) cannot be used. However we are trying to come up with a formula for Prob(X is Prime)and we seemed to have failed so far. === Subject: Re: Probability(X is Prime) > How about the following statement that using PNT pi(x)=x/log x > suggests the > Prob(x is prime) = 1/log (x). If this statement is to be accurate we > should have Sum (1/log x) over x=3,5,7,9,11.... should be 1 in order > to define a probability distribution on the integers. I dont know what > this sum actually is but my guess is that it is not 1. Does it > converge? > Only when the events are mutually exclsive the total probability is the sum of individual probabilities but a number can be divisible by many primes simultaneously. By Mertens theorem :(For N ---> inf.) Prob.(N is prime) = (1-1/2)(1-1/3)x....x(1-1/p)x e^g = 1/log(N) p = prime prox. inferior to N e = 2.718281828.. ; g = Euler's Constant = .577215... === Subject: Re: Probability(X is Prime) > How about the following statement that using PNT pi(x)=x/log x > suggests the Prob(x is prime) = 1/log (x). No, it doesn't, not unless you take a lot of care in setting up your definitions. > If this statement is to be accurate we should have Sum (1/log x) over > x=3,5,7,9,11.... should be 1 No, it shouldn't, it should sum to the number of primes, which it does. > in order to define a probability distribution on the integers. I dont > know what this sum actually is but my guess is that it is not 1. Does > it converge? No. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Probability(X is Prime) > How about the following statement that using PNT pi(x)=x/log x > suggests the Prob(x is prime) = 1/log (x). If a randomly select a 20-digit number, what is the probability that it is prime? Answer: approximately 1/log(10^20) = 0.0217147 . The actual value is (according to Maple) (pi(10^20)-pi(10^19))/(10^20-10^19) = 662253978428191411/30000000000000000000 = 0.02207513261 Actually, a better approximation is 1/(log(0.5*10^20)) = 0.02204655 === Subject: Re: Probability(X is Prime) > How about the following statement that using PNT pi(x)=x/log x > suggests the Prob(x is prime) = 1/log (x). -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Make money easily truely works X-Library: Indy 9.00.10 Comments Follow the directions below and in two weeks you'll have up to $20000.00 in your PayPal account. There is a very high rate of participation in the program because of its low investment and high rate of return. Just $5.00 to one person! THAT'S ALL !!! If you are a skeptic and don't think the program will work, I urge you to give it a try anyway! It REALLY WORKS! 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Become a part of the donation program and help people help people. This program is about helping each other! Success is a journey - Not a destination! Start Your Journey TODAY!!!! === Subject: simple pde question.. For some reason I can't solve one of the simplest equations.. Is there a solution? The equation: B = curl(A) Or in coordinate form: B_i = e_ijk* d/dx_j ( A_k ) (3 dimensions OK for now, e_ijk is anti-symmetric unit tensor or Levi-Civita tensor) The problem is to solve for the vector field A given a vector field B. I know that if A is a solution, so is A' = A + grad(f) where f is any function of the coordinates. I also know that there is no solution if div(B) != 0. But how can I write some (any) solution A in terms of divergenceless B? === Subject: Re: simple pde question.. Given div(B)=0, how can one solve the equation curl(A)=B? In addition to the answers prior posted, here is another direct approach. Let antigrad be an R^3 into R^3 mapping defined by: (antigrad_1(G))x = Int{y3=0 to x_3: G_2(x_1,x_2,y3)} - Int{y2=0 to x_2: G_3(x_1, y2, 0)} (antigrad_2(G))x = - Int{y3=0 to x_3: G_1(x_1,x_2,y3)} (antigrad_3(G))x = 0 Hint: These formulae will look better if manually rewritten, with y3 changed to x_3 bar, etc. ------------------------------------ We now have the following result: If div(B)=0 then [A=anticurl(B) is a solution of: curl(A)=B] Alternatively; If div(B)=0 then curl(anticurl(B))=B Notes: 1) The term anticurl is meant to be suggestive rather than literal -- it indicates that under a special condition (but not universally), the anticurl is the right inverse of the curl. 2) The proof of the above given result follows from direct computation of curl(anticurl(B)). David Ziskind ziskind@ntplx.net === Subject: Re: simple pde question.. Given div(B)=0, how can one solve the equation curl(A)=B? Sorry, there is a mis-writing in my first reply: the four instances of antigrad SHOULD BE anticurl. Please disregard my first reply, In addition to the answers prior posted, here is another direct approach. Let anticurl be an R^3 into R^3 mapping defined by: (anticurl_1(G))x = Int{y3=0 to x_3: G_2(x_1,x_2,y3)} - Int{y2=0 to x_2: G_3(x_1, y2, 0)} (anticurl_2(G))x = - Int{y3=0 to x_3: G_1(x_1,x_2,y3)} (anticurl_3(G))x = 0 Hint: These formulae will look better if manually rewritten, with y3 changed to x_3 bar, etc. ------------------------------------ We now have the following result: If div(B)=0 then [A=anticurl(B) is a solution of: curl(A)=B] Alternatively; If div(B)=0 then curl(anticurl(B))=B Notes: 1) The term anticurl is meant to be suggestive rather than literal -- it indicates that under a special condition (but not universally), the anticurl is the right inverse of the curl. 2) The proof of the above given result follows from direct computation of curl(anticurl(B)). David Ziskind ziskind@ntplx.net === Subject: Re: simple pde question.. > Given div(B)=0, how can one solve the equation curl(A)=B? > Sorry, there is a mis-writing in my first reply: the four instances of > ?antigrad? SHOULD BE ?anticurl?. Please disregard my first reply, > In addition to the answers prior posted, here is another direct > approach. > Let anticurl be an R^3 into R^3 mapping defined by: > (anticurl_1(G))x = Int{y3=0 to x_3: G_2(x_1,x_2,y3)} > - Int{y2=0 to x_2: G_3(x_1, y2, 0)} > (anticurl_2(G))x = - Int{y3=0 to x_3: G_1(x_1,x_2,y3)} > (anticurl_3(G))x = 0 > Hint: These formulae will look better if manually rewritten, with y3 > changed to x_3 bar, etc. > ------------------------------------ > We now have the following result: > If div(B)=0 then > [A=anticurl(B) is a solution of: curl(A)=B] > Alternatively; > If div(B)=0 then > curl(anticurl(B))=B > Notes: > 1) The term ?anticurl? is meant to be suggestive rather than literal -- > it indicates that under a special condition (but not universally), the > anticurl is the right inverse of the curl. > 2) The proof of the above given result follows from direct computation > of curl(anticurl(B)). I was indeed after an anticurl operator, and a few of you gave me specific solutions, as well as good general considerations to make life easier including defining a current (doesn't have to be electric) and using symmetry. I should tell you my application. It is ßuid mechanics.. I have been looking at the vorticity equation. If I come up with a solution for the vorticity of a ßuid.. i.e. I know curl(v) (v = bulk ßuid velocity).. then I was having trouble finding what the velocity is, and still more particularly what the divergence of the velocity is. Clearly (now) I need to include some boundary conditions on the ßow to uniquely determine the velocity from the vorticity.. and the velocity may not have a closed form. Author-Supplied-Address: bds ipp mpg de === Subject: Re: simple pde question.. Mail-To-News-Contact: abuse@dizum.com shevek asked: |> For some reason I can't solve one of the simplest equations.. |> Is there a solution? |> The equation: |> B = curl(A) |> The problem is to solve for the vector field A given a vector field B. |> I know that if A is a solution, so is A' = A + grad(f) where f is any |> function of the coordinates. I also know that there is no solution if |> div(B) != 0. |> But how can I write some (any) solution A in terms of divergenceless |> B? If you're willing to buy a divergenceless A, then you can do one further curl operation, leaving - Laplacian A = J = curl B Then you solve this elliptic equation given the appropriate boundary conditions. -- cu, Bruce drift wave turbulence: http://www.rzg.mpg.de/~bds/ === Subject: Re: simple pde question.. shevek: >For some reason I can't solve one of the simplest equations.. >Is there a solution? >The equation: >B = curl(A) >Or in coordinate form: >B_i = e_ijk* d/dx_j ( A_k ) >(3 dimensions OK for now, e_ijk is anti-symmetric unit tensor or >Levi-Civita tensor) >The problem is to solve for the vector field A given a vector field B. >I know that if A is a solution, so is A' = A + grad(f) where f is any >function of the coordinates. I also know that there is no solution if >div(B) != 0. >But how can I write some (any) solution A in terms of divergenceless >B? It depends upon what you are given _explicitly_. One way, is to take the curl of B = curl A to get J, curl V = curl curl A = grad (div A) - grad^2 A = (4pi/c) J Use the gauge freedom to set div A = 0, then solve the three poisson equations for the components of A, -grad^2 A = (4pi/c) J Another way is to exploit some symmetry and solve for the line integral of A, B = curl A => integral B.dS = integral (curl A).dS Using the identity, integral (curl A).dS = integral A.dl, I get, integral A.dl = integral B.dS If you can write down A in a coordinate system that allows you to take A outside the integral sign in order to evaluate the line integral easily, then this will work. === Subject: Re: simple pde question.. >For some reason I can't solve one of the simplest equations.. >Is there a solution? >The equation: >B = curl(A) >Or in coordinate form: >B_i = e_ijk* d/dx_j ( A_k ) >(3 dimensions OK for now, e_ijk is anti-symmetric unit tensor or >Levi-Civita tensor) >The problem is to solve for the vector field A given a vector field B. >I know that if A is a solution, so is A' = A + grad(f) where f is any >function of the coordinates. I also know that there is no solution if >div(B) != 0. >But how can I write some (any) solution A in terms of divergenceless If B is defined on all of R^3, then one answer is: A(r) = - int_0^1 du u r x B(u r), i.e. A_x = int_0^1 du [u z B_y(u x,u y,u z) - u y B_z(u x,u y,u z)], A_y = int_0^1 du [u x B_z(u x,u y,u z) - u z B_x(u x,u y,u z)], A_z = int_0^1 du [u y B_x(u x,u y,u z) - u x B_y(u x,u y,u z)]. David ----- === Subject: Re: simple pde question.. >B = curl(A) >The problem is to solve for the vector field A given a vector field B. >I know that if A is a solution, so is A' = A + grad(f) where f is any >function of the coordinates. I also know that there is no solution if >div(B) != 0. >But how can I write some (any) solution A in terms of divergenceless In other words, you want to construct a vector potential for the solenoidal vector field B. In general, you can do this on a domain D such that every closed surface in D is the boundary of a domain contained in D (i.e. D has no holes). I suppose your domain is all of R^3. The trick is to choose gauge conditions to reduce the amount of arbitrariness in the solution. One possible gauge condition is A_3 = 0. Then you want dA_2/dx_3 = -B_1 dA_1/dx_3 = B_2 dA_2/dx_1 - dA_1/dx_2 = B_3 The first and second conditions determine A_1 and A_2 up to functions of x_1 and x_2, i.e. A_1 = int B_2 dx_3 + a_1(x_1,x_2) A_2 = -int B_1 dx_3 + a_2(x_1,x_2) and then the third equation says da_1/dx_1 - da_2/dx_2 = g(x_1,x_2) for a certain function g (exercise: show that if div(B) = 0, this is a function of x_1 and x_2 only). Now you could e.g. take a_2 = 0 and a_1 = int g(x_1,x_2) dx_1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: The real numbers, and general comments Perhaps I should try to explain again my argument. (Provided that you haven't just decided to run away from me.) R is normally conceptualised as an infinite geometric line. I say that what we call a number is a point on this line, and so many points as we can locate can be called numbers. But, it is not a valid inference from this to state that the line is made up of numbers! This is also why hyperbolic geometry is inconsistent; it requires an uncountable number of (rational) numbers. Andrew Usher === Subject: Re: The real numbers, and general comments > > Perhaps I should try to explain again my argument. (Provided that you > haven't just decided to run away from me.) > R is normally conceptualised as an infinite geometric line. I say that > what we call a number is a point on this line, and so many points as > we can locate can be called numbers. But, it is not a valid inference > from this to state that the line is made up of numbers! > This is also why hyperbolic geometry is inconsistent; it requires an > uncountable number of (rational) numbers. > Andrew Usher Why on earth is hyperbolic geometry inconsistent? and how does it require an uncountable number of rational numbers? It doesn't and there aren't. === Subject: Re: The real numbers, and general comments > You can specify any one Dedekind cut, but you can't give the rule to > construct them all. A finite definition is acceptable, of course, as: > > N = {0, 1, 2, ...} > > is understood by all of us. > > But is it a finite definition? I mean it only gives the first 3. > Since the rationals are countable, they must be allowable, and since > sequences involve a countable number of terms they must be acceptable, so > are cauchy sequences of rationals not acceptable. It gives the rule to construct all the other terms, therefore it's a finite definition. Andrew Usher === Subject: Re: The real numbers, and general comments > You can specify any one Dedekind cut, but you can't give the rule to > construct them all. A finite definition is acceptable, of course, as: > > N = {0, 1, 2, ...} > > is understood by all of us. > > > But is it a finite definition? I mean it only gives the first 3. > Since the rationals are countable, they must be allowable, and since > sequences involve a countable number of terms they must be acceptable, so > are cauchy sequences of rationals not acceptable. > It gives the rule to construct all the other terms, therefore it's a > finite definition. > Andrew Usher And it is a way of defining the real numbers contradicting your original claim, I think. I don't have the original post anymore and can't be === Subject: Re: The real numbers, and general comments > You can specify any one Dedekind cut, but you can't give the rule to > construct them all. A finite definition is acceptable, of course, as: > > N = {0, 1, 2, ...} > > is understood by all of us. > Z = { ... -3, -2, -1, 0, 1, 2, 3, ...