mm-334 === So I am at a very akward stage now: if give me undergraduate math toread, I>feel too easy and boring, if give me difficult math to read as those in>Information Theory Transactions, I feel dizzy...Can anybody recommend some procedures/reference books to give me atreatment>for my current sickness? Can anybody give a booklist that any big guy inthe>field would recommend as a must-read as a systematic treatment to mathfor>researchers in math-rela engineering/science area? Graham, Knuth, and Patashnik's CONCRETE MATHEMATICS comes to mind.> Also, Oppenheim's SIGNALS AND SYSTEMS. --gregbo> gds at best dot comWe had studied Oppenheim's book in undergraduate, maybe I did not study itwell or fully understand it... :=)But Knuth's book seems a good one, I will go and take a look!Thank you!=== === Subject: : Re: Please advise! what kind of training I am lackfor math? I must say that your problem is very familiar to me too! So don't be> disheartened. Me too. Early on, math came with great difficulty to me> also and being in an electrical engineering program I was> constantly challenged. It takes perserverence but the> facility comes with usage. I am amazed today at how easily> I can move into a new mathematical area and how quickly it> makes sense. I'da never thought. Sometimes difficulty is a sign that you need total> understanding and perspective to carry on and that can make> it seem harder than it might to someone who can settle for> its utility. You might think you are in the latter school> when you are in fact in the former. That's a good thing in> the long run. The idea that this is simply language is a particularly powerful idea.The> problem is that the language part you're having trouble with is thatit's> shorthand! So you try to read at your own normal rate and theshorthand> just doesn't allow that - it's too compact. Yet, it's still languagewith a> few new terms thrown in. But with a big difference. To me, visualization was an> essential component in learning to swim in that ocean. I'd> never make a theoretical or mathematical physicist. Finally, understand this: many papers are not well written for clarityor> to be understood. So, in addition to there being issues of language and> shorthand, some stuff is just crap from a presentation point of view and> maybe from a real content point of view as well. Skip those papersunless> someone tells you they're really important. On the other hand, much of my learning was in puzzling> through exactly such papers and filling in the gaps in them> and in my own knowledge to come to an understanding of the> topic that was sufficient for implementation or variation.> I'd say don't skip a paper that you feel you should be able> to understand but don't. Rather, buckle down and work it> through. I can think of some papers that I spent months> coming back to, setting aside, looking at things that might> clarify them until the final aha moment that made it all> worth it.I think we agree. My point was in being choosy about which papers to spendtime on.Fred=== === Subject: : Re: Please advise! what kind of training I am lackfor math? On the other hand, much of my learning was in puzzling> through exactly such papers and filling in the gaps in them> and in my own knowledge to come to an understanding of the> topic that was sufficient for implementation or variation.> I'd say don't skip a paper that you feel you should be able> to understand but don't. Rather, buckle down and work it> through. I can think of some papers that I spent months> coming back to, setting aside, looking at things that might> clarify them until the final aha moment that made it all> worth it.I think we agree. My point was in being choosy about which papers to spend> time on.For sure. I'm just recommending that obscurity not be acriterion in and of itself. It can be difficult toanticipate that there may be a nugget at the end of thetrail but one develops a sense of it.I'm looking at one such now: http://www.atvs.diac.upm.es/publicaciones/docs/Bot00a.pdfI've got some serious work to do to know whether thiscontains an answer to my current problem but my sense isthat it does. The question now is whether it may be tooconcise in its discussion of the noise reduction part of theapproach that I won't be able to use it even when I'veaccumula the background. I know several authors that goto pains to develop a formulation of the problem to thenjust tell you that they've solved it. Bob-- Things should be described as simply as possible, but nosimpler. A. Einstein=== === Subject: : Re: Please advise! what kind of training I am lackfor math?> I'm looking at one such now: http://www.atvs.diac.upm.es/publicaciones/docs/Bot00a.pdf>Hi Bob,I have got the feeling that 4 page paper can just tell me what the authorshave done and it leaves me to journal paper to fully understand the author'smethod most of the time. I don't the other people have the same feeling as Ido...-Walala=== === Subject: : Re: Please advise! what kind of training I am lackfor math? (snipped)I'm looking at one such now: http://www.atvs.diac.upm.es/publicaciones/docs/Bot00a.pdf> (snipped)Hi Bob, just for giggles, I peeked at the first page of that paper. I was puzzled about how the authors could say, in the abstract, that a signal is decomposed in its allpass and minimum-phase components ...What the heck is the allpass component of a signal? You don't have to answer that for me Bob. I realize it's just careless use of terminology. They should have said: a system's *frequency response* is decomposed into its allpass and minimum-phase components ...Now if the rest of the text is carelessly-written, your job gets *much* tougher.Hope ya' find your nugget of gold.Good Luck,[-Rick-]=== === Subject: : Re: Please advise! what kind of training I am lackfor math?> (snipped)>I'm looking at one such now:>> http://www.atvs.diac.upm.es/publicaciones/docs/Bot00a.pdf (snipped)Hi Bob,> just for giggles, I peeked at the first page > of that paper. I was puzzled about how the > authors could say, in the abstract, that a > signal is decomposed in its allpass> and minimum-phase components ...What the heck is the allpass component of > a signal? You don't have to answer that for > me Bob. I realize it's just careless use of > terminology. They should have said:> a system's *frequency response* is > decomposed into its allpass> and minimum-phase components ...Now if the rest of the text is carelessly-> written, your job gets *much* tougher.Hope ya' find your nugget of gold.Good Luck,> [-Rick-]And compare their to to your into. Peeple who don't rite so good arehard to unnerstand.Jerry-- Engineering is the art of making what you want from things you can get.[OSlash][OSlash][OSlash][OSlash]=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityExpires: 28 days>I am pretty convinced now that source dependency would make virtually no> difference to the operation of a sagnac.>>Pretty convinced indeed. Of course you are.>>No experimental evidence can shatter your blind irrational>>faith in source dependency.>>You can twist anything if you have enough faith.>>And you do.>>Paul Are you suggesting that light DOES NOT move at c relative to its source? No. What everybody has been suggesting to moron > quantumoids for the last 110 yeais that > *photons* move at C relative to their source.> You freaking moronic, born-again followers> of the true eight-fold path to New York stupid.Why Zed?Henri Wilson. www.users.bigpond.com/hewn/index.htm=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityExpires: 28 days>Sorry Henri, that's wrong. Only _inertial_ instruments>>measure the speed as c according to SR.>> Yes. OK. You have to include the great SRian escape route. inertial.Try this experiment: stand up, turn round then sit down.>Now consider a flash of light moving away from a recent>supernova in the Andromeda galaxy. It is 2 million light>years away so in the frame that co-rota with you as you>turned, that light moved 2*pi*R or a distance of 13 million>light years in just a few seconds. Obviously the speed of>light in a rotating frame cannot possibly be c.I doubt if the one-way speed of light is ever c.But anyway, you have put forward a good case for scrapping rotating framesaltogether. I certainly agree with that. >> In this case, the speed of the instrument is the same as the speed of>the> source/receiver.> Analysis using the co-rotating frame shows that fringe shift should not>>occur> according to SR.>>A co-rotating instrument is not inertial. Do the analysis>>properly and you will find it predicts the observed shift.>> I said a co-rotating frame. You were the one who first introduced such to>this>> argument remember.No, you replied to Paul Anderson:>>I am stating the fact that in a ring laser, the speed of light is>isotropic in>>the inertial frame, EVEN when the source (lasing gas) is moving in this>frame.>>This is proven by the mere fact that the ring laser gyro works.>>This falsifies the ballistic light theory.>> SR says that the speed of light is c relative to its source. Therefore if>a>> ring laser is analysed in the co-rotating frame, no fringe shifts should>be>> expec according to SR.>> Sagnac clearly refutes SR.>SR does NOT imply that the speed of light is c relative to>its source in the co-rotating frame, only that it is isotropic>in the inertial frame as Paul sta.The point I made to Paul was that if the 'co-rotaing frame' argument was usedagainst source dependency then it would also apply to SR since SR is based onsource dependency.> I have done the analysis and it shows that there will be a fringe shift>with>> source dependency.That analysis is incorrect, see my other reply.> That means that there will also be one according to SR which>> clearly states that light moves at c relative to its source.You are right, there will be a shift according to SR but>you have been arguing the opposite.I was being a little facetious.The point is, the SR explanation is the same as source dependency. > Source and observer are the same in a ring gyro.Both are rotating and accelerating, hardly inertial.Well if the whole thing is not inertial, SR cannot apply at all - so how thehell can you even suggest that SR can provide any kind of explanation forsagnac effect when it is entirely beyond its scope?>In the 'lab' frame SR predicts the speed of the light>>will be c while the ballistic theory predicts it will be>> c + v' = v * cos(a) + c * cos(b)>> Not so. If the source is moving in the direction v, then light speed in>the>> horizontal direction will be c+[v*cos(a)]>Obviously they are not equivalent.Either way, SR says it is c in the inertial frame so they>are not equivalent. See my other post for the details.I don't care one iota what SR says. Why should I. None of it is proven.>> Are you trying to tell me that it wont detect rotation?>>No, I said that SR predicts the time taken for the beams>>to travel from source to detector is not the same in the>>two directions .. and varies with the speed of rotation.>>so I said SR predicts that it _will_ detect the rotation.>> You also explained the R 'explanation' by refering to a 'lab frame'.>> I have cp,letely removed the lab frame. Where is your explanation now?It remains the same, I said:>> SR says the speed will be c only as measured in the inertial>> (lab) frame.SR says the speed is c in any inertial frame, whether there>is a 'lab' handy to define that or not. You have removed the>lab but not the inertial frame.And what if the ring is also moving inertially in a direction perpendicular toits plane?>Perhaps you should read my posts more carefully, it is the>>_Ritzian_ theory that predicts that it won't be detec.> You people are really funny.>>Well you're the one pushing a theory that predicts there>>will be no shift.>> There will be a shift under source dependency.See my other post where I have shown that there is no speed>change on the reflections and the path lengths vary as (v/c)^2>for v<c as well.> You are becoming quite confused here George.Well you have been saying SR doesn't predict a shift but above>you said ... there will be a fringe shift with source>dependency. That means that there will also be one according>to SR .. and in the previous post you asked Are you trying to>tell me that it wont detect rotation? when I had just said it>would.Like I said. I don't think the sagnac effect has anything to do with thepreachings of SR.> Remember the source and observer are the same in a ring gyro. Therefore>> according to both you (and me), the light travels at the same speed 'c'>> relative to both.No, remember SR says the light travels at c ONLY in an inertial>frame. The co-rotating frame is not inertial. Ballistic theory>on the other hand says the light is emit a c relative to the>source and each mirror on subsequent reflections.No it doesn't. You cannot assume that it even arrives at each mirror at c.> I really don't see where SR comes into this at all.It didn't until you introduced it in your reply to Paul. You>said SR says that the speed of light is c relative to its>source. which it appears you know is untrue, SR says it is c>in any inertial frame but the rotating table is not inertial.>When you bear that in mind you will find SRFrankly, I don't want to 'find' SR. It is completely illogical from start tofinish and there is absolutely no evidence to back it up.George>Henri Wilson. www.users.bigpond.com/hewn/index.htm=== === Subject: : Re: Final Rout of Synchronization Clocks in Relativity>Sorry Henri, that's wrong. Only _inertial_ instruments>>measure the speed as c according to SR.>> Yes. OK. You have to include the great SRian escape route. inertial.Try this experiment: stand up, turn round then sit down.>Now consider a flash of light moving away from a recent>supernova in the Andromeda galaxy. It is 2 million light>years away so in the frame that co-rota with you as you>turned, that light moved 2*pi*R or a distance of 13 million>light years in just a few seconds. Obviously the speed of>light in a rotating frame cannot possibly be c. I doubt if the one-way speed of light is ever c. But anyway, you have put forward a good case for scrapping rotating frames> altogether. I certainly agree with that.How does this look in a rotating frame? is a perfectlyvalid question one can ask of any experiment so we have tobe able to answer it. After all, the Earth rotates so in thelimit all actual experiments have to take that into account.The point is that it is obvious from the above that SR doesnot say that the speed of light is c in a rotating frame.>> ... You were the one who first introduced such to this>> argument remember.No, you replied to Paul Anderson:...>> SR says that the speed of light is c relative to its source. Thereforeif a>> ring laser is analysed in the co-rotating frame, no fringe shiftsshould be>> expec according to SR.>> Sagnac clearly refutes SR.SR does NOT imply that the speed of light is c relative to>its source in the co-rotating frame, only that it is isotropic>in the inertial frame as Paul sta. The point I made to Paul was that if the 'co-rotaing frame' argument wasused> against source dependency then it would also apply to SR since SR is basedon> source dependency.The point I make is that you introduced SR, not me.In addition SR is based on source independence, specificallythe opposite of what you say.>> That means that there will also be one according to SR which>> clearly states that light moves at c relative to its source.You are right, there will be a shift according to SR but>you have been arguing the opposite. I was being a little facetious.OK, but you cannot then say I am confused when you arecontradicting yourself, glass houses and all that, what?> The point is, the SR explanation is the same as source dependency.Nope, they are opposites. Ballistic theory says the lightis emit at c in the rotating frame and you have tocalculate its speed in the inertial frame. SR says itwill be c in the inertial frame and you have to calculatewhat it will be in the rotating frame.>> Source and observer are the same in a ring gyro.Both are rotating and accelerating, hardly inertial. Well if the whole thing is not inertial, SR cannot apply at all - so howthe> hell can you even suggest that SR can provide any kind of explanation for> sagnac effect when it is entirely beyond its scope?You only know that the speed is c in the inertial frameso you calculate it based on that. Since the speed isisotropic but the mirrors move, there is a predicfringe shift. In Ritzian theory you only know the speedis c relative to the source at emission so have tocalculate in the co-rotating frame. In that the mirrorsare not moving or rotating so there is no fringe shift.>SR says the speed is c in any inertial frame, whether there>is a 'lab' handy to define that or not. You have removed the>lab but not the inertial frame. And what if the ring is also moving inertially in a directionperpendicular to> its plane?in _any_ inertial frame>See my other post where I have shown that there is no speed>change on the reflections and the path lengths vary as (v/c)^2>for v<c as well.It does but we know the shift is linear in (v/c) for small vand it is easier to calculate. In practice at high v, themirror will have moved too far and the light beam will missit so you get a crude variation of the toothed wheel methodfor measuring the speed of light.>No, remember SR says the light travels at c ONLY in an inertial>frame. The co-rotating frame is not inertial. Ballistic theory>on the other hand says the light is emit a c relative to the>source and each mirror on subsequent reflections. No it doesn't. You cannot assume that it even arrives at each mirror at c.I don't assume it, this is the calculation that demonstratesit http://www.briar.demon.co.uk/Henri/speed.gif>> I really don't see where SR comes into this at all.It didn't until you introduced it in your reply to Paul. You>said SR says that the speed of light is c relative to its>source. which it appears you know is untrue, SR says it is c>in any inertial frame but the rotating table is not inertial.>When you bear that in mind you will find SR Frankly, I don't want to 'find' SR. It is completely illogical from startto> finish and there is absolutely no evidence to back it up.Oops, that should have said .. you will find SR predictsa fringe shift. No doubt you can accuse me of treating SRas a religion now ;-)George=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityExpires: 28 daysSorry Henri, that's wrong. Only _inertial_ instruments>measure the speed as c according to SR.>> Yes. OK. You have to include the great SRian escape route. inertial.>>Try this experiment: stand up, turn round then sit down.>>Now consider a flash of light moving away from a recent>>supernova in the Andromeda galaxy. It is 2 million light>>years away so in the frame that co-rota with you as you>>turned, that light moved 2*pi*R or a distance of 13 million>>light years in just a few seconds. Obviously the speed of>>light in a rotating frame cannot possibly be c.>> I doubt if the one-way speed of light is ever c.>> But anyway, you have put forward a good case for scrapping rotating frames>> altogether. I certainly agree with that.How does this look in a rotating frame? is a perfectly>valid question one can ask of any experiment so we have to>be able to answer it. After all, the Earth rotates so in the>limit all actual experiments have to take that into account.The point is that it is obvious from the above that SR does>not say that the speed of light is c in a rotating frame.>> ... You were the one who first introduced such to this> argument remember.>>No, you replied to Paul Anderson:> SR says that the speed of light is c relative to its source. Therefore>if a> ring laser is analysed in the co-rotating frame, no fringe shifts>should be> expec according to SR.> Sagnac clearly refutes SR.>>SR does NOT imply that the speed of light is c relative to>>its source in the co-rotating frame, only that it is isotropic>>in the inertial frame as Paul sta.>> The point I made to Paul was that if the 'co-rotaing frame' argument was>used>> against source dependency then it would also apply to SR since SR is based>on>> source dependency.The point I make is that you introduced SR, not me.In addition SR is based on source independence, specifically>the opposite of what you say.George, if you are in a remote inertial spaceship and send a ray of light outin front of the ship, what do you expect its speed will be relative to theship. If you reflect the light back with a mirror on a long pole, how long willit take to return? 2D/c. If you send it around a square path using three fixedmirrohow long will it take to return? 4D/c. The speed of the light isobviously c relative to its source.If you rotate the spaceship in the plane of the mirrothe time taken willNOT be 4D/c for reasons whicj I have explained.SR doesn't know what it says.>> That means that there will also be one according to SR which> clearly states that light moves at c relative to its source.>>You are right, there will be a shift according to SR but>>you have been arguing the opposite.>> I was being a little facetious.OK, but you cannot then say I am confused when you are>contradicting yourself, glass houses and all that, what?> The point is, the SR explanation is the same as source dependency.Nope, they are opposites. Ballistic theory says the light>is emit at c in the rotating frame and you have to>calculate its speed in the inertial frame. SR says it>will be c in the inertial frame and you have to calculate>what it will be in the rotating frame.You also have to calculate its direction.>> Source and observer are the same in a ring gyro.>>Both are rotating and accelerating, hardly inertial.>> Well if the whole thing is not inertial, SR cannot apply at all - so how>the>> hell can you even suggest that SR can provide any kind of explanation for>> sagnac effect when it is entirely beyond its scope?You only know that the speed is c in the inertial frame>so you calculate it based on that. Since the speed is>isotropic but the mirrors move, there is a predic>fringe shift. In Ritzian theory you only know the speed>is c relative to the source at emission so have to>calculate in the co-rotating frame. In that the mirrors>are not moving or rotating so there is no fringe shift.That is an oversimplification and incorrect.>SR says the speed is c in any inertial frame, whether there>>is a 'lab' handy to define that or not. You have removed the>>lab but not the inertial frame.>> And what if the ring is also moving inertially in a direction>perpendicular to>> its plane?in _any_ inertial frame>See my other post where I have shown that there is no speed>>change on the reflections and the path lengths vary as (v/c)^2>>for v<> You cannot assume v<c as well.It does but we know the shift is linear in (v/c) for small v>and it is easier to calculate. In practice at high v, the>mirror will have moved too far and the light beam will miss>it so you get a crude variation of the toothed wheel method>for measuring the speed of light.Would the c beam in the MMX do the same?And it does that whether or not the light speed is source dependent.>No, remember SR says the light travels at c ONLY in an inertial>>frame. The co-rotating frame is not inertial. Ballistic theory>>on the other hand says the light is emit a c relative to the>>source and each mirror on subsequent reflections.>> No it doesn't. You cannot assume that it even arrives at each mirror at c.I don't assume it, this is the calculation that demonstrates>it http://www.briar.demon.co.uk/Henri/speed.gifSorry George. Your diagram doesn't make any sense.>> I really don't see where SR comes into this at all.>>It didn't until you introduced it in your reply to Paul. You>>said SR says that the speed of light is c relative to its>>source. which it appears you know is untrue, SR says it is c>>in any inertial frame but the rotating table is not inertial.>>When you bear that in mind you will find SR>> Frankly, I don't want to 'find' SR. It is completely illogical from start>to>> finish and there is absolutely no evidence to back it up.Oops, that should have said .. you will find SR predicts>a fringe shift. No doubt you can accuse me of treating SR>as a religion now ;-)Earth centricism acurately predicts that the moon rotates around the sun, too,George.George>Henri Wilson. www.users.bigpond.com/hewn/index.htm=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityExpires: 28 days> Answer the question Paul.>> Consider yourself flying through remote space with only a ring laser, an oxygen>> bottle and a tin of sardines.>> If you rotate the gyro, does it work?>> If so, why, - since there is no lab frame to be used as a reference.I have got it now, Henry.