mm-462 === Subject: : Oneness of a numberI am trying to build fractals generated by the principal of the oneness of anumber.For the definition of the oneness of a number, view the internet, or below.*I have been unseccesful in creating a pattern around the oneness although Ihave struck on some intriguing details. These mainly resolve around theissue of, if we take the oneness of the oneness of a number, and keep doingthat, how long does it take 'til we reach 1? (note that for 5 this neverhappens) I think I might be able to find a pattern in that since, if I callthis the recursive oneness loop length (if anybody can think a bettername, tell me), then, the recursive oneness loop length of a number like20000000 is still only 15. I am trying to make a large-scale 2d map of it,but so far it is still pure chaos, even if the range is so limited.Does anybody know any other numerical iterations like the oneness thatcreate a chaotic result? I think this is very interesting..* For the oneness of a number 'n' :if(n is even) n = n/2;if(n is odd) n = 3*n + 1;Keep doing this until n = 1. The amount of necessary steps for this iscalled the oneness of n.-- Quaternion=== === Subject: : Re: JSH: Difficult social problemcongratulation!> 7/x = 1 requires that x=7.> It's an intriguing problem.> http://mathforprofit.blogspot.com/--ils duces d'Enron!http://www.movisol.org/http://members.tripod.com/~ american_almanac/=== === Subject: : Re: JSH: Difficult social problem>And believe me, if you don't speak any French, the >waiters in France won't bring you soup.Have you considered the possibility that the reason waiters don'tbring you soup might *not* be that you don't speak French?-- Richard-- Spam filter: to mail me from a .com/.net site, put my surname in the headers.FreeBSD rules!=== === Subject: : Re: derivate of x^xprimefinder escribi.97 en el mensaje> Ok, I've probably asked this question before, but I've forgot theanswer...> What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ?> I guess asking for the integral of these functions would be foolish?Use logarithmicdrivative:f(x) = x^x ===> Ln(f(x)) = x*Ln(x) ===f'(x)/f(x) = 1*Ln(x) + x *(1/x) = 1 + Ln(x) ===f'(x) = x^x(1 + Ln(x))For the other, aply that twice.-- === === Subject: : Re: NOVA strings and branesit's hearsay, about the US casualty estimates. now,the Japanese were negotiating through the VAtican Special Officeand the OSS, and MacArthur's air & sea blockade was very successful;the fact is that he was not consulted! the tie by the parliament was a matter of saving face;damn right, it was broken by the emperor. you point is also taken that it was no worsethan the German situation circa '45, becausethey were also victim of the British terror-bombings. thus was the Nuclear Age begun.The next day, a political and diplomatic storm erupted. Truman,who had previewed the speech and approved it on Sept. 11, lied andtold the press that Wallace never showed him the speech. Secretaryof State Byrnes and the press went ballistic, and on Sept. 20,Truman asked for Wallace's resignation and got it. Truman promptlyappointed Averell Harriman in Wallace's place. For the next two decades, Wallace continued to battlefor national policy direction as he saw it. That is a storyfor another telling.References> Now consider that, if anything, the Japanese had a (well earned) > reputation of keeping fighting even in situations where the Germans > surrendered and draw your own conclusions.> The last large scale military operation in the Pacific preceding > Hiroshima was the battle of Okinawa. Okinawa was defended by about > 100000 Japanese troops, with practically no heavy weaponry and no > close air or naval support. The island is small enough so that every > part of could've been covered by direct fire from US navy ships (and, > of course, air support from the carriers). Yet, it still took nearly > 3 months of fighting and close to 50000 US casualties (about 12000 > Japanese forces, less than 10% were taken prisoner. In addition, some > 50000-100000 civilians died.> That was a rather small island. In Japan itself, the Japanese > military had ten times as many troops (granted, poorly equipped, but > so were those in Okinawa) and many millions of civilians who were > ready to fight to the end. Based on the Okinawan experience you can > begin to estimate the cost, in casualties and destruction that would > result from landing in Japan.--ils duces d'Enron!=== === Subject: : Re: Factorization dispute> It turns out that I can isolate the current dispute easily enough by> focusing on the factorizations:> Consider [snipped]Crank Information http://www.crank.net/harris.html http://www.crank.net/usenet.html http://www.google.com/search?q=harris+site%3Awww.crank.net http://www.google.com/search?q=%22james+harris%22+site% 3Ausers.pandora.be> Readers should please check out *all* the links Sam the Worm listed.> As for the math, notice that with> (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> no other factorization works as long as 7 is not a factor of 22.You have only demonstrated anything for the case x=0, not 'x' in general.--There are two things you must never attempt to prove: the unprovable -- and the obvious.----http://www.crbond.com=== === Subject: : Re: Oneness of a number===> === Subject: : Oneness of a number>Message-id: I am trying to build fractals generated by the principal of the oneness of anumber.>For the definition of the oneness of a number, view the internet, or below.*It's more commonly called the Collatz Conjecture or the 3x+1 problem.>I have been unseccesful in creating a pattern around the oneness although I>have struck on some intriguing details. These mainly resolve around the>issue of, if we take the oneness of the oneness of a number, and keep doing>that, how long does it take 'til we reach 1? Depends on how many binary digits the number has (which is related to themagnitude of the number) AND ALSO on the pattern of 1s and 0s in the binaryrepresentaion (which is NOT related to the magniyude of the number). >(note that for 5 this never happens) It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1>I think I might be able to find a pattern in that since, if I call>this the recursive oneness loop length (if anybody can think a better>name, tell me),It's sometimes refered to as the stopping time. Others make up their ownterminology.> then, the recursive oneness loop length of a number like>20000000 is still only 15. I am trying to make a large-scale 2d map of it,How are you plotting this in 2D? >but so far it is still pure chaos, even if the range is so limited.Maybe you just need to look at the chaos from the right perspective, say bylooking at the forest instead of the trees:http://members.aol.com/mensanator666/collatz/bang0010. htm>Does anybody know any other numerical iterations like the oneness that>create a chaotic result? I think this is very interesting..>* For the oneness of a number 'n' :>if(n is even) n = n/2;>if(n is odd) n = 3*n + 1;>Keep doing this until n = 1. The amount of necessary steps for this is>called the oneness of n.>-- >Quaternion--MensanatorAce of Clubs=== === Subject: : Re: JSH: Difficult social problemAnd believe me, if you don't speak any French, thewaiters in France won't bring you soup.> Have you considered the possibility that the reason waiters don't> bring you soup might *not* be that you don't speak French?This is unthinkable if you have ever been to Quebec or France. The onlyway I could get a cup of coffee in Quebec City was for my wife to orderit for me. :-)Chuck-- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net=== === Subject: : Re: NOVA strings and branes> than the German situation circa '45, because> they were also victim of the British terror-bombings.> thus was the Nuclear Age begun.Terror bombings? The Germans started bombing cities first. They simply were getting as good (or bad) as they gave. Arthur Bomber Harris said something to the effect that the Germans had sown the wind, and in due course would reap the whirlwind. So true.Bob Kolker=== === Subject: : Re: Factorization dispute> Let's take 'em out.> James HarrisTake 'em out? What do you mean by that? Are you back to your previousthreat of calling out the military or the CIA and the FBI, or are youadvocating the formation of a private militia?> I think James wants his vast silent audience to treat mathematicians to> dinner and a movie. I suspect that there's an ulterior motive.Physics students are a cool crowd and I know because I was once aphysics undergrad, and I'd get into all kinds of fascinatingdiscussions, as physics is that kind of field.When you deal with crackpots and cranks, which mathematicians are now,who have been backed into a logical corner, you go ahead and deliverthe coup de grace of the irrefutable facts.That is, you take 'em out.Now then, the crackpots may still believe their nonsense, butcivilized society is once again protected by the vigillance of activeand highly intelligent minds, as the bulk of society can move on withthe truth.I've isolated mathematicians by a logical argument, having made amajor discovery--that's what physics people do--in what they probablythink of as their area alone, and caught them trying to dodge theresult and its implications.Remember mathematicians have claimed that pure math was for thebenefit of humanity, but I've shown them trying to hide one of themost astounding results in math history (one of three, mind you).So it's time for the physics people to look at the facts, stand forlogic over social needs, and take 'em out.=== === Subject: : Re: Oneness of a number===> === Subject: : Oneness of a number>Message-id: I am trying to build fractals generated by the principal of the oneness of>a> number.>For the definition of the oneness of a number, view the internet, or>below.*> It's more commonly called the Collatz Conjecture or the 3x+1 problem.>I have been unseccesful in creating a pattern around the oneness although>I have struck on some intriguing details. These mainly resolve around the>issue of, if we take the oneness of the oneness of a number, and keep>doing that, how long does it take 'til we reach 1?> Depends on how many binary digits the number has (which is related to the> magnitude of the number) AND ALSO on the pattern of 1s and 0s in the> binary representaion (which is NOT related to the magniyude of the> number).>(note that for 5 this never happens)> It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1No, you didn't listen well, I take the oneness of the oneness for therecursive oneness looplength.The oneness of 5 is 5.the oneness of 5 is 5, etc..>I think I might be able to find a pattern in that since, if I call>this the recursive oneness loop length (if anybody can think a better>name, tell me),> It's sometimes refered to as the stopping time. Others make up their own> terminology.> then, the recursive oneness loop length of a number like>20000000 is still only 15. I am trying to make a large-scale 2d map of it,> How are you plotting this in 2D?1. Using complex number representation and messing around with that.2. Using y as a period of x (y = x/WIDTH, so to speak)-- Quaternion===>If you saw>(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2)>with the c's algebraic integers, I think few of you would have a>problem realizing that only two of the c's have 7 as a factor.No - the problem here is that this polynomial doesnot factor at all in the algebraic integers in the form you have shown. Its factorization is necessarily of the form 49*(x - r1)*(x - r2)*(x - r3),where r1, r2, and r3 are nonunit *algebraic integers* whichare the roots of x^3 + 5*x^2 + 3*x + 2 = 0, and are such that -r1*r2*r3 = 2.Knowing this, you can distribute the 49 among the factors in many different ways. However, *none* of thoseways will end up having constant terms of 7, 7, and 2.Why? Because 7 and 2 are coprime in the algebraic integers. It is not possible to write 2, which you wantas the constant term of the third factor, as a product of a nonunit factor of 7 and a nonunit factor of 2.Andrzej.>But, of course, you're looking at *functions* of x, as you have >f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, >so I could also write it as>(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2).>Notice that dividing both sides by 49 gives >(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2>as long as you're in a ring where 7 is not a factor of 22.>I want to emphasize that point as notice there's only *one* way to>divide through by 49 if 7 is not a factor of 22.>Usually you can *see* the other factors of 7, but I want you to>abstract, and generalize.>Please pay careful attention to that example.>You may see people who reply claiming that the word polynomial has>some significance, as if it's a mystical thing which refutes basic>logic, so if something isn't polynomial it no longer behaves>logically.>Now then, in my advanced factorization work, I just use functions of x>that are a lot more complicated than f_1(x) = c_1 x, and unfortunately>there are people who can use an unfamiliar leap in complexity to>confuse others.>Some of you have learned various advanced math topics, now imagine if>in your classrooms there were some hecklers who continually hollered>out at your teacher, or otherwise disrupted the class?>What if when there were difficult concepts those hecklers would try to>confuse everyone as they sought to discredit the mathematics?>If you find that hard to imagine, imagine me in your class with you>questioning the professor and calling him names.>How much would you have learned?>I need those of you interested in mathematics to focus on the basics,>so that you can understand the advanced.>James Harris>http://mathforprofit.blogspot.com/=== === Subject: : Re: derivate of x^x>Ok, I've probably asked this question before, but I've forgot the answer...>What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ?Logarithmic differentiation.Doug=== === Subject: : Re: Factorization dispute, again> Notice,http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.netStupid is as stupid does. There is no fixing stupid because it is notbroken. Harris would be much happier as a priest proclaiming godamdist heretics and buggering altar boys to His greater glory.Hey Harris, your village called: Its idiot is missing.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net!=== === Subject: : Re: Usenet Posting Guide?> With all respect, dear man, you strike me as that type more enamored> with process rather than function. Most of us out here have far better> things to do than screw around for weeks configuring newsreaders and> the like, even if we were so inclined, which most of us clearly are> not. I suppose we could memorize the phone book, too, but would that> help us communicate our ideas any better?> Some of us prefer to view the forest rather than count the trees down> there. Better view, too.Well, I find the technical aspects of how USENET functions much moreinteresting than most of the discussions that take place on it. By thesame token, I am much more interested in the hardware and operating systemsoftware of the systems I administer than in any of the applicationsfor which they are used.Once I hoped that the growing popularity of personal computers meantthat nearly everyone would learn to think like programmers. It neveroccurred to me that, sadly, the opposite would happen: That computerswould be designed to be used by people who *can't* program them.> Dear man, that is all fine and good, but you'd do well to examine your> thinking here a little more closely. You're seeing the world from a> very narrow vantage point. Technology and inventive genius exist to> serve those who use and take advantage of them, not merely those who> design and invent, otherwise you might just as easily make the> argument that the world was a far better place when mostly car> builders and those who tinkered all day with Model T Fords were out> there driving on the roads. Certainly the roads were safer then, and> that those driving them generally knew their cars inside and out, but> that ignores the fundamental purpose behind that of automobiles, which> is to facilitate transportation.> The same exact argument can be made for telephones, television,> airplanes and a whole host of other modern conveniences we now take> for granted and which represent fundamental shifts in the way we> communicate and get around. They were all once the domain of a select> few tinkerers and inventors who had absolutely no idea of what they> were about to unleash on the world.> With all respect, sir, I can guess your politics right across the> board. Just for openers, you're a Sierra Club, save-the-whales type.> Gotcha, huh?I think you're entrenching him. He said it was cooler when programmersused 'em. You said it's cooler when people who knew vehicles insideout were the ones driving them.You guys are agreeing.=== === Subject: : Re: Skeptickal Inquirer UFOI hope you're not speaking poorly about my Venus lizard folk, Catharsand all.This may be getting some folks a wee bit off topic from group focus,or that of my intended agenda upon Venus life (lizard folk and all),but according to some fairly recent feedback, I've learned a thing ortwo about our nasty moon, as a place that I believe we've long neededto establish a lunar space elevator (LSE) in order to be efficientlygetting ourselves off to visiting the wizard of Oz at Venus L2, aswell as for reaching out to those thoroughly irradiated to deathsnowman/snowwoman situated on Mars:http://guthvenus.tripod.com/gv-cm-ccm-01.htmHere's a little typical feedback of supposed facts from: Jay Windley(webmaster@clavius.org)Highnergy cosmic rays do not come from the sun. They come fromoutside the solar system, and our sun is the primary defense againstlower energy and thus their secondary effects in the ambient areminimal.My thoughts: http://guthvenus.tripod.com/gv-moon-radiation.htmUnfortunately, I may have incorrectly utilized the terminology ofcosmic rays, though fortunately, I never specified upon any specificspectrum of highnergy cosmic rays, just pointing out that our sunis certainly capable of tossing out its fair share of far worse thingsthan visible photons plus IR worth of BTUs and of those nasty UVs.Obviously a supernovae is worth a thousand fold in terms of beingnasty, thereby from the far off generated galactic influx must offer ameasurable degree of such events, and of the secondary radiation givenoff by all that infamous clumping lunar dirt should become a fairlydarn good as well as unobstructed indicator of whatevercosmic/galactic influx. Seems fairly odd that shuch measurementsaren't common place, as where's the justification for not otherwiseproviding this level of information?The assertions or premise offered by the likes of Jay Windley, that ofour moon not only lacking an atmosphere but also without a Van Allenbuffer zone is not such a bad thing if you're out and about on thelunar surface, seems somewhat risky if not downright lethal. I mighthave come into that understanding if we're referring to an earthshineilluminated lunar surface, but I'm not going so far if that's of anyfully solar illuminated environment while wearing a mostly syntheticmoon suit because, we're not talking about avoiding a 270 nm UV sunburn.Sorry about all my make-due reverse engineering logic, or lackthereof. I was simply trying to establish upon the amount of solarradiation that becomes hard X-Ray class. So exactly how much is it ona typical lunar day, or how about on a good day as well as on a trulyHighnergy cosmic rays do not come from the sun ???Do we suppose that happens to include the likes of the last couple ofweeks of horrific solar flak?Seems there should be some specific knowledge (excluding Apollo) ofwhat's what pertaining to the solar illuminated surface, as opposed tothe absolute lunar nighttime environment and, of something specificpertaining to whatever earthshine contributes.This task of obtaining knowledge is somewhat like my getting a graspupon the applied energy (thrust or torque) involved in acceleratingsomething the size and mass of the moon, so that it accelerates andthereby recedes form Earth at 38 mm/year.As worthy feedback provided from: Ami Silberman (silber@mitre.org)The mechanisms for the lunar recession have been well understood fordecades. In a nutshell, tides cause friction between the oceans andthe ocean floors, which transfers energy from the solid part of theearth to the oceans. One of the effects of this friction is that thetidal bulge is off-center, and is located eastward of the moon. (Sothe high tide actually occurs when the moon is west of overhead.) Theresult of the tidal bulge being off center is that there is a torgueeffect placed on the moon, and this in turn transfers energy from theearth to the moon. The earth's spin rate slows, the moon is speeded inits orbit and therefor moves further away from the earth. (Thistransfer of energy is essentially a transfer of angular momentum,which is a conserved quantity.) The historical (over geological eras)rate of recession has varied due to varying amounts of tidal frictiondue to shallower or deeper oceans, and the positions of thecontinents.For the benefit of all my loyal critics, I've conceded that there's adarn good chance that the likes of Tim Thompson has more than a fewvalid points as to his version of what's what. This following page isjust another example of my learning from the pros, of accepting otherinput, which may even including the likes of what Ami Silberman justpresented, that I'd not be calling flak, as there actually seems to besome considerable worth to at least Tim's version of the lunarrecession, if I don't say so myself.http://guthvenus.tripod.com/earth-moonnergy.htm=== === Subject: : Re: Skeptical Inquirer