} > Q = { m/n | m in Z, n in Z, n <> 0} These are fine ... > R = { | A union B = Q; for all a in A, for all b in B, a is not in A} > For a in R, if GLB(B) in B, then is identified with GLB(B) > which is in Q. > How's that? but this is not. It doesn't specify a procedure for producing these sets. As you know, I believe that quantify over all subsets of an infinite set is illegitimate. > I would found analysis on any countable dense set in R, and use Cauchy > sequences where necessary. It is obvious that analysis over any such > countable dense set is equivalent to any other, and to analysis over > R, as long as all functions are piecewise continuous. > There's a simple problem with that: Let f be the following function: > f(x) = 0 when x is irrational, f(x) = 1 when x is rational. This is not piecewise continuous. (Note the restriction above.) > You are suggesting we just use the rationals, but the integral of f(x) > from 0 to 1 is actually 0. True, the above integral taken over Q gives 1. But who really cares about this function anyway? >No, you can't diagonalise on the computables, S, because S itself is >not computable. Hence you can't construct a number that isn't >computable. I see no reason to consider a thing that can't be >constructed to be a number. Also, this Ôprocedure' can't show that >there are uncountably many Ônumbers'. > >S is enumerable, which is sufficient to demonstrate a number that is >enumerable but not computable. > > Such a number is provably not constructible. Even if you accept > non-constructibles as Ônumbers', you can't show that there are > uncountably many Ônumbers'. > not constructible -> not computable. Diagonal arguments to show that > there are uncountably many reals, combined with only countably many > computables, implies that there are uncountably many non-computables. Yes, R is uncountable. But you haven't justified considering all of R to be numbers. If you use the common sense definition of a Ônumber' as something you can calculate with, then obviously all numbers are in S. >Euclidean geometry naturally embeds in projective geometry. I perceive >metric and projective techniques is simply different methods of >proving results in geometry, not as different fields. > Use the extended parallel postulate: > > P. Every pair of lines meet in a point; this point is at infinity iff > the interior angles make two right angles. > > This includes both Euclidean and projective geometry. > You are modifying the set of postulates for Euclidean geometry when you > do that. As soon as you add a postulate, you get *different* results, > which means you are no longer in Euclidean geometry. It is a geometry > that is clearly very similar, but it is not the same. What different results? As far as I can see, adopting my postulate means that all the results of Euclidean and of projective geometry hold good. >No, math is not a science but is closer allied to philosophy. I >wouldn't doubt that modern philosophy suffers from the same defect. > >Having studied philosophy, I can't agree with this statement, either. >It does offer insight into your approach to math, however. > > Would you mind expanding on that last? > The defect that philosophy suffers, from what I've seen, is that whoever > can string the best looking line of BS gets famous, regardless of how > clearly the philosopher doesn't believe it. > Solipsism would be a great example. If you don't believe anything exists > besides yourself, why would you care if someone attacks you, hurts you, > etc? If everything is in your mind and nothing is real, then don't > worry about anything. Well, not quite. Even if the real world has no independent existence, you still can logically want to act so as to avoid being hurt. Also, solipsism may be ridiculous, but the consideration of the possibility increasing our understanding of knowledge. > Math, in contrast, is about finding logicly consistent axioms and > finding their conclusions. As a result, math is much like the other > sciences in that it seeks to find truth. Philosophy, on the other hand, > may claim to seek truth but seems to do anything but that. I see both mathematics and philosophy as attempting to find absolute truth (outside of the physical world). To the extent either deviates from that, I consider it wrong. Andrew Usher === Subject: Speculative, but at least interesting Long Standing Math Puzzle May be Solves http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html === Subject: Re: Speculative, but at least interesting > Long Standing Math Puzzle May be Solves > http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html Quoth the Grauniad: 1 Birch and Swinnerton-Dyer conjecture Euclid geometry for the 21st century, involving things called abelian points and zeta functions and both finite and infinite answers to algebraic equations Oh dear:-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Speculative, but at least interesting > I'd just like to ask what is the thinking behind the apparent > fear among experts that a proof of RH would endanger internet > security? Keith Devlin comments on this in his book > _The Millennium Problems_ as follows: > Keith Devlin is a fine mathematician; probably better than me. > He is not, however, an expert on breaking public key crypto. > When you say among experts, to which experts do you refer? > Certainly factoring experts have no such fear. I was just going by what Devlin says below about the encryption community. The word experts referred to members of that community. > Rather, the fear among the encryption community is that the > _methods_ used to prove the hypothesis will involve new insights > into the pattern of the primes that will lead to better factoring > method. (p. 50) > I can't imagine how. Certainly some new technique discovered while > proving RH *might* lead to a new factoring technique. But this is > pure speculation. It might also be the case that some new technique > unrelated to RH will lead to a better factoring method. The ÔRH' > part is a red-herring. As far as anyone currently knows, factoring > algorithms do NOT depend on gaps between primes. As I understand it, the RH is equivalent to the statement that the probability of a natural number (that cannot be divided by a square) having an even or odd number of prime factors is 50-50. It's one of the few interpretations of the RH that makes it seem intuititvely plausible. I take it then that facts of that sort would also be of no help in designing factoring algorithms? === Subject: Re: Speculative, but at least interesting > if somebody really has cracked the so-called Riemann hypothesis, > financial disaster might follow. Suddenly all cryptic codes could be > breakable. No internet transaction would be safe. > Let's start here. A proof of R.H. would have ZERO, repeat ZERO > effect on cryptography. With the proviso that my knowledge of cryptography is limited to an elementary discussion of RSA public-key cryptography in Kenneth Rosen's _Elementary Number Theory and its Applications_ (pp. 259-271; I re-read this section recently because of the brouhaha over de Branges's supposed proof of the RH), I'd just like to ask what is the thinking behind the apparent fear among experts that a proof of RH would endanger internet security? Keith Devlin comments on this in his book _The Millennium Problems_ as follows: Since the Riemann hypothesis tells us so much about the primes, a proof of that conjecture might well lead to a major breakthrough in factoring techniques. Not because we will then know the hypothesis is true. Suspecting that it is true, mathematicians have been investigating its consequences for years. Indeed, some factoring methods work on the _assumption_ that it's true. Rather, the fear among the encryption community is that the _methods_ used to prove the hypothesis will involve new insights into the pattern of the primes that will lead to better factoring method. (p. 50) This is rather vague. Can anyone explain how a proof of the RH might allow someone to break the RSA codes? Inquiring minds want to know... === Subject: Re: Speculative, but at least interesting > Long Standing Math Puzzle May be Solves > http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html if somebody really has cracked the so-called Riemann hypothesis, financial disaster might follow. Suddenly all cryptic codes could be breakable. No internet transaction would be safe. Assume the Riemann hypothesis is true, then crack every code. A valid proof would alter nothing here. NSA can brute force crack any Officially sanctioned encryption right now. Being able in theory to factor the product of large primes isn't the same as actually doing it. Eudora passwords are trivially cracked - your ISP account is naked. PKZIP encryptions are demonstrated crackable if you know some contained text. MS Word and Word Perfect encryptions are easily crackable. The DES is an NSA joke. Recent versions of PGP are rumored to include an NSA back door. What would change? -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: Speculative, but at least interesting > Assume the Riemann hypothesis is true, then crack every code. A valid > proof would alter nothing here. NSA can brute force crack any > Officially sanctioned encryption right now. Steganography, however, is like Quantum Physics. You can't crack it until you see it. === Subject: Re: Speculative, but at least interesting > Assume the Riemann hypothesis is true, then crack every code. A valid > proof would alter nothing here. NSA can brute force crack any > Officially sanctioned encryption right now. > Steganography, however, is like Quantum Physics. You can't crack it > until you see it. Sure! The best encryption is not to be discovered in plain sight - the Mexican pushing a wheelbarrow of dirt over the border every morning. He wasn't smuggling drugs, data... He was smuggling wheelbarrows. I use the same strategy when bringing products home from Canadian research. They hand-search the luggage, the assholes. So? Take a long pop-top shipping can. Put the stuff in vials in the bottom then tighty fill with empty vials. Nobody wants to put their hands down through the razor edge from the pop top. Smile at them and give helpful hints all the while they are fondling your underwear. A dirty labcoat takes care of the working dogs. Methyl benzoate is cocaine pseudoscent for training the doggies. Idiots. If you want to kill the plane (and yourself), load a deodorant can with your chemical agent. No problem. Homeland Severity is a Federal river of money to political cronies. It is wholly impotent (there is that word again) to stop any meaningful attack. It's girly crap against a masculine enemy. Steganography affects the least significant bits of an image. If you are in the business of being curious, adequate diagnostics are SOP. Now you can deal with the shear volume of data - as in individual frames of a DvD movie. What if the encryption merely ßags the frame number, and that forms the code? If you are up to your ass in alligators, you properly look for the hatchery. If you want to protect your country against bad people, purse snatchers to whatever, you arm your citizens and let them defend themselves. Uncle Al says, Vote for Kerry! Pull America's plug and have done with it. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: Speculative, but at least interesting Discussion, linux) > Long Standing Math Puzzle May be Solves > http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html ,---- | Mathematicians could be on the verge of solving two separate million | dollar problems. If they are right - still a big if - and somebody | really has cracked the so-called Riemann hypothesis, financial | disaster might follow. Suddenly all cryptic codes could be | breakable. No internet transaction would be safe. | | [...] | | The Riemann hypothesis would explain the apparently random pattern | of prime numbers - numbers such as 3, 17 and 31, for instance, are | all prime numbers: they are divisible only by themselves and | one. Prime numbers are the atoms of arithmetic. They are also the | key to internet cryptography: in effect they keep banks safe and | credit cards secure. `---- *Would* the Riemann hypothesis explain the pattern of primes? If so, would this really threaten cryptography? Sorry, I've always been fuzzy on this Riemann stuff. But the connection to primes is just about the value of pi(x) as x goes to infinity, right? Are there any actual results showing that a proof of the Riemann hypothesis would yield a fast means of factoring large primes[1]? Footnotes: [1] If Bill Gates says that's the issue, then that's the issue. Don't argue. ,---- | ``The obvious mathematical breakthrough would be development of an | easy way to factor large prime numbers.'' -Bill Gates, The Road | Ahead, pg. 265 `---- -- Jesse F. Hughes One is not superior merely because one sees the world as odious. -- Chateaubriand (1768-1848) === Subject: Re: Speculative, but at least interesting >If so, would this really threaten cryptography? It seems unlikely to me. We currently proceed pretty much as though the Riemann hypothesis were known to be true. Having a proof would presumably provide some further insight, but the possibility that it would be the kind that would let someone factor numbers faster seems like a kind of cute daydream to me. At one time, not so long ago, it wasn't known for certain whether testing a number for being prime or not could be done in time bounded by a polynomial in the number of digits. There were various methods that came very close to doing so: there was a probabilistic test that would fail for non-primes with probability of at least 1/2, and could be repeated however many times you wanted. In practice, this test would almost never fail to detect that a number was composite. There were as I recall also methods that for primes would nearly always provide you with a rigorous proof that the prime was prime. It had also been known that if the Riemann hypothesis was true, then for some constant C applying one of these other tests to the numbers from 1 to C(log n)^2 would be good enough. It was in that sense that the Riemann hypothesis was linked to the existence of an efficient algorithm for primality testing. But primality testing and factoring are very different problems. Moreover, the dependence on the Riemann hypothesis is gone now. A polynomial-time algorithm known to work regardless of whether the Riemann hypothesis is true was found, by an amateur, which was sort of neat. Keith Ramsay === Subject: Re: Speculative, but at least interesting > Long Standing Math Puzzle May be Solves > http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html > ,---- > | Mathematicians could be on the verge of solving two separate million > | dollar problems. If they are right - still a big if - and somebody > | really has cracked the so-called Riemann hypothesis, financial > | disaster might follow. Suddenly all cryptic codes could be > | breakable. No internet transaction would be safe. > | [...] > | The Riemann hypothesis would explain the apparently random pattern > | of prime numbers - numbers such as 3, 17 and 31, for instance, are > | all prime numbers: they are divisible only by themselves and > | one. Prime numbers are the atoms of arithmetic. They are also the > | key to internet cryptography: in effect they keep banks safe and > | credit cards secure. > `---- > *Would* the Riemann hypothesis explain the pattern of primes? > If so, would this really threaten cryptography? > Sorry, I've always been fuzzy on this Riemann stuff. But the > connection to primes is just about the value of pi(x) as x goes to > infinity, right? Are there any actual results showing that a proof of > the Riemann hypothesis would yield a fast means of factoring large > primes[1]? > Footnotes: > [1] If Bill Gates says that's the issue, then that's the issue. > Don't argue. > ,---- > | ``The obvious mathematical breakthrough would be development of an > | easy way to factor large prime numbers.'' -Bill Gates, The Road > | Ahead, pg. 265 > `---- :) nice quote. I wonder about this too, because my extremely limited understanding is that currently the general *belief* amongst mathematicians who know anything much about the Riemann Hypothesis is that it's almost certainly true, and just awaiting the technicality of a proof. Indeed I believe there's a substantial body of Ôresults' which assume RH as a premise. I don't think any of them are new fast ways to factor large numbers. -- Larry Lard Replies to group please === Subject: Re: Speculative, but at least interesting format=ßowed; charset=iso-8859-1; reply-type=response > Long Standing Math Puzzle May be Solves > http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html Louis de Branges sounds like their version of James Harris.... blimey! I wonder if that IS James Harris using a pseudonym! === Subject: Re: Speculative, but at least interesting > Long Standing Math Puzzle May be Solves > http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html > Louis de Branges sounds like their version of James Harris.... blimey! I > wonder if that IS James Harris using a pseudonym! No, based on his work on the Bieberbach Conjecture, de Branges is VERY legit. He has, in the past, published erroneous proofs of the Riemann hypothesis, but so have other mathematicians. Myxococcus xanthus === Subject: Re: Speculative, but at least interesting > Long Standing Math Puzzle May be Solves > http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html > Louis de Branges sounds like their version of James Harris.... blimey! I > wonder if that IS James Harris using a pseudonym! Not quite. It's true that he has in common with James an obsessive interest in one of the great unsolved problems. However, de Branges has actually contributed. He is on the faculty at Purdue, and is credited with solving another unsolved problem, the Bierbach conjecture. His proof has not been dismissed outright but is getting serious attention. If he's gone off the deep end late in his mathematical career, at least he had a distinguished mathematical career first. - Randy === Subject: Re: Speculative, but at least interesting >Long Standing Math Puzzle May be Solves > http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html > Louis de Branges sounds like their version of James Harris.... blimey! I >wonder if that IS James Harris using a pseudonym! > Not quite. It's true that he has in common with James an obsessive > interest in one of the great unsolved problems. However, de Branges > has actually contributed. He is on the faculty at Purdue, and is > credited with solving another unsolved problem, the Bierbach conjecture. > His