>You claim that gyros doesn't work because there is no ether with>an 'arbitrary inertial frame', no lab or any other reference frame.No Paul. George led me to believe that that was the SRian view.What is YOUR explanation?Paul>Henri Wilson. www.users.bigpond.com/hewn/index.htm=== === Subject: : Re: Final Rout of Synchronization Clocks in Relativity> Answer the question Paul.>> Consider yourself flying through remote space with only a ring laser, an oxygen>> bottle and a tin of sardines.>> If you rotate the gyro, does it work?>> If so, why, - since there is no lab frame to be used as a reference.I have got it now, Henry.>You claim that gyros doesn't work because there is no ether with>an 'arbitrary inertial frame', no lab or any other reference frame. No Paul. George led me to believe that that was the SRian view. What is YOUR explanation?Explanation of what?You have been told what SR say about the Sacnac many times,so why the hell do you keep asking over and over again?I have certainly explained it MANY times to you.George Disman explained it a couple of postings ago in this thread.Are you so senile that you don't remember from one postingto the next? If that the case, why don't you go back an readthe previous postings again?Print the following out, and pin it up over your computer:REMEMBER - I HAVE BEEN TOLD: According to SR, the speed of light is isotropic c when the Sagnac ring is rotating, the light will use different times to go around the ring in the two opposite directions, because the light path obviously is shorter in one direction than the other when measured in the inertial frame where the light moves at the speed c.You have obviously no rational argument against this explanation,so you try to ridcilule the expression 'inertial' by vague statementslike this:| That is a perfect example of why SR is just a disguised aether theory.| An aether with an 'arbitrary inertial frame'.| I'm not impressed!| Yes. OK. You have to include the great SRian escape route. inertial.What the hell are you implying with these remarks, if it isn'tthat there is no such thing as an inertial frame, becausethat implies the existence of an ether, which you claimdoesn't exist?Note this. YOU state something which cannot - if it meansanything at all - be interpre otherwise than a claim thatthere is no such thing as an inertial frame.And then YOU states that: George led me to believe that that was the SRian view.Say - how confised can you get?I am challenging you to answer my questions below:----------------------------------Do you really insist that inertial frames does not exist?In that case - how do gyros work?What is the reference for the rotation they measure?The fact is of course that the concept inertial framea system of co-ordinates in which the equationsof Newtonian mechanics hold good.?Paul=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityExpires: 28 days> No Paul. George led me to believe that that was the SRian view.>> What is YOUR explanation?Explanation of what?>You have been told what SR say about the Sacnac many times,>so why the hell do you keep asking over and over again?>I have certainly explained it MANY times to you.>George Disman explained it a couple of postings ago in this thread.>Are you so senile that you don't remember from one posting>to the next? If that the case, why don't you go back an read>the previous postings again?Print the following out, and pin it up over your computer:REMEMBER - I HAVE BEEN TOLD:> According to SR, the speed of light is isotropic c> when the Sagnac ring is rotating, the light will> use different times to go around the ring in the two> opposite directions, because the light path obviously> is shorter in one direction than the other when measured> in the inertial frame where the light moves at the speed c.You have obviously no rational argument against this explanation,>so you try to ridcilule the expression 'inertial' by vague statements>like this:HaHa!!!But all you have given is the aether solution.You don't even accept an aether. (or so you say Hahahah!)Besides, you don't have any evidence that the speed of light is isotropic in aninertial frame so your explanation is no more than circular speculation. It isno more credible than claiming that the fairies cause the fringes to shift.| That is a perfect example of why SR is just a disguised aether theory.>| An aether with an 'arbitrary inertial frame'.>| I'm not impressed!| Yes. OK. You have to include the great SRian escape route. inertial.What the hell are you implying with these remarks, if it isn't>that there is no such thing as an inertial frame, because>that implies the existence of an ether, which you claim>doesn't exist?Of course inertial frames exist - but in an instantaneous universe they are alleffectively the same so we only need one.Note this. YOU state something which cannot - if it means>anything at all - be interpre otherwise than a claim that>there is no such thing as an inertial frame.Bull.I'm simply pointing out that there is no evidence that light speed isisotropic. Of course it is not.And then YOU states that:> George led me to believe that that was the SRian view.Say - how confised can you get?I am challenging you to answer my questions below:>---------------------------------->Do you really insist that inertial frames does not exist?>In that case - how do gyros work?>What is the reference for the rotation they measure?Gyros work for the same reason that a three mirror sagnac works.Light beams moving on opposite directions are deflec by different amountswhen reflecting from the moving mirrors. The fact is of course that the concept inertial frame>a system of co-ordinates in which the equations>of Newtonian mechanics hold good.?Because he knew NM was correct - if not complete. Paul>Henri Wilson. www.users.bigpond.com/hewn/index.htm=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityExpires: 28 days> SR says that the speed of light is c relative to its source. Therefore if a>> ring laser is analysed in the co-rotating frame, no fringe shifts should be>> expec according to SR.>> Sagnac clearly refutes SR.>>We both know that you know nothing about and don't understand SR.>It wasn't really necessary to remind us.>> Do you only pull out the above statement as a last resort, Paul?>>The above demonstration of your understanding of SR>>make it obvious that it is a simple statement of fact.>>Paul>> Well come on Paul. Tell me what SR says about ring lasers.Do you pretend to have forgotten, or are you really senile?No, I just want a precise statement from YOU.Paul>Henri Wilson. www.users.bigpond.com/hewn/index.htm=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityExpires: 28 days>> For instance, can it be assumed that a ray, which is aligned with the>centre of> the first mirror, strikes that mirror at exactly the same point when>the> apparatus is rotating. If not, I don't think your interpretation of>what> happens in the co-rotating frame can be correct.>>You have to remember that a 'ray' is just the path formed by>>the normal to the wavefront. In reality any light beam has a>>finite width.>> And it isn't perfectly parallel.No, but it is the fact that the wavefront is normal to the>ray that answers your question.>> I understand your point but it is hardly convincing. You have not>considered> where the rays will strike the moving mirrors.>>The rays must hit the mirrors at exactly the same points as>>in the non-rotating case because, in the co-rotating frame,>>the mirrors are static and the light eventually has to hit>>the same receiver (telescope, screen or whatever.) The point>>of reflection could move if the angles turned along the curved>>paths differed from leg to leg, but since they must be>>symmetrical, there can be no change.>> None of that is obvious. I don't know how you can make those claims.Sometimes things may not be obvious but they may still be>correct. The symmetry about the midpoint of a path ensures>the angles of approach at each end of a light path from>mirror to mirror are the same. That the wavefront is normal>to the tangent to the path ensures the angle of incidence is>equal the angle of reflection. Once you know those, it means>the point where the ray hits the mirror has to be the same.But the angle of incidence is NOT equal to the angle of reflection from amoving mirror.>>The travel time is barely affec but more importantly it is>increased by exactly the same amount for both rays since it>depends on v^2, not v, so there should be _no_ path difference.>However, the observed fringe shift is exactly as if the speed>was c-v and c+v in the co-rotating frame.>> Again this is hardy convincing. When you say the time depends on v^2>(actually> 1/v^2) you are forced to include a constant (k) that is not>dimensionless,> (T/L^2). I'm not sure what that implies.>>Actually it depends on (v/c)^2 which is dimensionless but that>>is beside the point. the key is that any curve based on even>>powers is symmetrical about zero hence the path length change>>for +v is the same as for -v and the path difference is>>therefore zero.>> Well, I'll have to think a lot more about this. The curve can be>symmetrical>> but need not strike the same point on the mirror for all rotational>speeds.It takes some thought but the symmetry argument does>give you an answer.>The rays must follow identical paths when the apparatus is>>stationary from basic optics, I'm sure you know that. Since>>the angles change symmetrically, they do not change the point>>of reflection.>> This is where the second order factor comes in. The travel times between>the>> mirrors is different for the two paths because, with source dependency the>rays>> get a velocity 'kick' each time they are reflec. The mirrors act as>sources.>> You are ignoring that.No, we have looked at the equations several times. I>have expanded the sketches to try to answer your next>point as well:>> You must also consider velocity variations that occur during reflection>at each> moving mirror. Each acts as a source.> Under source dependency, the velocity of light leaving the source>appears to be> (c+vsin45).> But is it?> Is it c+vsin(45+x) for the red ray and c+vsin(45-x) for the green one?>>The angle 'a' in this diagram>> http://www.briar.demon.co.uk/Henri/speed.gif>>is 45+x for the red ray at the top right mirror in this diagram:>> http://www.briar.demon.co.uk/Henri/paths.gif>>The trick though is that you than have to find the speed at>>which the light approaches the mirror at the top left. The>>angle there is also 45+x so when you subtract the motion of>>the mirror which is moving away from the light, the nett>>result is a relative incident speed of exactly c.>> That is true only for the first mirror.>>The same applies to the green ray but you need to take a>>mirror image of the speed.gif sketch and the angle then, as>>you say, is 45-x at both ends but again since they are the>>same the contributions from v cancel.You drew this:>> www.users.bigpond.com/hewn/sagnac.jpgI have copied that and added two small blue lines at the>mid-point of two paths. The paths are S->A->B->C->S' and>S->C'->B->A'->S'http://www.briar.demon.co.uk/Henri/ lab.gifThe diagram on the right concentrates on the A/B section and>has the two aligned on the mid-point. This confirms what you>were asking about whether the angles were exactly 45 degrees.>For the green path, the angle turned by the ray is slightly>more than 90 degrees at each end so the angle between the>mirror and the horizontal ray is slightly less than 45>degrees. For the red path, the angle is slightly more at both>ends.Why do you think the two lines meet at point B?I haven't had time to really think about this but I would think that the greenand red lines strike the B mirror at different points.You said in another post:>[GD]>> c + v' = v * cos(a) + c * cos(b)>[HW]>> Not so. If the source is moving in the direction v, then light speed in>the>> horizontal direction will be c+[v*cos(a)]I have added a bit more to this diagram:http://www.briar.demon.co.uk/Henri/speed.gifObviously for the normal situation of v<is negligible but this is just simple vector addition.although less than 45 degrees for the green path and more for>the red path.Regardless of that, it should be clear that v*cos(a) adds at>one end and subtracts at the other and the factor of cos(b)>multiplies at one and divides at the other so the apparent>speed of the light at the left end of the diagram must be>exactly c regardless of angle a. The result is that since,>in your source-dependent theory, the light is re-emit>at c, there is no a velocity kick in this case. I am not>ignoring this aspect, I have calcula it and found it to>be zero.The same diagram can be applied when considering the leg>from the source or to the detector.I'm sorry George but I cannot see what you are trying to say with this diagram.What are all those lines supposd to mean?I cannot see any relationship to sagnac. Please explain.>>Why not? you seemed happy with it a few posts ago.>> I was happy with the diagram, not its significance.http://www.briar.demon.co.uk/Henri/paths.gifThe intention was to show that the two rays turned in the>same direction hence the angle between changes by the>difference, not the sum. What you cannot get from the>diagram is that the derivative of the angle for each ray>at v=0 must be the same for both rays so the angle between>them must be unchanging at v=0.I will try to spend more time on this. Please be patient.George>Henri Wilson. www.users.bigpond.com/hewn/index.htm=== === Subject: : Re: Final Rout of Synchronization Clocks in Relativity[lots trimmed to get to the gist] That the wavefront is normal>to the tangent to the path ensures the angle of incidence is>equal the angle of reflection. Once you know those, it means>the point where the ray hits the mirror has to be the same. But the angle of incidence is NOT equal to the angle of reflection from a> moving mirror.Draw out the wavefronts and you will find it is but themirror appears very slightly curved since it rotates inthe time it takes the wavefront to reach all parts ofthe mirror. However, at any point on the mirror, theangles are still equal.>You drew this:>> www.users.bigpond.com/hewn/sagnac.jpgI have copied that and added two small blue lines at the>mid-point of two paths. The paths are S->A->B->C->S' and>S->C'->B->A'->S'http://www.briar.demon.co.uk/Henri/ lab.gifThe diagram on the right concentrates on the A/B section and>has the two aligned on the mid-point. This confirms what you>were asking about whether the angles were exactly 45 degrees.>For the green path, the angle turned by the ray is slightly>more than 90 degrees at each end so the angle between the>mirror and the horizontal ray is slightly less than 45>degrees. For the red path, the angle is slightly more at both>ends. Why do you think the two lines meet at point B?> I haven't had time to really think about this but I would think that thegreen> and red lines strike the B mirror at different points.There may be a slight misunderstanding between us. Thinkabout it in the non-rotating frame and it should beobvious that since the paths start and end at the sameplace, the direct ray must hit all the mirrors at thesame point _on_the_mirror. That is not to say that theyhit at the same place 'in the lab'. As we have both shown,the points A and A' differ and the same for C and C'. TheB and B' points only coincide in the lab if it happens tobe exactly half way round.>http://www.briar.demon.co.uk/Henri/speed.gif...> although less than 45 degrees for the green path and more for>the red path.Regardless of that, it should be clear that v*cos(a) adds at>one end and subtracts at the other and the factor of cos(b)>multiplies at one and divides at the other so the apparent>speed of the light at the left end of the diagram must be>exactly c regardless of angle a. The result is that since,>in your source-dependent theory, the light is re-emit>at c, there is no a velocity kick in this case. I am not>ignoring this aspect, I have calcula it and found it to>be zero.The same diagram can be applied when considering the leg>from the source or to the detector. I'm sorry George but I cannot see what you are trying to say with thisdiagram.> What are all those lines supposd to mean?> I cannot see any relationship to sagnac. Please explain.http://www.briar.demon.co.uk/Henri/paths.gifThe intention was to show that the two rays turned in the>same direction hence the angle between changes by the>difference, not the sum. What you cannot get from the>diagram is that the derivative of the angle for each ray>at v=0 must be the same for both rays so the angle between>them must be unchanging at v=0. I will try to spend more time on this. Please be patient.OK, I'll go through the steps in order. We have jumpedabout between various aspects over the posts and thediscussion has become a bit fragmen so a summary inone place may help.The theory considers that light is emit at c relative'velocity kick', I take it you consider the light alsoleaves the mirror leaves a mirror at c relative to themirror regardless of the incident speed.In the lab frame, the paths look like this www.users.bigpond.com/hewn/sagnac.jpgBy rotating the drawing (not the equipment!), itbecomes easier to see that the mirror reflectionsare at slightly different angles for the red andgreen rays. The small blue lines indicate thealignment: http://www.briar.demon.co.uk/Henri/lab.gifNow think about the red ray from reflection A to B.http://www.briar.demon.co.uk/Henri/paths.gifThe magenta line indicates the mirror which is almostbut not quite at 45 degrees, call that angle 'a'. Themirror is moving with velocity v shown by the magentaarrow. The light is emit from the mirror at c (andthe same is true for the source) as shown by the greenarrow. I have offset the green line from the reflectionpoint in the usual way to show addition of vectors byforming a parallelogram. The vector addition of v and cthen produces a speed along the A-B line of c*cos(b) + v*cos(a) where c*sin(b) = v*sin(a)That is the speed of the light in the inertial frame.(In the rotating frame it would just be c of course.)When the light hits the mirror at B, we want to knowthe speed of incidence. The angles marked a are thesame through symmetry so since c*sin(b) = v*sin(a)the angles marked b must also be the same. The mirroris moving at v in the direction shown, that is withthe same component of speed to the left in the diagrambut with the component up/down the diagram in theopposite sense from the speed at A. We need to subtractvector v from c*cos(b) + v*cos(a) to find the speed thelight approaches the mirror. By the symmetry of thedrawing, it should be clear that this must be exactly c.If the light approaches the mirror B at c and leaves atthe same speed, there is no 'velocity kick' in this case.You should be able to imagine the same sketch for thegreen ray which would go from left to right hence themirror speed subtracts at B and adds at A. The speedin the inertial frame is then c*cos(b) - v*cos(a)and the speed of incidence on the mirror at A is againexactly c.You can also see the same thing in the rotating framewhere the light paths are curvedhttp://www.briar.demon.co.uk/Henri/paths.gifThe light is emit at c and the curves are symmetricalbecause the rate of rotation of the platform is constant.In this frame, the mirrors are neither rotating nortranslating so all the normal rules of optics apply. Inparticular the rays must trace the same paths when notrotating hence the points where the rays hit the mirrorsmust be identical. The curvature applies equally to bothrays so does not change the points where the direct ray the mirrors. The path lengths are a minimum for v=0(straight lines) so must vary as (v/c)^2 hence the pathdifference is unchanging for v< necessity to determine, a specification of any Clock. However, out of any> falling along any amalgam, which it can takes to an attempt along thefinal> confusion.> The define as the definite reason, for what, for instance, to know, that> the sun along its highest point in the sky, is absolutely a matter of> metamorphosis along its variations over an orbit as over its rotation.> Something, which it does allows the time of a being just an average along> any appearance among any solar days as along an orbital year. Therefore, there is a definitely also, the ultimate Clock, which is alongan> utilisation of an aic Clock, among a several other clocks. However, the> aic Clock, which is usually along any measure of any vibration alongany> energy over any cesium along any a. That, usually it is used along an> addition of a fraction of a seconds every more or less decades. However, a finally, there is an universal time, used as known> internationally as a Greenwich time. Therefore, basically used along the> rotation of the earth, whether, it does a definitely have as a majorpillar,> which is the aic Clock. Therefore, from that point to arrive todetermine> any Synchronization along any Clock. It would be as it should be,something,> which it could takes to an infinite calculations, which it would as should> always and a definitely takes to the departure case, and this is what isall> about, definitely as a matter a fact!!!!!!!!!!!!!!!!!!...............What this thread is all about is the Sagnac Experimentand there are no clocks in that. Otherwise, I have noidea what you intended to say, sorry.George=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityObviously!?........ ...Then, basically, what it would be if it could beSynchronized.Could it be if it would be your own upstairs?????!!!!!?????..................-- Ahmed Ouahi, ArchitectNice To See You Too!George Dishman kirjoitti viestiss.8a ......... ...First of all, it would be as it should be an essential> necessity to determine, a specification of any Clock. However, out ofany> falling along any amalgam, which it can takes to an attempt along the> final> confusion.> The define as the definite reason, for what, for instance, to know,that> the sun along its highest point in the sky, is absolutely a matter of> metamorphosis along its variations over an orbit as over its rotation.> Something, which it does allows the time of a being just an averagealong> any appearance among any solar days as along an orbital year. Therefore, there is a definitely also, the ultimate Clock, which isalong> an> utilisation of an aic Clock, among a several other clocks. However,the> aic Clock, which is usually along any measure of any vibration along> any> energy over any cesium along any a. That, usually it is used along an> addition of a fraction of a seconds every more or less decades. However, a finally, there is an universal time, used as known> internationally as a Greenwich time. Therefore, basically used along the> rotation of the earth, whether, it does a definitely have as a major> pillar,> which is the aic Clock. Therefore, from that point to arrive to> determine> any Synchronization along any Clock. It would be as it should be,> something,> which it could takes to an infinite calculations, which it would asshould> always and a definitely takes to the departure case, and this is what is> all> about, definitely as a matter a fact!!!!!!!!!!!!!!!!!!............... What this thread is all about is the Sagnac Experiment> and there are no clocks in that. Otherwise, I have no> idea what you intended to say, sorry. George=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityMessage-Id: <1075577329.46340.0@iris.uk.clara.net> kirjoitti viestiss.8a What this thread is all about is the Sagnac Experiment> and there are no clocks in that. Otherwise, I have no> idea what you intended to say, sorry.> Obviously!? ........ ...Then, basically, what it would be if it could be> Synchronized.> Could it be if it would be your own upstairs?????!!!!!?????...............If what could be synchronised? There is no clockin the Sagnac experiment.Don't be misled by the thread subject line, theconversation moved on a long time ago.George=== === Subject: : Re: Final Rout of Synchronization Clocks in Relativity.......... ...Do talk about!!!!!!!!!!!....... ...-- Ahmed Ouahi, ArchitectDefinitely!George Dishman kirjoitti viestiss.8a> George Dishman kirjoitti viestiss.8a What this thread is all about is the Sagnac Experiment> and there are no clocks in that. Otherwise, I have no> idea what you intended to say, sorry. Obviously!? ........ ...Then, basically, what it would be if it could be> Synchronized.