mm-418 Subject: Oneness of a number 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.* 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? (note that for 5 this never happens) 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), 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, 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 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 === Subject: Re: JSH: Difficult social problem congratulation! > 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't bring you soup might *not* be that you don't speak French? -- === Subject: Re: derivate of x^x primefinder escribi.97 en el mensaje > 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)) ? > 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 branes it's hearsay, about the US casualty estimates. now, the Japanese were negotiating through the VAtican Special Office and 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 worse than the German situation circa '45, because they 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 and told the press that Wallace never showed him the speech. Secretary of State Byrnes and the press went ballistic, and on Sept. 20, Truman asked for Wallace's resignation and got it. Truman promptly appointed Averell Harriman in Wallace's place. For the next two decades, Wallace continued to battle for national policy direction as he saw it. That is a story for 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. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Oneness of a number === >Subject: Oneness of a number >Message-id: 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 >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? >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 by looking 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 -- Mensanator Ace of Clubs === 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? This is unthinkable if you have ever been to Quebec or France. The only way I could get a cup of coffee in Quebec City was for my wife to order it 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. > Take 'em out? What do you mean by that? Are you back to your previous > threat of calling out the military or the CIA and the FBI, or are you > advocating 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 a physics undergrad, and I'd get into all kinds of fascinating discussions, 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 deliver the coup de grace of the irrefutable facts. That is, you take 'em out. Now then, the crackpots may still believe their nonsense, but civilized society is once again protected by the vigillance of active and highly intelligent minds, as the bulk of society can move on with the truth. I've isolated mathematicians by a logical argument, having made a major discovery--that's what physics people do--in what they probably think of as their area alone, and caught them trying to dodge the result and its implications. Remember mathematicians have claimed that pure math was for the benefit of humanity, but I've shown them trying to hide one of the most astounding results in math history (one of three, mind you). So it's time for the physics people to look at the facts, stand for logic 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 -> 1 No, you didn't listen well, I take the oneness of the oneness for the recursive 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 does not 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* which are 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 those ways 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 want as 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. 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.net Stupid is as stupid does. There is no fixing stupid because it is not broken. Harris would be much happier as a priest proclaiming god amdist heretics and buggering altar boys to His greater glory. Hey Harris, your village called: Its idiot is missing. -- Uncle Al http://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 more > interesting than most of the discussions that take place on it. By the > same token, I am much more interested in the hardware and operating system > software of the systems I administer than in any of the applications > for which they are used. Once I hoped that the growing popularity of personal computers meant > that nearly everyone would learn to think like programmers. It never > occurred to me that, sadly, the opposite would happen: That computers > would 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 programmers used 'em. You said it's cooler when people who knew vehicles inside out were the ones driving them. You guys are agreeing. === Subject: Re: Skeptickal Inquirer UFO I hope you're not speaking poorly about my Venus lizard folk, Cathars and 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 or two about our nasty moon, as a place that I believe we've long needed to establish a lunar space elevator (LSE) in order to be efficiently getting ourselves off to visiting the wizard of Oz at Venus L2, as well as for reaching out to those thoroughly irradiated to death snowman/snowwoman situated on Mars: http://guthvenus.tripod.com/gv-cm-ccm-01.htm Here's a little typical feedback of supposed facts from: Jay Windley (webmaster@clavius.org) High-energy cosmic rays do not come from the sun. They come from outside the solar system, and our sun is the primary defense against lower energy and thus their secondary effects in the ambient are minimal. My thoughts: http://guthvenus.tripod.com/gv-moon-radiation.htm Unfortunately, I may have incorrectly utilized the terminology of cosmic rays, though fortunately, I never specified upon any specific spectrum of high-energy cosmic rays, just pointing out that our sun is certainly capable of tossing out its fair share of far worse things than visible photons plus IR worth of BTUs and of those nasty UVs. Obviously a supernovae is worth a thousand fold in terms of being nasty, thereby from the far off generated galactic influx must offer a measurable degree of such events, and of the secondary radiation given off by all that infamous clumping lunar dirt should become a fairly darn good as well as unobstructed indicator of whatever cosmic/galactic influx. Seems fairly odd that shuch measurements aren't common place, as where's the justification for not otherwise providing this level of information? The assertions or premise offered by the likes of Jay Windley, that of our moon not only lacking an atmosphere but also without a Van Allen buffer zone is not such a bad thing if you're out and about on the lunar surface, seems somewhat risky if not downright lethal. I might have come into that understanding if we're referring to an earthshine illuminated lunar surface, but I'm not going so far if that's of any fully solar illuminated environment while wearing a mostly synthetic moon suit because, we're not talking about avoiding a 270 nm UV sun burn. Sorry about all my make-due reverse engineering logic, or lack thereof. I was simply trying to establish upon the amount of solar radiation that becomes hard X-Ray class. So exactly how much is it on a typical lunar day, or how about on a good day as well as on a truly High-energy cosmic rays do not come from the sun ??? Do we suppose that happens to include the likes of the last couple of weeks of horrific solar flak? Seems there should be some specific knowledge (excluding Apollo) of what's what pertaining to the solar illuminated surface, as opposed to the absolute lunar nighttime environment and, of something specific pertaining to whatever earthshine contributes. This task of obtaining knowledge is somewhat like my getting a grasp upon the applied energy (thrust or torque) involved in accelerating something the size and mass of the moon, so that it accelerates and thereby 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 for decades. In a nutshell, tides cause friction between the oceans and the ocean floors, which transfers energy from the solid part of the earth to the oceans. One of the effects of this friction is that the tidal bulge is off-center, and is located eastward of the moon. (So the high tide actually occurs when the moon is west of overhead.) The result of the tidal bulge being off center is that there is a torgue effect placed on the moon, and this in turn transfers energy from the earth to the moon. The earth's spin rate slows, the moon is speeded in its orbit and therefor moves further away from the earth. (This