proof has not been dismissed outright but is getting serious > attention. > If he's gone off the deep end late in his mathematical career, at > least he had a distinguished mathematical career first. > - Randy Reminds me of Abian. Alexander Abian apparently still made useful contributions to the mathematics literature even after he evolved into a crackpot. http://www.math.ucdavis.edu/~suh/abian/abian-list.html === Subject: Re: Speculative, but at least interesting Discussion, linux) > Long Standing Math Puzzle May be Solves > http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html > Louis de Branges sounds like their version of James Harris.... blimey! I > wonder if that IS James Harris using a pseudonym! > Not quite. It's true that he has in common with James an obsessive > interest in one of the great unsolved problems. However, de Branges > has actually contributed. He is on the faculty at Purdue, and is > credited with solving another unsolved problem, the Bierbach conjecture. > His proof has not been dismissed outright but is getting serious > attention. > If he's gone off the deep end late in his mathematical career, at > least he had a distinguished mathematical career first. But when he proved the Bierbach conjecture, it took some time for mathematicians to accept the proof. Similarly, once again, it has taken time for mathematicians to really look at his proposed proof of Riemann (partly because it's not the first time he's announced a proof of the Riemann hypothesis). paint such a rosy picture of mathematicians' reactions to de Branges and it also doesn't show him as having a distinguished mathematical career free of controversy prior to this. Note: I'm not saying whether mathematicians' reluctance to study his proposed proof is regrettable or not. I'm not really familiar with a fuller picture of the situation. -- Jesse Hughes But nothing's being Dr. Ullrich is a particular case of something's being such that nothing is it: (Ex)~(Ey)(y = x) -- John Correy on the failings of first order logic === Subject: Re: Speculative, but at least interesting |But when he proved the Bierbach conjecture, it took some time for |mathematicians to accept the proof. Similarly, once again, it has |taken time for mathematicians to really look at his proposed proof of |Riemann (partly because it's not the first time he's announced a proof |of the Riemann hypothesis). If I remember correctly, the proof of the Bieberbach conjecture came at the end of a book on related material. I think those of us who have read proofs of any complexity can appreciate why the specialists in the fields closest to De Branges' might be reluctant to take the time to read a whole book, especially if it's relatively technical. Any error in it could well be difficult to find, too. I think we might have to wait for awhile for a verdict. Keith Ramsay === Subject: Re: Speculative, but at least interesting > |But when he proved the Bierbach conjecture, it took some time for > |mathematicians to accept the proof. Similarly, once again, it has > |taken time for mathematicians to really look at his proposed proof of > |Riemann (partly because it's not the first time he's announced a proof > |of the Riemann hypothesis). > If I remember correctly, the proof of the Bieberbach conjecture came > at the end of a book on related material. > I think those of us who have read proofs of any complexity can > appreciate why the specialists in the fields closest to > De Branges' might be reluctant to take the time to read > a whole book, especially if it's relatively technical. Any error in it > could well be difficult to find, too. I think we might have to wait for > awhile for a verdict. Furthermore, de Branges is sloppy and nonstandard. Not non-standard, but not following established standards. He's just an independent so-and-so. Some people are like that. It took people a while to accept his proof of Bieberbach for that reason, and I suspect people will be loathe to look at his proof of Riemann for the same reason. Jon Miller === Subject: Re: Speculative, but at least interesting |Furthermore, de Branges is sloppy and nonstandard. Not non-standard, but |not following established standards. He's just an independent so-and-so. |Some people are like that. It took people a while to accept his proof of |Bieberbach for that reason, and I suspect people will be loathe to look at |his proof of Riemann for the same reason. I was an undergraduate when his proof of Bieberbach was accepted, and I went to a talk he gave. Early on, he made a remark about some people not liking the order in which he arranged his proof (what was it, that he started by doing operator theory?), but that that was the way he did it. Keith Ramsay === Subject: Re: Speculative, but at least interesting format=ßowed; charset=iso-8859-1; > Long Standing Math Puzzle May be Solves > http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html > Louis de Branges sounds like their version of James Harris.... blimey! I > wonder if that IS James Harris using a pseudonym! > Not quite. It's true that he has in common with James an obsessive > interest in one of the great unsolved problems. However, de Branges > has actually contributed. He is on the faculty at Purdue, and is > credited with solving another unsolved problem, the Bierbach conjecture. > His proof has not been dismissed outright but is getting serious > attention. > If he's gone off the deep end late in his mathematical career, at > least he had a distinguished mathematical career first. Cool, just checking.. === Subject: Re: Definition of Axiom of Choice? > A corollary to the Axiom of Choice is that if you have a single > non-empty set, then it is possible to construct a new set containing > only one of that set's elements. In other words, you can pick a > singleton out of the original set. This does *not* require the axiom of choice. That is, the following statement is true in ZF: For all x and y, if y is in x, then there exists z such that for all w, w is in z if and only if w = y. -SJH === Subject: Re: Eigenfunctions of the Laplacian >Consider the Laplacian operator Delta on some bounded two-dimensional >domain G. There is a L^2 orthonormal set of eigenfunctions of Delta >on G. >If G is rectangular and if a function f on G is smooth up to the >boundary then the (generalised) Fourier series of f (in terms of the >eigenfunctions of Delta) is uniformly convergent on G. >Under which conditions does that yield for more general domains G? >Tobias Nahring You don't mention the boundary conditions -- that's a necessary part of formulating the eigenvalue problem. Let's take Dirichlet conditions (so the eigenfunctions are products of certain sine functions when G is a rectangle, which = 0 on the boundary). So the eigenfunctions all are zero on the boundary, and there is no way a series of them can converge uniformly to f unless f too is zero on the boundary. /dan === Subject: Weird Trigonometric Identity After reading another post (unfortunately, I can't remember which) I was playing around with the function f(n) = 2arctan( cot(2n)) + 4n and it seems that for n integer, f(n) is always an odd multiple of pi. Does anyone know if this is true, and if so why? Andrew === Subject: Re: Weird Trigonometric Identity Earlier,I was also playing around with inverse trigonometric functions and got traingular/saw tooth after two functionally annulling inverse operations: For any trig function f and inverse function g with period P, g[f(x)] = x for |x|< P/4 if f is odd, and, |x|< P/2 if f is even. Periodicity is obtaned by straight line co-catenation of f(x) if you will, joining maximum and minimum points by straight lines. It will be a good exercise for you to establish under which conditions this is true for ANY peridic function. > I was playing around with the function f(n) = 2arctan( cot(2n)) + 4n > and it seems that for n integer, f(n) is always an odd multiple of > pi. === Subject: Re: Weird Trigonometric Identity > After reading another post (unfortunately, I can't remember which) I > was playing around with the function > f(n) = 2arctan( cot(2n)) + 4n > and it seems that for n integer, f(n) is always an odd multiple of > pi. > Does anyone know if this is true, and if so why? A simpler related identity is arctan(cot(t)) = pi/2 - t, for -pi/2 < t <= pi/2 The odd multiples come from the branches of arctan (just add 2pi), sort of an artifact of forcing a computation. Why is the simpler identity true? the proof by picture (using words!) is that cot and tan are reciprocals, so are reßections of each other over the line x==y. Which is exactly what pi/2 - t does. I'm sure you could prove it formally by converting to exponentials and using tedious algebra, but it's way too early in the morning for me to bother. -- Mitch Harris (remove q to reply) === Subject: Re: Weird Trigonometric Identity > After reading another post (unfortunately, I can't remember which) I > was playing around with the function > > f(n) = 2arctan( cot(2n)) + 4n > > and it seems that for n integer, f(n) is always an odd multiple of > pi. > > Does anyone know if this is true, and if so why? > A simpler related identity is > arctan(cot(t)) = pi/2 - t, > for -pi/2 < t <= pi/2 > The odd multiples come from the branches of arctan (just add 2pi), > sort of an artifact of forcing a computation. > Why is the simpler identity true? the proof by picture (using > words!) is that cot and tan are reciprocals, so are reßections of > each other over the line x==y. Which is exactly what pi/2 - t does. I can't say that I agree with your argument that cot and tan being _reciprocals_ implies that they are reßections of each other over the line x = y. I would prove the simpler identity as follows: tan(t)tan( pi/2 - t) = (sin(t)/cos(t))*(sin(pi/2 - t)/cos(pi/2 - t)) = (sin(t)/cos(t))*(cos(t)/sin(t)) = 1, t != 0 Thus cot(t) = tan(pi/2 -t) The rest follows from what you all told me. Andrew === Subject: Re: Weird Trigonometric Identity schrieb Andrew Duncan : > After reading another post (unfortunately, I can't remember which) I > was playing around with the function > f(n) = 2arctan( cot(2n)) + 4n > and it seems that for n integer, f(n) is always an odd multiple of > pi. > Does anyone know if this is true, and if so why? Try to express arctan in terms on arccot; then use 2arccot( cot(2n)) = 4n. === Subject: Re: Weird Trigonometric Identity >schrieb Andrew Duncan : > After reading another post (unfortunately, I can't remember which) I > was playing around with the function > f(n) = 2arctan( cot(2n)) + 4n > and it seems that for n integer, f(n) is always an odd multiple of > pi. >Try to express arctan in terms on arccot; >then use > 2arccot( cot(2n)) = 4n. That should be 2 arccot(cot(2n)) = 4n + 2k pi where k is an integer. 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: Weird Trigonometric Identity >schrieb Andrew Duncan : > After reading another post (unfortunately, I can't remember which) I > was playing around with the function > f(n) = 2arctan( cot(2n)) + 4n > and it seems that for n integer, f(n) is always an odd multiple of > pi. >Try to express arctan in terms on arccot; >then use > 2arccot( cot(2n)) = 4n. > That should be 2 arccot(cot(2n)) = 4n + 2k pi where k is an integer. How embarrassing! - I got that mixed up with 4cot(arccot(2n))=4n. Thomas === Subject: Fokker-Plank and Diffusive equations Let us compare Fokker-Plank and Diffusive equations in case of no drift (for simplicity). Fokker-Plank dn/dt = 1/2*ddQn/dxx Diffusive dn/dt = 1/2*dQdn/dxx n- density of distribution Q- diffusive coefficient Question: Why position of Q is different? ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: Time, the need for the lack of it. > There are other forces,true, but all are actions that require other > forces or reactions, to me this leaves the only force with enourh > ability to jump start the BIG BANG, gravity. For one small moment, > gravity was the only game in town. Who or what caused gravity can only > be Prayed to. As the first force,it, gravity, needs no other force to > exist. If it, gravity, does not require time, speed, energy,or mass > ,to be,then if you remove, these other forces from considerion in > determing the outcome of any problem, in communication or movement by > or of ones self across vast distances(more than 2 feet) the solution > will be correct. The conventional/current view of gravity is that it is due to bends in space-time. Einstein discovered that when people are travelling at different velocities time and space get mixed up. As an example, suppose Captain Kirk put clocks all along the outside of Enterprise from front to back, set them all to the same time, then measured the distance between them. The Klingons coming towards him at half light speed would be laughing because they would see that the clocks were all progressively reading different times from front to back of the ship. They would laugh even more as they watched Kirk measuring distances, he would be putting his ruler down at a slant so that the two ends would never touch the ship when the clocks read the same instant. The Klingons would communicate with Kirk telling him that he should measure lengths with both ends of the ruler applied at the same time. If Science Officer Spock took the call he would tell the Klingons that neither space nor time exist, they are inextricably bound together as Ôspace-time'. One man's space is another Klingon's time. If you put synchronised clocks along a ßag pole you would find that they all read very, very, very slightly different times. It turns out that mass changes the distribution of space and time in space-time. If you fell to earth from a few thousand miles up your acceleration would actually be adjusting your velocity to take account of the changing rate of clocks as you got nearer to earth. The changes in clock rate with velocity and height are very tiny but have huge effects on energy and hence the forces occurring in nature because mass changes slightly with velocity and e=mc^2 where c is an enormous number. Here is a conundrum. How could you see a movement or hear a word if you only exist for the present instant, if you are at the durationless boundary between the past and future? Best Wishes Alex Green === Subject: Re: Time, the need for the lack of it. > > There are other forces,true, but all are actions that require other > forces or reactions, to me this leaves the only force with enourh > ability to jump start the BIG BANG, gravity. For one small moment, > gravity was the only game in town. Who or what caused gravity can only > be Prayed to. As the first force,it, gravity, needs no other force to > exist. If it, gravity, does not require time, speed, energy,or mass > ,to be,then if you remove, these other forces from considerion in > determing the outcome of any problem, in communication or movement by > or of ones self across vast distances(more than 2 feet) the solution > will be correct. > The conventional/current view of gravity is that it is due to bends in > space-time. > Einstein discovered that when people are travelling at different > velocities time and space get mixed up. Einstein discovered no such thing. He merely re-discovered that when you let mathematicians use the words Space, Time, Energy and Velocity, the result is total recurring gibberish. Which unfortunately for Einstein, that result was very previously known, by one of his predecessors named Noah. And it is *not* the current view of gravity that it is *due* to bends in space-time. It is the current-view that SpaceTime (AKA QM, it's Science Fiction, it's low-IQ Biologists, and it's Scientically-derived, Zippy-The-Pinhead Photons) are due to bends in gravity. As an example, suppose Captain > Kirk put clocks all along the outside of Enterprise from front to > back, set them all to the same time, then measured the distance > between them. The Klingons coming towards him at half light speed > would be laughing because they would see that the clocks were all > progressively reading different times from front to back of the ship. > They would laugh even more as they watched Kirk measuring distances, > he would be putting his ruler down at a slant so that the two ends > would never touch the ship when the clocks read the same instant. The > Klingons would communicate with Kirk telling him that he should > measure lengths with both ends of the ruler applied at the same time. > If Science Officer Spock took the call he would tell the Klingons that > neither space nor time exist, they are inextricably bound together as > Ôspace-time'. One man's space is another Klingon's time. > If