> Could it be if it would be your ownupstairs?????!!!!!?????............... If what could be synchronised? There is no clock> in the Sagnac experiment. Don't be misled by the thread subject line, the> conversation moved on a long time ago. George=== === Subject: : Humpty Dumpty's M TheorySarfatti Commentary onFROM SLOWDOWN to SPEEDUPBy Adam G. Riess and Michael S. TurnerUntil recently, astronomers fully expec to see gravity slowing down the expansion of the cosmos. In 1998, however, researchers discovered the repulsive side of gravity. By carefully observing distant supernovaestellar explosions that for a brief time shine as brightly as 10 billion sunsastronomers found that they were fainter than expec. The most plausible explanation for the discrepancy is that the light from the supernovae, which exploded billions of years ago, traveled a greater distance than theorists had predic. And this explanation, in turn, led to the conclusion that the expansion of the universe is actually speeding up, not slowing down. general relativity also allows for the possibility of forms of energy with strange properties that produce repulsive gravity The discovery of accelerating rather than decelerating expansion has apparently revealed the presence of such an energy form, referred to as dark energy. The leading candidate to explain dark energys effects is vacuum energy, which is mathematically equivalent to the cosmological constant that Einstein inven in 1917. But to produce the observed acceleration of the universe, the constants density would have to be twice that of matter. Where could this energy density come from? The uncertainty principle of quantum borrowed time and energy, popping in and out of existence. But when theorists try to compute the energy density associa with the quantum vacuum, they come up with values that are at least 55 orders of magnitude too largeThis is the problem I claim I have solved with the idea of vacuum coherence that is in fact the inflation field in disguise. Einsteins gravity together with both dark energy and dark matter all emerge together from the self-organizing creative information-rich ripples in our conscious holographic universes phase and intensity of this vacuum coherence field. Stephen Hawking has glimpsed this through the glass darkly calling it The Mind of God. This subterranean giant tsunami virtual tidal wave of inflation decreases the randomness of the zero point vacuum fluctuations of all the quantum fields. I have provided a very simple dynamical reason based on Diracs theory of the negative energy virtual electron-positron pair quantum vacuum why this happens in an intrinsic instability of flat spacetime without gravity that is very like what happens when a normal metal undergoes a phase transition, an emergent More is different metamorphosis to a superconductor where electric currents freely flow without the annoying resistive heating of the circuits.This discrepancy has been called the worst embarrassment in all of theoretical physics, but it may actually be the sign of a great opportunity.Indeed it is. The really great opportunity that President Bushs science pundits are missing in their new vision for the American manned space exploration program is the metric engineering of super cosmos enabling us, like caterpillars to butterflies in a metamorphosis of Men Like Gods to fulfill our manifest super cosmic destiny to harness the repulsive dark energy. Indeed, anti-gravitating universally repelling dark energy is The Right Stuff enabling us to time travel effectively faster than light without actually moving faster than light with a free floating weightless warp drive through traversable star gate wormholes to distant places in our own universe in our cosmic present, past and future and to the other parallel universes next door. As General las MacArthur warned the cadets at West Point in his Duty, Honor, Country Farewell Address, we are not the first intelligent species to do this as the UFO flying saucer phenomenon is, in all likelihood, showing us, nor will we be the last. My esteemed colleagues, all honorable men, the Pundits of respectable physics and cosmology have put on blinders rejecting this evidence, which is, in my opinion, the missing link, the last piece needed to complete the jig saw puzzle.Although it is possible that new attempts to estimate the vacuum energy density may yield just the right amount to explain cosmic acceleration, many theorists believe that a correct calculation, incorporating a new symmetry principle, will lead to the conclusion that the energy associa with the quantum vacuum is zero. (Even quantum nothingnessweighs nothing!) If this is true, something else must be causing the expansion of the universe to speed up.Jack: No, this is simply confused thinking. What you see below Is All The Kings Men grasping wildly willy-nilly at straws like the Sufi Story where The Pundits shine strong light in the wrong part of Platos Dark Cave.Theorists have proposed a variety of ideas, ranging from the influence of extra, hidden dimensions Perhaps the most radical idea is that there is no dark energy at all but rather that Einsteins theory of gravity must be modified.Jack: This is a very bad idea. It is a false idol, a Golden Calf. It is Fools Gold. It is a wrong turn off the path through Dantes Inferno. The following however is in complete exact agreement with my theory that I published in 2002 my two books Destiny Matrix and Space-Time and Beyond II on file in the catalog of the Library of Congress.IN NEWTONS THEORY, gravity is always attractive and its strength depends on the mass of the attracting object. The twist in Einsteins theory is that the strength of the gravitational pull exer by an object also depends on its composition. Physicists characterize the composition of a substance by its internal pressure. An objects gravity is proportional to its energy density plus three times the pressure. Our sun, for example, is a hot sphere of gas with positive (outward) pressure; because gas pressure rises with temperature, the suns gravitational pull is slightly greater than that of a cold ball of matter of equivalent mass. On the other hand, a gas of photons has a pressure that is equal to one third its energy density, so its gravitational pull should be twice that of an equivalent mass of cold matter. Dark energy is characterized by negative pressure.(Elastic objectsfor instance, a rubber sheetalso have negative, or inward, pressure.) If the pressure falls below 1?3 times the energy density, then the combination of energy plus three times the pressure is negative and the gravitational force is repulsive. The quantum vacuum has a pressure that is 1 times its energy density, so the gravity of a vacuum is very repulsive. Only theories stipulating large variations in dark energy density have been ruled out ... The only way to forecast our cosmic future is to figure out the nature of dark energy.Amen. When gravity operates over microscopic distances for instance, at the center of a black hole, where a huge mass is packed into a subaic volumethe bizarre quantum properties of matter come into play, and string theory describes how the law of gravity changes. Over greater distances, string theorists have generally assumed that quantum effects are unimportant. Yet the cosmological discoveries of the past several years have encouraged researchers to reconsider. Four years ago my colleagues and I asked whether string theory would change the law of gravity not just on the smallest scales but also on the largest ones. The feature of string theory that could bring about this revision is its theory adds six or seven dimensions to the usual three.Theorists today are willing to pay any price to avoid signal nonlocality even extra dimensions. Well, perhaps, that is not too high a price after all?In the past, string theorists have argued that the extra dimensions are too small for us to see or move in. But recent progress reveals that some or all of the new dimensions could actually be infinite in size. They are hidden from view not because they are small but because the force of gravity, and as a result, the law of gravity changes.There is one nice feature about this idea of enormous branes in that it easily explains why there is no significant amount of antimatter in our universe. According to John Archibald Wheelers Geometrodynamics lepto-quarks are in reality David Bohms hidden or extra variable that are tiny wormholes in three dimensional space with quantized gauge force fluxes threading them like we see in Type II superconducting vortices in condensed matter physics near absolute zero temperature. Wheeler called this Mass without mass with Charge without charge and Spin without spin. Wheelers idea did not work back then almost fifty years ago because gravity was thought to be too weak. In other words, spacetime geometry was simply too stiff to bend easily. Spacetimes string tension was too large. This implied that Wheelers wormholes the electron. Indeed, this is just like the speed of sound barrier was mistakenly thought to be before the era of jet planes and Chuck Yeagers that Andrei Sakharovs metric elasticity for emergent gravity out of zero point vacuum energy fluctuations, or what Ed Witten, (The Big Cheese of M Theory) calls alpha can be a scale-dependent variable not an immutable constant. Wittens alpha is proportional to the reciprocal of the string tension, that origina in the observed universal gravity Regge parallel trajectories slope of the hadronic resonances of (1 Gev)^-2 unceremoniously kicked up to the quantum gravity scale of (10^19 Gev)^-2 is a variable that gets less stiff at smaller scales of higher energy transfer in scattering experiments. Indeed, the effective strong short-range gravity, sugges by Abdus Salam in the early 1970s, solves more than one hitherto unsolved fundamental conceptual problems in high-energy physics. Strong gravity at the scale of 1 fermi (10^-13 cm) is 40 powers of ten stronger than Newtons gravity. This means Wheelers wormholes are just the right size. Its like those size-changing drugs that Alice takes in Lewis Carrolls Wonderland. First of all it solves the mystery of the missing anti-matter called broken charge conjugation symmetry or C-violation. The missing antimatter is in a parallel brane world next door ac a thin barrier of extra dimensional hyperspace. Imagine a wormhole with two mouths or ends attached to two different brane worlds with infinitely large uncompactified extra space dimension as Dvali explains. The Faraday flux lines of electro-weak-strong gauge forces must form closed loops. Flux lines coming out of one mouth of the wormhole must enter the other mouth of the same wormhole. We know from Gausss divergence theorem that the sign of the effective charge changes from + to depending on whether the flux lines leave or enter the wormhole mouths that you can picture as little 2-dimensional spherical surfaces in ordinary 3-dimensional space. These spheres may themselves not be simply connec but may have little wormhole handles on them as well. It may be Ezekials vision of wormholes on wormholes. This might mean only a single electromagnetic flux in which the weak and strong forces are simply topological variations of the same theme. That would be a nice unification hardly less speculative than what you see on Brian Greenes Elegant Universe on NOVA. One famous Dead Physicist said Elegance is for tailors. If NOVA was produced by BBC they might call it Saville Row physics very fashionable. Therefore, there are equal amounts of matter and anti-matter but the corresponding wormhole mouths with equal and opposite quantized flux charges are attached to different parallel brane universe next door to each other! There is still another problem solved here. It is the problem that Richard Feynman tried and failed and All The Kings Men have not solved it to this day.The problem is the infinite self-energy of the point electron. Why a point electron you ask? Because, there was, hitherto, no way to prevent the extended electric charge distribution of a non-point electron from exploding! This unsolved problem is more than one hundred years old! Its solved now. The Lorentz-Abraham stress is simply the strongly short-range attractive exotic vacuum dark matter zero point fluctuation positive pressure from negative zero point energy density where the throats of Wheelers wormhole geons are like the vortex cores of superfluids where the vacuum coherence field drops to zero on the scale of a coherence length. Note that the trapped near field quantized Faraday flux lines of virtual spin 1 quanta in coherent states are contained in the Meissner effects penetration depth that is no more than one over the square root of two of the coherence length like in a Type II superconductor.WHEN ASTRONOMERS ENCOUNTERED the cosmic acceleration, their first reaction was to attribute it to the so-called cosmological constant. Notoriously introduced and then retrac by Einstein, the constant represents the energy inherent in spaceMaybe cosmic acceleration isnt caused by dark energy after all but by an inexorable leakage of gravity out of our world.Dont bet on that ENRON stock.Although our standard model of cosmology has been confirmed by recent observations, it still has a gaping hole: nobody knows why the expansion of the universe is accelerating.Speak for yourself George.=== === Subject: : analysis problem........show that f(x) = x sin(x) is not uniformly continuous--------------------i want to choose x_1,x_2,epsilon.thusi want to show|x_1 - x_2| |f(x_1)-f(x_2)|>=e_0but i can't..........=== === Subject: : Re: analysis problem........>show that f(x) = x sin(x) is not uniformly continuous--------------------i want to choose x_1,x_2,epsilon.thusi want to show|x_1 - x_2| |f(x_1)-f(x_2)|>=e_0but i can't.......... Whenever you talk about continuity (or uniform continuity), specify the open set on which thefunction is defined. Let's assume this is R, the real numbers. Suppose I claim f is uniform continuous butyou are challenging my claim. The definitionof uniform continuity says that for any epsilon > 0,there exists an epsilon > 0 such that|f(x1) - f(x2)| < epsilon whenever x1, x2 in R and |x1 - x2| < delta. If f is not uniform continuous, then some epsilon > 0 existssuch that no suitable delta can be found.You, the challenger, are supposed to produce an epsilon which will fail.You might pick epsilon = 1 or epsilon = 10^(-100) or epsilon = pi/sqrt(2).but must make your choice known. Your disproof should similarly start with something like Let epsilon = 0.5. Having heard your choice of epsilon, it is my jobto produce a corresponding delta which will work foryour epsilon. Perhaps I pick delta = MIN(1, epsilon/100).This will be positive whenever epsilon > 0Observe that my choice of delta is allowedto depend on your epsilon but not on x1, x2. You, the challenger, now show I am lying.Find a way to choose x1 and x2, using my delta,such that |x1 - x2| < delta but |f(x1) - f(x2)| is large.You are showing that, no matter how small my delta is,you can find two close points x1, x2 such thatf(x1) and f(x2) are at least epsilon apart. It may help to graph f.-- John Adams served two terms as Vice President and one as President, but lostreelection. Later his son became President despite losing the popular vote.That son lost his reelection attempt badly. Now history is repeating itself.pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California=== === Subject: : Re: analysis problem........> show that f(x) = x sin(x) is not uniformly continuous--------------------i want to choose x_1,x_2,epsilon.thusi want to show|x_1 - x_2| |f(x_1)-f(x_2)|>=e_0but i can't..........That isn't what you need to show.Given d, e_0 there exists x_1,x_2 such that ...-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland=== === Subject: : What is normalized surface area?What does normalized surface area mean?you divide the authalic radius squared by a^(1/3)*b^(2/3) youwill have the normalized surface area radius.So for the WGS84 spheroid, a= 6378.137,b=6356.7523 and (Surface area radius) Sr= 6371.00718, 6371.00718^2/(6378.137^(1/3)*6356.7523^(2/3))=++ 6378.14980383181.What is that?-------------- .b3[CapitalYAcute] .bd[Paragraph] KORNET -------------=== === Subject: : Re: simple equation solving, with a twist?> This is a problem which origina in chemistry, but being that math lover> that I am I quickly tried to find a way to turn it into a pure math problem.> I succeeded, but now I don't know how to solve it (other than trial and> error or similar... you'll understand what I mean), so any help would be> apprecia. I am particularly interes in whether this can be programmed> into a calculator--I have a TI-83+--so that would also be of interest.Say we have some equations with a number of variables. For example:5x - 3y = 212> 3y - z = 7Here's the twist: we don't want to solve the set of equations (infact, in> this case they are not solvable). Instead, we want to find what, say, 5x - z> is equal to. Now, in this case it is easy, we can simply add the bot to> the top, and we get 5x - z = 219. We can do similar things for other numbers> etc. In general, you are allowed to multiply any equation by a constant, add> any two equations, and subtract any two equations.But, with, say, 5 equations and 7 variables, it can get tricky. Each> equation doesn't have every variable in it; most of them have 3 or 4, which> may make it easier. But, any hints on how to solve this in a general case> (even though this isn't necessarily solvable for any expression)?One idea I had for programming my calculator for doing it was using a> matrix, e.g. we could represent the above example by5 -3 0 212> 0 3 -1 7We can add another row if we want, for the expression we want to get...5 0 -1 ??But I don't know what to do with it after that.Any hints etc. would be greatly apprecia.> If you have five equations in seven unknowns, you will have twodegrees of freedom. This means that usually five of the variables canat least be solved in terms of the other two, the values of which areleft open. This is referred to as underdetermination. So it's just aquestion of which of the variables you choose to keep open and thensolve for everything else in terms of them. What this actually meansis that there will be an infinite number of independent solutions,depending on what values the open variables are assigned. Equationsof this type never have unique solutions.=== === Subject: : how do circular combination/permutation different from normal combination/permutation?Dear all,I know the normal combination and permutation formular, suppose we select mfrom n,then they can be represen as C(n, m) and P(n, m), where n should be ontop of m;But how about circular combination and permutation? Can anybody tell me theequations?Thank you,=== === Subject: : Re: how do circular combination/permutation different from normal combination/permutation?>I know the normal combination and permutation formular, suppose we select m>from n,>then they can be represen as C(n, m) and P(n, m), where n should be on>top of m;But how about circular combination and permutation? Can anybody tell me the>equations?A circular permutation (i.e. a way to arrange objects in a circle, with (ABCD) considered the same as (BCDA) but not (DCBA)) becomes an ordinary permutation if you pick one of the chosen items to be first. So the number of circular permutations of n distinct objects taken m at a time is just P(n,m)/m.Since order doesn't matter at all in combinations, I'm not surewhat you mean by a circular combination.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Permutation GroupsI'm stuck on two problems involving the ORDER of permutation groups. I knowwhat to do when they are written in disjoint cycle form but not when they arewritten like this. (1 2 3 5)(2 4 5 6 7)The answer is 12and(3 4 5)(2 4 5)The answer is 2=== === Subject: : Re: Pure abstract demonstration by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i122RNg07786;>> The biggest possibility is in some kind of Your mistake.>> However lets check Your revelations for n=3:>>You claim Your parameters:>> T=3^u *abt once A=a^3 and B=b^3 or 3^(3u-1)*b^3>>Already it is known for n=3, that 3 should divide Z;X or Y >>so lets take 3/X ; X=T+B ; B=3^(3u-1) b^3>>also does Your parameters fits ? Ls=(T+A+B)^3=(T+A)^3 +(T+B)^3=Rs ? >> Ls= T^3 +3T^2(A+B) +3T(A+B)^2 +(A+B)^3>> Rs= 2T^3 +3T^2(A+B) +3T(A^2 +B^2) +A^3 +B^3>>then Rs-Ls= T^3 -6TAB -3AB(A+B) = 0 ; T^3 -3AB(2T+A+B) = 0>>3^3u a^3 b^3 t^3 -3*a^3 3^(3u-1)*b^3 [2*3^u abt +a^3 +3^(3u-1) b^3]=0>>and really axtracting 3^3u a^3 b^3 we'll have:>> t^3 - (2*3^u abt +a^3 +3^(3u-1) b^3] = 0>> What is just Your f(t) for n=3 and for the fall X/3>>Now You suggest Y^2 -YX +X^2 = s^3, what is also correct,>>but how to express everything in a;b;t;p parameters?>> I can see, that X+Y in this case should be some cube too:>> X+Y = T+B+T+A = 2T+A+B = 2*3^u abt +a^3 +3^(3u-1) b^3>>also Your t^3 = X+Y and once Y^2 -YX +X^2 = s^3 so Z = t*s>> now t*s = T+A+B; t^3 = 2T +A+B: T = t^3 -ts = t(t^2 -s);>>but also once T = 3^u abt = t(t^2 -s): t^2 -s = 3^u ab>>and finally s = t^2 -3^u ab *****(how smart are those numbers)>> What You sugess then before:>> Ls= s^3 = X^2 -XY +Y^2 =Rs>>Ls=(t^2 -3^u ab)^3 =>> = t^6 -3*t^4 3^u ab +3*t^2 3^2u a^2 b^2 - 3^3u a^3 b^3>>Rs={3^u abt +3^(3u-1) b^3}^2 -[3^u abt +3^(3u-1) b^3][3^u abt +a^3] +>> +{3^u abt + a^3}^2 ......>> I can feel, that there is really some more accounts as one web>>page could hold so I'll do it on some paper and come back>> very soon>> Juan C.C.> Dear Mr. Juan and anybody else, I should just accent some methodical mistake: My parameters creating relatively simple primary equation f(t): with some falls: 1.) for X;Y;Z not divided: t^n - n^u abpt - a^n - b^n = 0 2.) for Z/n : n^(nu-1) t^n - n^u abpt - a^n - b^n = 0 3.) for X/n : t^n - n^nu abpt - a^n - n^(nu-1) b^n = 0 ( for Y/n see only inversed values ) If we succed to get some t as natural number so we'll get for sure s^n as s = t^(n-1) - n^u abp ...............(*) My method involves double cing check of that same relation. It should be next to be check the p^5 , however again from (*) n^u abp = t^(n-1) - s and so on p should be dissolved as natural number. So on, the only really obstacles are with 1.);2.);3.) falls and with their equations. I hope to be more carefull and correct in my future targets but with Eisenstein criterion at the moment I can only simply assure u parameter bigger or equal to 2. Thank You for Your attention Ro === === Subject: : Re: intriguing numbers. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i122ROo07815;>i just read something in a magazine (forgot what title.) a small>true or not. because it seems very interesting. two numbers intrigued>me. one is the number (psi) and they said that its the so-called>last finite number. the number right before infinity. and a very>short mention on something called the end number. the end number is>the highest in the kingdom of numbers and makes absolute infinity a>dust on its shoulder. nothing is higher than the end number because by>theory it is the last number. now i was very intrigued and suddenly it>became my favorite number. so please verify this for me.Oh my God I beg of you to find the magazine title and issue number; please, I *need* that magazine :)Really, this message all by itself might make it to a top 100 math humor list :)=== === Subject: : Constructions of a HeptagonSee the construction of a Heptagon:http://hptgn.tripod.com/=== === Subject: : Re: need help in understanding Torkel's ZFC comment5. Is ZFC an attempt to formalize Set Theory (which also avoids the> paradoxes)? Where have authors been able to construct formal proofs> in ZFC?Get a grip, idiot:You know, you'd make a really lousy teacher. (In case - or rathersince - you haven't figured it out already, I do already know theanswers to my 6 questions. And, in spite of what you might have heardelsewhere, it IS a good idea to ask a question that you know theanswer to, to see where the other person's head is at - or to smoke'em out.)> Inspired by Whitehead and Russell's monumental Principia Mathematica,> the Metamath Proof Explorer has over 3,000 completely worked out proofs> in logic and [ZFC] set theory. > Essentially everything that is possible to know in mathematics can be> derived from a handful of axioms known as Zermelo-Fraenkel set theory.If you believe that, then I have a bridge that I'd like to show you(on sale).> which is the culmination of many years of effort to isolate the> essential nature of mathematics and is one of the most profound> achievements of mankind. Great! Because my system is a lot better. Wow - one of the mostprofound achievements of mankind (and outdone.)*blush*PS Where can I find quotes about how great ZFC is? I want to includesystem fixes its mistakes and is a lot better.> Taken from theMetamath Proof Explorer Home Page > http://metamath.planetmirror.com/mpegif/mmset.html F.