transfer 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 friction due to shallower or deeper oceans, and the positions of the continents. For the benefit of all my loyal critics, I've conceded that there's a darn good chance that the likes of Tim Thompson has more than a few valid points as to his version of what's what. This following page is just another example of my learning from the pros, of accepting other input, which may even including the likes of what Ami Silberman just presented, that I'd not be calling flak, as there actually seems to be some considerable worth to at least Tim's version of the lunar recession, if I don't say so myself. http://guthvenus.tripod.com/earth-moon-energy.htm === Subject: Re: Skeptical Inquirer UFO Unfortunately Scott, this sting/ruse is way so much bigger than even you can imagine. I've posted this folowing in at least a half dozen topics 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 or two about our nasty moon, as a place that I believe we've long needed to establish a lunar space elevator (LSE) in order to be efficiently getting ourselves off to visiting the wizard of Oz at Venus L2, as well as for reaching out to those thoroughly irradiated to death snowman/snowwoman situated on Mars: http://guthvenus.tripod.com/gv-cm-ccm-01.htm Here's a little typical feedback of supposed facts from: Jay Windley (webmaster@clavius.org) High-energy cosmic rays do not come from the sun. They come from outside the solar system, and our sun is the primary defense against lower energy and thus their secondary effects in the ambient are minimal. My thoughts: http://guthvenus.tripod.com/gv-moon-radiation.htm Unfortunately, I may have incorrectly utilized the terminology of cosmic rays, though fortunately, I never specified upon any specific spectrum of high-energy cosmic rays, just pointing out that our sun is certainly capable of tossing out its fair share of far worse things than visible photons plus IR worth of BTUs and of those nasty UVs. Obviously a supernovae is worth a thousand fold in terms of being nasty, thereby from the far off generated galactic influx must offer a measurable degree of such events, and of the secondary radiation given off by all that infamous clumping lunar dirt should become a fairly darn good as well as unobstructed indicator of whatever cosmic/galactic influx. Seems fairly odd that shuch measurements aren't common place, as where's the justification for not otherwise providing this level of information? The assertions or premise offered by the likes of Jay Windley, that of our moon not only lacking an atmosphere but also without a Van Allen buffer zone is not such a bad thing if you're out and about on the lunar surface, seems somewhat risky if not downright lethal. I might have come into that understanding if we're referring to an earthshine illuminated lunar surface, but I'm not going so far if that's of any fully solar illuminated environment while wearing a mostly synthetic moon suit because, we're not talking about avoiding a 270 nm UV sun burn. Sorry about all my make-due reverse engineering logic, or lack thereof. I was simply trying to establish upon the amount of solar radiation that becomes hard X-Ray class. So exactly how much is it on a typical lunar day, or how about on a good day as well as on a truly High-energy cosmic rays do not come from the sun ??? Do we suppose that happens to include the likes of the last couple of weeks of horrific solar flak? Seems there should be some specific knowledge (excluding Apollo) of what's what pertaining to the solar illuminated surface, as opposed to the absolute lunar nighttime environment and, of something specific pertaining to whatever earthshine contributes. This task of obtaining knowledge is somewhat like my getting a grasp upon the applied energy (thrust or torque) involved in accelerating something the size and mass of the moon, so that it accelerates and thereby 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 for decades. In a nutshell, tides cause friction between the oceans and the ocean floors, which transfers energy from the solid part of the earth to the oceans. One of the effects of this friction is that the tidal bulge is off-center, and is located eastward of the moon. (So the high tide actually occurs when the moon is west of overhead.) The result of the tidal bulge being off center is that there is a torgue effect placed on the moon, and this in turn transfers energy from the earth to the moon. The earth's spin rate slows, the moon is speeded in its orbit and therefor moves further away from the earth. (This transfer 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 friction due to shallower or deeper oceans, and the positions of the continents. For the benefit of all my loyal critics, I've conceded that there's a darn good chance that the likes of Tim Thompson has more than a few valid points as to his version of what's what. This following page is just another example of my learning from the pros, of accepting other input, which may even including the likes of what Ami Silberman just presented, that I'd not be calling flak, as there actually seems to be some considerable worth to at least Tim's version of the lunar recession, if I don't say so myself. http://guthvenus.tripod.com/earth-moon-energy.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.math anyway) do have an Achilles heel - a four-digit year at the end of the title. If your newsreader supports wildcard titles in kill rules, all one 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 occurs in any group: given a group element x and an integer n we compute the power x^n in the group; certainly (x^n)^m = (x^m)^n, which is what you appear to request in your last line. And yes, it's often difficult to take an element in a cyclic group and decide what power of the generator it is. Your initial example is of this type, where the group is (Z/kZ)^* , the multiplicative group mod k (where k is any fixed integer). Another example commonly used for one-way functions is to use the group of points in an elliptic curve mod p. === Subject: Re: division by general integer using register shifts > Hi all, > 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 be done > by distributing the operation across numbers that are powers of two and sum > 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. Can someone > tell me how (or is it possible) to do division by a general integer using > shifts? > -- > Best wishes, > Allen You could try not working binary but convert to the divisor as radix instead, 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, <3c65f87.0311080557.3c8f97f1@posting.google.com>: > 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 have a chance of not using trig, but it would take some work. I can tell you that the roots, as reported by Pari/GP, are approximately: -4.424076847035404679966063971 -0.2879615764822976600169680144 - 0.6075770270241740255965589263*I -0.2879615764822976600169680144 + 0.6075770270241740255965589263*I but this doesn't help much as algebraic units are provably dense around (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 giggles the 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 that x^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 both units (as is easily verified by multiplying them together), they factor little in the analysis. I can just as easily posit d_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 the equation y^3 - 4 = 0. (d_3 solves this equation too.) d_1' ? Well, lessee; the most obvious equation is 2 * y^3 - 343 = 0 which is clearly not going to lead to anything resembling an algebraic integer. Since d_1' requires a cube root no equation of lesser degree will do. But I may need those roots to your equation, and not just approximations, either. If one posits N(x) = x^3 + 5*x^2 + 3*x + 2 then, if one substitutes x = y-5/3, one can rewrite this as: N(y-5/3) = y^3 - 16/3 * x + 169/27 Now substitute y = w + 16/(9*w) (Vi.8fta's substitution, with p = -16/3) and get N(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 divide by 3^3/3^3 (or sqrt(3^6)/3^3), resulting in w^3 = (-169 ± sqrt(12177)) / (2 * 27) or w^3 = (-169 ± 3*sqrt(1353)) / (2 * 27) I'm going to take the + sign; the - sign will yield slightly identical 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 as x1 = -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 solution w^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 a Lost Cause(tm), although we know the answer is yes since they