you put synchronised clocks along a ßag pole you would find that > they all read very, very, very slightly different times. It turns out > that mass changes the distribution of space and time in space-time. If > you fell to earth from a few thousand miles up your acceleration would > actually be adjusting your velocity to take account of the changing > rate of clocks as you got nearer to earth. The changes in clock rate > with velocity and height are very tiny but have huge effects on energy > and hence the forces occurring in nature because mass changes slightly > with velocity and e=mc^2 where c is an enormous number. > Here is a conundrum. How could you see a movement or hear a word if > you only exist for the present instant, if you are at the durationless > boundary between the past and future? > Best Wishes > Alex Green === Subject: Re: Time, the need for the lack of it. > It has become increasingly ovious that the sceince comunity has > gotten it totaly wrong. Quite aside from your apparent delusions of grandeur, the fact that you are unwilling to even check your spelling/grammar and have not even spelt Ôscience' correctly resulted in me giving up reading this far through your post. If you're not going to take what you write seriously, why do you expect us to? alex === Subject: Re: Time, the need for the lack of it. > It has become increasingly ovious that the sceince comunity has > gotten it totaly wrong. > Quite aside from your apparent delusions of grandeur, the fact that you are > unwilling to even check your spelling/grammar and have not even spelt Ôscience' > correctly resulted in me giving up reading this far through your post. > If you're not going to take what you write seriously, why do you expect us to? > alex `Had I known I was to be dealing with perfection, not only would I have invested my life savings in a spell checker, but I would even put on clean drawers, so you could kiss my ass without care or bother of skid marks. Had I visions of granderur, I would mirical you ass to a far warmer climate. Some people, like myself, do not have the ability or learing to afford but a good effort at proper spelling, sorry about that. === Subject: theorems/problems with lots of quantifiers Most mathematical theorems and conjectures seem to have low quantifier depth, that is, the nesting of quantifiers is not very deep. For example, if you formalize the fundamental thm of arithmetic, you get something like for all integers n, there exists a unique factorization into primes. However, I would like to have some examples of not too obscure, fairly accessible problems (conjectures or thms) where it -is- nested deeply. For example, the pumping lemma for regular languages (notorious to students of automata/computability) has (at least) 4 quantifiers. I suppose I care more about natural problems rather than er...unmotivated ones. (I suppose a related notion is algorithmic problems with running time a higher degree polynomial: most of the algorithms we care about are degree 3 or less. what are some high degree ones? like the latest n^12 primality testing, or some of those n^6 group theory algorithms) -- Mitch Harris (remove q to reply) === Subject: Re: theorems/problems with lots of quantifiers > Most mathematical theorems and conjectures seem to have low quantifier > depth, that is, the nesting of quantifiers is not very deep. > For example, if you formalize the fundamental thm of arithmetic, you > get something like for all integers n, there exists a unique > factorization into primes. Do you take prime and factorization as primitive notions ? If not, you'll have to introduce more quantifiers as you expand them into primitive notation. Also, just your there exists a unique has a universal quantifier lurking: E! x F(x) means Ex Ay F(y) iff y=x > However, I would like to have some examples of not too obscure, fairly > accessible problems (conjectures or thms) where it -is- nested deeply. Well, in calculus the notion of continuity at a point has three alternating quantifiers: A epsilon E delta A y ... Try stating the theorem that a function continuous on a closed bounded interval is uniformly continuous. -- pa at panix dot com === Subject: Re: theorems/problems with lots of quantifiers > (I suppose a related notion is algorithmic problems with running time > a higher degree polynomial: most of the algorithms we care about are > degree 3 or less. what are some high degree ones? like the latest n^12 > primality testing, or some of those n^6 group theory algorithms) I like the O(n^42) algorithm to partition orthogonal polygons into fat rectangles. http://www.cccg.ca/proceedings/2002/04.ps Therese === Subject: Re: theorems/problems with lots of quantifiers > Most mathematical theorems and conjectures seem to have low quantifier > depth, that is, the nesting of quantifiers is not very deep. > For example, if you formalize the fundamental thm of arithmetic, you > get something like for all integers n, there exists a unique > factorization into primes. > However, I would like to have some examples of not too obscure, fairly > accessible problems (conjectures or thms) where it -is- nested deeply. > For example, the pumping lemma for regular languages (notorious to > students of automata/computability) has (at least) 4 quantifiers. > I suppose I care more about natural problems rather than > er...unmotivated ones. How about, there is a winning strategy for white in two moves at some game This is There is some white move so that For all black moves White wins Replace two moves by fifty moves, or if you like omega moves. === Subject: Re: theorems/problems with lots of quantifiers > Most mathematical theorems and conjectures seem to have low quantifier > depth, that is, the nesting of quantifiers is not very deep. > However, I would like to have some examples of not too obscure, fairly > accessible problems (conjectures or thms) where it -is- nested deeply. You need to specify what sort of language you wish these conjectures or theorems to be expressed in or alternatively what sort of objects do you accept quantification over. All arithmetical problems, for example, can be expressed using just one universal quantifier binding a function variable and one existential quantifier binding a number variable. Obviously this is not what you're after. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: theorems/problems with lots of quantifiers > Most mathematical theorems and conjectures seem to have low quantifier > depth, that is, the nesting of quantifiers is not very deep. > For example, if you formalize the fundamental thm of arithmetic, you > get something like for all integers n, there exists a unique > factorization into primes. > However, I would like to have some examples of not too obscure, fairly > accessible problems (conjectures or thms) where it -is- nested deeply. > For example, the pumping lemma for regular languages (notorious to > students of automata/computability) has (at least) 4 quantifiers. > I suppose I care more about natural problems rather than > er...unmotivated ones. > (I suppose a related notion is algorithmic problems with running time > a higher degree polynomial: most of the algorithms we care about are > degree 3 or less. what are some high degree ones? like the latest n^12 > primality testing, or some of those n^6 group theory algorithms) If x and the a_i's are integers it is true that: (Ax) ((Ea_1)((Ea_2)((Ea_3)((Ea_4)((Ea_5)((Ea_6)((Ea_7)((Ea_8)((Ea_ 9) ((Ea_10)((Ea_11)((Ea_12)((Ea_13)((Ea_14)((Ea_15)((Ea_16)((Ea_ 17) ((Ea_18)((Ea_19)((Ea_20)((Ea_21)((Ea_22)((Ea_23)((Ea_24)((Ea_ 25) ((Ea_26)((Ea_27)((Ea_28)((Ea_29)((Ea_30)((Ea_31)((Ea_32)((Ea_ 33) ((Ea_34)((Ea_35)((Ea_36)((Ea_37) (x=a_1^5+a_2^5+a_3^5+a_4^5+a_5^5+a_6^5+a_7^5+a_8^5+a_9^5+ a_10^5+a_11^5+a_12^5+a_13^5+a_14^5+a_15^5+a_16^5+a_17^5+ a_18^5+a_19^5+a_20^5+a_21^5+a_22^5+a_23^5+a_24^5+a_25^5+ a_26^5+a_27^5+a_28^5+a_29^5+a_30^5+a_31^5+a_32^5+a_33^5+ a_34^5+a_35^5+a_36^5+a_37^5) ))))))))))))))))))))))))))))))))))))) -- Clive Tooth http://www.clivetooth.dk === Subject: Re: theorems/problems with lots of quantifiers >However, I would like to have some examples of not too obscure, fairly >accessible problems (conjectures or thms) where it -is- nested deeply. > If x and the a_i's are integers it is true that: > (Ax) > ((Ea_1)((Ea_2)((Ea_3)((Ea_4)((Ea_5)((Ea_6)((Ea_7)((Ea_8)((Ea_ 9) > ((Ea_10)((Ea_11)((Ea_12)((Ea_13)((Ea_14)((Ea_15)((Ea_16)((Ea_ 17) > ((Ea_18)((Ea_19)((Ea_20)((Ea_21)((Ea_22)((Ea_23)((Ea_24)((Ea_ 25) > ((Ea_26)((Ea_27)((Ea_28)((Ea_29)((Ea_30)((Ea_31)((Ea_32)((Ea_ 33) > ((Ea_34)((Ea_35)((Ea_36)((Ea_37) > (x=a_1^5+a_2^5+a_3^5+a_4^5+a_5^5+a_6^5+a_7^5+a_8^5+a_9^5+ > a_10^5+a_11^5+a_12^5+a_13^5+a_14^5+a_15^5+a_16^5+a_17^5+ > a_18^5+a_19^5+a_20^5+a_21^5+a_22^5+a_23^5+a_24^5+a_25^5+ > a_26^5+a_27^5+a_28^5+a_29^5+a_30^5+a_31^5+a_32^5+a_33^5+ > a_34^5+a_35^5+a_36^5+a_37^5) > ))))))))))))))))))))))))))))))))))))) Hm. OK, Waring's problem. so you can get arbitrarily large #'s of quantifiers. OK, how about the extra rule (which I conveniently forgot to mention) that the quantifiers need to -alternate- (because in a logical sense, a string of there exists really just collapses to one there exists with a possibly nasty encoding of the variables. -- Mitch Harris (remove q to reply) === Subject: Re: theorems/problems with lots of quantifiers |OK, how about the extra rule (which I conveniently forgot to mention) |that the quantifiers need to -alternate- (because in a logical sense, |a string of there exists really just collapses to one there exists |with a possibly nasty encoding of the variables. you can ask things like can the first player force a win in go-moku within 30 moves?. probably not of really great mathematical interest, but some such questions can be rather challenging. -- [e-mail address jdolan@math.ucr.edu] === Subject: Localization without using Wavelet One of the merits using Wavelet is a localization. But in return, the correspondence between frequency and basis becomes a bit complicated. Are there any methods which bring localization with keeping the simple relation between frequency and basis like Fourier analysis? I tried to construct such basis in the following site: http://139.134.5.123/tiddler2/fbt/fbt.htm === Subject: Re: The multiplicative groups Z*_n; One more time... === > Subject: Re: The multiplicative groups Z*_n; One more time... > Use > http://mathquest.com/discuss/sci.math > http://mathforum.org/epigone/sci.math > 2 <= m, g generates (Z_p^m)^*, g^phi(p^m) /= 1 (mod p^(m+1)) >You assume this is true, right? If so, why do you need m => 2? > ==> g generates (Z_p^(m+1))^*, g^phi(p^(m+1)) /= 1 (mod p^(m+2)) > let b = g^phi(p^m); if b^p = g^phi(p^(m+1)) = 1 (mod p^(m+2)): > (b - 1)(b^(p-1) +..+ 1) = b^p - 1 = 0 (mod p^(m+2)) > p^m | b-1; not p^(m+1) | b-1; p^2 | b^(p-1) +..+ 1 > b = 1 (mod p^m); b = 1 (mod p^2); b^j = 1 (mod p^2) > 2 <= m is needed for this step^^^^^^^ > 0 = b^(p-1) +..+ 1 = p (mod p^2); p^2 | p, whoops >So g^phi(p^(m+1)) /= 1 (mod p^(m+2)), right? >This is what you are showing here--do I have that right? > Yes, the indented statements after if....: make contradiction. > g generates Z_p^* ==> g or g + p generates (Z_p^2)^* > if (g + p)^(p-1) = 1 = g^(p-1) (mod p^2) >(g + p)^(p-1) = [g^(p-1) - pg^(p-2)] > Huh? > 1 = g^(p-1) + (p-1) g^(p-2) p + ... = 1 - p.g^(p-2) (mod p^2) >g^(p-1) /= 1 mod p^2 ; g^(p-1) = 1 mod p ==> >g^(p-1) = 1 + ap ==> (g + p)^(p-1) = 1 + p[a - g^(p-2)]. > Where does a come from? It is just an adjustable constant I introduced from g^(p-1) = 1 mod p ==> p | g^(p-1) - 1 ==> g^(p-1) = 1 + ap ==> (g + p)^(p-1) = 1 + p[a - g^(p-2)]. The following is much like your case of p = 3, g = 2 below. ----------- Consider p = 5, g = 2. 2^4 = 16 = 1 + 3.5 ==> a = 3 In Z*_p^2 = Z*_25, g^10 = -1, so g generates Z*_25. g + p = 7, (g+p)^4 = 7^4 = 1 mod 25, so order(g+p) = p-1, as one can see from (g + p)^(p-1) = 1 + p[a - g^(p-2)] with a = 3, 2^3 = 8, p[a - g^(p-2)] = 5[3 - 8] mod 25 = 0. I have lost track of what I was doing. ------------- I think you are correct in saying that either g or g + p generates Z*_p^2. When I did this part, I said that one can find some a so that o(g + ap) = p - 1, and o(p+1) = p, so x = (g + ap)(p+1) generates Z*p^2. >I think its best to use g + kp and then choose k so that >either o(g) = p-1 or o(g) = p(p-1) in Z*_p^2, i.e we can always >find a k so that g generates Z*_p^2. > That's what I said, k = 0 or k = 1 in paticular. > p.g^(p-2) = 0 (mod p^2) > p = p.g^(p-1) = 0 (mod p^2); p^2 | p, whoops > > g generates (Z_p^2)^* ==> g + p^2 generates (Z_p^2)^*, Do you mean g + p rather than g + p^2 ? >I don't understand. In Z*_p^2, g + p^2 = g. > 2 generates Z_3^*, so does 5 and 8. 5 = 2, 5^2 = 1 (mod 3) > In Z_3^2, 2^2 = 4, 5^2 = 7, 8^2 = 1 (mod 9) > Thus, 2 and 5 but not 8 generate (Z_3^2)^* as from above: Right. You must have meant g + p. OK. > g generates Z_p^*, g^(p-1) /= 1 (mod p^2) ==> g generates (Z_p^2)^* > g generates Z_p^* ==> g or g + p generates (Z_p^2)^* > g + p generates Z_p^*, thus g + p or g + 2p generates (Z_p^2)^* If you really mean g + p^2 below, I don't understand. > (g + p^2)^phi(p^2) = 1 (mod p^3). > and (g + p^2)^phi(p^2) /= 1 (mod p^3). > otherwise: let b = (g + p^2)^p > 1 = b^(p-1) + (p-1) b^(p-2) p^2 + ... = 1 - b^(p-2) p^2 (mod p^3) > (only time p > 2 is used); b^(p-2) p^2 = 0 (mod p^3) > p^2 = b^(p-1) p^2 = 0 (mod p^3); p^3 | p^2 whoops >I will go over this last again. > Don't bother. Instead find for all odd prime p, > some g generates (Z_p^2)^* with g^(p-1) /= 1 (mod p^3). OK. Do you mean try g, g + p^2, and find one such that g^(p-1) /= 1 (mod p^3). Is this why you talk of g + p^2 ? >I am not clear what you are doing here. > Me neither. > ---- === Subject: Re: The multiplicative groups Z*_n; One more time... > g generates Z_p^* ==> g or g + p generates (Z_p^2)^* > if (g + p)^(p-1) = 1 = g^(p-1) (mod p^2) >(g + p)^(p-1) = [g^(p-1) - pg^(p-2)] > Huh? > 1 = g^(p-1) + (p-1) g^(p-2) p + ... = 1 - p.g^(p-2) (mod p^2) >g^(p-1) /= 1 mod p^2 ; g^(p-1) = 1 mod p ==> >g^(p-1) = 1 + ap ==> (g + p)^(p-1) = 1 + p[a - g^(p-2)]. > Where does a come from? > It is just an adjustable constant I introduced from > g^(p-1) = 1 mod p ==> p | g^(p-1) - 1 ==> g^(p-1) = 1 + ap ==> Then say so: ==> some a with g^(p-1) = 1 + ap. > (g + p)^(p-1) = 1 + p[a - g^(p-2)]. -- > The following is much like your case of p = 3, g = 2 below. > Consider p = 5, g = 2. 