UFOUnfortunately Scott, this sting/ruse is way so much bigger than evenyou can imagine. I've posted this folowing in at least a half dozentopics that I believe are related to obtaining the truth.This may be getting some folks a wee bit off topic from group focus,or that of my intended agenda upon Venus life (lizard folk and all),but according to some fairly recent feedback, I've learned a thing ortwo about our nasty moon, as a place that I believe we've long neededto establish a lunar space elevator (LSE) in order to be efficientlygetting ourselves off to visiting the wizard of Oz at Venus L2, aswell as for reaching out to those thoroughly irradiated to deathsnowman/snowwoman situated on Mars:http://guthvenus.tripod.com/gv-cm-ccm-01.htmHere's a little typical feedback of supposed facts from: Jay Windley(webmaster@clavius.org)Highnergy cosmic rays do not come from the sun. They come fromoutside the solar system, and our sun is the primary defense againstlower energy and thus their secondary effects in the ambient areminimal.My thoughts: http://guthvenus.tripod.com/gv-moon-radiation.htmUnfortunately, I may have incorrectly utilized the terminology ofcosmic rays, though fortunately, I never specified upon any specificspectrum of highnergy cosmic rays, just pointing out that our sunis certainly capable of tossing out its fair share of far worse thingsthan visible photons plus IR worth of BTUs and of those nasty UVs.Obviously a supernovae is worth a thousand fold in terms of beingnasty, thereby from the far off generated galactic influx must offer ameasurable degree of such events, and of the secondary radiation givenoff by all that infamous clumping lunar dirt should become a fairlydarn good as well as unobstructed indicator of whatevercosmic/galactic influx. Seems fairly odd that shuch measurementsaren't common place, as where's the justification for not otherwiseproviding this level of information?The assertions or premise offered by the likes of Jay Windley, that ofour moon not only lacking an atmosphere but also without a Van Allenbuffer zone is not such a bad thing if you're out and about on thelunar surface, seems somewhat risky if not downright lethal. I mighthave come into that understanding if we're referring to an earthshineilluminated lunar surface, but I'm not going so far if that's of anyfully solar illuminated environment while wearing a mostly syntheticmoon suit because, we're not talking about avoiding a 270 nm UV sunburn.Sorry about all my make-due reverse engineering logic, or lackthereof. I was simply trying to establish upon the amount of solarradiation that becomes hard X-Ray class. So exactly how much is it ona typical lunar day, or how about on a good day as well as on a trulyHighnergy cosmic rays do not come from the sun ???Do we suppose that happens to include the likes of the last couple ofweeks of horrific solar flak?Seems there should be some specific knowledge (excluding Apollo) ofwhat's what pertaining to the solar illuminated surface, as opposed tothe absolute lunar nighttime environment and, of something specificpertaining to whatever earthshine contributes.This task of obtaining knowledge is somewhat like my getting a graspupon the applied energy (thrust or torque) involved in acceleratingsomething the size and mass of the moon, so that it accelerates andthereby recedes form Earth at 38 mm/year.As worthy feedback provided from: Ami Silberman (silber@mitre.org)The mechanisms for the lunar recession have been well understood fordecades. In a nutshell, tides cause friction between the oceans andthe ocean floors, which transfers energy from the solid part of theearth to the oceans. One of the effects of this friction is that thetidal bulge is off-center, and is located eastward of the moon. (Sothe high tide actually occurs when the moon is west of overhead.) Theresult of the tidal bulge being off center is that there is a torgueeffect placed on the moon, and this in turn transfers energy from theearth to the moon. The earth's spin rate slows, the moon is speeded inits orbit and therefor moves further away from the earth. (Thistransfer of energy is essentially a transfer of angular momentum,which is a conserved quantity.) The historical (over geological eras)rate of recession has varied due to varying amounts of tidal frictiondue to shallower or deeper oceans, and the positions of thecontinents.For the benefit of all my loyal critics, I've conceded that there's adarn good chance that the likes of Tim Thompson has more than a fewvalid points as to his version of what's what. This following page isjust another example of my learning from the pros, of accepting otherinput, which may even including the likes of what Ami Silberman justpresented, that I'd not be calling flak, as there actually seems to besome considerable worth to at least Tim's version of the lunarrecession, if I don't say so myself.http://guthvenus.tripod.com/earth-moonnergy.htm=== === Subject: : Re: Vito Rizzuto - February 21st 1946> But DSK regularly changes his email address thus frustrating> those of us who try to killfile him.As you've no doubt noticed though Robin, his posts (to sci.mathanyway) do have an Achilles heel - a four-digit year at the endof the title.If your newsreader supports wildcard titles in kill rules, allone needs is a filter on titles of the form /(19|20)dd$/(in Perl wildcard syntax).Cheers------------------------------------------------ ---------------------------John R Ramsden (jr@adslate.com)---------------------------------------------- -----------------------------Eternity is a long time, especially towards the end. Woody Allen=== === Subject: : Re: Usenet Posting Guide?>But that's always been the case. We made a lot of money doing>work that our customers didn't want or have to do. What we>didn't do (when we weren't ing idiots) was hide all the>warts under pretty pictures.I'm sorry, WHAT did you say you were doing to the idiots?!=== === Subject: : Re: associative one-way function>Does anyone know if there is any potential one-way function>which is associative other than modular exponentiation?How is this associative? Is (a^b)^c = a^(b^c)? I think not!(And therefor I am not?)>That is, given a function f(x,y) (which is easy to compute),>- given z and x, it is difficult to find y such that z = f(x,y)>- f(f(x, y1), y2) = f(f(x, y2), y1)That doesn't look like the associative law, either.Setting the words aside, it sort of looks like what you want occursin any group: given a group element x and an integer n we computethe power x^n in the group; certainly (x^n)^m = (x^m)^n, which iswhat you appear to request in your last line. And yes, it's oftendifficult to take an element in a cyclic group and decide what power ofthe generator it is.Your initial example is of this type, where the group is (Z/kZ)^* , themultiplicative group mod k (where k is any fixed integer). Anotherexample commonly used for one-way functions is to use the group ofpoints in an elliptic curve mod p.dave=== === Subject: : Re: division by general integer using register shifts> ,> As most here are already aware (but just in case some aren't),> if> you shift the digits of a number to the left or right one time then it's> like multiplying or dividing the number by its radix. So if you perform a> binary left shift on say:> 00011011 (27 base 10)> you get:> 00110110 (54 base 10 or 27 * 2)> So you can multiply a number by any power of two very quickly (for a> computer, register shifts are much faster than mult/div. instructions) by> shifting. What about numbers that are not powers of two? They can bedone> by distributing the operation across numbers that are powers of two andsum> to the number you're trying to multiply by. Say you want to multiply by> 800. 800 is not a power of two, but 32, 256, and 512 are(32+256+512=800).> So:> n*800 = (n*32)+(n*256)+(n*512) = (n<<5)+(n<<8)+(n<<9)> The same is true for division. Binary right shifting is the same as> dividing by two. I can't seem to work out the algebra though. Cansomeone> tell me how (or is it possible) to do division by a general integer using> shifts?> --> Best wishes,> AllenYou could try not working binary but convert to the divisor as radixinstead, and then reconvert to binary.Does this help?=== === Subject: : differentiable...problem...if f is differentialbe on (0, infinite)and lim [f(x) +f'(x)] = L (x->infinite)show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite)------------------------------------it seems to be trivial. but i no touch.....oops........help.....me......please......===In sci.math, James Harris:> If you saw> (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2)> with the c's algebraic integers, I think few of you would have a> problem realizing that only two of the c's have 7 as a factor.Not clear at all without knowing precisely what the roots are.Since two of the roots are complex in this case I might havea chance of not using trig, but it would take some work.I can tell you that the roots, as reported by Pari/GP, areapproximately:-4.424076847035404679966063971- 0.2879615764822976600169680144 - 0.6075770270241740255965589263*I- 0.2879615764822976600169680144 + 0.6075770270241740255965589263*Ibut this doesn't help much as algebraic units are provably densearound (0 + 0i) -- sqrt(n+1) - sqrt(n) is a unit.> But, of course, you're looking at *functions* of x, as you have > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > so I could also write it as> (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2).> Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2> as long as you're in a ring where 7 is not a factor of 22.Only an issue modulo 3 or 5, AFAICT.... :-)I can posit a simpler example. Let's assume for gigglesthe following equation, similar to yours:(d_1 * x + 7) * (d_2 * x + 7) * (d_3 * x + 2) = 49 * (x^3 + 2)and carry through your logic. I know the roots to this equation. :-)We know thatx^3 + 2 = (x + 2^(1/3)) *(x + 2^(1/3) * (-1/2 + sqrt(3) * I/2)) *(x + 2^(1/3) * (-1/2 - sqrt(3) * I/2))(GP/Pari verifies this nicely, as it turns out).Therefore, one can pick d_1 = 7 / (2^(1/3)),d_2 = 7 / (2^(1/3) * (-1/2 + sqrt(3) * I/2)),d_3 = 2 / (2^(1/3) * (-1/2 - sqrt(3) * I/2)).Since -1/2 + sqrt(3) * I/2 and -1/2 - sqrt(3) * I/2 are bothunits (as is easily verified by multiplying them together),they factor little in the analysis. I can just as easilypositd_1' = d_2' = 7 / (2^(1/3))d_3' = 2 / (2^(1/3)) = 2^(2/3)Now d_3' turns out to be an algebraic integer, solving theequation y^3 - 4 = 0. (d_3 solves this equation too.)d_1' ? Well, lessee; the most obvious equation is2 * y^3 - 343 = 0which is clearly not going to lead to anything resembling analgebraic integer. Since d_1' requires a cube root no equationof lesser degree will do.But I may need those roots to your equation, and not justapproximations, either. If one positsN(x) = x^3 + 5*x^2 + 3*x + 2then, if one substitutes x = y-5/3, one can rewrite this as:N(y-5/3) = y^3 - 16/3 * x + 169/27Now substitutey = w + 16/(9*w)(Vi.8fta's substitution, with p = -16/3) and getN(w+16/(9*w)-5/3) = (729*w^6 + 4563*w^3 + 4096)/(729*w^3)Surprise: w^3 = (-4563 sqrt(4563^2-4*729*4096)) / (2 * 729)= (-4563 sqrt(8877033)) / (2 * 729) , by our old friend,the quadratic formula.Since 8877033 = 3^9*11*41 and 4563 = 3^3 * 13^2, we can divideby 3^3/3^3 (or sqrt(3^6)/3^3), resulting in w^3 = (-169 sqrt(12177)) / (2 * 27)orw^3 = (-169 3*sqrt(1353)) / (2 * 27)I'm going to take the + sign; the - sign will yield slightlyidentical results, as it turns out.Because GP/Pari does odd (but somewhat logical)things with (-1)^3, I'm going to take the sign out now.The results can be written:x1 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) - 16/(3*((169 - 3*sqrt(1353)) / 2)^(1/3))x2 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 - sqrt(3) * I/2) - 16/(3*((169 - 3*sqrt(1353)) / 2)^(1/3)) * (-1/2 + sqrt(3) * I/2)x3 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 + sqrt(3) * I/2) - 16/(3*((169 - 3*sqrt(1353)) / 2)^(1/3)) * (-1/2 - sqrt(3) * I/2)It's worth noting that(169 - 3*sqrt(1353)) * (169 + 3*sqrt(1353)) = 16384 = 2^14,so one can rewrite the above asx1 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) - 1/3 * ((169+3*sqrt(1353))/2)^(1/3)x2 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 - sqrt(3) * I/2) - 1/3 * ((169+3*sqrt(1353))/2)^(1/3) * (-1/2 + sqrt(3) * I/2)x3 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 + sqrt(3) * I/2) - 1/3 * ((169+3*sqrt(1353))/2)^(1/3) * (-1/2 - sqrt(3) * I/2)If you look carefully you'll see that using the alternate solutionw^3 = (-169 - 3*sqrt(1353)) / (2 * 27)will merely swap x2 and x3 in the final list.In this form, figuring out whether xi is an algebraic integer is aLost Cause(tm), although we know the answer is yes since theyall solve the equation x^3 + 5*x^2 + 3*x + 2 = 0, as one canverify if one multiplies(x-x1)*(x-x2)*(x-x3)in GP/Pari.However, there's a different route, making the exactroots entirely irrelevant. (Sorry, guys -- but it *was*an interesting solution using Vi.8fta's substitution. :-) )We are hypothesizing(c_1*x + 7)(c_2*x + 7)( c_3*x + 2) = 49(x^3 + 5 x^2 + 3x + 2).for some algebraic numbers ci. This hypothesis turns out to bequite valid, as it turns out -- but what are the ci?Since we knowx^3 + 5x^2 + 3x + 2 = (x-x1)*(x-x2)*(x-x3)we can deduce that c1 = -7/x1, c2 = -7/x2, and c3 = -2/x3, for somerotation of x1, x2, x3. Are these algebraic integers?We note that, if xi solvesx^3 + 5 x^2 + 3x + 2 = 0,then 1/xi has to solve2*z^3 + 3*z^2 + 5*z + 1 = 0,7/xi has to solve2*z^3 + 3*7*z^2 + 5*7^2*z + 7^3 = 0,and -7/xi has to solve2*z^3 - 3*7*z^2 + 5*7^2*z - 7^3 = 0.-7/xi is clearly *not* an algebraic integer (as the aboveis not a reducible polynomial over Q).Fortunately, -2/xi is:2*z^3 - 3*2*z^2 + 5*2^2*z - 2^3 = 0orz^3 - 3*z^2 + 10*z - 4 = 0Well, 1 out of 3 ain't bad.This counter-proof generalizes quite nicely for most cubic polynomialsx^3 + a_2 * x^2 + a_1 * x + a_0with non-unit a_0 not divisible by 7.I'm afraid you may be out of luck, James. :-)[rest snipped]-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Uncle Al is Sadistic .In sci.math, Nicolas Le Roux:> amanda grava .88 la saucisse et au marteau:> [SNIP]> Could you *please* the lines you are not responding to.Erm...are you sure you didn't inadvertantly snip the word snip?Pleasing the lines sounds impossible; I've never quite figuredout whether chocolates or flowers will make the sentence This statement is falsehappy.:-) ;-) :-)> Otherwise, this> is very hard and unpleasant to read and, besides, it takes more time to> download (and it's significative with a low-speed modem).I can relate to that to some extent; I've a 56k myself. Fortunately,I also use leafnode (fetchnews) -- though that doesn't do a thingfor having to pay by the connection minute.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: torque T = r x F and basic tensors Q( a x b ) = 1/(detQ) ( Qa x Qb ) where Q is the matrix of the change of basis, but I'm not quite seeing how that matches the A' = Q^{-1} A Q form.> As you say, under the change of orthonormal basis, the vectors a and b> go to a' = Qa and b' = Qb. In component form,> a_i = (sum over m) Q_(im) a_m and b_i = (sum over n) Q_(in) b_n> I've used different summation indices in order that the substitutions> below makes sense.> Because both the new and old bases are orthonormal, Q is an orthogonal> matrix, i.e., Q^T = Q^(-1), i.e if M = Q^(-1), M_(ij) = Q_(ji). Thus,> det Q = +/- 1. If in addition, right-handed orthonormal bases are> taken to right-handed orthonormal bases, then det Q = 1, and Q is called> a special orthogonal matrix. The set of all special orthogonal matrices> forms the group SO(3), where the notation is fairly selfxplanatory -> S for special (det +1), O for orthogonal (inverse = transpose), 3 for> 3 by 3.> Define matrix A by A_(ij) = a_i b_j - a_j b_i. Under a transformation,> A'_ij = a'_i b'_j - a'_j b'_i> = (sum over m,n) [Q_(im) a_m Q_(jn) b_n> - Q_(jn) a_n Q_(im) b_m]> = (sum over m,n) [Q_(im) a_m b_n Q_(jn)> - Q_(im) a_n b_m Q_(jn)]> = (sum over m,n) [Q_(im) (a_m b_n - a_n b_m) Q_jn]> = (sum over m,n) [Q_(im) A_(mn) Q^(-1)_nj]> A = Q A Q^(-1)> This is a little different than A = Q A Q^(-1) because, the way Feynman> defines things, a'_i = (sum over n) Q^(-1)_(in) a_n. It's a good> exercise to show this.sure what you were meaning to show here, if you will note the last 4 linesof your posting.that is inconsistent with how I specified the change of basis, it shouldhave been A' = Q A Q^{-1} and that is what you have.I also note that Feynman's matrix object T_ij = x_i F_j - x_j F_i doesindeed transform as T' = Q T Q^{-1}, as you have shown. I am still mystified as to what real value there is in taking a niceaxial vector object like Torque and making the anti-symmetric matrix out of it...=== === Subject: : Re: Uncle Al is Sadistic . grava .88 la saucisse et au marteau:Could you *please* the lines you are not responding to.> Erm...are you sure you didn't inadvertantly snip the word snip?Actually, as far as I remember, I think it was the word delete.-- NicolasWow, THAT'S an interessant discussion=== === Subject: : Re: differentiable...problem...Content-transferncoding: 8bit> if f is differentialbe on (0, infinite)> and lim [f(x) +f'(x)] = L (x->infinite)> show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite)Well. Perhaps L'Hopital's Rule has some use, after all.--Ron Bruck=== === Subject: : question on finding an isogeny for two curvesCan the direct isogeny be found between these two elliptic curves?It is known that they are in a set of 8 isogenous curves, but whatis the exact transformation that takes x <--> X, i.e. points on (1)to points on (2)?(1) y^2 = (x + 366)*(x^2 - 366*x + 2625489)(2) Y^2 = (X - 1551)*(X^2 + 1551*X + 497214)Both curves have conductor 210.More generally, how might one find the isogeny betweeny^2 = x^3+ax+b and Y^2 = X^3+AX+B, when the curves havethe same conductor? (a,b,A,B are known)=== === Subject: : Re: Uncle Al is Sadistic .> amanda grava .88 la saucisse et au marteau:> [SNIP]> Could you *please* the lines you are not responding to. Otherwise, this> is very hard and unpleasant to read and, besides, it takes more time to> download (and it's significative with a low-speed modem).Well..when I used to do that (not in this group though), I was accusedof leaving out the info to mislead the reader, etc. Anyway, I am very sorry. I will do that from now on.=== === Subject: : Re: Continuumfluently presume at least that you are presumably fluent in what passes forEnglish in the state that gave the world Hollywood . It's sheer fun to seehow all those quick to stomp into dirt some kid asking for homework help,wouldn't find a pick better then notational for my babbling. Know ye,philistines, that Z stands for Ze Any Set With Cardinality of ZeContinuum.amateurBesides, to comment on youawareness, what languages do you fluently read in except english ?> You presume that I am fluent in English?=== === Subject: : Re: Continuumfluently presume at least that you are presumably fluent in what passes forEnglish in the state that gave the world Hollywood . It's sheer fun to seehow all those quick to stomp into dirt some kid asking for homework help,wouldn't find a pick better then notational for my babbling. Know ye,philistines, that Z stands for Ze Any Set With Cardinality of ZeContinuum.amateurBesides, to comment on youawareness, what languages do you fluently read in except english ?> You presume that I am fluent in English?=== === Subject: : [JSH] Re: My [Crank] research [Yeah Right], publication [in Crank Net] announcement [Joke]> There's more to my work than just arguing on Usenet, so I'd like to> point out that my paper Advanced Polynomial Factorization is slated> to be published:> See http://www.megasociety.net/NoesisHighlights.html> The Mega Foundation is an organization of high IQ people, and I'm glad> to be associated with them. To learn further about the organization> you can use Google, or see:> Why a group like the Mega Foundation?> http://www.ultrahiq.org/Mega/WhyMega.htmMr. Harris,to me, this seems equivalent to the copyright you had on the web page youcreated for your famed Proof of FLT.Now, you are once again hell-bent on crying, cursing, cheating and trying todo anything to propagate your flawed lies.You also have some sort of greatness disorder that makes you think you areabove every person you come in contact with. IIRC, this is a NarcissisticPersonality Disorder.Here is excerpt from the web site: http://www.suite101.com/welcome.cfm/npdNarcissists attract abuse. Haughty, exploitative, demanding, insensitive,and quarrelsome - they tend to draw opprobrium and provoke anger and evenhatred. Sorely lacking in interpersonal skills, devoid of empathy, andsteeped in irksome grandiose fantasies - they invariably fail to mitigatethe irritation and revolt that they induce in others.Here is a clinical definition:http://www.behavenet.com/capsules/disorders/ narcissisticpd.htm***Diagnostic criteria for 301.81 Narcissistic Personality Disorder (cautionarystatement)A pervasive pattern of grandiosity (in fantasy or behavior), need foradmiration, and lack of empathy, beginning by early adulthood and present ina variety of contexts, as indicated by five (or more) of the following:(1) has a grandiose sense of self-importance (e.g., exaggerates achievementsand talents, expects to be recognized as superior without commensurateachievements)(2) is preoccupied with fantasies of unlimited success, power, brilliance,beauty, or ideal love(3) believes that he or she is special and unique and can only beunderstood by, or should associate with, other special or high-status people(or institutions)(4) requires excessive admiration(5) has a sense of entitlement, i.e., unreasonable expectations ofespecially favorable treatment or automatic compliance with his or herexpectations(6) is interpersonally exploitative, i.e., takes advantage of others toachieve his or her own ends(7) lacks empathy: is unwilling to recognize or identify with the feelingsand needs of others(8) is often envious of others or believes that others are envious of him orher(9) shows arrogant, haughty behaviors or attitudesReprinted with permission from the Diagnostic and Statistical Manual ofMental Disorders, fourth Edition. Copyright 1994 American PsychiatricAssociation***Perhaps instead of spewing your useless nonsense, you should study thisdisorder, go see a doctor and then drug and drink yourself to death.Think about, eh, let's wack you!