all solve the equation x^3 + 5*x^2 + 3*x + 2 = 0, as one can verify if one multiplies (x-x1)*(x-x2)*(x-x3) in GP/Pari. However, there's a different route, making the exact roots 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 be quite valid, as it turns out -- but what are the ci? Since we know x^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 some rotation of x1, x2, x3. Are these algebraic integers? We note that, if xi solves x^3 + 5 x^2 + 3x + 2 = 0, then 1/xi has to solve 2*z^3 + 3*z^2 + 5*z + 1 = 0, 7/xi has to solve 2*z^3 + 3*7*z^2 + 5*7^2*z + 7^3 = 0, and -7/xi has to solve 2*z^3 - 3*7*z^2 + 5*7^2*z - 7^3 = 0. -7/xi is clearly *not* an algebraic integer (as the above is not a reducible polynomial over Q). Fortunately, -2/xi is: 2*z^3 - 3*2*z^2 + 5*2^2*z - 2^3 = 0 or z^3 - 3*z^2 + 10*z - 4 = 0 Well, 1 out of 3 ain't bad. This counter-proof generalizes quite nicely for most cubic polynomials x^3 + a_2 * x^2 + a_1 * x + a_0 with non-unit a_0 not divisible by 7. I'm afraid you may be out of luck, James. :-) [rest snipped] -- #191, ewill3@earthlink.net It'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 figured out whether chocolates or flowers will make the sentence This statement is false happy. :-) ;-) :-) > 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 thing for having to pay by the connection minute. -- #191, ewill3@earthlink.net It'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 self-explanatory - > 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 lines of your posting. that is inconsistent with how I specified the change of basis, it should have 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 does indeed 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 nice axial 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. -- Nicolas Wow, THAT'S an interessant discussion === Subject: Re: differentiable...problem... Content-transfer-encoding: 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 curves Can the direct isogeny be found between these two elliptic curves? It is known that they are in a set of 8 isogenous curves, but what is 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 between y^2 = x^3+ax+b and Y^2 = X^3+AX+B, when the curves have the 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 accused of leaving out the info to mislead the reader, etc. Anyway, I am very sorry. I will do that from now on. === Subject: Re: Continuum fluently presume at least that you are presumably fluent in what passes for English in the state that gave the world Hollywood . It's sheer fun to see how 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, istines, that Z stands for Ze Any Set With Cardinality of Ze Continuum. amateur > Besides, to comment on you > awareness, what languages do you fluently read in except english ? > You presume that I am fluent in English? === Subject: Re: Continuum fluently presume at least that you are presumably fluent in what passes for English in the state that gave the world Hollywood . It's sheer fun to see how 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, istines, that Z stands for Ze Any Set With Cardinality of Ze Continuum. amateur > Besides, to comment on you > awareness, 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.htm Mr. Harris, to me, this seems equivalent to the copyright you had on the web page you created for your famed Proof of FLT. Now, you are once again hell-bent on crying, cursing, cheating and trying to do anything to propagate your flawed lies. You also have some sort of greatness disorder that makes you think you are above every person you come in contact with. IIRC, this is a Narcissistic Personality Disorder. Here is excerpt from the web site: http://www.suite101.com/welcome.cfm/npd Narcissists attract abuse. Haughty, exploitative, demanding, insensitive, and quarrelsome - they tend to draw opprobrium and provoke anger and even hatred. Sorely lacking in interpersonal skills, devoid of empathy, and steeped in irksome grandiose fantasies - they invariably fail to mitigate the 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 (cautionary statement) A pervasive pattern of grandiosity (in fantasy or behavior), need for admiration, and lack of empathy, beginning by early adulthood and present in a variety of contexts, as indicated by five (or more) of the following: (1) has a grandiose sense of self-importance (e.g., exaggerates achievements and talents, expects to be recognized as superior without commensurate achievements) (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 be understood 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 of especially favorable treatment or automatic compliance with his or her expectations (6) is interpersonally exploitative, i.e., takes advantage of others to achieve his or her own ends (7) lacks empathy: is unwilling to recognize or identify with the feelings and needs of others (8) is often envious of others or believes that others are envious of him or her (9) shows arrogant, haughty behaviors or attitudes Reprinted with permission from the Diagnostic and Statistical Manual of Mental Disorders, fourth Edition. Copyright 1994 American Psychiatric Association *** Perhaps instead of spewing your useless nonsense, you should study this disorder, 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 tensors X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 05:01 PM, jjensen14@hotmail.com (J Jensen) said: >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 map skew-symmetric tensors into vectors. For more than 3 dimensions, you have to explicitly treat the torque as a skew symmetric tensor or a 2-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 need to treat it as a vector and pay close attention to your weights. Either way, you need to be careful to distinguish covariant indices from contravariant. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Hard tensor question (kst). <3fa668f5$9$fuzhry+tra$mr2ice@news.patriot.net> <2202379a.0311052323.77555cbc@posting.google.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 11:23 PM, dynamics@vianet.on.ca (Ken S. Tucker) said: >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 polar coordinates 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 JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: [JSH] [LUNATIC FRINGE] Re: My [Flawed] research, [JOKE] publication announcement Mr. Harris, Psychopathology ------------------------ Narcissism: A pattern of traits and behaviors which signify infatuation and obsession with one's self to the exclusion of all others and the egotistic and ruthless pursuit of one's gratification, dominance and ambition. Perhaps research in to your delusional fits of grandeur would be a more appropriate 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 extraneous inessentials. 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 probability argument, as versions of the Berry-Esseen bounds just cannot be done strictly probabilistically; the bounds on the difference of sums of iid random variables with a finite third moment and sums of iid normal random variables with the same variance, but the sequences dependent, does not increase with the number. Both of these proofs go over almost verbatim for martingales. The characteristic function proof does go over, but not as easily. -- 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 === > 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 examined from the view point of calculus. Well, I know that a derivative measures change and I also know that when the instantaneous 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 finding f'(x). f'(x)=cos(x) We know that a function has a critical point at a point where the derivative is 0. cos(x)=0 at x=pi/2 and 3pi/2 Ok, we know where the critical points are, but we don't know what happens at those points. So take the 2nd derivative. f''(x)=-sin(x) When x=pi/2, f''(x)=-1. When x=3pi/2 f''(x)=1 Therefore I can now analyze my results. When x=pi/2, the 2nd derivative is negative, which means it's decreasing so pi/2 must be a maximum. When x=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 if you examine it from a calculus viewpoint, which may not be a bad idea for you 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. 