2^4 = 16 = 1 + 3.5 ==> a = 3 ... > I have lost track of what I was doing. End of story, I'm not a mind reader. -- > Don't bother. Instead find for all odd prime p, > some g generates (Z_p^2)^* with g^(p-1) /= 1 (mod p^3). > OK. Do you mean try g, g + p^2, and find one such that No, I say don't bother with that approach. As you see, I've resolved all the issues of this second method of showing (Z_p^n)^* for odd prime p, except for finding generators as noted above. So put your mind to finding them, I haven't found a way. What can you come up with? === Subject: Do you know the counter example? I have been doing battle with this problem trying to prove it but I cannot make the proof work so I am wondering if its even true. Let X be a topological space and A is a Sub set of X. Show that, The interior of the boundary of A is equal to null set. . ( B(A)^o = O/ ) It feels as if it should be true but i just cannot show it so any ideas would be welcome stephen === Subject: Re: Do you know the counter example? It is false. Next would be a modification of the problem. If A is an open or closed subset of X = R^n, then the interior of the boundary of A is null set. > I have been doing battle with this problem trying to prove it but I cannot > make the proof work so I am wondering if its even true. > Let X be a topological space and A is a Sub set of X. Show that, > The interior of the boundary of A is equal to null set. . ( > B(A)^o = O/ ) > It feels as if it should be true but i just cannot show it so any ideas > would be welcome > stephen === Subject: Re: Do you know the counter example? >I have been doing battle with this problem trying to prove it but I cannot >make the proof work so I am wondering if its even true. >Let X be a topological space and A is a Sub set of X. Show that, >The interior of the boundary of A is equal to null set. . ( >B(A)^o = O/ ) >It feels as if it should be true but i just cannot show it so any ideas >would be welcome It is false actually: the boundary of Q in R is the whole set R KP -- E-MAIL: K.P.Hart@EWI.TUDelft.NL PAPER: Faculty EWI PHONE: +31-15-2784572 TU Delft FAX: +31-15-2786178 Postbus 5031 URL: http://fa.its.tudelft.nl/~hart 2600 GA Delft the Netherlands === Subject: Co-re set with no infinite r.e. subset Is there an easy example of an infinite set of naturals that has no infinite r.e. subset? It would be better if it were co-r.e., but any example would -- Daryl McCullough Ithaca, NY === Subject: Re: Co-re set with no infinite r.e. subset > Is there an easy example of an infinite set of naturals that has no infinite > r.e. subset? It would be better if it were co-r.e., but any example would Generic sets, as in Boolos and Jeffreys, chapter 20 ? Because their exercise 20.5 is Show that all subsets of a generic set that are definable in arithmetic are finite. Recursively enumerable implies definable in arithmetic, so any subset of a generic that is r.e. must be finite. -- pa at panix dot com === Subject: Re: Co-re set with no infinite r.e. subset charset=iso-8859-1 > Is there an easy example of an infinite set of naturals that has no infinite > r.e. subset? It would be better if it were co-r.e., but any example would These are called Ôimmune' sets. An (infinite) r.e. complement of such a set is called Ôsimple.' Immune sets exist (and were named by Dekker), simple sets by Post. The existence is given as an exercise by Hartley Rogers (5.8), and discussed in his section 8.2. I won't type in the construction. Dennis === Subject: Re: Co-re set with no infinite r.e. subset format=ßowed; charset=Windows-1252; > Is there an easy example of an infinite set of naturals that has no > infinite > r.e. subset? It would be better if it were co-r.e., but any example would The set of indices of the TMs that do not halt on a blank tape? --r.e.s. === Subject: Re: Co-re set with no infinite r.e. subset | > Is there an easy example of an infinite set of naturals that has no | > infinite | > r.e. subset? It would be better if it were co-r.e., but any example would | | The set of indices of the TMs that do not halt on a blank tape? No, there are lots of infinite r.e. subsets of that. That's known as a complete co-r.e. set and they always do. For example, take the machines that have no transitions to the halt state, or ones that always move to the right and output 1s, and so on. What Daryl wants is known as an immune set. If it's the complement of an r.e. set, then it's the complement of what's called a simple set. Post proved there exists such a thing in 1944. Keith Ramsay === Subject: Re: Co-re set with no infinite r.e. subset Keith Ramsay says... >What Daryl wants is known as an immune set. If it's the complement >of an r.e. set, then it's the complement of what's called a simple >set. Post proved there exists such a thing in 1944. >Keith Ramsay of a simple set. This relates to the thread about Chaitin's Omega. As I understand it, if we let ONE = { n | the nth bit of Omega is 1 } ZERO = { n | the nth bit of Omega is 0 } then both ONE and ZERO are immune sets. At first, I thought that Omega was equivalent in some sense to the real HALT whose nth bit is 1 if the nth computer program halts on input n, and 0 otherwise. But it is possible to determine infinitely many bits of HALT, so Omega is more uncomputable than HALT, in some sense. -- Daryl McCullough Ithaca, NY === Subject: Re: Co-re set with no infinite r.e. subset > Is there an easy example of an infinite set of naturals that has no > infinite > r.e. subset? It would be better if it were co-r.e., but any example would > The set of indices of the TMs that do not halt on a blank tape? That set is obviously not r.e., but why can it not have an infinite r.e. subset? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Co-re set with no infinite r.e. subset format=ßowed; charset=Windows-1252; > Is there an easy example of an infinite set of naturals that has no > infinite > r.e. subset? It would be better if it were co-r.e., but any example > would > The set of indices of the TMs that do not halt on a blank tape? > That set is obviously not r.e., but why can it not have an infinite r.e. > subset? That's the part I wondered about myself (hence the ?). It does seem to be a bogus example, though, by the following hand-wavy construction: A set of indices of TMs corresponding to the programs {while true: n=n}, n=1,2,3,..., can be enumerated by some TM, since the index of an arbitrary TM is supposed to be computable, and can presumably be made to depend straighforwardly on n. I hope someone does post a valid example (or shows that there are none). --r.e.s. === Subject: Re: Co-re set with no infinite r.e. subset > Is there an easy example of an infinite set of naturals that has no > infinite > r.e. subset? It would be better if it were co-r.e., but any example > would > The set of indices of the TMs that do not halt on a blank tape? > That set is obviously not r.e., but why can it not have an infinite r.e. > subset? > That's the part I wondered about myself (hence the ?). > It does seem to be a bogus example, though, by the > following hand-wavy construction: A set of indices of > TMs corresponding to the programs {while true: n=n}, > n=1,2,3,..., can be enumerated by some TM, since the > index of an arbitrary TM is supposed to be computable, > and can presumably be made to depend straighforwardly > on n. > I hope someone does post a valid example (or shows > that there are none). David Ullrich has already posted an example. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Co-re set with no infinite r.e. subset >Is there an easy example of an infinite set of naturals that has no infinite >r.e. subset? not sure how easy it has to be. here's a proof that there is such a thing that can't be -too- hard, since i didn't know the answer a minute ago: let s[1], s[2], ... be an enumeration of the infinite re sets. choose n[1] in s[1]. choose m[1] > n[1]. choose n[2] > m[1] with n[2] in s[2]. choose m[2] > n[2]. etc. let s be the set of all the m[k]'s. then s is infinite, and s[k] is not a subset of s since n[k] is not in s. >It would be better if it were co-r.e., but any example would ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: On the partial sums of reciprocals of primes by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i87CBIw09302; >Could we say that if >f(t) = o(g(t)), then >? int_{a}^{x} f(t) dt = o (int_{a}^{x} g(t) dt) ? >We suppose that f(t) could be negative or positive (we don't know), >while g(t) is positive for all sufficiently large positive t. >For example, if f(t) = t and g(t) = e^t, then >t = o(e^t). >And >lim x to infty int_{a}^{x} t dt / int_{a}^{x} e^t dt >= lim x ((x^2 - a^2)/2)/(e^x - e^a) = 0. >Using the hypothesis, seems to me that we could prove the question >in the positive easily. >Let me outline the proof: >Set >int_{a}^{x} f(t) dt / int_{a}^{x} g(t) dt. >By hypothesis, for any epsilon > 0, there exists M such that >for all x > M, >0 < |f(t)/g(t)| < epsilon < 1. (1) >Thus we could write >int_{a}^{x} f(t) dt / int_{a}^{x} g(t) dt >= [ int_{a}^{M} f(t) dt + int_{M}^{x}f(t) dt ] / int_{a}^{x} g(t) dt >= [ constant(M) + int_{M}^{x}f(t) dt ]/int_{a}^{x} g(t) dt >= o(1) + int_{M}^{x} f(t) dt/int_{a}^{x} g(t) dt. (2) >By (1), >|f(t)| = epsilon g(t). (3) >Integrating both sides of (3) over [a, x], we get No; instead, over [M, x]. Below are corrected; int_{M}^{x} |f(t)| dt = epsilon int_{M}^{x} g(t) dt or int_{M}^{x} |f(t)| dt / int_{M}^{x} g(t) dt = epsilon. (4) Using (4) in (2), we obtain o(1) + int_{M}^{x} f(t) dt/int_{a}^{x} g(t) dt < o(1) + int_{M}^{x} |f(t)| dt / int_{M}^{x} g(t) dt = o(1) + epsilon. (a < M) erdos fan === Subject: Re: A Non Linear Trigonometric Expression for P i . by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i87CBHk09257; >For any positive real number X: >Pi= 4*ArcTanX + 2*ArcTan[(1-X^2)/2X] , Working in Radians, >[ or , >180= 4*ArcTanX + 2*ArcTan[(1-X^2)/2X] , Working in Degrees ]. ADDENDUM -------- It is worth adding to the above formula ,my previously presented (similar in past discussions)formula : Pi/4 =ArcTAN[X] - ArcTan[(X-1)/(X+1)] , working in Radians , For X positive real number Copyright P. Stefanides >Panagiotis Stefanides >http://www.stefanides.gr P.stefanides http://www.stefanides.gr === Subject: Re: Are logarithms still useful? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i87CBJf09333; >We learn that logarithms were invented to facilitate arithmetic with big >numbers. Electronic calculators have made them unecessary for this >purpose. In what ways are logarithms still useful in mathematics? Of course computers , still, cannot give this answer. Nevertheless my theory and work on the ETIMOLOGY of the word LOGARITHM gives a meaning that it is the ratio of two numbers,and particularly the ratio of two angles, e.g: Ln 5=1.609437912 , and this is the ratio of 144.8494121..Deg /90 Deg. Ln 2=0.693147181 , this is the ratio of 62.38324625..Deg / 90 Deg,etc. Similarly ,the same applies to other ,than e, base. [Ref: http://www.stefanides.gr/why_logarithm.htm http://www.stefanides.gr/logarithm.htm http://www.stefanides.gr/nautilus.htm The above ,is still under examination and hope some care should be taken for acceptance . Concluding ,logarithms are needed and are as good as geometry trigonometry ,algebra etc., and computer cannot replace their use. Panagiotis Stefanides http://www.stefanides.gr === Subject: Re: Are logarithms still useful? >I , DO NOT KNOW IF LOGARITHMS WERE INVENTED to facilitate arithmetic >GIVE THE ETIMOLOGYOF THIS TERMINOLOGY. >Of course computers , still, cannot give this answer. The etymology of the word logarithm is as follows, according to the OED: [ad. mod.L. logarithm-us (Napier, 1614), f. Gr. logos word, proportion, ratio + arithmos number. Napier does not explain his view of the literal meaning of logarithmus. It is commonly taken to mean ratio-number, and as thus interpreted it is not inappropriate, though its fitness is not obvious without explanation. Perhaps, however, Napier may have used logos merely in the sense of Ôreckoning', Ôcalculation' (cf. LOGISTIC).] My computer did give this answer, or at least look it up at http://www.oed.com. >Nevertheless my theory and work on the ETIMOLOGY of the word LOGARITHM >gives a meaning that it is the ratio of two numbers,and particularly >the ratio of two angles, e.g: >Ln 5=1.609437912 , and this is the ratio of 144.8494121..Deg /90 Deg. >Ln 2=0.693147181 , this is the ratio of 62.38324625..Deg / 90 Deg,etc. Any real number can be written as a ratio of some number of degrees to 90 degrees. So what? Is there anything special about 144.8494121... other than that it is 90 ln(5)? 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: Infinite cardinal number Let m be an infinite cardinal number. Many proofs of m+m=m that I saw are colloraries of m x m=m. Does anyone know a direct proof ? === Subject: Re: Infinite cardinal number >Let m be an infinite cardinal number. Many proofs of >m+m=m that I saw are colloraries of m x m=m. >Does anyone know a direct proof ? With Axiom of Choice, it is straightforward. Without Axiom of Choice, it is true in some models, and false in others. On the other hand m x m = m implies the Axiom of choice. So let X be a set of infinite cardinality m. Then there is a limit ordinal alpha such that alpha has cardinality m. So we need to show that there is a 1-1 mapping of 2 x alpha onto alpha. Just use the ordinal product; this maps omega x h + k, k finite, onto omega x h + {2k, 2k+1}. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Math Cobweb Game I combine elements of the following games of mine, plus a little Poker, into one game. Tangle: The Game: Game involving an n-gon & lines (revised): Integer-Alterations On a Grid (fun): First, draw a regular r-gon on a piece of paper, where r is a multiple of the number of players playing this game. Each player then makes up math-functions, the same number of functions per player for a total of r functions, which transform the integers m and n (n is a positive integer) into an integer, new m. For example, m <= m + n, m <= ßoor(|m|/n), m <= d(|m| + n^2), d is number-of-divisors function, m <= ceiling(sqrt(m*n)), m <= numerator of reduced |m|/n, m <= Fibonacci(|m|) - n, etc. (The functions can be appropriate for any level of math-ability.) The functions are written around the perimeter of the r-gon, one function per side. it down without showing it to anybody else. (This rule is subject to change.) m begins with the value of 1. Player 1 starts at any vertex of the r-gon and draws a straight line-segment, using a straight-edge if necessary, to any side. m is modified by the function at that side and by n (which equals 1 on first move; see below on how n is calculated). Players alternate drawing line-segments, starting each segment where last player finished their segment, drawing the segment in any direction (with restrictions), and finishing the segment at the first previously drawn segment or edge of the r-gon encountered. Segments coming off of a segment hitting another player-drawn segment must pass the earlier segment, continuing on the opposite side from the most-previously drawn segment. Segments coming off of a segment drawn to the edge of the r-gon must Ôbounce' (in either direction) back into the r-gon's interior. Do not draw any segments to any vertexes (of the r-gon or where previously drawn segments intersect). (See the link above about the game Tangle for a similar set of segment-drawing rules, plus for some ascii diagrams.) Now n is equal to the number of segments in the previous stretch. (A stretch is the the collection of segments between any two sides of the r-gon which are immediately connected by a path of segments {without connecting in the mean-time to any other sides of the r-gon}.) Every time a side of the r-gon is reached, m is modified by the previous m and by n and by the function at that side. Also: A stretch may intersect any particular segment at most once. Each side of the r-gon can be visited any number of times, but the game is complete when the last-to-be-visited side is finally reached. The winner is the player(s) whose secret integer is closest to the final value of m. Alternative scoring: Maybe players can make up a different situation for scoring/winning, such as: (for 2 player game) player 1 wins if the number of distinct primes dividing the final m is odd, player 2 wins if this number of primes is even. Leroy Quet === Subject: Re: Probability(X is Prime) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i87EGdl21188; >However we are trying to Who is we? >come up with a formula for Prob(X is Prime)and we seemed to have >failed so far. You will not succeed at all until you define your terminology. A formula in the sense you ask is impossible; The probability, for any X is either 0 or 1. You need to define a pdf before you can begin talking about probability. One such is the following: Let S = {x | (1-eps)X < x < (1+eps) X} x in Z+ for some X. and (say) 1/2 < eps < 1 Thus, S is the set of all integers x, near X. If we select x uniformly at random from the set S, then the probability that x is prime is approximately 1/log(X). You can take x uniformly from all of Z+; no such density function exists. You have to take some bounded subset of Z+. === Subject: Re: Probability(X is Prime) > > Who is we? Looks like we are the people writing to this thread. > You need to define a pdf before you can begin talking about > probability. One such is the following: > Let S = {x | (1-eps)X < x < (1+eps) X} x in Z+ for some X. > and (say) 1/2 < eps < 1 > Thus, S is the set of all integers x, near X. > If we select x uniformly at random from the set S, then the > probability that x is prime is approximately 1/log(X). > You can[Ôt] take x uniformly from all of Z+; no such density function > exists. You have to take some bounded subset of Z+. So using your example as a good starting point for further discussion, what values of eps and X make the statement not approximate but exact? === Subject: Re: Probability(X is Prime) > Let S = {x | (1-eps)X < x < (1+eps) X} x in Z+ for some X. > and (say) 1/2 < eps < 1 > > Thus, S is the set of all integers x, near X. > > If we select x uniformly at random from the set S, then the > probability that x is prime is approximately 1/log(X). > > You can[Ôt] take x uniformly from all of Z+; no such density function > exists. You have to take some bounded subset of Z+. > So using your example as a good starting point for further discussion, what > values of eps and X make the statement not approximate but exact? You want the probability to be exactly 1 / log X ? This will be tricky, since the probability is going to be some rational number, so you're going to have to choose X so that its natural logarithm is rational. This will make X transcendental, so you'll be talking about the probability of primality in the neighborhood of a transcendental number, whereas people generally ask for stuff like the probability of primality in a neighborhood of, say, some particular power of 10. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Probability(X is Prime) > Let S = {x | (1-eps)X < x < (1+eps) X} x in Z+ for some X. > and (say) 1/2 < eps < 1 > > Thus, S is the set of all integers x, near X. > > If we select x uniformly at random from the set S, then the > probability that x is prime is approximately 1/log(X). > > You can[Ôt] take x uniformly from all of Z+; no such density function > exists. You have to take some bounded subset of Z+. > > So using your example as a good starting point for further discussion, what > values of eps and X make the statement not approximate but exact? > You want the probability to be exactly 1 / log X ? > This will be tricky, since the probability is going to be > some rational number, Why must the probability be rational? >so you're going to have to choose X > so that its natural logarithm is rational. This will make X > transcendental, so you'll be talking about the probability > of primality in the neighborhood of a transcendental number, > whereas people generally ask for stuff like the probability > of primality in a neighborhood of, say, some particular > power of 10. === Subject: Differential Equations and Boundary Value Problems... I just found my old college DiffEQ text book. Its called Differential Equations and Boundary Value Problems. I am just curious but could someone explain the difference between ordinary diffeqs and boundary value problems? === Subject: Re: Differential Equations and Boundary Value Problems... > I just found my old college DiffEQ text book. Its called Differential > Equations and Boundary Value Problems. I am just curious but could > someone explain the difference between ordinary diffeqs and boundary > value problems? boundary value problems are often partial differential equations. === Subject: Least Square Estimation problem Hi all, here is my problem. I'd like to find the minimum of that expression and not the roots alpha_k like in the classical case: parallel mathcal{D} (n+1)- sum _{k=0} ^{n} alpha_k mathcal{D}(k) parallel =parallel gamma parallel where in my pratical case: $mathcal{D} (k)$ is a matrix 8x8 is known. $alpha_k$ are the roots of the equation, unknown. $gamma$ is what I want so unknown. I don't want to find roots which minimize gamma but just the minimum of gamma. Do I really need to compute the roots to get the mimimum of gamma or is there a technique to just get the mimimum ? Any helpful suggestion or reference is appreciated. Yannick === Subject: a series divergence problem I have been failed to solve this problem. b_0 = 1. b_(n+1) = ( b_n / 6 ) * ( (b_n)^2 - 3 b_n + 6 ) . Prove that the sum b_0 + b_1 + b_2 + ... diverges to infinity. What I know is that { b_n } is a decreasing sequence and limit_(n->infinity) b_n = 0. The ratio test gives : b_(n+1) / b_n --> 1 as n --> infinity. === Subject: Re: a series divergence problem >I have been failed to solve this problem. >b_0 = 1. >b_(n+1) = ( b_n / 6 ) * ( (b_n)^2 - 3 b_n + 6 ) . >Prove that the sum b_0 + b_1 + b_2 + ... diverges to infinity. The given recurrence implies b_n >= 6/(4n+6) by induction. And 6/(4n+6) is a divergent series. Mike Guy === Subject: Re: a series divergence problem >I have been failed to solve this problem. >b_0 = 1. >b_(n+1) = ( b_n / 6 ) * ( (b_n)^2 - 3 b_n + 6 ) . >Prove that the sum b_0 + b_1 + b_2 + ... diverges to infinity. >What I know is that { b_n } is a decreasing sequence and limit_(n->infinity) >b_n = 0. >The ratio test gives : >b_(n+1) / b_n --> 1 as n --> infinity. I have a rather messy proof: . . . . . . Spoiler . . . . . . b_{n + 1} = f(b_n), where f(x) = x^3/6 - x^2/2 + x is strictly increasing for all real x, because f'(x) = [(x - 1)^2 + 1]/2 > 0. If n > 0, then f(1/n) = (6n^2 - 3n + 1)/(6n^3) > 1/(n + 1), because (n + 1)(6n^2 - 3n + 1) - 6n^3 = 3n^2 - 2n + 1 = 2n^2 + (n - 1)^2 > 0. Therefore by induction on n, for all n >= 0, b_n >= 1/(n + 1), and since sum 1/(n + 1) diverges, sum b_n also diverges. -- Angus Rodgers (angus_prune@ eats spam; reply to angusrod@) Contains mild peril === Subject: Charting transformations Hi there, My maths is extremely rusty given that I haven't been in a math class for 16 years so please forgive my ignorance. I'm doing some charting and I need to transform coordinates from one chart into a second so that I can plot the point on a computer. Rather than try to explain precisely what I'm try to do I've created a graphic that illustrates it. It's available at http://192.168.120.15/transformation.gif Please take a look. If anyone can offer me the solution to this problem I would be extremely grateful. Mat P.S. Please alo reply by email to mat_guthrie@hotmail.com === Subject: Re: Charting transformations > Hi there, > My maths is extremely rusty given that I haven't been in a math class > for 16 years so please forgive my ignorance. > I'm doing some charting and I need to transform coordinates from one > chart into a second so that I can plot the point on a computer. Rather > than try to explain precisely what I'm try to do I've created a > graphic that illustrates it. It's available at > http://192.168.120.15/transformation.gif That address is a local IP address; the file you have put there will not be accessible via that address to anyone outside your local network. You will need to find some webspace somewhere to host your image to show us it. Your ISP may well provide you with some such public webspace. -- Larry Lard Replies to group please === Subject: n! = a! b! It seems to be a folk theorem that the equation in the subject has only one nontrivial solution: n = 10, a = 7, b = 6. Here we exclude as trivial solutions where a or b is 0 or 1 and the collection n = (k + 1)!, a = (k + 1)! - 1, b = k + 1. I haven't been able to find a published proof--has anyone come across one? Rick === Subject: Re: n! = a! b! What if n = r! for some r? Then n! = r!(n-1)! There are an infinite number of solutions. Kevin > It seems to be a folk theorem that the equation in the subject > has only one nontrivial solution: n = 10, a = 7, b = 6. Here we > exclude as trivial solutions where a or b is 0 or 1 and the > collection n = (k + 1)!, a = (k + 1)! - 1, b = k + 1. I haven't > been able to find a published proof--has anyone come across one? > Rick === Subject: Re: n! = a! b! > What if n = r! for some r? Then n! = r!(n-1)! > There are an infinite number of solutions. > Kevin > It seems to be a folk theorem that the equation in the subject > has only one nontrivial solution: n = 10, a = 7, b = 6. Here we > exclude as trivial solutions where a or b is 0 or 1 and the > collection n = (k + 1)!, a = (k + 1)! - 1, b = k + 1. I haven't > been able to find a published proof--has anyone come across one? > > > > Rick That would be the collection n = (k + 1)!, a = (k + 1)! - 1, b = k + 1 k = r - 1 -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: n! = a! b! These solutions are called trivial in the original post. >What if n = r! for some r? Then n! = r!(n-1)! >There are an infinite number of solutions. >Kevin >It seems to be a folk theorem that the equation in the subject >has only one nontrivial solution: n = 10, a = 7, b = 6. Here we >exclude as trivial solutions where a or b is 0 or 1 and the >collection n = (k + 1)!, a = (k + 1)! - 1, b = k + 1. I haven't >been able to find a published proof--has anyone come across one? >Rick > === Subject: Re: n! = a! b! > It seems to be a folk theorem that the equation in the subject > has only one nontrivial solution: n = 10, a = 7, b = 6. Here we > exclude as trivial solutions where a or b is 0 or 1 and the > collection n = (k + 1)!, a = (k + 1)! - 1, b = k + 1. I haven't > been able to find a published proof--has anyone come across one? In Problem B23 of the 2nd edition of Unsolved Problems in Number Theory, Guy notes 9! = 7! 3! 3! 2!, 10! = 7! 6! = 7! 5! 3!, 16! = 14! 5! 2!, and the trivial example a_1 = a_2! ... a_r! - 1, n = a_2! ... a_r! giving n! = a_1! ... a_r!. He says nothing else has been found up to n = 18160. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: n! = a! b! > It seems to be a folk theorem that the equation in the subject > has only one nontrivial solution: n = 10, a = 7, b = 6. Here we > exclude as trivial solutions where a or b is 0 or 1 and the > collection n = (k + 1)!, a = (k + 1)! - 1, b = k + 1. I haven't > been able to find a published proof--has anyone come across one? > In Problem B23 of the 2nd edition of Unsolved Problems in Number > Theory, Guy notes 9! = 7! 3! 3! 2!, 10! = 7! 6! = 7! 5! 3!, > 16! = 14! 5! 2!, and the trivial example a_1 = a_2! ... a_r! - 1, > n = a_2! ... a_r! giving n! = a_1! ... a_r!. He says nothing else > has been found up to n = 18160. Huh! Too bad they gave up at that point. 18161! = 11590! x 7908! Tracie. === Subject: Re: n! = a! b! >Huh! Too bad they gave up at that point. >18161! = 11590! x 7908! Nobody has come out and said it yet, so I will. There can be no primes in the interval [max(a,b)+1,n]. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: n! = a! b! > > It seems to be a folk theorem that the equation in the subject > has only one nontrivial solution: n = 10, a = 7, b = 6. Here we > exclude as trivial solutions where a or b is 0 or 1 and the > collection n = (k + 1)!, a = (k + 1)! - 1, b = k + 1. I haven't > been able to find a published proof--has anyone come across one? > > In Problem B23 of the 2nd edition of Unsolved Problems in Number > Theory, Guy notes 9! = 7! 3! 3! 2!, 10! = 7! 6! = 7! 5! 3!, > 16! = 14! 5! 2!, and the trivial example a_1 = a_2! ... a_r! - 1, > n = a_2! ... a_r! giving n! = a_1! ... a_r!. He says nothing else > has been found up to n = 18160. > Huh! Too bad they gave up at that point. > 18161! = 11590! x 7908! I know that was a joke, but there are probably some wide-eyed readers on this group which might be led astray by the above. To those I ask Does the prime 18149 divide both sides?. Phil -- They no longer do my traditional winks tournament lunch - liver and bacon. It's just what you need during a winks tournament lunchtime to replace lost === Subject: Re: n! = a! b! > It seems to be a folk theorem that the equation in the subject > has only one nontrivial solution: n = 10, a = 7, b = 6. Here we > exclude as trivial solutions where a or b is 0 or 1 and the > collection n = (k + 1)!, a = (k + 1)! - 1, b = k + 1. I haven't > been able to find a published proof--has anyone come across one? > In Problem B23 of the 2nd edition of Unsolved Problems in Number > Theory, Guy notes 9! = 7! 3! 3! 2!, 10! = 7! 6! = 7! 5! 3!, > 16! = 14! 5! 2!, and the trivial example a_1 = a_2! ... a_r! - 1, > n = a_2! ... a_r! giving n! = a_1! ... a_r!. He says nothing else > has been found up to n = 18160. > Huh! Too bad they gave up at that point. > 18161! = 11590! x 7908! > I know that was a joke, but there are probably some wide-eyed readers > on this group which might be led astray by the above. To those I ask > Does the prime 18149 divide both sides?. For the n!=a!b! conjecture, this looks like a good start, by the way, showing that a must be greater than about n-ln n , so b must be quite small. Alas, it doesn't seem to lead anywhere :-( > Phil === Subject: Re: n! = a! b! > It seems to be a folk theorem that the equation in the subject > has only one nontrivial solution: n = 10, a = 7, b = 6. Here we > exclude as trivial solutions where a or b is 0 or 1 and the > collection n = (k + 1)!, a = (k + 1)! - 1, b = k + 1. I haven't > been able to find a published proof--has anyone come across one? > In Problem B23 of the 2nd edition of Unsolved Problems in Number > Theory, Guy notes 9! = 7! 3! 3! 2!, 10! = 7! 6! = 7! 5! 3!, > 16! = 14! 5! 2!, and the trivial example a_1 = a_2! ... a_r! - 1, > n = a_2! ... a_r! giving n! = a_1! ... a_r!. He says nothing else > has been found up to n = 18160. > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) The third edition of Guy's book (same section) adds that Chris Caldwell has by Caldwell The Diophantine equation A! B! = C!, J. Recreational Math., 26(1994) 128-133. === Subject: Questions on the Wiener algebra The Wiener algebra W is the set of all functions on the unit circle f(t) = Sum_{n=-infty}{infty} a_n t^n, where t = e^{I theta}, (0 <= theta < 2 Pi) which have absolutely summable Fourier coefficients: Sum |a_n| < infty It is a famous result of Wiener that: If f(t) has no zeros on the unit circle, then 1/f(t) is also in W. Also, another known result (perhaps Gelfand?) is that, under the same conditions, f(t) = e^{g(t)}, where g(t) is also in W. My sources quote the results but not the proofs. My question: can anyone provide (i) a proof of either assertion or (ii) good (english) refs. which contains the same? alan === Subject: Re: Questions on the Wiener algebra Originator: grubb@lola >It is a famous result of Wiener that: >If f(t) has no zeros on the unit circle, >then 1/f(t) is also in W. The standard proof of this uses Banach algebras. The multiplicative linear functionals on W are easily seen to be point evalutations. Hence, the spectrum of an element is just the range of that element thought of as a function on the circle. If the spectrum doesn't contain zero, the element is invertible in the algebra (by definition of the spectrum). Look at ÔAn Introduction to Harmonic Analysis' by Yitzak Katznelson in the chapter on Commutative Banach algebras. >Also, another known result (perhaps Gelfand?) is that, under the same >conditions, >f(t) = e^{g(t)}, where g(t) is also in W. Hmmm...what about f(t)=t? This has no zero on the circle but is not e^g(t) for any continuous function g on the circle. On the other hand, if f(t) has a continuous logarithm on the circle, it has one in W. Again, this is a specific case of a result for commutative Banach algebras, the Arens-Royden theorem. See ÔBanach Algebras and Several Complex Variables' by john Wermer. >My sources quote the results but not the proofs. >My question: can anyone provide >(i) a proof of either assertion or >(ii) good (english) refs. which contains the same? --Dan Grubb === Subject: Re: Questions on the Wiener algebra >It is a famous result of Wiener that: >If f(t) has no zeros on the unit circle, >then 1/f(t) is also in W. > The standard proof of this uses Banach algebras. The multiplicative > linear functionals on W are easily seen to be point evalutations. > Hence, the spectrum of an element is just the range of that element > thought of as a function on the circle. If the spectrum doesn't contain > zero, the element is invertible in the algebra (by definition of the > spectrum). > Look at ÔAn Introduction to Harmonic Analysis' by Yitzak Katznelson > in the chapter on Commutative Banach algebras. >Also, another known result (perhaps Gelfand?) is that, under the same >conditions, >f(t) = e^{g(t)}, where g(t) is also in W. > Hmmm...what about f(t)=t? This has no zero on the circle but is not > e^g(t) for any continuous function g on the circle. On the other hand, > if f(t) has a continuous logarithm on the circle, it has one in W. Again, > this is a specific case of a result for commutative Banach algebras, the > Arens-Royden theorem. See ÔBanach Algebras and Several Complex Variables' > by john Wermer. Yes, Dan, you're right -- I forgot to add the required continuous log. condition (winding number = 0) for the second result. alan === Subject: Re: Questions on the Wiener algebra |My sources quote the results but not the proofs. |My question: can anyone provide | |(i) a proof of either assertion or |(ii) good (english) refs. which contains the same? I gather the original proof is somewhat technical. At least the original Wiener lemma is proven nicely in: Thierry Coquand and Gabriel Stolzenberg. The Wiener lemma and certain of its generalizations. Bull. Amer. Math. Soc. (N.S.) 24 (1991), no. 1, 1-9.(42B05) Keith Ramsay === Subject: Analytic interpolation I have an infinite family of sequences a_{m,n}. Three, (a_{m,5},a_{m,6} and a_{m,7}, m in N), are shown here: http://users.forthnet.gr/ath/jgal/math/figs/seq.gif All the sequences eventually become constant (to the right). I need to interpolate for m in N, using real analytic functions in the region between the first term and the first constant term (that's the S shaped region). I also have a recursive relationship for all the terms. (A recursive relationship for ALL terms of ALL sequences). Questions: 1) Do there exist real analytic functions that can interpolate them? (In the sequences only the initial terms are different. All the rest are given by a recursion). (The sequences seem to beg for some sort of variant of the exp function, but all my numerical analysis books are away from home). I know that polynomials can be used, but I am rather fond of the exp function. 