=== === Subject: : Re: torque T = r x F and basic tensorsX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzXcrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith 1. Why is there a cross product in the definition of torque T = r x>F ?Because you're doing Physics in 3 dimensions, you can mapskew-symmetric tensors into vectors. For more than 3 dimensions, youhave to explicitly treat the torque as a skew symmetric tensor or a2-form.>2. How is torque a tensor? Because the definition of cross product makes it transform like one.If you don't want to treat it as a tensor or a 2-form, than you needto treat it as a vector and pay close attention to your weights.Either way, you need to be careful to distinguish covariant indicesfrom contravariant.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : Re: Hard tensor question (kst). <3fa668f5$9$fuzhry+tra$mr2ice@news.patriot.net> <2202379a.0311052323.77555cbc@posting.google.comX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzXcrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith As I understand, the indices forming the relatively>defined tensor K are meaningless after the definition,>but I have some uncertainty.I'm not sure what you're trying to say, but change to polarcoordinates and observe what happens.>The relative tensor K_uv is actually a 4th rank>(counting weight) like the Riemann-Christoffel >Curvature tensor. The weight has nothing to do with the rank.>It is very evident the Kronecker delta (d^u_v) >serves as an orthogonal metric,No it doesn't.>relativity of K_uv?-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to spamtrap@library.lspace.org=== === Subject: : [JSH] [LUNATIC FRINGE] Re: My [Flawed] research, [JOKE] publication announcementMr. Harris,Psychopathology------------------------Narcissism:A pattern of traits and behaviors which signify infatuation and obsessionwith one's self to the exclusion of all others and the egotistic andruthless pursuit of one's gratification, dominance and ambition.Perhaps research in to your delusional fits of grandeur would be a moreappropriate vocation for you?Still pissing out your research?=== === Subject: : Re: Why is math so difficult for some people?>Similarly, Cramer's proof of the Levy-Cramer theorem that>the sum of two independent random variables is not normal>unless they both are is probably the only easy proof, but>it likewise obscures the ideas. Any time characteristic>functions are used to prove a probability theorem, the>concepts are not even present.Do you happen to know of a probabilistic proof of the Cramr-Lvy theorem?> There are entropy proofs. Any published?-- A.=== === Subject: : Re: Why is math so difficult for some people?> Many of the so-called elegant proofs completely [sic] hide the concepts.>Elegance is characterized by the propensity toward dispensing with>extraneous inessentials.> Proving the Central Limit Theorem for sums of> independent identical distributions is certainly the most> elegant, but it [sic] hides virtually everything; it has no> probability in it at all.But the characteristic function proof includes extraneousinessentials. The Lindeberg proof and the Skorokhod embedding use different strictly probability approaches,although they are not completely probability arguments;I do not believe there can be a completely probabilityargument, as versions of the Berrysseen bounds just cannot be done strictly probabilistically; the bounds onthe difference of sums of iid random variables with a finite third moment and sums of iid normal random variableswith the same variance, but the sequences dependent, doesnot increase with the number.Both of these proofs go over almost verbatim for martingales. The characteristic function proof does go over, but not aseasily. -- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558===> If you saw> (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) => 49(x^3 + 5x^2 + 3x + 2)> with the c's algebraic integers, I think few of you would have a> problem realizing that only two of the c's have 7 as a factor.> But, of course, you're looking at *functions* of x, as you have> f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x,> so I could also write it as> (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2).> Notice that dividing both sides by 49 gives> (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2> as long as you're in a ring where 7 is not a factor of 22.> I want to emphasize that point as notice there's only *one* way to> divide through by 49 if 7 is not a factor of 22.> Usually you can *see* the other factors of 7, but I want you to> abstract, and generalize.> Please pay careful attention to that example.> You may see people who reply claiming that the word polynomial has> some significance, as if it's a mystical thing which refutes basic> logic, so if something isn't polynomial it no longer behaves> logically.All functions that I know of, all of them behave logically when examinedfrom the view point of calculus.Well, I know that a derivative measures change and I also know that when theinstantaneous change is 0, there is some sort of a critical point(determined by the 2nd derivative).Let's say we want to describe the graph of f(x) in words. Begin by findingf'(x).f'(x)=cos(x)We know that a function has a critical point at a point where the derivativeis 0.cos(x)=0 at x=pi/2 and 3pi/2Ok, we know where the critical points are, but we don't know what happens atthose points. So take the 2nd derivative. f''(x)=-sin(x)When x=pi/2, f''(x)=-1. When x=3pi/2 f''(x)=1Therefore I can now analyze my results. When x=pi/2, the 2nd derivative isnegative, which means it's decreasing so pi/2 must be a maximum. Whenx=3pi/2, the 2nd derivative is positive so 3pi/2 must be a minimum.In fact, if I graph this, my predictions are correct.My point is that all elementary functions that I know of behave logically ifyou examine it from a calculus viewpoint, which may not be a bad idea foryou to do.David Moran> Now then, in my advanced factorization work, I just use functions of x> that are a lot more complicated than f_1(x) = c_1 x, and unfortunately> there are people who can use an unfamiliar leap in complexity to> confuse others.> Some of you have learned various advanced math topics, now imagine if> in your classrooms there were some hecklers who continually hollered> out at your teacher, or otherwise disrupted the class?> What if when there were difficult concepts those hecklers would try to> confuse everyone as they sought to discredit the mathematics?> If you find that hard to imagine, imagine me in your class with you> questioning the professor and calling him names.> How much would you have learned?> I need those of you interested in mathematics to focus on the basics,> so that you can understand the advanced.> James Harris> http://mathforprofit.blogspot.com/===> You may see people who reply claiming that the word polynomial has> some significance, as if it's a mystical thing which refutes basic> logic, so if something isn't polynomial it no longer behaves> logically.No both polynomial and non-polynomial functions behave logically.It is just that polynomials have some properties that arenot true in general.Let P(x) by a polynomial divisible by p, and let g_1(x) andg_2(x) be polynomials such that P(x) = g_1(x) * g_2(x)Then p divides at least one of g_1(x) and g_2(x).If g_1(x) and g_2(x) are not polynomials, this is no longer true.Counterexample.Let P(x) = 15-3x. Then P(x) is divisible by 3.Let g_1(x)=4-sqrt(1+3x) and g_2(x)=4+sqrt(1+3x).Then P(x) = g1(x)*g2(x)list a few valuesP( 0) = 15, g_1( 0) = 3, g_2( 0) = 5P( 1) = 12, g_1( 1) = 2, g_2( 1) = 6P( 5) = 0, g_1( 5) = 0. g_2( 5) = 8P( 8) = -9, g_1( 8) = -1. g_2( 8) = 9P(16) = -33, g_1(16) = -3. g_2(16) = 11Note how the factor of 3 switches back and forthbetween g_1 and g_2 depending on the value of x.Thus neither g_1(x) nor g_2(x) is divisible by3 for all x. In particular the fact that g_1(0)is divisible by 3 does not mean that g_1(1) isdivisible by 3. - William Hughes=== > If you saw > (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2) > with the c's algebraic integers, I think few of you would have a > problem realizing that only two of the c's have 7 as a factor.How do you know that this is possible with algebraic integer c1 to c3?I do not think this is a valid factorisation in the algebraic integers.Could you provide the monic polynomial with integer coefficients ofwhich the c's are the roots? More to the point, x^3 + 5x^3 + 3x + 2factors as (x - r1)(x - r2)(x - r3) in the algebraic integers. Wherethe r's are the roots of the polynomial. We can multiply the factorsby 7/r1, 7/r2 and 2/r3 to get your factorisation. So we get: (7/r1 x + 7)(7/r2 x + 7)(2/r3 x + 2).So c1 = 7/r1, c2 = 7/r2, c3 = 2/r3.Of these only c3 is an algebraic integer (r3 is a divisor of 2).Neither 7/r1, nor 7/r2 are algebraic integers. Unless you claimthat r1 and r2 are units (they are divisors of 2). But they arenot, because the constant term of a polynomial (see the correctusage here, James) must be 1 for a root to be a unit.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/=== === Subject: : Re: differentiable...problem...> if f is differentiable on (0, infinite)> and lim [f(x) + f'(x)] = L (x->infinite)> show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite)Hint: L'HopitalNote: this classic problem is well-known due to the factthat it appeared in Hardy's classic calculus textbook witha solution more awkward than the elegant L'Hopital approach.For further details and references on the L'Hospital-based approach see my prior post [1], and [2].-Bill Dubuque[1] http://google.com/groups?threadm=y8zognpn1wa.fsf% 40berne.ai.mit.edu[2] Martin D. Landau; William R. Jones. A Hardy Old Problem.Math. Magazine 56 (1983) 230-232.=== === Subject: : Re: differentiable...problem... >if f is differentialbe on (0, infinite) >and lim [f(x) +f'(x)] = L (x->infinite) >show that lim f(x) = L (x->infinite) and >lim f'(x) = 0 (x->infinite)Bill Dubuque (wgd@berne.ai.mit.edu) > f e^x (f + f') e^x > lim f + f' = L => lim f = lim ----- = lim ------------ = L >x->oo x->oo x->oo e^x x->oo e^xProvided f e^x -> +-oo, in which case 0 = L - L = lim f+f' - lim f = lim f'However when f e^x -> k; then f = fe^x e^-x -> 0 f e^-x = f e^x e^-2x -> 0L = lim f+f' - lim 2f = lim f'-f0 = lim f = lim f e^-x / e^-x = (f' - f)e^-x / ^-x = lim f-f' = -L0 = lim f+f' - lim f = lim f'So if lim f = k and lim f' exists, then lim(x->oo) f+f' = k + lim f' k = lim(x->oo) f = k + lim f'; lim f' = 0If lim f' doesn't exist, then for n >= 2 f_n(x) = sin(x^n)/x -> 0 (f_n)'(x) = nx^(n-1) cos (x^n)/x - (sin x^n)/x^2 -> oscillation----=== === Subject: : Re: differentiable...problem...> if f is differentialbe on (0, infinite)> and lim [f(x) +f'(x)] = L (x->infinite)> show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite)Let's suppose L=0 (substitute f by f-L).We must prove that if f(x)+f'(x)->0 (x->inf), then f(x)->0 (x->inf)f(x)+f'(x)=o(1) (x->inf)multiply by e^(x) :e^(x)[f(x)+f'(x)]=o(e(x)) (x->inf)integrate from 0 to t : (the derivative of e^(x)f(x) is e^(x)[f(x)+f'(x)])e^(t)f(t)-f(0)=o(e(t)) (t->inf)Divide by e^t :f(t)=o(1) which means that f(t)->0 (t->inf)(excuse my english)=== === Subject: : if x has normally distributed digits, does 1/x?Is there a simple proof, or counterexample, to the propositionthat if we consider the expansion of the fractional partof some (irrational) number x in some base, and know that it is normal,then it follows that the expansion of 1/x is also normal?I'd appreciate a pointer to any literature in this area.Rob Shaw=== === Subject: : Re: if x has normally distributed digits, does 1/x?>Is there a simple proof, or counterexample, to the proposition>that if we consider the expansion of the fractional part>of some (irrational) number x in some base, and know that it is normal,>then it follows that the expansion of 1/x is also normal?>I'd appreciate a pointer to any literature in this area.What do you mean by normally distributed digits? The normal distribution is a continuous one.Nor would it make sense to speak of x - floor(x) being normally distributed, since the normal dsitrubtion is supported by the entire real line.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: if x has normally distributed digits, does 1/x?Is there a simple proof, or counterexample, to the propositionthat if we consider the expansion of the fractional partof some (irrational) number x in some base, and know that it is normal,then it follows that the expansion of 1/x is also normal?I'd appreciate a pointer to any literature in this area.> What do you mean by normally distributed digits? The normal > distribution is a continuous one.> Nor would it make sense to speak of x - floor(x) being normally > distributed, since the normal dsitrubtion is supported by the entire > real line.A real number is said to be normal (to the base 10) if for all n every string of digits of length n appears with frequency 10^(-n) in its decimal expansion. This has nothing to do with the (Gaussian) normal distribution that you are thinking of. As others have noted, it seems unlikely that x normal implies 1/x normal, but I don't know what's in the literature. Let y be normal to base 2, and let x be the number you get by making believe y is a decimal expansion. Then x isn't normal - it has only the digits 0 and 1 - but I'd expect 1/x to be normal (though I wouldn't be able to prove it).-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)=== === Subject: : Re: if x has normally distributed digits, does 1/x?>Is there a simple proof, or counterexample, to the proposition>that if we consider the expansion of the fractional part>of some (irrational) number x in some base, and know that it is normal,>then it follows that the expansion of 1/x is also normal?I seriously doubt it. Here's an indication of why:Let's talk about base 2. Start with x = .01010101... .Of course x is not normal, but it does contain the rightnumber of 0's and 1's; it's weakly normal in base 2, ina sense. But x = 1/3, so 1/x = three = 11.000... , which isfar from normal.I wouldn't be surprised if you could actually constructa counterexample along these lines - you'd need to verifya few things...>I'd appreciate a pointer to any literature in this area.>Rob Shaw=== === Subject: : Re: if x has normally distributed digits, does 1/x?Is there a simple proof, or counterexample, to the propositionthat if we consider the expansion of the fractional partof some (irrational) number x in some base, and know that it is normal,then it follows that the expansion of 1/x is also normal?> I seriously doubt it. Here's an indication of why:> Let's talk about base 2. Start with x = .01010101... .> Of course x is not normal, but it does contain the right> number of 0's and 1's; it's weakly normal in base 2, in> a sense. But x = 1/3, so 1/x = three = 11.000... , which is> far from normal.You're in the land of rationals. Rationals and normality don't mix.(Take that out of context, man!)Phil-- Unpatched IE vulnerability: window.open search injectionDescription: cross-domain scripting, cookie/data/identity theft, command executionReference: http://safecenter.net/liudieyu/WsFakeSrc/ WsFakeSrc-Content.HTMExploit: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm= == === Subject: : Re: if x has normally distributed digits, does 1/x?>Is there a simple proof, or counterexample, to the proposition>that if we consider the expansion of the fractional part>of some (irrational) number x in some base, and know that it is normal,>then it follows that the expansion of 1/x is also normal?> I seriously doubt it. Here's an indication of why:> Let's talk about base 2. Start with x = .01010101... .> Of course x is not normal, but it does contain the right> number of 0's and 1's; it's weakly normal in base 2, in> a sense. But x = 1/3, so 1/x = three = 11.000... , which is> far from normal.>You're in the land of rationals. >Rationals and normality don't mix.>(Take that out of context, man!)Well, that's amusing enough to make it seem possible that youwere just trying to be funny. In case you were also trying tomake a serious point:I certainly didn't mean that this was a counterexample, orsomething that would lead to a counterexample with no newideas required. But what I had in mind was something_vaguely_ like this: Suppose we could find x_n such thateach n-bit sequence in the binary expansion of x_n occurswith the right frequency, but 1/x_n has a terminatingbinary expansion. Then let x consist of a long stretchof the bits of x_1, followed by a much longer stretchof the bits of x_2, etc. For suitable values of longand longer x will be normal, but it doesn't seem sounlikely that 1/x would turn out to be abnormal.>Phil>-- >Unpatched IE vulnerability: window.open search injection>Description: cross-domain scripting, cookie/data/identity > theft, command execution>Reference: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-Content.HTM >Exploit: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm= == === Subject: : Image of a straight line in a convex cylindrical mirrorWhat is the reflection in a cylindrical mirror x^2+y^2 = a^2 of astraight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c)? Is it a catenary ? If not, what is it?=== === Subject: : Re: Image of a straight line in a convex cylindrical mirrorWhen is the homework due?> What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a> straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c)> ? Is it a catenary ? If not, what is it?=== === Subject: : Re: Image of a straight line in a convex cylindrical mirror> When is the homework due?:) this is my curiosity about curvarture of lines when the virtualimage is distorted in reflection.I observed clear images of square lines of floor mosaic in a polishedcylindrical flowerpot in Twelve Oak Mall, Detroit. What is commonlyknown is the optics: image-object distances are connected by standardformula (u-f)(v-f)=f^2 where f=a/2 or focal length, u,v are object &image distances etc. in reflection. I did not do much of image / raytracing in 3D before .I believed that cos(slope angle) in the x-z projection is inverselyproportional to z-coordinate, as it is for a catenary z=A cosh(x/A).I asked this question expecting that there could be elementarygeometrical operations with straight lines, cylinders, reflection &c.My imagination surrounded vague thoughts about geodesic deviation, butI right now I would thank in advance for any ensuing pointers andclarifications.=== === Subject: : Re: Image of a straight line in a convex cylindrical mirror> What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a> straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c)> ? Is it a catenary ? If not, what is it?What have you done so far?=== === Subject: : another idea about HarrisThis guy is totally out of touch with reality. So whatwould be the harm in simply ignoring him from thisday forward and forever, and permitting him to retirein the triumphant knowledge (as far as he's concerned)that he is absolutely and totally correct in everythinghe's ever said? He would be quite happy, and then youall could have a lot more fun communicating with realmathematicians. You all must know by now that youare wasting your time. I mean the ones of you whoare responding in a thoughtful reasoned way to James.The others, who are simply calling him a horse's ass,should join another group, maybe themutualasskicking.math group.=== === Subject: : Re: another idea about Harris> This guy is totally out of touch with reality. So what> would be the harm in simply ignoring him from this> day forward and forever, and permitting him to retire> in the triumphant knowledge (as far as he's concerned)> that he is absolutely and totally correct in everything> he's ever said?he wouldn't retire. he is a world-class crackpot and troll.have you seen his picture?> He would be quite happy, and then you> all could have a lot more fun communicating with real> mathematicians. You all must know by now that you> are wasting your time. I mean the ones of you who> are responding in a thoughtful reasoned way to James.> The others, who are simply calling him a horse's ass,> should join another group, maybe the> mutualasskicking.math group.=== === Subject: : Re: another idea about Harris> This guy is totally out of touch with reality. So what> would be the harm in simply ignoring him from this> day forward and forever, and permitting him to retire> in the triumphant knowledge (as far as he's concerned)> that he is absolutely and totally correct in everything> he's ever said? He would be quite happy, and then you> all could have a lot more fun communicating with real> mathematicians. You all must know by now that you> are wasting your time. I mean the ones of you who> are responding in a thoughtful reasoned way to James.> The others, who are simply calling him a horse's ass,> should join another group, maybe the> mutualasskicking.math group.Besides being a crank, he is a master troll. His posts are irresistable. You yourself have added yet another JSH-related post to this newsgroup.