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 are not true in general. Let P(x) by a polynomial divisible by p, and let g_1(x) and g_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 values P( 0) = 15, g_1( 0) = 3, g_2( 0) = 5 P( 1) = 12, g_1( 1) = 2, g_2( 1) = 6 P( 5) = 0, g_1( 5) = 0. g_2( 5) = 8 P( 8) = -9, g_1( 8) = -1. g_2( 8) = 9 P(16) = -33, g_1(16) = -3. g_2(16) = 11 Note how the factor of 3 switches back and forth between g_1 and g_2 depending on the value of x. Thus neither g_1(x) nor g_2(x) is divisible by 3 for all x. In particular the fact that g_1(0) is divisible by 3 does not mean that g_1(1) is divisible 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 of which the c's are the roots? More to the point, x^3 + 5x^3 + 3x + 2 factors as (x - r1)(x - r2)(x - r3) in the algebraic integers. Where the r's are the roots of the polynomial. We can multiply the factors by 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 claim that r1 and r2 are units (they are divisors of 2). But they are not, because the constant term of a polynomial (see the correct usage here, James) must be 1 for a root to be a unit. -- 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: 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'Hopital Note: this classic problem is well-known due to the fact that it appeared in Hardy's classic calculus textbook with a 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]. - 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) 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^x Provided 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 -> 0 L = lim f+f' - lim 2f = lim f'-f 0 = lim f = lim f e^-x / e^-x = (f' - f)e^-x / -e^-x = lim f-f' = -L 0 = 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' = 0 If 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 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. 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 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. 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 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. I wouldn't be surprised if you could actually construct a counterexample along these lines - you'd need to verify a 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 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!) -- 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: 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 you were just trying to be funny. In case you were also trying to make a serious point: I certainly didn't mean that this was a counterexample, or something that would lead to a counterexample with no new ideas required. But what I had in mind was something _vaguely_ like this: Suppose we could find x_n such that each n-bit sequence in the binary expansion of x_n occurs with the right frequency, but 1/x_n has a terminating binary expansion. Then let x consist of a long stretch of the bits of x_1, followed by a much longer stretch of the bits of x_2, etc. For suitable values of long and longer x will be normal, but it doesn't seem so unlikely that 1/x would turn out to be abnormal. -- >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 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? === Subject: Re: Image of a straight line in a convex cylindrical mirror When 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 virtual image is distorted in reflection. I observed clear images of square lines of floor mosaic in a polished cylindrical flowerpot in Twelve Oak Mall, Detroit. What is commonly known is the optics: image-object distances are connected by standard formula (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 / ray tracing in 3D before . I believed that cos(slope angle) in the x-z projection is inversely proportional to z-coordinate, as it is for a catenary z=A cosh(x/A). I asked this question expecting that there could be elementary geometrical operations with straight lines, cylinders, reflection &c. My imagination surrounded vague thoughts about geodesic deviation, but I right now I would thank in advance for any ensuing pointers and clarifications. === 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 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. === 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 eight years of claiming to have a proof of FLT, he doesn't know an algebraic integer 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 him nothing to cross-post to other rooms from which he would still get his arse kicked. The JSH thing is tedious most of the time, and morbid all the time, but still good for the occasional laugh. Try Google with these keywords: guffaw idiot polynomial LH === Subject: 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. 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 Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de === Subject: Re: Analysis problem... Content-transfer-encoding: 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 in int_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: 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 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) what I ordered...two cups of milk. Marvin Sebourn osugeography@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 main course. A better example would be ordering cold water in Italy. Do not be surprised when you get hot water. -- 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: 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 left side: x^2 - y^2 = (x - y)(x + y) Now, let A = x - y, and B = x + y. What can you say about the relationship between A and B? What sorts of numbers can occur as A and B? For instance, you can probably see that B - A = 2y. What does that say about A and B (thought of together, not just one at a time)? Having decided what numbers A and B that can appear in this factorization, now look at all the factorizations of 66. What sorts of numbers do you get from those factorizations? If this isn't enough of a hint, you're trying too hard. The answer is really pretty simple. === Subject: Re: help me, please! ha scritto nel messaggio > Is there a fast way to determine X and Y both integer such that: > X^2 - Y^2 = A > 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. 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 process then to learn what's to be know, solve for integers x^2 - y^2 = 4972 > 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. No. The number of factors has nothing to do with it. === 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 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 mistaken I 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. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === 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 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 mistaken Carlos -- === 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) if b<1/(4a). Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === 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. === 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. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === 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 Greek ancestry, or to be from those of other groups, mainly not Egyptians, who joined the Greeks. The Ptolemies made no attempt to submerge the other cultures, and the Greeks were sufficiently xenophobic (barbarian was the Greek for those who did not speak Greek, basically) that they made no attempt to do so. When the Seleucids did, they ran into open revolts. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: 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 modern English speaker, but Herodotus for example seems to use it in a quite neutral sense to mean non-Greeks. -- === Subject: Re: Who contributed most to mathematics? charset=iso-8859-7 Ì Herman Rubin ó.8d.98.87.8b.8c .97.99.95 .92.86.94.9d.92.87 > , 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 like nonsensical 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 Galidakis http://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 write a 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 publish much, he corresponded with many important mathematicians, so he was quite 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 of any 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 was to discover the idea of mathematical proof. Nearly all Western mathematicians until very recent times got their first taste of mathematics from studying Euclid. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Who contributed most to mathematics? > >Please forgive me my ignorance, I missed a few classes in college >thanks to some heavily hopped Czechish Pilsner, but could you >elaborate on this list especially regarding Czechoslovakia and >Taiwan? 