2) If anyone can provide some online links and/or references, I'd be greatful. -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: Uncountable sets in CZF? |The Platonic attitude that there is exactly one set R of real numbers is |not fully appropriate. I disagree. Even while wearing my constructivist hat I still disagree. If you're going to adopt the attitude that there isn't one, which is up to you, you still are left with the fact that as far as ZF is concerned, there exists a unique set R of real numbers. So if you're in the middle of doing some reasoning inside ZF, your reasoning should respect that. If you are not in the middle of doing some reasoning based on ZF, then a few words about what you are doing instead might be in order. I think trying to think of the reals as not being unique can easily lead to confusion if you don't carefully partition such beliefs off from the mathematics one is trying to do, if it's in a system that metatheory in which the unique existence of R is not a theorem, but the results from model theory you want to use are, it should be named. |One cannot assume that there is exactly one set R |that must be common to all models of ZF, Of course. |or that models of ZF are ßawed |on the basis that R can be altered by taking a generic extension. It depends on what you mean by ßawed. Certainly it's possible for a model of ZF to have the real line as its R. (If you like, think of it as a theorem of ZF. ZF doesn't have a theorem saying there are standard models of ZF, but adding an assumption such as the existence of a measurable cardinal, then there exists a model M of ZF where the set of reals relative to M is R.) Such a model might be considered more perfect. |For myself, the easiest way to understand generic extensions is through |Boolean-valued models. Nice of you to go to the trouble of explaining that in your posting. It looks like you tersely covered a good fraction of the key points of the development of boolean-valued model as it's done in textbooks. A Boolean-valued model M^G can be arranged to behave the way you describe; we can wind up with an element f in M^G such that |f:Z^->R^ is a bijection| = 1, where Z^ and R^ are the counterparts of Z and R in M^G, although R^ is not what M^G considers to be R. This f is essentially a function from ZxR to G. The most obvious choice of f in this particular case is symmetrical under permutations of Z and permutations of R (which both act on G). The model can get away with thinking of for each n, there exists an r such that f(n)=r as having a truth-value of 1 because the value of there exists an r such that f(n)=r is the least upper bound of the values of f(n)=r ranging over all r in R. I think the problem arises when you try to apply the equivalence of the Boolean-valued approach (which doesn't need to construct an actual model of ZF) with a forcing approach that does. |A set G (not necessarily an element of V) is called a generic set if | | (1) G is a subset of F (i.e. all the elements of G are forcing | conditions), | | (2) if p is an element of G, then any condition q weaker than p | is an element of G, | | (3) if p and q are elements of G, then p and q are compatible, | | (4) if D is an element of V and D is a dense set of conditions, | then G and D have nonempty intersection, i.e. G intersects | every dense subset of F. | |For all x in V^B, define i_G(x) by induction on x by | | i_G(x) = { i_G(y) : y in dom(x), G intersects x(y) }, | |and define V[G] to be the class with elements i_G(x) for x in V^B. I wish you hadn't chosen to use the letter V for your model, since V is the usual notation for the class of all sets. It requires a little care to apply the construction to a proper class, but it can be done. One useful trick is to realize that when we're working with proper classes, we're really dealing with the predicates that define them. So with V, really we're working with the predicate that's always true, like 0=0. If we apply this process to V, V^G can be a boolean-valued model that fails to satisfy various statements true in V. If we reduce back to a 0-1-valued model using such a mapping i_G, though, we go right back to V. It doesn't leave us with any new elements than what we started with. In what sense are none of the elements new? In the sense that they are in V, i.e. that 0=0. That's just a matter of definition. It's not a metaphysical claim about the real world of sets or anything. The whole discussion still can be regarded as an informal version of an argument to be formalized in the language of ZF. |Then V[G] is the smallest model of ZFC which contains V as a submodel and |contains G as an element. V[G] has the same ordinals as V, and the same |constructible sets as V. In the case where we use V to mean the class of all sets, G already was an element of V. |For a formula phi(x_1,...,x_k) with free variables x_1,...,x_k, and for |elements y_1,...,y_k, of V^B, V[G] satisfies phi(i_G(y_1),...,i_G(y_k)) |iff there exists a condition p in G such that p forces phi(y_1,...,y_k). In our main example, all conditions force f to be a bijection, but i_G(R^)=R and i_G(Z^)=Z, and there is no bijection between Z and R. I believe there is a problem with preserving quantifiers. When we are evaluating a formula with a quantifier in V^B, we take the least upper bound (for existential quantifiers) or the greatest lower bound (for universal quantifiers). For this to be preserved under reduction to a 0-1-valued model, it needs to be the case that when each member of the set of elements of B whose greatest lower bound we are taking is in the filter, then so is their greatest lower bound. |Define G' in V^B by dom(G') = { hat p : p in F }, and G'(p) = e(p) for |all conditions p in F, then ||G' is a generic set|| = 1, and i_G(G') = G. |It follows that if ZFC is consistent, then the existence of a generic set |is consistent with ZFC. | |The connection between generic ultrafilters and generic sets is given as |follows. | |If G is a generic set, then | | U = { A a regular open set in F : A intersects G } | |is the corresponding generic ultrafilter, and V[U] = V[G]. | |If U is a generic ultrafilter, then | | G = { p in F : e(p) in U } | |is the corresponding generic set, and V[G] = V[U]. I think it's only the case that we get back a 0-1-valued model of ZF satisfying the sentences whose values in V^B lie in U, if some other condition is satisfied. Perhaps we need to use Lowenheim-Skolem to cut down to a countable model, like Cohen did in his original forcing papers. Keith Ramsay === Subject: Re: Uncountable sets in CZF? |As others have said, it is hard to reconcile Platonism and first-order |logic. No, just with first-order-itis, in which you go around thinking that the things you can say in first-order language about a structure are all the things you can say about it. Keith Ramsay === Subject: Re: Uncountable sets in CZF? >This is essentially the same as my beliefs. If a bijection can be >constructed between N in R[V] in some model V', then given that the >size of N, and of R[V], has not been changed, that would seem to say >that R[V] is countable, only unprovably so in V. Since we know that >the real R is not countable, that would seem to say that ZFC is >incomplete, or wrong. > >But that is already well known for a long time. >There is no way to characterize Ôthe real model of ZFC' (whatever that >means) with new axioms or so. It was once known as the Skolem-Loewenheim >paradox. > > > I see, ZFC is incomplete. Is this related to Godel's theorem? > The Skolem-Loewenheim theorem/paradox is older (early 1920s). I meant, is it proved by the same type of argument? If it is, it confirms my sentiment that logic is incapable of talking about the uncountable. Andrew Usher === Subject: Re: Uncountable sets in CZF? >This is essentially the same as my beliefs. If a bijection can be >constructed between N in R[V] in some model V', then given that the >size of N, and of R[V], has not been changed, that would seem to say >that R[V] is countable, only unprovably so in V. Since we know that >the real R is not countable, that would seem to say that ZFC is >incomplete, or wrong. > >But that is already well known for a long time. >There is no way to characterize Ôthe real model of ZFC' (whatever that >means) with new axioms or so. It was once known as the Skolem-Loewenheim >paradox. >I see, ZFC is incomplete. Is this related to Godel's theorem? >The Skolem-Loewenheim theorem/paradox is older (early 1920s). > I meant, is it proved by the same type of argument? No, not at all. In the form any consistent theory has a countable model, it shows more resemblance with the ordinary completeness theorem of FOL. If it is, it > confirms my sentiment that logic is incapable of talking about the > uncountable. FOL is incapable of distinguishing between uncountable and countable models. That doesn't mean, however, that there exist no FOL-theories that are capable of talking Ôabout' the uncountable, in the way ZFC does. (FYI: your Ôsentiment' was well known in the 1920s, but is currently no longer generally held. The S-L Ôparadox' simply is no longer felt to be paradoxical anymore. Set-domain models of ZFC are only interesting in so far as they allow one to establish the consistency of certain theories (independence of certain assumptions from others). They are no realistic models anyway, as the Ôreal' universe of sets has a proper class for its domain.) -- Herman Jurjus === Subject: Re: Uncountable sets in CZF? > >By plain set theory I meant theory where the only primary objects >were sets as opposed to a theory where numbers are primary objects. >ZF and ZFC are therefore what you call plain set theories. In ZF and >ZFC, numbers are defined AS SETS. I wonder where you got the erroneous >idea that numbers were primary objects in ZF and ZFC??? >A natural number is defined to be either the empty set or a successor >ordinal whose elements are all either empty or successor ordinals, a >set n is a natural number if >n is empty or [n is a successor ordinal and for all m in n (m is empty >or m is a successor ordinal)]. >Equivalently, a natural number is an element of the smallest limit >ordinal. Equivalently, a natural number is an element of the unique smallest inductive set, where a set x is called inductive if the empty set is an element of x and for all y in x exists z in x for all t (t in z iff (t in y or t = y)). Sufficient axioms for ZF are: (1) exists x (x in x) - this asserts the existence of at least one set, so it can be called the Axiom of Existence; (2) for all x for all y (for all z (z in x <=> z in y) => x = y); (3) for all x exists y for all z (exists t (z in t and t in x) => z in y); (4) for all x exists y for all z (for all t (t in z => t in x) => z in y); (5) for any formula , P(u,v), in ZF, and for arbitrary values for all free variables in P(u,v), except for u and v, for all u for all v for all w ((P(u,v) and P(u,w)) => v = w) => for all x exists y for all v (v in y <=> exists u (u in x and P(u,v))); (6) exists x (empty set in x and for all y in x exists z in x for all t (t in z <=> (t in y or t = y))) - Axioms 1 and 5 guarantee the existence of an empty set, and Axiom 2 guarantees its uniqueness; (7) for all x (exists y (y in x) => exists y in x for all z not (z in x and z in y)). For ZFC, this is augmented by another Axiom: (8) for all x ([exists y (y in x) and for all y in x [exists t (t in y) and for all z in x (y = z or for all t not (t in y and t in z))]] => exists u for all y in x exists v for all t ((t in y and t in u) <=> t = v)). David ----- === Subject: Re: Uncountable sets in CZF? Regarding IZF, [...] | But most especially, there can't be a surjection from the whole of the | naturals to the reals. |Cantor-Schroeder-Bernstein: it works both ways. | |What that means is that one of the reasons that people call the reals |uncountable is because they've figured out a bijection between the |reals and the powerset of the naturals, thus they reason that there |are no bijections between the reals and the naturals, because |Cantor-Schroeder-Bernstein says the existence of a surjection either |way between two sets is proof of the existence of a bijection between |those two sets. I thought the theorem was about sets A and B where there exist injections from A to B and from B to A. I don't know why one would use either result (either Cantor-Bernstein or the analogous result about surjections) to argue against the existence of a bijection between the natural numbers and the reals. It's so much more direct to just use one of the two proofs Cantor had to show that there is no surjection from the naturals to the reals. I see no point in complicating the issue. |That is to say, the existence of a surjection from A to B and from B |to A implies that A and B are equivalent, and as well from A to B to C |and C to B to A through composition. Moreover, in the context of IZF, you can't use Cantor-Schroeder- Bernstein; it's not a theorem of IZF. The bijection described in the proof has to be defined by cases. As often is the case, when a function has been defined by cases, the law of excluded middle is implicitly invoked to claim that the domain of the function is the whole original domain. That is, if X is a subset of A and we say f=g on X and f=h on A-X, then the domain of f is the set of elements of A that either are members of X or are not members of X. And of course IZF does not have the law of excluded middle. |That implies it is not a mathematical fact and to promote the other |view as gospel, immutable, written in stone, etcetera, would thus be |deceitful. I'm rather angered that you would suggest the acceptance |of a mathematical falsehood as mathematical fact. Wouldn't you be? Not always, no. I already have plenty of experience with seeing people who are confused like you are offering nonsense as mathematical fact. If I were angry about it, I would be angry quite a lot of the time. Of course, it's worse when the person who is confused is so confused, like you are, that he not only thinks that what he's saying is correct, but he's really *sure* of its being correct, and that hence everyone who disagrees with it is wrong. But worse still are the people like you who not only are confused and sure they're right, but angry at other people who are sure they're right too. |Reexamine the claim about there being only one or none proper classes. Nobody claims there's exactly one proper class. The ordinals are a proper class, and so is class of all sets. They may believe that there is no such thing as a proper class, but if they agree there are any, they agree there are many. I think you're giving a confused version of the claim that any two proper classes can be put into one-to-one correspondence. | Consider why that demands dual representation of the ur-element as |both zero and Ord, regardless of whether Ord is N. (Steve, infinite |sets are equivalent.) | |Unitize the analog! It's better to have a tool, even a primitive |tool, to measure sparse points of the continuum, than none, and bar |further consideration of the matter by fiat. | |Why don't you consider that the direct sum of infinitely many copies |of N is zero? I can't make any sense out of this. |union of countable sets, You should try to write more precisely (not that I expect you to), especially when you're allegedly stating what someone else said. This sounds like you're describing him as saying that each uncountable set is a countable union of countable sets, which he surely didn't something about its being consistent with ZF for there to exist an uncountable set that's a countable union of countable sets. |to that the constructivist powerset mapping |argument is shoddy or fails, How about getting rid of some of the shoddy that you keep distributing? |to these metatheoretical hand-wavings of |yes and no and ignorance fo the excluded middle, we've seen some of |what I would call progress in the state of discussion of these issues |here on sci.math, an open public forum where the participants are |often professional mathematicians and as such loathe to promote |untruth, particularly in the stark, concrete world of mathematics, |that is already a paradise with no need for the transfinite: |inconsistent and thus hellish. The only appearance of inconsistency is because you're confused! [...] |You're intelligent and understand the basic tenets of rational |discourse, and by now you're