=== === Subject: : Re: another idea about Harris>This guy is totally out of touch with reality. So what>would be the harm in simply ignoring him from this>day forward and forever, and permitting him to retire>in the triumphant knowledge (as far as he's concerned)>that he is absolutely and totally correct in everything>he's ever said? He would be quite happy, and then you>all could have a lot more fun communicating with real>mathematicians. You all must know by now that you>are wasting your time. I mean the ones of you who>are responding in a thoughtful reasoned way to James.>The others, who are simply calling him a horse's ass,>should join another group, maybe the>mutualasskicking.math group.> Besides being a crank, he is a master troll. His posts are > irresistable. You yourself have added yet another JSH-related post to > this newsgroup.And you another ... and me too.Gib=== === Subject: : Re: another idea about Harris ( [JSH] for killfiles)Francis Harrington> This guy is totally out of touch with reality. So what> would be the harm in simply ignoring him from this> day forward and forever, and permitting him to retire> in the triumphant knowledge (as far as he's concerned)> that he is absolutely and totally correct in everything> he's ever said? He would be quite happy, and then you> all could have a lot more fun communicating with real> mathematicians. You all must know by now that you> are wasting your time. I mean the ones of you who> are responding in a thoughtful reasoned way to James.> The others, who are simply calling him a horse's ass,> should join another group, maybe the> mutualasskicking.math group.Talking to Harris _about mathematics_ is a waste of time, yes. After eightyears of claiming to have a proof of FLT, he doesn't know an algebraicinteger is, or what a ring extension is. Conclusion? He doesn't _care_whether his statements are true or false.Even if he had no audience at sci.math, would he stop posting? It costs himnothing to cross-post to other rooms from which he would still get his arsekicked.The JSH thing is tedious most of the time, and morbid all the time, butstill good for the occasional laugh. Try Google with these keywords: guffaw idiot polynomialLH=== === Subject: : Analysis problem...I'm having a hard time with the following problem:Let f be a continuous function on [1,oo) such thatf^4 is Lebesgue integrable. Prove that if a > 4 then |integral of f(x^a)| < oo wherethe integral is also over [1,oo).Also, give an example where the integral of f(x^4) is infinite.Any hints or input on this would be greatly appreciated.Hugh=== === Subject: : Re: Analysis problem...> I'm having a hard time with the following problem:> Let f be a continuous function on [1,oo) such that> f^4 is Lebesgue integrable. > Prove that if a > 4 then |integral of f(x^a)| < oo where> the integral is also over [1,oo).> Also, give an example where the integral of f(x^4) is infinite.> Any hints or input on this would be greatly appreciated.Can you write the integral of f(x^a) differently?Do you know Hoelders inequality?HTH,Michael.-- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&Dr. Michael UlmFB Mathematik, Universitaet Rostockmichael.ulm@mathematik.uni-rostock.de=== === Subject: : Re: Analysis problem...Content-transferncoding: 8bit[[ This message was both posted and mailed: see the To, Cc, and Newsgroups headers for details. ]]> I'm having a hard time with the following problem:> Let f be a continuous function on [1,oo) such that> f^4 is Lebesgue integrable. > Prove that if a > 4 then |integral of f(x^a)| < oo where> the integral is also over [1,oo).> Also, give an example where the integral of f(x^4) is infinite.I'll give you a hint for a = 5: consider a change of variable inint_1^infty f(x^5)^4 x^4 dx,then think Holder.--Ron Bruck=== === Subject: : Re: JSH: Difficult social problem===> === Subject: : Re: JSH: Difficult social problem>Message-id: And believe me, if you don't speak any French, the >waiters in France won't bring you soup.Have you considered the possibility that the reason waiters don'tbring you soup might *not* be that you don't speak French?>I think that when I order soup in France (in English), I would get soup.>As the French have the word soupe which is some form of soup. E.g.bouillabaisse is a soupe. But in that case it will be the main>course.>A better example would be ordering cold water in Italy. Do not be>surprised when you get hot water.I ordered two lattes at an AutoGrill in Italy, and got exactly (and only) whatI ordered...two cups of milk.Marvin Sebournosugeography@aol.com>dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,>+31205924131>home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/=== === Subject: : Re: JSH: Difficult social problem >And believe me, if you don't speak any French, the >waiters in France won't bring you soup. > Have you considered the possibility that the reason waiters don't > bring you soup might *not* be that you don't speak French?I think that when I order soup in France (in English), I would get soup.As the French have the word soupe which is some form of soup. E.g.bouillabaisse is a soupe. But in that case it will be the maincourse.A better example would be ordering cold water in Italy. Do not besurprised when you get hot water.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/=== === Subject: : Re: help me, please!> William Elliot ha scritto nel messaggio>Is there a fast way to determine X and Y both integer such that:>X^2 - Y^2 = A>Of course the answer to your question is yes.>Indeed, you phrased your question incorrectly.>Factor x^2 - y^2 = (x-y)(x+y) = a>Thus x-y and x+y are factors of a>For each factorization of a = nm you have>x-y = n; x+y = m>solve for x and y. Beware,>if the solution turns out not be an integer, discard it.>Do this for all factorizations of a.>Collect the solutions you find.>To make sure you understand the process>solve x^2 - y^2 = 12 for integers, x,y.>First list all the ways of factoring 12.>Let's see what you try.>What are your answers?>x^2 - y^2 = 66 has no solution. Why?> bacause it has tre factors.Think of the form of the factorization of the leftside: x^2 - y^2 = (x - y)(x + y)Now, let A = x - y, and B = x + y. What can you sayabout the relationship between A and B? What sorts ofnumbers can occur as A and B? For instance, you canprobably see that B - A = 2y. What does that say aboutA and B (thought of together, not just one at a time)?Having decided what numbers A and B that can appear inthis factorization, now look at all the factorizationsof 66. What sorts of numbers do you get from thosefactorizations?If this isn't enough of a hint, you're trying too hard.The answer is really pretty simple.> Thanx> marcoDale.=== === Subject: : Re: help me, please! William Elliot ha scritto nel messaggio> Is there a fast way to determine X and Y both integer such that:> X^2 - Y^2 = AFactor x^2 - y^2 = (x-y)(x+y) = aThus x-y and x+y are factors of aFor each factorization of a = nm you havex-y = n; x+y = msolve for x and y. Beware,if the solution turns out not be an integer, discard it.Do this for all factorizations of a.Collect the solutions you find.To make sure you understand the processsolve x^2 - y^2 = 12 for integers, x,y.Clearly you skipped this step because of the question below you asked.So show us some work and what answers you get.If you just guess instead of following the processthen to learn what's to be know, solve for integers x^2 - y^2 = 4972First list all the ways of factoring 12.Let's see what you try.What are your answers?x^2 - y^2 = 66 has no solution. Why?> bacause it has tre factors.No. The number of factors has nothing to do with it.=== === Subject: : Re: sum of random variablesSaying that one random variable isequal to another one means that the value of the first is, with probability 1, equal to the value of the second.> Huh? I think you're understating. Saying that one random variable> is equal to another is stronger than what you're stating -- it means> that *every instance* of the value of the first variable is equal to> the value of the second one. (yes, what I just said implies that> the probability is one -- but probability one does not imply what I> said).> Please correct me if I'm mistakenI won't insist on this point. I think it depends on your point of view. E.g. in analysis L^1 functions are not really functions, but rather equivalence classes for equality almost everywhere.=== === Subject: : Re: sum of random variables> Saying that one random variable is> equal to another one means that the value of the first is, with > probability 1, equal to the value of the second.Huh? I think you're understating. Saying that one random variableis equal to another is stronger than what you're stating -- it meansthat *every instance* of the value of the first variable is equal tothe value of the second one. (yes, what I just said implies thatthe probability is one -- but probability one does not imply what Isaid).Please correct me if I'm mistakenCarlos--=== === Subject: : Re: Solution to an inequation>Here is an important equation I am encountering.>b <= x * ( (1-x)^a )Inequality, not equation.> I accept.>Solve for x in terms of a and b.Unlikely to have a closed-form solution in general.> Okay, for my purposes, a is an even integer and b is a ratioanl number less> than one.> I am looking for a solution in the range (0,1).> Particularly, I am interested in the smallest value of x which will satisfy> the inequality. If you can't get me the smallest value, can you get me> something as small as possible in a closed form solution. Right now the> smallest I have is the place where x * ( (1-x)^a ) is maximised.... that place being x_0 = 1/(1+a). I assume b < f(x_0) = a^a/(1+a)^(1+a), where f(x) = x (1-x)^a. It seems that f'''(x) > 0 for 0 < x < 1/(1+a), so if x_1 is any point with 0 <= x_1 <= x_0 with f(x_1) < b, f(x) > f(x_1) + f'(x_1) (x-x_1) + 1/2 f''(x_1) (x - x_1)^2.So if x_1 is not itself a solution, solve the quadratic f(x_1) + f'(x_1) (x - x_1) + 1/2 f''(x_1) (x - x_1)^2 = b;any solution x_2 of this with x_1 < x_2 < x_0 will have f(x_2) > b.In particular, with x_1 = 0 you get x_2 = (1 - sqrt(1-4ab))/(2a) ifb<1/(4a).=== === Subject: : Re: Solution to an inequationHere is an important equation I am encountering.b <= x * ( (1-x)^a )> Inequality, not equation.I accept.Solve for x in terms of a and b.> Unlikely to have a closed-form solution in general.Okay, for my purposes, a is an even integer and b is a ratioanl number lessthan one.I am looking for a solution in the range (0,1).Particularly, I am interested in the smallest value of x which will satisfythe inequality. If you can't get me the smallest value, can you get mesomething as small as possible in a closed form solution. Right now thesmallest I have is the place where x * ( (1-x)^a ) is maximised.=== === Subject: : Re: Who contributed most to mathematics?>but the Greeks started it.Just about every thing the Greeks did was rediscovered by othersas it was needed in their time. Much of Archemedies work was for War efforts as they were needed. Sure they have being first name recognition but played no more role than India and other countries to thought today.> I have heard that Euclid's Elements is the book with more editions> published than any other except the Bible. But wasn't his school in> Alexandria? So perhaps he should be assigned to Egypt (even if he was> Greek by culture and ancestry).And in fact the MacTutor site does assign Euclid to Egypt, althoughthere's no reliable information about his birthplace or ancestry.=== === Subject: : Re: Who contributed most to mathematics?>but the Greeks started it.> Just about every thing the Greeks did was rediscovered by others> as it was needed in their time. Much of Archemedies work was for War > efforts as they were needed. Sure they have being first name > recognition but played no more role than India and other countries > to thought today.> I have heard that Euclid's Elements is the book with more editions> published than any other except the Bible. But wasn't his school in> Alexandria? So perhaps he should be assigned to Egypt (even if he was> Greek by culture and ancestry).>And in fact the MacTutor site does assign Euclid to Egypt, although>there's no reliable information about his birthplace or ancestry.The Greeks in Alexandria seem to have been mainly of Greekancestry, or to be from those of other groups, mainly notEgyptians, who joined the Greeks. The Ptolemies made noattempt to submerge the other cultures, and the Greeks weresufficiently xenophobic (barbarian was the Greek for thosewho did not speak Greek, basically) that they made no attemptto do so. When the Seleucids did, they ran into open revolts.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== === Subject: : Re: Who contributed most to mathematics?>and the Greeks were>sufficiently xenophobic (barbarian was the Greek for those>who did not speak Greek, basically) Use of the term barbarian might suggest xenophobia in a modernEnglish speaker, but Herodotus for example seems to use it in aquite neutral sense to mean non-Greeks.-- Richard-- Spam filter: to mail me from a .com/.net site, put my surname in the headers.FreeBSD rules!=== === Subject: : Re: Who contributed most to mathematics? charset=iso-8859-7 Herman Rubin > , and the Greeks were> sufficiently xenophobic (barbarian was the Greek for those> who did not speak Greek, basically)Right, but the word THEN, did not have a derogative meaning.To native Greeks of the time, foreign languages sounded very much likenonsensical babling similar to var-var-var noises, thus the adjective.(b in Greek is pronounced as v in English).> -- > This address is for information only. I do not claim that these views> are those of the Statistics Department or of Purdue University.> Herman Rubin, Department of Statistics, Purdue University> hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------- -----------------------Eventually, _everything_ is understandable=== === Subject: : Re: Who contributed most to mathematics?>I was just trying to see who in fact borned the most mathematicians. >My SWAG list was not intended to be complete but just a start. Ok, I >blew it after France at least according to MacTutor link below. >However I was close with Italy, Germany and former Russia but totally >underestimated England and the U.S. and way over estimated India and>Greece. This is a quantative analysis. The next step is qualitative>analysis which is why I asked you guys the question.I doubt that anybody would claim that being in the MacTutor list is all that significant as an indication of mathematical importance: it's mainly a question of which people somebody has bothered to writea biography of. >There sure have been a lot of important French mathematicians;>Fourier, Galois, d'Alembert...>Fermat - but his last theorem was published after his death. >So during his life time it did not play a major role in mathematical>developement.Fermat's importance in mathematics does not rest on his Last Theorem.He made many contributions in various areas. Although he didn't publishmuch, he corresponded with many important mathematicians, so he wasquite influential in his lifetime.> The same can be said of Da Vinci and other Greats whose>certain works were suppressed during their lifetime.Although da Vinci did study a lot of mathematics, I'm not aware ofany really significant contributions he made to mathematics. >but the Greeks started it.>Just about every thing the Greeks did was rediscovered by others>as it was needed in their time. Much of Archemedies work was for War >efforts as they were needed. Sure they have being first name >recognition but played no more role than India and other countries >to thought today.This is just silliness. The main contribution of the Greeks wasto discover the idea of mathematical proof. Nearly all Western mathematicians until very recent times got their first taste ofmathematics from studying Euclid.=== === Subject: : Re: Who contributed most to mathematics?Please forgive me my ignorance, I missed a few classes in collegethanks to some heavily hopped Czechish Pilsner, but could youelaborate on this list especially regarding Czechoslovakia andTaiwan?> Others have addressed some of the other countries. My recollection is> that Hamilton was from Ireland and that Chern was from Taiwan.Shiingshen Chern. Born 26 Oct 1911 in Chia-hsing (or Jiaxing),Chekiang province(now Zhejiang), China.You, too, need some well-hopped Pilsner to refresh your memory ...=== === Subject: : Re: Who contributed most to mathematics?but the Greeks started it.> Just about every thing the Greeks did was rediscovered by others> as it was needed in their time. Much of Archemedies work was for War > efforts as they were needed. Sure they have being first name > recognition but played no more role than India and other countries > to thought today.I have heard that Euclid's Elements is the book with more editionspublished than any other except the Bible. But wasn't his school inAlexandria? So perhaps he should be assigned to Egypt (even if he wasGreek by culture and ancestry).=== === Subject: : Re: [Probability] Convergence in probability metrizable?> There are lots of metrics. In fact, if d is any bounded> metric, and there are lots of bounded metrics equivalent> to any one, take D(f,g) = int d(f(x),g(x)) dP(x). This> is equivalent to the topology you specified.Thank you very much.Noel.=== === Subject: : Re: [Probability] Convergence in probability metrizable?>I am given a probability space (W,F,P) and a seperable metric space E.>I consider the set M(E) of all maps f:W->E which are Borel measurable.>I consider on M(E) the topology generated by the base {B(f,a,e): f in>M(E), a,e>0} where:>B(f,a,e)={g in M(E): P(d(f,g)>=a)(fn) in M(E) and f in M(E), the convergence fn->f with respect to this>topology is equivalent to the convergence in probability, i.e.:>For all a>0, P(d(fn,f)>=a)->0 as n->+oo>I was trying to find a metric for this topology and am starting to>suspect it may not be metrizable. Any hint is appreciated.There are lots of metrics. In fact, if d is any boundedmetric, and there are lots of bounded metrics equivalentto any one, take D(f,g) = int d(f(x),g(x)) dP(x). Thisis equivalent to the topology you specified.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== === Subject: : Re: Question on 7d geometryThank you Tobias. Your analysis is very clear, and I appreciate it.-StevenHello list,This question may be too simple, but it is important to me.In 7D space, I have two hyper planes:(*1):(x1 x2 x3 x4 x5 x6 x7)*(a1 a2 a3 a4 a5 a6 a7)T = 0(*2):(x1 x2 x3 x4 x5 x6 x7)*(b1 b2 b3 b4 b5 b6 b7)T = 0where T is the transpose sign, * is the matrix multiplication sign.Now I first get the subspace(sub-hyperplane) S1 which is theintersection of (*1) and (*2). Then I combine the subspace S1 and avectorv1 = (a1 a2 a3 a4 a5 a6 a7)T to form a hyper plane (*3), so that thisnew hyperplane is the span of S1 and v1, and S1, v1 are on thehyperplane (*3). The questions what is the formula of this newhyperplane?> Let u be a vector such that the induced hyperplane x*(u^T)=0 (like in (*1)> and (*2)) contains S1. Then u is a linear combination of a and b.> Proof: Every linear combination of a and b clearly has this property, this> gives a 2D vector space. On the other hand, the orthogonal complement of S!> (that is the set of all such u) only is 2D because S1 is 5D and 2+5=7. qed.> So we make the ansatz u=s*a+t*b with unknown coefficients s and t. Because> v1 also lies in the hyperplane (*3) we have v1*(u^T)=0 which gives a linear> condition on s and t. The result is 1D solution space for s and t. It is 1D> because every multiple of any solution u also is a solution. Just pick any> value where not both s and t are zero and calculate u.=== === Subject: : Re: Question on 7d geometryX-ID: XVJlQTZVreMTQOGiFYTIke+6LeVTmFh1nZep7Ih-GuseThIA7ehOcI> Hello list,> This question may be too simple, but it is important to me.> In 7D space, I have two hyper planes:> (*1):> (x1 x2 x3 x4 x5 x6 x7)*(a1 a2 a3 a4 a5 a6 a7)T = 0> (*2):> (x1 x2 x3 x4 x5 x6 x7)*(b1 b2 b3 b4 b5 b6 b7)T = 0> where T is the transpose sign, * is the matrix multiplication sign.> Now I first get the subspace(sub-hyperplane) S1 which is the> intersection of (*1) and (*2). Then I combine the subspace S1 and a> vector> v1 = (a1 a2 a3 a4 a5 a6 a7)T to form a hyper plane (*3), so that this> new hyperplane is the span of S1 and v1, and S1, v1 are on the> hyperplane (*3). The questions what is the formula of this new> hyperplane?Let u be a vector such that the induced hyperplane x*(u^T)=0 (like in (*1)and (*2)) contains S1. Then u is a linear combination of a and b.Proof: Every linear combination of a and b clearly has this property, thisgives a 2D vector space. On the other hand, the orthogonal complement of S!(that is the set of all such u) only is 2D because S1 is 5D and 2+5=7. qed.So we make the ansatz u=s*a+t*b with unknown coefficients s and t. Becausev1 also lies in the hyperplane (*3) we have v1*(u^T)=0 which gives a linearcondition on s and t. The result is 1D solution space for s and t. It is 1Dbecause every multiple of any solution u also is a solution. Just pick anyvalue where not both s and t are zero and calculate u.-- reverse my forename for mail!=== === Subject: : Re: Maximizing and minimizing in optimization>Is it possible to convert a problem where we are trying to maximize a>quantity (profit) into a problem where we are trying to minimize a>(different) quantity (loss)?Maximizing f is the same as minimizing -f. Or perhaps what you have inmind is the equivalence in Linear Programming between solving the primalproblem (a max problem) and the dual problem (a min problem).> Yes. The equivalence in Linear programming was what I was thinking> about.