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 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). === 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 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. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Question on 7d geometry Thank you Tobias. Your analysis is very clear, and I appreciate it. -Steven 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, 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 geometry X-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, 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. -- 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 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 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 equivalent > to 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 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). Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === 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 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 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 equivalent > to 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 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. Best regards, Pradip === 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 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 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 equivalent > to 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? 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 recursive oneness 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 will cause a loop. The oneness of 32 is 5. Very few numbers have a oneness of 5. I'm not sure but something tells me the only next one of which ROLL(n) becomes infinite is 2^32. (which causes the 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 than 16. This only happens regularily when n exceeds 2000000000 (2 ion) You just gave me a cool idea btw. Suppose you have a number 's' and you call N(s) the amount of numbers for which the oneness is 's'. I'll try that later. There must be, for each 's', at least one even and at least one odd number that reaches them? Could those two systems form two two-dimensional axises? Also, I made another recursive loop that does S(n) S(S(n)) S(S(S(n))) Until this reaches 'n', not one. Let's call this looplength the recursive oneness identity loop length, ROILL. Note that this only happens for 1,2,3,4,5,7, and 16 Also note that in my definition ROLL(1) = ROILL(1) = 6. Since S(1) = 3 S(3) = 7 S(7) = 16 S(16) = 4 S(4) = 2 S(2) = 1 I think this set of numbers might be interesting, in intuitive terms 3 and 7 might be seen as the 'prime' counterparts of the 2-powers 16 and 4, and 2 as 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 with written language. I did say what is necessary : I've tried to visualize any possible patterns in the one-dimensional series, let's leave it at that because I've tried many things. As for the complex numbers, I tried dividing by 2+2i when the complex and real parts are even, and multiplying with 3+3i and adding 1+1i when they are odd. I'm still experimenting with that.. -- Quaternion === Subject: Re: Oneness of a number Does 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->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 === 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 = 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 = 1 Damn let me take that in tonight, it's quite complex for me to understand what you mean to represent with 'C_a1,a2,a3..,an' right now and how it can fight the logic that if you follow the iteration scheme you get a bijection between n and the amount of iterations that are needed to transform it to 1.. -- 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 higher aspect of Mathematics beyond the typical undergraduate study of Calculus? At a simpler level, it seems that you could treat any single group of a given quantity as ONE WHOLE; so a grouped unit is one whole set. This allows us to find 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: 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? 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 else interesting 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 -- Mensanator Ace of Clubs === Subject: existence of envelope singular solutions What 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 and del F/ del c = 0 ; perhaps it is a condition using second order partial derivative. In the crank - connecting rod example, we have a crank of radius R centered 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 connecting rod length L ; It is found that an envelope exists in c limits + - ArcTan(L/R), and there is no envelope outside these limits. === Subject: Re: parallelizability of manifolds X-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 ends together 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 a top. space Y together with an epimorphism f:X->Y, can we say that Y is simply connected? 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)=s so that we can prove that Y has to be simply connected. Can we turn this idea into a proof? According to your post, the answer seems to be no. So can you give an 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 continuous map X->Y. Of course it need not be an epimorphism, so the conclusion was wrong 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 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.) >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, then the right phrase is homeomorphic image, and then yes, the homeomorphic 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 usual example is the same f I gave above, where X is the real line and Y 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)=s That'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 that you can lift loops; that is, if s is a loop in Y, then it's also a path, and so it lifts to a path r if f is a covering (say); but this lifted path r need not be a loop, i.e., there's no reason you must have r(0)=r(1). If you can't see this with formulas, draw a picture of a helix (a spring) X, casting a circular shadow. The shadow, Y is a circle and the projection is the map f. Draw a loop around Y, and then 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 lift paths does not by itself guarantee that you can lift other maps such as homotopies (maps from [0,1]x[0,1] into Y). If you _are_ in the context of covering spaces, there is a pretty characterization of what can be lifted: a map r : Z --> Y induces a homomorphism r* : 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)) must be contained in the image f*(pi_1(X)). If f is a covering, then this condition is also sufficient. In particular, any map from a contractible space Z into a space Y will lift to a map from Z into any cover X of Y. I know you never said you interested only in covering maps, but these things put your confusion into stark relief. You might want to think about some of the classical examples of these. Start with any simply-connected space X and a discrete group G of self-homeomorphisms of 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. === Subject: Re: parallelizability of manifolds X-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 continuous map) f:X->Y is called an epimorphism iff there is a morphism g:Y->X such that fg=id_Y. 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. 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. 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 -- 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 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 manifolds X-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's Algebra, and he defines epimorphism and monomorphism only for abelian categories. My notion is much stronger; it implies the other, but if we take a set X and let 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 an epimorphism, but does not have a right inverse. In the context of covering spaces (where 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 consider this case. -- reverse my forename for mail! Tobias Fritz Student of Mathematics and Physics University 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 is 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. 