=== === Subject: : Re: need help in understanding Torkel's ZFC comment>So are my questions not well-defined? Unclear? Too hard? Too>embarrassing? Why can't anyone answer a few simple questions about>ZFC? People have - the fact that you don't believe the answers doesn't> mean that the questions have not been answered. (For example,> you continue to ask what the inference rules of ZFC are, after> a detailed explanation of why the question makes no sense;> why ZFC does not contain inference rules, any more than> there are inference rules of group theory.)Not so. You only responded to one or two of them, which includes theabove response that only avoids the question. To say that thequestion makes no sense basically contradicts what you said earlier:In what seems to me to be the most common formalizations there are asmall number of purely logical rules of inference - sometimes justmodus ponens, sometimes modus ponens plus universal generalization,depending on how one prefers to set up ones formal system. In otherwords, authors have to add them ad hoc. That is true. So why doesn'tit make sense to ask if it would be better to have them in ZFC per seinstead of having to add them in different ways in differenttreatments?These are 6 fairly simple questions with mostly yes/no answers. Youhave avoided them by ignoring them or declaring them not validquestions. It would take only a minute to answer all 6. Why not justinsert the answers after the questions? (anyone)1. Would ZFC be better if it included its rules of inference?2. Does ZFC violates Occam's Razor to have axiom schemes when goodold rules of inference will do?3. Is it better to replace a set of axioms with a smaller set ofaxioms and rules of inference that can be used to derive them astheorems?4. And what if these same rules of inference could be used to generatetheorems from the Theory of Computation and to auatically constructcomputer programs that implement functions from number theory? Wouldthat enhance its usefulness?5. Is ZFC an attempt to formalize Set Theory (which also avoids theparadoxes)? Where have authors been able to construct formal proofsin ZFC? Would it be better if they did?6. Where do people talk about using rules of inference instead ofaxiom schemes in ZFC, or in general?> ************************=== === Subject: : Re: need help in understanding Torkel's ZFC commentX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>>So are my questions not well-defined? Unclear? Too hard? Too>>embarrassing? Why can't anyone answer a few simple questions about>>ZFC? People have - the fact that you don't believe the answers doesn't>> mean that the questions have not been answered. (For example,>> you continue to ask what the inference rules of ZFC are, after>> a detailed explanation of why the question makes no sense;>> why ZFC does not contain inference rules, any more than>> there are inference rules of group theory.)Not so. You only responded to one or two of them, which includes the>above response that only avoids the question. To say that the>question makes no sense basically contradicts what you said earlier:>In what seems to me to be the most common formalizations there are a>small number of purely logical rules of inference - sometimes just>modus ponens, sometimes modus ponens plus universal generalization,>depending on how one prefers to set up ones formal system. No it doesn't. You _really_ need to learn to read much more carefully.In fact ZFC does not contain any rules of inference. That does notcontradict the previous paragraph - there I'm not talking about ZFC,I'm talking about common _formalizations of_ ZFC.The disctinction has been explained to you several times. Someonewho understands logic better than the rest of us should _really_be a little quicker.>In other>words, authors have to add them ad hoc. That is true. So why doesn't>it make sense to ask if it would be better to have them in ZFC per se>instead of having to add them in different ways in different>treatments?Because people studying set theory are interes in _set theory_,not in proof theory. (Of course someone could be interes inboth, but they're not the same topic.) Set theory is a mathematicaltheory, and like all mathematical theories it takes the notion oflogical consequence of as already understood.>These are 6 fairly simple questions with mostly yes/no answers. You>have avoided them by ignoring them or declaring them not valid>questions. It would take only a minute to answer all 6. Why not just>insert the answers after the questions? (anyone)1. Would ZFC be better if it included its rules of inference?Have you asked this question before? Last I recall you were stillinsisting that ZFC _does_ include inference rules...>2. Does ZFC violates Occam's Razor to have axiom schemes when good>old rules of inference will do?Whether it violates Occam's razor strikes everyone but you as a sillyquestion. More important is the question of why it's better to haveaxioms schemes instead of doing the same thing with inferencerules. The answer is that we want to separate the _logic_ fromthe _set theory_ - if the set theory included inference rulesthen we couldn't study the set theory in an (equivalent!)formal logical system with different inference rules.>3. Is it better to replace a set of axioms with a smaller set of>axioms and rules of inference that can be used to derive them as>theorems?No.(You want one-word answeyou got one. An explanation for theno is above - doesn't really matter, since you've given no reasonfor a yes.)>4. And what if these same rules of inference could be used to generate>theorems from the Theory of Computation and to auatically construct>computer programs that implement functions from number theory? Would>that enhance its usefulness?What if.>5. Is ZFC an attempt to formalize Set Theory (which also avoids the>paradoxes)? Yes.> Where have authors been able to construct formal proofs>in ZFC? Would it be better if they did?Why would anyone care? People studying set theory areinteres in what does and what does not follow from theaxioms, not in formal proofs.>6. Where do people talk about using rules of inference instead of>axiom schemes in ZFC, or in general?Again, why would anyone care. Recasting the axiom schemesas inference rules seems like a bad idea.************************************************=== === Subject: : Re: need help in understanding Torkel's ZFC comment5. Is ZFC an attempt to formalize Set Theory (which also avoids the> paradoxes)? Where have authors been able to construct formal proofs> in ZFC?> Get a grip, idiot:Inspired by Whitehead and Russell's monumental Principia Mathematica,> the Metamath Proof Explorer has over 3,000 completely worked out proofs> in logic and [ZFC] set theory, interconnec with more than 200,000> hyperlinked c-references. . . .> Taken from theMetamath Proof Explorer Home Page > http://metamath.planetmirror.com/mpegif/mmset.htmlF.What does that prove?Inspired by the work of Andrew Wiles, the C-B System solves theGoldbach Conjecture, the Twin Prime Conjecture, and a whole host ofpreviously unsolved problems of mathematics...Taken from: The Charlie-Boo System Manual=== === Subject: : Re: need help in understanding Torkel's ZFC comment> There's a link to download a copy in the View or Download section of> the page.Thanks. I'll check it out.Ok, I read the paper, Mechanizing Set Theory - Cardinal Arithmeticand the Axiom of Choice and it is like so many papers onformalizing/auating mathematics: There's no examples! It givesformal definitions, which is fine, but who's to say they mean anythingor that they did anything with them? They also give what they say aretheorems derived by the system. Same comment goes. But they don'teven pretend to give the proof that the system supposedly generates(or verifies)!This kind of nonsense is way too common in computer sciencepublications. (It also explains why these systems are never availableto the public - except the simple Resolution proof findewhich ofcourse are very limi in scope though fun to play with.)BTW My formalizations tend to be pretty short (see my earlier postingswhere I give expressions for some of ZFC), in contrast to what thispaper gives as definitions. If you really do formalize somethinglike ZFC (or Recursion Theory, Theory of Computation, ProgramSynthesis, et. al.) you quickly find that there are common themes thatyou can abstract out and the resulting expressions are often a lotshorter than the equivalent expressed in semi-formal set notation. For example, much of ZFC is of the form There is a set X such thatfor all Y, Y is an element of X iff predicate(Y) which is primitiveto my whole formalization. I use the notation P(x)[Q(a,b)] where P isthe predicate and Q is the base, either a set or formula, or even aclass, Turing Machine, English sentence, etc. I use it throughout myformalization of computer science and logic.Any more BS papers for me to read? :)=== === Subject: : Re: need help in understanding Torkel's ZFC comment> But if the system does verify them, then fine - that's half of the> job.Right. Metamath is a proof checker, not a theorem prover.> But I don't see offhand how to get a copy or access it over the> internet to try it out.There are source files and executables for Windows, OS X, and Unix athttp://au.metamath.org/#downloads> And when I looked for a proof using ZFC axioms, I saw tons of other> rules being used.The development uses 24 axiom schemas: 3 for propositional calculus, 14for predicate calculus with equality, and 7 for ZFC. http://au.metamath.org/mpegif/mmset.html#axioms> How about if you point me to the best example or two of proofs using> ZFC that is given - preferably not just proving part of ZFC with> another part?The square root of 2 is irrational: http://au.metamath.org/mpegif/sqr2irr.html=== === Subject: : Re: need help in understanding Torkel's ZFC commentHow about if you point me to the best example or two of proofs using> ZFC that is given - preferably not just proving part of ZFC with> another part? The square root of 2 is irrational:> http://au.metamath.org/mpegif/sqr2irr.html Q: What does that have to do with ZFC? There are 42 lines and> offhand it doesn't look like any of them are from ZFC.A: This theorem was proved from axioms: ax-1 ax-2 ax-3 ax-mp ax-4 > ax-5 ax-6 ax-7 ax-gen ax-8 ax-9 ax-10 ax-11 ax-12 ax-13 ax-14 > ax-15 ax-16 ax-17 ax-ext ax-rep ax-un ax-pow ax-reg ax-inf.> (http://au.metamath.org/mpegif/sqr2irr.html)As I said elsewhere, 19 of these are axioms outside of ZFC, and Idon't think you really need ZFC to prove that sqrt(2) is irrationalanyway. This is NOT a good example of a formal use of ZFC, IMEO, andif anything shows that ZFC alone doesn't suffice.(I will look more closely now that I can concentrate on the bestexample.)F.=== === Subject: : Re: need help in understanding Torkel's ZFC comment> How about if you point me to the best example or two of proofs using> ZFC that is given - preferably not just proving part of ZFC with> another part?The square root of 2 is irrational:> http://au.metamath.org/mpegif/sqr2irr.htmlWhat does that have to do with ZFC? There are 42 lines and offhand itdoesn't look like any of them are from ZFC.=== === Subject: : Re: need help in understanding Torkel's ZFC comment> What does that have to do with ZFC? There are 42 lines and offhand it> doesn't look like any of them are from ZFC.In the Metamath Proof Explorer we derive the postulates or axioms ofcomplex arithmetic as theorems of ZFC set theory. [1]The bot of sqr2irr.html reads: This theorem was proved from axioms:[...] ax-ext ax-rep ax-un ax-pow ax-reg ax-inf. Those are Megill'snames for the axioms of ZFC (minus ax-ac, the Axiom of Choice). Youhave to drill down a bit to see how they're actually used.sqr2irr depends on ax0re, which depends on 0r, ..., which depends on ax-inf:THEOREM DESCRIPTION------- -----------sqr2irr The square root of 2 is irrational.ax0re 0 is a real number.0r The constant 0_R is a signed real.1pr The positive real number 'one'.1q The positive fraction 'one'.enqex The equivalence relation for positive fractions exists.niex The class of positive integers is a set.omex The existence of omega (the class of natural numbers).zfinf A standard version of the axiom of infinity (using definitions)inf4 A standard version of the axiom of infinity (expanded to primitives)axinf Axiom of Infinity (with the fewest number of different variables)ax-inf Axiom of Infinity. An axiom of Zermelo-Fraenkel set theory.[1] http://us.metamath.org/mpegif/mmcomplex.html=== === Subject: : Re: need help in understanding Torkel's ZFC comment> What does that have to do with ZFC? There are 42 lines and offhand it> doesn't look like any of them are from ZFC.In the Metamath Proof Explorer we derive the postulates or axioms of> complex arithmetic as theorems of ZFC set theory. [1]Ax-1 et. al. are derived from ZFC? Where?> sqr2irr depends on ax0re, which depends on 0r, ..., which depends on ax-inf.I just don't see that they in fact all start with ZFC, but show meotherwise if you can. Can you take Ax-1 for starters?Looking at this proof (the square root of 2 is irrational) alone, Isee that:1. It uses 19 axioms (their term) outside of ZFC, confirming mycontention that ZFC alone doesn't do it. (I don't think ZFC is reallyneeded for this proof at all, actually.)2. Some of the axioms don't actually capture the semantics sta andneeded. For example, Ax-17 is |-(P => all x P) with descriptionAxiom to quantify a variable over a formula in which it does notoccur. But how does |-(P => all x P) signify that x does not occurwithin P? How can this proof checker know to check something in theuse of an axiom that is not specified? (They show all of thesepresumably valid proofs, but it would be instructive to type in aninvalid proof and see what happens!)I avoid low level problems such as this by using a higher levelformalization (as I say, abstracting out common principles),resulting in a shorter, simpler formalization. Trying to stay within100 year old syntax and semantics is a noble (and sentimental)gesture, but, IMHO, is pointless. (Just the simple proof of thesquare root of 2 being irrational becomes unwieldy.)After generations of mathematicians trying to come up with theultimate system, Turing et. al. showed us that the most generalmathematical system possible (that can be carried out) is a verysimple programming language. I really think that people should moveforward with that idea, to develop shorter, simpler, more powerfulformalizations. That is central to my research.For example, I think that set theory and logic should have beencombined long ago. There is so much redundancy (overlap) betweenthem. In my own software, they are one and the same. There is onlyone DeMorgan's Rule (which also applies to solving problems in theTheory of Computation, Program Synthesis, and other branches oftheoretical Computer Science as well.)You know, computer programmers are very accused to learning fromexperience in the development of a new system, and how this experiencetranslates into replacing the earlier design with a much better one. So much, in fact, that we use fast prototyping: We assume that thefirst design will be thrown out, and take various shortcuts to more orless simulate the use of the first attempt. Then, after we discoverits strengths and weaknesses, the real, long term design isimplemen.Trying to use 100 year old set theory is like using software thathasn't been upda in 100 years - and we can all relate to that!http://www.arxiv.org/html/cs.lo/0003071> [1] http://us.metamath.org/mpegif/mmcomplex.html=== === Subject: : Re: need help in understanding Torkel's ZFC comment> In any case, the 6> questions that I ask don't actually refer to my alternative. Why not> address them? They are of no apparent interest. If you want to learn about ZFC,why not study the subject?=== === Subject: : Re: need help in understanding Torkel's ZFC commentIn any case, the 6> questions that I ask don't actually refer to my alternative. Why not> address them? They are of no apparent interest.You're not interes in whether there are shortcomings or possibleimprovements to a system that you use so much?> If you want to learn about ZFC, why not study the subject?I'm doing that right now.=== === Subject: : Re: need help in understanding Torkel's ZFC comment> You're not interes in whether there are shortcomings or possible> improvements to a system that you use so much? I am indeed interes in possible improvements of the publictransport system, but not in any active way.> I'm doing that right now. OK, so I take it that the news postings are a form of relaxation.=== === Subject: : Re: need help in understanding Torkel's ZFC commentYou're not interes in whether there are shortcomings or possible> improvements to a system that you use so much? I am indeed interes in possible improvements of the public> transport system, but not in any active way.Do you specialize in ZFC more than you specialize in use of the publictransportation system? Do you see a big difference? Or do you thinkthat's a fair analogy?(To quantify: # of people as involved in ZFC as you vs. # of peopleinvolved in public transportation as you. Approximately the same?)> I'm doing that right now. OK, so I take it that the news postings are a form of relaxation.Relaxation? Well yeah, they are relaxing, actually. Don't you thinkposting to news groups is relaxing?=== === Subject: : Re: need help in understanding Torkel's ZFC comment> Do you specialize in ZFC more than you specialize in use of the public> transportation system? Specialize how?> (To quantify: # of people as involved in ZFC as you vs. # of people> involved in public transportation as you. Involved how?> Relaxation? Well yeah, they are relaxing, actually. Don't you think> posting to news groups is relaxing? Right, so your study of ZFC takes place outside news.=== === Subject: : Re: need help in understanding Torkel's ZFC commentCharlie-Boo says...>1. Would ZFC be better if it included its rules of inference?For mathematical purposes, the theory *is* its set of theorems. Howyou describe those theorems as axioms/axiom schemas/rules of inferencedoesn't really matter much. The standard view is to try to separatethe rules of *logic*, which are the same for any first-order theory,from the *axioms*, which are different for different theories. Theadvantage is that you can prove facts about all first-order theories(such as compactness) and then they auatically apply to ZFC, orto PA, or to GNB, etc.But for most purposes, it doesn't really matter how you describe thetheory.--Daryl McCulloughIthaca, NY=== === Subject: : Re: need help in understanding Torkel's ZFC comment> Charlie-Boo says...1. Would ZFC be better if it included its rules of inference?For mathematical purposes, the theory *is* its set of theorems. How> you describe those theorems as axioms/axiom schemas/rules of inference> doesn't really matter much.But if different authors use different rules of inference, thetheorems (and thus the theory) may vary. And if some are laterdiscovered to be inconsistent - oops. Why reinvent the wheel? Thoseare just some of the problems caused by using an incomplete system andeach author having to fill in the gaps.> The standard view is to try to separate> the rules of *logic*, which are the same for any first-order theory,> from the *axioms*, which are different for different theories. The> advantage is that you can prove facts about all first-order theories> (such as compactness) and then they auatically apply to ZFC, or> to PA, or to GNB, etc.Yes, if the rules of inference are given. But Zermelo et. al. nevergave rules of inference (although unintentionally including somemisconstrued as axioms.)> But for most purposes, it doesn't really matter how you describe the> theory.There's also the question of efficiency - minimizing the number ofaxioms (dropping redundant ones). There are plenty of ramificationsto using a poorly designed system!=== === Subject: : Re: need help in understanding Torkel's ZFC comment|> Charlie-Boo says...|> |>1. Would ZFC be better if it included its rules of inference?|> |> For mathematical purposes, the theory *is* its set of theorems. How|> you describe those theorems as axioms/axiom schemas/rules of inference|> doesn't really matter much.||But if different authors use different rules of inference, the|theorems (and thus the theory) may vary.The logic associa with ZFC is classical first-order logic. There aremultiple formalizations of classical first-order logic. Since it doesn'tmatter which one is used, one doesn't assume that any one of them is theset of inference rules for ZFC.|And if some are later|discovered to be inconsistent - oops.So you're worried that first-order logic might turn out not to have beencorrectly formalized? Or that it might turn out to be incoherent?|Why reinvent the wheel?Why indeed? First-order logic is an off the shelf component, andrestating it each time you want to present an axiom system would be acase of reinventing the wheel. So would taking your rules of inferenceto include extra stuff besides first-order logic (when you could just aswell simply use a standard set of rules of inference).|Those|are just some of the problems caused by using an incomplete system and|each author having to fill in the gaps.I've read Cohen's book on the continuum hypothesis and the axiom ofchoice. I've seen various papers referring to ZFC in journals. I don'tremember any of them having to fill in gaps. I think you're stillthinking that ZFC should include some stuff that simply isn't consideredpart of it.|> The standard view is to try to separate|> the rules of *logic*, which are the same for any first-order theory,|> from the *axioms*, which are different for different theories. The|> advantage is that you can prove facts about all first-order theories|> (such as compactness) and then they auatically apply to ZFC, or|> to PA, or to GNB, etc.||Yes, if the rules of inference are given. But Zermelo et. al. never|gave rules of inference (although unintentionally including some|misconstrued as axioms.)For most purposes, giving rules of inference for first-order logic isunnecessary. Zermelo wasn't trying to implement ZFC on a computer oranything like that, so he didn't need formal rules. He wasn't trying todo syntactic proof-theory. He just needed an ordinary mathematician'sunderstanding of the quantifiers and connectives, just like the way wewould reason about any other first-order theory.Axiom schemes like the axiom scheme of replacement are coun as axiomsbecause they have content. They restrict which structures can count asmodels. Ordinary rules of inference in first-order logic, in order to bevalid, need to be valid for all possible models, whether they are modelsof ZFC or not.|> But for most purposes, it doesn't really matter how you describe the|> theory.||There's also the question of efficiency - minimizing the number of|axioms (dropping redundant ones). There are plenty of ramifications|to using a poorly designed system!Efficiency for what purpose? I can't think of many purposes for whichhaving fewer axioms would always help in efficiency. More compact storageof the axiom set is not a very important consideration.=== === Subject: : Re: need help in understanding Torkel's ZFC commentX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>> Charlie-Boo says...1. Would ZFC be better if it included its rules of inference? For mathematical purposes, the theory *is* its set of theorems. How>> you describe those theorems as axioms/axiom schemas/rules of inference>> doesn't really matter much.But if different authors use different rules of inference, the>theorems (and thus the theory) may vary. No. You really haven't been paying attention:The theorems are the logical consequences of the axioms.There are various different ways to formalize the notion oflogical consequence, which have different inference rules,but which are all _equivalent_ - if A is a logical consequenceof B in one (sound and complete) formal proof system thenit is also a logical consequence of B in any other (soundand complete) formal proof system.It's remarkable how many times one needs to explain thesethings to you. (I mean it's remarkable, given your assertionthat you understand logic so much better than the restof us - if you didn't make cracks like that people mightbe a little more patient with their explanations...)>And if some are later>discovered to be inconsistent - oops. Why reinvent the wheel? Those>are just some of the problems caused by using an incomplete system and>each author having to fill in the gaps.> The standard view is to try to separate>> the rules of *logic*, which are the same for any first-order theory,>> from the *axioms*, which are different for different theories. The>> advantage is that you can prove facts about all first-order theories>> (such as compactness) and then they auatically apply to ZFC, or>> to PA, or to GNB, etc.Yes, if the rules of inference are given. But Zermelo et. al. never>gave rules of inference (although unintentionally including some>misconstrued as axioms.)For the fourth time: This is because ZFC is a mathematical theory,not a formal proof system. Mathematical theories do not containinference rules.