> If yes, can we do the transform in linear>time? In particular, is it possible to transform a Network flow>problem where we try to maximize the flow between the source and sink>into a shortest path problem where we are trying to find the shortest>between two points (say source and sink)?Maximum flow is equivalent to minimal cut. If it's on a planar graph,and the source and sink are on the same face, then this is equivalentto a shortest-path problem in the dual graph.> Here, I want to know if the Maximum flow in a network can be> formulated as a Linear programming problem. Also, can you please> elaborate as to why the Maximal flow is equivalent to the> shortest-path problem _only_ for planar graphs? Certainly maximum-flow (with source s and sink t) is a linear programming problem:maximize fsubject to sum_j (x_{sj} - x_{js}) = f sum_j (x_{tj) - x_{jt)) = -f sum_j (x_{ij) - x_{ji}) = 0 for all other i 0 <= x_{ij} <= c_{ij} for all i,j where c_{ij} is the capacity of the arc i -> j, and x_{ij} the flow in that arc (make x_{ij} = 0 if there is no arc i -> j). === === Subject: : Re: Maximizing and minimizing in optimizationIs it possible to convert a problem where we are trying to maximize a>quantity (profit) into a problem where we are trying to minimize a>(different) quantity (loss)?> Maximizing f is the same as minimizing -f. Or perhaps what you have in> mind is the equivalence in Linear Programming between solving the primal> problem (a max problem) and the dual problem (a min problem).Yes. The equivalence in Linear programming was what I was thinkingabout.> If yes, can we do the transform in linear>time? In particular, is it possible to transform a Network flow>problem where we try to maximize the flow between the source and sink>into a shortest path problem where we are trying to find the shortest>between two points (say source and sink)?> Maximum flow is equivalent to minimal cut. If it's on a planar graph,> and the source and sink are on the same face, then this is equivalent> to a shortest-path problem in the dual graph.> Here, I want to know if the Maximum flow in a network can beformulated as a Linear programming problem. Also, can you pleaseelaborate as to why the Maximal flow is equivalent to theshortest-path problem _only_ for planar graphs? > Certainly maximum-flow (with source s and sink t) is a linear programming problem:> maximize f> subject to> sum_j (x_{sj} - x_{js}) = f> sum_j (x_{tj) - x_{jt)) = -f> sum_j (x_{ij) - x_{ji}) = 0 for all other i> 0 <= x_{ij} <= c_{ij} for all i,j> where c_{ij} is the capacity of the arc i -> j, and x_{ij} the flow in > that arc (make x_{ij} = 0 if there is no arc i -> j). Thank you for the clarification. Pradip=== === Subject: : Re: Maximizing and minimizing in optimizationIs it possible to convert a problem where we are trying to maximize aquantity (profit) into a problem where we are trying to minimize a(different) quantity (loss)?> Maximizing f is the same as minimizing -f. Or perhaps what you have in> mind is the equivalence in Linear Programming between solving the primal> problem (a max problem) and the dual problem (a min problem).Yes. The equivalence in Linear programming was what I was thinkingabout.If yes, can we do the transform in lineartime? In particular, is it possible to transform a Network flowproblem where we try to maximize the flow between the source and sinkinto a shortest path problem where we are trying to find the shortestbetween two points (say source and sink)?> Maximum flow is equivalent to minimal cut. If it's on a planar graph,> and the source and sink are on the same face, then this is equivalent> to a shortest-path problem in the dual graph.Here, I want to know if the Maximum flow in a network can beformulated as a Linear programming problem. Also, can you pleaseelaborate as to why the Maximal flow is equivalent to theshortest-path problem _only_ for planar graphs? Pradip.=== === Subject: : Re: Oneness of a number===> === Subject: : Re: Oneness of a number>Message-id: === Subject: : Oneness of a number>Message-id: I am trying to build fractals generated by the principal of the oneness>of a> number.>For the definition of the oneness of a number, view the internet, or>below.*> It's more commonly called the Collatz Conjecture or the 3x+1 problem.> >I have been unseccesful in creating a pattern around the oneness>although I have struck on some intriguing details. These mainly resolve>around the issue of, if we take the oneness of the oneness of a number,>and keep doing that, how long does it take 'til we reach 1?> Depends on how many binary digits the number has (which is related to> the magnitude of the number) AND ALSO on the pattern of 1s and 0s in the> binary representaion (which is NOT related to the magniyude of the> number).>(note that for 5 this never happens)> It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1>No, you didn't listen well, I take the oneness of the oneness for the>recursive oneness looplength.> That is a bit hard to follow.>The oneness of 5 is 5.>the oneness of 5 is 5, etc..> Let me see if I got this straight. If S(n) is the stopping time, you are> then finding the stopping time of the stopping time?> S(n)> S(S(n))> S(S(S(n)))> etc.> until that reaches 1?Yes, and the amount of recursive calls needed for that I call the recursiveoneness loop length. Let's call this ROLL(n)> So if the stopping time of n is n (as with 5) you get a loop?Yeah,> I haven't considered that before, any other loops other than 5? Anything> else interesting about doing that?5 is not the only one that loops, any number of which the oneness is 5 willcause a loop. The oneness of 32 is 5.Very few numbers have a oneness of 5. I'm not sure but something tells methe only next one of which ROLL(n) becomes infinite is 2^32. (which causesthe trap 2^32->32->5->5->5->... of course)And very few numbers n cause ROLL(n) to become infinite.Again like I said, it takes a long while for ROLL(n) to become larger than16. This only happens regularily when n exceeds 2000000000 (2 billion)You just gave me a cool idea btw. Suppose you have a number 's' and you callN(s) the amount of numbers for which the oneness is 's'. I'll try thatlater. There must be, for each 's', at least one even and at least one oddnumber that reaches them? Could those two systems form two two-dimensionalaxises?Also, I made another recursive loop that doesS(n)S(S(n))S(S(S(n)))Until this reaches 'n', not one. Let's call this looplength the recursiveoneness identity loop length, ROILL.Note that this only happens for 1,2,3,4,5,7, and 16Also note that in my definition ROLL(1) = ROILL(1) = 6.Since S(1) = 3S(3) = 7S(7) = 16S(16) = 4S(4) = 2S(2) = 1I think this set of numbers might be interesting, in intuitive terms 3 and 7might be seen as the 'prime' counterparts of the 2-powers 16 and 4, and 2as the main axis, being as well prime as the smallest power of 2.Well perhaps you are into intuitive stuff and care to think about it.Again note that this is my definition, some definitions would say S(1) = 0.I'm not a believer in that :)>I think I might be able to find a pattern in that since, if I call>this the recursive oneness loop length (if anybody can think a better>name, tell me),> It's sometimes refered to as the stopping time. Others make up their own> terminology.> then, the recursive oneness loop length of a number like>20000000 is still only 15. I am trying to make a large-scale 2d map of>it,> How are you plotting this in 2D?>1. Using complex number representation and messing around with that.>2. Using y as a period of x (y = x/WIDTH, so to speak)> That doesn't help much. Can you be more specific?That's because called brainstorm coding and it's a bit hard to express withwritten language. I did say what is necessary : I've tried to visualize anypossible patterns in the one-dimensional series, let's leave it at thatbecause I've tried many things.As for the complex numbers, I tried dividing by 2+2i when the complex andreal parts are even, and multiplying with 3+3i and adding 1+1i when theyare odd. I'm still experimenting with that..-- Quaternion=== === Subject: : Re: Oneness of a numberDoes two have more oneness than one?=== === Subject: : Re: Oneness of a number> Does two have more oneness than one?In my defintion, the oneness of one is 3.1->4->2->1takes three iterationsIt could be that some definitions say that the oneness of 1 is 0, since itstarts at the stopping time. It doesn't matter that much, and I prefer itthis way, where S(1)=3. And thus larger than S(2)-- Quaternion=== === Subject: : Re: Oneness of a numberDoes two have more oneness than one?> In my defintion, the oneness of one is 3.> 1->4->2->1> takes three iterations> It could be that some definitions say that the oneness of 1 is 0, since it> starts at the stopping time. It doesn't matter that much, and I prefer it> this way, where S(1)=3. And thus larger than S(2)> --> QuaternionIf you use an index to represent the collatz-transformations,like C_0 = 1 C_1 = 1*2^1 C_2 = 1*2^2 C_a = 1*2^a C_2,0 = (C_2 -1)/3 C_a,0 = (C_a -1)/3 C_a,b = C_a,0 * 2^b with the periodic property that C_0 = C_2,0 = C_2,2,0 = C_2,2,2,2,0 = C_2,...2,0 = 1 (which reflects the loop 1-4-2-1....) and so on, then this indicates in a very natural way the oneness and stopping time of a number: it's just the sum of all indexes plus the number of them: C_4,0 = 5 Stopping-time = (4+0) + (1) = 5 C_4,1,0 = 3 Stopping-time = (4+1+0) + (2) = 7 Because of the loop at 1 it is also C_4,0 = C_2,4,0 = C_2,2,....,2,4,0 = 5 so the stopping time is not exactly unique. With the additional restriction of allowing only the smallest number of indexes the definition is unique then. For 1 the stopping time is then also unique, namely 0, since the representation with the shortest index is C_0 = 1 Gottfried Helms=== === Subject: : Re: Oneness of a number> > Does two have more oneness than one?> In my defintion, the oneness of one is 3.> 1->4->2->1> takes three iterations> It could be that some definitions say that the oneness of 1 is 0, since> it starts at the stopping time. It doesn't matter that much, and I prefer> it this way, where S(1)=3. And thus larger than S(2)> --> Quaternion> If you use an index to represent the collatz-transformations,> like> C_0 = 1> C_1 = 1*2^1> C_2 = 1*2^2> C_a = 1*2^a> C_2,0 = (C_2 -1)/3> C_a,0 = (C_a -1)/3> C_a,b = C_a,0 * 2^b> with the periodic property that> C_0 = C_2,0 = C_2,2,0 = C_2,2,2,2,0 = C_2,...2,0 = 1> (which reflects the loop 1-4-2-1....)> and so on, then this indicates in a very natural way the oneness and> stopping time of a number: it's just the sum of all indexes plus the> number of them:> C_4,0 = 5 Stopping-time = (4+0) + (1) = 5> C_4,1,0 = 3 Stopping-time = (4+1+0) + (2) = 7> Because of the loop at 1 it is also> C_4,0 = C_2,4,0 = C_2,2,....,2,4,0 = 5> so the stopping time is not exactly unique. With the additional> restriction of allowing only the smallest number of indexes the> definition is unique then. For 1 the stopping time is then also unique,> namely 0, since the representation with the shortest index is> C_0 = 1Damn let me take that in tonight, it's quite complex for me to understandwhat you mean to represent with 'C_a1,a2,a3..,an' right now and how it canfight the logic that if you follow the iteration scheme you get a bijectionbetween n and the amount of iterations that are needed to transform it to1..-- Quaternion=== === Subject: : Re: Oneness of a number> Damn let me take that in tonight, :-)) So I wish, that the nightmares won't catch you... Gottfried(P.S. I had them already, just some weeks ago :-)=== === Subject: : Re: Oneness of a number>Does two have more oneness than one?That seems vague, and depends on context. Is this ONENESS part of some higheraspect of Mathematics beyond the typical undergraduate study of Calculus? At a simpler level, it seems that you could treat any single group of a givenquantity as ONE WHOLE; so a grouped unit is one whole set. This allows us tofind a fraction of any number as well as find the fraction of 1. G C=== === Subject: : Re: Oneness of a number===>Subject: Re: Oneness of a number>Message-id: === Subject: : Oneness of a number>Message-id: I am trying to build fractals generated by the principal of the oneness of>a> number.>For the definition of the oneness of a number, view the internet, or>below.*> It's more commonly called the Collatz Conjecture or the 3x+1 problem.> >I have been unseccesful in creating a pattern around the oneness although>I have struck on some intriguing details. These mainly resolve around the>issue of, if we take the oneness of the oneness of a number, and keep>doing that, how long does it take 'til we reach 1?> Depends on how many binary digits the number has (which is related to the> magnitude of the number) AND ALSO on the pattern of 1s and 0s in the> binary representaion (which is NOT related to the magniyude of the> number).>(note that for 5 this never happens)> It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1>No, you didn't listen well, I take the oneness of the oneness for the>recursive oneness looplength.That is a bit hard to follow.>The oneness of 5 is 5.>the oneness of 5 is 5, etc..Let me see if I got this straight. If S(n) is the stopping time, you are thenfinding the stopping time of the stopping time?S(n)S(S(n))S(S(S(n)))etc.until that reaches 1?So if the stopping time of n is n (as with 5) you get a loop?I haven't considered that before, any other loops other than 5? Anything elseinteresting about doing that?>I think I might be able to find a pattern in that since, if I call>this the recursive oneness loop length (if anybody can think a better>name, tell me),> It's sometimes refered to as the stopping time. Others make up their own> terminology.> then, the recursive oneness loop length of a number like>20000000 is still only 15. I am trying to make a large-scale 2d map of it,> How are you plotting this in 2D?>1. Using complex number representation and messing around with that.>2. Using y as a period of x (y = x/WIDTH, so to speak)That doesn't help much. Can you be more specific?>-- >Quaternion--MensanatorAce of Clubs=== === Subject: : existence of envelope singular solutionsWhat conditions should be satisfied so that a singular solution envelope for F(x,y,c) = 0 exists for parametric variation of c ?It is known that singular solution should satisfy the above F anddel F/ del c = 0 ; perhaps it is a condition using second orderpartial derivative.In the crank - connecting rod example, we have a crank of radius Rcentered on origin, angle c with x-axis. The end [R cos(c), R sin(c)]connects to a point on x-axis (reciprocating piston) with connectingrod length L ;It is found that an envelope exists in c limits + - ArcTan(L/R), andthere is no envelope outside these limits.=== === Subject: : Re: parallelizability of manifoldsX-ID: TFvqfyZHreIVQWjEdgHkcQCc+laPMsvXjdnB5Legpor1MHktX1Vf4m>so every continuous image of it also is simply connected.> No! Take a balloon (S^2), deflate it to a disk, roll it up like a crepe> or something to get an interval, and then join the ends together> to get a circle, which is not simply connected.What do you mean with roll it up like a crepe? And joining the endstogether is no continuous operation!?> Sort of at the basis of the theory of covering spaces is the> observation that the continuous image of a simply-connected space is> not necessarily simply connected.Yes, I see that I was wrong. But given a simply connected top. space X and atop. space Y together with an epimorphism f:X->Y, can we say that Y issimply connected? My idea was that any continuous curve s:[0,1]->Y can bepulled back to a continuous curve r:[0,1]->X such that f(r)=s so that wecan prove that Y has to be simply connected. Can we turn this idea into aproof? According to your post, the answer seems to be no. So can you givean example where this pull-back does not work?In the present case, we would have X=S^2, Y=R^2{(0,0)} and f any continuousmap X->Y. Of course it need not be an epimorphism, so the conclusion waswrong even if the proposition mentioned above was correct.I'll think about it further, but this is it for now.Thank you for your help,Tobias-- reverse my forename for mail!=== === Subject: : Re: parallelizability of manifolds> so every continuous image of it also is simply connected.> No! Take a balloon (S^2), deflate it to a disk, roll it up like a crepe> or something to get an interval, and then join the ends together> to get a circle, which is not simply connected.> What do you mean with roll it up like a crepe? Sorry, too many math for poets lectures. That really just means to takea projection to an interval. Everything to this point is just theprojection pi : R^3 --> R^1, restricted to S^2. (That's continuous.)>And joining the ends together is no continuous operation!?Sure it is; it's the map f : t |--> ( cos(2pi t), sin(2pi t) ).Perhaps you're thinking of continuous maps _with continuous inverses_,which this isn't. If that's what you mean by continuous image, thenthe right phrase is homeomorphic image, and then yes, thehomeomorphic image of a simply-connected space is also simply-connected.> Sort of at the basis of the theory of covering spaces is the> observation that the continuous image of a simply-connected space is> not necessarily simply connected.> Yes, I see that I was wrong. But given a simply connected top. space X and a> top. space Y together with an epimorphism f:X->Y, can we say that Y is> simply connected? No, that is still exactly the setting for covering spaces. The usualexample is the same f I gave above, where X is the real line andY is the circle.> My idea was that any continuous curve s:[0,1]->Y can be> pulled back to a continuous curve r:[0,1]->X such that f(r)=sThat's called the path-lifting property; it's a consequence of f being a covering map. (It's also possessed by other maps, such as fibrations.)> so that we can prove that Y has to be simply connected.No! First of all, the fact that you can lift paths does not mean thatyou can lift loops; that is, if s is a loop in Y, then it's alsoa path, and so it lifts to a path r if f is a covering (say); butthis lifted path r need not be a loop, i.e., there's no reason youmust have r(0)=r(1). If you can't see this with formulas, draw apicture of a helix (a spring) X, casting a circular shadow. The shadow, Yis a circle and the projection is the map f. Draw a loop around Y, andthen lift it back to X and you'll see you don't get a loop.Also you want to be a little careful : the fact that you can liftpaths does not by itself guarantee that you can lift other maps suchas homotopies (maps from [0,1]x[0,1] into Y). If you _are_ in thecontext of covering spaces, there is a pretty characterization ofwhat can be lifted: a map r : Z --> Y induces a homomorphismr* : pi_1(Z) --> pi_1(Y), from which you can show that a _necessary_condition for liftability is that the subgroup r*(pi_1(Z)) mustbe contained in the image f*(pi_1(X)). If f is a covering, thenthis condition is also sufficient. In particular, any map from acontractible space Z into a space Y will lift to a map from Zinto any cover X of Y.I know you never said you interested only in covering maps, but thesethings put your confusion into stark relief. You might want to thinkabout some of the classical examples of these. Start with anysimply-connected space X and a discrete group G of self-homeomorphismsof X and then let Y be the quotient X/G . You could take(X,G) to be (R, Z), (R^2, Z^2), (S^2, Z/2Z), (complex half-plane, modular group), (S^3, binary icosahedral group), etc.dave=== === Subject: : Re: parallelizability of manifoldsX-ID: E2GuPqZToeZepxr7wCJzde28vTkgTMjK+4i1VRxfNz9ztXgbMvSDEM> Sorry, too many math for poets lectures. That really just means to take> a projection to an interval. Everything to this point is just the> projection pi : R^3 --> R^1, restricted to S^2. (That's continuous.)> Sure it is; it's the map f : t |--> ( cos(2pi t), sin(2pi t) ).Ah, now your example is clear. Nice!> Perhaps you're thinking of continuous maps _with continuous inverses_,> which this isn't. If that's what you mean by continuous image, then> the right phrase is homeomorphic image, and then yes, the> homeomorphic image of a simply-connected space is also simply-connected.> Yes, I see that I was wrong. But given a simply connected top. space X> and a top. space Y together with an epimorphism f:X->Y, can we say that Y> is simply connected?In the meantime, I could figure this out. A morphism (here: a continuousmap) f:X->Y is called an epimorphism iff there is a morphism g:Y->X suchthat fg=id_Y. Using this definition, and any closed curve s:[0,1]->Y, wecan easily lift it to gs:[0,1]->X and still it is closed: gs(0)=gs(1).Because X is simply connected, there is continuous c:[0,1]^2->X withc(0,t)=gs(t) and c(1,t)=x for some x in X. Applying f gives usfc:[0,1]^2->Y with fc(0,t)=fgs(t)=s(t) and fc(1,t)=f(x). So Y is simplyconnected.But what does it mean for f to be an epimorphism? fg=id_Y is equivalent to:for every S subset Y we have g-(f-(S))=S (where the - sign denotespre-images; this is just set-theoretic). Because of continuity reasons, anecessary condition isS open <=> f-(S) open (*)For surjective f with the property (*), you can take any set-theoreticg:Y->X with fg=id_Y to prove the other direction of the following theorem:f is an epimorphism in the category of topological spaces, iff it issurjective and has property (*).This is equivalent to that f be a quotient map. So being an epimorphism ispretty special, so I agree with you that continuous images ofsimply-connected spaces need not be simply-connected.Tobias-- reverse my forename for mail!=== === Subject: : Re: parallelizability of manifolds>Yes, I see that I was wrong. But given a simply connected top. space X>and a top. space Y together with an epimorphism f:X->Y, can we say that Y>is simply connected?