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. === 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} 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. -Tyler === Subject: Re: v sp over field {0,1} 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. === 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 about the inner product (sesquilinear form) you're using. As others have pointed out, you really need to take into account whether the form is degenerate. For example, if X has a non-zero radical R then dim(R) + dim(R^perp) = dim(R) + dim(X) > dim(X). -- === 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 are canonical inner products in R^n and in C^n, but I don't know if there is 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 you require of an inner product? In some books, an inner product is required to be positive definite (i.e., =0, and =0 if and only if v = 0)... === 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 like the 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 solution to a system of linear equations. Do you know how to compute the dimension of a nullspace (kernel)? Note that the result is _false_ if for example =0 for all u and v. You'll need to look to see what properties you're assuming about the inner product before you try to prove something about it! (Nondegeneracy, for starters.) === 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. -Tyler It is best to think of K^perp not as a subspace of X but as a subspace of the dual space X*. (Now via your pairing , it is true that X* is identified with X, but that just confuses things.) -- 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. Best regards, Jose Carlos Santos === Subject: Re: v sp over field {0,1} 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. === 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. === 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 This one is in Sloane's online database of offbeat sequences, number A008847. Go to http://www.research.att.com/~njas/sequences/ and enter 1, 9, 20, 180, 1306 and click Search. The search will cough up more terms of the table, and some 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 67 313 104855 117831 5 67 313 132682 187131 2 11 37 163 198534 347529 2 3 7 29 163 247863 347529 3 7 11 29 37 292374 479787 [non-primitive - 9*32486] 300876 503347 2 2 3 25073 312374 414309 2 313 499 313929 436107 3 3 3 7 11 151 334330 496713 2 5 67 499 345807 478401 [non-primitive - 9*38423] 376095 503347 3 5 25073 428184 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 19 107 498892 696787 2 2 191 653 contains 1306=2*653 623615 696787 5 191 653 754956 1296141 [non-primitive - 9*83884] 768460 1348221 [non-primitive - 20*38423] 787127 828723 11 163 439 943695 1296141 [non-primitive - 9*104855] 985369 1125579 7 11 67 191 1194138 2058441 [non-primitive - 9*132682] 1276880 2106853 2 2 2 2 5 11 1451 1378608 2650557 2 2 2 2 3 7 11 11 373 1547052 2728713 2 2 3 13 47 211 1606281 1993719 3 29 37 499 1676904 2854579 2 2 2 3 107 653 contains 1306=2*653 1754039 1903629 7 83 3091 1933815 2728713 3 5 13 47 211 2015094 3310671 2 3 29 37 313 2034423 2501877 3 3 3 151 499 2156730 3969147 2 3 5 29 37 67 2181409 2322957 19 29 37 107 2452440 4657471 2 2 2 3 5 107 191 2552202 4154493 2 3 3 3 151 313 2731590 4980801 2 3 3 3 5 67 151 2811366 4557399 [non-primitive - 9*312374] 3008970 5463843 [non-primitive - 9*334330] 3043401 3779139 3 19 107 499 3817974 6275451 2 3 19 107 313 4086330 7523607 2 3 5 19 67 107 4490028 7664657 [non-primitive - 9*498892] 4957260 10773399 [non-primitive - 20*247863] 5147241 6540807 3 11 61 2557 5545617 7536711 3 7 11 24007 5570344 8390001 2 2 2 13 19 2819 5612535 7664657 [non-primitive - 9*623615] 5643638 8786379 2 7 11 13 2819 5805336 10287381 2 2 2 3 19 29 439 5881722 10773399 2 3 7 11 29 439 6278580 13519317 [non-primitive - 20*313929] 6385703 6457269 67 191 499 6403664 9346701 2 2 2 2 29 37 373 6416984 10185273 2 2 2 7 19 37 163 6916140 14830431 [non-primitive - 9*20*38423] 7084143 9115953 [non-primitive - 9*787127 7542184 11178921 2 2 2 13 47 1543 7642712 12516231 2 2 2 7 11 19 653 contains 1306=652*2 8010922 10722621 2 67 191 313 8868321 12381369 [non-primitive - 9*985369] 8934096 15205827 2 2 2 2 3 373 499 9392460 204212197 [non-primitive - 20*469623] 9821396 14335671 2 2 13 67 2819 Some 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 67 67 3 31 7^2 (313) => can cancel 31 107 7 13 127 191 7 31 13^2 313 3 181^2 => can cancel 3 373 3 13 73 7^2 439 3 67 31^2 499 3 7 109^2 653 7 13^2 19^2 => can cancel 7 2819 61 19^4 => can cancel 61 And some other primes: 3 13 => can cancel 13 (313&373) => can cancel 73 11 7 19 (653) => can cancel 19 17 307 => can cancel 307 19 3 127 (313) => can cancel 127 23 7 79 (653) => can cancel 79 31 3 331 (313) => can cancel 331 41 1723 => can cancel 1723 43 3 631 => can cancel 631 47 37 61 (2819) => can cancel 37 53 7 409 (653) => can cancel 409 61 3 13 97 (3&313) => can cancel 97 79 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, the similarity with QS is massive, as you'd want to sieve over a 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 with pretty massive numbers of factors. I'll leave that for someone else. -- Unpatched IE vulnerability: Basic Authentication URL spoofing Description: Spoofing the URL displayed in the Address bar Reference: 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.html Needless to say, I had the last laugh. 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 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. does this equality actually always hold, and is there a pleasant bijective proof of it? also, i tried the obvious greedy algorithm for creating such circular words (start with the all-zeros word of length n and then start appending letters to it, preferably 1 but if that causes a repeated word then 0 instead), and it seemed to work; for example 00000111110111001101011000101001. does this greedy algorithm actually always work, and if so then why? also, are there are any particular nice generalizations of any or all of this? (by the way, just in case any of these questions are answered in on numbers and games, i did try to get it out of the library but it was in 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 who don'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 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. Ah, then it's a good thing I presuaded him to state the Word explicitly. It's a word I really would not have thought of 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 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 . . . . . . . . . . . . . . . . . . . . . . . PROBABILITY Michael === 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, for example. Took another minute to see the actual solution: there are no uncountable chains of _rationals_. Very clever - I _really_ never woulda thought of applying that word in a problem about logic. >Michael === Subject: Re: Two questions in propositional logic >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 that false 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 was trying to prove. (Regardless, it was thinking about open subsets of the middle-thirds Cantor set that led me to realize what I was 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,
>> 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. 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. 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? J === Subject: Re: Greek Alphebet >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? 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 -- good point, except that the contemporary Greek word for cow (AGELADA, from AGELI = 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 the eponymous theorem; the story survives anywhere from Diogenes Laertius (4th century 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 >>
 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
>>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 the
cardinality of the integers, the reals, functions on the
reals, and functions on that.  These were constants, not
variables.  At this time, aleph_z is the z-th smallest
transfinite cardinal of a well-ordered set, where z is an
ordinal number, starting from 0, and omega_z is the
corresponding initial ordinal.  At least with the Axiom
of Choice, beth_z is the z-th least power transfinite
cardinal; with the GCH, beth_z is the cardinality of the
power set of a set with cardinality aleph_z.  
I have seen zayin introduced as the opposite of aleph in
models 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 without
exception, they use Latin and Greek letters as we do.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558