>> But for most purposes, it doesn't really matter how you describe the>> theory.There's also the question of efficiency - minimizing the number of>axioms (dropping redundant ones). Which of the standard axioms for ZFC follows from the others?>There are plenty of ramifications>to using a poorly designed system!************************=== === Subject: : showing sequence convergesSuppose you want to show lim(n -> oo) |f-f_n| = 0. Also suppose you knowthat lim(k,n -> oo) |f_n(k) - f_n| = 0 (i.e. f_n is Cauchy) for n(k) anincreasing sequence, and also that lim(k -> oo) |f-f_n(k)| = 0 also. Can youthen conclude thatlim (n -> oo) |f-f_n| = 0 since |f-f_n| <= |f-f_n(k)| + |f_n(k) - f_n|. Thisseems like it should work, but on the far right-hand side you need k -> ooin addition to n -> oo (so I'm not entirely sure if it's legit). Thanks!=== === Subject: : question about strictly increasing integer sequencesHello.Let S be the set of functions from N to N which are strictlyincreasing, that is to say, if f is a function in S then f(n+1)>f(n)for all n>1 in N.If S' is an arbitrary subset of S, is it necessarily true that thereis a least element of S' (in the lexicographical sense)?(This is not homework. Thanks very much in advance for help)=== === Subject: : Re: question about strictly increasing integer sequences>Let S be the set of functions from N to N which are strictly>increasing, that is to say, if f is a function in S then f(n+1)>f(n)>for all n>1 in N.>If S' is an arbitrary subset of S, is it necessarily true that there>is a least element of S' (in the lexicographical sense)?Hint: construct a set of functions that does not contain f(n)=n, butcontains functions that agree with it up to any given natural number.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : want to improve my math skillsI was wondering if anyone could point me to a website, or book (withthe answers) that would test me on problem solving. Like word problemsimprove my problem solving skills but for some weird reason I can'tfind anything on the web. PLEASE HELP!!!! please feel free to email mewith your responses as well. Thank you very much.=== === Subject: : Re: The Lost Proof of FermatEgads! I am under TROLL ATTACK!!!Calling all trollsCalling all trollsWake up and smell the logic!x^3 + y^3 = (x+y)*[(x+y)^2 - 3*x*y ]3^3 + 4^3 = (3+4)*[(3+4)^2 - 3*3*4 ]3^3 + 4^3 = 91 = 7*[49 - 36 ] = 7*13 = 91Wake up ya @#$%% **&&^ TROLLS!!!!Are there any REAL mathematicians here???=== === Subject: : Re: The Lost Proof of FermatUtilizing the generalized equation (x+y)*[(x+y)^(p-1) + [{(x^p+y^p)/(x+y)} - (x+y)^(p-1)] ] becomes:x^3 + y^3 = (x+y)*[(x+y)^2 - 3xy] x^5 + y^5 = (x+y)*[(x+y)^4 -5x^3 y -5(xy)^2 -5x y^3 ] x^7 + y^7 = (x+y)*[(x+y)^6 -7yx^5 -14x^4 y^2 -21(xy)^3 -14x^2 y^4 - 7xy^5 ] In general, x^p + y^p = (x+y)*[(x+y)^(p-1) - p*f(x,y) ]=== === Subject: : Re: The Lost Proof of Fermat (x+y)*[(x+y)^(p-1) + [{(x^p+y^p)/(x+y)} - (x+y)^(p-1)] ] > becomes:x^3 + y^3 = (x+y)*[(x+y)^2 - 3xy] Let's see, for x=2 and y=3, left side is 8 + 27 = 35 and the right side is (5)*[5^2 - 12] = 5*[13] = 65.I don't think it works.> In general, x^p + y^p = (x+y)*[(x+y)^(p-1) - p*f(x,y) ]Multiply through your right side by your (x+y) factor. You get (x+y)^p - p*(x+y)f(x,y). Obviously, if you try to expand (x+y)^p, you will get an x^p term and a y^p term, which is what you have on the left side... then you magically subtract off an f(x,y) which you don't really describe.Perhaps you want to know the factorizationx^p + y^p = (x+y)( x^p - x^(p-1)y + x^(p-2)y^2 - x^(p-3)y^3 + ... + y^p )(prime p only).It's clean, easy to prove, no messy coefficients... and fairly well-known.J=== === Subject: : Re: The Lost Proof of Fermat===>>B > A>>A^x + B^x>>[A^x + B^x]^[1/x]>>L'Hopital's Rule:>>Limit f(x)/g(x) = Limit f'(x)/g'(x)> Aha! What is lim_{x->1} x^2/(x + 1)?> It is lim_{x->1} 2x/1 = 2. Very neat> Ok but the above representation of L'Hopitals rule is over-simplified. It would be more accurate to say that IF lim f'(x)/g'(x) x->a = c in R then lim f(x)/g(x) x->a = c.=== === Subject: : Re: The Lost Proof of Fermat Ok but the above representation of L'Hopitals rule is over-simplified.> It would be more accurate to say that IF lim f'(x)/g'(x) x->a = c in R> then lim f(x)/g(x) x->a = c.Not sure whether you're deliberately trying to mess with people, or ifyou're just stupid, but you can *EASILY* look this stuff up.=== === Subject: : Re: The Lost Proof of Fermat===>Ok but the above representation of L'Hopitals rule is over-simplified.>>It would be more accurate to say that IF lim f'(x)/g'(x) x->a = c in R>>then lim f(x)/g(x) x->a = c.> Not sure whether you're deliberately trying to mess with people, or if> you're just stupid, but you can *EASILY* look this stuff up.I agree. However even though everyone can easily look it up, it is also very easy to misinterpret what it says. The most common mistakes (besides from forgetting to verify that the preconditions are satisfied) seems to be people thinking that lim f(x)/g(x) = lim f'(x)/g'(x) (for x->a), sometimes leading to the false conclusion that if f'(x)/g'(x), x->a doesn't converge then the same is true for f(x)/g(x).=== === Subject: : Re: The Lost Proof of Fermat>>L'Hopital's Rule:>>Limit f(x)/g(x) = Limit f'(x)/g'(x)>Ok but the above representation of L'Hopitals rule is over-simplified. >It would be more accurate to say that IF lim f'(x)/g'(x) x->a = c in R >then lim f(x)/g(x) x->a = c.Yes, if by more accurate you mean wrong.-- I'm not interes in mathematics that might have anythingto do with reality. -- Russell Easterly, in sci.math=== === Subject: : Re: The Lost Proof of Fermat===>L'Hopital's Rule:>>Limit f(x)/g(x) = Limit f'(x)/g'(x)>Ok but the above representation of L'Hopitals rule is over-simplified. >>It would be more accurate to say that IF lim f'(x)/g'(x) x->a = c in R >>then lim f(x)/g(x) x->a = c.> Yes, if by more accurate you mean wrong.> In my reply I assumed the reader was familiar with the rule (considering f(x) / g(x) = lim f'(x) / g'(x) (for x->a). Therefore I didn't write the Let f and g be real functions defined near a point a, and assume that f(x) -> 0 for x -> a and g(x) -> 0 for x -> a. If the fraction f'(x) / g'(x) -> c in R for x -> a then f(x) / g(x) -> c for x -> a.=== === Subject: : Re: The Lost Proof of Fermat>Let f and g be real functions defined near a point a, and assume that >f(x) -> 0 for x -> a and g(x) -> 0 for x -> a. If the fraction f'(x) / >g'(x) -> c in R for x -> a then f(x) / g(x) -> c for x -> a.Or if f(x) -> oo and g(x) -> oo. L'Hospital gets misunderstood enoughto warrant not quoting it wrong whenever somebody might read the falsestatement and arrive to misconceptions such as displayed by theoriginal poster.-- I'm not interes in mathematics that might have anythingto do with reality. -- Russell Easterly, in sci.math === Subject: : re:The Lost Proof of Fermat===The general equation IS correct...x^p + y^p equals (x+y)*[(x+y)^(p-1) + [{(x^p+y^p)/(x+y)} - (x+y)^(p-1)] ] for all p and n >= 1(x+y)*[(x+y)^2 + {[(x^3+y^3)/(x+y)} - (x+y)^2] ](x^3+y^3)/(x+y) = (x+y)*[x^2 -xy +y^2]/(x+y) = x^2 -xy + y^2so the equation becomes:(x+y)*[(x+y)^2 + (x^2 -xy +y^2) -(x+y)^2] which becomes(x+y)*[x^2 -xy + y^2]this equals:x^3 + y^3 :lol: :lol: :lol:----== Pos via Newsfeed.Com - Unlimi-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =---=== === Subject: : Re: The Lost Proof of Fermatx^n + y^n = z^n x^3 + y^3 = (x+y)[x^2-xy+y^2] A more general identity for primes p, and numben: A*B means A times B A/B means A divided by B + and - , we all know... follow the order of operations: x^p + y^p equals (x+y)*[(x+y)^(p-1) + [{(x^p+y^p)/(x+y)} - (x+y)^(p-1)] ] for all p and n > 1=== === Subject: : Re: The Lost Proof of Fermat> x^p + y^p> equals> (x+y)*[(x+y)^(p-1) + [{(x^p+y^p)/(x+y)} - (x+y)^(p-1)] ]No. No, it doesn't. My biggest hint was when you added we all know.=== === Subject: : Re: The Lost Proof of Fermat x^3 + y^3 = (x+y)[x^2-xy+y^2] > A more general identity for primes p, and numben: > x^p + y^p > equals > (x+y)*[(x+y)^(p-1) + [{(x^p+y^p)/(x+y)} - (x+y)^(p-1)] ] > for all p and n > 1 Unfortunately, that equation is NOT more general than the sum of subes formula. What happens when you plug in p=3 in your general equation? (If it were truly a generalization, it would reduce to the sum of cubes formula.) Your equation says x^3 + y^3 = (x+y) [(x+y)^2 + [ (x^3+y^3)/(x+y) ] - (x+y)^2 ]which is not the same thing. I'm not sure why would would have (x+y)^2 - (x+y)^2in the expression, since those go to 0. The rest of the equation just says that x^3+y^3 = (x+y) * (x^3+y^3) / (x+y) , which, again, we can just cancel. [[ Essentially, it is saying something like 10 = 7(5 + 10/7 - 5) ]]J=== === Subject: : Re: LISP routines to do symbolic differentiation> apologies to 2nd followup of my own post. What does> the right syntax is arctan(sin(x))... and that DOES> give an answer different from same as input.> For Maple 7, the> integral of arctan(sin(x))+arctan(1/sin(x)) comes out> (-1/2 pi)*x.> I think this is wrong.Yes. The necessary simplification is not done auatically.One has to use int(combine(arctan(sin(x))+arctan(1/sin(x))),x);-- Thomas RichardMaple SupportScientific Computers GmbHhttp://www.scientific.de=== === Subject: : waq LISP routines to do symbolic differentiation> Daniel Lichtblau (private email) politely questioned what> I had written about what I had done in Mathematica last night. Mathematica actually gets a fine answer to the question> posed below, essentially the same step-function S, times x.> (By contrast, Commercial Macsyma gives atan2(0,0) genera and (free)> Maxima returns the input expression unchanged. Maple 7 also> declines to do it. So Mathematica is the winner here.)What I tried to integrate was D, the DERIVATIVE of S wrt x,> which is zero except in those infinitely-many places where> sin(x)=0, when D should not be zero. (But Mathematica thinks it is).> More precisely, D is 0 except at a denombrable infinite numberof values of x where it is not defined. Now if you do theconverse step of integrating the derivative of S, then youmust have a theory of integration in mind: forexample for Lebesgue integration of functions (and not distributions),since D(x) is 0 except at a 0-measure set you will get 0+constant.Therefore I don't see any problem if your prefered CAS returns 0.=== === Subject: : Re: Running Matlab from Maple> I am trying to run Matlab from Maple (version 9). Though Matlab> and the path variables have been setup properly, I have the following> error message while trying to use ode45(). Can somebody help me??`Matlab/valid_os`();> true> with(Matlab);> [chol, closelink, defined, det, dimensions, eig, evalM, fft, getvar,> inv, lu, ode45, openlink, qr, setvar, size, square, transpose](T, Y) := ode45(rigidode, 0..12, [0, 1, 1], `tol`=.001);> Error, (in ode45) evaluation failed ode45(rigidode, 0 .. 12, [0, 1,> 1], tol = 0.1e-2)Which Matlab version is that?Notice that Maple 9 supports only Matlab 6.5 (R13) (which I don't have).Your input looks correct to me. It works with older combinations of bothtools.-- Thomas RichardMaple SupportScientific Computers GmbHhttp://www.scientific.de=== === Subject: : Running Matlab from Maple by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i150HqZ18916;===The Matlab version that I have is 6.5.0.180913a Release 13.madhu=== === Subject: : Re: AlgebraThanks to whomever replied to me.I found a good site that helps.www.algebrahelp.com> Hi everyone I'm trying to remember Algebra & I'm not succeding too well. Any good web> sites to look at? Thanks for your help. Dan=== === Subject: : Order of the complex numbers and its consequencesIn this work we define the complex number order and the power of complexnumber base and real number index by means of the limit of a monotonictrigonometric function;and from these we define the pi number.In: http://www.telecable.es/personales/carloman/=== === Subject: : Mass matrix for ode15s from simulink block by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i150HrT18934;=== I have an S-function block in simulink wherein I am solving someset of DAEs. Since the system of ODEs are stiff, I am using ode15s asthe solver, with variable step size. But I also need to pass the massmatrix to ode15s. Can somebody tell me how one can set the optionsin ode15s(@filename,tspan,x0,options,...) when using a simulink block.madhu=== === Subject: : Middle school mathI am currently a University student majoring in elementary education. I am very interes in teaching math in the middle schools; grades 6through 8. If anyone happens to know about what topics of math thisage group learns about, I would love to know. I have tutored a fewstudents from various cities, and it seems as though the topicscovered differ greatly depending on what school district you are in. Any information on general math knowledge in these grades, as well asany good websites or lesson plans I could look at, would me very muchapprecia.-- tml=== === Subject: : Re: Middle school mathTry a webinternet search for Mathematics Standards for your State. The topics should include Number Sense, Measurement, Probablility and Data(maybe measurement and data), Algebra, Geometry. The topics will probablyEXCLUDE Intermediate Algebra, Trigonometry, and Calculus. G CTrakakaka1@aol.com wants to know:>I am currently a University student majoring in elementary education. >I am very interes in teaching math in the middle schools; grades 6>through 8. If anyone happens to know about what topics of math this>age group learns about, I would love to know. I have tutored a few>students from various cities, and it seems as though the topics>covered differ greatly depending on what school district you are in. >Any information on general math knowledge in these grades, as well as>any good websites or lesson plans I could look at, would me very much>apprecia.-- tml=== === Subject: : Re: Middle school mathKaija asked,> I am very interes in teaching math in the middle schools; grades 6> through 8. If anyone happens to know about what topics of math this> age group learns about, I would love to know. I have tutored a few> students from various cities, and it seems as though the topics> covered differ greatly depending on what school district you are in.Nothing can cure the variance from school district to school district in theUni States, but you can prepare now to teach middle school mathematicsWELL, wherever you end up teaching.A good source of reasonable standards for different age ranges ishttp://www.cde.ca.gov/cdepress/standards-pdfs/ mathematics.pdfwhich is based on reasonable international standards.> Any information on general math knowledge in these grades, as well as> any good websites or lesson plans I could look at, would me very much> apprecia.sitehttp://math.berkeley.edu/~wuas soon as possible. That will lead you to more good reading.Hope this helps!-- tml=== === Subject: : Re: Help...Never seen this beforeI believe you are referring to a product symbol (not sure of the exact> terminology for it).> My Harper Collins DICTIONARY OF MATHEMATICS, by Borowski and Borwein,uses the phrase continued product while Mathworld says:http://mathworld.wolfram.com/Product.html(To all interes: For examples, look to number theory, wherecontinued products are common. The totient function or Euler's phifunction [one of the most basic functions in number theory and one ofmy favorites] is a continued product.)Paul-- tml=== === Subject: : Re: Help...Never seen this before> Sheila, I'm wondering if Potterwasp isn't referring to a summation,> uppercase sigma. Either that, or educate me as to where this sort of> operation, a progressive product might be used.Please. > -- > charlie dickThe right to be left alone -- the most comprehensive > of rights, and the right most valued by a free people. - Justice Louis Brandeis, Olmstead v. U.S. (1928). ****************************************A nice usage for the capital Pi, product notation is in elementary numbertheory.For instance, let p_1, p_2, ... represent the infinite sequence of primenumbers.Every positive integer m can then be written in the form m = Pi(i=1 toinfinity) p_i^(e_i) where the e_i are nonnegative integer exponents, onlyfinitely many of which are nonzero.If n = Pi(i=1 to infinity) p_i^(f_i) is another positive integer, then wecan writegcd(m, n) = Pi(i=1 to infinity) p_i^(min{e_i, f_i})and lcm(m, n) = Pi(i=1 to infinity) p_i^(max{e_i, f_i}).Please excuse the Pi(i=1 to infinity) expression; in mathematicaltypography this would be written entirely similar to capital-sigmanotation, with the sigma replaced by capital Pi, of course.--- Joe-- Delete the second o to e-mail me.-- tml=== === Subject: : TouchMoneyI am a special education teacher in Boulder, CO. and am interes ingetting my hands on a copy of Touch Money. Can you steer me in theright direction? Holly-- tml=== === Subject: : times tablesthere aren't any strategies per say just have them stand in class andsay them other than rewarding them with a trinket it is just somethingthey need to do. You do them a great disservice by not fullyimpressing on them the joy in knowing these facts. Rote is not a badword. Practice and repition are not evil concepts. -- tml=== === Subject: : Prove that 7^203 ends in 07this ia my math homework. been trying to prove it but failed.basically, they told us that the last two digit of 7^2 => 497^3 => 437^4 => 017^5 => 07now i have to prove that the last two digits of 7^201 is 07help!-thanks -- tml=== === Subject: : Re: Prove that 7^203 ends in 07> this ia my math homework. been trying to prove it but failed.basically, they told us that the last two digit of 7^2 => 49> 7^3 => 43> 7^4 => 01> 7^5 => 07now i have to prove that the last two digits of > 7^201 is 07Hint: what does 7^6 end with? 7^7? Is there a pattern? Can you provethat the pattern continues?-- Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karpluslife member (LAB, Adventure Cycling, American Youth Hostels)Effective Cycling Instructor #218-ck (lapsed)Professor of Computer Engineering, University of California, Santa CruzUndergraduate and Graduate Director, BioinformaticsAffiliations for identification only.-- tml=== === Subject: : Re: order of ops> Man Abe! Give the guy a break.What level do you teach? Even in high school the students mess up the> order of operations because of PEMDAS. Mnics don't work for> everyone. Even after the teacher logically shows the students the> right way to handle simplification of expressions involving multiple> inline addition/substraction or multiplication/division they still> make a very common mistake. They do what they want to do first> because they see addition and multiplication as being easier than> subtraction and division. It is a good idea to give another mnic device...how about People> Enjoy McDonald's SAndwiches...? Notice the capitalization. Both> sequences should be presen and the differences should be emphasized> to show the equal level of addition and subtraction. That being said,> it is also important that the students get beyond the mnic device.> Order of operations should be second nature whether you plan on doing> exponential stuff or not.This emphasizes an important point: That PEMDAS, although yieldingcorrect results, can confuse students about the relative priority ofall of these operations. With PEMDAS, there is no qualitativedifference in the precedence that exponentiation has overmultiplication vs. the precedence that multiplication has overdivision, or that addition has over subtraction for that matter. Iagree that multiplication and division should be understood as groupedtogether, and addition and subtraction should be understood as groupedtogether. The acronym I gave earlier, PEMA, where multiplicationsubsumes division and addition subsumes subtraction via thedefinitions of division and subtraction, gives students a better viewof the relative priorities of all of these operations.Unlike the priorities of exponentiation over multiplication and ofmultiplication over addition, the priorities of multiplication overdivision and of addition over subtraction in the left-to-right contextare entirely rote in nature, devoid of providing any understanding.These priorities in this left-to-right context are just ways to applydivision and subtraction without having to understand theirdefinitions.This lack of understanding these definitions is a fundamental problemand more widespread than many might want to believe. I've seen anumber of high school students who have come to me not understandingthat an expression of the form (a/b)*c is equivalent to (ac)/b. Theynever learned the definitions of division and subtraction to fluency(or they never learned them at all). They just don't understand whatthey write. Instilling these definitions is the only fix.Paul-- tml=== === Subject: : Re: order of opsAs earlier sta, this is a difficult discussion to carry. Fluency here, isknowing what to do with an equation or an expression depending on what iswan from the equation or expression; and without having to formally rememberand recite the relevant properties; and without needing to look for therelevant properties in a book. They just must be understood as if they wereacquired language.Again; should I give an example?G C-- tml=== === Subject: : Re: order of opsDisappointing if true that This lack of understanding these definitions is afundamental problemand more widespread than many might want to believe;Are these the same students who learned so well about dealing with fractionsand decimals, and then have great trouble in introductory algebra? The order of operations should become auatic to the student, and mnicdevices like PEMA, or PEMDA, or PEMDAS should be nothing more than some retarding crutch. Basically, at the level of this introductory algebra, after4 or 6 weeks, the student moves from the properties of equality, inequality,and the arithmetic operations, to just knowing what they see when it is on thepage (understanding the expressions upon looking at them each for a fewseconds). If this discussion seems unclear, ask, and I could give an example.G C>This emphasizes an important point: That PEMDAS, although yielding>correct results, can confuse students about the relative priority of>all of these operations. With PEMDAS, there is no qualitative>difference in the precedence that exponentiation has over>multiplication vs. the precedence that multiplication has over>division, or that addition has over subtraction for that matter. I>agree that multiplication and division should be understood as grouped>together, and addition and subtraction should be understood as grouped>together. The acronym I gave earlier, PEMA, where multiplication>subsumes division and addition subsumes subtraction via the>definitions of division and subtraction, gives students a better view>of the relative priorities of all of these operations.Unlike the priorities of exponentiation over multiplication and of>multiplication over addition, the priorities of multiplication over>division and of addition over subtraction in the left-to-right context>are entirely rote in nature, devoid of providing any understanding.>These priorities in this left-to-right context are just ways to apply>division and subtraction without having to understand their>definitions.This lack of understanding these definitions is a fundamental problem>and more widespread than many might want to believe. I've seen a>number of high school students who have come to me not understanding>that an expression of the form (a/b)*c is equivalent to (ac)/b. They>never learned the definitions of division and subtraction to fluency>(or they never learned them at all). They just don't understand what>they write. Instilling these definitions is the only fix.Paul>-- tml=== === Subject: : Fraction StripsTry www.teachervision.com and click on printables-- tml=== === Subject: : Re: General Factorization ToolI'm posting now to make several corrections.> Consider n functions f_1(x), f_2(x),..., f_n(x), wheref_1(x) f_2(x)...f_n = F(x), where F(x) is a polynomial where F(0) = 0,That should be f_1(x) f_2(x)...f_n = KF(x), Also the f functions each must equal 0 when x=0.> and numbers g_1, ..., g_n, k_1, ..., k_n, G and K wherek_1 g_1...k_n g_n = KG, where also(f_1(x) + k_1 g_1)...(f_n(x) + k_n g_n) = K[F(x) + H(x) + G]and H(x) is a polynomial where H(0) = 0.Here you have the factorization k_1 g_1...k_n g_n = KG, balanced against the factorization(f_1(x) + k_1 g_1)...