> In the meantime, I could figure this out. A morphism (here: a continuous> map) f:X->Y is called an epimorphism iff there is a morphism g:Y->X such> that fg=id_Y. No - that's _not_ the usual (category-theoretic) definition of epimorphism. Instead, f: X --> Y is an epimorphism iff whenever g_1,g_2 : Y --> Z are morphisms such that g_1 o f =g_2 o f, it follows that g_1 = g_2. (Your notion might be called a split epimorphism ... )> Using this definition, and any closed curve s:[0,1]->Y, we> can easily lift it to gs:[0,1]->X and still it is closed: gs(0)=gs(1).> Because X is simply connected, there is continuous c:[0,1]^2->X with> c(0,t)=gs(t) and c(1,t)=x for some x in X. Applying f gives us> fc:[0,1]^2->Y with fc(0,t)=fgs(t)=s(t) and fc(1,t)=f(x). So Y is simply> connected. In the context of covering spaces (where Dave Rusin was explaininf stuff), the only covering maps f: X --> Y with X simply connected and f an epimorphism in your sense are _homeomorphisms_ ... so Y _would_ be simply connected in this (fairly uninteresting) case.> But what does it mean for f to be an epimorphism? fg=id_Y is equivalent to:> for every S subset Y we have g-(f-(S))=S (where the - sign denotes> pre-images; this is just set-theoretic). Because of continuity reasons, a> necessary condition is> S open <=> f-(S) open (*)> For surjective f with the property (*), you can take any set-theoretic> g:Y->X with fg=id_Y to prove the other direction of the following theorem:> f is an epimorphism in the category of topological spaces, iff it is> surjective and has property (*).> This is equivalent to that f be a quotient map. ... Yep ...> So being an epimorphism is> pretty special, so I agree with you that continuous images of> simply-connected spaces need not be simply-connected.> Tobias=== === Subject: : Re: parallelizability of manifoldsX-ID: T873pTZOQe-SmC5727w4cf2y+dJtZB-ykAqnFtohVoVyMy-6VP0GYp> No - that's _not_ the usual (category-theoretic) definition> of epimorphism. Instead, f: X --> Y is an epimorphism iff> whenever g_1,g_2 : Y --> Z are morphisms such that g_1 o f =g_2 o f,> it follows that g_1 = g_2.> (Your notion might be called a split epimorphism ... )Sorry for this one, the only reference I have for category theory is Lang'sAlgebra, and he defines epimorphism and monomorphism only for abeliancategories.My notion is much stronger; it implies the other, but if we take a set X andlet X_1 be X with the discrete topology, X_2 with the indiscrete topology,and define f:X_1->X_2 as the identity, then f is continuous. It is anepimorphism, but does not have a right inverse.> In the context of covering spaces (where Dave Rusin was explaininf> stuff), the only covering maps f: X --> Y with X simply connected and> f an epimorphism in your sense are _homeomorphisms_ ... so Y _would_> be simply connected in this (fairly uninteresting) case.Maybe, but since I am not familiar with covering maps, I did not considerthis case.-- reverse my forename for mail!Tobias FritzStudent of Mathematics and PhysicsUniversity of Heidelberg, Germany=== === Subject: : Re: parallelizability of manifolds>Indeed, algebraic topology seems to be an extremely powerful tool: Yes!>S^2 is simply connected, Yes!>so every continuous image of it also isNo! Take a balloon (S^2), deflate it to a disk, roll it up like a crepeor something to get an interval, and then join the ends togetherto get a circle, which is not simply connected.Sort of at the basis of the theory of covering spaces is theobservation that the continuous image of a simply-connected space isnot necessarily simply connected.dave=== === Subject: : v sp over field {0,1}I learned all of my linear algebra over the reals or complexes,so this seemingly simply problem has baffled me.X is a vector space over the field F_2 = {0,1} offinite dimension. K is a subspace and K^perp is itsorthogonal complement,K^perp := {x in X with =0 for all k in K}.The equation I want to prove isdim K + dim (K^perp) = dim X.I've tried on my own, and noticed many discouraging things.Normally (in R^n, say), we have that K and K^perp onlyintersect at zero. But this is no longer necessarily true.I don't think that you can extend any orthonormal basis ofa subspace to an orthonormal basis of the entire space. Forone thing, Gram-Schmidt certainly doesn't work any more.Many of these troubles probably stem from the fact that thereare many nonzero vectors v for which =0. -Tyler=== === Subject: : Re: v sp over field {0,1}I think I understand now. The equalitydim(Null(A)) + dim(Range(A)) = dim(X)is still true,where A is a linear transformation in the finite-dimensional space X. I say still true becauseI was not certain that it would hold for vectorspaces over arbitrary fields. (I reviewed theproof to confirm it still works.)We want to check that dim(F) + dim(F^perp) = dim(X).Choose a basis of F, and let A have these elementsas rows. Then Null(A) = F^perp; and the row rankof A is clearly dim(F). Since row rank = col rank= dim(Range(A)), we are done.=== === Subject: : Re: v sp over field {0,1}in message <43b20b30.0311091614.90ffa88@posting.google.com>:> I think I understand now. The equality> dim(Null(A)) + dim(Range(A)) = dim(X)> is still true,> where A is a linear transformation in the finite-> dimensional space X. I say still true because> I was not certain that it would hold for vector> spaces over arbitrary fields. (I reviewed the> proof to confirm it still works.)> We want to check that dim(F) + dim(F^perp) = dim(X).> Choose a basis of F, and let A have these elements> as rows. Then Null(A) = F^perp; and the row rank> of A is clearly dim(F). Since row rank = col rank> = dim(Range(A)), we are done.You can't assume Null(A) = F^perp without knowing more aboutthe inner product (sesquilinear form) you're using. As othershave pointed out, you really need to take into account whetherthe form is degenerate. For example, if X has a non-zero radicalR then dim(R) + dim(R^perp) = dim(R) + dim(X) > dim(X).-- Jim Heckman=== === Subject: : Re: v sp over field {0,1} Adjunct Assistant Professor at the University of Montana.>I learned all of my linear algebra over the reals or complexes,>so this seemingly simply problem has baffled me.>X is a vector space over the field F_2 = {0,1} of>finite dimension. K is a subspace and K^perp is its>orthogonal complement,What is the inner product for the vector space, though? There arecanonical inner products in R^n and in C^n, but I don't know if thereis one for vector spaces over F_2...>K^perp := {x in X with =0 for all k in K}.>The equation I want to prove is>dim K + dim (K^perp) = dim X.>I've tried on my own, and noticed many discouraging things.>Normally (in R^n, say), we have that K and K^perp only>intersect at zero. But this is no longer necessarily true.>I don't think that you can extend any orthonormal basis of>a subspace to an orthonormal basis of the entire space. For>one thing, Gram-Schmidt certainly doesn't work any more.>Many of these troubles probably stem from the fact that there>are many nonzero vectors v for which =0.What is your definition of inner product? What are the axioms yourequire of an inner product? In some books, an inner product isrequired to be positive definite (i.e., =0, and =0 if andonly if v = 0)...=====Arturo Magidinmagidin@math.berkeley.edu=== === Subject: : Re: v sp over field {0,1}>X is a vector space over the field F_2 = {0,1} of>finite dimension. K is a subspace and K^perp is its>orthogonal complement,>K^perp := {x in X with =0 for all k in K}.>The equation I want to prove is>dim K + dim (K^perp) = dim X.As you noted, inner products don't always behave likethe standard one in R^n. But this result still holds if,for example, is still defined as the dot product.In that case, you can describe K^perp as the solutionto a system of linear equations. Do you know how to computethe dimension of a nullspace (kernel)?Note that the result is _false_ if for example =0 forall u and v. You'll need to look to see what propertiesyou're assuming about the inner product before you try toprove something about it! (Nondegeneracy, for starters.)dave=== === Subject: : Re: v sp over field {0,1}> I learned all of my linear algebra over the reals or complexes,> so this seemingly simply problem has baffled me.> X is a vector space over the field F_2 = {0,1} of> finite dimension. K is a subspace and K^perp is its> orthogonal complement,> K^perp := {x in X with =0 for all k in K}.> The equation I want to prove is> dim K + dim (K^perp) = dim X.> I've tried on my own, and noticed many discouraging things.> Normally (in R^n, say), we have that K and K^perp only> intersect at zero. But this is no longer necessarily true.> I don't think that you can extend any orthonormal basis of> a subspace to an orthonormal basis of the entire space. For> one thing, Gram-Schmidt certainly doesn't work any more.> Many of these troubles probably stem from the fact that there> are many nonzero vectors v for which =0.> -TylerIt is best to think of K^perp not as a subspace of X but as a subspaceof the dual space X*. (Now via your pairing , it is true that X*is identified with X, but that just confuses things.)-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/=== === Subject: : Re: v sp over field {0,1}> I learned all of my linear algebra over the reals or complexes,> so this seemingly simply problem has baffled me.> X is a vector space over the field F_2 = {0,1} of> finite dimension. K is a subspace and K^perp is its> orthogonal complement,> K^perp := {x in X with =0 for all k in K}.> The equation I want to prove is> dim K + dim (K^perp) = dim X.What do you mean by ? You have not defined it.Jose Carlos Santos=== === Subject: : Re: v sp over field {0,1}Sorry.Fix a basis of X (suppose dim X = m). Then let thecoordinates/coefficients of x and k bex1...xm and k1...km respectively. Now we candefine :=x1k1 + x2k2 + ... + xmkm,in a manner analogous to the standard innerproduct. It is linear and symmetric, but doesnot really give a norm.=== === Subject: : Re: v sp over field {0,1} Adjunct Assistant Professor at the University of Montana.>Sorry.>Fix a basis of X (suppose dim X = m). Then let the>coordinates/coefficients of x and k be>x1...xm and k1...km respectively. Now we can>define :=x1k1 + x2k2 + ... + xmkm,>in a manner analogous to the standard inner>product. It is linear and symmetric, but does>not really give a norm.So? Not all inner products are positive definite.=====Arturo Magidinmagidin@math.berkeley.edu=== === Subject: : Re: Extremely bizarre number theoretical property> Well, I discovered something rather very odd (at least to me) today.> Let n be a positive integer, and let s(n) denote the sum of all the> divisors of n. So, for example, s(6) = 1+2+3+6 = 12.> It turns out there is a class of numbers with the property that> a) n is a perfect square> b) s(n) is a perfect squareThis one is in Sloane's online database of offbeat sequences, numberA008847. Go tohttp://www.research.att.com/~njas/sequences/and enter1, 9, 20, 180, 1306and click Search. The search will cough up more terms of the table, andsome references in the lit.Larry=== === Subject: : Re: Extremely bizarre number theoretical property> Well, I discovered something rather very odd (at least to me) today. > Let n be a positive integer, and let s(n) denote the sum of all the> divisors of n. So, for example, s(6) = 1+2+3+6 = 12.> It turns out there is a class of numbers with the property that > a) n is a perfect square> b) s(n) is a perfect square> Here's some examples:annotated, with factors of sqrt(n):> s(9^2) = 11^2 3 3> s(20^2) = 31^2 2 2 5> s(180^2) = 341^2 [non-primitive - 9*20]> s(1306^2) = 1729^2 2 653> s(1910^2) = 2821^2 2 5 191> s(11754^2) = 19019^2 [non-primitive - 9*1306]> s(17190^2) = 31031^2 [non-primitive - 9*1910]> s(32486^2) = 43617^2 2 37 439> s(38423^2) = 43491^2 7 11 499> s(47576^2) = 68961^2 2 2 2 19 313> s(48202^2) = 72219^2 2 7 11 313> s(50920^2) = 82677^2 2 2 2 5 19 67> s(51590^2) = 86583^2 2 5 7 11 67> s(83884^2) = 117831^2 2 2 67 313104855 117831 5 67 313132682 187131 2 11 37 163198534 347529 2 3 7 29 163247863 347529 3 7 11 29 37292374 479787 [non-primitive - 9*32486]300876 503347 2 2 3 25073312374 414309 2 313 499313929 436107 3 3 3 7 11 151334330 496713 2 5 67 499345807 478401 [non-primitive - 9*38423]376095 503347 3 5 25073428184 758571 [non-primitive - 9*47576]433818 794409 [non-primitive - 9*48202]458280 909447 [non-primitive - 9*50920]464310 952413 [non-primitive - 9*51590]469623 658749 3 7 11 19 107498892 696787 2 2 191 653 contains 1306=2*653623615 696787 5 191 653754956 1296141 [non-primitive - 9*83884]768460 1348221 [non-primitive - 20*38423]787127 828723 11 163 439943695 1296141 [non-primitive - 9*104855]985369 1125579 7 11 67 1911194138 2058441 [non-primitive - 9*132682]1276880 2106853 2 2 2 2 5 11 14511378608 2650557 2 2 2 2 3 7 11 11 3731547052 2728713 2 2 3 13 47 2111606281 1993719 3 29 37 4991676904 2854579 2 2 2 3 107 653 contains 1306=2*6531754039 1903629 7 83 30911933815 2728713 3 5 13 47 2112015094 3310671 2 3 29 37 3132034423 2501877 3 3 3 151 4992156730 3969147 2 3 5 29 37 672181409 2322957 19 29 37 1072452440 4657471 2 2 2 3 5 107 1912552202 4154493 2 3 3 3 151 3132731590 4980801 2 3 3 3 5 67 1512811366 4557399 [non-primitive - 9*312374]3008970 5463843 [non-primitive - 9*334330]3043401 3779139 3 19 107 4993817974 6275451 2 3 19 107 3134086330 7523607 2 3 5 19 67 1074490028 7664657 [non-primitive - 9*498892]4957260 10773399 [non-primitive - 20*247863]5147241 6540807 3 11 61 25575545617 7536711 3 7 11 240075570344 8390001 2 2 2 13 19 28195612535 7664657 [non-primitive - 9*623615]5643638 8786379 2 7 11 13 28195805336 10287381 2 2 2 3 19 29 4395881722 10773399 2 3 7 11 29 4396278580 13519317 [non-primitive - 20*313929]6385703 6457269 67 191 4996403664 9346701 2 2 2 2 29 37 3736416984 10185273 2 2 2 7 19 37 1636916140 14830431 [non-primitive - 9*20*38423]7084143 9115953 [non-primitive - 9*7871277542184 11178921 2 2 2 13 47 15437642712 12516231 2 2 2 7 11 19 653 contains 1306=652*28010922 10722621 2 67 191 3138868321 12381369 [non-primitive - 9*985369]8934096 15205827 2 2 2 2 3 373 4999392460 204212197 [non-primitive - 20*469623]9821396 14335671 2 2 13 67 2819Some prime, such as 37, 67, 107, 191, 313, 373, 439, 499, 653 are a factor of n quite frequently? I wonder why.Their squares' sigma contributions are as follows:37 3 7 67 (313&653) => can cancel 6767 3 31 7^2 (313) => can cancel 31107 7 13 127191 7 31 13^2313 3 181^2 => can cancel 3373 3 13 73 7^2439 3 67 31^2499 3 7 109^2 653 7 13^2 19^2 => can cancel 72819 61 19^4 => can cancel 61And some other primes:3 13 => can cancel 13 (313&373) => can cancel 7311 7 19 (653) => can cancel 1917 307 => can cancel 30719 3 127 (313) => can cancel 12723 7 79 (653) => can cancel 7931 3 331 (313) => can cancel 33141 1723 => can cancel 172343 3 631 => can cancel 63147 37 61 (2819) => can cancel 3753 7 409 (653) => can cancel 40961 3 13 97 (3&313) => can cancel 9779 3 43 7^2 (313) => can cancel 43(2 gives 7 more easily than 653, and 5 gives 31 more easily than 67.)I guess that when looked at in the simplest possible terms you can just create a binary matrix of non-square terms in sigma(p^2) for prime p, and find combinations that sum to zero in exactly the same way as you do in QS (in fact, thesimilarity with QS is massive, as you'd want to sieve overa quadratic expression to form the vectors in the first place).I'd conjecture that there is an infinite quantity of such numbers, and that it's not hard to construct one withpretty massive numbers of factors.I'll leave that for someone else.Phil-- Unpatched IE vulnerability: Basic Authentication URL spoofingDescription: Spoofing the URL displayed in the Address barReference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/15.html=== === Subject: : Re: Algebraic Closure> |There's an interesting description in Conway's On Numbers and Games.> |(You take nim addition/multiplication on the ordinals: the ordinals> |below a certain one form algebraic closure of F_2.)> a somewhat related question:> from fooling around with sloane's encyclopedia of integer sequences,> it appears that the number of:> field structures on the vector space [f_2]^[2^[n-1]] for which the> frobenius automorphism is the obvious rotation operator> is equal to the number of:> circular words of circumference 2^n in the alphabet {0,1}> containing every word of length n as a contiguous sub-word.Such circular words are called de Bruijn sequences.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times)=== === Subject: : Re: Algebraic Closure|There's an interesting description in Conway's On Numbers and Games.|(You take nim addition/multiplication on the ordinals: the ordinals|below a certain one form algebraic closure of F_2.)a somewhat related question:from fooling around with sloane's encyclopedia of integer sequences,it appears that the number of: field structures on the vector space [f_2]^[2^[n-1]] for which the frobenius automorphism is the obvious rotation operatoris equal to the number of: circular words of circumference 2^n in the alphabet {0,1} containing every word of length n as a contiguous sub-word.does this equality actually always hold, and is there a pleasantbijective proof of it?also, i tried the obvious greedy algorithm for creating such circularwords (start with the all-zeros word of length n and then startappending letters to it, preferably 1 but if that causes a repeatedword then 0 instead), and it seemed to work; for example00000111110111001101011000101001. does this greedy algorithm actuallyalways work, and if so then why?also, are there are any particular nice generalizations of any or allof this?(by the way, just in case any of these questions are answered in onnumbers and games, i did try to get it out of the library but it wasin circulation.)-- [e-mail address jdolan@math.ucr.edu]=== === Subject: : Re: Two questions in propositional logic> >Anyway, as I said in the other post I think I've got this sorted out now >(although I'll want to write my own proof of it before I'm fully happy >with it).>If you're interested, John's proof went roughly as follows:>Define an equivalence relation on C by p ~ q if there is a bijection f >from the set of primitive propositions to itself, such that q = p with >all the p_i in p replaced with f(p_i).>He then showed that there can be at most one element of each equivalence >class in C, and that there are countably many equivalence classes >greater than p (because you can take some countable subset of the whole >set of primitive propositions and represent every formula in C as a >formula using only this countable subset and the primitives that appear >in p).>He then went on to show there were only countably many elements less >than p in a similar manner. I pointed out that this was rather easier, >as you could just reverse the chain by negation and use the previous >result. I found this the most natural approach. However (and IBL will probably> kill me for saying anything at all about this sheet in public) you> should also be aware there is a one word answer to this question. That> is, there is one particular word (familiar to everyone) which> instantly makes this question a triviality. I'll leave you to guess> what it is.>Hmm.>...>Oh hell. Is it the one-word that you always use to prove obvious >statements? It is, isn't it... ::sketches proof in his head:: Excuse me >while I go sulk.>(To anyone reading this who doesn't understand that line, it's a >Leaderism. :)Come on. For those of us who aren't getting any of this, and whodon't even know what a Leaderism is: what the heck is the one word?> By the way you didn't ask about the last question (if the set of> primitive propositions is allowed to be uncountable is it true given a> set S of propositions that you can find an independent set of> propositions equivalent to S); does that mean you've solved it? I> still don't have it, one year after taking the course. No hints> please!>Well... yes and no. I thought I had a solution. There was a tiny problem >in it contained right at the end which I think I can fix (but it may be >a way bigger problem than I thought it was. :)>That being said, my solution to the last question was only one line. My >solution to the previous question did not assume countability. I was >lazy and didn't feel like proving it twice...>David>(E-mail address spam-blocked in the obvious way)=== === Subject: : Re: Two questions in propositional logic>Anyway, as I said in the other post I think I've got this sorted out now >(although I'll want to write my own proof of it before I'm fully happy >with it).>If you're interested, John's proof went roughly as follows:>Define an equivalence relation on C by p ~ q if there is a bijection f >from the set of primitive propositions to itself, such that q = p with >all the p_i in p replaced with f(p_i).>He then showed that there can be at most one element of each equivalence >class in C, and that there are countably many equivalence classes >greater than p (because you can take some countable subset of the whole >set of primitive propositions and represent every formula in C as a >formula using only this countable subset and the primitives that appear >in p).>He then went on to show there were only countably many elements less >than p in a similar manner. I pointed out that this was rather easier, >as you could just reverse the chain by negation and use the previous >result.>I found this the most natural approach. However (and IBL will probably>kill me for saying anything at all about this sheet in public) you>should also be aware there is a one word answer to this question. That>is, there is one particular word (familiar to everyone) which>instantly makes this question a triviality. I'll leave you to guess>what it is.>Hmm.>...>Oh hell. Is it the one-word that you always use to prove obvious >statements? It is, isn't it... ::sketches proof in his head:: Excuse me >while I go sulk.