probably familiar with the rather few |arguments of transfinite set theory: Set theory is a rather huge subject. Keith Ramsay === Subject: Re: Uncountable sets in CZF? Discussion, linux) > |Reexamine the claim about there being only one or none proper classes. > Nobody claims there's exactly one proper class. Ross does. -- But remember, as long as one human being follows the rules of mathematics, then mathematics as a human discipline survives. Right now I'm that one human being, so mathematics survives. -- James S. Harris === Subject: Re: Uncountable sets in CZF? David, I want to compliment you, those are some very informative posts and that post should be recommended reading. There are notions of theories where numbers are primary objects in the theory and are not sets. Keith presented a statement that he could map a proper subset of the naturals bijectively to the reals. What's the deal with that? That reminds me that the direct sum of infinitely many copies of N is empty, but not, by exceptional definition. Then again having {} as an element of a set implies regularity by many naive constructions of that predicate. I think the ur-element is zero, and sometimes infinity, and one. We've probably all had calculus instruction, or at least the pre-calculus instruction about limits and the definition of a of the antiderivative, the integral. Familiarity with a strict and perhaps overly strict concept of limit is a given, where that is a relatively modern response to the vagaries of infinitesimals of Newton and Leibniz' functional, useful, empirical result driving infinitesimal analysis: the integral calculus. An even more modern response arose last century in the consideration of the infinitesimals and the establishment of definitions to allow the consideration of hyperreals. The hyperreals as a set contain no elements that are not elements of the reals, and they do include infinitesimals. Now before everyone hauls out umpteen proofs that 0.999... = 1, they are perhaps true in a similar manner as to how some geometrical results are true (sound, valid, correct), in the Euclidean geometry, and not true. McAnally, I don't know how long you lurked on sci.math before beginning your voluminous well-informed postings, you may well know that I have since shortly after discovering this unmediated forum argued the notion that infinite sets are equivalent, ignorantly, naively. My response to the declaration that there were not mappings from the naturals to the reals was the definition of the Natural/Unit Equivalency Function. It's defined in a way that the domain is the naturals and the range is the unit interval of reals. EF(0) = 0 lim EF(oo) = 1 EF(n+1) > EF(n) lim (EF(n+1)-EF(n)) = 0 People ask then what is EF(1) and I say iota and they say what is (EF(0)+EF(1))/2 and I say undefined and then (EF(0)+EF(2))/2 = EF(1) and then (EF(0)+EF(n))/m is defined only when m divides into n. Anyways that leads to the consideration of iota, the least positive real infinitesimal, with properties of an x-transcendental real, and the vague fugue. Claims are made that no function bijects between the naturals and reals for a) the antidiagonal argument, b) the nested intervals argument, and c) Cantor-Bernstein transitivity. The antidiagonal argument and nested intervals argument are shown to not apply, for dual representation and enforced list order, and for functions with properties similar to EF in monotonically mapping. The Cantor-Schroeder-Bernstein theorem(s), about surjection either way implying a bijection, presents an inconsistency in mapping between the reals and naturals and reals and powerset of the naturals. To that end I described a model (obviously part of the theory in disguise) of ubiquitous ordinals or naturals. Another alternate consideration even to that has the ur-element represented as zero and infinity. In the model of ubiquitous ordinals, the order type is the successor is the powerset, and a a set X and P(X) f(P(X)) = f(X)+1, and f(x)=x+1 for x E X is a bijective mapping, with S = {}, which while being zero when zero={} is also the ur-element and equal to X+1. About as looking from outside, metatheories/metamodels, extensions, etcetera, you have in them a set N, the set of all naturals integers, in one of those models mapping bijectively to a set R, the set of all real numbers. Ross F. === Subject: Re: Uncountable sets in CZF? |Keith presented a statement that he could map a proper subset of the |naturals bijectively to the reals. What's the deal with that? Do you think you're quoting me at all accurately here? You do agree that you should attempt to, don't you? Keith Ramsay === Subject: Re: Uncountable sets in CZF? > |Keith presented a statement that he could map a proper subset of the > |naturals bijectively to the reals. What's the deal with that? > Do you think you're quoting me at all accurately here? You do agree > that you should attempt to, don't you? > Keith Ramsay Yes, I think you claimed there was a surjection from some proper subset of N onto R, and through Cantor-Schroeder-Bernstein as there is a trivial surjection from R onto any subset of N there is a bijection. You say specifically that it doesn't follow that there is a bijection. Yet, it necessarily does, until you present some disproof or negation of the Cantor-Schroeder-Bernstein theorem in that context. I don't base my arguments (that the reals and naturals are equivalent) upon what you said, I haven't seen your explanation of a surjection from some proper subset of the naturals to the reals, and I have my own explanations for why the naturals can biject with some proper subset of the reals. Apparently, so do you. I think that's good, and progress. Ross F. === Subject: Re: Uncountable sets in CZF? |> |Keith presented a statement that he could map a proper subset of the |> |naturals bijectively to the reals. What's the deal with that? |> Do you think you're quoting me at all accurately here? You do agree |> that you should attempt to, don't you? | |Yes, I think you claimed there was a surjection from some proper |subset of N onto R, and through Cantor-Schroeder-Bernstein as there is |a trivial surjection from R onto any subset of N there is a bijection. Well I didn't, and it doesn't. |I've seen proofs that it's consistent with various constructive formal |systems to assume that each function from N to N is computable. |If I'm not mistaken the same sort of proof works for IZF. | |Assuming that each function from N to N is computable has a |number of consequences that are liable to be unfamiliar. Since it's |contradictory with the law of excluded middle, you have to be ready |to work with intuitionist logic. | |Here, though, the relevant consequence is that the function mapping |those Turing machines that compute real numbers to the real numbers |they compute would be a SURJECTION from a SUBSET of the natural |numbers to the real numbers. Please, now that we've gone through this |more than once, either don't try to quote the result, or remember these |key details. It doesn't follow that there's a bijection. Do not quote people as claiming things that they expressly inform you that they are not saying, just because you think the thing they refused to say follows from what they did say. Even if the person is simply confused, it's not accurate to describe them as having said they agree with your conclusion. There's no reason why you can't say in these cases, X says that Y In this case, I specifically asked you not to try again to quote this result without getting the details straight. 1. I didn't say these consequences of the assumption that each function from N to N is computable were *true*. All I've ever said is that it is *consistent with IZF* that they are true. There's a big difference between claiming something is true and claiming that it's consistent to assume it. 2. I didn't say it was consistent for there to be a bijection, just a surjection from the subset of N. 3. Cantor-Bernstein doesn't say that if there are surjections from A to B and from B to A, then there's a bijection between A and B. It says it for injections. (Cantor-Bernstein isn't a theorem of IZF. It doesn't require the axiom of choice, but it does require the law of excluded middle, which we had to drop in order to work with the assumption that each function from N to N is recursive.) 4. After a bit of searching, I found a paper on the web about the dual Cantor-Bernstein theorem, which is the statement that if there are surjections from A to B and from B to A, then there's a bijection between A and B. This is the result you're trying to apply in the context of IZF. It can't be proven without the axiom of choice, let alone without using the law of excluded middle, so it's very far from being a theorem of IZF. The assumption that each function from N to N is computable is of course contradictory with it. 5. The existence of a surjection from R to Z isn't a theorem of IZF! The existence of non-constant functions from R to Z isn't a theorem of IZF! You probably are thinking of some function like the greatest integer function [x] = max {n in Z: n<=x}. But this is only defined for all reals if we are able to say that for each real x, either x<1 or x>=1 (and likewise x=n for each integer n). That's not a constructive result. If you are given a real number x, in the form of a sequence of rational approximations, there's nothing to guarantee that you can tell whether x<1 or not. (If x is not <1, then x>=1.) |You say specifically that it doesn't follow that there is a |bijection. Yet, it necessarily does, until you present some disproof |or negation of the Cantor-Schroeder-Bernstein theorem in that context. No, even if you were right, you're not entitled to pretend I agree with you until I show you what's wrong with it. It's up to you to make sure that you're right, not up to other people to correct you when you're mistaken. |I don't base my arguments (that the reals and naturals are equivalent) |upon what you said, I haven't seen your explanation of a surjection |from some proper subset of the naturals to the reals, I said it was a consequence, in IZF, of assuming that all functions from N to N are computable. If each function from N to N is computable, then each real is computable. Each computable real can be represented by a machine that computes a sequence of approximations to it. So the mapping from those integers that represent machines that compute reals, to the reals that they compute, is from a subset of the natural numbers onto the computable reals. It's not one-to-one, because many different machines compute the same real number. Even when I'm thinking in constructive terms, I don't regard the statement that each function from N to N is computable as a fact. (And nonconstructively, it's false, of course.) It's just something that can be consistently assumed, much like the continuum hypothesis. In the Bishop tradition of constructive mathematics, such statements are used to help us know what to try to prove, and what not to bother trying to prove. |and I have my |own explanations for why the naturals can biject with some proper |subset of the reals. Well that they can do! But I think you mean the reverse, which is a different story. If there were a bijection between the reals and a subset of the natural numbers, then equality between reals would be decidable, i.e. either r=s or r<>s for each r and s. That's perfectly natural for classical mathematics, following from the law of excluded middle, but isn't appropriate constructively. Classically, then this bijection is plainly impossible; constructively it's at least not appropriate. |Apparently, so do you. I think that's good, and progress. You're mistaken. Keith Ramsay P.S. Cantor's theorem is a fact of mathematics! === Subject: Re: Uncountable sets in CZF? > ... > Do not quote people as claiming things that they expressly inform > you that they are not saying, just because you think the thing they > refused to say follows from what they did say. Even if the person is > simply confused, it's not accurate to describe them as having said > they agree with your conclusion. > ... > No, even if you were right, you're not entitled to pretend I agree > with you until I show you what's wrong with it. It's up to you to > make sure that you're right, not up to other people to correct you > when you're mistaken. > ... Hi Keith, That helps to clear up that misunderstanding. It also is quite obfuscatory. So now I will agree that you definitely claim that no set bijects with its own powerset. Instead, you just claim that it's consistent for the naturals to surject onto the reals in this IZF or some other framework with no excluded middle, and that if you assume you can compute those functions NxN then you claim that there is a surjection from N onto R. For a surjection from N onto R, that is an injection from R into N. By your notion, Cantor-Bernstein demands excluded middle, thus that while it might be consistent with IZF or what-have-you, as it (C-B) is probably pretty definitely not inconsistent, you claim that it is not itself a theorem but would demand something else to make the same claim that a surjection either way or an injection either way between two sets implies bijection. Indeed, I think the point to make clear is that the set of all reals numbers is the set of all real numbers, and the set of all naturals numbers or finite ordinals is the set of all natural numbers, and those sets contain the same elements in each of these models. The set R of real numbers is totally ordered and obeys trichotomy, for any two real numbers x and y it is exactly one of that xy. This is where parallel lines do not intersect, etcetera, yet as well where they do and the principle value of a real number is the value. So you are claiming that it is not inconsistent in, say, IZF, for there to be a surjection from N onto R because that doesn't necessarily imply a bijection because Cantor-Bernstein is not a theorem without excluded middle. Now I'm trying to think about each function from N to N vis-a-vis NxN and NxNxNx.... For each function F from N to N, for y=F(x) (x,y) is an element of NxN. That's pretty simple, just saying that each possible 2-tuple of naturals is contained with NxN. About the set of all functions from N to N, then that would be Nx{0,1}xN, no? I think it is obvious that it is a subset of NxNxN. Plainly to represent a single function it is a proper subset of NxN. How about all the functions from NxN to NxN? What are those? Anyways we agree to disagree, I wish you l in unitizing the analog, figuring out why there could ever or never be another proper class, if there is one, explaining why N<->R is consistent, why a set containing only all the real numbers is the set of real numbers, why all real numbers are computable, etcetera. Ross F. === Subject: Re: Differential Equations and Boundary Value Problems... > I just found my old college DiffEQ text book. Its called Differential > Equations and Boundary Value Problems. I am just curious but could > someone explain the difference between ordinary diffeqs and boundary > value problems? You have to specify the boundary conditions to get a particular solution. It is the constant of integration. e.g., F = ma = m d^2(x)/dt^2 gives the motion of a ptl. with mass m, the boundary conditions are the initial position x(t_o) = x_o and initial velocity dx/dt(t_o) = v_o. === === Subject: Re: Do you know the counter example? > It is false. > Next would be a modification of the problem. > If A is an open or closed subset of X = R^n, then the interior of the > boundary of A is null set. > I have been doing battle with this problem trying to prove it but I cannot > make the proof work so I am wondering if its even true. > Let X be a topological space and A is a Sub set of X. Show that, > The interior of the boundary of A is equal to null set. . ( > B(A)^o = O/ ) > It feels as if it should be true but i just cannot show it so any ideas > would be welcome > stephen Is this related to b^2 = 0 from differential geometry--the boundary of a boundary = 0 ? === Subject: Re: Do you know the counter example? > It is false. > Next would be a modification of the problem. > If A is an open or closed subset of X = R^n, then the interior of the > boundary of A is null set. >Is this related to b^2 = 0 from differential geometry--the boundary of >boundary = 0 ? Er, no... the boundary of the boundary of an open or closed set A is the boundary of A. 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: Speculative, but at least interesting by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i87H0Ot04339;