(f_n(x) + k_n g_n) = K[F(x) + H(x) + G],should bef_1(x) f_2(x)...f_n = KF(x) > where the point is that it's *unknown* how factors of K split between> f_1(x) through f_n(x).So you can think of my method as a solution for an unknown> factorization.Here the solutions vary dependent on the ring.For instance, in the ring of algebraic integers dividing K from both> sides *must* giveg_1...g_n = G.I wasn't thinking about unit factors and gave a strong condition whichis unnecessary.The appropriate thing to note is that the factorization exists in thering. > That may seem trivial, but let n=2, k_1 = 3, k_2 = 1, and consider> what can happen in some other ring, like the field of algebraic> numbers:(sqrt(3) g_1)(g_2/sqrt(3)) = G.Notice there are an *infinity* of solutions in the field of algebraic> numbebut just one in the ring of algebraic integers.That's incorrect as there are an infinity of solutions in the ring ofalgebraic integers as well.What's important here is that the factorization shown while availablein the field of algebraic numbers is not available in the ring ofalgebraic integers. > Now then given that you have(f_1(x) + k_1 g_1)...(f_n(x) + k_n g_n) = K[F(x) + H(x) + G],it's forced that f_a(x) *should* have g_a, where 'a' is a counting> number such that 1<=a<=n, as a factor in any ring where you have the> single factorizationg_1...g_n = G.I'll call such a ring a complete ring.It turns out that you need an additional condition, which is that withj and k some counting numbers less than or equal to n, iff_j(x)/f_k(x) is in the ring for some value of x, then it must be inthe ring for all possible values of x within that ring.It's a consistency requirement between the factors. === === Subject: : Re: Complexity of solving N non-linear equationsContent-Length: 1460Originator: rusin@vesuvius>I must admit I know a little about it, and I really nead some help.>>Can anyone tell me what is the complexity of solving a set of N>>non-linear equations (simultaneous equations - having all variables in>>each equation). The equations themselves are very complex - contains>>integrals of a sample maximum functions.In any great degree of generality, it is not computable at all. > The solution to Hilbert's 10th problem says that there is no> algorithm to solve diophantine equations in several variables: given > a polynomial P(x_1,...,x_n) with integer coefficients, there is no > way to tell whether there are positive integers x_1,...,x_n such that > P(x_1,...,x_n) = 0. > This is equivalent to saying there is no way to tell whether there are > real numbers t_1,...,t_n such that P(t_1^2, ..., t_n^2)^2 + sum_{j=1}^n sin^2(pi t_j^2) = 0This needs a qualification;'no way' = 'no algorithm that decides in each instance'Once you bound the range of the variables (which is usuallyreasonable in applications), the problem becomes finite and(in principle) tractable.Moreover, the question was most likely about solutions inreal numbeand the result needs to be given only to acertain accuracy, and the equations are of a different type -all reasons for making the diophantine result not applicable.Arnold Neumaier=== === Subject: : Re: Complexity of solving N non-linear equationsContent-Length: 1521Originator: rusin@vesuviusI must admit I know a little about it, and I really nead some help.> Can anyone tell me what is the complexity of solving a set of N> non-linear equations (simultaneous equations - having all variables in> each equation). The equations themselves are very complex - contains> integrals of a sample maximum functions.Depends. I'll assume you're talking about real values (not just integers).If you are just talking about multivariate polynomials -without-any fancy integration or optimization constraints, the best you canhope for is Grobner Bases computation (like generalized gaussian elimination but deals with nonlinear terms). This problem is EXPSPACE-complete (definitely takes exponential time, most likely will take exponential space) (Kuehnle and Mayr '96). So, it's bad.But if you're adding in integration (I suppose multivariate integration) then this is equivalent to solving a PDE (right?),which I've heard is (in general) is undecidable. So it's worse than bad.But you may have restrictions on your system (low bounded degree, integration on one variable at a time, lots of symmetry (despite it being dense)). So it may not be so bad.But if all you care about is an approximation, then I'm sure a bunch of numerical analysis techniques can be thrown at it to give reasonable answewhere you can take more time to get better answers. as mentioned elsewhere see a numerical analysis (or computational sciences) text.-- Mitch Harris(remove q to reply)=== === Subject: : Norm-one complemen Banach spacesContent-Length: 333Originator: rusin@vesuvius friends.Where can I find a proof of the (classical) factasserting that every dual Banach spaceis norm-one complemen in its bidual?To be more precise, the fact is as follows:For every normed space X, there is a projectionof norm 1 from X''' onto (the natural image of) X'.--Alexander E. Gutmangutman@math.nsc.ru=== === Subject: : Re: Norm-one complemen Banach spacesContent-Length: 578Originator: rusin@vesuvius> friends.Where can I find a proof of the (classical) fact> asserting that every dual Banach space> is norm-one complemen in its bidual?To be more precise, the fact is as follows:> For every normed space X, there is a projection> of norm 1 from X''' onto (the natural image of) X'.Let J: X to X'' be the natural (isometric) injection.Then J^*: X''' to X' is a (norm 1) projection. QED (modulo checking the algebraic details).ZTW=== === Subject: : Behaviour of Entire FunctionsEpigone-thread: skonblunstrimContent-Length: 180Originator: rusin@vesuviusIf f is an entire function such that for each z either |f(z)| is less than or equal to 1 or |f'(z)| is less than or equal to 1, then mustf be a constant or a linear polynomial ?=== === Subject: : Re: Behaviour of Entire FunctionsContent-Length: 896Originator: rusin@vesuvius>If f is an entire function such that for each z either |f(z)| is less >than or equal to 1 or |f'(z)| is less than or equal to 1, then must>f be a constant or a linear polynomial ?If |f(z)| > 1 everywhere, then |f'(z)| <= 1 everywhere, and that caseis trivial by Liouville's Theorem. Otherwise we may assume wlog that |f(0)| <= 1. Now for any complexnumber z, let s = max { t in [0,1]: |f(tz)| <= 1}. We havef(z) = f(sz) + int_L f'(w) dw where L is the straight line segmentfrom sz to z, and since |f(sz)| <= 1 and |f'(w)| <= 1 for w in L,|f(z)| <= 1 + |z|. Cauchy's estimates complete the proof.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Erdos Number Project updateContent-Length: 1370Originator: rusin@vesuviusThe bi-annual update of the Erdos Number Project web site has beencomple. The URL is http://www.oakland.edu/~gman/erdoshp.htmlBriefly, this project studies the subject of collaboration inmathematical research in general and the collaborations of Paul Erdos(1913-1996) in particular. On the web site are the list of the 509people who have written joint papers with Erdos (they have Erdos number1) and the lists of their other coauthors (a total of 6984 of them --the people with Erdos number 2). The data are organized in severalways, and various statistics are summarized. We also include an updateto the complete bibliography of Erdos (numbering more than 1500 papers),preprints of papers about this subject, and dozens of links to relamaterial, including the flurry of recent serious research activity oncollaboration graphs.Further information is contained in the README file, available at thesite.As always, we want to know about corrections, additions, and relainformation.-- Jerrold W. Gman, Professor VOICE: (248) 370-3443 Department of Mathematics and Statistics FAX: (248) 370-4184 Oakland University FLESH: 346 SEB Rochester, MI 48309-4485 E-MAIL: gman@oakland.edu WEB: http://www.oakland.edu/~gman/=== === Subject: : sums of binominalsContent-Length: 729Originator: rusin@vesuviusI'm interes in formulas of sums like: S(n,k) := sum_{i=0}^nbinom{k}{i} forall n,kn N with nleq kLudwig Schlaefli inven the so called Schlaefli-Triangle induced by thefact that S(0,k) = 1 and S(n,k) = S(n-1,k-1)+S(n,k-1) 1 1 2 1 3 4 1 4 7 81 5 11 15 16I'm looking for equations or upper and lower bounds of S(n,k) assuming n isvariable. I.e. from kryptology we know S(n,k) < 2^{k(-n/k*log_2(n/k)-(1-n/k)*log_2(1-n/k))-1}.Thanks for any suggestions=== === Subject: : Re: Complexity of solving non-linear equationsContent-Length: 2958Originator: rusin@vesuvius>> In any great degree of generality, it is not computable at all.>> The solution to Hilbert's 10th problem says that there is no>> algorithm to solve diophantine equations in several variables: given>> a polynomial P(x_1,...,x_n) with integer coefficients, there is no>> way to tell whether there are positive integers x_1,...,x_n such that>> P(x_1,...,x_n) = 0.>> This is equivalent to saying there is no way to tell whether there are>> real numbers t_1,...,t_n such that>> P(t_1^2, ..., t_n^2)^2 + sum_{j=1}^n sin^2(pi t_j^2) = 0A more sophistica construction (see Wang, The Undecidability of theExistence of Zeros of Real Elementary Functions, Journal of the ACM21(4), 586-589 (1974), available from) shows thatthere is no algorithm to tell whether a function f(x) of one variable(expressed in terms of x, rational numbepi and the sine function,using addition, multiplication and composition) has a zero.>This needs a qualification;>'no way' = 'no algorithm that decides in each instance'Yes. In fact there is a particular, rather complica, polynomialP(x,y,z_1,...,z_n) for which there is no algorithm such that, givenpositive integers x and y, the algorithm will decide correctly in finitetime whether there exist positive integers z_1,...,z_n such thatP(x,y,z_1,...,z_n) = 0.>Once you bound the range of the variables (which is usually>reasonable in applications), the problem becomes finite and>(in principle) tractable.One has to be careful with this. If you can't tell whether F(x) hasa zero in (-infinity, infinity), you can't tell whether F(tan(x)) hasa zero in (-pi/2, pi/2).>Moreover, the question was most likely about solutions in>real numbeand the result needs to be given only to a>certain accuracy, and the equations are of a different type ->all reasons for making the diophantine result not applicable.In practical terms, I think the key here is the only to acertain accuracy - i.e. you'd be satisfied with an algorithmsuch that for a function f (in some suitable class) on a compactregion R, and given 0 < epsilon_1 < epsilon_2:1) if there is x in R where |f(x)| < epsilon_1, the algorithm willsupply an x in R where |f(x)| < epsilon_22) if there is no x in R where |f(x)| < epsilon_2, the algorithmwill return no solution3) if neither 1) nor 2) is true, the algorithm is allowed to eithersupply an x with |f(x)| < epsilon_2, or return no solution.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, BC, Canada V6T 1Z2=== === Subject: : Pairs of convex bodies orderContent-Length: 535Originator: rusin@vesuviusIt is well known(Hormander:Sur la fonction dappui des esemblesconvexes dans un espace localment convexe, Theorem 9(pag.185),Arkivfor Matematik Band #3 nr#12 1954) that the space H_o of differences ofsupport functions is a lattice, is this lattice (sigma) complete?.i.e if0 Ooookaaaay..... berserker attacks Soldier #1 with +3 battle axe: 6 HP damage!> Soldier #1 is wounded.> berserker attacks Soldier #1 with +3 battle axe: 4 HP damage> Soldier #1 is gravely wounded.> Soldier #2 comes up.> Soldier #2 attacks berserker with M-16. *critical hit*, 8 HP damage!!> berserker is gravely wounded.> Medical Attendant #1 comes up.> Medical Attendant attacks Solder #1 with bandage: -1 HP damage.I'm not sure where you thought you were going with this. But I think youproved his point.=== === Subject: : Re: first deriviate with FFT, indexing problemsHey,> ...> then force the n/2 mode to be real.> ...This is exactly the step I wan to avoid!Because it is altering the data, I'm introducing an error in theresults. It might be small, but bugs me :)My solution at the moment is to loose hermitianity at this particularsample, and receive therefore some phaseshifts after the inversefouriertransform. I can get rid of this phaseshifts more or less easily, butit introduces additional computation I wan to avoid as well.If there is no better solution than this, I would realy be interesin a good, maybe formal explanation.Maybe Gordon Sande is allready close to the solution.anyway thanks for your effort!Martin> With n input points you only resolve n/2 fourier modes.if f is the result of a forward fft on the n samples of the> fcn to be differentia then the multiplication with ik is as> follows (depends on the way the fft you use arranges the> spectrum output, usually the modes are ordered as> -n/2+1,-n/2+2,...,-1,0,1,2,...n/2 ):for modes k=0,...,n/2> multiply f(k) with ikfor modes k=-n/2+1,...,-1> multiply f(k) with ikthen force the n/2 mode to be real.A backward fft of the array f will contain the> derivative up to a scale factor of 1/n (for most> fft programs).PeterHey there,I have n evenly spaced sample values taken from a bandlimi function.> I assume that sampling has been done properly (nyquist criteria).I like to calculate the first derivative of this function by using> the Derivative Theorem of the Fourier transform:> df[n]/dt <=> -jnF[n]- First I do a Fourier transform. As I only have real values, the> resulting function is hermitian:> f(x) = f*(-x) f* means conjugate complex- Then the multiplication with -j (-sqrt(-1)) is a rotation of 90deg around> the time axis. Easy saying:> re_new = im_old> im_new = -re_old- than I multiply the result values with their index n for example size 8:> index: 0 1 2 3 4 5 6 7 > factor 0 1 2 3 k -3 -2 -1 and there is the problem! what to do at index 4? in order to keep the hermitian property f(4)=f*(4)> I would have to force the imaginary component to 0, but that would alter > the data and therefore lead to a less precise results.> The reason why I want to keep the hermitian property is, that I want> a real function after the inverse Fourier transform. Fortunately this is only a problem when I have an even number of> samples, so I can always zero pad to a odd number.> But that is not very satisfying, and I would appreciate a cleaner solution.> Maybe someone can help me out here!> Thanks a lotMartin=== === Subject: : Re: first deriviate with FFT, indexing problems> Because it is altering the data, I'm introducing an error in the> results. It might be small, but bugs me :)No, you are not introducing any error. The forward FFT outputsthe modes -n/2+1,...,0,...,n/2. The Fourier series, whose coefficientsthe FFT is computing, is a sum of the modes -n/2,...,0,...,n/2.The -n/2 and n/2 modes are real so by forcing f(n/2) to be real youare removing any error that introduced a small complex value at thatindex.Peter=== === Subject: : Re: first deriviate with FFT, indexing problems>Hey,> ...>> then force the n/2 mode to be real.>> ...>This is exactly the step I wan to avoid!>Because it is altering the data, I'm introducing an error in the>results. It might be small, but bugs me :)If you can't set the Nyquist frequency term to zero, then your resultsare probably wrong anyway because of aliasing.When you use FFT, you assume that your signal contains no energyat and above n/2 df. Otherwise, you are not allowed to multiplythe (n/2-1) term by (n/2-1) df, you should distinguish which part of it comes from (n/2-1) df in the signal, and which part from(n/2+1) df, multiply them separately accordingly, and add them.But of course, you need to resample your signal at a higher frequencyto be able to make the distinction.My solution at the moment is to loose hermitianity at this particular>sample, and receive therefore some phaseshifts after the inverse>fourier>transform. I can get rid of this phaseshifts more or less easily, but>it introduces additional computation I wan to avoid as well.If there is no better solution than this, I would realy be interes>in a good, maybe formal explanation.Maybe Gordon Sande is allready close to the solution.anyway thanks for your effort!>Martin> With n input points you only resolve n/2 fourier modes. if f is the result of a forward fft on the n samples of the>> fcn to be differentia then the multiplication with ik is as>> follows (depends on the way the fft you use arranges the>> spectrum output, usually the modes are ordered as>> -n/2+1,-n/2+2,...,-1,0,1,2,...n/2 ): for modes k=0,...,n/2>> multiply f(k) with ik for modes k=-n/2+1,...,-1>> multiply f(k) with ik then force the n/2 mode to be real. A backward fft of the array f will contain the>> derivative up to a scale factor of 1/n (for most>> fft programs). Peter Hey there, I have n evenly spaced sample values taken from a bandlimi function.>> I assume that sampling has been done properly (nyquist criteria). I like to calculate the first derivative of this function by using>> the Derivative Theorem of the Fourier transform:>> df[n]/dt <=> -jnF[n] - First I do a Fourier transform. As I only have real values, the>> resulting function is hermitian:>> f(x) = f*(-x) f* means conjugate complex - Then the multiplication with -j (-sqrt(-1)) is a rotation of 90deg around>> the time axis. Easy saying:>> re_new = im_old>> im_new = -re_old - than I multiply the result values with their index n for example size 8:>> index: 0 1 2 3 4 5 6 7 >> factor 0 1 2 3 k -3 -2 -1 and there is the problem! what to do at index 4? in order to keep the hermitian property f(4)=f*(4)>> I would have to force the imaginary component to 0, but that would alter >> the data and therefore lead to a less precise results. The reason why I want to keep the hermitian property is, that I want>> a real function after the inverse Fourier transform. Fortunately this is only a problem when I have an even number of>> samples, so I can always zero pad to a odd number.>> But that is not very satisfying, and I would appreciate a cleaner solution. >> Maybe someone can help me out here!>> Thanks a lot Martin=== === Subject: : Integration and the equipartition formulaHey,In Kreyzig it says that the integral from 0 to infinity of[x^2*exp[-b*x^2] is the same as the integral from 0 to infinity of1/2b*exp[-b*x^2] but it doesn't say why. Does anybody know themathematical instincts which are behind this claim?[Also Does anyone know where I could find a derivation for theequipartition formula? I tried deriving it but got into mess.]thank you,Jon=== === Subject: : Re: cycloic polynomialsX-AUTHid: minutsil> Hi there!!!> SOS!!! Help me !!!! I Know nothing about cycloic polynomials but I> have to make a small research for a course in number theory. Can> anyone give me some help? Any sites I could study for this?> Thanks!!!The fastest way you can get an answer is if you do a google search.I've found http://mathworld.wolfram.com/CycloicPolynomial.htmlWolfram is the company that produces Mathematica, and you can find a wealth of information at their page www.wolfram.comand at http://mathworld.wolfram.com.However, this is a classical topic in algebra, well understood,so a good book might explain it better. Any graduate textbookin algebra will have it (or advanced undergraduate textbook).Lookup e.g. Lang, or Hungerford.Here's a brief explanation though. You want to solve the equation x^n=1. If you're looking for solution over the set of real numbethen it's trivial,x=1 if n is odd, and x=+/-1 if n is even.Over the set of complex numbers things are more complica.Any poly of degree n, has precisely n roots (counting multiplicities).So x^n-1 has n roots, 1,w_1,...w_n-1. It turns out that theyform a cyclic group, so the roots can be expressed as 1,w,w^2,...w_n-1 where w is a certain w_k. Obviously, w^n=1 (in fact w_k^n=1 for each k, because w_k arethe roots of x^n-1), but maybe there's a poly of smaller degreewhich has w as a root. The minimal polynomial of w is called the n-th cycloic polynomial. I.e. it is the polynomial of smallest degree of whichw is a root of.Ex: n=2: x^2-1 has roots 1 and w=-1. The min poly of -1 is x+1.n=3: x^3-1 has roots 1, w=cos(2*pi/3)+i*sin(2*pi/3) and w^2.The min poly of w is x^2+x+1. Note that x^3-1=(x-1)(x^2+x+1).n=4: x^4-1 has roots 1,i,i^2=-1,i^3=-i, so if we take as generatorw=i, the min poly is x^2+1. Again, this is a factor of x^4-1:x^4-1=(x-1)(x+1)(x^2+1).etc.=== === Subject: : The FFT derivation threadHi fellows,I read the recent post on FFT derivation. I've been trying to get thesame thing working, too - but with less than perfect results (why elsewould I post here :)I follow the treatise taken by Numerical Recipes. The signal in timedomain is in N (N is even) discrete values, h[k] (0<=k<=N-1). They aresampled from a continuous function f such that: h[k] = f(stepsize*k) = f(k) // we use a stepsize of oneNow the discrete Fourier transform is: H[n] = sum(k=1,N-1) h[k]*e^(2*PI*i*k*n/N) // 0<=n<=N-1Then the transformation backwards is simply: h[k] = 1/N * sum(n=1,N-1) H[n]*e^(-2*PI*i*k*n/N) // 0<=k<=N-1Now, my first idea for derivation (computing h'[k]) was to derivativethe backward transformation formula wrt. k: h'[k] = 1/N * sum(n=1,N-1) H[n]*e^(-2*PI*i*k*n/N)*(-2*PI*n/N) //0<=k<=N-1This worked somewhat, but gave some nasty errors: the derivative wasabout 20 times larger than reference derivative (compu usingcentral differences) and it was negative (so it had to be nega byhand). To date, the reasons for these errors is unclear to me!The second candidate idea is the idea introduced in the recent post onFFT derivation on this newsgroup. The idea was to multiply each H[n]by -jn if nN/2 (where j^2=-1) (and dosomething when n==N/2 :) I tried this method, too, but the resultingderivative was wildly oscillating at ends and well... it just wasn'tthe derivative :)So I thought: should the method proposed in the recent post also workfor the transformation formulas in Numerical Recipes?Any description of how to get this FFT derivation thing *working* (andexplanations of what I did wrong) would be nice :) - Mikko KauppilaPS. I know the transformation formulas above are O(N^2) when applieddirectly. I just want to get this derivation thing to *work* beforegoing to the O(N log N) algorithm.=== === Subject: : derived tensor productNNTP-Posting-User: [FlpdK7CkhjkTHuPrQ86bX+xi7sq7GbYm]Cancel-Lock: sha1:sxx8qRig/YFU7b59RbbGq56lHcg=Originator: israel@math.ubc.ca (Robert Israel)Let R be a commutative ring. Does there exist a nonzero object X inthe (unbounded) derived category for R so that the derived tensorproduct of X with itself is zero? (I think that this is impossible ifR is noetherian, but what if it isn't?) For any n>=2, does thereexist an object X so that the n-fold derived tensor product of X withitself is zero, but the (n-1)-fold derived tensor product is not?Here's a rela non-example: notice that as Z-modules, Q/Z tensor Q/Zis zero. In the derived category of Z, though, I think that thederived tensor product of Q/Z with itself is isomorphic to Q/Z indegree 1 (or degree -1, depending on how you grade things). This iscorrect, right? (I'm not very good with the derived tensor product,but it's legitimate to compute it using flat resolutions, isn't it?)-- J. H. PalmieriDept of Mathematics, Box 354350 mailto:palmieri@math.washington.eduUniversity of Washington http://www.math.washington.edu/~palmieri/Seattle, WA 98195-4350=== === Subject: : Re: derived tensor productOriginator: israel@math.ubc.ca (Robert Israel)Let R be a commutative ring. Does there exist a nonzero object X in> the (unbounded) derived category for R so that the derived tensor> product of X with itself is zero? (I think that this is impossible if> R is noetherian, but what if it isn't?) For any n>=2, does there> exist an object X so that the n-fold derived tensor product of X with> itself is zero, but the (n-1)-fold derived tensor product is not?Here's a rela non-example: notice that as Z-modules, Q/Z tensor Q/Z> is zero. In the derived category of Z, though, I think that the> derived tensor product of Q/Z with itself is isomorphic to Q/Z in> degree 1 (or degree -1, depending on how you grade things). This is> correct, right? (I'm not very good with the derived tensor product,> but it's legitimate to compute it using flat resolutions, isn't it?)--> J. H. Palmieri> Dept of Mathematics, Box 354350 mailto:palmieri@math.washington.edu> University of Washington http://www.math.washington.edu/~palmieri/> Seattle, WA 98195-4350If nobody else does i try a sketch based on my dusty Hartshorne [H]Residues and Duality, Chap II 4 and 5.For derived tensors (or local hyperTor) i think you need some leftboundedness (as you need flat resolutions). Tor^i(F,F) is thehomology of the complex F@F (@=tensor product) and beeing 0 meansit is (homotop to) exact. Now taking F0 the left-most homology ofF itself, my idea would be (sorry, forget to say: i have no exactproof) to use a spectral sequence argument (like [H] p.94 = EGA)to show: that assertion implies that the usual Tor^i(F0,F0) (i ein the category of modules) is 0 for all i. Which means your Q is transla to the usual case. May be for thatyou find something sound in Matsumura. Otherwise i would try touse adjointness of Ext and Tor with morphisms to k = R/m, as inthe noetherian case one may assume R to be local. Hmmm.The second (on n-1 etc) i do not quite understand.For the last i would again apply (derived) adjointness of Homand @ as 0 -> Z -> Q -> Q/Z -> 0 is an injective resolution ofZ (i e: Q and Q/Z are injective modules). May be to get yourdesired identity you need that all usual Ext^i and Tor^i overZ vanish for i < 2.Hopefully the raesoning is not to odd.