>(To anyone reading this who doesn't understand that line, it's a >Leaderism. :)> Come on. For those of us who aren't getting any of this, and who> don't even know what a Leaderism is: what the heck is the one word?Well, the one word that I thought it was is induction. I thought I saw a really simple proof by induction on the length of a proof of q from p. Unfortunately it didn't work. :) This happens to me a lot with Dr Leader's example sheets (Dr Leader is the man who lectures the course lecturing method, which are referred to as leaderisms.)>By the way you didn't ask about the last question (if the set of>primitive propositions is allowed to be uncountable is it true given a>set S of propositions that you can find an independent set of>propositions equivalent to S); does that mean you've solved it? I>still don't have it, one year after taking the course. No hints>please!>Well... yes and no. I thought I had a solution. There was a tiny problem >in it contained right at the end which I think I can fix (but it may be >a way bigger problem than I thought it was. :)>That being said, my solution to the last question was only one line. My >solution to the previous question did not assume countability. I was >lazy and didn't feel like proving it twice...>David>(E-mail address spam-blocked in the obvious way)> === === Subject: : Re: Two questions in propositional logicAnyway, as I said in the other post I think I've got this sorted out now >(although I'll want to write my own proof of it before I'm fully happy >with it).>If you're interested, John's proof went roughly as follows:>Define an equivalence relation on C by p ~ q if there is a bijection f from the set of primitive propositions to itself, such that q = p with all the p_i in p replaced with f(p_i).>He then showed that there can be at most one element of each equivalence >class in C, and that there are countably many equivalence classes >greater than p (because you can take some countable subset of the whole >set of primitive propositions and represent every formula in C as a >formula using only this countable subset and the primitives that appear >in p).>He then went on to show there were only countably many elements less >than p in a similar manner. I pointed out that this was rather easier, >as you could just reverse the chain by negation and use the previous >result.>I found this the most natural approach. However (and IBL will probably>kill me for saying anything at all about this sheet in public) you>should also be aware there is a one word answer to this question. That>is, there is one particular word (familiar to everyone) which>instantly makes this question a triviality. I'll leave you to guess>what it is.>Hmm.>...>Oh hell. Is it the one-word that you always use to prove obvious >statements? It is, isn't it... ::sketches proof in his head:: Excuse me >while I go sulk.>(To anyone reading this who doesn't understand that line, it's a >Leaderism. :)> > Come on. For those of us who aren't getting any of this, and who> don't even know what a Leaderism is: what the heck is the one word?>Well, the one word that I thought it was is induction. Ah, then it's a good thing I presuaded him to state theWord explicitly. It's a word I really would not have thoughtof applying here, and it really does give a one-line solution.(I imagine you've already seen the word in his reply to me.Incredibly keen argument.)>I thought I saw a >really simple proof by induction on the length of a proof of q from p. >Unfortunately it didn't work. :) This happens to me a lot with Dr >Leader's example sheets (Dr Leader is the man who lectures the course >lecturing method, which are referred to as leaderisms.)>By the way you didn't ask about the last question (if the set of>primitive propositions is allowed to be uncountable is it true given a>set S of propositions that you can find an independent set of>propositions equivalent to S); does that mean you've solved it? I>still don't have it, one year after taking the course. No hints>please!>Well... yes and no. I thought I had a solution. There was a tiny problem >in it contained right at the end which I think I can fix (but it may be >a way bigger problem than I thought it was. :)>That being said, my solution to the last question was only one line. My >solution to the previous question did not assume countability. I was >lazy and didn't feel like proving it twice...>David>(E-mail address spam-blocked in the obvious way)> > === === Subject: : Re: Two questions in propositional logic>Anyway, as I said in the other post I think I've got this sorted out now >(although I'll want to write my own proof of it before I'm fully happy >with it).>If you're interested, John's proof went roughly as follows:>Define an equivalence relation on C by p ~ q if there is a bijection f >from the set of primitive propositions to itself, such that q = p with >all the p_i in p replaced with f(p_i).>He then showed that there can be at most one element of each equivalence >class in C, and that there are countably many equivalence classes >greater than p (because you can take some countable subset of the whole >set of primitive propositions and represent every formula in C as a >formula using only this countable subset and the primitives that appear >in p).>He then went on to show there were only countably many elements less >than p in a similar manner. I pointed out that this was rather easier, >as you could just reverse the chain by negation and use the previous >result.>I found this the most natural approach. However (and IBL will probably>kill me for saying anything at all about this sheet in public) you>should also be aware there is a one word answer to this question. That>is, there is one particular word (familiar to everyone) which>instantly makes this question a triviality. I'll leave you to guess>what it is.Hmm....Oh hell. Is it the one-word that you always use to prove obvious statements? It is, isn't it... ::sketches proof in his head:: Excuse me while I go sulk.(To anyone reading this who doesn't understand that line, it's a Leaderism. :)> Come on. For those of us who aren't getting any of this, and who> don't even know what a Leaderism is: what the heck is the one word?I really should have just e-mailled my clue. I didn't want to post theone word because it is a fun problem and it will likely be set infuture years as well. However as I've given away the fact there is aone word solution I suppose it is unfair on people who won't find outotherwise. So I'll post it and hope to goodness that no-one else doingthe course happens to read this group.The one word is.......................PROBABILITYMichael=== === Subject: : Re: Two questions in propositional logic> >Anyway, as I said in the other post I think I've got this sorted out now >(although I'll want to write my own proof of it before I'm fully happy >with it).>If you're interested, John's proof went roughly as follows:>Define an equivalence relation on C by p ~ q if there is a bijection f >from the set of primitive propositions to itself, such that q = p with >all the p_i in p replaced with f(p_i).>He then showed that there can be at most one element of each equivalence >class in C, and that there are countably many equivalence classes >greater than p (because you can take some countable subset of the whole >set of primitive propositions and represent every formula in C as a >formula using only this countable subset and the primitives that appear >in p).>He then went on to show there were only countably many elements less >than p in a similar manner. I pointed out that this was rather easier, >as you could just reverse the chain by negation and use the previous >result. I found this the most natural approach. However (and IBL will probably> kill me for saying anything at all about this sheet in public) you> should also be aware there is a one word answer to this question. That> is, there is one particular word (familiar to everyone) which> instantly makes this question a triviality. I'll leave you to guess> what it is.>Hmm.>...>Oh hell. Is it the one-word that you always use to prove obvious >statements? It is, isn't it... ::sketches proof in his head:: Excuse me >while I go sulk.>(To anyone reading this who doesn't understand that line, it's a >Leaderism. :)> Come on. For those of us who aren't getting any of this, and who> don't even know what a Leaderism is: what the heck is the one word?>I really should have just e-mailled my clue. I didn't want to post the>one word because it is a fun problem and it will likely be set in>future years as well. However as I've given away the fact there is a>one word solution I suppose it is unfair on people who won't find out>otherwise. So I'll post it and hope to goodness that no-one else doing>the course happens to read this group.>The one word is>[one word snipped]Took only another minute to realize I wasn't sure why the solution made sense, since there _are_ uncountable chains of reals, forexample.Took another minute to see the actual solution: there are no uncountable chains of _rationals_.Very clever - I _really_ never woulda thought of applying thatword in a problem about logic.>Michael=== === Subject: : Re: Two questions in propositional logic>Unless you're using some extra property about the p_i, q_i in saying >this that I'm not seeing (which is very possible, on account of it being >Early here and I just got up. :), that needn't be true.>For example take the following sequence:>p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3.> > Pointed out a few minutes ago that this is not a counterexample.> Of course what I said is wrong - a counterexample is> p_1 & (p_2), p_1 & (p_2 v p_3), ... p_1.>Oops. Should have read this post before my last one. That's exactly the >counterexample I had in mind when I made the (wrong) counterexample this >morning, I was just half asleep and misremembering. :)> (Realized I wanted to show that the intersection of a strictly> decreasing sequence of open subsets of the Cantor set had> empty interior. Realized that that was false. Now, I bet> that there's no strictly decreasing omega_1-sequence of> subsets of the Cantor set, but never mind, the proof you> already have is simpler.)>Are you sure that the cantor set approach works? Also, do you mean >{0,1}^X for some possibly uncountable X rather than the cantor set?When I was thinking along these lines I was trying to prove thatfalse fact, in which only countably many formulas appear...>(which is homeomorphic when X is countable). Because I tried that, and I >couldn't come to the conclusion that you could get the association to >work both ways - you could get a clopen subset of {0, 1}^X for every >proposition, but I wasn't sure you could go the other way... it looked >like if you chose some sufficiently nasty clopen set it wouldn't >correspond to a proposition. (I didn't prove that you couldn't, but it >looked like it was going to go badly wrong if I tried to prove that you >could, so I gave up on that approach).No, not every clopen set corresponds to a proposition, at least I don't think it does. That doesn't matter to the false thing I wastrying to prove. (Regardless, it was thinking about open subsetsof the middle-thirds Cantor set that led me to realize what Iwas trying to prove was obviously false.)>David>(E-mail address spam-blocked in the obvious way)=== === Subject: : Re: Two questions in propositional logic> Well... yes and no. I thought I had a solution. There was a tiny problem > in it contained right at the end which I think I can fix (but it may be > a way bigger problem than I thought it was. :)Hmm... Yes. It does indeed appear to be a bigger problem than I initially anticipated. There's a surprise. ::grumble grumble evil frigging example sheets grumble::David(E-mail address spam-blocked in the obvious way)=== === Subject: : Re: Greek Alphebet
 In my studies on
triginometry, and calculus, I am comming across many>symbols
which appear to be the Greek Alphebet, as there was a chart
of>the Alphebet in the front of the book. I am wondering if
someone could>tell me what they all mean, or point me to a
site where I could find>out.>Thank you,>TomYou need this rule:
The symbols mean whatever the author says they mean.Take each
one in the context where it is discussed. Somewhere else,
thesame symbol is very likely to mean something different.Even
the venerable PI does not necessarily mean the constant
3.1415...,although that is probably the meaning in the context
of the subjects youmentioned.Lynn KillingbeckDo the
Greeks use Latin and Greek letters in a similar way to us
Latin>alphabet users? How about the Russians, do they spurn
their own>alphabet and use Latin and Greek letters?Others have
answered your question about the Russians, and I can assure you
that Greeks use Greek letters the same way everybody else does:
they are not Greek to them :-) [Curious thread, this one, and I
have probably missed some good stuff :-)I liked the comment
about an inverted A looking like an ox head -- goodpoint,
except that the contemporary Greek word for cow (AGELADA,
fromAGELI = herd) first shows up in Greek around the 10th
century *AD* ... whereas the ancient Greek word for ox starts
with, alas, a B (BOUS*, modern Greek BODI)...]*Speaking of
oxen, there is an apocryphal story, attributed to Apollodorus,
about Pythagoras having sacrificed one hundred of them upon
discovering theeponymous theorem; the story survives anywhere
from Diogenes Laertius (4thcentury AD) to Veniamin of Lesbos
(early 19th century AD) -- who disputes it on the grounds of
Pythagoras having been a metempsychosis believer :-)[I
recently read this in
http://195.134.75.8/0709011144070000/main.htm(Veniamin of
Lesbos' rendering of The Elements into modern Greek), p. 6]
baloglouAToswego.edu===
===
Subject:
: Re: Greek Alphebet>
 In
my studies on triginometry, and calculus, I am comming across
many> symbols which appear to be the Greek Alphebet, as there
was a chart of> the Alphebet in the front of the book. I am
wondering if someone could> tell me what they all mean, or
point me to a site where I could find> out.> Thank you,>
Tom>You need this rule: The symbols mean whatever the author
says they mean.>Take each one in the context where it is
discussed. Somewhere else, the>same symbol is very likely to
mean something different.>Even the venerable PI does not
necessarily mean the constant 3.1415...,>although that is
probably the meaning in the context of the subjects
you>mentioned.>Lynn Killingbeck>
As has been discussed
here, the Greek alphabet is well used by>mathematics. I have
sometimes wondered why further alphabets have not>been
borrowed.>Aleph from the Hebrew is used and seems to have a
meaning even more>fixed than pi (but not so well known). But
it appears to stop there. >I have not seen beth, gimmel used
in maths.Cantor introduced aleph, beth, gimel, and daleth as
thecardinality of the integers, the reals, functions on
thereals, and functions on that. These were constants,
notvariables. At this time, aleph_z is the z-th
smallesttransfinite cardinal of a well-ordered set, where z is
anordinal number, starting from 0, and omega_z is
thecorresponding initial ordinal. At least with the Axiomof
Choice, beth_z is the z-th least power transfinitecardinal;
with the GCH, beth_z is the cardinality of thepower set of a
set with cardinality aleph_z. I have seen zayin introduced as
the opposite of aleph inmodels without choice in one
paper.>The Russian alphabet would blend in well but I have not
seen it used. >Many letters would be hard to distinguish from
Latin and Greek but>many other would be usable.I have not seen
Russian letters used for mathematical constants or variables in
any Russian mathematics papers.>And of course there are many,
many more alphabets available. >Even more extreme we could use
Chinese and have thousands of symbols>available.>I have also
wondered what mathematics in Greek and Russian look like. >Do
the Greeks use Latin and Greek letters in a similar way to us
Latin>alphabet users? How about the Russians, do they spurn
their own>alphabet and use Latin and Greek letters?I have not
seen Greek papers, but I have seen papers in Russian and
Chinese and other languages. Almost withoutexception, they use
Latin and Greek letters as we do.-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue
Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
(765)494-0558===
===
Subject:
: Re: Greek Alphebet>Even more extreme
we could use Chinese and have thousands of
symbols>available.Tohoku Journal, in which he used a Japanese
character for the nameof a field. I cannot recall ever seeing
an arabic character used(apart from Arabic numerals, of
course), nor a Norse rune, nor anyof the south Asian alphabets
(e.g. Thai) nor the old English thorn, nor an accented Latin
character from one of the modern European alphabets, though
Cech (co)homology is usually indicated by the use of that
diacritical mark (it's a hachek or something like that)on the
H. The German Fraktur lettering is a font, really, not
analphabet, but those gylphs have been used heavily in the
past,especially in algebraic number theory. There are zillions
of othersymbols used, some of which appear to be letters, but I
don't thinkthey really are -- nabla and pe come to mind. There
are some ancientGreek letters which I don't think I have ever
seen used in math(including qoppa or something like that; I
think the letters werenever used in post-Homeric Greek). There
are of course many morealphabets from which one can draw; take
a look at the full Unicode suite.And that doesn't include
Klingon! (although it WAS proposed!)>I have also wondered what
mathematics in Greek and Russian look like. >Do the Greeks use
Latin and Greek letters in a similar way to us Latin>alphabet
users? How about the Russians, do they spurn their
own>alphabet and use Latin and Greek letters?Well, research
mathematics is usually communicated in one of just a
fewlanguages, so people whose native language is, say,
Cherokee, don't usuallywrite much high-level mathematics in
that language -- not if they wanttheir results to get much
exposure. Before the (first) October revolution,that included
Russia; the preferred language for the St Petersburg
societywas French. But during the 20th century there was quite
a bit ofresearch mathematics published in Russian, and indeed
the text would bein Russian (using Cyrillic, of course) and
the mathematics using exactlythe same notation used in the
west. I don't recall even seeing a pointin a geometric diagram
ever labelled with a Cyrillic character.(The fact that the
typesetting of mathematics was done using Latin andGreek
characters must have driven the Soviet printers crazy, but it
wasa boon to the translators into western languages, who would
typicallyjust cut and paste the displays. This was occasionally
confusing since,for example, the names of the trig functions
are not the same in allwestern languages.)There is not always
a clear distinction between research monograph andgraduate
text, so university materials followed the same conventions.I
am not sure whether the same is true in elementary-school
texts;obviously the bulk of the text would be in the native
language. If I had to write the primary-grade materials for
them, I guess I wouldopt for using letters which appear
identical in both alphabets, atleast for a couple of years
until the youngsters start the study ofEuropean languages in
earnest.The same situation occurs in the Chinese texts I have
seen.This helps a lot when I have to assess the courses an
internationalstudent has taken in the past. They think I'm
kidding when I say,show me the book you used, but it's pretty
easy to recognize whatwas in a course from the equations which
appear.Just imagine the joys of writing mathematics in, say,
Hebrew, wherethe bulk of the writing is done right-to-left but
the equations arewritten left-to-right. Back in the typewriter
days, that meant havingto guess how much space would be
required, jumping ahead (i.e. left),and proceeding to type (to
the right).Please don't ask me what typewriter
means.dave===
===
Subject:
: Re: Greek Alphebet> Aleph from the
Hebrew is used and seems to have a meaning even more> fixed
than pi (but not so well known). But it appears to stop there.
> I have not seen beth, gimmel used in maths.Beth is used in
set theory for cardinal arithmetic.Russian letter Sha is used
for the Shafarevich group.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re: Greek
AlphebetJohn Savard But in
other contexts, the letters of the Greek alphabet are
fraught>with meaning!>Thus, alpha comes from aleph, which
means Ox.References please? The _name_ alpha might be a
degenerate form of theHebrew aleph, but the letter itself
started AS the symbol for an ox, inGreek. It actually comes
from the upside down capital alpha, which wastheactual symbol
for the head of an ox. (Just picture an upside down A).Who
says it's not the other way around and aleph is not a
degenerateformof alpha?> With all due respect to the Greeks,
the Semitic alphabet> was well established before the Greeks
considered using> an alphabet; they started with it, and the
Semitic alphabet> already had many versions.And just when do
you think the Greeks were using the Cadmian alphabet.1438 BC
is when. This is 50 years before Abraham.There were already 5
varianets of the Cadmian script by 900 BC.===
===
Subject:
: Re:
Greek Alphebet> With all due respect to the Greeks, the
Semitic alphabet> was well established before the Greeks
considered using> an alphabet[...]>And just when do you think
the Greeks were using the Cadmian alphabet.>1438 BC is when.
This is 50 years before Abraham.The Greeks couldn't have
invented an alphabet. The language has thewrong structure.