===

Subject: Re: 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 name
of a field. I cannot recall ever seeing an arabic character used
(apart from Arabic numerals, of course), nor a Norse rune, nor any
of 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 an
alphabet, but those gylphs have been used heavily in the past,
especially in algebraic number theory. There are zillions of other
symbols used, some of which appear to be letters, but I don't think
they really are -- nabla and pe come to mind. There are some ancient
Greek letters which I don't think I have ever seen used in math
(including qoppa or something like that; I think the letters were
never used in post-Homeric Greek). There are of course many more
alphabets 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 few
languages, so people whose native language is, say, Cherokee, don't usually
write much high-level mathematics in that language -- not if they want
their results to get much exposure. Before the (first) October revolution,
that included Russia; the preferred language for the St sburg society
was French. But during the 20th century there was quite a bit of
research mathematics published in Russian, and indeed the text would be
in Russian (using Cyrillic, of course) and the mathematics using exactly
the same notation used in the west. I don't recall even seeing a point
in a geometric diagram ever labelled with a Cyrillic character.
(The fact that the typesetting of mathematics was done using Latin and
Greek characters must have driven the Soviet printers crazy, but it was
a boon to the translators into western languages, who would typically
just cut and paste the displays. This was occasionally confusing since,
for example, the names of the trig functions are not the same in all
western languages.)
There is not always a clear distinction between research monograph and
graduate 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 would
opt for using letters which appear identical in both alphabets, at
least for a couple of years until the youngsters start the study of
European 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 international
student 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 what
was in a course from the equations which appear.
Just imagine the joys of writing mathematics in, say, Hebrew, where
the bulk of the writing is done right-to-left but the equations are
written left-to-right. Back in the typewriter days, that meant having
to 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.