---remove the no for mail=== === Subject: : Enriched categoriesOriginator: israel@math.ubc.ca (Robert Israel)I'm looking for information on n-categories and categories withHom-sets enriched in topological, metric and Hilbert spaces. Whatbooks (or papee.g. arXiv) do you recommend for learning thisstuff?-I=== === Subject: : Re: Enriched categoriesOriginator: israel@math.ubc.ca (Robert Israel)>I'm looking for information on n-categories and categories with>Hom-sets enriched in topological, metric and Hilbert spaces. What>books (or papee.g. arXiv) do you recommend for learning this>stuff?-IFor n-categories probably you can do no better than to turn to theonline book by T. Leinster (and references therein)Higher Operads, Higher Categories math.CT/0305049For enriched category theory you can turn to Borceux's 3-volume bookon category theory. I think the one you're interes is volume II.There is also the older (1970?)Dubuc - Kan extensions in enriched category theoryLet me just remark that the (usual) category of topological spaces isnot cartesian closed, so it's not very convenient for enrichedcategory theory. There are replacements, though, such as the categoryof compactly genera spaces, that are enough for, homotopy theory,for example. Much more so with the category of Hilbert spaces andbounded linear operators. Restricting to Hilbert-Schmidt operatorsdoes not work because the identity is not Hilbert-Schmidt.With my best G. Rodrigues=== === Subject: : Re: Enriched categoriesOriginator: israel@math.ubc.ca (Robert Israel)> I'm looking for information on n-categories and categories with> Hom-sets enriched in topological, metric and Hilbert spaces. What> books (or papee.g. arXiv) do you recommend for learning this> stuff?-IFor n-categories try Jon Baez's homepage. www.math.ucr.edu follow thelinks (sorry, being lazy). There are a couple of expository papers there,and he says he wants to start writing some free to internet book. Alsosearch through his weekly finds for more handwaving arguments.=== === Subject: : This week in the mathematics arXiv (26 Jan - 30 Jan)Content-Disposition: inlineOriginator: israel@math.ubc.ca (Robert Israel)Here are this week's titles in the mathematics arXiv, available at: http://front.math.ucdavis.edu/ http://front.math.ucdavis.edu/submissionsThis week in the mathematics arXiv may be freely redistribuwith attribution and without modification.Titles in the mathematics arXiv (26 Jan - 30 Jan)-------------------------------------------------AC: Commutative Algebra-----------------------math.AC/0401306 Alberto Corso, Craig Huneke, Daniel Katz, Wolmer Vasconcelos: Integral closure of ideals and annihilators of homologyAG: Algebraic Geometry----------------------math.AG/0401409 Alexander Braverman: Instanton counting via affine Lie algebras I: Equivariant J-functions of (affine) flag manifolds and Whittaker vectorsmath.AG/0401405 Giorgio Ottaviani, Elena Rubei: Resolutions of homogeneous bundles on P^2math.AG/0401403 Amit Khetan, Carlos D'Andrea: Implicitization of rational surfaces using toric varietiesmath.AG/0401401 Jaroslaw Wlodarczyk: Simple Hironaka resolution in characteristic zeromath.AG/0401396 Dirk Siersma, Mihai Tibar: Singularity exchange at infinitymath.AG/0401394 Marian Aprodu: Green-Lazarsfeld's Gonality Conjecture for a Generic Curve of Odd Genusmath.AG/0401378 J. P. Pridham: The structure of the pro-l-unipotent group of a smooth varietymath.AG/0401375 Mireille Martin-Deschamps: Caract`eres num'eriquesmath.AG/0401374 Willem Veys: Arc spaces, motivic integration and stringy invariantsmath.AG/0401367 Chien-Hao Liu, Kefeng Liu, Shing-Tung Yau: $S^1$-fixed-points in hyper-Quot-schemes and an exact mirror formula for flag manifolds from the extended mirror principle diagrammath.AG/0401366 Holger Brenner: There is no Bogomolov type restriction theorem for strong semistability in positive characteristicmath.AG/0401364 Donu Arapura: Partial Regularity and Amplitudemath.AG/0401354 A. B. Goncharov: Euclidean scissor congruence groups and mixed Tate motives over dual numbersmath.AG/0401344 J. P. Pridham: The deformation theory of representations of the fundamental group of a smooth varietyAP: Analysis of PDEs--------------------math.AP/0401410 Kari Astala, Matti Lassas, Lassi Paivarinta: Calderon's inverse problem for anisotropic conductivity in the planemath.AP/0401397 Claudia Garetto, Guenther Hoermann: Microlocal analysis of generalized functions: pseudodifferential techniques and propagation of singularitiesmath.AP/0401395 S. Coriasco, E. Schrohe, J. Seiler: Realizations of Differential Operators on Conic Manifolds with Boundarymath.AP/0401384 Alessandro Giacomini: Size effects on quasistatic growth of fracturesmath.AP/0401383 Alessandro Giacomini, Marcello Ponsiglione: Discontinuous finite element approximation of quasistatic crack growth in finite elasticitymath.AP/0401376 Katrin Wehrheim: Banach space valued Cauchy-Riemann equations with totally real boundary conditionsmath.AP/0401355 J. Ginibre, G. Velo: Scattering Theory for the Schrodinger Equation in some external time dependent magnetic fieldsmath.AP/0401352 Axel Gruenrock, Hartmut Pecher: Bounds in time for the Klein-Gordon-Schroedinger and the Zakharov systemmath.AP/0401315 Andreas Wannebo: Removability: First order Sobolev space, PDE solutions, holomorphic functions. - Sobolev space construc in a new wayAT: Algebraic Topology----------------------math.AT/0401424 Boris Chorny: A generalization of Quillen's small object argumentmath.AT/0401346 Randy McCarthy, Vahagn Minasian: On Triples, Operads, and Generalized Homogeneous FunctorsCA: Classical Analysis and ODEs-------------------------------math.CA/0401418 Yuan Xu: On discrete orthogonal polynomials of several variablesmath.CA/0401417 Yuan Xu: Generalized translation operator and approximation in several variablesmath.CA/0401416 Yuan Xu: On polynomials of least deviation from zero in several variablesmath.CA/0401382 Y. Chen, J.C. Griffin, M.E.H. Ismail: Generalizations of Chebyshev Polynomials and Polynomial Mappingsmath.CA/0401348 Alexei Yu. Karlovich, Andrei K. Lerner: Commutators of singular integrals on generalized $L^p$ spaces with variable exponentmath.CA/0401310 Ilia Krasikov: New bounds on the Hermite polynomialsCO: Combinatorics-----------------math.CO/0401404 Nathan Reading: Lattice congruences of the weak ordermath.CO/0401398 Y. Caro, R. Yuster: A Tur'{a}n Type Problem Concerning the Powers of the Degrees of a Graph (revised)math.CO/0401379 Serkan Hosten, Seth Sullivant: A finiteness theorem for Markov bases of hierarchical modelsmath.CO/0401373 Anders Bjorner, Irena Peeva, Jessica Sidman: Subspace arrangements defined by products of linear formsmath.CO/0401363 Frank Harary, Wolfgang Slany, Oleg Verbitsky: On the Lengths of Symmetry Breaking-Preserving Games on Graphsmath.CO/0401361 Oleg Verbitsky: The First Order Definability of Graphs with Separators via the Ehrenfeucht Gamemath.CO/0401350 Michael Huber: The classification of flag-transitive Steiner 3-designsmath.CO/0401342 Eric H. Kuo: Viterbi Sequences and Polytopesmath.CO/0401339 Pierre Lalonde: Alternating sign matrices with one -1 under vertical reflectionmath.CO/0401313 Alexander V. Karzanov: Integer concave cocirculations and honeycombsCT: Category Theory-------------------math.CT/0401347 Victor Ostrik: Tensor categories (after P. Deligne)CV: Complex Variables---------------------math.CV/0401413 Keizo Hasegawa: Four-dimensional compact solvmanifolds with and without complex structuresmath.CV/0401359 Genevra Neumann: Valence of complex-valued planar harmonic functionsDG: Differential Geometry-------------------------math.DG/0401420 Camille Laurent-Gengoux, Jean-Louis Tu, Ping Xu: Chern-Weil map for principal bundles over groupoidsmath.DG/0401419 Naichung Conan Leung, Xiaowei Wang: Intersection theory of coassociative submanifolds in G_(2)-manifolds and Seiberg-Witten invariantsmath.DG/0401412 Iskander A. Taimanov: Surfaces in the four-space and the Davey--Stewartson equationsmath.DG/0401407 M. Saralegi-Aranguren, R. Wolak: The BIC of a singular foliation defined by an abelian group of isometriesgr-qc/0401112 Antonio N. Bernal, Miguel Sanchez: Smoothness of time functions and the metric splitting of globally hyperbolic spacetimesmath.DG/0401393 David Marin, Manuel de Leon: Classification of Material G-structuresmath.DG/0401386 Piotr T. Chrusciel, Erwann Delay, John M. Lee, Dale N. Skinner: Boundary regularity of conformally compact Einstein metricshep-th/0401206 Jos'e Figueroa-O'Farrill, Joan Sim'on: Supersymmetric Kaluza-Klein reductions of AdS backgroundsmath.DG/0401381 Burt Totaro: The curvature of a Hessian metricmath.DG/0401380 Jorge Cortes, Alexandre M. Vinogradov: Hamiltonian theory of constrained impulsive motionmath.DG/0401372 Henri Anciaux, Ildefonso Castro, Pascal Romon: Lagrangian submanifolds folia by (n-1)-spheres in R^2nmath.DG/0401365 R. R. Isangulov: Isospectral flat 3-manifoldsmath.DG/0401320 Vestislav Apostolov, David M. J. Calderbank, Paul Gauduchon, Christina W. Tonneson-Friedman: Hamiltonian 2-forms in Kahler geometry, II Global ClassificationFA: Functional Analysis-----------------------math.FA/0401411 Matthias Lesch: The uniqueness of the spectral flow on spaces of unbounded self--adjoint Fredholm operatorsmath.FA/0401337 Omran Kouba: On The Interpolation of Injective or Projective Tensor Products of Banach Spacesmath.FA/0401336 Omran Kouba: $H^1$-Projective Banach Spacesmath.FA/0401335 Omran Kouba: L'Application Canonique $J:H^2(X) otimes H^2(X)->H^1(Xotimes X)$ n'est pas Surjective en G'en'eralmath.FA/0401333 Josh Isralowitz: A Characterization Of Compactness in Finite Lp SpacesGR: Group Theory----------------math.GR/0401423 : Capable groups of prime exponent and class twomath.GR/0401357 Kevin Wortman: Quasi-isometric rigidity of higher rank S-arithmetic latticesmath.GR/0401349 Valerij Bardakov, Leonid Bokut, Andrei Vesnin: Twis conjugacy in free groups and Makanin's questionmath.GR/0401328 Edith Adan-Bante: Products of Characters and Finite $p$-groups IIImath.GR/0401324 David Bessis: A dual braid monoid for the free groupmath.GR/0401312 Peter Kropholler, Peter Linnell, Wolfgang Lueck: Groups of small homological dimension and the Atiyah Conjecturemath.GR/0401308 Gadde A. Swarup: Delzant's variation on Scott complexityGT: Geometric Topology----------------------math.GT/0401421 David Futer: Involutions of knots that fix unknotting tunnelsmath.GT/0401399 Saul Schleimer: The disjoint curve propertymath.GT/0401360 Kevin Wortman: Quasiflats with holes in reductive groupsmath.GT/0401351 Meirav Amram, David Garber, Mina Teicher: Local braid monodromies and local fundamental groups of tangen conic-line arrangementsmath.GT/0401345 Sergey Finashin: Refinement of the Seiberg-Witten and the Ozsvath-Szabo 4-dimensional invariants along embedded surfacesmath.GT/0401311 Richard Evan Schwartz: Modular circle quotients and PL limit setsKT: K-Theory and Homology-------------------------math.KT/0401414 Francis Clarke, Martin Cley, Sarah Whitehouse: Algebras of operations in K-theoryLO: Logic---------math.LO/0401343 J. D. Monk: Concerning problems about cardinal invariants on Boolean algebrasmath.LO/0401307 Oleg Pikhurko, Joel Spencer, Oleg Verbitsky: Succinct Definitions in the First Order Theory of GraphsMP: Mathematical Physics------------------------math-ph/0401055 C.Klein, D.Korotkin, V.Shramchenko: Ernst equation, Fay identities and variational formulas on hyperelliptic curvesmath-ph/0401054 Arturo Ramos: Poisson structures for reduced non-holonomic systemsmath-ph/0401053 Remi Carles, Peter A. Markowich, Christof Sparber: Semiclassical asymptotics for weakly nonlinear Bloch wavesmath-ph/0401052 Yuly Billig: Magnetic hydrodynamics with asymmetric stress tensormath-ph/0401051 M. Lorente: Integrable systems on the lattice and orthogonal polynomials of discrete variablehep-th/0401166 B.Eynard, A.Kokotov, D.Korotkin: $1/N^2$ correction to free energy in hermitian two-matrix modelmath-ph/0401050 Armand Wirgin: The inverse crimemath-ph/0401049 K. Pankrashkin: On semiclassical dispersion relations of Harper-like operatorsmath-ph/0401048 Masato Sakakibara: On the Differential equations of the characters for the Renormalization Groupmath-ph/0401047 Frederic Helein, Joseph Kouneiher: The notion of observable in the covariant Hamiltonian formalism for the calculus of variations with several variablesmath-ph/0401046 Frederic Helein, Joseph Kouneiher: Covariant Hamiltonian formalism for the calculus of variations with several variables: Lepage--Dedecker versus de Donder--Weylmath-ph/0401045 V.M. Mekhitarian, V.E. Mkrtchian: A new approach in theoretical physics based on the Einstein covariance principlehep-th/0401198 Branko Dragovich, Zoran Rakic: Path Integral Approach to Noncommutative Quantum Mechanicshep-th/0311150 Carlos Leiva, Mikhail S. Plyushchay: Nonlinear superconformal symmetry of a fermion in the field of a Dirac monopolemath-ph/0401044 A. Cervellino, S. Ciccariello: The Algebraic Approach to the Phase Problem for Neutron Scatteringmath-ph/0401043 Marius Mantoiu, Radu Purice: The Magnetic Weyl Calculuscond-mat/0401440 E D Belokolos, V Z Enolskii, M Salerno: Wannier functions of elliptic one-gap potentialsmath-ph/0401042 Maria J. Esteban: Erratum : Existence of 3D Skyrmions. Complete versionhep-th/0401176 D.Bashkirov, G.Sardanshvily: The BRST extension of gauge non-invariant Lagrangianshep-th/0401172 Thomas Thiemann: The LQG -- String: Loop Quantum Gravity Quantization of String Theory I. Flat Target Spacehep-th/0401157 F. Girelli, T. Krajewski, P. Martinetti: An algebraic Birkhoff decomposition for the continuous renormalization groupquant-ph/0401139 Nevena Ilieva, Heide Narnhofer, Walter Thirring: Finite Supersymmetry Transformationsnlin.PS/0401032 Ioana Bena, Avinash Khare, Avadh Saxena: Soliton Lattice and Single Soliton Solutions of the Associa Lam'e and Lam'e Potentialsmath-ph/0401041 B. Bagchi, C. Quesne: Conditionally exactly solvable potential and dual transformation in quantum mechanicsmath-ph/0401040 H. C. Rosu, O. Cornejo-Perez: Factorizations of ODEs with polynomial nonlinearitiesNA: Numerical Analysis----------------------math.NA/0401415 P. Vertesi, Yuan Xu: Mean convergence of orthogonal Fourier series and interpolating polynomialsmath.NA/0401369 Teijo Arponen, Ben Leimkuhler: An efficient geometric integrator for thermostat anti-/ferromagnetic modelsNT: Number Theory-----------------math.NT/0401406 Jonathan Sondow: A Faster Product for Pi and a New Integral for ln(Pi/2)math.NT/0401392 Simon Kristensen: A metric theorem for restric Diophantine approximation in positive characteristicmath.NT/0401371 Simon Kristensen: Metric Diophantine approximation with respect to planar distance functionsmath.NT/0401362 Pietro Corvaja, Umberto Zannier: On the length of continued fractions for values of quotients of power sumsmath.NT/0401358 Nikolaj Glazunov: Critical lattices, elliptic curves and their possible dynamicsmath.NT/0401356 Joseph H. Silverman: Common divisors of a^n-1 and b^n-1 over function fieldsmath.NT/0401341 Yann Bugeaud, Florian Luca: On the period of the continued fraction expansion of ${sqrt {2^{2n+1}+1}}$math.NT/0401334 Kimberly Spears: Bounds on discriminants with one class per genusmath.NT/0401319 Michael E. Hoffman: Quasi-symmetric functions and mod p multiple harmonic sumsOA: Operator Algebras---------------------math.OA/0401408 Marius Ionescu: Operator Algebras and Mauldin Williams Graphsmath.OA/0401340 S. Walters: Periodic Integral Transforms and C*-algebrashep-th/0401168 Varghese Mathai, Jonathan Rosenberg: T-duality for torus bundles via noncommutative topologyPR: Probability Theory----------------------math.PR/0401422 D.A. Dawson, L.G. Gorostiza, A. Wakolbinger: Degrees of transience and recurrence and hierarchical random walksmath.PR/0401402 Hans-Otto Georgii, Hyun Jae Yoo: Conditional Intensity and Gibbsianness of Determinantal Point Processesmath.PR/0401390 Uwe Franz, Naofumi Muraki: Markov property of monotone L'evy processesmath.PR/0401389 Antar Bandyopadhyay: Bivariate Uniqueness in the Logistic Recursive Distributional Equationmath.PR/0401388 Antar Bandyopadhyay, David J. Aldous: A Survey of Max-Type Recursive Distributional Equationsmath.PR/0401385 Serguei Popov: A note on random Bulgarian solitairemath.PR/0401370 E. Lytvynov: The square of white noise as a Jacobi fieldmath.PR/0401368 Sever Silvestru Dragomir: A General Divergence Measure for Monotonic Functions and Applications in Information Theorymath.PR/0401353 Nicolas Lanchier: Multitype Contact Process with Frozen statesmath.PR/0401309 Panki Kim: Relative Fatou's Theorem for $(-Delta)^{alpha/2}$-harmonic Functions in Bounded $kappa$-fat Open SetQA: Quantum Algebra-------------------math.QA/0401400 Boris Feigin, Andrey Losev, Boris Shoikhet: Riemann-Roch-Hirzebruch theorem and Topological Quantum MechanicsRT: Representation Theory-------------------------math.RT/0401387 Fr'ed'eric Latour: Representations of trigonometric Cherednik algebras of rank 1 in positive characteristicmath.RT/0401332 Harsh Pittie, Arun Ram: A Pieri-Chevalley formula for K(G/B)math.RT/0401331 Harsh Pittie, Arun Ram: A Pieri-Chevalley formula in the K-theory of a G/B-bundlemath.RT/0401330 Halverson, Arun Ram: q-rook monoid algebras, Hecke algebras and Schur-Weyl dualitymath.RT/0401329 Arun Ram: Affine Hecke algebras and generalized standard Young tableauxmath.RT/0401327 Arun Ram: Representations of rank two affine Hecke algebrasmath.RT/0401326 Arun Ram: Skew shape representations are irreduciblemath.RT/0401325 Arun Ram: Standard Young tableaux for finite root systemsmath.RT/0401323 Arun Ram: Calibra representations of affine Hecke algebrasmath.RT/0401322 Arun Ram, Jacqui Ramagge: Affine Hecke algebras, cycloic Hecke algebras and Clifford theorymath.RT/0401318 Persi Diaconis, Arun Ram: Analysis of systematic scan Metropolis algorithms using Iwahori-Hecke algebra techniquesmath.RT/0401317 Rosa Orellana, Arun Ram: Affine braids, Markov traces and the category Omath.RT/0401316 ippe Caldero, Frederic Chapoton, Ralf Schiffler: Quivers with relations arising from clusters (A_n case)math.RT/0401314 Halverson, Arun Ram: Partition AlgebrasSG: Symplectic Geometry-----------------------math.SG/0401377 Katrin Wehrheim: Anti-self-dual instantons with Lagrangian boundary conditions II: Bubblingmath.SG/0401338 Fan Ding, Hansjorg Geiges, Andr'as I. Stipsicz: Lutz twist and contact surgerymath.SG/0401321 Ricardo Casta~no-Bernard: Symplectic invariants of some families of Lagrangian T^3 fibrationsSP: Spectral Theory-------------------math.SP/0401391 J. Bellissard, J. Geronimo, A. Volberg, P.Yuditskii: If they are limit periodic?-- / Greg Kuperberg (UC Davis) / / Visit the Math ArXiv Front at http://front.math.ucdavis.edu/ / * All the math that's fit to e-print *=== === Subject: : Question about Group T.Originator: israel@math.ubc.ca (Robert Israel)Dear all: Let's say G is an h-group if for any natural number n thereexists at most a finite number of subgroups(=sbgps.) of G of indexn (for example finitely genera groups are h-groups). It's easy to show that G is an h-group iff for any normal sbgp. K of G of finite index (=f.i.) and for any finite simple group S there exists only a finite number of normal sbgps. L of K s.t. K/L is isomorphic to S.[1] So let be given an h-group G, a f.i. normal sbgp. K of G and a finite simple group S, and let L1,..., Lr be all the normal sbgps. of K s.t. K/Li isomorphic to S. We can see at once that for all pairof different indexes 1 <= i,j <= r we have that LiLj = K.My question is: can we bound somehow, probably by imposing someadditional conditions on G, or on K, the number r in terms of n?For example, we could demmand that K/Li be a chief factor of G (thus Li is a normal sbgp. OF G which is maximal with respect to being contained in K), and if G is say solvable, or even locally solvable, then S would be abelian...or we could begin by supposing that n is the least natural number for which there is a normal sbgp. K with some quotient isomorphic to S; we could even begin at firstconsidering sbgps. of index AT MOST n, for some natural n, or we could consider only sbgps. Li which are normal IN G, etc.If, for example, we take K = G and we ask how many different normal sbgps. Li of G are there s.t. G/Li is isom. to S, then auatically G/Li is a chief factor of G and thus we should, probably, ask about these factors under so and so conditions.The above question popped up pretty surprisingly during my PhD research and I didn't pay much attention to it back then. Anyinsight, hint, recommendation, quote, etc. will be much apprecia.SaludosTonio[1] Wilson, John S., Groups Satisfying the Maximal Condition forNormal Subgroups, Math. Z. 118, 107-114 (1970), Lemma 1.=== === Subject: : Re: Question about Group T.Originator: israel@math.ubc.ca (Robert Israel)>Dear all:> Let's say G is an h-group if for any natural number n there>exists at most a finite number of subgroups(=sbgps.) of G of index>n (for example finitely genera groups are h-groups). It's easy to >show that G is an h-group iff for any normal sbgp. K of G of finite >index (=f.i.) and for any finite simple group S there exists only a >finite number of normal sbgps. L of K s.t. K/L is isomorphic to S.[1] So let be given an h-group G, a f.i. normal sbgp. K of G and a >finite simple group S, and let L1,..., Lr be all the normal sbgps. >of K s.t. K/Li isomorphic to S. We can see at once that for all pair>of different indexes 1 <= i,j <= r we have that LiLj = K.>My question is: can we bound somehow, probably by imposing some>additional conditions on G, or on K, the number r in terms of n?What exactly is n here please?The only n I can see is in the first papragraph, where it isquantified by a `for all'.I cannot tell whether it is |G:K|, |G:Li|, or something else.Derek Holt.>For example, we could demmand that K/Li be a chief factor of G (thus >Li is a normal sbgp. OF G which is maximal with respect to being >contained in K), and if G is say solvable, or even locally solvable, >then S would be abelian...or we could begin by supposing that n is >the least natural number for which there is a normal sbgp. K with >some quotient isomorphic to S; we could even begin at first>considering sbgps. of index AT MOST n, for some natural n, or we >could consider only sbgps. Li which are normal IN G, etc.If, for example, we take K = G and we ask how many different normal >sbgps. Li of G are there s.t. G/Li is isom. to S, then auatically >G/Li is a chief factor of G and thus we should, probably, ask about >these factors under so and so conditions.>The above question popped up pretty surprisingly during my PhD >research and I didn't pay much attention to it back then. Any>insight, hint, recommendation, quote, etc. will be much apprecia.>Saludos>Tonio[1] Wilson, John S., Groups Satisfying the Maximal Condition for>Normal Subgroups, Math. Z. 118, 107-114 (1970), Lemma 1.