They would have [or: already had [1]] invented a syllabry.Only
a Semitic language has the vowel-consonant factorization in
itsmorphology to enable an illiterate person to surmise the
segmentationof morphemes at the phonemic level. Without the
syntactic redundancyof vowels an illiterate person isn't going
to see below the syllabiclevel. Most newly developed writing
systems in the world were pictographicor syllabic. In fact,
I'm not even sure there ARE any alphabetic systemsoutside the
Greek/Roman/Phoenician/Arabic/Hebrew lineage.The sequencing
and names are best understood as probably arising fromthe
names and constellations of a lunar zodiac, which were once
prevalentthroughout the old world. It's easy to see a
situation where theconstellations in a Semitic language are,
as a mnemonic device, areeffectively labelled by consonants in
a Abe, Baker, Charlie, Daniel, etc.type system. In later times,
the icons that stand for the constellationscome to also denote
the consonants themselves.There's a passage in Psalms that
might be a latter-day survival of thattradition, in which the
representations and names of the alphabet symbolsare embedded
in mnemonic verses.Note:[1] I believe Linear B (or was it A?)
was syllabic, as a case in point.===
===
Subject:
: Re: Greek
Alphebet> With all due respect to the Greeks, the Semitic
alphabet> was well established before the Greeks considered
using> an alphabet[...]>And just when do you think the Greeks
were using the Cadmian alphabet.>1438 BC is when. This is 50
years before Abraham.>The Greeks couldn't have invented an
alphabet. The language has the>wrong structure. They would
have [or: already had [1]] invented a syllabry.>Only a Semitic
language has the vowel-consonant factorization in
its>morphology to enable an illiterate person to surmise the
segmentation>of morphemes at the phonemic level. Without the
syntactic redundancy>of vowels an illiterate person isn't
going to see below the syllabic>level. Most newly developed
writing systems in the world were pictographic>or syllabic. In
fact, I'm not even sure there ARE any alphabetic
systems>outside the Greek/Roman/Phoenician/Arabic/Hebrew
lineage.There are; the languages of India use an alphabet
partly derived from the ancient Semitic alphabet. Also,
ancientPersian used an alphabet not sufficiently close to
theSemitic alphabet, but they knew about it.The Hangul
alphabet for Korean is not derived from theSemitic, although
it is probable that the existence of analphabet was known to
the designers. A Korean characterhas two to four alphabetic
characters.-- This address is for information only. I do not
claim that these viewsare those of the Statistics Department
or of Purdue University.Herman Rubin, Department of
Statistics, Purdue Universityhrubin@stat.purdue.edu Phone:
(765)494-6054 FAX: (765)494-0558===
===
Subject:
: Re: Greek
Alphebet>With all due respect to the Greeks, the Semitic
alphabet>was well established before the Greeks considered
using>an alphabet[...]And just when do you think the Greeks
were using the Cadmian alphabet.1438 BC is when. This is 50
years before Abraham.[ZAP!] Lies and *CHEAP* PropagandaWHAT A
BUNCH OF IDIOCIES!!!!You are out of your tiny little mind
bloke!! No need to comment*ANYTHING* of your CHEAP
UNBELIEVABLE lies and propaganda!!!! Yor liesare *ENORMOUS*,
you little bwoy. Is Goebbels your spiritualfather??? The
bigger the lie it is, the more believable it is. ??But I can
explain your Ignorance AND your cause if you are Jewish.Are
you Jewish bwoy? That could explain a
lot!propagandist.-Hades===
===
Subject:
: Re: Greek Alphebet>With all
due respect to the Greeks, the Semitic alphabet>was well
established before the Greeks considered using>an
alphabet[...]And just when do you think the Greeks were using
the Cadmian alphabet.1438 BC is when. This is 50 years before
Abraham.> The Greeks couldn't have invented an alphabet. The
language has the> wrong structure. They would have [or:
already had [1]] invented asyllabry.> Only a Semitic language
has the vowel-consonant factorization in its> morphology to
enable an illiterate person to surmise the segmentation> of
morphemes at the phonemic level. Without the syntactic
redundancy> of vowels an illiterate person isn't going to see
below the syllabic> level. Most newly developed writing
systems in the world werepictographic> or syllabic. In fact,
I'm not even sure there ARE any alphabetic systems> outside
the Greek/Roman/Phoenician/Arabic/Hebrew lineage.TWADDLE.The
Greeks were already using 4 scripts before the arrival of
Cadmus, GreekLinear A (Aeolic and Ionic), Cypriot Linear A
(Mycenaean), Linaer B(Mycenaean), and Phaistos Script (Ionic).
The Akkadian-Phoenician scribesknew absolutely nothing about
vowels. It was the Greeks who invented vowelsboth in Linear A
and B and in Cadmian script. Up until 1400 BC which is thedate
of the Ras Sharma tablets the Phoenicians were still using
cuneiform onclay.Cadmus arrival in Greece in 1438 indicates
that it was in Greece that theCadmian script was developed.
According to Phoenician historical recordstranslated by Philo
of Byblis the so-called 3 letters were invented byEisirius
(Celix) who was a foreigner and the brother of Phoenix (Chna)
andby implication the brother of Cadmus as well.The divergence
between Cadmian script and Canaanite script took
placeimmediately on Cadmus arrival in Greece. The Greeks added
vowels to thealphabet but the Phoenicians did not.> The
sequencing and names are best understood as probably arising
from> the names and constellations of a lunar zodiac, which
were once prevalent> throughout the old world. It's easy to
see a situation where the> constellations in a Semitic
language are, as a mnemonic device, are> effectively labelled
by consonants in a Abe, Baker, Charlie, Daniel,etc.> type
system. In later times, the icons that stand for the
constellations> come to also denote the consonants
themselves.PREPOSTEROUS. The names of the Phoenician
constellations are known and theydo not go A, B, C. Besides
which both Greek and Hebrew start of as A, B, Gand there are
no zodiacal constellations in that order either, nor enoughfor
an alphablet.An upside down Alpha looks more like a horses had
than that of an ox so itis more likely that is was derived
from the Greek Alogo.> There's a passage in Psalms that might
be a latter-day survival of that> tradition, in which the
representations and names of the alphabet symbols> are
embedded in mnemonic verses.> Note:> [1] I believe Linear B
(or was it A?) was syllabic, as a case in point.Wrong. Linear
A and B had separate individual symbols for VOWELS as well
asbi-syllabic consonant-vowel and tri-syllabic
combinations.===
===
Subject:
: Re: Greek Alphebet>And just when do
you think the Greeks were using the Cadmian alphabet.>1438 BC
is when. This is 50 years before Abraham.>There were already 5
varianets of the Cadmian script by 900 BC.And just where do you
get these dates from? If you believe the Bible,your date for
Abraham is too late. If you don't believe the Bible,why do you
think there was an Abraham at all? As for the Greek
alphabet,various dates are suggested for it, but I'm not aware
of any archaeological evidence for it before the mid 8th
century BCE.===
===
Subject:
: Re: Greek AlphebetAnd just when do you
think the Greeks were using the Cadmian alphabet.1438 BC is
when. This is 50 years before Abraham.There were already 5
varianets of the Cadmian script by 900 BC.> And just where do
you get these dates from? If you believe the Bible,The
Traditional Chronology as recorded by Diodorus, Castor and
Apollonius etal, and used by Tatian, Eusebius and Clement. The
siege of Troy began 1193BC so add up the figures and
generations.> your date for Abraham is too late. If you don't
believe the Bible,If you knew how to do some simple arithmetic
and statistics you would knowthat the numbers in the bible are
all the numerators of
fractions.http://www.enthymia.co.uk/myths/bible/
EstablishChronology.htm(the above research is copyrighted)>
why do you think there was an Abraham at all? As for the Greek
alphabet,> various dates are suggested for it, but I'm not
aware of any> archaeological evidence for it before the mid
8th century BCE.Then you know nothing about the subject. 5
different scripts are known tohave existed in 900 BC. The only
way there could have been so many is if allof them derived from
a common root long before the Ionian mitigation of 1071BC. 1200
BC is the latest possible date for the use of Cadmian script by
theGreeks and the start of its divergence into 5 independent
Greek scripts.Herodotus records an inscription made in the
time of Amphytrion and Laiuswhich firmly dates to Amhytrions
victory over the Teleboans which is datedto 1286 BC by the
occurrence of a total solar eclipse over the Aegean
whichoccurred on February 10 and is confirmed by the
traditional chronology.> Robert Israel israel@math.ubc.ca>
Department of Mathematics http://www.math.ubc.ca/~israel>
University of British Columbia> Vancouver, BC, Canada V6T
1Z2===
===
Subject:
: Re: Greek Alphebet>If you knew how to do some
simple arithmetic and statistics you would know>that the
numbers in the bible are all the numerators of fractions.Yes,
all the integers used in the bible are the numerators of
fractions.As are all the integers that are not used in the
bible. So?===
===
Subject:
: Re: Greek AlphebetIf you knew how to do
some simple arithmetic and statistics you wouldknowthat the
numbers in the bible are all the numerators of fractions.>
Yes, all the integers used in the bible are the numerators of
fractions.> As are all the integers that are not used in the
bible. So?Do the
maths.http://www.enthymia.co.uk/myths/bible/
EstablishChronology.htm> Robert Israel israel@math.ubc.ca>
Department of Mathematics http://www.math.ubc.ca/~israel>
University of British Columbia> Vancouver, BC, Canada V6T
1Z2===
===
Subject:
: Re: Greek Alphebet charset=iso-8859-7Yes, all
the integers used in the bible are the numerators of
fractions.As are all the integers that are not used in the
bible. So?> Do the maths.Agamemnon,I think you may want to be
a little more courteous with Robert. He cancertainly do the
maths much easier than anyone here, so you'll probably bevery
lucky if you can convince him to read your webpage.And you
never know....If he does, you may become famous :*)>
http://www.enthymia.co.uk/myths/bible/
EstablishChronology.htmRobert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israelUniversity of British
ColumbiaVancouver, BC, Canada V6T 1Z2--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: Greek Alphebet> Yes, all the
integers used in the bible are the numerators offractions.> As
are all the integers that are not used in the bible. So?Do the
maths.> Agamemnon,> I think you may want to be a little more
courteous with Robert. He can> certainly do the maths much
easier than anyone here, so you'll probablybe> very lucky if
you can convince him to read your webpage.If he can do the
maths then he should have no problem in confirming that Iam
right.> And you never know....If he does, you may become
famous
:*)http://www.enthymia.co.uk/myths/bible/
EstablishChronology.htm> Robert Israel israel@math.ubc.ca>
Department of Mathematics http://www.math.ubc.ca/~israel>
University of British Columbia> Vancouver, BC, Canada V6T 1Z2>
--> Ioannis Galidakis> http://users.forthnet.gr/ath/jgal/>
------------------------------------------> Eventually,
_everything_ is understandable===
===
Subject:
: Re: Greek
AlphebetAgamamnonas is a person who is a man of all trades but
master of none. He isinfamously famous, notoriously famous
already. He has absolutely NO need ofmere professors of
maths!Why, Agamamnonas the linguist has already told us that
Dolmades is fromthe Greek word gemistes or gamistes or
something like that and not fromthe Turkish root word
dol/dolma
meaningfill/stuff/filling/stuffing/filled/stuffed.Why don't we
all get together to tell Agamemnonas to go, get stuffed.Pigenai
na gemithis, re Agamemnona!Gia sou re Agamemnona;-) Ean esei
mou stelleis loukanika, kai egw na soustellw Ntpolmades? 8a
kanomai allagei/trampa, re. Entaksi?Entaksi, Mparataksi!--
choro-nik********> Yes, all the integers used in the bible are
the numerators offractions.> As are all the integers that are
not used in the bible. So?Do the maths.> Agamemnon,> I think
you may want to be a little more courteous with Robert. He
can> certainly do the maths much easier than anyone here, so
you'll probablybe> very lucky if you can convince him to read
your webpage.> And you never know....If he does, you may
become famous
:*)http://www.enthymia.co.uk/myths/bible/
EstablishChronology.htm> Robert Israel israel@math.ubc.ca>
Department of Mathematics http://www.math.ubc.ca/~israel>
University of British Columbia> Vancouver, BC, Canada V6T 1Z2>
--> Ioannis Galidakis> http://users.forthnet.gr/ath/jgal/>
------------------------------------------> Eventually,
_everything_ is understandable===
===
Subject:
: Re: Greek Alphebet>
Agamamnonas is a person who is a man of all trades but master
of none. Heis> infamously famous, notoriously famous already.
He has absolutely NO needof> mere professors of maths!> Why,
Agamamnonas the linguist has already told us that Dolmades is
from> the Greek word gemistes or gamistes or something like
that and notfrom> the Turkish root word dol/dolma meaning>
fill/stuff/filling/stuffing/filled/stuffed.I said it was from
the Latin word Tomaculum meaning sausage and Toomentummeaning
stuffing. There is no Altaic root for Dolma.> Why don't we all
get together to tell Agamemnonas to go, get stuffed.> Pigenai
na gemithis, re Agamemnona!> Gia sou re Agamemnona;-) Ean esei
mou stelleis loukanika, kai egw na sou> stellw Ntpolmades? 8a
kanomai allagei/trampa, re. Entaksi?Mpikse angkouri mes to
kolo sou.> Entaksi, Mparataksi!> --> choro-nik> ********> Yes,
all the integers used in the bible are the numerators of>
fractions.> As are all the integers that are not used in the
bible. So?> Do the maths.Agamemnon,I think you may want to be
a little more courteous with Robert. He cancertainly do the
maths much easier than anyone here, so you'llprobably> bevery
lucky if you can convince him to read your webpage.And you
never know....If he does, you may become famous :*)>
http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm Robert Israel israel@math.ubc.ca> Department of Mathematics
http://www.math.ubc.ca/~israel> University of British
Columbia> Vancouver, BC, Canada V6T 1Z2--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: Greek AlphebetAgamamnonas is a
person who is a man of all trades but master of none.He>
isinfamously famous, notoriously famous already. He has
absolutely NO need> ofmere professors of maths!Why,
Agamamnonas the linguist has already told us that Dolmades
isfromthe Greek word gemistes or gamistes or something like
that and not> fromthe Turkish root word dol/dolma
meaningfill/stuff/filling/stuffing/filled/stuffed.> I said it
was from the Latin word Tomaculum meaning sausage and
Toomentum> meaning stuffing. There is no Altaic root for
Dolma.Absolutely brilliant! What's wrong with you, Agamemnona?
Can't even find aword in a dictionary? Oh, I know why. It is
because it is Turkish. The rootword is dol with the verb being
doldurmak meaning to fill. Dolma isthat which has been stuffed.
Dolgu is a tooth filling. Doldur is whatyou say at the petrol
pump if you want to say Fill her up.Tomaculum and Toomentum my
foot!!!!-- choro-nik********Why don't we all get together to
tell Agamemnonas to go, get stuffed.Pigenai na gemithis, re
Agamemnona!Gia sou re Agamemnona;-) Ean esei mou stelleis
loukanika, kai egw na soustellw Ntpolmades? 8a kanomai
allagei/trampa, re. Entaksi?> Mpikse angkouri mes to kolo
sou.Entaksi, Mparataksi!--choro-nik********> > Yes, all the
integers used in the bible are the numerators offractions.> As
are all the integers that are not used in the bible. So?> Do
the maths.> Agamemnon,> I think you may want to be a little
more courteous with Robert. He can> certainly do the maths
much easier than anyone here, so you'll> probablybe> very
lucky if you can convince him to read your webpage.> And you
never know....If he does, you may become famous :*)>
http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm Robert Israel israel@math.ubc.ca> Department of Mathematics
http://www.math.ubc.ca/~israel> University of British Columbia>
Vancouver, BC, Canada V6T 1Z2> --> Ioannis Galidakis>
http://users.forthnet.gr/ath/jgal/>
------------------------------------------> Eventually,
_everything_ is understandable> ===
===
Subject:
: Re: Greek
Alphebet> Agamamnonas is a person who is a man of all trades
but master of none.> Heis> infamously famous, notoriously
famous already. He has absolutely NOneedof> mere professors of
maths!> Why, Agamamnonas the linguist has already told us that
Dolmades is> from> the Greek word gemistes or gamistes or
something like that and notfrom> the Turkish root word
dol/dolma meaning>
fill/stuff/filling/stuffing/filled/stuffed.I said it was from
the Latin word Tomaculum meaning sausage andToomentumTomentum
rather.meaning stuffing. There is no Altaic root for Dolma.>
Absolutely brilliant! What's wrong with you, Agamemnona? Can't
even find a> word in a dictionary? Oh, I know why. It is
because it is Turkish. Theroot> word is dol with the verb
being doldurmak meaning to fill. Dolma is> that which has been
stuffed. Dolgu is a tooth filling. Doldur is what> you say at
the petrol pump if you want to say Fill her up.POPPYCOCK.The
Turks got the word from the Greeks. The Turkish word Dol is a
corruptionof the Latin Tom.> Tomaculum and Toomentum my
foot!!!!> --> choro-nik> ********> Why don't we all get
together to tell Agamemnonas to go, getstuffed.> Pigenai na
gemithis, re Agamemnona!> Gia sou re Agamemnona;-) Ean esei
mou stelleis loukanika, kai egw nasou> stellw Ntpolmades? 8a
kanomai allagei/trampa, re. Entaksi?Mpikse angkouri mes to
kolo sou.> Entaksi, Mparataksi!> --> choro-nik> ********> >
Yes, all the integers used in the bible are the numerators of>
fractions.> As are all the integers that are not used in the
bible. So?> Do the maths.> Agamemnon,> I think you may
want to be a little more courteous with Robert. Hecan>
certainly do the maths much easier than anyone here, so
you'llprobably> be> very lucky if you can convince him to read
your webpage.> And you never know....If he does, you may
become famous :*)>
http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm Robert Israel israel@math.ubc.ca> Department of Mathematics
http://www.math.ubc.ca/~israel> University of British Columbia>
Vancouver, BC, Canada V6T 1Z2> --> Ioannis Galidakis>
http://users.forthnet.gr/ath/jgal/>
------------------------------------------> Eventually,
_everything_ is understandable> ===
===
Subject:
: Re: Greek
AlphebetNo, Agamemnona, the Turkish word Dol is a corruption
of the Greek TaVgaleis Ap'ton Kolosou!Ma nomizeis oth monon
esei esai Karpasitis?-- choro-nik********> Agamamnonas is a
person who is a man of all trades but master ofnone.He> is>
infamously famous, notoriously famous already. He has
absolutely NO> need> of> mere professors of maths!> Why,
Agamamnonas the linguist has already told us that Dolmades
isfrom> the Greek word gemistes or gamistes or something like
that andnot> from> the Turkish root word dol/dolma meaning>
fill/stuff/filling/stuffing/filled/stuffed.> I said it was
from the Latin word Tomaculum meaning sausage and> Toomentum>
Tomentum rather.> meaning stuffing. There is no Altaic root
for Dolma.Absolutely brilliant! What's wrong with you,
Agamemnona? Can't even findaword in a dictionary? Oh, I know
why. It is because it is Turkish. The> rootword is dol with
the verb being doldurmak meaning to fill. Dolmaisthat which
has been stuffed. Dolgu is a tooth filling. Doldur iswhatyou
say at the petrol pump if you want to say Fill her up.>
POPPYCOCK.> The Turks got the word from the Greeks. The
Turkish word Dol is acorruption> of the Latin Tom.Tomaculum
and Toomentum my foot!!!!--choro-nik********> Why don't we
all get together to tell Agamemnonas to go, get> stuffed.>
Pigenai na gemithis, re Agamemnona!> Gia sou re
Agamemnona;-) Ean esei mou stelleis loukanika, kai egw na>
sou> stellw Ntpolmades? 8a kanomai allagei/trampa, re.
Entaksi?> Mpikse angkouri mes to kolo sou.> Entaksi,
Mparataksi!> --> choro-nik> ********> > > Yes, all the
integers used in the bible are the numerators of> fractions.>
As are all the integers that are not used in the bible. So?>
Do the maths.> Agamemnon,> I think you may want to be a
little more courteous with Robert. He> can> certainly do the
maths much easier than anyone here, so you'll> probably> be>
very lucky if you can convince him to read your webpage.>
And you never know....If he does, you may become famous :*)>
http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm Robert Israelisrael@math.ubc.ca> Department of
Mathematicshttp://www.math.ubc.ca/~israel> University of
British Columbia> Vancouver, BC, Canada V6T 1Z2> --> Ioannis
Galidakis> http://users.forthnet.gr/ath/jgal/>
------------------------------------------> Eventually,
_everything_ is understandable> > > ===
===
Subject:
: Re: Greek
Alphebet> Thus, alpha comes from aleph, which means Ox.>
References please?> Alpha does not come from aleph, they
simply have a common source.>Punic (Phonecian) alphabet.
Hebrew, Aramaic and Arabic are all >descendants of the
Urguritic -language-. I am not sure what the original >scripts
were (assuming there was any scripts).>Bob KolkerThe Ugaritic
alphabet is a 32 character cuneiform modification of the early
Semitic alphabets; therewere lots of them. The earliest known
Semiticalphabetic writing is from about 1800 BCE, foundon a
cliff wall in Egypt.Hebrew and Ugaritic are almost identical
as languages;Aramaic is similar, but different, and these
probablydiverged before alphabetic writing; Arabic is froma
different branch of the Semitic family.-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue
Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
(765)494-0558