===

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.
-- 
                           
http://www.math.ohio-state.edu/~edgar/

===

Subject: Re: Greek Alphebet
> John 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 
the
>Hebrew aleph, but the letter itself started AS the symbol for an ox, 
in
>Greek. It actually comes from the upside down capital alpha, which 
was
the
>actual 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 degenerate
form
>of  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 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.
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.
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.

===

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, ancient
Persian used an alphabet not sufficiently close to the
Semitic alphabet, but they knew about it.
The Hangul alphabet for Korean is not derived from the
Semitic, although it is probable that the existence of an
alphabet was known to the designers.  A Korean character
has two to four alphabetic characters.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558

===

Subject: Re: 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* Propaganda
WHAT 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 lies
are *ENORMOUS*, you little bwoy. Is Goebbels your spiritual
father??? 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 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.
TWADDLE.
The Greeks were already using 4 scripts before the arrival of Cadmus, Greek
Linear A (Aeolic and Ionic), Cypriot Linear A (Mycenaean), Linaer B
(Mycenaean), and Phaistos Script (Ionic). The Akkadian-Phoenician scribes
knew absolutely nothing about vowels. It was the Greeks who invented vowels
both in Linear A and B and in Cadmian script. Up until 1400 BC which is the
date of the Ras Sharma tablets the Phoenicians were still using cuneiform 
on
clay.
Cadmus arrival in Greece in 1438 indicates that it was in Greece that the
Cadmian script was developed. According to Phoenician historical records
translated by o of Byblis the so-called 3 letters were invented by
Eisirius (Celix) who was a foreigner and the brother of Phoenix (Chna) and
by implication the brother of Cadmus as well.
The divergence between Cadmian script and Canaanite script took place
immediately on Cadmus arrival in Greece. The Greeks added vowels to the
alphabet 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 they
do not go A, B, C. Besides which both Greek and Hebrew start of as A, B, G
and there are no zodiacal constellations in that order either, nor enough
for an alphablet.
An upside down Alpha looks more like a horses had than that of an ox so it
is 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 as
bi-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.
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
 

===

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,
The Traditional Chronology as recorded by Diodorus, Castor and Apollonius 
et
al, and used by Tatian, Eusebius and Clement. The siege of Troy began 1193
BC 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 know
that 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 to
have existed in 900 BC. The only way there could have been so many is if 
all
of them derived from a common root long before the Ionian mitigation of 
1071
BC. 1200 BC is the latest possible date for the use of Cadmian script by 
the
Greeks and the start of its divergence into 5 independent Greek scripts.
Herodotus records an inscription made in the time of Amphytrion and Laius
which firmly dates to Amhytrions victory over the Teleboans which is dated
to 1286 BC by the occurrence of a total solar eclipse over the Aegean which
occurred on February 10 and is confirmed by the traditional chronology.
> 
> 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?
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?
Do the maths.
http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
> 
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2

===

Subject: Re: Greek Alphebet
        charset=iso-8859-7
> 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
> 
> 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
> 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.
If he can do the maths then he should have no problem in confirming that I
am right.
> And you never know....If he does, you may become famous :*)
> http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
> 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. 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 is from
the Greek word gemistes or gamistes or something like that and not 
from
the Turkish root word dol/dolma meaning
fill/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 sou
stellw Ntpolmades? 8a kanomai allagei/trampa, re. Entaksi?
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
> 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. 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 is 
from
> the Greek word gemistes or gamistes or something like that and 
not
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
meaning 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 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
 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.
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 is
from
> the Greek word gemistes or gamistes or something like that and 
not
> 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
> 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. The root
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.
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
> 
> 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.
> 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 
is
> from
> the Greek word gemistes or gamistes or something like that and 
not
> 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 find 
a
> word in a dictionary? Oh, I know why. It is because it is Turkish. The
root
> 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 
corruption
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
> 
> 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
No, Agamemnona, the Turkish word Dol is a corruption of the Greek Ta
Vgaleis 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 of
none.
> 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 
is
> from
> the Greek word gemistes or gamistes or something like that 
and
not
> 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 
find
a
> word in a dictionary? Oh, I know why. It is because it is Turkish. The
> root
> 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
corruption
> 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 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
>>  > 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 Kolker
The Ugaritic alphabet is a 32 character cuneiform 
modification of the early Semitic alphabets; there
were lots of them.  The earliest known Semitic
alphabetic writing is from about 1800 BCE, found
on a cliff wall in Egypt.
Hebrew and Ugaritic are almost identical as languages;
Aramaic is similar, but different, and these probably
diverged before alphabetic writing; Arabic is from
a different branch of the Semitic family.
-- 
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