>=== === Subject: : approximation theoremIn-reply-to: Epigone-thread: phalzhongskixOriginator: israel@math.ubc.ca (Robert Israel)Am I right, when I'm assuming that the field K is analgebraic function field of one variable over the fieldk?Depending on what you assume, when saying that in Kthe product formula holds, this is a consequence ofthis product formula.If K|k is an algebraic function field your statementabout the constants c_K and c_R is correct. It can beproved using the so-called Strong Approximation Theorem.The dimension of the vector space T in principal can becompu using the Theorem of Riemann-Roch. Howeverwithout further assumptions the Riemann-Roch formulawill contain quantities that may not be known in yoursituation.You find both theorems mentioned in books specializedon algebraic curves resp. algebraic function fieldsof one variable.H === === Subject: : Re: mathematics departments directoryOriginator: israel@math.ubc.ca (Robert Israel) Some other good directories are:> http://www.ams.org/mathweb/mi-depts.html> http://www.math.fsu.edu/Science/Servers/> http://archives.math.utk.edu/math_dept.html> http://www.numbertheory.org/math_depts.html> http://directory.google.com/Top/Science/Math/Academic_ Departments/>Another interesting directory ishttp://csm.msuiit.edu.ph/math/links/index.htm .I would like to thank members of this groupfor valuable suggestions concerning possibleadditions to the mathematical department directoryhttp://www.math.hr/~duje/mathdept.html .Futher suggestions are very welcomed.By the way, I was motiva to make this directorywhen I prepared the link sectionhttp://www.math.hr/~mathe/linkovi.htmlfor recently established online mathematical journal math.ehttp://www.math.hr/~mathe/for secondary school and undergraduate students.published by Croatian Mathematical Society.Matrix transformations of the plane;Vigenere cipher;Algebraic method for solving constructive problems;Mathematicians on postage stamps.Any comments are very welcomed.Andrej Dujella=== === Subject: : Re: mathematics departments directoryOriginator: israel@math.ubc.ca (Robert Israel)>> The page with links to mathematics departments in the world>> http://www.math.hr/~duje/mathdept.html>> now contains links to 1375 departments from 117 countries. Any comments and suggestions are very welcomed.How does your directory compare with the one at Penn State>http://www.math.psu.edu/MathLists/Contents.html ?In a completely unscientific test (only two universities tried. The first one (www.math.hr) was correct on both, the second(www.math.psu.edu) was correct on only one.Dan-- Dan Luecking Department of Mathematical SciencesUniversity of Arkansas Fayetteville, Arkansas 72701luecking at uark dot edu=== === Subject: : subgroups of linear algebraic groupsOriginator: israel@math.ubc.ca (Robert Israel)Let F be a field. Does there exist a subgroup G of GL(n,F) with thefollowing properties?1) G acts transitively on the set of lines in F^n (i.e. on P^{n-1}(F))2) Denote by D the subgroup of diagonal matrices, and Z the subgroupof scalar matrices (i.e. the centre of GL(n,F)). Then G intersects Donly in Z.In other words, I want a group which acts transitively on lines, butwhose diagonal elements act trivially on lines. Does such a groupexist?=== === Subject: : Universal property of tensor product of linear maps?Originator: israel@math.ubc.ca (Robert Israel)Does the tensor product of linear maps have a universal property likethe tensor product of modules does? I am preparing a document in whichif U and W are modules over some commutative unital ring, thenUotimes W is defined to be a universal arrow from the singleton inthe category _Set_ to (the functor B(U,W;-) from the category ofmodules over the ring to _Set_), and would like a definition of theformer which is 'in spirit'. (The object function of B(U,W;-) takesthe module V to the set of bilinear maps from UxW to V)Appreciatively,Jeremy Martin=== === Subject: : universal set theory with 3 valued logicEpigone-thread: vermsmixblerOriginator: israel@math.ubc.ca (Robert Israel)this is an attempt at axiomatizing a universal set using three valuedlogic to avoid russell's paradox.i have changed the subsets axiom so that if binary logic is applied,then it remains the same axiom.i have changed the foundation axiom so that U can be an element of Uyet if U did not exist, the change to the foundation axiom would notexist (it is formula as IF there is a universal set then it MAYcontain itself).some results:cantor's diagonal argument fails to prove that U does not map ontoP(U).U=P(U), where = means equals, not in bijection with.if U<=x for any x, where this means domination, then U~x where ~ meansis bijection with. thus U is the largest set.it immediately follows that the set of all functions from U to U is inbijection with U and so OMEGA to the OMEGA is OMEGA.this also shows that the OMEGA product of OMEGA is OMEGA.i then dabble into an investigation of U as a boolean ring and discusssome more elementarily derivable properties. i discuss the concept ofa universal limit and convergence without reference to a topology ormetric.this is still a rough draft and all feedback is apprecia so that ican publish it one day.thank youbrian=== === Subject: : right chain ring that is not a chain ringOriginator: israel@math.ubc.ca (Robert Israel)Does anybody know where can be found some example of a right chainring that is not a chain ring? I'd really appreciate such a ringwithout a Krull dimension. Thanks for answers if any.Pavel=== === Subject: : Re: Bernoullis equation?I wonder if there is anyone who could help me solve this equation step by step.y(x) = y(x)/x + 3(y(x))^2I think that this is Bernoullis equation, But not sure.JaakkoIt is a Bernouilli equation. Look it up in any book onODE's. For this one, substitute v = 1/y .-- ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything and thereby take away our free will and that share of glory that rightfully belongs to us. -- N. Machiavelli, The Prince.=== === Subject: : set. Combinations/permutations questionCombinations/permutations from multiply setsYou have 3 sets each have 2 membershow many doubles can be selec from these 3 sets, if each doublecannot have 2 members from the same set??any help would be greatthank you=== === Subject: : Re: Real Integral using complex integral> Hello everyone.Since x / (x^3-8) is not an even funtion,Yes. If you plug in x=-x the integrand becomesx / (x^3+8)But, as x->oo (x^3 + 8) -> x^3Therefore for large enough x, the integrand is approximately even.Hope that helps,John=== === Subject: : Re: Is there a set of numbers between the rationals and reals?> If the real numbers consist of the following:1) rationals> 2) numbers that can be represen by finite algorithms (example: e> and pi)> 3) numbers that are neither rational nor representable by a finite> algorithmThe last item would include numbers like this:0.123245223643792134....(infinite sequence of digits that are> impossible to represent with any algorithm.)Why is there not a well known set of numbers that does not include #3?There's more than one. For instance, the so-called algebraic numbers(numbers that are roots of polynomials with integer coefficients)lie between #1 and #2. Your #2 is itself a well known set, generallycalled the computable numbers.> It would be more dense than the rationals but less than than the> reals.That sounds plausible, but actually (for most meanings of dense)it's wrong, because in a very strong sense almost all numbersare in set 3.> What is the purpose of describing numbers like #3?I don't understand that question, so I can't answer it. Sorry!-- Gareth McCaughan.sig under construc=== === Subject: : diferencial equation. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1GFTaV26976;===Do you know how to solve this equation?y''-a y^4 + b = 0I would like to know if there is a solution, or how to solve itnumerically.boundary conditions y(x=0)= T y'(x=0)= cHow can I solve that in mathematica?=== === Subject: : Re: diferencial equation.> Do you know how to solve this equation?> y''-a y^4 + b = 0I would like to know if there is a solution, or how to solve it> numerically.boundary conditions y(x=0)= T> y'(x=0)= c> How can I solve that in mathematica?Especially because you have two conditions at y = 0, I would solve itas a pair of first order ODEs:Let v = y'Then you have the system of equations:v' = -b + ay^4y' = vy(0) = Tv(0) = cThere are many methods for solving systems of first order differentialequations. Pick your favorite!Hope that helps,John=== === Subject: : Impact of windowing data on confidence intervalsI'm working with a time series of data and have the end goal ofproducing plots of spectral energy density with confidence limits. Itis well known that each spectral estimation is a chi-squared functionwith 2 degrees of freedom. And if we had a record that had N=1000points of data, and divided this into sub-groups of M=100 points, wewould have K=N/M=10 seperate realizations of the spectral estimate. And if we averaged these results together, the resultant spectralestimate would have v=2K=20 degrees of freedom.With v in hand we could find the confidence limits through referenceto the chi-squared distribution, for 20 DOF and whichever confidencelimit we wan, say 95%.If the raw data is initally passed through a windowing process(Hamming, Hanning, etc.) one must modify the estimate of the DOF byfinding the Equivalent Degrees of Freedom. For common windows, thecalculation of the EDOF is given by Priestly (1981) and isstraightforward.Furthermore it is common to overlap adjacent segments of the originaldata when windowing, at times with up to 50% overlap. For the exampleabove, we could start at the 50th data point of the record and startagain constructing sub-groups of M=100. This would result is 9 newrecords that we could add to the spectral realization. Now finally myquestion:What effect does such overlapping have on the EDOF calculation. Atfirst I though that such an overlap would not result in any increasein EDOF, as the 9 new records are not statistically independant fromthe original 10. However Emery & Thomson (2001) state that ...by notoverlapping adjacent segments, we could be overly conservative in ourestimate of the number of degrees of freedom. ... The effect of thewindow is to reduce the effective length of each segment in the timedomain so that, for some sharply defined windows such as theKaiser-Bessel window, even adjoining segments with 50% overlap can beconsidered independant time series for spectral analysis.Now this makes a lot of sense, and I'm left wondering how one would goabout determining the overlap/window combinations which could still beconsidered independant, and hence yield an increase in EDOF andfinally lead to an accurate estimate of confidence intervals.Thanks for listening and for any insights you might have to offer.MarkEmery & Thomson, 2001 Data Analysis Methods in Phys. Oceanography2nd. Ed.Priestley 1981 Spectral Analysis and Times Series=== === Subject: : Re: Impact of windowing data on confidence intervals>I'm working with a time series of data and have the end goal of>producing plots of spectral energy density with confidence limits. It>is well known that each spectral estimation is a chi-squared function>with 2 degrees of freedom. And if we had a record that had N=1000>points of data, and divided this into sub-groups of M=100 points, we>would have K=N/M=10 seperate realizations of the spectral estimate. >And if we averaged these results together, the resultant spectral>estimate would have v=2K=20 degrees of freedom.With v in hand we could find the confidence limits through reference>to the chi-squared distribution, for 20 DOF and whichever confidence>limit we wan, say 95%.If the raw data is initally passed through a windowing process>(Hamming, Hanning, etc.) one must modify the estimate of the DOF by>finding the Equivalent Degrees of Freedom. For common windows, the>calculation of the EDOF is given by Priestly (1981) and is>straightforward.Furthermore it is common to overlap adjacent segments of the original>data when windowing, at times with up to 50% overlap. For the example>above, we could start at the 50th data point of the record and start>again constructing sub-groups of M=100. This would result is 9 new>records that we could add to the spectral realization. Now finally my>question:What effect does such overlapping have on the EDOF calculation. At>first I though that such an overlap would not result in any increase>in EDOF, as the 9 new records are not statistically independant from>the original 10. However Emery & Thomson (2001) state that ...by not>overlapping adjacent segments, we could be overly conservative in our>estimate of the number of degrees of freedom. ... The effect of the>window is to reduce the effective length of each segment in the time>domain so that, for some sharply defined windows such as the>Kaiser-Bessel window, even adjoining segments with 50% overlap can be>considered independant time series for spectral analysis.Now this makes a lot of sense, and I'm left wondering how one would go>about determining the overlap/window combinations which could still be>considered independant, and hence yield an increase in EDOF and>finally lead to an accurate estimate of confidence intervals.Thanks for listening and for any insights you might have to offer.MarkEmery & Thomson, 2001 Data Analysis Methods in Phys. Oceanography>2nd. Ed.>Priestley 1981 Spectral Analysis and Times SeriesOverlapping windows increase the number of degrees of freedombecause you then take into account correlations betweenvalues that would otherwise be some at the end of a windowand the others at the beginning of the next (the spectrumis the Fourier transform of the autocorrelation).The increment in the number of degrees of freedom for eachnew window processed is then 2/(1 + 2 ovl_1^2 + 2 ovl_2^2 + ...)where ovl_1 is overlap between a window and the next,ovl_2 between a window and the second next, etc.Interaction with tapering is discussed in Harris (1978)On the use of windows for harmonic analysis with the discreteFourier transform, Proc. IEEE, vol 66, No 1 pp.51-84.My understanding is that good taper windows make the estimates from consecutive windows almost independent up to 50% overlap.=== === Subject: : Re: Kernel for cubic interpolation>> You can find the Lagrange interpolation formula in Abramowitz & Stegun,>> Handbook of Mathematical Functions. It is just a simple way to write a>> polynomial of degree N that goes through N+1 points of a function.>> For example, suppose you know f(x) at the points x0, x1 and x2. Call>> these values f0, f1 and f2. Then you write> f0 * (x-x1)*(x-x2) f1 * (x-x0)*(x-x2) f2 * (x-x0)*(x-x1)>f(x) ~ ------------------ + ------------------ + ------------------> (x0-x1)*(x0-x2) (x1-x0)*(x1-x2) (x2-x0)*(x2-x1)>> You see this is quadratic in x. Higher-order polynomials areessentially>> the same.>>This formula directly gives the value of f at any x. What I'm actually>after is the kernel. Ie. | g(x) 0 <= |x| <= 1>k(x) = | h(x) 1 <= |x| <= 2> | 0 |x| >= 2>What is g and h? Can you measure the impulse response of an interpolation formula?> Is there a relationship between the impulse response and the convolution> kernal?The impulse response is really the same as the convolution kernel, thoughthe impulse response may be specified tabularly (a list of points) whereasthe OP is looking for a formula.=== === Subject: : Re: Kernel for cubic interpolation You can find the Lagrange interpolation formula in Abramowitz & Stegun,>> Handbook of Mathematical Functions. It is just a simple way to write a>> polynomial of degree N that goes through N+1 points of a function. For example, suppose you know f(x) at the points x0, x1 and x2. Call>> these values f0, f1 and f2. Then you write > f0 * (x-x1)*(x-x2) f1 * (x-x0)*(x-x2) f2 * (x-x0)*(x-x1)>f(x) ~ ------------------ + ------------------ + ------------------> (x0-x1)*(x0-x2) (x1-x0)*(x1-x2) (x2-x0)*(x2-x1) You see this is quadratic in x. Higher-order polynomials are essentially>> the same. >> This formula directly gives the value of f at any x. What I'm actually>after is the kernel. Ie. | g(x) 0 <= |x| <= 1>k(x) = | h(x) 1 <= |x| <= 2> | 0 |x| >= 2>What is g and h?Can you measure the impulse response of an interpolation formula?> Is there a relationship between the impulse response and the convolution> kernal?> I'm not sure whether you're asking me or being rhetorical (ie. whatI'm asking is dumb)!! :) If the former, I don't know! If the latter, Iagree!At least in the case of linear interpolation the Lagrange formula isf(x) = f0 * (x-x1) + f1 * (x-x0) ----------- ----------- -1 +1and the kernel is | 1-x |x| <= 1k(x) = | | 0 |x| >= 1What the relationship is... I can't see. The cubic, quintic, septicwould also be interesting.=== === Subject: : Q. On Crout's Algorithm For Matrix LU DecompositionGreetings.I'm a computer science grad student, trying to write a softwarelibrary for handling matrices that, among other things, findsdeterminants efficiently. The best way to do this is by LUdecomposition, and the best way to do that (that I know of) is Crout'salgorithm.I stumbled upon Crout's algorithm in _Numerical Recipes In C: The Artof Scientific Computing_, which has a pretty good discussion. Iunderstand most of it.Where I get stuck is the concept of the pivoting (Swapping rows,when appropriate. This results in a row-wise permutation of thethe stability of Crout's method. Calculation of the subdiagonalelements (alpha[i][j]) involves dividing by a pivot element. Themathematical discussion of the algorithm just uses the non-unitydiagonal elements (beta[j][j]), without regard to pivoting. I workedthrough it by hand with a random 3x3 matrix, and it seemed to workjust fine. However, the example C code goes out of its way to choosemaximum pivot element, and exchange rows when it becomes necessary.So...my burning question, in three closely-rela parts, are: 1) Why is pivoting necessary? Is it a safeguard againstdivide-by-zero? 2) If I were to skip pivoting altogether, why would thisde-stabilize the algorithm? 3) What scenarios could come up that would fall apart in the absenceof pivoting?Thanks in advance,ff=== === Subject: : Re: Q. On Crout's Algorithm For Matrix LU DecompositionGreetings.I'm a computer science grad student, trying to write a software> library for handling matrices that, among other things, finds> determinants efficiently. The best way to do this is by LU> decomposition, and the best way to do that (that I know of) is Crout's> algorithm.I stumbled upon Crout's algorithm in _Numerical Recipes In C: The Art> of Scientific Computing_, which has a pretty good discussion. I> understand most of it.Where I get stuck is the concept of the pivoting (Swapping rows,> when appropriate. This results in a row-wise permutation of the> the stability of Crout's method. Calculation of the subdiagonal> elements (alpha[i][j]) involves dividing by a pivot element. The> mathematical discussion of the algorithm just uses the non-unity> diagonal elements (beta[j][j]), without regard to pivoting. I worked> through it by hand with a random 3x3 matrix, and it seemed to work> just fine. However, the example C code goes out of its way to choose> maximum pivot element, and exchange rows when it becomes necessary.So...my burning question, in three closely-rela parts, are: 1) Why is pivoting necessary? Is it a safeguard against> divide-by-zero?> 2) If I were to skip pivoting altogether, why would this> de-stabilize the algorithm?> 3) What scenarios could come up that would fall apart in the absence> of pivoting?Thanks in advance,ffHave a look at A. Ralston, A first course in numerical analysis 2nd ed(McGraw-Hill, New York, 1978). It discusses the pivoting question insickening detail.Pivoting is necessary to reduce roundoff error. Any of the iterativeelimination schemes for solving linear equations or inverting matricesinvolves multiplying a row by a constant and subtracting it from anotherrow. It is the subtraction step that hurts because it frequently happensthat you subtract numbers that are close in magnitude. The difference hasmany fewer significant figures of precision than either of the numbersyou were subtracting. By pivoting and normalizing to the largest elementin a row one reduces this (unavoidable) loss of precision. The bestform of pivoting involves permuting both rows and columns, but the resultingalgorithm is far mor complex than row-wise pivoting alone. Fortunately,in most practical cases row-wise pivoting is adequate.-- ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything, and thereby take away our free wiil and that share of glory that rightfully belongs to ourselves. -- N. Machiavelli, The Prince.=== === Subject: : Bessel rela integralX-ID: Zj+LTrZJQebFql-HnM2lFS-Tu+fNCLtJnuwYAOW3VspO9tA0pH8krAI want exp(c*x)*BesselK(1,a*(b^2+x^2)^(1/2))/(b^2+x^2)^(1/2)to be integra from a some m to infinity (or what becomesequivalent: over the positive reals but with a locationparameter xi+m instead of x).Looking through Abramowitz and as well Gradsteyn there seemsto exist no closed analytic solution in terms of known fcts.This is a probability density function and surely one do theintegration through some Gauss method or similar with a goodaccuracy (or through FFT as the char Fct is known).What i would like to have is a fast one with medium accuracy(say 8 digits?) through a polynomial or rational approximationas it exists for the cdf normal (i intend to use it in Excel,within Maple it is of course not a problem).Due to the additional parameters i have no idea whether onecan do that. Any hints or solutions for that or a 'no, thereis no reasonable way'?=== === Subject: : Re: Bessel rela integralI want exp(c*x)*BesselK(1,a*(b^2+x^2)^(1/2))/(b^2+x^2)^(1/2)> to be integra from a some m to infinity (or what becomes> equivalent: over the positive reals but with a location> parameter xi+m instead of x).Looking through Abramowitz and as well Gradsteyn there seems> to exist no closed analytic solution in terms of known fcts.This is a probability density function and surely one do the> integration through some Gauss method or similar with a good> accuracy (or through FFT as the char Fct is known).What i would like to have is a fast one with medium accuracy> (say 8 digits?) through a polynomial or rational approximation> as it exists for the cdf normal (i intend to use it in Excel,> within Maple it is of course not a problem).Due to the additional parameters i have no idea whether one> can do that. Any hints or solutions for that or a 'no, there> is no reasonable way'?Have you looked at Watson, Theory of Bessel Functions? Mine ishome so I can't consult it, but you might be able to employ anintegral representation of the modified Bessel function. Or youshould try the Bateman Manuscript Project. I suspect the resultmight be a Legendre function of the second kind.BTW, I trust that a > 0 and a > c ;-)If you are willing to do it numerically, look at Gauss-Laguerreintegration. It should be pretty rapidly convergent since K_1is a monotonically decreasing function of its argument.-- ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything, and thereby take away our free wiil and that share of glory that rightfully belongs to ourselves. -- N. Machiavelli, The Prince.=== === Subject: : Re: Bessel rela integral> I want exp(c*x)*BesselK(1,a*(b^2+x^2)^(1/2))/(b^2+x^2)^(1/2)> to be integra from a some m to infinity (or what becomes> equivalent: over the positive reals but with a location> parameter xi+m instead of x). Looking through Abramowitz and as well Gradsteyn there seems> to exist no closed analytic solution in terms of known fcts. This is a probability density function and surely one do the> integration through some Gauss method or similar with a good> accuracy (or through FFT as the char Fct is known). What i would like to have is a fast one with medium accuracy> (say 8 digits?) through a polynomial or rational approximation> as it exists for the cdf normal (i intend to use it in Excel,> within Maple it is of course not a problem). Due to the additional parameters i have no idea whether one> can do that. Any hints or solutions for that or a 'no, there> is no reasonable way'?Have you looked at Watson, Theory of Bessel Functions? Mine is> home so I can't consult it, but you might be able to employ an> integral representation of the modified Bessel function. Or you> should try the Bateman Manuscript Project. I suspect the result> might be a Legendre function of the second kind.BTW, I trust that a > 0 and a > c ;-)If you are willing to do it numerically, look at Gauss-Laguerre> integration. It should be pretty rapidly convergent since K_1> is a monotonically decreasing function of its argument. Yes, there are restrictions, but i did not want to write down thecomplete normal inveres Gauss distribution (Barndorff-Nielsen) ...No, as former algebraist i do not have them. I found somethinglooking close and incidential there is a proof in Lebedev whichuses integral representation. But sadly it is BesselJ instead ofexp and whithout location - that damned exp does not fit in theapparatus, i tried (x/2)^n*exp(-t-x^2/4/t)*t^(-n-1) dt / 2 overthe positive reals.At least i can cut off integration very soon. But some (rational)approximation i would still prefer=== === Subject: : Still conjecture? on polynomials by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1GLcuj24590;===Hi all,reading a book on polynomials few year ago (sorry I don't remember thetitle) I took a note on a strange conjecture:Let p be a polynomial of degree n, and let sc(p) denote the numberof sign changes in the sequence of nonzero coefficients of p. Thenholdssc(p(x+1)) + sc((x+1)^n * p(1/(x+1))) <= sc(p(x))Has anyone came ac this conjecture or any attempt to prove it?Does anyone see any possibility or probability to prove it?John Vassiliou=== === Subject: : Re: 'erf' function in C>> Are there any C compilers which have the erf function (from>> probability) as part of their math libraries? Or are there any math>> libraries available to download which implement this function? The GNU Scientific Library (http://www.gnu.org/software/gsl/) has erf.